A 543920
}
ARCHIMEDES
WORKS
i
QA
.A673]
Z.e.
1897

ARTES
LIBRARY
18371
VERITAS
SCIENTIA
OF THE
UNIVERSITY OF MICHIGAN
TUEBOR
31-QUÆRIS PENINSULAM AMENAM
CIRCUMSPICE
1
.
THE WORKS
OF
24087
ARCHIMEDES.
London: C. J. CLAY AND SONS,
CAMBRIDGE UNIVERSITY PRESS WAREHOUSE,
AVE MARIA LANE.
Glasgow: 263, ARGYLE STREET.

Leipzig: F. A. BROCKHAUS.
New York: THE MACMILLAN COMPANY.
1
THE WORKS
醋
​OF
67918
ARCHIMEDES
EDITED IN MODERN NOTATION
WITH INTRODUCTORY CHAPTERS
BY
T. L. HEATH, Sc.D.,
✔
SOMETIME FELLOW OF TRINITY COLLEGE, CAMBRIDGE.
CAMBRIDGE:
AT THE UNIVERSITY PRESS.
1897
[All Rights reserved.]
५
5°3
ر بیسیم مجو
>
PREFACE.
THIS book is intended to form a companion volume to my
édition of the treatise of Apollonius on Conic Sections
lately published. If it was worth while to attempt to make the
work of "the great geometer" accessible to the mathematician
of to-day who might not be able, in consequence of its length
and of its form, either to read it in the original Greek or in a
Latin translation, or, having read it, to master it and grasp the
whole scheme of the treatise, I feel that I owe even less of an
apology for offering to the public a reproduction, on the same
lines, of the extant works of perhaps the greatest mathematical
genius that the world has ever seen.
Michel Chasles has drawn an instructive distinction between
the predominant features of the geometry of Archimedes and
of the geometry which we find so highly developed in Apollo-
nius. Their works may be regarded, says Chasles, as the origin
and basis of two great inquiries which seem to share between
them the domain of geometry. Apollonius is concerned with
the Geometry of Forms and Situations, while in Archimedes
we find the Geometry of Measurements dealing with the quad-
rature of curvilinear plane figures and with the quadrature
and cubature of curved surfaces, investigations which "
"gave
birth to the calculus of the infinite conceived and brought
to perfection successively by Kepler, Cavalieri, Fermat, Leibniz,
and Newton." But whether Archimedes is viewed as the
man who, with the limited means at his disposal, nevertheless
succeeded in performing what are really integrations for the
purpose of finding the area of a parabolic segment and a
110
N
67918
vi
PREFACE.
spiral, the surface and volume of a sphere and a segment
of a sphere, and the volume of any segments of the solids
of revolution of the second degree, whether he is seen finding
the centre of gravity of a parabolic segment, calculating
arithmetical approximations to the value of π, inventing a
system for expressing in words any number up to that which
we should write down with 1 followed by 80,000 billion
ciphers, or inventing the whole science of hydrostatics and at
the same time carrying it so far as to give a most complete
investigation of the positions of rest and stability of a right
segment of a paraboloid of revolution floating in a fluid, the
intelligent reader cannot fail to be struck by the remarkable
range of subjects and the mastery of treatment. And if these
are such as to create genuine enthusiasm in the student of
Archimedes, the style and method are no less irresistibly
attractive. One feature which will probably most impress the
mathematician accustomed to the rapidity and directness secured
by the generality of modern methods is the deliberation with
which Archimedes approaches the solution of any one of his
main problems. Yet this very characteristic, with its incidental
effects, is calculated to excite the more admiration because the
method suggests the tactics of some great strategist who
foresees everything, eliminates everything not immediately
conducive to the execution of his plan, masters every position
in its order, and then suddenly (when the very elaboration of
the scheme has almost obscured, in the mind of the spectator,
its ultimate object) strikes the final blow. Thus we read in
Archimedes proposition after proposition the bearing of which is
not immediately obvious but which we find infallibly used later
on; and we are led on by such easy stages that the difficulty of
the original problem, as presented at the outset, is scarcely
appreciated. As Plutarch says, "it is not possible to find in
geometry more difficult and troublesome questions, or more
simple and lucid explanations." But it is decidedly a rhetorical
exaggeration when Plutarch goes on to say that we are deceived
PREFACE.
vii
(C
by the easiness of the successive steps into the belief that anyone
could have discovered them for himself. On the contrary, the
studied simplicity and the perfect finish of the treatises involve
at the same time an element of mystery. Though each step
depends upon the preceding ones, we are left in the dark as to
how they were suggested to Archimedes. There is, in fact,
much truth in a remark of Wallis to the effect that he seems
as it were of set purpose to have covered up the traces of his
investigation as if he had grudged posterity the secret of his
method of inquiry while he wished to extort from them assent
to his results." Wallis adds with equal reason that not only
Archimedes but nearly all the ancients so hid away from
posterity their method of Analysis (though it is certain that
they had one) that more modern mathematicians found it easier
to invent a new Analysis than to seek out the old. This is no
doubt the reason why Archimedes and other Greek geometers
have received so little attention during the present century and
why Archimedes is for the most part only vaguely remembered
as the inventor of a screw, while even mathematicians scarcely
know him except as the discoverer of the principle in hydro-
statics which bears his name. It is only of recent years that
we have had a satisfactory edition of the Greek text, that of
Heiberg brought out in 1880-1, and I know of no complete
translation since the German one of Nizze, published in 1824,
which is now out of print and so rare that I had some difficulty
in procuring a copy.
The plan of this work is then the same as that which I
followed in editing the Conics of Apollonius. In this case,
however, there has been less need as well as less opportunity for
compression, and it has been possible to retain the numbering
of the propositions and to enunciate them in a manner more
nearly approaching the original without thereby making the
enunciations obscure. Moreover, the subject matter is not so
complicated as to necessitate absolute uniformity in the notation
used (which is the only means whereby Apollonius can be made
viii
PREFACE.
even tolerably readable), though I have tried to secure as much
uniformity as was fairly possible. My main object has been to
present a perfectly faithful reproduction of the treatises as they
have come down to us, neither adding anything nor leaving out
anything essential or important. The notes are for the most
part intended to throw light on particular points in the text or
to supply proofs of propositions assumed by Archimedes as
known; sometimes I have thought it right to insert within
square brackets after certain propositions, and in the same type,
notes designed to bring out the exact significance of those
propositions, in cases where to place such notes in the Intro-
duction or at the bottom of the page might lead to their being
overlooked.
Much of the Introduction is, as will be seen, historical; the
rest is devoted partly to giving a more general view of certain
methods employed by Archimedes and of their mathematical
significance than would be possible in notes to separate propo-
sitions, and partly to the discussion of certain questions arising
out of the subject matter upon which we have no positive
historical data to guide us. In these latter cases, where it is
necessary to put forward hypotheses for the purpose of explaining
obscure points, I have been careful to call attention to their
speculative character, though I have given the historical evidence
where such can be quoted in support of a particular hypothesis,
my object being to place side by side the authentic information
which we possess and the inferences which have been or may
be drawn from it, in order that the reader may be in a position
to judge for himself how far he can accept the latter as probable.
Perhaps I may be thought to owe an apology for the length of
one chapter on the so-called vevoeis, or inclinationes, which goes
somewhat beyond what is necessary for the elucidation of
Archimedes; but the subject is interesting, and I thought it
well to make my account of it as complete as possible in
order to round off, as it were, my studies in Apollonius and
Archimedes.
PREFACE.
ix
I have had one disappointment in preparing this book for
the press. I was particularly anxious to place on or opposite
the title-page a portrait of Archimedes, and I was encouraged
in this idea by the fact that the title-page of Torelli's edition
bears a representation in medallion form on which are endorsed
the words Archimedis effigies marmorea in veteri anaglypho
Romae asservato. Caution was however suggested when I
found two more portraits wholly unlike this but still claiming to
represent Archimedes, one of them appearing at the beginning
of Peyrard's French translation of 1807, and the other in
Gronovius' Thesaurus Graecarum Antiquitatum; and I thought
it well to inquire further into the matter. I am now informed
by Dr A. S. Murray of the British Museum that there does
not appear to be any authority for any one of the three, and
that writers on iconography apparently do not recognise an
Archimedes among existing portraits. I was, therefore, re-
luctantly obliged to give up my idea.
The proof sheets have, as on the former occasion, been read
over by my brother, Dr R. S. Heath, Principal of Mason College,
Birmingham; and I desire to take this opportunity of thanking
him for undertaking what might well have seemed, to any one
less genuinely interested in Greek geometry, a thankless task.
March, 1897.
T. L. HEATH.
•
LIST OF THE PRINCIPAL WORKS CONSULTED.
JOSEPH TORELLI, Archimedis quae supersunt omnia cum Eutocii Asca-
lonitae commentariis. (Oxford, 1792.)
ERNST NIZZE, Archimedes von Syrakus vorhandene Werke aus dem
griechischen übersetzt und mit erläuternden und kritischen Anmerk-
ungen begleitet. (Stralsund, 1824.)
J. L. HEIBERG, Archimedis opera omnia cum commentariis Eutocii.
(Leipzig, 1880–1.)
J. L. HEIBERG, Quaestiones Archimedeae. (Copenhagen, 1879.)
F. HULTSCH, Article Archimedes in Pauly-Wissowa's Real-Encyclopädie der
classischen Altertumswissenschaften. (Edition of 1895, II. 1, pp.
507-539.)
C. A. BRETSCHNEIDER, Die Geometrie und die Geometer vor Euklides.
(Leipzig, 1870.)
M. CANTOR, Vorlesungen über Geschichte der Mathematik, Band 1, zweite
Auflage. (Leipzig, 1894.)
G. FRIEDLEIN, Procli Diadochi in primum Euclidis elementorum librum
commentarii. (Leipzig, 1873.)
JAMES GOW, A short history of Greek Mathematics. (Cambridge, 1884.)
SIEGMUND GÜNTHER, Abriss der Geschichte der Mathematik und der
Naturwissenschaften im Altertum in Iwan von Müller's Handbuch der
klassischen Altertumswissenschaft, v. 1.
HERMANN HANKEL, Zur Geschichte der Mathematik in Alterthum und
Mittelalter. (Leipzig, 1874.)
J. L. HEIBERG, Litterargeschichtliche Studien über Euklid. (Leipzig,
1882.)
J. L. HEIBERG, Euclidis elementa. (Leipzig, 1883–8.)
F. HULTSCH, Article Arithmetica in Pauly-Wissowa's Real-Encyclopädie,
II. 1, pp. 1066–1116.
xii
LIST OF PRINCIPAL WORKS CONSULTED.
F. HULTSCH, Heronis Alexandrini geometricorum et stereometricorum
reliquiae. (Berlin, 1864.)
F. HULTSCH, Pappi Alexandrini collectionis quae supersunt. (Berlin,
1876-8.)
GINO LORIA, Il periodo aureo della geometria greca. (Modena, 1895.)
MAXIMILIEN MARIE, Histoire des sciences mathématiques et physiques,
Tome I. (Paris, 1883.)
J. H. T. MÜLLER, Beiträge zur Terminologie der griechischen Mathematiker.
(Leipzig, 1860.)
G. H. F. NESSELMANN, Die Algebra der Griechen. (Berlin, 1842.)
F. SUSEMIHL, Geschichte der griechischen Litteratur in der Alexandrinerzeit,
Band I. (Leipzig, 1891.)
P. TANNERY, La Géométrie grecque, Première partie, Histoire générale de la
Géométrie élémentaire. (Paris, 1887.)
H. G. ZEUTHEN, Die Lehre von den Kegelschnitten im Altertum. (Copen-
hagen, 1886.)
H. G. ZEUTHEN, Geschichte der Mathematik im Altertum und Mittelalter.
(Copenhagen, 1896.)
ร
ARCHIMEDES.
xvii
catapults so ingeniously constructed as to be equally serviceable.
at long or short ranges, machines for discharging showers of
missiles through holes made in the walls, and others consisting
of long moveable poles projecting beyond the walls which either
dropped heavy weights upon the enemy's ships, or grappled the
prows by means of an iron hand or a beak like that of a crane,
then lifted them into the air and let them fall again*. Marcellus
is said to have derided his own engineers and artificers with the
words, "Shall we not make an end of fighting against this geo-
metrical Briareus who, sitting at ease by the sea, plays pitch and
toss with our ships to our confusion, and by the multitude of
missiles that he hurls at us outdoes the hundred-handed giants of
mythology?+"; but the exhortation had no effect, the Romans being
in such abject terror that "if they did but see a piece of rope
or wood projecting above the wall, they would cry 'there it is
again,' declaring that Archimedes was setting some engine in motion
against them, and would turn their backs and run away, insomuch
that Marcellus desisted from all conflicts and assaults, putting all
his hope in a long siege‡."
If we are rightly informed, Archimedes died, as he had lived,
absorbed in mathematical contemplation. The accounts of the
exact circumstances of his death differ in some details. Thus
Livy says simply that, amid the scenes of confusion that followed
the capture of Syracuse, he was found intent on some figures which
he had drawn in the dust, and was killed by a soldier who did
not know who he was §. Plutarch gives more than one version in
the following passage.
"Marcellus was most of all afflicted at
the death of Archimedes; for, as fate would have it, he was intent
on working out some problem with a diagram and, having fixed
his mind and his eyes alike on his investigation, he never noticed
the incursion of the Romans nor the capture of the city. And
when a soldier came up to him suddenly and bade him follow to
* Polybius, Hist. VIII. 7—8; Livy xxiv. 34; Plutarch, Marcellus, 15—17.
+ Plutarch, Marcellus, 17.
+ ibid.
§ Livy xxv. 31. Cum multa irae, multa auaritiae foeda exempla ederentur,
Archimedem, memoriae proditum est in tanto tumultu, quantum pauor captae
urbis in discursu diripientium militum ciere poterat, intentum formis, quas in
puluere descripserat, ab ignaro milite quis esset interfectum; aegre id Marcellum
tulisse sepulturaeque curam habitam, et propinquis etiam inquisitis honori
praesidioque nomen ac memoriam eius fuisse.
H. A.
xviii
INTRODUCTION.
Marcellus, he refused to do so until he had worked out his problem
to a demonstration; whereat the soldier was so enraged that he
drew his sword and slew him. Others say that the Roman ran
up to him with a drawn sword offering to kill him; and, when
Archimedes saw him, he begged him earnestly to wait a short time
in order that he might not leave his problem incomplete and
unsolved, but the other took no notice and killed him. Again
there is a third account to the effect that, as he was carrying to
Marcellus some of his mathematical instruments, sundials, spheres,
and angles adjusted to the apparent size of the sun to the sight, some
soldiers met him and, being under the impression that he carried
gold in the vessel, slew him *.” The most picturesque version of the
story is perhaps that which represents him as saying to a Roman
soldier who came too close, "Stand away, fellow, from my diagram,”
whereat the man was so enraged that he killed him†. The addition
made to this story by Zonaras, representing him as saying mapà
κεφαλὰν καὶ μὴ παρὰ γραμμάν, while it no doubt recalls the second
version given by Plutarch, is perhaps the most far-fetched of the
touches put to the picture by later hands.
Archimedes is said to have requested his friends and relatives
to place upon his tomb a representation of a cylinder circumscribing
a sphere within it, together with an inscription giving the ratio
which the cylinder bears to the sphere; from which we may
infer that he himself regarded the discovery of this ratio [On the
Sphere and Cylinder, 1. 33, 34] as his greatest achievement. Cicero,
when quaestor in Sicily, found the tomb in a neglected state and
restored it§.
Beyond the above particulars of the life of Archimedes, we
have nothing left except a number of stories, which, though perhaps
not literally accurate, yet help us to a conception of the personality
of the most original mathematician of antiquity which we would
not willingly have altered. Thus, in illustration of his entire
preoccupation by his abstract studies, we are told that he would
forget all about his food and such necessities of life, and would
be drawing geometrical figures in the ashes of the fire, or, when
* Plutarch, Marcellus, 19.
+ Tzetzes, Chil. 11. 35, 135; Zonaras IX. 5.
‡ Plutarch, Marcellus, 17 ad fin.
§ Cicero, Tusc. v. 64 sq.
ARCHIMEDES.
xix
anointing himself, in the oil on his body. Of the same kind is
the well-known story that, when he discovered in a bath the
solution of the question referred to him by Hieron as to whether
a certain crown supposed to have been made of gold did not in
reality contain a certain proportion of silver, he ran naked through
the street to his home shouting εύρηκα, εὕρηκατ.
Eucka
་
According to Pappus it was in connexion with his discovery
of the solution of the problem To move a given weight by a given
force that Archimedes uttered the famous saying,
"Give me a
place to stand on, and I can move the earth (dós μoɩ toû σtŵ kai
KIVÔ Tην YŶv)." Plutarch represents him as declaring to Hieron
that any given weight could be moved by a given force, and
boasting, in reliance on the cogency of his demonstration, that, if
he were given another earth, he would cross over to it and move
this one.
"And when Hieron was struck with amazement and asked
him to reduce the problem to practice and to give an illustration
of some great weight moved by a small force, he fixed upon a ship
of burden with three masts from the king's arsenal which had
only been drawn up with great labour and many men'; and loading
her with many passengers and a full freight, sitting himself the
while far off, with no great endeavour but only holding the end
of a compound pulley (roλúσraσTos) quietly in his hand and pulling
at it, he drew the ship along smoothly and safely as if she were
moving through the sea§." According to Proclus the ship was one
which Hieron had had made to send to king Ptolemy, and, when all
the Syracusans with their combined strength were unable to launch
it, Archimedes contrived a mechanical device which enabled Hieron
to move it by himself, insomuch that the latter declared that
"from that day forth Archimedes was to be believed in every-
thing that he might say." While however it is thus established
that Archimedes invented some mechanical contrivance for moving
a large ship and thus gave a practical illustration of his thesis,
it is not certain whether the machine used was simply a compound
* Plutarch, Marcellus, 17.
+ Vitruvius, Architect. ix. 3. For an explanation of the manner in which
Archimedes probably solved this problem, see the note following On floating
bodies, 1. 7 (p. 259 sq.).
+
Pappus VIII. p. 1060.
§ Plutarch, Marcellus, 14.
|| Proclus, Comm. on Eucl. 1., p. 63 (ed. Friedlein).
b2
XX
INTRODUCTION.
pulley (ToλúσTaσTоs) as stated by Plutarch; for Athenaeus*, in
describing the same incident, says that a helix was used. This
term must be supposed to refer to a machine similar to the koɣλías
described by Pappus, in which a cog-wheel with oblique teeth
moves on a cylindrical helix turned by a handlet. Pappus, how-
ever, describes it in connexion with the ẞapovλkós of Heron, and,
while he distinctly refers to Heron as his authority, he gives no
hint that Archimedes invented either the Bapovλkós or the par-
ticular koxías; on the other hand, the ToλúσTaσтos is mentioned
by Galen ‡, and the τρíσπаσros (triple pulley) by Oribasius§, as one
of the inventions of Archimedes, the rpíσTаσтоs being so called
either from its having three wheels (Vitruvius) or three ropes
(Oribasius). Nevertheless, it may well be that though the ship
could easily be kept in motion, when once started, by the тpí-
σπαστος οι πολύσπαστος, Archimedes was obliged to use an appliance
similar to the Koxλías to give the first impulse.
The name of yet another instrument appears in connexion with
the phrase about moving the earth. Tzetzes' version is, "Give
me a place to stand on (a Bo), and I will move the whole earth
with a xapioríwv"; but, as in another passage he uses the word
iρíσnaσтos, it may be assumed that the two words represented one
and the same thing**.
χαριστίων ||
It will be convenient to mention in this place the other
mechanical inventions of Archimedes. The best known is the
>
* Athenaeus v. 207 a-b, κατασκευάσας γὰρ ἕλικα τὸ τηλικοῦτον σκάφος εἰς τὴν
θάλασσαν κατήγαγε πρῶτος δ' Αρχιμήδης εὗρε τὴν τῆς ἕλικος κατασκευήν. Τo the
same effect is the statement of Eustathius ad Il. III. p. 114 (ed. Stallb.) AéYETαL
δὲ ἕλιξ καί τι μηχανῆς εἶδος, ὁ πρῶτος εὑρὼν ὁ ᾿Αρχιμήδης εὐδοκίμησέ, φασι, δι' αὐτοῦ.
+ Pappus VIII. pp. 1066, 1108 sq.
Galen, in Hippocr. De artic., Iv. 47 (=XVIII. p. 747, ed. Kühn).
§ Oribasius, Coll. med., XLIX. 22 (ïv. p. 407, ed. Bussemaker), 'Aπe\\ídous ✯
Αρχιμήδους τρίσπαστον, described in the same passage as having been invented
πρὸς τὰς τῶν πλοίων καθολκάς.
|| Tzetzes, Chil. 11. 130.
[ Ibid., III. 61, ὁ γῆν ἀνασπῶν μηχανῇ τῇ τρισπάστῳ βοῶν· ὅπα βῶ καὶ σαλεύσω
τὴν χθόνα.
** Heiberg compares Simplicius, Comm. in Aristot. Phys. (ed. Diels, p. 1110,
1. 2), ταύτῃ δὲ τῇ ἀναλογίᾳ τοῦ κινοῦντος καὶ τοῦ κινουμένου καὶ τοῦ διαστήματος
τὸ σταθμιστικὸν ὄργανον τὸν καλούμενον χαριστίωνα συστήσας ὁ ᾿Αρχιμήδης ὡς
μέχρι παντὸς τῆς ἀναλογίας προχωρούσης ἐκόμπασεν ἐκεῖνο τὸ πᾶ βῶ καὶ κινῶ τὰν
γᾶν.
CONTENTS.
INTRODUCTION.
CHAPTER I.
ARCHIMEDES
PAGE
XV
CHAPTER II. MANUSCRIPTS AND PRINCIPAL EDITIONS—ORDER
OF COMPOSITION—DIALECT-LOST WORKS
xxiii
CHAPTER III.
§ 1.
RELATION OF ARCHIMEDES TO HIS PREDECESSORS
Use of traditional geometrical methods
xxxix
xl
•
§ 2.
Earlier discoveries affecting quadrature and
cubature.
xlvii
§ 3.
Conic Sections
lii
§ 4. Surfaces of the second degree
liv
§ 5.
Two mean proportionals in continued propor-
tion .
lxvii
CHAPTER IV.
ARITHMETIC IN ARCHIMEDES
lxviii
§ 1.
Greek numeral system
lxix
•
§ 2.
Addition and subtraction
lxxi
•
§ 3. Multiplication
lxxii
•
§ 4.
Division
lxxiii
§ 5.
Extraction of the square root
lxxiv
•
rables
§ 6. Early investigations of surds or incommensu-
J
§ 7. Archimedes' approximations to √3
§ 8. Archimedes' approximations to the square roots
of large numbers which are not complete
squares
Note on alternative hypotheses with regard to
the approximations to √3
lxxvii
lxxx
lxxxiv
XC
'xiv
CONTENTS.
CHAPTER V.
§ 1.
Nevoeis referred to by Archimedes
§ 2.
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ
Mechanical constructions: the conchoid of Nico-
medes
§ 3. Pappus' solution of the veûois referred to in
Props. 8, 9 On Spirals.
PAGE
هرية
CV
cvii
§ 4.
The problem of the two mean proportionals
CX
§ 5.
The trisection of an angle
cxi
§ 6.
On certain plane νεύσεις
cxiii
CHAPTER VI.
CUBIC EQUATIONS
cxxiii
CHAPTER VII. ANTICIPATIONS BY ARCHIMEDES OF THE INTE-
GRAL CALCULUS
CHAPTER VIII. THE TERMINOLOGY OF ARCHIMEDES
THE WORKS OF ARCHIMEDES.
ON THE SPHERE AND CYLINDER, BOOK I.
""
レ
​MEASUREMENT OF A CIRCLE
ON CONOIDS AND SPHEROIDS
ON SPIRALS
cxlii
clv
1
BOOK II.
56
91
99
151
189
BOOK II.
203
221-
233
253
263
301
319
ON THE EQUILIBRIUM OF PLANES, BOOK I.
""
""
THE SAND-RECKONER
QUADRATURE OF THE PARABOLA
ON FLOATING BODIES, BOOK I.
BOOK OF LEMMAS .
THE CATTLE-PROBLEM
BOOK II.
•
INTRODUCTION.
CHAPTER I.
ARCHIMEDES.
A LIFE of Archimedes was written by one Heracleides*, but
this biography has not survived, and such particulars as are known
have to be collected from many various sources †. According to
Tzetzes he died at the age of 75, and, as he perished in the sack
of Syracuse (B.C. 212), it follows that he was probably born about
287 B.C. He was the son of Pheidias the astronomer§, and was
on intimate terms with, if not related to, king Hieron and his
* Eutocius mentions this work in his commentary on Archimedes' Measure-
ment of the circle, ὥς φησιν Ηρακλείδης ἐν τῷ ᾿Αρχιμήδους βίῳ. He alludes to it
again in his commentary on Apollonius' Conics (ed. Heiberg, Vol. 1. p. 168),
where, however, the name is wrongly given as 'Hрákλelos. This Heracleides is
perhaps the same as the Heracleides mentioned by Archimedes himself in the
preface to his book On Spirals.
+ An exhaustive collection of the materials is given in Heiberg's Quaestiones
Archimedeae (1879). The preface to Torelli's edition also gives the main points,
and the same work (pp. 363-370) quotes at length most of the original
references to the mechanical inventions of Archimedes. Further, the article
Archimedes (by Hultsch) in Pauly-Wissowa's Real-Encyclopädie der classischen
Altertumswissenschaften gives an entirely admirable summary of all the available
information. See also Susemihl's Geschichte der griechischen Litteratur in der
Alexandrinerzeit, 1. pp. 723–733.
*‡ Tzetzes, Chiliad., 11. 35, 105.
§ Pheidias is mentioned in the Sand-reckoner of Archimedes, тŵν πротéρшv
ἀστρολόγων Εὐδόξου...Φειδία δὲ τοῦ ἁμοῦ πατρὸς (the last words being the correction
of Blass for Toû 'AкоÚπαтроs, the reading of the text). Cf. Schol. Clark. in
Gregor. Nazianz. Or. 34, p. 355 a Morel. Þeidías rò μèv Yévos v Zvрakóσlos
ἀστρολόγος ὁ ᾿Αρχιμήδους πατήρ.
xvi
INTRODUCTION.
son Gelon. It appears from a passage of Diodorus* that he spent
a considerable time at Alexandria, where it may be inferred that
he studied with the successors of Euclid. It may have been at
Alexandria that he made the acquaintance of Conon of Samos
(for whom he had the highest regard both as a mathematician
and as a personal friend) and of Eratosthenes. To the former
he was in the habit of communicating his discoveries before their
publication, and it is to the latter that the famous Cattle-problem
purports to have been sent. Another friend, to whom he dedicated
several of his works, was Dositheus of Pelusium, a pupil of Conon,
presumably at Alexandria though at a date subsequent to Archi-
medes' sojourn there.
After his return to Syracuse he lived a life entirely devoted
to mathematical research. Incidentally he made himself famous
by a variety of ingenious mechanical inventions. These things
were however merely the "diversions of geometry at play †," and
he attached no importance to them. In the words of Plutarch, “he
possessed so high a spirit, so profound a soul, and such treasures
of scientific knowledge that, though these inventions had obtained
for him the renown of more than human sagacity, he yet would
not deign to leave behind him any written work on such subjects,
but, regarding as ignoble and sordid the business of mechanics
and every sort of art which is directed to use and profit, he placed
his whole ambition in those speculations in whose beauty and
subtlety there is no admixture of the common needs of life ‡.
fact he wrote only one such mechanical book, On Sphere-makings,
to which allusion will be made later.
""
In
Some of his mechanical inventions were used with great effect
against the Romans during the siege of Syracuse. Thus he contrived

* Diodorus v. 37, 3, οὓς [τοὺς κοχλίας] ᾿Αρχιμήδης ὁ Συρακόσιο
παρέβαλεν εἰς Αἴγυπτον.
+ Plutarch, Marcellus, 14.
‡ ibid. 17.
§ Pappus VIII. p. 1026 (ed. Hultsch). Κάρπος δὲ πού φησιν ὁ ᾿Αντιοχεὺς
᾿Αρχιμήδη τὸν Συρακόσιον ἓν μόνον βιβλίον συντεταχέναι μηχανικὸν τὸ κατὰ τὴν
σφαιροποιΐαν, τῶν δὲ ἄλλων οὐδὲν ἠξιωκέναι συντάξαι. καίτοι παρὰ τοῖς πολλοῖς ἐπὶ
μηχανικῇ δοξασθεὶς καὶ μεγαλοφυής τις γενόμενος ὁ θαυμαστὸς ἐκεῖνος, ὥστε διαμεῖναι
παρὰ πᾶσιν ἀνθρώποις ὑπερβαλλόντως ὑμνούμενος, τῶν τε προηγουμένων γεωμετρικῆς
καὶ ἀριθμητικῆς ἐχομένων θεωρίας τὰ βραχύτατα δοκοῦντα εἶναι σπουδαίως συνέγραφεν
ὃς φαίνεται τὰς εἰρημένας επιστήμας οὕτως ἀγαπήσας ὡς μηδὲν ἔξωθεν ὑπομένειν
αὐταῖς ἐπεισάγειν.
ARCHIMEDES.
xxi
*
water-screw (also called κoxías) which was apparently invented
by him in Egypt, for the purpose of irrigating fields. It was
also used for pumping water out of mines or from the hold of
ships.
Another invention was that of a sphere constructed so as to
imitate the motions of the sun, the moon, and the five planets
in the heavens. Cicero actually saw this contrivance and gives a
description of it†, stating that it represented the periods of the
moon and the apparent motion of the sun with such accuracy that
it would even (over a short period) show the eclipses of the sun
and moon.
Hultsch conjectures that it was moved by water ‡.
We know, as above stated, from Pappus that Archimedes wrote
a book on the construction of such a sphere (πeρì σpαɩрожоias),
and Pappus speaks in one place of "those who understand the
making of spheres and produce a model of the heavens by means
of the regular circular motion of water." In any case it is certain
that Archimedes was much occupied with astronomy. Livy calls
him "unicus spectator caeli siderumque." Hipparchus says §,
"From these observations it is clear that the differences in the
years are altogether small, but, as to the solstices, I almost
think (ovê åteλTíйw) that both I and Archimedes have erred to
the extent of a quarter of a day both in the observation and in the
deduction therefrom." It appears therefore that Archimedes had
considered the question of the length of the year, as Ammianus
also states ||. Macrobius says that he discovered the distances of
the planets T. Archimedes himself describes in the Saru-reckoner
the apparatus by which he measured the apparent diameter of the
sun, or the angle subtended by it at the eye.
The story that he set the Roman ships on fire by an arrange-
ment of burning-glasses or concave mirrors is not found in any
* Diodorus 1. 34, v. 37; Vitruvius x. 16 (11); Philo 1. p. 330 (ed. Pfeiffer);
Strabo xvII. p. 807; Athenaeus v. 208 f.
+ Cicero, De rep., 1. 21–22; Tusc., 1. 63; De nat. deor., II. 88. Cf. Ovid,
Fasti, vi. 277; Lactantius, Instit., II. 5, 18; Martianus Capella, 11. 212, vI.
583 sq.; Claudian, Epigr. 18; Sextus Empiricus, p. 416 (ed. Bekker).
‡ Zeitschrift f. Math. u. Physik (hist. litt. Abth.), xxII. (1877), 106 sq.
§ Ptolemy, σúvražis, I. p. 153.
σύνταξις, Ι.
|| Ammianus Marcell., xxvI. i. 8.
¶ Macrobius, in Somn. Scip., 11. 3.
f
+
7
xxii
INTRODUCTION.
authority earlier than Lucian*; and the so-called loculus Archi-
medius, which was a sort of puzzle made of 14 pieces of ivory of
different shapes cut out of a square, cannot be supposed to be his
invention, the explanation of the name being perhaps that it was
only a method of expressing that the puzzle was cleverly made,
in the same way as the πρόβλημα ᾿Αρχιμήδειον came to be simply
a proverbial expression for something very difficult †.
* The same story is told of Proclus in Zonaras XIV. 3. For the other
references on the subject see Heiberg's Quaestiones Archimedeae, pp. 39–41.
+ Cf. also Tzetzes, Chil. xΙΙ. 270, τῶν ᾿Αρχιμήδους μηχανών χρείαν ἔχω.
CHAPTER II.
MANUSCRIPTS AND PRINCIPAL EDITIONS-ORDER OF
COMPOSITION-DIALECT-LOST WORKS.
THE sources of the text and versions are very fully described
by Heiberg in the Prolegomena to Vol. III. of his edition of Archi-
medes, where the editor supplements and to some extent amends
what he had previously written on the same subject in his dis-
sertation entitled Quaestiones Archimedeae (1879). It will there-
fore suffice here to state briefly the main points of the discussion.
The MSS. of the best class all had a common origin in a MS.
which, so far as is known, is no longer extant. It is described
in one of the copies made from it (to be mentioned later and dating
from some time between A.D. 1499 and 1531) as 'most ancient'
(TaλαιoτάTOν), and all the evidence goes to show that it was written
as early as the 9th or 10th century. At one time it was in the
possession of George Valla, who taught at Venice between the
years 1486 and 1499; and many important inferences with regard
to its readings can be drawn from some translations of parts of
Archimedes and Eutocius made by Valla himself and published
in his book entitled de expetendis et fugiendis rebus (Venice, 1501).
It appears to have been carefully copied from an original belonging
to some one well versed in mathematics, and it contained figures
drawn for the most part with great care and accuracy, but there
was considerable confusion between the letters in the figures and
those in the text. This MS., after the death of Valla in 1499,
became the property of Albertus Pius Carpensis (Alberto Pio,
prince of Carpi). Part of his library passed through various hands
and ultimately reached the Vatican; but the fate of the Valla
MS. appears to have been different, for we hear of its being in
the possession of Cardinal Rodolphus Pius (Rodolfo Pio), a nephew
of Albertus, in 1544, after which it seems to have disappeared.
xxiv
INTRODUCTION.
The three most important MSS. extant are:
F (= Codex Florentinus bibliothecae Laurentianae Mediceae
plutei XXVIII. 4to.).
B (=Codex Parisinus 2360, olim Mediceus).
C (Codex Parisinus 2361, Fonteblandensis).
Of these it is certain that B was copied from the Valla MS.
This is proved by a note on the copy itself, which states that the
archetype formerly belonged to George Valla and afterwards to
Albertus Pius. From this it may also be inferred that B was
written before the death of Albertus in 1531; for, if at the date
of B the Valla MS. had passed to Rodolphus Pius, the name of
the latter would presumably have been mentioned. The note re-
ferred to also gives a list of peculiar abbreviations used in the
archetype, which list is of importance for the purpose of com-
parison with F and other MSS.
From a note on C it appears that that MS. was written by
one Christophorus Auverus at Rome in 1544, at the expense of
Georgius Armagniacus (Georges d'Armagnac), Bishop of Rodez,
then on a mission from King Francis I. to Pope Paul III. Further,
a certain Guilelmus Philander, in a letter to Francis I. published
in an edition of Vitruvius (1552), mentions that he was allowed,
by the kindness of Cardinal Rodolphus Pius, acting at the instance
of Georgius Armagniacus, to see and make extracts from a volume
of Archimedes which was destined to adorn the library founded
by Francis at Fontainebleau. He adds that the volume had been
the property of George Valla. We can therefore hardly doubt
that C was the copy which Georgius Armagniacus had made in
order to present it to the library at Fontainebleau.
Now F, B and C all contain the same works of Archimedes
and Eutocius, and in the same order, viz. (1) two Books de sphaera
et cylindro, (2) de dimensione circuli, (3) de conoidibus, (4) de
lineis spiralibus, (5) de planis aeque ponderantibus, (6) arenarius,
(7) quadratura parabolae, and the commentaries of Eutocius on
(1) (2) and (5). At the end of the quadratura parabolae both
F and B give the following lines:
εὐτυχοίης λέον γεωμετρα
πολλοὺς εἰς λυκάβαντας ἴοις πολὺ φίλτατε μούσαις.
F and C also contain mensurae from Heron and two fragments
περὶ σταθμῶν and περὶ μέτρων, the order being the sane in both
MANUSCRIPTS.
XXV
and the contents only differing in the one respect that the last
fragment Teρì μérpwv is slightly longer in F than in C.
A short preface to C states that the first page of the archetype
was so rubbed and worn with age that not even the name of
Archimedes could be read upon it, while there was no copy at
Rome by means of which the defect could be made good, and
further that the last page of Heron's de mensuris was similarly
obliterated. Now in F the first page was apparently left blank
at first and afterwards written in by a different hand with many
gaps, while in B there are similar deficiencies and a note attached
by the copyist is to the effect that the first page of the archetype
was indistinct. In another place (p. 4 of Vol. III., ed. Heiberg)
all three MSS. have the same lacuna, and the scribe of B notes
that one whole page or even two are missing.
Now C could not have been copied from F because the last
page of the fragment Tepi péтpov is perfectly distinct in F; and,
on the other hand, the archetype of F must have been illegible
at the end because there is no word réλos at the end of F, nor any
other of the signs by which copyists usually marked the completion
of their task. Again, Valla's translations show that his MS. had
certain readings corresponding to correct readings in B and C
instead of incorrect readings given by F. Hence F cannot have
been Valla's MS. itself.
The positive evidence about F is as follows. Valla's trans-
lations, with the exception of the few readings just referred to,
agree completely with the text of F. From a letter written at
Venice in 1491 by Angelus Politianus (Angelo Poliziano) to Lau-
rentius Mediceus (Lorenzo de' Medici), it appears that the former
had found a MS. at Venice containing works by Archimedes and
Heron and proposed to have it copied. As G. Valla then lived
at Venice, the MS. can hardly have been any other but his, and
no doubt F was actually copied from it in 1491 or soon after.
Confirmatory evidence for this origin of F is found in the fact
that the form of most of the letters in it is older than the 15th
century, and the abbreviations etc., while they all savour of an
ancient archetype, agree marvellously with the description which
the note to B above referred to gives of the abbreviations used
in Valla's MS. Further, it is remarkable that the corrupt passage
corresponding to the illegible first page of the archetype just takes
up one page of F, no more and no less.
2
xxvi
INTRODUCTION.
!
The natural inference from all the evidence is that F, B and
C all had their origin in the Valla MS.; and of the three F is
the most trustworthy. For (1) the extreme care with which the
copyist of F kept to the original is illustrated by a number of
mistakes in it which correspond to Valla's readings but are cor-
rected in B and C, and (2) there is no doubt that the writer of
B was somewhat of an expert and made many alterations on his
own authority, not always with success.
Passing to other MSS., we know that Pope Nicholas V. had
a MS. of Archimedes which he caused to be translated into Latin.
The translation was made by Jacobus Cremonensis (Jacopo Cas-
siani*), and one copy of this was written out by Joannes Regio-
montanus (Johann Müller of Königsberg, near Hassfurt, in Fran-
conia), about 1461, who not only noted in the margin a number
of corrections of the Latin but added also in many places Greek
readings from another MS. This copy by Regiomontanus is pre-
served at Nürnberg and was the source of the Latin translation
given in the editio princeps of Thomas Gechauff Venatorius (Basel,
1544); it is called Nb by Heiberg. (Another copy of the same
translation is alluded to by Regiomontanus, and this is doubtless
the Latin MS. 327 of 15th c. still extant at Venice.) From the
fact that the translation of Jacobus Cremonensis has the same
lacuna as that in F, B and C above referred to (Vol. III., ed.
Heiberg, p. 4), it seems clear that the translator had before him
either the Valla MS. itself or (more likely) a copy of it, though
the order of the books in the translation differs in one respect
from that in our MSS., viz. that the arenarius comes after instead
of before the quadratura parabolae.
It is probable that the Greek MS. used by Regiomontanus was V
(= Codex Venetus Marcianus cccv. of the 15th c.), which is still extant
and contains the same books of Archimedes and Eutocius with the
same fragment of Heron as F has, and in the same order. If the
above conclusion that F dates from 1491 or thereabouts is correct,
then, as V belonged to Cardinal Bessarione who died in 1472, it
cannot have been copied from F, and the simplest way of accounting
for its similarity to F is to suppose that it too was derived from
Valla's MS.
* Tiraboschi, Storia della Letteratura Italiana, Vol. vi. Pt. 1 (p. 358 of the
edition of 1807). Cantor (Vorlesungen üb. Gesch. d. Math., II. p. 192) gives the
full name and title as Jacopo da S. Cassiano Cremonese canonico regolare.
MANUSCRIPTS.
xxvii
Regiomontanus mentions, in a note inserted later than the
rest and in different ink, two other Greek MSS., one of which he
calls "exemplar vetus apud magistrum Paulum." Probably the
monk Paulus (Albertini) of Venice is here meant, whose date was
1430 to 1475; and it is possible that the "exemplar vetus" is
the MS. of Valla.
The two other inferior MSS., viz. A (= Codex Parisinus 2359,
olim Mediceus) and D (= Cod. Parisinus 2362, Fonteblandensis),
owe their origin to V.
It is next necessary to consider the probabilities as to the MSS.
used by Nicolas Tartaglia for his Latin translation of certain of
the works of Archimedes. The portion of this translation published
at Venice in 1543 contained the books de centris gravium vel de
aequerepentibus I-II, tetragonismus [parabolae], dimensio circuli
and de insidentibus aquae I; the rest, consisting of Book II de
insidentibus aquae, was published with Book I of the same treatise,
after Tartaglia's death in 1557, by Troianus Curtius (Venice, 1565).
Now the last-named treatise is not extant in any Greek MS. and,
as Tartaglia adds it, without any hint of a separate origin, to the
rest of the books which he says he took from a mutilated and
almost illegible Greek MS., it might easily be inferred that the
Greek MS. contained that treatise also. But it is established, by
a letter written by Tartaglia himself eight years later (1551) that
he then had no Greek text of the Books de insidentibus aquae, and
it would be strange if it had disappeared in so short a time without
leaving any trace. Further, Commandinus in the preface to his
edition of the same treatise (Bologna, 1565) shows that he had
never heard of a Greek text of it. Hence it is most natural to
suppose that it reached Tartaglia from some other source and in the
Latin translation only*.
The fact that Tartaglia speaks of the old MS. which he used
as "fracti et qui vix legi poterant libri," at practically the same
time as the writer of the preface to C was giving a similar de-
scription of Valla's MS., makes it probable that the two were
* The Greek fragment of Book I., περὶ τῶν ὕδατι ἐφισταμένων ἢ περὶ τῶν
¿xovμévwv, edited by A. Mai from two Vatican MSS. (Classici auct. 1. p. 426–30;
Vol. II. of Heiberg's edition, pp. 356-8), seems to be of doubtful authenticity.
Except for the first proposition, it contains enunciations only and no proofs.
Heiberg is inclined to think that it represents an attempt at retranslation into
Greek made by some mediaeval scholar, and he compares the similar attempt
made by Rivault.
*
1
xxviii
INTRODUCTION.
fr
!
identical; and this probability is confirmed by a considerable agree-
ment between the mistakes in Tartaglia and in Valla's versions.
But in the case of the quadratura parabolae and the dimensio
circuli Tartaglia adopted bodily, without alluding in any way to
the source of it, another Latin translation published by Lucas
Gauricus "Iuphanensis ex regno Neapolitano" (Luca Gaurico of
Gifuni) in 1503, and he copied it so faithfully as to reproduce most
obvious errors and perverse punctuation, only filling up a few
gaps and changing some figures and letters. This translation by
Gauricus is seen, by means of a comparison with Valla's readings
and with the translation of Jacobus Cremonensis, to have been
made from the same MS. as the latter, viz. that of Pope Nicolas V.
Even where Tartaglia used the Valla MS. he does not seem
to have taken very great pains to decipher it when it was
not easily legible-it may be that he was unused to deciphering
MSS.—and in such cases he did not hesitate to draw from other
sources. In one place (de planor. equilib. 11. 9) he actually
gives as the Archimedean proof a paraphrase of Eutocius some-
what retouched and abridged, and in many other instances he
has inserted corrections and interpolations from another Greek
MS. which he once names. This MS. appears to have been a copy
made from F, with interpolations due to some one not unskilled
in the subject-matter; and this interpolated copy of F was ap-
parently also the source of the Nürnberg MS. now to be mentioned.
Na (= Codex Norimbergensis) was written in the 16th century
and brought from Rome to Nürnberg by Wilibald Pirckheymer.
It contains the same works of Archimedes and Eutocius, and in
the same order, as F, but was evidently not copied from F direct,
while, on the other hand, it agrees so closely with Tartaglia's
version as to suggest a common origin. Na was used by Vena-
torius in preparing the editio princeps, and Venatorius corrected
many mistakes in it with his own hand by notes in the margin
or on slips attached thereto; he also made many alterations in
the body of it, erasing the original, and sometimes wrote on it
directions to the printer, so that it was probably actually used
to print from. The character of the MS. shows it to belong to
the same class as the others; it agrees with them in the more
important errors and in having a similar lacuna at the beginning.
Some mistakes common to it and F alone show that its source was
F, though at second hand, as above indicated.
EDITIONS AND TRANSLATIONS.
xxix
It remains to enumerate the principal editions of the Greek
text and the published Latin versions which are based, wholly or
partially, upon direct collation of the MSS. These are as follows,
in addition to Gaurico's and Tartaglia's translations.
1. The editio princeps published at Basel in 1544 by Thomas
Gechauff Venatorius under the title Archimedis opera quae quidem
exstant omnia nunc primum graece et latine in lucem edita. Adiecta
quoque sunt Eutocii Ascalonitae commentaria item graece et latine
nunquam antea excusa. The Greek text and the Latin version in
this edition were taken from different sources, that of the Greek
text being Na, while the translation was Joannes Regiomontanus'
revised copy (N) of the Latin version made by Jacobus Cremo-
nensis from the MS. of Pope Nicolas V. The revision by
Regiomontanus was effected by the aid of (1) another copy of
the same translation still extant, (2) other Greek MSS., one of
which was probably V, while another may have been Valla's MS.
itself.
2. A translation by F. Commandinus (containing the following
works, circuli dimensio, de lineis spiralibus, quadratura parabolae,
de conoidibus et sphaeroidibus, de arenae numero) appeared at
Venice in 1558 under the title Archimedis opera nonnulla in
latinum conversa et commentariis illustrata. For this translation
several MSS. were used, among which was V, but none preferable
to those which we now possess.
3. D. Rivault's edition, Archimedis opera quae exstant graece
et latine novis demonstr. et comment. illustr. (Paris, 1615), gives
only the propositions in Greek, while the proofs are in Latin and
somewhat retouched. Rivault followed the Basel editio princeps
with the assistance of B.
4. Torelli's edition (Oxford, 1792) entitled 'Apxundovs rà σw-
ζόμενα μετὰ τῶν Εὐτοκίου Ασκαλωνίτου ὑπομνημάτων, Archimedis
quae supersunt omnia cum Eutocii Ascalonitae commentariis ex
recensione J. Torelli Veronensis cum nova versione latina. Acced-
unt lectiones variantes ex codd. Mediceo et Parisiensibus. Torelli
followed the Basel editio princeps in the main, but also collated
V. The book was brought out after Torelli's death by Abram
Robertson, who added the collation of five more MSS., F, A, B, C, D,
with the Basel edition. The collation however was not well done,
and the edition was not properly corrected when in the press.
3
2
XXX
1
INTRODUCTION.
5. Last of all comes the definitive edition of Heiberg (Archi-
medis opera omnia cum commentariis Eutocii. E codice Florentino
recensuit, Latine uertit notisque illustrauit J. L. Heiberg. Leipzig,
1880-1).
The relation of all the MSS. and the above editions and trans-·
lations is well shown by Heiberg in the following scheme (with
the omission, however, of his own edition):
Codex Uallae saec. IX-X
[
Cod. Nicolai V
c. 1453
F
c. 1491
Tartalea
a. 1543
V
B
saec. XV
c. 1500
C
a. 1514

Cod. Tartaleae II
I
Na saec. XVI
Ed. Basil. 1544
Ed. Riualti
a. 1615
A, D
saec. XVI
Commandinus
1558
Torellius 1792
I
Gauricus
Cremonensis c. 1460
Cod. Uenet. 327 Nb, c. 1461
saec. XV
The remaining editions which give portions of Archimedes in
Greek, and the rest of the translations of the complete works or
parts of them which appeared before Heiberg's edition, were not
based upon any fresh collation of the original sources, though some
excellent corrections of the text were made by some of the editors,
notably Wallis and Nizze. The following books may be mentioned.
Joh. Chr. Sturm, Des unvergleichlichen Archimedis Kunstbücher,
übersetzt und erläutert (Nürnberg, 1670). This translation em-
braced all the works extant in Greek and followed three years
after the same author's separate translation of the Sand-reckoner.
It appears from Sturm's preface that he principally used the edition
of Rivault.
Is. Barrow, Opera Archimedis, Apollonii Pergaei conicorum libri,
Theodosii sphaerica methodo novo illustrata et demonstrata (London,
1675).
Wallis, Archimedis arenarius et dimensio circuli, Eutocii in hanc
commentarii cum versione et notis (Oxford, 1678), also given
in Wallis' Opera, Vol. 111. pp. 509–546.
Karl Friedr. Hauber, Archimeds zwei Bücher über Kugel und
Cylinder. Ebendesselben Kreismessung. Uebersetzt mit Anmerkungen
u. s. w. begleitet (Tübingen, 1798).
TRANSLATIONS-ORDER OF WORKS.
xxxi
F. Peyrard, Œuvres d'Archimède, traduites littéralement, avec
un commentaire, suivies d'un mémoire du traducteur, sur un nouveau
miroir ardent, et d'un autre mémoire de M. Delambre, sur l'arith-
métique des Grecs. (Second edition, Paris, 1808.)
Ernst Nizze, Archimedes von Syrakus vorhandene Werke, aus dem
Griechischen übersetzt und mit erläuternden und kritischen Anmer-
kungen begleitet (Stralsund, 1824).
The MSS. give the several treatises in the following order.
1. περὶ σφαίρας καὶ κυλίνδρου α' β', two Books On the Sphere
and Cylinder.
2. KÚKλOV μÉтpηois*, Measurement of a Circle.
3. περὶ κωνοειδέων καὶ σφαιροειδέων, On Conoids and Spheroids.
4. περὶ ἑλίκων, On Spirals.
5. éππédшv iσopроruv a B't, two Books On the Equilibrium
of Planes.
6. чaμμíτns, The Sand-reckoner.
7. τετραγωνισμός παραβολῆς (a name substituted later for that
given to the treatise by Archimedes himself, which must
undoubtedly have been τετραγωνισμὸς τῆς τοῦ ὀρθογωνίου
KÚVOV TOµĤs‡), Quadrature of the Parabola.
To these should be added
8. Teρì ỏxovμévov §, the Greek title of the treatise On floating
bodies, only preserved in a Latin translation.
* Pappus alludes (1. p. 312, ed. Hultsch) to the kúkλov µétρìjʊ…. ¡n the words
ἐν τῷ περὶ τῆς τοῦ κύκλου περιφερείας.
+ Archimedes himself twice alludes to properties proved in Book 1. as
demonstrated ev Toîs µnxavikoîs (Quadrature of the Parabola, Props. 6, 10).
Pappus (VIII. p. 1034) quotes тà'Apxiµýdovs teρì iσopporɩŵv. The beginning of
Book 1. is also cited by Proclus in his Commentary on Eucl. 1., p. 181, where the
reading should be τοῦ ἃ ἰσορροπιών, and not τῶν ἀνισορροπιών (Hultsch).
‡ The name ‘parabola' was first applied to the curve by Apollonius. Archi-
medes always used the old term 'section of a right-angled cone.' Cf. Eutocius
(Heiberg, vol. III., p. 342) δέδεικται ἐν τῷ περὶ τῆς τοῦ ὀρθογωνίου κώνου τομῆς.
§ This title corresponds to the references to the book in Strabo I. p. 54
(Αρχιμήδης ἐν τοῖς περὶ τῶν ὀχουμένων) and Pappus VIII. p. 1024 (ὡς ᾿Αρχιμήδης
¿xovμévois). The fragment edited by Mai has a longer title, Teρì Tŵv vdaтɩ
ἐφισταμένων ἢ περὶ τῶν ὀχουμένων, where the first part corresponds to Tartaglia's
version, de insidentibus aquae, and to that of Commandinus, de iis quae vehun-
tur in aqua. But Archimedes intentionally used the more general word vypóv
(fluid) instead of vowp; and hence the shorter title repì ¿xovµévwv, de iis quae
in humido vehuntur (Torelli and Heiberg), seems the better.
•
1
1
xxxii
INTRODUCTION.
I
4
I
I
1
The books were not, however, written in the above order; and
Archimedes himself, partly through his prefatory letters and partly
by the use in later works of properties proved in earlier treatises,
gives indications sufficient to enable the chronological sequence
to be stated approximately as follows:
On the equilibrium of planes, I.
- 2. Quadrature of the Parabola.
-3. On the equilibrium of planes, II.
•
On the Sphere and Cylinder, I, II.
-5. On Spirals.
.6. On Conoids and Spheroids.
- 7. On floating bodies, I, II.
- 8. Measurement of a circle.
9. The Sand-reckoner.
It should however be observed that, with regard to (7), no
more is certain than that it was written after (6), and with regard
to (8) no more than that it was later than (4) and before (9).
In addition to the above we have a collection of Lemmas (Liber
Assumptorum) which has reached us through the Arabic. The
collection was first edited by S. Foster, Miscellanea (London, 1659),
and next by Borelli in a book published at Florence, 1661, in
which the title is given as Liber assumptorum Archimedis interprete
Thebit ben Kora et exponente doctore Almochtasso Abilhasan. The
Lemmas cannot, however, have been written by Archimedes in
their present form, because his name is quoted in them more than
once. The probability is that they were propositions collected by
some Greek writer* of a later date for the purpose of elucidating
some ancient work, though it is quite likely that some of the
propositions were of Archimedean origin, e.g. those concerning
the geometrical figures called respectively apßnλost (literally
It would seem that the compiler of the Liber Assumptorum must have
drawn, to a considerable extent, from the same sources as Pappus. The
number of propositions appearing substantially in the same form in both
collections is, I think, even greater than has yet been noticed. Tannery (La
Géométrie grecque, p. 162) mentions, as instances, Lemmas 1, 4, 5, 6; but it
will be seen from the notes in this work that there are several other coin-
cidences.
+ Pappus gives (p. 208) what he calls an 'ancient proposition' (ȧpxala
πρότασις) about the same figure, which he describes as χωρίον, δ δὴ καλοῦσιν
ǎрßηλov. Cf. the note to Prop. 6 (p. 308). The meaning of the word is gathered
WORKS ASCRIBED TO ARCHIMEDES.
xxxiii
'shoemaker's knife') and σáλivov (probably a 'salt-cellar'*), and
Prop. 8 which bears on the problem of trisecting an angle.
from the Scholia to Nicander, Theriaca, 423: äpßnλoi Xéyovrai tà kukλotepĤ
σιδήρια, οἷς οἱ σκυτοτόμοι τέμνουσι καὶ ξύουσι τὰ δέρματα. Cf. Hesychius,
ἀνάρβηλα, τὰ μὴ ἐξεσμένα δέρματα· ἄρβηλοι γὰρ τὰ σμιλία.
* The best authorities appear to hold that in any case the name σáλivov was
not applied to the figure in question by Archimedes himself but by some later
writer. Subject to this remark, I believe σáλwov to be simply a Graecised
form of the Latin word salinum. We know that a salt-cellar was an essential
part of the domestic apparatus in Italy from the early days of the Roman
Republic. "All who were raised above poverty had one of silver which
descended from father to son (Hor., Carm. 11. 16, 13, Liv. xxvi. 36), and
was accompanied by a silver patella which was used together with the salt-
cellar in the domestic sacrifices (Pers. III. 24, 25). These two articles of
silver were alone compatible with the simplicity of Roman manners in the
early times of the Republic (Plin., H. N. xxxIII. § 153, Val. Max. iv. 4, § 3).
...In shape the salinum was probably in most cases a round shallow bowl”
[Dict. of Greek and Roman Antiquities, article salinum]. Further we have
in the early chapters of Mommsen's History of Rome abundant evidence
of similar transferences of Latin words to the Sicilian dialect of Greek. Thus
(Book 1., ch. xiii.) it is shown that, in consequence of Latino-Sicilian com-
merce, certain words denoting measures of weight, libra, triens, quadrans,
sextans, uncia, found their way into the common speech of Sicily in the third
century of the city under the forms λίτρα, τριᾶς, τετρᾶς, ἑξᾶς, οὐγκία. Similarly
Latin law-terms (ch. xi.) were transferred; thus mutuum (a form of loan)
became μoîтov, carcer (a prison) κáρкαрov. Lastly, the Latin word for lard,
arvina, became in Sicilian Greek åpßívn, and patina (a dish) warávŋ. The last
word is as close a parallel for the supposed transfer of salinum as could be
wished. Moreover the explanation of σáλwov as salinum has two obvious
advantages in that (1) it does not require any alteration in the word, and
4
་
ľ
*
』

(2) the resemblance of the lower curve to an ordinary type of salt-cellar is
evident. I should add, as confirmation of my hypothesis, that Dr A. S. Murray,
of the British Museum, expresses the opinion that we cannot be far wrong in
accepting as a salinum one of the small silver bowls in the Roman ministerium
H. A.
C
xxxiv
INTRODUCTION.
Archimedes is further credited with the authorship of the
Cattle-problem enunciated in the epigram edited by Lessing in
1773. According to the heading prefixed to the epigram it was
communicated by Archimedes to the mathematicians at Alexandria
in a letter to Eratosthenes*. There is also in the Scholia to Plato's
Charmides 165 E a reference to the problem "called by Archimedes
the Cattle-problem” (τὸ κληθὲν ὑπ᾽ ᾿Αρχιμήδους βοεικόν πρόβλημα).
The question whether Archimedes really propounded the problem,
or whether his name was only prefixed to it in order to mark the
extraordinary difficulty of it, has been much debated. A complete
account of the arguments for and against is given in an article
by Krumbiegel in the Zeitschrift für Mathematik und Physik
(Hist. litt. Abtheilung) xxv. (1880), p. 121 sq., to which Amthor
added (ibid. p. 153 sq.) a discussion of the problem itself. The
general result of Krumbiegel's investigation is to show (1) that
at the Museum which was found at Chaourse (Aisne) in France and is of a
section sufficiently like the curve in the Salinon.
The other explanations of σáλvov which have been suggested are as follows.
(1) Cantor connects it with σáλos, "das Schwanken des hohen Meeres,"
and would presumably translate it as wave-line. But the resemblance is
not altogether satisfactory, and the termination -wov would need explanation.
(2) Heiberg says the word is "sine dubio ab Arabibus deprauatum," and
suggests that it should be σéλivov, parsley (“ex similitudine frondis apii").
But, whatever may be thought of the resemblance, the theory that the word is
corrupted is certainly not supported by the analogy of äpßŋλos which is correctly
reproduced by the Arabs, as we know from the passage of Pappus referred to in
the last note.
(3) Dr Gow suggests that σálivov may be a ‘sieve,' comparing σáλağ. But
this guess is not supported by any evidence.
* The heading is, Πρόβλημα ὅπερ ᾿Αρχιμήδης ἐν ἐπιγράμμασιν εὑρὼν τοῖς ἐν
᾿Αλεξανδρείᾳ περὶ ταῦτα πραγματευομένοις ζητεῖν ἀπέστειλεν ἐν τῇ πρὸς Ερατοσθένην
τὸν Κυρηναῖον ἐπιστολῇ. Heiberg translates this as "the problem which
Archimedes discovered and sent in an epigram...in a letter to Eratosthenes.”
He admits however that the order of words is against this, as is also the use of
the plural ἐπιγράμμασιν. It is clear that to take the two expressions ἐν
ἐπιγράμμασιν and ἐν ἐπιστολῇ as both following ἀπέστειλεν is very awkward. In
fact there seems to be no alternative but to translate, as Krumbiegel does, in
accordance with the order of the words, "a problem which Archimedes found
among (some) epigrams and sent...in his letter to Eratosthenes "; and this sense
is certainly unsatisfactory. Hultsch remarks that, though the mistake πραγ
ματουμένοις for πραγματευομένοις and the composition of the heading as a whole
betray the hand of a writer who lived some centuries after Archimedes, yet he
must have had an earlier source of information, because he could hardly have
invented the story of the letter to Eratosthenes.
12
WORKS ASCRIBED TO ARCHIMEDES.
XXXV
the epigram can hardly have been written by Archimedes in its
present form, but (2) that it is possible, nay probable, that the
problem was in substance originated by Archimedes. Hultsch* has
an ingenious suggestion as to the occasion of it. It is known that
Apollonius in his wкUTÓKLOV had calculated a closer approximation to
the value of π than that of Archimedes, and he must therefore have
worked out more difficult multiplications than those contained in
the Measurement of a circle. Also the other work of Apollonius
on the multiplication of large numbers, which is partly preserved
in Pappus, was inspired by the Sand-reckoner of Archimedes; and,
though we need not exactly regard the treatise of Apollonius as
polemical, yet it did in fact constitute a criticism of the earlier
book. Accordingly, that Archimedes should then reply with a
problem which involved such a manipulation of immense numbers
as would be difficult even for Apollonius is not altogether outside
the bounds of possibility. And there is an unmistakable vein of
satire in the opening words of the epigram "Compute the number
of the oxen of the Sun, giving thy mind thereto, if thou hast a
share of wisdom," in the transition from the first part to the
second where it is said that ability to solve the first part would
entitle one to be regarded as "not unknowing nor unskilled in
numbers, but still not yet to be numbered among the wise,” and
again in the last lines. Hultsch concludes that in any case the
problem is not much later than the time of Archimedes and dates
from the beginning of the 2nd century B.C. at the latest.
Of the extant books it is certain that in the 6th century A.D.
only three were generally known, viz. On the Sphere and Cylinder,
the Measurement of a circle, and On the equilibrium of planes. Thus
Eutocius of Ascalon who wrote commentaries on these works only
knew the Quadrature of the Parabola by name and had never seen
it nor the book On Spirals. Where passages might have been
elucidated by references to the former book, Eutocius gives ex-
planations derived from Apollonius and other sources, and he
speaks vaguely of the discovery of a straight line equal to the
circumference of a given circle "by means of certain spirals,"
whereas, if he had known the treatise On Spirals, he would have
quoted Prop. 18. There is reason to suppose that only the three
treatises on which Eutocius commented were contained in the
* Pauly-Wissowa's Real-Encyclopädie, 11. 1, pp. 534, 5.
}
c 2
xxxvi
INTRODUCTION.
ordinary editions of the time such as that of Isidorus of Miletus,
the teacher of Eutocius, to which the latter several times alludes.
In these circumstances the wonder is that so many more books
have survived to the present day. As it is, they have lost to a
considerable extent their original form. Archimedes wrote in the
Doric dialect*, but in the best known books (On the Sphere and
Cylinder and the Measurement of a circle) practically all traces
of that dialect have disappeared, while a partial loss of Doric forms
has taken place in other books, of which however the Sand-
reckoner has suffered least. Moreover in all the books, except the
Sand-reckoner, alterations and additions were first of all made by
an interpolator who was acquainted with the Doric dialect, and
then, at a date subsequent to that of Eutocius, the book On the
Sphere and Cylinder and the Measurement of a circle were completely
recast.
Of the lost works of Archimedes the following can be identified.
1. Investigations relating to polyhedra are referred to by
Pappus who, after alluding (v. p. 352) to the five regular polyhedra,
gives a description of thirteen others discovered by Archimedes
which are semi-regular, being contained by polygons equilateral
and equiangular but not similar.
2. A book of arithmetical content, entitled apxaí Principles
and dedicated to Zeuxippus. We learn from Archimedes himself
that the book dealt with the naming of numbers (Kaтovóμatis TŵV
apiμov)† and expounded a system of expressing numbers higher
* Thus Eutocius in his commentary on Prop. 4 of Book 1. On the Sphere
and Cylinder speaks of the fragment, which he found in an old book and which
appeared to him to be the missing supplement to the proposition referred to,
as "preserving in part Archimedes' favourite Doric dialect" (èv µépei dè Tĥv
᾿Αρχιμήδει φίλην Δωρίδα γλῶσσαν ἀπέσωζον). From the use of the expression ἐν
μέρει
μépeɩ Heiberg concludes that the Doric forms had by the time of Eutocius
begun to disappear in the books which have come down to us no less than in
the fragment referred to.
+ Observing that in all the references to this work in the Sand-reckoner
Archimedes speaks of the naming of numbers or of numbers which are named or have
their names (ἀριθμοὶ κατωνομασμένοι, τὰ ὀνόματα ἔχοντες, τὰν κατονομαξίαν ἔχοντες),
Hultsch (Pauly-Wissowa's Real-Encyclopädie, 11. 1, p. 511) speaks of katovó-
μαξις τῶν ἀριθμών as the name of the work; and he explains the words τινὰς τῶν
ἐν ἀρχαῖς < ἀριθμῶν > τῶν κατονομαξίαν ἐχόντων as meaning some of the
numbers mentioned at the beginning which have a special name," where "at
the beginning" refers to the passage in which Archimedes first mentions Tŵv
L
LOST WORKS.
xxxvii
than those which could be expressed in the ordinary Greek no-
tation. This system embraced all numbers up to the enormous
figure which we should now represent by a 1 followed by 80,000
billion ciphers; and, in setting out the same system in the Sand-
reckoner, Archimedes explains that he does so for the benefit of
those who had not had the opportunity of seeing the earlier work
addressed to Zeuxippus.
3. πeρì (vyŵv, On balances or levers, in which Pappus says (VIII.
p. 1068) that Archimedes proved that "greater circles overpower
(Kaтaкρaтovσi) lesser circles when they revolve about the same
centre." It was doubtless in this book that Archimedes proved
the theorem assumed by him in the Quadrature of the Parabola,
Prop. 6, viz. that, if a body hangs at rest from a point, the centre
of gravity of the body and the point of suspension are in the same
vertical line.
4. KEVтρоẞαρiká, On centres of gravity. This work is mentioned
by Simplicius on Aristot. de caelo 11. (Scholia in Arist. 508 a 30).
Archimedes may be referring to it when he says (On the equilibrium
of planes 1. 4) that it has before been proved that the centre of
gravity of two bodies taken together lies on the line joining the
centres of gravity of the separate bodies. In the treatise On
floating bodies Archimedes assumes that the centre of gravity of a
segment of a paraboloid of revolution is on the axis of the segment
at a distance from the vertex equal to 3rds of its length. This
may perhaps have been proved in the KEVтpoßaρiká, if it was
not made the subject of a separate work.
Doubtless both the περὶ ζυγῶν and the κεντροβαρικά preceded
the extant treatise On the equilibrium of planes.
5. κатожтριкά, an optical work, from which Theon (on Ptolemy,
Synt. 1. p. 29, ed. Halma) quotes a remark about refraction.
Cf. Olympiodorus in Aristot. Meteor., II. p. 94, ed. Ideler.
ὑφ' ἡμῶν κατωνομασμένων ἀριθμῶν καὶ ἐνδεδομένων ἐν τοῖς ποτὶ Ζεύξιππον γεγραμ-
μévois. But év ȧpxaîs seems a less natural expression for "at the beginning"
than ἐν ἀρχῇ οι κατ᾿ ἀρχάς would have been. Moreover, there being no
participial expression except κατονομαξίαν ἐχόντων to be taken with ἐν ἀρχαῖς in
this sense, the meaning would be unsatisfactory; for the numbers are not
named at the beginning, but only referred to, and therefore some word like
eipnμévwv should have been used. For these reasons I think that Heiberg,
Cantor and Susemihl are right in taking åpɣal to be the name of the treatise.
t
1
xxxviii
INTRODUCTION.
6. Teρì σpaιpoжоtas, On sphere-making, a mechanical work on
the construction of a sphere representing the motions of the
heavenly bodies as already mentioned (p. xxi).
7. ¿pódiov, a Method, noticed by Suidas, who says that Theo-
dosius wrote a commentary on it, but gives no further information
about it.
8. According to Hipparchus Archimedes must have written
on the Calendar or the length of the year (cf. p. xxi).
Some Arabian writers attribute to Archimedes works (1) On
a heptagon in a circle, (2) On circles touching one another, (3) On
parallel lines, (4) On triangles, (5) On the properties of right-
angled triangles, (6) a book of Data; but there is no confirmatory
evidence of his having written such works. A book translated
into Latin from the Arabic by Gongava (Louvain, 1548) and en-
titled antiqui scriptoris de speculo comburente concavitatis parabolae
cannot be the work of Archimedes, since it quotes Apollonius.
Į
མ་
CHAPTER III.
THE RELATION OF ARCHIMEDES TO HIS PREDECESSORS.
An extraordinarily large proportion of the subject matter of
the writings of Archimedes represents entirely new discoveries of
his own. Though his range of subjects was almost encyclopaedic,
embracing geometry (plane and solid), arithmetic, mechanics, hydro-
statics and astronomy, he was no compiler, no writer of text-
books; and in this respect he differs even from his great successor
Apollonius, whose work, like that of Euclid before him, largely
consisted of systematising and generalising the methods used, and
the results obtained, in the isolated efforts of earlier geometers.
There is in Archimedes no mere working-up of existing materials ;
his objective is always some new thing, some definite addition to
the sum of knowledge, and his complete originality cannot fail
to strike any one who reads his works intelligently, without any
corroborative evidence such as is found in the introductory letters
prefixed to most of them. These introductions, however, are emi-
nently characteristic of the man and of his work; their directness
and simplicity, the complete absence of egoism and of any effort
to magnify his own achievements by comparison with those of
others or by emphasising their failures where he himself succeeded:
all these things intensify the same impression. Thus his manner
is to state simply what particular discoveries made by his pre-
decessors had suggested to him the possibility of extending them
in new directions; e.g. he says that, in connexion with the efforts
of earlier geometers to square the circle and other figures, it
occurred to him that no one had endeavoured to square a parabola,
and he accordingly attempted the problem and finally solved it.
In like manner, he speaks, in the preface of his treatise On the
xl
INTRODUCTION.
{
1
1
Sphere and Cylinder, of his discoveries with reference to those
solids as supplementing the theorems about the pyramid, the cone
and the cylinder proved by Eudoxus. He does not hesitate to
say that certain problems baffled him for a long time, and that
the solution of some took him many years to effect; and in one
place (in the preface to the book On Spirals) he positively insists,
for the sake of pointing a moral, on specifying two propositions
which he had enunciated and which proved on further investigation
to be wrong. The same preface contains a generous eulogy of
Conon, declaring that, but for his untimely death, Conon would
have solved certain problems before him and would have enriched
geometry by many other discoveries in the meantime.
In some of his subjects Archimedes had no fore-runners, e.g.
in hydrostatics, where he invented the whole science, and (so
far as mathematical demonstration was concerned) in his me-
chanical investigations. In these cases therefore he had, in laying
the foundations of the subject, to adopt a form more closely re-
sembling that of an elementary textbook, but in the later parts
he at once applied himself to specialised investigations.
Thus the historian of mathematics, in dealing with Archimedes'
obligations to his predecessors, has a comparatively easy task before
him. But it is necessary, first, to give some description of the use
which Archimedes made of the general methods which had found
acceptance with the earlier geometers, and, secondly, to refer to
some particular results which he mentions as having been previously
discovered and as lying at the root of his own investigations, or
which he tacitly assumes as known.
§ 1. Use of traditional geometrical methods.
In my edition of the Conics of Apollonius*, I endeavoured,
following the lead given in Zeuthen's work, Die Lehre von den
Kegelschnitten im Altertum, to give some account of what has been
fitly called the geometrical algebra which played such an important
part in the works of the Greek geometers. The two main methods
included under the term were (1) the use of the theory of pro-
portions, and (2) the method of application of areas, and it was
shown that, while both methods are fully expounded in the Elements
of Euclid, the second was much the older of the two, being
attributed by the pupils of Eudemus (quoted by Proclus) .to the
Apollonius of Perga, pp. ci sqq.
RELATION OF ARCHIMEDES TO HIS PREDECESSORS.
xli
Pythagoreans. It was pointed out that the application of areas,
as set forth in the second Book of Euclid and extended in the
sixth, was made by Apollonius the means of expressing what he
takes as the fundamental properties of the conic sections, namely
the properties which we express by the Cartesian equations
y² = px,
y² = px + = x²,
2x²,
referred to any diameter and the tangent at its extremity as axes;
and the latter equation was compared with the results obtained in the
27th, 28th and 29th Props. of Euclid's Book VI, which are equivalent
to the solution, by geometrical means, of the quadratic equations
b
ax + − x² = D.
с
It was also shown that Archimedes does not, as a rule, connect his
description of the central conics with the method of application of
areas, as Apollonius does, but that Archimedes generally expresses
the fundamental property in the form of a proportion
y²
y's
X.X1
x.x
y³
b2
X.X1
α
a²,
and, in the case of the ellipse,
where x, x, are the abscissae measured from the ends of the diameter
of reference.
It results from this that the application of areas is of much less
frequent occurrence in Archimedes than in Apollonius. It is
however used by the former in all but the most general form. The
simplest form of "applying a rectangle" to a given straight line
which shall be equal to a given area occurs e.g. in the proposition On
the equilibrium of Planes II. 1; and the same mode of expression
is used (as in Apollonius) for the property y=px in the parabola,
px being described in Archimedes' phrase as the rectangle "applied
tο” (πаρаñíπтоν Taρá) a line equal to p and "having at its width”
(πλáros ĕxov) the abscissa (x). Then in Props. 2, 25, 26, 29 of the
book On Conoids and Spheroids we have the complete expression
which is the equivalent of solving the equation
ax + x³ = b²,
"let a rectangle be applied (to a certain straight line) exceeding by
xlii
INTRODUCTION.
a square figure (παραπεπτωκέτω χωρίον ὑπερβάλλον εἴδει τετραγώνῳ)
and equal to (a certain rectangle)." Thus a rectangle of this sort
has to be made (in Prop. 25) equal to what we have above called
x.x, in the case of the hyperbola, which is the same thing as
x (α + x) or ax+x³, where a is the length of the transverse axis.
But, curiously enough, we do not find in Archimedes the application
of a rectangle "falling short by a square figure," which we should
obtain in the case of the ellipse if we substituted x (α-x) for x. x₁-
In the case of the ellipse the area x.x is represented (On Conoids
and Spheroids, Prop. 29) as a gnomon which is the difference
between the rectangle h. h (where h, h₁ are the abscissae of the
ordinate bounding a segment of an ellipse) and a rectangle applied
to h₁-h and exceeding by a square figure whose side is h - x; and
the rectangle h. h₁ is simply constructed from the sides h, h. Thus
Archimedes avoids the application of a rectangle falling short by a
square, using for x.x, the rather complicated form
h.h₁ - {(h₂-h) (h − x) + (h−x)}.
-
It is easy to see that this last expression is equal to x. x₁, for it
reduces to
h. h₁ — {h¸ (h — x) − x (h − x)}
= x (h₁+ h) − x²,
= ax − x², since h₁+h=a,
-
= x.X1·
It will readily be understood that the transformation of rectangles
and squares in accordance with the methods of Euclid, Book II, is
just as important to Archimedes as to other geometers, and there is
no need to enlarge on that form of geometrical algebra.
The theory of proportions, as expounded in the fifth and sixth
Books of Euclid, including the transformation of ratios (denoted by
the terms componendo, dividendo, etc.) and the composition or
multiplication of ratios, made it possible for the ancient geometers
to deal with magnitudes in general and to work out relations
between them with an effectiveness not much inferior to that of
modern algebra. Thus the addition and subtraction of ratios could
be effected by procedure equivalent to what we should in algebra
* The object of Archimedes was no doubt to make the Lemma in Prop. 2
(dealing with the summation of a series of terms of the form a .rx+ (rx)², where r
successively takes the values 1, 2, 3,...) serve for the hyperboloid of revolution
and the spheroid as well.
4
RELATION OF ARCHIMEDES TO HIS PREDECESSORS. xliii
call bringing to a common denominator. Next, the composition or
multiplication of ratios could be indefinitely extended, and hence
the algebraical operations of multiplication and division found easy
and convenient expression in the geometrical algebra. As a par-
ticular case, suppose that there is a series of magnitudes in continued
proportion (i.e. in geometrical progression) as do, α₁, αz, ... αn, so that
αo
01
An-1
а1
a2
an
We have then, by multiplication,
ап
ао
n
(a)",
а1
or
αo
It is easy to understand how powerful such a method as that of
proportions would become in the hands of an Archimedes, and a few
instances are here appended in order to illustrate the mastery with
which he uses it.
1. A good example of a reduction in the order of a ratio after
the manner just shown is furnished by On the equilibrium of Planes
II. 10. Here Archimedes has a ratio which we will call a³/b³, where
a²/b²=c/d; and he reduces the ratio between cubes to a ratio
between straight lines by taking two lines x, y such that
It follows from this that
or
X
d
18
018
()
8170
с
d
C
х
Y
a²
b2
•
and hence
a³
b3
3
с
x d
C
•
х y Y
2. In the last example we have an instance of the use of
auxiliary fixed lines for the purpose of simplifying ratios and
thereby, as it were, economising power in order to grapple the more
successfully with a complicated problem. With the aid of such
auxiliary lines or (what is the same thing) auxiliary fixed points in
a figure, combined with the use of proportions, Archimedes is able to
effect some remarkable eliminations.
Thus in the proposition On the Sphere and Cylinder II. 4 he obtains
three relations connecting three as yet undetermined points, and
xliv
+
1
INTRODUCTION.
proceeds at once to eliminate two of the points, so that the problem
is then reduced to finding the remaining point by means of one
equation. Expressed in an algebraical form, the three original
relations amount to the three equations
За X
Y
2α- x
x
a + x
Ꮳ
2a
2
א
2α- x
Y
m
N
Z
and the result, after the elimination of y and z, is stated by
Archimedes in a form equivalent to
m + n a + x
n
4a2
α (2α — x)²°
Again the proposition On the equilibrium of Planes II. 9 proves
by the same method of proportions that, if a, b, c, d, x, y, are straight
lines satisfying the conditions
α
b
с
b
d
с
X
(a>b>c>d)
a - d˜¯¯ 3 ( a − c ) '
2a + 4b + 6c+ 3d
Y
and
then
5a + 106 + 10c + 5d
x + y = za.
a
The proposition is merely brought in as a subsidiary lemma to the
proposition following, and is not of any intrinsic importance; but a
glance at the proof (which again introduces an auxiliary line) will
show that it is a really extraordinary instance of the manipulation
of proportions.
3. Yet another instance is worth giving here. It amounts to
the proof that, if
then
x²
y²
+
a²
2a + x
a + x
y² (a− x) +
b2
2α-x
1,
'—— * • y² (a + x) = 4ab².
a-x
A, A' are the points of contact of two parallel tangent planes to a
spheroid; the plane of the paper is the plane through AA' and the
RELATION OF ARCHIMEDES TO HIS PREDECESSORS.
xlv
axis of the spheroid, and PP' is the intersection of this plane with
another plane at right angles to it (and therefore parallel to the
tangent planes), which latter plane divides the spheroid into two
segments whose axes are AN, A’N. Another plane is drawn through
B
P

c x
Y
A
N
the centre and parallel to the tangent plane, cutting the spheroid
into two halves. Lastly cones are drawn whose bases are the
sections of the spheroid by the parallel planes as shown in the
figure.
Archimedes' proposition takes the following form [On Conoids
and Spheroids, Props. 31, 32].
APP' being the smaller segment of the two whose common base
is the section through PP', and x, y being the coordinates of P,
he has proved in preceding propositions that
and
(volume of) segment APP'
(volume of) cone APP'
2α + x
a + x
half spheroid ABB'
cone ABB'
=
2;...
and he seeks to prove that
segment A'PP'
cone A'PP'
2α - x
α
х
The method is as follows.
We have
cone ABB'
α
b2
cone APP'
a - x˚ y²
α
α
α
a²
- xa² - x²·
Z
α
If we suppose
the ratio of the cones becomes
α
α
za
a² — x²°
(a),
(B),
(Y),
xlvi
INTRODUCTION.
Next, by hypothesis (a),
cone APP'
a + x
segmt. APP'— 2a + x
Therefore, ex aequali,
cone ABB'
It follows from (B) that
κα
segmt. APP' ~ (a − x) (2a + x) *
spheroid
4za
segmt. APP' (α − x) (2α + x)'

whence
segmt. A'PP
4za — (a − x) (2a +x)
segmt. APP'
(a− x) (2α + x)
z (2a−x)+(2a+x) ( z − a − x
(a− x) (2α + x)
Now we have to obtain the ratio of the segment A'PP' to the cone
A'PP', and the comparison between the segment APP' and the cone
A'PP' is made by combining two ratios ex aequali. Thus
segmt. APP'
cone APP'
2a + x
"
by (a),
a + x
cone APP'
α
-
XC
and
cone A'PP'= a + x
Thus combining the last three proportions, ex aequali, we have
since
α
segmt. A'PP' __ ≈ (2α − x) + (2α + x) (≈ − a − x
cone A'PP'
a² + 2αx + x²
-a-x)
≈ (2α − x)+(2a + x) (z — a
z (a − x) + (2a + x) x
a² = z (a − x), by (y).
[The object of the transformation of the numerator and denominator
of the last fraction, by which z (2a - x) and ≈ (a−x) are made the
first terms, is now obvious, because
2a х
——
a — x
x.
is the fraction which
Archimedes wishes to arrive at, and, in order to prove that the
required ratio is equal to this, it is only necessary to show that
2α- x
z − (a− x)
a-x
XC
•]
RELATION OF ARCHIMEDES TO HIS PREDECESSORS. xlvii
Now
so that
2α- x
a - x
= 1 +
α
α
-
XC
= 1 + 2, by (y),
a
a+z
α
≈ — (α − x)
х
(dividendo),
segmt. A'PP'
2a
XC
cone A'PP'
а
XC
4. One use by Euclid of the method of proportions deserves
mention because Archimedes does not use it in similar circumstances.
Archimedes (Quadrature of the Parabola, Prop. 23) sums a particular
geometric series
a + a ( 1 ) + a ( 1 )² +
...
+ a ( 1 ) n − ¹
in a manner somewhat similar to that of our text-books, whereas
Euclid (1x. 35) sums any geometric series of any number of terms by
means of proportions thus.
Suppose a, a, ... an, an+1 to be (n + 1) terms of a geometric
series in which an+1 is the greatest term.
Then
An+1
an
an
an_1 An-2
An-1
a2
α1
Therefore
An+1- an An - An-1
ап
ап
An-1
—
а1
Adding all the antecedents and all the consequents, we have
an+1-α1
A2-A1
A1 + A2 + Az +
...
+ an
1
which gives the sum of n terms of the series.
§ 2. Earlier discoveries affecting quadrature and cuba-
ture.
Archimedes quotes the theorem that circles are to one another as
the squares on their diameters as having being proved by earlier
geometers, and he also says that it was proved by means of a certain
lemma which he states as follows: "Of unequal lines, unequal
surfaces, or unequal solids, the greater exceeds the less by such a
magnitude as is capable, if added [continually] to itself, of exceeding
-
xlviii
INTRODUCTION.
any given magnitude of those which are comparable with one another
(τῶν πρὸς ἄλληλα λεγομένων).” We know that Hippocrates of Chios
proved the theorem that circles are to one another as the squares on
their diameters, but no clear conclusion can be established as to the
method which he used. On the other hand, Eudoxus (who is
mentioned in the preface to The Sphere and Cylinder as having
proved two theorems in solid geometry to be mentioned presently)
is generally credited with the invention of the method of exhaustion
by which Euclid proves the proposition in question in XII. 2. The
lemma stated by Archimedes to have been used in the original proof
is not however found in that form in Euclid and is not used in the
proof of XII. 2, where the lemma used is that proved by him in
x. 1, viz. that “Given two unequal magnitudes, if from the greater
[a part] be subtracted greater than the half, if from the remainder
[a part] greater than the half be subtracted, and so on continually,
there will be left some magnitude which will be less than the lesser
given magnitude." This last lemma is frequently assumed by
Archimedes, and the application of it to equilateral polygons in-
scribed in a circle or sector in the manner of XII. 2 is referred to as
having been handed down in the Elements*, by which it is clear
that only Euclid's Elements can be meant. The apparent difficulty
caused by the mention of two lemmas in connexion with the theorem
in question can, however, I think, be explained by reference to
the proof of x. 1 in Euclid. He there takes the lesser magnitude
and says that it is possible, by multiplying it, to make it some time
exceed the greater, and this statement he clearly bases on the 4th
definition of Book v. to the effect that "magnitudes are said to bear
a ratio to one another, which can, if multiplied, exceed one another."
Since then the smaller magnitude in x. 1 may be regarded as the
difference between some two unequal magnitudes, it is clear that the
lemma first quoted by Archimedes is in substance used to prove the
lemma in x. 1 which appears to play so much larger a part in the in-
vestigations in quadrature and cubature which have come down to us.
The two theorems which Archimedes attributes to Eudoxus
by namet are
(1) that any pyramid is one third part of the prism which has
the same base as the pyramid and equal height, and
* On the Sphere and Cylinder, 1. 6.
+ ibid. Preface.
୮
RELATION OF ARCHIMEDES TO HIS PREDECESSORS.
xlix
(2) that any cone is one third part of the cylinder which has
the same base as the cone and equal height.
The other theorems in solid geometry which Archimedes quotes
as having been proved by earlier geometers are*:
(3) Cones of equal height are in the ratio of their bases, and
conversely.
(4) If a cylinder be divided by a plane parallel to the base,
cylinder is to cylinder as axis to axis.
(5) Cones which have the same bases as cylinders and equal
height with them are to one another as the cylinders.
(6) The bases of equal cones are reciprocally proportional to
their heights, and conversely.
(7) Cones the diameters of whose bases have the same ratio as
their axes are in the triplicate ratio of the diameters of their bases.
In the preface to the Quadrature of the Parabola he says
that earlier geometers had also proved that
(8) Spheres have to one another the triplicate ratio of their
diameters; and he adds that this proposition and the first of those
which he attributes to Eudoxus, numbered (1) above, were proved
by means of the same lemma, viz. that the difference between
any two unequal magnitudes can be so multiplied as to exceed
any given magnitude, while (if the text of Heiberg is right) the
second of the propositions of Eudoxus, numbered (2), was proved
by means of "a lemma similar to that aforesaid." As a matter
of fact, all the propositions (1) to (8) are given in Euclid's twelfth
Book, except (5), which, however, is an easy deduction from (2);
and (1), (2), (3), and (7) all depend upon the same lemma [x. 1]
as that used in Eucl. XII. 2.
The proofs of the above seven propositions, excluding (5), as
given by Euclid are too long to quote here, but the following sketch
will show the line taken in the proofs and the order of the propo-
sitions. Suppose ABCD to be a pyramid with a triangular base,
and suppose it to be cut by two planes, one bisecting AB, AC,
AD in F, G, E respectively, and the other bisecting BC, BD, BA
in H, K, F respectively. These planes are then each parallel to
one face, and they cut off two pyramids each similar to the original
* Lemmas placed between Props. 16 and 17 of Book 1. On the Sphere and
Cylinder.
H. A.
d
i
1
INTRODUCTION.
pyramid and equal to one another, while the remainder of the
pyramid is proved to form two equal prisms which, taken together,
A

B
H
F
K
G
E
D
are greater than one half of the original pyramid [XII. 3]. It is
next proved [XII. 4] that, if there are two pyramids with triangular
bases and equal height, and if they are each divided in the
manner shown into two equal pyramids each similar to the whole
and two prisms, the sum of the prisms in one pyramid is to the
sum of the prisms in the other in the ratio of the bases of the
whole pyramids respectively. Thus, if we divide in the same
manner the two pyramids which remain in each, then all
the pyramids which remain, and so on continually, it follows
on the one hand, by x. 1, that we shall ultimately have
pyramids remaining which are together less than any assigned
solid, while on the other hand the sums of all the prisms
resulting from the successive subdivisions are in the ratio of
the bases of the original pyramids. Accordingly Euclid is able
to use the regular method of exhaustion exemplified in XII. 2,
and to establish the proposition [XII. 5] that pyramids with the
same height and with triangular bases are to one another as their
bases. The proposition is then extended [XII. 6] to pyramids with the
same height and with polygonal bases. Next [XII. 7] a prism with
a triangular base is divided into three pyramids which are shown
to be equal by means of XII. 5; and it follows, as a corollary, that
any pyramid is one third part of the prism which has the same
base and equal height. Again, two similar and similarly situated
pyramids are taken and the solid parallelepipeds are completed,
which are then seen to be six times as large as the pyramids
respectively; and, since (by XI. 33) similar parallelepipeds are in
the triplicate ratio of corresponding sides, it follows that the same
RELATION OF ARCHIMEDES TO HIS PREDECESSORS.
li
is true of the pyramids [x11. 8]. A corollary gives the obvious
extension to the case of similar pyramids with polygonal bases.
The proposition [XII. 9] that, in equal pyramids with triangular
bases, the bases are reciprocally proportional to the heights is
proved by the same method of completing the parallelepipeds and
using XI. 34; and similarly for the converse. It is next proved
[XII. 10] that, if in the circle which is the base of a cylinder a
square be described, and then polygons be successively described
by bisecting the arcs remaining in each case, and so doubling the
number of sides, and if prisms of the same height as the cylinder
be erected on the square and the polygons as bases respectively,
the prism with the square base will be greater than half the
cylinder, the next prism will add to it more than half of the
remainder, and so on. And each prism is triple of the pyramid with
the same base and altitude. Thus the same method of exhaustion
as that in XII. 2 proves that any cone is one third part of the
cylinder with the same base and equal height. Exactly the same
method is used to prove [xII. 11] that cones and cylinders which
have the same height are to one another as their bases, and
[XII. 12] that similar cones and cylinders are to one another in
the triplicate ratio of the diameters of their bases (the latter
proposition depending of course on the similar proposition XII. 8
for pyramids). The next three propositions are proved without
fresh recourse to x. 1. Thus the criterion of equimultiples laid
down in Def. 5 of Book v. is used to prove [XII. 13] that, if a
cylinder be cut by a plane parallel to its bases, the resulting
cylinders are to one another as their axes. It is an easy deduction
[XII. 14] that cones and cylinders which have equal bases are
proportional to their heights, and [XII. 15] that in equal cones
and cylinders the bases are reciprocally proportional to the heights,
and, conversely, that cones or cylinders having this property are
equal. Lastly, to prove that spheres are to one another in the
triplicate ratio of their diameters [XII. 18], a new procedure is
adopted, involving two preliminary propositions. In the first of
these [XII. 16] it is proved, by an application of the usual lemma
x. 1, that, if two concentric circles are given (however nearly
equal), an equilateral polygon can be inscribed in the outer circle
whose sides do not touch the inner; the second proposition [XII. 17]
uses the result of the first to prove that, given two concentric
spheres, it is possible to inscribe a certain polyhedron in the outer
2
[
d 2
1
2
.
į
ཟ*
lii
INTRODUCTION.
so that it does not anywhere touch the inner, and a corollary adds
the proof that, if a similar polyhedron be inscribed in a second
sphere, the volumes of the polyhedra are to one another in the
triplicate ratio of the diameters of the respective spheres. This
last property is then applied [XII. 18] to prove that spheres are
in the triplicate ratio of their diameters.
3. Conic Sections.
In my edition of the Conics of Apollonius there is a complete
account of all the propositions in conics which are used by Archi-
medes, classified under three headings, (1) those propositions
which he expressly attributes to earlier writers, (2) those which
are assumed without any such reference, (3) those which appear to
represent new developments of the theory of conics due to Archi-
medes himself. As all these properties will appear in this
volume in their proper places, it will suffice here to state only
such propositions as come under the first heading and a few under
the second which may safely be supposed to have been previously
known.
Archimedes says that the following propositions "are proved
in the elements of conics," i.e. in the earlier treatises of Euclid
and Aristaeus.
1. In the parabola
(a) if PV be the diameter of a segment and QVq the
chord parallel to the tangent at P, then QV = Vq ;
(b) if the tangent at Q meet VP produced in T, then
PV=PT;
(c) if two chords QVq, Q'V'q' each parallel to the tangent
at P meet the diameter PV in V, V' respectively,
PV: PV'=QV° : Q'V".
2. If straight lines drawn from the same point touch any
conic section whatever, and if two chords parallel to the respective
tangents intersect one another, then the rectangles under the
segments of the chords are to one another as the squares on the
parallel tangents respectively.
3. The following proposition is quoted as proved "in the conics.”
If in a parabola pa be the parameter of the principal ordinates,
RELATION OF ARCHIMEDES TO HIS PREDECESSORS.
liii
QQ'any chord not perpendicular to the axis which is bisected in V
by the diameter PV, p the parameter of the ordinates to PV, and
if QD be drawn perpendicular to PV, then
QV²: QD²=p: Pa
[On Conoids and Spheroids, Prop. 3, which see.]
The properties of a parabola, PN² = Pɑ. AN, and QV²=p.PV,
were already well known before the time of Archimedes. In fact
the former property was used by Menaechmus, the discoverer of
conic sections, in his duplication of the cube.
It may be taken as certain that the following properties of the
ellipse and hyperbola were proved in the Conics of Euclid.
and
1. For the ellipse
PN² : AN. A'N=P'N'² : AN' . A'N' = CB² : CA²
2
QV2: PV. P'V=Q'V' : PV'. P'V' = CD: CP.
(Either proposition could in fact be derived from the proposition
about the rectangles under the segments of intersecting chords
above referred to.)
and
2. For the hyperbola
2
PN: AN. A'N=P'N' : AN'. A'N'
2
QV²: PV. P'V=Q'V' : PV'. P'V',
though in this case the absence of the conception of the double
hyperbola as one curve (first found in Apollonius) prevented Euclid,
and Archimedes also, from equating the respective ratios to those
of the squares on the parallel semidiameters.
3. In a hyperbola, if P be any point on the curve and PK,
PL be each drawn parallel to one asymptote and meeting the
other,
PK.PL (const.)
=
This property, in the particular case of the rectangular hyperbola,
was known to Menaechmus.
It is probable also that the property of the subnormal of the
parabola (NG= |Pa) was known to Archimedes' predecessors. It
is tacitly assumed, On floating bodies, II. 4, etc.
From the assumption that, in the hyperbola, AT <AN (where
N is the foot of the ordinate from P, and T the point in which the
i
+
liv
INTRODUCTION.
tangent at P meets the transverse axis) we may perhaps infer
that the harmonic property
TP : TP' = PV : P'V,
or at least the particular case of it,
TA: TA' AN: A'N,
was known before Archimedes' time.
Lastly, with reference to the genesis of conic sections from
cones and cylinders, Euclid had already stated in his Phaenomena
that, "if a cone or cylinder be cut by a plane not parallel to the
base, the resulting section is a section of an acute-angled cone
[an ellipse] which is similar to a Oupeós." Though it is not probable
that Euclid had in mind any other than a right cone, the statement
should be compared with On Conoids and Spheroids, Props. 7, 8, 9.
§ 4. Surfaces of the second degree.
Prop. 11 of the treatise On Conoids and Spheroids states without
proof the nature of certain plane sections of the conicoids of revo-
lution. Besides the obvious facts (1) that sections perpendicular
to the axis of revolution are circles, and (2) that sections through
the axis are the same as the generating conic, Archimedes asserts
the following.
1. In a paraboloid of revolution any plane section parallel to
the axis is a parabola equal to the generating parabola.
2. In a hyperboloid of revolution any plane section parallel
to the axis is a hyperbola similar to the generating hyperbola.
In a hyperboloid of revolution a plane section through the
vertex of the enveloping cone is a hyperbola which is not similar
to the generating hyperbola.
4. In any spheroid a plane section parallel to the axis is an
ellipse similar to the generating ellipse.
Archimedes adds that "the proofs of all these propositions
are manifest (pavepaí)." The proofs may in fact be supplied as
follows.
1. Section of a paraboloid of revolution by a plane parallel
to the axis.
RELATION OF ARCHIMEDES TO HIS PREDECESSORS.
lv
P
Suppose that the plane of the paper represents the plane section
through the axis AN which intersects the given plane section at right
angles, and let A'O be the line of intersection.
Let POP' be any double ordinate to AN in the
section through the axis, meeting A'O and AN
at right angles in O, N respectively. Draw A'M
perpendicular to AN.
Suppose a perpendicular drawn from 0 to
A'O in the plane of the given section parallel to
the axis, and let y be the length intercepted by
the surface on this perpendicular.
Then, since the extremity of y is on the
circular section whose diameter is PP',

y= PO. OP'.
A
A
M
N
P
If A'0=x, and if p is the principal parameter of the generating
parabola, we have then
y²=PN° - ON-
= PN² – A'M² \*.
=p (AN—AM)
px,
1
so that the section is a parabola equal to the generating parabola.
2. Section of a hyperboloid of revolution by a plane parallel to
the axis.
Take, as before, the plane section through the axis which intersects

C'
A
P
C
A M
IN
P
the given plane section at right angles in A'O. Let the hyperbola
lvi
INTRODUCTION.
PAP' in the plane of the paper represent the plane section through
the axis, and let C be the centre (or the vertex of the enveloping
cone). Draw CC' perpendicular to CA, and produce ОA' to meet it
in C'. Let the rest of the construction be as before.
Suppose that
CA = a, C'A' a', C'O=x,
and let y have the same meaning as before.
Then
y² = PO. OP' = PN² – A'M².
And, by the property of the original hyperbola,
PN: CN-CA2 A'M2: CM2-CA2 (which is constant).
2
www
Thus A'M²: CM² - CA²
PN2: CN² – CA2
2
– PN² – A'M²: CN2 - CM²
=
2
= y²x²- a'²,
whence it appears that the section is a hyperbola similar to the
original one.
3. Section of a hyperboloid of revolution by a plane passing
through the centre (or the vertex of the enveloping cone).
I think there can be no doubt that Archimedes would have proved
his proposition about this section by means of the same general
property of conics which he uses to prove Props. 3 and 12-14 of
the same treatise, and which he enunciates at the beginning of
Prop. 3 as a known theorem proved in the "elements of conics," viz.
that the rectangles under the segments of intersecting chords are as
the squares of the parallel tangents.
Let the plane of the paper represent the plane section through
the axis which intersects the given plane passing through the
centre at right angles. Let CA'O be the line of intersection, C
being the centre, and A' being the point where CA'O meets the
surface. Suppose CAMN to be the axis of the hyperboloid, and
POp, P'O'p' two double ordinates to it in the plane section through
the axis, meeting CA'O in O, O' respectively; similarly let A'M be
the ordinate from A'. Draw the tangents at A and A' to the
section through the axis meeting in T, and let QOq, Q'O'q' be the
two double ordinates in the same section which are parallel to the
tangent at A' and pass through O, O' respectively.
Suppose, as before, that y, y' are the lengths cut off by the
RELATION OF ARCHIMEDES TO HIS PREDECESSORS.
lvii
surface from the perpendiculars at O and O' to OC in the plane of
the given section through CA'O, and that
CO=x, CO' = ', CA = a, CA = a.
Q'

P
P
C'
A'
T
A
C
M
N
N'
q
q
સ
P
P
Then, by the property of the intersecting chords, we have, since
Q0=0q,
Also
PO . Op : QO² = TA³ : TA'²
= P'O'. O'p': Q'O'².
y² = PO. Op, y'²=P'O' . O'p',
and, by the property of the hyperbola,
2
12
α
2
QO²x² - a¹²=Q'O'² : x'² — a' 2.
It follows, ex aequali, that
y² : x² — a'² = y'² : x'² a
and therefore that the section is a hyperbola.
12
.(a),
To prove that this hyperbola is not similar to the generating
hyperbola, we draw CC' perpendicular to CA, and C'A' parallel to
CA meeting CC' in C' and Pp in U.
If then the hyperbola (a) is similar to the original hyperbola, it
must by the last proposition be similar to the hyperbolic section
made by the plane through C'A'U at right angles to the plane of
the paper.
and
Now
CO2-CA' (C'U³ — C'A'³) + (CC' + OU)² -- CC12 Simlen 25.
>C′U² — C'A'²,
PO. Op<PU. Up.
F
lviii
INTRODUCTION.
2
Therefore PO. Op : CO² – CA'² <PU . Up : C'U² — C'A'³,
and it follows that the hyperbolas are not similar*.
4. Section of a spheroid by a plane parallel to the axis.
That this is an ellipse similar to the generating ellipse can of
course be proved in exactly the same way as theorem (2) above
for the hyperboloid.
* I think Archimedes is more likely to have used this proof than one on the
lines suggested by Zeuthen (p. 421). The latter uses the equation of the
hyperbola simply and proceeds thus. If y have the same meaning as above,
and if the coordinates of P referred to CA, CC' as axes be z, x, while those of O
referred to the same axes are z, x', we have, for the point P,
x² = k (z² — a²),
where x is constant.
K
Also, since the angle A'CA is given, x'=az, where a is constant.
Thus
y²=x² — x²² = (k − a²) z² — ka².
CO
Now z is proportional to CO, being in fact equal to
becomes
√1+a²,
and the equation
y2=
K-a²
1+a2°
.(1),
which is clearly a hyperbola, since a²K.
Now, though the Greeks could have worked out the proof in a geometrical
form equivalent to the above, I think that it is alien from the manner in which
Archimedes regarded the equations to central conics. These he always expressed
in the form of a proportion
and never in the
Apollonius, viz.
y2
y'2
x² ~ a²x²² ~ a²
[=
b2
a²
in the case of the ellipse],
form of an equation between areas like that used by
y² = px± 2x².
Moreover the occurrence of the two different constants and the necessity
of expressing them geometrically as ratios between areas and lines respectively
would have made the proof very long and complicated; and, as a matter of fact,
Archimedes never does express the ratio y²/(x² - a²) in the case of the hyperbola
in the form of a ratio between constant areas like b²/a2. Lastly, when the
equation of the given section through CA'O was found in the form (1), assuming
that the Greeks had actually found the geometrical equivalent, it would still
have been held necessary, I think, to verify that
C¿¹² — × (1+ª²), a²,
CA'
K-a2
before it was finally pronounced that the hyperbola represented by the equation
and the section made by the plane were one and the same thing.
RELATION OF ARCHIMEDES TO HIS PREDECESSORS.
lix
We are now in a position to consider the meaning of Archimedes'
remark that "the proofs of all these properties are manifest." In
the first place, it is not likely that "manifest" means "known” as
having been proved by earlier geometers; for Archimedes' habit is
to be precise in stating the fact whenever he uses important
propositions due to his immediate predecessors, as witness his
references to Eudoxus, to the Elements [of Euclid], and to the
"elements of conics." When we consider the remark with reference
to the cases of the sections parallel to the axes of the surfaces
respectively, a natural interpretation of it is to suppose that
Archimedes meant simply that the theorems are such as can easily
be deduced from the fundamental properties of the three conics now
expressed by their equations, coupled with the consideration that
the sections by planes perpendicular to the axes are circles. But I
think that this particular explanation of the "manifest" character
of the proofs is not so applicable to the third of the theorems
stating that any plane section of a hyperboloid of revolution
through the vertex of the enveloping cone but not through the axis
is a hyperbola. This fact is indeed no more "manifest" in the
ordinary sense of the term than is the like theorem about the
spheroid, viz. that any section through the centre but not through
the axis is an ellipse. But this latter theorem is not given along
with the other in Prop. 11 as being "manifest"; the proof of it is
included in the more general proposition (14) that any section of a
spheroid not perpendicular to the axis is an ellipse, and that parallel
sections are similar. Nor, seeing that the propositions are essen-
tially similar in character, can I think it possible that Archimedes
wished it to be understood, as Zeuthen suggests, that the proposition
about the hyperboloid alone, and not the other, should be proved
directly by means of the geometrical equivalent of the Cartesian
equation of the conic, and not by means of the property of the
rectangles under the segments of intersecting chords, used earlier
[Prop. 3] with reference to the parabola and later for the case of
the spheroid and the elliptic sections of the conoids and spheroids
generally. This is the more unlikely, I think, because the proof
by means of the equation of the conic alone would present much
more difficulty to the Greek, and therefore could hardly be called
"manifest."
It seems necessary therefore to seek for another explanation,
and I think it is the following. The theorems, numbered 1, 2, and
lx
INTRODUCTION.
4 above, about sections of conoids and spheroids parallel to the axis
are used afterwards in Props. 15–17 relating to tangent planes ;
whereas the theorem (3) about the section of the hyperboloid by a
plane through the centre but not through the axis is not used in
connexion with tangent planes, but only for formally proving that a
straight line drawn from any point on a hyperboloid parallel to any
transverse diameter of the hyperboloid falls, on the convex side of
the surface, without it, and on the concave side within it. Hence
it does not seem so probable that the four theorems were collected
in Prop. 11 on account of the use made of them later, as that they
were inserted in the particular place with special reference to the
three propositions (12-14) immediately following and treating of the
elliptic sections of the three surfaces. The main object of the whole
treatise was the determination of the volumes of segments of the
three solids cut off by planes, and hence it was first necessary to
determine all the sections which were ellipses or circles and therefore
could form the bases of the segments. Thus in Props. 12-14
Archimedes addresses himself to finding the elliptic sections, but,
before he does this, he gives the theorems grouped in Prop. 11 by
way of clearing the ground, so as to enable the propositions about
elliptic sections to be enunciated with the utmost precision. Prop.
11 contains, in fact, explanations directed to defining the scope of
the three following propositions rather than theorems definitely
enunciated for their own sake; Archimedes thinks it necessary to
explain, before passing to elliptic sections, that sections perpen-
dicular to the axis of each surface are not ellipses but circles, and
that some sections of each of the two conoids are neither ellipses nor
circles, but parabolas and hyperbolas respectively. It is as if he had
said, "My object being to find the volumes of segments of the three
solids cut off by circular or elliptic sections, I proceed to consider
the various elliptic sections; but I should first explain that sections
at right angles to the axis are not ellipses but circles, while sections.
of the conoids by planes drawn in a certain manner are neither
ellipses nor circles, but parabolas and hyperbolas respectively. With
these last sections I am not concerned in the next propositions, and
I need not therefore cumber my book with the proofs; but, as some
of them can be easily supplied by the help of the ordinary properties
of conics, and others by means of the methods illustrated in the
propositions now about to be given, I leave them as an exercise for
the reader." This will, I think, completely explain the assumption
RELATION OF ARCHIMEDES TO HIS PREDECESSORS.
lxi
of all the theorems except that concerning the sections of a spheroid
parallel to the axis; and I think this is mentioned along with the
others for symmetry, and because it can be proved in the same way
as the corresponding one for the hyperboloid, whereas, if mention of
it had been postponed till Prop. 14 about the elliptic sections of a
spheroid generally, it would still require a proposition for itself, since
the axes of the sections dealt with in Prop. 14 make an angle with
the axis of the spheroid and are not parallel to it.
At the same time the fact that Archimedes omits the proofs of
the theorems about sections of conoids and spheroids parallel to the
axis as
"manifest" is in itself sufficient to raise the presumption
that contemporary geometers were familiar with the idea of three
dimensions and knew how to apply it in practice. This is no matter
for surprise, seeing that we find Archytas, in his solution of the
problem of the two mean proportionals, using the intersection of a
certain cone with a curve of double curvature traced on a right
circular cylinder*. But, when we look for other instances of early
investigations in geometry of three dimensions, we find practically
nothing except a few vague indications as to the contents of a lost
treatise of Euclid's consisting of two Books entitled Surface-loci
(τόποι πρὸς ἐπιφανεία) t. This treatise is mentioned by Pappus
among other works by Aristaeus, Euclid and Apollonius grouped
as forming the so-called τóños avaλvóμevos‡. As the other works in
the list which were on plane subjects dealt only with straight lines,
circles and conic sections, it is a priori likely that the surface-loci of
* Cf. Eutocius on Archimedes (Vol. I. pp. 98–102), or Apollonius of Perga,
pp. xxii.-xxiii.
A
""
+ By this term we conclude that the Greeks meant "loci which are surfaces"
as distinct from loci which are lines. Cf. Proclus' definition of a locus as
“a position of a line or a surface involving one and the same property'
(γραμμῆς ἢ ἐπιφανείας θέσις ποιοῦσα ἓν καὶ ταὐτὸν σύμπτωμα), p. 394. Pappus
(pp. 660—2) gives, quoting from the Plane Loci of Apollonius, a classification of
loci according to their order in relation to that of which they are the loci. Thus,
he says, loci are (1) epektɩkol, i.e. fixed, e.g. in this sense the locus of a point is
a point, of a line a line, and so on; (2) dieĝodikoi or moving along, a line being in
this sense the locus of a point, a surface of a line, and a solid of a surface;
(3) ȧvaσтpopikol, turning backwards, i.e., presumably, moving backwards and
forwards, a surface being in this sense the locus of a point, and a solid of a line.
Thus a surface-locus might apparently be either the locus of a point or the
locus of a line moving in space.
Pappus, pp. 634, 636.
lxii
INTRODUCTION.
Euclid included at least such loci as were cones, cylinders and
spheres. Beyond this, all is conjecture based upon two lemmas
given by Pappus in connexion with the treatise.
First lemma to the Surface-loci of Euclid*.
The text of this lemma and the attached figure are not satisfac-
tory as they stand, but they have been explained by Tannery in a
way which requires a change in the figure, but only the very slightest
alteration in the text, as follows†.
"If AB be a straight line and CD be parallel to a straight line
given in position, and if the ratio AD. DB: DC² be [given], the
point C lies on a conic section.
If now AB be no longer given in
position and A, B be no longer
given but lie on straight lines
AE, EB given in position, the
point C raised above [the plane
containing AE, EB] is on a
surface given in position. And
this was proved."
A
E

C
B
D
According to this interpretation, it is asserted that, if AB moves
with one extremity on each of the lines AE, EB which are fixed,
while DC is in a fixed direction and AD. DB: DC2 is constant,
then lies on a certain surface. So far as the first sentence is
concerned, AB remains of constant length, but it is not made
precisely clear whether, when AB is no longer given in position, its
length may also vary§. If however AB remains of constant length
for all positions which it assumes, the surface which is the locus of
C would be a complicated one which we cannot suppose that Euclid
could have profitably investigated. It may, therefore, be that
Pappus purposely left the enunciation somewhat vague in order to
make it appear to cover several surface-loci which, though belonging
to the same type, were separately discussed by Euclid as involving
* Pappus, p. 1004.
+ Bulletin des sciences math., 2º Série, vi. 149.
# The words of the Greek text are γένηται δὲ πρὸς θέσει εὐθεῖα ταῖς ΑΕ, ΕΒ,
and the above translation only requires ev@elais instead of eveîa. The figure in
the text is so drawn that ADB, AEB are represented as two parallel lines, and
CD is represented as perpendicular to ADB and meeting AEB in E.
§ The words are simply "if AB be deprived of its position (σTEρηēĥ TĤs
Béσews) and the points A, B be deprived of their [character of] being given"
(στερηθῇ τοῦ δοθέντος είναι).
RELATION OF ARCHIMEDES TO HIS PREDECESSORS. lxiii
in each case somewhat different sets of conditions limiting the
generality of the theorem.
It is at least open to conjecture, as Zeuthen has pointed out*,
that two cases of the type were considered by Euclid, namely, (1)
that in which AB remains of constant length while the two fixed
straight lines on which A, B respectively move are parallel instead
of meeting in a point, and (2) that in which the two fixed straight
lines meet in a point while AB moves always parallel to itself
and varies in length accordingly.
(1) In the first case, where the length of AB is constant and
the two fixed lines parallel, we should have a surface described by a
conic moving bodily t. This surface would be a cylindrical surface,
though it would only have been called a "cylinder" by the ancients
in the case where the moving conic was an ellipse, since the essence
of a "cylinder" was that it could be bounded between two parallel
circular sections. If then the moving conic was an ellipse, it would
not be difficult to find the circular sections of the cylinder; this
could be done by first taking a section at right angles to the axis,
after which it could be proved, after the manner of Archimedes,
On Conoids and Spheroids, Prop. 9, first that the section is an ellipse
or a circle, and then, in the former case, that a section made by
a plane drawn at a certain inclination to the ellipse and passing
through, or parallel to, the major axis is a circle. There was
nothing to prevent Euclid from investigating the surface similarly
generated by a moving hyperbola or parabola; but there would
be no circular sections, and hence the surfaces might perhaps not
have been considered as of very great importance.
(2) In the second case, where AE, BE meet at a point and
AB moves always parallel to itself, the surface generated is of
course a cone. Some particular cases of this sort may easily have
been discussed by Euclid, but he could hardly have dealt with the
general case, where DC has any direction whatever, up to the
point of showing that the surface was really a cone in the sense
in which the Greeks understood the term, or (in other words)
of finding the circular sections. To do this it would have been
necessary to determine the principal planes, or to solve the dis-
* Zeuthen, Die Lehre von den Kegelschnitten, pp. 425 sqq.
+ This would give a surface generated by a moving line, διεξοδικὸς γραμμῆς
as Pappus has it.
ļ
INTRODUCTION.
lxiv
my to
}
4
፡
J
}
+
criminating cubic, which we cannot suppose Euclid to have done.
Moreover, if Euclid had found the circular sections in the most
general case, Archimedes would simply have referred to the fact
instead of setting himself to do the same thing in the particular
case where the plane of symmetry is given. These remarks apply
to the case where the conic which is the locus of C is an ellipse;
there is still less ground for supposing that Euclid could have
proved the existence of circular sections where the conic was a
hyperbola, for there is no evidence that Euclid even knew that
hyperbolas and parabolas could be obtained by cutting an oblique
circular cone.
Second lemma to the Surface-loci.
+
In this Pappus states, and gives a complete proof of the propo-
sition, that the locus of a point whose distance from a given point
is in a given ratio to its distance from a fixed line is a conic
section, which is an ellipse, a parabola, or a hyperbola according
as the given ratio is less than, equal to, or greater than unity*.
Two conjectures are possible as to the application of this theorem
by Euclid in the treatise referred to.
(1) Consider a plane and a straight line meeting it at any angle.
Imagine any plane drawn at right angles to the straight line and
meeting the first plane in another straight line which we will call
X. If then the given straight line meets the plane at right angles
to it in the point S, a conic can be described in that plane with
S for focus and X for directrix; and, as the perpendicular on X
from any point on the conic is in a constant ratio to the per-
pendicular from the same point on the original plane, all points
on the conic have the property that their distances from S are in
a given ratio to their distances from the given plane respectively.
Similarly, by taking planes cutting the given straight line at right
angles in any number of other points besides S, we see that the locus
of a point whose distance from a given straight line is in a given
ratio to its distance from a given plane is a cone whose vertex is
the point in which the given line meets the given plane, while the
plane of symmetry passes through the given line and is at right
angles to the given plane. If the given ratio was such that the
guiding conic was an ellipse, the circular sections of the surface
* See Pappus, pp. 1006-1014, and Hultsch's Appendix, pp. 1270–1273; or
cf. Apollonius of Perga, pp. xxxvi.—xxxviii.
5
L
L
RELATION OF ARCHIMEDES TO HIS PREDECESSORS.
lxv
could, in that case at least, be found by the same method as
that used by Archimedes (On Conoids and Spheroids, Prop. 8) in
the rather more general case where the perpendicular from the
vertex of the cone on the plane of the given elliptic section does
not necessarily pass through the focus.
(2) Another natural conjecture would be to suppose that, by
means of the proposition given by Pappus, Euclid found the locus
of a point whose distance from a given point is in a given ratio
to its distance from a fixed plane. This would have given surfaces
identical with the conoids and spheroids discussed by Archimedes
excluding the spheroid generated by the revolution of an ellipse
about the minor axis. We are thus brought to the same point as
Chasles who conjectured that the Surface-loci of Euclid dealt with
surfaces of revolution of the second degree and sections of the
same*.
Recent writers have generally regarded this theory as
improbable. Thus Heiberg says that the conoids and spheroids
were without any doubt discovered by Archimedes himself; other-
wise he would not have held it necessary to give exact definitions
of them in his introductory letter to Dositheus; hence they could
not have been the subject of Euclid's treatise†. I confess I think
that the argument of Heiberg, so far from being conclusive against
the probability of Chasles' conjecture, is not of any great weight.
To suppose that Euclid found, by means of the theorem enunciated
and proved by Pappus, the locus of a point whose distance from
a given point is in a given ratio to its distance from a fixed plane
does not oblige us to assume either that he gave a name to the
loci or that he investigated them further than to show that sections
through the perpendicular from the given point on the given plane
were conics, while sections at right angles to the same perpendicular
were circles; and of course these facts would readily suggest them-
selves. Seeing however that the object of Archimedes was to
find the volumes of segments of each surface, it is not surprising
that he should have preferred to give a definition of them which
would indicate their form more directly than a description of them
as loci would have done; and we have a parallel case in the dis-
tinction drawn between conics as such and conics regarded as loci,
which is illustrated by the different titles of Euclid's Conics and
the Solid Loci of Aristaeus, and also by the fact that Apollonius,
Aperçu historique, pp. 273, 4.
+ Litterargeschichtliche Studien über Euklid, p. 79.
H. A.
e
lxvi
INTRODUCTION.
though he speaks in his preface of some of the theorems in his
Conics as useful for the synthesis of 'solid loci' and goes on to
mention the 'locus with respect to three or four lines,' yet enun-
ciates no proposition stating that the locus of such and such a point
is a conic. There was a further special reason for defining the
conoids and spheroids as surfaces described by the revolution of
a conic about its axis, namely that this definition enabled Archi-
medes to include the spheroid which he calls 'flat' (èmiλaTÙ
σpaɩpoeɩdés), i.e. the spheroid described by the revolution of an
ellipse about its minor axis, which is not one of the loci which
the hypothesis assumes Euclid to have discovered. Archimedes'
new definition had the incidental effect of making the nature of
the sections through and perpendicular to the axis of revolution
even more obvious than it would be from Euclid's supposed way
of treating the surfaces; and this would account for Archimedes'
omission to state that the two classes of sections had been known
before, for there would have been no point in attributing to Euclid
the proof of propositions which, with the new definition of the
surfaces, became self-evident. The further definitions given by
Archimedes may be explained on the same principle. Thus the
axis, as defined by him, has special reference to his definition of
the surfaces, since it means the axis of revolution, whereas the
axis of a conic is for Archimedes a diameter. The enveloping cone
of the hyperboloid, which is generated by the revolution of the
asymptotes about the axis, and the centre regarded as the point
of intersection of the asymptotes were useful to Archimedes' dis-
cussion of the surfaces, but need not have been brought into
Euclid's description of the surfaces as loci. Similarly with the
axis and vertex of a segment of each surface. And, generally, it
seems to me that all the definitions given by Archimedes can be
explained in like manner without prejudice to the supposed dis-
covery of three of the surfaces by Euclid.
I think, then, that we may still regard it as possible that
Euclid's Surface-loci was concerned, not only with cones, cylinders
and (probably) spheres, but also (to a limited extent) with three
other surfaces of revolution of the second degree, viz. the paraboloid,
the hyperboloid and the prolate spheroid. Unfortunately however
we are confined to the statement of possibilities; and certainty
can hardly be attained unless as the result of the discovery of
fresh documents.
RELATION OF ARCHIMEDES TO HIS PREDECESSORS. lxvii
§ 5. Two mean proportionals in continued proportion.
Archimedes assumes the construction of two mean proportionals
in two propositions (On the Sphere and Cylinder II. 1, 5). Perhaps
he was content to use the constructions given by Archytas,
Menaechmus*, and Eudoxus. It is worth noting, however, that
Archimedes does not introduce the two geometric means where
they are merely convenient but not necessary; thus, when (On the
Sphere and Cylinder 1. 34) he has to substitute for a ratio.
where ẞ>y, a ratio between lines, and it is sufficient for his
purpose that the required ratio cannot be greater than
(2)
(9)*,
but
may be less, he takes two arithmetic means between ß, y, as d, e,
and then assumes† as a known result that
ß³ß
83 γ
<
* The constructions of Archytas and Menaechmus are given by Eutocius
[Archimedes, Vol. г. pp. 92—102]; or see Apollonius of Perga, pp. xix–xxiii.
The proposition is proved by Eutocius; see the note to On the Sphere
and Cylinder 1. 34 (p. 42).
"
e 2
ร
叔
​CHAPTER IV.
ARITHMETIC IN ARCHIMEDES.
Two of the treatises, the Measurement of a circle and the
Sand-reckoner, are mostly arithmetical in content. Of the Sand-
reckoner nothing need be said here, because the system for expressing
numbers of any magnitude which it unfolds and applies cannot be
better described than in the book itself; in the Measurement of a
circle, however, which involves a great deal of manipulation of
numbers of considerable size though expressible by means of the
ordinary Greek notation for numerals, Archimedes merely gives the
results of the various arithmetical operations, multiplication, extrac-
tion of the square root, etc., without setting out any of the operations
themselves. Various interesting questions are accordingly involved,
and, for the convenience of the reader, I shall first give a short
account of the Greek system of numerals and of the methods by
which other Greek mathematicians usually performed the various
operations included under the general term λoyiσTIKÝ (the art of
calculating), in order to lead up to an explanation (1) of the way in
which Archimedes worked out approximations to the square roots of
large numbers, (2) of his method of arriving at the two approximate
values of √3 which he simply sets down without any hint as to how
they were obtained*.
* In writing this chapter I have been under particular obligations to Hultsch's
articles Arithmetica and Archimedes in Pauly-Wissowa's Real-Encyclopädie, 11.
1, as well as to the same scholar's articles (1) Die Näherungswerthe irrationaler
Quadratwurzeln bei Archimedes in the Nachrichten von der kgl. Gesellschaft der
Wissenschaften zu Göttingen (1893), pp. 367 sqq., and (2) Zur Kreismessung des
Archimedes in the Zeitschrift für Math. u. Physik (Hist. litt. Abtheilung) xxxix.
(1894), pp. 121 sqq. and 161 sqq. I have also made use, in the earlier part
of the chapter, of Nesselmann's work Die Algebra der Griechen and the histories
of Cantor and Gow.
ARITHMETIC IN ARCHIMEDES.
lxix
§ 1. Greek numeral system.
It is well known that the Greeks expressed all numbers from 1
to 999 by means of the letters of the alphabet reinforced by the
addition of three other signs, according to the following scheme, in
which however the accent on each letter might be replaced by a
short horizontal stroke above it, as a.
a', B', y', d', é', ', ', n', ' are
í', k', X', µ', v', §', o',
';
' „,
p', o', T', v', ', X', y', w', I'„,
τ', ύ, Χ, Ψ', ω,
1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
10, 20, 30,
90
""
100, 200, 300,......900
""
Thus
Intermediate numbers were expressed by simple juxtaposition
(representing in this case addition), the largest number being placed
on the left, the next largest following it, and so on in order.
the number 153 would be expressed by pvý or pvy.
sign for zero, and therefore 780 was π', and 306 τs' simply.
There was no
Thousands (xiλiádes) were taken as units of a higher order, and
1,000, 2,000, ... up to 9,000 (spoken of as xíλioi, dioxidwi, K.T.λ.) were
represented by the same letters as the first nine natural numbers
but with a small dash in front and below the line; thus e.g. ' was
4,000, and, on the same principle of juxtaposition as before, 1,823 was
expressed by awкy' or awkɣ, 1,007 by a', and so on.
Above 9,999 came a myriad (µvpiás), and 10,000 and higher
numbers were expressed by using the ordinary numerals with the
substantive µvpiádes taken as a new denomination (though the words
μύριοι, δισμύριοι, τρισμύριοι, κ.τ.λ. are also found, following the
analogy of χίλιοι, δισχίλιοι and so on). Various abbreviations were
used for the word μvpiás, the most common being M or Mu; and,
where this was used, the number of myriads, or the multiple of
10,000, was generally written over the abbreviation, though some-
times before it and even after it. Thus 349,450 was M0vv'*.
*
λδ
Fractions (λeπrá) were written in a variety of ways. The most
usual was to express the denominator by the ordinary numeral with
two accents affixed. When the numerator was unity, and it was
therefore simply a question of a symbol for a single word such as
Diophantus denoted myriads followed by thousands by the ordinary signs
for numbers of units, only separating them by a dot from the thousands. Thus
for 3,069,000 he writes 75.0, and Xy. ayos for 331,776. Sometimes myriads
were represented by the ordinary letters with two dots above, as p=100 myriads
(1,000,000), and myriads of myriads with two pairs of dots, as i for 10 myriad-
myriads (1,000,000,000).
"
F
7
lxx
a"
INTRODUCTION.
9
5)
TρíTOV,, there was no need to express the numerator, and the
symbol was y'"; similarly s"=, te″=1, and so on. When the
numerator was not unity and a certain number of fourths, fifths,
etc., had to be expressed, the ordinary numeral was used for the
numerator; thus 0′ la″ = 11, l' oa". In Heron's Geometry the
denominator was written twice in the latter class of fractions; thus
ξ (δύο πέμπτα) was β'ε'ε", 33 λεπτά τριακοστότριτα κγ' or εἰκοσιτρία
23
трiaкоσTÓтρiтa) was kỳ λy" Ay". The sign for , pov, is in
τριακοστότριτα) κγ'
Archimedes, Diophantus and Eutocius ", in Heron C or a sign
similar to a capital S*.
5
ß'e"
A favourite way of expressing fractions with numerators greater
than unity was to separate them into component fractions with
numerator unity, when juxtaposition as usual meant addition. Thus
was written "8" =+1; 15 was C8"n"15" = 1 + 1 + 1 + 1/8 ;
Eutocius writes "8" or + for 32, and so on.
½ 1
Sometimes the
same fraction was separated into several different sums; thus in
Heron (p. 119, ed. Hultsch) 16 is variously expressed as
and
(a)
+1+ 34
64
224
16
1
1/2 + 1/4 + 1/1/14 + 112 + 224
1
(b) 1/2 + 1/3 + 1/1/8 + 3/2 + 112,
글 ​10
1
(c) + + + 12 + 214.
1/2 1/2 21
1
Sexagesimal fractions. This system has to be mentioned because
the only instances of the working out of some arithmetical operations
which have been handed down to us are calculations expressed in
terms of such fractions; and moreover they are of special interest
as having much in common with the modern system of decimal
fractions, with the difference of course that the submultiple is 60
instead of 10. The scheme of sexagesimal fractions was used by the
Greeks in astronomical calculations and appears fully developed in
the σúvraέis of Ptolemy. The circumference of a circle, and along
with it the four right angles subtended by it at the centre, are
divided into 360 parts (τµýμaτα or μoîpai) or as we should say degrees,
each poîpa into 60 parts called (first) sixtieths, (πρŵтα) ¿έŋкоσtá,
or minutes (λeπtά), each of these again into deúrepa éέnkoσtá (seconds),
and so on.
A similar division of the radius of the circle into 60
* Diophantus has a general method of expressing fractions which is the
exact reverse of modern practice; the denominator is written above the
γ
ΚΕ
α. ωις
numerator, thus e=5/3, κα = 21/25, and ρ§. &§n=1,270,568/10,816. Some-
times he writes down the numerator and then introduces the denominator
with ἐν μορίῳ or μορίου, e.g. rs . θ μορ. λγ. αψος = 3,069,000/331,776.
ARITHMETIC IN ARCHIMEDES.
lxxi
parts (Tunμara) was also made, and these were each subdivided into
sixtieths, and so on. Thus a convenient fractional system was
available for general arithmetical calculations, expressed in units of
any magnitude or character, so many of the fractions which we
should represent by, so many of those which we should write
(1)³, (1)³, and so on to any extent. It is therefore not surprising
that Ptolemy should say in one place "In general we shall use the
method of numbers according to the sexagesimal manner because of
the inconvenience of the [ordinary] fractions." For it is clear that
the successive submultiples by 60 formed a sort of frame with fixed
compartments into which any fractions whatever could be located,
and it is easy to see that e.g. in additions and subtractions the
sexagesimal fractions were almost as easy to work with as decimals
are now, 60 units of one denomination being equal to one unit of
the next higher denomination, and "carrying" and "borrowing"
being no less simple than it is when the number of units of one
denomination necessary to make one of the next higher is 10 instead
of 60. In expressing the units of the circumference, degrees, poîpai
or the symbolů was generally used along with the ordinary numeral
which had a stroke above it; minutes, seconds, etc. were expressed
by one, two, etc. accents affixed to the numerals. Thus ů B=2°,
μopŵv µš µß' µ” = 47° 42′ 40″. Also where there was no unit in any
particular denomination O was used, signifying οὐδεμία μοῖρα, οὐδὲν
¿έŋkoσтóv and the like; thus Ō a' ß″′ 0″″ = 0° 1′ 2″′ 0″. Similarly, for
the units representing the divisions of the radius the word Tµńμata
or some equivalent was used, and the fractions were represented as
before; thus Tunμáτwv ¿d ve" = 67 (units) 4' 55".
νε
§ 2. Addition and Subtraction.
There is no doubt that, in writing down numbers for these
purposes, the several powers of 10 were kept separate in a manner
corresponding practically to our system of numerals, and the
hundreds, thousands, etc., were written in separate vertical rows.
The following would therefore be a typical form of a sum in addition;
ανκ δ' = 1424
ρ γ
103
α
Μ βσπ α'
a
12281
M
Μλ'
30030
8
Mywλ n' 43838
lxxii
INTRODUCTION.
and the mental part of the work would be the same for the Greek as
for us.
Similarly a subtraction would be represented as follows:
A
Μιγχλς' - 93636
β
My O' 23409
5
Μ σκζ' 70227
§ 3. Multiplication.
A number of instances are given in Eutocius' commentary on
the Measurement of a circle, and the similarity to our procedure is
just as marked as in the above cases of addition and subtraction.
The multiplicand is written first, and below it the multiplier preceded
by èπí (= "into"). Then the highest power of 10 in the multiplier
is taken and multiplied into the terms containing the separate
multiples of the successive powers of 10, beginning with the highest
and descending to the lowest; after which the next highest power
of 10 in the multiplier is multiplied into the various denominations
in the multiplicand in the same order. The same procedure is
followed where either or both of the numbers to be multiplied
contain fractions. Two instances from Eutocius are appended from
which the whole procedure will be understood.
(1)
ψπ'
ἐπὶ ψπ!
με ε
MM¸s'
Ε
780
× 780
490000 56000
M₁5,5v'
56000
6400
§
ὁμοῦ Μην
sum 608400
(2)
vey' L"S"
ἐπὶ γιγ' κ"δ"
γ
ΜΜ θ αφήν
Y
Μπλέ β' "
α
„OλO'a' L" L″S"
αφ'
ap'é'a' "S"n"
Yv′ß′ L″ L″8″n" 15"
30131130133]
× 301311
9,000,000 30,000 9,000 1500
750
30,000 100
30
5 21/2
9,000
30
9
1 1/2. 12 + 1
1,500
5
1/1/
1 1/
750
21/
1/2 + 1/2
1 1
8 16
λη
[ὁμοῦ] Μβχπθις" [9,041,250 + 30,137, +9,041
+ 1506 + 1 + 1 +
+ 753 + 1 + 1/1/ 1
+ 16
= 9,082,689, 18.
ARITHMETIC IN ARCHIMEDES.
lxxiii
One instance of a similar multiplication of numbers involving
fractions may be given from Heron (pp. 80, 81). It is only one of
many, and, for brevity, the Greek notation will be omitted. Heron
has to find the product of 433 and 782, and proceeds as follows:
4.7 = 28,
6
4.82=248,
64
64
64
33.7=281,
649
62 1
64 64
33.82=2046.1 = 31+ 82.74.
6464
The result is accordingly
62
1
28 +510 + 82.1=28+7 +82 +82.34
64
62
64 64
62
62
= 35 +82 +82.74.
64 64 64
The multiplication of 37° 4′ 55″ (in the sexagesimal system) by
itself is performed by Theon of Alexandria in his commentary on
Ptolemy's σúvraέıs in an exactly similar manner.
§ 4. Division.
The operation of dividing by a number of one digit only was
easy for the Greeks as for us, and what we call "long division" was
with them performed, mutatis mutandis, in the same way as now
with the help of multiplication and subtraction. Suppose, for
instance, that the operation in the first case of multiplication given
ૐ
above had to be reversed and that M¸nv' (608,400) had to be divided
by ' (780). The terms involving the different powers of 10 would
be mentally kept separate as in addition and subtraction, and the
first question would be, how many times will 7 hundreds go into 60
myriads, due allowance being made for the fact that the 7 hundreds
have 80 behind them and that 780 is not far short of 8 hundreds ?
The answer is 7 hundreds or ', and this multiplied by the divisor
νδ
ૐ
π′ (780) would give Ms' (546,000) which, subtracted from Mu'
5
(608,400), leaves the remainder M,ßv (62,400). This remainder has
then to be divided by 780 or a number approaching 8 hundreds, and
8 tens or 'would have to be tried. In the particular case the
result would then be complete, the quotient being ' (780), and
there being no remainder, since ' (80) multiplied by ′ (780) gives
S
the exact figure M,ẞv' (62,400).
lxxiv
INTRODUCTION.
An actual case of long division where the dividend and divisor
contain sexagesimal fractions is described by Theon. The problem
is to divide 1515 20′ 15″ by 25 12′ 10″, and Theon's account of the
process comes to this.
Divisor
25 12′ 10″
Dividend
1515 20' 15"
Quotient
First term 60
25.60 = 1500
Remainder
15 = 900'
Sum
920'
12'.60 =
720'
Remainder
200'
10″. 60 =
10'
Remainder
190'
Second term 7'
25.7' =
175'
15′ = 900"
Sum
915"
12'.7'
84″
Remainder
831"
10". 7'
1" 10"
Remainder
829" 50"
Third term 33″
25.33"
825"
Remainder
12′.33"
હું
4″ 50″″ = 290"
396""
(too great by) 106″
Thus the quotient is something less than 60 7′ 33″. It will be
observed that the difference between this operation of Theon's and
that followed in dividing Myv′ (608,400) by ч' (780) as above is
that Theon makes three subtractions for one term of the quotient,
whereas the remainder was arrived at in the other case after one
subtraction. The result is that, though Theon's method is quite
clear, it is longer, and moreover makes it less easy to foresee what
will be the proper figure to try in the quotient, so that more time
would be apt to be lost in making unsuccessful trials.
§ 5. Extraction of the square root.
We are now in a position to see how the operation of extracting
the square root would be likely to be attacked. First, as in the case
of division, the given whole number whose square root is required
would be separated, so to speak, into compartments each containing
ARITHMETIC IN ARCHIMEDES.
lxxv
such and such a number of units and of the separate powers of 10.
Thus there would be so many units, so many tens, so many hundreds,
etc., and it would have to be borne in mind that the squares of
numbers from 1 to 9 would lie between 1 and 99, the squares of
numbers from 10 to 90 between 100 and 9900, and so on. Then the
first term of the square root would be some number of tens or
hundreds or thousands, and so on, and would have to be found in
much the same way as the first term of a quotient in a "long
division," by trial if necessary. If A is the number whose square
root is required, while a represents the first term or denomination of
the square root and x the next term or denomination still to be
found, it would be necessary to use the identity (a + x)² = a² + 2ax + x²
and to find x so that 2ax + x² might be somewhat less than the
remainder A-a². Thus by trial the highest possible value of x
satisfying the condition would be easily found. If that value were
b, the further quantity 2ab+b² would have to be subtracted from
the first remainder A – a³, and from the second remainder thus left
a third term or denomination of the square root would have to be
derived, and so on. That this was the actual procedure adopted is
clear from a simple case given by Theon in his commentary on the
σύνταξις.
σúvragis. Here the square root of 144 is in question, and it is
obtained by means of Eucl. II. 4. The highest possible denomina-
tion (i.e. power of 10) in the square root is 10; 102 subtracted from
144 leaves 44, and this must contain not only twice the product of
10 and the next term of the square root but also the square of that
next term itself. Now, since 2.10 itself produces 20, the division
of 44 by 20 suggests 2 as the next term of the square root; and
this turns out to be the exact figure required, since
2. 20+ 2² = 44.
The same procedure is illustrated by Theon's explanation of
Ptolemy's method of extracting square roots according to the
sexagesimal system of fractions. The problem is to find approxi-
mately the square root of 4500 μoîpai or degrees, and a geometrical
figure is used which makes clear the essentially Euclidean basis of
the whole method. Nesselmann gives a complete reproduction of
the passage of Theon, but the following purely arithmetical represen-
tation of its purport will probably be found clearer, when looked at
side by side with the figure.
Ptolemy has first found the integral part of √4500 to be 67.
t
し
​I
den
г
lxxvi
INTRODUCTION.
Now 67² = 4489, so that the remainder is 11. Suppose now that
the rest of the square root is expressed by means of the usual
sexagesimal fractions, and that we may therefore put
√4500= √672 + 11 = 67 +
X Y
+
60 602,
2.67x
where x, y are yet to be found. Thus x must be such that 60
is somewhat less than 11, or x must be somewhat less than
or
330
67
11.60
2.67
which is at the same time greater than 4. On trial, it
turns out that 4 will satisfy the conditions of the problem, namely
2
4
that (67
67 +
60
must be less than 4500, so that a remainder will
be left by means of which y may be found.
α
n
K
δ

67⁰
4'
55"
4489
268'
3688" 40"
€
مد
4'
268'
16"
0
λ
55"
3688" 40"
β
γ
2.67.4
4\2
Now 11
60
60
is the remainder, and this is equal to
11.60º-2.67.4.60-16
602
7424
60°
Thus we must suppose that 2 (67 +
that 2 (67
4
Y
60 602 approximates to
7424
60º,
or that 8048y is approximately equal to 7424. 60.
ARITHMETIC IN ARCHIMEDES.
lxxvii
4 \ 55
60/60°
+
Therefore y is approximately equal to 55. We have then to
subtract
2 (67 +11)
(209)
2
442640 3025
or
,
+
603
6049
from the remainder
7424
above found.
602
The subtraction of
442640
603
7424
2800 46
from
40
gives
602
603,
or
602
60³ ;
but Theon does not go further and subtract the remaining
3025
6049
instead of which he merely remarks that the
55
square of
602
approximates to +
46 40
602 60ª•
As a matter of fact, if we deduct the
3025
from
604
2800
603,
SO as to obtain the correct remainder, it is
164975
found to be
604
To show the power of this method of extracting square roots by
means of sexagesimal fractions, it is only necessary to mention that
103 55 23
Ptolemy gives +
60 602 60³
+ as an approximation to √3, which
approximation is equivalent to 1-7320509 in the ordinary decimal
notation and is therefore correct to 6 places.
But it is now time to pass to the question how Archimedes
obtained the two approximations to the value of √3 which he
assumes in the Measurement of a circle. In dealing with this
subject I shall follow the historical method of explanation adopted
by Hultsch, in preference to any of the mostly a priori theories
which the ingenuity of a multitude of writers has devised at
different times.
§ 6. Early investigations of surds or incommensurables.
*
From a passage in Proclus' commentary on Eucl. 1. we learn
that it was Pythagoras who discovered the theory of irrationals
(ʼn tŵv ådóywv πрayμarcíα). Further Plato says (Theaetetus 147 D),
"On square roots this Theodorus [of Cyrene] wrote a work in
p. 65 (ed. Friedlein).
lxxviii
INTRODUCTION.
which he proved to us, with reference to those of 3 or 5 [square] feet
that they are incommensurable in length with the side of one square
foot, and proceeded similarly to select, one by one, each [of the other
incommensurable roots] as far as the root of 17 square feet, beyond
which for some reason he did not go." The reason why √2 is not
mentioned as an incommensurable square root must be, as Cantor
says, that it was before known to be such. We may therefore
conclude that it was the square root of 2 which was geometrically
constructed by Pythagoras and proved to be incommensurable with
the side of a square in which it represented the diagonal. A clue
to the method by which Pythagoras investigated the value of √2
is found by Cantor and Hultsch in the famous passage of Plato
(Rep. VIII. 546 B, C) about the 'geometrical' or 'nuptial' number.
Thus, when Plato contrasts the ῥητή and ἄρρητος διάμετρος τῆς
TЄμжάdos, he is referring to the diagonal of a square whose side
contains five units of length; the appητos diáμerpos, or the irrational
diagonal, is then √50 itself, and the nearest rational number is
√50-1, which is the nr diapeтpos. We have herein the
explanation of the way in which Pythagoras must have made the
first and most readily comprehensible approximation to √2; he
must have taken, instead of 2, an improper fraction equal to it but
such that the denominator was a square in any case, while the
numerator was as near as possible to a complete square. Thus
50
Pythagoras chose and the first approximation to √2 was
25'
accordingly it being moreover obvious that √2>7. Again,
5'
5
Pythagoras cannot have been unaware of the truth of the
proposition, proved in Eucl. 11. 4, that (a + b) = a + 2ab+b², where
a, b are any two straight lines, for this proposition depends solely
upon propositions in Book I. which precede the Pythagorean
proposition 1. 47 and which, as the basis of 1. 47, must necessarily
have been in substance known to its author. A slightly different
geometrical proof would give the formula (a - b)² = a² — 2ab + b²,
which must have been equally well known to Pythagoras. It could
not therefore have escaped the discoverer of the first approximation
√50-1 for √50 that the use of the formula with the positive sign
would give a much nearer approximation, viz. 7 +
which is only
1
14'
ARITHMETIC IN ARCHIMEDES.
lxxix
2
greater than √50 to the extent of
Thus we may properly
assign to Pythagoras the discovery of the fact represented by
7 1/14 > √50 > 7.
3 doc.pi
The consequential result that √2> √50-1 is used by
Aristarchus of Samos in the 7th proposition of his work On the
size and distances of the sun and moon*.
With reference to the investigations of the values of √3, √5,
√6,......17 by Theodorus, it is pretty certain that √3 was
geometrically represented by him, in the same way as it appears
* Part of the proof of this proposition was a sort of foretaste of the first part
of Prop. 3 of Archimedes' Measurement of a
circle, and the substance of it is accordingly
appended as reproduced by Hultsch.
ABEK is a square, KB a diagonal, HBE
= 4KBE, LFBE=3°, and AC is perpendicu-
lar to BF so that the triangles ACB, BEF are
similar.
Aristarchus seeks to prove that
AB: BC > 18: 1.

If R denote a right angle, the angles KBE,
HBE, FBE are respectively 80R, 15R, ¿R.
Then HE : FE > LHBE: LFBE.
A
B
K
H
D
F
E
[This is assumed as a known lemma by Aristarchus as well as Archimedes.]
Therefore
HE: FE > 15 : 2……………..
Now, by construction,
BK2=2BE2.
Also [Eucl. vi. 3]
BK: BE=KH : HE;
..(a).
whence
And, since
KH=√2HE.
√2 >
50-1
25
so that
KH: HE > 7: 5,
KE : EH > 12 : 5 ......
.(B).
From (a) and (B), ex aequali,
KE : FE > 18 : 1.
Therefore, since
BF > BE (or KE),
BF : FE > 18 : 1,
so that, by similar triangles,
AB: BC > 18: 1.
lxxx
INTRODUCTION.
1
afterwards in Archimedes, as the perpendicular from an angular
point of an equilateral triangle on the opposite side. It would
thus be readily comparable with the side of the "1 square foot"
mentioned by Plato. The fact also that it is the side of three
square feet (rpírovs dúvapus) which was proved to be incommensurable
suggests that there was some special reason in Theodorus' proof for
specifying feet, instead of units of length simply; and the ex-
planation is probably that Theodorus subdivided the sides of his
triangles in the same way as the Greek foot was divided into
halves, fourths, eighths and sixteenths. Presumably therefore,
50
exactly as Pythagoras had approximated to √2 by putting 25
48
for 2, Theodorus started from the identity 3
be clear that
It would then
16*
48 + 1
7
√3<
i.e.
>
16
4°
To investigate √48 further, Theodorus would put it in the form
√49–1, as Pythagoras put √50 into the form √49 + 1, and the
result would be
1
√/48 (= √49 − 1) <7
— —
14'
We know of no further investigations int incommensurable
square roots until we come to Archimedes.
§ 7. Archimedes' approximations to √3.
Seeing that Aristarchus of Samos was still content to use the
first and very rough approximation to √2 discovered by Pythagoras,
it is all the more astounding that Aristarchus' younger contemporary
Archimedes should all at once, without a word of explanation, give
out that
1351
780
265
> √3: >
153'
as he does in the Measurement of a circle.
In order to lead up to the explanation of the probable steps by
which Archimedes obtained these approximations, Hultsch adopts
the same method of analysis as was used by the Greek geometers in
solving problems, the method, that is, of supposing the problem
solved and following out the necessary consequences.
To compare
ARITHMETIC IN ARCHIMEDES.
lxxxi
the two fractions
265
153
and
1351
780
we first divide both denominators
into their smallest factors, and we obtain
780 2.2.3.5.13,
153 = 3.3.17.
We observe also that 2.2.13 52, while 3. 1751, and we may
therefore show the relations between the numbers thus,
780
3.5.52,
1533.51.
For convenience of comparison we multiply the numerator and
265
denominator of by 5; the two original fractions are then
153
1351
1325
and
15.52
15.51'
so that we can put Archimedes' assumption in the form
1351
52
> 15√√3>
1325
51,
and this is seen to be equivalent to
1
26-
12 > 15 √3 > 26-
52
51'
1
2
Now 26
-
52
26º — 1 + (533)
and the latter expression
}
is an approximation to √26 - 1.
1
We have then
26-
52>√√26º — 1.
1
52
As 26- was compared with 15/3, and we want an ap-
proximation to √3 itself, we divide by 15 and so obtain
that
1
15
But √26ª – 1
1
15
(26 - 5712):
1
√√26-1.
676-1
225
√
675
=
√3, and it follows
225
1
15 (26 - 512) > √/5.
The lower limit for √3 was given by
H. A.
-
√3 > 1—15 (26 − 1),
51
f
:
A
lxxxii
INTRODUCTION.
¡
and a glance at this suggests that it may have been arrived at by
simply substituting (52–1) for 52.
Now as a matter of fact the following proposition is true. If
a³±b is a whole number which is not a square, while a² is the nearest
square number (above or below the first number, as the case may be),
then
b
= 2 > √ a² + b > a ±
a± 2 a
b
2a+1'
Hultsch proves this pair of inequalities in a series of propositions
formulated after the Greek manner, and there can be little doubt
that Archimedes had discovered and proved the same results in
substance, if not in the same form. The following circumstances
confirm the probability of this assumption.
(1) Certain approximations given by Heron show that he
knew and frequently used the formula
√a²±b~a±
b
2a'
(where the sign ~ denotes "is approximately equal to ").
Thus he gives
(2) The formula
1
√50~7+
14'
1
√63~8
16'
11
√75~8+
16*
b
a+ba+
is used by the Arabian
2α + 1
Alkarkhi (11th century) who drew from Greek sources (Cantor,
p. 719 sq.).
It can therefore hardly be accidental that the formula
b
2α
a+ >√√a³±b>a±
b
2a + 1
gives us what we want in order to obtain the two Archimedean
approximations to √3, and that in direct connexion with one
another*.
* Most of the a priori theories as to the origin of the approximations are
open to the serious objection that, as a rule, they give series of approximate
values in which the two now in question do not follow consecutively, but are
separated by others which do not appear in Archimedes. Hultsch's explanation
is much preferable as being free from this objection. But it is fair to say that
the actual formula used by Hultsch appears in Hunrath's solution of the puzzle
ARITHMETIC IN ARCHIMEDES.
lxxxiii
We are now in a position to work out the synthesis as follows.
From the geometrical representation of √3 as the perpendicular
from an angle of an equilateral triangle on the opposite side we
obtain √2-1 = √3 and, as a first approximation,
2–1 > √3.
4
Using our formula we can transform this at once into
25
,
√3>2
1
or 2
?
4-1
Archimedes would then square (2-1)
which he would compare with 3, or
,
1
3'
5
or and would obtain
27
9
3'
; i.e. he would put
25+2
√3 =
and would obtain
9
5+
3
1 ( 5 + 1 ) > √3, i.e.
26
15
> √√3.
15/',
√3
To obtain a still nearer approximation, he would proceed in the
same márner and compare
would appear that
26\² 676
225'
675
with 3, or
whence it
225'
262-1
or
225
1
and therefore that
that is,
15 (26 - 512) > √3,
780
The application of the formula would then give the result
1351
> √3.
1
√3>
15
(26
1
52–
:1),
1326-1
265
that is,
√√3>
or
15.51
153'
The complete result would therefore be
•
1351
> √3>
780
265
153*
(Die Berechnung irrationaler Quadratwurzeln vor der Herrschaft der Decimal-
brüche, Kiel, 1884, p. 21; cf. Ueber das Ausziehen der Quadratwurzel bei
Griechen und Indern, Hadersleben, 1883), and the same formula is implicitly
used in one of the solutions suggested by Tannery (Sur la mesure du cercle
d'Archimède in Mémoires de la société des sciences physiques et naturelles de
Bordeaux, 2ª série, iv. (1882), p. 313-337).
ƒ2
lxxxiv
INTRODUCTION.
Thus Archimedes probably passed from the first approximation
4 3'
7 5
5 26
to from to
3 15'
26
and from directly to
1351
780
the closest
15
approximation of all, from which again he derived the less close
approximation
265
153'
nearer approximation than
The reason why he did not proceed to a still
1351
780
is probably that the squaring of
this fraction would have brought in numbers much too large to be
conveniently used in the rest of his calculations. A similar reason
will account for his having started from instead of 7; if he had
5
3
4
used the latter, he would first have obtained, by the same method,
97
or >√3; the squaring
56
√3 =
√
49-1
16
7-1/4
and thence
7—*> √3,
4
97
√972-1
and the corresponding
where again the numbers
56.194'
of would have given √3=
56
approximation would have given
56
18817
are inconveniently large for his purpose.
§ 8.
numbers.
Approximations to the square roots of large
Archimedes gives in the Measurement of a circle the following
approximate values:
301339082321,
(1)
+
(2)
18383380929,
(3)
100911018405,
(4)
2017√4069284,
(5)
(6)
(7)
5911<√349450,
11723<√137394333,
23391◄√5472132.
There is no doubt that in obtaining the integral portion
of the square root of these numbers Archimedes used the method
based on the Euclidean theorem (a + b)² = a² + 2ab+b² which has
ARITHMETIC IN ARCHIMEDES.
lxxxv
already been exemplified in the instance given above from Theon,
where an approximation to 4500 is found in sexagesimal fractions.
The method does not substantially differ from that now followed; but
whereas, to take the first case, √9082321, we can at once see what
will be the number of digits in the square root by marking off pairs
of digits in the given number, beginning from the end, the absence
of a sign for 0 in Greek made the number of digits in the square
root less easy to ascertain because, as written in Greek, the number
Mẞтka' only contains six signs representing digits instead of seven.
Even in the Greek notation however it would not be difficult to see
that, of the denominations, units, tens, hundreds, etc. in the square
root, the units would correspond to ka' in the original number, the
tens to ẞr, the hundreds to M, and the thousands to M. Thus it
would be clear that the square root of 9082321 must be of the form
an
1000x + 100y+ 10z +w,
where x, y, z, w can only have one or other of the values 0, 1, 2, ….. 9.
Supposing then that x is found, the remainder N-(1000x)², where
N is the given number, must next contain 2.1000x. 100y and
(100y)³, then 2 (1000x + 100y). 10% and (10%), after which the
remainder must contain two more numbers similarly formed.
In the particular case (1) clearly x=3. The subtraction of
(3000)* leaves 82321, which must contain 2.3000. 100y. But, even
if y is as small as 1, this product would be 600,000, which is greater
than 82321. Hence there is no digit representing hundreds in the
square root. To find z, we know that 82321 must contain
2.3000.10% + (10%)³,
and z has to be obtained by dividing 82321 by 60,000. Therefore
=1. Again, to find w, we know that the remainder
(82321-2.3000. 10-10²),
or 22221, must contain 2.3010w+w², and dividing 22221 by
2.3010 we see that w=3. Thus 3013 is the integral portion of
the square root, and the remainder is 22221 –(2.3010.3 +3²), or
4152.
The conditions of the proposition now require that the approxi-
mate value to be taken for the square root must not be less than
lxxxvi
INTRODUCTION.
} -
the real value, and therefore the fractional part to be added to 3013
must be if anything too great. Now it is easy to see that the
fraction to be added is greater than
1
because 2.3013.
2
3 · 1/2 + ( 1 ) ²
is
less than the remainder 4152.
required (which is nearer to 3014 than to 3013) is 3014-
Suppose then that the number
p
and has to be if anything too small.
q
Now (3014) (3013)+2.3013+1 (3013)² + 6027
whence
=
= 9082321 – 4152 + 6027,
9082321 = (3014)³ – 1875.
By applying Archimedes' formula √a³±b<a±
P
b
we obtain
2α'
3014-
1875
2.3014
9082321.
1875
The required value has therefore to be not greater than
Չ
It remains to be explained why Archimedes put for the value
P
q
6028
1
4
1507
which is equal to
In the first place, he evidently preferred
6028
fractions with unity for numerator and some power of 2 for
denominator because they contributed to ease in working, e.g. when
two such fractions, being equal to each other, had to be added.
9
(The exceptions, the fractions and 10
11
are to be explained by
exceptional circumstances presently to be mentioned.) Further, in
the particular case, it must be remembered that in the subsequent
work 2911 had to be added to 3014-? and the sum divided by 780,
q
or 2.2.3.5.13. It would obviously lead to simplification if a
factor could be divided out, e.g. the best for the purpose, 13. Now,
dividing 2911 + 3014, or 5925, by 13, we obtain the quotient 455,
and a remainder 10, so that 10-2 remains to be divided by 13.
q
Therefore has to be so chosen that 10q -p is divisible by 13, while
P
g
p
approximates to, but is not greater than,
Չ
1875
6028'
The solution
p=1, q=4 would therefore be natural and easy.
ARITHMETIC IN ARCHIMEDES.
lxxxvii
(2) √3380929.
The usual process for extraction of the square root gave as the
integral part of it 1838, and as the remainder 2685. As before, it
was easy to see that the exact root was nearer to 1839 than to 1838,
and that
√33809291838² + 2685 = 1839 - 2.1838-1 +2685
=
= 1839º — 992.
The Archimedean formula then gave
1839
992
2.1839
> √3380929.
1
It could not have escaped Archimedes that 4
was a near approxima-
tion to
or
992 1984
3678 7356'
since
1 1839
4 7356
1
; and
would have satisfied
4
+
the necessary condition that the fraction to be taken must be less
than the real value. Thus it is clear that, in taking
2
11
as the
approximate value of the fraction, Archimedes had in view the
simplification of the subsequent work by the elimination of a factor.
If the fraction be denoted by p
3662
p
the sum of 1839. and 1823, or
-2,
q
had to be divided by 240, i.e. by 6.40.
Չ
Division of 3662
be so chosen that
զ
by 40 gave 22 as remainder, and then p, q had to
22-P
q
was conveniently divisible by 40, while P
approximately equal to
q
was less than but
992
3678'
The solution p=2, q= 11 was easily
seen to satisfy the conditions.
(3) √1018405.
The usual procedure gave 1018405 = 1009²+324 and the ap-
proximation
324
2018
1009 > √1018405.
324
It was here necessary that the fraction to replace
should be
2018
1
greater but approximately equal to it, and satisfied the conditions,
6
while the subsequent work did not require any change in it.
Į
lxxxviii
INTRODUCTION.
(4) √4069284.
36
The usual process gave 4069284 2017 + 995; it followed
that
2017 +
36.995 + 1
36.2.2017
>√40692848
69
and 2017 was an obvious value to take as an approximation
somewhat greater than the left side of the inequality.
(5) √349450.
In the case of this and the two following roots an approximation
had to be obtained which was less, instead of greater, than the true
value. Thus Archimedes had to use the second part of the formula
a±
b
2α
> √a³±b>a±
b
2a+1'
In the particular case of √349450 the integral part of the root is
591, and the remainder is 169. This gave the result
169
591 + > √349450>591 +
2.591
169
2.591+1'
and since 16913, while 2.591+1=7.139, it resulted without
further calculation that
√349450 > 5914.
Why then did Archimedes take, instead of this approximation,
another which was not so close, viz. 5911? The answer which the
subsequent working and the other approximations in the first part of
the proof suggest is that he preferred, for convenience of calculation,
to use for his approximations fractions of the form 1 only. But he
2n
could not have failed to see that to take the nearest fraction of this
1
form,
8'
1
instead of might conceivably affect his final result and
7
make it less near the truth than it need be. As a matter of fact,
as Hultsch shows, it does not affect the result to take 591 and to
work onwards from that figure. Hence we must suppose that
Archimedes had satisfied himself, by taking 5914 and proceeding on
that basis for some distance, that he would not be introducing any
appreciable error in taking the more convenient though less accurate
approximation 5911.
ARITHMETIC IN ARCHIMEDES.
lxxxix
(6)
137394338.
In this case the integral portion of the root is 1172, and the
remainder 35933.
Thus, if R denote the root,
R> 1172+
35933
2.1172+1
64
> 1172 +
359
2.1172+1'
a fortiori.
W
Now 2.1172+1=2345; the fraction accordingly becomes
1 359
2513
359.
2345'
and
satisfies the necessary conditions, viz. that it must
be approximately equal to, but not greater than, the given fraction.
Here again Archimedes would have taken 11724 as the approximate
value but that, for the same reason as in the last case, 11721 was
more convenient.
(7) √5472132.
The integral portion of the root is here 2339, and the remainder
1211, so that, if R is the exact root,
R>2339+
16
12111
2.2339 +1
> 23391, a fortiori.
A few words may be added concerning Archimedes' ultimate
reduction of the inequalities
3+
667/
46731
2841
>>3+
20171
1
10
to the simpler result
3
>>3
T
7
71°
1
6671/
As a matter of fact
so that in the first fraction it was
7- 467212
>
only necessary to make the small change of diminishing the de-
1
nominator by 1 in order to obtain the simple 3.
As regards the lower limit for π, we see that
2841 1137
20171 8069
and
Hultsch ingeniously suggests the method of trying the effect of
increasing the denominator of the latter fraction by 1. This
XC
INTRODUCTION.
1137 379
produces or ; and, if we divide 2690 by 379, the quotient
8070 2690
is between 7 and 8, so that
1
379 1
>
>
7 2690 8
Now it is a known proposition (proved in Pappus VII. p. 689)
α C
that, if > then
b d'
α a + c
>
ō b+d'
Similarly it may be proved that
a + c с
b+d² d'
It follows in the above case that
379
2690
379 +1
1
>
2690+88'
which exactly gives
10 1
71 8'
10.
379
71
and is very much nearer to
2690
than f
1
is.
Note on alternative hypotheses with regard to the
approximations to √3.
For a description and examination of all the various theories put
forward, up to the year 1882, for the purpose of explaining Archimedes'
approximations to √3 the reader is referred to the exhaustive paper by
Dr Siegmund Günther, entitled Die quadratischen Irrationalitäten der Alten
und deren Entwickelungsmethoden (Leipzig, 1882). The same author gives
further references in his Abriss der Geschichte der Mathematik und der Natur-
wissenschaften im Altertum forming an Appendix to Vol. v. Pt. 1 of Iwan von
Müller's Handbuch der klassischen Altertums-wissenschaft (München, 1894).
Günther groups the different hypotheses under three general heads :
(1) those which amount to a more or less disguised use of the
method of continued fractions and under which are included the solutions
of De Lagny, Mollweide, Hauber, Buzengeiger, Zeuthen, P. Tannery (first
solution), Heilermann;
(2) those which give the approximations in the form of a series
of fractions such as a + + + +...; under this class come the
1
1
1
21 9192 919293
solutions of Radicke, v. Pessl, Rodet (with reference to the Çulvasutras),
Tannery (second solution);
•
ARITHMETIC IN ARCHIMEDES.
xci
(3) those which locate the incommensurable surd between a greater
and lesser limit and then proceed to draw the limits closer and closer.
This class includes the solutions of Oppermann, Alexejeff, Schönborn,
Hunrath, though the first two are also connected by Günther with the
method of continued fractions.
Of the methods so distinguished by Günther only those need be here
referred to which can, more or less, claim to rest on a historical basis
in the sense of representing applications or extensions of principles laid
down in the works of Greek mathematicians other than Archimedes which
have come down to us. Most of these quasi-historical solutions connect
themselves with the system of side- and diagonal-numbers (λevpikoì and
diaμeтρikoì ápiμoi) explained by Theon of Smyrna (c. 130 A.D.) in a work
which was intended to give so much of the principles of mathematics as
was necessary for the study of the works of Plato.
The side- and diagonal-numbers are formed as follows. We start with
two units, and (a) from the sum of them, (b) from the sum of twice
the first unit and once the second, we form two new numbers; thus
1.1+1=2,
2.1+1=3.
Of these numbers the first is a side- and the second a diagonal-number
respectively, or (as we may say)
Ag=2,
dq=3.
In the same way as these numbers were formed from a₁=1, d₁=1, suc-
cessive pairs of numbers are formed from a₂, d2, and so on, in accordance
with the formula
whence we have
an+1=αn+dn,
ag=1.2+3=5,
a₁=1.5+7=12,
dn+1=2αn+dn,
dz=2.2+3=7,
d=2.5+7=17,
and so on.
Theon states, with reference to these numbers, the general proposition
which we should express by the equation
dr²=2a,2±1.
The proof (no doubt omitted because it was well-known) is simple. For
we have
d2-2a,²=(2αn-1+dn-1)-2(an-1+dn-1)²
་
=2an-12-dn-12
=-(d₂-12-2αn-12)
=
+(dn-22-2αn-22), and so on,
2
while d₁² — 2a₁² = -1; whence the proposition is established.
Cantor has pointed out that any one familiar with the truth of this
proposition could not have failed to observe that, as the numbers were
successively formed, the value of d²/an² would approach more and more
nearly to 2, and consequently the successive fractions dn/an would give
2 2
α
xcii
INTRODUCTION.
nearer and nearer approximations to the value of √2, or in other words that
1 3 7 17 41
1' 2' 5' 12' 29
are successive approximations to √2. It is to be observed that the third
7
5'
of these approximations, is the Pythagorean approximation which
appears to be hinted at by Plato, while the above scheme of Theon,
amounting to a method of finding all the solutions in positive integers of
the indeterminate equation
2x2 y2±1,
A
and given in a work designedly introductory to the study of Plato,
distinctly suggests, as Tannery has pointed out, the probability that even
in Plato's lifetime the systematic investigation of the said equation had
already begun in the Academy. In this connexion Proclus' commentary
on Eucl. I. 47 is interesting. It is there explained that in isosceles
right-angled triangles "it is not possible to find numbers corresponding to
the sides; for there is no square number which is double of a square
except in the sense of approximately double, e.g. 72 is double of 52 less 1."
When it is remembered that Theon's process has for its object the finding
of any number of squares differing only by unity from double the squares
of another series of numbers respectively, and that the sides of the two
sets of squares are called diagonal- and side-numbers respectively, the
conclusion becomes almost irresistible that Plato had such a system in
mind when he spoke of pnrn diάuerpos (rational diagonal) as compared
with äppηtos diáμerpos (irrational diagonal) τŷs ñeµñádos (cf. p. lxxviii above).
One supposition then is that, following a similar line to that by which
successive approximations to √2 could be obtained from the successive
solutions, in rational numbers, of the indeterminate equations 2x² - y²= ±1,
Archimedes set himself the task of finding all the solutions, in rational
numbers, of the two indeterminate equations bearing a similar relation
to √/3, viz.
x² — 3y²=1,
x² — 3y² — — 2.
Zeuthen appears to have been the first to connect, eo nomine, the ancient
approximations to √3 with the solution of these equations, which are also
made by Tannery the basis of his first method. But, in substance, the
same method had been used as early as 1723 by De Lagny, whose
hypothesis will be, for purposes of comparison, described after Tannery's
which it so exactly anticipated.
Zeuthen's solution.
After recalling the fact that, even before Euclid's time, the solution
of the indeterminate equation x²+y²=z² by means of the substitutions
x=mn, y=
m² - n²
2
2=
m²+n²
2
ARITHMETIC IN ARCHIMEDES.
xciii
was well known, Zeuthen concludes that there could have been no
difficulty in deducing from Eucl. II. 5 the identity
2
2
3 (mn)² + (m² = 3n³)² = (m² + 3n³) ³
-
2
2
"
from which, by multiplying up, it was easy to obtain the formula
3 (2mn)²+(m² - 3n²)²= (m²+3n²)².
If therefore one solution m²-3n²=1 was known, a second could at once
be found by putting
x=m²+3n²,
y=2mn.
m² — 3n² = 1
Now obviously the equation
is satisfied by the values m=2, n=1; hence the next solution of the
equation
is
x²-3y²=1
x₁=22+3.1=7, Y₁=2.2.1=4;
and, proceeding in like manner, we have any number of solutions as
X2=72+3.42=97, Y2=2.7.4 56,
x3=97²+3.56²=18817,
and so on.
y3=2.97.56=10864,
Next, addressing himself to the other equation
Zeuthen uses the identity
x² — 3y² — — 2,
(m+3n)2-3 (m+n)²= -2 (m² - 3n²).
Thus, if we know one solution of the equation m² — 3n²=1, we can proceed
to substitute
x=m+3n, y=m+n.
Suppose m=2, n=1, as before; we then have
x₁=5,
Y₁ = 3.
If we put x2=x₁+3y₁=14, y₂=x₁+y₁=8, we obtain
X2 14 7
Y2 8 4
(and m=7, n=4 is seen to be a solution of m² — 3n²= 1).
Starting again from x₂, y₂, we have
and
X3=38,
Y3=22,
X3
23 = 118
19
Y3
(m=19, n=11 being a solution of the equation m²-3n²= − 2);
whence
X4=104,
Y4=60,
X4
26
Y4
15
1
xciv
INTRODUCTION.
เ
(and m=26, n=15 satisfies m² - 3n² = 1),
or
Xg=284,
s=164,
X5 71
Y5 41
Similarly
'6
97 X7
Yo 56' YT
Ход
265
153'
and so on.
This method gives all the successive approximations to √3, taking
account as it does of both the equations
Tannery's first solution.
x²-3y²=1,
x²-3y²=-2.
Tannery asks himself the question how Diophantus would have set
about solving the two indeterminate equations. He takes the first equation
in the generalised form
x²-ay2=1,
and then, assuming one solution (p, q) of the equation to be known, he
supposes
Then
P1=mx-P, q1=x+q.
2
p₁²-aq₁²=m²x² - 2mpx+p²-ax²-2aqx-aq²=1,
whence, since p²-aq2=1, by hypothesis,
mp+aq
x=2.
m² - a
so that
P1=
(m²+a) p+2amq
m² - a
91
2mp+(m²+a) q
m² - a
2
and p₁²- aq₁²=1.
The values of P₁, q₁ so found are rational but not necessarily integral ;
if integral solutions are wanted, we have only to put
P₁ = (u²+ av²) p+2auvq, 91=2puv+(u²+av²) q,
where (u, v) is another integral solution of x2 — ay²=1.
Generally, if (p, q) be a known solution of the equation
x²-ay²=r,
suppose p₁=ap+ßq, q₁=yp+dq, and "il suffit pour déterminer a, ß, y, d de
connaître les trois groupes de solutions les plus simples et de résoudre
deux couples d'équations du premier degré à deux inconnues." Thus
(1) for the equation
the first three solutions are
whence
so that
x²-3y2=1,
(p=1, q=0), (p=2, q=1), (p=7, q=4),
2=a) 7=2a+Br
and
1=y√ 4=2y+df
a=2, B=3, y=1, 8=2,
,
ARITHMETIC IN ARCHIMEDES.
XCV
and it follows that the fourth solution is given by
(2) for the equation
p=2.7+3.4=26,
q=1.7+2.4=15;
x²-3y²= -2,
the first three solutions being (1, 1), (5, 3), (19, 11), we have
5=a+B) 19=5a+38)
and
3=y+df 11=5y+38)'
whence a=2, B=3, y=1, d=2, and the next solution is given by
p=2.19+3.11=71,
q=1.19+2.11=41,
and so on.
Therefore, by using the two indeterminate equations and proceeding as
shown, all the successive approximations to √3 can be found.
Of the two methods of dealing with the equations it will be seen that
Tannery's has the advantage, as compared with Zeuthen's, that it can be
applied to the solution of any equation of the form x² — ay²=r.
De Lagny's method.
The argument is this. If √3 could be exactly expressed by an im-
proper fraction, that fraction would fall between 1 and 2, and the square of
its numerator would be three times the square of its denominator. Since
this is impossible, two numbers have to be sought such that the square of
the greater differs as little as possible from 3 times the square of the
smaller, though it may be either greater or less. De Lagny then evolved
the following successive relations,
viz.
2²=3.1²+1, 52=3.32-2, 7²=3.4²+1, 192=3.112-2,
26²=3.15²+1, 71²=3.41²— 2, etc.
From these relations were derived a series of fractions greater than √3,
etc., and another series of fractions less than √3, viz.
2 7 26
1' 4' 15'
5 19 71
etc. The law of formation was found in each case to be that, if
3'11' 41'
P
Չ
was one fraction in the series and 2 the next, then
This led to the results
p' _ 2p+3q
q'・ p+2q
?>
26
97
362 1351
>
>√√3,
15 56 209 780
5 19
71
265 989 3691
and
<√3;
41
153 571 2131
xcvi
INTRODUCTION.
while the law of formation of the successive approximations in each series
is precisely that obtained by Tannery as the result of treating the two
indeterminate equations by the Diophantine method.
Heilermann's method.
This method needs to be mentioned because it also depends upon a
generalisation of the system of side- and diagonal-numbers given by Theon
of Smyrna.
Theon's rule of formation was
Sn=-1+Dn-1, Du=2n-1+Dn-1;
and Heilermann simply substitutes for 2 in the second relation any
arbitrary number a, developing the following scheme,
S₁ =So+Do D₁=aSo+Do,
It follows that
S₂=S₁+D1,
Dq=aS₁+D1,
S3=S₂+D2,
D3=aS₂+D2,
Sn=Sn-1+Dn-1, Dn=aSn-1+Dn−1·
aS₁²=aS₂-12+2aSn-1Dn-1+aDn-12,
D₂²=a2Sn-12+2aSn-1Dn-1+D-12.
2
2
By subtraction, D₁² - aSm² = (1 − a) (Dn-1² — αSn-12)
=(1-α)² (D₂-22-aS-22), similarly,
=(1-a)" (D2-aS2).
This corresponds to the most general form of the "Pellian" equation
x²-ay2= (const.).
If now we put Do=So=1, we have
D₂2
(1 − a)n+1
=a+
S₁₂2
Sn²
2
9
from which it appears that, where the fraction on the right-hand side
Dn
approaches zero as n increases, is an approximate value for a.
Clearly in the case where a=3, D。=2, S。=1 we have
2
4
52 26
Do_2 D1_5 D₂_14 7 D3_19 DA
So I' S₁ 3' S2 8 4' S 11' S4 30 15'
D5_71 D6 194 97 Dr 265
S 41' S
112
56' S 153'
and so on.
=
ARITHMETIC IN ARCHIMEDES.
xcvii
1
But the method is, as shown by Heilermann, more rapid if it is used to
find, not √a, but b√a, where b is so chosen as to make b²a (which takes
the place of a) somewhat near to unity. Thus suppose a=
3
√a= √3, and we then have (putting D₁=√。=1)
52
S₁=2, D₁= and √3.26
25'
27
so that
25'
or
3'25
26
15'
102
54+52 106
5 106
S₂-
D₂=
25 '
,
and 1/3 2
265
25
25
3'102' 153'
or
208
102.27 106
5404
S3=
D3=
25
+
25.25 25 25.25'
and
5404 5
√3~
or
25.208'3
1351
780
This is one of the very few instances of success in bringing out the two
Archimedean approximations in immediate sequence without any foreign
values intervening. No other methods appear to connect the two values
in this direct way except those of Hunrath and Hultsch depending on the
formula
a±
b
2α
b
>√a²±b>a±2α±1°
We now pass to the second class of solutions which develops the
approximations in the form of the sum of a series of fractions, and under
this head comes
Tannery's second method.
This may be exhibited by means of its application (1) to the case of the
square root of a large number, e.g. √349450 or 5712+23409, the first of
the kind appearing in Archimedes, (2) to the case of /3.
(1) Using the formula
b
√a²+b~a+ 2α'
we try the effect of putting for √√/5712+23409 the expression
571+
23409
1142
It turns out that this gives correctly the integral part of the root, and we
now suppose the root to be
Squaring and regarding
1
m²
571+20+
1
m
as negligible, we have
1142 40
5712+400+22840+ + =571²+23409,
m
m
H. A.
g
xcviii
INTRODUCTION.
1
whence
1182
= 169,
m
and
so that
1 169 1
m 1182
√/349450>591.
(2) Bearing in mind that
b
√a²+b∞a+
2a+1'
2
we have
√/3=√/1²+2~1+2.1+1
2
witz, or
5
3
Assuming then that √3 = (3+), squaring and neglecting
25 10
+ = 3,
9 3m
whence m=15, and we get as the second approximation
We have now
5 1
26
+ or
3 15' 15°
262-3.15²=1,
1
m²,
we obtain
and can proceed to find other approximations by means of Tannery's first
method.
1\2
=3,
put (1+++)² = 3,
Or we can also put
15
1
and, neglecting we get
n2,
262 52
+
152 15n
=3,
whence n=-15.52— — 780, and
√3 ~ ( 1 + 3 3 +
2
1 1
1351
15 780 780
1351
It is however to be observed that this method only connects
with
780
26
and not with the intermediate approximation
15
265
153'
to obtain which
Tannery implicitly uses a particular case of the formula of Hunrath and
Hultsch.
Rodet's method was apparently invented to explain the approximation
in the Çulvasūtras*
1
1
1
3.4
3.4.34;
* See Cantor, Vorlesungen über Gesch. d. Math. p. 600 sq.
ARITHMETIC IN ARCHIMEDES.
xcix
4
but, given the approximation the other two successive approximations
3'
indicated by the formula can be obtained by the method of squaring just
described without such elaborate work as that of Rodet, which, when
applied to √3, only gives the same results as the simpler method.
Lastly, with reference to the third class of solutions, it may be
mentioned
(1) that Oppermann used the formula
a+b
2 >Nab>
2ab
a+b'
which gave successively
2>√/3>300
2'
12
3>
7
97
168
56
√3>
97'
but only led to one of the Archimedean approximations, and that by
combining the last two ratios, thus
97 +168
56+97
265
153'
(2) that Schönborn came somewhat near to the formula successfully used
by Hunrath and Hultsch when he proved + that
b
2α
a±>√ a²±b>a±
Ъ
* Cantor had already pointed this out in his first edition of 1880.
+ Zeitschrift für Math. u. Physik (Hist. litt. Abtheilung) xXVIII. (1883),
p. 169 sq.
Vor M 22
F
31
2
CHAPTER V.
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ.
THE word veûois, commonly inclinatio in Latin, is difficult to
translate satisfactorily, but its meaning will be gathered from some
general remarks by Pappus having reference to the two Books of
Apollonius entitled veúσes (now lost). Pappus says, "A line is
said to verge (vevew) towards a point if, being produced, it reach the
point," and he gives, among particular cases of the general form of
the problem, the following.
"Two lines being given in position, to place between them a
straight line given in length and verging towards a given point."
"If there be given in position (1) a semicircle and a straight
line at right angles to the base, or (2) two semicircles with their
bases in a straight line, to place between the two lines a straight
line given in length and verging towards a corner (yovíav) of a
semicircle."
Thus a straight line has to be laid across two lines or curves so
that it passes through a given point and the intercept on it between
the lines or curves is equal to a given length†.
§ 1. The following allusions to particular vevores are found in
Archimedes. The proofs of Props. 5, 6, 7 of the book On Spirals
use respectively three particular cases of the general theorem that,
Pappus (ed. Hultsch) vII. p. 670.
+ In the German translation of Zeuthen's work, Die Lehre von den
Kegelschnitten im Altertum, veûoɩs is translated by "Einschiebung," or as we
might say "insertion," but this fails to express the condition that the required
line must pass through a given point, just as inclinatio (and for that matter the
Greek term itself) fails to express the other requirement that the intercept on
the line must be of given length.
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ.
ci
if A be any point on a circle and BC any diameter, it is possible to
draw through A a straight line, meeting the circle again in P and
BC produced in R, such that the intercept PR is equal to any given
A

P
B
D
C
R
length. In each particular case the fact is merely stated as true
without any explanation or proof, and
(1) Prop. 5 assumes the case where the tangent at A is parallel
to BC,
1
(2) Prop. 6 the case where the points A, P in the figure are
interchanged,.
(3) Prop. 7 the case where A, P are in the relative positions
shown in the figure.
B
P
A
R D
Again, (4) Props. 8 and 9 each assume (as before, without proof,
and without giving any solution of the
implied problem) that, if AE, BC be two
chords of a circle intersecting at right
angles in a point D such that BD> DC,
then it is possible to draw through A
another line ARP, meeting BC in R and
the circle again in P, such that PR = DE.
Lastly, with the assumptions in Props.
5, 6, 7 should be compared Prop. 8 of the
Liber Assumptorum, which may well be
due to Archimedes, whatever may be said of the composition of the
whole book. This proposition proves that, if in the first figure
APR is so drawn that PR is equal to the radius OP, then the arc
AB is three times the arc PC. In other words, if an arc AB of a
circle be taken subtending any angle at the centre O, an arc equal
to one-third of the given arc can be found, i.e. the given angle can be
trisected, if only APR can be drawn through A in such a manner
E
L

cii
INTRODUCTION.
Į
that the intercept PR between the circle and BO produced is equal to
the radius of the circle. Thus the trisection of an angle is reduced to
a vevois exactly similar to those assumed as possible in Props. 6, 7
of the book On Spirals.
The vevσes so referred to by Archimedes are not, in general,
capable of solution by means of the straight line and circle alone,
as may be easily shown. Suppose in the first figure that x
represents the unknown length OR, where O is the middle point
of BC, and that k is the given length to which PR is to be equal;
also let OD = a, AD = b, BC = 2c. Then, whether BC be a diameter
or (more generally) any chord of the circle, we have
and therefore
AR.RP=BR. RC,
k√b² + (x − a)² = x² - c².
The resulting equation, after rationalisation, is an equation of the
fourth degree in x; or, if we denote the length of AR by y, we have,
for the determination of x and y, the two equations
y² = (x − a)² + b³
ky = x² - c²
}....
........
In other words, if we have a rectangular system of coordinate
axes, the values of x and y satisfying the conditions of the problem
can be determined as the coordinates of the points of intersection of
a certain rectangular hyperbola and a certain parabola.
In one particular case, that namely in which D coincides with O
the middle point of BC, or in which A is one extremity of the
diameter bisecting BC at right angles, a=0, and the equations
reduce to the single equation
y² — ky = b² + c²,
which is a quadratic and can be geometrically solved by the
traditional method of application of areas; for, if u be substituted
for y-k,
y – k, so that u = AP, the equation becomes
u (k + u) = b² + c³,
and we have simply "to apply to a straight line of length k a
rectangle exceeding by a square figure and equal to a given
area (b²+ c²)."
The other vevous referred to in Props. 8 and 9 can be solved in
the more general form where k, the given length to which PR
is to be equal, has any value within a certain maximum and is not
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ.
ciii
necessarily equal to DE, in exactly the same manner; and the two
equations corresponding to (a) will be for the second figure
y² = ( a − x)² + b²
ky = c² — x²
}.....
..(B).
Here, again, the problem can be solved by the ordinary method
of application of areas in the particular case where AE is the
diameter bisecting BC at right angles; and it is interesting to note
that this particular case appears to be assumed in a fragment
of Hippocrates' Quadrature of lunes preserved in a quotation
by Simplicius* from Eudemus' History of Geometry, while Hippo-
crates flourished probably as early as 450 B.C.
Accordingly we find that Pappus distinguishes different classes
of veúσes corresponding to his classification of geometrical problems
in general. According to him, the Greeks distinguished three kinds
of problems, some being plane, others solid, and others linear. He
proceeds thus: "Those which can be solved by means of a straight
line and a circumference of a circle may properly be called plane
(emímeda); for the lines by means of which such problems are
(ἐπίπεδα)
solved have their origin in a plane. Those however which are
solved by using for their discovery (eupeow) one or more of the
sections of the cone have been called solid (σTepeά); for the
construction requires the use of surfaces of solid figures, namely,
those of cones. There remains a third kind of problem, that
which is called linear (ypaµµukóv); for other lines [curves] besides
those mentioned are assumed for the construction whose origin
is more complicated and less natural, as they are generated from
more irregular surfaces and intricate movements." Among other
instances of the linear class of curves Pappus mentions spirals, the
curves known as quadratrices, conchoids and cissoids. He adds
that "it seems to be a grave error which geometers fall into
whenever any one discovers the solution of a plane problem by
means of conics or linear curves, or generally solves it by means of
a foreign kind, as is the case, for example, (1) with the problem in
the fifth Book of the Conics of Apollonius relating to the parabola‡,
* Simplicius, Comment. in Aristot. Phys. pp. 61-68 (ed. Diels). The whole
quotation is reproduced by Bretschneider, Die Geometrie und die Geometer vor
Euklides, pp. 109–121. As regards the assumed construction see particularly
p. 64 and p. xxiv of Diels' edition; cf. Bretschneider, pp. 114, 115, and Zeuthen,
Die Lehre von den Kegelschnitten im Altertum, pp. 269, 270.
+ Pappus IV. pp. 270–272.
‡ Cf. Apollonius of Perga, pp. cxxviii. cxxix.
1
civ
INTRODUCTION.
5
and (2) when Archimedes assumes in his work on the spiral a
veûσis of a solid character with reference to a circle; for it is
possible without calling in the aid of anything solid to find the
[proof of the] theorem given by the latter [Archimedes], that is, to
prove that the circumference of the circle arrived at in the first
revolution is equal to the straight line drawn at right angles to the
initial line to meet the tangent to the spiral."
The “solid vevois" referred to in this passage is that assumed to
be possible in Props. 8 and 9 of the book On Spirals, and is mentioned
again by Pappus in another place where he shows how to solve the
problem by means of conics*. This solution will be given later, but,
when Pappus objects to the procedure of Archimedes as unorthodox,
the objection appears strained if we consider what precisely it is that
Archimedes assumes. It is not the actual solution which is assumed,
but only its possibility; and its possibility can be perceived without
any use of conics. For in the particular case it is only necessary,
as a condition of possibility, that DE in the second figure above
should not be the maximum length which the intercept PR could
have as APR revolves about A from the position ADE in the
direction of the centre of the circle; and that DE is not the
maximum length which PR can have is almost self-evident. In
fact, if P, instead of moving along the circle, moved along the
straight line through E parallel to BC, and if ARP moved from the
position ADE in the direction of the centre, the length of PR would
continually increase, and a fortiori, so long as P is on the arc of the
circle cut off by the parallel through E to BC, PR must be greater
in length than DE; and on the other hand, as ARP moves further
in the direction of B, it must sometime intercept a length PR
equal to DE before P reaches B, when PR vanishes. Since, then,
Archimedes' method merely depends upon the theoretical possibility
of a solution of the vevous, and this possibility could be inferred
from quite elementary considerations, he had no occasion to use
conic sections for the purpose immediately in view, and he cannot
fairly be said to have solved a plane problem by the use of conics.
At the same time we may safely assume that Archimedes
was in possession of a solution of the veûois referred to. But there
is no evidence to show how he solved it, whether by means of conics,
or otherwise. That he would have been able to effect the solution,
* Pappus IV. p. 298 sq.
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ.
CV
as Pappus does, by the use of conics cannot be doubted. A precedent
for the introduction of conics where a "solid problem" had to be
solved was at hand in the determination of two mean proportionals
between two unequal straight lines by Menaechnius, the inventor of
the conic sections, who used for the purpose the intersections of a
parabola and a rectangular hyperbola. The solution of the cubic
equation on which the proposition On the Sphere and Cylinder 11. 4
depends is also effected by means of the intersections of a parabola
with a rectangular hyperbola in the fragment given by Eutocius
and by him assumed to be the work of Archimedes himself *.
Whenever a problem did not admit of solution by means of the
straight line and circle, its solution, where possible, by means of
conics was of the greatest theoretical importance. First, the
possibility of such a solution enabled the problem to be classified
as a "solid problem"; hence the importance attached by Pappus
to solution by means of conics. But, secondly, the method had
other great advantages, particularly in view of the requirement that
the solution of a problem should be accompanied by a Stopioµós
giving the criterion for the possibility of a real solution. Often too
the diopɩoμós involved (as frequently in Apollonius) the determination
of the number of solutions as well as the limits for their possibility.
Thus, in any case where the solution of a problem depended on the
intersections of two conics, the theory of conics afforded an effective
means of investigating διορισμοί.
§ 2. But though the solution of "solid problems" by means of
conics had such advantages, it was not the only method open to
Archimedes. An alternative would be the use of some mechanical
construction such as was often used by the Greek geometers and is
recognised by Pappus himself as a legitimate substitute for conics,
which are not easy to draw in a planet. Thus in Apollonius'
solution of the problem of the two mean proportionals as given by
Eutocius a ruler is supposed to be moved about a point until the
points at which the ruler crosses two given straight lines at right
angles are equidistant from a certain other fixed point; and the
same construction is also given under Heron's name. Another
version of Apollonius' solution is that given by Ioannes Philoponus,
which assumes that, given a circle with diameter OC and two
* See note to On the Sphere and Cylinder, 11. 4.
Pappus III. p. 54.
豐
​cvi
INTRODUCTION.
straight lines OD, OE through O and at right angles to one
another, a line can be drawn through C, meeting the circle again
in F and the two lines in D, E respectively, such that the in-
tercepts CD, FE are equal. This solution was no doubt discovered
by means of the intersection of the circle with a rectangular hyper-
bola drawn with OD, OE as asymptotes and passing through C;
and this supposition accords with Pappus' statement that Apollonius
solved the problem by means of the sections of the cone*. The
equivalent mechanical construction is given by Eutocius as that
of Philo Byzantinus, who turns a ruler about C until CD, FE are
equal†.
νεσις.
►
Now clearly a similar method could be used for the purpose of
effecting a vevous. We have only to suppose a ruler (or any object
with a straight edge) with two marks made on it at a distance
equal to the given length which the problem requires to be
intercepted between two curves by a line passing through the
fixed point; then, if the ruler be so moved that it always passes
through the fixed point, while one of the marked points on it follows
the course of one of the curves, it is only necessary to move the
ruler until the second marked point falls on the other curve. Some
such operation as this may have led Nicomedes to the discovery of
his curve, the conchoid, which he introduced (according to Pappus)
into his doubling of the cube, and by which he also trisected an
angle (according to the same authority). From the fact that
Nicomedes is said to have spoken disrespectfully of Eratosthenes'
mechanical solution of the duplication problem, and therefore must
have lived later than Eratosthenes, it is concluded that his date
must have been subsequent to 200 B.C., while on the other hand
he must have written earlier than 70 B.C., since Geminus knew the
name of the curve about that date; Tannery places him between
Archimedes and Apolloniust. While therefore there appears to
be no evidence of the use, before the time of Nicomedes, of such
a mechanical method of solving a vevous, the interval between
Archimedes and the discovery of the conchoid can hardly have
been very long. As a matter of fact, the conchoid of Nicomedes
can be used to solve not only all the revσes mentioned in Archimedes
but any case of such a problem where one of the curves is a straight
* Pappus III. p. 56.
+ For fuller details see Apollonius of Perga, pp. cxxv—cxxvii.
‡ Bulletin des Sciences Mathématiques, 2º série vïI. p. 284,
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ.
cvii
line. Both Pappus and Eutocius attribute to Nicomedes the inven-
tion of a machine for drawing his conchoid. AB is supposed to be
A
B
1

E
a ruler with a slot in it parallel to its length, FE a second ruler at
right angles to the first with a fixed peg in it, C. This peg moves
in a slot made in a third ruler parallel to its length, while this
ruler has a fixed peg on it, D, in a straight line with the slot in
which C moves; and the peg D can move along the slot in AB. If
then the ruler PD moves so that the peg D describes the length of
the slot in AB on each side of F, the extremity of the ruler, P,
describes the curve which is called a conchoid. Nicomedes called
the straight line AB the ruler (kavov), the fixed point C the pole
(Tóλos), and the length PD the distance (Siάornua); and the
fundamental property of the curve, which in polar coordinates
would now be denoted by the equation r = a + b sec 0, is that, if
any radius vector be drawn from C to the curve, as CP, the length
intercepted on the radius vector between the curve and the straight
line AB is constant. Thus any veûous in which one of the two
given lines is a straight line can be solved by means of the
intersection of the other line with a certain conchoid whose pole
is the fixed point to which the required straight line must verge
(veúewv). In practice Pappus tells us that the conchoid was not
always actually drawn, but that "some," for greater convenience,
moved the ruler about the fixed point until by trial the intercept
was made equal to the given length*.
§ 3. The following is the way in which Pappus applies
conic sections to the solution of the vevois referred to in Props. 8, 9
of the book On Spirals. He begins with two lemmas.
* Pappus IV. p. 246.
(
}-
2
cviii
INTRODUCTION.
(1) If from a given point A any straight line be drawn meeting
a straight line BC given in position in R, and if RQ be drawn
perpendicular to BC and bearing a given ratio to AR, the locus of
Q is a hyperbola.

A"
A
B
OR
D
A'
N'
P
N
K
E
For draw AD perpendicular to BC, and on AD produced take A'
such that
QR: RA=A'D: DA = (the given ratio).
Measure DA" along DA equal to DA'.
or
Then, if QN be perpendicular to AN,
(AR² – AD²): (QR² – A'D²) = (const.),
QN° : A'N. A″N = (const.)
(2) If BC be given in length, and if RQ, a straight line drawn
at right angles to BC from any point R on it, be such that
BR. RC = k. RQ,
where k is a straight line of given length, then the locus of Q is a
parabola.
Let ◇ be the middle point of BC, and let OK be drawn at right
angles to it and of such length that
OC² = k. KO.
Draw QN' perpendicular to OK.
Then
QN'2 = OR² = OC² – BR. RC
= k. (KO - RQ), by hypothesis,
=
- k. KN'.
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ.
cix
In the particular case referred to by Archimedes (with the slight
generalisation that the given length k to which PR is to be equal is
not necessarily equal to DE) we have
(1) the given ratio RQ: AR is unity, or RQAR, whence A"
coincides with A, and, by the first, lemma,
QN2 AN. A'N,
=
so that Q lies on a rectangular hyperbola.
(2) BR. RC=AR. RP=k. AR=k. RQ, and, by the second
lemma, lies on a certain parabola.
y,
If now we take O as origin, OC as axis of x and OK as axis of
and if we put OD=a, AD=b, BC= 2c, the hyperbola and parabola
determining the position of Q are respectively denoted by the
equations
(a — x)² = y² — b²,
c² — x² = ky,
which correspond exactly to the equations (B) above obtained by
purely algebraical methods.
Pappus says nothing of the Stopioμós which is necessary to the
complete solution of the generalised problem, the duopioμós namely
which determines the maximum value of k for which the solution is
possible. This maximum value would of course correspond to the
case in which the rectangular hyperbola and the parabola touch one
another. Zeuthen has shown* that the corresponding value of k can
be determined by means of the intersection of two other hyperbolas or
of a hyperbola and a parabola, and there is no doubt that Apollonius,
with his knowledge of conics, and in accordance with his avowed
object in giving the properties useful and necessary for diopipoi,
would have been able to work out this particular dopioμós by means
of conies; but there is no evidence to show that Archimedes investi-
gated it by the aid of conics, or indeed at all, it being clear, as shown
above, that it was not necessary for his immediate purpose.
This chapter may fitly conclude with a description of (1) some
important applications of vevores given by Pappus, and (2) certain
particular cases of the same class of problems which are plane, that
is, can be solved by the aid of the straight line and circle only, and
which were (according to Pappus) shown by the Greek geometers to
be of that character.
* Zeuthen, Die Lehre von den Kegelschnitten im Altertum, pp. 273—5.
#
}
CX
INTRODUCTION.
•
§ 4. One of the two important applications of 'solid' vevσes was
discovered by Nicomedes, the inventor of the conchoid, who intro-
duced that curve for solving a vevous to which he reduced the problem
of doubling the cube* or (what amounts to the same thing) the finding
of two mean proportionals between two given unequal straight lines.
Let the given unequal straight lines be placed at right angles as
CL, LA. Complete the parallelogram ABCL, and bisect AB at D,
and BC at E. Join LD and produce it to meet CB produced in H.
From E draw EF at right angles to BC, and take a point F on EF
such that CF is equal to AD. Join HF, and through C draw CG
parallel to HF. If we produce BC to K, the straight lines CG, CK
M

A
D
E
H
K
B
F
form an angle, and we now draw from the given point F a straight
line FGK, meeting CG, CK in G, K respectively, such that the
intercept GK is equal to AD or FC. (This is the veûous to which
the problem is reduced, and it can be solved by means of a conchoid
with F as pole.)
Join KL and produce it to meet BA produced in M.
Then shall CK, AM be the required mean proportionals between
CL, LA, or
CL: CK=CK: AMAM: AL.
We have, by Eucl. II. 6,
BK . KC + CE² – EK².
If we add EF2 to each side,
BK. KC + CF² = FK².
Now, by parallels,
MA: AB= ML : LK
· BC: CK;
* Pappus IV. p. 242 sq. and III. p. 58 sq.; Eutocius on Archimedes, On the
Sphere and Cylinder, 11. 1 (Vol. I. p. 114 sq.)
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ.
cxi
and, since AB = 2AD, and BC = 1HC,
MA: AD = HC: CK
whence, componendo,
= FG: GK, by parallels,
MD: AD FK : GK.
=
But GK = AD; therefore MD = FK, and MD² – FK².
Again,
and
while
therefore
Hence
that is,
MD² = BM . MA + AD³,
FK² = BK . KC + CF², from above,
MD2 FK, and AD2 = CF2;
=
BM. MA=BK . KC.
CK: MA=BM : BK
= MA: AL, by parallels,
LC: CK},
=
LC : CK=CK : MA = MA : AL.
§ 5. The second important problem which can be reduced to
a 'solid' vevoɩs is the trisection of any angle. One method of
reducing it to a vevơis has been mentioned above as following from
Prop. 8 of the Liber Assumptorum. This method is not mentioned
by Pappus, who describes (Iv. p. 272 sq.) another way of effecting
the reduction, introducing it with the words, "The earlier
geometers, when they sought to solve the aforesaid problem about
the [trisection of the] angle, a problem by nature solid,' by
'plane' methods, were unable to discover the solution; for they
were not yet accustomed to the use of the sections of the cone,
and were for that reason at a loss. Later, however, they trisected
an angle by means of conics, having used for the discovery of it
the following νεῦσις.”
The vevois is thus enunciated: Given a rectangle ABCD, let it
be required to draw through A a straight line AQR, meeting CD in
Q and BC produced in R, such that the intercept QR is equal to a
given length, k suppose.
Suppose the problem solved, QR being equal to k. Draw DP
parallel to QR and RP parallel to CD, meeting in P. Then, in the
parallelogram DR, DP = QR = k.
Hence P lies on a circle with centre D and radius k.
Again, by Eucl. 1. 43 relating to the complements of the
parallelograms about the diagonal of the complete parallelogram,
BC. CD BR. QD
= ᏢᎡ . ᎡᏴ ;
1
cxii
INTRODUCTION.
and, since BC. CD is given, it follows that P lies on a rectangular
hyperbola with BR, BA as asymptotes and passing through D.
B
C
A
D
R

P
Therefore, to effect the construction, we have only to draw this
rectangular hyperbola and the circle with centre D and radius equal
to k. The intersection of the two curves gives the point P, and R
is determined by drawing PR parallel to DC. Thus AQR is found.
[Though Pappus makes ABCD a rectangle, the construction
applies equally if ABCD is any parallelogram.]
Now suppose ABC to be any acute angle which it is required to
trisect. Let AC be perpendicular to BC. Complete the parallelo-
gram ADBC, and produce DA.
Suppose the problem solved, and let the angle CBE be one-third
of the angle ABC. Let BE meet AC in E and DA produced in F.
Bisect EF in H, and join AH.
Then, since the angle ABE is equal to twice the angle EBC and,
by parallels, the angles EBC, EFA are equal,
Therefore
and
LABE = 2 ▲ AFH = L AHB.
L
AB=AH= HF,
EF=2HF
= 2AB.
=
D
A

E
B
C
H
Hence, in order to trisect the angle ABC, we have only to solve
the following veûois: Given the rectangle ADBC whose diagonal
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ.
cxiii
is AB, to draw through B a straight line BEF, meeting AC in E and
DA produced in F, such that EF may be equal to twice AB; and this
vevσis is solved in the manner just shown.
These methods of doubling the cube and trisecting any acute
angle are seen to depend upon the application of one and the same
veûσis, which may be stated in its most general form thus. Given
any two straight lines forming an angle and any fixed point
which is not on either line, it is required to draw through the
fixed point a straight line such that the portion of it intercepted
between the fixed lines is equal to a given length. If AE, AC be

D
A
R
E
B
C
P
the fixed lines and B the fixed point, let the parallelogram ACBD
be completed, and suppose that BQR, meeting CA in Q and AE in
R, satisfies the conditions of the problem, so that QR is equal to
the given length. If then the parallelogram CQRP is completed,
we may regard P as an auxiliary point to be determined in order
that the problem may be solved; and we have seen that P can be
found as one of the points of intersection of (1) a circle with centre
C and radius equal to k, the given length, and (2) the hyperbola
which passes through C and has DE, DB for its asymptotes.
It remains only to consider some particular cases of the problem
which do not require conics for their solution, but are 'plane'
problems requiring only the use of the straight line and circle.
§ 6. We know from Pappus that Apollonius occupied him-
self, in his two Books of vevσes, with problems of that type
which were capable of solution by 'plane' methods. As a matter
of fact, the above vevois reduces to a 'plane' problem in the
particular case where B lies on one of the bisectors of the angle
between the two given straight lines, or (in other words) where the
parallelogram ACBD is a rhombus or a square. Accordingly we
find Pappus enunciating, as one of the 'plane' cases which had
H. A.
h
cxiv
INTRODUCTION.
been singled out for proof on account of their greater utility for
many purposes, the following *: Given a rhombus with one side
produced, to fit into the exterior angle a straight line given in
length and verging to the opposite angle; and he gives later on, in
his lemmas to Apollonius' work, a theorem bearing on the problem
with regard to the rhombus, and (after a preliminary lemma)
a solution of the vevous with reference to a square.
The question therefore arises, how did the Greek geometers
discover these and other particular cases, where a problem which
is in general 'solid,' and therefore requires the use of conics (or a
mechanical equivalent), becomes 'plane'? Zeuthen is of opinion that
they were probably discovered as the result of a study of the general
solution by means of conicst. I do not feel convinced of this, for
the following reasons.
(1) The authenticated instances appear to be very rare in
which we should be justified in assuming that the Greeks used
the properties of conics, in the same way as we should combine
and transform two Cartesian equations of the second degree, for
the purpose of proving that the intersections of two conics also
lie on certain circles or straight lines. It is true that we may
reasonably infer that Apollonius discovered by a method of this sort
his solution of the problem of doubling the cube where, in place
of the parabola and rectangular hyperbola used by Menaechmus,
he employs the same hyperbola along with the circle which passes
through the points common to the hyperbola and parabola‡; but
in the only propositions contained in his conics which offer an
opportunity for making a similar reduction§, Apollonius does not
make it, and is blamed by Pappus for not doing so.
In the pro-
positions referred to the feet of the normals to a parabola drawn
from a given point are determined as the intersections of the
parabola with a certain rectangular hyperbola, and Pappus objects
* Pappus vII. p. 670.
+ "Mit dieser selben Aufgabe ist nämlich ein wichtiges Beispiel dafür
verknüpft, dass man bemüht war solche Fälle zu entdecken, in denen Aufgaben,
zu deren Lösung im allgemeinen Kegelschnitte erforderlich sind, sich mittels
Zirkel und Lineal lösen lassen. Da nun das Studium der allgemeinen Lösung
durch Kegelschnitte das beste Mittel gewährt solche Fälle zu entdecken, so ist
es ziemlich wahrscheinlich, dass man wirklich diesen Weg eingeschlagen hat."
Zeuthen, op. cit. p. 280.
‡ Apollonius of Perga, p. cxxv, cxxvi.
§ Ibid. p. cxxviii and pp. 182, 186 (Conics, v. 58, 62
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ.
CXV
to this method as an instance of discovering the solution of a
'plane' problem by means of conics*, the objection having reference
to the use of a hyperbola where the same points could be obtained
as the intersections of the parabola with a certain circle. Now the
proof of this latter fact would present no difficulty to Apollonius,
and Pappus must have been aware that it would not; if therefore
he objects in the circumstances to the use of the hyperbola, it is at
least arguable that he would equally have objected had Apollonius
brought in the hyperbola and used its properties for the purpose
of proving the problem to be 'plane' in the particular case.
(2) The solution of the general problem by means of conics
brings in the auxiliary point P and the straight line CP. We
should therefore naturally expect to find some trace of these in the
particular solutions of the vevous for a rhombus and square; but
they do not appear in the corresponding demonstrations and figures
given by Pappus.
Zeuthen considers that the vevous with reference to a square was
probably shown to be 'plane' by means of the same investigation
which showed that the more general case of the rhombus was also
capable of solution with the help of the straight line and circle
only, i.e. by a systematic study of the general solution by means of
conics. This supposition seems to him more probable than the view
that the discovery of the plane construction for the square may have
been accidental; for (he says) if the same problem is treated solely
by the aid of elementary geometrical expedients, the discovery that
it is 'plane' is by no means a simple mattert. Here, again, I am
not convinced by Zeuthen's argument, as it seems to me that a
simpler explanation is possible of the way in which the Greeks were
led to the discovery that the particular vevoels were plane. They
knew in the first place that the trisection of a right angle was a
'plane' problem, and therefore that half a right angle could be
trisected by means of the straight line and circle. It followed
Pappus IV. p. 270. Cf. p. ciii above.
+ "Die Ausführbarkeit kann dann auf die zuerst angedeutete Weise gefunden
sein, die den allgemeinen Fall, wo der Winkel zwischen den gegebenen Geraden
beliebig ist, in sich begreift. Dies scheint mir viel wahrscheinlicher als die
Annahme, dass die Entdeckung dieser ebenen Konstruction zufällig sein sollte;
denn wenn man dieselbe Aufgabe nur mittels rein elementar-geometrischer
Hülfsmittel behandelt, so liegt die Entdeckung, dass sie eben ist, ziemlich fern."
Zeuthen, op. cit. p. 282.
h 2
cxvi
INTRODUCTION.
therefore that the corresponding veûous, i.e. that for a square, was
a 'plane' problem in the particular case where the given length
to which the required intercept was to be equal was double of
the diagonal of the square. This fact would naturally suggest
the question whether the problem was still plane if k had
any other value; and, when once this question was thoroughly
investigated, the proof that the problem was 'plane,' and the
solution of it, could hardly have evaded for long the pursuit of
geometers so ingenious as the Greeks. This will, I think, be
clear when the solution given by Pappus and reproduced below
is examined. Again, after it had been proved that the veûous with
reference to a square was 'plane,' what more natural than the further
inquiry as to whether the intermediate case between that of the
square and parallelogram, that of the rhombus, might perhaps be a
'plane' problem?
As regards the actual solution of the plane vevσes with respect
to the rhombus and square, i.e. the cases in general where the fixed
point B lies on one of the bisectors of the angles between the two
given straight lines, Zeuthen says that only in one of the cases have
we a positive statement that the Greeks solved the veûσis by means
of the circle and ruler, the case, namely, where ACBD is a square*.
This appears to be a misapprehension, for not only does Pappus
mention the case of the rhombus as one of the plane vevσes which
the Greeks had solved, but it is clear, from a proposition given by
him later, how it was actually solved. The proposition is stated
by Pappus to be "involved" (Taрalewрovμevov, meaning presumably
"the subject of concurrent investigation") in the 8th problem of
Apollonius' first Book of veúσes, and is enunciated in the following
formt. Given a rhombus AD with diameter BC produced to E, if EF
be a mean proportional between BE, EC, and if a circle be described
with centre E and radius EF cutting CD in K and AC produced in
H, BKH shall be a straight line. The proof is as follows.
Let the circle cut AC in L,
meet BC in M.
and join HE, KE, LE. Let LK
* "Indessen besitzen wir doch nur in einem einzelnen hierher gehörigen
Falle eine positive Angabe darüber, dass die Griechen die Einschiebung mittels
Zirkel und Lineal ausgeführt haben, wenn nämlich die gegebenen Geraden
zugleich rechte Winkel bilden, AIBC also ein Quadrat wird." Zeuthen, op. cit.
p. 281.
+ Pappus VII. p. 778.
F
1
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ,
cxvii
Since, from the property of the rhombus, the angles LCM, KCM
are equal, and therefore CL, CK make equal angles with the diameter
FG of the circle, it follows that CL = CK.

E
A
L
C
M
F
K
N
G
H
B
D
Also EK = EL, and CE is common to the triangles ECK, ECL.
Therefore the said triangles are equal in all respects, and
or
▲ CKE=LCLE =▲ CHE.
Now, by hypothesis,
EB: EF = EF : EC,
EB : EK = EK : EC
(since EF = EK),
and the angle CEK is common to the triangles BEK, KEC; there-
fore the triangles BEK, KEC are similar, and
Again,
LCBK = LCKE
=LCHE, from above.
▲ HCE = L ACB = ▲ BCK.
Thus in the triangles CBK, CHE two angles are equal re-
spectively;
therefore
¿CEH=LCKB.
But, since CKEL CHE, from above, the points K, C, E, H
are concyclic.
Hence
Accordingly, since
▲ CEH + ¿CKH = (two right angles).
LCEH = LCKB,
▲ CKH +▲ CKB = (two right angles),
and BKH is a straight line.
cxviii
INTRODUCTION.
Now the form of the proposition at once suggests that, in the
8th problem referred to, Apollonius had simply given a construction
involving the drawing of a circle cutting CD and AC produced in
the points K, H respectively, and Pappus' proof that BKH is a
straight line is intended to prove that HK verges towards B, or (in
other words) to verify that the construction given by Apollonius
solves a certain veûois requiring BKH to be drawn so that KH is
equal to a given length.
The analysis leading to the construction must have been worked
out somewhat as follows.
Suppose BKH drawn so that KH is equal to the given length k.
Bisect KH at N, and draw NE at right angles to KH meeting BC
produced in E.
Draw KM perpendicular to BC and produce it to meet CA in L.
Then, from the property of the rhombus, the triangles KCM, LCM
are equal in all respects.
Therefore KM = ML; and accordingly, if MN be joined, MN,
LH are parallel.
Now, since the angles at M, N are right, a circle can be described
about EMKN.
Therefore
LCEK = LMNK, in the same segment,
=LCHK, by parallels.
Hence a circle can be described about CEHK. It follows that
▲ BCD = L CEK + ≤ CKE
= _CHK + ▲ CHE
=
L EHK = LEKH.
Therefore the triangles EKH, DBC are similar.
Lastly,
▲ CKN = ▲ CBK + ▲ BCK;
and, subtracting from these equals the equal angles EKN, BCK
respectively, we have
L EKC = L EBK.
Hence the triangles EBK, EKC are similar, and
BE: EK = EK : EC,
or
BE. ECEK².
But, by similar triangles, EK: KH = DC: CB,
and the ratio DC: CB is given, while KH is also given (= k).
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ.
cxix
Therefore EK is given, and, in order to find E, we have only, in
the Greek phrase, to "apply to BC a rectangle exceeding by a square
figure and equal to the given area EK².”
Thus the construction given by Apollonius was clearly the
following*.
If k be the given length, take a straight line p such that
p: k = AB : BC.
Apply to BC a rectangle exceeding by a square figure and equal to
the area p². Let BE. EC be this rectangle, and with E as centre and
radius equal to p describe a circle cutting AC produced in H and
CD in K.
HK is then equal to k, and verges towards B, as proved by
Pappus; the problem is therefore solved.
The construction used by Apollonius for the 'plane' veûois with
reference to the rhombus having been thus restored by means of the
theorem given by Pappus, we are enabled to understand the purpose
* This construction was suggested to me by a careful examination of
Pappus' proposition without other aid; but it is no new discovery.
Samuel Horsley gives the same construction in his restoration of Apollonii
Pergaei Inclinationum libri duo (Oxford, 1770); he explains, however, that
he went astray in consequence of a mistake in the figure given in the мss.,
and was unable to deduce the construction from Pappus's proposition until he
was recalled to the right track by a solution of the same problem by Hugo
d'Omerique. This solution appears in a work entitled, Analysis geometrica, sive
nova et vera methodus resolvendi tam problemata geometrica quam arithmeticas
quaestiones, published at Cadiz in 1698. D'Omerique's construction, which is
practically identical with that of Apollonius, appears to have been evolved by
means of an independent analysis of his own, since he makes no reference to
Pappus, as he does in other cases where Pappus is drawn upon (e.g. when giving
the construction for the case of the square attributed by Pappus to one
Heraclitus). The construction differs from that given above only in the fact
that the circle is merely used to determine the point K, after which BK is joined
and produced to meet AC in H. Of other solutions of the same problem two
may here be mentioned. (1) The solution contained in Marino Ghetaldi's
posthumous work De Resolutione et Compositione Mathematica Libri quinque
(Rome, 1630), and included among the solutions of other problems all purporting
to be solved "methodo qua antiqui utebantur,” is, though geometrical, entirely
different from that above given, being effected by means of a reduction of the
problem to a simpler plane veûσis of the same character as that assumed by
Hippocrates in his Quadrature of lunes. (2) Christian Huygens (De circuli
magnitudine inventa; accedunt problematum quorundam illustrium constructiones,
Lugduni Batavorum, 1654) gave a rather complicated solution, which may be
described as a generalisation of Heraclitus' solution in the case of a square.
CXX
INTRODUCTION.
for which Pappus, while still on the subject of the "8th problem "
of Apollonius, adds a solution for the particular case of the square
(which he calls a "problem after Heraclitus") with an introductory
lemma. It seems clear that Apollonius did not treat the case of the
square separately from the rhombus because the solution for the
rhombus was equally applicable to the square, and this supposition
is confirmed by the fact that, in setting out the main problems
discussed in the veúoes, Pappus only mentions the rhombus and not
the square. Being however acquainted with a solution by one
Heraclitus of the vevois relating to a square which was not on the
same lines as that of Apollonius, while it was not applicable to the
case of the rhombus, Pappus adds it as an alternative method for
the square which is worth noting*. This is no doubt the explanation
of the heading to the lemma prefixed to Heraclitus' problem which
Hultsch found so much difficulty in explaining and put in brackets
as an interpolation by a writer who misunderstood the figure
and the object of the theorem. The words mean "Lemma useful
for the [problem] with reference to squares taking the place
of the rhombus" (literally "having the same property as the
rhombus”), i.e. a lemma useful for Heraclitus' solution of the
H
A
C
B
D
* This view of the matter receives strong support from the following
facts. In Pappus' summary (p. 670) of the contents of the veúσels of Apollonius
"two cases" of the veûous with reference to the rhombus are mentioned last
among the particular problems given in the first of the two Books. As we have
seen, one case (that given above) was the subject of the "8th problem" of
Apollonius, and it is equally clear that the other case was dealt with in the
“9th problem.” The other case is clearly that in which
the line to be drawn through B, instead of crossing the
exterior angle of the rhombus at C, lies across the angle
C itself, i.e. meets CA, CD both produced. In the former
case the solution of the problem is always possible what-
ever be the length of k; but in the second case clearly
the problem is not capable of solution if k, the given
length, is less than a certain minimum. Hence the
problem requires a duopioμós to determine the minimum
length of k. Accordingly we find Pappus giving, after
the interposition of the case of the square, a "lemma useful for the diopiouós of
the 9th problem,” which proves that, if CH=CK and B be the middle point of
HK, then HK is the least straight line which can be drawn through B to meet
CH, CK. Pappus adds that the diopioμós for the rhombus is then evident; if
HK be the line drawn through B perpendicular to CB and meeting CA, CD
produced in H, K, then, in order that the problem may admit of solution, the
given length k must be not less than HK.

K
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ.
cxxi
vevσis in the particular case of a square*. The lemma is as
follows.
ABCD being a square, suppose BHE drawn so as to meet CD in
H and AD produced in E, and let EF be drawn perpendicular to BE
meeting BC produced in F. To prove that
CF2 = BC² + HE².
Suppose EG drawn parallel to DC meeting CF in G. Then
since BEF is a right angle, the angles HBC, FEG are equal.

A
D
E
H
or
B
C
G
F.
Therefore the triangles BCH, EGF are equal in all respects, and
Now
EF= BH.
BF² – BE² + EF²,
BC. BF + BF. FC : BH. BE + BE. EH + EF2.
But, the angles HCF, HEF being right, the points C, H, E, F
are concyclic, and therefore
BC. BFBH. BE.
Subtracting these equals, we have
BF. FC -- BE. EH + EF²
·BE. EH + BH²
· BH. HE + EH² + BH²
= EB. BH + EH²
= FB. BC + EH².
* Hultsch translates the words λῆμμα χρήσιμον εἰς τὸ ἐπὶ τετραγώνων ποιούντων
TÀ AνTÀ Tŵ ¿Óµßy (p. 780) thus, "Lemma utile ad problema de quadratis quorum
summa rhombo aequalis est," and has a note in his Appendix (p. 1260) explaining
what he supposes to be meant. The 'squares' he takes to be the given square
and the square on the given length of the intercept, and the rhombus to be one
for which he indicates a construction but which is not shown in Pappus' figure.
Thus he is obliged to translate re jóµßy as "a rhombus," which is one objec-
tion to his interpretation, while "whose squares are equal" scarcely seems a
possible rendering of ποιούντων τὰ αὐτά.
cxxii
INTRODUCTION.
Take away the common part BC . CF, and
CFBC + EH².
Heraclitus' analysis and construction are now as follows.
Suppose that we have drawn BHE so that HE has a given
length k.
Since
CF² = BC² + EH², or BC² + k²,
and BC and k are both given,
CF is given, and therefore BF is given.
Thus the semicircle on BF as diameter is given, and therefore
also E, its intersection with the given line ADE; hence BE is
given.
To effect the construction, we first find a square equal to the
sum of the given square and the square on k. We then produce
BC to F so that CF is equal to the side of the square so found. If
a semicircle be now described on BF as diameter, it will pass above
D (since CF> CD, and therefore BC . CF > CD²), and will therefore
meet AD produced in some point E.
Join BE meeting CD in H.
Then HE=k, and the problem is solved.
1
CHAPTER VI.
CUBIC EQUATIONS.
Ir has often been explained how the Greek geometers were able
to solve geometrically all forms of the quadratic equation which give
positive roots; while they could take no account of others because
the conception of a negative quantity was unknown to them. The
quadratic equation was regarded as a simple equation connecting
areas, and its geometrical expression was facilitated by the methods
which they possessed of transforming any rectilineal areas whatever
into parallelograms, rectangles, and ultimately squares, of equal
area; its solution then depended on the principle of application of
areas, the discovery of which is attributed to the Pythagoreans.
Thus any plane problem which could be reduced to the geometrical
equivalent of a quadratic equation with a positive root was at once
solved. A particular form of the equation was the pure quadratic,
which meant for the Greeks the problem of finding a square equal
to a given rectilineal area. This area could be transformed into a
rectangle, and the general form of the equation thus became x² = ab,
so that it was only necessary to find a mean proportional between ɑ
and b. In the particular case where the area was given as the
sum of two or more squares, or as the difference of two squares,
an alternative method depended on the Pythagorean theorem of
Eucl. 1. 47 (applied, if necessary, any number of times successively).
The connexion between the two methods is seen by comparing
Eucl. vi. 13, where the mean proportional between a and b is
found, and Eucl. II. 14, where the same problem is solved without
the use of proportions by means of I. 47, and where in fact the
formula used is
α
2
x² = ab = ('
(a + b)² - (a - b) ².
2
ー​ド
​2
}
cxxiv
INTRODUCTION.
The choice between the two methods was equally patent when the
equation to be solved was x2 = pa², where p is any integer; hence
x²
the 'multiplication' of squares was seen to be dependent on the
finding of a mean proportional. The equation a² = 2a was the
simplest equation of the kind, and the discovery of a geometrical
construction for the side of a square equal to twice a given square
was specially important, as it was the beginning of the theory of
incommensurables or 'irrationals' (åλóywv πpayμateía) which was
invented by Pythagoras. There is every reason to believe that this
successful doubling of the square was what suggested the question
whether a construction could not be found for the doubling of the
cube, and the stories of the tomb erected by Minos for his son and
of the oracle bidding the Delians to double a cubical altar were no
doubt intended to invest the purely mathematical problem with an
element of romance. It may then have been the connexion between
the doubling of the square and the finding of one mean proportional
which suggested the reduction of the doubling of the cube to the
problem of finding two mean proportionals between two unequal
straight lines. This reduction, attributed to Hippocrates of Chios,
showed at the same time the possibility of multiplying the cube
Thus, if x, y are two mean proportionals between
by any ratio.
a, b, we have
and we derive at once
a : x = x : y = y : b,
a: b = a³ : x³,
whence a cube (2³) is obtained which bears to a³ the ratio b : a,
while any fraction can be transformed into a ratio between lines
P
9
of which one (the consequent) is equal to the side a of the given
cube. Thus the finding of two mean proportionals gives the solution
of any pure cubic equation, or the equivalent of extracting the cube
root, just as the single mean proportional is equivalent to extracting
the square root. For suppose the given equation to be = bcd.
We have then only to find a mean proportional a between c and d,
and the equation becomes x³ = a². b = a³ . - which is exactly, the
a
b
a
multiplication of a cube by a ratio between lines which the two
mean proportionals enable us to effect.
As a matter of fact, we do not find that the great geometers
were in the habit of reducing problems to the multiplication of the
J
CUBIC EQUATIONS.
CXXV
cube eo nomine, but to the equivalent problem of the two mean
proportionals; and the cubic equation a³=a2b is not usually stated
in that form but as a proportion. Thus in the two propositions On
the Sphere and Cylinder 11. 1, 5, where Archimedes uses the two
mean proportionals, it is required to find x where
a² : x² = x : b ;
he does not speak of finding the side of a cube equal to a certain
parallelepiped, as the analogy of finding a square equal to a given
rectangle might have suggested. So far therefore we do not find
any evidence of a general system of adding and subtracting solids
by transforming parallelepipeds into cubes and cubes into parallel-
epipeds which we should have expected to see in operation if the
Greeks had systematically investigated the solution of the general
form of the cubic equation by a method analogous to that of the
application of areas employed in dealing with quadratic equations.
The question then arises, did the Greek geometers deal thus
generally with the cubic equation
x²+ax²+ Bx+T=0,
±
which, on the supposition that it was regarded as an independent
problem in solid geometry, would be for them a simple equation
between solid figures, x and a both representing linear magnitudes,
B an area (a rectangle), and T a volume (a parallelepiped)? And
was the reduction of a problem of an order higher than that which
could be solved by means of a quadratic equation to the solution of
a cubic equation in the form shown above a regular and recognised
method of dealing with such a problem? The only direct evidence
pointing to such a supposition is found in Archimedes, who reduces
the problem of dividing a sphere by a plane into two segments
whose volumes are in a given ratio (On the Sphere and Cylinder 11. 4)
to the solution of a cubic equation which he states in a form
equivalent to
4a³ : x² = (3α- x) :
m
m + n
α
……………..(1),
where a is the radius of the sphere, m:n the given ratio (being a
ratio between straight lines of which m>n), and x the height of the
greater of the required segments. Archimedes explains that this is
a particular case of a more general problem, to divide a straight
line (a) into two parts (x, a - x) such that one part (a−x) is to an-
other given straight line (c) as a given area (which for convenience'
cxxvi
INTRODUCTION.
J
sake we suppose transformed into a square, b²) is to the square on
the other part (x²), i.e. so that
(aα − x): c = b² : x².
·(2).
He further explains that the equation (2) stated thus generally
requires a dɩopioµós, i.e. that the limits for the possibility of a real
solution, etc., require to be investigated, but that the particular case
(with the conditions obtaining in the particular proposition) requires
no dɩopioμós, i.e. the equation (1) will always give a real solution.
He adds that "the analysis and synthesis of both these problems
will be given at the end." That is, he promises to give separately a
complete investigation of the equation (2), which is equivalent to the
cubic equation
x² (α − x) = b²c ...
and to apply it to the particular case (1).
.(3)
Wherever the solution was given, it was temporarily lost, having
apparently disappeared even before the time of Dionysodorus and
Diocles (the latter of whom lived, according to Cantor, not later
than about 100 B.C.); but Eutocius describes how he found an
old fragment which appeared to contain the original solution of
Archimedes, and gives it in full. It will be seen on reference to
Eutocius' note (which I have reproduced immediately after the
proposition to which it relates, On the Sphere and Cylinder II. 4)
that the solution (the genuineness of which there seems to be no
reason to doubt) was effected by means of the intersection of a
parabola and a rectangular hyperbola whose equations may re-
spectively be written thus,
b2
x2
y,
a
The Siopioμós takes the
(a − x) y = ac.
form of investigating the maximum
possible value of x² (a-x), and it is proved that this maximum
value for a real solution is that corresponding to the value x =
2
ğa.
4
This is established by showing that, if b³c=
a², the curves touch
27
2
4
at the point for which x=
a. If on the other hand b³c <
3
a², it
27
is proved that there are two real solutions. In the particular case
(1) it is clear that the condition for a real solution is satisfied, for
CUBIC EQUATIONS.
cxxvii
the expression in (1) corresponding to b'c in (2) is
is only necessary that
m
4
4a³
m + n
27 (3a)³, or 4a³,
which is obviously true.
m
m + n
4a³, and it
Hence it is clear that not only did Archimedes solve the cubic
equation (3) by means of the intersections of two conics, but he also
discussed completely the conditions under which there are 0, 1 or 2
roots lying between 0 and a. It is to be noted further´that the
Suopioμós is similar in character to that by which Apollonius
investigates the number of possible normals that can be drawn
to a conic from a given point*. Lastly, Archimedes' method is
seen to be an extension of that used by Menaechmus for the solution
of the pure cubic equation. This can be put in the form
a³: x³: = a:b,
which can again be put in Archimedes' form thus,
a² : x² = x : b,
and the conics used by Menaechmus are respectively
x² = ay, xy = ab,
which were of course suggested by the two mean proportionals
satisfying the equations.
a : x = x : y = y : b.
The case above described is not the only one where we may
assume Archimedes to have solved a problem by first reducing it
to a cubic equation and then solving that. At the end of the
preface to the book On Conoids and Spheroids he says that the
results therein obtained may be used for discovering many theorems
and problems, and, as instances of the latter, he mentions the
following, "from a given spheroidal figure or conoid to cut off,
by a plane drawn parallel to a given plane, a segment which shall
be equal to a given cone or cylinder, or to a given sphere." Though
Archimedes does not give the solutions, the following considerations
may satisfy us as to his method.
(1) The case of the 'right-angled conoid' (the paraboloid of
revolution) is a 'plane' problem and therefore does not concern us
here.
* Cf. Apollonius of Perga, p. 168 sqq.
L
cxxviii
INTRODUCTION.
(2) In the case of the spheroid, the volume of the whole
spheroid could be easily ascertained, and, by means of that, the
ratio between the required segment and the remaining segment;
after which the problem could be solved in exactly the same way
as the similar one in the case of the sphere above described,
since the results in On Conoids and Spheroids, Props. 29-32,
correspond to those of On the Sphere and Cylinder II. 2. Or
Archimedes may have proceeded in this case by a more direct
method, which we may represent thus. Let a plane be drawn
through the axis of the spheroid perpendicular to the given
plane (and therefore to the base of the required segment). This
plane will cut the elliptical base of the segment in one of its
axes, which we will call 2y. Let x be the length of the axis
of the segment (or the length intercepted within the segment
of the diameter of the spheroid passing through the centre of the
base of the segment). Then the area of the base of the segment will
vary as y² (since all sections of the spheroid parallel to the given
plane must be similar), and therefore the volume of the cone which
has the same vertex and base as the required segment will vary as
y³x. And the ratio of the volume of the segment to that of the
cone is (On Conoids and Spheroids, Props. 29–32) the ratio
(3α - x): (2α-x), where 2a is the length of the diameter of the
spheroid which passes through the vertex of the segment. There-
fore
y³x.
За - х
2α- x
1
C,
where C is a known volume. Further, since x, y are the coordinates
of a point on the elliptical section of the spheroid made by the plane
through the axis perpendicular to the cutting plane, referred to a
diameter of that ellipse and the tangent at the extremity of the
diameter, the ratio y²: x (2a - x) is given. Hence the equation
can be put in the form
x² (3α- x) = b³c,
and this again is the same equation as that solved in the fragment
given by Eutocius. A duopioμós is formally necessary in this case,
though it only requires the constants to be such that the volume
to which the segment is to be equal must be less than that of the
whole spheroid.
(3) For the 'obtuse-angled conoid' (hyperboloid of revolution)
it would be necessary to use the direct method just described for
CUBIC EQUATIONS.
cxxix
the spheroid, and, if the notation be the same, the corresponding
equations will be found, with the help of On Conoids and Spheroids,
Props. 25, 26, to be
y²x.
3a + x
2a + x
C,
and, since the ratio y² : x (2a + x) is constant,
x² (3a + x) = b²c.
If this equation is written in the form of a proportion like the
similar one above, it becomes
b² : x³ = (3a + x): c.
There can be no doubt that Archimedes solved this equation as
well as the similar one with a negative sign, i.e. he solved the two
equations
x³ + ax² + b²c = 0,
obtaining all their positive real roots. In other words, he solved
completely, so far as the real roots are concerned, a cubic equation
in which the term in x is absent, although the determination of the
positive and negative roots of one and the same equation meant for
him two separate problems. And it is clear that all cubic equations
can be easily reduced to the type which Archimedes solved.
We possess one other solution of the cubic equation to which
the division of a sphere into segments bearing a given ratio to one
another is reduced by Archimedes. This solution is by Dionysodorus,
and is given in the same note of Eutocius*. Dionysodorus does not
generalise the equation, however, as is done in the fragment quoted
above; he merely addresses himself to the particular case,
m
2
4a² : x³ = (3α − x) :
m + n
thereby avoiding the necessity for a duopioμós. The curves which he
uses are the parabola
m
a (3a - x) = y³
2
m + n
and the rectangular hyperbola
m
2a² = xy.
m + n
When we turn to Apollonius, we find him emphasising in his
* On the Sphere and Cylinder II. 4 (note at end).
H. A.
i
1
T
CXXX
.: INTRODUCTION.
[
preface to Book IV. of the Conics* the usefulness of investigations
of the possible number of points in which conics may intersect one
another or circles, because "they at all events afford a more ready
means of observing some things, e.g. that several solutions are
possible, or that they are so many in number, and again that no
solution is possible"; and he shows his mastery of this method
of investigation in Book v., where he determines the number of
normals that can be drawn to a conic through any given point, the
condition that two normals through it coincide, or (in other words)
that the point lies on the evolute of the conic, and so on. For these
purposes he uses the points of intersection of a certain rectangular
hyperbola with the conic in question, and among the cases we find
.(v. 51, 58, 62) some which can be reduced to cubic equations, those
namely in which the conic is a parabola and the axis of the parabola
is parallel to one of the asymptotes of the hyperbola. Apollonius
however does not bring in the cubic equation; he addresses himself
to the direct geometrical solution of the problem in hand without
reducing it to another. This is after all only natural, because the
solution necessitated the drawing of the rectangular hyperbola in
the actual figure containing the conic in question; thus, e.g. in the
case of the problem leading to a cubic equation, Apollonius can, so
to speak, compress two steps into one, and the introduction of the
cubic as such would be mere surplusage. The case was different
with Archimedes, when he had no conic in his original figure; and
the fact that he set himself to solve a cubic somewhat more general
than that actually involved in the problem made separate treatment
with a number of new figures necessary. Moreover Apollonius was
at the same time dealing, in other propositions, with cases which did
not reduce to cubics, but would, if put in an algebraical form, lead
to biquadratic equations, and these, expressed as such, would have
had no meaning for the Greeks; there was therefore the less reason
in the simpler case to introduce a subsidiary problem.
As already indicated, the cubic equation, as a subject of syste-
matic and independent study, appears to have been lost sight of
within a century or so after the death of Archimedes. Thus Diocles,
the discoverer of the cissoid, speaks of the problem of the division of
the sphere into segments in a given ratio as having been reduced
by Archimedes "to another problem, which he does not solve in
his work on the sphere and cylinder"; and he then proceeds to
Apollonius of Perga, p. lxxiii.
*
CUBIC EQUATIONS.
cxxxi
solve the original problem directly, without in any way bringing
in the cubic. This circumstance does not argue any want of
geometrical ability in Diocles; on the contrary, his solution of the
original problem is a remarkable instance of dexterity in the use of
conics for the solution of a somewhat complicated problem, and it
proceeds on independent lines in that it depends on the intersection
of an ellipse and a rectangular hyperbola, whereas the solutions of
the cubic equation have accustomed us to the use of the parabola
and the rectangular hyperbola. I have reproduced Diocles' solution
in its proper place as part of the note of Eutocius on Archimedes'
proposition; but it will, I think, be convenient to give here its
equivalent in the ordinary notation of analytical geometry, in
accordance with the plan of this chapter. Archimedes had proved
[On the Sphere and Cylinder 11. 2] that, if k be the height of a
segment cut off by a plane from a sphere of radius a, and if h be
the height of the cone standing on the same base as that of the
segment and equal in volume to the segment, then
(3α −k): (2a - k) = h : k.
Also, if h' be the height of the cone similarly related to the
remaining segment of the sphere,
(a + k) k=h': (2a-k).
From these equations we derive
and
(hk): ka : (2a – k),
(h' – 2a + k) : (2a − k) = a : k.
Slightly generalising these equations by substituting for a in the
third term of each proportion another length b, and adding the
condition that the segments (and therefore the cones) are to bear to
each other the ratio mn, Diocles sets himself to solve the three
equations
and
(h − k) : k=b : (2a – k)
(h' — 2a + k) : (2a – k) = b : k
h : h' = m : n
…………….(A).
Suppose m>n, so that k>a. The problem then is to divide a
straight line of length 2a into two parts k and (2a-k) of which k is
the greater, and which are such that the three given equations are
all simultaneously satisfied.
Imagine two coordinate axes such that the origin is the middle
point of the given straight line, the axis of y is at right angles to it,
i2
cxxxii
INTRODUCTION.
and x is positive when measured along that half of the given straight
line which is to contain the required point of division. Then the
conics drawn by Diocles are
(1) the ellipse represented by the equation
n
(y + a − x)² = {(a + b)² — x²},
m
and (2) the rectangular hyperbola
(x + a) (y + b) = 2ab.
One intersection between these conics gives a value of x between 0
and a, and leads to the solution required. Treating the equations
algebraically, and eliminating y by means of the second equation
which gives
У
we obtain from the first equation
b
a 20
a + x
•
b,
n
(a− x)º ( 1 + — — — )² = 1/2 {(a + b)ª --未"},
a + x
m
m
that is, (a + x)² (α + b − x):
=
(a − x)² (a + b + x) ..………………. (B).
N
In other words Diocles' method is the equivalent of solving a
complete cubic equation containing all the three powers of x and a
constant, though no mention is made of such an equation.
To verify the correctness of the result we have only to remember
that, x being the distance of the point of division from the middle
point of the given straight line,
k = a + x,
2a-ka-x.
Thus, from the first two of the given equations (A) we obtain
respectively
h = a + x +
а + x
a-x
b,
α
X
h' = a −x +
.b,
a + x
whence, by means of the third equation, we derive
m
(a + x)² (α + b −x)
(α − x)² (α + b + x),
n
which is the same equation as that found by elimination above (B).
CUBIC EQUATIONS.
cxxxiii
I have purposely postponed, until the evidence respecting the
Greek treatment of the cubic equation was complete, any allusion
to an interesting hypothesis of Zeuthen's* which, if it could be
accepted as proved, would explain some difficulties involved in
Pappus' account of the orthodox classification of problems and loci.
I have already quoted the passage in which Pappus distinguishes
the problems which are plane (¿πíñeda), those which are solid (σrepeά)
and those which are linear (ypaµµiká)†. Parallel to this division of
problems into three orders or classes is the distinction between three
classes of loci. The first class consists of plane loci (róñoɩ étíñedol)
which are exclusively straight lines and circles, the second of solid
loci (tótoɩ σtepeoí) which are conic sections§, and the third of
linear loci (Tómo yраμμiкoí). It is at the same time clearly implied
by Pappus that problems were originally called plane, solid or linear
respectively for the specific reason that they required for their
solution the geometrical loci which bore the corresponding names.
But there are some logical defects in the classification both as
regards the problems and the loci.
*
(1) Pappus speaks of its being a serious error on the part of
geometers to solve a plane problem by means of conics (i.e. solid
loci ') or 'linear' curves, and generally to solve a problem "by means
of a foreign kind” (¿§ ȧvoikeíov yévovs). If this principle were
applied strictly, the objection would surely apply equally to the
solution of a 'solid' problem by means of a 'linear' curve.
though e.g. Pappus mentions the conchoid and the cissoid as being
'linear' curves, he does not object to their employment in the
solution of the problem of the two mean proportionals, which is a
'solid' problem.
Yet,
(2) The application of the term 'solid loci' to the three conic
sections must have reference simply to the definition of the curves
as sections of a solid figure, viz. the cone, and it was no doubt in
contrast to the 'solid locus' that the 'plane locus' was so called.
This agrees with the statement of Pappus that 'plane' problems may
* Die Lehre von den Kegelschnitten, p. 226 sqq.
+ p. ciii.
Pappus VII. pp. 652, 662.
§ It is true that Proclus (p. 394, ed. Friedlein) gives a wider definition of
“solid lines" as those which arise "from some section of a solid figure, as the
cylindrical helix and the conic curves"; but the reference to the cylindrical
helix would seem to be due to some confusion.
¿xxxiv
INTRODUCTION.
properly be so called because the lines by means of which they are
solved "have their origin in a plane." But, though this may be
regarded as a satisfactory distinction when 'plane' and 'solid' loci
are merely considered in relation to one another, it becomes at once
logically defective when the third or linear' class is also brought
in. For, on the one hand, Pappus shows how the 'quadratrix' (a
"linear' curve) can be produced by a construction in three
dimensions ("by means of surface-loci," dia rav прòs éπipaveíais
TÓTTOV); and, on the other hand, other linear' loci, the conchoid
and cissoid, have their origin in a plane. If then Pappus' account
of the origin of the terms 'plane' and 'solid' as applied to problems
and loci is literally correct, it would seem necessary to assume that
the third name of 'linear' problems and loci was not invented until
a period when the terms 'plane' and 'solid loci' had been so long
recognised and used that their origin was forgotten.
To get rid of these difficulties, Zeuthen suggests that the terms
'plane' and 'solid' were first applied to problems, and that they
came afterwards to be applied to the geometrical loci which were
used for the purpose of solving them. On this interpretation, when
problems which could be solved by means of the straight line and
circle were called 'plane,' the term is supposed to have had reference,
not to any particular property of the straight line or circle, but to
the fact that the problems were such as depend on an equation of a
degree not higher than the second. The solution of a quadratic
equation took the geometrical form of application of areas, and the
term 'plane' became a natural one to apply to the class of problems
so soon as the Greeks found themselves confronted with a new class
of problems to which, in contrast, the term 'solid' could be applied.
This would happen when the operations by which problems were
reduced to applications of areas were tried upon problems which
depend on the solution of a cubic equation. Zeuthen, then,
supposes that the Greeks sought to give this equation a similar
shape to that which the reduced 'plane' problem took, that is, to
form a simple equation between solids corresponding to the cubic
equation
x³ + αx² + Bx +T=0;
,
the term 'solid' or 'plane' being then applied according as it had
been reduced, in the manner indicated, to the geometrical equivalent
of a cubic or a quadratic equation.
Zeuthen further explains the term 'linear problem' as having
CUBIC EQUATIONS.
CXXXV
been invented afterwards to describe the cases which, being
equivalent to algebraical equations of an order higher than the
third, would not admit of reduction to a simple relation between
lengths, areas and volumes, and either could not be reduced to an
equation at all or could only be represented as such by the use of
compound ratios. The term 'linear' may perhaps have been applied
because, in such cases, recourse was had to new classes of curves,
directly and without any intermediate step in the shape of an
equation. Or, possibly, the term may not have been used at all
until a time when the original source of the names ‘plane' and
'solid' problems had been forgotten.
On these assumptions, it would still be necessary to explain how
Pappus came to give a more extended meaning to the term 'solid;
problem,' which according to him equally includes those problems
which, though solved by the same method of conics as was used to
solve the equivalent of cubics, do not reduce to cubic equations but
to biquadratics. This is explained by the supposition that, the
cubic equation having by the time of Apollonius been obscured
from view owing to the attention given to the method of solution
by means of conics and the discovery that the latter method was.
one admitting of wider application, the possibility of solution by
means of conics came itself to be regarded as the criterion deter-
mining the class of problem, and the name 'solid problem' came
to be used in the sense given to it by Pappus through a natural
misapprehension. A similar supposition would account, in Zeuthen's
view, for a circumstance which would otherwise seem strange, viz.
that Apollonius does not use the expression 'solid problem,' though
it might have been looked for in the preface to the fourth Book
of the Conics. The term may have been avoided by Apollonius
because it then had the more restricted meaning attributed to it by
Zeuthen and therefore would not have been applicable to all the
problems which Apollonius had in view.
It must be admitted that Zeuthen's hypothesis is in several
respects attractive. I cannot however feel satisfied that the
positive evidence in favour of it is sufficiently strong to outweigh
the authority of Pappus where his statements tell the other way.
To make the position clear, we have to remember that Menaechmus,
the discoverer of the conic sections, was a pupil of Eudoxus who
flourished about 365 B.C.; probably therefore we may place the
discovery of conics at about 350 B.C. Now Aristaeus 'the elder'
cxxxvi
INTRODUCTION.
wrote a book on solid loci (σrepeoì Tóπoi) the date of which Cantor
concludes to have been about 320 B.C. Thus, on Zeuthen's hypo-
thesis, the 'solid problems' the solution of which by means of conics
caused the latter to be called 'solid loci' must have been such as
had been already investigated and recognised as solid problems
before 320 B.C., while the definite appropriation, so to speak, of the
newly discovered curves to the service of the class of problems must
have come about in the short period between their discovery and
the date of Aristaeus' work. It is therefore important to consider
what particular problems leading to cubic equations appear to have
been the subject of speculation before 320 B.C. We have certainly
no ground for assuming that the cubic equation used by Archimedes
(On the Sphere and Cylinder II. 4) was one of these problems; for
the problem of cutting a sphere into segments bearing a given ratio
to one another could not have been investigated by geometers who
had not succeeded in finding the volume of a sphere and a segment
of a sphere, and we know that Archimedes was the first to discover
this. On the other hand there was the duplication of the cube, or
the solution of a pure cubic equation, which was a problem dating
from very early times. Also it is certain that the trisection of an
angle had long exercised the minds of the Greek geometers. Pappus
says that "the ancient geometers" considered this problem and first
tried to solve it, though it was by nature a solid problem (πρóßλŋja
τῇ φύσει στερεὸν ὑπάρχον), by means of plane considerations (διὰ τῶν
érɩrédov) but failed; and we know that Hippias of Elis invented,
about 420 B.C., a transcendental curve which was capable of being
used for two purposes, the trisection of an angle, and the quadrature
of a circle*. This curve came to be called the Quadratrix†, but, as
Deinostratus, a brother of Menaechmus, was apparently the first to
apply the curve to the quadrature of the circle‡, we may no doubt
conclude that it was originally intended for the purpose of trisecting
* Proclus (ed. Friedlein), p. 272.
+ The character of the curve may be described as follows. Suppose there
are two rectangular axes Oy, Ox and that a straight line OP of a certain length
(a) revolves uniformly from a position along Oy to a position along Ox, while a
straight line remaining always parallel to Ox and passing through P in its
original position also moves uniformly and reaches Ox in the same time as the
moving radius OP. The point of intersection of this line and OP describes the
Quadratrix, which may therefore be represented by the equation
+ Pappus IV. pp. 250—2.
y/a=20/π.
CUBIC EQUATIONS.
cxxxvii
an angle. Seeing therefore that the Greek geometers had used their
best efforts to solve this problem before the invention of conics, it
may easily be that they had succeeded in reducing it to the geo-
metrical equivalent of a cubic equation. They would not have been
unequal to effecting this reduction by means of the figure of the
vevσis given above on p. cxii. with a few lines added. The proof
would of course be the equivalent of eliminating x between the two
equations
xy = ab
−
−
=
(x − a)² + (y — b)² = 4 (a² + b²)
where x = DF, y=FP = EC, a = DA, b = DB.
The second equation gives
a=DA,
})
(x + a) (x − 3a) = (y + b) (3b − y).
From the first equation it is easily seen that
…………..(a)
(x + α) : (y + b) = a : y,
and that
(x − 3a) y = a (b − 3y) ;
we have therefore
[or
a² (b − 3y) = y² (3b — y)…..
a³b
y³ — 3by² - 3a³y + a2b= 0].
.(B)
If then the trisection of an angle had been reduced to the geo-
metrical equivalent of this cubic equation, it would be natural for
the Greeks to speak of it as a solid problem. In this respect it
would be seen to be similar in character to the simpler problem of
the duplication of the cube or the equivalent of a pure cubic
equation; and it would be natural to see whether the transformation
of volumes would enable the mixed cubic to be reduced to the form
of the pure cubic, in the same way as the transformation of areas
enabled the mixed quadratic to be reduced to the pure quadratic.
The reduction to the pure cubic would soon be seen to be impossible,
and the stereometric line of investigation would prove unfruitful
and be abandoned accordingly.
The two problems of the duplication of the cube and the
trisection of an angle, leading in one case to a pure cubic equation
and in the other to a mixed cubic, are then the only problems
leading to cubic equations which we can be certain that the Greeks
had occupied themselves with up to the time of the discovery of the
conic sections. Menaechmus, who discovered these, showed that
they could be successfully used for finding the two mean propor-
tionals and therefore for solving the pure cubic equation, and the
?
cxxxviii
INTRODUCTION.
next question is whether it had been proved before the date of
Aristaeus' Solid Loci that the trisection of an angle could be
effected by means of the same conics, either in the form of the
vevois above described directly and without the reduction to a cubic
equation, or in the form of the subsidiary cubic (B). Now (1) the
solution of the cubic would be somewhat difficult in the days when
conics were still a new thing. The solution of the equation (B) as
such would involve the drawing of the conics which we should
represent by the equations
xy = a²,
α,
bx = 3a³ + 3by − y³,
and the construction would be decidedly more difficult than that
used by Archimedes in connexion with his cubic, which only requires
the construction of the conics
x²
b2
y,
α
(ax) y = ac;
hence we can hardly assume that the trisection of an angle in the
form of the subsidiary cubic equation was solved by means of conics
before 320 B.C. (2) The angle may have been trisected by means
of conics in the sense that the vevois referred to was effected by
drawing the curves (a), i.e. a rectangular hyperbola and a circle.
This could easily have been done before the date of Aristaeus; but
if the assignment of the name 'solid loci' to conics had in view their
applicability to the direct solution of the problem in this manner
without any reference to the cubic equation, or simply because
the problem had been before proved to be 'solid' by means of the
reduction to that cubic, then there does not appear to be any
reason why the Quadratrix, which had been used for the same
purpose, should not at the time have been also regarded as a 'solid
locus,' in which case Aristaeus could hardly have appropriated the
latter term, in his work, to conics alone. (3) The only remaining
alternative consistent with Zeuthen's view of the origin of the
name 'solid locus' appears to be to suppose that conics were so
called simply because they gave a means of solving one 'solid
problem,' viz. the doubling of the cube, and not a problem of the
more general character corresponding to a mixed cubic equation, in
which case the justification for the general name 'solid locus' could
only be admitted on the assumption that it was adopted at a time
CUBIC EQUATIONS.
cxxxix
i
when the Greeks were still hoping to be able to reduce the general
cubic equation to the pure form. I think however that the
traditional explanation of the term is more natural than this
would be. Conics were the first curves of general interest for
the description of which recourse to solid figures was necessary as
distinct from the ordinary construction of plane figures in a plane*;
hence the use of the term 'solid locus' for conics on the mere ground
of their solid origin would be a natural way of describing the new
class of curves in the first instance, and the term would be likely
to remain in use, even when the solid origin was no longer thought
of, just as the individual conics continued to be called "sections of
a right-angled, obtuse-angled, and acute-angled" cone respectively.
4
While therefore, as I have said, the two problems mentioned
might naturally have been called 'solid problems' before the dis-
covery of 'solid loci,' I do not think there is sufficient evidence
to show that 'solid problem' was then or later a technical term
for a problem capable of reduction to a cubic equation in the sense
of implying that the geometrical equivalent of the general cubic
equation was investigated for its own sake, independently of its
applications, and that it ever occupied such a recognised position
in Greek geometry that a problem would be considered solved so
soon as it was reduced to a cubic equation. If this had been so,
and if the technical term for such a cubic was 'solid problem,' I
find it hard to see how Archimedes could have failed to imply some-
thing of the kind when arriving at his cubic equation. Instead of
this, his words rather suggest that he had attacked it as res integra.
Again, if the general cubic had been regarded over any length of
time as a problem of independent interest which was solved by
means of the intersections of conics, the fact could hardly have been
unknown to Nicoteles who is mentioned in the preface to Book IV.
of the Conics of Apollonius as having had a controversy with Conon
respecting the investigations in which the latter discussed the maxi-
mum number of points of intersection between two conics. Now
Nicoteles is stated by Apollonius to have maintained that no use
01
* It is true that Archytas' solution of the problem of the two mean propor-
tionals used a curve of double curvature drawn on a cylinder; but this was not
such a curve as was likely to be investigated for itself or even to be regarded as
a locus, strictly speaking; hence the solid origin of this isolated curve would
not be likely to suggest objections to the appropriation of the term 'solid locus'
to conics.
exl
INTRODUCTION.
could be made of the discoveries of Conon for diopioμoí; but it seems
incredible that Nicoteles could have made such a statement, even for
controversial purposes, if cubic equations then formed a recognised
class of problems for the discussion of which the intersections of
conics were necessarily all-important.
I think therefore that the positive evidence available will not
justify us in accepting the conclusions of Zeuthen except to the
following extent.
1. Pappus' explanation of the meaning of the term 'plane
problem' (éπíπedоν πρóßλnμα) as used by the ancients can hardly
be right. Pappus says, namely, that "problems which can be
solved by means of the straight line and circle may properly be
called plane (λέγοιτ' ἂν εἰκότως ἐπίπεδα); for the lines by means of
which such problems are solved have their origin in a plane." The
words "may properly be called" suggest that, so far as plane
problems were concerned, Pappus was not giving the ancient
definition of them, but his own inference as to why they were
called 'plane.' The true significance of the term is no doubt, as
Zeuthen says, not that straight lines and circles have their origin
in a plane (which would be equally true of some other curves), but
that the problems in question admitted of solution by the ordinary
plane methods of transformation of areas, manipulation of simple
equations between areas, and in particular the application of areas.
In other words, plane problems were those which, if expressed
algebraically, depend on equations of a degree not higher than the
second.
2. When further problems were attacked which proved to be
beyond the scope of the plane methods referred to, it would be
found that some of such problems, in particular the duplication
of the cube and the trisection of an angle, were reducible to simple
equations between volumes instead of equations between areas; and
it is quite possible that, following the analogy of the distinction
existing in nature between plane figures and solid figures (an analogy
which was also followed in the distinction between numbers as 'plane'
and 'solid' expressly drawn by Euclid), the Greeks applied the term
'solid problem' to such a problem as they could reduce to an
equation between volumes, as distinct from a 'plane problem
reducible to a simple equation between areas.
3. The first 'solid problem' in this sense which they succeeded
CUBIC EQUATIONS.
cxli
in solving was the multiplication of the cube, corresponding to the
solution of a pure cubic equation in algebra, and it was found that
this could be effected by means of curves obtained by making plane
sections of a solid figure, namely the cone. Thus curves having a
solid origin were found to solve one particular solid problem, which
could not but seem an appropriate result; and hence the conic, as
being the simplest curve so connected with a solid problem, was
considered to be properly termed a 'solid locus,' whether because of
its application or (more probably) because of its origin.
4. Further investigation showed that the general cubic equation
could not be reduced, by means of stereometric methods, to the
simpler form, the pure cubic; and it was found necessary to try
the method of conics directly either (1) upon the derivative cubic
equation or (2) upon the original problem which led to it. In
practice, as e.g. in the case of the trisection of an angle, it was
found that the cubic was often more difficult to solve in that
manner than the original problem was. Hence the reduction of
it to a cubic was dropped as an unnecessary complication, and
the geometrical equivalent of a cubic equation stated as an in-
dependent problem never obtained a permanent footing as the
'solid problem' par excellence.
་
5. It followed that solution by conics came to be regarded as
the criterion for distinguishing a certain class of problem, and, as
conics had retained their old name of 'solid loci,' the corresponding
term 'solid problem' came to be used in the wider sense in which
Pappus interprets it, according to which it includes a problem
depending on a biquadratic as well as a problem reducible to a
cubic equation.
6. The terms 'linear problem' and 'linear locus' were then
invented on the analogy of the other terms to describe respectively
a problem which could not be solved by means of straight lines,
circles, or conics, and a curve which could be used for solving such
a problem, as explained by Pappus.
**
}
CHAPTER VII.
ANTICIPATIONS BY ARCHIMEDES OF THE INTEGRAL CALCULUS.
It has been often remarked that, though the method of exhaustion
exemplified in Euclid XII. 2 really brought the Greek geometers face
to face with the infinitely great and the infinitely small, they
never allowed themselves to use such conceptions. It is true that
Antiphon, a sophist who is said to have often had disputes with
Socrates, had stated* that, if one inscribed any regular polygon,
say a square, in a circle, then inscribed an octagon by constructing
isosceles triangles in the four segments, then inscribed isosceles
triangles in the remaining eight segments, and so on, "until the
whole area of the circle was by this means exhausted, a polygon
would thus be inscribed whose sides, in consequence of their small-
ness, would coincide with the circumference of the circle.” But as
against this Simplicius remarks, and quotes Eudemus to the same
effect, that the inscribed polygon will never coincide with the
circumference of the circle, even though it be possible to carry
the division of the area to infinity, and to suppose that it would
is to set aside a geometrical principle which lays down that magni-
tudes are divisible ad infinitum†. The time had, in fact, not come
for the acceptance of Antiphon's idea, and, perhaps as the result of
the dialectic disputes to which the notion of the infinite gave rise,
the Greek geometers shrank from the use of such expressions as
infinitely great and infinitely small and substituted the idea of things
greater or less than any assigned magnitude. Thus, as Hankel says‡,
they never said that a circle is a polygon with an infinite number of
* Bretschneider, p. 101.
+ Bretschneider, p. 102.
+ Hankel, Zur Geschichte der Mathematik im Alterthum und Mittelalter,
p. 123.
ARCHIMEDES' ANTICIPATIONS OF THE INTEGRAL CALCULUS. cxliii
infinitely small sides; they always stood still before the abyss of the
infinite and never ventured to overstep the bounds of clear con-
ceptions. They never spoke of an infinitely close approximation or
a limiting value of the sum of a series extending to an infinite
number of terms. Yet they must have arrived practically at such
a conception, e.g., in the case of the proposition that circles are to
one another as the squares on their diameters, they must have been
in the first instance led to infer the truth of the proposition by the
idea that the circle could be regarded as the limit of an inscribed
regular polygon with an indefinitely increased number of corre-
spondingly small sides. They did not, however, rest satisfied with
such an inference; they strove after an irrefragable proof, and this,
from the nature of the case, could only be an indirect one. Ac-
cordingly we always find, in proofs by the method of exhaustion,
a demonstration that an impossibility is involved by any other
assumption than that which the proposition maintains. Moreover
this stringent verification, by means of a double reductio ad ab-
surdum, is repeated in every individual instance of the use of the
method of exhaustion; there is no attempt to establish, in lieu of
this part of the proof, any general propositions which could be
simply quoted in any particular case.
The above general characteristics of the Greek method of
exhaustion are equally present in the extensions of the method
found in Archimedes. To illustrate this, it will be convenient,
before passing to the cases where he performs genuine integrations,
to mention his geometrical proof of the property that the area of a
parabolic segment is four-thirds of the triangle with the same base
and vertex. Here Archimedes exhausts the parabola by continually
drawing, in each segment left over, a triangle with the same base
and vertex as the segment. If A be the area of the triangle so
inscribed in the original segment, the process gives a series of areas
▲, 1ª, (1)³A,
and the area of the segment is really the sum of the infinitę series.
A {1 + 1 + (1)² + (7)³ + ... }.
But Archimedes does not express it in this way. He first proves
that, if A1, A„…..A„ be any number of terms of such a series, so that
4A, ..., then
1
A₁ = 442, 4,
=
2
or
A₁ + A2 + Ag + ... + A„ + §¹„ = 341,
n-1
(1)~−1 3
n
A {1 + 1 + (1)² + ... + (7)~~¹ + § (1)"−¹} = {4.
cxliv
INTRODUCTION.
Having obtained this result, we should nowadays suppose n to
increase indefinitely and should infer at once that (1)"-¹ becomes
indefinitely small, and that the limit of the sum on the left-hand side
is the area of the parabolic segment, which must therefore be equal
to 4. Archimedes does not avow that he inferred the result in
this way; he merely states that the area of the segment is equal
to 4, and then verifies it in the orthodox manner by proving that
it cannot be either greater or less than 4.
I pass now to the extensions by Archimedes of the method
of exhaustion which are the immediate subject of this chapter. It
will be noticed, as an essential feature of all of them, that
Archimedes takes both an inscribed figure and a circumscribed
figure in relation to the curve or surface of which he is investigating
the area or the solid content, and then, as it were, compresses the
two figures into one so that they coincide with one another and
with the curvilinear figure to be measured; but again it must
be understood that he does not describe his method in this way or
say at any time that the given curve or surface is the limiting form
of the circumscribed or inscribed figure. I will take the cases
in the order in which they come in the text of this book.
1. Surface of a sphere or spherical segment.
The first step is to prove (On the Sphere and Cylinder 1. 21, 22)
that, if in a circle or a segment of a circle there be inscribed
polygons, whose sides AB, BC, CD, are all equal, as shown
in the respective figures, then
(a) for the circle
...
(BB' + CC' + ...) : AA' = A'B : BA,
(b) for the segment
(BB' + CC' + ... + KK' + LM): AM = A'B: BA.
Next it is proved that, if the polygons revolve about the
diameter AA', the surface described by the equal sides of the
polygon in a complete revolution is [1. 24, 35]
(a) equal to a circle with radius √AB (BB' + CC' +
... + YY')
or (b) equal to a circle with radius √AB (BB' + CC' + ... + LM).
Therefore, by means of the above proportions, the surfaces
described by the equal sides are seen to be equal to
ARCHIMEDES' ANTICIPATIONS OF THE INTEGRAL CALCULUS. cxlv
(a) a circle with radius AA'. A'B,
and (b) a circle with radius √AM.A'B;
they are therefore respectively [1. 25, 37] less than
(a) a circle with radius AA',
(b) a circle with radius AL.
Archimedes now proceeds to take polygons circumscribed to the
circle or segment of a circle (supposed in this case to be less than a
semicircle) so that their sides are parallel to those of the inscribed
polygons before mentioned (cf. the figures on pp. 38, 51); and he
proves by like steps [1. 30, 40] that, if the polygons revolve about the
diameter as before, the surfaces described by the equal sides during
a complete revolution are greater than the same circles respectively.
Lastly, having proved these results for the inscribed and
circumscribed figures respectively, Archimedes concludes and proves
[1. 33, 42, 43] that the surface of the sphere or the segment of the
sphere is equal to the first or the second of the circles respectively.
In order to see the effect of the successive steps, let us express
the several results by means of trigonometry. If, in the figures on
pp. 33, 47 respectively, we suppose 4n to be the number of sides in
the polygon inscribed in the circle and 2n the number of the equal
sides in the polygon inscribed in the segment, while in the latter
case the angle AOL is denoted by a, the proportions given above
are respectively equivalent to the formulae*
П
2π
sin + sin
2n
...
+ + sin (2n − 1)
Π
π
cot
2n
2n
4n'
α
2a
α
2 sin
+ sin
-
n
n
n
and
1 - cos a
α
cot
2n
+ ... + sin (n − 1) 2} + sin a
Thus the two proportions give in fact a summation of the series
sin 0 + sin 20 + ... + sin (n − 1) 0
both generally where n✪ is equal to any angle a less than π, and in
the particular case where n is even and 0 = π/~.
Again, the areas of the circles which are equal to the surfaces
described by the revolution of the equal sides of the inscribed
* These formulae are taken, with a slight modification, from Loria, Il periodo
aureo della geometria greca, p. 108.
H. A.
k
cxlvi
INTRODUCTION.
polygons are respectively (if a be the radius of the great circle
of the sphere)
Π
4πα sin
4n
and
{sin 2n
α
+ ... + sin
π
2π
+ sin
2n
+ sin (2n-1)} or 4πα cos
2+ sin
2a
a],
παρ. 2 cos
Ta². 2 sin [2 {sin
πα
2n
π
4n'
or
n
α
+ ... + sin (n−1) a} + sin a
2n (1 − cos a).
The areas of the circles which are equal to the surfaces described
by the equal sides of the circumscribed polygons are obtained from
the areas of the circles just given by dividing them by cos π/4n and
cos' a/2n respectively.
2
Thus the results obtained by Archimedes are the same as would
be obtained by taking the limiting value of the above trigonometri-
cal expressions when n is indefinitely increased, and when therefore
cos π/4n and cos a/2n are both unity.
But the first expressions for the areas of the circles are (when n
is indefinitely increased) exactly what we represent by the
integrals
and
πα.
π
4πα.
sin e de, or 4a³,
2 sin ◊ de, or 2πа² (1 − cos a).
-
Thus Archimedes' procedure is the equivalent of a genuine
integration in each case.
2. Volume of a sphere or a sector of a sphere.
The method does not need to be separately set out in detail here,
because it depends directly on the preceding case.
The investiga-
tion proceeds concurrently with that of the surface of a sphere or a
segment of a sphere. The same inscribed and circumscribed figures
are used, the sector of a sphere being of course compared with the
solid figure made up of the figure inscribed or circumscribed to the
segment and of the cone which has the same base as that figure and
has its vertex at the centre of the sphere. It is then proved,
(1) for the figure inscribed or circumscribed to the sphere, that its
volume is equal to that of a cone with base equal to the surface of
the figure and height equal to the perpendicular from the centre of
the sphere on any one of the equal sides of the revolving polygon,
(2) for the figure inscribed or circumscribed to the sector, that the
ARCHIMEDES' ANTICIPATIONS OF THE INTEGRAL CALCULUS. cxlvii
volume is equal to that of a cone with base equal to the surface of
the portion of the figure which is inscribed or circumscribed to the
segment of the sphere included in the sector and whose height is the
perpendicular from the centre on one of the equal sides of the
polygon.
Thus, when the inscribed and circumscribed figures are, so
to speak, compressed into one, the taking of the limit is practically
the same thing in this case as in the case of the surfaces, the
resulting volumes being simply the before-mentioned surfaces
multiplied in each case by a.
3. Area of an ellipse.
This case again is not strictly in point here, because it does
not exhibit any of the peculiarities of Archimedes' extensions of
the method of exhaustion. That method is, in fact, applied in
the same manner, mutatis mutandis, as in Eucl. XII. 2. There
is no simultaneous use of inscribed and circumscribed figures, but
only the simple exhaustion of the ellipse and auxiliary circle by
increasing to any desired extent the number of sides in polygons
inscribed to each (On Conoids and Spheroids, Prop. 4).
4. Volume of a segment of a paraboloid of revolution.
Archimedes first states, as a Lemma, a result proved incidentally
in a proposition of another treatise (On Spirals, Prop. 11), viz. that,
if there be n terms of an arithmetical progression h, 2h, 3h, ….., then
and
h+2h+ 3h+
+nh>]nh
{
h+2h+ 3h+ ... + (n − 1) h < n²h)
………..(a).
Next he inscribes and circumscribes to the segment of the
paraboloid figures made up of small cylinders (as shown in the figure
of On Conoids and Spheroids, Props. 21, 22) whose axes lie along
the axis of the segment and divide it into any number of equal
parts. If c is the length of the axis AD of the segment, and if
there are n cylinders in the circumscribed figure and their axes are
each of length h, so that c=nh, Archimedes proves that
cylinder CE
n²h
...
+ (n − 1) h
inscribed fig. h + 2h + 3h+
> 2, by the Lemma,
(1)
cylinder CE
and (2)
n²h
circumscribed fig. ¯ h+2h+ 3h+ ... + nh
¯¯
< 2.
t
k2
cxlviii
INTRODUCTION.
*
Meantime it has been proved [Props. 19, 20] that, by increasing
n sufficiently, the inscribed and circumscribed figure can be made
to differ by less than any assignable volume. It is accordingly
concluded and proved by the usual rigorous method that
so that
(cylinder CE) = 2 (segment),
(segment ABC) = 3 (cone ABC).
The proof is therefore equivalent to the assertion, that if h is
indefinitely diminished and n indefinitely increased, while nh remains
equal to c,
limit of h {h + 2h + 3h + ... + (n − 1) h} = {c² ;
that is, in our notation,
x dx = {c².
.
Thus the method is essentially the same as ours when we
express the volume of the segment of the paraboloid in the form
K
* [° y dax,
K
where κ is a constant, which does not appear in Archimedes' result
for the reason that he does not give the actual content of the
segment of the paraboloid but only the ratio which it bears to the
circumscribed cylinder.
5. Volume of a segment of a hyperboloid of revolution.
The first step in this case is to prove [On Conoids and Spheroids,
Prop. 2] that, if there be a series of n terms,
ah+h², a. 2h+ (2h)², a.3h+(3h)², ... a.nh + (nh)²,
and if (ah+h³) + {a . 2h + (2h)²} + ... + {a. nh + (nh)²}
+{a .nh+(nh)} = Sn
a
Sn,
then
n{a .nh+(nh)}/n< (a + nh)/ (1
+ ng
nh
3
.(B).
and
n{a .nh+ (nh)} Sn_z > (a +
α
nh
ni)/( + )
2 3
Next [Props. 25, 26] Archimedes draws inscribed and circum-
scribed figures made up of cylinders as before (figure on p. 137), and
ARCHIMEDES' ANTICIPATIONS OF THE INTEGRAL CALCULUS. cxlix
proves that, if AD is divided into n equal parts of length h, so that
nh = AD, and if AA' = a, then
cylinder EB'
inscribed figure
cylinder EB'
and
circumscribed fig.
n{a .nh+(nh)"}
Sn-1
α
2
+
nh),
"
n{a .nh+(nh)?}
Sn
α nh
3
<(a + n) ( + ).
The conclusion, arrived at in the same manner as before, is that
cylinder EB'
segment ABB'= (a + nh) / (
α nh
+
2 3
This is the same as saying that, if nhb, and if h be indefinitely
diminished while n is indefinitely increased,
or
limit of n (ab + b²)/Sn = (a + b)
= / (
b
limit of - S₂ = b²
N
(2
α
+
+
Now Sn=a (h+2h + ... + nh) + {h² + (2h)² + ... + (nh)³},
so that hSn=ah (h + 2h + +nh) + h {h² + (2h)² + +(nh)²}.
...
...
The limit of the last expression is what we should write as
1°
(ax + x²) dx,
α
which is equal to
b2
+
2
3
and Archimedes has given the equivalent of this integration.
6. Volume of a segment of a spheroid.
Archimedes does not here give the equivalent of the integration
S˚ (αx − x0³),
presumably because, with his method, it would have required yet
another lemma corresponding to that in which the results (8) above
are established.
cl
INTRODUCTION.
Suppose that, in the case of a segment less than half the spheroid
(figure on p. 142), AA' = a, CD=c, AD=b; and let AD be divided
into n equal parts of length h.
The gnomons mentioned in Props. 29, 30 are then the differences
between the rectangle cb + b² and the successive rectangles
ch+h², c.2h+ (2h)³, c. (n-1) h + {(n - 1) h},
...
and in this case we have the conclusions that (if S₂ be the sum of
n terms of the series representing the latter rectangles)
cylinder EB'
inscribed figure
n (cb + b²)
n (cb + b²) – Sn
cylinder EB'
and
circumscribed fig.
> (c + b) / (1 + 24).
3
n (cb + b²)
n (cb + b³) — Sn−1
and in the limit
cylinder EB'
segment ABB'
26
<
(c + b) +
3
26
b)
3
(0+ 6)/ (
6)/ (1 + 25).
Accordingly we have the limit taken of the expression
n (cb + b²) - Sn
n (cb + b²)
or 1
Sn
n (cb + b²) ›
and the integration performed is the same as that in the case of the
hyperboloid above, with c substituted for a.
Archimedes discusses, as a separate case, the volume of half a
spheroid [Props. 27, 28]. It differs from that just given in that c
vanishes and b = a, so that it is necessary to find the limit of
2
h² + (2h)² + (3h)* + ... + (nh)²
n(nh)
;
and this is done by means of a corollary to the lemma given on
pp. 107-9 [On Spirals, Prop. 10] which proves that
and
h²
h* + (2h)® + ... + (nh) > }n(nh),
h² + (2h)² + ... + {(n − 1) h}² < }n (nh)³.
The limit of course corresponds to the integral
b
x²dx = {b³.
ARCHIMEDES' ANTICIPATIONS OF THE INTEGRAL CALCULUS. cli
7. Area of a spiral.
(1) Archimedes finds the area bounded by the first complete
turn of a spiral and the initial line by means of the proposition just
quoted, viz.
h® +(2h)® + ... + (nh) >]n (nhi
+{(n−1)h} <
h² + (2h)² + ... + {(n − 1) h}² < }n (nh)².
He proves [Props. 21, 22, 23] that a figure consisting of similar
sectors of circles can be circumscribed about any arc of a spiral such
that the area of the circumscribed figure exceeds that of the spiral
by less than any assigned area, and also that a figure of the same
kind can be inscribed such that the area of the spiral exceeds that
of the inscribed figure by less than any assigned area. Then, lastly,
he circumscribes and inscribes figures of this kind [Prop. 24]; thus
e.g. in the circumscribed figure, if there are n similar sectors, the
radii will be n lines forming an arithmetical progression, as h, 2h,
3h, nh, and nh will be equal to a, where a is the length inter-
cepted on the initial line by the spiral at the end of the first turn.
Since, then, similar sectors are to one another as the square of their
radii, and ʼn times the sector of radius nh or a is equal to the circle
with the same radius, the first of the above formulae proves that
(circumscribed fig.) > }πа².
A similar procedure for the inscribed figure leads, by the use of the
second formula, to the result that
(inscribed fig.) < πα².
The conclusion, arrived at in the usual manner, is that
(area of spiral) = }πα²;
and the proof is equivalent to taking the limit of
or of
π
= [h² + (2h)² + ... + {(n − 1) h}³]
n
πλ
[h² + (2h)² + ... + {(n − 1) h}³],
a
which last limit we should express as
Π
a
πα
#fxdx = fra®
a o
clii
INTRODUCTION.
[It is clear that this method of proof equally gives the area
bounded by the spiral and any radius vector of length b not being
greater than a; for we have only to substitute πb/ɑ for π, and to
remember that in this case nh=b. We thus obtain for the area
618
a
•b
[* 2ªdx, or }#b¾a.]
a
(2) To find the area bounded by an arc on any turn of the
spiral (not being greater than a complete turn) and the radii
vectores to its extremities, of lengths b and c say, where c>b,
Archimedes uses the proposition that, if there be an arithmetic
progression consisting of the terms
b, b+h, b+ 2h, ... b+ (n − 1) h,
and if S₁₂ = b² + (b + h)² + (b + 2h)² + ... + {b + (n − 1) h}²,
N
then
(n−1){b+(n−1)h}²
Sp-b²
<
{b + (n-1) h}²
{b+(n−1)h}b+}{(n−1)h}² '
and
(n − 1){b+(n−1)h}²
Sp-1
>
{b + (n-1) h}²
{b+(n−1)h}b+} {(n−1)h}² *
[On Spirals, Prop. 11 and note.]
Then in Prop. 26 he circumscribes and inscribes figures consisting
of similar sectors of circles, as before. There are n-1 sectors in
each figure and therefore n radii altogether, including both b and c,
so that we can take them to be the terms of the arithmetic progres-
sion given above, where {b+ (n-1) h}=c. It is thus proved, by
means of the above inequalities, that
sector OB'C
{b + (n − 1) h}²
circumscribed fig. ¯ {b+(n−1)h}b+} {(n − 1)h}²
and it is concluded after the usual manner that
sector OB'C
{b + (n − 1) h}²
2
<
sector OB'C
inscr. fig.
spiral OBC ~ {b+(n−1)h}b+}{(n−1) h}²
c²
cb + § (c − b)² °
Remembering that n−1 = (c-b)/h, we see that the result is the
1
ARCHIMEDES' ANTICIPATIONS OF THE INTEGRAL CALCULUS. cliii
.
same thing as proving that, in the limit, when n becomes indefinitely
great and h indefinitely small, while b + (n − 1) h = c,
2
limit of h [b² + (b + h)² + ... + {b + (n − 2) h}³]
= (c-b) {cb + (c-b)"}
=
(c³ - b³);
that is, with our notation,
S
x² dx = 1} (c³ — b³).
b
•
(3) Archimedes works out separately [Prop. 25], by exactly
the same method, the particular case where the area is that described
in any one complete turn of the spiral beginning from the initial
line. This is equivalent to substituting (n-1) a for b and na for c,
where a is the radius vector to the end of the first complete turn of
the spiral.
It will be observed that Archimedes does not use the result
corresponding to
dx
x²
·b
[® at du - ["a" da = ['" at da.
ľ
8. Area of a parabolic segment.
Of the two solutions which Archimedes gives of the problem of
squaring a parabolic segment, it is the mechanical solution which
gives the equivalent of a genuine integration. In Props. 14, 15 of
the Quadrature of the Parabola it is proved that, of two figures
inscribed and circumscribed to the segment and consisting in each
case of trapezia whose parallel sides are diameters of the parabola,
the inscribed figure is less, and the circumscribed figure greater,
than one-third of a certain triangle (EqQ in the figure on p. 242).
Then in Prop. 16 we have the usual process which is equivalent to
taking the limit when the trapezia become infinite in number and
their breadth infinitely small, and it is proved that
(area of segment)
EqQ.
The result is the equivalent of using the equation of the parabola
referred to Qq as axis of x and the diameter through Q as axis of
y, viz.
py = x (2α − x),
which can, as shown on p. 236, be obtained from Prop. 4, and finding
•2a
ydx,
cliv
INTRODUCTION.
where Y has the value in terms of x given by the equation; and of
course
1 2a
½ [th (2ax − xº) dx =
−
P Jo
4a³
3p
The equivalence of the method to an integration can also be
seen thus. It is proved in Prop. 16 (see figure on p. 244) that, if
qE be divided into n equal parts and the construction of the
proposition be made, Qq is divided at O₁, O2, ... into the same
number of equal parts. The area of the circumscribed figure is then
easily seen to be the sum of the areas of the triangles
QqF,
01,
QR,F₁, QR,F2,
that is, of the areas of the triangles
QqF, QO₁R₁, QO₂D1,
...
Suppose now that the area of the triangle QqF is denoted by ▲, and
it follows that
(circumscribed fig.) = A
1
fig.) = ▲ {1 +
(n − 1)², (n − 2)²
+
n2
...
+ +
n²
J
1
n² A²
2
2
▲ {A² + 2³A² + ... + n²A²}.
Similarly we obtain
1
(inscribed fig.) =
▲ {A² + 2²A² + ... + (n − 1)³▲³}.
n² A 2
Taking the limit we have, if A denote the area of the triangle EqQ,
so that A = nA,
1
A
(area of segment)=
A2dA
A2
= 4.
14.
If the conclusion be regarded in this manner, the integration is
the same as that which corresponds to Archimedes' squaring of the
spiral.
CHAPTER VIII.
+
THE TERMINOLOGY OF ARCHIMEDES.
So far as the language of Archimedes is that of Greek geometry
in general, it must necessarily have much in common with that of
Euclid and Apollonius, and it is therefore inevitable that the
present chapter should repeat many of the explanations of terms of
general application which I have already given in the corresponding
chapter of my edition of Apollonius' Conics*. But I think it will
be best to make this chapter so far as possible complete and self-
contained, even at the cost of some slight repetition, which will
however be relieved (1) by the fact that all the particular phrases
quoted by way of illustration will be taken from the text of
Archimedes instead of Apollonius, and (2) by the addition of a large
amount of entirely different matter corresponding to the great
variety of subjects dealt with by Archimedes as compared with the
limitation of the work of Apollonius to the one subject of conics.
One element of difficulty in the present case arises out of the
circumstance that, whereas Archimedes wrote in the Doric dialect,
the original language has been in some books completely, and in others.
partially, transformed into the ordinary dialect of Greek. Uni-
formity of dialect cannot therefore be preserved in the quotations
about to be made; but I have thought it best, when explaining
single words, to use the ordinary form, and, when illustrating their
use by quoting phrases or sentences, to give the latter as they appear
in Heiberg's text, whether in Doric or Attic in the particular case.
Lest the casual reader should imagine the paroxytone words eveíaι,
διαμέτροι, πεσείται, πεσούνται, ἐσσείται, δυνάνται, ἁπτέται, καλείσθαι,
Kelolaι and the like to be misprints, I add that the quotations in
Doric from Heiberg's text have the unfamiliar Doric accents.
I shall again follow the plan of grouping the various technical
Apollonius of Perga, pp. clvii—clxx.
T
clvi
INTRODUCTION.
terms under certain general headings, which will enable the Greek
term corresponding to each expression in the ordinary mathematical
phraseology of the present day to be readily traced wherever such
a Greek equivalent exists.
Points and lines.
A point is σημεῖον, the point Β τὸ Β σημεῖον or τὸ B simply; a
point on (a line or curve) onueîov érí (with gen.) or ev; a point
raised above (a plane) onμéîov μeréwpov; any two points whatever
being taken δύο σημείων λαμβανομένων ὁποιωνοῦν.
At a point (e.g. of an angle) após (with dat.), having its vertex at
the centre of the sphere κορυφὴν ἔχων πρὸς τῷ κέντρῳ τῆς σφαίρας; of
lines meeting in a point, touching or dividing at a point, etc., katá
κατά
(with acc.), thus AE is bisected at Z is à AE díxa teµvétai katà tò Z;
of a point falling on or being placed on another ẻπí or κará (with
acc.), thus Z will fall on T, rò µèv Z ènì tò г teσeirai, so that E lies
on Δ, ὥστε τὸ μὲν Ε κατὰ τὸ Δ κείσθαι.
Particular points are extremity répas, vertex коpvpn, centre
κέντρον, point of division διαίρεσις, point of meeting σύμπτωσις, point
of section τομή, point of bisection διχοτομία, the middle point τὸ
μέσον; the points of division Η, Ι, Κ, τὰ τῶν διαιρεσίων σαμεῖα τὰ Η,
I, K; let B be its middle point pérov dè avrâs eσTO Tò В; the point of
section in which (a circle) cuts à тоµá, кal av тéµvel.
V
A line is γραμμή, a curved line καμπύλη γραμμή, a straight line
εὐθεῖα with or without γραμμή. The straight line ΘΙΚΛ, ά ΙΚΛ
evoeîa; but sometimes the older expression is used, the straight line
on which (èπí with gen. or dat. of the pronoun) are placed certain
letters, thus let it be the straight line M, oтw ¿p' ₫ Tò M, other
straight lines K, Λ, ἄλλαι γραμμαί, ἐφ᾽ ἂν τὰ Κ, Λ. The straight
lines between the points αἱ μεταξὺ τῶν σημείων εὐθεῖαι, of the lines
which have the same extremities the straight line is the least tŵv Tà
αὐτὰ πέρατα ἐχουσῶν γραμμῶν ἐλαχίστην εἶναι τὴν εὐθεῖαν, straight lines
cutting one another εὐθείαι τεμνούσαι ἀλλάλας.
For points in relation to lines we have such expressions as the
following: the points T,, M are on a straight line in eveías ẻorì
тà T',, M σaμeia, the point of bisection of the straight line containing
the centres of the middle magnitudes & διχοτομία τᾶς εὐθείας τᾶς
ἐχούσας τὰ κέντρα τῶν μέσων μεγεθέων. A very characteristic phrase
for at a point which divides the straight line in such a proportion
that.. is ἐπὶ τᾶς εὐθείας διαιρεθείσας ὥστε...; similarly ἐπὶ τᾶς ΧΕ
THE TERMINOLOGY OF ARCHIMEDES.
clvii
τμαθείσας οὕτως, ὥστε. A certain point will be on the straight line...
dividing it so that... ἐσσείται ἐπὶ τᾶς εὐθείας...διαιρέον οὕτως τὰν
εἰρημέναν εὐθεῖαν, ὥστε....
The middle point of a line is often elegantly denoted by an
adjective in agreement; thus at the middle point of the segment πì
ἐπὶ
μέσου τοῦ τμάματος, (α line) drawn from Γ to the middle point of
ΕΒ, ἀπὸ τοῦ Γ ἐπὶ μέσαν τὰν ΕΒ ἀχθεῖσα, drawn to the middle point of
the base ἐπὶ μέσαν τὴν βάσιν ἀγομένα.
A straight line produced is the (straight line) in the same straight
line with it ἡ ἐπ᾿ εὐθείας αὐτῇ. In the same straight line with the
ααὶς ἐπὶ τᾶς αὐτᾶς εὐθείας τῷ ἄξονι. Of a straight line falling on
another line κατά (with gen.) is used, e.g. πίπτουσι κατ' αὐτῆς; ἐπί
(with acc.) is also used of a straight line placed on another, thus if
EH be placed on ΒΔ, τεθείσας τᾶς ΕΗ ἐπὶ τὴν ΒΔ.
ΤΟ
For lines passing through points we find the following ex-
pressions: will pass through N, ἥξει διὰ τοῦ Ν ; will pass through the
centre διὰ τοῦ κέντρου πορεύσεται, will fall through Θ πεσείται διὰ τοῦ
©, verging towards Β νεύουσα ἐπὶ τὸ Β, pass through the same point
ἐπὶ τὸ αὐτὸ σαμεῖον ἐρχόνται; the diagonals of the parallelogram fall
(i.e. meet) at 5, κατὰ δὲ τὸ Θ αἱ διαμέτροι τοῦ παραλληλογράμμου
πίπτοντι; ΕΖ (passes) through the points bisecting ΑΒ, ΓΔ, ἐπὶ δὲ τὰν
διχοτομίαν τῶν ΑΒ, ΓΔ & ΕΖ. The verb εἰμί is also used of passing
through, thus ἐσσείται δὴ αὐτὰ διὰ τοῦ Θ.
For lines in relation to other lines we have perpendicular to
κάθετος ἐπί (with acc.), parallel to παράλληλος with dat. or παρά
(with acc.); let ΚΛ be (drawn) from K parallel to ΓΔ, ἀπὸ τοῦ Κ
παρὰ τὴν ΓΔ ἔστω ἁ ΚΛ.
Lines meeting one another συμπίπτουσαι ἀλλήλαις; the point in
which ΖΗ, ΜΝ produced meet one another and ΑΓ, τὸ σημεῖον, καθ᾿ ὃ
συμβάλλουσιν ἐκβαλλόμεναι αἱ ΖΗ, ΜΝ ἀλλήλαις τε καὶ τῇ ΑΓ ; so as
to meet the tangent ὥστε ἐμπεσεῖν τῷ ἐπιψανούσα, let straight lines be
drawn parallel to ΑΓ to meet the section of the cone ἄχθων εὐθείαι
παρὰ τὰν ΑΓ ἔστε ποτὶ τὰν τοῦ κώνου τομάν, to draw a straight line to
meet its circumference ποτὶ τὰν περιφέρειαν αὐτοῦ ποτιβαλεῖν εὐθεῖαν,
the line drawn to meet & ποτιπεσοῦσα, let ΑΕ, ΑΔ be drawn from the
point A to meet the spiral and produced to meet the circumference of
the circle ποτιπιπτόντων ἀπὸ τοῦ Α σαμείου ποτὶ τὰν ἕλικα αἱ ΑΕ, ΑΔ
καὶ ἐκπιπτόντων ποτὶ τὰν τοῦ κύκλου περιφέρειαν ; until it meets ΘΑ in
Ο, ἔστε κα συμπέσῃ τῇ ΘΑ κατὰ τὸ 0 (of a circle).
clviii
INTRODUCTION.
(The straight line) will fall outside (i.e. will extend beyond) P,
ἐκτὸς τοῦ Ρ πεσεῖται ; will fall within the section of the figure ἐντὸς
πεσούνται τᾶς τοῦ σχήματος τομᾶς.
The (perpendicular) distance between (two parallel lines) AZ, BH,
τὸ διάστημα τῶν ΑΖ, ΒΗ. Other ways of expressing distances are the
following: the magnitudes equidistant from the middle one τὰ ἴσον
ἀπέχοντα ἀπὸ τοῦ μέσου μεγέθεα, are at equal distances from one
another ἴσα ἀπ᾿ ἀλλάλων διέστακεν ; the segments (lengths) on ΛΗ
equal to N, τὰ ἐν τῇ ΜΗ τμάματα ισομεγέθεα τῷ N; greater by one
segment ἑνὶ τμάματι μείζων.
The word eveîa itself is also often used in the sense of distance;
cf. the terms πρώτη εὐθεῖα etc. in the book On Spirals, also & εὐθεῖα
ἁ μεταξὺ τοῦ κέντρου τοῦ ἁλίου καὶ τοῦ κέντρου τᾶς γᾶς the distance
between the centre of the sun and the centre of the earth.
The word for join is ἐπιζευγνύω or ἐπιζεύγνυμι ; the straight line
joining the points of contact & τὰς ἁφὰς ἐπιζευγνύουσα εὐθεῖα, ΒΔ when
joined & ΒΔ ἐπιζευχθεῖσα; let EZ join the points of bisection of ΑΔ,
ΒΓ, ὁ δὲ ΕΖ ἐπιζευγνυέτω τὰς διχοτομίας τῶν ΑΔ, ΒΓ. In one case
the word seems to be used in the sense of drawing simply, εἴ κα
εὐθεῖα ἐπιζευχθῇ γραμμὰ ἐν ἐπιπέδῳ.
Angles.
An angle is γωνία, the three kinds of angles are right ὀρθή, acute
ὀξεία, obtuse ἀμβλεία; right-angled etc. ὀρθογώνιος, ὀξυγώνιος, ἀμβλυ-
γώνιος ; equiangular ισογώνιος; with an even number of angles
ἀρτιόγωνος or ἀρτιογώνιος.
At right angles to ὀρθὸς πρός (with acc.) or πρὸς ὀρθάς (with dat.
following); thus if a line be erected at right angles to the plane γραμμᾶς
ανεστακούσας ὀρθᾶς ποτὶ τὸ ἐπίπεδον, the planes are at right angles to
one another ὀρθὰ ποτ᾽ ἄλλαλά ἐντι τὰ ἐπίπεδα, being at right angles
to ΑΒΓ, πρὸς ὀρθὰς ὢν τῷ ΑΒΓ; ΚΓ, ΕΛ are at right angles to one
another ποτ' ὀρθάς ἐντι ἀλλήλαις αἱ ΚΓ, ΕΛ, to cut at right angles
τέμνειν πρὸς ὀρθάς. The expression making right angles with is also
used, e.g. ὀρθὰς ποιοῦσα γωνίας ποτὲ τὰν ΑΒ.
å
The complete expression for the angle contained by the lines AH,
ΑΓ is & γωνία ἁ περιεχομένα ὑπὸ τῶν ΑΗ, ΑΓ; but there are a great
variety of shorter expressions, ywvía itself being often understood;
thus the angles Δ, Ε, Α, Β, αἱ Δ, Ε, Α, Β γωνίαι ; the angle at Θ, ά ποτὶ
τῷ Θ; the angle contained by ΑΔ, ΔΖ, ἡ γωνία ἡ ὑπὸ τῶν ΑΔ, ΔΖ; the
angle ΔΗΓ, ἡ ὑπὸ τῶν ΔΗΓ γωνία, ἡ ὑπὸ ΔΗΓ (with or without γωνία).
å
•
THE TERMINOLOGY OF ARCHIMEDES.
clix
e
Making the angle K equal to the angle ©, γωνίαν ποιοῦσα τὸν Κ
ἴσαν τᾷ Θ ; the angle into which the sun fits and which has its vertex
at the eye γωνία, εἰς ἂν ὁ ἅλιος ἐναρμόζει τὴν κορυφὰν ἔχουσαν ποτὶ τῷ
ὄψει; of the sides subtending the right angle (hypotenuses) τἂν ὑπὸ
τὰν ὀρθὰν γωνίαν ὑποτεινουσαν, they subtend the same angle ἐντὶ ὑπὸ
τὰν αὐτὰν γωνίαν.
If a line through an angular point of a polygon divides it
exactly symmetrically, the opposite angles of the polygon, αἱ ἀπεναντίον
γωνίαι τοῦ πολυγώνου, are those answering to each other on each side
of the bisecting line.
Planes and plane figures.
A plane ἐπίπεδον ; the plane through ΒΔ, τὸ ἐπίπεδον τὸ κατὰ
τὴν ΒΔ, or τὸ διὰ τῆς ΒΔ, plane of the base ἐπίπεδον τῆς βάσεως, plane
(i.e. base) of the cylinder ἐπίπεδον τοῦ κυλίνδρου ; cutting plane ἐπί-
πεδον τέμνον, tangent plane ἐπίπεδον ἐπιψαῦον; the intersection of
planes is their common section κοινή τομή.
In the same plane as the circle ἐν τῷ αὐτῷ ἐπιπέδῳ τῷ κύκλῳ.
Let a plane be erected on IIZ at right angles to the plane in which
ΑΒ, ΓΔ are ἀπὸ τὰς ΠΖ ἐπίπεδον ἀνεστακέτω ὀρθὸν ποτὶ τὸ ἐπίπεδον τό,
ἐν ᾧ ἐντι αἱ ΑΒ, ΓΔ.
C
The plane surface ἡ ἐπίπεδος (ἐπιφάνεια), a plane segment ἐπίπεδον
τμήμα, a plane figure σχῆμα ἐπίπεδον.
A rectilineal figure ευθύγραμμον (σχήμα), a side πλευρά, perimeter
ἡ περίμετρος, similar ὅμοιος, similarly situated ὁμοίως κείμενος.
To coincide with (when one figure is applied to another),
ἐφαρμόζειν followed by the dative or ἐπί (with acc.); one part
coincides with the other ἐφαρμόζει τὸ ἕτερον μέρος ἐπὶ τὸ ἕτερον; the
plane through NZ coincides with the plane through ΑΓ, τὸ ἐπίπεδον τὸ
κατὰ τὰν ΝΖ ἐφαρμόζει τῷ ἐπιπέδῳ τῷ κατὰ τὰν ΑΓ. The passive is
also used; if equal and similar plane figures coincide with one another
τῶν ἴσων καὶ ὁμοίων σχημάτων ἐπιπέδων ἐφαρμοζομένων ἐπ᾽ ἄλλαλα.
Triangles.
A triangle is τρίγωνον, the triangles bounded by their three
sides) τὰ περιεχόμενα τρίγωνα ὑπὸ τῶν Α right-angled triangle
A
τρίγωνον ὀρθογώνιον, one of the sides about the right angle μία τῶν περὶ
τὴν ὀρθήν. The triangle through the axis of a cone) τὸ διὰ τοῦ ἄξονος
τρίγωνον.
clx
INTRODUCTION.
Quadrilaterals.
A quadrilateral is a four-sided figure (τετράπλευρον) as dis-
tinguished from a four-angled figure, τετράγωνον, which means a
square. A trapezium, τραπέζιον, is in one place more precisely
described as a trapezium having its two sides parallel τραπέζιον τὰς
δύο πλευρὰς ἔχον παραλλάλους ἀλλάλαις.
A parallelogram παραλληλόγραμμον; for a parallelogram on a
straight line as base èπí (with gen.) is used, thus the parallelograms
on them are of equal height ἐστὶν ἰσούψη τὰ παραλληλόγραμμα τὰ ἐπ᾿
αὐτῶν. A diagonal of a parallelogram is διάμετρος, the opposite sides
of the parallelogram αἱ κατ' ἐναντίον τοῦ παραλληλογράμμου πλευραί.
Rectangles.
The word generally used for a rectangle is χωρίον (space or area)
without any further description. As in the case of angles, the
rectangles contained by straight lines are generally expressed more
shortly than by the phrase τὰ περιεχόμενα χωρία υπό ; either χωρίον
may be omitted or both χωρίον and περιεχόμενον, thus the rectangle
ΑΓ, ΓΕ may be any of the following, τὸ ὑπὸ τῶν ΑΓ, ΓΕ, τὸ ὑπὸ
ΑΓ, ΓΕ, τὸ ὑπὸ ΑΓΕ, and the rectangle under ΘΚ, ΑΗ is τὸ ὑπὸ τῆς
ΘΚ καὶ τῆς ΑΗ. Rectangles Θ, Ι, Κ, Λ, χωρία ἐν οἷς τὰ (or ἐφ᾽ ὧν
ἕκαστον τῶν Θ, Ι, Κ, Λ.
To apply a rectangle to a straight line (in the technical sense) is
παραβάλλειν, and παραπίπτω is generally used in place of the passive;
the participle παρακείμενος is also used in the sense of applied to. In
each case applying to a straight line is expressed by mapá (with acc.).
Examples are, areas which we can apply to a given straight line (i.e.
which we can transform into a rectangle of the same area) χωρία, ἃ
δυνάμεθα παρὰ τὰν δοθεῖσαν εὐθεῖαν παραβαλείν, let a rectangle be
applied to each of them παραπεπτωκέτω παρ᾽ ἑκάσταν αὐτῶν χωρίον ;
if there be applied to each of them a rectangle exceeding by a square
figure, and the sides of the excesses exceed each other by an equal
amount (i.e. form an arithmetical progression) εἴ κα παρ᾽ ἑκάσταν
αὐτῶν παραπέσῃ τι χωρίον ὑπερβάλλον εἴδει τετραγώνῳ, ἔωντι δὲ αἱ
πλευραὶ τῶν ὑπερβλημάτων τῷ ἴσῳ ἀλλάλαν ὑπερεχούσαι.
The rectangle applied is παράβλημα.
Squares.
A square is τετράγωνον, a square on a straight line is a square
(erected) from it (από). The square on ΓΞ, τὸ ἀπὸ τᾶς ΓΕ τετράγωνον,
THE TERMINOLOGY OF ARCHIMEDES.
clxi
is shortened into τὸ ἀπὸ τᾶς ΓΞ, or τὸ ἀπὸ ΓΞ simply. The square
next in order to it (when there are a number of squares in a row) is
τὸ παρ᾽ αὐτῷ τετράγωνον or τὸ ἐχόμενον τετράγωνον.
;
With reference to squares, a most important part is played by
the word δύναμις and the various parts of the verb δύναμαι. δύναμις
expresses a square (literally a power); thus in Diophantus it is used
throughout as the technical term for the square of the unknown
quantity in an algebraical equation, i.e. for x². In geometrical
language it is the dative singular Suváμe which is mostly used
thus a straight line is said to be potentially equal, δυνάμει ἴσα, to a
certain rectangle where the meaning is that the square on the straight
line is equal to the rectangle; similarly for the square on BA is less
than double the square on AK we have ἡ ΒΑ ἐλάσσων ἐστὶν ἢ διπλασίων
δυνάμει της ΑΚ. The verb δύνασθαι (with or without ἴσον) has the
sense of being δυνάμει ἴσα, and, when δύνασθαι is used alone, it is
followed by the accusative; thus the square (on a straight line) is
equal to the rectangle contained by... is (εὐθεῖα) ἴσον δύναται τῷ
περιεχομένῳ ὑπό...; let the square on the radius be equal to the
rectangle ΒΔ, ΔΖ, ἡ ἐκ τοῦ κέντρου δυνάσθω τὸ ὑπὸ τῶν ΒΔΖ, (the
difference) by which the square on ZT is greater than the square on
half the other diameter ᾧ μεῖζον δυνάται ἡ ΖΓ τᾶς ἡμισείας τᾶς ἑτέρας
διαμέτρου.
A gnomon is γνώμων, and its breadth (πλάτος) is the breadth of
each end; a gnomon of breadth. equal to BI, γνώμων πλάτος ἔχων ἴσον
Tâ BI, (a gnomon) whose breadth is greater by one segment than the
breadth of the gnomon last taken away οὗ πλάτος ἑνὶ τμάματι μείζον
τοῦ πλάτεος τοῦ πρὸ αὐτοῦ ἀφαιρουμένου γνώμονος.
Polygons.
A polygon is πολύγωνον, an equilateral polygon is ισόπλευρον,
a polygon of an even number of sides or angles αρτιόπλευρον οι
ἀρτιόγωνον ; a polygon with all its sides equal except ΒΔ, ΔΑ, ἴσας
ἔχον τὰς πλευρὰς χωρὶς τῶν ΒΔΑ; a polygon with its sides, excluding
the base, equal and even in number τὰς πλευρὰς ἔχον χωρὶς τῆς βάσεως
ἴσας καὶ ἀρτίους ; an equilateral polygon the number of whose sides is
measured by four πολύγωνον ἰσόπλευρον, οὗ αἱ πλευραὶ ὑπὸ τετράδος
μετροῦνται, let the number of its sides be measured by four τὸ πλῆθος
τῶν πλευρῶν μετρείσθω ὑπὸ τετράδος. A chiliagon χιλιάγωνον.
The straight lines subtending two sides of the polygon (i.e. joining
angles next but one to each other) αἱ ὑπὸ δύο πλευρὰς τοῦ πολυγώνου
H. A.
7
clxii
INTRODUCTION.
VπоTEίVOVσαι, the straight line subtending one less than half the
number of the sides ἡ ὑποτείνουσα τὰς μιᾷ ἐλάσσονας τῶν ἡμίσεων.
Circles.
A circle is κύκλος, the circle Ψ is ὁ Ψ κύκλος or ὁ κύκλος ἐν ᾧ τὸ Ψ,
let the given circle be that drawn below ἔστω ὁ δοθεὶς κύκλος ὁ
ὑποκείμενος.
The centre is κέντρον, the circumference περιφέρεια, the former
word having doubtless been suggested by something stuck in and
the latter by something, e.g. a cord stretched tight, carried round
the centre as a fixed point and describing a circle with its other
extremity. Accordingly repɩpépaa is used for a circular arc as well
as for the whole circumference; thus the arc BA is BA πepipépeia,
the (part of the) circumference of the circle cut off by the same
(straight line) ή του κύκλου περιφέρεια ἡ ὑπὸ τῆς αὐτῆς ἀποτεμνομένη.
Though the circumference of a circle is also sometimes called its
perimeter (Tepíuerpos) in the treatises On the Sphere and Cylinder
and on the Measurement of a Circle, the word does not seem to have
been used by Archimedes himself in this sense; he speaks, however,
in the Sand-reckoner of the perimeter of the earth (TEρíµETρOS тâs yâs).
The radius is † EK TOû KÉVтpov simply, and this expression
without the article is used as a predicate as if it were one word;
thus the circle whose radius is ΘΕ is ὁ κύκλος οἱ ἐκ τοῦ κέντρου ἁ
ΘΕ; ΒΕ is a radius of the circle ἡ δὲ ΒΕ ἐκ τοῦ κέντρου ἐστὶ τοῦ κύκλου.
A diameter is διάμετρος, the circle on ΔΕ as diameter ὁ περὶ
διάμετρον τὴν ΔΕ κύκλος.
For drawing a chord of a circle there is no special technical
term, but we find such phrases as the following: éav eis Tòv KÚKλOV
evéîa yрaµµÝ µπéon if in a circle a straight line be placed, and the
chord is then the straight line so placed ureσovoa, or quite
commonly ἡ ἐν τῷ κύκλῳ (εὐθεῖα) simply. For the chord subtending
one 656th part of the circumference of a circle we have the following
interesting phrase, ἡ ὑποτείνουσα ἓν τμᾶμα διαιρεθείσας τᾶς τοῦ ΑΒΓ
κύκλου περιφερείας ἐς χνς'.
A segment of a circle is tµîµa kúkλov; sometimes, to distinguish
it from a segment of a sphere, it is called a plane segment
τμῆμα ἐπίπεδον. A semicircle is ἡμικύκλιον ; a segment less than a
semicircle cut of by AB, τμῆμα ἔλασσον ἡμικυκλίου ὃ ἀποτέμνει
ý AB. The segments on AE, EB (as bases) are Tà ¿Tì Tŵv
AE, EB τµńμаτα; but the semicircle on ZH as diameter is ro
THE TERMINOLOGY OF ARCHIMEDES.
clxiii
or
ἡμικύκλιον τὸ περὶ διάμετρον τὰν ΖΗ οι τὸ ἡμικύκλιον τὸ περὶ τὰν ΖΗ
simply. The expression the angle of the semicircle, á тov nμuкvкλίOV
(ywvía), is used of the (right) angle contained by the diameter and
the arc (or tangent) at one extremity of it.
A sector of a circle is roμcús or, when it is necessary to
distinguish it from what Archimedes calls a 'solid sector,' èπíñedos
ToμEÙS Kúkλov a plane sector of a circle. The sector including the
right angle (at the centre) is ὁ τομεὺς ὁ τὴν ὀρθὰν γωνίαν περιέχων.
Either of the radii bounding a sector is called a side of it, πλevpá;
each of the sectors (is) equal to the sector which has a side common
(with it) ἕκαστος τῶν τομέων ἴσος τῷ κοινὰν ἔχοντι πλευρὰν τομεῖ; α
sector is sometimes regarded as described on one of the bounding
radii as a side, thus similar sectors have been described on all (the
straight lines) αναγεγραφάται ἀπὸ πασᾶν ὁμοίοι τομέες.
Of polygons inscribed in or circumscribed about a circle éyypápe
εἰς or ev and περιγράφειν περί. (with acc.) are used we also find
περιγεγραμμένος used with the simple dative, thus τὸ περιγε-
γραμμένον σχημα τῷ τομεῖ is the figure circumscribed to the sector.
A polygon is said to be inscribed in a segment of a circle when
the base of the segment is one side and the other sides subtend
arcs making up the circumference; thus let a polygon be inscribed
on ΑΓ in the segment ΑΒΓ, ἐπὶ τῆς ΑΓ πολύγωνον ἐγγεγράφθω
εἰς τὸ ΑΒΓ
eis Tò ABT τμñμa. A regular polygon is said to be inscribed in
a sector when the two radii are two of the sides and the other sides
are all equal to one another, and a similar polygon is said to be
circumscribed about a sector when the equal sides are formed by the
tangents to the arc which are respectively parallel to the equal
sides of the inscribed polygon and the remaining two sides are the
bounding radii produced to meet the adjacent tangents. Of a
circle circumscribed to a polygon πepidaµßáveiv is also used; thus
πολύγωνον κύκλος περιγεγραμμένος περιλαμβανέτω περὶ τὸ αὐτὸ κέντρον
yivóμevos, as we might say let a circumscribed circle be drawn with
the same centre going round the polygon. Similarly the circle ABTA
containing the polygon ὁ ΑΒΓΔ κύκλος ἔχων τὸ πολύγωνον.
When a polygon is inscribed in a circle, the segments left over
between the sides of the polygon and the subtended arcs are
περιλειπόμενα τμήματα; when a polygon is circumscribed to the
circle, the spaces between the two are variously called τà TEρi-
λειπόμενα τῆς περιγραφῆς τμήματα, τὰ περιλειπόμενα σχήματα, τὰ
περιλείμματα οι τὰ ἀπολείμματα.
12
clxiv
INTRODUCTION.
U
Spheres, etc.
or
In connexion with a sphere (opaîpa) a number of terms are
used on the analogy of the older and similar terms connected with
the circle. Thus the centre is κέντρον, the radius ἡ ἐκ τοῦ κέντρου,
the diameter ἡ διάμετρος. Two segments, τμήματα σφαίρας οι
τμήματα σφαιρικά, are formed when a sphere is cut by a plane;
a hemisphere is ημισφαίριον ; the segment of the sphere at Γ, τὸ κατὰ τὸ
Γ τμῆμα τῆς σφαίρας ; the segment on the side of ΑΒΓ, τὸ ἀπὸ ΑΒΓ
τμημα; the segment including the circumference ΒΑΔ, τὸ κατὰ τὴν ΒΑΔ
περιφέρειαν τμῆμα. The curved surface of a sphere or segment
is ἐπιφάνεια ; thus of spherical segments bounded by equal surfaces the
hemisphere is greatest is τῶν τῇ ἴσῃ ἐπιφανείᾳ περιεχομένων σφαιρικών
τμημάτων μεῖζόν ἐστι τὸ ἡμισφαίριον. The terms base (βάσις), vertex
(κορυφή) and height (ύψος) are also used with reference to a segment
of a sphere.
Another term borrowed from the geometry of the circle is the
word sector (τομεύς) qualified with the adjective στερεός (solid).
A solid sector (τομεὺς στερεός) is defined by Archimedes as the
figure bounded by a cone which has its vertex at the centre of
a sphere and the part of the surface of the sphere within the cone.
The segment of the sphere included in the sector is τὸ τμῆμα τῆς
σφαίρας τὸ ἐν τῷ τομεῖ or τὸ κατὰ τὸν τομέα.
A great circle of a sphere is ὁ μέγιστος κύκλος τῶν ἐν τῇ σφαίρα
and often ὁ μέγιστος κύκλος alone.
Let a sphere be cut by a plane not through the centre τετμήσθω
σφαίρα μὴ διὰ τοῦ κέντρου ἐπιπέδῳ; a sphere cut by a plane through
the centre in the circle ΕΖΗΘ, σφαῖρα ἐπιπέδω τετμημένη διὰ τοῦ
κέντρου κατὰ τὸν ΕΖΗΘ κύκλον.
Prisms and pyramids.
A prism is πρίσμα, & pyramid πυραμίς. As usual, ἀναγράφειν ἀπό
is used of describing a prism or pyramid on a rectilineal figure
as base; thus let a prism be described on the rectilineal figure
(as base) αναγεγράφθω ἀπὸ τοῦ εὐθυγράμμου πρίσμα, on the polygon
circumscribed about the circle A let a pyramid be set up ἀπὸ τοῦ περὶ
τὸν Α κύκλον περιγεγραμμένου πολυγώνου πυραμὶς ἀνεστάτω αναγεγραμ-
μένη. A pyramid with an equilateral base ABT is πυραμὶς ἰσόπλευρον
ἔχουσα βάσιν τὸ ΑΒΓ.
The surface is, as usual, ἐπιφάνεια and, when any particular face
or a base is excluded, some qualifying phrase has to be used.
·
THE TERMINOLOGY OF ARCHIMEDES.
clxv
Thus the surface of the prism consisting of the parallelograms
(i.e. excluding the bases) ή ἐπιφάνεια τοῦ πρίσματος ἡ ἐκ τῶν·
παρaλλn λopáμμov σvykeμévη; the surface (of a pyramid) excluding
the base or the triangle ΑΕΓ, ἡ ἐπιφάνεια χωρὶς τῆς βάσεως or τοῦ
ΑΕΠ τριγώνου.
The triangles bounding the pyramid τὰ περιέχοντα τρίγωνα τὴν
Tupaμída (as distinct from the base, which may be polygonal).
Cones and solid rhombi.
The Elements of Euclid only introduce right cones, which are
simply called cones without the qualifying adjective. A cone is
there defined as the surface described by the revolution of a right-
angled triangle about one of the sides containing the right angle.
Archimedes does not define a cone, but generally describes a. right
cone as an isosceles cone (kŵvos iσookeλýs), though once he calls it
right (oplós). J. H. T. Müller rightly observes that the term
isosceles applied to a cone was suggested by the analogy of the
isosceles triangle, but I doubt whether such a cone was thought of
(as he supposes) as one which could be described by making an
isosceles triangle revolve about the perpendicular from the vertex
on the base; it seems more natural to connect it with the use of
the word side (πλeupá) by which Archimedes designates a generator
of the cone, a right cone being thus directly regarded as a cone having
all its legs equal. The latter supposition would also accord better
with the term scalene cone (kŵvos σkaλŋvós) by which Apollonius
denotes an oblique circular cone; such a cone could not of course
be described by the revolution of a scalene triangle. An oblique
circular cone is simply a cone for Archimedes, and he does not
define it; but, while he speaks of finding a cone with a given
vertex and passing through every point on a given 'section of an
acute-angled cone' [ellipse], he regards the finding of the cone as
being equivalent to finding the circular sections, and we may
therefore conclude that he would have defined the cone in
practically the same way as Apollonius does, namely as the surface
described by a straight line always passing through a fixed point
and moving round the circumference of any circle not in the same
plane with the point.
The vertex of a cone is, as usual, kopupń, the base ẞáois, the axis
ἄξων and the height ὕψος; the cones are of the same height εἰσὶν οἱ
κῶνοι ὑπὸ τὸ αὐτὸ ὕψος. A generator is called a side (πλευρά); if α
I
}
clxvi
INTRODUCTION. ·
cone be cut by a plane meeting all the generators of the cone ei ka
κῶνος ἐπιπέδω τμαθῇ συμπίπτοντι πάσαις ταῖς τοῦ κώνου πλευραῖς.
The surface of the cone excluding the base ἡ ἐπιφάνεια τοῦ κώνου
Xwpis Tηs Báσews; the conical surface between (two generators) AA, AB,
κωνικὴ ἐπιφάνεια ἡ μεταξὺ τῶν ΑΔΒ.
There is no special name for what we call a frustum of a cone
or the portion intercepted between two planes parallel to the base;
the surface of such a frustum is simply the surface of the cone
between the parallel planes ἡ ἐπιφάνεια τοῦ κώνου μεταξὺ τῶν
παραλλήλων ἐπιπέδων.
A curious term is segment of a cone (åñóτµаµa kávov), which is
used of the portion of any circular cone, right or oblique, cut off
towards the vertex by any plane which makes an elliptic and not a
circular section. With reference to a segment of a cone the axis
(ağwv) is defined as the straight line drawn from the vertex of the
cone to the centre of the elliptic base.
As usual, avaypápe anò is used of describing a cone on a circle
as base.
Similarly, a very common phrase is ἀπὸ τοῦ κύκλου κῶνος
cσrw let there be a cone on the circle (as base).
او
A solid rhombus (póµßos σtepeós) is the figure made up of two
cones having their base common, their vertices on opposite sides of
it, and their axes in one straight line. A rhombus made up of
isosceles cones ῥόμβος ἐξ ἰσοσκελῶν κώνων συγκείμενος, and the two
cones are spoken of as the cones bounding the rhombus oi kŵvoɩ oi
περιέχοντες τὸν ῥόμβον.
Cylinders.
A right cylinder is κύλινδρος ὀρθός, and the following terms
apply to the cylinder as to the cone: base Báois, one base or the
other n érépa Báois, of which the circle AB is a base and TA opposite
to it οὗ βάσις μὲν ὁ ΑΒ κύκλος, ἀπεναντίον δὲ ὁ ΓΔ ; axis ἄξων, height
vos, generator Tλevpά. The cylindrical surface cut off by (two
generators) ΑΓ, ΒΔ, ἡ ἀποτεμνομένη κυλινδρικὴ ἐπιφάνεια ὑπὸ τῶν ΑΓ,
BA; the surface of the cylinder adjacent to the circumference ABT, y
ἐπιφάνεια τοῦ κυλίνδρου ἡ κατὰ τὴν ΑΒΓ περιφέρειαν denotes the
surface of the cylinder between the two generators drawn through
the extremities of the arc.
A frustum of a cylinder τόμος κυλίνδρου is a portion of a
cylinder intercepted between two parallel sections which are elliptic
and not circular, and the axis (ägwv) of it is the straight line
THE TERMINOLOGY OF ARCHIMEDES.
clxvii
•
joining the centres of the two sections, which is in the same straight
line with the axis of the cylinder.
Conic Sections.
General terms are κωνικά στοιχεία, elements of conics, τὰ κωνικά
(the theory of) conics. Any conic section KÁVOV TOμỲ Öπоιαοûν.
Chords are simply εὐθείαι ἐν τῇ τοῦ κώνου τομᾷ ἀγμέναι. Archimedes
never uses the word axis (aέwv) with reference to a conic; the axes
are with him diameters (diáµetpoi), and diáμeтpos, when it has
reference to a complete conic, is used in this sense exclusively. A
tangent is ἐπιψαύουσα οι ἐφαπτομένη (with gen.).
The separate conic sections are still denoted by the old names;
a parabola is a section of a right-angled cone ὀρθογωνίου κώνου τομή,
a hyperbola a section of an obtuse-angled cone außλvywvíov kávov
τομή, and an ellipse a section of an acute-angled cone ὀξυγωνίου κώνου
τομή.
The parabola.
Only the axis of a complete parabola is called a diameter, and
the other diameters are simply lines parallel to the diameter. Thus
parallel to the diameter or itself the diameter is πapà tàv diáµetpov †
αὐτὰ διάμετρος; ΔΖ is parallel to the diameter & ΔΖ παρὰ τὰν
Siάμeтρóv éσтi. Once the term principal or original (diameter) is
used, αρχικά (sc. διάμετρος).
A segment of a parabola is tµîµa, which is more fully described
as the segment bounded by a straight line and a section of a right-
angled cone τμᾶμα τὸ περιεχόμενον ὑπό τε εὐθείας καὶ ὀρθογωνίου κώνου
τομᾶς. The word diáμeтpos is again used with reference to a
segment of a parabola in the sense of our word axis; Archimedes
defines the diameter of any segment as the line bisecting all the
straight lines (chords) drawn parallel to its base Tàv díxa réμvovσáv
τὰς εὐθείας πάσας τὰς παρὰ τὴν βάσιν αὐτοῦ ἀγομένας.
The part of a parabola included between two parallel chords is
called a frustum τόμος (ἀπὸ ὀρθογωνίου κώνου τομᾶς ἀφαιρούμενος),
the two chords are its lesser and greater base (λáoσwv and μeížov
Báois) respectively, and the line joining the middle points of the
two chords is the diameter (Siάµerpos) of the frustum.
What we call the latus rectum of a parabola is in Archimedes
the line which is double of the line drawn as far as the axis å diλaσíα
τᾶς μέχρι τοῦ ἄξονος. In this expression the axis (ἄξων) is the axis
clxviii
INTRODUCTION.
αν
of the right-angled cone from which the curve was originally derived
by means of a section perpendicular to a generator*. Or, again, the
equivalent of our word parameter (παρ᾽ ἂν δυνάνται αἱ ἀπὸ τᾶς τομᾶς)
is used by Archimedes as by Apollonius, meaning the straight line
to which the rectangle which has its breadth equal to the abscissa
of a point and is equal to the square of the ordinate must be
applied as base. The full phrase states that the ordinates have
their squares equal to the rectangles applied to the line equal to N (or
the parameter) which have as their breadth the lines which they (the
ordinates) cut off from AZ (the diameter) towards the extremity ▲,
δυνανται τὰ παρὰ τὰν ἴσαν τᾷ Ν παραπίπτοντα πλάτος ἔχοντα, ἃς αὐταὶ
ἀπολαμβάνοντι ἀπὸ τᾶς ΔΖ ποτὶ τὸ Δ πέρας.
Ordinates are the lines drawn from the section to the diameter
(of the segment) parallel to the base of the segment) αἱ ἀπὸ τᾶς τομᾶς
ἐπὶ τὰν ΔΖ ἀγομέναι παρὰ τὴν ΑΕ, or simply αἱ ἀπὸ τᾶς τομᾶς. Once
also the regular phrase drawn ordinate-wise τεταγμένως κατηγμένη is
used to describe an ordinate, as in Apollonius.
The hyperbola.
What we call the asymptotes (αἱ ἀσύμπτωτοι in Apollonius) are
in Archimedes the lines (approaching) nearest to the section of the
obtuse-angled cone αἱ ἔγγιστα τᾶς τοῦ ἀμβλυγωνίου κώνου τομᾶς.
The centre is not described as such, but it is the point at which
the lines nearest (to the curve) meet το σαμεῖον, καθ᾿ ὃ αἱ ἔγγιστα
συμπίπτοντι.
This is a property of the sections of obtuse-angled cones τοῦτο γάρ
ἐστιν ἐν ταῖς τοῦ ἀμβλυγωνίου κώνου τομαῖς σύμπτωμα.
The ellipse.
The major and minor axes are the greater and lesser diameters
μείζων and ἐλάσσων διάμετρος. Let the greater diameter be ΑΓ,
διάμετρος δὲ αὐτᾶς) ἃ μὲν μείζων ἔστω ἐφ᾽ ἃς τὰ Α, Γ. The rectangle
contained by the diameters (axes) τὸ περιεχόμενον ὑπὸ τῶν διαμέτρων.
One axis is called conjugate (συζυγής) to the other: thus let the
straight line N be equal to half of the other diameter which is
conjugate to AB, ὁ δὲ Ν εὐθεῖα ἴσα ἔστω τᾷ ἡμισείᾳ τᾶς ἑτέρας διαμέτρου,
ἅ ἐστι συζυγὴς τα ΑΒ.
α
The centre is here κέντρον.
* Cf. Apollonius of Perga, pp. xxiv, xxv.
THE TERMINOLOGY OF ARCHIMEDES.
clxix
.
•
Conoids and Spheroids.
There is a remarkable similarity between the language in which
Archimedes describes the genesis of his solids of revolution and that
used by Euclid in defining the sphere. Thus Euclid says: when, the
diameter of a semicircle remaining fixed, the semicircle revolves and
returns to the same position from which it began to move, the included
figure is a sphere σφαῖρά ἐστιν, ὅταν ἡμικυκλίου μενούσης τῆς διαμέτρου
περιενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο
φέρεσθαι, τὸ περιληφθὲν σχῆμα; and he proceeds to state that the
axis of the sphere is the fixed straight line about which the semicircle
turns ἄξων δὲ τῆς σφαίρας ἐστὶν ἡ μένουσα εὐθεῖα, περὶ ἣν τὸ ἡμικύκλιον
στρέφεται.
OTρéperαL Compare with this e.g. Archimedes' definition of the
right-angled conoid (paraboloid of revolution): if a section of a
right-angled cone, with its diameter (axis) remaining fixed, revolves
and returns to the position from which it started, the figure included
by the section of the right-angled cone is called a right-angled conoid,
and its axis is defined as the diameter which has remained fixed,
εἴ κα ὀρθογωνίου κώνου τομὰ μενούσας τᾶς διαμέτρου περιενεχθεῖσα
ἀποκατασταθῇ πάλιν, ὅθεν ὥρμασεν, τὸ περιλαφθὲν σχῆμα ὑπὸ τᾶς τοῦ
ὀρθογωνίου κώνου τομᾶς ὀρθογώνιον κωνοειδὲς καλείσθαι, καὶ ἄξονα
μὲν αὐτοῦ τὰν μεμενακοῦσαν διάμετρον καλείσθαι, and it will be seen
that the several phrases used are practically identical with those of
Euclid, except that ώρμασεν takes the place of ἤρξατο φέρεσθαι; and
even the latter phrase occurs in Archimedes' description of the
genesis of the spiral later on.
EL
The words conoid κωνοειδές (σχήμα) and spheroid σφαιροειδὲς
(σχήμα) are simply adapted from κώνος and σφαίρα, meaning that
the respective figures have the appearance (cidos) of, or resemble,
cones and spheres; and in this respect the names are perhaps more
satisfactory than paraboloid, hyperboloid and ellipsoid, which can
only be said to resemble the respective conics in a different sense.
But when kwvoedés is qualified by the adjective right-angled
ὀρθογώνιον to denote the paraboloid of revolution, and by ἀμβλυ
yúviov obtuse-angled to denote the hyperboloid of revolution, the
expressions are less logical, as the solids do not resemble right-
angled and obtuse-angled cones respectively; in fact, since the
angle between the asymptotes of the generating hyperbola may be
acute, a hyperboloid of revolution would in that case more resemble
an acute-angled cone. The terms right-angled and obtuse-angled
clxx
INTRODUCTION.
were merely transferred to the conoids from the names for the
respective conics without any more thought of their meaning.
It is unnecessary to give separately the definition of each
conoid and spheroid; the phraseology is in all cases the same
as that given above for the paraboloid. But it may be remarked
that Archimedes does not mention the conjugate axis of a hyperbola
or the figure obtained by causing a hyperbola to revolve about that
axis; the conjugate axis of a hyperbola first appears in Apollonius,
who was apparently the first to conceive of the two branches of a
hyperbola as one curve. Thus there is only one obtuse-angled
conoid in Archimedes, whereas there are two kinds of spheroids
according as the revolution takes place about the greater diameter
(axis) or lesser diameter of the generating section of an acute-
angled cone (ellipse); the spheroid is in the former case oblong
(παραμάκες σφαιροειδές) and in the latter case fat (ἐπιπλατύ
σφαιροειδές).
·
A special feature is, however, to be observed in the description
of the obtuse-angled conoid (hyperboloid of revolution), namely that
the asymptotes of the hyperbola are supposed to revolve about the
axis at the same time as the curve, and Archimedes explains that
they will include an isosceles cone (κῶνον ἰσοσκελέα περιλαψούνται),
which he thereupon defines as the cone enveloping the conoid
(περιέχων τὸ κωνοειδές). Also in a spheroid the term diameter
(diáμeтpos) is appropriated to the straight line drawn through
the centre at right angles to the axis (ἡ διὰ τοῦ κέντρου ποτ᾽ ὀρθὰς
ἀγομένα τῷ ἄξονι). The centre of a spheroid is the middle point of
the axis τὸ μέσον τοῦ ἄξονος.
The following terms are used of all the conoids and spheroids.
The vertex (kopvoý) is the point at which the axis meets the surface tò
σαμεῖον, καθ᾿ ὃ ἁπτέται ὁ ἄξων τᾶς ἐπιφανείας, the spheroid having of
course two vertices. A segment (7µâµ¤) is a part cut off by a plane,
and the base (Báois) of the segment is defined as the plane (figure)
included by the section of the conoid (or spheroid) in the cutting
plane τὸ ἐπίπεδον τὸ περιλαφθὲν ὑπὸ τᾶς τοῦ κωνοειδέος (or σφαιροειδέος)
τομᾶς ἐν τῷ ἀποτέμνοντι ἐπιπέδῳ. The vertex of a segment is the point
at which the tangent plane parallel to the base of the segment meets
the surface, το σαμεῖον, καθ᾿ ὃ ἁπτέται τὸ ἐπίπεδον τὸ ἐπιψαῦον (τοῦ
Kwvoeιdéos). The axis (aswv) of a segment is differently defined for
the three surfaces; (a) in the paraboloid it is the straight line cut off
within the segment from the line drawn through the vertex of the
THE TERMINOLOGY OF ARCHIMEDES.
clxxi.
segment parallel to the axis of the conoid ὁ ἐναπολαφθεῖσα εὐθεῖα ἐν τῷ
τμάματι ἀπὸ τῆς ἀχθείσας διὰ τᾶς κορυφᾶς τοῦ τμάματος παρὰ τὸν
ἄξονα τοῦ κωνοειδέος, (β) in the hyperboloid it is the straight line cut
off within the segment from the line drawn through the vertex of the
segment and the vertex of the cone enveloping the conoid ἀπὸ τᾶς
αχθείσας διὰ τᾶς κορυφᾶς τοῦ τμάματος καὶ τᾶς κορυφᾶς τοῦ κώνου τοῦ
περιέχοντος τὸ κωνοειδές, (c) in the spheroid it is the part similarly
cut off from the straight line joining the vertices of the two segments
into which the base divides the spheroid, ἀπὸ τῆς εὐθείας τᾶς τὰς
κορυφὰς αὐτῶν (τῶν τμαμάτων) ἐπιζευγνυούσας.
Archimedes does not use the word centre with respect to the
hyperboloid of revolution, but calls the centre the vertex of the
enveloping cone. Also the axis of a hyperboloid or a segment is
only that part of it which is within the surface. The distance
between the vertex of the hyperboloid or segment and the vertex
of the enveloping cone is the line adjacent to the axis & ποτεοῦσα
τῷ ἄξονι.
The following are miscellaneous expressions. The part inter-
cepted within the conoid of the intersection of the planes ὁ ἐναπο-
λαφθεῖσα ἐν τῷ κωνοειδεῖ τᾶς γενομένας τομᾶς τῶν ἐπιπέδων, (the plane)
will have cut the spheroid through its axis τετμακὸς ἐσσείται τὸ
σφαιροειδές διὰ τοῦ ἄξονος, so that the section it makes will be a
conic section ὥστε τὰν τομὰν ποιήσει κώνου τομάν, let two segments be
cut of in any manner αποτετμάσθω δύο τμάματα ὡς ἔτυχεν or by
planes drawn in any manner ἐπιπέδοις ὁπωσοῦν ἀγμένοις.
Half the spheroid τὸ ἁμίσεον τοῦ σφαιροειδέος, half the line
joining the vertices of the segments (of a spheroid), i.e. what we should
call a semi-diameter, ἡ ἡμισέα αὐτᾶς τῆς ἐπιζευγνυούσας τὰς κορυφὰς
τῶν τμαμάτων.
The spiral.
We have already had, in the conoids and spheroids, instances of
the evolution of figures by the motion of curves about an axis. The
same sort of motion is used for the construction of solid figures
inscribed in and circumscribed about a sphere, a circle and an
inscribed or circumscribed polygon being made to revolve about
a diameter passing through an angular point of the polygon and
dividing it and the circle symmetrically. In this case, in Archimedes'
phrase, the angular points of the polygon will move along the circum-
ferences of circles, αἱ γωνίαι κατὰ κύκλων περιφερειῶν ἐνεχθήσονται (or
clxxii
INTRODUCTION.
οἰσθήσονται) and the sides will move on certain cones, or on the surface
of a cone κατά τινων κώνων ενεχθήσονται or κατ' ἐπιφανείας κώνου ; and
sometimes the angular points or the points of contact of the sides of
a circumscribed polygon are said to describe circles γράφουσι κύκλους.
The solid figure so formed is τὸ γενηθὲν στερεὸν σχημα, and let the
sphere by its revolution make a figure περιενεχθεῖσα ἡ σφαῖρα ποιείτω
σχημά τι.
For the construction of the spiral, however, we have a new
element introduced, that of time, and we have two different uniform
motions combined; if a straight line in a plane turn uniformly
about one extremity which remains fixed, and return to the position
from which it started and if, at the same time as the line is revolving,
a point move at a uniform rate along the line starting from the fixed
extremity, the point will describe a spiral in the plane, εἴ κα εὐθεῖα...ἐν
ἐπιπέδῳμένοντος τοῦ ἑτέρου πέρατος αὐτᾶς ἰσοταχέως περιενεχθεῖσα
ἀποκατασταθῇ πάλιν, ὅθεν ὥρμασεν, ἅμα δὲ τῇ γραμμᾷ περιαγομένα
φερήται τι σαμεῖον ἰσοταχέως αὐτὸ ἑαυτῷ κατὰ τᾶς εὐθείας ἀρξάμενον ἀπὸ
τοῦ μένοντος πέρατος, τὸ σαμεῖον ἕλικα γράψει ἐν τῷ ἐπιπέδῳ.
The spiral (described in the first, second, or any turn is ὁ ἕλιξ ὁ ἐν
τα πρώτα, δευτέρᾳ, or ὁποιμοῦν περιφορᾷ γεγραμμένα, and the turns
other than any particular ones are the other spirals αἱ ἄλλαι ἕλικες.
The distance traversed by the point along the line in any time is
ἡ εὐθεῖα ἡ διανυσθεῖσα, and the times in which the point moved over the
distances οἱ χρόνοι, ἐν οἷς τὸ σαμεῖον τὰς γραμμὰς ἐπορεύθη ; in the time
in which the revolving line reaches ΑΓ from AB, ἐν ᾧ χρόνῳ ἁ περιαγομένα
γραμμὰ ἀπὸ τᾶς ΑΒ ἐπὶ τὴν ΑΓ ἀφικνείται.
The origin of the spiral is ἀρχὰ τῆς ἕλικος, the initial line ἀρχὰ τᾶς
περιφοράς. The distance described by the point along the line in
the first complete revolution is εὐθεῖα πρώτα (first distance), that
described during the second revolution the second distance ev0۔a
δευτέρα, and so on, the distances being called by the number of the
revolutions ὁμωνύμως ταῖς περιφοραῖς. The first area, χωρίον πρώτον,
is the area bounded by the spiral described in the first revolution and
by the first distance' τὸ χωρίον τὸ περιλαφθὲν ὑπό τε τᾶς ἕλικος τᾶς ἐν
τα πρώτα περιφορᾷ γραφείσας καὶ τᾶς εὐθείας, ἃ ἐστιν πρώτα; the second
area is that bounded by the spiral in the second turn and the 'second
distance,' and so on. The area added by the spiral in any turn is rò
χωρίον τὸ ποτιλαφθὲν ὑπὸ τῆς ἕλικος ἔν τινι περιφορᾷ.
α
The first circle, κύκλος πρῶτος, is the circle described with the
'first distance' as radius and the origin as centre, the second circle
THE TERMINOLOGY OF ARCHIMEDES.
clxxiii
that with the origin as centre and twice the 'first distance' as
radius, and so on.
Together with as many times the whole of the circumference of the
circle as (is represented by) the number less by one than (that of)
the revolutions μεθ᾽ ὅλας τᾶς τοῦ κύκλου περιφερείας τοσαυτάκις λαμ-
βανομένας, ὅσος ἐστὶν ὁ ἑνὶ ἐλάσσων ἀριθμὸς τῶν περιφορᾶν, the circle
called by the number corresponding to that of the revolutions å kúkλos
ὁ κατὰ τὸν αὐτὸν ἀριθμὸν λεγόμενος ταῖς περιφοραῖς.
With reference to any radius vector, the side which is in the
direction of the revolution is forward τὰ προηγούμενα, the other
backward τὰ ἑπόμενα.
Tangents, etc.
Though the word åлтоμαι is sometimes used in Archimedes of a
line touching a curve, its general meaning is not to touch but simply
to meet; e.g. the axis of a conoid or spheroid meets (άπτεται) the
surface in the vertex. (The word is also often used elsewhere than
in Archimedes of points lying on a locus; e.g. in Pappus, p. 664, the
point will lie on a straight line given in position ἅψεται τὸ σημεῖον
θέσει δεδομένης εὐθείας.)
To touch a curve or surface is generally ἐφάπτεσθαι or ἐπιψαύειν
(with gen.). A tangent is ἐφαπτομένη or ἐπιψαύουσα (sc. εὐθεία) and
a tangent plane ἐπιψανον ἐπίπεδον. Let tangents be drawn to the circle
ΑΒΓ, τοῦ ΑΒΓ κύκλου εφαπτόμεναι ἤχθωσαν; if straight lines be drawn
touching the circles ἐὰν ἀχθωσίν τινες ἐπιψαύουσαι τῶν κύκλων. The
full phrase of touching without cutting is sometimes found in
Archimedes; if a plane touch (any of) the conoidal figures
without cutting the conoid εἴ κα τῶν κωνοειδέων σχημάτων ἐπίπεδον
ἐφαπτήται μὴ τέμνον τὸ κωνοειδές. The simple word ψαύειν is
occasionally used participially), the tangent planes τὰ ἐπίπεδα τὰ
ψαύοντα.
To touch at a point is expressed by κατά (with acc.); the points
at which the sides...touch (or meet) the circle σημεία, καθ᾽ ἃ ἅπτονται
τοῦ κύκλου αἱ πλευραί.... Let them touch the circle at the middle
points of the circumferences cut off by the sides of the inscribed
polygon ἐπιψαυέτωσαν τοῦ κύκλου κατὰ μέσα τῶν περιφερειῶν τῶν
ἀποτεμνομένων ὑπὸ τοῦ ἐγγεγραμμένου πολυγώνου πλευρών.
The distinction between ἐπιψαύειν and ἅπτομαι is well brought
out in the following sentence; but that the planes touching the
spheroid meet its surface at one point only we shall prove ὅτι δὲ
,
clxxiv
INTRODUCTION.
τὰ ἐπιψαύοντα ἐπίπεδα τοῦ σφαιροειδέος καθ' ἓν μόνον ἁπτόνται σαμεῖον
τᾶς ἐπιφανείας αὐτοῦ δειξοῦμες.
The point of contact ἡ ἁφή.
Tangents drawn from (a point) åyμévai ảπó; we find also the
elliptical expression ἀπὸ τοῦ Ξ ἐφαπτέσθω ἡ ΟΕΤ, let ΟΞΠ be the
tangent from E, where, in the particular case, E is on the circle.
Constructions.
The richness of the Greek language in expressions for con-
structions is forcibly illustrated by the variety of words which
may be used (with different shades of meaning) for drawing a
line. Thus we have in the first place ayw and the compounds
Stayw (of drawing a line through a figure, with eis or ev following,
of producing a plane beyond a figure, or of drawing a line in a
plane), karáуw (used of drawing an ordinate down from a point on
a conic), Tрooáyw (of drawing a line to meet another). As an
alternative to προσάγω, προσβάλλω is also used; and προσπίπτω
may take the place of the passive of either verb. To produce is
Kẞáλλw, and the same word is also used of a plane drawn through a
point or through a straight line; an alternative for the passive is
supplied by ἐκπίπτω. Moreover πρόσκειμαι is an alternative word
for being produced (literally being added).
In the vast majority of cases constructions are expressed by the
elegant use of the perfect imperative passive (with which may be
classed such forms as γεγονέτω from γίγνομαι, ἔστω from εἰμί, and
κείσθω from κείμαι), or occasionally the aorist imperative passive.
The great variety of the forms used will be understood from the
following specimens. Let BT be made (or supposed) equal to A,
κείσθω τῷ Δ ἴσον τὸ ΒΓ; let it be drawn ἤχθω, let a straight line be
drawn in it (a chord of a circle) dińxow tɩs eis autòv evßeîa, let KM be
drawn equal to... ἴση κατήχθω ἡ KM, let it be joined επεζεύχθω, let
ΚΛ be drawn to meet προσβεβλήσθω ἡ ΚΛ, let them be produced
ἐκβεβλήσθωσαν, suppose them found εὑρήσθωσαν, let a circle be set out
ἐκκείσθω κύκλος, let it be taken εἰλήφθω, let K, Η be taken ἔστωσαν
ειλημμέναι αἱ Κ, H, let a circle Ψ be taken λελάφθω κύκλος ἐν ᾧ τὸ Ψ, let
it be cut τετμήσθω, let it be divided διαιρήσθω (διῃρήσθω) ; let one cone be
cut by a plane parallel to the base and produce the section EZ, TµNOŃTW 8
ἕτερος κῶνος ἐπιπέδῳ παραλλήλῳ τῇ βάσει καὶ ποιείτω τομὴν τὴν ΕΖ, let
TZ be cut off aroλeλáp0w ȧ TZ; let (such an angle) be left and let it
δε ΝΗΓ, λελείφθω καὶ ἔστω ἡ ὑπὸ ΝΗΓ, let a figure be made γεγενήσθω
THE TERMINOLOGY OF ARCHIMEDES.
clxxv
σχημα, let the sector be made ἔστω γεγενημένος ὁ τομεύς, let cones be
described on the circles (as bases) αναγεγράφθωσαν ἀπὸ τῶν κύκλων
κῶνοι, ἀπὸ τοῦ κύκλου κῶνος ἔστω, let it be inscribed or circumscribed
ἐγγεγράφθω (or ἐγγεγραμμένον ἔστω), περιγεγράφθω ; let an area (equal
to that) of AB be applied to ΛΗ, παραβεβλήσθω παρὰ τὰν ΛΗ τὸ χωρίον
τοῦ ΑΒ ; let a segment of a circle be described on ΘΚ, ἐπὶ τῆς ΘΚ
κύκλου τμῆμα ἐφεστάσθω, let the circle be completed αναπεπληρώσθω ὁ
κύκλος, let N= (α parallelogram) be completed συμπεπληρώσθω τὸ ΝΕ,
let it be made πεποιήσθω, let the rest of the construction be the same as
before τὰ ἄλλα κατεσκευάσθω τὸν αὐτὸν τρόπον τοῖς πρότερον. Suppose
it done γεγονέτω.
Another method is to use the passive imperative of voéw (let it be
conceived). Let straight lines be conceived to be drawn νοείσθωσαν
εὐθεῖαι ἠγμέναι, let the sphere he conceived to be cut νοείσθω ἡ σφαίρα
τετμημένη, let a figure (generated from the inscribed polygon be
conceived as inscribed in the sphere ἀπὸ τοῦ πολυγώνου τοῦ ἐγγραφο-
μένου νοείσθω τι εἰς τὴν σφαῖραν ἐγγραφὲν σχῆμα. Sometimes the
participle for drawn is left out; thus ἀπ' αὐτοῦ νοείσθω ἐπιφάνεια let
a surface be conceived (generated) from it.
The active is much more rarely used; but we find (1) ἐὰν with
subjunctive, if we cut ἐὰν τέμωμεν, if we draw ἐὰν ἀγάγωμεν, if you
produce ἐὰν ἐκβαλῇς; (2) the participle, it is possible to inscribe...and
(ultimately) to leave δυνατόν ἐστιν ἐγγράφοντα...λείπειν, if we con-
tinually circumscribe polygons, bisecting the remaining circumferences
and drawing tangents, we shall (ultimately) leave ἀεὶ δὴ περιγράφοντες
πολύγωνα δίχα τεμνομένων τῶν περιλειπομένων περιφερειῶν καὶ ἀγομένων
ἐφαπτομένων λείψομεν, it is possible, if we take the area..., to inscribe
λαβόντα (or λαμβάνοντα) τὸ χωρίον...δυνατόν ἐστιν...ἐγγράψαι; (3) the
first person singular, I take two straight lines λαμβάνω δύο εὐθείας,
I took a straight line ἔλαβόν τινα εὐθεῖαν ; I draw @M from – parallel
to ΑΖ, ἄγω ἀπὸ τοῦ Θ τὰν ΘΜ παράλληλον τα AZ, having drawn ΓΚ
perpendicular, I cut off AK equal to TK ȧyaywv káletov tàv TK tậ
ΓΚ ἴσαν ἀπέλαβον τὰν ΑΚ, Ι inscribed a solid figure...and circum-
scribed another ἐνέγραψα σχῆμα στερεόν...καὶ ἄλλο περιέγραψα.
The genitive of the passive participle is used absolutely,
εὑρεθέντος δή it being supposed found, ἐγγραφέντος δή (the figure)
being inscribed.
To make a figure similar to one (and equal to another) ὁμοιώσαι,
to find experimentally ὀργανικώς λαβεῖν, to cut into unequal parts εἰς
ἄνισα τέμνειν.
:
clxxvi
INTRODUCTION.
Operations (addition, subtraction, etc.).
1. Addition, and sums, of magnitudes.
To add is προστίθημι, for the passive of which πρόσκειμαι is often
used; thus one segment being added ἑνὸς τμάματος ποτιτεθέντος, the
added (straight line) & ποτικειμένα, let the common HA, ΖΓ be added
κοιναὶ προσκείσθωσαν αἱ ΗΑ, ΖΓ; the words are generally followed
by πρός (with acc. of the thing added to), but sometimes by the
dative, that to which the addition was made ᾧ ποτετέθη.
For being added together we have συντίθεσθαι; thus being added
to itself συντιθέμενον αὐτὸ ἑαυτῷ, added together ἐς τὸ αὐτὸ συντεθέντα,
added to itself (continually) ἐπισυντιθέμενον ἑαυτῷ.
Sums are commonly expressed for two magnitudes by συναμφό
τερος used in the following different ways; the sum of BΑ, ΑΛ
συναμφότερος ἡ ΒΑΛ, the sum of ΔΓ, ΓΒ συναμφότερος ἡ ΔΓ, ΓΒ, the
sum of the area and the circle τὸ συναμφότερον ὅ τε κύκλος καὶ τὸ
χωρίον. Again for sums in general we have such expressions as the
line which is equal to both the radii ἡ ἴση ἀμφοτέραις ταῖς ἐκ τοῦ
κέντρου, the line equal to the sum of all the lines joining ἡ ἴση
πάσαις ταῖς ἐπιζευγνυούσαις. Also all the circles οἱ πάντες κύκλοι
means the sum of all the circles; and σύγκειται ἐκ is used for is
equal to the sum of (two other magnitudes).
To denote plus μετά (with gen.) and σύν are used; together with
the bases μετὰ τῶν βάσεων, together with half the base of the segment
σὺν τῇ ἡμισείᾳ τῆς τοῦ τμήματος βάσεως; τε and καί also express the
same thing, and the participle of προσλαμβάνω gives another way of
describing having something added to it; thus the squares on (all)
the lines equal to the greatest together with the square on the greatest...
is τὰ τετράγωνα τὰ ἀπὸ τῶν ἰσᾶν τῇ μεγίστᾳ ποτιλαμβάνοντα τό τε ἀπὸ
τᾶς μεγίστας τετράγωνον....
2. Subtraction and differences.
To subtract from is ἀφαιρεῖν ἀπό; if (the rhombus) be conceived as
taken away ἐὰν νοηθῇ ἀφηρημένος, let the segments be subtracted
ἀφαιρεθέντων τα τμήματα. Terms common to each side in an
equation are κοινά ; the squares are common to both sides) κοινά ἐντι
ἑκατέρων τὰ τετράγωνα. Then let the common area be subtracted
is κοινὸν ἀφῃρήσθω τὸ χωρίον, and so on; the remainder is denoted
by the adjective λοιπός, e.g. the conical surface remaining λοιπὴ ἡ
κωνικὴ ἐπιφάνεια.
The difference or excess is ὑπεροχή, or more fully the excess by
THE TERMINOLOGY OF ARCHIMEDES.
clxxvii
W
•
which (one magnitude) exceeds (another) vπeрoɣń, ʼn vπepéɣeɩ….. or
ὑπεροχά, μείζων
veроɣá, a μεíčшv orí.... The excess is also expressed by means of
the verb vτepéɣew alone; let the difference by which the said triangles
exceed the triangle ΑΔΓ be ®, ὦ δὴ ὑπερέχει τὰ εἰρημένα τρίγωνα τοῦ
ΑΔΓ τριγώνου ἔστω τὸ Θ, to exceed by less than the excess of the cone
Ψ over the half of the spheroid ὑπερέχειν ἐλάσσονι ἢ ᾧ (or ἁλίκῳ)
ὑπερέχει ὁ Ψ κώνος τοῦ ἡμίσεος τοῦ σφαιροειδέος (where ᾧ ὑπερέχει may
also be omitted). Again the excess may be μeilwv čoτí. The
opposite to υπερέχει is λείπεται (with gen.).
Equal to twice a certain excess loa dvoìv vπepoɣaîs, with which
equal to one excess, loa pia vπeроxâ, is contrasted.
The following sentence practically states the equivalent of an
algebraical equation; the rectangle under ZH, EA exceeds the rect-
angle under ZE, EA by the (sum of) the rectangle contained by EA,
EH and the rectangle under ZE, ΞΕ, ὑπερέχει τὸ ὑπὸ τῶν ΖΗ, ΕΔ τοῦ
vñò tâv ZE, EA tậ te vñò tâv EA, EH πepieɣoµévų kaì tê vπò tâv ZE,
EE. Similarly twice PH together with II is (equal to) the sum of
ΣΡ, ΡΠ, δύο μὲν αἱ ΡΗ μετὰ τᾶς ΠΣ συναμφότερός ἐστιν ὁ ΣΡΠ.
3. Multiplication.
To multiply is Tоλλaλασráйw; multiply one another (of numbers)
πολλαπλασιάζειν αλλάλους; to multiply by a number is expressed by
the dative ; let – be multiplied by @ πεπολλαπλασιάσθω ὁ Δ τῷ Θ.
Multiplied into is sometimes èπí (with acc.); thus the rectangle
HO, OA into OA (i.e. a solid figure) is rò væò tŵv HO, ☺A ¿πì
τὴν ΘΑ.
4. Division.
To divide diapeîv; let it be divided into three equal parts at the
points K, Θ, διηρήσθω εἰς τρία ἴσα κατὰ τὰ Κ, Θ ταμεία; to be divisible
by μετρείσθαι ὑπό.
Proportions.
A ratio is λóyos, proportional is expressed by the phrase in
proportion ἀνάλογον, and a proportion is ἀναλογία. We find in
Archimedes some uses of the verb λéyw which seem to throw light
on the definition found in Euclid of the relation or ratio between
two magnitudes. One passage (On Conoids and Spheroids, Prop. 1)
says if the terms similarly placed have, two and two, the same ratio
and the first magnitudes are taken in relation to some other mag-
nitudes in any ratios whatever εἴ κα κατὰ δύο τὸν αὐτὸν λόγον ἔχωντι
H. A.
m
•
clxxviii
INTRODUCTION.
τὰ ὁμοίως τεταγμένα, λεγήται δὲ τὰ πρῶτα μεγέθεα ποτί τινα ἄλλα
μεγέθεα...ἐν λόγοις ὁποιοισοῦν, ἐξ Α, Β... be in relation to N, E... but
Z be not in relation to anything (i.e. has no term corresponding to
it) εἴ κα... τὰ μὲν Α, Β,... λεγωνται ποτὶ τὰ Ν, Ξ, τὸ δὲ Ζ μηδὲ
ποθ' ἓν λεγήται.
A mean proportional between is μέση ἀνάλογον των….., is a mean
proportional between μέσον λόγον ἔχει τῆς...καὶ τῆς..., two mean pro-
portionals δύο μέσαι. ἀνάλογον with or without κατὰ τὸ συνεχές in
continued proportion.
If three straight lines be proportional ἐὰν τρεῖς εὐθεῖαι ἀνάλογον
ὦσι, a fourth proportional τετάρτα ἀνάλογον, if four straight lines be
proportional in continued proportion εἴ κα τέσσαρες γραμμαὶ ἀνάλογον
ἔωντι ἐν τῷ συνεχεί αναλογίᾳ, at the point dividing (the line) in the
said proportion κατὰ τὴν ἀνάλογον τομὰν τῷ εἰρημένα.
The ratio of one straight line to another is eg. ὁ τῆς ΡΛ πρὸς ΛΧ
λόγος or ὁ (λόγος), ὃν ἔχει ἡ ΡΛ πρὸς τὴν ΛΧ ; the ratio of the bases ὁ
τῶν βασίων λόγος; has the ratio of 5 to 2 λόγον ἔχει, ὃν πέντε πρὸς
δύο.
For having the same ratio as we find the following constructions.
Have the same ratio to one another as the bases τὸν αὐτὸν ἔχοντι λόγον
ποτ᾽ ἀλλάλους ταῖς βάσεσιν, as the squares on the radii ὃν αἱ ἐκ τῶν
κέντρων δυνάμει; ΤΔ has to PZ the linear ratio which the square on
ΤΔ has to the square on H, ὃν ἔχει λόγον ἡ ΤΔ πρὸς τὴν Η δυνάμει,
τοῦτον ἔχει τὸν λόγον ἡ ΤΔ πρὸς ΡΖ μήκει. Ιs divided in the same
Is
ratio εἰς τὸν αὐτὸν λόγον τέτμηται, or simply ὁμοίως; will divide the
diameter in the proportion of the successive odd numbers, unity
corresponding to the (part) adjacent to the vertex of the segment ràv
διάμετρον τεμοῦντι εἰς τοὺς τῶν ἑξῆς περισσῶν ἀριθμῶν λόγους, ἑνὸς
λεγομένου ποτὶ τῷ κορυφᾷ τοῦ τμήματος.
α
To have a less (or greater) ratio than is ἔχειν λόγον ἐλάσσονα (or
μείζονα) with the genitive of the second ratio or a phrase introduced
by ; to have a less ratio than the greater magnitude has to the less,
ἔχειν λόγον ἐλάσσονα ἢ τὸ μεῖζον μέγεθος πρὸς τὸ ἔλασσον.
For duplicate, triplicate etc. ratios we have the following
expressions : has the triplicate ratio of the same ratio τριπλασίονα
λόγον ἔχει τοῦ αὐτοῦ λόγου, has the duplicate ratio of ΕΛ to ΑΚ
διπλασίονα λόγον ἔχει ἤπερ ἡ ΕΛ προς ΑΚ, are in the triplicate ratio
of the diameters in the bases ἐν τριπλασίονι λόγῳ εἰσὶ τῶν ἐν ταῖς
βάσεσι διαμέτρων, sesquialterate ratio ἡμιόλιος λόγος. With these
expressions must be contrasted the use of double, quadruple etc.
THE TERMINOLOGY OF ARCHIMEDES.
clxxix
ratio in the sense of a simple multiple by 2, 4 etc., e.g. if any
number of areas be placed in order, each being four times the next el
κα χωρία τεθέωντι ἑξῆς ὁποσαοῦν ἐν τῷ τετραπλασίονι λόγῳ.
The ordinary expression for a proportion is as A is to B so is r
to A y A.
το Δ, ως ή Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ. Let ΔΕ be made so
that AE is to TE as the sum of OA, AE is to AE, Teжоýσłw, ws
συναμφότερος ἡ ΘΑ, ΑΕ πρὸς τὴν ΑΕ, οὕτως ἡ ΔΕ προς ΓΕ. The
antecedents are τὰ ἡγούμενα, the consequents τὰ ἑπόμενα.
For reciprocally proportional the parts of avτInéπovОa. are used;
the bases are reciprocally proportional to the heights åvriæeñóvlaσiv
αἱ βάσεις ταῖς ὕψεσιν, to be reciprocally in the same proportion
ἀντιπεπονθέμεν κατὰ τὸν αὐτὸν λόγον.
A ratio compounded of is λόγος συνημμένος (or συγκείμενος) ἔκ τε
TOÛ…..Kaì TOû…….; the ratio of PA to AX is equal to that compounded of
ὁ τῆς ΡΛ πρὸς Λ.Χ λόγος συνῆπται ἐκ…. Two other expressions for
compounded ratios are ὁ τοῦ ἀπὸ ΑΘ πρὸς τὸ ἀπὸ ΒΘ καὶ ὁ (or
προσλαβὼν τὸν τῆς ΑΘ πρὸς Β, the ratio of the square on ΑΘ to
the square on B® multiplied by the ratio of A® to ®B.
The technical terms for transforming such a proportion as
a: b = c :d are as follows:
1. évadλáέ alternately (usually called permutando or alternando)
means transforming the proportion into a :cb: d.
2. ávánaλiv reversely (usually invertendo), b: a=d: c.
3. σύνθεσις λόγου is composition of a ratio by which the ratio
a: b becomes a+b:b. The corresponding Greek term to com-
ponendo is σvvlévri, which means no doubt literally "to one who
has compounded," i.e. "if we compound," the ratios. Thus ovv0évti
denotes the inference that a+b: b=c+d: d. Karà σúveσɩ is also
used in the same sense by Archimedes.
кат
4. διαίρεσις λόγον signifies the division of a ratio in the sense of
separation or subtraction by which a : b becomes a-b: b. Similarly
διελόντι (or κατὰ διαίρεσιν) denotes the inference that a - b : 6 -
c-d: d. The translation dividendo is therefore somewhat mis-
leading.
5. ἀναστροφή λόγου conversion of a ratio and ἀναστρέψαντι
correspond respectively to the ratio a: a b and to the inference
that a: a b c : c-d.
=
—
•
I
1
clxxx
INTRODUCTION.
6. Si toov ex aequali (sc. distantia) is applied e.g. to the
inference from the proportions
that
abcd etc. = A : B : C : D etc.
a: d= A: D.
When this dividing-out of ratios takes place between proportions
with corresponding terms placed crosswise, it is described as di trov
ἐν τῇ τεταραγμένῃ ἀναλογίᾳ, ex aequali in disturbed proportion or
ἀνομοίως τῶν λόγων τεταγμένων the ratios being dissimilarly placed ;
this is the case e.g. when we have two proportions
ab=B: C,
b : c = A : B,
and we infer that
Arithmetical terms.
a: c⇒ A : C.
Whole multiples of any magnitude are generally described as the
double of, the triple of etc., ô diñλáσios, d. tpiñdáoios K.T.A., following
the gender of the particular magnitude; thus the (surface which is)
four times the greatest circle in the sphere ἡ τετραπλασία του μεγίστου
kúkλov tŵv év tŷ opaípa; five times the sum of AB, BE together with
ten times the sum of ΓΒ, ΒΔ, & πενταπλασία συναμφοτέρου τᾶς ΑΒ, ΒΕ
μετὰ τᾶς δεκαπλασίας συναμφοτέρου τᾶς ΓΒ, ΒΔ. The same multiple
å
ας τοσαυταπλασίων... ὁσαπλασίων ἐστί, οι ισάκις πολλαπλασίων... καί.
The general word for a multiple of is πολλαπλάσιος or πολλαπλασίων,
which may be qualified by any expression denoting the number of
times multiplied; thus multiplied by the same number oλλαπλάσιos
Tậ avτô ảρiðµê, multiples according to the successive numbers
πολλαπλάσια κατὰ τοὺς ἑξῆς ἀριθμούς.
Another method is to use the adverbial forms twice Sís, thrice
тpís, etc., which are either followed by the nominative, e.g. twice E▲
Sis ʼn EA, or constructed with a participle, e.g. twice taken dis laµ-
βανόμενος οι δὶς εἰρημένος; together with twice the whole circumference
of the circle μεθ᾽ ὅλας τὰς τοῦ κύκλου περιφερείας δὶς λαμβανομένας.
Similarly the same number of times (the said circumference) as is
expressed by the number one less than (that of) the revolutions
τοσαυτάκις λαμβανομένας, ὅσος ἐστὶν ὁ ἑνὶ ἐλάσσων ἀριθμὸς τῶν
Teρipoρâv. An interesting phrase is the following, as many times as
the line TA is contained (literally added together) in AA, so many times
let the time ZH be contained in the time ΛΗ, όσάκις συγκείται ἡ ΓΔ
THE TERMINOLOGY OF ARCHIMEDES.
clxxxi
γραμμὰ ἐν τῷ ΑΔ, τοσαυτάκις συγκείσθω ὁ χρόνος ὁ ΖΗ ἐν τῷ χρόνῳ τῷ
ΛΗ.
Submultiples are denoted by the ordinal number followed by
μέρος ; one-seventh is ἕβδομον μέρος and so on, one-half being however
ἥμισυς. When the denominator is a large number, a circumlocutory
phrase is used; thus less than τith part of a right angle ἐλάττων ἢ
διαιρεθείσας τᾶς ὀρθᾶς εἰς ρξδ' τούτων ἓν μέρος.
1
When the numerator of a fraction is not unity, it is expressed
by the ordinal number, and the denominator by a compound
substantive denoting such and such a submultiple; e.g. two-thirds
δύο τριταμόρια, three-fifths τρία πεμπταμόρια.
There are two improper fractions which have special names,
thus one-and-a-half of is ἡμιόλιος, one-and-a-third of ἐπίτριτος.
Where a number is partly integral and partly fractional, the integer
is first stated and the fraction follows introduced by καὶ ἔτι or καί
and besides. The phrases used to express the fact that the cir-
cumference of a circle is less than 34 but greater than 310 times its
diameter deserve special notice; (1) παντὸς κύκλου ἡ περίμετρος τῆς
διαμέτρου τριπλασίων ἐστί, καὶ ἔτι ὑπερέχει ἐλάσσονι μὲν ἢ ἑβδόμῳ μέρει
τῆς διαμέτρου, μείζονι δὲ ἢ δέκα ἑβδομηκοστομόνοις, and (2) τριπλασίων
ἐστὶ καὶ ἐλάσσονι μὲν ἢ ἑβδόμῳ μέρει, μείζονι δὲ ἢ ι' οα" μείζων. We
also have the phrase for the first part ἐλάσσων ἢ τριπλασίων καὶ
ἑβδόμῳ μέρει μείζων.
To measure μετρεῖν, common measure κοινὸν μέτρον, commensurable,
incommensurable σύμμετρος, ἀσύμμετρος.
Mechanical terms.
Mechanics τὰ μηχανικά, weight βάρος; centre of gravity κέντρον
τοῦ βάρεος with another genitive of the body or magnitude; in the
plural we have either τὰ κέντρα αὐτῶν τοῦ βάρεος or τὰ κέντρα τῶν
βαρέων. κέντρον is also used alone.
A lever ζυγός or ζύγιον, the horizon ὁ ὁρίζων ; in a vertical line is
represented by perpendicularly κατὰ κάθετον, thus the point of
suspension and the centre of gravity of the body suspended are in a
vertical line κατὰ κάθετόν ἐστι τό τε σαμεῖον τοῦ κρεμαστοῦ καὶ τὸ
κέντρον τοῦ βάρεος τοῦ κρεμαμένου. Of suspension from or at ἐκ or
κατά (with acc. is used. Let the triangle be suspended from the
points Β, Γ, κρεμάσθω τὸ τρίγωνον ἐκ τῶν Β, Γ σαμείων; if the
suspension of the triangle BAT at B, I be set free, and it be suspended
at E, the triangle remains in its position εἴ κα τοῦ ΒΔΓ τριγώνου &
clxxxii
INTRODUCTION.
μὲν κατὰ τὰ Β, Γ κρέμασις λυθῇ, κατὰ δὲ τὸ Ε κρεμασθῇ, μένει τὸ
τρίγωνον, ὡς νῦν ἔχει
Το incline towards ῥέπειν ἐπί (acc.); to be in equilibrium
ισορροπεῖν, they will be in equilibrium with s held fast κατεχομένου
τοῦ Δ ἰσορροπήσει, they will be in equilibrium at Δ (i.e. will balance
about Δ) κατὰ τὸ Δ ισορροπησοῦντι; ΑΒ is too great to balance Γ
μεῖζόν ἐστι τὸ ΑΒ ἢ ὥστε ισορροπεῖν τῷ Γ. The adjective for in
equilibrium is ἰσορρεπής; let it be in equilibrium with the triangle
ΓΔΗ, ισορρεπὲς ἔστω τῷ ΓΔΗ τριγώνῳ. Το balance at certain
distances (from the point of support or the centre of gravity of a
system) is από τινων μακέων ισορροπεῖν.
Theorems, problems, etc.
A theorem θεώρημα (from θεωρεῖν to investigate); a problem
πρόβλημα, with which the following expressions may be compared,
the questions) propounded concerning the figures τὰ προβεβλημένα
περὶ τῶν σχημάτων, these things are propounded for investigation
προβαλλεται τάδε θεωρήσαι; also πρόκειμαι takes the place of the
passive, which it was proposed (or required) to find ὅπερ προέκειτο
εὑρεῖν.
Another similar word is ἐπίταγμα, direction or requirement ;
thus the theorems and directions necessary for the proofs of them rà
θεωρήματα καὶ τὰ ἐπιτάγματα τὰ χρείαν ἔχοντα εἰς τὰς ἀποδειξίας αὐτῶν,
in order that the requirement may be fulfilled όπως γένηται τὸ ἐπι-
ταχθέν (or ἐπίταγμα). Το satisfy the requirement is ποιεῖν τὸ ἐπίταγμα
(either e.g. of lines in a figure, or of the person solving the
problem).
After the setting out (ἔκθεσις) in any proposition there follows
the short statement of what it is required to prove or to do. In
the former case (that of a theorem) Archimedes uses one of three
expressions δεικτέον it is required to prove, λέγω or φαμὶ δή I assert
or say; and in the second case (that of a problem) δεῖ δή it is
required (to do so and so).
In a problem the analysis ανάλυσις and synthesis σύνθεσις are
distinguished, the latter being generally introduced with the words
the synthesis of the problem will be as follows συντεθήσεται τὸ
πρόβλημα οὕτως. The parts of the verb ἀναλύειν are similarly
used; thus the analysis and synthesis of each of these (problems) will
be given at the end ἑκάτερα δὲ ταῦτα ἐπὶ τέλει ἀναλυθήσεταί τε καὶ
συντεθήσεται.
THE TERMINOLOGY OF ARCHIMEDES.
clxxxiii
A notable term in connexion with problems is the Stoptopós
(determination), which means the determination of the limits within
which a solution is possible*. If a solution is always possible, the
problem does not involve a διορισμός, οὐκ ἔχει διορισμόν ; otherwise
it does involve it, ἔχει διορισμόν.
Data and hypotheses.
For given some part of the verb dídwμi is used, generally the
participle δοθείς, but sometimes δεδομένος and once or twice διδόμενος.
Let a circle be given dedóσow KÚKλos, given two unequal magnitudes
δύο μεγεθῶν ἀνίσων δοθέντων, each of the two lines ΓΔ, ΕΖ is given
ἐστὶν δοθεῖσα ἑκατέρα τῶν ΓΔ, ΕΖ, the same ratio as the given one
λόγος ὁ αὐτὸς τῷ δοθέντι. Similar expressions are the assigned ratio
ὁ ταχθεὶς λόγος, the given area τὸ προτεθὲν (or προκείμενον) χωρίον.
Given in position θέσει simply (sc. δεδομένη).
Of hypotheses the parts of the verb vrоríbeμai and (for the
passive) υπόκειμαι are used; with the same suppositions τῶν αὐτῶν
ὑποκειμένων, let the said suppositions be made ὑποκείσθω τὰ εἰρημένα,
we make these suppositions υποτιθέμεθα τάδε.
Where in a reductio ad absurdum the original hypothesis is
referred to, and generally where an earlier step is quoted, the past
tense of the verb is used; but it was not (so) ovê îv dé, for it was less
ἦν γὰρ ἐλάσσων, they were proved equal ἀπεδείχθησαν ἴσοι, for this has
been proved to be possible δεδείκται γὰρ τοῦτο δυνατὸν ἐόν. Where a
hypothesis is thus quoted, the past tense of vπókeμαι has various
constructions after it, (1) an adjective or participle, AZ, BH were
supposed equal loaɩ vwékeɩvto ai AZ, BH, it is by hypothesis a tangent ·
útékeito éπifaúovoa, (2) an infinitive, for by hypothesis it does not
cut ὑπέκειτο γὰρ μὴ τέμνειν, the axis is by hypothesis not at right
angles to the parallel planes ὑπέκειτο ὁ ἄξων μὴ εἶμεν ὀρθὸς ποτὶ τὰ
πaρádλada éπíñeda, (3) the plane is supposed to have been drawn
through the centre τὸ ἐπίπεδον ὑπόκειται διὰ τοῦ κέντρου ἄχθαι.
Supposing it found evρelévтos absolutely.
γεγονέτω.
Suppose it done
The usual idiomatic use of ei dè µý after a negative statement
may be mentioned; it will not meet the surface in another point,
otherwise... οὐ γὰρ ἁψέται κατ᾽ ἄλλο σαμεῖον τῆς ἐπιφανείας· εἰ δὲ
μή....
* Cf. Apollonius of Perga, p. lxx, note,
•
clxxxiv
INTRODUCTION.
Inferences, and adaptation to different cases.
The usual equivalent for therefore is ἄρα; οὖν and τοίνυν are
generally used in a somewhat weaker sense to mark the starting-
point of an argument, thus ἐπεὶ οὖν may be translated as since, then.
Since is ἐπεί, because διότι.
πολλῷ μᾶλλον much more then is apparently not used in Archi-
medes, who has Toλ alone; thus much less then is the ratio of the
circumscribed figure to the inscribed than that of K to Η πολλῷ
ἄρα τὸ περιγραφὲν πρὸς τὸ ἐγγραφὲν ἐλάσσονα λόγον ἔχει τοῦ, ὃν ἔχει ἡ
Κ προς Η.
Siά with the accusative is a common way of expressing the
reason why; because the cone is isosceles διὰ τὸ ἰσοσκελῆ εἶναι τὸν
κώνον, for the same reason διὰ ταὐτά.
Siά with the genitive expresses the means by which a proposition
is proved ; by means of the construction διὰ τῆς κατασκευῆς, by the .
same means διὰ τῶν αὐτῶν, by the same method διὰ τοῦ αὐτοῦ τρόπου.
Whenever this is the case, the surface is greater ὅταν τοῦτο ᾖ,
μείζων γίνεται ἡ ἐπιφάνεια..., if this is the case, the angle ΒΑΘ is
equal..., εἰ δὲ τοῦτο, ἴσα ἐστὶν ἡ ὑπὸ ΒΑΘ γωνία..., which is the same
thing as showing that... ὃ ταὐτόν ἐστι τῷ δεῖξαι, ὅτι....
Similarly for the sector ὁμοίως δὲ καὶ ἐπὶ τοῦ τομέως, the proof
is the same as that used to show that & αὐτὰ ἀπόδειξις ἅπερ καὶ ὅτι,
the proof that...is the same ἡ αὐτὰ ἀπόδειξίς ἐντι καὶ διότι..., the same
argument holds for all rectilineal figures inscribed in the segments in
the recognised manner (see p. 204) ἐπὶ πάντων εὐθυγράμμων τῶν
ἐγγραφομένων ἐς τὰ τμάματα γνωρίμως ὁ αὐτὸς λόγος ; it will be possible,
having proved it for a circle, to transfer the same argument in
the case of the sector ἔσται ἐπὶ κύκλου δείξαντα μεταγαγεῖν τὸν ὅμοιον
λόγον καὶ ἐπὶ τοῦ τομέως ; the rest will be the same, but it will be the
lesser of the diameters which will be intercepted within the spheroid
instead of the greater) τὰ μὲν ἄλλα τὰ αὐτὰ ἐσσείται, τῶν δὲ διαμέτρων
ὁ ἐλάσσων ἐσσείται ἡ ἐναπολαφθεῖσα ἐν τῷ σφαιροειδεῖ; it will make
no difference whether...or... διοίσει δὲ οὐδέν, εἴτε... εἴτε....
Conclusions.
The proposition is therefore obvious, or is proved δῆλον οὖν ἐστι
(οι δέδεικται) τὸ προτεθέν ; similarly φανερὸν οὖν ἐστιν, ὃ ἔδει δείξαι,
and ἔδει δὲ τοῦτο δείξαι. Which is absurd, or impossible ὅπερ ἄτοπον,
or ἀδύνατον.
A curious use of two negatives is contained in the following:
THE TERMINOLOGY OF ARCHIMEDES.
clxxxv
οὐκ ἄρα οὐκ ἐστι κέντρον τοῦ βάρεος τοῦ ΔΕΖ τριγώνου τὸ Ν σαμεῖον.
ἔστιν ἄρα, therefore it is not possible that the point N should not be the
centre of gravity of the triangle ▲EZ. It must therefore be so.
Thus a rhombus will have been formed ἔσται δὴ γεγονως ῥόμβος ;
two unequal straight lines have been found satisfying the requirement
εὑρημέναι εἰσὶν ἄρα δύο εὐθεῖαι ἄνισοι ποιοῦσαι τὸ ἐπίταγμα.
Direction, concavity, convexity.
In the same direction ἐπὶ τὰ αὐτά, in the other direction ἐπὶ
τὰ ἕτερα, concave in the same direction ἐπὶ τὰ αὐτὰ κοίλη ; in the same
direction as ἐπὶ τὰ αὐτά with the dative or ἐφ᾽ ἅ, thus in the same
direction as the vertex of the cone ἐπὶ τὰ αὐτὰ τᾷ τοῦ κώνου κορυφᾷ,
drawn in the same direction as (that of) the convex side of it ἐπὶ τὰ
αὐτὰ ἀγομέναι, ἐφ᾽ ἃ ἐντι τὰ κυρτὰ αὐτοῦ. For on the same side of ἐπὶ
τὰ αὐτά is followed by the genitive, they fall on the same side of the
line ἐπὶ τὰ αὐτὰ πίπτουσι τῆς γραμμής.
On each side of ἐφ' ἑκάτερα (with gen.); on each side of the plane
of the base ἐφ' ἑκάτερα τοῦ ἐπιπέδου τῆς βάσεως.
Miscellaneous.
Property σύμπτωμα. Proceeding thus continually, ἀεὶ τοῦτο
ποιοῦντες, ἀεὶ τούτου γενομένου, or τούτου ἑξῆς γινομένου. In the
elements ἐν τῇ στοιχειώσει.
One special difference between our terminology and the Greek is
that whereas we speak of any circle, any straight line and the like,
the Greeks say every circle, every straight line, etc.
Thus any
pyramid is one third part of the prism with the same base as the
pyramid and equal height πᾶσα πυραμὶς τρίτον μέρος ἐστὶ τοῦ πρίσματος
τοῦ τὴν αὐτὴν βάσιν ἔχοντος τῷ πυραμίδι καὶ ὕψος ἴσον. I define the
diameter of any segment as διάμετρον καλέω παντὸς τμάματος. Το
exceed any assigned (magnitude) of those which are comparable with
one another υπερέχειν παντὸς τοῦ προτεθέντος τῶν πρὸς ἄλληλα
λεγομένων.
Another noteworthy difference is illustrated in the last sentence.
The Greeks did not speak as we do of a given area, a given ratio
etc., but of the given area, the given ratio, and the like. Thus It is
possible...to leave certain segments less than a given area δυνατόν
ἐστιν...λείπειν τινα τμήματα, ἅπερ ἔσται ἐλάσσονα τοῦ προκειμένου
Xwpíov; to divide a given sphere by a plane so that the segments have
to one another an assigned ratio τὰν δοθεῖσαν σφαῖραν ἐπιπέδῳ τεμεῖν,
ὥστε τὰ τμάματα αὐτᾶς ποτ᾽ ἄλλαλα τὸν ταχθέντα λόγον ἔχειν.
Η. Α.
n
clxxxvi
INTRODUCTION.
Magnitudes in arithmetical progression are said to exceed each
other by an equal (amount); if there be any number of magnitudes in
arithmetical progression εἴ κα ἔωντι μεγέθεα ὁποσαοῦν τῷ ἴσῳ ἀλλάλων
ὑπερέχοντα. The common diference is the excess υπεροχα, and the
terms collectively are spoken of as the magnitudes exceeding by the
equal (diference) τὰ τῷ ἴσῳ ὑπερέχοντα. The least term is τὸ ἐλάχιστον,
the greatest term τὸ μέγιστον. The sum of the terms is expressed by
πάντα τὰ τῷ ἴσῳ· ὑπερέχοντα,
Terms of a geometrical progression are simply in (continued)
proportion ἀνάλογον, the series is then ἡ ἀναλογία, the proportion,
and a term of the series is τὶς τῶν ἐν τῷ αὐτῷ ἀναλογίᾳ. Numbers in
geometrical progression beginning from unity are ἀριθμοὶ ἀνάλογον
ἀπὸ μονάδος. Let the term Λ of the progression be taken which
is distant the same number of terms from ℗ as ▲ is distant from
unity λελάφθω ἐκ τᾶς ἀναλογίας ὁ Λ ἀπέχων ἀπὸ τοῦ Θ τοσούτους, ὅσους
ὁ Δ ἀπὸ μονάδος ἀπέχει.
.
1
THE WORKS OF
ARCHIMEDES.
!
F
الله
**
ON THE SPHERE AND CYLINDER.
BOOK I.
ARCHIMEDES to Dositheus greeting.
On a former occasion I sent you the investigations which
I had up to that time completed, including the proofs, showing
that any segment bounded by a straight line and a section of a
right-angled cone [a parabola] is four-thirds of the triangle
which has the same base with the segment and equal height.
Since then certain theorems not hitherto demonstrated (ave-
λéуKTOV) have occurred to me, and I have worked out the proofs
of them. They are these: first, that the surface of any sphere
is four times its greatest circle (του μεγίστου κύκλου); next,
that the surface of any segment of a sphere is equal to a circle
whose radius (ἡ ἐκ τοῦ κέντρου) is equal to the straight line
drawn from the vertex (xopupń) of the segment to the circum-
ference of the circle which is the base of the segment; and,
further, that any cylinder having its base equal to the greatest
circle of those in the sphere, and height equal to the diameter
of the sphere, is itself [i.e. in content] half as large again as the
sphere, and its surface also [including its bases] is half as large
again as the surface of the sphere. Now these properties were
all along naturally inherent in the figures referred to (avTŷ TỘ
φύσει προυπῆρχεν περὶ τὰ εἰρημένα σχήματα), but remained
unknown to those who were before my time engaged in the
study of geometry. Having, however, now discovered that the
properties are true of these figures, I cannot feel any hesitation
H. A.
1
2
1
7
5
F
}
1
#
"
2
ARCHIMEDES
in setting them side by side both with my former investiga-
tions and with those of the theorems of Eudoxus on solids
which are held to be most irrefragably established, namely,
that any pyramid is one third part of the prism which has the
same base with the pyramid and equal height, and that any
cone is one third part of the cylinder which has the same
base with the cone and equal height. For, though these
properties also were naturally inherent in the figures all along,.
yet they were in fact unknown to all the many able geometers
who lived before Eudoxus, and had not been observed by any
one. Now, however, it will be open to those who possess the
requisite ability to examine these discoveries of mine. They
ought to have been published while Conon was still alive,
for I should conceive that he would best have been able to
grasp them and to pronounce upon them the appropriate
verdict; but, as I judge it well to communicate them to those
who are conversant with mathematics, I send them to you with
the proofs written out, which it will be open to mathematicians
to examine. Farewell.
I first set out the axioms* and the assumptions which I
have used for the proofs of my propositions.
DEFINITIONS.
1. There are in a plane certain terminated bent lines
(καμπύλαι γραμμαὶ πεπερασμέναι), which either lie wholly on
the same side of the straight lines joining their extremities, or
have no part of them on the other side.
2. I apply the term concave in the same direction
to a line such that, if any two points on it are taken, either
all the straight lines connecting the points fall on the same
side of the line, or some fall on one and the same side while
others fall on the line itself, but none on the other side.
* Though the word used is åğıwµara, the “axioms” are more of the nature
of definitions; and in fact Eutocius in his notes speaks of them as such (öpoɩ).
† Under the term bent line Archimedes includes not only curved lines of
continuous curvature, but lines made up of any number of lines which may be
either straight or curved.
*
ON THE SPHERE AND CYLINDER I.
3
3. Similarly also there are certain terminated surfaces, not
themselves being in a plane but having their extremities in a
plane, and such that they will either be wholly on the same
side of the plane containing their extremities, or have no part
of them on the other side.
4. I apply the term concave in the same direction
to surfaces such that, if any two points on them are taken, the
straight lines connecting the points either all fall on the same
side of the surface, or some fall on one and the same side of
it while some fall upon it, but none on the other side.
I use the term solid sector, when a cone cuts a sphere,
and has its apex at the centre of the sphere, to denote the
figure comprehended by the surface of the cone and the surface
of the sphere included within the cone.
6. I apply the term solid rhombus, when two cones with
the same base have their apices on opposite sides of the plane
of the base in such a position that their axes lie in a straight
line, to denote the solid figure made up of both the cones.
ASSUMPTIONS.
1.
Of all lines which have the same extremities the straight
line is the least*.
* This well-known Archimedean assumption is scarcely, as it stands, a
definition of a straight line, though Proclus says [p. 110 ed. Friedlein] "Archi-
medes defined (wpíσaro) the straight line as the least of those [lines] which have
the same extremities. For because, as Euclid's definition says, ¿§ loov Keîtai ToÛS
ép èavrîs onμelous, it is in consequence the least of those which have the same
extremities." Proclus had just before [p. 109] explained Euclid's definition,
which, as will be seen, is different from the ordinary version given in our text-
books; a straight line is not "that which lies evenly between its extreme points,"
but that which ἐξ ἴσου τοῖς ἐφ' ἑαυτῆς σημείοις κεῖται.” The words of Proclus
roîs éavrîs
are, "He [Euclid] shows by means of this that the straight line alone [of all
lines] occupies a distance (kaтéxeiv diáornua) equal to that between the points
on it. For, as far as one of its points is removed from another, so great is the
length (μéyelos) of the straight line of which the points are the extremities;
and this is the meaning of τὸ ἐξ ἴσου κεῖσθαι τοῖς ἐφ' ἑαυτῆς σημείοις. But, if you
take two points on a circumference or any other line, the distance cut off
between them along the line is greater than the interval separating them; and
this is the case with every line except the straight line." It appears then from
this that Euclid's definition should be understood in a sense very like that of
J
1-2
4
ARCHIMEDES
2. Of other lines in a plane and having the same extremi-
ties, [any two] such are unequal whenever both are concave in
the same direction and one of them is either wholly included
between the other and the straight line which has the same
extremities with it, or is partly included by, and is partly
common with, the other; and that [line] which is included is
the lesser [of the two].
3. Similarly, of surfaces which have the same extremities,
if those extremities are in a plane, the plane is the least [in
area].
4. Of other surfaces with the same extremities, the ex-
tremities being in a plane, [any two] such are unequal when-
ever both are concave in the same direction and one surface
is either wholly included between the other and the plane which
has the same extremities with it, or is partly included by, and
partly common with, the other; and that [surface] which is
included is the lesser [of the two in area].
5. Further, of unequal lines, unequal surfaces, and unequal
solids, the greater exceeds the less by such a magnitude as,
when added to itself, can be made to exceed any assigned
magnitude among those which are comparable with [it and
with] one another*.
These things being premised, if a polygon be inscribed in a
circle, it is plain that the perimeter of the inscribed polygon is
less than the circumference of the circle; for each of the sides
of the polygon is less than that part of the circumference of the
circle which is cut off by it."
Archimedes' assumption, and we might perhaps translate as follows, "A straight
line is that which extends equally (eg tσov keîral) with the points on it," or, to
follow Proclus' interpretation more closely, "A straight line is that which
represents equal extension with [the distances separating] the points on it."
* With regard to this assumption compare the Introduction, chapter III. § 2.
ON THE SPHERE AND CYLINDER I.
5
10
Proposition 1.
A
If a polygon be circumscribed about a circle, the perimeter
of the circumscribed polygon is greater
than the perimeter of the circle.
Let any two adjacent sides, meet-
ing in A, touch the circle at P, Q
respectively.
Then [Assumptions, 2]
PA+AQ>(arc PQ).
A similar inequality holds for each
angle of the polygon; and, by ad-
dition, the required result follows.

Proposition 2.
P
Given two unequal magnitudes, it is possible to find two un-
equal straight lines such that the greater straight line has to the
less a ratio less than the greater magnitude has to the less.
Let AB, D represent the two unequal magnitudes, AB being
the greater.
Suppose BC measured along BA equal to D, and let GH be
any straight line.
Then, if CA be added to itself a sufficient
number of times, the sum will exceed D. Let
AF be this sum, and take E on GH produced
such that GH is the same multiple of HE that
AF is of AC.
Thus
-
EH: HG AC: AF.
But, since AF > D (or CB),
AC: AF< AC : CB.
Therefore, componendo,
EG: GH <AB: D.
H
E
B
Hence EG, GH are two lines satisfying the given condition.
D
"
6
ARCHIMEDES
口
​Proposition 3.
Given two unequal magnitudes and a circle, it is possible to
inscribe a polygon in the circle and to describe another about it
so that the side of the circumscribed polygon may have to the side
of the inscribed polygon a ratio less than that of the greater
magnitude to the less.
Let A, B represent the given magnitudes, A being the
greater.
Find [Prop. 2] two straight lines F, KL, of which F is the
greater, such that
T
N
D
SC.
H
G
F:KL<A: B ……..
K
E
A F B
……………..(1).


Draw LM perpendicular to LK and of such length that
KM = F.
In the given circle let CE, DG be two diameters at right
angles. Then, bisecting the angle DOC, bisecting the half
again, and so on, we shall arrive ultimately at an angle (as
NOC) less than twice the angle LKM.
Join NC, which (by the construction) will be the side of a
regular polygon inscribed in the circle. Let OP be the radius
of the circle bisecting the angle NOC (and therefore bisecting
NC at right angles, in H, say), and let the tangent at P meet
OC, ON produced in S, T respectively.
Now, since
▲ CON < 2 ▲ LKM,
<HOC<< LKM,
;
ON THE SPHERE AND CYLINDER I.
7
and the angles at H, L are right;
Hence
therefore MK: LK > OC : OH
> OP.: OH.
ST: CNMK: LK
therefore, a fortiori, by (1),
<F: LK;
ST : CN <A : B.
Thus two polygons are found satisfying the given condition.
Proposition 4.
Again, given two unequal magnitudes and a sector, it is
possible to describe a polygon about the sector and to inscribe
another in it so that the side of the circumscribed polygon may
have to the side of the inscribed polygon a ratio less than the
greater magnitude has to the less.
[The "inscribed polygon" found in this proposition is one
which has for two sides the two radii bounding the sector, while
the remaining sides (the number of which is, by construction,
some power of 2) subtend equal parts of the arc of the sector;
the "circumscribed polygon" is formed by the tangents parallel
to the sides of the inscribed polygon and by the two bounding
radii produced.]
S
P
H
T
N
A
F B
E
K
£
1
f


L
M
G
In this case we make the same construction as in the last
proposition except that we bisect the angle COD of the sector,
instead of the right angle between two diameters, then bisect
the half again, and so on. The proof is exactly similar to the
preceding one.
}
•
8
ARCHIMEDES
Proposition 5.
Given a circle and two unequal magnitudes, to describe a
polygon about the circle and inscribe another in it, so that the
circumscribed polygon may have to the inscribed a ratio less than
the greater magnitude has to the less.
Let A be the given circle and B, C the given magnitudes, B
being the greater.

D-
E
F
A
B
C
Take two unequal straight lines D, E, of which D is the
greater, such that D: E<B: C [Prop. 2], and let F be a mean
proportional between D, E, so that D is also greater than F.
Describe (in the manner of Prop. 3) one polygon about the
circle, and inscribe another in it, so that the side of the former
has to the side of the latter a ratio less than the ratio D: F.
Thus the duplicate ratio of the side of the former polygon
to the side of the latter is less than the ratio D²: F2.
But the said duplicate ratio of the sides is equal to the
ratio of the areas of the polygons, since they are similar;
therefore the area of the circumscribed polygon has to the
area of the inscribed polygon a ratio less than the ratio D² : F²,
or D: E, and a fortiori less than the ratio B: C.
ON THE SPHERE AND CYLINDER I.
9
Proposition 6.
Similarly we can show that, given two unequal magnitudes
and a sector, it is possible to circumscribe a polygon about the
sector and inscribe in it another similar one so that the circum-
scribed may have to the inscribed a ratio less than the greater
magnitude has to the less.
And it is likewise clear that, if a circle or a sector, as well
as a certain area, be given, it is possible, by inscribing regular
polygons in the circle or sector, and by continually inscribing
such in the remaining segments, to leave segments of the circle or
sector which are [together] less than the given area. For this is
proved in the Elements [Eucl. XII. 2].
But it is yet to be proved that, given a circle or sector and
an area, it is possible to describe a polygon about the circle or
sector, such that the area remaining between the circumference
and the circumscribed figure is less than the given area.”


B
A
The proof for the circle (which, as Archimedes says, can be
equally applied to a sector) is as follows.
Let A be the given circle and B the given area.
Now, there being two unequal magnitudes A + B and A, let
a polygon (C) be circumscribed about the circle and a polygon
(I) inscribed in it [as in Prop. 5], so that
C: I<A+B: A
The circumscribed polygon (C) shall be that required.
.(1).
10
ARCHIMEDES
!
-སྙ -"- : ན་
I
!
**
For the circle (A) is greater than the inscribed polygon (I).
Therefore, from (1), a fortiori,
C: A <A+B: A,
whence
or
C< A + B,
C-A<B
Proposition 7.
If in an isosceles cone [i.e. a right circular cone] a pyramid
be inscribed having an equilateral base, the surface of the
pyramid excluding the base is equal to a triangle having its
base equal to the perimeter of the base of the pyramid and its
height equal to the perpendicular drawn from the apex on one
side of the base.
Since the sides of the base of the pyramid are equal, it
follows that the perpendiculars from the apex to all the sides
of the base are equal; and the proof of the proposition is
obvious.
Proposition 8.
If a pyramid be circumscribed about an isosceles cone, the
surface of the pyramid excluding its base is equal to a triangle
having its base equal to the perimeter of the base of the pyramid
and its height equal to the side [i.e. a generator] of the cone.
The base of the pyramid is a polygon circumscribed about
the circular base of the cone, and the line joining the apex of
the cone or pyramid to the point of contact of any side of the
polygon is perpendicular to that side. Also all these perpen-
diculars, being generators of the cone, are equal; whence the
proposition follows immediately.
ON THE SPHERE AND CYLINDER I.
11
Proposition 9.
If in the circular base of an isosceles cone a chord be placed,
and from its extremities straight lines be drawn to the apex of
the cone, the triangle so formed will be less than the portion of
the surface of the cone intercepted between the lines drawn to the
apex.
Let ABC be the circular base of the cone, and O its apex.
Draw a chord AB in the circle, and join OA, OB. Bisect
the arc ACB in C, and join AC, BC, OC.
Then
AOAC + A OBC > A OAB.

E
G
HK
F

L
D
A
B
Let the excess of the sum of the first two triangles over the
third be equal to the area D.
Then D is either less than the sum of the segments AEC,
CFB, or not less.
I. Let D be not less than the sum of the segments referred
to.
We have now two surfaces
(1) that consisting of the portion OAEC of the surface
of the cone together with the segment AEC, and
(2) the triangle OAC;
and, since the two surfaces have the same extremities (the
perimeter of the triangle OAC), the former surface is greater
than the latter, which is included by it [Assumptions, 3 or 4].
12
ARCHIMEDES
Hence (surface OAEC)+(segment AEC) >▲ OAC.
Similarly (surface OCFB) + (segment CFB) > ▲ OBC.
Therefore, since D is not less than the sum of the segments,
we have, by addition,
(surface OAECFB)+D>^ OAC +▲ OBC
>▲ OAB+D, by hypothesis.
Taking away the common part D, we have the required
result.
II. Let D be less than the sum of the segments AEC,
CFB.
If now we bisect the arcs AC, CB, then bisect the halves,
and so on, we shall ultimately leave segments which are
together less than D.
[Prop. 6]
Let AGE, EHC, CKF, FLB be those segments, and join
OE, OF.
+
and
Then, as before,
(surface OAGE)+(segment AGE) > ▲ OAE
(surface OEHC)+(segment EHC) > ▲ OEC.
Therefore (surface OAGHC) + (segments AGE, EHC)
>ΔΟΑΕ+ΔΟΕΣ
> AOAC, a fortiori.
Similarly for the part of the surface of the cone bounded by
OC, OB and the arc CFB.
Hence, by addition,
(surface OAGEHCKFLB)+(segments AGE, EHC, CKF, FLB)
>AOAC+A OBC
>^OAB+ D, by hypothesis.
But the sum of the segments is less than D, and the re-
quired result follows.
+
ON THE SPHERE AND CYLINDER I.
13
Proposition 10.
If in the plane of the circular base of an isosceles cone two
tangents be drawn to the circle meeting in a point, and the points
of contact and the point of concourse of the tangents be respectively
joined to the apex of the cone, the sum of the two triangles
formed by the joining lines and the two tangents are together
greater than the included portion of the surface of the cone.
Let ABC be the circular base of the cone, O its apex, AD,
BD the two tangents to the circle meeting in D. Join OA,
OB, OD.
Let ECF be drawn touching the circle at C, the middle
point of the arc ACB, and therefore parallel to AB. Join
OE, OF.
Then
ED+DF>EF,
and, adding AE+ FB to each side,
AD+DB>AE+ EF + FB.
Now OA, OC, OB, being generators of the cone, are equal,
and they are respectively perpendicular to the tangents at A,
C, B.
D

E
Q CR
F
H
K
P
S
A
B
G
8
14
ARCHIMEDES
སྣ
It follows that
▲ OAD+▲ ODB>▲ OAE+^OEF+^ OFB.
Let the area G be equal to the excess of the first sum over
the second.
G is then either less, or not less, than the sum of the spaces
EAHC, FCKB remaining between the circle and the tangents,
which sum we will call L.
I. Let G be not less than L.
We have now two surfaces
(1) that of the pyramid with apex 0 and base AEFB,
excluding the face OAB,
(2) that consisting of the part OACB of the surface of the
cone together with the segment ACB.
These two surfaces have the same extremities, viz. the
perimeter of the triangle OAB, and, since the former includes
the latter, the former is the greater [Assumptions, 4].
That is, the surface of the pyramid exclusive of the face
OAB is greater than the sum of the surface OACB and the
segment ACB.
Taking away the segment from each sum, we have
▲ ОAE+▲ ОEF+▲ OFB+L> the surface OAHCKB.
And G is not less than L.
It follows that
ΔΟΑΕ+Δ0EF+Δ 0FB+ G,
which is by hypothesis equal to ▲OAD+▲ ODB, is greater
than the same surface.
II. Let G be less than L.
If we bisect the arcs AC, CB and draw tangents at their
middle points, then bisect the halves and draw tangents, and
so on, we shall lastly arrive at a polygon such that the sum
of the parts remaining between the sides of the polygon and
the circumference of the segment is less than G.
ON THE SPHERE AND CYLINDER I.
15
Let the remainders be those between the segment and the
polygon APQRSB, and let their sum be, M. Join OP, OQ,
etc.
Then, as before,
ΔΟΑΕ+ ΔΟEF +Δ0FB>ΔΟΑΡ+ΔΟΡΟ+...+Δ OSB
Also, as before,
(surface of pyramid OAPQRSB excluding the face OAB)
> the part OACB of the surface of the
cone together with the segment ACB.
Taking away the segment from each sum,
▲ OAP + ▲ OPQ+...+M> the part OACB of the
Hence, a fortiori,
surface of the cone.
ΔΟΑΕ + ΔΟEF + Δ0FB+ G,
which is by hypothesis equal to
AOAD +▲ ODB,
is greater than the part OACB of the surface of the cone.
Proposition 11.
If a plane parallel to the axis of a right cylinder cut the
cylinder, the part of the surface of the cylinder cut off by the
plane is greater than the area of the parallelogram in which the
plane cuts it.
Proposition 12.
If at the extremities of two generators of any right cylinder
tangents be drawn to the circular bases in the planes of those
bases respectively, and if the pairs of tangents meet, the
parallelograms formed by each generator and the two corre-
sponding tangents respectively are together greater than the
included portion of the surface of the cylinder between the two
generators.
[The proofs of these two propositions follow exactly the
methods of Props. 9, 10 respectively, and it is therefore un-
necessary to reproduce them.]
16
ARCHIMEDES
J
J|
From the properties thus proved it is clear (1) that, if a
pyramid be inscribed in an isosceles cone, the surface of the
pyramid excluding the base is less than the surface of the cone
[excluding the base], and (2) that, if a pyramid be circumscribed
about an isosceles cone, the surface of the pyramid excluding the
base is greater than the surface of the cone excluding the base.
"It is also clear from what has been proved both (1) that,
if a prism be inscribed in a right cylinder, the surface of the
prism made up of its parallelograms [i.e. excluding its bases] is
less than the surface of the cylinder excluding its bases, and
(2) that, if a prism be circumscribed about a right cylinder, the
surface of the prism made up of its parallelograms is greater
than the surface of the cylinder excluding its bases."
Proposition 13.
The surface of any right cylinder excluding the bases is equal
to a circle whose radius is a mean proportional between the side
[i.e. a generator] of the cylinder and the diameter of its base.
Let the base of the cylinder be the circle A, and make CD
equal to the diameter of this circle, and EF equal to the height
of the cylinder.

M
D
K
B
N
E
F
H
A
ON THE SPHERE AND CYLINDER I.
17
Let H be a mean proportional between CD, EF, and B
a circle with radius equal to H.
Then the circle B shall be equal to the surface of the
cylinder (excluding the bases), which we will call S.
For, if not, B must be either greater or less than S.
I. Suppose B< S.
Then it is possible to circumscribe a regular polygon about
B, and to inscribe another in it, such that the ratio of the
former to the latter is less than the ratio S : B.
Suppose this done, and circumscribe about A a polygon
similar to that described about B; then erect on the polygon
about A a prism of the same height as the cylinder. The
prism will therefore be circumscribed to the cylinder.
Let KD, perpendicular to CD, and FL, perpendicular to
EF, be each equal to the perimeter of the polygon about A.
Bisect CD in M, and join MK.
Then
Also
Δ
▲ KDM = the polygon about A.
□ EL = surface of prism (excluding bases).
Produce FE to N so that FE = EN, and join NL.
Now the polygons about A, B, being similar, are in the
duplicate ratio of the radii of A, B.
Thus
▲ KDM : (polygon about B) = MD³ : H
2
= MD2: CD. EF
= MD: NF
=▲ KDM : ▲ LFN
Therefore (polygon about B) = ^ LFN
But
H. A.
=□ EL
(since DK = FL).
=(surface of prism about A),
from above.
"(polygon about B): (polygon in B) < § : B.
2
18
ARCHIMEDES
Therefore
(surface of prism about A): (polygon in B) <S: B,
and, alternately,
(surface of prism about A): S<(polygon in B) :B;
which is impossible, since the surface of the prism is greater
than S, while the polygon inscribed in B is less than B.
Therefore
II. Suppose B >S.
B&S.
Let a regular polygon be circumscribed about B and another
inscribed in it so that
(polygon about B): (polygon in B) < B: S.
Inscribe in A a polygon similar to that inscribed in B, and
erect a prism on the polygon inscribed in A of the same height
as the cylinder.
Again, let DK, FL, drawn as before, be each equal to the
perimeter of the polygon inscribed in A.
Then, in this case,
▲ KDM >(polygon inscribed in A)
(since the perpendicular from the centre on a side of the
polygon is less than the radius of A).
Also ▲ LFN=□ EL = surface of prism (excluding bases).
Now
(polygon in A): (polygon in B) = MD² : H²,
And
▲KDM: ^LFN, as before.
AKDM > (polygon in A).
Therefore
▲ LFN, or (surface of prism) > (polygon in B).
But this is impossible, because
so that
(polygon about B): (polygon in B) < B: S,
<(polygon about B): S, a fortiori,
(polygon in B) > S,
>(surface of prism), a fortiori.
Hence B is neither greater nor less than S, and therefore
B = S.
ON THE SPHERE AND CYLINDER I.
19
Proposition 14.
The surface of any isosceles cone excluding the base is equal
to a circle whose radius is a mean proportional between the side
of the cone [a generator] and the radius of the circle which is the
base of the cone.
Let the circle A be the base of the cone; draw C equal to
the radius of the circle, and D equal to the side of the cone, and
let E be a mean proportional between C, D.
E
D
A

B
Draw a circle B with radius equal to E.
Then shall B be equal to the surface of the cone (excluding
the base), which we will call S.
If not, B must be either greater or less than S.
I. Suppose B< S.
Let a regular polygon be described about B and a similar
one inscribed in it such that the former has to the latter a ratio
less than the ratio S: B.
Describe about A another similar polygon, and on it set up
a pyramid with apex the same as that of the cone.
Then (polygon about A): (polygon about B)
12
C¹²: E²
= C: D
=
(polygon about A): (surface of pyramid excluding base).
2-2
20
ARCHIMEDES
Therefore
(surface of pyramid) = (polygon about B).
Now (polygon about B): (polygon in B) < S: B.
Therefore
(surface of pyramid): (polygon in B)<S : B,
which is impossible, (because the surface of the pyramid is
greater than S, while the polygon in B is less than B).
Hence
II. Suppose B > S.
B & S.
Take regular polygons circumscribed and inscribed to B such
that the ratio of the former to the latter is less than the ratio
B: S.
Inscribe in A a similar polygon to that inscribed in B, and
erect a pyramid on the polygon inscribed in A with apex the
same as that of the cone.
In this case
(polygon in 4): (polygon in B) = C² : E²
= C: D
>(polygon in A): (surface of pyramid excluding base).
This is clear because the ratio of C to D is greater than the
ratio of the perpendicular from the centre of A on a side of the
polygon to the perpendicular from the apex of the cone on the
same side*.
Therefore
(surface of pyramid) > (polygon in B).
But (polygon about B): (polygon in B)< B: S.
Therefore, a fortiori,
(polygon about B): (surface of pyramid) < B: S;
which is impossible.
Since therefore B is neither greater nor less than S,
B = S.
* This is of course the geometrical equivalent of saying that, if a, ß be two
angles each less than a right angle, and a>ß, then sin a>sin ß.
ON THE SPHERE AND CYLINDER I.
21
Proposition 15.
The surface of any isosceles cone has the same ratio to its
base as the side of the cone has to the radius of the base.
By Prop. 14, the surface of the cone is equal to a circle
whose radius is a mean proportional between the side of the
cone and the radius of the base.
Hence, since circles are to one another as the squares of
their radii, the proposition follows.
Proposition 16.
If an isosceles cone be cut by a plane parallel to the base, the
portion of the surface of the cone between the parallel planes is
equal to a circle whose radius is a mean proportional between (1)
the portion of the side of the cone intercepted by the parallel
planes and (2) the line which is equal to the sum of the radii of
the circles in the parallel planes.
Let OAB be a triangle through the axis of a cone, DE its
intersection with the plane cutting off the
frustum, and OFC the axis of the cone.
Then the surface of the cone OAB is
equal to a circle whose radius is equal to
√OA.AC.
[Prop. 14.]
Similarly the surface of the cone ODE
is equal to a circle whose radius is equal
to √OD.DF.

E
F
A
C
B
And the surface of the frustum is
equal to the difference between the two circles.
Now
OA.AC-OD.DF=DA. AC+ OD.AC- OD.DF.
But
since
OD. AC=OA.DF,
OA: AC OD: DF.
=
22
ARCHIMEDES
Hence OA. AC-OD.DFDA. AC + DA.DF
-
= DA . (AC + DF).
And, since circles are to one another as the squares of their
radii, it follows that the difference between the circles whose.
radii are √OA.AC, √OD.DF respectively is equal to a circle
whose radius is √DA. (AC + DF).
Therefore the surface of the frustum is equal to this circle.
te
Lemmas.
"1. Cones having equal height have the same ratio as their
bases; and those having equal bases have the same ratio as their
heights*.
2. If a cylinder be cut by a plane parallel to the base, then,
as the cylinder is to the cylinder, so is the axis to the axis †.
3. The cones which have the same bases as the cylinders [and
equal height] are in the same ratio as the cylinders.
4. Also the bases of equal cones are reciprocally proportional
to their heights; and those cones whose bases are reciprocally
proportional to their heights are equal‡.
5. Also the cones, the diameters of whose bases have the same
ratio as their axes, are to one another in the triplicate ratio of the
diameters of the bases §.
And all these propositions have been proved by earlier
geometers."
* Euclid XII. 11. "Cones and cylinders of equal height are to one another
as their bases."
Euclid XII. 14. "Cones and cylinders on equal bases are to one another as
their heights."
+ Euclid XII. 13. "If a cylinder be cut by a plane parallel to the opposite
planes [the bases], then, as the cylinder is to the cylinder, so will the axis be
to the axis."
Euclid XII. 15. "The bases of equal cones and cylinders are reciprocally
proportional to their heights; and those cones and cylinders whose bases are
reciprocally proportional to their heights are equal."
§ Euclid XII. 12. "Similar cones and cylinders are to one another in the
triplicate ratio of the diameters of their bases.”
ON THE SPHERE AND CYLINDER I.
23
Proposition 17.
If there be two isosceles cones, and the surface of one cone be
equal to the base of the other, while the perpendicular from the
centre of the base [of the first cone] on the side of that cone is
equal to the height [of the second], the cones will be equal.
Let OAB, DEF be triangles through the axes of two cones
respectively, C, G the centres of the respective bases, GH the

A
C
O
G
E

H
F
·D
B
perpendicular from G on FD; and suppose that the base of the
cone OAB is equal to the surface of the cone DEF, and
that OC=GH.
Then, since the base of OAB is equal to the surface of
DEF,
(base of cone OAB): (base of cone DEF)
=
=
= (surface of DEF): (base of DEF)
·DF : FG
[Prop. 15]
= DG : GH, by similar triangles,
= DG : OC.
=
Therefore the bases of the cones are reciprocally propor-
tional to their heights; whence the cones are equal. [Lemma
4.]
५
24
ARCHIMEDES
༔
Proposition 18.
Any solid rhombus consisting of isosceles cones is equal to
the cone which has its base equal to the surface of one of the
cones composing the rhombus and its height equal to the perpen-
dicular drawn from the apex of the second cone to one side of
the first cone.
Let the rhombus be OABD consisting of two cones with
apices O, D and with a common base (the circle about AB as
diameter).
A
F
H
G
K
D
M
C
B
P
E
3
[



N
Let FHK be another cone with base equal to the surface of
the cone OAB and height FG equal to DE, the perpendicular
from D on OB.
Then shall the cone FHK be equal to the rhombus.
Construct a third cone LMN with base (the circle about
MN) equal to the base of OAB and height LP equal to OD.
Then, since
and
LP=OD,
LP:CD=0D:CD.
But [Lemma 1] OD: CD = (rhombus OADB): (cone DAB),
LP: CD = (cone LMN): (cone DAB).
It follows that
(rhombus (ADB) = (cone LMN)…………..
(1).
ON THE SPHERE AND CYLINDER I.
25
Again, since AB= MN, and
(surface of OAB) = (base of FHK),
(base of FHK): (base of LMN)
=(surface of OAB): (base of OAB)
= OB: BC
[Prop. 15]
=OD: DE, by similar triangles,
=LP: FG, by hypothesis.
Thus, in the cones FHK, LMN, the bases are reciprocally
proportional to the heights.
Therefore the cones FHK, LMN are equal,
and hence, by (1), the cone FHK is equal to the given
solid rhombus.
Proposition 19.
If an isosceles cone be cut by a plane parallel to the base,
and on the resulting circular section a cone be described having
as its apex the centre of the base [of the first cone], and if the
rhombus so formed be taken away from the whole cone, the part
remaining will be equal to the cone with base equal to the surface
of the portion of the first cone between the parallel planes and
with height equal to the perpendicular drawn from the centre of
the base of the first cone on one side of that cone.
Let the cone OAB be cut by a plane parallel to the base in
the circle on DE as diameter. Let C be the centre of the base
of the cone, and with C as apex and the circle about DE as base
describe a cone, making with the cone ODE the rhombus
ODCE.
Take a cone FGH with base equal to the surface of the
frustum DABE and height equal to the perpendicular (CK)
from Con AO.
Then shall the cone FGH be equal to the difference between
the cone OAB and the rhombus ODCE.
Take (1) a cone LMN with base equal to the surface of the
cone OAB, and height equal to CK,
26
ARCHIMEDES
(2) a cone PQR with base equal to the surface of the cone
ODE and height equal to CK.

D
E
G
K
A
C
B
L
P
F
LL



M
N
R
Now, since the surface of the cone OAB is equal to the
surface of the cone ODE together with that of the frustum
DABE, we have, by the construction,
(base of LMN)= (base of FGH)+(base of PQR)
and, since the heights of the three cones are equal,
(cone LMN)=(cone FGH) + (cone PQR).
But the cone LMN is equal to the cone OAB [Prop. 17],
and the cone PQR is equal to the rhombus ODCE [Prop. 18].
Therefore (cone OAB) = (cone FGH)+(rhombus ODCE),
and the proposition is proved.
Proposition 20.
If one of the two isosceles cones forming a rhombus be cut
by a plane parallel to the base and on the resulting circular
section a cone be described having the same apex as the second
cone, and if the resulting rhombus be taken from the whole
rhombus, the remainder will be equal to the cone with base equal
to the surface of the portion of the cone between the parallel
planes and with height equal to the perpendicular drawn from
the apex of the second cone to the side of the first cone.
*
* There is a slight error in Heiberg's translation "prioris coni” and in the
corresponding note, p. 93. The perpendicular is not drawn from the apex of
the cone which is cut by the plane but from the apex of the other.
ON THE SPHERE AND CYLINDER I.
27
K
A
Let the rhombus be OACB, and let the cone OAB be cut
by a plane parallel to its base in the circle about DE as diameter.
With this circle as base and Cas apex describe a cone, which
therefore with ODE forms the rhombus ODCE.

F
D
E
G
H
M
B
P
R
Q
N
Take a cone FGH with base equal to the surface of the
frustum DABE and height equal to the perpendicular (CK)
from Con OA.
The cone FGH shall be equal to the difference between the
rhombi OACB, ODCE.
For take (1) a cone LMN with base equal to the surface of
OAB and height equal to CK,
(2) a cone PQR, with base equal to the surface of ODE,
and height equal to CK.
Then, since the surface of OAB is equal to the surface of
ODE together with that of the frustum DABE, we have, by
construction,
(base of LMN)=(base of PQR) + (base of FGH),
and the three cones are of equal height;
therefore (cone LMN)=(cone PQR)+(cone FGH).
But the cone LMN is equal to the rhombus OACB, and the
cone PQR is equal to the rhombus ODCE [Prop. 18].
Hence the cone FGH is equal to the difference between the
two rhombi OACB, ODCE.
28
ARCHIMEDES
Proposition 21.
A regular polygon of an even number of sides being inscribed
in a circle, as ABC...A'...C'B'A, so that AA' is a diameter,
if two angular points next but one to each other, as B, B′, be
joined, and the other lines parallel to BB' and joining pairs
of angular points be drawn, as CC', DD'..., then
(BB' +CC' + ...) : AA′ = A'B :BA.
Let BB', CC', DD',... meet AA' in F, G, H,...; and let
CB', DC',... be joined meeting AA' in K, L,... respectively.
B
D

E
A
F/K
GLH
NI
B'
C
D'
E'
AB.
Then clearly CB', DC',... are parallel to one another and to
Hence, by similar triangles,
BF: FA=B'F: FK
= CG: GK
=C'G: GL
- E'I: IA':
=
ON THE SPHERE AND CYLINDER I.
29
and, summing the antecedents and consequents respectively, we
have
(BB' + CC' + ...): AA' = BF: FA
= A'B : BA.
Proposition 22.
If a polygon be inscribed in a segment of a circle LAL' so
that all its sides excluding the base are equal and their number
even, as LK...A...K'L', A being the middle point of the segment,
and if the lines BB', CC',... parallel to the base LL' and joining
pairs of angular points be drawn, then
(BB'+CC' + ... + LM): AM = A'B : BA,
where M is the middle point of LL' and AA' is the diameter
through M.
A
B
B'
D

K
C
F/P G
A
o
૭
H R M
1
Ľ
C
K'
D'
Joining CB', DC',...LK', as in the last proposition, and
supposing that they meet AM in P, Q,...R, while BB', CC',...,
KK' meet AM in F, G,... H, we have, by similar triangles,
BF: FA-B'F: FP
= CG: PG
= CG: GQ
=LM : RM ;
30
ARCHIMEDES
and, summing the antecedents and consequents, we obtain
(BB'+CC'+...+LM): AM=BF: FA
Proposition 23.
= A'B: BA.
Take a great circle ABC... of a sphere, and inscribe in it
a regular polygon whose sides are a multiple of four in number.
Let AA', MM' be diameters at right angles and joining
opposite angular points of the polygon.
M

A
B
B'
N
N'
Y
Yo
A
,
M'
Then, if the polygon and great circle revolve together about
the diameter AA', the angular points of the polygon, except A,
A', will describe circles on the surface of the sphere at right
angles to the diameter AA'. Also the sides of the polygon
will describe portions of conical surfaces, e.g. BC will describe
a surface forming part of a cone whose base is a circle about
CC' as diameter and whose apex is the point in which CB,
C'B' produced meet each other and the diameter AA'.
Comparing the hemisphere MAM' and that half of the
figure described by the revolution of the polygon which is
included in the hemisphere, we see that the surface of the
hemisphere and the surface of the inscribed figure have the
same boundaries in one plane (viz. the circle on MM' as
ON THE SPHERE AND CYLINDER I.
31
diameter), the former surface entirely includes the latter, and
they are both concave in the same direction.
Therefore [Assumptions, 4] the surface of the hemisphere
is greater than that of the inscribed figure; and the same is
true of the other halves of the figures.
Hence the surface of the sphere is greater than the surface
described by the revolution of the polygon inscribed in the great
circle about the diameter of the great circle.
Proposition 24.
If a regular polygon AB...A'...B'A, the number of whose
sides is a multiple of four, be inscribed in a great circle of a
sphere, and if BB' subtending two sides be joined, and all the
other lines parallel to BB' and joining pairs of angular points
be drawn, then the surface of the figure inscribed in the sphere
by the revolution of the polygon about the diameter AA' is equal
to a circle the square of whose radius is equal to the rectangle
BA (BB' + CC′ + ...).
The surface of the figure is made up of the surfaces of parts
of different cones.
A
B
B
M

N
C'
N'
M'
A
Now the surface of the cone ABB' is equal to a circle whose
radius is √BA. ¿BB'.
[Prop. 14]
32
ARCHIMEDES
I
}
The surface of the frustum BB'C'C is equal to a circle of
radius √BC.¿(BB′+CC′),
and so on.
[Prop. 16]
It follows, since_ BA = BC = ..., that the whole surface is
equal to a circle whose radius is equal to
√ BA (BB' + CC' + ... + MM' + ...
Proposition 25.
+YY').
The surface of the figure inscribed in a sphere as in the last
propositions, consisting of portions of conical surfaces, is less than
four times the greatest circle in the sphere.
Let AB...A...B'A be a regular polygon inscribed in a
great circle, the number of its sides being a multiple of four.
À
B
BI
M

C
N
C'
N'
MI
A'
As before, let BB' be drawn subtending two sides, and
CC',...YY' parallel to BB'.
to
Let R be a circle such that the square of its radius is equal
AB(BB' + CC' + ... + YY'),
so that the surface of the figure inscribed in the sphere is equal
to R.
[Prop. 24]
ON THE SPHERE AND CYLINDER I.
33
}
Now
whence
(BB' + CC' + ... + YY') : AA' = A'B: AB, [Prop. 21]
Hence
AB (BB' + CC' + ... + YY') = AA'. A'B.
(radius of R)² = AA'. A'B
<AA".
Therefore the surface of the inscribed figure, or the circle R,
is less than four times the circle AMA'M'.
Proposition 26.
The figure inscribed as above in a sphere is equal [in volume]
to a cone whose base is a circle equal to the surface of the figure
inscribed in the sphere and whose height is equal to the
perpendicular drawn from the centre of the sphere to one side of
the polygon.
Suppose, as before, that AB...A'...B'A is the regular
polygon inscribed in a great circle, and let BB', CC',... be
joined.
T
A
B
B
M

N
C'
M'
Y'
A'
With apex O construct cones whose bases are the circles
on BB', CC',... as diameters in planes perpendicular to AA'.
H. A.
3
34
ARCHIMEDES
Then OBAB' is a solid rhombus, and its volume is equal to
a cone whose base is equal to the surface of the cone ABB′ and
whose height is equal to the perpendicular from O on AB
[Prop. 18]. Let the length of the perpendicular be p.
Again, if CB, C'B' produced meet in T, the portion of the
solid figure which is described by the revolution of the triangle
BOC about AA' is equal to the difference between the rhombi
OCTC' and OBTB', i.e. to a cone whose base is equal to the
surface of the frustum BB'C'C and whose height is p [Prop. 20].
Proceeding in this manner, and adding, we prove that, since
cones of equal height are to one another as their bases, the
volume of the solid of revolution is equal to a cone with height
p and base equal to the sum of the surfaces of the cone BAB',
the frustum BB'C'C, etc., i.e. a cone with height p and base
equal to the surface of the solid.
Proposition 27.
The figure inscribed in the sphere as before is less than
four times the cone whose base is equal to a great circle of
the sphere and whose height is equal to the radius of the
sphere.
By Prop. 26 the volume of the solid figure is equal to a cone
whose base is equal to the surface of the solid and whose height
is p, the perpendicular from O on any side of the polygon. Let
R be such a cone.
Take also a cone S with base equal to the great circle, and
height equal to the radius, of the sphere.
Now, since the surface of the inscribed solid is less than four
times the great circle [Prop. 25], the base of the cone R is less
than four times the base of the cone S.
Also the height (p) of R is less than the height of S.
Therefore the volume of R is less than four times that of S;
and the proposition is proved.
1
ON THE SPHERE AND CYLINDER I.
35
Proposition 28.
Let a regular polygon, whose sides are a multiple of four in
number, be circumscribed about a great circle of a given
sphere, as AB...A'...B'A; and about the polygon describe
another circle, which will therefore have the same centre as the
great circle of the sphere. Let AA' bisect the polygon and
cut the sphere in a, a'.
M

m
A
a
K
B
B'
K'
m²
M'
a'
'A'
If the great circle and the circumscribed polygon revolve
together about AA', the great circle will describe the surface
of a sphere, the angular points of the polygon except A, A′ will
move round the surface of a larger sphere, the points of contact
of the sides of the polygon with the great circle of the inner
sphere will describe circles on that sphere in planes perpen-
dicular to AA', and the sides of the polygon themselves will
describe portions of conical surfaces. The circumscribed figure
will thus be greater than the sphere itself.
Let any side, as BM, touch the inner circle in K, and let K'
be the point of contact of the circle with B'M'.
Then the circle described by the revolution of KK' about
AA' is the boundary in one plane of two surfaces
(1) the surface formed by the revolution of the circular
segment KaK', and
>
3-2
36
ARCHIMEDES
(2) the surface formed by the revolution of the part
KB…..A...B'K' of the polygon.
Now the second surface entirely includes the first, and they
are both concave in the same direction;
therefore [Assumptions, 4] the second surface is greater
than the first.
The same is true of the portion of the surface on the opposite
side of the circle on KK' as diameter.
Hence, adding, we see that the surface of the figure
circumscribed to the given sphere is greater than that of the
sphere itself.
Proposition 29.
In a figure circumscribed to a sphere in the manner shown
in the previous proposition the surface is equal to a circle the
square on whose radius is equal to AB (BB' + CC' + ...).
For the figure circumscribed to the sphere is inscribed in a
larger sphere, and the proof of Prop. 24 applies.
Proposition 30.
The surface of a figure circumscribed as before about a sphere
is greater than four times the great circle of the sphere.
B
P
A
a
B'
M

C
m
O
C'
m'
M'
Α'
a'
ON THE SPHERE AND CYLINDER I.
37
Let AB...A'...B'A be the regular polygon of 4n sides
which by its revolution about AA' describes the figure circum-
scribing the sphere of which ama'm' is a great circle. Suppose
aa', AA' to be in one straight line.
Let R be a circle equal to the surface of the circumscribed
solid.
Now (BB'+CC' + ...): AA' = A'B: BA, [as in Prop. 21]
AB (BB' + CC′ + ...) = AA'. A'B.
so that
Hence
(radius of R)=√AA'. A'B
> A'B.
[Prop. 29]
But A'B=20P, where P is the point in which AB touches
the circle ama'm'.
Therefore (radius of R)> (diameter of circle ama'm');
whence R, and therefore the surface of the circumscribed solid,
is greater than four times the great circle of the given sphere.
Proposition 31.
The solid of revolution circumscribed as before about a sphere
is equal to a cone whose base is equal to the surface of the solid
and whose height is equal to the radius of the sphere.
The solid is, as before, a solid inscribed in a larger sphere;
and, since the perpendicular on any side of the revolving polygon
is equal to the radius of the inner sphere, the proposition is
identical with Prop. 26.
COR. The solid circumscribed about the smaller sphere is
greater than four times the cone whose base is a great circle
of the sphere and whose height is equal to the radius of the
sphere.
For, since the surface of the solid is greater than four times
the great circle of the inner sphere [Prop. 30], the cone whose
base is equal to the surface of the solid and whose height is the
radius of the sphere is greater than four times the cone of
the same height which has the great circle for base. [Lemma 1.]
Hence, by the proposition, the volume of the solid is greater
than four times the latter cone.
7
388
ARCHIMEDES
1
Proposition 32.
If a regular polygon with 4n sides be inscribed in a great
circle of a sphere, as ab...a'...b'a, and a similar polygon
AB...A'...B'A be described about the great circle, and if the
polygons revolve with the great circle about the diameters aa',
AA' respectively, so that they describe the surfaces of solid
figures inscribed in and circumscribed to the sphere respectively,
then
(1) the surfaces of the circumscribed and inscribed figures
are to one another in the duplicate ratio of their sides, and
(2) the figures themselves [i.e. their volumes] are in the
triplicate ratio of their sides.
(1) Let AA', aa' be in the same straight line, and let
MmOm'M' be a diameter at right angles to them.
M

m
B
A
a
B'
m!
M'
A'
u'
Join BB', CC',... and bb', cc', ... which will all be parallel
to one another and MM'.
Suppose R, S to be circles such that
R=(surface of circumscribed solid),
S=(surface of inscribed solid).
ON THE SPHERE AND CYLINDER I.
39
Then
AB(BB'+CC'+...)
(radius of R)² = AB (BB' + CC′+...) [Prop. 29]
(radius of S)² = ab (bb′ + cc' + ...).
==
[Prop. 24]
And, since the polygons are similar, the rectangles in these
two equations are similar, and are therefore in the ratio of
Hence
AB² : ab².
(surface of circumscribed solid): (surface of inscribed solid)
= AB² : ab².
(2) Take a cone V whose base is the circle R and whose
height is equal to Oa, and a cone W whose base is the circle S
and whose height is equal to the perpendicular from O on ab,
which we will call p.
Then V, W are respectively equal to the volumes of the
circumscribed and inscribed figures.
Now, since the polygons are similar,
AB: ab
Oa: P
=
[Props. 31, 26]
(height of cone V): (height of cone W);
and, as shown above, the bases of the cones (the circles R, S)
are in the ratio of AB² to ab².
Therefore
V: WAB: abs.
Proposition 33.
The surface of any sphere is equal to four times the greatest
circle in it.
Let C be a circle equal to four times the great circle.
Then, if C is not equal to the surface of the sphere, it must
either be less or greater.
I. Suppose C less than the surface of the sphere.
It is then possible to find two lines ẞ, y, of which ẞ is the
greater, such that
By<(surface of sphere): C.
[Prop. 2]
Take such lines, and let & be a mean proportional between
them.
40
ARCHIMEDES
C
*
Suppose similar regular polygons with 4n sides circum-
scribed about and inscribed in a great circle such that the ratio
of their sides is less than the ratio : S.
[Prop. 3]
A
a
B'
B
M

m
A'
a
β. δ γ
જ
m'
M'
Let the polygons with the circle revolve together about
a diameter common to all, describing solids of revolution as
before.
Then (surface of outer solid): (surface of inner solid)
=
= (side of outer)²: (side of inner)2 [Prop. 32]
< ß² : S², or B: y where 82=38
< (surface of sphere) : C, a fortiori.
But this is impossible, since the surface of the circum-
scribed solid is greater than that of the sphere [Prop. 28], while
the surface of the inscribed solid is less than C [Prop. 25].
Therefore C is not less than the surface of the sphere.
II. Suppose C greater than the surface of the sphere.
Take lines ẞ, y, of which ẞ is the greater, such that
B: y<C: (surface of sphere).
Circumscribe and inscribe to the great circle similar regular
polygons, as before, such that their sides are in a ratio less than
that of ẞ to 8, and suppose solids of revolution generated in the
usual manner.
ON THE SPHERE AND CYLINDER I.
41
Then, in this case,
(surface of circumscribed solid): (surface of inscribed solid)
<C: (surface of sphere).
But this is impossible, because the surface of the circum-
scribed solid is greater than C [Prop. 30], while the surface of
the inscribed solid is less than that of the sphere [Prop. 23].
Thus C is not greater than the surface of the sphere.
Therefore, since it is neither greater nor less, C is equal to
the surface of the sphere.
Proposition 34.
Any sphere is equal to four times the cone which has its base
equal to the greatest circle in the sphere and its height equal
to the radius of the sphere.
Let the sphere be that of which ama'm' is a great circle.
If now the sphere is not equal to four times the cone
described, it is either greater or less.
I. If possible, let the sphere be greater than four times the
cone.
Suppose V to be a cone whose base is equal to four times
the great circle and whose height is equal to the radius of the
sphere.
Then, by hypothesis, the sphere is greater than V; and two
lines B, y can be found (of which ẞ is the greater) such that
B: y<(volume of sphere): V.
Between ẞ and y place two arithmetic means 8, e.
As before, let similar regular polygons with sides 4n in
number be circumscribed, about and inscribed in the great
circle, such that their sides are in a ratio less than ß: S.
Imagine the diameter aa' of the circle to be in the same
straight line with a diameter of both polygons, and imagine
the latter to revolve with the circle about aa', describing the
42
ARCHIMEDES
2
- ມ
surfaces of two solids of revolution. The volumes of these solids
are therefore in the triplicate ratio of their sides. [Prop. 32]
Thus (vol. of outer solid): (vol. of inscribed solid)
< ß³: 8³, by hypothesis,
< ß: y, a fortiori (since ß: y> ß³ : 8³)*,
<(volume of sphere): V, a fortiori.
But this is impossible, since the volume of the circumscribed
M

A
a
B
b
m
u
A'
m
β δε γ
M'
* That B: >83: 6 is assumed by Archimedes.
property in his commentary as follows.
Eutocius proves the
Take a such that
Thus
X
and, since ẞ>d, ß-d>d-x.
But, by hypothesis,
Therefore
B:d=8x.
B-d:ß=d-x: 8
B-d=8-ε.
d-€>d-x,
x>Є.
or
Again, suppose
d:x=x:y,
and, as before, we have
d-x>x-y,
so that, a fortiori,
Therefore
and, since x>€, Y>Y.
¤−y>x-y;
Now, by hypothesis, ß, d, x, y are in continued proportion;
therefore
p³ : 8³=ß: y
<B: 7.
ON THE SPHERE AND CYLINDER I.
43
solid is greater than that of the sphere [Prop. 28], while the
volume of the inscribed solid is less than V [Prop. 27].
Hence the sphere is not greater than V, or four times the
cone described in the enunciation.
II. If possible, let the sphere be less than V.
In this case we take ß, y (ß being the greater) such that
B: y<V: (volume of sphere).
The rest of the construction and proof proceeding as before,
we have finally
(volume of outer solid): (volume of inscribed solid)
< V : (volume of sphere).
But this is impossible, because the volume of the outer
solid is greater than V [Prop. 31, Cor.], and the volume of the
inscribed solid is less than the volume of the sphere.
Hence the sphere is not less than V.
Since then the sphere is neither less nor greater than V, it
is equal to V, or to four times the cone described in the enun-
ciation.
O
COR. From what has been proved it follows that every
cylinder whose base is the greatest circle in a sphere and whose
height is equal to the diameter of the sphere is of the sphere,
and its surface together with its bases is of the surface of the
sphere.
For the cylinder is three times the cone with the same
base and height [Eucl. XII. 10], i.e. six times the cone with
the same base and with height equal to the radius of the
sphere.
But the sphere is four times the latter cone [Prop. 34].
Therefore the cylinder is of the sphere.
Again, the surface of a cylinder (excluding the bases) is
equal to a circle whose radius is a mean proportional between
the height of the cylinder and the diameter of its base
[Prop. 13].
0
:
F
44
ARCHIMEDES
In this case the height is equal to the diameter of the base
and therefore the circle is that whose radius is the diameter of
the sphere, or a circle equal to four times the great circle of
the sphere.
Therefore the surface of the cylinder with the bases is equal
to six times the great circle.
And the surface of the sphere is four times the great circle
[Prop. 33]; whence
(surface of cylinder with bases) = §. (surface of sphere).
Proposition 35.
If in a segment of a circle LAL' (where A is the middle
point of the arc) a polygon LK...A...K'L' be inscribed of which
LL' is one side, while the other sides are 2n in number and all
equal, and if the polygon revolve with the segment about the
diameter AM, generating a solid figure inscribed in a segment of
a sphere, then the surface of the inscribed solid is equal to a
circle the square on whose radius is equal to the rectangle
AB ( BB' + CC' + ... + KK' +
LL'
2
K
B
A

M
B'
K'
A'
The surface of the inscribed figure is made up of portions of
surfaces of cones.
ON THE SPHERE AND CYLINDER I.
45
·
If we take these successively, the surface of the cone BAB'
is equal to a circle whose radius is
√AB. BB'.
[Prop. 14]
The surface of the frustum of a cone BCC'B' is equal to
a circle whose radius is

AB.
BB' + CC'
2
1;
[Prop. 16]
and so on.
Proceeding in this way and adding, we find, since circles
are to one another as the squares of their radii, that the
surface of the inscribed figure is equal to a circle whose radius
is

AB (BB' + CC' + ... + KK' +
LL'
2
Proposition 36.
The surface of the figure inscribed as before in the segment
of a sphere is less than that of the segment of the sphere.
This is clear, because the circular base of the segment is a
common boundary of each of two surfaces, of which one, the
segment, includes the other, the solid, while both are concave
in the same direction [Assumptions, 4].
Proposition 37.
The surface of the solid figure inscribed in the segment of the
sphere by the revolution of LK...A...K'L' about AM is less than
a circle with radius equal to AL.
Let the diameter AM meet the circle of which LAL' is a
segment again in A'. Join A'B.
As in Prop. 35, the surface of the inscribed solid is equal to
a circle the square on whose radius is
AB (BB' + CC' + ... + KK' + LM).
46
ARCHIMEDES
}
But this rectangle
B
= A'B.AM
=
< A'A. AM
<AL
A
B'
[Prop. 22]

K
ட்
M
C'
K'
Α'
Hence the surface of the inscribed solid is less than the
circle whose radius is AL.
Proposition 38.
The solid figure described as before in a segment of a sphere
less than a hemisphere, together with the cone whose base is the
base of the segment and whose apex is the centre of the sphere,
is equal to a cone whose base is equal to the surface of the
inscribed solid and whose height is equal to the perpendicular
from the centre of the sphere on any side of the polygon.
Let O be the centre of the sphere, and p the length of the
perpendicular from 0 on AB.
Suppose cones described with O as apex, and with the
circles on BB', CC',... as diameters as bases.
Then the rhombus OBAB' is equal to a cone whose base is
equal to the surface of the cone BAB', and whose height is p.
[Prop. 18]
Again, if CB, C'B' meet in T, the solid described by the
triangle BOC as the polygon revolves about AO is the difference
ON THE SPHERE AND CYLINDER I.
47
between the rhombi OCTC' and OBTB', and is therefore equal
to a cone whose base is equal to the surface of the frustum
BCC'B' and whose height is p.
[Prop. 20]
Similarly for the part of the solid described by the triangle
COD as the polygon revolves; and so on.
T

K
¡A
B
B'
C
K
L'
Hence, by addition, the solid figure inscribed in the segment
together with the cone OLL' is equal to a cone whose base is
the surface of the inscribed solid and whose height is p.
COR. The cone whose base is a circle with radius equal to
AL and whose height is equal to the radius of the sphere is
greater than the sum of the inscribed solid and the cone OLL'.
For, by the proposition, the inscribed solid together with
the cone OLL' is equal to a cone with base equal to the surface
of the solid and with height p.
This latter cone is less than a cone with height equal to OA
and with base equal to the circle whose radius is AL, because
the height p is less than OA, while the surface of the solid is
less than a circle with radius AL.
[Prop. 37]
Proposition 39.
Let lal' be a segment of a great circle of a sphere, being less
than a semicircle. Let O be the centre of the sphere, and join
Ol, Ol'. Suppose a polygon circumscribed about the sector Olal'
such that its sides, excluding the two radii, are 2n in number
48
ARCHIMEDES
and all equal, as LK,... BA, AB', ... K'L'; and let OA be that
radius of the great circle which bisects the segment lal'.
The circle circumscribing the polygon will then have the
same centre O as the given great circle.
Now suppose the polygon and the two circles to revolve
together about OA. The two circles will describe spheres, the
A

a
B
B'
'M
'In.
|
い
​angular points except A will describe circles on the outer
sphere, with diameters BB' etc., the points of contact of the
sides with the inner segment will describe circles on the inner
sphere, the sides themselves will describe the surfaces of cones
or frusta of cones, and the whole figure circumscribed to the
segment of the inner sphere by the revolution of the equal
sides of the polygon will have for its base the circle on LL'
as diameter.
The surface of the solid figure so circumscribed about the
sector of the sphere [excluding its base] will be greater than that
of the segment of the sphere whose base is the circle on ll' as
diameter.
For draw the tangents IT, IT" to the inner segment at l, l'.
These with the sides of the polygon will describe by their
revolution a solid whose surface is greater than that of the
segment [Assumptions, 4].
But the surface described by the revolution of IT is less
than that described by the revolution of LT, since the angle TIL
is a right angle, and therefore LT >IT.
Hence, a fortiori, the surface described by LK...A...K'L'
is greater than that of the segment.
1
ON THE SPHERE AND CYLINDER I.
49
COR. The surface of the figure so described about the sector
of the sphere is equal to a circle the square on whose radius
is equal to the rectangle
AB (BB' + CC' + ... + KK' + ½ LL').
For the circumscribed figure is inscribed in the outer sphere,
and the proof of Prop. 35 therefore applies.
Proposition 40.
The surface of the figure circumscribed to the sector as before
is greater than a circle whose radius is equal to al.
Let the diameter AɑO meet the great circle and the circle
circumscribing the revolving polygon again in a', A'. Join
A'B, and let ON be drawn to N, the point of contact of AB
with the inner circle.
A

a
N
B
B'
ՊՈՆ
MO
Α'
a'
Now, by Prop. 39, Cor., the surface of the solid figure
circumscribed to the sector OlAl' is equal to a circle the square
on whose radius is equal to the rectangle
(BB'
AB BB'+CC'+...+KK'+
LL
2
But this rectangle is equal to A'B. AM [as in Prop. 22].
H. A.
"
+
4
1
*
1
50
ARCHIMEDES
Next, since AL', al' are parallel, the triangles AML', amľ
are similar. And AL'>al'; therefore AM> am.
Also
Therefore
A′B = 20N = aa′.
A'B.AM> am. aa'
> al'.
Hence the surface of the solid figure circumscribed to the
sector is greater than a circle whose radius is equal to al', or al.
COR. 1. The volume of the figure circumscribed about the
sector together with the cone whose apex is O and base the circle
on LL' as diameter, is equal to the volume of a cone whose base
is equal to the surface of the circumscribed figure and whose
height is ON.
For the figure is inscribed in the outer sphere which has the
same centre as the inner. Hence the proof of Prop. 38 applies.
COR. 2. The volume of the circumscribed figure with the cone
OLL' is greater than the cone whose base is a circle with radius
equal to al and whose height is equal to the radius (Oa) of the
inner sphere.
For the volume of the figure with the cone OLL' is equal to
a cone whose base is equal to the surface of the figure and
whose height is equal to ON.
And the surface of the figure is greater than a circle with
radius equal to al [Prop. 40], while the heights Oa, ON are
equal.
Proposition 41.
Let lal' be a segment of a great circle of a sphere which is
less than a semicircle.
Suppose a polygon inscribed in the sector Olal' such that
the sides lk,.... ba, ab',...k'l' are 2n in number and all equal.
Let a similar polygon be circumscribed about the sector so that
its sides are parallel to those of the first polygon; and draw
the circle circumscribing the outer polygon.
Now let the polygons and circles revolve together about
OaA, the radius bisecting the segment lal'.
ON THE SPHERE AND CYLINDER I.
51
Then (1) the surfaces of the outer and inner solids of revolution
so described are in the ratio of AB² to ab², and (2) their volumes
together with the corresponding cones with the same base and
with apex O in each case are as AB³ to ab³.
A

a
B
n
B'
M
m
(1) For the surfaces are equal to circles the squares on
whose radii are equal respectively to
and
AB (BB' + CC' + ... + KK' +
LL'
ab (bb′+cc′ + ... + kk' + 2).
2
[Prop. 39, Cor.]
[Prop. 35]
But these rectangles are in the ratio of AB² to ab². Therefore
so are the surfaces.
(2) Let OnN be drawn perpendicular to ab and AB; and
suppose the circles which are equal to the surfaces of the outer
and inner solids of revolution to be denoted by S, s respectively.
Now the volume of the circumscribed solid together with
the cone OLL' is equal to a cone whose base is S and whose
height is ON [Prop. 40, Cor. 1].
And the volume of the inscribed figure with the cone Oll' is
equal to a cone with base s and height On [Prop. 38].
and
But
S : s = AB² : ab²,
ON: On = AB: ab.
Therefore the volume of the circumscribed solid together with
the cone OLL' is to the volume of the inscribed solid together
with the cone Oll' as AB³ is to ab³ [Lemma 5].
4-2
52
ARCHIMEDES
Proposition 42.
If lal' be a segment of a sphere less than a hemisphere and
Oa the radius perpendicular to the base of the segment, the
surface of the segment is equal to a circle whose radius is equal
to al.
Let R be a circle whose radius is equal to al. Then the
surface of the segment, which we will call S, must, if it be not
equal to R, be either greater or less than R.
B
b
A

a
B'
.
K
ん
​I. Suppose, if possible, S> R.
Let lal' be a segment of a great circle which is less than a
semicircle. Join Ol, Ol', and let similar polygons with 2n equal
sides be circumscribed and inscribed to the sector, as in the
previous propositions, but such that
(circumscribed polygon): (inscribed polygon)<S : R.
[Prop. 6]
Let the polygons now revolve with the segment about OɑA,
generating solids of revolution circumscribed and inscribed to
the segment of the sphere.
Then
(surface of outer solid) : (surface of inner solid)
– AB² : ab²
[Prop. 41]
(circumscribed polygon): (inscribed polygon)
<S: R, by hypothesis.
But the surface of the outer solid is greater than S [Prop. 39].
ON THE SPHERE AND CYLINDER I.
53
Therefore the surface of the inner solid is greater than R;
which is impossible, by Prop. 37.
II. Suppose, if possible, S<R.
In this case we circumscribe and inscribe polygons such that
their ratio is less than R: S; and we arrive at the result that
(surface of outer solid): (surface of inner solid)
<R: S.
But the surface of the outer solid is greater than R [Prop. 40].
Therefore the surface of the inner solid is greater than S: which
is impossible [Prop. 36].
Hence, since S is neither greater nor less than R,
S=R.
Proposition 43.
Even if the segment of the sphere is greater than a hemisphere,
its surface is still equal to a circle whose radius is equal to al.
For let lal'a' be a great circle of the sphere, aa' being the
diameter perpendicular to l'; and let
la'l' be a segment less than a semi-
circle.
Then, by Prop. 42, the surface of
the segment la'l' of the sphere is
equal to a circle with radius equal to
al.
Also the surface of the whole
sphere is equal to a circle with radius
equal to aa' [Prop. 33].
I
a'

a
But aa'² - a'l² = al³, and circles are to one another as the
squares on their radii.
Therefore the surface of the segment lal', being the difference
between the surfaces of the sphere and of la'l', is equal to a
circle with radius equal to al.
3
J
54
ARCHIMEDES
Proposition 44.
The volume of any sector of a sphere is equal to a cone whose
base is equal to the surface of the segment of the sphere included
in the sector, and whose height is equal to the radius of the
sphere.
Let R be a cone whose base is equal to the surface of the
segment lal' of a sphere and whose height is equal to the radius
of the sphere; and let S be the volume of the sector Olal'.
A

a
B
B'
b
B-
δ
E-
γ
Then, if S is not equal to R, it must be either greater or
less.
I. Suppose, if possible, that S> R.
Find two straight lines B, y, of which ẞ is the greater, such
that
B: Y<S:R;
and let 8, e be two arithmetic means between ß, y.
Let lal' be a segment of a great circle of the sphere.
Join Ol, Ol', and let similar polygons with 2n equal sides be
circumscribed and inscribed to the sector of the circle as before,
but such that their sides are in a ratio less than B: 8.
[Prop. 4].
ON THE SPHERE AND CYLINDER I.
55
.
Then let the two polygons revolve with the segment about
OaA, generating two solids of revolution.
Denoting the volumes of these solids by V, v respectively,
we have
Now
(V+cone OLL') : (v +cone Oll') = AB³ : ab³ [Prop. 41]
Therefore also
< B³: 89
<B: y, a fortiori*,
<S: R, by hypothesis.
(V+cone OLL') > S.
(v + cone Oll')> R.
But this is impossible, by Prop. 38, Cor. combined with Props.
42, 43.
Hence
SR.
II. Suppose, if possible, that S < R.
In this case we take ẞ, y such that
B: y < R: S,
and the rest of the construction proceeds as before.
We thus obtain the relation
Now
Therefore
(V+ cone OLL') : (v + cone Ol') < R : S.
(v + cone Oll') < S.
(V + cone OLL') < R ;
which is impossible, by Prop. 40, Cor. 2 combined with Props.
42, 43.
Since then S is neither greater nor less than R,
S = R.
* Cf. note on Prop. 34, p. 42..
J
+
ON THE SPHERE AND CYLINDER.
BOOK II.
“ARCHIMEDES to Dositheus greeting.
On a former occasion you asked me to write out the proofs of
the problems the enunciations of which I had myself sent to
Conon. In point of fact they depend for the most part on the
theorems of which I have already sent you the demonstrations,
namely (1) that the surface of any sphere is four times the
greatest circle in the sphere, (2) that the surface of any
segment of a sphere is equal to a circle whose radius is equal
to the straight line drawn from the vertex of the segment to
the circumference of its base, (3) that the cylinder whose base
is the greatest circle in any sphere and whose height is equal
to the diameter of the sphere is itself in magnitude half as
large again as the sphere, while its surface [including the two
bases] is half as large again as the surface of the sphere, and
(4) that any solid sector is equal to a cone whose base is the
circle which is equal to the surface of the segment of the sphere
included in the sector, and whose height is equal to the radius
of the sphere. Such then of the theorems and problems as
depend on these theorems I have written out in the book
which I send herewith; those which are discovered by means
of a different sort of investigation, those namely which relate
to spirals and the conoids, I will endeavour to send you soon.
ON THE SPHERE AND CYLINDER II.
57
The first of the problems was as follows: Given a sphere, to
find a plane area equal to the surface of the sphere.
The solution of this is obvious from the theorems aforesaid.
For four times the greatest circle in the sphere is both a plane
area and equal to the surface of the sphere.
The second problem was the following."
Proposition 1. (Problem.)
Given a cone or a cylinder, to find a sphere equal to the cone
or to the cylinder.
If V be the given cone or cylinder, we can make a cylinder
equal to V. Let this cylinder be the cylinder whose base
is the circle on AB as diameter and whose height is OD.
Now, if we could make another cylinder, equal to the
cylinder (OD) but such that its height is equal to the diameter
of its base, the problem would be solved, because this latter
cylinder would be equal to V, and the sphere whose diameter
is equal to the height (or to the diameter of the base) of the
same cylinder would then be the sphere required [I. 34, Cor.].
G
E
F
M
N
D


A
B
Suppose the problem solved, and let the cylinder (CG) be
equal to the cylinder (OD), while EF, the diameter of the base,
is equal to the height CG.
58
ARCHIMEDES
Then, since in equal cylinders the heights and bases are
reciprocally proportional,
AB2: EF2 CG: OD
Suppose MN to be such a line that
- EF : OD.......
..(1).
EF2AB.MN.
(2).
Hence
AB: EFEF: MN,
and, combining (1) and (2), we have
AB: MN=EF: OD,
or
Therefore
AB: EF MN: OD.
AB: EF = EF: MN = MN: OD,
and EF, MN are two mean proportionals between AB, OD.
The synthesis of the problem is therefore as follows. Take
two mean proportionals EF, MN between AB and OD, and
describe a cylinder whose base is a circle on EF as diameter
and whose height CG is equal to EF.
Then, since
AB: EF EF: MN MN: OD,
=
EF2 AB. MN,
----
and therefore
AB2: EF² = AB : MN
= EF: OD
= CG: OD;
whence the bases of the two cylinders (OD), (CG) are recipro-
cally proportional to their heights.
Therefore the cylinders are equal, and it follows that
cylinder (CG) = {V.
The sphere on EF as diameter is therefore the sphere
required, being equal to V.
ON THE SPHERE AND CYLINDER II.
59
Proposition 2.
If BAB′ be a segment of a sphere, BB′ a diameter of the
base of the segment, and O the centre of the sphere, and if AA'
be the diameter of the sphere bisecting BB' in M, then the volume
of the segment is equal to that of a cone whose base is the same
as that of the segment and whose height is h, where
h: AMOA' + A'M : A'M.
Measure MH along MA equal to h, and MH' along MA'
equal to h', where
h': A'M = OA+AM: AM.
Suppose the three cones constructed which have 0, H
H' for their apices and the base (BB′) of the segment for their
common base. Join AB, A'B.

H'
C
B
A
H
M
A
B'
Let C be a cone whose base is equal to the surface of the
segment BAB′ of the sphere, i.e. to a circle with radius equal
to AB [I. 42], and whose height is equal to OA.
Then the cone C is equal to the solid sector OBAB′ [I. 44].
Now, since HM: MAOA' + A'M: A'M,
dividendo,
HA: AMOA : A'M,
and, alternately, HA: AO AM: MA',
so that
=
HO: OA = AA' : A'M
= AB² : BM²
(base of cone C): (circle on BB' as diameter).
60
ARCHIMEDES
But OA is equal to the height of the cone C; therefore, since
cones are equal if their bases and heights are reciprocally
proportional, it follows that the cone C (or the solid sector
OBAB') is equal to a cone whose base is the circle on BB' as
diameter and whose height is equal to OH.
And this latter cone is equal to the sum of two others
having the same base and with heights OM, MH, i.e. to the
solid rhombus OBHB'.
Hence the sector OBAB' is equal to the rhombus OBHB'.
Taking away the common part, the cone OBB',
the segment BAB' = the cone HBB'.
Similarly, by the same method, we can prove that
the segment BA'B' the cone H'BB'.
=
Alternative proof of the latter property.
Suppose D to be a cone whose base is equal to the surface
of the whole sphere and whose height is equal to OA.
Thus D is equal to the volume of the sphere.
Now, since OA' + A'M : A'M = HM : MA,
dividendo and alternando, as before,
[I. 33, 34]
Again, since
OA : AH = A'M : MA.
H'M: MA' = 0A + AM : AM,
H'A': OA = A'M : MA
Componendo,
=OA: AH, from above.
H'O: OA OH: HA......………………….
=
Alternately,
H'O: OH = OA : AH ....
(1).
.(2),
and, componendo,
HH': HO OH: HA,
=
= H'O: OA, from (1),
whence
HH'.0A = H'O.OH....
(3).
Next, since
H'O: OH=0A: AH, by (2),
=
A'M: MA,
(H'0+0H)² : H'O.OH = (A'M + MA)² : A'M. MA,
ON THE SPHERE AND CYLINDER II.
61
whence, by means of (3),
2
HH": HH'. OA
AA": A'M. MA,
or
HH': 0A = AA" : BM".
Now the cone D, which is equal to the sphere, has for its base
a circle whose radius is equal to AA', and for its height a line
equal to OA.
Hence this cone D is equal to a cone whose base is the circle
on BB' as diameter and whose height is equal to HH';
the cone D = the rhombus HBH'B',
the rhombus HBH'B' = the sphere.
therefore
or
But
the segment BAB' = the cone HBB';
therefore the remaining segment BA'B'
the cone H'BB'.
COR. The segment BAB' is to a cone with the same base and
equal height in the ratio of OA' + A'M to A'M.
Proposition 3. (Problem.)
To cut a given sphere by a plane so that the surfaces of the
segments may have to one another a given ratio.
Suppose the problem solved. Let AA' be a diameter of a
great circle of the sphere, and suppose that a plane perpendicular
to AA' cuts the plane of the great circle in the straight
B
Ан

A
B
M
A
line BB', and AA' in M, and that it divides the sphere so that
the surface of the segment BAB' has to the surface of the
segment BA'B' the given ratio.
62
ARCHIMEDES
2
Now these surfaces are respectively equal to circles with
radii equal to AB, A'B [I. 42, 43].
Hence the ratio AB² : A'B² is equal to the given ratio, i.e.
AM is to MA' in the given ratio.
Accordingly the synthesis proceeds as follows.
If H : K be the given ratio, divide AA' in M so that
AM: MA' H: K.
Then AM: MA'= AB² : A'B²
=
(circle with radius AB): (circle with radius A'B)
(surface of segment BAB'): (surface of segment BA'B').
Thus the ratio of the surfaces of the segments is equal to
the ratio H : K.
Proposition 4. (Problem.)
To cut a given sphere by a plane so that the volumes of the
segments are to one another in a given ratio.
Suppose the problem solved, and let the required plane cut
the great circle ABA' at right angles in the line BB'. Let
AA' be that diameter of the great circle which bisects BB' at
right angles (in M), and let O be the centre of the sphere.
·

He
B
D
A'
MA EH
B
Take H on OA produced, and H' on OA' produced, such
that
OA' + A'M: A'M
HM: MA,....
..(1),
and
OA+AM: AM=H'M : MA' ……….
..(2).
Join BH, B'H, BH', B'H'.
ON THE SPHERE AND CYLINDER II.
63
Then the cones HBB', H'BB' are respectively equal to the
segments BAB', BA'B' of the sphere [Prop. 2].
Hence the ratio of the cones, and therefore of their altitudes,
is given, i.e.
HM : H'M = the given ratio……………….
.(3).
We have now three equations (1), (2), (3), in which there
appear three as yet undetermined points M, H, H'; and it is
first necessary to find, by means of them, another equation in
which only one of these points (M) appears, i.e. we have, so to
speak, to eliminate H, H'.
Now, from (3), it is clear that HH': H'M is also a given
ratio; and Archimedes' method of elimination is, first, to find
values for each of the ratios A'H' : H'M and HH': H'A' which
are alike independent of H, H', and then, secondly, to equate
the ratio compounded of these two ratios to the known value
of the ratio HH': H'M.
(a) To find such a value for A'H': H'M.
It is at once clear from equation (2) above that
A'H' : H'M = OA : OA + AM ……………………………….(4).
(b) To find such a value for HH': A'H'.
From (1) we derive
A'M: MAOA' + A'M: HM
ОА': АН ..
and, from (2),
A'M: MAH'M: OA+AM
= A'H': OA
HA: AO
OA': A'H',
………….(5);
.(6).
Thus
whence
or
It follows that
or
OH : OA'OH': A'H',
OH OH' OA': A'H'.
=
HH': OH' = OH' : A'H',
HH'. H'A' = OH”.
12
Therefore HH' : H'A' = OH” : H'A'
12
= AA" : A'M², by means of (6)
64
ARCHIMEDES
Hu
(c) To express the ratios A'H': H'M and HH': H'M more
simply we make the following construction. Produce OA to D
so that OA=AD. (D will lie beyond H, for A'M > MA, and
therefore, by (5), OA >AH.)
A'H': H'M = OA: OA+AM
Then
***
· AD: DM....
Now divide AD at E so that
HH' : H'M = AD : DE……...
DE.........
·(7).
.(8).
Thus, using equations (8), (7) and the value of HH': H'A'
above found, we have
AD: DE = HH': H'M
But
Therefore
AD:
=
·(HH': H'A'). (A'H': H'M)
· (AA" : A'M²).(AD : DM).
DE=(DM: DE). (AD: DM).
MD: DE = AA” : A'M²
And D is given, since AD=OA.
to HH': H'M) is a given ratio.
Also AD: DE (being equal
Therefore DE is given.
Hence the problem reduces itself to the problem of dividing
A'D into two parts at M so that
MD : (a given length) = (a given area) : A'M².
Archimedes adds: "If the problem is propounded in this
general form, it requires a duopioμós [i.e. it is necessary to
investigate the limits of possibility], but, if there be added the
conditions subsisting in the present case, it does not require a
διορισμός.”
In the present case the problem is:
Given a straight line A'A produced to D so that A'A = 2AD,
and given a point E on AD, to cut AA' in a point M so that
AA”: A'M² – MD : DE.
=
"And the analysis and synthesis of both problems will be
given at the end*.
The synthesis of the main problem will be as follows. Let
RS be the given ratio, R being less than S. AA' being a
* See the note following this proposition.
T
ON THE SPHERE AND CYLINDER II.
65
diameter of a great circle, and O the centre, produce OA to D
so that OA = AD, and divide AD in E so that
AE : ED=R : S.
Then cut AA' in M so that
MD: DE AA” : A'M².
=
Through M erect a plane perpendicular to AA'; this plane
will then divide the sphere into segments which will be to one
another as R to S.
Take H on A'A produced, and H' on AA' produced, so that
OA' + A'M : A'M = HM : MA,………………………………….(1),
OA+AM: AM = H'M : MA'…..
We have then to show that
or
HM: MH' R: S, or AE: ED.
=
(a) We first find the value of HH': H'A' as follows.
As was shown in the analysis (b),
HH'. H'A' OH",
=
>
HH': H'A' = OH": H'A"
(B) Next we have
Therefore
=AA": A'M2
.(2).
= MD: DE, by construction.
H'A': H'MOA: OA+AM
= AD : DM.
=
HH': H'M=(HH': H'A').(H'A': H'M)
= (MD: DE). (AD:DM)
= AD: DE,
whence
HM : MH' = AE : ED
=R: S
Q. E. D.
Note. The solution of the subsidiary problem to which the
original problem of Prop. 4 is reduced, and of which Archimedes
promises a discussion, is given in a highly interesting and
important note by Eutocius, who introduces the subject with
the following explanation.
H. A.
5
66 ·
ARCHIMEDES
"He [Archimedes] promised to give a solution of this
problem at the end, but we do not find the promise kept in any
of the copies. Hence we find that Dionysodorus too failed to
light upon the promised discussion and, being unable to grapple
with the omitted lemma, approached the original problem in a
different way, which I shall describe later. Diocles also ex-
pressed in his work Teρì Tuρiwv the opinion that Archimedes
made the promise but did not perform it, and tried to supply
the omission himself. His attempt I shall also give in its
order. It will however be seen to have no relation to the
omitted discussion but to give, like Dionysodorus, a construction
arrived at by a different method of proof. On the other hand,
as the result of unremitting and extensive research, I found in
a certain old book some theorems discussed which, although the
reverse of clear owing to errors and in many ways faulty as
regards the figures, nevertheless gave the substance of what I
sought, and moreover to some extent kept to the Doric dialect
affected by Archimedes, while they retained the names familiar in
old usage, the parabola being called a section of a right-angled
cone, and the hyperbola a section of an obtuse-angled cone;
whence I was led to consider whether these theorems might
not in fact be what he promised he would give at the end. For
this reason I paid them the closer attention, and, after finding
great difficulty with the actual text owing to the multitude of
the mistakes above referred to, I made out the sense gradually
and now proceed to set it out, as well as I can, in more familiar
and clearer language. And first the theorem will be treated
generally, in order that what Archimedes says about the limits
of possibility may be made clear; after which there will follow
the special application to the conditions stated in his analysis.
of the problem."
The investigation which follows may be thus reproduced.
The general problem is:
Given two straight lines AB, AC and an area D, to divide
AB at M so that
1
AM: AC=D: MB².
ON THE SPHERE AND CYLINDER II.
67
Analysis.
Suppose M found, and suppose AC placed at right angles to
AB. Join CM and produce it. Draw EBN through B parallel
to AC meeting CM in N, and through C draw CHE parallel to
AB meeting EBN in E. Complete the parallelogram CENF,
and through M draw PMH parallel to AC meeting FN in P.
Measure EL along EN so that
CE. EL (or AB. EL) = D.
Then, by hypothesis,
AM: ACCE. EL: MB2.
P

N
And
AM: AC CE: EN,
A
B
M
by similar triangles,
=CE.EL:EL.EN.
C
H
It follows that PN2 = MB" = EL.EN.
E
Hence, if a parabola be described with vertex E, axis EN, and
parameter equal to EL, it will pass through P; and it will be
given in position, since EL is given.
Therefore P lies on a given parabola.
Next, since the rectangles FH, AE are equal,
FP.PH = AB.BE.
Hence, if a rectangular hyperbola be described with CE, CF
as asymptotes and passing through B, it will pass through P.
And the hyperbola is given in position.
Therefore P lies on a given hyperbola.
Thus P is determined as the intersection of the parabola
and hyperbola. And since P is thus given, M is also given.
διορισμός.
Now, since
AM: AC=D: MB²,
AM. MB² = AC.D.
But AC. D is given, and it will be proved later that the maximum
value of AM.MB² is that which it assumes when BM=2AM.
5-2
68
ARCHIMEDES
Hence it is a necessary condition of the possibility of a
solution that AC. D must not be greater than AB.(§AB)², or
AB².
Synthesis.
If O be such a point on AB that BO=2A0, we have seen
that, in order that the solution may be possible,
AC.DAO.OB².
Thus AC.D is either equal to, or less than, AO.OB².
(1) If AC.D=AO. OB³, then the point O itself solves the
problem.
(2) Let AC.D be less than AO.OB².
Place AC at right angles to AB. Join CO, and produce it
to R. Draw EBR through B parallel to AC meeting CO in R,
and through C draw CE parallel
to AB meeting EBR in E. Com-
plete the parallelogram CERF,
and through O draw QOK parallel
to AC meeting FR in Q and CE
in K.

F
Q Q'
R
P
G
N
A
B
M
Then, since
AC.D< AO.OB²,
C
K
H
E
measure RQ' along RQ so that
AC.D=AO.QʻR²,
AO: AC=D: QʻR².
or
Measure EL along ER so that
Now, since
and
it follows that
D=CE. EL (or AB. EL).
AO: AC=D: Q'R², by hypothesis,
=CE.EL: Q'R²,
AO: AC CE: ER, by similar triangles,
2
=
=
CE.EL:EL.ER,
Q'R' = EL.ER.
ON THE SPHERE AND CYLINDER II.
69
Describe a parabola with vertex E, axis ER, and parameter
equal to EL. This parabola will then pass through Q'.
Again,
rect. FK: =rect. AE,
or
FQ.QKAB.BE;
and, if we describe a rectangular hyperbola with asymptotes
CE, CF and passing through B, it will also pass through Q.
Let the parabola and hyperbola intersect at P, and through
P draw PMH parallel to AC meeting AB in M and CE
in H, and GPN parallel to AB meeting CF in G and ER
in N.
Then shall M be the required point of division.
Since
PG.PH=AB.BE,
rect. GM = rect. ME,
and therefore CMN is a straight line.
Thus
AB.BE = PG. PHAM.EN
…………………………
(1).
Again, by the property of the parabola,
PN² = EL.EN,
or
=
MB2 EL. EN
·(2).
From (1) and (2)
or
Alternately,
or
AM: EL AB. BE: MB2,
=
AM.AB: AB.EL=AB.AC:MB².
AM. AB: AB. AC-AB. EL: MB2,
AM: AC=D: MB2.
Proof of διορισμός.
It remains to be proved that, if AB be divided at O so that
BO = 2A0, then AO. OB² is the maximum value of AM.MB²,
or
AO. OB²>AM.MB²,
where M is any point on AB other than 0.
710
ARCHIMEDES
F
Suppose that AO: AC CE. EL' : OB³,
so that
=
AO.OB² = CE.EL'.AC.
Join CO, and produce it to N;
draw EBN through B parallel
to AC, and complete the paral-
lelogram CENF.
Through O draw POH
parallel to AC meeting FN
in P and CE in H.

W
F
With vertex E, axis EN,
and parameter EL', describe
a parabola. This will pass
through P, as shown in the
analysis above, and beyond P G
will meet the diameter CF of
the parabola in some point.
Next draw a rectangular
hyperbola with asymptotes CE,
CF and passing through B.
This hyperbola will also pass
through P, as shown in the
analysis.
Produce NE to T so that
TE=EN. Join TP meeting
CE in Y, and produce it to
meet CF in W. Thus TP will
touch the parabola at P.
Then, since
A
N
P
g'
M
R
B
E
H
K
Y
T
BO = 2A0,
TP=2PW.
And
TP=2PY.
Therefore
PW=PY.
Since, then, WY between the asymptotes is bisected at P, the
point where it meets the hyperbola,
WY is a tangent to the hyperbola.
Hence the hyperbola and parabola, having a common tangent
at P, touch one another at P.
ON THE SPHERE AND CYLINDER II.
71
Now take any point M on AB, and through M draw QMK
parallel to AC meeting the hyperbola in Q and CE in K.
Lastly, draw GqQR through Q parallel to AB meeting CF in G,
the parabola in q, and EN in R.
Then, since, by the property of the hyperbola, the rectangles
GK, AE are equal, CMR is a straight line.
By the property of the parabola,
qR2= EL'. ER,
so that
Suppose
and we have
or
Therefore
QR² < EL'. ER.
QR² = EL. ER,
AM: ACCE: ER
= CE.EL: EL.ER
=
= CE. EL: QRª
=CE.EL: MB²,
AM.MB²=CE. EL.AC.
AM.MB² < CE.EL'.AC
< AO. OB².
If AC.D< AO. OB², there are two solutions because there
will be two points of intersection between the parabola and the
hyperbola.
For, if we draw with vertex E and axis EN a parabola
whose parameter is equal to EL, the parabola will pass through
the point Q (see the last figure); and, since the parabola meets
the diameter CF beyond Q, it must meet the hyperbola again
(which has CF for its asymptote).
[If we put AB≈α, BM −x, AC = c, and D= b², the pro-
portion
AM: AC=D: MB²
is seen to be equivalent to the equation
x² (ax) = b²c,
being a cubic equation with the term containing a omitted.
Now suppose EN, EC to be axes of coordinates, EN being
the axis of y.
72
ARCHIMEDES
Then the parabola used in the above solution is the
parabola
b2
•y,
a
and the rectangular hyperbola is
y (ax) = ac.
Thus the solution of the cubic equation and the conditions
under which there are no positive solutions, or one, or two
positive solutions are obtained by the use of the two conics.]
[For the sake of completeness, and for their intrinsic interest,
the solutions of the original problem in Prop. 4 given by
Dionysodorus and Diocles are here appended.
Dionysodorus' solution.
Let AA' be a diameter of the given sphere. It is required
to find a plane cutting AA' at right angles (in a point M,
suppose) so that the segments into which the sphere is divided
are in a given ratio, as CD: DE.
Produce A'A to F so that AF OA, where O is the centre
of the sphere.

K'
P
R
K
B
H
F
A
M
A'
G
B'
D
C
E
Draw AH perpendicular to AA' and of such length that
FA: AHCE: ED,
ON THE SPHERE AND CYLINDER II.
73
and produce AH to K so that
AK² = FA. AH….
.(a).
With vertex F, axis FA, and parameter equal to AH
describe a parabola. This will pass through K, by the equa-
tion (a).
Draw A'K' parallel to AK and meeting the parabola in K';
and with A'F, A'K' as asymptotes describe a rectangular
hyperbola passing through H. This hyperbola will meet the
parabola at some point, as P, between K and K'.
Draw PM perpendicular to AA' meeting the great circle in
B, B', and from H, P draw HL, PR both parallel to AA' and
meeting A'K' in L, R respectively.
i.e.
or
and
i.e.
or
Then, by the property of the hyperbola,
PR.PM = AH.HL,
PM.MA' HA. AA',
=
PM: AH-AA': A'M,
12
PM: AH² = AA”² : A'M².
Also, by the property of the parabola,
PM² = FM.AH,
FM:PM=PM : AH,
FM: AH= PM² : AH²
2
= AA" : A'M², from above.
Thus, since circles are to one another as the squares of their
radii, the cone whose base is the circle with A'M as radius and
whose height is equal to FM, and the cone whose base is the
circle with AA' as radius and whose height is equal to AH,
have their bases and heights reciprocally proportional.
Hence the cones are equal; i.e., if we denote the first cone
by the symbol c (A'M), FM, and so on,
c(A'M), FM = c(AA′), AH.
Now c(AA'), FA : c(AA'), AH=FA:AH
= CE : ED, by construction.
*
74
ARCHIMEDES
A
Therefore
But (1)
c(AA'), FA : c(A'M), FM = CE : ED ………………..(B).
c (AA′), FA = the sphere.
[I. 34]
(2) c(A'M), FM can be proved equal to the segment of
the sphere whose vertex is A' and height A'M.
For take G on AA' produced such that
GM : MA' – FM : MA
= OA+AM: AM.
Then the cone GBB′ is equal to the segment A'BB′ [Prop. 2].
And
Therefore
FM: MG=AM: MA', by hypothesis,
= BM2: A'M².
(circle with rad. BM): (circle with rad. A'M)
= FM: MG,
so that
c(A'M), FM = c (BM), MG
= the segment A'BB'.
We have therefore, from the equation (8) above,
(the sphere): (segmt. A'BB') = CE : ED,
whence (segmt. ABB') : (segmt. A'BB') = CD: DE.
Diocles' solution.
Diocles starts, like Archimedes, from the property, proved in
Prop. 2, that, if the plane of section cut a diameter AA' of the
sphere at right angles in M, and if H, H' be taken on OA, OA'
produced respectively so that
OA' + A'M : A'M = HM : MA,
OA+AM : AM = H'M : MA',
then the cones HBB', H'BB' are respectively equal to the
segments ABB', A'BB'.
ON THE SPHERE AND CYLINDER II.
75
Then, drawing the inference that
HA: AMOA': A'M,
H'A' : A'M = 0A : AM,

B
HA
M
O
A'
H'
B'
he proceeds to state the problem in the following form, slightly
generalising it by the substitution of any given straight line for
OA or ОA':
Given a straight line AA', its extremities A, A', a ratio C: D,
and another straight line as AK, to divide AA' at M and to find
two points H, H' on A'A and AA' produced respectively so that
the following relations may hold simultaneously,
Analysis.
C: D= HM: MH'
HA: AM = AK : A'M
H'A' : A'M = AK : AM
(a),
..(B),
(x).
Suppose the problem solved and the points M, H, H' all
found.
Place AK at right angles to AA', and draw A'K' parallel
and equal to AK. Join KM, K'M, and produce them to meet
K'A', KA respectively in E, F. Join KK', draw EG through
E parallel to A'A meeting KF in G, and through M draw QMN
parallel to AK meeting EG in Q and KK' in N.
Now
whence
Similarly
Next,
HA : AM = A'K' : A'M, by (B),
= FA: AM, by similar triangles,
HA=FA.
H'A' A'E.
FA+AM : A'K' + A'M = AM : A'M
=AK+AM : EA' + A'M, by similar triangles.
76
ARCHIMEDES
Therefore
(FA + AM). (EA' + A'M) = (KA + AM). (K'A' + A'M).
Take AR along AH and A'R' along A'H' such that
AR=A'R' = AK.
Then, since FA+AM=HM, EA' + A'M = MH', we have
HM.MH' = RM.MR'
·(8).
(Thus, if R falls between A and H, R' falls on the side of H'
remote from A', and vice versa.)

G
F
PI
E
R
H-
A
M
Α'
P
V
K
N
K
主
​R'
H'
Now
C : D=HM: MH', by hypothesis,
= HM.MH' : MH'
= RM. MR': MH", by (8).
=
Measure MV along MN so that MV A'M. Join A'V and
produce it both ways. Draw RP, R'P' perpendicular to RR'
meeting A'V produced in P, P' respectively. Then, the angle
MA'V being half a right angle, PP' is given in position, and,
since R, R' are given, so are P, P'.
And, by parallels,
P'V: PV = R'M : MR.
ON THE SPHERE AND CYLINDER II.
77
Therefore
But
Therefore
PV.P'V: PV²= RM. MR' : RM².
PV² = 2RM².
And it was shown that
PV. P'V = 2RM.MR'.
RM.MR': MH" = C: D.
But
Hence
Therefore
PV.P'V: MH2 = 20: D.
MH' = A'M + A'E = VM + MQ = QV.
QV² : PV.P'V=D: 2C, a given ratio.
Thus, if we take a line p such that
D: 2Cp: PP'*,
and if we describe an ellipse with PP' as a diameter and p as
the corresponding parameter [= DD/PP' in the ordinary
notation of geometrical conics], and such that the ordinates to
PP' are inclined to it at an angle equal to half a right angle,
i.e. are parallel to QV or AK, then the ellipse will pass
through Q.
Hence Q lies on an ellipse given in position.
Again, since EK is a diagonal of the parallelogram GK',
GQ. QN = AA'. A'K'.
If therefore a rectangular hyperbola be described with KG,
KK' as asymptotes and passing through A', it will also pass
through Q.
Hence Q lies on a given rectangular hyperbola.
Thus Q is determined as the intersection of a given ellipse
* There is a mistake in the Greek text here which seems to have escaped the
notice of all the editors up to the present. The words are éàv äpa toińowμev, ws
τὴν Δ πρὸς τὴν διπλασίαν τῆς Γ, οὕτως τὴν ΤΥ πρὸς ἄλλην τινὰ ὡς τὴν Φ, i.e. (with
the lettering above) "If we take a length p such that D : 2C = PP' : p." This
cannot be right, because we should then have
QV2: PV. P'V = PP' : p,
whereas the two latter terms should be reversed, the correct property of the
ellipse being
QV²: PV. P'V=p: PP'.
[Apollonius I. 21]
The mistake would appear to have originated as far back as Eutocius, but I
think that Eutocius is more likely to have made the slip than Diocles himself,
because any intelligent mathematician would be more likely to make such a slip
in writing out another man's work than to overlook it if made by another.
"
*
78
ARCHIMEDES
L
and a given hyperbola, and is therefore given. Thus M is
given, and H, H' can at once be found.
Synthesis.
Place AA', AK at right angles, draw A'K' parallel and
equal to AK, and join KK'.
Make AR (measured along A'A produced) and A'R'
(measured along AA' produced) each equal to AK, and
through R, R' draw perpendiculars to RR'.
Then through A' draw PP' making an angle (AA’P) with
AA' equal to half a right angle and meeting the perpendiculars
just drawn in P, P' respectively.
Take a length p such that
D: 2C-p: PP'*,
and with PP' as diameter and p as the corresponding parameter
describe an ellipse such that the ordinates to PP' are inclined
to it at an angle equal to AA'P, i.e. are parallel to AK.
With asymptotes KA, KK' draw a rectangular hyperbola
passing through A'.
Let the hyperbola and ellipse meet in Q, and from Q draw
QMVN perpendicular to AA' meeting AA' in M, PP' in V
and KK' in N. Also draw GQE parallel to AA' meeting AK,
A'K' respectively in G, E.
Produce KA, K'M to meet in F.
Then, from the property of the hyperbola,
GQ.QN=AA'. A'K',
and, since these rectangles are equal, KME is a straight line.
Measure AH along AR equal to AF, and A'H' along A'R'
equal to A'E.
From the property of the ellipse,
QV²: PV.P'V=p: PP'
=D : 2C.
* Here too the Greek text repeats the same error as that noted on p. 77.
ON THE SPHERE AND CYLINDER II.
79
or
And, by parallels,
PV : P'V=RM : R'M,
PV.P'V:P'V² = RM.MR' : R'M²,
while P'V2R'M', since the angle RA'P is half a right
angle.
PV.P'V=2RM. MR',
Therefore
whence
QV2: 2RM. MR' =D: 2C.
But
QV=EA' + A'M = MH'.
Therefore
RM.MR': MH" = C: D.
Again, by similar triangles,
Therefore
FA+AM: K'A'+A'M=AM: A'M
=KA+AM: EA' + A'M.
or
(FA+AM).(EA' + A'M)=(KA+AM). (K'A' + A'M)
It follows that
or
Also
and
HM.MH' = RM.MR'.
HM.MH': MH" = C: D,
HM: MH'-C: D
HA:AM=FA : AM,
(a).
= A'K': A'M, by similar
triangles...(B),
H'A' : A'M = EA' : A'M
=AK:
ᎪᏦ : ᎪᎷ ....
(y).
Hence the points M, H, H' satisfy the three given
relations.]
Proposition 5. (Problem.)
To construct a segment of a sphere similar to one segment
and equal in volume to another.
Let ABB' be one segment whose vertex is A and whose
base is the circle on BB' as diameter; and let DEF be another
segment whose vertex is D and whose base is the circle on EF
80
ARCHIMEDES
as diameter. Let AA', DD' be diameters of the great circles
passing through BB', EF respectively, and let O, C be the
respective centres of the spheres.
Suppose it required to draw a segment similar to DEF and
equal in volume to ABB'.
Analysis. Suppose the problem solved, and let def be the
required segment, d being the vertex and ef the diameter of
the base. Let dd' be the diameter of the sphere which bisects
ef at right angles, c the centre of the sphere.
E
K
D
G
F
H
A
ФФФ



d'
R
B
M
A'
Let M, G, g be the points where BB', EF, ef are bisected
at right angles by AA', DD', dd' respectively, and produce OA,
CD, cd respectively to H, K, k, so that
OA' + A'M: A'M=HM: MA
CD' + D'G : D'G=KG:GD
cd'+d'g: d'g=kg: gd
and suppose cones formed with vertices H, K, k and with the
same bases as the respective segments. The cones will then be
equal to the segments respectively [Prop. 2].
Therefore, by hypothesis,
´the cone HBB' = the cone kef.
B'
ON THE SPHERE AND CYLINDER II.
81
Hence
(circle on diameter BB'): (circle on diameter ef)= kg: HM,
so that
BB" : ef² = kg : HM ....
(1).
But, since the
segments DEF, def are similar, so are the
Therefore
KG: EF-kg: ef.
cones KEF, kef
And the ratio KG: EF is given. Therefore the ratio kg : ef
is given.
Suppose a length R taken such that
Thus R is given.
kg : ef = HM ;R ………..
(2).
2
Again, since kg: HM=BB'² : ef² = ef : R, by (1) and (2),
suppose a length S taken such that
or
Thus
ef² = BB'. S,
BB" : ef² = BB': S.
BB' : ef = ef : S = S : R,
and ef, S are two mean proportionals in continued proportion
between BB', R.
Synthesis. Let ABB', DEF be great circles, AA', DD'
the diameters bisecting BB', EF at right angles in M, G
respectively, and O, C the centres.
Take H, K in the same way as before, and construct the
cones HBB', KEF, which are therefore equal to the respective
segments ABB', DEF.
Let R be a straight line such that
KG: EF=HM: R,
and between BB', R take two mean proportionals ef, S.
On ef as base describe a segment of a circle with vertex d
and similar to the segment of a circle DEF. Complete the
circle, and let dd' be the diameter through d, and c the centre.
Conceive a sphere constructed of which def is a great circle,
and through ef draw a plane at right angles to dd'.
H. A.
6
82
ARCHIMEDES
Then shall def be the required segment of a sphere.
For the segments DEF, def of the spheres are similar, like
the circular segments DEF, def.
Produce cd to k so that
cd'+d'g: d'g=kg: gd.
The cones KEF, kef are then similar.
Therefore
whence
kg: ef= KG: EF=HM: R,
kg: HM=ef: R.
But, since BB', ef, S, R are in continued proportion,
BB": ef² = BB' : S
=ef: R
= kg: HM.
Thus the bases of the cones HBB', kef are reciprocally
proportional to their heights. The cones are therefore equal,
and def is the segment required, being equal in volume to the
cone kef.
[Prop. 2]
Proposition 6. (Problem.)
Given two segments of spheres, to find a third segment of a
sphere similar to one of the given segments and having its
surface equal to that of the other.
Let ABB' be the segment to whose surface the surface of
the required segment is to be equal, ABA'B' the great circle
whose plane cuts the plane of the base of the segment ABB' at
right angles in BB'. Let AA' be the diameter which bisects
BB' at right angles.
Let DEF be the segment to which the required segment
is to be similar, DED'F the great circle cutting the base of the
segment at right angles in EF. Let DD' be the diameter
bisecting EF at right angles in G.
Suppose the problem solved, def being a segment similar
to DEF and having its surface equal to that of ABB'; and
ON THE SPHERE AND CYLINDER II.
83
complete the figure for def as for DEF, corresponding points
being denoted by small and capital letters respectively.
D

E
G
F
A
D'
d
B'
g
B
M
A'
Join AB, DF, df.
d'
Now, since the surfaces of the segments def, ABB' are equal,
so are the circles on df, AB as diameters ;
that is,
df = AB.
[I. 42, 43]
From the similarity of the segments DEF, def we obtain
and
whence
or
d'd: dg
=
D'D: DG,
dg: df = DG: DF;
d'd: df = D'D : DF,
d'd: ABD'D: DF.
therefore d'd is given.
But AB, D'D, DF are all given;
Accordingly the synthesis is as follows.
Take d'd such that
d'd: ABD'D: DF..
………..(1).
Describe a circle on d'd as diameter, and conceive a sphere
constructed of which this circle is a great circle.
6-2
84
ARCHIMEDES
Divide d'd at g so that
d'g: gd
=
D'G: GD,
and draw through g a plane perpendicular to d'd cutting off
the segment def of the sphere and intersecting the plane of the
great circle in ef. The segments def, DEF are thus similar,
dg : df = DG : DF.
and
But from above, componendo,
d'd: dg = D'D: DG.
Therefore, ex aequali,
d'd: df= D'D: DF,
whence, by (1), df = AB.
Therefore the segment def has its surface equal to the
surface of the segment ABB′ [I. 42, 43], while it is also similar
to the segment DEF.
Proposition 7. (Problem.)
From a given sphere to cut off a segment by a plane so that
the segment may have a given ratio to the cone which has the same
base as the segment and equal height.
Let AA′ be the diameter of a great circle of the sphere.
It is required to draw a plane at right angles to AA' cutting
off a segment, as ABB', such that the segment ABB' has to
the cone ABB' a given ratio.
Analysis.
Suppose the problem solved, and let the plane of section
cut the plane of the great circle in BB', and the diameter
AA' in M. Let O be the centre of the sphere.

8
D
A
3
B'
H
A
Produce OA to H so that
OA' + A'M : A'M = HM : MA……………….
་་་
.(1).
ON THE SPHERE AND CYLINDER II.
85
Thus the cone HBB' is equal to the segment ABB'. [Prop. 2]
Therefore the given ratio must be equal to the ratio of the
cone HBB' to the cone ABB', i.e. to the ratio HM : MA.
Hence the ratio OA'+A'M : A'M is given; and therefore
A'M is given.
διορισμός
Now
so that
OA': A'M>OA' : A'A,
OA' + A'M: A'M > OA' + A'A : A'A
> 3:2.
Thus, in order that a solution may be possible, it is a
necessary condition that the given ratio must be greater than
3:2.
The synthesis proceeds thus.
Let AA' be a diameter of a great circle of the sphere, O the
centre.
Take a line DE, and a point F on it, such that DE: EF is
equal to the given ratio, being greater than 3: 2.
Now, since
so that
OA' + A'A: A'A = 3 : 2,
3:2,
DE: EF >OA' + A'A : A'A,
DF: FE> OA': A'A.
Hence a point M can be found on AA' such that
=
DF: FE OA': A'M.
(2).
Through M draw a plane at right angles to AA' intersecting
the plane of the great circle in BB', and cutting off from the
sphere the segment ABB'.
As before, take H on OA produced such that
OA' + A'M : A'M = HM : MA.
Therefore HM: MA – DE : EF, by means of (2).
=
It follows that the cone HBB', or the segment ABB', is to
the cone ABB' in the given ratio DE: EF.
1
86
ARCHIMEDES
1
Proposition 8.
If a sphere be cut by a plane not passing through the centre
into two segments A’BB′, ABB′, of which A'BB' is the greater,
then the ratio
(segmt. A'BB'): (segmt. ABB')
< (surface of A'BB')² : (surface of ABB′)²
but > (surface of A'BB')³ : (surface of ABB′)**.
Let the plane of section cut a great circle A'BAB' at right
angles in BB', and let AA' be the diameter bisecting BB' at
right angles in M.
Let O be the centre of the sphere.
Join A'B, AB.

B
H'
H
K
Α
MN
ΝΑ
B'
As usual, take H on OA produced, and H' on OA' produced,
so that
OA' + A'M : A'M = HM : MA….....
OA+AM: AM = H'M : MA'………...
…..(1),
(2),
and conceive cones drawn each with the same base as the two
segments and with apices H, H' respectively. The cones are
then respectively equal to the segments [Prop. 2], and they
are in the ratio of their heights HM, H'M.
Also
(surface of A'BB'): (surface of ABB') = A'B² : AB² [I. 42, 43]
- A'M : AM.
* This is expressed in Archimedes' phrase by saying that the greater seg-
ment has to the lesser a ratio "less than the duplicate (diλáocov) of that which
the surface of the greater segment has to the surface of the lesser, but greater
than the sesquialterate (nóλcov) [of that ratio]."
ON THE SPHERE AND CYLINDER II.
87
We have therefore to prove
(a) that
H'M : MH < A'M² : MA²,
(b) that
H'M: MH> A'M¹: MA¹.
(a) From (2) above,
A'M: AM = H'M : OA+AM
- H'A': OA', since OA = OA'.
Since A'M> AM, H'A' > OA'; therefore, if we take K on
H'A' so that OA' A'K, K will fall between H' and A'.
And, by (1),
Thus
Therefore
=
A'M : AM=KM : MH.
KM: MH = H'A': A'K, since A'K OA',
> H'M: MK.
H'M.MH < KM².
or
or
It follows that
H'M.MH: MH < KM²: MH²,
H'M :MH<KM² : MH²
< A'M² : AM", by (1).
(b) Since OA' = OA,
A'M. MA < A'0.0A,
A'M: OA'< 0A: AM
Therefore
<H'A': A'M, by means of (2).
A'M² <H'A'. OA'
< H'A'. A'K.
Take a point N on A'A such that
A'N² = H'A'. A'K.
Thus
Also
H'A' : A'K = A'N² : A'K”
..(3).
H'A': A'N = A'N : A'K,
and, componendo,
H'N: A'N = NK : A'K,
whence
A'N² : A'K² = H'N² : NK².
•
88
ARCHIMEDES
[
Therefore, by (3),
H'A': A'K=H'N²: NK².
Now
H'M: MK>H'N: NK.
Therefore
H'M²: MK2>H'A': A'K
> H'A': OA'
> A'M: MA, by (2), as above,
> OA' + A'M: MH, by (1),
> KM: MH.
Hence
H'M² : MH² = (H'M² : MK²). (KM² : MH²)
It follows that
>(KM : MH).(KM²: MH²).
H'M : MH>KM* : MH*
> A'M* : AM³, by (1).
[The text of Archimedes adds an alternative proof of this
proposition, which is here omitted because it is in fact neither
clearer nor shorter than the above.]
Proposition 9.
Of all segments of spheres which have equal surfaces the
hemisphere is the greatest in volume.
Let ABA'B' be a great circle of a sphere, AA′ being
a diameter, and O the centre. Let the sphere be cut by
a plane, not passing through O, perpendicular to AA' (at M),
and intersecting the plane of the great circle in BB'. The
segment ABB' may then be either less than a hemisphere as
in Fig. 1, or greater than a hemisphere as in Fig. 2.
Let DED'E' be a great circle of another sphere, DD'
being a diameter and C the centre. Let the sphere be cut by
a plane through C perpendicular to DD' and intersecting the
plane of the great circle in the diameter EE'.
ON THE SPHERE AND CYLINDER II.
89
Suppose the surfaces of the segment ABB' and of the
hemisphere DEE' to be equal.
K-
B

B
K
H
A'
MRO
A
B'
H
E

A'
о
R M
A
B'
Since the surfaces are equal, AB=DE.
Now, in Fig. 1, AB² >2AM² and <2A0²,
and, in Fig. 2,
AB² <2AM² and >2A0².
Hence, if R be taken on AA' such that
AR² =†AB²,
E'
F
C
D
[I. 42, 43]
R will fall between 0 and M.
Also, since AB²=DE², AR= CD.
Produce OA' to K so that OA' = A'K, and produce A'A to
H so that
A'K: A'M=HA: AM,
or, componendo, A'K+A'M: A'M=HM: MA….....
...(1).
[Prop. 2]
Thus the cone HBB' is equal to the segment ABB'.
Again, produce CD to F so that CD=DF, and the cone
FEE' will be equal to the hemisphere DEE'.
Now
and
AR.RA'>AM.MA',
AR² = †AB² = †AM. AA' = AM.A’K.
[Prop. 2]
}
90
ARCHIMEDES
}
+
or
Hence
AR. RA' + RA' > AM. MA' + AM. A'K,
AA'. AR>AM. MK
> HM. A'M, by (1).
Therefore
AA': A'M> HM: AR,
or
AB² : BM² > HM: AR,
i.e.
AR: BM2>HM: 2AR, since AB² = 2AR2,
> HM: CF.
Thus, since AR = CD, or CE,
(circle on diam. EE'): (circle on diam. BB')> HM: CF.
It follows that
(the cone FEE')> (the cone HBB'),
and therefore the hemisphere DEE' is greater in volume than
the segment ABB'.
>
MEASUREMENT OF A CIRCLE.
Proposition 1.
The area of any circle is equal to a right-angled triangle in
which one of the sides about the right angle is equal to the radius,
and the other to the circumference, of the circle.
Let ABCD be the given circle, K the triangle described.

T
A
G.
D
K
F
E
B

Then, if the circle is not equal to K, it must be either
greater or less.
I. If possible, let the circle be greater than. K.
Inscribe a square ABCD, bisect the arcs AB, BC, CD, DA,
then bisect (if necessary) the halves, and so on, until the sides
of the inscribed polygon whose angular points are the points of
division subtend segments whose sum is less than the excess of
the area of the circle over K.
92
ARCHIMEDES
Thus the area of the polygon is greater than K.
Let AE be any side of it, and ON the perpendicular on AE
from the centre 0.
Then ON is less than the radius of the circle and therefore
less than one of the sides about the right angle in K. Also the
perimeter of the polygon is less than the circumference of the
circle, i.e. less than the other side about the right angle in K.
Therefore the area of the polygon is less than K; which is
inconsistent with the hypothesis.
Thus the area of the circle is not greater than K.
II. If possible, let the circle be less than K.
Circumscribe a square, and let two adjacent sides, touching
the circle in E, H, meet in T. Bisect the arcs between adjacent
points of contact and draw the tangents at the points of
bisection. Let A be the middle point of the arc EH, and FAG
the tangent at A.
Then the angle TAG is a right angle.
Therefore
TG> GA
> GH
It follows that the triangle FTG is greater than half the area
TEAH.
Similarly, if the arc AH be bisected and the tangent at the
point of bisection be drawn, it will cut off from the area GAH
more than one-half.
Thus, by continuing the process, we shall ultimately arrive
at a circumscribed polygon such that the spaces intercepted
between it and the circle are together less than the excess of
K over the area of the circle.
Thus the area of the polygon will be less than K.
Now, since the perpendicular from O on any side of the
polygon is equal to the radius of the circle, while the perimeter
of the polygon is greater than the circumference of the circle,
it follows that the area of the polygon is greater than the
triangle K; which is impossible.
?
MEASUREMENT OF A CIRCLE.
93
Therefore the area of the circle is not less than K.
Since then the area of the circle is neither greater nor less
than K, it is equal to it.
Proposition 2.
The area of a circle is to the square on its diameter as 11
to 14.
[The text of this proposition is not satisfactory, and Archi-
medes cannot have placed it before Proposition 3, as the
approximation depends upon the result of that proposition.]
Proposition 3.
The ratio of the circumference of any circle to its diameter
is less than 34 but greater than 310.
[In view of the interesting questions arising out of the
arithmetical content of this proposition of Archimedes, it is
necessary, in reproducing it, to distinguish carefully the actual
steps set out in the text as we have it from the intermediate
steps (mostly supplied by Eutocius) which it is convenient to
put in for the purpose of making the proof easier to follow.
Accordingly all the steps not actually appearing in the text
have been enclosed in square brackets, in order that it may be
clearly seen how far Archimedes omits actual calculations and
only gives results. It will be observed that he gives two
fractional approximations to √3 (one being less and the other
greater than the real value) without any explanation as to how
he arrived at them; and in like manner approximations to the
square roots of several large numbers which are not complete
squares are merely stated. These various approximations and
the machinery of Greek arithmetic in general will be found
discussed in the Introduction, Chapter IV.]
I. Let AB be the diameter of any circle, O its centre, AC
the tangent at A; and let the angle AOC be one-third of a
right angle.
#
f
1
94
ARCHIMEDES
г
D
E
F
H
Then
and
in D.
OA : AC[=√3 : 1]> 265 : 153...
OC: CA [21]=306: 153...………………….
(1),
(2).
First, draw OD bisecting the angle AOC and meeting AC
Now
so that
CO: OA = CD: DA,
[CO+ OA : OA = CA: DA, or]
b.
[Eucl. VI. 3]
CO+OA: CA = OA: AD.
Therefore [by (1) and (2)] M
Hence
OA: AD> 571 : 153
2
OD²: AD² [=(OA² + AD²) : AD²
>(571² + 153²): 153²]
> 349450 : 23409,
so that
OD: DA > 591: 153
(3).
(4).

Secondly, let OE bisect the angle AOD, meeting AD in E.
[Then
so that
DO: OA=DE : EA,
DO+OA: DA= OA : AE.]
Therefore OA : AE[>(591 + 571) : 153, by (3) and (4)]
> 1162: 153.............(5).
B
MEASUREMENT OF A CIRCLE.
95
[It follows that
Thus
OE² : EA²> {(1162)²+153²) : 153²
> (135053483+23409): 23409
> 137394388: 23409.]
OE: EA > 1172½ : 153……….
(6).
Thirdly, let OF bisect the angle AOE and meet AE in F.
We thus obtain the result [corresponding to (3) and (5)
above] that
OA : AF[> (11623 +11721): 153]
> 23341: 153….....
[Therefore OF² : FA² > {(23341)²+153²} : 153²
·(7).
Thus
> 5472132: 23409.]
OF: FA > 23394: 153................
.(8).
Fourthly, let OG bisect the angle AOF, meeting AF in G.
We have then
OA: AG[>(23344 + 23391) : 153, by means of (7) and (8)]
> 46731 : 153.
Now the angle AOC, which is one-third of a right angle,
has been bisected four times, and it follows that
▲ AOG = (a right angle).
48
Make the angle AOH on the other side of OA equal to the
angle AOG, and let GA produced meet OH in H.
Then
▲ GOH = 14 (a right angle).
24
Thus GH is one side of a regular polygon of 96 sides cir-
cumscribed to the given circle.
And, since
while
it follows that
OA: AG > 4673: 153,
AB=20A, GH = 2AG,
AB: (perimeter of polygon of 96 sides) [>46731 : 153 × 96]
> 46731: 14688.
96
ARCHIMEDES
1
14688
667
But
3+
46731
4673/1/
[<3+
667
< 34.
46721
Therefore the circumference of the circle (being less than
the perimeter of the polygon) is a fortiori less than 34 times
the diameter AB.
II. Next let AB be the diameter of a circle, and let AC,
meeting the circle in C, make the angle CAB equal to one-third
of a right angle. Join BC.
Then
AC: CB[=√3 : 1]<1351 : 780.
First, let AD bisect the angle BAC and meet BC in d and
the circle in D. Join BD.
Then
▲ BAD = ▲ dAC
= <dBD,
and the angles at D, C are both right angles.
It follows that the triangles ADB, [ACd], BDd are similar.

or
F
G
B
E
D
d
Therefore
о
AD: DB=BD: Dd
[= AC: Cd]
= AB: Bd
= AB+ AC: Bd + Cd
=AB+ AC: BC
BA+AC : BC=AD : DB.
[Eucl. VI. 3]
MEASUREMENT OF A CIRCLE.
97
[But
while
Therefore
[Hence
Thus
AC: CB<1351 780, from above,
:
BA: BC 2:1
=1560: 780.]
AD: DB <2911: 780....
AB² : BD² < (2911²+780²): 780*
<9082321: 608400.]
AB: BD <3013: 780.......
..(1).
(2).
Secondly, let AE bisect the angle BAD, meeting the circle
in E; and let BE be joined.
Then we prove, in the same way as before, that
AE : EB[=BA + AD : BD
< (30132 + 2911): 780, by (1) and (2)]
<59243 780
<59242 × 4:780 ×
<1823 : 240
(3).
[Hence
AB2: BE² < (18232 + 2402): 2402
< 3380929: 57600.]
Therefore
AB: BE <1838, 240..........
.(4).
Thirdly, let AF bisect the angle BAE, meeting the circle
in F.
Thus
AF : FB[= BA + AE : BE
< 36619: 240, by (3) and (4)]
11
< 3661 × 11: 240 × 11
[It follows that
Therefore
< 1007: 66.......
AB² : BF¹² < (1007² + 66²) : 662
<1018405 : 4356.]
AB: BF < 1009½ : 66………………….
(5).
..(6).
Fourthly, let the angle BAF be bisected by AG meeting the
circle in G.
Then
H. A.
AG : GB [=BA + AF : BF]
< 20161 : 66, by (5) and (6).
7
98
ARCHIMEDES
[And
AB² ; BG² < {(20161)² + 66²} : 66²
< 4069284: 4356.]
Therefore
AB: BG
2017: 66,
whence
BG : AB > 66 : 20171.
(7).
[Now the angle BAG which is the result of the fourth bisection
of the angle BAC, or of one-third of a right angle, is equal to
one-fortyeighth of a right angle.
Thus the angle subtended by BG at the centre is
24
(a right angle).]
Therefore BG is a side of a regular inscribed polygon of 96
sides.
It follows from (7) that
(perimeter of polygon): AB [> 96 × 66 : 20171]
And
> 6336: 20171.
6336
> 379.
20171
Much more then is the circumference of the circle greater than
310 times the diameter.
Thus the ratio of the circumference to the diameter
< 34 but > 349.
ON
CONOIDS AND SPHEROIDS.
Introduction*.
"ARCHIMEDES to Dositheus greeting.
In this book I have set forth and send you the proofs of the
remaining theorems not included in what I sent you before, and
also of some others discovered later which, though I had often
tried to investigate them previously, I had failed to arrive at
because I found their discovery attended with some difficulty.
And this is why even the propositions themselves were not
published with the rest. But afterwards, when I had studied
them with greater care, I discovered what I had failed in
before.
Now the remainder of the earlier theorems were propositions
concerning the right-angled conoid [paraboloid of revolution];
but the discoveries which I have now added relate to an obtuse-
angled conoid [hyperboloid of revolution] and to spheroidal
figures, some of which I call oblong (πapaµáкea) and others flat
(ἐπιπλατέα).
I. Concerning the right-angled conoid it was laid down
that, if a section of a right-angled cone [a parabola] be made to
revolve about the diameter [axis] which remains fixed and
* The whole of this introductory matter, including the definitions, is trans-
lated literally from the Greek text in order that the terminology of Archimedes
may be faithfully represented. When this has once been set out, nothing will
be lost by returning to modern phraseology and notation. These will accordingly
be employed, as usual, when we come to the actual propositions of the treatise.
7-2
JorM
100
ARCHIMEDES
return to the position from which it started, the figure compre-
hended by the section of the right-angled cone is called a right-
angled conoid, and the diameter which has remained fixed
is called its axis, while its vertex is the point in which the
axis meets (äπTETαι) the surface of the conoid. And if a plane
touch the right-angled conoid, and another plane drawn parallel
to the tangent plane cut off a segment of the conoid, the base
of the segment cut off is defined as the portion intercepted by
the section of the conoid on the cutting plane, the vertex
[of the segment] as the point in which the first plane touches
the conoid, and the axis [of the segment] as the portion cut
off within the segment from the line drawn through the vertex
of the segment parallel to the axis of the conoid.
The questions propounded for consideration were
(1) why, if a segment of the right-angled conoid be cut off
by a plane at right angles to the axis, will the segment so cut
off be half as large again as the cone which has the same base
as the segment and the same axis, and
(2) why, if two segments be cut off from the right-angled
conoid by planes drawn in any manner, will the segments so cut
off have to one another the duplicate ratio of their axes.
II. Respecting the obtuse-angled conoid we lay down the
following premisses. If there be in a plane a section of an
obtuse-angled cone [a hyperbola], its diameter [axis], and the
nearest lines to the section of the obtuse-angled cone [i.e. the
asymptotes of the hyperbola], and if, the diameter [axis]
remaining fixed, the plane containing the aforesaid lines be
made to revolve about it and return to the position from which
it started, the nearest lines to the section of the obtuse-angled
cone [the asymptotes] will clearly comprehend an isosceles cone
whose vertex will be the point of concourse of the nearest lines
and whose axis will be the diameter [axis] which has remained
fixed. The figure comprehended by the section of the obtuse-
angled cone is called an obtuse-angled conoid [hyperboloid of
revolution], its axis is the diameter which has remained fixed,
and its vertex the point in which the axis meets the surface
ON CONOIDS AND SPHEROIDS.
101
of the conoid. The cone comprehended by the nearest lines to
the section of the obtuse-angled cone is called [the cone]
enveloping the conoid (περιέχων τὸ κωνοειδές), and the
straight line between the vertex of the conoid and the vertex
of the cone enveloping the conoid is called [the line] adjacent
to the axis (Tотeоûσа тâ ¤§ovi). And if a plane touch the
obtuse-angled conoid, and another plane drawn parallel to the
tangent plane cut off a segment of the conoid, the base of
the segment so cut off is defined as the portion intercepted by
the section of the conoid on the cutting plane, the vertex [of
the segment] as the point of contact of the plane which touches
the conoid, the axis [of the segment] as the portion cut off
within the segment from the line drawn through the vertex of
the segment and the vertex of the cone enveloping the conoid;
and the straight line between the said vertices is called
adjacent to the axis.
Right-angled conoids are all similar; but of obtuse-angled
conoids let those be called similar in which the cones enveloping
the conoids are similar.
The following questions are propounded for consideration,
(1) why, if a segment be cut off from the obtuse-angled
conoid by a plane at right angles to the axis, the segment so
cut off has to the cone which has the same base as the segment
and the same axis the ratio which the line equal to the sum
of the axis of the segment and three times the line adjacent
to the axis bears to the line equal to the sum of the axis of
the segment and twice the line adjacent to the axis, and
(2) why, if a segment of the obtuse-angled conoid be cut
off by a plane not at right angles to the axis, the segment so
cut off will bear to the figure which has the same base as
the segment and the same axis, being a segment of a cone *
(ἀπότμαμα κώνου), the ratio which the line equal to the sum
of the axis of the segment and three times the line adjacent
to the axis bears to the line equal to the sum of the axis of the
segment and twice the line adjacent to the axis.
* A segment of a cone is defined later (p. 104).
102
ARCHIMEDES
III. Concerning spheroidal figures we lay down the follow-
ing premisses. If a section of an acute-angled cone [ellipse] be
made to revolve about the greater diameter [major axis] which
remains fixed and return to the position from which it started,
the figure comprehended by the section of the acute-angled
cone is called an oblong spheroid (παραμᾶκες σφαιροειδές).
But if the section of the acute-angled cone revolve about the
lesser diameter [minor axis] which remains fixed and return
to the position from which it started, the figure comprehended
by the section of the acute-angled cone is called a flat spheroid
(ἐπιπλατὺ σφαιροειδές). In either of the spheroids the axis
is defined as the diameter [axis] which has remained fixed, the
vertex as the point in which the axis meets the surface of the
spheroid, the centre as the middle point of the axis, and the
diameter as the line drawn through the centre at right angles
to the axis. And, if parallel planes touch, without cutting,
either of the spheroidal figures, and if another plane be drawn
parallel to the tangent planes and cutting the spheroid, the
base of the resulting segments is defined as the portion inter-
cepted by the section of the spheroid on the cutting plane, their
vertices as the points in which the parallel planes touch the
spheroid, and their axes as the portions cut off within the
segments from the straight line joining their vertices. And
that the planes touching the spheroid meet its surface at one
point only, and that the straight line joining the points of
contact passes through the centre of the spheroid, we shall
prove. Those spheroidal figures are called similar in which
the axes have the same ratio to the 'diameters.' And let
segments of spheroidal figures and conoids be called similar if
they are cut off from similar figures and have their bases
similar, while their axes, being either at right angles to the
planes of the bases or making equal angles with the corre-
sponding diameters [axes] of the bases, have the same ratio
to one another as the corresponding diameters [axes] of the
bases.
The following questions about spheroids are propounded for
consideration,
(1) why, if one of the spheroidal figures be cut by a plane
ON CONOIDS AND SPHEROIDS.
103
through the centre at right angles to the axis, each of the
resulting segments will be double of the cone having the same
base as the segment and the same axis; while, if the plane of
section be at right angles to the axis without passing through
the centre, (a) the greater of the resulting segments will bear
to the cone which has the same base as the segment and the
same axis the ratio which the line equal to the sum of half the
straight line which is the axis of the spheroid and the axis of
the lesser segment bears to the axis of the lesser segment, and
(b) the lesser segment bears to the cone which has the same
base as the segment and the same axis the ratio which the line
equal to the sum of half the straight line which is the axis
of the spheroid and the axis of the greater segment bears to the
axis of the greater segment;
(2) why, if one of the spheroids be cut by a plane passing
through the centre but not at right angles to the axis, each of
the resulting segments will be double of the figure having the
same base as the segment and the same axis and consisting of a
segment of a cone*.
(3) But, if the plane cutting the spheroid be neither
through the centre nor at right angles to the axis, (a) the
greater of the resulting segments will have to the figure
which has the same base as the segment and the same axis
the ratio which the line equal to the sum of half the line
joining the vertices of the segments and the axis of the lesser
segment bears to the axis of the lesser segment, and (b) the
lesser segment will have to the figure with the same base
as the segment and the same axis the ratio which the line
equal to the sum of half the line joining the vertices of the
segments and the axis of the greater segment bears to the axis
of the greater segment. And the figure referred to is in these
cases also a segment of a cone*.
When the aforesaid theorems are proved, there are dis-
covered by means of them many theorems and problems.
Such, for example, are the theorems
(1) that similar spheroids and similar segments both of
* See the definition of a segment of a cone (áñóтµaµa kávov) on p. 104.
104
ARCHIMEDES
spheroidal figures and conoids have to one another the triplicate
ratio of their axes, and
(2) that in equal spheroidal figures the squares on the
'diameters' are reciprocally proportional to the axes, and, if in
spheroidal figures the squares on the 'diameters' are reciprocally
proportional to the axes, the spheroids are equal.
Such also is the problem, From a given spheroidal figure
or conoid to cut off a segment by a plane drawn parallel to a
given plane so that the segment cut off is equal to a given cone
or cylinder or to a given sphere.
After prefixing therefore the theorems and directions (èπɩ-
Táyμaтa) which are necessary for the proof of them, I will
then proceed to expound the propositions themselves to you.
Farewell.
DEFINITIONS.
If a cone be cut by a plane meeting all the sides [generators]
of the cone, the section will be either a circle or a section of an
acute-angled cone [an ellipse]. If then the section be a circle,
it is clear that the segment cut off from the cone towards the
same parts as the vertex of the cone will be a cone. But, if
the section be a section of an acute-angled cone [an ellipse], let
the figure cut off from the cone towards the same parts as the
vertex of the cone be called a segment of a cone. Let the
base of the segment be defined as the plane comprehended by
the section of the acute-angled cone, its vertex as the point
which is also the vertex of the cone, and its axis as the straight
line joining the vertex of the cone to the centre of the section
of the acute-angled cone.
And if a cylinder be cut by two parallel planes meeting all
the sides [generators] of the cylinder, the sections will be either
circles or sections of acute-angled cones [ellipses] equal and
similar to one another. If then the sections be circles, it is
clear that the figure cut off from the cylinder between the
parallel planes will be a cylinder. But, if the sections be
sections of acute-angled cones [ellipses], let the figure cut off
from the cylinder between the parallel planes be called a
frustum (TóμOs) of a cylinder. And let the bases of the
ON CONOIDS AND SPHEROIDS.
105
frustum be defined as the planes comprehended by the sections
of the acute-angled cones [ellipses], and the axis as the straight
line joining the centres of the sections of the acute-angled
cones, so that the axis will be in the same straight line with
the axis of the cylinder."
Lemma.
If in an ascending arithmetical progression consisting of the
magnitudes A1, A2, ... An the common difference be equal to the
least term A,, then
and
n. An< 2(A₂+ A₂+ ... + An),
> 2 (A1 + A2 + ... + An-1).
A₂+
[The proof of this is given incidentally in the treatise On
Spirals, Prop. 11. By placing lines side by side to represent
the terms of the progression and then producing each so as to
make it equal to the greatest term, Archimedes gives the
equivalent of the following proof.
If
we have also
And
Therefore
whence
and
Sn = A₂+A₂+ ... + An−1 + An,
2
Sn = An+ An-1 + An-2 + ... + A₁.
A₁+ An-1 A2 + An~2 = ... = An·
=
2Sn = (n+1) An,
n. An < 2Sn,
n. An > 2Sn-1"
Thus, if the progression is a, 2a, ... na,
and
but
Sn
n (n + 1)
2
α,
n'a < 2Sn
> 2Sn-1.]
Proposition 1.
2
If A1, B1, C1, ...K, and A2, B2, C2, ...K₂ be two series of
magnitudes such that
1
A₁ : B₁ = A½ : B2,
B₁ : C₁ = B₂ : C2, and so on
}
(a),
106
ARCHIMEDES
F
4
and if A3, B3, С3, ...K, and A4, B4, C₁, ...K₁ be two other series
such that
A₁: A₂ = A₂: A4,
3
2
B₁: B₁ = B₂ : B₁, and so on
}.....
then (A₁+B₁+ C₁ + ... + K₁) : (Ag + B3 + C₂+ ... + K3)
3
(B),
= (A₂ + B₂+ C₂ + ... + K₂): (A + B₁ + ... + K₁).
2
The proof is as follows.
Since
and
while
A3: A₁ = A₁: A2,
1
A₁: B₁ = A₂: B2,
B₁ : B3 = B₂ : B4,
we have, ex aequali,
Similarly
2
A₂ B₂ = A₁: B₁.
A3: B3
4.
B: C3 B₁ C4, and so on
=
Again, it follows from equations (a) that
1
2
A₁: A₁ = B₁: B₂ = C₁: C₂ = ....
2
Therefore
}
......
(7).
A₁ : A₂ = (A¸ + B₁ + C₁ + ... + K₁) : (A₂+ B₂ + ... + K₂),
1
2
2
or (A₁ + B₁ + C₁ + ... + K₁) : A₁ = (A₂ + B₂+ C₂ + ... + K₂) : A₂;
and
1
2
A₁: A¸= A₂: A4,
3
2
while from equations (y) it follows in like manner that
A3: (A3+ B3 + C3 + ... + Kg) = A4 : (A¸ + В₁ + С₁ + ... + K₁).
By the last three equations, ex aequali,
4
(A₁ + B₁ + C₁ + ... + K₁) : (A¸ + B3 + Сs + ... + Kg)
3
= (A₂+B₂ + C₂ + ... + K₂) : (A + B₁ + C₁ + ... + K₁).
4
COR. If any terms in the third and fourth series corre-
sponding to terms in the first and second be left out, the
result is the same. For example, if the last terms K,, K, are
absent,
(A₁ + B₁ + C₂ + ... + K₂): (A + B₂+ C₂+ ... + I3)
3
3
3
4
= (A₂+ B₂+ C₂ + ... + K₂) : (A + В₁ + С₁ + ... + I4),
where I immediately precedes K in each series.
ON CONOIDS. AND SPHEROIDS.
107
If A1,
arithmetic
to the lea
Lemma to Proposition 2.
3,
[On Spirals, Prop. 10.]
A2, A¸, ...An be n lines forming an ascending
progression in which the common difference is equal
41, then
(n + 1) An²+ Aï(A₁ + A₂+ Ag + ... + An)

2
3
2
2
= 3 (A₁²+A₂²+ Ag² + ... + An²).
2
1
A1 A2
An-sAn-2 An-1
An An-1 An-2
A3 A2 A1
1
Let the lines An, An-1, An-2, ...A, be placed in a row
from left to right. Produce An-1, An-2, …..A, until they are
each equal to An, so that the parts produced are respectively
equal to A1, A2, …..An-1.
Taking each line successively, we have
2An² = 2An²,
2
(A₁+ An−1)² = A₁² + A³n−1 + 2A1 . An−1,
2
(A2 + An-2)² = A½³ + A³n-2 + 2A,. An−2,
2
2
(An-1+ A₁)² = A²n−1 + A1² + 2An−1 · A₁.
108
ARCHIMEDES
:
L.
And, by addition,
2
2
2
(n + 1)An² = 2 (A₂² + A₂² + ... + A„²)
+2A₁. An-1+2A₂. An-2 + ... + 2An−1 · A₁.
.
Therefore, in order to obtain the required result, we have to
prove that
2(A₁.An−1+A2. An-2+...+An-1• A₁)+A₁(A₁+A₂+Ag+...+An)
= A₂² + A½² + ... + An²
+ An² ……………………………. (a).
=
2
2
Now 242. An-2= A₁. 4An-2, because A,
=
2A1,
2A3. An-3= A1. 6A-s, because A₁ = 341,
2An−1. A₁ = A₁. 2(n − 1)A¸.
It follows that
3
2(A₁.An-1+A2. An-2+...+An-1 · A₁) + A₁(A₁ + A₂+ ... + An)
1
2
+(2n − 1) A₂}.
A₁{An+3An−1 + 5An-2 + ... + (2n
And this last expression can be proved to be equal to
2
2
A₁² + A₂² + ... + An².
For
2
A = A1(n. An)
N
=
· A₁{An + (n − 1)An}
= A₁{An+2(An−1 + An~2 + ... + A₁)},
because (n - 1) An = An−1 + A1
+ An−2+ A₂
+
1
+ A₁ + An−1·
2
Similarly A³n-1 = A₁ {An−1 + 2(An~2 + An-s + ... + A₁)},
2
A₂² = A‚(A½+2A₁),
2
A₁² = A₁. A₁;
whence, by addition,
2
A¸² + A2 + Ag² + ... + An²
==
A₁{A
n+3An~1 + 5An−2 + ... + (2n − 1) A,}.
ON CONOIDS AND SPHEROIDS.
109
that
Thus the equation marked (a) above is true; and it follows
2
2
(n+1)An²+A₁(A₁+A₂+ ¸+...+An) = 3(A¸² + A¸²+...+An²).
COR. 1. From this it is evident that
Also
so that
and therefore
n. A„² <3(A‚²+ A2² + ... + An²) .
(1).
An² = A₁{An+2(An−1 + An-2+...+A₁)}, as above,
An² > A₁(An+An−1 + ... + A₁),
2
An²+A₁(A1+A₂+ ... + An) < 2A„².
It follows from the proposition that
n. An² >3(A‚²+A₂² + ... + A³n−1) ………….
(2).
COR. 2. All these results will hold if we substitute similar
figures for squares on all the lines; for similar figures are in the
duplicate ratio of their sides.
[In the above proposition the symbols A1, A2, …..An have
been used instead of a, 2a, 3a, ...na in order to exhibit the
geometrical character of the proof; but, if we now substitute
the latter terms in the various results, we have (1)
(n + 1) n²a³ + a (a + 2a + ... + na)
Therefore
Also (2)
and (3)
=3 {a²+(2a)² +(3a)* + ... + (na)³}.
a² + (2a)² + (3a)² + ... + (na)²
n (n
- a² {(n + 1) n² + " (" + 1)}
α
3
= a²
2
n (n + 1) (2n + 1)
6
n<3(1*+2*+3*+...+n*),
3
n³ > 3 (1² + 2² +32 + ... + n − 1).]
-
110
ARCHIMEDES
then
and
and
and
Proposition 2.
Iƒ A1, A2……. An be any number of areas such that*
A₁
= ax + x²,
1
A₂ = a. 2x + (2x)²,
Ag=a.3x + (3x)²,
An = a.nx + (nx)²,
n. An : (A1 + Á₂+ ... + An)<(a + nx) : ('
nx
+
3
α nx\
n. An : (A1 + A2 + ... + An−1) > (a + nx) : (2/
+
For, by the Lemma immediately preceding Prop. 1,
n. anx 2 (ax+ a. 2x+...+a.nx),
> 2 (ax+a. 2x + ... + a. n 1x).
+a.n−1x).
Also, by the Lemma preceding this proposition,
n. (nx)³ <3 {x² + (2x)² + (3x)* + ... + (nx)*}
>3 {x² + (2x)² + ... + (n − 1 x)²}.
Hence
an²x n(n)
3
+
2
3
<[(ax+x²)+{a. 2x + (2x)²} + ... + {a. nx+(nx)²}],
and
>[(ax+x²)+{a. 2x + (2x)²} + ... + {a. n − 1 x + (n− 1 x)²}],
an²x n(n)
or
and
+
2
3
1
< A₂+ A₂+ ... + An,
1
> A₂+ A₂+ ... + An-1.
It follows that
(an²x
1
2
n. An: (A₂+A₂+ ... + An) < n {a. nx+(nx)*} :·
n(n)
+
2
3
пх
or
2
n. An: (A₂+ A₂+ ... + An) < (a+nx):
+
;
2
3
α
also
n. An : (A1 + A2 + ... + An-1) > (a + nx) : ( ~ + næ²).
(A₁+A₂+
2 3
* The phraseology of Archimedes here is that associated with the traditional
method of application of areas: εἴ κα...παρ' ἑκάσταν αὐτῶν παραπέσῃ τι χωρίον
ὑπερβάλλον εἴδει τετραγώνῳ, "if to each of the lines there be applied a space
[rectangle] exceeding by a square figure." Thus A, is a rectangle of height x ap-
plied to a line a but overlapping it so that the base extends a distance x beyond a.
ON CONOIDS AND SPHEROIDS.
111
Proposition 3.
(1) If TP, TP' be two tangents to any conic meeting in T,
and if Qq, Q'q' be any two chords parallel respectively to TP,
TP' and meeting in O, then
(C
QO. Oq: Q'O. Oq′ = TP² : TP²².
And this is proved in the elements of conics*.”
(2) If QQ' be a chord of a parabola bisected in V by the
diameter PV, and if PV be of constant length, then the areas of
the triangle PQQ' and of the segment PQQ' are both constant
whatever be the direction of QQ'.
Let ABB' be the particular segment of the parabola whose
vertex is A, so that BB' is bisected perpendicularly by the axis
at the point H, where AH=PV.
Draw QD perpendicular to PV.

B
E P
D
ν
M' O
Pa P
TAW
H
M
B'
p
Pa
Let pa be the parameter of the principal ordinates, and let
be another line of such length that
QV² : QD²=p: Pa;
it will then follow that p is equal to the parameter of the ordi-
nates to the diameter PV, i.e. those which are parallel to QV.
* i.e. in the treatises on conics by Aristaeus and Euclid.
112
ARCHIMEDES
"For this is proved in the conics*."
Thus
QV² = p.PV.
And BH² = pa.AH, while AH=PV.
Therefore
QV²: BH²=p: Pa·
1-
But
hence
Thus
and therefore
QV² : QD"=p : Pa;
BH = QD.
BH.AH=QD.PV,
ΔΑΒΒ' = ΔΡQQ';
that is, the area of the triangle PQQ' is constant so long as PV
is of constant length.
Hence also the area of the segment PQQ' is constant under
the same conditions; for the segment is equal to APQQ'.
[Quadrature of the Parabola, Prop. 17 or 24.]
* The theorem which is here assumed by Archimedes as known can be
proved in various ways.
(1) It is easily deduced from Apollonius I. 49 (cf. Apollonius of Perga,
pp. liii, 39). If in the figure the tangents at A and P be drawn, the former
meeting PV in E, and the latter meeting the axis in T, and if AE, PT meet
at C, the proposition of Apollonius is to the effect that
CP:PE=p:2PT,
where Ρ is the parameter of the ordinates to PV.
(2) It may be proved independently as follows,
Let QQ' meet the axis in O, and let QM, QʻʼM', PN be ordinates to the axis.
Then
AM: AM'=QM²: Q'M²²=OM² : OM¹²,
whence
so that
That is to say,
or
AM: MM'-OM²: OM² – OM'
2
=0M²: (OM – OM'). MM',
OM²=AM. (OM – OM').
(AM-AO)2=AM. (AM+AM' -2A0),
A02 AM.AM'.
And, since QM²=Pa. AM, and Q'M'²= Pa. AM',
it follows that
Now
QM.Q'M'=Pa.AO.
:
QM+Q'M' \ ²
2
QV² : QD²=QV² ; ( QM
?M')
\
+QM. Q'M'
=QV²: (QM - Q'M' )²
2
=QV²: (PN²+QM. QʻM')
=p.PV: Pa. (AN+40), by (a).
PV=TO=AN+AO.
But
Therefore
QV²: QD²=p: Pa•
......
(a).
ON CONOIDS AND SPHEROIDS.
113
趣
​Proposition 4.
The area of any ellipse is to that of the auxiliary circle as
the minor axis to the major.
Let AA' be the major and BB' the minor axis of the
ellipse, and let BB' meet the auxiliary circle in b, b'.
Suppose to be such a circle that
(circle AbA'b') : 0 – CA : CB.
Then shall O be equal to the area of the ellipse.
For, if not, O must be either greater or less than the
ellipse.
I. If possible, let O be greater than the ellipse.
We can then inscribe in the circle O an equilateral polygon
of 4n sides such that its area is greater than that of the ellipse.
[cf. On the Sphere and Cylinder, I. 6.]
f
b


B
F
G
h
e
E
H
A
M
N
C
B'
A'
b'
Let this be done, and inscribe in the auxiliary circle of the
ellipse the polygon AefbghA'... similar to that inscribed in 0.
Let the perpendiculars eM, ƒN,... on AA′ meet the ellipse in
E, F,... respectively. Join AE, EF, FB,……….
Suppose that P' denotes the area of the polygon inscribed
in the auxiliary circle, and P that of the polygon inscribed in
the ellipse.
H. A.
8
114
ARCHIMEDES
Then, since all the lines eM, ƒN,... are cut in the same
proportions at E, F,…….,
i.e.
eM: EM=ƒN: FN... = bc: BC,
eM:EM=fN:FN=..
the pairs of triangles, as eAM, EAM, and the pairs of trapeziums,
as eMNf, EMNF, are all in the same ratio to one another
as bC to BC, or as CA to CB.
Therefore, by addition,
P': P-CA: CB.
Now P': (polygon inscribed in 0)
=
= (circle AbA'b') : 0
=
CA: CB, by hypothesis.
Therefore P is equal to the polygon inscribed in 0.
But this is impossible, because the latter polygon is by
hypothesis greater than the ellipse, and a fortiori greater
than P.
Hence O is not greater than the ellipse.
II. If possible, let O be less than the ellipse.
In this case we inscribe in the ellipse a polygon P with 4n
equal sides such that P>0.
Let the perpendiculars from the angular points on the
axis AA' be produced to meet the auxiliary circle, and let the
corresponding polygon (P') in the circle be formed.
Inscribe in O a polygon similar to P'.
Then
P': P-CA: CB
=
· (circle AbA′b′): O, by hypothesis,
- P': (polygon inscribed in O).
Therefore the polygon inscribed in O is equal to the
polygon P; which is impossible, because P>0.
Hence 0, being neither greater nor less than the ellipse, is
equal to it; and the required result follows.
ON CONOIDS AND SPHEROIDS.
115
Proposition 5.
If AA', BB' be the major and minor axis of an ellipse
respectively, and if d be the diameter of any circle, then
(area of ellipse): (area of circle) = AA'. BB' : d².
For
(area of ellipse): (area of auxiliary circle) = BB': AA' [Prop. 4]
· AA'. BB' : AA”.
And
=
(area of aux. circle): (area of circle with diam. d) AA": d².
Therefore the required result follows ex aequali.
Proposition 6.
The areas of ellipses are as the rectangles under their axes.
This follows at once from Props. 4, 5.
COR. The areas of similar ellipses are as the squares of
corresponding axes.
Proposition 7.
Given an ellipse with centre C, and a line CO drawn per-
pendicular to its plane, it is possible to find a circular cone
with vertex O and such that the given ellipse is a section of it
[or, in other words, to find the circular sections of the cone with
vertex O passing through the circumference of the ellipse].
Conceive an ellipse with BB' as its minor axis and lying in
a plane perpendicular to that of the paper. Let CO be drawn
perpendicular to the plane of the ellipse, and let O be the
vertex of the required cone. Produce OB, OC, OB', and in the
same plane with them draw BED meeting OC, OB' produced
in E, D respectively and in such a direction that
BE. ED : E0² = CA² : CO²,
where CA is half the major axis of the ellipse.
8-2
116
ARCHIMEDES
And this is possible, since
BE.ED: E0"> BC. CB': CO2."
[Both the construction and this proposition are assumed as
known.]

H
F
A
B
B'
CN
E
G
Q
K
M
D
Now conceive a circle with BD as diameter lying in a plane
at right angles to that of the paper, and describe a cone with
this circle for its base and with vertex 0.
We have therefore to prove that the given ellipse is a
section of the cone, or, if P be any point on the ellipse, that P
lies on the surface of the cone.
Draw PN perpendicular to BB'. Join ON and produce it
to meet BD in M, and let MQ be drawn in the plane of the
circle on BD as diameter perpendicular to BD and meeting the
circle in Q. Also let FG, HK be drawn through E, M respec-
tively parallel to BB'.
We have then
QM²: HM.MK = BM. MD: HM.MK
= BE. ED: FE. EG
=
(BE.ED : E0²).(E0²:FE.EG)
(CA² : CO²). (CO² : BC .CB')
= CA² : CB2
= PN2: BN.NB'.
ON CONOIDS AND SPHEROIDS.
117
Therefore QM² : PN² – HM.MK: BN.NB'
=
= OM2: ON²;
whence, since PN, QM are parallel, OPQ is a straight line.
But Q is on the circumference of the circle on BD as
diameter; therefore OQ is a generator of the cone, and hence
P lies on the cone.
Thus the cone passes through all points on the ellipse.
Proposition 8.
Given an ellipse, a plane through one of its axes AA' and
perpendicular to the plane of the ellipse, and a line CO drawn
from C, the centre, in the given plane through AA' but not
perpendicular to AA', it is possible to find a cone with vertex O
such that the given ellipse is a section of it [or, in other words,
to find the circular sections of the cone with vertex O whose
surface passes through the circumference of the ellipse].
By hypothesis, OA, OA' are unequal. Produce OA' to D so
that OA=OD. Join AD, and draw FG through C parallel to it.
The given ellipse is to be supposed to lie in a plane per-
pendicular to the plane of the paper. Let BB' be the other
axis of the ellipse.
Conceive a plane through AD perpendicular to the plane
of the paper, and in it describe either (a), if CB² = FC. CG, a
circle with diameter AD, or (b), if not, an ellipse on AD as
axis such that, if d be the other axis,
d2: AD² = CB2: FC. CG.
Take a cone with vertex O whose surface passes through
the circle or ellipse just drawn. This is possible even when the
curve is an ellipse, because the line from 0 to the middle point
of AD is perpendicular to the plane of the ellipse, and the
construction is effected by means of Prop. 7.
Let P be any point on the given ellipse, and we have only
to prove that P lies on the surface of the cone so described.
L
* A
118
ARCHIMEDES
Draw PN perpendicular to AA'. Join ON, and produce it
to meet AD in M. Through M draw HK parallel to A´A.

F
A'
N
H
Q
D
M
B
G
A
K
Lastly, draw MQ perpendicular to the plane of the paper
(and therefore perpendicular to both HK and AD) meeting the
ellipse or circle about AD (and therefore the surface of the cone)
in Q.
Then
QM² : HM.MK = (QM²: DM. MA). (DM. MA: HM.MK)
= (d² : AD³). (FC.CG : A'C.CA)
=
(CB2: FC. CG). (FC. CG: A'C.CA)
= CB2: CA"
= PN² : A'N.NA.
Therefore, alternately,
QM² : PN² – HM.MK: A'N.NA
=
2
= OM² : ON².
Thus, since PN, QM are parallel, OPQ is a straight line;
and, Q being on the surface of the cone, it follows that P is also
on the surface of the cone.
Similarly all points on the ellipse are also on the cone, and
the ellipse is therefore a section of the cone.
ON CONOIDS AND SPHEROIDS.
119
Proposition 9.
Given an ellipse, a plane through one of its axes and perpen-
dicular to that of the ellipse, and a straight line CO drawn from
the centre C of the ellipse in the given plane through the axis but
not perpendicular to that axis, it is possible to find a cylinder
with axis OC such that the ellipse is a section of it [or, in other
words, to find the circular sections of the cylinder with axis OC
whose surface passes through the circumference of the given
ellipse].
Let AA' be an axis of the ellipse, and suppose the plane
of the ellipse to be perpendicular to that of the paper, so that
OC lies in the plane of the paper.

D
B
M'
O
M
E
P
A
C
N
Draw AD, A'E parallel to CO, and let DE be the line
through O perpendicular to both AD and A'E.
We have now three different cases according as the other
axis BB' of the ellipse is (1) equal to, (2) greater than, or
(3) less than, DE.
(1) Suppose BB' – DE.
=
Draw a plane through DE at right angles to OC, and in
this plane describe a circle on DE as diameter. Through this
circle describe a cylinder with axis OC.
This cylinder shall be the cylinder required, or its surface
shall pass through every point P of the ellipse.
For, if P be any point on the ellipse, draw PN perpendicular
to AA'; through N draw NM parallel to CO meeting DE in
M, and through M, in the plane of the circle on DE as diameter,
draw MQ perpendicular to DE, meeting the circle in Q.
J
120
Then, since
And
Therefore
since AD, NM, CO, A'E are parallel.
PN² = DM.ME
=
= QM²,
PN²: AN.NA' = DO² : AC. CA'.
DM.ME: AN. NA' = DO² : AC²,
ARCHIMEDES
DE = BB',
190
عد
не
~M
An
NA!
AN
by the property of the circle.
Hence, since PN, QM are equal as well as parallel, PQ is
parallel to MN and therefore to CO. It follows that PQ is a
generator of the cylinder, whose surface accordingly passes
through P.
(2) If BB'> DE, we take E' on A'E such that DE' = BB'
and describe a circle on DE' as diameter in a plane perpen-
dicular to that of the paper; and the rest of the construction
and proof is exactly similar to those given for case (1).
(3) Suppose BB' < DE.
Take a point K on CO produced such that
DO² - CB² = 0K”.
From K draw KR perpendicular to the plane of the paper
and equal to CB.
Thus
2
OR² = OK² + CB² = OD².
D
P
R
K
H
M
E
A
C
N
Α
In the plane containing DE, OR describe a circle on DE as
diameter. Through this circle (which must pass through R)
draw a cylinder with axis OC.
1

ON CONOIDS AND SPHEROIDS.
121
We have then to prove that, if P be any point on the given
ellipse, P lies on the cylinder so described.
Draw PN perpendicular to AA', and through N draw NM
parallel to CO meeting DE in M. In the plane of the circle on
DE as diameter draw MQ perpendicular to DE and meeting
the circle in Q.
Lastly, draw QH perpendicular to NM produced. QH will
then be perpendicular to the plane containing AC, DE, i.e. the
plane of the paper.
Now
QH² : QM² = KR2 : OR², by similar triangles.
And QM² AN. NA' DM. ME: AN. NA'
=
=OD2 CA.
Hence, ex aequali, since OR = OD,
Thus QH-PN.
QH": AN. NA' KR: CA'
AN.NA'
=
= CB² : CA2
2
= PN² : AN. NA'.
And QH, PN are also parallel. Accordingly
PQ is parallel to MN, and therefore to CO, so that PQ is a
generator, and the cylinder passes through P.
Proposition 10.
It was proved by the earlier geometers that any two cones
have to one another the ratio compounded of the ratios of their
bases and of their heights*. The same method of proof will
show that any segments of cones have to one another the ratio
compounded of the ratios of their bases and of their heights.
The proposition that any 'frustum' of a cylinder is triple
of the conical segment which has the same base as the frustum
and equal height is also proved in the same manner as the
proposition that the cylinder is triple of the cone which has
the same base as the cylinder and equal height†.
* This follows from Eucl. XII. 11 and 14 taken together. Cf. On the Sphere
and Cylinder 1, Lemma 1.
+ This proposition was proved by Eudoxus, as stated in the preface to On
the Sphere and Cylinder 1. Cf. Eucl. XII. 10.
122
ARCHIMEDES
Proposition 11.
(1) If a paraboloid of revolution be cut by a plane through,
or parallel to, the axis, the section will be a parabola equal to the
original parabola which by its revolution generates the paraboloid.
And the axis of the section will be the intersection between the
cutting plane and the plane through the axis of the paraboloid
at right angles to the cutting plane.
If the paraboloid be cut by a plane at right angles to its
axis, the section will be a circle whose centre is on the axis.
(2) If a hyperboloid of revolution be cut by a plane through
the axis, parallel to the axis, or through the centre, the section
will be a hyperbola, (a) if the section be through the axis, equal,
(b) if parallel to the axis, similar, (c) if through the centre,
not similar, to the original hyperbola which by its revolution
generates the hyperboloid. And the axis of the section will be
the intersection of the cutting plane and the plane through the
axis of the hyperboloid at right angles to the cutting plane.
Any section of the hyperboloid by a plane at right angles to
the axis will be a circle whose centre is on the axis.
(3) If any of the spheroidal figures be cut by a plane through
the axis or parallel to the axis, the section will be an ellipse,
(a) if the section be through the axis, equal, (b) if parallel to the
axis, similar, to the ellipse which by its revolution generates the
figure. And the axis of the section will be the intersection of the
cutting plane and the plane through the axis of the spheroid
at right angles to the cutting plane.
If the section be by a plane at right angles to the axis of the
spheroid, it will be a circle whose centre is on the axis.
(4) If any of the said figures be cut by a plane through the
axis, and if a perpendicular be drawn to the plane of section
from any point on the surface of the figure but not on the section,
that perpendicular will fall within the section.
"And the proofs of all these propositions are evident.”*
* Cf. the Introduction, chapter 111. § 4.
ON CONOIDS AND SPHEROIDS.
123
Proposition 12.
If a paraboloid of revolution be cut by a plane neither parallel
nor perpendicular to the axis, and if the plane through the axis
perpendicular to the cutting plane intersect it in a straight line
of which the portion intercepted within the paraboloid is RR',
the section of the paraboloid will be an ellipse whose major axis
is RR' and whose minor axis is equal to the perpendicular
distance between the lines through R, R' parallel to the axis
of the paraboloid.
Suppose the cutting plane to be perpendicular to the plane
of the paper, and let the latter be the plane through the axis
ANF of the paraboloid which intersects the cutting plane at
right angles in RR'. Let RH be parallel to the axis of the
paraboloid, and R'H perpendicular to RH.
Let Q be any point on the section made by the cutting
plane, and from Q draw QM perpendicular to RR'. QM will
therefore be perpendicular to the plane of the paper.
Through M draw DMFE perpendicular to the axis ANF
meeting the parabolic section made by the plane of the paper
in D, E. Then QM is perpendicular to DE, and, if a plane be
drawn through DE, QM, it will be perpendicular to the axis
and will cut the paraboloid in a circular section.
R2
AN
T

P
H
E
F
LL.
D
M
R
Since Q is on this circle,
QM² = DM.ME:
Again, if PT be that tangent to the parabolic section in the
124
ARCHIMEDES
plane of the paper which is parallel to RR', and if the tangent
at A meet PT in O, then, from the property of the parabola,
DM.ME: RM. MR' = A02: OP2
[Prop. 3 (1)]
Therefore
2
= A0²: OT², since ANAT.
QM² : RM.MR′ = A0² : OT²
= R'H': RR",
by similar triangles.
Hence Qlies on an ellipse whose major axis is RR' and
whose minor axis is equal to R'H.
Propositions 13, 14.
If a hyperboloid of revolution be cut by a plane meeting all
the generators of the enveloping cone, or if an 'oblong' spheroid
be cut by a plane not perpendicular to the axis*, and if a plane
through the axis intersect the cutting plane at right angles in a
straight line on which the hyperboloid or spheroid intercepts
a length RR', then the section by the cutting plane will be an
ellipse whose major axis is RR'.
Suppose the cutting plane to be at right angles to the
plane of the paper, and suppose the latter plane to be that
T
A
P
K'
N
R'
T

A
R'
N

E
D
F M
B
D
C
BR
E
IF
M
R
K
* Archimedes begins Prop. 14 for the spheroid with the remark that, when the
cutting plane passes through or is parallel to the axis, the case is clear (òîλov).
Cf. Prop. 11 (3).
ON CONOIDS AND SPHEROIDS.
125
through the axis ANF which intersects the cutting plane
at right angles in RR'. The section of the hyperboloid or
spheroid by the plane of the paper is thus a hyperbola or ellipse
having ANF for its transverse or major axis.
Take any point on the section made by the cutting plane,
as Q, and draw QM perpendicular to RR'. QM will then
be perpendicular to the plane of the paper.
Through M draw DFE at right angles to the axis ANF
meeting the hyperbola or ellipse in D, E; and through QM,
DE let a plane be described. This plane will accordingly be
perpendicular to the axis and will cut the hyperboloid or
spheroid in a circular section.
Thus
QM² = DM.ME.
Let PT be that tangent to the hyperbola or ellipse which
is parallel to RR', and let the tangent at A meet PT in 0.
or
Then, by the property of the hyperbola or ellipse,
2
DM.ME: RM.MR' = OA² : OP²,
2
QM: RM. MR' = OA²: OP².
Now (1) in the hyperbola OA < OP, because AT< AN*, and
accordingly OT<OP, while OA < OT,
(2) in the ellipse, if KK' be the diameter parallel to RR',
and BB' the minor axis,
BC.CB': KC.CK'
OA: OP²;
and BC.CB' < KC. CK', so that OA<OP.
Hence in both cases the locus of Q is an ellipse whose major
axis is RR'.
COR. 1. If the spheroid be a 'flat' spheroid, the section will
be an ellipse, and everything will proceed as before except that
RR' will in this case be the minor axis.
COR. 2. In all conoids or spheroids parallel sections will be
similar, since the ratio OA2: OP" is the same for all the
parallel sections.
OA²
* With reference to this assumption cf. the Introduction, chapter III. § 3.
126
ARCHIMEDES
1
Proposition 15.
(1) If from any point on the surface of a conoid a line be
drawn, in the case of the paraboloid, parallel to the axis, and, in
the case of the hyperboloid, parallel to any line passing through
the vertex of the enveloping cone, the part of the straight line
which is in the same direction as the convexity of the surface will
fall without it, and the part which is in the other direction
within it.
For, if a plane be drawn, in the case of the paraboloid,
through the axis and the point, and, in the case of the hyperbo-
loid, through the given point and through the given straight
line drawn through the vertex of the enveloping cone, the
section by the plane will be (a) in the paraboloid a parabola
whose axis is the axis of the paraboloid, (b) in the hyperboloid
a hyperbola in which the given line through the vertex of the
enveloping cone is a diameter *.
[Prop. 11]
Hence the property follows from the plane properties of the
conics.
(2) If a plane touch a conoid without cutting it, it will
touch it at one point only, and the plane drawn through the
point of contact and the axis of the conoid will be at right
angles to the plane which touches it.
For, if possible, let the plane touch at two points. Draw
through each point a parallel to the axis. The plane passing
through both parallels will therefore either pass through, or be
parallel to, the axis. Hence the section of the conoid made by
this plane will be a conic [Prop. 11 (1), (2)], the two points
will lie on this conic, and the line joining them will lie within
the conic and therefore within the conoid. But this line
will be in the tangent plane, since the two points are in it.
Therefore some portion of the tangent plane will be within
the conoid; which is impossible, since the plane does not
cut it.
* There seems to be some error in the text here, which says that "the
diameter" (i.e. axis) of the hyperbola is "the straight line drawn in the conoid
from the vertex of the cone." But this straight line is not, in general, the
axis of the section.
ON CONOIDS AND SPHEROIDS.
127
Therefore the tangent plane touches in one point only.
That the plane through the point of contact and the axis is
perpendicular to the tangent plane is evident in the particular
case where the point of contact is the vertex of the conoid.
For, if two planes through the axis cut it in two conics, the
tangents at the vertex in both conics will be perpendicular
to the axis of the conoid. And all such tangents will be in the
tangent plane, which must therefore be perpendicular to the
axis and to any plane through the axis.
D
P
A
N
E
If the point of contact P is not the vertex, draw the plane
passing through the axis AN and the point P.
It will cut the conoid in a conic whose axis is
AN and the tangent plane in a line DPE
touching the conic at P. Draw PNP' perpen-
dicular to the axis, and draw a plane through it
also perpendicular to the axis. This plane will
make a circular section and meet the tangent
plane in a tangent to the circle, which will
therefore be at right angles to PN. Hence the
tangent to the circle will be at right angles to the plane
containing PN, AN; and it follows that this last plane is
perpendicular to the tangent plane.

Proposition 16.
P
(1) If a plane touch any of the spheroidal figures without
cutting it, it will touch at one point only, and the plane through
the point of contact and the axis will be at right angles to the
tangent plane.
This is proved by the same method as the last proposition.
(2) If any conoid or spheroid be cut by a plane through the
axis, and if through any tangent to the resulting conic a plane be
erected at right angles to the plane of section, the plane so erected
will touch the conoid or spheroid in the same point as that in
which the line touches the conic.
For it cannot meet the surface at any other point. If it
did, the perpendicular from the second point on the cutting
128
ARCHIMEDES
plane would be perpendicular also to the tangent to the conic
and would therefore fall outside the surface. But it must fall
within it.
[Prop. 11 (4)]
(3) If two parallel planes touch any of the spheroidal
figures, the line joining the points of contact will pass through
the centre of the spheroid.
If the planes are at right angles to the axis, the proposition
is obvious. If not, the plane through the axis and one point of
contact is at right angles to the tangent plane at that point.
It is therefore at right angles to the parallel tangent plane, and
therefore passes through the second point of contact. Hence
both points of contact lie on one plane through the axis, and
the proposition is reduced to a plane one.
Proposition 17.
If two parallel planes touch any of the spheroidal figures,
and another plane be drawn parallel to the tangent planes and
passing through the centre, the line drawn through any point of
the circumference of the resulting section parallel to the chord
of contact of the tangent planes will fall outside the spheroid.
This is proved at once by reduction to a plane proposition.
Archimedes adds that it is evident that, if the plane
parallel to the tangent planes does not pass through the
centre, a straight line drawn in the manner described will
fall without the spheroid in the direction of the smaller
segment but within it in the other direction.
Proposition 18.
Any spheroidal figure which is cut by a plane through the
centre is divided, both as regards its surface and its volume, into
two equal parts by that plane.
To prove this, Archimedes takes another equal and similar
spheroid, divides it similarly by a plane through the centre, and
then uses the method of application.
ON CONOIDS AND SPHEROIDS.
129
Propositions 19, 20.
Given a segment cut off by a plane from a paraboloid or
hyperboloid of revolution, or a segment of a spheroid less than
half the spheroid also cut off by a plane, it is possible to inscribe
in the segment one solid figure and to circumscribe about it
another solid figure, each made up of cylinders or 'frusta' of
cylinders of equal height, and such that the circumscribed figure
exceeds the inscribed figure by a volume less than that of any
given solid.
Let the plane base of the segment be perpendicular to the
plane of the paper, and let the plane of the paper be the plane
through the axis of the conoid or spheroid which cuts the base
of the segment at right angles in BC. The section in the plane
of the paper is then a conic BAC.
[Prop. 11]
Let EAF be that tangent to the conic which is parallel to
BC, and let A be the point of contact. Through EAF draw
a plane parallel to the plane through BC bounding the
segment. The plane so drawn will then touch the conoid
or spheroid at A.
[Prop. 16]
(1) If the base of the segment is at right angles to the
axis of the conoid or spheroid, A will be the vertex of the
conoid or spheroid, and its axis AD will bisect BC at right
angles.
(2) If the base of the segment is not at right angles to the
axis of the conoid or spheroid, we draw AD
(a) in the paraboloid, parallel to the axis,
(b) in the hyperboloid, through the centre (or the vertex of
the enveloping cone),
(c) in the spheroid, through the centre,
and in all the cases it will follow that AD bisects BC in D.
Then A will be the vertex of the segment, and AD will be
its axis.
Further, the base of the segment will be a circle or an
ellipse with BC as diameter or as an axis respectively, and
with centre D. We can therefore describe through this circle
H. A.
9
130
ARCHIMEDES
or ellipse a cylinder or a 'frustum' of a cylinder whose axis is
AD.
E
A
G
F
L
P
Q
/M
H
K
Ο
D
[Prop. 9]

B
Dividing this cylinder or frustum continually into equal
parts by planes parallel to the base, we shall at length arrive
at a cylinder or frustum less in volume than any given solid.
Let this cylinder or frustum be that whose axis is OD, and
let AD be divided into parts equal to OD, at L, M,.... Through
L, M,... draw lines parallel to BC meeting the conic in P, Q,……….,
and through these lines draw planes parallel to the base of the
segment. These will cut the conoid or spheroid in circles or
similar ellipses. On each of these circles or ellipses describe
two cylinders or frusta of cylinders each with axis equal to OD,
one of them lying in the direction of A and the other in the
direction of D, as shown in the figure.
Then the cylinders or frusta of cylinders drawn in the
direction of A make up a circumscribed figure, and those in
the direction of D an inscribed figure, in relation to the
segment.
Also the cylinder or frustum PG in the circumscribed figure
is equal to the cylinder or frustum PH in the inscribed figure,
QI in the circumscribed figure is equal to QK in the inscribed
figure, and so on.
Therefore, by addition,
(circumscribed fig.) = (inscr. fig.)
+(cylinder or frustum whose axis is OD).
But the cylinder or frustum whose axis is OD is less than
the given solid figure; whence the proposition follows.
Having set out these preliminary propositions, let us
proceed to demonstrate the theorems propounded with reference
to the figures."
:
ON CONOIDS AND SPHEROIDS.
131
:
Propositions 21, 22.
Any segment of a paraboloid of revolution is half as large
again as the cone or segment of a cone which has the same base
and the same axis.
Let the base of the segment be perpendicular to the plane of
the paper, and let the plane of the paper be the plane through
the axis of the paraboloid which cuts the base of the segment
at right angles in BC and makes the parabolic section BAC.
Let EF be that tangent to the parabola which is parallel to
BC, and let A be the point of contact.
Then (1), if the plane of the base of the segment is
perpendicular to the axis of the paraboloid, that axis is the
line AD bisecting BC at right angles in D.
(2) If the plane of the base is not perpendicular to the
axis of the paraboloid, draw AD parallel to the axis of the
paraboloid. AD will then bisect BC, but not at right angles.
Draw through EF a plane parallel to the base of the seg-
ment. This will touch the paraboloid at A, and A will be
the vertex of the segment, AD its axis.
The base of the segment will be a circle with diameter BC
or an ellipse with BC as major axis.
Accordingly a cylinder or a frustum of a cylinder can be
found passing through the circle or ellipse and having AD for
its axis [Prop. 9]; and likewise a cone or a segment of a cone
can be drawn passing through the circle or ellipse and having
A for vertex and AD for axis.
[Prop. 8]
Suppose to be a cone equal to 3 (cone or segment of
cone ABC). The cone X is therefore equal to half the cylinder
or frustum of a cylinder EC.
[Cf. Prop. 10]
We shall prove that the volume of the segment of the
paraboloid is equal to X.
If not, the segment must be either greater or less than X.
I. If possible, let the segment be greater than X.
We can then inscribe and circumscribe, as in the last
9-2
132
ARCHIMEDES
proposition, figures made up of cylinders or frusta of cylinders
with equal height and such that
(circumscribed fig.) - (inscribed fig.)<(segment) – X.
Let the greatest of the cylinders or frusta forming the
circumscribed figure be that whose base is the circle or ellipse
about BC and whose axis is OD, and let the smallest of them be
that whose base is the circle or ellipse about PP' and whose
axis is AL.
Let the greatest of the cylinders forming the inscribed
figure be that whose base is the circle or ellipse about RR' and
whose axis is OD, and let the smallest be that whose base is
the circle or ellipse about PP' and whose axis is LM.
R'
C
F
A
E

P
P
L
Q
M
D
N
S
R
H
T
B
Produce all the plane bases of the cylinders or frusta to
meet the surface of the complete cylinder or frustum EC.
Now, since
(circumscribed fig.) — (inscr. fig.)<(segment) — X,
it follows that
(inscribed figure) > X ....
.(a).
Next, comparing successively the cylinders or frusta with
heights equal to OD and respectively forming parts of the
complete cylinder or frustum EC and of the inscribed figure,
we have
(first cylinder or frustum in EC): (first in inscr. fig.)
= BD² : RO²
=AD: AO
= BD: TO, where AB meets OR in T.
And (second cylinder or frustum in EC): (second in inscr. fig.)
- HO: SN, in like manner,
and so on.
ON CONOIDS AND SPHEROIDS.
133
Hence [Prop. 1] (cylinder or frustum EC): (inscribed figure)
= (BD+HO+...): (TO + SN+...),
where BD, HO,... are all equal, and BD, TO, SN,……. diminish in
arithmetical progression.
But [Lemma preceding Prop. 1]
Therefore
or
BD+HO+... > 2 (TO + SN+...).
2(TO+SN
(cylinder or frustum EC) > 2 (inscribed fig.),
X>(inscribed fig.);
which is impossible, by (a) above.
II. If possible, let the segment be less than X.
In this case we inscribe and circumscribe figures as before,
but such that
(circumscr. fig.) (inscr. fig.)<X - (segment),
whence it follows that
(circumscribed figure)< X
..(B).
And, comparing the cylinders or frusta making up the
complete cylinder or frustum CE and the circumscribed figure
respectively, we have
(first cylinder or frustum in CE): (first in circumscr. fig.)
and so on.
= BD² : BD²
=
BD: BD.
(second in CE): (second in circumscr. fig.)
Hence [Prop. 1]
= H02: R02
=
· AD: AO
= HO: TO,
(cylinder or frustum CE): (circumscribed fig.)
= (BD + HO + ...) : (BD+TO+ ...),
< 2:1,
and it follows that
[Lemma preceding Prop. 1]
X < (circumscribed fig.);
which is impossible, by (B).
Thus the segment, being neither greater nor less than X, is
equal to it, and therefore to § (cone or segment of cone ABC).
134
ARCHIMEDES
**
i
Proposition 23.
If from a paraboloid of revolution two segments be cut off,
one by a plane perpendicular to the axis, the other by a plane not
perpendicular to the axis, and if the axes of the segments are
equal, the segments will be equal in volume.
Let the two planes be supposed perpendicular to the plane
of the paper, and let the latter plane be the plane through the
axis of the paraboloid cutting the other two planes at right
angles in BB', QQ' respectively and the paraboloid itself in the
parabola QPQ'B'.
Let AN, PV be the equal axes of the segments, and A, P
their respective vertices.
L

B
P
D
V
A
N
K
Q'
B'
Draw QL parallel to AN or PV and Q'L perpendicular
to QL.
Now, since the segments of the parabolic section cut off by
BB', QQ' have equal axes, the triangles ABB', PQQ' are equal
[Prop. 3]. Also, if QD be perpendicular to PV, QD = BN (as
in the same Prop. 3).
Conceive two cones drawn with the same bases as the
segments and with A, P as vertices respectively. The height
of the cone PQQ' is then PK, where PK is perpendicular to
QQ'.
1
ON CONOIDS AND SPHEROIDS.
135
Now the cones are in the ratio compounded of the ratios of
their bases and of their heights, i.e. the ratio compounded of
(1) the ratio of the circle about BB' to the ellipse about QQ',
and (2) the ratio of AN to PK.
That is to say, we have, by means of Props. 5, 12,
(cone ABB'): (cone PQQ') = (BB'² : QQ' . QʻL).(AN : PK).
And BB' = 2BN = 2QD = Q'L, while QQ' = 2QV.
Therefore
(cone ABB'): (cone PQQ') = (QD : QV).(AN: PK)
= (PK : PV). (AN: PK)
= AN : PV.
Since ANPV, the ratio of the cones is a ratio of equality;
and it follows that the segments, being each half as large again
as the respective cones [Prop. 22], are equal,
Proposition 24.
If from a paraboloid of revolution two segments be cut off by
planes drawn in any manner, the segments will be to one another
as the squares on their axes.
For let the paraboloid be cut by a plane through the axis
in the parabolic section P'PApp', and let the axis of the
parabola and paraboloid be ANN'.
p'
Measure along ANN' the lengths AN, AN' equal to the
respective axes of the given segments,
and through N, N' draw planes perpen-
dicular to the axis, making circular
sections on Pp, P'p' as diameters re-
spectively. With these circles as bases
and with the common vertex A let two
cones be described.
Now the segments of the paraboloid
whose bases are the circles about Pp,
P'p' are equal to the given segments
respectively, since their respective axes
are equal [Prop. 23]; and, since the
segments APP, AP'p' are half as large

A
N
N
P
2
136
ARCHIMEDES
again as the cones APP, AP'p' respectively, we have only
to show that the cones are in the ratio of AN² to AN'².
But
(cone APp): (cone AP'p') = (PN² : P'N'²). (AN : AN')
thus the proposition is proved.
= (AN: AN'). (AN: AN')
2
= AN²: AN'";
Propositions 25, 26.
In any hyperboloid of revolution, if A be the vertex and AD
the axis of any segment cut off by a plane, and if CA be the
semidiameter of the hyperboloid through A (CA being of course
in the same straight line with AD), then
(segment) : (cone with same base and axis)
=(AD+3CA): (AD + 2CA).
Let the plane cutting off the segment be perpendicular to
the plane of the paper, and let the latter plane be the plane
through the axis of the hyperboloid which intersects the cutting
plane at right angles in BB', and makes the hyperbolic
segment BAB'. Let C be the centre of the hyperboloid (or
the vertex of the enveloping cone).
Let EF be that tangent to the hyperbolic section which is
parallel to BB'. Let EF touch at A, and join CA. Then CA
produced will bisect BB' at D, CA will be a semi-diameter of
the hyperboloid, A will be the vertex of the segment, and AD
its axis. Produce AC to A' and H, so that AC=CA' = A'H.
Through EF draw a plane parallel to the base of the seg-
ment. This plane will touch the hyperboloid at A.
Then (1), if the base of the segment is at right angles to the
axis of the hyperboloid, A will be the vertex, and AD the axis,
of the hyperboloid as well as of the segment, and the base of the
segment will be a circle on BB' as diameter.
ON CONOIDS AND SPHEROIDS.
137
(2) If the base of the segment is not perpendicular to the
axis of the hyperboloid, the base will be an ellipse on BB' as
major axis.
H
[Prop. 13]

b (AD)
B
E
M
Q
N
P
P
D
Ρ
q
འ
L
a
a
a
a
(AA')
A
A
F
し
​B


❤
Then we can draw a cylinder or a frustum of a cylinder
EBB'F' passing through the circle or ellipse about BB' and
having AD for its axis; also we can describe a cone or a
segment of a cone through the circle or ellipse and having A
for its vertex.
We have to prove that
(segment ABB'): (cone or segment of cone ABB')= HD : A'D.
138
ARCHIMEDES
Let V be a cone such that
V: (cone or segment of cone ABB') = HD : A'D„…………….(a)
and we have to prove that V is equal to the segment.
Now
(cylinder or frustum EB'): (cone or segmt. of cone ABB')=3:1.
Therefore, by means of (a),
(cylinder or frustum EB') : V = A'D :
HD
3
..(B).
If the segment is not equal to V, it must either be greater
or less.
I. If possible, let the segment be greater than V.
Inscribe and circumscribe to the segment figures made up
of cylinders or frusta of cylinders, with axes along AD and all
equal to one another, such that
whence
(circumscribed fig.) — (inscr. fig.)<(segmt.) — V,
(inscribed figure) > V........(Y).
Produce all the planes forming the bases of the cylinders or
frusta of cylinders to meet the surface of the complete cylinder
or frustum EB'.
Then, if ND be the axis of the greatest cylinder or frustum
in the circumscribed figure, the complete cylinder will be
divided into cylinders or frusta each equal to this greatest
cylinder or frustum.
Let there be a number of straight lines a equal to AA' and
as many in number as the parts into which AD is divided by
the bases of the cylinders or frusta. To each line a apply a
rectangle which shall overlap it by a square, and let the greatest
of the rectangles be equal to the rectangle AD. A'D and the
least equal to the rectangle AL. AʼL; also let the sides of the
overlapping squares b, p, q,...l be in descending arithmetical
progression. Thus b, p, q,...l will be respectively equal to AD,
AN, AM,...AL, and the rectangles (ab+b²), (ap+p²),...(al+l³)
will be respectively equal to AD. A'D, AN . A'N,...AL . AʼL.
ON CONOIDS AND SPHEROIDS.
139
Suppose, further, that we have a series of spaces S each
equal to the largest rectangle AD. A'D and as many in number
as the diminishing rectangles.
Comparing now the successive cylinders or frusta (1) in the
complete cylinder or frustum EB' and (2) in the inscribed
figure, beginning from the base of the segment, we have
(first cylinder or frustum in EB'): (first in inscr. figure)
Again
· BD² : PN²
=AD. A'D: AN. A'N, from the hyperbola,
=S: (ap+ p³).
(second cylinder or frustum in EB'): (second in inscr. fig.)
and so on.
=
= BD² : QM²
= AD.A'D: AM.A'M
=
= S: (aq + q³),
The last cylinder or frustum in the complete cylinder or
frustum EB' has no cylinder or frustum corresponding to it in
the inscribed figure.
Combining the proportions, we have
(cylinder or frustum EB'): (inscribed figure)
[Prop. 1]
= (sum of all the spaces S): (ap + p²)+(aq + q³) + ...
> (a+b): (a
+
[Prop. 2]
3
HD
> A'D:
since a = AA', b=AD,
3
Hence
>(EB'): V, by (8) above.
(inscribed figure) < V.
But this is impossible, because, by (y) above, the inscribed
figure is greater than V.
140
ARCHIMEDES
II. Next suppose, if possible, that the segment is less
than V.
In this case we circumscribe and inscribe figures such that
(circumscribed fig.) — (inscribed fig.) < V — (segment),
whence we derive
V> (circumscribed figure).
-
……..(8).
We now compare successive cylinders or frusta in the
complete cylinder or frustum and in the circumscribed figure;
and we have
(first cylinder or frustum in EB'): (first in circumscribed fig.)
=S: S
=S: (ab+b²),
(second in EB') : (second in circumscribed fig.)
and so on.
Hence [Prop. 1]
=S: (ap+p³),
(cylinder or frustum EB'): (circumscribed fig.)
= (sum of all spaces S) : (ab + b²) + (ap + p²) + ...
α
b
< (a + b) : (2 + 1)
[Prop. 2]
HD
<A'D:
3
<(EB'): V, by (B) above.
Hence the circumscribed figure is greater than V; which is
impossible, by (8) above.
Thus the segment is neither greater nor less than V, and is
therefore equal to it.
Therefore, by (a),
(segment ABB'): (cone or segment of cone ABB')
= (AD+3CA): (AD+2CA).
ON CONOIDS AND SPHEROIDS.
141
Propositions 27, 28, 29, 30.
(1) In any spheroid whose centre is C, if a plane meeting
the axis cut off a segment not greater than half the spheroid and
having A for its vertex and AD for its axis, and if A'D be the
axis of the remaining segment of the spheroid, then
(first segmt.): (cone or segmt. of cone with same base and axis)
= CA+A'D: A'D
[=3CA - AD: 2CA – AD].
(2) As a particular case, if the plane passes through the
centre, so that the segment is half the spheroid, half the spheroid
is double of the cone or segment of a cone which has the same
vertex and axis.
Let the plane cutting off the segment be at right angles to
the plane of the paper, and let the latter plane be the plane
through the axis of the spheroid which intersects the cutting
plane in BB' and makes the elliptic section ABA'B'.
Let EF, E'F" be the two tangents to the ellipse which are
parallel to BB', let them touch it in A, A', and through the
tangents draw planes parallel to the base of the segment.
These planes will touch the spheroid at A, A', which will
be the vertices of the two segments into which it is divided.
Also AA' will pass through the centre C and bisect BB'
in D.
Then (1) if the base of the segments be perpendicular to
the axis of the spheroid, A, A' will be the vertices of the
spheroid as well as of the segments, AA' will be the axis
of the spheroid, and the base of the segments will be a circle on
BB' as diameter;
(2) if the base of the segments be not perpendicular to the
axis of the spheroid, the base of the segments will be an
ellipse of which BB' is one axis, and AD, A'D will be the
axes of the segments respectively.
142
ARCHIMEDES
We can now draw a cylinder or a frustum of a cylinder
EBB'F' through the circle or ellipse about BB' and having AD
for its axis; and we can also draw a cone or a segment of
a cone passing through the circle or ellipse about BB' and
having A for its vertex.
a
ไป
E
A
F

L
Q
M
P
N
B'
B
D
C
E'
A'
FI
સ
X
a
H
a'

a
a
d
d
d
d
с
b
We have then to show that, if CA' be produced to H so
that CA' A'H,
=
(segment ABB'): (cone or segment of cone ABB') = HD: A'D.
Let V be such a cone that
V: (cone or segment of cone ABB') = HD: A'D ... (a);
and we have to show that the segment ABB′ is equal to V.
ON CONOIDS AND SPHEROIDS.
143
But, since
(cylinder or frustum EB'): (cone or segment of cone ABB')
=3:1,
we have, by the aid of (a),
HD
(cylinder or frustum EB'): V=A'D:
(B).
3
Now, if the segment ABB' is not equal to V, it must
be either greater or less.
I. Suppose, if possible, that the segment is greater
than V.
Let figures be inscribed and circumscribed to the segment
consisting of cylinders or frusta of cylinders, with axes along
AD and all equal to one another, such that
(circumscribed fig.) — (inscribed fig.)<(segment) — V,
whence it follows that
(inscribed fig.) > V
(7).
Produce all the planes forming the bases of the cylinders or
frusta to meet the surface of the complete cylinder or frustum
EB'. Thus, if ND be the axis of the greatest cylinder or
frustum of a cylinder in the circumscribed figure, the complete
cylinder or frustum EB' will be divided into cylinders or frusta
of cylinders each equal to the greatest of those in the circum-
scribed figure.
Take straight lines da' each equal to A'D and as many in
number as the parts into which AD is divided by the bases of
the cylinders or frusta, and measure da along da' equal to AD.
It follows that aa' = 2CD.
Apply to each of the lines a'd rectangles with height equal
to ad, and draw the squares on each of the lines ad as in
the figure. Let S denote the area of each complete rectangle.
From the first rectangle take away a gnomon with breadth
equal to AN (i.e. with each end of a length equal to AN);
take away from the second rectangle a gnomon with breadth
equal to AM, and so on, the last rectangle having no gnomon
taken from it.
144
ARCHIMEDES
T
Then
the first gnomon = A'D. AD-ND. (A'D — AN)
Similarly,
= A'D. AN+ND. AN
= AN.A'N.
the second gnomon = AM. A'M,
and so on.
And the last gnomon (that in the last rectangle but one) is
equal to AL. AʼL.
2
Also, after the gnomons are taken away from the successive
rectangles, the remainders (which we will call R₁, R₂,... Rn,
where n is the number of rectangles and accordingly R₂ = S)
are rectangles applied to straight lines each of length aa' and
exceeding by squares" whose sides are respectively equal
to DN, DM,... DA.
CC
For brevity, let DN be denoted by x, and aa' or 2CD by c,
so that R₁ = cx + x², R₂ = c. 2x + (2x)³‚…….
2x+(2x)³,...
2
Then, comparing successively the cylinders or frusta of
cylinders (1) in the complete cylinder or frustum EB' and
(2) in the inscribed figure, we have
(first cylinder or frustum in EB'): (first in inscribed fig.)
: BD2 : PN2
= AD. A'D: AN.A'N
=S: (first gnomon);
(second cylinder or frustum in EB'): (second in inscribed fig.)
=S: (second gnomon),
and so on.
The last of the cylinders or frusta in the cylinder or
frustum EB' has none corresponding to it in the inscribed
figure, and there is no corresponding gnomon.
Combining the proportions, we have [by Prop. 1]
(cylinder or frustum EB'): (inscribed fig.)
= (sum of all spaces S): (sum of gnomons).
ON CONOIDS AND SPHEROIDS.
145
Now the differences between S and the successive gnomons
are R₁, R₂,... Rn, while
R₁ = cx + x²,
R₂ = c. 2x + (2x)³,
2
Rn cb+b²=S,
=
where b
= nx = AD.
Hence [Prop. 2]
:
(sum of all spaces §) : (R¸ + R₂ + ... + Run) < (c + b) : ( §2 + 2 ) .
It follows that
2
(sum of all spaces S): (sum of gnomons) > (c+b):
C
2
HD
> A'D:
3
+
3
26
3
Thus (cylinder or frustum EB'): (inscribed fig.)
> A'D:
HD
3
from (8) above.
Therefore
>(cylinder or frustum EB'): V,
(inscribed fig.) < V ;
which is impossible, by (y) above.
Hence the segment ABB' is not greater than V.
II. If possible, let the segment ABB' be less than V.
We then inscribe and circumscribe figures such that
(circumscribed fig.) (inscribed fig.)< V- (segment),
whence
V>(circumscribed fig.).....
·(8).
In this case we compare the cylinders or frusta in (EB')
with those in the circumscribed figure.
Thus
(first cylinder or frustum in EB'): (first in circumscribed fig.)
=S: S;
(second in EB'): (second in circumscribed fig.)
and so on.
H. A.
=S: (first gnomon),
10
146
ARCHIMEDES
!
Lastly (last in EB'): (last in circumscribed fig.)
Now
=S: (last gnomon).
{S+(all the gnomons)} = nS − (R₁ + R₂+ ... + Rn−1).
And nS : R₁ + R₂ + ... + Rm-1>(c+b): (+), [Prop. 2]
so that
nS: {S+(all the gnomons)} < (c+b):
+
3
26).
It follows that, if we combine the above proportions as in
Prop. 1, we obtain
(cylinder or frustum EB'): (circumscribed fig.)
< (c+b):
с 26
+
2 3
HD
< A'D:
3
<(EB'): V, by (B) above.
Hence the circumscribed figure is greater than V; which is
impossible, by (8) above.
Thus, since the segment ABB' is neither greater nor less
than V, it is equal to it; and the proposition is proved.
(2) The particular case [Props. 27, 28] where the segment
is half the spheroid differs from the above in that the distance
CD or c/2 vanishes, and the rectangles cb + b² are simply squares
(b²), so that the gnomons are simply the differences between b²
and a², b² and (2x)², and so on.
Instead therefore of Prop. 2 we use the Lemma to Prop. 2,
Cor. 1, given above [On Spirals, Prop. 10], and instead of the
we obtain the ratio 3 : 2, whence
ratio (c +b): (+26)
3
(segment ABB'): (cone or segment of cone ABB') = 2 : 1.
[This result can also be obtained by simply substituting
CA for AD in the ratio (3CA – AD) : (2CA – AD).]
-
ON CONOIDS AND SPHEROIDS.
147
Propositions 31, 32.
If a plane divide a spheroid into two unequal segments, and
if AN, A'N be the axes of the lesser and greater segments
respectively, while C is the centre of the spheroid, then
(greater segmt.): (cone or segmt. of cone with same base and axis)
= CA+AN: AN.
Let the plane dividing the spheroid be that through PP'
perpendicular to the plane of the paper, and let the latter plane
be that through the axis of the spheroid which intersects the
cutting plane in PP' and makes the elliptic section PAP'A'.

P
B
-H'
Α'
K
N
A
H
P/
B'
Β
Draw the tangents to the ellipse which are parallel to PP';
let them touch the ellipse at A, A', and through the tangents
draw planes parallel to the base of the segments. These planes
will touch the spheroid at A, A', the line AA' will pass
through the centre C and bisect PP' in N, while AN, A'N will
be the axes of the segments.
Then (1) if the cutting plane be perpendicular to the axis
of the spheroid, AA' will be that axis, and A, A' will be the
vertices of the spheroid as well as of the segments. Also the
sections of the spheroid by the cutting plane and all planes
parallel to it will be circles.
(2) If the cutting plane be not perpendicular to the axis,
10-2
148
ARCHIMEDES
the base of the segments will be an ellipse of which PP' is an
axis, and the sections of the spheroid by all planes parallel
to the cutting plane will be similar ellipses.
Draw a plane through C parallel to the base of the segments
and meeting the plane of the paper in BB'.
Construct three cones or segments of cones, two having A
for their common vertex and the plane sections through PP',
BB' for their respective bases, and a third having the plane
section through PP' for its base and A' for its vertex.
Produce CA to H and CA' to H' so that
AH = A'H' = CA.
We have then to prove that
(segment A'PP'): (cone or segment of cone A'PP')
=
CA + AN: AN
=NH:AN.
Now half the spheroid is double of the cone or segment of a
cone ABB' [Props. 27, 28]. Therefore
(the spheroid) = 4 (cone or segment of cone ABB').
But
(cone or segmt. of cone ABB'): (cone or segmt. of cone APP')
= (CA: AN).(BC²: PN³)
= (CA : AN). (CA.CA': AN.A'N)…..(a).
If we measure AK along AA' so that
we have
AK: AC=AC: AN,
AK.A'N AC. A'N CA: AN,
:
and the compound ratio in (a) becomes
i.e.
Thus
=
(AK. AʼN : CA. AʼN). (CA. CA' : AN. AʼN),
AK.CA': AN.A'N.
(cone or segmt. of cone ABB'): (cone or segmt. of cone APP')
=AK.CA': AN.A'N.
ON CONOIDS AND SPHEROIDS.
149
But (cone or segment of cone APP'): (segment APP')
Therefore, ex aequali,
=AN:NH
[Props. 29, 30]
=
AN. A'N: AN.NH'.
so that
(cone or segment of cone ABB'): (segment APP')
=AK.CA': AN. NH',
(spheroid): (segment APP')
= HH'. AK: AN. NH',
since
HH' = 4CA'.
Hence
(segment A'PP'): (segment APP')
=(HH'.AK-AN.NH):AN.NH
(AK.NH+NH'.NK):AN.NH.
and
Further,
(segment APP'): (cone or segment of cone APP')
-NH':AN
=AN. NH': AN. A'N,
(cone or segmt. of cone APP'): (cone or segmt. of cone A'PP')
AN A'N
:
=AN.A'N : A'N'.
From the last three proportions we obtain, ex aequali,
(segment A'PP'): (cone or segment of cone A'PP')
=(AK.NH+NH'.NK):AN
=
(AK.NH+NH'.NK):(CA*+NH'.CN)
=(AK.NH+NH'.NK):(AK.AN+NH'.CN)...(3).
But
AK.NH: AK. AN = NH: AN
= CA+AN: AN
=AK+CA: CA
HK: CA
(since AK: AC AC: AN)
= HK – NH: CA — AN
=NK: CN
=NH'.NK:NH'.CN.
=
150
ARCHIMEDES.
Hence the ratio in (B) is equal to the ratio
Therefore
AK.NH:AK.AN, or NH:AN.
(segment A'PP'): (cone or segment of cone A'PP')
=
NH:AN
= CA+AN: AN.
[If (x, y) be the coordinates of P referred to the conjugate
diameters AA', BB' as axes of x, y, and if 2a, 2b be the lengths
of the diameters respectively, we have, since
(spheroid) - (lesser segment)=(greater segment),
2a + x
2a XC
4. ab²
• y² (a — x) =
y² (a + x);
a + x
α XC
and the above proposition is the geometrical proof of the truth
of this equation where x, y are connected by the equation
y²
=
x²
+
α b2
a²
1.]
ON SPIRALS.
“ARCHIMEDES to Dositheus greeting.
Of most of the theorems which I sent to Conon, and of
which you ask me from time to time to send you the proofs, the
demonstrations are already before you in the books brought to
you by Heracleides; and some more are also contained in that
which I now send you. Do not be surprised at my taking a
considerable time before publishing these proofs. This has
been owing to my desire to communicate them first to persons
engaged in mathematical studies and anxious to investigate
them. In fact, how many theorems in geometry which have
seemed at first impracticable are in time successfully worked out!
Now Conon died before he had sufficient time to investigate
the theorems referred to; otherwise he would have discovered
and made manifest all these things, and would have enriched
geometry by many other discoveries besides. For I know well
that it was no common ability that he brought to bear on
mathematics, and that his industry was extraordinary. But,
though many years have elapsed since Conon's death, I do not
find that any one of the problems has been stirred by a single
person. I wish now to put them in review one by one,
particularly as it happens that there are two included among
them which are impossible of realisation* [and which may
serve as a warning] how those who claim to discover every-
thing but produce no proofs of the same may be confuted as
having actually pretended to discover the impossible.
* Heiberg reads τέλος δὲ ποθεσόμενα, but F has τέλους, so that the true reading
is perhaps Téλous dè totideóμeva. The meaning appears to be simply 'wrong.'
152
ARCHIMEDES
What are the problems I mean, and what are those of which
you have already received the proofs, and those of which the
proofs are contained in this book respectively, I think it proper
to specify. The first of the problems was, Given a sphere, to find
a plane area equal to the surface of the sphere; and this was
first made manifest on the publication of the book concerning the
sphere, for, when it is once proved that the surface of any sphere
is four times the greatest circle in the sphere, it is clear that it
is possible to find a plane area equal to the surface of the sphere.
The second was, Given a cone or a cylinder, to find a sphere
equal to the cone or cylinder; the third, To cut a given sphere
by a plane so that the segments of it have to one another an
assigned ratio; the fourth, To cut a given sphere by a plane so
that the segments of the surface have to one another an assigned
ratio; the fifth, To make a given segment of a sphere similar to
a given segment of a sphere*; the sixth, Given two segments of
either the same or different spheres, to find a segment of a sphere
which shall be similar to one of the segments and have its
surface equal to the surface of the other segment. The seventh
was, From a given sphere to cut off a segment by a plane so
that the segment bears to the cone which has the same base as
the segment and equal height an assigned ratio greater than
that of three to two. Of all the propositions just enumerated
Heracleides brought you the proofs. The proposition stated
next after these was wrong, viz. that, if a sphere be cut by a
plane into unequal parts, the greater segment will have to the
less the duplicate ratio of that which the greater surface has to
the less. That this is wrong is obvious by what I sent you
before; for it included this proposition: If a sphere be cut into
unequal parts by a plane at right angles to any diameter in the
sphere, the greater segment of the surface will have to the less
the same ratio as the greater segment of the diameter has
to the less, while the greater segment of the sphere has to the
less a ratio less than the duplicate ratio of that which the
* τὸ δοθὲν τμᾶμα σφαίρας τῷ δοθέντι τμάματι σφαίρας ὁμοιώσαι, i.e. to make a
segment of a sphere similar to one given segment and equal in content to
another given segment. [Cf. On the Sphere and Cylinder, II. 5.]
ON SPIRALS.
153
greater surface has to the less, but greater than the sesqui-
alterate* of that ratio. The last of the problems was also wrong,
viz. that, if the diameter of any sphere be cut so that the square
on the greater segment is triple of the square on the lesser
segment, and if through the point thus arrived at a plane be
drawn at right angles to the diameter and cutting the sphere,
the figure in such a form as is the greater segment of the sphere
is the greatest of all the segments which have an equal surface.
That this is wrong is also clear from the theorems which I
before sent you. For it was there proved that the hemisphere
is the greatest of all the segments of a sphere bounded by an
equal surface.
After these theorems the following were propounded con-
cerning the cone†. If a section of a right-angled cone [a
parabola], in which the diameter [axis] remains fixed, be made to
revolve so that the diameter [axis] is the axis [of revolution],
let the figure described by the section of the right-angled cone
be called a conoid. And if a plane touch the conoidal figure
and another plane drawn parallel to the tangent plane cut off
a segment of the conoid, let the base of the segment cut off be
defined as the cutting plane, and the vertex as the point in which
the other plane touches the conoid. Now, if the said figure be
cut by a plane at right angles to the axis, it is clear that the
section will be a circle; but it needs to be proved that the
segment cut off will be half as large again as the cone which has
the same base as the segment and equal height. And if two
segments be cut off from the conoid by planes drawn in any
manner, it is clear that the sections will be sections of acute-
angled cones [ellipses] if the cutting planes be not at right
angles to the axis; but it needs to be proved that the
segments will bear to one another the ratio of the squares on
the lines drawn from their vertices parallel to the axis to meet
the cutting planes. The proofs of these propositions are not
yet sent to you.
After these came the following propositions about the spiral,
(λόγον) μείζονα ἢ ἡμιόλιον τοῦ, ὃν ἔχει κ.τ.λ., i.e. a ratio greater than (the
ratio of the surfaces). See On the Sphere and Cylinder, II. 8.
C
+ This should be presumably the conoid,' not 'the cone.'
154
ARCHIMEDES
which are as it were another sort of problem having nothing
in common with the foregoing; and I have written out the
proofs of them for you in this book. They are as follows. If a
straight line of which one extremity remains fixed be made to
revolve at a uniform rate in a plane until it returns to the
position from which it started, and if, at the same time as the
straight line revolves, a point move at a uniform rate along the
straight line, starting from the fixed extremity, the point will
describe a spiral in the plane. I say then that the area
bounded by the spiral and the straight line which has returned
to the position from which it started is a third part of the circle
described with the fixed point as centre and with radius the
length traversed by the point along the straight line during the
one revolution. And, if a straight line touch the spiral at the
extreme end of the spiral, and another straight line be drawn at
right angles to the line which has revolved and resumed its
position from the fixed extremity of it, so as to meet the
tangent, I say that the straight line so drawn to meet it is
equal to the circumference of the circle. Again, if the revolving
line and the point moving along it make several revolutions
and return to the position from which the straight line started,
I say that the area added by the spiral in the third revolution
will be double of that added in the second, that in the fourth
three times, that in the fifth four times, and generally the areas
added in the later revolutions will be multiples of that added in
the second revolution according to the successive numbers,
while the area bounded by the spiral in the first revolution is a
sixth part of that added in the second revolution. Also, if on
the spiral described in one revolution two points be taken and
straight lines be drawn joining them to the fixed extremity of
the revolving line, and if two circles be drawn with the fixed
point as centre and radii the lines drawn to the fixed extremity
of the straight line, and the shorter of the two lines be produced,
I say that (1) the area bounded by the circumference of the
greater circle in the direction of (the part of) the spiral included
between the straight lines, the spiral (itself) and the produced
straight line will bear to (2) the area bounded by the circum-
ference of the lesser circle, the same (part of the) spiral and the
ON SPIRALS.
155
straight line joining their extremities the ratio which (3) the
radius of the lesser circle together with two thirds of the excess
of the radius of the greater circle over the radius of the lesser
bears to (4) the radius of the lesser circle together with one
third of the said excess.
The proofs then of these theorems and others relating to the
spiral are given in the present book. Prefixed to them, after the
manner usual in other geometrical works, are the propositions
necessary to the proofs of them. And here too, as in the books
previously published, I assume the following lemma, that, if
there be (two) unequal lines or (two) unequal areas, the excess
by which the greater exceeds the less can, by being [continually]
added to itself, be made to exceed any given magnitude among
those which are comparable with [it and with] one another."
Proposition 1.
If a point move at a uniform rate along any line, and two
lengths be taken on it, they will be proportional to the times of
describing them.
Two unequal lengths are taken on a straight line, and two
lengths on another straight line representing the times; and
they are proved to be proportional by taking equimultiples of
each length and the corresponding time after the manner of
Eucl. V. Def. 5.
Proposition 2.
If each of two points on different lines respectively move along
them each at a uniform rate, and if lengths be taken, one on each
line, forming pairs, such that each pair are described in equal
times, the lengths will be proportionals.
This is proved at once by equating the ratio of the lengths
taken on one line to that of the times of description, which
must also be equal to the ratio of the lengths taken on the other
line.
156
ARCHIMEDES
Proposition 3.
Given any number of circles, it is possible to find a straight
line greater than the sum of all their circumferences.
For we have only to describe polygons about each and then
take a straight line equal to the sum of the perimeters of the
polygons.
Proposition 4.
Given two unequal lines, viz. a straight line and the circum-
ference of a circle, it is possible to find a straight line less than
the greater of the two lines and greater than the less.
For, by the Lemma, the excess can, by being added a sufficient
number of times to itself, be made to exceed the lesser line.
Thus e.g., if c>l (where c is the circumference of the circle
and 7 the length of the straight line), we can find a number n
such that
Therefore
and
n (cl) > l.
c-l>
し
​N
c> l + - > l.
N
Hence we have only to divide l into n equal parts and add
one of them to l. The resulting line will satisfy the condition.
Proposition 5.
Given a circle with centre O, and the tangent to it at a point
A, it is possible to draw from O a straight line OPF, meeting the
circle in P and the tangent in F, such that, if c be the circum-
ference of any given circle whatever,
FP: OP< (arc AP): c.
Take a straight line, as D, greater than the circumference c.
[Prop. 3]
ON SPIRALS.
157
Through O draw OH parallel to the given tangent, and
draw through A a line APH, meeting the circle in P and OH
A

P
D
in H, such that the portion PH intercepted between the circle
and the line OH may be equal to D*. Join OP and produce
it to meet the tangent in F.
Then
FP: OPAP : PH, by parallels,
= AP: D
<< (arc AP): c.
Proposition 6.
Given a circle with centre 0, a chord AB less than the
diameter, and OM the perpendicular on AB from O, it is possible
to draw a straight line OFP, meeting the chord AB in F and the
circle in P, such that
FP:PB=D : E,
where D: E is any given ratio less than BM: MO.
Draw OH parallel to AB, and BT perpendicular to BO
meeting OH in T.
Then the triangles BMO, OBT are similar, and therefore
BM: MO=0B: BT,
whence
D: E< OB: BT.
* This construction, which is assumed without any explanation as to how it
is to be effected, is described in the original Greek thus: "let PH be placed
(kelow) equal to D, verging (veúovoa) towards A." This is the usual phraseology
used in the type of problem known by the name of νεῦσις.
158
ARCHIMEDES
that
Suppose that a line PH (greater than BT) is taken such
P
D:E=OB: PH,

A
B
F
M
T
H
E
D
and let PH be so placed that it passes through B and P lies on
the circumference of the circle, while H is on the line OH*.
(PH will fall outside BT, because PH>BT.) Join OP meeting
AB in F.
We now have
FP: PB OP: PH
=
-
ОВ :PH
:D: E.
Proposition 7.
Given a circle with centre 0, a chord AB less than the
diameter, and OM the perpendicular on it from O, it is possible
to draw from O a straight line OPF, meeting the circle in P and
AB produced in F, such that
FP: PB D: E,
where D: E is any given ratio greater than BM : MO.
Draw OT parallel to AB, and BT perpendicular to BO
meeting OT in T.
* The Greek phrase is "let PH be placed between the circumference and the
straight line (OH) through B." The construction is assumed, like the similar
one in the last proposition.
ON SPIRALS.
159
In this case,
D:E>BM : MO
A
M
> OB: BT, by similar triangles.

B
D
E
P
H
T
Take a line PH (less than BT) such that
D: E=0B: PH,
and place PH so that P, H are on the circle and on OT respec-
tively, while HP produced passes through B*.
Then
FP: PB=0P : PH
=D: E.
Proposition 8.
Given a circle with centre O, a chord AB less than the
diameter, the tangent at B, and the perpendicular OM from O
on AB, it is possible to draw from O a straight line OFP,
meeting the chord AB in F, the circle in P and the tangent in G,
such that
FP: BG =D: E,
where D: E is any given ratio less than BM: MO.
If OT be drawn parallel to AB meeting the tangent at B in T,
BM: MO = OB : BT,
so that
D: E< OB : BT.
Take a point C on TB produced such that
whence
D: E=OB: BC,
BC > BT.
* PH is described in the Greek as vevovσav éπì (verging to) the point B. As
before the construction is assumed.
160
ARCHIMEDES
Through the points O, T, C describe a circle, and let OB be
produced to meet this circle in K.

C
G
A
M
P
B
F
૩
K
D
E
Then, since BC > BT, and OB is perpendicular to CT, it is
possible to draw from O a straight line OGQ, meeting CT in G
and the circle about OTC in Q, such that GQ = BK*.
Let OGQ meet AB in F and the original circle in P.
Now
and
so that
It follows that
or
CG.GT OG. GQ;
=
OF: OG = BT : GT,
OF. GT=OG.BT.
CG.GT: OF. GT = OG. GQ : OG. BT,
CG OF GQ : BT
=
= BK: BT, by construction,
=
BC: OB
= BC: OP.
Hence
and therefore
OP OF BC: CG,
or
PF: OP BG: BC,
=
PF: BGOP: BC
= OB : BC
D: E.
* The Greek words used are: "it is possible to place another [straight line]
GQ equal to KB verging (veúovσav) towards 0." This particular veûous is
discussed by Pappus (p. 298, ed. Hultsch). See the Introduction, chapter v.
To
ON SPIRALS.
161
Proposition 9.
Given a circle with centre 0, a chord AB less than the
diameter, the tangent at B, and the perpendicular OM from O
on AB, it is possible to draw from O a straight line OPGF,
meeting the circle in P, the tangent in G, and AB produced in F,
such that
FP: BGD: E,
where D: E is any given ratio greater than BM: MO.
in T
Let OT be drawn parallel to AB meeting the tangent at B
Then
D:E>BM : MO
Produce TB to C so that
whence
>OB : BT, by similar triangles.
D: E=OB: BC,
BC < BT.

C
B
A
F
M
G
P
K
T
D
E
Describe a circle through the points O, T, C, and produce OB
to meet this circle in K.
Then, since TB > BC, and OB is perpendicular to CT, it is
possible to draw from O a line OGQ, meeting CT in G, and the
H. A.
11
1
162
ARCHIMEDES
circle about OTC in Q, such that GQ BK*. Let OQ meet
==
the original circle in P and AB produced in F.
We now prove, exactly as in the last proposition, that
CG: OF = BK : BT
= BC: OP.
Thus, as before,
OP: OF
=
BC: CG,
and
OP: PFBC: BG,
whence
PF: BG-OP: BC
=OB: BC
D: E
Proposition 10.
Iƒ A1, A2, A3, …..An be n lines forming an ascending arith-
metical progression in which the common difference is equal
to A₁, the least term, then
2
1
2
2
(n + 1) An² + A₁ (A1 + A2 + ... + An) = 3 (A‚² + A₂² + ... + An³).
[Archimedes' proof of this proposition is given above, p. 107-
9, and it is there pointed out that the result is equivalent to
n (n + 1) (2n + 1)
12+22+32 + ... + n²
6
∙1).]
COR. 1. It follows from this proposition that
and also that
2
2
n. An² <3 (A,² + A₂² + ... + An²),
A„²
2
2
2
n. An² > 3 (A₂² + A₂² + ... + An-1²).
[For the proof of the latter inequality see p. 109 above.]
COR. 2. All the results will equally hold if similar figures
are substituted for squares.
* See the note on the last proposition.
ON SPIRALS.
163
Proposition 11.
Iƒ A1, A2,…….An be n lines forming an ascending arith-
metical progression [in which the common difference is equal to
the least term A₁]*, then
:
2
(n − 1) An² (An² + An-1² + ... + A₂²)
but
< An² : {An. A1 + } (An − A₁)³} ;
2
2
(n − 1) An²: (An-1² + An-2² + ... + A₂³)
-
> An² : {An· A₁ + § (An − A₁)²}.
CH
=
TU
[Archimedes sets out the terms side by side in the manner
shown in the figure, where BC= An, DE = An-1,...RS A1, and
produces DE, FG,...RS until they are
respectively equal to BC or An, so that
EH, GI,...SU in the figure are re-
spectively equal to A1, A2...An-1. He
further measures lengths BK, DL,
FM,...PV along BC, DE, FG, ...PQ re-
spectively each equal to RS.
The figure makes the relations
between the terms easier to see with
the eye, but the use of so large a
number of letters makes the proof
somewhat difficult to follow, and it
E
4
BD F
may be more clearly represented as follows.]
It is evident that (An A₁) = An-1⋅
Q+
-M
V
S
PR
The following proportion is therefore obviously true, viz.
2
(n − 1) An² : (n − 1) (An. A, +} An–1²)
1
= An² : {An⋅ A₂+} (An− A₁)³}.
•
* The proposition is true even when the common difference is not equal to
A₁, and is assumed in the more general form in Props. 25 and 26. But, as
Archimedes' proof assumes the equality of 4, and the common difference, the
words are here inserted to prevent misapprehension.
11-2
164
ARCHIMEDES
In order therefore to prove the desired result, we have only
to show that
2
(n − 1) An. A₁+} (n − 1) An-1² < (An² + An-1² + ... + A₂²)
>(An-1² + An-2² + ... + A₂²).
but
I. To prove the first inequality, we have
(n − 1) An. A₁ + § (n − 1) An−1²
And
1
2
2
2
= (n−1) A¸² + (n − 1) A₁. An−1 + § (n − 1) An—1² …..(1).
2
An³ + An-1²+ ... + ½²
2
=(An−1 + A₁)² + (An~2 + A₂)² + ... + (A₂+ A₁)²
= (An-2² + An-2² + ... + A₂²)
2
+ (n − 1) A¸²
+2A₁ (An-1+An-2 + ... + A₁)
= (An-1² + An-2² + ... + A1²)
2
+ (n − 1) A¸²
1
+ A₁ {An~2+ An-2 + An-3 + ... + A₁
1
2
+A₁ +A₂ + ... + An-2 + An-1}
= (An-1² + An-2² + ... + A₂³)
2
+ (n − 1) A¸²
+nA₁. An-1
(2).
Comparing the right-hand sides of (1) and (2), we see that
(n-1) A² is common to both sides, and
1
2
(n − 1) A₁. An-1 <n A₁. An-1,
while, by Prop. 10, Cor. 1,
2
} (n − 1) An-1² < An-1² + An-2² + ... + A₂².
It follows therefore that
2
2
1.
(n − 1) An. A₂+} (n − 1) An-1² < (An² + An-1² + ... + A₂²);
1
and hence the first part of the proposition is proved.
II. We have now, in order to prove the second result, to
show that
2
2
(n − 1) An. A₁+ } (n − 1) An-1² > (An-1² + An-2² + ... +A₁²).
ON SPIRALS.
165
The right-hand side is equal to
(An−2+ A₁)³+ (An−s + A₁)² + ... + (A₁ + A₁)² + A‚²
2
= An-2² + An-3² + ... + A₂²
2
+(n − 1) A₂²
2
+2A1 (An~2 + An-s + ... + A1)
2
-3
= (An-2² + An-3² + ... + A₂²)
2
+ (n − 1) A¸²
1
+A₁S An-2+An-s+ ... + A₁
\+A₁ +A₂ + ... + An-2)
1
2
= (An-2² + An-3³ + ... + A₂³)
+(n-1) A,2
+ (n − 2) A¸ . An-1…….
(3).
Comparing this expression with the right-hand side of (1) above,
we see that (n - 1) A," is common to both sides, and
2
(n − 1) A₁. An-1 > (n − 2) A¸ . An−1,
while, by Prop. 10, Cor. 1,
Hence
2
} (n − 1) An-1² > (An-2² + An-3² + ... + A₂³).
(n − 1) An. A₁ + } (n − 1) An-1² > (An-1² + An-2² + ... + A‚ª³);
and the second required result follows.
COR. The results in the above proposition are equally true if
similar figures be substituted for squares on the several lines.
DEFINITIONS.
1. If a straight line drawn in a plane revolve at a uniform
rate about one extremity which remains fixed and return to
the position from which it started, and if, at the same time as
the line revolves, a point move at a uniform rate along the
straight line beginning from the extremity which remains fixed,
the point will describe a spiral (λığ) in the plane.
2. Let the extremity of the straight line which remains
166
ARCHIMEDES
fixed while the straight line revolves be called the origin*
(ápɣá) of the spiral.
3. And let the position of the line from which the straight
line began to revolve be called the initial line* in the
revolution (åpxà tâs tepidopâs).
4. Let the length which the point that moves along the
straight line describes in one revolution be called the first
distance, that which the same point describes in the second
revolution the second distance, and similarly let the distances
described in further revolutions be called after the number of
the particular revolution.
5. Let the area bounded by the spiral described in the
first revolution and the first distance be called the first area,
that bounded by the spiral described in the second revolution
and the second distance the second area, and similarly for the
rest in order.
6. If from the origin of the spiral any straight line be
drawn, let that side of it which is in the same direction as that
of the revolution be called forward (πpoayouμeva), and that
which is in the other direction backward (éπóμeva).
7. Let the circle drawn with the origin as centre and the
first distance as radius be called the first circle, that drawn
with the same centre and twice the radius the second circle,
and similarly for the succeeding circles.
Proposition 12.
If any number of straight lines drawn from the origin to
meet the spiral make equal angles with one another, the lines will
be in arithmetical progression.
[The proof is obvious.]
* The literal translation would of course be the "beginning of the spiral "
and "the beginning of the revolution" respectively. But the modern names
will be more suitable for use later on, and are therefore employed here.
ON SPIRALS.
167
Proposition 13.
If a straight line touch the spiral, it will touch it in one point
only.
Let O be the origin of the spiral, and BC a tangent to it.
If possible, let BC touch the spiral in two points P, Q.
Join OP, OQ, and bisect the angle POQ by the straight line OR
meeting the spiral in R.

Q
R
P
K
B
A
Then [Prop. 12] OR is an arithmetic mean between OP and
OQ, or
OP + OQ = 20R.
But in any triangle POQ, if the bisector of the angle POQ
meets PQ in K,
OP + OQ > 20K*.
Therefore OK < OR, and it follows that some point on BC
between P and Q lies within the spiral. Hence BC cuts the
spiral; which is contrary to the hypothesis.
Proposition 14.
If O be the origin, and P, Q two points on the first turn of
the spiral, and if OP, OQ produced meet the 'first circle'
AKP'Q' in P', Q' respectively, OA being the initial line, then
OP: 0Q= (arc AKP'): (arc AKQ').
For, while the revolving line OA moves about O, the point
A on it moves uniformly along the circumference of the circle
* This is assumed as a known proposition; but it is easily proved.
168
ARCHIMEDES
.
AKP'Q', and at the same time the point describing the spiral
moves uniformly along OA.
p'

Q'
Q
P
A
K
Thus, while A describes the arc AKP', the moving point on
OA describes the length OP, and, while A describes the arc
AKQ', the moving point on OA describes the distance OQ.
Hence OP: 0Q=(arc AKP'): (arc AKQ').
[Prop. 2]
Proposition 15.
If P, Q be points on the second turn of the spiral, and OP,
OQ meet the 'first circle' AKP'Q' in P', Q', as in the last
proposition, and if c be the circumference of the first circle, then
OP : OQ = c + (arc AKP') : c + (arc AKQ').
For, while the moving point on OA describes the distance
OP, the point A describes the whole of the circumference of
the 'first circle' together with the arc AKP'; and, while the
moving point on OA describes the distance OQ, the point A
describes the whole circumference of the 'first circle' together
with the arc AKQ'.
COR. Similarly, if P, Q are on the nth turn of the spiral,
OP : OQ = (n − 1) c + (arc AKP') : (n − 1) c + (arc AKQ').
ON SPIRALS.
169
Propositions 16, 17.
If BC be the tangent at P, any point on the spiral, PC being
the 'forward' part of BC, and if OP be joined, the angle OPC
is obtuse while the angle OPB is acute.
I. Suppose P to be on the first turn of the spiral.
Let OA be the initial line, AKP' the 'first circle.' Draw
the circle DLP with centre O and radius OP, meeting OA in
D. This circle must then, in the 'forward' direction from P,

B
PI
C
R
R
P
Q
K
D
A
fall within the spiral, and in the 'backward' direction outside
it, since the radii vectores of the spiral are on the 'forward' side
greater, and on the 'backward' side less, than OP. Hence the
angle OPC cannot be acute, since it cannot be less than the
angle between OP and the tangent to the circle at P, which is
a right angle.
It only remains therefore to prove that OPC is not a right
angle.
If possible, let it be a right angle. BC will then touch
the circle at P.
Therefore [Prop. 5] it is possible to draw a line OQC
meeting the circle through P in Q and BC in C, such that
CQ: 0Q< (arc PQ): (arc DLP)............(1).
170
ARCHIMEDES
Suppose that OC meets the spiral in R and the 'first circle'
in R'; and produce OP to meet the 'first circle' in P'.
From (1) it follows, componendo, that
CO: OQ (arc DLQ): (arc DLP)
<(arc AKR'): (arc AKP')
< OR: OP.
[Prop. 14]
But this is impossible, because OQ= OP, and OR < OC.
Hence the angle OPC is not a right angle. It was also
proved not to be acute.
Therefore the angle OPC is obtuse, and the angle OPB
consequently acute.
II. If P is on the second, or the nth turn, the proof is the
same, except that in the proportion (1) above we have to
substitute for the arc DLP an arc equal to (p+arc DLP) or
(n−1.p+arc DLP), where p is the perimeter of the circle
C

R
Q
P
R'
P
B
K
L
A
4
DLP through P. Similarly, in the later steps, p or (n − 1) p
will be added to each of the arcs DLQ and DLP, and c or
(n-1)c to each of the arcs AKR', AKP', where c is the
circumference of the 'first circle' AKP'.
ON SPIRALS.
171
Propositions 18, 19.
I. If OA be the initial line, A the end of the first turn of
the spiral, and if the tangent to the spiral at A be drawn, the
straight line OB drawn from O perpendicular to OA will meet
the said tangent in some point B, and OB will be equal to the
circumference of the 'first circle.'
II. If A' be the end of the second turn, the perpendicular
OB will meet the tangent at A' in some point B', and OB' will
be equal to 2 (circumference of' second circle').
III. Generally, if An be the end of the nth turn, and OB
meet the tangent at An in Bn, then
OBn
= ncn,
where on is the circumference of the 'nth circle.'
<
I. Let AKC be the first circle.' Then, since the 'back-
ward' angle between OA and the tangent at A is acute [Prop.
16], the tangent will meet the 'first circle' in a second point C.
And the angles CAO, BOA are together less than two right
angles; therefore OB will meet AC produced in some point B.
oàù
p/
FA
Q'
Q
F

B
E
E
D
K
Then, if c be the circumference of the first circle, we have
to prove that
OB = c
If not, OB must be either greater or less than c.
172
ARCHIMEDES
(1) If possible, suppose OB > c.
Measure along OB a length OD less than OB but greater
than c.
We have then a circle AKC, a chord AC in it less than
the diameter, and a ratio AO:OD which is greater than the
ratio AO: OB or (what is, by similar triangles, equal to it) the
ratio of AC to the perpendicular from O on AC. Therefore
[Prop. 7] we can draw a straight line OPF, meeting the circle
in P and CA produced in F, such that
FP: PA AO: OD.
Thus, alternately, since AO
=
PO,
FP: POPA: OD
< (arc PA): c,
since (arc PA) > PA, and OD > c.
FO: PO< (c + arc PA) : c
Componendo,
< OQ: OA,
where OF meets the spiral in Q.
[Prop. 15]
Therefore, since OA = OP, FO < OQ; which is impossible.
Hence
OBc.
(2) If possible, suppose OB < c.
Measure OE along OB so that OE is greater than OB but
less than c.
In this case, since the ratio AO: OE is less than the ratio
AO:OB (or the ratio of AC to the perpendicular from 0
on AC), we can [Prop. 8] draw a line OF'P'G, meeting AC in
F", the circle in P', and the tangent at A to the circle in G,
such that
F'P' : AG = AO : OE.
Let OP'G cut the spiral in Q'.
Then we have, alternately,
F'P': P'0 AG: OE
=
> (arc AP'): c,
because AG > (arc AP'), and OE < c.
ON SPIRALS.
173
Therefore
F'0: P'0 < (arc AKP'): c
<OQ' : OA.
[Prop. 14]
But this is impossible, since OA = OP', and OQ' < OF".
Hence
OB &c.
OB = c.
Since therefore OB is neither greater nor less than c,
II. Let A'K'C' be the second circle,' A'C' being the
tangent to the spiral at A' (which will cut the second circle,
since the backward' angle OA'C' is acute). Thus, as before,
the perpendicular OB' to OA' will meet A'C' produced in some
point B'.
If then c' is the circumference of the 'second circle,' we
have to prove that OB' = 2c'.
E'
B'
F

A'
C'
Q
P
A
K
K'
For, if not, OB' must be either greater or less than 2c'.
(1) If possible, suppose OB'>2c'.
Measure OD' along OB' so that OD' is less than OB' but
greater than 2c'.
Then, as in the case of the first circle' above, we can draw
a straight line OPF meeting the 'second circle' in P and C'A'
produced in F, such that
FP: PA' A'O: OD'.
=
174
ARCHIMEDES
Let OF meet the spiral in Q.
We
e now have, since A'O = PO,
FP: POPA' : OD'
< (arc A'P): 2c,
because (arc A'P) > A'P and OD′ > 2c'.
Therefore
FO: PO < (2c+arc A'P): 2c'
<OQ : OA'.
Hence FO < 0Q; which is impossible.
Thus
OB' 2c'.
[Prop. 15, Cor.]
Similarly, as in the case of the 'first circle', we can prove that
Therefore
OB' ☀ 2c'.
OB' = 2c'.
III. Proceeding, in like manner, to the 'third' and suc-
ceeding circles, we shall prove that
OBn = ncn•
Proposition 20.
I. If P be any point on the first turn of the spiral and OT
be drawn perpendicular to OP, OT will meet the tangent at P tỏ
the spiral in some point T; and, if the circle drawn with centre
O and radius OP meet the initial line in K, then OT is equal to
the arc of this circle between K and P measured in the 'forward'
direction of the spiral.
II. Generally, if P be a point on the nth turn, and the
notation be as before, while p represents the circumference of the
circle with radius OP,
OT = (n − 1) p + arc KP (measured ‘forward').
I. Let P be a point on the first turn of the spiral, OA the
initial line, PR the tangent at P taken in the 'backward'
direction.
Then [Prop. 16] the angle OPR is acute. Therefore PR
ON SPIRALS.
175
meets the circle through P in some point R; and also OT will
meet PR produced in some point T.
If now OT is not equal to the arc KRP, it must be either
greater or less.
U
F

P
R
Q
K
A
(1) If possible, let OT be greater than the arc KRP.
Measure OU along OT less than OT but greater than the
arc KRP.
Then, since the ratio PO: OU is greater than the ratio
PO: OT, or (what is, by similar triangles, equal to it) the
ratio of PR to the perpendicular from O on PR, we can draw
a line OQF, meeting the circle in Q and RP produced in F,
such that
=
FQ: PQ PO: OU
[Prop. 7]
Let OF meet the spiral in Q'.
We have then
FQ: QO = PQ : OU
Componendo,
<(arc PQ): (arc KRP), by hypothesis.
FO: QO< (arc KRQ): (arc KRP)
But QO = OP.
<OQ': OP.
[Prop. 14]
176
ARCHIMEDES
:
Therefore FO< OQ'; which is impossible.
Hence
OT
(arc KRP).
(2) The proof that OT (arc KRP) follows the method of
Prop. 18, I. (2), exactly as the above follows that of Prop. 18,
I. (1).
Since then OT is neither greater nor less than the arc KRP,
it is equal to it.
II. If P be on the second turn, the same method shows
that
OT=p+(arc KRP);
and, similarly, we have, for a point P on the nth turn,
OT = (n-1) p+(arc KRP).
Propositions 21, 22, 23.
Given an area bounded by any arc of a spiral and the lines
joining the extremities of the arc to the origin, it is possible to
circumscribe about the area one figure, and to inscribe in it
another figure, each consisting of similar sectors of circles, and
such that the circumscribed figure exceeds the inscribed by less
than any assigned area.
For let BC be any arc of the spiral, O the origin. Draw
the circle with centre O and radius OC, where C is the 'forward'
end of the arc.
Then, by bisecting the angle BOC, bisecting the resulting
angles, and so on continually, we shall ultimately arrive at
an angle COr cutting off a sector of the circle less than any
assigned area. Let COr be this sector.
Let the other lines dividing the angle BOC into equal parts
meet the spiral in P, Q, and let Or meet it in R. With O as
centre and radii OB, OP, OQ, OR respectively describe arcs of
circles Bp', bBq', pQr', qRc', each meeting the adjacent radii as
shown in the figure. In each case the arc in the 'forward'
direction from each point will fall within, and the arc in the
'backward' direction outside, the spiral.
ON SPIRALS.
177
We have now a circumscribed figure and an inscribed figure
each consisting of similar sectors of circles. To compare their
areas, we take the successive sectors of each, beginning from OC,
and compare them.

B
R
C
p
q'
C'
O
The sector OCr in the circumscribed figure stands alone.
And
(sector ORq) = (sector ORc'),
(sector OQp) = (sector OQr'),
(sector OPb) = (sector OPq'),
while the sector OBp' in the inscribed figure stands alone.
Hence, if the equal sectors be taken away, the difference be-
tween the circumscribed and inscribed figures is equal to the
difference between the sectors OCr and OBp'; and this difference
is less than the sector OCr, which is itself less than any
assigned area.
The proof is exactly the same whatever be the number of
angles into which the angle BOC is
B
b

P
P
Q
divided, the only difference being
that, when the arc begins from the
origin, the smallest sectors OPb, OPq'
in each figure are equal, and there is
therefore no inscribed sector standing
by itself, so that the difference
between the circumscribed and in-
scribed figures is equal to the sector
Ocr itself.
H. A.
C'
R
12
178
ARCHIMEDES
Thus the proposition is universally true.
COR. Since the area bounded by the spiral is intermediate
in magnitude between the circumscribed and inscribed figures,
it follows that
(1) a figure can be circumscribed to the area such that it
exceeds the area by less than any assigned space,
(2) a figure can be inscribed such that the area exceeds it by
less than any assigned space.
Proposition 24.
The area bounded by the first turn of the spiral and the
initial line is equal to one-third of the 'first circle' [= }π (2πα)³,
where the spiral is r=a0].
[The same proof shows equally that, if OP be any radius
vector in the first turn of the spiral, the area of the portion of
the spiral bounded thereby is equal to one-third of that sector of
the circle drawn with radius OP which is bounded by the initial
line and OP, measured in the 'forward' direction from the
initial line.]
Let O be the origin, OA the initial line, A the extremity of
the first turn.
Draw the 'first circle,' i.e. the circle with O as centre and
OA as radius.
Then, if C₁ be the area of the first circle, R, that of the first
turn of the spiral bounded by OA, we have to prove that
R₁ = {С₁.
Ꭱ
For, if not, R₁ must be either greater or less than C₁.
I. If possible, suppose R₁ < ¿C₁.
We can then circumscribe a figure about R₂ made up of
similar sectors of circles such that, if F be the area of this
figure,
whence F<C₁.
F— R₁< ¿C₁ — R₁,
ON SPIRALS.
179
Let OP, OQ,... be the radii of the circular sectors, beginning
from the smallest. The radius of the largest is of course OA.
The radii then form an ascending arithmetical progression
in which the common difference is equal to the least term OP.
If n be the number of the sectors, we have [by Prop. 10, Cor. 1]
n. OA² < 3 (OP² + OQ² + ... + OA³);
2

Q
q
a
A
P
Ρ
and, since the similar sectors are proportional to the squares on
their radii, it follows that
or
C₁ <3F,
F > ¿C₁.
But this is impossible, since F was less than C₁.
Therefore
R₁ JC₁.
II. If possible, suppose R₁ > C₁.
We can then inscribe a figure made up of similar sectors of
circles such that, if ƒ be its area,
whence ƒ> C₁.
R₁ − ƒ < R₁ − ¿C₁,
If there are (n - 1) sectors, their radii, as OP, OQ,..., form
an ascending arithmetical progression in which the least term
is equal to the common difference, and the greatest term, as
OY, is equal to (n-1) OP.
12-2
180
ARCHIMEDES
Thus [Prop. 10, Cor. 1]
n. OA² > 3 (OP² + OQ² + ... + OY³),
whence
or
C₁> 3f,
f< 30₁;
which is impossible, since ƒ> ¿C₁.
Therefore
R₁ 101.
Since then R₁ is neither greater nor less than C₁,
R₁ = {C₁.
[Archimedes does not actually find the area of the spiral
cut off by the radius vector OP, where P is any point on the
first turn; but, in order to do this, we have only to substitute

P
E
K
A
L
in the above proof the area of the sector KLP of the circle
drawn with O as centre and OP as radius for the area C₁ of
the 'first circle', while the two figures made up of similar sectors
have to be circumscribed about and inscribed in the portion
OEP of the spiral. The same method of proof then applies
exactly, and the area of OEP is seen to be (sector KLP).
We can prove also, by the same method, that, if P be a
point on the second, or any later turn, as the nth, the complete
area described by the radius vector from the beginning up to
the time when it reaches the position OP is, if C denote the
area of the complete circle with O as centre and OP as radius,
(C+ sector KLP) or (n−1. C+ sector KLP) respectively.
The area so described by the radius vector is of course not
the same thing as the area bounded by the last complete turn
ON SPIRALS.
181
of the spiral ending at P and the intercepted portion of the
radius vector OP. Thus, suppose R₁ to be the area bounded
by the first turn of the spiral and OA, (the first turn ending at
A₁ on the initial line), R₂ the area added to this by the second
complete turn ending at A, on the initial line, and so on. R, has
then been described twice by the radius vector when it arrives
at the position OA₂; when the radius vector arrives at the
position OA,, it has described R₁ three times, the ring R, twice,
and the ring R, once; and so on.
Thus, generally, if Cn denote the area of the 'nth circle,' we
shall have
‡nCn = Rn +2Rn−1 +3Rn−2+...+nR1,
while the actual area bounded by the outside, or the complete
nth, turn and the intercepted portion of OA, will be equal to
Rn+ Rn−1 + Rn-2 + ... + R₁.
It can now be seen that the results of the later Props. 25
and 26 may be obtained from the extension of Prop. 24 just
given.
To obtain the general result of Prop. 26, suppose BC to be
an arc on any turn whatever of the spiral, being itself less than
a complete turn, and suppose B to be beyond An the extremity
of the nth complete turn, while C is forward' from B.
Let
P
q
be the fraction of a turn between the end of the nth
turn and the point B.
Then the area described by the radius vector up to the
position OB (starting from the beginning of the spiral) is
equal to
↓ (n+2) (circle with rad. OB).
Also the area described by the radius vector from the beginning
up to the position OC is
+ {(n+2) (circle with rad. OC) + (sector B'MC)}.
182
ARCHIMEDES
The area bounded by OB, OC and the portion BEC of the
spiral is equal to the difference between these two expressions;
and, since the circles are to one another as OB² to OC², the
difference may be expressed as
OB2
2) (105) (circle with rad.
(circle with rad. OC) + (sector B′M
3 {(n + 2) (1
But, by Prop. 15, Cor.,
B'MO)}.
(n + 2) (circle B'MO) : {{(n + 2) (circle B'MO) + (sector B'MC)
= OB: OC,

M
so that
B
E
L
An
C
C
B
n+2) (circle B'MC): (sector B'MC) = OB : (OC — OB).
Thus
area BEC
sector B'MC
>
= 1.
OB
{(00-0B) (1-08)+1)
OB
OB(OC +OB) + 00²
00
OC.OB+(OC — OB)²
00
12
The result of Prop. 25 is a particular case of this, and the
result of Prop. 27 follows immediately, as shown under that
proposition.]
ON SPIRALS.
183
Propositions 25, 26, 27.
[Prop. 25.] If A, be the end of the second turn of the spiral,
the area bounded by the second turn and OA, is to the area
of the 'second circle' in the ratio of 7 to 12, being the ratio of
{r₂r₁+} (r₂ − r₁)²} to r2, where r₁, r₂ are the radii of the 'first'
and ‘second' circles respectively.
1
2
[Prop. 26.] If BC be any arc measured in the 'forward
direction on any turn of a spiral, not being greater than the
complete turn, and if a circle be drawn with O as centre and OC
as radius meeting OB in B', then
(area of spiral between OB, OC): (sector OB'C)
=
= {OC. OB+ † (OC — OB)²} : OC².
[Prop. 27.] If R₁ be the area of the first turn of the spiral
bounded by the initial line, R₂ the area of the ring added by the
second complete turn, Rs that of the ring added by the third turn,
and so on, then
Also
R₂.
R₁ = 2R, R₁ =3R2, R5 = 4R2, ..., R₂ = (n − 1) R2
R₂ = 6R₁.
[Archimedes' proof of Prop. 25 is, mutatis mutandis, the
same as his proof of the more general Prop. 26. The latter
will accordingly be given here, and applied to Prop. 25 as a
particular case.]
Let BC be an arc measured in the 'forward' direction on
any turn of the spiral, CKB' the circle drawn with O as centre
and OC as radius.
Take a circle such that the square of its radius is equal
to OC.OB+(OC – OB)", and let σ be a sector in it whose
central angle is equal to the angle BOC.
Thus σ : (sector OB'C) = {OC. OB+} (OC – OB)³} : OC²,
and we have therefore to prove that
(area of spiral OBC)= σ.
For, if not, the area of the spiral OBC (which we will call S)
must be either greater or less than σ.
184
ARCHIMEDES
I. Suppose, if possible, S< σ.
Circumscribe to the area S a figure made up of similar
sectors of circles, such that, if F be the area of the figure,
whence
F-S<σ-S,
F<σ.
Let the radii of the successive sectors, starting from OB,
be OP, OQ,... OC. Produce OP, OQ,... to meet the circle
CKB',...

B'
B
Y
Q
P
૧
σ
K
If then the lines OB, OP, OQ,... OC be n in number, the
number of sectors in the circumscribed figure will be (n − 1),
and the sector OB'C will also be divided into (n-1) equal
sectors. Also OB, OP, OQ,... OC will form an ascending
arithmetical progression of n terms.
Therefore [see Prop. 11 and Cor.]
(n − 1) OC² : (OP² + OQ² + ... + OC³)
< OC² : {OC. OB + {(OC − OB)²}
<(sector OB'C): σ, by hypothesis.
0,
Hence, since similar sectors are as the squares of their radii,
so that
(sector OB'C): F< (sector OB'C): σ,
F>σ.
But this is impossible, because F< σ.
Therefore
So.
ON SPIRALS.
185
II. Suppose, if possible, S> a.
Inscribe in the area S a figure made up of similar sectors of
circles such that, if ƒ be its area,
whence
S−ƒ<S− σ,
f>o.
Suppose OB, OP,... OY to be the radii of the successive
sectors making up the figure ƒ, being (n − 1) in number.
We shall have in this case [see Prop. 11 and Cor.]
(n − 1) OC²: (OB² + OP² + ... + OY²)
00²
whence
so that
-
> 00²: {00.OB + { (OC — OB)³},
(sector OB'C): ƒ> (sector OB'C) : 0,
f<σ.
But this is impossible, because f>σ.
Therefore
S$0.
Since then S is neither greater nor less than σ, it follows that
S=σ.
In the particular case where B coincides with A₁, the end
of the first turn of the spiral, and C with A₂, the end of the
second turn, the sector OB'C becomes the complete 'second
circle,' that, namely, with OA, (or r₂) as radius.
Thus
(area of spiral bounded by OA₂): ('second circle').
= {r₂r₁ + } (r₂ − r₁)³} : r²
=
2
·
2
(since r₁ = 2r₁)
=(2+1): 4 (since r½
= 7:12.
2
Again, the area of the spiral bounded by OA, is equal to
R₁+R (ie. the area bounded by the first turn and OA1,
together with the ring added by the second turn). Also the
'second circle' is four times the 'first circle,' and therefore
equal to 12 R₂.
Hence
or
7:12,
(R₁ + R₁): 12R₁ = 7: 12,
R₁ + R₂ = 7R₁.
Thus
R₂ = 6R₁………….
186
ARCHIMEDES
.
Next, for the third turn, we have
and
(R₂+ R₂+R₂) : ('third circle') = {r¸r₂+ } (r; − r₂)²} : r¸²
therefore
= (3.2+}): 3º
= 19: 27,
19:27,
('third circle') = 9 ('first circle')
= 27R₁;
R₁ + R₂+ R₂ = 19R₁,
and, by (1) above, it follows that
2
R₁ = 12R₁
2R....
.(2),
and so on.
Generally, we have
2
(R₂+ R₂+ ... + Rn): (nth circle) = {rnºn−1 + § (˜n − ˜n−1)²} : In²,
(R₁ + R₂+ ... + Rn-1): (n-1th circle)
and
Therefore
=
2
· {˜n−1 ˜n−2 + {} (˜n—1 − ˜n-2)²} : rn-1²,
(nth circle): (n-1th circle) = r²: rn-1.
(R₁ + R₂ + ... + R₂) : (R₂ + R₂ + ... + Run-1)
Dirimendo,
= {n (n − 1) + }}
{(n − 1) (n − 2) + }}
-
{3n (n − 1) + 1} : {3 (n − 1) (n − 2) + 1}.
-
Rn: (R₂+ R₂+ ... + Run-1)
Rn−1)
= 6 (n − 1): {3 (n − 1) (n − 2) + 1}
(a).
Similarly
R₂-1: (R₂+ R₂+ ... + Rn-2) = 6 (n − 2): {3 (n − 2) (n − 3) + 1},
from which we derive
:
Rn-1 (R₂+ R₂+ ... + Rn-1)
6 (n − 2) : {6 (n − 2) + 3 (n − 2) (n − 3) + 1}
= 6 (n − 2) : {3 (n − 1) (n − 2) + 1} ……………………………. (B).
ON SPIRALS.
187
Combining (a) and (B), we obtain
Thus
Rn: Rn-1 = (n − 1): (n − 2).
R2, R3, R4,... Rn are in the ratio of the successive numbers
1, 2, 3 ... (n - 1).
Proposition 28.
If O be the origin and BC any arc measured in the forward'
direction on any turn of the spiral, let two circles be drawn
(1) with centre 0, and radius OB, meeting OC in C', and
(2) with centre O and radius OC, meeting OB produced in B'.
Then, if E denote the area bounded by the larger circular arc
B'C, the line B'B, and the spiral BC, while F denotes the area
bounded by the smaller arc BC', the line CC' and the spiral BC,
E : F = {OB+ } (OC – OB)} : {OB+{(OC − OB)}.
Leto denote the area of the lesser sector OBC′; then the
larger sector OB'C is equal to σ + F + E.

E
F
B'
C'
σ
B
Thus [Prop. 26]
-
(σ + F) : (0 + F + E) = {OC. OB + † (OC – OB)*} : OC²...(1),
whence
E: (o+F) = {0C (OC – OB) – † (OC — OB)²}
-
:{OC.OB+}(OC — (B)³}
= {OB (OC — OB) + } (OC — OB)²}
: {OC. OB+(OC — OB)}... ..(2).
188
Again
ARCHIMEDES ON SPIRALS.
(σ+F+E): σ = 002: OB2.
Therefore, by the first proportion above, ex aequali,
(o + F) : o = {OC . OB + § (OC – OB)²} : OB²,
whence
(o+F): F= {0C. OB+(OC-OB)³}
: {OB(OC - OB) + † (OC — OB)²}.
Combining this with (2) above, we obtain
E:F={OB(OC — OB) + § (OC — OB)²}
: {OB(OC — OB) + } (OC — OB)³}
= {OB+(OC - OB)}: {OB+(OC – OB)}.
=
I
ON THE EQUILIBRIUM OF PLANES
OR
THE CENTRES OF GRAVITY OF PLANES.
1
BOOK I.
“I POSTULATE the following:
1. Equal weights at equal distances are in equilibrium,
and equal weights at unequal distances are not in equilibrium
but incline towards the weight which is at the greater distance.
2. If, when weights at certain distances are in equilibrium,
something be added to one of the weights, they are not in
equilibrium but incline towards that weight to which the
addition was made.
3. Similarly, if anything be taken away from one of the
weights, they are not in equilibrium but incline towards the
weight from which nothing was taken.
4. When equal and similar plane figures coincide if applied
to one another, their centres of gravity similarly coincide.
5. In figures which are unequal but similar the centres of
gravity will be similarly situated. By points similarly situated
in relation to similar figures I mean points such that, if straight
lines be drawn from them to the equal angles, they make equal
angles with the corresponding sides.
190
ARCHIMEDES
Lop
Lofic
Lot༦
sures to
talk geometric
geometrically
6. If magnitudes at certain distances be in equilibrium,
(other) magnitudes equal to them will also be in equilibrium at
the same distances.
7. In any figure whose perimeter is concave in (one and)
the same direction the centre of gravity must be within the
figure."
Proposition 1.
Weights which balance at equal distances are equal.
For, if they are unequal, take away from the greater the
difference between the two. The remainders will then not
balance [Post. 3]; which is absurd.
Therefore the weights cannot be unequal.
Proposition 2.
Unequal weights at equal distances will not balance but will
incline towards the greater weight.
For take away from the greater the difference between the
two. The equal remainders will therefore balance [Post. 1].
Hence, if we add the difference again, the weights will not
balance but incline towards the greater [Post. 2].
Proposition 3.
Unequal weights will balance at unequal distances, the greater
weight being at the lesser distance.
Let A, B be two unequal weights (of which A is the
greater) balancing about C at distances AC, BC respectively.

A
B
Then shall AC be less than BC. For, if not, take away
from A the weight (A – B.) The remainders will then incline
ON THE EQUILIBRIUM OF PLANES. I.
191
towards B [Post. 3]. But this is impossible, for (1) if AC = CB,
the equal remainders will balance, or (2) if AC> CB, they will
incline towards A at the greater distance [Post. 1].
Hence AC<CB.
Conversely, if the weights balance, and AC< CB, then
A> B.
Proposition 4. novial

If two equal weights have not the same centre of gravity, the
centre of gravity of both taken together is at the middle point of
the line joining their centres of gravity.
[Proved from Prop. 3 by reductio ad absurdum. Archimedes
assumes that the centre of gravity of both together is on the
straight line joining the centres of gravity of each, saying that
this had been proved before (πpodéseikтai). The allusion is no
doubt to the lost treatise On levers (πepì (vyŵv).]
Proposition 5.
If three equal magnitudes have their centres of gravity on a
straight line at equal distances, the centre of gravity of the
system will coincide with that of the middle magnitude.
[This follows immediately from Prop. 4.]
COR 1. The same is true of any odd number of magnitudes
if those which are at equal distances from the middle one are
equal, while the distances between their centres of gravity are
equal.
COR. 2. If there be an even number of magnitudes with
their centres of gravity situated at equal distances on one straight
line, and if the two middle ones be equal, while those which are
equidistant from them (on each side) are equal respectively, the
centre of gravity of the system is the middle point of the line
joining the centres of gravity of the two middle ones.
192
ARCHIMEDES
:
Propositions 6, 7.
Two magnitudes, whether commensurable [Prop. 6] or in-
commensurable [Prop. 7], balance at distances reciprocally
proportional to the magnitudes.
I. Suppose the magnitudes A, B to be commensurable,
and the points A, B to be their centres of gravity. Let DE be
a straight line so divided at C that
A:B=DC : CE.
We have then to prove that, if A be placed at E and B at
D, C is the centre of gravity of the two taken together.


A
B
По

C
D
H
K
N
Since A, B are commensurable, so are DC, CE. Let N be
a common measure of DC, CE. Make DH, DK each equal to
CE, and EL (on CE produced) equal to CD. Then. EH = CD,
since DHCE. Therefore LH is bisected at E, as HK is
bisected at D.
Thus LH, HK must each contain N an even number of
times.
Take a magnitude O such that O is contained as many
times in A as N is contained in LH, whence
But
A:0 LH: N.
=
B:A=CE : DC
=
= HK: LH.
Hence, ex aequali, B: 0 HK: N, or 0 is contained in B as
many times as N is contained in HK.
Thus O is a common measure of A, B.
ON THE EQUILIBRIUM OF PLANES I.
193
Divide LH, HK into parts each equal to N, and A, B into
parts each equal to 0. The parts of A will therefore be equal
in number to those of LH, and the parts of B equal in number
to those of HK. Place one of the parts of A at the middle
point of each of the parts N of LH, and one of the parts of B
at the middle point of each of the parts N of HK.
Then the centre of gravity of the parts of A placed at equal
distances on LH will be at E, the middle point of LH [Prop. 5,
Cor. 2], and the centre of gravity of the parts of B placed at
equal distances along HK will be at D, the middle point of HK.
Thus we may suppose A itself applied at E, and B itself
applied at D.
But the system formed by the parts O of A and B together
is a system of equal magnitudes even in number and placed at
equal distances along LK. And, since LE = CD, and EC = DK,
LC=CK, so that C is the middle point of LK. Therefore C is
the centre of gravity of the system ranged along LK.
Therefore A acting at E and B acting at D balance about
the point C.
II. Suppose the magnitudes to be incommensurable, and
let them be (4+a) and B respectively. Let DE be a line
divided at C so that
D
(A+a): B=DC: CE
C
E

B
a
A
Then, if (A+a) placed at E and B placed at D do not
balance about C', (A + a) is either too great to balance B, or not
great enough.
Suppose, if possible, that (A +a) is too great to balance B.
Take from (A+a) a magnitude a smaller than the deduction
which would make the remainder balance B, but such that the
remainder A and the magnitude B are commensurable.
H. A.
13
194
ARCHIMEDES
}
Then, since A, B are commensurable, and
A: B<DC: CE,
A and B will not balance [Prop. 6], but D will be depressed.
But this is impossible, since the deduction a was an
insufficient deduction from (A+a) to produce equilibrium, so
that E was still depressed.
Therefore (4+ a) is not too great to balance B; and
similarly it may be proved that B is not too great to balance
(A + a).
Hence (A+a), B taken together have their centre of
gravity at C.
Proposition 8.
If AB be a magnitude whose centre of gravity is C, and AD
a part of it whose centre of gravity is F, then the centre of
gravity of the remaining part will be a point G on FC produced
such that
GC: CF=(AD) : (DE).
A
F
D
E

H.
B
For, if the centre of gravity of the remainder (DE) be not
G, let it be a point H. Then an absurdity follows at once from
Props. 6, 7.
Proposition 9.
The centre of gravity of any parallelogram lies on the
straight line joining the middle points of opposite sides.
Let ABCD be a parallelogram, and let EF join the middle
points of the opposite sides AD, BC.
If the centre of gravity does not lie on EF, suppose it to be
H, and draw HK parallel to AD or BC meeting EF in K.
ON THE EQUILIBRIUM OF PLANES I.
195
Then it is possible, by bisecting ED, then bisecting the
halves, and so on continually, to arrive at a length EL less
A
E L

K
H
B
than KH. Divide both AE and ED into parts each equal
to EL, and through the points of division draw parallels to AB
or CD.
We have then a number of equal and similar parallelograms,
and, if any one be applied to any other, their centres of gravity
coincide [Post. 4]. Thus we have an even number of equal
magnitudes whose centres of gravity lie at equal distances along
a straight line. Hence the centre of gravity of the whole
parallelogram will lie on the line joining the centres of gravity
of the two middle parallelograms [Prop. 5, Cor. 2].
But this is impossible, for H is outside the middle
parallelograms.
Therefore the centre of gravity cannot but lie on EF.
Proposition 10.
The centre of gravity of a parallelogram is the point of
intersection of its diagonals.
For, by the last proposition, the centre of gravity lies on
each of the lines which bisect opposite sides. Therefore it
is at the point of their intersection; and this is also the
point of intersection of the diagonals.
Alternative proof.
Let ABCD be the given parallelogram, and BD a diagonal.
Then the triangles ABD, CDB are equal and similar, so that
[Post. 4], if one be applied to the other, their centres of gravity
will fall one upon the other.
13-2
196
ARCHIMEDES
:
+
L
Suppose F to be the centre of gravity of the triangle ABD.
Let G be the middle point of BD.
Join FG and produce it to H, so
that FGGH.
If we then apply the triangle
ABD to the triangle CDB so that
AD falls on CB and AB on CD, the
point F will fall on H.
B
A
F
H
C
D

But [by Post. 4] F will fall on the centre of gravity of
CDB. Therefore H is the centre of gravity of CDB.
Hence, since F, H are the centres of gravity of the two
equal triangles, the centre of gravity of the whole parallelogram
is at the middle point of FH, i.e. at the middle point of BD,
which is the intersection of the two diagonals.
Proposition 11.
If abc, ABC be two similar triangles, and g, G two points in
them similarly situated with respect to them respectively, then, if
g
be the centre of gravity of the triangle abc, G must be the centre
of gravity of the triangle ABC.
Suppose ab: bc: ca= AB: BC : CA.
a
H
A


B
The proposition is proved by an obvious reductio ad
absurdum. For, if G be not the centre of gravity of the
triangle ABC, suppose H to be its centre of gravity.
Post. 5 requires that g, H shall be similarly situated with
respect to the triangles respectively; and this leads at once
to the absurdity that the angles HAB, GAB are equal.
ON THE EQUILIBRIUM OF PLANES I.
197
Proposition 12.
Given two similar triangles abc, ABC, and d, D the middle
points of bc, BC respectively, then, if the centre of gravity of abc
lie on ad, that of ABC will lie on AD.
Let g be the point on ad which is the centre of gravity
of abc.

a
G
A

b
d
Take G on AD such that
B
D
ad: ag = AD: AG,
and join gb, gc, GB, GC.
Then, since the triangles are similar, and bd, BD are the
halves of bc, BC respectively,
ab: bd AB: BD,
and the angles abd, ABD are equal.
Therefore the triangles abd, ABD are similar, and

Also
while, from above,
bad=BAD.
ba: ad=BA : AD,
ad ag=AD: AG.
adag:
Therefore ba: ag = BA: AG, while the angles bag, BAG
are equal.
Hence the triangles bag, BAG are similar, and
▲ abg=▲ ABG.
And, since the angles abd, ABD are equal, it follows that
<gbd=GBD.
In exactly the same manner we prove that
<gac=4GAC,
Zacg=<ACG,
gcd=GCD.
198
ARCHIMEDES
medium
Therefore g, G are similarly situated with respect to the
triangles respectively; whence [Prop. 11] G is the centre of
gravity of ABC.
Proposition 13.
In any triangle the centre of gravity lies on the straight line
joining any angle to the middle point of the opposite side.
Let ABC be a triangle and D the middle point of BC.
Join AD. Then shall the centre of gravity lie on AD.
For, if possible, let this not be the case, and let H be the
centre of gravity. Draw HI parallel to CB meeting AD in I.
Then, if we bisect DC, then bisect the halves, and so on,
we shall at length arrive at a length, as DE, less than HI.
A

M
N
R
L
S
P
H
O
Q
K
T
X
B
F
DE
Divide both BD and DC into lengths each equal to DE, and
through the points of division draw lines each parallel to DA
meeting BA and AC in points as K, L, M and N, P, Q
respectively.
Join MN, LP, KQ, which lines will then be each parallel
to BC.
We have now a series of parallelograms as FQ, TP, SN,
and AD bisects opposite sides in each. Thus the centre
of gravity of each parallelogram lies on AD [Prop. 9], and
therefore the centre of gravity of the figure made up of them
all lies on AD.
ON THE EQUILIBRIUM OF PLANES I.
199
Let the centre of gravity of all the parallelograms taken
together be 0. Join OH and produce it; also draw CV
parallel to DA meeting OH produced in V.
Now, if n be the number of parts into which AC is divided,
AADC: (sum of triangles on AN, NP,...)
Similarly
And
12
= AC²: (AN² + NP² + ...)
= n²:n 2
=n: 1
?
= AC: AN.
▲ ABD : (sum of triangles on AM, ML, ...) = AB : AM.
It follows that
AC: AN AB: AM.
A ABC (sum of all the small As) = CA: AN
Suppose OV produced to X so that
> VO: OH, by parallels.
▲ ABC: (sum of small As) = X0:0H,
whence, dividendo,
(sum of parallelograms): (sum of small ▲s)=XH : HO.
Since then the centre of gravity of the triangle ABC is at H,
and the centre of gravity of the part of it made up of the
parallelograms is at O, it follows from Prop. 8 that the centre
of gravity of the remaining portion consisting of all the small
triangles taken together is at X.
no
But this is impossible, since all the triangles are on one side
of the line through X parallel to AD.
Therefore the centre of gravity of the triangle cannot but
lie on AD.
Alternative proof.
Suppose, if possible, that H, not lying on AD, is the centre
of gravity of the triangle ABC. Join AH, BH, CH. Let
E, F be the middle points of CA, AB respectively, and join
DE, EF, FD. Let EF meet AD in M.
200
ARCHIMEDES
L
Draw FK, EL parallel to AH meeting BH, CH in K,
respectively. Join KD, HD, LD, KL. Let KL meet DH in
N, and join MN.
A

F
E
M
K
B
Since DE is parallel to AB, the triangles ABC, EDC are
similar.
And, since CEEA, and EL is parallel to AH, it follows
that CL = LH. And CD = DB. Therefore BH is parallel
to DL.
Thus in the similar and similarly situated triangles ABC,
EDC the straight lines AH, BH are respectively parallel to
EL, DL; and it follows that H, L are similarly situated with
respect to the triangles respectively.
But His, by hypothesis, the centre of gravity of ABC.
Therefore L is the centre of gravity of EDC. [Prop. 11]
Similarly the point K is the centre of gravity of the
triangle FBD.
And the triangles FBD, EDC are equal, so that the centre
of gravity of both together is at the middle point of KL, i.e. at
the point N.
The remainder of the triangle ABC, after the triangles FBD,
EDC are deducted, is the parallelogram AFDE, and the centre
of gravity of this parallelogram is at M, the intersection of its
diagonals.
It follows that the centre of gravity of the whole triangle
ABC must lie on MN; that is, MN must pass through H, which
is impossible (since MN is parallel to AH).
Therefore the centre of gravity of the triangle ABC cannot
but lie on AD.
ON THE EQUILIBRIUM OF PLANES I.
201
Proposition 14.
It follows at once from the last proposition that the centre
of gravity of any triangle is at the intersection of the lines drawn
from any two angles to the middle points of the opposite sides
respectively.
Proposition 15.
If AD, BC be the two parallel sides of a trapezium ABCD,
AD being the smaller, and if AD, BC be bisected at E, F
respectively, then the centre of gravity of the trapezium is at a
point G on EF such that
GE: GF=(2BC+AD) : (2AD+BC).
Produce BA, CD to meet at 0. Then FE produced will
also pass through 0, since AE = ED, and BF = FC.

A
RK
M
S
G
P
Q
W
H
B
Now the centre of gravity of the triangle OAD will lie on
OE, and that of the triangle OBC will lie on OF. [Prop. 13]
It follows that the centre of gravity of the remainder, the
trapezium ABCD, will also lie on OF.
[Prop. 8]
Join BD, and divide it at L, M into three equal parts.
Through L, M draw PQ, RS parallel to BC meeting BA in
P, R, FE in W, V, and CD in Q, S respectively.
Join DF, BE meeting PQ in H and RS in K respectively.
Now, since
BL=1BD,
FH=1FD.
་
202
ARCHIMEDES.
Therefore H is the centre of gravity of the triangle DBC*.
Similarly, since EK = BE, it follows that K is the centre
of gravity of the triangle ADB.
Therefore the centre of gravity of the triangles DBC, ADB
together, i.e. of the trapezium, lies on the line HK.
But it also lies on OF.
Therefore, if OF, HK meet in G, G is the centre of gravity
of the trapezium.
Hence [Props. 6, 7]
But
Therefore
It follows that
A DBC: A ABD = KG: GH
=VG: GW.
▲ DBC: ▲ ABD = BC : AD.
› BC:AD=VG:GW.
FC: EAD:G:
(2BC+AD) : (2AD+BC)=(2VG+GW) : (2GW +VG)
= EG : GF.
Q. E. D.
* This easy deduction from Prop. 14 is assumed by Archimedes without
proof.
ON THE EQUILIBRIUM OF PLANES.
BOOK II.
Proposition 1.
If P, P' be two parabolic segments and D, E their centres
of gravity respectively, the centre of gravity of the two segments
taken together will be at a point C on DE determined by the
relation
P: P' CE: CD*.
In the same straight line with DE measure EH, EL each
equal to DC, and DK equal to DH; whence it follows at once
that DK = CE, and also that KC = CL.

(P)
x 3
K
D
H
C
E
N
(P')
* This proposition is really a particular case of Props. 6, 7 of Book I. and
is therefore hardly necessary. As, however, Book II. relates exclusively to
parabolic segments, Archimedes' object was perhaps to emphasize the fact
that the magnitudes in I. 6, 7 might be parabolic segments as well as
rectilinear figures. His procedure is to substitute for the segments rect-
angles of equal area, a substitution which is rendered possible by the results
obtained in his separate treatise on the Quadrature of the Parabola.
204
ARCHIMEDES
Apply a rectangle MN equal in area to the parabolic
segment P to a base equal to KH, and place the rectangle so
that KH bisects it, and is parallel to its base.
Then D is the centre of gravity of MN, since KD=DH.
Produce the sides of the rectangle which are parallel to KH,
and complete the rectangle NO whose base is equal to HL.
Then E is the centre of gravity of the rectangle NO.
Now
(MN): (NO)= KH : HL
= DH: EH
=CE: CD
=P: P'.
But
Therefore
(MN)=P.
(NO)=P'.
Also, since C is the middle point of KL, C is the centre
of gravity of the whole parallelogram made up of the two
parallelograms (MN), (NO), which are equal to, and have the
same centres of gravity as, P, P' respectively.
Hence C is the centre of gravity of P, P' taken together.
Definition and lemmas preliminary to Proposition 2.
"If in a segment bounded by a straight line and a section
of a right-angled cone [a parabola] a triangle be inscribed
having the same base as the segment and equal height, if again
triangles be inscribed in the remaining segments having the
same bases as the segments and equal height, and if in the
remaining segments triangles be inscribed in the same manner,
let the resulting figure be said to be inscribed in the
recognised manner (yvwpíμws éyypápeσbai) in the segment.
And it is plain
(1) that the lines joining the two angles of the figure so inscribed
which *are nearest to the vertex of the segment, and the next
ON THE EQUILIBRIUM OF PLANES II.
205
pairs of angles in order, will be parallel to the base of the
segment,
(2) that the said lines will be bisected by the diameter of the
segment, and
(3) that they will cut the diameter in the proportions of the
successive odd numbers, the number one having reference to [the
length adjacent to] the vertex of the segment.
And these properties will have to be proved in their proper
places (ἐν ταῖς τάξεσιν).”
[The last words indicate an intention to give these pro-
positions in their proper connexion with systematic proofs; but
the intention does not appear to have been carried out, or at
least we know of no lost work of Archimedes in which they
could have appeared. The results can however be easily
derived from propositions given in the Quadrature of the
Parabola as follows.
(1) Let BRQPApqrb be a figure inscribed in the recog-
nised manner' in the parabolic segment BAb of which Bb is
the base, A the vertex and AO the diameter.
B

P
R
G
A
W
X
e
P
q
E
Bisect each of the lines BQ, BA, QA, Aq, Ab, qb, and
through the middle points draw lines parallel to AO meeting
Bb in G, F, E, e, f, g respectively.
206
ARCHIMEDES
1
1
These lines will then pass through the vertices R, Q, P,
p, q, r of the respective parabolic segments [Quadrature of the
Parabola, Prop. 18], i.e. through the angular points of the
inscribed figure (since the triangles and segments are of equal
height).
Also BG = GF=FE = EO, and Oe=ef=fg = gb. But
BO = Ob, and therefore all the parts into which Bb is divided
are equal.
If now AB, RG meet in L, and Ab, rg in l, we have
whence GL=gl.
BG: GL = BO: OA, by parallels,
Again [ibid., Prop. 4]
=b0: OA
=bg: gl,
GL : LR=BO : OG
=b0: Og
= gl: lr ;
and, since GL = gl, LR= lr.
Therefore GR, gr are equal as well as parallel.
Hence GRrg is a parallelogram, and Rr is parallel to Bb.
Similarly it may be shown that Pp, Qq are each parallel
to Bb.
(2) Since RGgr is a parallelogram, and RG, rg are
parallel to AO, while GO Og, it follows that Rr is bisected
by AO.
=
And similarly for Pp, Qq.
(3) Lastly, if V, W, X be the points of bisection of Pp,
Qq, Rr,
AV: AW: AX : A0 = PV² : QW²: RX²: BO²
=1:49:16,
whence
AV: VW: WX:X0=1:35:7.]
ON THE EQUILIBRIUM OF PLANES II.
207
Proposition 2.
If a figure be 'inscribed in the recognised manner' in a
parabolic segment, the centre of gravity of the figure so inscribed
will lie on the diameter of the segment.
For, in the figure of the foregoing lemmas, the centre of
gravity of the trapezium BRrb must lie on XO, that of the
trapezium RQqr on WX, and so on, while the centre of gravity
of the triangle PAp lies on AV.
Hence the centre of gravity of the whole figure lies on AO.
Proposition 3.
If BAB', bab' be two similar parabolic segments whose
diameters are AO, ao respectively, and if a figure be inscribed
in each segment ‘in the recognised manner,' the number of sides
in each figure being equal, the centres of gravity of the inscribed
figures will divide AO, ao in the same ratio.
[Archimedes enunciates this proposition as true of similar
segments, but it is equally true of segments which are not
similar, as the course of the proof will show.]
Suppose BRQPAP'Q'R'B', brqpap'q'r'b' to be the two
figures inscribed in the recognised manner.' Join PP', QQ',
RR' meeting AO in L, M, N, and pp', qq', rr' meeting ao
in l, m, n.
Then [Lemma (3)]
AL: LM: MN: NO
=1:3:5:7
=al: lm : mn: no,
so that AO, ao are divided in the same proportion.
Also, by reversing the proof of Lemma (3), we see that
PP': pp'=QQ' : qq′ = RR' : rr' = BB' : bb'.
Since then RR' : BB' = rr' : bb', and these ratios respec-
tively determine the proportion in which NO, no are divided
208
ARCHIMEDES
=
*
by the centres of gravity of the trapezia BRR'B', brr′b′ [1. 15],
it follows that the centres of gravity of the trapezia divide NO,
no in the same ratio.
go
P
A
Im
L
n
M
N
R
B

R

B'
Similarly the centres of gravity of the trapezia RQQʻR',
rqqʻr' divide MN, mn in the same ratio respectively, and so on.
Lastly, the centres of gravity of the triangles PAP', pap'
divide AL, al respectively in the same ratio.
Moreover the corresponding trapezia and triangles are, each
to each, in the same proportion (since their sides and heights
are respectively proportional), while AO, ao are divided in
the same proportion.
Therefore the centres of gravity of the complete inscribed
figures divide AO, ao in the same proportion.
Proposition 4.
The centre of gravity of any parabolic segment cut off by a
straight line lies on the diameter of the segment.
Let BAB' be a parabolic segment, A its vertex and AO its
diameter.
Then, if the centre of gravity of the segment does not lie on
AO, suppose it to be, if possible, the point F. Draw FE
parallel to AO meeting BB' in E.
ON THE EQUILIBRIUM OF PLANES II.
209
Inscribe in the segment the triangle ABB' having the same
vertex and height as the segment, and take an area S such
that
AABB': S=BE: EO.

K
B
F
E
A
O
H
A
B'
We can then inscribe in the segment 'in the recognised
manner' a figure such that the segments of the parabola left
over are together less than S. [For Prop. 20 of the Quadrature
of the Parabola proves that, if in any segment the triangle with
the same base and height be inscribed, the triangle is greater
than half the segment; whence it appears that, each time that
we increase the number of the sides of the figure inscribed 'in
the recognised manner,' we take away more than half of the
remaining segments.]
Let the inscribed figure be drawn accordingly; its centre
of gravity then lies on AO [Prop. 2]. Let it be the point H.
Join HF and produce it to meet in K the line through B
parallel to AO.
Then we have
(inscribed figure): (remainder of segmt.)>▲ABB' : S
> BE: EO
> KF: FH.
Suppose I taken on HK produced so that the former ratio is
equal to the ratio LF: FH.
H. A.
14
'210
ARCHIMEDES
籬
​Then, since H is the centre of gravity of
the inscribed
figure, and F that of the segment, I must be the centre
of gravity of all the segments taken together which form the
remainder of the original segment.
[I. 8]
But this is impossible, since all these segments lie on one
side of the line drawn through L parallel to AO [Cf. Post. 7].
Hence the centre of gravity of the segment cannot but lie
on AO.
Proposition 5.
If in a parabolic segment a figure be inscribed in the
recognised manner,' the centre of gravity of the segment is nearer
to the vertex of the segment than the centre of gravity of the
inscribed figure is.
Let BAB' be the given segment, and AO its diameter.
First, let ABB' be the triangle in-
scribed in the recognised manner.'
Divide AO in F so that AF=2F0;
F is then the centre of gravity of the
triangle ABB'.
Q H
K
A
O
E F
H'
B

Bisect AB, AB' in D, D' respec-
tively, and join DD' meeting 40 in E.
Draw DQ, D'Q' parallel to OA to meet
the curve. QD, Q'D' will then be the
diameters of the segments whose bases
are AB, AB', and the centres of gravity
of those segments will lie respectively
on QD, Q'D' [Prop. 4]. Let them be H, H', and join HH'
meeting AO in K.
B'
Now QD, Q'D' are equal*, and therefore the segments of
which they are the diameters are equal [On Conoids and
Spheroids, Prop. 3].
* This may either be inferred from Lemma (1) above (since QQ', DD' are
both parallel to BB'), or from Prop. 19 of the Quadrature of the Parabola, which
applies equally to Q or Q'.

ON THE EQUILIBRIUM OF PLANES II.
211 ·
Also, since QD, Q'D' are parallel*, and DE = ED', K is the
middle point of HH'.
Hence the centre of gravity of the equal segments AQB,
AQ'B' taken together is K, where K lies between E and A.
And the centre of gravity of the triangle ABB' is F.
It follows that the centre of gravity of the whole segment
BAB' lies between K and F, and is therefore nearer to the
vertex A than F is.
Secondly, take the five-sided figure BQAQ'B' inscribed 'in
the recognised manner,' QD, Q'D' being, as before, the diameters
of the segments AQB, AQ'B'.
Then, by the first part of this proposition, the centre of
gravity of the segment AQB (lying of course on QD) is nearer
to Q than the centre of gravity of the triangle AQB is. Let
the centre of gravity of the segment be H, and that of the
triangle I.
Similarly let H' be the centre of gravity of the segment
AQ'B', and I' that of the triangle AQ'B'.
It follows that the centre of gravity
of the two segments AQB, AQ'B' taken
together is K, the middle point of HH',
and that of the two triangles AQB, AQ‘B'
is L, the middle point of II'.
If now the centre of gravity of the
triangle ABB' be F, the centre of gravity
of the whole segment BAB' (i.e. that of
the triangle ABB' and the two segments
AQB, AQ'B' taken together) is a point
G on KF determined by the proportion
A
QHI
HT
D
K
L GFF
D'
B'
B

(sum of segments AQB, AQ'B'): ^ABB' = FG : GK. [I. 6, 7]
* There is clearly some interpolation in the text here, which has the words
καὶ ἐπεὶ παραλληλόγραμμόν ἐστι τὸ ΘΖΗΙ. It is not yet proved that HDDH is
a parallelogram; this can only be inferred from the fact that H, H' divide QD,
Q'D' respectively in the same ratio. But this latter property does not appear
till Prop. 7, and is then only enunciated of similar segments. The interpolation
must have been made before Eutocius' time, because he has a note on the
phrase, and explains it by gravely assuming that H, H' divide QD, Q'D' respec-
tively in the same ratio.
14-2
212
ARCHIMEDES
TA
And the centre of gravity of the inscribed figure BQAQ'B'
is a point F' on LF determined by the proportion
[Hence
or
(AAQB+^AQ'B') : ▲ ABB' = FF' : F'L. [I. 6, 7]
and, componendo,
FG: GK >FF": FL,
GK: FGF'L: FF",
FK: FG < FL: FF", while FK > FL.]
Therefore FG> FF", or G lies nearer than F" to the vertex A.
Using this last result, and proceeding in the same way,
we can prove the proposition for any figure inscribed in the
recognised manner.’
Proposition 6.
Given a segment of a parabola cut off by a straight line, it is
possible to inscribe in it 'in the recognised manner' a figure such
that the distance between the centres of gravity of the segment and
of the inscribed figure is less than any assigned length.
Let BAB' be the segment, AO its diameter, G its centre
of gravity, and ABB' the triangle inscribed in the recognised
manner.'
Let D be the assigned length and S an area such that
AG: D-AABB': S.
In the segment inscribe 'in the recognised manner' a figure
such that the sum of the segments left over is less than S.
Let F be the centre of gravity of the inscribed figure.
We shall prove that FG< D.
For, if not, FG must be either equal to, or greater than, D.
And clearly
(inscribed fig.): (sum of remaining segmts.)
>▲ABB' : S
> AG: D
> AG: FG, by hypothesis (since FG & D).
ON THE EQUILIBRIUM OF PLANES II.
213
Let the first ratio be equal to the ratio KG: FG (where K
lies on GA produced); and it follows that K is the centre of
gravity of the small segments taken together.
[I. 8]
B

K
10
A
GF
B'
S
But this is impossible, since the segments are all on the
same side of a line drawn through K parallel to BB'.
Hence FG cannot but be less than D.
Proposition 7.
If there be two similar parabolic segments, their centres of
gravity divide their diameters in the same ratio.
[This proposition, though enunciated of similar segments
only, like Prop. 3 on which it depends, is equally true of
any segments. This fact did not escape Archimedes, who
uses the proposition in its more general form for the proof of
Prop. 8 immediately following.]
Let BAB', bab' be the two similar segments, AO, ao their
diameters, and G, g their centres of gravity respectively.
Then, if G, g do not divide AO, ao respectively in the same
ratio, suppose H to be such a point on 40 that
AH: HO=ag : go ;
214
ARCHIMEDES
:
and inscribe in the segment BAB' 'in the recognised manner'
a figure such that, if F be its centre of gravity,
GF GH.
B
[Prop. 6]

a
A
gf
GFH
b

B'
Inscribe in the segment bab' 'in the recognised manner' a
similar figure; then, if ƒ be the centre of gravity of this figure,
ag <af.
[Prop. 5]
And, by Prop. 3,
af: fo= AF: FO.
But
AF: FO<AH: HO
<ag : go, by hypothesis.
Therefore
af: fo<ag: go; which is impossible.
It follows that G, g cannot but divide AO, ao in the same
ratio.
Proposition 8.
If AO be the diameter of a parabolic segment, and G its
centre of gravity, then
AG=& GO.
Let the segment be BAB'. Inscribe the triangle ABB' ‘in
the recognised manner,' and let F be its centre of gravity.
Bisect AB, AB' in D, D', and draw DQ, D'Q' parallel to OA
to meet the curve, so that QD, Q'D' are the diameters of the
segments AQB, AQ'B' respectively.
Let H, H' be the centres of gravity of the segments AQB,
AQ'B' respectively. Join QQ', HH' meeting AO in V, K
respectively.
ON THE EQUILIBRIUM OF PLANES II.
215
K is then the centre of gravity of the two segments AQB,
AQ'B' taken together.
B

Now
whence
AG: GO=QH : HD,
[Prop. 7]
H
D
AO: OG=QD : HD.
But A0=4QD [as is easily proved
by means of Lemma (3), p. 206].
Therefore
and, by subtraction,
A
V/LK
GF
OG=4HD;
D'
H'
AG=4QH.
Also, by Lemma (2), QQ' is paral-
lel to BB' and therefore to DD'. It
B'
follows from Prop. 7 that HH' is also parallel to QQ' or DD',
and hence
Therefore
and
QH=VK.
AG=4VK,
AV+KG=3VK.
Measuring VL along VK so that VL AV, we have
Again
whence
Now, by I. 6, 7,
KG=3LK..
.(1).
A0=4AV
[Lemma (3)]
=3AL, since AV-3VL,
AL=3A0=0F …..
(2).
and
▲ABB': (sum of segmts. AQB, AQ'B') = KG : GF,
▲ABB' = 3 (sum of segments AQB, AQ'B')
[since the segment ABB' is equal to ₫ ▲ ABB' (Quadrature of
the Parabola, Props. 17, 24)].
Hence
But
Therefore
KG=3GF.
KG = 3LK, from (1) above.
LF = LK+KG+GF
= 5GF.
216
ARCHIMEDES
And, from (2),
LF (AO-AL-OF)=1A0=OF.
Therefore
OF=5GF,
and
OG=6GF.
But
A0=30F=15GF.
Therefore, by subtraction,
AG=9GF
= & GO.
Proposition 9 (Lemma).
If a, b, c, d be four lines in continued proportion and in
descending order of magnitude, and if
d: (a–d) = x: § (a−c),
and (2a+4b+6c+3d): (5a+10b+10c + 5d)=y: (a−c),
it is required to prove that
x+y=fa.
A
T
[The following is the proof given by Archimedes, with
the only difference that it is set out in
algebraical instead of geometrical notation.
This is done in the particular case simply in
order to make the proof easier to follow.
Archimedes exhibits his lines in the figure
reproduced in the margin, but, now that it is
possible to use algebraical notation, there is
no advantage in using the figure and the more
cumbrous notation which only obscures the course
of the proof. The relation between Archimedes'
figure and the letters used below is as follows;
AB
H
A
E
Z
= α, TB=b, AB = c, EB =d, ZH = x, HO=y, AO = z.]
We have
(1),
B
α
b
C
b
C
α ·b b-c
whence
b
C
c- d
d
a-b
b-c
α
b
C
and therefore
11
b-c
C
d b
(2).
C
Now
2c
2(a+b)_a+b_a+b b
α
c b - c a
C
C
bc
b - cc - d
c - d'
ON THE EQUILIBRIUM OF PLANES II.
217
And, in like manner,
b+c b+c c
d
α
C
C d
c-d'
It follows from the last two relations that
α C 2a+3b+c
C
d
2c+ d
(3).
Suppose z to be so taken that
so that z<(c–d).
Therefore
2a +46 + 4c+2d
2c+d
a-c+z
α-c
And, by hypothesis,
a - c
(4),
2
2a + 4b+6c+3d
2 (a + d) + 4 (b+c) *
c _ 5 (a + d) + 10 (b + c)
2a + 4b+ 6c+ 3d
α
Y
a-c+z
so that
Y
5 (a + d) + 10 (b + c)
2 (a+d)+4(b+c)
10
5
(5).
2
Again, dividing (3) by (4) crosswise, we obtain
2
2a+3b+c
c-d 2(a+d) + 4 (b+c) '
whence
But, by (2),
c-d-z
b+3c+2d
c-d
2 (a + d) + 4 (b + c)
(6).
c - d_a−b_3 (b−c)_2 (c-d)
_
d
b
b
so that
3c
2d
c_d__ (a - b) + 3 (b −c) + 2 (c–d)
d
b+3c+2d
Combining (6) and (7), we have
c - d − z
(a−b) + 3 (b −c) + 2 (c–d)
(7).
whence
d
And, since [by (1)]
2(c
2 (a + d) + 4 (b + c)
C
Z
3a + 6b+3c
.(8).
d
2 (a + d) + 4 (b + c)
c-d_b-c
a-b
c+d¯¯¯ b + c
a + b '
218
ARCHIMEDES
1
we have
c - d
c+d
>
a C b+c+a+b
a+2b+2c+d__ 2 (a + d) + 4 (b+c)
2 (a + d) + 4 (b + c ) ..... (9).
2 (a + c) + 4b
2 (u + d) + 4 (b+c)
a-d
whence
a
C
a + 2b + c
a - d
Thus
% (a - c)
d
X
C
But, by (8),
ď
and therefore, by hypothesis,
3
{2(a+c)+4b}
2 (a + d) + 4 (b + c)
§ {2 (a + c) + 4b}
3a+6b+3c
2 (a + d) + 4 (b+c)
and it follows, ex aequali, that
And, by (5),
Therefore
or
c - z
&
OC
}
M 103
3(a+c)+6b 5 3
{2(a+c)+4b} 32
α c+ z 5
5
20 121
Y
α
x + y
x+y=za.
2
Proposition 10.
11
5
10
ல்
2'
If PP'B'B be the portion of a parabola intercepted between
two parallel chords PP', BB' bisected respectively in N, O by
the diameter ANO (N being nearer than O to A, the vertex
of the segments), and if NO be divided into five equal parts of
which LM is the middle one (L being nearer than M to N), then,
if G be a point on LM such that
LG : GM = BO². (2PN+BO) : PN². (2B0+PN),
G will be the centre of gravity of the area PP'B' B.
Take a line ao equal to AO, and an on it equal to AN. Let
p, q be points on the line ao such that
ao : aq=aq : an
ao : an= aq: ap
(1),
(2),
[whence ao : aq= aq : an = an : ap, or ao, aq, an, ap are lines in
continued proportion and in descending order of magnitude].
Measure along GA a length GF such that
op: ap = OL: GF…………..
…………..(3)..
ON THE EQUILIBRIUM OF PLANES II.
219
Then, since PN, BO are ordinates to ANO,
so that
and
B02: PN2=AO: AN
= ao : an
ao² : aq², by (1),
B0: PN = αo : aq
3
BO³: PN³ = αo³: aq³
= (ao : ag). (aq: an).(an : ap)
=
P
= ao : ap
B
A
HN FLOM
α
p n
P
心
​0
(4),
(5).

B'
Thus (segment BAB'): (segment PAP')
whence
= ABAB': APAP
= BO³ : PN³
= ao : ap,
(area PP'B'B): (segment PAP') = op: ap
=OL: GF, by (3),
=gON : GF ……………………. (6).
:GF...
Now B0². (2PN+BO) : B0³ = (2PN+BO) : BO
and
= (2aq+ao): ao, by (4),
BO³ : PN³=ao : ap, by (5),
PN³: PN². (2B0+PN)=PN : (2B0+PN)
=aq: (2ao+aq), by (4),
= ap: (2an+ap), by (2).
220
ARCHIMEDES.
}
}
[
L
Hence, ex aequali,
B0². (2PN+BO) : PN². (2B0+PN)=(2aq+ao) : (2an+ap),
so that, by hypothesis,
LG : GM = (2aq + ao): (2an+ap).
Componendo, and multiplying the antecedents by 5,
ON : GM = {5 (ao+ap) + 10 (aq + an)} : (2an+ap).
But ON: OM 5:2
=
{5 (ao+ap) + 10 (aq +an)} : {2 (ao+ap)+4(aq +an)}.
It follows that
ON : OG= {5 (ao + ap) + 10 (aq + an)} : (2ao + 4aq+6an+3ap).
Therefore
(2ao+4aq+6an+3ap): {5 (ao+ap) + 10 (aq + an)} = OG : ON
And
ap: (ao — ap) = ap: op
-
OG: on.
GF: OL, by hypothesis,
=
GF: on,
while ao, aq, an, ap are in continued proportion.
Therefore, by Prop. 9,
GF+ OG = OFαo=OA.
Thus F is the centre of gravity of the segment BAB'. [Prop. 8]
Let H be the centre of gravity of the segment PAP', so
that AH=&AN.
And, since
AF=3AO,
HF=&ON.
we have, by subtraction, HF = %ON.
But, by (6) above,
(area PP'B'B) : (segment PAP')=§ON:GF
=HF: FG.
Thus, since F, H are the centres of gravity of the segments
BAB', PAP' respectively, it follows [by I. 6, 7] that G is the
centre of gravity of the area PP'B'B.
THE SAND-RECKONER.
'THERE are some, king Gelon, who think that the number
of the sand is infinite in multitude; and I mean by the sand
not only that which exists about Syracuse and the rest of Sicily
but also that which is found in every region whether inhabited
or uninhabited. Again there are some who, without regarding
it as infinite, yet think that no number has been named which
is great enough to exceed its multitude. And it is clear that
they who hold this view, if they imagined a mass made up of
sand in other respects as large as the mass of the earth, in-
cluding in it all the seas and the hollows of the earth filled up
to a height equal to that of the highest of the mountains,
would be many times further still from recognising that any
number could be expressed which exceeded the multitude of
the sand so taken. But I will try to show you by means of
geometrical proofs, which you will be able to follow, that, of the
numbers named by me and given in the work which I sent to
Zeuxippus, some exceed not only the number of the mass of
sand equal in magnitude to the earth filled up in the way
described, but also that of a mass equal in magnitude to the
universe. Now you are aware that 'universe' is the name
given by most astronomers to the sphere whose centre is the
centre of the earth and whose radius is equal to the straight
line between the centre of the sun and the centre of the earth.
This is the common account (rà ypaþóµeva), as you have heard
from astronomers. But Aristarchus of Samos brought out a
222
ARCHIMEDES
book consisting of some hypotheses, in which the premisses lead
to the result that the universe is many times greater than that
now so called. His hypotheses are that the fixed stars and the
sun remain unmoved, that the earth revolves about the sun in
the circumference of a circle, the sun lying in the middle of the
orbit, and that the sphere of the fixed stars, situated about the
same centre as the sun, is so great that the circle in which he
supposes the earth to revolve bears such a proportion to the
distance of the fixed stars as the centre of the sphere bears to
its surface. Now it is easy to see that this is impossible; for,
since the centre of the sphere has no magnitude, we cannot
conceive it to bear any ratio whatever to the surface of the
sphere. We must however take Aristarchus to mean this:
since we conceive the earth to be, as it were, the centre of
the universe, the ratio which the earth bears to what we
describe as the 'universe' is the same as the ratio which the
sphere containing the circle in which he supposes the earth to
revolve bears to the sphere of the fixed stars. For he adapts
the proofs of his results to a hypothesis of this kind, and in
particular he appears to suppose the magnitude of the sphere
in which he represents the earth as moving to be equal to what
we call the 'universe.'
I say then that, even if a sphere were made up of the sand,
as great as Aristarchus supposes the sphere of the fixed stars
to be, I shall still prove that, of the numbers named in the
Principles*, some exceed in multitude the number of the
sand which is equal in magnitude to the sphere referred to,
provided that the following assumptions be made.
1. The perimeter of the earth is about 3,000,000 stadia and
not greater.
It is true that some have tried, as you are of course aware,
to prove that the said perimeter is about 300,000 stadia. But
I go further and, putting the magnitude of the earth at ten
times the size that my predecessors thought it, I suppose its
perimeter to be about 3,000,000 stadia and not greater.
* 'Apxaí was apparently the title of the work sent to Zeuxippus. Cf. the
note attached to the enumeration of lost works of Archimedes in the Introduction,
Chapter II., ad fin.
THE SAND-RECKONER.
223
2. The diameter of the earth is greater than the diameter of
the moon, and the diameter of the sun is greater than the diameter
of the earth.
In this assumption I follow most of the earlier astronomers.
3. The diameter of the sun is about 30 times the diameter of
the moon and not greater.
It is true that, of the earlier astronomers, Eudoxus declared
it to be about nine times as great, and Pheidias my father*
twelve times, while Aristarchus tried to prove that the diameter
of the sun is greater than 18 times but less than 20 times the
diameter of the moon. But I go even further than Aristarchus,
in order that the truth of my proposition may be established
beyond dispute, and I suppose the diameter of the sun to be
about 30 times that of the moon and not greater.
4. The diameter of the sun is greater than the side of the
chiliagon inscribed in the greatest circle in the (sphere of the)
universe.
I make this assumption+ because Aristarchus discovered
that the sun appeared to be about th part of the circle of
the zodiac, and I myself tried, by a method which I will now
describe, to find experimentally (ỏpyavikŵs) the angle sub-
tended by the sun and having its vertex at the eye (Tàv Ywvíav,
εἰς ἂν ὁ ἅλιος ἐναρμόζει τὴν κορυφὰν ἔχουσαν ποτὶ τῇ ὄψει).”
[Up to this point the treatise has been literally translated
because of the historical interest attaching to the ipsissima
verba of Archimedes on such a subject. The rest of the work
can now be more freely reproduced, and, before proceeding to
the mathematical contents of it, it is only necessary to remark
that Archimedes next describes how he arrived at a higher and
a lower limit for the angle subtended by the sun. This he did
* τοῦ ἁμοῦ πατρὸς is the correction of Blass for τοῦ ᾿Ακούπατρος (Jahrb. f.
Philol. CXXVII. 1883).
+ This is not, strictly speaking, an assumption; it is a proposition proved
later (pp. 224-6) by means of the result of an experiment about to be
described.
224
ARCHIMEDES
by taking a long rod or ruler (kavóv), fastening on the end of it
a small cylinder or disc, pointing the rod in the direction of the
sun just after its rising (so that it was possible to look directly
at it), then putting the cylinder at such a distance that it just
concealed, and just failed to conceal, the sun, and lastly measur-
ing the angles subtended by the cylinder. He explains also the
correction which he thought it necessary to make because "the
eye does not see from one point but from a certain area” (èπEì
αἱ ὄψιες οὐκ ἀφ' ἑνὸς σαμείου βλέποντι, ἀλλὰ ἀπό τινος
μεγέθεος).]
1
The result of the experiment was to show that the angle
subtended by the diameter of the sun was less than 4th part,
and greater than th part, of a right angle.
To prove that (on this assumption) the diameter of the sun
is greater than the side of a chiliagon, or figure with 1000 equal
sides, inscribed in a great circle of the universe.'
Suppose the plane of the paper to be the plane passing
through the centre of the sun, the centre of the earth and the
eye, at the time when the sun has just risen above the horizon.
Let the plane cut the earth in the circle EHL and the sun
in the circle FKG, the centres of the earth and sun being C, O
respectively, and E being the position of the eye.
Further, let the plane cut the sphere of the 'universe' (i.e.
the sphere whose centre is C and radius CO) in the great
circle AOB.
Draw from E two tangents to the circle FKG touching it
at P, Q, and from C draw two other tangents to the same circle
touching it in F, G respectively.
Let CO meet the sections of the earth and sun in H, K
respectively; and let CF, CG produced meet the great circle
AOB in A, B.
Join EO, OF, OG, OP, OQ, AB, and let AB meet CO in M.
Now CO > EO, since the sun is just above the horizon.
Therefore
< PEQ><FCG.
THE SAND-RECKONER.
225
but
And PEQ> 20 where R represents a right angle.
< TRS

L
E
H
K
PP
M
G
B
A
Thus
<FCG <TR, a fortiori,
and the chord AB subtends an arc of the great circle which is
less thanth of the circumference of that circle, i.e.
AB<(side of 656-sided polygon inscribed in the circle).
Now the perimeter of any polygon inscribed in the great
circle is less than 44CO. [Cf. Measurement of a circle, Prop. 3.]
AB: CO < 11: 1148,
Therefore
and, a fortiori,
...(α).
Again, since CA=CO, and AM is perpendicular to CO,
while OF is perpendicular to CA,
·AM = OF.
Therefore
AB = 2AM = (diameter of sun).
Thus
and, a fortiori,
(diameter of sun) <CO, by (a),
(diameter of earth)<¸¡¿CO.
H. A.
[Assumption 2]
15
•
226
ARCHIMEDES
Hence
so that
or
And
while
Therefore
CH+OK <ro00,
TooCO,
HK > 99 CO,
100
CO: HK < 100 : 99.
CO > CF,
HK < EQ.
CF: EQ< 100 99..
..(B).
Now in the right-angled triangles CFO, EQO, of the sides
about the right angles,
Therefore
OFOQ, but EQ < CF (since EO < CO).
ZOEQ: OCF > CO: EO,
< CF: EQ*.
but
Doubling the angles,
< PEQ : ≤ ACB < CF : EQ
< 100 : 99, by (B) above.
But
Therefore
< PEQ>R, by hypothesis.
< ACB > 20000 R
> 203 R.
It follows that the arc AB is greater than 12th of the circum-
ference of the great circle AOB.
Hence, a fortiori,
AB> (side of chiliagon inscribed in great circle),
and AB is equal to the diameter of the sun, as proved above.
The following results can now be proved:
(diameter of 'universe') < 10,000 (diameter of earth),
and (diameter of' universe') < 10,000,000,000 stadia.
* The proposition here assumed is of course equivalent to the trigonometrical
formula which states that, if a, ẞ are the circular measures of two angles, each
less than a right angle, of which a is the greater, then
tan a
>
α sin a
>
tan ßßsin ß
+
1
THE SAND-RECKONER.
227
(1) Suppose, for brevity, that du represents the diameter
of the 'universe,' d, that of the sun, de that of the earth, and dm
that of the moon.
By hypothesis,
and
therefore
ds 30dm
de > dm;
ds < 30de..
[Assumption 3]
[Assumption 2]
Now, by the last proposition,
so that
ds> (side of chiliagon inscribed in great circle),
(perimeter of chiliagon) < 1000ds
< 30,000de.
But the perimeter of any regular polygon with more sides
than 6 inscribed in a circle is greater than that of the inscribed
regular hexagon, and therefore greater than three times the
diameter. Hence
(perimeter of chiliagon) > 3du.
It follows that
du 10,000de.
(2) (Perimeter of earth) 3,000,000 stadia.
[Assumption 1]
and
Therefore
(perimeter of earth) > 3de.
de < 1,000,000 stadia,
whence
du< 10,000,000,000 stadia.
Assumption 5.
Suppose a quantity of sand taken not greater than a poppy-
seed, and suppose that it contains not more than 10,000 grains.
Next suppose the diameter of the poppy-seed to be not less
than 4th of a finger-breadth.
Orders and periods of numbers.
I. We have traditional names for numbers up to a
myriad (10,000); we can therefore express numbers up to a
myriad myriads (100,000,000). Let these numbers be called
numbers of the first order.
Suppose the 100,000,000 to be the unit of the second order,
and let the second order consist of the numbers from that unit
up to (100,000,000).
15-2
228
ARCHIMEDES
Let this again be the unit of the third order of numbers
ending with (100,000,000); and so on, until we reach the
100,000,000th order of numbers ending with (100,000,000)100,000,000,
which we will call P.
II. Suppose the numbers from 1 to P just described to
form the first period.
Let P be the unit of the first order of the second period, and
let this consist of the numbers from P up to 100,000,000 P.
Let the last number be the unit of the second order of the
second period, and let this end with (100,000,000)* P.
We can go on in this way till we reach the 100,000,000th order
of the second period ending with (100,000,000)100,000,000 P, or P².
III. Taking P² as the unit of the first order of the third
period, we proceed in the same way till we reach the
100,000,000th order of the third period ending with P³.
IV.. Taking P³ as the unit of the first order of the fourth
period, we continue the same process until we arrive at the
100,000,000th order of the 100,000,000th period ending with
P100,000,000. This last number is expressed by Archimedes as "a
myriad-myriad units of the myriad-myriad-th order of the
myriad-myriad-th period (ai pvpiакioμνριοσтâs πeρiódov μvρia-
κισμυριοστῶν ἀριθμῶν μυρίαι μυριάδες),” which is easily seen
to be 100,000,000 times the product of (100,000,000) 99,999,999 and
P99,999,999, i.e. P100,000,000
[The scheme of numbers thus described can be exhibited
more clearly by means of indices as follows.
FIRST PERIOD.
First order. Numbers from 1 to 108.
་
Second order.
108 to 1016.
Third order.
1016 to 1024.
""
""
(108)th order.
33
35
108.(108-1) to 108.108 (P, say).
THE SAND-RECKONER.
229
·
SECOND PERIOD.
Numbers from P.1 to P.108.
First order.
Second order.
"
""
P.108 to P.1016
(108)th order.
(10%)TH PERIOD.
First order.
Second order.
",
P.108.(108-1) to
P.108.108 (or P²).
""
P108-1.1 to P108-1.108.
""
33
P108-1.108 to P108-1.1016.
""
(108)th order.
""
""
P108-1.108. (108—1) to
P108-1.108.108 (i.e. P108).
The prodigious extent of this scheme will be appreciated
when it is considered that the last number in the first period
would be represented now by 1 followed by 800,000,000 ciphers,
while the last number of the (108)th period would require
100,000,000 times as many ciphers, i.e. 80,000 million millions
of ciphers.]
Octads.
Consider the series of terms in continued proportion of
which the first is 1 and the second 10 [i.e. the geometrical
progression 1, 10¹, 102, 10³, ...]. The first octad of these terms
[¿.e. 1, 10², 10², ... 107] fall accordingly under the first order
of the first period above described, the second octad [i.e.
10°, 10°,... 10¹³] under the second order of the first period, the
first term of the octad being the unit of the corresponding
order in each case. Similarly for the third octad, and so on.
We can, in the same way, place any number of octads.
Theorem.
If there be any number of terms of a series in continued
proportion, say A1, A2, A¸‚……… Am‚……. An,……. Am+n-1,... of which
A₁ = 1, A, = 10 [so that the series forms the geometrical pro-
gression 1, 101, 102,...10m-1,... 10-,... 10m+n2,...], and if any
two terms as Am, An be taken and multiplied, the product
1
I
230
ARCHIMEDES
Am. An will be a term in the same series and will be as many
terms distant from An as Am is distant from ▲₁; also it will be
distant from A, by a number of terms less by one than the sum
of the numbers of terms by which Am and An respectively are
distant from A₁.
Take the term which is distant from An by the same
number of terms as Am is distant from A₁. This number of
terms is m (the first and last being both counted). Thus the
term to be taken is m terms distant from An, and is therefore
the term Am+n-1.
We have therefore to prove that
Am⋅ An = Am+n−1·
Now terms equally distant from other terms in the con-
tinued proportion are proportional.
Thus
But
Therefore
A1
Am ___ Am+n−1
An
Am= Am. A₁, since A₁ = 1.
Am+n−1 = Am. An …………….
(1).
The second result is now obvious, since Am is m terms
distant from A₁, An is n terms distant from A₁, and Am+n-ı is
(m + n − 1) terms distant from A₁.
Application to the number of the sand.
By Assumption 5 [p. 227],
(diam. of poppy-seed)
(finger-breadth);
and, since spheres are to one another in the triplicate ratio
of their diameters, it follows that
(sphere of diam. 1 finger-breadth)
64,000 poppy-seeds
64,000 × 10,000
$640,000,000
6 units of second grains
order + 40,000,000
units of first order
(a fortiori) < 10 units of second
order of numbers.
of
sand.
THE SAND-RECKONER.
231
We now gradually increase the diameter of the supposed
sphere, multiplying it by 100 each time. Thus, remembering
that the sphere is thereby multiplied by 100³ or 1,000,000, the
number of grains of sand which would be contained in a sphere
with each successive diameter may be arrived at as follows.
Diameter of sphere.
(1) 100 finger-breadths
(2) 10,000 finger-breadths
(3) 1 stadium
(4) 100 stadia
Corresponding number of grains of sand.
<1,000,000 × 10 units of second order
<(7th term of series) x (10th term of
series)
16th term of series
[i.e. 1015]
<<[107 or] 10,000,000 units of the second
order.
<1,000,000 × (last number)
<(7th term of series) x (16th term)
<22nd term of series
[i.e. 1021]
<[105 or] 100,000 units of third order.
<100,000 units of third order.
(<10,000 finger-breadths)
<1,000,000 × (last number)
<<(7th term of series) × (22nd term)
<28th term of series
[1027]
<[10³ or] 1,000 units of fourth order.
(5) 10,000 stadia
<1,000,000 x (last number)
<(7th term of series) × (28th term)
<34th term of series
[1033]
<10 units of fifth order.
(6) 1,000,000 stadia
<(7th term of series) × (34th term)
<40th term
[1039]
(7) 100,000,000 stadia
<[107 or] 10,000,000 units of fifth order.
<(7th term of series) × (40th term)
<46th term
[1045]
(8) 10,000,000,000 stadia
<[105 or] 100,000 units of sixth order.
<(7th term of series) × (46th term)
<52nd term of series
[1051]
<[103 or] 1,000 units of seventh order.
But, by the proposition above [p. 227],
(diameter of 'universe') < 10,000,000,000 stadia.
Hence the number of grains of sand which could be contained
in a sphere of the size of our universe' is less than 1,000 units
of the seventh order of numbers [or 10º¹].
232
ARCHIMEDES
1
From this we can prove further that a sphere of the size
attributed by Aristarchus to the sphere of the fixed stars would
contain a number of grains of sand less than 10,000,000 units
of the eighth order of numbers [or 1056+7 = 106].
For, by hypothesis,
(earth): ('universe ') = (' universe'): (sphere of fixed stars).
And [p. 227]
whence
(diameter of universe') < 10,000 (diam. of earth);
(diam. of sphere of fixed stars) < 10,000 (diam. of ‘universe').
Therefore
(sphere of fixed stars) <(10,000)³. (‘universe ').
It follows that the number of grains of sand which would be
contained in a sphere equal to the sphere of the fixed stars
<(10,000)³ × 1,000 units of seventh order
< (13th term of series) × (52nd term of series)
<64th term of series
[i.e. 1063]
< [10' or] 10,000,000 units of eighth order of numbers.
Conclusion.
"I conceive that these things, king Gelon, will appear
incredible to the great majority of people who have not studied
mathematics, but that to those who are conversant therewith
and have given thought to the question of the distances and
sizes of the earth the sun and moon and the whole universe the
proof will carry conviction. And it was for this reason that
I thought the subject would be not inappropriate for your
consideration.”
QUADRATURE OF THE PARABOLA.
“ARCHIMEDES to Dositheus greeting.
"When I heard that Conon, who was my friend in his life-
time, was dead, but that you were acquainted with Conon and
withal versed in geometry, while I grieved for the loss not only
of a friend but of an admirable mathematician, I set myself the
task of communicating to you, as I had intended to send to
Conon, a certain geometrical theorem which had not been
investigated before but has now been investigated by me, and
which I first discovered by means of mechanics and then
exhibited by means of geometry. Now some of the earlier
geometers tried to prove it possible to find a rectilineal area
equal to a given circle and a given segment of a circle; and
after that they endeavoured to square the area bounded by the
section of the whole cone* and a straight line, assuming lemmas
not easily conceded, so that it was recognised by most people
that the problem was not solved. But I am not aware that
any one of my predecessors has attempted to square the
segment bounded by a straight line and a section of a right-
angled cone [a parabola], of which problem I have now dis-
covered the solution. For it is here shown that every segment
bounded by a straight line and a section of a right-angled cone
[a parabola] is four-thirds of the triangle which has the same base
and equal height with the segment, and for the demonstration
* There appears to be some corruption here: the expression in the text is
τᾶς ὅλου τοῦ κώνου τομᾶς, and it is not easy to give a natural and intelligible
meaning to it. The section of the whole cone' might perhaps mean a section
cutting right through it, i.e. an ellipse, and the 'straight line' might be an axis
or a diameter. But Heiberg objects to the suggestion to read râs ỏğvywvlov
κώνου τομᾶς, in view of the addition of καὶ εὐθείας, on the ground that the former
expression always signifies the whole of an ellipse, never a segment of it
(Quaestiones Archimedeae, p. 149).
→
234
ARCHIMEDES
of this property the following lemma is assumed: that the
excess by which the greater of (two) unequal areas exceeds
the less can, by being added to itself, be made to exceed any
given finite area. The earlier geometers have also used this
lemma; for it is by the use of this same lemma that they have
shown that circles are to one another in the duplicate ratio of
their diameters, and that spheres are to one another in the
triplicate ratio of their diameters, and further that every
pyramid is one third part of the prism which has the same base
with the pyramid and equal height; also, that every cone is
one third part of the cylinder having the same base as the cone
and equal height they proved by assuming a certain lemma
similar to that aforesaid. And, in the result, each of the afore-
said theorems has been accepted* no less than those proved
without the lemma. As therefore my work now published has
satisfied the same test as the propositions referred to, I have
written out the proof and send it to you, first as investigated
by means of mechanics, and afterwards too as demonstrated by
geometry. Prefixed are, also, the elementary propositions in
conics which are of service in the proof (στοιχεία κωνικὰ χρείαν
ἔχοντα ἐς τὰν ἀπόδειξιν). Farewell.”
Proposition 1.
If from a point on a para-
bola a straight line be drawn
which is either itself the axis or
parallel to the axis, as PV, and
if QQ' be a chord parallel to
the tangent to the parabola at P
and meeting PV in V, then

QV = VQ'.
Conversely, if QV = VQ', the
chord QQ' will be parallel to the
tangent at P.
*
P
The Greek of this passage is: συμβαίνει δὲ τῶν προειρημένων θεωρημάτων
ἕκαστον μηδὲν ἧσσον τῶν ἄνευ τούτου τοῦ λήμματος ἀποδεδειγμένων πεπιστευκέναι.
Here it would seem that TETOTEUкévaι must be wrong and that the passive
should have been used.
QUADRATURE OF THE PARABOLA.
235
Proposition 2.
If in a parabola QQ' be a chord parallel to the tangent at P,
and if a straight line be drawn through P which is either itself
the axis or parallel to the axis, and which meets QQ' in V and
the tangent at Q to the parabola in T, then
PV=PT.
{
T
P
V
Proposition 3.
If from a point on a parabola a straight line be drawn
which is either itself the axis or parallel to the axis, as PV,
and if from two other points Q, Q' on the parabola straight
lines be drawn parallel to the tangent at P and meeting PV in
V, V' respectively, then
PV : PV' = QV² : Q'V'½.
"And these propositions are proved in the elements of conics.*"
Proposition 4.
If Qq be the base of any segment of a parabola, and P the
vertex of the segment, and if the diameter through any other point
R meet Qq in O and QP (produced if necessary) in F, then
QV: VO OF: FR.
=
Draw the ordinate RW to PV, meeting QP in K.
* i.e. in the treatises on conics by Euclid and Aristaeus.
·

236
ARCHIMEDES
Then
whence, by parallels,
PV: PW = QV: RW';
PQ : PK = PQ² : PF².


Q
R
F
K
P
P
W
W
R
q
In other words, PQ, PF, PK are in continued proportion;
therefore
PQ: PF = PF: PK
Hence, by parallels,
= PQ ± PF : PF ± PK
= QF: KF.
QV: VO OF: FR.
==
[It is easily seen that this equation is equivalent to a change of
axes of coordinates from the tangent and diameter to new axes
consisting of the chord Qq (as axis of x, say) and the diameter
through Q (as axis of y).
a²
For, if QV = a,
= a, PV
=
p
ordinates to PV.
where p is the parameter of the
Thus, if QO = x, and RO=y, the above result gives
1
α
OF
X
a
OF-y
α
X .
a
OF
whence
2a
2α-x
Y
y
or
py=x (2α − x).]
•
QUADRATURE OF THE PARABOLA.
237
Proposition 5.
If Qq be the base of any segment of a parabola, P the vertex
of the segment, and PV its diameter, and if the diameter of the
parabola through any other point R meet Qq in O and the
tangent at Q in E, then
QO: Oq = ER: RO.
Let the diameter through R meet QP in F.

E
T
P
F
Then, by Prop. 4,
QV: VO OF: FR.
Since QV = Vq, it follows that
=
QV : qO = OF: OR...
Also, if VP meet the tangent in T,
PT=PV, and therefore EF = OF.
Accordingly, doubling the antecedents in (1), we have
whence
Qq: qO = OE: OR,
QỞ : 0q = ER : RO.
00- FOIRC
.(1).
238
ARCHIMEDES
Propositions 6, 7*.
Suppose a lever AOB placed horizontally and supported at
its middle point O. Let a triangle BCD in which the angle C is
right or obtuse be suspended from B and O, so that C is attached
to O and CD is in the same vertical line with O. Then, if P be
such an area as, when suspended from A, will keep the system in
equilibrium,
P = }^BCD.
Take a point E on OB such that BE=20E, and draw EFH
parallel to OCD meeting BC, BD in F, H respectively. Let G
be the middle point of FH.

A
P
E
B
F
C
G
H
D
Then G is the centre of gravity of the triangle BCD.
Hence, if the angular points B, C be set free and the
triangle be suspended by attaching F to E, the triangle will
hang in the same position as before, because EFG is a vertical
straight line. "For this is proved†."
or
Therefore, as before, there will be equilibrium.
Thus
P: ABCD OE: AO
=
=1:3,
P = {^BCD.
* In Prop. 6 Archimedes takes the separate case in which the angle BCD of
the triangle is a right angle so that C coincides with O in the figure and F with
E. He then proves, in Prop. 7, the same property for the triangle in which
BCD is an obtuse angle, by treating the triangle as the difference between two
right-angled triangles BOD, BOC and using the result of Prop. 6. I have com-
bined the two propositions in one proof, for the sake of brevity. The same
remark applies to the propositions following Props. 6, 7.
+ Doubtless in the lost book weρi Švyŵv. Cf. the Introduction, Chapter II.,
ad fin.
QUADRATURE OF THE PARABOLA.
239
Propositions 8, 9.
Suppose a lever AOB placed horizontally and supported at
its middle point 0. Let a triangle BCD, right-angled or obtuse-
angled at C, be suspended from the points B, E on OB, the
angular point C being so attached to E that the side CD is in the
same vertical line with E. Let Q be an area such that
AO: OE-ABCD: Q.
Then, if an area P suspended from A keep the system in
equilibrium,
P<ABCD but > Q.
Take G the centre of gravity of the triangle BCD, and draw
GH parallel to DC, i.e. vertically, meeting BO in H.

A
P
C
D
E
H
Q
B
We may now suppose the triangle BCD suspended from H,
and, since there is equilibrium,
ABCD: PAO: OH...............
..(1),
whence
Also
P < ABCD.
ABCD: Q = AO- OE.
Therefore, by (1), ABCD: Q> ABCD: P,
and
P> Q.
Propositions 10, 11.
Suppose a lever AOB placed horizontally and supported at 0,
its middle point. Let CDEF be a trapezium which can be so
placed that its parallel sides CD, FE are vertical, while C is
vertically below 0, and the other sides CF, DE meet in B. Let
EF meet BO in H, and let the trapezium be suspended by attaching
F to H and C to 0. Further, suppose Q to be an area such that
AO: OH = (trapezium CDEF) : Q.
240
ARCHIMEDES
+
J
Then, if P be the area which, when suspended from A, keeps the
system in equilibrium,
P<Q.
The same is true in the particular case where the angles at
C, F are right, and consequently C, F coincide with O, H.
respectively.
Divide OH in K so that
(2CD + FE): (2FE + CD) HK: KO.
=
A
P
ок
H
B

F
C
G
E
Q
Draw KG parallel to OD, and let G be the middle point of
the portion of KG intercepted within the trapezium. Then G
is the centre of gravity of the trapezium [On the equilibrium of
planes, I. 15].
Thus we may suppose the trapezium suspended from K, and
the equilibrium will remain undisturbed.
Therefore
AO: OK = (trapezium CDEF) : P,
and, by hypothesis,
AO: OH (trapezium CDEF) : Q.
=
Since OK < OH, it follows that
P<Q.
Propositions 12, 13.
If the trapezium CDEF be placed as in the last propositions,
except that CD is vertically below a point L on OB instead of
being below O, and the trapezium is suspended from L, H,
suppose that Q, R are areas such that
and
AO: OH = (trapezium CDEF): Q,
AO: OL= (trapezium CDEF): R.
}
QUADRATURE OF THE PARABOLA.
241
If then an area P suspended from A keep the system in
equilibrium,
P> R but <Q.
Take the centre of gravity G of the trapezium, as in the
last propositions, and let the line through G parallel to DC
meet OB in K.
A
a
Q
R
C
L K
H
B

IF
G
E
Then we may suppose the trapezium suspended from K,
and there will still be equilibrium.
Therefore (trapezium CDEF): P = A0 : OK.
Hence
but
(trapezium CDEF) : P >(trapezium CDEF) : Q,
< (trapezium CDEF): R.
It follows that
PQ but > R.
Propositions 14, 15.
Let Qq be the base of any segment of a parabola. Then, if
two lines be drawn from Q, q, each parallel to the axis of the
parabola and on the same side of Qq as the segment is, either
(1) the angles so formed at Q, q are both right angles, or
(2) one is acute and the other obtuse. In the latter case let
the angle at q be the obtuse angle.
Divide Qq into any number of equal parts at the points
01, 02,... On. Draw through q, 01, 02, ... On diameters of the
parabola meeting the tangent at Q in E, E₁, E2, ….. En and the
parabola itself in q, R1, R2, Rn. Join QR₁, QR2,
meeting qE, 0,Е₁, О₂E₂, ... On-En-1 in F, F1, F2, ... Fn-1.
2,
QR n
H. A.
.
16
242
ARCHIMEDES
L
Let the diameters Eq, E101,... EnOn meet a straight line
QOA drawn through Q perpendicular to the diameters in the
points 0, H1, H2, ... H₂ respectively. (In the particular case
where Qq is itself perpendicular to the diameters q will coincide
with 0, 0, with H₁, and so on.)
It is required to prove that
(1) ▲ EqQ<3(sum of trapezia FO₁, F102,…….Fn-1On and ▲ EnOnQ),
(2) ▲ EqQ>3(sum of trapezia R₁O2, R203, ……. Rn-10n and ▲ RnOnQ).
A
Hi H2 Hn-1 Hn
On

P1
P2
Pnt!
01
LL
F
R1
F1
R 2
E
>
E1
E2
Fn-1
En-1
Rn
En
Suppose 40 made equal to OQ, and conceive QOA as a
lever placed horizontally and supported at 0. Suppose the
triangle EqQ suspended from OQ in the position drawn, and
suppose that the trapezium EO, in the position drawn is
balanced by an area P, suspended from A, the trapezium E₁0₂
in the position drawn is balanced by the area P, suspended
1
2
QUADRATURE OF THE PARABOLA.
243
from A, and so on, the triangle EnOnQ being in like manner
balanced by Pn+1•
Then P₁+P₂+...+P+1 will balance the whole triangle
EqQ as drawn, and therefore
Again
P₁+ P₂+...+Pn+1={▲EqQ.
AO: OH₁ =Q0 : 0H₁
1
[Props. 6, 7]
Qq: qO1
whence [Props. 10, 11]
Next
while
E₁0₁: 0₁R₁ [by means of Prop. 5]
= (trapezium EO₁): (trapezium FO₁);
(FO₁) > P₁.
AO: OH₁ = E₁0₁ : 0₁R₁
1
= (E₂0₂½) : (R₂O₂) ……….. ……………….(a),
=
AO: OH₂ = Е202 : 0₂R₂
2
= (E₁0₂) : (F₁0₂) ..
.(B);
and, since (a) and (B) are simultaneously true, we have, by
Props. 12, 13,
(F₁0₂) > P₂> (R₂O₂).
2
(F₂03) > P3> (R₂O3),
Similarly it may be proved that
and so on.
Lastly [Props. 8, 9]
▲ EnOnQ > Pn+1 > ▲ RnOnQ.
By addition, we obtain
(1) (FO₁)+(F₂O₂)+...+(Fn-1On)+▲ EnOnQ> P₁+P₂+
> } ▲ EqQ,
or ▲ EqQ<3 (FO₂+ F102 + ... + Fn-10n'+ ▲ EnOnQ).
+Pn+1
(2) (R₁O₂)+(R₂03)+...+(Rn-10n)+^Rn0nQ<P₂+P3+...+Pn+1
or
< P₁+ P₂+ ... + Pn+1, a fortiori,
1
2
< ▲ EqQ,
▲ EqQ>3 (R₂O₂+ R₂O3 + ... + Rn-10n+▲ RnOnQ).
2
16-2
244
ARCHIMEDES
T
Proposition 16.
Suppose Qq to be the base of a parabolic segment, q being
not more distant than Q from the vertex of the parabola. Draw
through q the straight line qE parallel to the axis of the parabola
to meet the tangent at Q in E. It is required to prove that
(area of segment) = { ▲ EqQ.
For, if not, the area of the segment must be either greater
or less than ▲EqQ.

I. Suppose the area of the
segment greater than ▲ EqQ.
Then the excess can, if con-
tinually added to itself, be
made to exceed AEqQ. And
it is possible to find a submul-
tiple of the triangle EqQ less
than the said excess of the
segment over & ▲EqQ.
Let the triangle FqQ be such
a submultiple of the triangle
EqQ. Divide Eq into equal
parts each equal to qF, and let
all the points of division in-
cluding F be joined to Q meet-
ing the parabola in R₁, R,,
Rn respectively. Through R₁, R2,
parabola meeting qQ in 01, 02,
Let O₁R₁ meet QR₂ in F₁.
F
E
Ο
On
En
R₁₂
R2
... Rn draw diameters of the
On respectively.
Оп
Let O₂R₂ meet QR, in D, and QR, in F₂.
2
Let OR meet QR₂ in D₂ and QR in F¸, and so on.
We have, by hypothesis,
▲FqQ< (area of segment) - AEqQ,
or
(area of segment) — ▲FqQ > } ▲ EqQ
(a).
1
QUADRATURE OF THE PARABOLA.
245
Now, since all the parts of qE, as qF and the rest, are equal,
0₁R₁ = R₁₂F₁, O₂D₁ = D₁R₂ = R₂F₂, and so on; therefore
2
2
▲ FqQ=(FO₂+R₂O₂+D₂O₂+…..)
But
1
2
=(FO₂+F₁D₁+F₂D₂+...+Fn-1Dn-1+▲ EnRnQ)...(B).
2
(area of segment) < (FO₁ + F₁02 +...+Fn-10n +▲ EnOnQ).
Subtracting, we have
(area of segment) — ^ FqQ< (R₁O₂ + R₂O3 + …..
whence, a fortiori, by (a),
2
2
+ Rn-10n+^RnOnQ),
▲ EqQ<(R₂O₂+ R₂O3 + ... + Rn-10n + ▲ RnOnQ).
But this is impossible, since [Props. 14, 15]
}^ EqQ>(R₂O₂+ R₂O3 + ... + Rn-10n+ ▲ RnOnQ).
Therefore
(area of segment)
▲ EqQ.
II. If possible, suppose the area of the segment less than
▲ EqQ.
Take a submultiple of the triangle EqQ, as the triangle
FqQ, less than the excess of AEqQ over the area of the
segment, and make the same construction as before.
Since ▲FqQ<ž▲EqQ – (area of segment),
it follows that
AFqQ+(area of segment) < ^EqQ
< (FO₁+ F₂O₂+ ... + Fn-10n + ▲EnOnQ).
2
Subtracting from each side the area of the
▲ FqQ<(sum of spaces qFR₁, R₁F₁R₂,
[Props. 14, 15]
segment, we have
En RnQ)
< (FO₁+ F₂D₁ + ... + Fn-Dn-1+ ▲ EnRnQ), a fortiori;
which is impossible, because, by (ß) above,
▲ FqQ=FO₂+ F₁D₁ + ... + Fn¬1Dn−1 + ▲ EnRnQ.
Hence
(area of segment) & §▲ EqQ.
Since then the area of the segment is neither less nor
greater than ▲ EqQ, it is equal to it.
246
ARCHIMEDES
+
1
J
Proposition 17.
It is now manifest that the area of any segment of a
parabola is four-thirds of the triangle which has the same base
as the segment and equal height.
Let Qq be the base of the segment, P its vertex. Then
PQq is the inscribed triangle with the
same base as the segment and equal
height.
Since P is the vertex* of the seg-
ment, the diameter through P bisects
Qq. Let V be the point of bisection.
Let VP, and qE drawn parallel to
it, meet the tangent at Q in T, E re-
spectively.

Then, by parallels,
P
T
qE=2VT,
and
PV = PT,
[Prop. 2]
so that
VT=2PV.
E
Hence
EqQ= 4^PQq.
But, by Prop. 16, the area of the segment is equal to ▲EqQ.
Therefore
(area of segment) = A PQq.
DEF. "In segments bounded by a straight line and any
curve I call the straight line the base, and the height the
greatest perpendicular drawn from the curve to the base of the
segment, and the vertex the point from which the greatest
perpendicular is drawn."
* It is curious that Archimedes uses the terms base and vertex of a segment
here, but gives the definition of them later (at the end of the proposition).
Moreover he assumes the converse of the property proved in Prop. 18.
QUADRATURE OF THE PARABOLA.
247
ነ
Proposition 18.
If Qq be the base of a segment of a parabola, and V the
middle point of Qq, and if the diameter through V meet the
curve in P, then P is the vertex of the segment.
Q

P
g
For Qq is parallel to the tangent at P [Prop. 1]. Therefore,
of all the perpendiculars which can be drawn from points on the
segment to the base Qq, that from P is the greatest. Hence,
by the definition, P is the vertex of the segment.
Proposition 19.
If Qq be a chord of a parabola bisected in V by the diameter
PV, and if RM be a diameter bisecting QV in M, and RW
be the ordinate from R to PV, then
PV = ‡RM.
Q

R
M
P
V
W
Չ
For, by the property of the parabola,
PV: PW=QV²: RW²
so that
whence
= 4RW2: RW2,
PV = 4PW,
PV = ‡RM.
248
ARCHIMEDES
Proposition 20.
If Qq be the base, and P the vertex, of a parabolic segment,
then the triangle PQq is greater than half the segment PQq.
છે
For the chord Qq is parallel to the tangent at P, and the
triangle PQq is half the parallelogram
formed by Qq, the tangent at P, and the
diameters through Q, q.
Therefore the triangle PQq is greater
than half the segment.
COR. It follows that it is possible
to inscribe in the segment a polygon such
that the segments left over are together
less than any assigned area.

P
V
q
Proposition 21.
If Qq be the base, and P the vertex, of any parabolic
segment, and if R be the vertex of the segment cut off by PQ,
then
APQq=8APRQ.
Q
The diameter through R will bisect the chord PQ, and
therefore also QV, where PV is the
diameter bisecting Qq. Let the dia-
meter through R bisect PQ in Y and
QV in M. Join PM.

By Prop. 19,
R
Y
M
PV = 4RM.
P
V
W
Also
PV=2YM.
Therefore
YM = 2RY,
20
and
ΔΡΟΜ=2ΔΡΕΟ
Hence
▲PQV=4▲PRQ,
q
and
▲PQq=8^ PRQ.
QUADRATURE OF THE PARABOLA.
249
Also, if RW, the ordinate from R to PV, be produced to
meet the curve again in r,
RW=rW,
and the same proof shows that
▲PQq = 8▲ Prq.
Proposition 22.
If there be a series of areas A, B, C, D, ... each of which is
four times the next in order, and if the largest, A, be equal to the
triangle PQq inscribed in a parabolic segment PQq and having
the same base with it and equal height, then
(A+B+C+D+...) < (area of segment PQq).
For, since APQg = 8^PRQ = 8▲Pqr, where R, r are the
vertices of the segments cut off by PQ,

Pq, as in the last proposition,
▲ PQq=4(APQR + APqr).
Therefore, since ▲PQq= A,
▲PQR+ ▲Pqr = B.
In like manner we prove that the
triangles similarly inscribed in the re-
maining segments are together equal to
the area C, and so on.
R
P
V
Therefore A+B+C+D+... is equal to the area of a
certain inscribed polygon, and is therefore less than the area of
the segment.
Proposition 23.
Given a series of areas A, B, C, D, ... Z, of which A is the
greatest, and each is equal to four times the next in order, then
A+B+C+...+Z+}Z= {A.
250
ARCHIMEDES
#
[
M
Take areas b, c, d, ... such that
b = {B,
c = {C,
d = 1D, and so on.
-
Then, since
b = {B,
and
B = 1A,
B+b= {A.
Similarly
Therefore
C+c={B.
B+C+D+...+Z+b+c+d+...+2=}(A+B+C+...+ Y).
But
b+c+d+ ... + y = {(B+C+D+ ... + Y).

A
Therefore, by subtraction,
or
B
C
D
b
d
C
B+C+D+...+Z+z=}A
A+B+C+...+Z+}Z=‡A.
QUADRATURE OF THE PARABOLA.
251
!
The algebraical equivalent of this result is of course
1 + 1 + (1)² + ... + (†)~~¹ = ƒ − } (†)"−1
1+1+(1)²+
1-(1)n
1-1
Proposition 24.
Every segment bounded by a parabola and a chord Qq is
equal to four-thirds of the triangle which has the same base as
the segment and equal height.
Suppose
K = &▲PQq,
where P is the vertex of the segment; and we have then to
prove that the area of the segment is
equal to K.
For, if the segment be not equal to
K, it must either be greater or less.
I. Suppose the area of the segment
greater than K.

P
1°
R
If then we inscribe in the segments
cut off by PQ, Pq triangles which have
the same base and equal height, i.e.
triangles with the same vertices R, r as
those of the segments, and if in the
remaining segments we inscribe triangles in the same manner,
and so on, we shall finally have segments remaining whose sum
is less than the area by which the segment PQq exceeds K.
Therefore the polygon so formed must be greater than the
area K; which is impossible, since [Prop. 23]
where
A+B+C+...+Z<§A,
A = ▲ PQq.
Thus the area of the segment cannot be greater than K.
II. Suppose, if possible, that the area of the segment is
less than K.
252
ARCHIMEDES.
L
I
If then APQq=A, B=1A, C=1B, and so on, until we
arrive at an area X such that X is less than the difference
between K and the segment, we have
A+B+C+...+X+3X=‡A
[Prop. 23]
= K.
Now, since K exceeds A+B+C+...+X by an area less
than X, and the area of the segment by an area greater than X,
it follows that
A+B+C+...+X> (the segment);
X>(the
which is impossible, by Prop. 22 above.
Hence the segment is not less than K.
Thus, since the segment is neither greater nor less than K,
(area of segment PQq)=K=÷APQq.
ON FLOATING BODIES.
BOOK I.
Postulate 1.
"Let it be supposed that a fluid is of such a character that,
its parts lying evenly and being continuous, that part which is
thrust the less is driven along by that which is thrust the
more; and that each of its parts is thrust by the fluid which is
above it in a perpendicular direction if the fluid be sunk in
anything and compressed by anything else.'
Proposition 1.
""
If a surface be cut by a plane always passing through a
certain point, and if the section be always a circumference [of a
circle] whose centre is the aforesaid point, the surface is that of
a sphere.
For, if not, there will be some two lines drawn from the
point to the surface which are not equal.
Suppose O to be the fixed point, and A, B to be two points
on the surface such that OA, OB are unequal. Let the surface
be cut by a plane passing through OA, OB. Then the section
is, by hypothesis, a circle whose centre is 0.
Thus OA=0B; which is contrary to the assumption.
Therefore the surface cannot but be a sphere.
254
ARCHIMEDES
}
>
Proposition 2.
The surface of any fluid at rest is the surface of a sphere
whose centre is the same as that of the earth.
Suppose the surface of the fluid cut by a plane through 0,
the centre of the earth, in the curve ABCD.
ABCD shall be the circumference of a circle.
For, if not, some of the lines drawn from 0 to the curve
will be unequal. Take one of them, OB, such that OB is
greater than some of the lines from 0 to the curve and less
than others. Draw a circle with OB as radius. Let it be EBF,
which will therefore fall partly within and partly without the
surface of the fluid.

H
G
B
EA
P
R
DF
Draw OGH making with OB an angle equal to the angle
EOB, and meeting the surface in H and the circle in G. Draw
also in the plane an arc of a circle PQR with centre O and
within the fluid.
Then the parts of the fluid along PQR are uniform and
continuous, and the part PQ is compressed by the part between
it and AB, while the part QR is compressed by the part
between QR and BH. Therefore the parts along PQ, QR will
be unequally compressed, and the part which is compressed the
less will be set in motion by that which is compressed the
more.
Therefore there will not be rest; which is contrary to the
hypothesis.
Hence the section of the surface will be the circumference
of a circle whose centre is 0; and so will all other sections by
planes through 0.
Therefore the surface is that of a sphere with centre O.
ON FLOATING BODIES I.
255
Proposition 3.
10. Same
Of solids those which, size for size, are of equal weight with
a fluid will, if let down into the fluid, be immersed so that they
do not project above the surface but do not sink lower.
If possible, let a certain solid EFHG of equal weight,
volume for volume, with the fluid remain immersed in it so
that part of it, EBCF, projects above the surface.
Draw through O, the centre of the earth, and through the
solid a plane cutting the surface of the fluid in the circle
ABCD.
Conceive a pyramid with vertex O and base a parallelogram
at the surface of the fluid, such that it includes the immersed
portion of the solid. Let this pyramid be cut by the plane of
ABCD in OL, OM. Also let a sphere within the fluid and
below GH be described with centre O, and let the plane of
ABCD cut this sphere in PQR.
A
F
E
CM
LB
S VN
GH
P
Q
R
D
Conceive also another pyramid in the fluid with vertex 0,
continuous with the former pyramid and equal and similar to
it. Let the pyramid so described be cut in OM, ON by the
plane of ABCD.
Lastly, let STUV be a part of the fluid within the second
pyramid equal and similar to the part BGHC of the solid, and
let SV be at the surface of the fluid.
Then the pressures on PQ, QR are unequal, that on PQ
being the greater. Hence the part at QR will be set in motion
density

256
ARCHIMEDES
by that at PQ, and the fluid will not be at rest; which is
contrary to the hypothesis.
Therefore the solid will not stand out above the surface.
Nor will it sink further, because all the parts of the fluid
will be under the same pressure.
Proposition 4.
A solid lighter than a fluid will, if immersed in it, not be
completely submerged, but part of it will project above the
surface.
In this case, after the manner of the previous proposition,
we assume the solid, if possible, to be completely submerged and
the fluid to be at rest in that position, and we conceive (1) a
pyramid with its vertex at 0, the centre of the earth, including
the solid, (2) another pyramid continuous with the former and
equal and similar to it, with the same vertex 0, (3) a portion of
the fluid within this latter pyramid equal to the immersed solid
in the other pyramid, (4) a sphere with centre O whose surface
is below the immersed solid and the part of the fluid in the
second pyramid corresponding thereto. We suppose a plane to
be drawn through the centre O cutting the surface of the
fluid in the circle ABC, the solid in S, the first pyramid in OA,
OB, the second pyramid in OB, OC, the portion of the fluid in
the second pyramid in K, and the inner sphere in PQR.
Then the pressures on the parts of the fluid at PQ, QR are
unequal, since S is lighter than K. Hence there will not be
rest; which is contrary to the hypothesis.
B
portree of

A
8
K
Q
P
R
C
Water
Therefore the solid S cannot, in a condition of rest, be
completely submerged.
ON FLOATING BODIES 1.
257
Proposition 5.
to
Any solid lighter than a fluid will, if placed in the fluid,
be so far immersed that the weight of the solid will be equal
the weight of the fluid displaced.
For let the solid be EGHF, and let BGHC be the portion
of it immersed when the fluid is at rest. As in Prop. 3,
conceive a pyramid with vertex O including the solid, and
another pyramid with the same vertex continuous with the
former and equal and similar to it. Suppose a portion of the
fluid STUV at the base of the second pyramid to be equal and
similar to the immersed portion of the solid; and let the con-
struction be the same as in Prop. 3.
F

E
CMS
LB
V
N
H
G
Q
P
R
A
D
Then, since the pressure on the parts of the fluid at PQ, QR
must be equal in order that the fluid may be at rest, it follows
that the weight of the portion STUV of the fluid must be
equal to the weight of the solid EGHF. And the former is
equal to the weight of the fluid displaced by the immersed
portion of the solid BGHC.
Proposition 6.
If a solid lighter than a fluid be forcibly immersed in it, the
solid will be driven upwards by a force equal to the difference
between its weight and the weight of the fluid displaced.
For let A be completely immersed in the fluid, and let G
represent the weight of A, and (G+H) the weight of an equal
volume of the fluid. Take a solid D, whose weight is H
H. A.
17
Far
Rey
wew
258 (ach)
| Brimet)
and add it to A.
Arch.
Alist.
не или
Pike.
ARCHIMEDES
Then the weight of (A+D) is less than
that of an equal volume of the fluid; and, if (A+D) is
immersed in the fluid, it will project so that its weight will
be equal to the weight of the fluid displaced. But its weight
is (G+ H).

D
G
B
A
C
H
Therefore the weight of the fluid displaced is (G+H), and
hence the volume of the fluid displaced is the volume of the
solid A. There will accordingly be rest with A immersed
and D projecting.
Thus the weight of D balances the upward force exerted by
the fluid on A, and therefore the latter force is equal to H,
which is the difference between the weight of A and the weight
of the fluid which A displaces.
Proposition 7.
A solid heavier than a fluid will, if placed in it, descend
to the bottom of the fluid, and the solid will, when weighed
in the fluid, be lighter than its true weight by the weight of the
fluid displaced.
(1) The first part of the proposition is obvious, since the
part of the fluid under the solid will be under greater pressure,
and therefore the other parts will give way until the solid
reaches the bottom.
(2) Let A be a solid heavier than the same volume of the
fluid, and let (G+H) represent its weight, while G represents
the weight of the same volume of the fluid.
ON FLOATING BODIES I.
259
Take a solid B lighter than the same volume of the fluid,
and such that the weight of B is G, while the weight of the
same volume of the fluid is (G+H).


A
G
B
H
Let A and B be now combined into one solid and immersed.
Then, since (A+B) will be of the same weight as the same
volume of fluid, both weights being equal to (G+H)+G, it
follows that (A+B) will remain stationary in the fluid.
Therefore the force which causes A by itself to sink must
be equal to the upward force exerted by the fluid on B by
itself. This latter is equal to the difference between (G+H)
and G [Prop. 6]. Hence A is depressed by a force equal to
H, i.e. its weight in the fluid is H, or the difference between
(G+H) and G.
[This proposition may, I think, safely be regarded as decisive
of the question how Archimedes determined the proportions of
gold and silver contained in the famous crown (cf. Introduction,
Chapter I.). The proposition suggests in fact the following
method.
Let W represent the weight of the crown, w₁ and w₂ the
weights of the gold and silver in it respectively, so that
W=W₁+W₂.
(1) Take a weight W of pure gold and weigh it in a fluid.
The apparent loss of weight is then equal to the weight of
the fluid displaced. If F, denote this weight, F, is thus known
as the result of the operation of weighing.
1
It follows that the weight of fluid displaced by a weight w₁
of gold is
W1
F1.
W
F
17-2
260
ARCHIMEDES
:
(2) Take a weight W of pure silver and perform the same
operation. If F₂ be the loss of weight when the silver is
weighed in the fluid, we find in like manner that the weight
2
W2
W
of fluid displaced by w, is . F
(3) Lastly, weigh the crown itself in the fluid, and let F be
the loss of weight. Therefore the weight of fluid displaced by
the crown is F.
or
W2
W1
It follows that
W
1
F₁+ F₁ = F,
whence
W
2
w₁ F₁+w₂ F₂ = (w₁ + w₂) F,
W1
W2
2
F-F
F-F₁
This procedure corresponds pretty closely to that described
in the poem de ponderibus et mensuris (written probably about
500 A.D.)* purporting to explain Archimedes' method. Ac-
cording to the author of this poem, we first take two equal
weights of pure gold and pure silver respectively and weigh
them against each other when both immersed in water; this
gives the relation between their weights in water and therefore
between their loss of weight in water. Next we take the
mixture of gold and silver and an equal weight of pure silver
and weigh them against each other in water in the same
manner.
The other version of the method used by Archimedes is
that given by Vitruviust, according to which he measured
successively the volumes of fluid displaced by three equal
weights, (1) the crown, (2) the same weight of gold, (3) the
same weight of silver, respectively. Thus, if as before the
weight of the crown is W, and it contains weights w₁ and w₂ of
gold and silver respectively,
(1) the crown displaces a certain quantity of fluid, V say.
(2) the weight W of gold displaces a certain volume of
* Torelli's Archimedes, p. 364; Hultsch, Metrol. Script. . 95 sq., and
Prolegomena § 118.
+ De architect. IX. 3.
ON FLOATING BODIES I.
261
fluid, V₁ say; therefore a weight w₁ of gold displaces a volume
W1
V₁ of fluid.
W
• 1
(3) the weight W of silver displaces a certain volume of
fluid, say V₂; therefore a weight w, of silver displaces a volume
W2 .V₂ of fluid.
W
It follows that
whence, since
WI
Ꮴ
W
V=W.V₂+ W
W2
Ꮴ
W₂ V2
W
W = W₁ + W₂,
W1
V2
V₁- V
W2
V – V,
and this ratio is obviously equal to that before obtained, viz.
F-F
2
F — F₁
•]
Postulate 2.
"Let it be granted that bodies which are forced upwards in
a fluid are forced upwards along the perpendicular [to the
surface] which passes through their centre of gravity.”
Proposition 8.
If a solid in the form of a segment of a sphere, and of a
substance lighter than a fluid, be immersed in it so that its base
does not touch the surface, the solid will rest in such a position
that its axis is perpendicular to the surface; and, if the solid be
forced into such a position that its base touches the fluid on one
side and be then set free, it will not remain in that position but
will return to the symmetrical position.
[The proof of this proposition is wanting in the Latin
version of Tartaglia. Commandinus supplied a proof of his
own in his edition.]
Proposition 9.
If a solid in the form of a segment of a sphere, and of a
substance lighter than a fluid, be immersed in it so that its base
is completely below the surface, the solid will rest in such a
position that its axis is perpendicular to the surface.
1
1
262
ARCHIMEDES
[The proof of this proposition has only survived in a
mutilated form. It deals moreover with only one case out of
three which are distinguished at the beginning, viz. that in
which the segment is greater than a hemisphere, while figures
only are given for the cases where the segment is equal to, or
less than, a hemisphere.]
Suppose, first, that the segment is greater than a hemisphere.
Let it be cut by a plane through its axis and the centre of the
earth; and, if possible, let it be at rest in the position shown
in the figure, where AB is the intersection of the plane with
the base of the segment, DE its axis, C the centre of the
sphere of which the segment is a part, O the centre of the
earth.

D
F
G
H
C
B
E
A
о
The centre of gravity of the portion of the segment outside
the fluid, as F, lies on OC produced, its axis passing through C.
Let G be the centre of gravity of the segment. Join FG,
and produce it to H so that
FG: GH = (volume of immersed portion): (rest of solid).
Join OH.
Then the weight of the portion of the solid outside the fluid
acts along FO, and the pressure of the fluid on the immersed
portion along OH, while the weight of the immersed portion
acts along HO and is by hypothesis less than the pressure of
the fluid acting along OH.
Hence there will not be equilibrium, but the part of the
segment towards A will ascend and the part towards B descend,
until DE assumes a position perpendicular to the surface of
the fluid.
ON FLOATING BODIES.
BOOK II.
+
Proposition 1.
If a solid lighter than a fluid be at rest in it, the weight of
the solid will be to that of the same volume of the fluid as the
immersed portion of the solid is to the whole.
Let (A+B) be the solid, B the portion immersed in the
fluid.
¿
Let (C+D) be an equal volume of the fluid, C being equal
in volume to A and B to D.
Further suppose the line E to represent the weight of the
solid (A+B), (F+G) to represent the weight of (C+D), and
G that of D.
1

A
E

B
C
F
D
G
Then
weight of (A + B) : weight of (C + D) = E : (F+G)...(1).
264
ARCHIMEDES
And the weight of (A+B) is equal to the weight of a
volume B of the fluid [I. 5], i.e. to the weight of D.
That is to say, E = G.
Hence, by (1),
weight of (A + B) : weight of (C + D) = G : F +G
=D:C+D
=
B: A+B
Proposition 2.
If a right segment of a paraboloid of revolution whose axis is
not greater than p (where p is the principal parameter of the
generating parabola), and whose specific gravity is less than that
of a fluid, be placed in the fluid with its axis inclined to the
vertical at any angle, but so that the base of the segment does not
touch the surface of the fluid, the segment of the paraboloid will
not remain in that position but will return to the position in
which its axis is vertical.
Let the axis of the segment of the paraboloid be AN, and
through AN draw a plane perpendicular to the surface of the
fluid. Let the plane intersect the paraboloid in the parabola
BAB', the base of the segment of the paraboloid in BB', and
the plane of the surface of the fluid in the chord QQ' of the
parabola.
Then, since the axis AN is placed in a position not perpen-
dicular to QQ', BB' will not be parallel to QQ'.
Draw the tangent PT to the parabola which is parallel to
QQ', and let P be the point of contact*.
[From P draw PV parallel to AN meeting QQ' in V.
Then PV will be a diameter of the parabola, and also the
axis of the portion of the paraboloid immersed in the fluid.
* The rest of the proof is wanting in the version of Tartaglia, but is given
in brackets as supplied by Commandinus.
ON FLOATING BODIES II.
265
Let C be the centre of gravity of the paraboloid BAB', and
F that of the portion immersed in the fluid. Join FC and
produce it to H so that H is the centre of gravity of the
remaining portion of the paraboloid above the surface.
Then, since
B
V
F
N
H
Q'
A
LPKM T
AN=AC*,
B'
and
it follows that
AN+ $p,
Ρ
AC >
Therefore, if CP be joined,
the angle CPT is acute†.
Hence, if CK be drawn perpendicular to PT, K will fall between
P and T. And, if FL, HM be drawn parallel to CK to meet
PT, they will each be perpendicular to the surface of the fluid.
Now the force acting on the immersed portion of the
segment of the paraboloid will act upwards along LF, while
the weight of the portion outside the fluid will act downwards
along HM.
Therefore there will not be equilibrium, but the segment
* As the determination of the centre of gravity of a segment of a paraboloid
which is here assumed does not appear in any extant work of Archimedes, or
in any known work by any other Greek mathematician, it appears probable that
it was investigated by Archimedes himself in some treatise now lost.
+ The truth of this statement is easily proved from the property of the sub-
normal. For, if the normal at P meet the axis in G, AG is greater than
except in the case where the normal is the normal at the vertex A itself. But
the latter case is excluded here because, by hypothesis, AN is not placed vertically.
Hence, P being a different point from A, AG is always greater than AC; and,
since the angle TPG is right, the angle TPC must be acute.
J

266
ARCHIMEDES
will turn so that B will rise and B' will fall, until AN takes
the vertical position.]
[For purposes of comparison the trigonometrical equivalent
of this and other propositions will be appended.
Suppose that the angle NTP, at which in the above figure
the axis AN is inclined to the surface of the fluid, is denoted
by 0.
Then the coordinates of P referred to AN and the tangent
at A as axes are
p
cot² 0, 2 cot 0,
4
E
where p
is the principal parameter.
Suppose that AN=h, PV-k.
If now be the distance from T of the orthogonal projection
of F on TP, and a the corresponding distance for the point C,
we have
x
2/2
P
2
cot³0. cos 0+cot 0. sin 0 + cos 0,
2
XC cot 0. cos 0+h cose,
4
2
whence a' - a = cos 02 (cot³ 0 + 2) — (h−k)} .
3
In order that the segment of the paraboloid may turn in
the direction of increasing the angle PTN, x' must be greater
than x, or the expression just found must be positive.
This will always be the case, whatever be the value of 0, if
or
p
2h
2
3
håp.]
Proposition 3.
If a right segment of a paraboloid of revolution whose axis
is not greater than &p (where p is the parameter), and whose
specific gravity is less than that of a fluid, be placed in the fluid
with its axis inclined at any angle to the vertical, but so that its
ON FLOATING BODIES II.
267
base is entirely submerged, the solid will not remain in that posi-
tion but will return to the position in which the axis is vertical.
Let the axis of the paraboloid be AN, and through AN
draw a plane perpendicular to the surface of the fluid inter-
secting the paraboloid in the parabola BAB', the base of the
segment in BNB', and the plane of the surface of the fluid in
the chord QQ' of the parabola.
Q
A
T MKPL

F
B'
H
N
B
Then, since AN, as placed, is not perpendicular to the
surface of the fluid, QQ' and BB' will not be parallel.
Draw PT parallel to QQ' and touching the parabola at P.
Let PT meet NA produced in T. Draw the diameter PV
bisecting QQ' in V. PV is then the axis of the portion of the
paraboloid above the surface of the fluid.
Let C be the centre of gravity of the whole segment of the
paraboloid, F that of the portion above the surface. Join FC
and produce it to H so that H is the centre of gravity of
the immersed portion.
Then, since AC2, the angle CPT is an acute angle, as in
the last proposition.
Hence, if CK be drawn perpendicular to PT, K will fall
between P and T. Also, if HM, FL be drawn parallel to CK,
they will be perpendicular to the surface of the fluid.
And the force acting on the submerged portion will act
upwards along HM, while the weight of the rest will act
downwards along LF produced.
Thus the paraboloid will turn until it takes the position
in which AN is vertical.
268
ARCHIMEDES
Proposition 4.
-
2
Given a right segment of a paraboloid of revolution whose
axis AN is greater than åp (where p is the parameter), and
whose specific gravity is less than that of a fluid but bears
to it a ratio not less than (AN — §p)² : AN², if the segment
of the paraboloid be placed in the fluid with its axis at any
inclination to the vertical, but so that its base does not touch
the surface of the fluid, it will not remain in that position but
will return to the position in which its axis is vertical.
Let the axis of the segment of the paraboloid be AN, and
let a plane be drawn through AN perpendicular to the surface
of the fluid and intersecting the segment in the parabola BAB',
the base of the segment in BB', and the surface of the fluid in
the chord QQ' of the parabola.
N
H
B
Q
C
R
F
K
A
PT
B'

Then AN, as placed, will not be perpendicular to QQ'.
Draw PT parallel to QQ' and touching the parabola at P.
Draw the diameter PV bisecting QQ' in V. Thus PV will be
the axis of the submerged portion of the solid.
Let C be the centre of gravity of the whole solid, F that of
the immersed portion. Join FC and produce it to H so that H
is the centre of gravity of the remaining portion.
Now, since
and
AN = &AC,
AN>3p,
it follows that
AC>
Ρ
2
ON FLOATING BODIES II.
269
Measure CO along C'A equal to 2, and OR along OC′ equal to
AO.
Then, since
ANAC,
and
AR=3AO,
we have, by subtraction,
NR = 30C.
That is,
or
AN-AR-300
=
= 3p,
AR= (AN-3p).
Thus (AN – ³p)² : AN” – AR² : AN²,
and therefore the ratio of the specific gravity of the solid to
that of the fluid is, by the enunciation, not less than the ratio
AR² : AN².
But, by Prop. 1, the former ratio is equal to the ratio
of the immersed portion to the whole solid, i.e. to the ratio
PV2: AN' [On Conoids and Spheroids, Prop. 24].
Hence
PV2: AN
AR²: AN²,
or
PV & AR.
It follows that
PF(= 3PV) }AR
AO.
If, therefore, OK be drawn from O perpendicular to OA, it will
meet PF between P and F.
Also, if CK be joined, the triangle KCO is equal and similar
to the triangle formed by the normal, the subnormal and the
ordinate at P (since CO = p or the subnormal, and KO is
equal to the ordinate).
Therefore CK is parallel to the normal at P, and therefore
perpendicular to the tangent at P and to the surface of the
fluid.
Hence, if parallels to CK be drawn through F, H, they will
be perpendicular to the surface of the fluid, and the force
acting on the submerged portion of the solid will act upwards
along the former, while the weight of the other portion will
act downwards along the latter.
1
270
ARCHIMEDES
!
Therefore the solid will not remain in its position but will
turn until AN assumes a vertical position.
[Using the same notation as before (note following Prop. 2),
we have
Sp
2
x = cos 0
{2 (cotº 0 + 2) — 3 (h
(h – k
- k)},
and the minimum value of the expression within the bracket,
for different values of 0, is
Ρ
2
1 - 3
2
(h — k),
corresponding to the position in which AM is vertical, or =
π
2
Therefore there will be stable equilibrium in that position only,
provided that
k✶ (h− åp),
or, if s be the ratio of the specific gravity of the solid to that of
the fluid (= k²/h² in this case),
2
s✶ (h − &p)³/h².]
Proposition 5.
Given a right segment of a paraboloid of revolution such that
its axis AN is greater than åp (where p is the parameter), and
its specific gravity is less than that of a fluid but in a ratio to
it not greater than the ratio {AN² – (AN — ³p)²³} : ANª, if the
segment be placed in the fluid with its axis inclined at any angle
to the vertical, but so that its base is completely submerged, it will
not remain in that position but will return to the position in
which AN is vertical.
Let a plane be drawn through AN, as placed, perpendicular
to the surface of the fluid and cutting the segment of the
paraboloid in the parabola BAB', the base of the segment in
BB', and the plane of the surface of the fluid in the chord
QQ' of the parabola.
Draw the tangent PT parallel to QQ', and the diameter
PV, bisecting QQ', will accordingly be the axis of the portion
of the paraboloid above the surface of the fluid.
1
ON FLOATING BODIES II.
271
Let F be the centre of gravity of the portion above the
surface, C that of the whole solid, and produce FC to H, the
centre of gravity of the immersed portion.
As in the last proposition, AC >2, and we measure CO along
CA equal to 2, and OR along OC equal to §40.
Then AN=3AC, and AR = §AO;
and we derive, as before,
Now, by hypothesis,
AR=(AN — ¾p).
(spec. gravity of solid): (spec. gravity of fluid)
> {AN² – (AN — ¾p)²} : AN²
>(AN² – AR²) : AN².
and
TP

A
K
R
C
Q
Q'
V
B'
H
N
B
Therefore
(portion submerged): (whole solid)
> (AN² – AR²) : AN²,
(whole solid): (portion above surface)
Thus
whence
and
AN²: AR2.
AN: PV² AN²: AR",
2
PVAR,
PFAR
40.
272
ARCHIMEDES
靠
​Therefore, if a perpendicular to AC be drawn from 0, it will
meet PF in some point K between P and F.
And, since CO = p, CK will be perpendicular to PT, as in
the last proposition.
Now the force acting on the submerged portion of the solid
will act upwards through H, and the weight of the other
portion downwards through F, in directions parallel in both
cases to CK; whence the proposition follows.
Proposition 6.
If a right segment of a paraboloid lighter than a fluid be
such that its axis AM is greater than åp, but AM : †p < 15 : 4,
and if the segment be placed in the fluid with its axis so inclined
to the vertical that its base touches the fluid, it will never remain
in such a position that the base touches the surface in one point
only.
Suppose the segment of the paraboloid to be placed in the
position described, and let the plane through the axis AM
perpendicular to the surface of the fluid intersect the segment
of the paraboloid in the parabolic segment BAB' and the plane
of the surface of the fluid in BQ.
Take C on AM such that AC = 2CM (or so that C is the
centre of gravity of the segment of the paraboloid), and measure
CK along CA such that
AM: CK = 15: 4.

B
M
C
F
K
R
L
N
H
Q
PT
Thus AM: CK > AM: p, by hypothesis; therefore CK <{p.
ON FLOATING BODIES II.
273
Measure CO along CA equal to p. Also draw KR per-
pendicular to AC meeting the parabola in R.
Draw the tangent PT parallel to BQ, and through P draw
the diameter PV bisecting BQ in V and meeting KR in I.
Then
PV: PI KM: AK,
"for this is proved."*
And
whence
or >
CK = 4AM=‡AC;
AK = AC — CK = {AC = }AM.
Thus
KM = &AM.
Therefore
KM = &AK.
It follows that
PV
or > & PI,
so that
PI2IV.
or <
1.
Let F be the centre of gravity of the immersed portion of
the paraboloid, so that PF2FV. Produce FC to H, the
centre of gravity of the portion above the surface.
Draw OL perpendicular to PV.
* We have no hint as to the work in which the proof of this proposition was
contained. The following proof is shorter than Robertson's (in the Appendix
to Torelli's edition).
Let BQ meet AM in U, and let PN be the ordinate from P to AM.
We have to prove that PV. AK or PI.KM, or in other words that
Now
>
(PV.AK-PI. KM) is positive or zero.
PV.AK-PI.KM=AK.PV - (AK- AN) (AM – AK)
(since AN-AT).
Now
Therefore
whence
= AK² — AK (AM+AN− PV)+AM.AN
— AK² — AK.UM+AM.AN,
UM: BM=NT: PN.
UM²: p. AM=4AN²:p.AN,
UM²=4AM.AN,
or
UM2
AM.AN=
4
Therefore
PV.AK-PI.KM=AK² – AK. UM+
2
AK
=(41
= (1x- UM)",
2
and accordingly (PV. AK - PI. KM) cannot be negative.
H. A.
UM2
4
18
274
ARCHIMEDES
Then, since CO = p, CL must be perpendicular to PT and
therefore to the surface of the fluid.
And the forces acting on the immersed portion of the
paraboloid and the portion above the surface act respectively
upwards and downwards along lines through F and H parallel
to CL.
Hence the paraboloid cannot remain in the position in which
B just touches the surface, but must turn in the direction of
increasing the angle PTM.
The proof is the same in the case where the point I is not
on VP but on VP produced, as in the second figure*.
B'

M
H
0
K
O
A
Q
B
ER
Pi
T
[With the notation used on p. 266, if the base BB' touch
the surface of the fluid at B, we have
BM= BV sin + PN,
and, by the property of the parabola,
Therefore
BV² = (p+4AN) PV
= pk (1+cot² 0).
Ρ
√ph = √pk + ½ cot ✪.
2
To obtain the result of the proposition, we have to eliminate
k between this equation and
x'
x = cos
4
(cot³ 0 + 2) − ( h −k)} .
2
3
* It is curious that the figures given by Torelli, Nizze and Heiberg are all
incorrect, as they all make the point which I have called I lie on BQ instead of
VP produced.
ON FLOATING BODIES II.
275
We have, from the first equation,
p
k = h − √ph cot 0 + cot² 0,
4
h-k=√ph cot 0 - 2 cot' 0.
2
4
or
Therefore
x − x = cos 0
{2 (cotº 0 + 2) — 3 (√ph cot 0 - 2 cot
cot³ 0)
4
10)}
= cos 0
:0}.
4
(§ coť² 0 + 2) — } √/ph cot
If then the solid can never rest in the position described,
but must turn in the direction of increasing the angle PTM,
the expression within the bracket must be positive whatever
be the value of 0.
or
Therefore
(3)² ph < žp²,
h< 15p.]
Proposition 7.
Given a right segment of a paraboloid of revolution lighter
than a fluid and such that its axis AM is greater than p, but
AM: p<15: 4, if the segment be placed in the fluid so that
its base is entirely submerged, it will never rest in such a position
that the base touches the surface of the fluid at one point only.
Suppose the solid so placed that one point of the base
only (B) touches the surface of the fluid. Let the plane
through B and the axis AM cut the solid in the parabolic
segment BAB' and the plane of the surface of the fluid in the
chord BQ of the parabola.
Let C be the centre of gravity of the segment, so that
AC = 2CM; and measure CK along CA such that
It follows that
AM: CK= 15 : 4.
CK< ½ p.
Measure CO along CA equal to 1p. Draw KR perpen-
dicular to AM meeting the parabola in R.
+
18-2
276
ARCHIMEDES
י
Let PT, touching at P, be the tangent to the parabola
which is parallel to BQ, and PV the diameter bisecting BQ, i.e.
the axis of the portion of the paraboloid above the surface.

H
A
L
K
M
R
V
B
and
B'
Then, as in the last proposition, we prove that
PV PI,
3
or >
PI or 2IV.
<
Let F be the centre of gravity of the portion of the solid
above the surface; join FC and produce it to H, the centre of
gravity of the portion submerged.
Draw OL perpendicular to PV; and, as before, since
CO=p, CL is perpendicular to the tangent PT. And the
lines through H, F parallel to CL are perpendicular to the
surface of the fluid; thus the proposition is established as
before.
The proof is the same if the point I is not on VP but on
VP produced.
Proposition 8.
3
Given a solid in the form of a right segment of a paraboloid
of revolution whose axis AM is greater than p, but such that
AM: ¿p<15 : 4, and whose specific gravity bears to that of a
fluid a ratio less than (AM-p): AM2, then, if the solid be
placed in the fluid so that its base does not touch the fluid and
its axis is inclined at an angle to the vertical, the solid will not
return to the position in which its axis is vertical and will not
ON FLOATING BODIES II.
277
remain in any position except that in which its axis makes with
the surface of the fluid a certain angle to be described.
Let am be taken equal to the axis AM, and let c be a point
on am such that ac2cm. Measure co along ca equal to p,
and or along oc equal to ao.

d
B
X
a b
r c
m
B'

M
H
F
N
A
PT
Q'
Let X + Y be a straight line such that
2
(spec. gr. of solid): (spec. gr. of fluid) = (X + Y): am³......(a),
and suppose X=2Y.
Now
ar = 3 ao = 3 (} am–1p)
= am - ³ p
-AM-ip.
Therefore, by hypothesis,
(X + Y)² : am² < ar²: am²,
whence (X + Y) < ar, and therefore X< ao.
>
Measure ob along oa equal to X, and draw bd perpendicular
to ab and of such length that
Join ad.
bd² = 1 co.ab.......
(B).
Now let the solid be placed in the fluid with its axis AM
inclined at an angle to the vertical. Through AM draw a
plane perpendicular to the surface of the fluid, and let this
278
ARCHIMEDES
plane cut the paraboloid in the parabola BAB' and the plane
of the surface of the fluid in the chord QQ' of the parabola.
Draw the tangent PT parallel to QQ', touching at P, and
let PV be the diameter bisecting QQ' in V (or the axis of the
immersed portion of the solid), and PN the ordinate from P.
Measure AO along AM equal to ao, and OC along OM
equal to oc, and draw OL perpendicular to PV.
But
I. Suppose the angle OTP greater than the angle dab.
PN² : NT² > db² : ba².
Thus
and
Therefore
or
whence
PN² : NT²=p:4AN
=co: NT,
db² : ba² = ½co : ab, by (B).
NT<2ab,
AN< ab,
NO>bo (since ao = A0)
> X.
2
Now (X + Y)²: am² = (spec. gr. of solid): (spec. gr. of fluid)
so that
or
But
and therefore
=(portion immersed): (rest of solid).
=PV2: AM2,
X+Y=PV.
PL (=NO)>X
> }(X + Y), since X = 2Y,
> } PV,
PV<3PL,
PL>2LV.
Take a point F on PV so that PF=2FV, i.e. so that Fis
the centre of gravity of the immersed portion of the solid.
Also AC=ac=&am=AM, and therefore C is the centre
of gravity of the whole solid.
Join FC and produce it to H, the centre of gravity of the
portion of the solid above the surface.
ON FLOATING BODIES II.
279
Now, since CO=1p, CL is perpendicular to the surface of
the fluid; therefore so are the parallels to CL through F and
H. But the force on the immersed portion acts upwards
through F and that on the rest of the solid downwards
through H.
Therefore the solid will not rest but turn in the direction of
diminishing the angle MTP.
II. Suppose the angle OTP less than the angle dab. In
this case, we shall have, instead of the above results, the
following,
Also
and therefore
AN>ab,
· NO<X.
PV>&PL,
PL<2LV.
B'

M
H
B
FKL
N
A
P
Make PF equal to 2FV, so that F is the centre of gravity
of the immersed portion.
And, proceeding as before, we prove in this case that the
solid will turn in the direction of increasing the angle MTP.
III. When the angle MTP is equal to the angle dab,
equalities replace inequalities in the results obtained, and L is
itself the centre of gravity of the immersed portion. Thus all
the forces act in one straight line, the perpendicular CL;
therefore there is equilibrium, and the solid will rest in the
position described.
280
ARCHIMEDES
[With the notation before used
2
x' — x = cos 0 {2 (cot' 0 + 2) — 3 (h−k)},
and a position of equilibrium is obtained by equating to zero the
expression within the bracket. We have then
p
2
2 cot' = (kk) - 22
4₁
3
(h
2
It is easy to verify that the angle Ø satisfying this equation
is the identical angle determined by Archimedes. For, in the
above proposition,
3X
2.
=PV=k,
whence
Also
It follows that
2
= /h
2
ab h − − k − (b −k) — 2.
=
1 (h
bd-2.ab.
cot dab = ab³/bd" = = {} (h — k) — 22}]
4 (2
p (3
Proposition 9.
3
4
3
Given a solid in the form of a right segment of a paraboloid
of revolution whose axis AM is greater than p, but such that
AM: p< 15:4, and whose specific gravity bears to that of a
fluid a ratio greater than {AM² — (AM – ¾ p)2}: AM², then, if
the solid be placed in the fluid with its axis inclined at an angle
to the vertical but so that its base is entirely below the surface,
the solid will not return to the position in which its axis is
vertical and will not remain in any position except that in which
its axis makes with the surface of the fluid an angle equal to that
described in the last proposition.
Take am equal to AM, and take c on am such that ac= 2cm.
Measure co along ca equal to p, and ar along ac such that
ar= ao.
ON FLOATING BODIES II.
281
;
Let X + Y be such a line that
(spec. gr. of solid): (spec. gr. of fluid) = {am² — (X + Y)²} : am²,
and suppose X=2Y.

d
++
a b
rc
m
ТР

N
FNL
V
H
M
'B'
X
Now
B
ar=&ao
Therefore, by hypothesis,
whence
= 3 (3 am — 1 p)
= AM-3p.
am² — ar² : am² < {am² − (X + Y)²} : am³,
and therefore
X + Y < ar,
X < ao.
Make ob (measured along oa) equal to X, and draw bd
perpendicular to ba and of such length that
Join ad.
bd² = ½co. ab.
Now suppose the solid placed as in the figure with its axis
AM inclined to the vertical. Let the plane through AM
perpendicular to the surface of the fluid cut the solid in the
parabola BAB' and the surface of the fluid in QQ'.
Let PT be the tangent parallel to QQ', PV the diameter
bisecting QQ'(or the axis of the portion of the paraboloid above
the surface), PN the ordinate from P.
· ad
282
ARCHIMEDES
I. Suppose the angle MTP greater than the angle dab.
Let AM be cut as before in C and O so that AC=2CM,
OC=1p, and accordingly AM, am are equally divided. Draw
OL perpendicular to PV.
Then, we have, as in the last proposition,
whence
and therefore
It follows that
PN² : NT² >db² : ba²,
co:NT>]co:ab,
AN< ab.
NO>bo
> X.
Again, since the specific gravity of the solid is to that of
the fluid as the immersed portion of the solid to the whole,
or
AM² - (X + Y)²: AM² = AM² – PV² : AM²,
That is,
And
(X + Y)² : AM² = PV² : AM².
X + Y=PV.
PL (or NO)>X
so that
>{PV,
PL>2LV.
Take F on PV so that PF=2FV. Then F is the centre
of gravity of the portion of the solid above the surface.
Also C is the centre of gravity of the whole solid. Join FC
and produce it to H, the centre of gravity of the immersed
portion.
Then, since CO = p, CL is perpendicular to PT and to the
surface of the fluid; and the force acting on the immersed
portion of the solid acts upwards along the parallel to CL
through H, while the weight of the rest of the solid acts down-
wards along the parallel to CL through F.
Hence the solid will not rest but turn in the direction of
diminishing the angle MTP.
II. Exactly as in the last proposition, we prove that, if the
angle MTP be less than the angle dab, the solid will not remain
ON FLOATING BODIES II.
283
in its position but will turn in the direction of increasing the
angle MTP.
T
P

A
N
V
B'
Ho
M
B
III. If the angle MTP is equal to the angle dab, the solid
will rest in that position, because L and F will coincide, and all
the forces will act along the one line CL..
Proposition 10.
Given a solid in the form of a right segment of a paraboloid
of revolution in which the axis AM is of a length such that
AM:p>15:4, and supposing the solid placed in a fluid
of greater specific gravity so that its base is entirely above the
surface of the fluid, to investigate the positions of rest.
(Preliminary.)
Suppose the segment of the paraboloid to be cut by a plane
through its axis AM in the parabolic segment BAB, of which
BB, is the base.
Divide AM at C so that AC = 2CM, and measure CK along
CA so that
AM: CK= 15: 4
whence, by the hypothesis, CK > †p.
.(a),
Suppose CO measured along CA equal to 1p, and take a
point R on AM such that MR = 3CO.
Thus
AR=AM – MR
= 3 (AC — CO)
= §AO.
284
ARCHIMEDES
I
!
Join BA, draw KA, perpendicular to AM meeting BA in A,
bisect BA in A,, and draw A¿M₂, AM, parallel to AM meeting
BM in M₂, M, respectively.
B
3

Mз M₂ D M B2
B1
བs
Qs
iR
|A³
Q2
10
K
A
A
On A,M₂, AM, as axes describe parabolic segments similar
to the segment BAB₁. (It follows, by similar triangles, that
BM will be the base of the segment whose axis is A,M, and
BB, the base of that whose axis is A,M₂, where BB₂ = 2BM2.)
The parabola BAB, will then pass through C.
2
[For
BM₂: M₂M = BM½ : A¿K
2
· KM: AK
= CM+ CK: AC-CK
= (} + {}) AM : (} − †) AM
=
5
9: 6.......
= MA: AC.
5
..(B)
•
ON FLOATING BODIES II.
285
Thus C is seen to be on the parabola BA,B, by the converse
of Prop. 4 of the Quadrature of the Parabola.]
2
Also, if a perpendicular to AM be drawn from O, it will
meet the parabola BA,B, in two points, as Q2, P2. Let Q1Q2QD
be drawn through Q₂ parallel to AM meeting the parabolas
BAB₁, BAM respectively in Q1, Q, and BM in D; and let
P₁PP be the corresponding parallel to AM through P. Let
the tangents to the outer parabola at P₁, Q, meet MA produced
in T₁, U respectively.
3
Then, since the three parabolic segments are similar and
similarly situated, with their bases in the same straight line
and having one common extremity, and since Q1Q2QD is a
diameter common to all three segments, it follows that
Now
And
Q1Q2 : Q2Qs = (B‚‚ : B‚B). (BM : MB₂)*.
BB₁ : B₁B = MM₂ : BM
2
= 2:5,
(dividing by 2)
by means of (B) above.
by means of (B),
BM : MB₂=BM : (2BM₂ – BM)
=5(65),
= 5 : 1.
* This result is assumed without proof, no doubt as being an easy deduction
from Prop. 5 of the Quadrature of the Parabola. It may be established
as follows.
First, since AAA,В is a straight line, and AN=AT with the ordinary
notation (where PT is the tangent at P and PN the ordinate), it follows, by
similar triangles, that the tangent at B to the outer parabola is a tangent to
each of the other two parabolas at the same point B.
Now, by the proposition quoted, if DQ3Q2Q1 produced meet the tangent BT
in E,
whence
Similarly
and
EQ3: Q¿D=BD : DM,
EQ3: ED=BD : BM.
EQ2: ED=BD : BB2,
EQ₁: ED=BD : BB₁·
The first two proportions are equivalent to
and
By subtraction,
Similarly
It follows that
EQ₂: ED=BD.BB2:BM.BB2,
3
EQ2: ED=BD.BM:BM.BB2.
Q2Q3: ED=BD. MB,: BM.BB2.
2
Q1Q2: ED=BD. B2B₁: BB2. BB1.
1
Q1Q2: Q2Q3=(BB₁ : B₁B). (BM : MB₂).
286
ARCHIMEDES
+
or
It follows that
Similarly
Also, since
Q1Q2: Q2Q3 = 2: 1,
Q1Q2 = 2Q2Q3.
P₁P₂ = 2P₂P3
2
MR = &CO = 3p,
AR = AM – MR
3•.
= AM - 3p.
Enunciation.)
If the segment of the paraboloid be placed in the fluid with
its base entirely above the surface, then
(I.) if
(spec. gr. of solid): (spec. gr. of fluid) ‡ AR² : AM²
[† (AM − 3p)² : AM²],
the solid will rest in the position in which its axis AM is vertical;
(II.) if
2
(spec. gr. of solid): (spec. gr. of fluid) < AR² : AM²
but > Q1Q32: AM²,
2.
the solid will not rest with its base touching the surface of the
fluid in one point only, but in such a position that its base does
not touch the surface at any point and its axis makes with the
surface an angle greater than U;
(III. a) if
་
(spec. gr. of solid): (spec. gr. of fluid) = Q1Q32 : AM²,
the solid will rest and remain in the position in which the base
touches the surface of the fluid at one point only and the axis
makes with the surface an angle equal to U;
(III. b) if
3
2
(spec. gr. of solid): (spec. gr. of fluid) = P₁P²: AM²,
the solid will rest with its base touching the surface of the fluid
at one point only and with its axis inclined to the surface at an
angle equal to T₁;
(IV.) if
(spec. gr. of solid) : (spec. gr. of fluid) > P₁P¸² :
2
AM²
3
but < Q1Q3² : AM²,
2
ON FLOATING BODIES II.
287
the solid will rest and remain in a position with its base more
submerged;
(V.) if
2
(spec. gr. of solid): (spec. gr. of fluid) < P₁P¸ª : AM²,
the solid will rest in a position in which its axis is inclined to the
surface of the fluid at an angle less than T₁, but so that the base
does not even touch the surface at one point.
(Proof.)
(I.) Since AM > p, and
(spec. gr. of solid): (spec. gr. of fluid) ✶ (AM – ¾p)² : AM³,
it follows, by Prop. 4, that the solid will be in stable equilibrium
with its axis vertical.
(II.) In this case
2
(spec. gr. of solid): (spec. gr. of fluid) < AR2: AM²
B
but > Q1Q32: AM².

M B2
B1
P
P
P
Q3
F'
21
O/R
Ο
'Q'
יד
N'
A
Suppose the ratio of the specific gravities to be equal to
so that < AR but > Q1Q3-
12: AM²,
Place P'V' between the two parabolas BAB₁, BP,Q,M equal
288
ARCHIMEDES
30
tol and parallel to AM*; and let P'V' meet the intermediate
parabola in F″.
Then, by the same proof as before, we obtain
P'F' = 2F'V'.
Let P'T', the tangent at P' to the outer parabola, meet
MA in T", and let P'N' be the ordinate at P'.
Join BV' and produce it to meet the outer parabola in Q'.
Let OQ,P₂ meet P'V' in I.
2
Now, since, in two similar and similarly situated parabolic
* Archimedes does not give the solution of this problem, but it can be
supplied as follows.
Let BR₁Q1, BRQ₂ be two similar and similarly situated parabolic segments
with their bases in the same straight line, and let BE be the common tangent
at B.
E
R
R
1
R₂
B

O
Q
Suppose the problem solved, and let ERRO, parallel to the axes, meet the
parabolas in R, R₁ and BQ₂ in O, making the intercept RR₁ equal to 1.
2
Then, we have, as usual,
ER₁: E0=B0 : BQ1
=BO.BQ2: BQ1. BQ2,
and
ER: E0=BO: BQ2
=BO.BQ1: BQ1. BQ2.
By subtraction,
RR₁: E0=B0. Q1Q2: BQ1. BQ2,
or
BO.OE=1.
BQ1.BQ2 which is known.
Q102
And the ratio BO: OE is known. Therefore B02, or OE2, can be found, and
therefore 0.
ON FLOATING BODIES II.
289
segments with bases BM, BB, in the same straight line, BV', BQ'
are drawn making the same angle with the bases,
so that
BV': BQ' = BM : BB₁*
= 1:2,
BV' = V'Q'.
Suppose the segment of the paraboloid placed in the fluid,
as described, with its axis inclined at an angle to the vertical,
and with its base touching the surface at one point B only.
Let the solid be cut by a plane through the axis and per-
M
H
V
C
B.
N
A
P
B'

pendicular to the surface of the fluid, and let the plane intersect
the solid in the parabolic segment BAB' and the plane of the
surface of the fluid in BQ.
Take the points C, O on AM as before described.
Draw
* To prove this, suppose that, in the figure on the opposite page, BR₁ is
produced to meet the outer parabola in R2.
We have, as before,
whence
ER₁ : E0=BO : BQ1,
ER: E0=B0 : BQ2,
ER₁: ER=BQ2 : BQ1•
And, since R₁ is a point within the outer parabola,
Hence
H. A.
ER: ER₁=BR₁ : BR₂, in like manner.
BQ1 : BQ2=BR₁ : BR½•
19
290
ARCHIMEDES
the tangent parallel to BQ touching the parabola in P and
meeting AM in T; and let PV be the diameter bisecting BQ
(i.e. the axis of the immersed portion of the solid).
Then
whence
ľ² : AM² = (spec. gr. of solid): (spec. gr. of fluid)
= (portion immersed): (whole solid)
=
= PV² : AM²,
P'V' l = PV.
Thus the segments in the two figures, namely BP'Q',
BPQ, are equal and similar.
PTN=zPTN.
Therefore
Also
AT-AT', ANAN', PNP'N'.
Now, in the first figure, P'I< 2IV'.
Therefore, if OL be perpendicular to PV in the second
figure,
PL<2LV.
Take F on LV so that PF2FV, i.e. so that F is the centre
of gravity of the immersed portion of the solid. And C is the
centre of gravity of the whole solid. Join FC and produce it to
H, the centre of gravity of the portion above the surface.
Now, since CO = p, CL is perpendicular to the tangent at
P and to the surface of the fluid. Thus, as before, we prove
that the solid will not rest with B touching the surface, but will
turn in the direction of increasing the angle PTN.
Hence, in the position of rest, the axis AM must make with
the surface of the fluid an angle greater than the angle U which
the tangent at Q makes with AM.
(III. a) In this case
2
(spec. gr. of solid): (spec. gr. of fluid) = Q₁Q;² : AM².
Let the segment of the paraboloid be placed in the fluid so
that its base nowhere touches the surface of the fluid, and its
axis is inclined at an angle to the vertical.
ON FLOATING BODIES II.
291
Let the plane through AM perpendicular to the surface of
the fluid cut the paraboloid in the parabola BAB' and the
B
M
H
F
B
M B2
P
P3
ar
P
Quiz
Β΄

Q'
O
N
A

9
Q1
N'
AU
T₁
B1
plane of the surface of the fluid in QQ'. Let PT be the tangent
parallel to QQ', PV the diameter bisecting QQ', PN the ordinate
at P.
Divide AM as before at C, O.
19-2
292
ARCHIMEDES
In the other figure let Q₁N' be the ordinate at Q1. Join
BQ and produce it to meet the outer parabola in q. Then
BQ = Q:q, and the tangent Q₁U is parallel to Bq. Now
2
Q₁Q²: AM² = (spec. gr. of solid): (spec. gr. of fluid)
3
= (portion immersed): (whole solid)
= PV2: AM².
Therefore Q1Q = PV; and the segments QPQ', BQıq of the
paraboloid are equal in volume. And the base of one passes
through B, while the base of the other passes through Q, a point
nearer to A than B is.
It follows that the angle between QQ' and BB' is less than
the angle B,Bq.
Therefore
whence
<U<< PTN,
AN'> AN,
and therefore
N'O (or Q1Q2) < PL,
where OL is perpendicular to PV.
It follows, since Q1Q2 = 2Q2Q3, that
PL>2LV.
Therefore F, the centre of gravity of the immersed portion
of the solid, is between P and L, while, as before, CL is perpen-
dicular to the surface of the fluid.
Producing FC to H, the centre of gravity of the portion of
the solid above the surface, we see that the solid must turn in
the direction of diminishing the angle PTN until one point B
of the base just touches the surface of the fluid.
When this is the case, we shall have a segment BPQ equal
and similar to the segment BQ19, the angle PTN will be equal
to the angle U, and AN will be equal to AN'.
Hence in this case PL = 2LV, and F, L coincide, so that F,
C, H are all in one vertical straight line.
Thus the paraboloid will remain in the position in which
one point B of the base touches the surface of the fluid, and the
axis makes with the surface an angle equal to U.
ON FLOATING BODIES II.
293
(III. b) In the case where
(spec. gr. of solid): (spec. gr. of fluid) = P₁P¸² : AM²,
we can prove in the same way that, if the solid be placed in the
fluid so that its axis is inclined to the vertical and its base does
not anywhere touch the surface of the fluid, the solid will take
up and rest in the position in which one point only of the base
touches the surface, and the axis is inclined to it at an angle
equal to T, (in the figure on p. 284).
(IV.) In this case
(spec. gr. of solid) : (spec. gr. of fluid) > P₁P² : AM²
2
but < Q1Q,² : AM².
Suppose the ratio to be equal to 12: AM², so that l is greater
than P₁P, but less than Q1Q..
3
Place P'V' between the parabolas BP₁Q₁, BPQ, so that
P'V' is equal to l and parallel to AM, and let P'V' meet the
intermediate parabola in F" and OQ,P₂ in I.
B
M B2
2

B₁
P
F'
Pi
IN'
Q1
IA
9
Join BV' and produce it to meet the outer parabola in q.
Then, as before, BV' =
P'T' at P' is parallel to Bq.
V'q, and accordingly the tangent
Let P'N' be the ordinate of P'.
+
294
ARCHIMEDES
1. Now let the segment be placed in the fluid, first, with
its axis so inclined to the vertical that its base does not
anywhere touch the surface of the fluid.
B'
1

M
H
B
V
F
N
P
A
T
Let the plane through AM perpendicular to the surface of
the fluid cut the paraboloid in the parabola BAB' and the
plane of the surface of the fluid in QQ'. Let PT be the
tangent parallel to QQ', PV the diameter bisecting QQ'.
Divide AM at C, O as before, and draw OL perpendicular to PV.
Then, as before, we have PV = l = P'V'.
Thus the segments BP'q, QPQ' of the paraboloid are equal
in volume; and it follows that the angle between QQ' and BB'
is less than the angle B,Bq.
Therefore
and hence
so that
i.e.
LPTN'</PTN,
AN' > AN,
NO > N'O,
PL> P'I
> P'F', a fortiori,
Thus PL>2LV, so that F, the centre of gravity of the
immersed portion of the solid, is between L and P, while CL
is perpendicular to the surface of the fluid.
ON FLOATING BODIES II.
295
If then we produce FC to H, the centre of gravity of the
portion of the solid above the surface, we prove that the solid
will not rest but turn in the direction of diminishing the
angle PTN.
2. Next let the paraboloid be so placed in the fluid that
its base touches the surface of the fluid at one point B only,
and let the construction proceed as before.
Then PV=P'V', and the segments BPQ, BP'q are equal
and similar, so that
It follows that
and therefore
whence
LPTN=¿P'T'N'.
AN-AN', NO = N'O,
P'I = PL,
PL>2LV.
B'

M
H
U
B
N
LKF
A
Thus F again lies between P and L, and, as before, the
paraboloid will turn in the direction of diminishing the angle
PTN, i.e. so that the base will be more submerged.
(V.) In this case
2
(spec. gr. of solid): (spec. gr. of fluid) < P₁P² : AM².
3
If then the ratio is equal to l² : AM³, 1 < P₂P¸. Place P'V'
between the parabolas BP,Q, and BP;Q; equal in length to l
1
296
ARCHIMEDES
and parallel to AM. Let P'V' meet the intermediate parabola
in F" and OP, in I.
Join BV' and produce it to meet the outer parabola in q.
Then, as before, BV' = V'q, and the tangent P'T' is parallel
to Bq.
B
M B2
V'P3
P
A
C
1. Let the paraboloid be so placed in the fluid that its
base touches the surface at one point only.
B'
M
V
B
H
A
+


ON FLOATING BODIES II.
297
Let the plane through AM perpendicular to the surface
of the fluid cut the paraboloid in the parabolic section BAB'
and the plane of the surface of the fluid in BQ.
Making the usual construction, we find
PV = l = P'V',
and the segments BPQ, BP,q are equal and similar.
Therefore
and
Therefore
LPTN =
P'TN,
ANAN', N'ONO.
whence it follows that
PL = P'I,
PL < 2LV.
Thus F, the centre of gravity of the immersed portion of the
solid, lies between L and V, while CL is perpendicular to the
surface of the fluid.
Producing FC to H, the centre of gravity of the portion
above the surface, we prove, as usual, that there will not be
rest, but the solid will turn in the direction of increasing the
angle PTN, so that the base will not anywhere touch the
surface.
2. The solid will however rest in a position where its axis
makes with the surface of the fluid an angle less than T₁.
B'

B
M
Ι
P
O
N
A
298
ARCHIMEDES
For let it be placed so that the angle PTN is not less
than T₁.
Then, with the same construction as before, PV=1= P'V'.
And, since
<T<T,
AN AN₁,
and therefore NO N₁0, where P₁N, is the ordinate of P₁.
Hence
But
1
PL P₁P2.
P₁P₂> P'F".
PL>&PV,
Therefore
so that F, the centre of gravity of the immersed portion of
the solid, lies between P and L.
Thus the solid will turn in the direction of diminishing
the angle PTN until that angle becomes less than T₁.
[As before, if x, x' be the distances from T of the orthogonal
projections of C, F respectively on TP, we have
2
x'
- x = cos
{2} (cot¹ 0 + 2) — 3 (h —
•
(1),
where h= AM, k = PV.
Also, if the base BB' touch the surface of the fluid at one
point B, we have further, as in the note following Prop. 6,
and
√ph = √ pk + 2 cot 0 .....
h− k = √ph cot 0 - 2 cot* 0
..
•
(2),
(3).
Therefore, to find the relation between h and the angle at
which the axis of the paraboloid is inclined to the surface of the
fluid in a position of equilibrium with B just touching the
surface, we eliminate k and equate the expression in (1) to
zero; thus
or
2 (cot¹ 0 + 2) — 2 (√ph cot 0 - 2 cot* 0) = 0,
5p cot³ 0 - 8√ph cot 0 + 6p = 0 ....
(4).
ON FLOATING BODIES II.
299
The two values of @ are given by the equations
5√p cot 0 = 4√h ± √16h – 30p ……..
(5).
The lower sign corresponds to the angle U, and the upper
sign to the angle T₁, in the proposition of Archimedes, as can
be verified thus.
In the first figure of Archimedes (p. 284 above) we have
AK = 3h,
-
M,D² = 3p. OK = } p (} h − h − p)
2
3p (4h
5
15 2
(一​).
If P₁PP, meet BM in D', it follows that
3
M,D
M₂D'S
M₂D ± M,M₂
3
/3p (4h - 1) + 10 √ph,
5 15 2
MD
and
=
MD'
MM₂ M₂D
2
2
√ph +
3p/4h
p
515
2/2
Now, from the property of the parabola,
cot U=2MD/p,
cot T₁ = 2MD'/p,
so that
el cot
{}
2
/ph =
5
3p/4h p
V5 15 2
>
or
5 √p cot
4 √h = √16h – 30p,
T₁
which agrees with the result (5) above.
To find the corresponding ratio of the specific gravities, or
k²/h², we have to use equations (2) and (5) and to express k in
terms of h and p.
300
ARCHIMEDES
Equation (2) gives, on the substitution in it of the value of
cote contained in (5),
√k = √ñ − † (4 √h ± √16h − 30p)
= § √h ↓ ↓ √16h — 30p,
whence we obtain, by squaring,
-
k = f f h − fo p = 3 √h (16h — 30p) ………………….. (6).
25
The lower sign corresponds to the angle U and the upper to
the angle T₁, and, in order to verify the results of Archimedes,
we have simply to show that the two values of k are equal to
Q1 Q3, P₁P, respectively.
3
Now it is easily seen that
Q1Qs=h/2 — MD²/p+2M,D²/p,
P₁P¸=h/2-MD'/p+2M,D'/p.
3
Therefore, using the values of MD, MD', MD, M,D' above
found, we have
Q1 Q3)
P₁P
h 3/4h
+
3
=
p
7h6 3h (4h p
2 5 15 2 50 5 N 5 15 2
-
= 18 h − + p ± 3√h (16h – 30p),
1 25
which are the values of k given in (6) above.]
BOOK OF LEMMAS.
Proposition 1.
If two circles touch at A, and if BD, EF be parallel diameters
in them, ADF is a straight line.
[The proof in the text only applies to the particular case
where the diameters are perpendicular to the radius to the
point of contact, but it is easily adapted to the more general
case by one small change only.]
Let O, C be the centres of the circles, and let OC be joined
and produced to A. Draw DH parallel to A0 meeting OF
in H.
E
A.

C
H
F
BI
Then, since
and
we have, by subtraction,
Therefore
OH = CD = CA,
OF OA,
=
HF = CO = DH.
HDF=HFD.
302
ARCHIMEDES
Thus both the triangles CAD, HDF are isosceles, and the
third angles ACD, DHF in each are equal. Therefore the
equal angles in each are equal to one another, and
<ADC=<DFH.
Add to each the angle CDF, and it follows that
2. ADC + ≤ CDF = / CDF +▲ DFH
(two right angles).
Hence ADF is a straight line.
The same proof applies if the circles touch externally*.
Proposition 2.
Let AB be the diameter of a semicircle, and let the tangents
to it at B and at any other point D on it meet in T. If now DE
be drawn perpendicular to AB, and if AT, DE meet in F,
DFFE.
Produce AD to meet BT produced in H. Then the angle
ADB in the semicircle is right; therefore the angle BDH is
also right. And TB, TD are equal.
A
F
E
B
H

T
Therefore T is the centre of the semicircle on BH as
diameter, which passes through D.
Hence
HT = TB.
And, since DE, HB are parallel, it follows that DF = FE.
* Pappus assumes the result of this proposition in connexion with the
äpßnλos (p. 214, ed. Hultsch), and he proves it for the case where the circles
touch externally (p. 840).
BOOK OF LEMMAS.
303
Proposition 3.
Let P be any point on a segment of a
circle whose base is
AB, and let PN be perpendicular to AB. Take D on AB so
If now PQ be an arc equal to the arc PA, and
that AN = ND.
ND.
BQ be joined,
BQ, BD shall be equal*.
Q

P
N
D
B
Join PA, PQ, PD, DQ.
* The segment in the figure of the мs. appears to have been a semicircle,
though the proposition is equally true of any segment. But the case where the
segment is a semicircle brings the proposition into close connexion with a
proposition in Ptolemy's μeyáλn oúvtağıs, I. 9 (p. 31, ed. Halma; cf. the repro-
duction in Cantor's Gesch. d. Mathematik, I. (1894), p. 389). Ptolemy's object is
to connect by an equation the lengths of the chord of an arc and the chord of half
the arc. Substantially his procedure is as follows. Suppose AP, PQ to be
equal arcs, AB the diameter through A; and let AP, PQ, AQ, PB, QB be joined.
Measure BD along BA equal to BQ. The perpendicular PN is now drawn, and
it is proved that PA=PD, and AN=ND.
Then
AN=}(BA−BD)= ½ (BA−BQ) = ½ (BA −√√ BA² – AQ²).
And, by similar triangles,
Therefore
AN: AP=AP: AB.
AP-AB.AN
= } (AB — √√ AB² — AQ²). AB.
This gives AP in terms of AQ and the known diameter AB. If we divide by
AB2 throughout, it is seen at once that the proposition gives a geometrical
proof of the formula
sin2
≥ (1 − cos a).
The case where the segment is a semicircle recalls also the method used by
Archimedes at the beginning of the second part of Prop. 3 of the Measurement
of a circle. It is there proved that, in the figure above,
AB+BQ: AQ=BP:PA,
or, if we divide the first two terms of the proposition by AB,
α
(1+cos a)/sin a=cot
cot 2.
304
ARCHIMEDES
and
Then, since the arcs PA, PQ are equal,
PA = PQ.
But, since AN = ND, and the angles at N are right,
Therefore
PA = PD.
PQ = PD,
·
< PQD = < PDQ.
Now, since A, P, Q, B are concyclic,
whence
< PAD+< PQB = (two right angles),
<PDA+<PQB = (two right angles)
=<PDA + PDB.
Therefore
▲ PQB = ¿ PDB;
and, since the parts, the angles PQD, PDQ, are equal,
and
+ BQD=zBDQ,
BQ = BD.
Proposition 4.
If AB be the diameter of a semicircle and N any point on AB,
and if semicircles be described within the first semicircle and
having AN, BN as diameters respectively, the figure included
between the circumferences of the three semicircles is “what
Archimedes called an apẞnλos*"; and its area is equal to the
ἄρβηλος
circle on PN as diameter, where PN is perpendicular to AB
and meets the original semicircle in P.
For
AB² = AN² + NB² + 2AN.NB
= AN² + NB² + 2PN².
But circles (or semicircles) are to one another as the squares of
their radii (or diameters).
ǎpẞnλos is literally 'a shoemaker's knife.' Cf. note attached to the remarks
on the Liber Assumptorum in the Introduction, Chapter II.
BOOK OF LEMMAS.
305
Hence
(semicircle on AB)=(sum of semicircles on AN, NB)
+ 2 (semicircle on PN).
A
P
N
B
7

That is, the circle on PN as diameter is equal to the
difference between the semicircle on AB and the sum of the
semicircles on AN, NB, i.e. is equal to the area of the apßnλos.
Proposition 5.
Let AB be the diameter of a semicircle, C any point on AB,
and CD perpendicular to it, and let semicircles be described
within the first semicircle and having AC, CB as diameters.
Then, if two circles be drawn touching CD on different sides
and each touching two of the semicircles, the circles so drawn
will be equal.
Let one of the circles touch CD at E, the semicircle on AB
in F, and the semicircle on AC in G.
Draw the diameter EH of the circle, which will accordingly
be perpendicular to CD and therefore parallel to AB.
.
Join FH, HA, and FE, EB. Then, by Prop. 1, FHA, FEB
are both straight lines, since EH, AB are parallel.
For the same reason AGE, CGH are straight lines.
Let AF produced meet CD in D, and let AE produced
meet the outer semicircle in I. Join BI, ID.
H. A.
20
I
306
ARCHIMEDES
Then, since the angles AFB, ACD are right, the straight
lines AD, AB are such that the perpendiculars on each from the
extremity of the other meet in the point E. Therefore, by the
properties of triangles, AE is perpendicular to the line joining
B to D.
H
F
D

E
A
But AE is perpendicular to BI.
Therefore BID is a straight line.
B
Now, since the angles at G, I are right, CH is parallel
to BD.
Therefore
so that
AB: BC= AD: DH
= AC: HE,
AC. CB AB. HE.
In like manner, if d is the diameter of the other circle, we can
prove that
AC. CB AB. d.
=
Therefore d = HE, and the circles are equal*.
* The property upon which this result depends, viz. that
AB: BC=AC: HE,
appears as an intermediate step in a proposition of Pappus (p. 230, ed. Hultsch)
which proves that, in the figure above,
AB: BC=CE2: HE².
The truth of the latter proposition is easily seen. For, since the angle CEH
is a right angle, and EG is perpendicular to CH,
CE2: EH2=CG: GH
=AC: HE.
BOOK OF LEMMAS.
307
[As pointed out by an Arabian Scholiast Alkauhi, this
proposition may be stated more generally. If, instead of one
point C on AB, we have two points C, D, and semicircles be
described on AC, BD as diameters, and if, instead of the
perpendicular to AB through C, we take the radical axis of the
two semicircles, then the circles described on different sides of
the radical axis and each touching it as well as two of the
semicircles are equal. The proof is similar and presents no
difficulty.]
Proposition 6.
Let AB, the diameter of a semicircle, be divided at C so that
AC = 3 CB [or in any ratio]. Describe semicircles within the
first semicircle and on AC, CB as diameters, and suppose a
circle drawn touching all three semicircles. If GH be the
diameter of this circle, to find the relation between GH and AB.
Let GH be that diameter of the circle which is parallel to
AB, and let the circle touch the semicircles on AB, AC, CB
in D, E, F respectively.
Join AG, GD and BH, HD. Then, by Prop. 1, AGD, BHD
are straight lines.

A
N
D
H
E
F
K
M
P
B
For a like reason AEH, BFG are straight lines, as also
are CEG, CFH.
Let AD meet the semicircle on AC in I, and let BD meet
the semicircle on CB in K. Join CI, CK meeting AE, BF
20-2
308
ARCHIMEDES
respectively in L, M, and let GL, HM produced meet AB in
N, P respectively.
Now, in the triangle AGC, the perpendiculars from A, C on
the opposite sides meet in L. Therefore, by the properties of
triangles, GLN is perpendicular to AC.
Similarly HMP is perpendicular to CB.
Again, since the angles at I, K, D are right, CK is parallel
to AD, and CI to BD.
Therefore
AC: CB AL: LH
=
= AN: NP,
and
Hence
BC: CA = BM : MG
= BP: PN.
AN:NP=NP: PB,
or AN, NP, PB are in continued proportion*.
Now, in the case where AC = 3 CB,
AN = NP = 2 PB,
½
whence BP:PN: NA:AB=4:6:9: 19.
Therefore
6.
GH = NP = & AB.
And similarly GH can be found when AC: CB is equal to
any other given ratio†.
* This same property appears incidentally in Pappus (p. 226) as an inter-
mediate step in the proof of the "ancient proposition" alluded to below.
+ In general, if AC: CB=λ: 1, we have
and
BP: PN: NA: AB=1:λ: λ²: (1+λ+λ²),
GH:AB=\: (1+λ+λ²).
It may be interesting to add the enunciation of the "ancient proposition "
stated by Pappus (p. 208) and proved by him after several auxiliary lemmas.

02
01
A
Ng
N2
C N₁
B
BOOK OF LEMMAS.
309
Proposition 7.
If circles be circumscribed about and inscribed in a square,
the circumscribed circle is double of the inscribed circle.
For the ratio of the circumscribed to the inscribed circle is
equal to that of the square on the diagonal to the square itself,
i.e. to the ratio 2 : 1.
Proposition 8.
If AB be any chord of a circle whose centre is O, and if AB
be produced to C so that BC is equal to the radius; if further CO
meet the circle in D and be produced to meet the circle a second
time in E, the arc AE will be equal to three times the arc BD.

A
B
C
E
F
D
Draw the chord EF parallel to AB, and join OB, OF.
Let an äрßnλos be formed by three semicircles on AB, AC, CB as diameters, and
let a series of circles be described, the first of which touches all three semicircles,
while the second touches the first and two of the semicircles forming one end
of the apẞnλos, the third touches the second and the same two semicircles, and
so on. Let the diameters of the successive circles be d₁, da, dg,... their centres
01, 02, Og,... and O₁N₁, ON½, ON¸,... the perpendiculars from the centres on
AB. Then it is to be proved that
O₁N₁=d₁,
O₂N₂=2dg,
ON₂=3dg,
OnNn=ndn•
310
ARCHIMEDES
Then, since the angles OEF, OFE are equal,
<COF=2/OEF
2 BCO, by parallels,
= 2 BOD, since BC BO.
-
Therefore
< BOF=32 BOD,
so that the arc BF is equal to three times the arc BD.
Hence the arc AE, which is equal to the arc BF, is equal to
three times the arc BD*.
Proposition 9.
If in a circle two chords AB, CD which do not pass through
the centre intersect at right angles, then
(arc AD)+(arc CB)=(arc AC)+(arc DB).
Let the chords intersect at O, and draw the diameter EF
parallel to AB intersecting CD in
H. EF will thus bisect CD at
right angles in H, and
(arc ED)=(arc EC).
Also EDF, ECF are semi-
circles, while
(arc ED)=(arc EA) + (arc AD).
Therefore
(sum of arcs CF, EA, AD) = (arc
of a semicircle).
And the arcs AE, BF are equal.
Therefore
A
D

B
F
E
H
C
(arc CB) + (arc AD) = (arc of a semicircle).
* This proposition gives a method of reducing the trisection of any angle,
i.e. of any circular arc, to a problem of the kind known as vevσels. Suppose that
AE is the arc to be trisected, and that ED is the diameter through E of the circle
of which AE is an arc. In order then to find an arc equal to one-third of AE,
we have only to draw through A a line ABC, meeting the circle again in B and
ED produced in C, such that BC is equal to the radius of the circle. For a
discussion of this and other vevσels see the Introduction, Chapter V.
BOOK OF LEMMAS
311
Hence the remainder of the circumference, the sum of the
arcs AC, DB, is also equal to a semicircle; and the proposition
is proved.
Proposition 10.
Suppose that TA, TB are two tangents to a circle, while TC
cuts it. Let BD be the chord through B parallel to TC, and let
AD meet TC in E. Then, if EH be drawn perpendicular to BD,
it will bisect it in H.
Let AB meet TC in F, and join BE.
Now the angle TAB is equal to the angle in the alternate
segment, i.e.
▲ TAB=ZADB
= AET, by parallels.
C
A
[m]
E
F
B
H
T
Hence the triangles EAT, AFT have one angle equal and
another (at 7) common. They are therefore similar, and
Therefore
FT: AT-AT: ET
ET. TF TA²
=TB².
It follows that the triangles EBT, BFT are similar.
Therefore
TEB=LTBF
= <TAB
But the angle TEB is equal to the angle EBD, and the
angle TAB was proved equal to the angle EDB.
1

312
ARCHIMEDES
Therefore
LEDB=LEBD.
And the angles at H are right angles.
It follows that
BH = HD*.
Proposition 11.
If two chords AB, CD in a circle intersect at right angles in
a point O, not being the centre, then
AO² + BO² + CO2 + DO² = (diameter)".
Draw the diameter CE, and join AC, CB, AD, BE.
Then the angle CAO is equal
to the angle CEB in the same seg-
ment, and the angles AOC, EBC
are right; therefore the triangles
AOC, EBC are similar, and

▲ ACO = 4 ECB.
It follows that the subtended
A
B
arcs, and therefore the chords AD,
BE, are equal.
D
E
* The figure of this proposition curiously recalls the figure of a problem
given by Pappus (pp. 836-8) among his lemmas to the first Book of the treatise
of Apollonius On Contacts (Teρì èπapŵv). The problem is, Given a circle and
two points E, F (neither of which is necessarily, as in this case, the middle
point of the chord of the circle drawn through E, F), to draw through E, F
respectively two chords AD, AB having a common extremity A and such that DB
is parallel to EF. The analysis is as follows. Suppose the problem solved, BD
being parallel to FE. Let BT, the tangent at B, meet EF produced in T. (T
is not in general the pole of AB, so that TA is not generally the tangent at A.)
Then
▲ TBF= ▲ BDA, in the alternate segment,
= LAET, by parallels.
Therefore A, E, B, T are coǹcyclic, and
EF.FT-AF.FB.
But, the circle ADB and the point F being given, the rectangle AF. FB is given.
Also EF is given.
Hence FT is known.
Thus, to make the construction, we have only to find the length of FT from
the data, produce EF to T so that FT has the ascertained length, draw the
tangent TB, and then draw BD parallel to EF. DE, BF will then meet in A on
the circle and will be the chords required.
1
BOOK OF LEMMAS.
J
313
Thus
(A0* + DO*)+(BO + CO*) = AD +BC*
= BE² + BC²
= CE³.
Proposition 12.
If AB be the diameter of a semicircle, and TP, TQ the
tangents to it from any point T, and if AQ, BP be joined
meeting in R, then TR is perpendicular to AB.
Let TR produced meet AB in M, and join PA, QB.
Since the angle APB is right,
¿PAB+¿ PBA = (a right angle)
= <AQB.
T

P
R
A
M
B
Add to each side the angle RBQ, and
But
< PAB+¿QBA = (exterior) ≤ PRQ.
<TPR=/ PAB, and
in the alternate segments;
therefore
TQR = ≤ QBA,
<TPR+< TQR = ≤ PRQ.
It follows from this that TP=TQ=TR.
[For, if PT be produced to O so that TO= TQ, we have
And, by hypothesis,
By addition,
< TOQ = < TQO.
<PRQ=<≤ TPR+TQR.
POQ+< PRQ= < TPR+OQR.
314
ARCHIMEDES
It follows that, in the quadrilateral OPRQ, the opposite
angles are together equal to two right angles. Therefore a
circle will go round OPQR, and T is its centre, because
TP=TO = TQ. Therefore TR = TP.]
Thus
▲ TRP = ▲ TPR=¿ PAM.
Adding to each the angle PRM,
¿PAM+¿PRM=/ TRP +▲ PRM
=
= (two right angles).
Therefore
< APR+<AMR = (two right angles),
whence
< AMR = (a right angle)*.
Proposition 13.
If a diameter AB of a circle meet any chord CD, not a
diameter, in E, and if AM, BN be drawn perpendicular to CD,
then
CN = DM†.
Let O be the centre of the
circle, and OH perpendicular to
CD. Join BM, and produce HO to
meet BM in K.
Then
CH= HD.
And, by parallels,
N
A

D
M
E
H
O
K
since
BO=OA,
BK = KM.
Therefore
NH=HM.
8
Accordingly
CN= DM.
* TM is of course the polar of the intersection of PQ, AB, as it is the line
joining the poles of PQ, AB respectively.
+ This proposition is of course true whether M, N lie on CD or on CD
produced each way. Pappus proves it for the latter case in his first lemma
(p. 788) to the second Book of Apollonius' vevσeis.
BOOK OF LEMMAS.
315
Proposition 14.
Let ACB be a semicircle on AB as diameter, and let AD,
BE be equal lengths measured along AB from A, B respectively.
On AD, BE as diameters describe semicircles on the side towards
C, and on DE as diameter a semicircle on the opposite side. Let
the perpendicular to AB through O, the centre of the first semi-
circle, meet the opposite semicircles in C, F respectively.
Then shall the area of the figure bounded by the circumferences
of all the semicircles (“ which Archimedes calls 'Salinon '"*) be
equal to the area of the circle on CF as diameter†.
By Eucl. II. 10, since ED is bisected at O and produced
to A,
and
EA²+AD² = 2 (E0² + OA²),
CF OA+OE-EA.
=
A
C
Д

D
F
O
E
B
* For the explanation of this name see note attached to the remarks on the
Liber Assumptorum in the Introduction, Chapter II. On the grounds there
given at length I believe σávov to be simply a Graecised form of the Latin
word salinum, 'salt-cellar.'
+ Cantor (Gesch. d. Mathematik, 1. p. 285) compares this proposition
with Hippocrates' attempt to square the circle by means of lunes, but
points out that the object of Archimedes may have been the converse of that
of Hippocrates. For, whereas Hippocrates wished to find the area of a circle
from that of other figures of the same sort, Archimedes' intention was possibly
to equate the area of figures bounded by different curves to that of a circle
regarded as already known.
316
ARCHIMEDES
1
Therefore
AB² + DE² = 4(E0² + OA²) = 2 (CF² + AD²).
But circles (and therefore semicircles) are to one another as
the squares on their radii (or diameters).
Therefore
(sum of semicircles on AB, DE)
=(circle on CF) + (sum of semicircles on AD, BE).
Therefore
(area of 'salinon')=(area of circle on CF as diam.).
Proposition 15.
Let AB be the diameter of a circle, AC a side of an in-
scribed regular pentagon, D the middle point of the arc AC.
Join CD and produce it to meet BA produced in E; join AC,
DB meeting in F, and draw FM perpendicular to AB. Then
CB.
EM=(radius of circle)*.
Let O be the centre of the circle, and join DA, DM, DO,
Now
and
whence
< ABC = (right angle),
< ABD=<DBC = (right angle),
< AOD=2(right angle).
* Pappus gives (p. 418) a proposition almost identical with this among the
lemmas required for the comparison of the five regular polyhedra. His enunci-
ation is substantially as follows. If DH be half the side of a pentagon inscribed
in a circle, while DH is perpendicular to the radius OHA, and if HM be made
equal to AH, then OA is divided at M in extreme and mean ratio, OM being the
greater segment.
In the course of the proof it is first shown that AD, DM, MO are all equal,
as in the proposition above.
Then, the triangles ODA, DAM being similar,
or (since AD=OM)
OA: AD=AD: AM,
OA: OM-OM: MA.
BOOK OF LEMMAS.
317
Further, the triangles FCB, FMB are equal in all respects.
Therefore, in the triangles DCB, DMB, the sides CB, MB
being equal and BD common, while the angles CBD, MBD are
equal,
< BCD=2BMD=§(right angle).

But
E
D
F
A H
H M
< BCD+2 BAD=(two right angles)
=¿BAD+< DAE
=<BMD+<DMA,
B
so that
and
Therefore
<DAE= < BCD,
▲ BAD=¿ AMD.
AD=MD.
Now, in the triangle DMO,
<MOD=(right angle),
<DMO= g(right angle).
Therefore ODM = (right angle) = AOD;
whence
Again
OM = MD.
▲ EDA = (supplement of ADC)
=▲ CBA
= (right angle)
=≤ODM.
318
ARCHIMEDES
I
1
Therefore, in the triangles EDA, ODM,
<EDA = <ODM,
<EAD=20MD,
and the sides AD, MD are equal.
Hence the triangles are equal in all respects, and
EA = MO.
Therefore
EM=AO.
=
Moreover DE DO; and it follows that, since DE is equal
to the side of an inscribed hexagon, and DC is the side of an
inscribed decagon, EC is divided at D in extreme and mean
ratio [i.e. EC: ED-ED: DC]; "and this is proved in the
book of the Elements." [Eucl. XIII. 9, "If the side of the
hexagon and the side of the decagon inscribed in the same
circle be put together, the whole straight line is divided in
extreme and mean ratio, and the greater segment is the
side of the hexagon."]
1
THE CATTLE-PROBLEM.
It is required to find the number of bulls and cows of each
of four colours, or to find 8 unknown quantities. The first
part of the problem connects the unknowns by seven simple
equations; and the second part adds two more conditions to
which the unknowns must be subject.
Let W, w be the numbers of white bulls and cows respectively,
X, x
""
black
""
""
Y, Y
Z, z
""
""
yellow
""
""
""
""
dappled
""
34
""
First part.
(I)
W = (} + {) X + Y ....
……..(a),
X = (1 + }) Z + Y ....
.(B),
Z = (¿ + 4) W + Y ...
..(y),
(II)
w=
(} + 1) (X + ∞)…………..
..(8),
¿c = (} + {) (Z + z) ………….
.(€),
2 = (ž + z) (Y + y) ……..
·(5),
y = (} + 4) (W +w)
..(n).
Second part.
W+ X = a square
Y+Z = a triangular number……
.(c).
320
ARCHIMEDES
[There is an ambiguity in the language which expresses the
condition (0). Literally the lines mean "When the white bulls
joined in number with the black, they stood firm (еµπedov)
with depth and breadth of equal measurement (ioóμerpoi eis
Bálos eis eûpós te); and the plains of Thrinakia, far-stretching
all ways, were filled with their multitude" (reading, with
Krumbiegel, πλήθους instead of πλίνθου). Considering that, if
the bulls were packed together so as to form a square figure,
the number of them need not be a square number, since a bull
is longer than it is broad, it is clear that one possible interpre-
tation would be to take the 'square' to be a square figure, and
to understand condition (0) to be simply
W+X= a rectangle (i.e. a product of two factors).
The problem may therefore be stated in two forms:
(1) the simpler one in which, for the condition (0), there is
substituted the mere requirement that
W+ X = a product of two whole numbers;
(2) the complete problem in which all the conditions have to
be satisfied including the requirement (0) that
W+ X = a square number.
The simpler problem was solved by Jul. Fr. Wurm and may
be called
Wurm's Problem.
The solution of this is given (together with a discussion of
the complete problem) by Amthor in the Zeitschrift für Math.
u. Physik (Hist. litt. Abtheilung), xxv. (1880), p. 156 sqq.
Multiply (a) by 336, (B) by 280, (y) by 126, and add; thus
and
297 W = 742Y, or 3³.11 W = 2.7.53Y.........(α).
Then from (y) and (B) we obtain
891Z=1580 Y, or 3.11Z = 2.5.79 Y.......(B′),
99X = 178Y, or 3².11X=2.89Y............(y').
Again, if we multiply (8) by 4800, (e) by 2800, (Y) by 1260,
(n) by 462, and add, we obtain
4657w=2800X + 1260Z +462Y+143 W;
1
=
321
THE CATTLE-PROBLEM.
and, by means of the values in (a′), (B′), (y′), we derive
or
297.4657w 2402120Y,
=
3º. 11. 4657w = 2³ . 5.7.23.373Y........(d').
Hence, by means of (n), (C), (e), we have
32.11.4657y = 13.46489Y.
..(€'),
38.4657% = 22.5.7.7617
………..(5'),
32.11.4657∞ = 2.17.15991 Y .....
…….(n´).
and
And, since all the unknowns must be whole numbers, we see
from the equations (a′), (B′), ….. (n') that Y must be divisible by
34.11.4657, i.e. we may put
Y = 34.11.4657n
=
4149387n.
Therefore the equations (a'), (B′),…..(ʼn′) give the following values
for all the unknowns in terms of n, viz.
W=2.3.7.53.4657n
X=2.32.89.4657n
Y = 3.11.4657n
Z=22.5.79.4657n
w=23.3.5.7.23.373n
x = 2.32.17.15991n
X
y=32.13.46489n
z=22.3.5.7.11.761m
10366482n
7460514n
= 4149387n
7358060n
.(A).
=
7206360n
4893246n
5439213n
=
3515820n,
If now n = 1, the numbers are the smallest which will satisfy
the seven equations (a), (B),...(n); and we have next to find
such an integral value for n that the equation () will be
satisfied also. [The modified equation (0) requiring that W + X
must be a product of two factors is then simultaneously
satisfied.]
Equation (1) requires that
Y+Z=2(+1)
Y + Z = ? (q + 1)
2
C
21
where q is some positive integer.
H. A.
!
322
ARCHIMEDES
Putting for Y, Z their values as above ascertained, we have
q (q+1)
2
=2471.4657n
=(3.11+22.5.79). 4657n
2s, or
= 7.353.4657n.
Now q is either even or odd, so that either q = 2s,
q=2s-1, and the equation becomes
s (2s ± 1) = 7.353.4657n.
As n need not be a prime number, we suppose nu. v, where
u is the factor in n which divides s without a remainder and v
the factor which divides 2s ±1 without a remainder; we then
have the following sixteen alternative pairs of simultaneous
equations:
2s+1=7.353.4657v,
(1)
S=
u,
(2)
S=
7u.
28 + 1 =
353.4657v,
(3)
s=
353u,
28 + 1 =
7.4657v,
(4)
S=
4657u,
28 + 1 =
7.353v,
(5)
S=
7.353u,
28 + 1 =
4657v,
(6)
S=
7.4657u,
28 + 1 =
353v,
(7)
S=
353.4657u,
2s + 1
=
7v,
(8)
s=7.353.4657u,
28 + 1 =
V.
In order to find the least value of n which satisfies all the
conditions of the problem, we have to choose from the various
positive integral solutions of these pairs of equations that
particular one which gives the smallest value for the product
uv or n.
If we solve the various pairs and compare the results, we
find that it is the pair of equations
s = 7u,
28-1353.4657v,
which leads to the solution we want; this solution is then
u = 117423,
v = 1,
so that
N = UV = 117423 = 33. 4349,
THE CATTLE-PROBLEM.
323
whence it follows that
and
Thus
s=7u= 821961,
q=28-1=1643921.
Y+Z=2471.4657n
= 2471.4657.117423
=
1351238949081
1643921.1643922
2
which is a triangular number, as required.
The number in equation (8) which has to be the product of
two integers is now
W+X=2.3.(7.53 +3.89). 4657n
= 23.3.11.29.4657n
= 22.3.11.29.4657.117423
= 22.34.11.29.4657.4349
=(22.34.4349). (11.29.4657)
=
1409076.1485583,
which is a rectangular number with nearly equal factors.
The solution is then as follows (substituting for n its value
117423):
W = 1217263415886
X = 876035935422
Y = 487233469701
Z= 864005479380
w = 846192410280
X = 574579625058
y= 638688708099
2
412838131860
and the sum = 5916837175686
324
ARCHIMEDES
[
1
3
The complete problem.
In this case the seven original equations (a), (B),…….(n) have
to be satisfied, and the following further conditions must hold,
W + X = a square number = p², say,
Y+Z = a triangular number
=
q (q + 1)
2 , say.
Using the values found above (A), we have in the first place
p² = 2.3.(7.53 +3.89). 4657n
= 22.3.11.29.4657n,
and this equation will be satisfied if
n = 3.11.29.46572 = 44567492,
where § is any integer.
Thus the first 8 equations (a), (B),…….(n), (0) are satisfied by
the following values:
W=2.32.7.11.29.53.46572.2
X=2.33.11.29.89.46572.§
Y = 35.112.29.46572.2
Z=22.3.5.11.29.79.46572.2
- 46200808287018. §²
33249638308986. §²
-2
= 18492776362863. §*
=32793026546940.§²
w=23.32.5.7.11.23.29.373.4657. §2=32116937723640. §²
x = 2.3.11.17.29.15991.4657. §²
y = 3³.11.13.29.46489.4657 . §²
2
2
= 21807969217254. §²
24241207098537. §²
2
=
15669127269180. §²
22.32.5.7.112.29.761.4657. §²
It remains to determine § so that
satisfied, i.e. so that
Ÿ+Z= · 9 (q + 1)
2
=
equation (1) may be
Substituting the ascertained values of Y, Z, we have
q (q + 1)
2
= 51285802909803.2
= 3.7.11.29.353.46572. §2.
?
THE CATTLE-PROBLEM.
Multiply by 8, and put
2q+1=t, 2.4657 . § = u,
and we have the "Pellian" equation
325
ť² − 1 = 2.3.7.11.29.353. u²,
that is,
ť² - 4729494 u² = 1.
Of the solutions of this equation the smallest has to be
chosen for which u is divisible by 2.4657.
When this is done,
us
૬
2.4657
and is a whole number;
whence, by substitution of the value of so found in the last
system of equations, we should arrive at the solution of the
complete problem.
It would require too much space to enter on the solution of
the "Pellian" equation
ť² – 4729494 u² = 1,
and the curious reader is referred to Amthor's paper itself.
Suffice it to say that he develops √4729494 in the form of a
continued fraction as far as the period which occurs after 91
convergents, and, after an arduous piece of work, arrives at the
conclusion that
W = 1598 206541
where 206541 represents the fact that there are 206541 more
digits to follow, and that, with the same notation,
the whole number of cattle = 7766 206541
One may well be excused for doubting whether Archimedes
solved the complete problem, having regard to the enormous
倡
​326
ARCHIMEDES
J
size of the numbers and the great difficulties inherent in the
work. By way of giving an idea of the space which would be
required for merely writing down the results when obtained,
Amthor remarks that the large seven-figured logarithmic tables
contain on one page 50 lines with 50 figures or so in each, say
altogether 2500 figures; therefore one of the eight unknown
quantities would, when found, occupy 82 such pages, and to
write down all the eight numbers would require a volume of
660 pages!]
CAMBRIDGE: PRINTED BY J. AND C. F. CLAY, AT THE UNIVERSITY PRESS.

UNIVERSITY OF MICHIGAN
3 9015 06388 8815