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GEOMETRIC EXERCISES IN PAPER-FOLDING. By T.
SUNDARA Row. Edited and revised by W. W. Beman and
D. E. Smith. With many half-tone engravings from pho-
tographs of actual exercises, and a package of papers for
folding. Pages, x, 148. Cloth, fi.oo net (45. 6d. net).
THE OPEN COURT PUBLISHING COMPANY
324 DEARBORN ST., CHICAGO.
LONDON: Kegan Paul, Trench, Trubner & Co., Ltd.
T. SUNDARA ROW'S
Geometric Exercises in
Paper Folding
Edited and Revised by
WOOSTER WOODRUFF BEMAN
PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF MICHIGAN
and
DAVID EUGENE SMITH
PROFESSOR OF MATHEMATICS IN TEACHERS* COLLEGE
OF COLUMBIA UNIVERSITY
WITH 87 ILLUSTRATIONS
CHICAGO
THE OPEN COURT PUBLISHING COMPANY
LONDON AGENTS
KEGAN PAUL, TRENCH, TRUBNER & Co., LTD.
IQOI
COPYRIGHT BY
THE OPEN COURT PUBLISHING Co.
1901
EDITORS' PREFACE.
OUR attention was first attracted to Sundara Row's Geomet-
rical Exercises in Paper Folding by a reference in Klein's Vor-
lesungen iiber ausgezvahlte Fragen der Elementar geometric.
An examination of the book, obtained after many vexatious delays,
convinced us of its undoubted merits and of its probable value to
American teachers and students of geometry. Accordingly we
sought permission of the author to bring out an edition in this
country, which permission was most generously granted.
The purpose of the book is so fully set forth in the author's
introduction that we need only to say that it is sure to prove of
interest to every wide-awake teacher of geometry from the graded
school to the college. The methods are so novel and the results
so easily reached that they cannot fail to awaken enthusiasm.
Our work as editors in this revision has been confined to some
slight modifications of the proofs, some additions in the way of
references, and the insertion of a considerable number of half-tone
reproductions of actual photographs instead of the line-drawings
of the original.
W. W. BEMAN.
D. E. SMITH.
104
CONTENTS.
PAGE
Introduction vii
I. The Square I
II. The Equilateral Triangle 9
III. Squares and Rectangle?; 14
IV. The Pentagon 30
V. The Hexagon 35
VI. The Octagon 39
VII. The Nonagon 45
VIII. The Decagon and the Dodecagon . 47
IX. The Pentedecagon 50
X. Series 52
XI. Polygons 67
XII. General Principles 82
XIII. The Conic Sections.
Section i. The Circle 102
Section n. The Parabola 115
Section in. The Ellipse 121
Section iv. The Hyperbola 126
XIV. Miscellaneous Curves 131
INTRODUCTION.
THE idea of this book was suggested to me by
Kindergarten Gift No. VIII. Paper-folding. The
gift consists of two hundred variously colored squares
of paper, a folder, and diagrams and instructions for
folding. The paper is colored and glazed on one side.
The paper may, however, be of self-color, alike on
both sides. In fact, any paper of moderate thickness
will answer the purpose, but colored paper shows the
creases better, and is more attractive. The kinder-
garten gift is sold by any dealers in school supplies ;
but colored paper of both sorts can be had from sta-
tionery dealers. Any sheet of paper can be cut into
a square as explained in the opening articles of this
book, but it is neat and convenient to have the squares
ready cut.
2. These exercises do not require mathematical
instruments, the only things necessary being a pen-
knife and scraps of paper, the latter being used for
setting off equal lengths. The squares are themselves
simple substitutes for a straight edge and a T square.
3. In paper-folding several important geometric
processes can be effected much more easily than with
viii INTR OD UC TION.
a pair of compasses and ruler, the only instruments
the use of which is sanctioned in Euclidean geom-
etry; for example, to divide straight lines and angles
into two or more equal parts, to draw perpendiculars
and parallels to straight lines. It is, however, not
possible in paper-folding to describe a circle, but a
number of points on a circle, as well as other curves,
may be obtained by other methods. These exercises
do not consist merely of drawing geometric figures
involving straight lines in the ordinary way, and fold-
ing upon them, but they require an intelligent appli-
cation of the simple processes peculiarly adapted to
paper-folding. This will be apparent at the very com-
mencement of this book.
4. The use of the kindergarten gifts not only affords
interesting occupations to boys and girls, but also
prepares their minds for the appreciation of science
and art. Conversely the teaching of science and art
later on can be made interesting and based upon
proper foundations by reference to kindergarten occu-
pations. This is particularly the case with geometry,
which forms the basis of every science and art. The
teaching of plane geometry in schools can be made
very interesting by the free use of the kindergarten
gifts. It would be perfectly legitimate to require pu-
pils to fold the diagrams with paper. This would
give them neat and accurate figures, and impress the
truth of the propositions forcibly on their minds. It
would not be necessary to take any statement on trust.
INTRODUCTION. ix
But what is now realised by the imagination and ideal-
isation of clumsy figures can be seen in the concrete.
A fallacy like the following would be impossible.
5. To prove that every triangle is isosceles. Let
ABC, Fig. 1, be any triangle. Bisect AB in Z, and
through Z draw ZO perpendicular to AB. Bisect the
angle ACS by CO.
Z B
Fig. i.
(1) If CO and ZO do not meet, they are parallel.
Therefore CO is at right angles to AB. Therefore
(2) If CO and ZO do meet, let them meet in O.
Draw OX perpendicular to BC and O Y perpendicular
to AC. Join OA, OB. By Euclid I, 26 (B. and S.,
88, cor. 7)* the triangles YOC and XOC are con-
* These references are to Beman and Smith's New Plane and Solid Geom-
etry, Boston, Ginn & Co., 1899.
V
x INTRO DUG TfON.
gruent; also by Euclid I, 47 and I, 8 (B. and S.,
156 and 79) the triangles AOY and BOX are con-
gruent. Therefore
AY+ YC=BX+XC,
i.e., AC=C.
Fig. 2 shows by paper-folding that, whatever tri-
angle be taken, CO and ZO cannot meet within the
triangle.
Fig. 2.
O is the mid-point of the arc A OB of the circle
which circumscribes the triangle ABC.
6. Paper-folding is not quite foreign to us. Fold-
ing paper squares into natural objects a boat, double
INTRODUCTION. xi
boat, ink bottle, cup-plate, etc., is well known, as
also the cutting of paper in symmetric forms for pur-
poses of decoration. In writing Sanskrit and Mah-
rati, the paper is folded vertically or horizontally to
keep the lines and columns straight. In copying let-
ters in public offices an even margin is secured by fold-
ing the paper vertically. Rectangular pieces of paper
folded double have generally been used for writing,
and before the introduction of machine-cut letter pa-
per and envelopes of various sizes, sheets of convenient
size were cut by folding and tearing larger sheets, and
the second half of the paper was folded into an envel-
ope inclosing the first half. This latter process saved
paper and had the obvious advantage of securing the
post marks on the paper written upon. Paper-folding
has been resorted to in teaching the Xlth Book of
Euclid, which deals with figures of three dimensions.*
But it has seldom been used in respect of plane fig-
ures.
7. I have attempted not to write a complete trea
tise or text-book on geometry, but to show how reg-
ular polygons, circles and other curves can be folded
or pricked on paper. I have taken the opportunity to
introduce to the reader some well known problems of
ancient and modern geometry, and to show how alge-
bra and trigonometry may be advantageously applied
to geometry, so as to elucidate each of the subjects
which are usually kept in separate pigeon-holes.
* See especially Beman and Smith's New Plane and Solid Geometry ', p. 287.
xii IN TR OD UC TION.
8. The first nine chapters deal with the folding of
the regular polygons treated in the first four books of
Euclid, and of the nonagon. The paper square of the
kindergarten has been taken as the foundation, and
the other regular polygons have been worked out
thereon. Chapter I shows how the fundamental square
is to be cut and how it can be folded into equal right-
angled isosceles triangles and squares. Chapter II
deals with the equilateral triangle described on one of
the sides of the square. Chapter III is devoted to
the Pythagorean theorem (B. and S., 156) and the
propositions of the second book of Euclid and certain
puzzles connected therewith. It is also shown how a
right-angled triangle with a given altitude can be de-
scribed on a given base. This is tantamount to find-
ing points on a circle with a given diameter.
9. Chapter X deals with the arithmetic, geometric,
and harmonic progressions and the summation of cer-
tain arithmetic series. In treating of the progressions,
lines whose lengths form a progressive series are ob-
tained. A rectangular piece of paper chequered into
squares exemplifies an arithmetic series. For the geo-
metric the properties of the right-angled triangle, that
the altitude from the right angle is a mean propor-
tional between the segments of the hypotenuse (B.
and S., 270), and that either side is a mean propor-
tional between its projection on the hypotenuse and
the hypotenuse, are made use of. In this connexion
the Delian problem of duplicating a cube has been
INTRODUCTION. xiii
explained.* In treating of harmonic progression, the
fact that the bisectors of an interior arid correspond-
ing exterior angle of a triangle divide the opposite
side in the ratio of the other sides of the triangle (B.
and S., 249) has been used. This affords an inter-
esting method of graphically explaining systems in
involution. The sums of the natural numbers and of
their cubes have been obtained graphically, and the
sums of certain other series have been deduced there-
from.
10. Chapter XI deals with the general theory of
regular polygons, and the calculation of the numerical
value of n. The propositions in this chapter are very
interesting.
11. Chapter XII explains certain general princi-
ples, which have been made use of in the preceding
chapters, congruence, symmetry, and similarity of
figures, concurrence of straight lines, and collinearity
of points are touched upon.
12. Chapters XIII and XIV deal with the conic
sections and other interesting curves. As regards
the circle, its harmonic properties among others are
treated. The theories of inversion and co-axial circles
are also explained. As regards other curves it is
shown how they can be marked on paper by paper-
folding. The history of some of the curves is given,
and it is shown how they were utilised in the solution
*See Beman and Smith's translation of Klein's Famous Problems of Ele-
mentary Geometry, Boston, 1897; also their translation of Fink's History of
Mathematics, Chicago, The Open Court Pub. Co., 1900.
xiv INTRODUCTION.
of the classical problems, to find two geometric means
between two given lines, and to trisect a given recti-
lineal angle. Although the investigation of the prop-
erties of the curves involves a knowledge of advanced
mathematics, their genesis is easily understood and is
interesting.
13. I have sought not only to aid the teaching of
geometry in schools and colleges, but also to afford
mathematical recreation to young and old, in an at-
tractive and cheap form. "Old boys" like myself
may find the book useful to revive their old lessons,
and to have a peep into modern developments which,
although very interesting and instructive, have been
ignored by university teachers.
T. SUNDARA Row.
MADRAS, INDIA, 1893.
I. THE SQUARE.
1. The upper side of a piece of paper lying flat
upon a table is a plane surface, and so is the lower
side which is in contact with the table.
2. The two surfaces are separated by the material
of the paper. The material being very thin, the other
sides of the paper do not present appreciably broad
surfaces, and the edges of the paper are practically
lines. The two surfaces though distinct are insepa-
rable from each other.
3. Look at the irregularly shaped piece of paper
shown in Fig. 3, and at this page which is rectangu-
lar. Let us try and shape the former paper like the
latter.
4. Place the irregularly shaped piece of paper
upon the table, and fold it flat upon itself. Let X' X
be the crease thus formed. It is straight. Now pass
a knife along the fold und separate the smaller piece.
We thus obtain one straight edge.
5. Fold the paper again as before along BY, so
that the edge X' X is doubled upon itself. Unfolding
the paper, we see that the crease HY\s> at right angles
to the edge X'X. It is evident by superposition that
GEOMETRIC EXERCISES
the angle YBX' equals the angle XBY, and that each
of these angles equals an angle of the page. Now pass
Fig. 3.
a knife as before along the second fold and remove
the smaller piece.
IN PAPER FOLDING
6. Repeat the above process and obtain the edges
CD and DA. It is evident by superposition that the
angles at A, B, C, D, are right angles, equal to one
another, and that the sides BC, CD are respectively
Fig. 4-
equal to DA, AB. This piece of paper (Fig. 3) is
similar in shape to the page.
7. It can be made equal in size to the page by
taking a larger piece of paper and measuring off AB
and BC equal to the sides of the latter.
4 GEOMETRIC EXERCISES
8. A figure like this is called a rectangle. By
superposition it is proved that (1) the four angles are
right angles and all equal, (2) the four sides are not
all equal, (3) but the two long sides are equal, and so
also are the two short sides.
9. Now take a rectangular piece of paper, A B' CD,
and fold it obliquely so that one of the short sides, CD,
Fig. 5-
falls upon one of the longer sides, DA', as in Fig. 4.
Then fold and remove the portion A' B'BA which
overlaps. Unfolding the sheet, we find that ABCD
is now square, i. e. , its four angles are right angles,
and all its sides are equal.
10. The crease which passes through a pair of the
IN PAPER FOLDING 5
opposite corners B, D, is a diagonal of the square.
One other diagonal is obtained by folding the square
through the other pair of corners as in Fig. 5
n. We see that the diagonals are at right angles
to each other, and that they bisect each other.
12. The point of intersection of the diagonals is
called the center of the square.
Fig. 6.
13. Each diagonal divides the square into two con-
gruent right-angled isosceles triangles, whose vertices
are at opposite corners.
14. The two diagonals together divide the square
into four congruent right-angled isosceles triangles,
whose vertices are at the center of the square.
6 GEOMETRIC EXERCISES
15. Now fold again, as in Fig. 6, laying one side
cf the square upon its opposite side. We get a crease
which passes through the center of the square. It is
at right angles to the other sides and (1) bisects them;
(2) it is also parallel to the first two sides ; (3) it is
itself bisected at the center ; (4) it divides the square
Fig. 7.
into two congruent rectangles, which are, therefore,
each half of it; (5) each of these rectangles is equal
to one of the triangles into which either diagonal
divides the square.
16. Let us fold the square again, laying the re-
maining two sides one upon the other. The crease
IN PAPER FOLDING 7
now obtained and the one referred to in 15 divide
the square into four congruent squares.
17. Folding again through the corners of the
smaller squares which are at the centers of the sides
of the larger square, we obtain a square which is in-
scribed in the latter. (Fig. 7.)
Fig. 8.
18. This square is half the larger square, and has
the same center.
19. By joining the mid-points of the sides of the
inner square, we obtain a square which is one-fourth
of the original square (Fig. 8). By repeating the pro-
cess, we can obtain any number of squares which are
to one another as
8 GEOMETRIC EXERCISES
1111 I L 1 L
"2* "4' "8 ' 16' 6tC '' r 2 ' 2 2 ' 2 3 ' 2 4 ' '
Each square is half of the next larger square,
i. e., the four triangles cut from each square are to-
gether equal to half of it. The sums of all these tri-
angles increased to any number cannot exceed the
original square, and they must eventually absorb the
whole of it.
Therefore -J- ^ + 03 ~^~ etc ' to ^ n ^ n ^y = *
20. The center of the square is the center of its
circumscribed and inscribed circles. The latter circle
touches the sides at their mid-points, as these are
nearer to the center than any other points on the
sides.
21. Any crease through the center of the square
divides it into two trapezoids which are congruent. A
second crease through the center at right angles to
the first divides the square into four congruent quadri-
laterals, of which two opposite angles are right angles.
The quadrilaterals are concyclic, i. e., the vertices of
each lie in a circumference.
II. THE EQUILATERAL TRIANGLE.
22. Now take this square piece of paper (Fig. 9),
and fold it double, laying two opposite edges one upon
the other. We obtain a crease which passes through
Fig. 9.
the mid-points of the remaining sides and is at right
angles to those sides. Take any point on this line,
fold through it and the two corners of the square which
10
GEOMETRIC EXERCISES
are on each side of it. We thus get isosceles triangles
standing on a side of the square.
23. The middle line divides the isosceles triangle
into two congruent right-angled triangles.
24. The vertical angle is bisected.
25. If we so take the point on the middle line, that
Fig. 10.
its distances from two corners of the square are equal
to a side of it, we shall obtain an equilateral triangle
(Fig. 10). This point is easily determined by turning
the base AB through one end of it, over A A', until the
other end, B, rests upon the middle line, as at C.
26. Fold the equilateral triangle by laying each
IN PAPER FOLDING n
of the sides upon the base. We thus obtain the
three altitudes of the triangle, viz.: A A', BB', CC,
(Fig. 11).
27. Each of the altitudes divides the triangle into
two congruent right-angled triangles.
28. They bisect the sides at right angles.
Fig. ii.
29. They pass through a common point.
30. Let the altitudes A A' and CC' meet in O.
Draw BO and produce it to meet AC in B '. BB'
will now be proved to be the third altitude. From
the triangles C'OA and COA', OC' = OA'. From
triangles OC'B and A' OB, / OBC =-. ^A'BO. Again
from triangles ABB' and CB'B, / AB'B = / BB 1 C,
12 GEOMETRIC EXERCISES
i. e., each of them is a right angle. That is, BOB' is
an altitude of the equilateral triangle ABC. It also
bisects AC in B '.
31. It can be proved as above that OA, OB, and
OC are equal, and that OA', OB', and OC are also
equal.
32. Circles can therefore be described with O as a
center and passing respectively through A, B, and C
and through A', B' , and C*. The latter circle touches
the sides of the triangle.
33. The equilateral triangle ABC is divided into
six congruent right-angled triangles which have one
set of their equal angles at O, and into three congru-
ent, symmetric, concyclic quadrilaterals.
34. The triangle A OC is double the triangle A'OC;
therefore, AO = 2OA'. Similarly, BO = 2OB' and
CO = 2 OC'. Hence the radius of the circumscribed
circle of triangle ABC is twice the radius of the in-
scribed circle.
35. The right angle A, of the square, is trisected
by the straight lines AO, AC. Angle BAC = \ of a
right angle. The angles C'AO and OAB' are each \
of a right angle. Similarly with the angles at B and C.
36. The six angles at O are each f of a right angle.
37. Fold through A'B', B' C , and C'A' (Fig. 12).
Then A'B'C' is an equilateral triangle. It is a fourth
of the triangle ABC.
38. A'B', B'C, C'A' are each parallel toAB, BC,
CA, and halves of them.
IN PAPER FOLDING
13
39. AC'A'B' is a rhombus. So are C'BA'B' and
CB'C'A'.
40. A'B', B'C', C'A' bisect the corresponding alti-
tudes.
41. CC
= 0.866.. .
Fig. 12.
42. The A ABC= rectangle of AC and CC, i. e.
\AB X Jl/3 -^^ ba \V*'A& = 0.433 .... X^^ 2 -
43. The angles of the triangle AC'C are in the
ratio of 1:2:3, and its sides are in the ratio of ]/ 1
: 1/3 : 1/4 .
III. SQUARES AND RECTANGLES.
44. Fold the given square as in Fig. 13. This
affords the well-known proof of the Pythagorean the-
Fig. 13-
orem. FGH being a right-angled triangle, the square
on FH equals the sum of the squares on FG and GH.
It is easily proved that FC is a square, and that
PAPER FOLDING 15
the triangles FGH, HBC, KDC, and FEK are con-
gruent.
If the triangles FGH and HBC are cut off from
the squares FA and D, and placed upon the other
two triangles, the square FHCK is made up.
IiAB = a, GA = b, and FH=c, then a> + P = <*.
Fig. 14.
45. Fold the given square as in Fig. 14. Here the
rectangles AF, BG, CH, and DE are congruent, as
also the triangles of which they are composed. EFGH
is a square as also KLMN.
Let AK=a, KB = b, and NK=c,
then + ^ = ^, i. e. uKLMN.
i6
GEOMETRIC EXERCISES
Now square ABCD overlaps the square KLMN
by the four triangles AKN t BLK, CML, and DNM.
But these four triangles are together equal to two
of the rectangles, i. e., to 2ab.
Therefore (a ~j- ^) 2 = a 2 -f P -f 2ab.
46. EF=a b, zu&uEFGH=(a b}' i .
The square EFGH is less than the square KLMN
by the four triangles ^WAT, A'Z, HLM, and EMN.
But these four triangles make up two of the rect-
angles, i. e., 2ab.
M
Fig. 15
47. The square ABCD overlaps the square EFGH
by the four rectangles AF t BG, CH, and DE.
48. In Fig. 15, the square ABCD=(a + ) 2 , and
IN PAPER FOLDING 17
the square EFGH=(a ) 2 . Also square AKGN
= square ELCM '= a 1 . Square KBLF =
square
Squares ABCD and EFGH are together equal to
the latter four squares put together, or to twice the
square AKGN and twice the square KBLF, that is,
(a _j_ )2 _j_ (0 _ )2 = 2a* -f 2 2 .
D _ Q _ N _ C
M
Fig. 16.
49. In Fig. 16 the rectangle PL is equal to (#
(<*)-
Because the rectangle EK = FM, therefore rect-
angle PL = square PK square A, i. e., (a-{-b)
50. If squares be described about the diagonal of
a given square, the right angle at one corner being
common to them, the lines which join this corner with
the mid-points of the opposite sides of the given
i8
GEOMETRIC EXERCISES
square bisect the corresponding sides of all the inner
squares. (Fig. 17.) For the angles which these lines
make with the diagonal are equal, and their magni-
tude is constant for all squares, as may be seen by
Fig. 17.
superposition. Therefore the mid-points of the sides
of the inner squares must lie on these lines.
51. ABCD being the given square piece of paper
(Fig. 18), it is required to obtain by folding, the point
X in AB, such that the rectangle AB-XB is equal to
the square on AX.
Double BC upon itself and take its mid-point E.
Fold through E and A.
Lay EB upon EA and fold so as to get EF, and
G such that EG = EB.
IN PAPER FOLDING
Then rectangle
Complete the rectangle BCHX and the square
AXKL.
Let Xffcut EA in M. Take FY=FB.
Then FB = FG = FY = XM and XM=\AX.
Fig. 18.
Now, because BY is bisected in F and produced
to A,
= AF*, by 49,
, by 44.
But ^^ 2 ^
20 GEOMETRIC EXERCISES
AY=XB,
AB is said to be divided in X in median section.*
Also
i. e., AB is also divided in Fin median section.
52. A circle can be described with F as a center,
its circumference passing through B, G, and K It
will touch EA at G, because FG is the shortest dis-
tance from F \.Q the line EGA.
53. Since
subtracting BK we have
rectangle XKNY= square CHRP,
i. e., AX'YX=AY' i ,
i. e., AX is divided in Fin median section.
Similarly HY'is divided in X in median section.
54. .- AB-XB^AX*
CD-CP
55. Rectangles Btfand YD being each = AB-XB,
rectangle HY+ square CK= AX* = AB- XB.
56. Hence rectangle HY^= rectangle BK, i. e.,
57. Hence rectangle HN=AX-XBBX < *.
*The term " golden section " is also used. See Beman and Smith's New
Plane and Solid Geometry, p. 196.
IN PAPER FOLDING 21
58. l^t\.AB = a, XB = x.
Then (a x}^ = ax, by 51.
= 3a*, by 54;
and ^ = ^(3 1/5).
... a _# = -?. (1/5 1) =0X0. 6180 ____
. . . (a _ #)2 = (3 _ |5) = ^ X 0. 3819 ____
A
The rect.
59. In the language of proportion
AB :AX=AX: XB.
The straight line AB is said to be divided " in ex-
treme and mean ratio."
60. Let AB be divided in X in median section.
Complete the rectangle CBXH (Fig. 19). Bisect the
rectangle by the line MNO. Find the point N by
laying XA over X so that A falls on MO, and fold
through JfTV, NB t and ^4. Then ^^V is an isos-
22
GEOMETRIC EXERCISES
celes triangle having its angles ABN and BNA double
the angle NAB.
Fig. 19.
= BN* BM* -f
IN PAPER FOLDING 23
.-.AN=AB
and /_NAB = ^ of a right angle.
61. The right angle at A can be divided into five
equal parts as in Fig. 20. Here N' is found as in
60. Then fold^^'<2; bisect / QAB by folding,
Fig. 20.
fold over the diagonal AC and thus get the point
Q', /".
62. To describe a right-angled triangle, given the
hypotenuse AB, and the altitude.
Fold EF (Fig. 21) parallel to AB at the distance
of the given altitude.
Take G the middle point of AB. Find JTby fold-
ing GB through G so that B may fall on EF.
GEOMETRIC EXERCISES
Fold through H and A, G, and B.
AHB is the triangle required.
Fig. 21.
63. ABCD (Fig. 22) is a rectangle. It is required
to find a square equal to it in area.
Q P
M R O B M
Fig. 22.
Mark off BM=BC.
Find O, the middle point of AM, by folding.
IN PAPER FOLDING 25
Fold OM, keeping O fixed and letting M fall on
line BC, thus finding/*, the vertex of the right-angled
triangle AMP.
Describe on PB the square BPQR.
The square is equal to the given rectangle.
For . BP = QP, and the angles are equal, triangle
BMP is evidently congruent to triangle QSP.
. . triangles DA T and QSP are congruent.
.-. PC=SR and triangles RSA and CPT are con-
gruent.
.-. tl^ABCD can be cut into three parts which
can be fitted together to form the square RBPQ.
Fig. 23.
64. Take four equal squares and cut each of them
into two pieces through the middle point of one of the
sides and an opposite corner. Take also another
26
GEOMETRIC EXERCISES
equal square. The eight pieces can be arranged round
the square so as to form a complete square, as in Fig.
23, the arrangement being a very interesting puzzle.
The fifth square may evidently be cut like the
others, thus complicating the puzzle.
65. Similar puzzles can be made by cutting the
squares through one corner and the trisection points
of the opposite side, as in Fig. 24.
Fig. 24.
66. If the nearer point is taken 10 squares are re-
quired, as in Fig. 24; if the remoter point is taken 13
squares are required, as in Fig. 25.
67. The puzzles mentioned in 65, 66, are based
upon the formulas
IN PAPER FOLDING
27
The process may be continued, but the number of
squares will become inconveniently large.
68. Consider again Fig. 13 in 44. If the four
triangles at the corners of the given square are re-
moved, one square is left. If the two rectangles FK
and KG are removed, two squares in juxtaposition
are left.
Fig. 25.
69. The given square may be cut into pieces which
can be arranged into two squares. There are various
ways of doing this. Fig. 23, in 65, suggests the
following elegant method : The required pieces are
(1) the square in the center, and (2) the four con-
gruent symmetric quadrilaterals at the corners, to-
gether with the four triangles. In this figure the lines
from the mid-points of the sides pass through the cor-
28
GEOMETRIC EXERCISES
ners of the given square, and the central square is.
one-fifth of it. The magnitude of the inner square
can be varied by taking other points on the sides in-
stead of the corners.
70. The given square can be divided as follows
(Fig. 26) into three equal squares :
Take ^G = hali the diagonal of the square.
Fig. 26.
Fold through C and G.
Fold BM perpendicular to CG.
Take MP, CN, and NL each = BM.
Fold PH, NK, LF at right angles to CG, as in
Fig. 26.
ag
Take NK = BM, and fold KE at right angles to
NK.
Then the pieces 1, 4, and 6, 3 and 5, and 2 and 7
form three equal squares.
Now C& = *B&,
and from the triangles GBC and CMB
BM BG
Letting J3C=a, we have
IV. THE PENTAGON.
71. To cut off a regular pentagon from the square
A BCD.
Divide BA in X in median section and take M
the mid-point of AX.
C
A M X N B
Fig. 27.
Then AB-AX=X&, and AM= MX.
Take BN=AM or MX.
Lay off NP and J/^? equal to MN, so that P and
R may lie on .Z?C and AD respectively.
PAPER FOLDING 31
Lay off RQ and PQ = MR and NP.
MNPQR is the pentagon required.
In Fig. 19, p. 22, AN, which is equal to AB, has
the point N on the perpendicular MO. If A be moved
on ^^ over the distance MB, then it is evident that
TV^will be moved on to BC, and X to M.
Therefore, in Fig. 27, NR = AB. Similarly MP =
AB. RP is also equal to AB and parallel to it.
= * of a rt. /.
= % of a rt /.
Similarly / PNM= f of a rt. / .
From triangles MNR and QRP, L NMR= / RQP
= | of a rt. /.
The three angles at M, JV, and Q of the pentagon
being each equal to f of a right angle, the remain-
ing two angles are together equal to J^ 2 - of a right
angle, and they are equal. Therefore each of them
is |^ of a right angle.
Therefore all the angles of the pentagon are equal.
The pentagon is also equilateral by construction.
72. The base MN of the pentagon is equal to XJ3,
i. e., to^?- (i/5 1)^^^x0.6180.. . 58.
z
The greatest breadth of the pentagon is AB.
73. If/ be the altitude,
GEOMETRIC EXERCISES
5 + 1/5
T/104- 2 v'5
= ^^ X 0.9510. . . . =AB cos 18
Fig. 28.
74. If be the radius of the circumscribed circle,
AB 2AB
2cosl8 c
= AX 0.5257....
IN PAPER FOLDING
33
75. If r be the radius of the inscribed circle, then
from Fig. 28 it is evident that
0.4253
76. The area of the pentagon is 5r X the base of
the pentagon, i. e.,
40
5 1/5
IF -^^ X 0.6571....
77- In Fig. 27 let PR be divided by MQ and NQ
in E and f.
Then . MN= 2- - (1/5 1) ... 72
and cos 36 =
34 GEOMETRIC EXERCISES
2)... (2)
RF: RE = RE: EF (by 51) (3)
1/5 1 : 3 1/5 = 3 1/5 : 2 (1/5 2) (4)
By 76 the area of the pentagon
5 5 1/5
= MN* 1 1/25 + 101/5,
snce =
.-. the area of the inner pentagon
=r^^ 2 * 1/25-1- 101/5
= ^^ 2 (1/5 2) 2 I - 1/25 + 101/5 .
The larger pentagon divided by the smaller
= 2 : (7 31/5)
= 1 : 0.145898.. ..
78. If in Fig. 27, angles <2^^ and Z^<2 are made
equal to ERQ or FQP, K, L being points on the sides
QR and gjP respectively, then EFL QK will be a reg-
ular pentagon congruent to the inner pentagon. Pen-
tagons can be similarly described on the remaining
sides of the inner pentagon. The resulting figure
consisting of six pentagons is very interesting.
V. THE HEXAGON.
79. To cut off a regular hexagon from a given
square.
Fig. 29.
Fold through the mid-points of the opposite sides,
and obtain the lines A OB and COD.
On both sides of AO and OB describe equilateral
triangles ( 25), AOE, AHO; BFO and BOG.
36 GEOMETRIC EXERCISES
Draw EF and HG.
AHGBFE is a regular hexagon.
It is unnecessary to give the proof.
The greatest breadth of the hexagon is AB.
80. The altitude of the hexagon is
= 0.866....
Fig. 30.
81. If R be the radius of the circumscribed circle,
82. If r be the radius of the inscribed circle,
r== ] ' 1 ^^==0.433
IN PAPER FOLDING
37
83. The area of the hexagon is 6 times the area of
the triangle HGO,
. . . . X AS*.
4 4
Also the hexagon = | AB - CD.
= 1 times the equilateral triangle on AB.
Fig. 31-
84. Fig. 30 is an example of ornamental folding
into equilateral triangles and hexagons.
85. A hexagon is formed from an equilateral tri-
angle by folding the three corners to the center.
The side of the hexagon is 1 of the side of the
equilateral triangle.
38 GEOMETRIC EXERCISES
The area of the hexagon = \ of the equilateral
triangle.
86. The hexagon can be divided into equal regular
hexagons and equilateral triangles as in Fig. 31 by
folding through the points of trisection of the sides.
VI. THE OCTAGON.
87. To cut off-a regular octagon from a given square.
Obtain the inscribed square by joining the mid-
points A, B, C, D of the sides of the given square.
Fig. 32-
Bisect the angles which the sides of the inscribed
square make with the sides of the other. Let the bi-
secting lines meet in E. F, G, and If.
4 o GEOMETRIC EXERCISES
AEBFCGDH is a regular octagon.
The triangles AEB, BFC, CGD, and DHA are
congruent isosceles triangles. The octagon is there-
fore equilateral.
The angles at the vertices, E, F, G, H of the same
four triangles are each one right angle and a half,
since the angles at the base are each one-fourth of a
right angle.
Therefore the angles of the octagon at A, B, C,
and D are each one right angle and a half.
Thus the octagon is equiangular.
The greatest breadth of the octagon is the side of
the given square, a.
88. If R be the radius of the circumscribed circle,
and a be the side of the original square,
/? a
= ~2'
89. The angle subtended at the center by each of
the sides is half a right angle.
90. Draw the radius OE and let it cut AB in K
(Fig. 33).
Then AK= OK= = -A=-
1/2 2v 7 2
KE=OA-OK=~ -- = " (2 V2).
2]/2 4
Now from triangle AEK,
IN PAPER FOLDING
(4-2J/2)
(2-1/2).
91. The altitude of the octagon is C (Fig. 33).
But C* = AC*
= a *- (2-1/2) = - (2 + v/2).
A
Fig. 33-
=v- 1/2 + 1/2.
92. The area of the octagon is eight times the tri-
angle AOE and
o
"2 2 1/2 ~~i/f'
42
GE OME TRIG EXER CISES
93. A regular octagon may also be obtained by
dividing the angles of the given square into four equal
parts.
2 Y
w K X
Fig. 34-
It is easily seen that EZ= WZ a, the side of the
square.
XE = WH= WK\
Now KZ* = a* + ai (1/2 I) 2 = * (4 2 1/2 )
.-. KZ=a 21/2.
Also GE = XZ2XE
= (2 1/2).
IN PAPER FOLDING 43
.-.ffO = ^ (2-1/2).
Again OZ=V2,
=al/2 1/2.
2 1/2
= 1/10 7 1/2.
= lzrA'=-| 1/107^2,
and HA = ^ 1/20 141/2.
94. The area of the octagon is eight times the
area of the triangle ffOA,
HO
1/2
^'2 i/2 -(6 41/2)
4)
= fl s -l/2-(i/2 I) 2 .
44 GEOMETRIC EXERCISES
95. This octagon : the octagon in 92
= (2 1/2 ) 2 : 1 or 2 : (1/2"+ I) 2 ;
and their bases are to one another as
1/2": 1/2"+ 1.
VII. THE NONAGON.
96. Any angle can be trisected fairly accurately by
paper folding, and in this way we may construct ap-
proximately the regular nonagon.
Fig- 35-
Obtain the three equal angles at the center of an
equilateral triangle. ( 25.)
46 GEOMETRIC EXERCISES
For convenience of folding, cut out the three
angles, AOF, FOC, and CO A.
Trisect each of the angles as in Fig. 35, and make
each of the arms = OA.
97. Each of the angles of a nonagon is -\, 4 - of a
right angle = 140.
The angle subtended by each side at the center is
f of a right angle or 40.
Half this angle is ^ of the angle of the nonagon.
98. OA = \a, where a is the side of the square ; it
is also the radius of the circumscribed circle, R.
The radius of the inscribed circle =R . cos 20
= \a cos 20
= ~ X 0.9396926
= aX 0.4698463.
The area of the nonagon is 9 times the area of
the triangle AOL
= f./? 2 -sin 40
Q,i 2
= 4r-X 0.6427876
o
= a 2 X 0.7231 36.
VIII. THE DECAGON AND THE DODECAGON
99. Figs. 36, 37 show how a regular decagon, and
a regular dodecagon, may be obtained from a penta-
gon and hexagon respectively.
Fig. 36.
The main part of the process is to obtain the
angles at the center.
In Fig. 36, the radius of the inscribed circle of
the pentagon is taken for the radius of the circum-
scribed circle of the decagon, in order to keep it
within the square.
48 GEOMETRIC EXERCISES
ioo. A regular decagon may also be obtained as
follows :
Obtain X, Y, (Fig. 38), as in 51, dividing AB in
median section.
Take J/the mid-point of AB.
Fold XC, MO, YD at right angles to AB.
Take O in MO such that YO = AY, or YO = XB.
Fig. 37-
Let YO, and XO produced meet XC, and YD in C
and D respectively.
Divide the angles XOC and DOY into four equal
parts by HOE, KOF, and LOG.
Take OH, OK, OL, OE, OF, and OG equal to
OYor OX.
Join X, H, K, L, C, D, E, F, G, and Y, in order.
IN PAPER FOLDING
As in 60,
/ YOX=\ of a rt. /=
49
Fig. 38.
By bisecting the sides and joining the points thus
determined with the center, the perigon is divided
into sixteen equal parts. A 16-gon is therefore easily
constructed, and so for a 32-gon, and in general a
regular 2 M -gon.
IX. THE PENTEDECAGON.
101. Fig. 39 shows how the pentedecagon is ob-
tained from the pentagon.
Let ABODE be the pentagon and O its center.
Draw OA, OB, OC, OD, and OE. Produce DO
to meet AB in K.
Take OF=\ of OD.
Fold GFH at right angles to OF. Make OG =
= OD.
PAPER FOLDING 51
Then GDH is an equilateral triangle, and the
angles DOG and HOD are each 120.
But angle DO A is 144; therefore angle GOA is
24.
That is, the angle EOA, which is 72, is trisected
by OG.
Bisect the angle EOG by OL, meeting EA in L,
and let OG cut EA in M; then
OL = OM.
In OA and OE take OP and 6>(? equal to OL or
0JT.
Then PM, ML, and Z<2 are three sides of the
pentedecagon.
Treating similarly the angles AOB, BOC, COD,
and DOE, we obtain the remaining sides of the pente-
decagon.
X. SERIES.
ARITHMETIC SERIES.
102. Fig. 40 illustrates an arithmetic series. The
horizontal lines to the left of the diagonal, including
the upper and lower edges, form an arithmetic series.
The initial line being a, and d the common difference,
the series is a, a -j- d, a-\- 2d, a -j- 3^/, etc.
103. The portions of the horizontal lines to the
right of the diagonal also form an arithmetic series,
PAPER FOLDING 53
but they are in reverse order and decrease with a
common difference.
104. In general, if / be the last term, and s the
sum of the series, the above diagram graphically
proves the formula
105. If a and c are two alternate terms, the middle
term is
a -j- c
~2~'
106. To insert n means between a and /, the ver-
tical line has to be folded into n-\- 1 equal parts. The
common difference will be
I a
107. Considering the reverse series and interchan-
ging a and /, the series becomes
a, a (n !) form a geometric series, the common rate
being OPi : OP 2 .
Fig. 41.
If OP\ be the unit of length, the series consists of
the natural powers of the common rate.
109. Representing the series by a, ar, ar 2 , ....
These lines also form a geometric series with the
common rate r.
no. The terms can also be reversed, in which case
the common rate will be a proper fraction. If OP*>
be the unit, OP is the common rate. The sum of
the series to infinity is
IN PAPER FOLDING
55
in. In the manner described in 108, one geo-
metric mean can be found between two given lines,
and by continuing the process, 3, 7, 15, etc., means
can be found. In general, 2* 1 means can be found,
n being any positive integer.
112. It is not possible to find two geometric means
between two given lines, merely by folding through
known points. It can, however, be accomplished in
the following manner : In Fig. 41, OP\ and OP being
given, it is required to find PI and P 3 . Take two rect-
angular pieces of paper and so arrange them, that their
outer edges pass through PI and P^ and two corners
lie on the straight lines OP', P' the perimeters of the in-
scribed and circumscribed polygons respectively of 2
sides, and A', B' their areas.
Then
p = n-AB, P = n-CD, p' = 2n-AE, P' = 2n-FG.
Because OF bisects / COE, and AB is parallel
to CD,
CFCO CO CD
FE ~ OE
72 GEOMETRIC EXERCISES
CE
~~AB '
n-CE n-
T
n-AB
2P _P+p
=
Again, from the similar triangles EIF and A HE,
Y _ EF
~AH~ ~~
or AE^^
or / = P.
Now,
The triangles A OH ^and AOE are of the same alti-
tude, AH,
A A OH OH
Similarly,
f^AOE OA
~~ ~OC'
Again because AB || CD,
&AOH &AOE
' ' t\AOE ~~
--- = -, ^
Now to find B'. Because the triangles COE and
IN PAPER FOLDING
73
FOE have the same altitude, and OF bisects the
angle OC,
A COE _ CE _ OC+OE
OE
&AOE
&FOE
and OE=OA,
, OC OE
and -- = ^rr =
&COE &AOE+ &AOH
' FO~E
&AOH
%B A' I A
From this equation we easily obtain 7 IT
2AB
A O
Fig. 49-
136. Given the radius It and apothem r of a reg-
ular polygon, to find the radius R and apothem r' of
a regular polygon of the same perimeter but of double
the number of sides.
Let AB be a side of the first polygon, O its center,
OA the radius of the circumscribed circle, and OD
the apothem. On OD produced take OC= OA or
OB. Draw AC, BC. Fold OA' and OB' perpen-
74 GEOMETRIC EXERCISES
dicular to AC and BC respectively, thus fixing the
points A', B'. Draw A' B' cutting OC in D '. Then
the chord A' B' is ha'f of AB, and the angle B'OA' is
half of BOA. OA' and OD' are respectively the ra-
dius R' and apothem r of the second polygon.
Now OD' is the arithmetic mean between OC and
OD, and OA' is the mean proportional between OC
and 0ZT.
and '
137. Now, take on 0(7, OE=OA' and draw ^'^.
Then ^4'Z>' being less than ^'C, and / D'A'C being
bisected by A'E,
ED' is less than \CD ', i. e., less than \CD
... 7?j_ n is less than \{R f}.
As the number of sides is increased, the polygon
approaches the circle of the same perimeter, and R
and r approach the radius of the circle.
That is,
-f R l n-h^ 2
= the diameter of the circle = .
n
Also,
RJ = Rn or R- 1 = tf l
and , and so on.
Multiplying both sides,
? -- - -- -- . . . . = the radius of the circle = -.
K\ /t2 -*M} ^7T
IN PAPER FOLDING 75
138. The radius of the circle lies between R n and
r Mf the sides of the polygon being 4-2" in number;
2 2
and TT lies between and --. The numerical value
r n &n
of n can therefore be calculated to any required de-
gree of accuracy by taking a sufficiently large number
of sides.
The following are the values of the radii and ap-
othems of the regular polygons of 4, 8, 16.... 2048
sides.
4-gon, r = 0-500000 R = n/ 1 Z = 0-7071 07
8-gon, n = 0-603553 ^ = 0-653281
2048-gon, r<) = 0-636620 7? 9 = 0-636620.
139. If R" be the radius of a regular isoperimetric
polygon of 4 sides
_
or in general
2
140. The radii R\, R^ .... successively diminish,
and the ratio Tr-is less than unity and equal to the
cosine of a certain angle ct.
a
76 GEOMETRIC EXERCISES
^+1- a
-#--- 'S^
multiplying together the different ratios, we get
a a a
The limit of cos a- cos ^ . cos _ i? when
ciri 2/y
is r - , a result known as Ruler's Formula.
141. It was demonstrated by Karl Friedrich Gauss*
(1777-1855) that besides the regular polygons of 2",
3-2", 5-2", 15-2" sides, the only regular polygons
which can be constructed by elementary geometry
are those the number of whose sides is represented
by the product of 2" and one or more different num-
bers of the form 2 w -f-l. We shall show here how
polygons of 5 and 17 sides can be described.
The following theorems are required :f
(1) If C and D are two points on a semi-circum-
ference A CDS, and if C' be symmetric to C with re-
spect to the diameter AB, and R the radius of the
circle,
.
AC'BC=R-CC ............. iii.
(2) Let the circumference of a circle be divided
into an odd number of equal parts, and let AO be the
*Beman and Smith's translation of Fink's History of Mathematics, p.
245 ; see also their translation of Klein's Famous Problems of Elementary
Geometry, pp. 16, 24, and their New Plane and Solid Geometry, p. 212.
t These theorems may be found demonstrated in Catalan's Theortmes et
Problemes de Gtometrie Eltmentaire.
IN PAPER FOLDING
77
diameter through one of the points of section A and
the mid-point O of the opposite arc. Let the points
of section on each side of the diameter be named A\,
A in G '. Draw Z>^ per-
pendicular to CD and take DF= OA.
Draw .F and FG '.
Take H in FG and ./if' in FG' produced so that
GH=EG and G'H' = G'D.
Then it is evident that
= M N,
IN PAPER FOLDING. 81
also
N, - . (DE + FIT) FH=
FH' = P, '.- (FH' DE') FH= DF* = JP.
Again in DF take K such that FK^FH.
Draw KL perpendicular to J}F and take L in KL
such that FL is perpendicular to DL.
Then FL^ = DF- FK=RN.
Again draw H' N perpendicular to FH' and take
H'N=FL. Draw NM perpendicular to NH' . Find
M in NM such that H'M is perpendicular to FM.
Draw MF' perpendicular to FH' .
Then
F'H' FF' = F'M* = FL*
But
XII. GENERAL PRINCIPLES.
147. In the preceding pages we have adopted sev-
eral processes, e. g., bisecting and trisecting finite
lines, bisecting rectilineal angles and dividing them
into other equal parts, drawing perpendiculars to a
given line, etc. Let us now examine the theory of
these processes.
148. The general principle is that of congruence.
Figures and straight lines are said to be congruent, if
they are identically equal, or equal in all respects.
In doubling a piece of paper upon itself, we ob-
tain the straight edges of two planes coinciding with
each other. This line may also be regarded as the
intersection of two planes if we consider their posi-
tion during the process of folding.
In dividing a finite straight line, or an angle into a
number of equal parts, we obtain a number of con-
gruent parts. Equal lines or equal angles are con-
gruent.
149. Let X'X be a given finite line, divided into
any two parts by A'. Take O the mid-point by doub-
ling the line on itself. Then OA' is half the difference
PAPER FOLDING 83
between A'X and X'A'. Fold X'X over O, and take
A in OX corresponding to A'. Then A A' is the differ-
ence between A'X and X'A' and it is bisected in O.
X- A' O A X
Fig. 52-
As A' is taken nearer O, A' O diminishes, and at the
same time A' A diminishes at twice the rate. This
property is made use of in finding the mid-point of a
line by means of the compasses.
150. The above observations apply also to an
angle. The line of bisection is found easily by the
compasses by taking the point of intersection cf two
circles.
151. In the line X'X, segments to the right of
O may be considered positive and segments to the
left of O may be considered negative. That is, a
point moving from O to A moves positively, and a
point moving in the opposite direction OA' moves
negatively.
AX= OX OA.
O A' = OX' A'X',
both members of the equation being negative.*
152. If OA, one arm of an angle A OP, be fixed and
OP be considered to revolve round O, the angles
which it makes with OA are of different magnitudes.
*See Beman and Smith's New Plane and Solid Geometry ', p. 56.
84 GEOMETRIC EXERCISES
All such angles formed by OP revolving in the direc-
tion opposite to that of the hands of a watch are re-
garded positive. The angles formed by OP revolving
in an opposite direction are regarded negative.*
153. After one revolution, OP coincides with OA.
Then the angle described is called a perigon, which
evidently equals four right angles. When OP has
completed half the revolution, it is in a line with
OAB. Then the angle described is called a straight
angle, which evidently equals two right angles. f
When OP has completed quarter of a revolution, it is
perpendicular to OA. All right angles are equal in
magnitude. So are all straight angles and all peri-
gons.
154. Two lines at right angles to each other form
four congruent quadrants. Two lines otherwise in-
clined form four angles, of which those vertically op-
posite are congruent.
155. The position of a point in a plane is deter-
mined by its distance from each of two lines taken as
above. The distance from one line is measured par-
allel to the other. In analytic geometry the proper-
ties of plane figures are investigated by this method.
The two lines are called axes ; the distances of the
point from the axes are called co-ordinates, and the
intersection of the axes is called the origin. This
* See Beman and Smith's New Plane and Solid Geometry, p. 56.
t/J.,p.5.
IN PAPER FOLDING, 85
method was invented by Descartes in 1637 A. D.* It
has greatly helped modern research.
156. If X'X, YY' be two axes intersecting at O,
distances measured in the direction of OX, i. e., to
the right of O are positive, while distances measured
to the left of O are negative. Similarly with reference
to YY', distances measured in the direction of O Fare
positive, while distances measured in the direction of
OY' are negative.
157. Axial symmetry is defined thus : If two fig-
ures in the same plane can be made to coincide by
turning the one about a fixed line in the plane through
a straight angle, the two figures are said to be sym-
metric with regard to that line as axis of symmetry. |
158. Central symmetry is thus defined : If two fig-
ures in the same plane can be made to coincide by
turning the one about a fixed point in that plane
through a straight angle, the two figures are said to
be symmetric with regard to that point as center of
symmetry. J
In the first case the revolution is outside the given
plane, while in the second it is in the same plane.
If in the above two cases, the two figures are halves
of one figure, the whole figure is said to be symmetric
with regard to the axis or center these are called axis
or center of symmetry or simply axis or center.
*Beman and Smith's translation of Fink's History of Mathematics, p. 230.
tBeman and Smith's New Plane and Solid Geometry, p. 26.
\ Ib., p. 183.
86 GEOMETRIC EXERCISES
159. Now, in the quadrant XOVmake a triangle
PQR. Obtain its image in the quadrant VOX' by
folding on the axis YY' and pricking through the
paper at the vertices. Again obtain images of the two
triangles in the fourth and third quadrants. It is seen
that the triangles in adjacent quadrants posses axial
Fig- 53-
symmetry, while the triangles in alternate quadrants
possess central symmetry.
160. Regular polygons of an odd number of sides
possess axial symmetry, and regular polygons of an
even number of sides possess central symmetry as
well.
IN PAPER FOLDING. 87
161. If a figure has two axes of symmetry at right
angles to each other, the point of intersection of the
axes is a center of symmetry. This obtains in reg-
ular polygons of an even number of sides and certain
curves, such as the circle, ellipse, hyperbola, and the
lemniscate ; regular polygons of an odd number of
Fig. 54-
sides may have more axes than one, but no two of
them will be at right angles to each other. If a sheet
of paper is folded double and cut, we obtain a piece
which has axial symmetry, and if it is cut fourfold, we
obtain a piece which has central symmetry as well, as
in Fig. 54.
88 GEOMETRIC EXERCISES
162. Parallelograms have a center of symmetry.
A quadrilateral of the form of a kite, or a trapezium
with two opposite sides equal and equally inclined to
either of the remaining sides, has an axis of sym-
metry.
163. The position of a point in a plane is also de-
termined by its distance from a fixed point and the
inclination of the line joining the two points to a fixed
line drawn through the fixed point.
If OA be the fixed line and P the given point, the
length OP and /_AOP, determine the position of P.
Fig. 55-
O is called the pole, OA the prime-vector, OP the
radius vector, and /_AOP the vectorial angle. OP
and /_AOP are called polar co-ordinates of P.
164. The image of a figure symmetric to the axis
OA may be obtained by folding through the axis OA.
The radii vectores of corresponding points are equally
inclined to the axis.
165. Let ABC be a triangle. Produce the sides
CA, AB, BC to D, E, F respectively. Suppose a
person to stand at A with face towards D and then to
IN PAPER FOLDING. 89
proceed from A to B, B to C, and C to A. Then he
successively describes the angles DAB, EBC, FCD.
Having come to his original position A, he has corn-
Fig. 56.
pleted a perigon, i. e., four right angles. We may
therefore infer that the three exterior angles are to-
gether equal to four right angles.
The same inference applies to any convex polygon.
161. Suppose the man to stand at A with his face
towards C, then to turn in the direction of AB and
proceed along AB, BC, and CA.
In this case, the man completes a straight angle,
i. e., two right angles. He successively turns through
the angles CAB, EBC, and FCA. Therefore /_EBF
+ / FCA -\- / CAB (neg. angle) = a straight angle.
This property is made use of in turning engines
on the railway. An engine standing upon DA with
its head towards A is driven on to CF, with its head
towards F. The motion is then reversed and it goes
backwards to EB. Then it moves forward along BA
on to AD. The engine has successively described
go GEOMETRIC EXERCISES
the angles ACB, CBA, and BAG. Therefore the
three interior angles of a triangle are together equal
to two right angles.
167. The property that the three interior angles of
a triangle are together equal to two right angles is
illustrated as follows by paper folding.
Fold CC' perpendicular to AB. Bisect C'B in N,
and AC' in M. Fold NA', MB' perpendicular to AB,
meeting BC and AC'm A' and B'. Draw A'C', B'C.
By folding the corners on NA' , MB' and A ' B ( ', we
find that the angles A, B, C of the triangle are equal
to the angles B'C'A, BC'A', and A' C'B' respectively,
which together make up two right angles.
168. Take any line ABC. Draw perpendiculars
to ABC at the points A, B, and C. Take points
D, E, P in the respective perpendiculars equidistant
IN PAPER FOLDING. 91
from their feet. Then it is easily seen by superposi-
tion and proved by equal triangles that DE is equal
to AB and perpendicular to AD and BE, and that
Fis equal to BC and perpendicular to BE and CF.
Now AB (=DE} is the shortest distance between the
lines AD and BE, and it is constant. Therefore AD
B
Fig. 58.
and BE can never meet, i. e., they are parallel. Hence
lines which are perpendicular to the same line are
parallel.
The two angles BAD and EBA are together equal
to two right angles. If we suppose the lines AD and
BE to move inwards about A and B, they will meet
and the interior angles will be less than two right
angles. They will not meet if produced backwards.
This is embodied in the much abused twelfth postulate
of Euclid's Elements.*
169. If AGffbe any line cutting BE in G and CF
in H, then
*For historical sketch see Beman and Smith's translation of Fink's
History of Mathematics, p. 270.
GEOMETRIC EXERCISES
/ GAD = the alternate /_AGB,
'.' each is the complement of /_BAG\ and
/_HGE= the interior and opposite / GAD.
.'. they are each =/ A GB.
Also the two angles GAD and EGA are together
equal to two right angles.
170. Take a line AX and mark off on it, from A,
equal segments AB, BC, CD, DE . . ..Erect perpen-
diculars to AE at B, C, D, E Let a line AF' cut
the perpendiculars in B', C', D', E' . . . . Then AB ',
B'C, CD', D'E' . . . . are all equal.
C D
Fig. 59-
If
, CD, DE be unequal, then
AB:BC=AB':B'C
BC: CD = B'C: CD', and so on.
171. If ABCDE . . . .be a polygon, similar polygons
may be obtained as follows.
Take any point O within the polygon, and draw
OA, OB, OC,....
Take any point A' in OA and draw A'B', B'C',
C'D, ____ parallel to AB, BC, CD ____ respectively.
IN PAPER FOLDING.
93
Then the polygon A'B'C'D'. . . . will be similar to
A BCD . . . . The polygons so described around a com-
mon point are in perspective. The point O may also
lie outside the polygon. It is called the center of per-
spective.
172. To divide a given line into 2, 3, 4, 5. . . .equal
parts. Let AB be the given line. Draw AC, BD
at right angles to AB on opposite sides and make
AC=BD. Draw CD cutting AB in /V Then
Now produce AC and take CE = EF= FG . . . .
= AC or BD. Draw DE, DF, DG ____ cutting AB
in />,, Pi, P 6 , ....
Then from similar triangles,
94 GEOMETRIC EXERCISES
= BD\ AF
= 1:3.
Similarly
PB\ A = l : 4,
and so on.
If AB = 1 9
But APi + PtP 9 + P B Pt + is ultimately = AB.
Or
2 ~l-
i 1
Adding
IN PAPER FOLDING. 95
.
1-22-3 (!)* rc
The limit of 1 -- when n is co is 1.
#
173. The following simple contrivance may be
used for dividing a line into a number of equal parts.
Take a rectangular piece of paper, and mark off n
equal segments on each or one of two adjacent sides.
Fold through the points of section so as to obtain
perpendiculars to the sides. Mark the points of sec-
tion and the corners 0, 1, 2, . . . .n. Suppose it is re-
quired to divide the edge of another piece of paper
AB into n equal parts. Now place AB so that A or
B may lie on 0, and B or A on the perpendicular
through n.
In this case AB must be greater than ON. But
the smaller side of the rectangle may be used for
smaller lines.
The points where AB crosses the perpendiculars
are the required points of section.
174. Center of mean position. If a line AB con-
tains (m -|- ri) equal parts, and it is divided at C so
that A C contains m of these parts and CB contains n
of them ; then if from the points A, C, B perpendicu-
lars AD, CF, BE be let fall on any line,
m-BE+ n-AD = (m-\-ri)- CF.
Now, draw BGH parallel to ED cutting CF'm G
and AD in H. Suppose through the points of divi-
sion AB lines are drawn parallel to BH. These lines
96 GEOMETRIC EXERCISES
will divide AH into (/ + ) equal parts and CG into
equal parts.
and since DH and .Z?^ are each = GF,
n-HD -f m-BE = f m -f
Hence, by addition
C is called the center of mean position, or the
mean center of A and B for the system of multiples
m and n.
The principle can be extended to any number of
points, not in a line. Then if P represent the feet of
the perpendiculars on any line from A, B, C, etc., if
a, b y c,... be the corresponding multiples, and if M
be the mean center
a-AP+b-BP+c-CP. ...
If the multiples are all equal to a, we get
a(AP+BP+CP+.. ..)=na-MP
n being the number of points.
175. The center of mean position of a number of
points with equal multiples is obtained thus. Bisect
the line joining any two points A, B in G, join G to a
third point C and divide GCinlf so that GH=\GC\
join H to a fourth point D and divide HD in ^T so that
HK\HD and so on: the last point found will be
the center of mean position of the system of points.
IN PAPER FOLDING. 97
176. The notion of mean center or center of mean
position is derived from Statics, because a system of
material points having their weights denoted by a, b,
c . . . ., and placed at A, B, C. . . .would balance about
the mean center M, if free to rotate about M under
the action of gravity.
The mean center has therefore a close relation to
the center of gravity of Statics.
177. The mean center of three points not in a line,
is the point of intersection of the medians of the tri-
angle formed by joining the three points. This is also
the center of gravity or mass center of a thin tri-
angular plate of uniform density.
178. If M is the mean center of the points A, B,
C, etc., for the corresponding multiples a, b, c, etc.,
and if P is any other point, then
b-BM* -f c- CM 2 + . . . .
Hence in any regular polygon, if O is the in-center
or circum-center and P is any point
AP* -f Bpt -f ____ = OA 2 -f OB* -f ____ -\-n- OP 2
= n (*-{- OP 2 ).
Now
Similarly
9 8 GEOMETRIC EXERCISES
Adding
179. The sum of the squares of the lines joining
the mean center with the points of the system is a
minimum.
If J/be the mean center and P any other point
not belonging to the system,
2PA* = 2MA*+2PM*, (where 2 stands for "the
sum of all expressions of the type").
.*. 2PA' 2 is the minimum when PM=Q, i. e.,
when P is the mean center.
180. Properties relating to concurrence of lines
and collinearity of points can be tested by paper fold-
ing.* Some instances are given below:
(1) The medians of a triangle are concurrent. The
common point is called the centroid.
(2) The altitudes of a triangle are concurrent
The common point is called the orthocenter.
(3) The perpendicular bisectors of the sides of a
triangle are concurrent. The common point is called
the circum-center.
(4) The bisectors of the angles of a triangle are
concurrent. The common point is called the in-center.
(5) Let ABCD be a parallelogram and P any
point. Through P draw Zf and EF parallel to BC
*For treatment of certain of these properties see Beman and Smith's
New Plane and Solid Geometry, pp. 84, 182.
IN PAPER FOLDING. 99
and AB respectively. Then the diagonals EG, HP,
and the line DB are concurrent.
(6) If two similar unequal rectineal figures are so
placed that their corresponding sides are parallel, then
the joins of corresponding corners are concurrent.
The common point is called the center of similarity.
(7) If two triangles are so placed that their corners
are two and two on concurrent lines, then their corre-
sponding sides intersect collinearly. This is known
as Desargues's theorem. The two triangles are said
to be in perspective. The point of concurrence and
line of collinearity are respectively called the center
and axis of perspective.
(8) The middle points of the diagonals of a com-
plete quadrilateral are collinear.
(9) If from any point on the circumference of the
circum-circle of a triangle, perpendiculars are dropped
on its sides, produced when necessary, the feet of
these perpendiculars are collinear. This line is called
Simson's line.
Simson's line bisects the join of the orthocenter
and the point from which the perpendiculars are
drawn.
(10) In any triangle the orthocenter, circum-center,
and centroid are collinear.
The mid-point of the join of the orthocenter and
circum-center is the center of the nine-points circle, so
called because it passes through the feet of the alti-
tudes and medians of the triangle and the mid-point
ioo GEOMETRIC EXERCISES
of that part of each altitude which lies between the
orthocenter and vertex.
The center of the nine-points circle is twice as far
from the orthocenter as from the centroid. This is
known as Poncelet's theorem.
(11) If A, B, C, D, , F, are any six points on a
circle which are joined successively in any order, then
the intersections of the first and fourth, of the second
and fifth, and of the third and sixth of these joins pro-
duced when necessary) are collinear. This is known
as Pascal's theorem.
(12) The joins of the vertices of a triangle with the
points of contact of the in-circle are concurrent. The
same property holds for the ex circles.
(13) The internal bisectors of two angles of a tri-
angle, and the external bisector of the third angle in-
tersect the opposite sides collinearly.
(14) The external bisectors of the angles of a tri-
angle intersect the opposite sides collinearly.
(15) If any point be joined to the vertices of a
triangle, the lines drawn through the point perpen-
dicular to those joins intersect the opposite sides of
the triangle collinearly.
(16) If on an axis of symmetry of the congruent
triangles ABC, A'B'C a point O be taken A'O, B' O,
and CO intersect the sides BC, CA and AB collin-
early.
(17) The points of intersection of pairs of tangents
to a circle at the extremities of chords which pass
IN PAPER FOLDING 101
through a given point are collinear. This line is called
the polar of the given point with respect to the circle.
(18) The isogonal conjugates of three concurrent
lines A X, BX, CX with respect to the three angles of
a triangle ABC are concurrent. (Two lines AX, AY
are said to be isogonal conjugates with respect to an
angle BAC, when they make equal angles with its
bisector.)
(19) If in a triangle ABC, the lines AA' , BB', CC
drawn from each of the angles to the opposite sides
are concurrent, their isotomic conjugates with respect
to the corresponding sides are also concurrent. (The
lines AA', AA" are said to be isotomic conjugates,
with respect to the side BC of the triangle ABC, when
the intercepts BA' and CA" are equal.)
(20) The three symmedians of a triangle are con-
current. (The isogonal conjugate of a median AM of
a triangle is called a symmedian.)
XIII. THE CONIC SECTIONS.
SECTION I. THE CIRCLE.
181. A piece of paper can be folded in numerous
ways through a common point. Points on each of the
lines so taken as to be equidistant from the common
point will lie on the circumference of a circle, of which
the common point is the center. The circle is the
locus of points equidistant from a fixed point, the
centre.
182. Any number of concentric circles can be
drawn. They cannot meet each other.
183. The center may be considered as the limit of
concentric circles described round it as center, the
radius being indefinitely diminished.
184. Circles with equal radii are congruent and
equal.
185. The curvature of a circle is uniform through-
out the circumference. A circle can therefore be made
to slide along itself by being turned about its center.
Any figure connected with the circle may be turned
about the center of the circle without changing its re-
lation to the circle.
PAPER FOLDING 103
186. A straight line can cross a circle in only two
points.
187. Every diameter is bisected at the center of
the circle. It is equal in length to two radii. All
diameters, like the radii, are equal.
188. The center of a circle is its center of sym-
metry, the extremities of any diameter being corre-
sponding points.
189. Every diameter is an axis of symmetry of the
circle, and conversely.
190. The propositions of 188, 189 are true for
systems of concentric circles.
191. Every diameter divides the circle into two
equal halves called semicircles.
192. Two diameters at right angles to each other
divide the circle into four equal parts called quadrants.
193. By bisecting the right angles contained by
the diameters, then the half right angles, and so on,
we obtain 2 n equal sectors of the circle. The angle
4
between the radii of each sector is ^ of a right angle
27T_ 7T
r W ~ 2^'
194. As shown in the preceding chapters, the right
angle can be divided also into 3, 5, 9, 10, 12, 15 and
17 equal parts. And each of the parts thus obtained
can be subdivided into 2" equal parts.
io 4 GEOMETRIC EXERCISES
195. A circle can be inscribed in a regular polygon,
and a circle can also be circumscribed round it. The
former circle will touch the sides at their mid-points.
196. Equal arcs subtend equal angles at the cen-
ter; and conversely. This can be proved by super-
position. If a circle be folded upon a diameter, the
two semicircles coincide. Every point in one semi-
circumference has a corresponding point in the other,
below it.
197. Any two radii are the sides of an isosceles tri-
angle, and the chord which joins their extremities is
the base of the triangle.
198. A radius which bisects the angle between two
radii is perpendicular to the base chord and also bi-
sects it.
199. Given one fixed diameter, any number of
pairs of radii may be drawn, the two radii of each set
being equally inclined to the diameter on each side of
it. The chords joining the extremities of each pair of
radii are at right angles to the diameter. The chords
are all parallel to one another.
200. The same diameter bisects all the chords as
well as arcs standing upon the chords, i. e., the locus
of the mid-points of a system of parallel chords is a
diameter.
201. The perpendicular bisectors of all chords of a
circle pass through the center.
IN PAPER FOLDING 105
202. Equal chords are equidistant from the center.
203. The extremities of two radii which are equally
inclined to a diameter on each side of it, are equi-
distant from every point in the diameter. Hence, any
number of circles can be described passing through
the two points. In other words, the locus of the cen-
ters of circles passing through two given points is the
straight line which bisects at right angles the join of
the points.
204. Let CC' be a chord perpendicular to the ra-
dius OA. Then the angles AOCand AOC' are equal.
Suppose both move on the circumference towards A
with the same velocity, then the chord CC' is always
parallel to itself and perpendicular to OA. Ultimately
the points C, A and C' coincide at A, and CAC' is
perpendicular to OA. A is the last point common to
the chord and the circumference. CAC' produced
becomes ultimately a tangent to the circle.
205. The tangent is perpendicular to the diameter
through the point of contact; and conversely.
206. If two chords of a circle are parallel, the arcs
joining their extremities towards the same parts are
equal. So are the arcs joining the extremities of either
chord with the diagonally opposite extremities of the
other and passing through the remaining extremities.
This is easily seen by folding on the diameter perpen-
dicular to the parallel chords.
io6
GEOMETRIC EXERCISES
207. The two chords and the joins of their extrem-
ities towards the same parts form a trapezoid which
has an axis of symmetry, viz., the diameter perpen-
dicular to the parallel chords. The diagonals of the
trapezoid intersect on the diameter. It is evident by
folding that the angles between each of the parallel
chords and each diagonal of the trapezoid are equal.
Also the angles upon the other equal arcs are equal.
208. The angle subtended at the center of a circle
by any arc is double the angle subtended by it at the
circumference.
Fig. 61.
Fig. 63.
An inscribed angle equals half the central angle
standing on the same arc.
Given
AVB an inscribed angle, and AOB the central angle
on the same arc AB.
To prove that / A VB = J / A OB.
Proof.
1. Suppose VO drawn through center <9, and pro-
duced to meet the circumference at X.
IN PAPER r>DING 107
Then _
2. And tXOB= /_XVB+ / VBO,
3. .-.
4. Similarly / A VX= J / A OX(each=zero in Fig. 62),
and .-. tAVB = ^/_AOB.
The proof holds for all three figures, point A hav-
ing moved to X (Fig. 62), and then through X (Fig.
63).*
209. The angle at the center being constant, the
angles subtended by an arc at all points of the cir-
cumference are equal.
210. The angle in a semicircle is a right angle.
211. If AB be a diameter of a circle, and DC a
chord at right angles to it, then ACBD is a quadri-
lateral of which AB is an axis of symmetry. The
angles BCA and ADB being each a right angle, the
remaining two angles DBC and CAD are together
equal to a straight angle. If A' and B' be any other
points on the arcs DAC and CBD respectively, the
/ CAD = L CA'Dand /_DBC=/_DB'C, and / CA'D
+DB'C = a straight angle. Therefore, also, /_B'CA'
+ / A'DB' = a straight angle.
Conversely, if a quadrilateral has two of its oppo-
site angles together equal to two right angles, it is
inscriptible in a circle.
*The above figures and proof are from Beman and Smith's New Plane
~:nd Solid Geometry, p. 129.
io8
GEOMETRIC EXERCISES
212. The angle between the tangent to a circle and
a chord which passes through the point of contact is
equal to the angle at the circumference standing upon
that chord and having its vertex on the side of it op-
posite to that on which the first angle lies.
Let AC be a. tangent to the circle at A and AB a
chord. Take O the center of the circle and draw OA,
OB. Draw OD perpendicular to AB.
Then BAC= AOD = J / BOA.
A
213. Perpendiculars to diameters at their extremi-
ties touch the circle at these extremities. (See Fig. 64).
The line joining the center and the point of intersection
IN PAPER FOLDING 109
of two tangents bisects the angles between the two
tangents and between the two radii. It also bisects
the join of the points of contact. The tangents are
equal.
This is seen by folding through the center and
the point of intersection of the tangents.
Let AC, AB be two tangents and ADEOF the
line through the intersection of the tangents A and
the center O, cutting the circle in D and F and BC
in E.
Then AC or AB is the geometric mean of AD and
AF\ AE is the harmonic mean; andAO the arith-
metic mean.
AD-AF 2AD-AF
OA AD + AF
Similarly, if any other chord through A be ob-
tained cutting the circle in P and R and BC in Q,
then AQ is the harmonic mean and AC the geometric
mean between AP and AR.
214. Fold a right-angled triangle OCB and CA
the perpendicular on the hypotenuse. Take D in AB
such that OD= OC (Fig. 65).
Then OA - OB = OC*= OP 2 ,
and OA : OC=OC: OB,
OA : OD=OD-. OB.
no
GEOMETRIC EXERCISES
A circle can be described with O as center and
OC or OD as radius.
The points A and B are inverses of each other
with reference to the center of inversion O and the
circle of inversion CDE.
Fig. 65.
Hence when the center is taken as the origin, the
foot of the ordinate of a point on a circle has for its
inverse the point of intersection of the tangent and
the axis taken.
215. Fold FBG perpendicular to OB. Then the
line FBG is called the polar of point A with reference
to the polar circle CDE and polar center O ; and A is
called the pole of FBG. Conversely B is the pole of
IN PAPER FOLDING m
CA and CA is the polar of B with reference to the
same circle.
216. Produce OC to meet FBG in F t and fold AH
perpendicular to OC.
Then F and H are inverse points.
AH\s> the polar of ^, and the perpendicular at F
to O/^ is the polar of H.
217. The points A, B, F, H, are concyclic.
That is, two points and their inverses are con-
cyclic ; and conversely.
Now take another point G on FBG. Draw OG,
and fold AK perpendicular to OG. Then K and G
are inverse points with reference to the circle CDE.
218. The points F, B, G are collinear, while their
polars pass through A.
Hence, the polars of collinear points are concur-
rent.
219. Points so situated that each lies on the polar
of the other are called conjugate points, and lines so
related that each passes through the pole of the other
are called conjugate lines.
A and ^are conjugate points, so are A and B, A
and G.
The point of intersection of the polars of two
points is the pole of the join of the points.
220. As A moves towards D, B also moves up to it.
Finally A and B coincide and FBG is the tangent at B.
ii2 GEOMETRIC EXERCISES
Hence the polar of any point on the circle is the
tangent at that point.
221. As A moves back to O, B moves forward to
infinity. The polar of the center of inversion or the
polar center is the line at infinity.
222. The angle between the polars of two points
is equal to the angle subtended by these points at the
polar center.
223. The circle described with B as a center and
BC as a radius cuts the circle CDE orthogonally.
224. Bisect AB in Z and fold LN perpendicular
to AB. Then all circles passing through A and B
will have their centers on this line. These circles cut
the circle CDE orthogonally. The circles circum-
scribing the quadrilaterals ABFH and ABGK are
such circles. AF and AG are diameters of the re-
spective circles. Hence if two circles cut orthogon-
ally the extremities of any diameter of either are con-
jugate points with respect to the other.
225. The points O, A, Zf and K are concyclic. H,
A, K being inverses of points on the line FBG, the
inverse of a line is a circle through the center of in-
version and the pole of the given line, these points
being the extremities of a diameter ; and conversely.
226. If DO produced cuts the circle CDE in D',
D and D' are harmonic conjugates of A and B. Sim-
IN PA PER FOLDING 1 1 3
ilarly, if any line through B cuts AC in A' and the
circle CDE in d and d { ', then d and d' are harmonic
conjugates of A' and B.
227. Fold any line LM=LB LA, and J/<7 per-
pendicular to LM meeting AB produced in O'.
Then the circle described with center O' and ra-
dius O'Mcuts orthogonally the circle described with
center Z and radius LM.
Now OL 2 = OE 1 + L&,
and
.-. LNis the radical axis of the circles O (OC)
and O'(O'M).
By taking other points in the semicircle AMB and
repeating the same construction as above, we get two
infinite systems of circles co-axial with O(OC} and
O\O'M}, viz., one system on each side of the radical
axis, LN. The point circle of each system is a point,
A or B, which may be regarded as an infinitely small
circle.
The two infinite systems of circles are to be re-
garded as one co-axial system, the circles of which
range from infinitely large to infinitely small the
radical axis being the infinitely large circle, and the
limiting points the infinitely small. This system of
co-axial circles is called the limiting point species.
If two circles cut each other their common chord
is their radical axis. Therefore all circles passing
ii4 GEOMETRIC EXERCISES
through A and B are co-axial. This system of co-
axial circles is called the common point species.
228. Take two lines OAB and OPQ, From two
points A and B in OAB draw AP, BQ perpendicular
to OPQ. Then circles described with A and B as
centers and AP and BQ as radii will touch the line
OPQ at P and Q.
Then OA:OB = AP:BQ.
This holds whether the perpendiculars are towards
the same or opposite parts. The tangent is in one
case direct, and in the other transverse.
In the first case, O is outside AB, and in the sec-
ond it is between A and B. In the former it is called
the external center of similitude and in the latter the
internal centre of similitude of the two circles.
229. The line joining the extremities of two par-
allel radii of the two circles passes through their ex-
ternal center of similitude, if the radii are in the same
direction, and through their internal center, if they
are drawn in opposite directions.
230. The two radii of one circle drawn to its points
of intersection with any line passing through either
center of similitude, are respectively parallel to the
two radii of the other circle drawn to its intersections
with the same line.
231. All secants passing through a center of simil-
itude of two circles are cut in the same ratio by the
circles.
IN PAPER FOLDING 115
232.. If 2?i, DI, and B
Let OX, OYbe the axes; divide the .right angle
VOX into a number of equal parts. Let XOA, A OB
130
GEOMETRIC EXERCISES
be two of the equal angles. Fold XB at right angles
to OX. Produce BO and take OF= OX. Fold OG
perpendicular to BF and find G in OG such that FGB
is a right angle. Take OA = OG. Then A is a point
on the curve.
Y
Fig. 75-
Now, the angles XOA and AOB being each 6,
And OA* = OG , =,
or <#, the origin is a
node, a cusp, or a con-
jugate point. The fig-
uref represents the case
when b a.
Nicomedes (c.
Fig. 77-
267. This curve also
was employed for finding
two geometric means, and for the trisection of an angle.
*See Beman and Smith's translation of Klein's Famous Problems of Ele-
mentary Geometry, p. 40.
tFroni Beman and Smith's translation of Klein's Famous Problems of
Elementary Geometry, p. 46.
134 GEOMETRIC EXERCISES
Let OA be the longer of the two lines of which
two geometric means are required.
Bisect OA in B\ with O as a center and OB as a
radius describe a circle. Place a chord BC in the
circle equal to the shorter of the given lines. Draw
AC and produce AC and BC to D and E, two points
collinear with O and such that DE OB, or BA.
Fig. 78.
Then ED and CE are the two mean proportionals
required.
Let OE cut the circles in E and G.
By Menelaus's Theorem,*
BC -ED OA=CE -OD-BA
... BC-OA=CE-OD
BC __
f C ~OA
BE __ OD + OA _ GE
' ' ~CE ~ OA ~~OA'
*See Beman and Smith's New Plane and Solid Geometry, p. 240.
IN PAPER FOLDING, 135
But GE-EF=BE'EC.
.-. OA -OD =
The position of E is found by the aid of the con-
choid of which AD is the asymptote, O the focus, and
DE the constant intercept.
268. The trisection of the angle is thus effected.
In Fig. 77, let < = / MOV, the angle to be trisected.
On OM lay off OM=b, any arbitrary length. With
M as a center and a radius b describe a circle, and
through M perpendicular to the axis of Jf with origin
O draw a vertical line representing the asymptote of
the conchoid to be constructed. Construct the con-
choid. Connect O with A, the intersection of the circle
and the conchoid. Then is / A OY one third of (p.*
THE WITCH.
269. If OQA (Fig. 79) be a semicircle and NQ an
ordinate of it, and NP be taken a fourth proportional
to ON, OA and QN t then the locus of P is the witch.
Fold AM at right angles to OA.
Fold through O, Q, and M.
Complete the rectangle NAMP.
PN\ QN=OM: OQ
= OA: ON.
*Beman and Smith's translation of Klein's Famous Problems of Elemen-
tary Geometry, p. 46.
136 GEOMETRIC EXERCISES
Therefore P is a point on the curve.
Its equation is,
xy* = a 2 (a .#).
Fig. 79.
This curve was proposed by a lady, Maria Gaetana
Agnesi, Professor of Mathematics at Bologna.
THE CUBICAL PARABOLA.
270. The equation to this curve is a^y = x z .
Let OX and OYbe the rectangular axes, OA = a,
and OX=x.
In the axis <9Ftake OB = x.
Draw BA and draw AC at right angles to AB cut-
ting the axis OY'm C.
IN PAPER FOLDING.
137
Draw CX, and draw XV at right angles to CX.
Complete the rectangle XOY.
P is a point on the curve.
Fig. 80.
THE HARMONIC CURVE OR CURVE OF SINES.
271. This is the curve in which a musical string
vibrates when sounded. The ordinates are propor-
tional to the sines of angles which are the same frac-
tions of four right angles that the corresponding ab-
scissas are of some given length.
Let AB (Fig. 81) be the given length. Produce BA
138
GEOMETRIC EXERCISES
to C and fold AD perpendicular to AB. Divide the
right angle DAC into a number of equal parts, say,
four. Mark on each radius a length equal to the am-
plitude of the vibration, AC=AP=AQ = AR = AD.
From points/ 5 , Q, J? fold perpendiculars to AC;
then PP', QQ', JZR', and DA are proportional to the
sines of the angles PAC, QAC, RAC, DAC.
Now, bisect AB in E and divide AE and EB into
twice the number of equal parts chosen for the right
C ?' Q'
S' T' U' V
angle. Draw the successive ordinates SS' , TT', UU',
VV, etc., equal to PP' , QQ', R', DA, etc. Then
S, T, U, V are points on the curve, and V is the
highest point on it. By folding on VV and pricking
through S, T, U, V, we get corresponding points on
the portion of the curve VE. The portion of the
curve corresponding to EB is equal to A VE but lies
on the opposite side of AB. The length from A to E
is half a wave length, which will be repeated from E
PAPER FOLDING
139
to B on the other side of AB. E is a point of inflec-
tion on the curve, the radius of curvature there be-
coming infinite.
THE OVALS OF CASSINI.
272. When a point moves in a plane so that the
product of its distances from two fixed points in the
plane is constant, it traces out one of Cassini's ovals.
The fixed points are called the foci. The equation of
M 'A
Fig. 82.
the curve is rr' &, where r and r' are the distances
of any point on the curve from the foci and k is a con-
stant.
Let F and F' be the foci. Fold through F and
F' . Bisect FF' in C, and fold BCB' perpendicular to
FF'. Find points B and B' such that FB and FB'
are each =k. Then B and B' are evidently points on
the curve.
1 4 o GEOMETRIC EXERCISES
Fold FK perpendicular to FF' and make FK=k,
and on FF' take CA and CA' each equal to CK. Then
A and A' are points on the curve.
For CA* = CK* = CF ti + 7^ 2 .
. C^ 2 C7^ 2 = 2 :=(a4 + CF)(CA CF}
= F'A-FA.
Produce .FA and take AT=FK. In A T take a
point M and draw J/A'. Fold KM' perpendicular to
MK meeting FA' in M'.
With the center F and radius FM, and with the
center F' and radius FM', describe two arcs cutting
each other in P. Then P is a point on the curve.
When a number of points between A and B are
found, corresponding points in the other quadrants
can be marked by paper folding.
When FF' = V2k and rr* = \k* the curve as-
sumes the form of a lemniscate. ( 279.)
When FF' is greater than 1/2 , the curve consists
of two distinct ovals, one about each focus.
THE LOGARITHMIC CURVE.
273. The equation to this curve \sy = a*.
The ordinate at the origin is unity.
If the abscissa increases arithmetically, the ordi-
nate increases geometrically.
The values of y for integral values of x can be ob-
tained by the process given in 108.
IN PAPER FOLDING. 141
The curve extends to infinity in the angular space
XOY.
If x be negative y= and approaches zero as x
increases numerically. The negative side of the axis
OX is therefore an asymptote to the curve.
THE COMMON CATENARY.
274. The catenary is the form assumed by a heavy
inextensible string freely suspended from two points
and hanging under the action of gravity.
The equation of the curve is
the axis of y being a vertical line through the lowest
point of the curve, and the axis of x a horizontal line
in the plane of the string at a distance c below the
lowest point ; c is the parameter of the curve, and e
the base of the natural system of logarithms.
When x = c, y=^(e l + e~ l )
when x 2,
Draw OD.
Fig. 85.
Then / A CB is \ of /_AOB.
For CD = DO =
/_ODB
THE LEMNISCATE OF BERNOULLI.
279. The polar equation to the curve is
Let O be the origin, and OA=a.
Produce AO, and draw OD at right angles to OA
Take the angle A OP =6 and AOB=--2d.
Draw AB perpendicular to OB.
In AO produced take OC=OB.
IN PAPER FOLDING
Find D in OD such that CDA is a right angle.
Take OP = OD.
P is a point on the curve.
r 2 = <9Z> 2 =<9C-<9^
145
As stated above, this curve is a particular case of
the ovals of Cassini.
Fig. 86.
It is the inverse of the rectangular hyperbola, with
reference to its center as center of inversion, and also
its pedal with respect to the center.
The area of the curve is a 2 .
THE CYCLOID.
280. The cycloid is the path described by a point
on the circumference of a circle which is supposed to
roll upon a fixed straight line.
Let A and A' be the positions of the generating
point when in contact with the fixed line after one
i 4 6
GEOMETRIC EXERCISES
complete revolution of the circle. Then AA' is equal
to the circumference of the circle.
The circumference of a circle may be obtained in
length in this way. Wrap a strip of paper round a
circular object, e. g., the cylinder in Kindergarten
gift No. II., and mark off two coincident points. Un-
fold the paper and fold through the points. Then the
straight line between the two points is equal to the
circumference corresponding to the diameter of the
cylinder.
By proportion, the circumference corresponding
to any diameter can be found arid vice versa.
A'
D G
Fig. 87.
Bisect AA' in D and draw DB at right angles to
AA', and equal to the diameter of the generating
circle.
Then A, A' and B are points on the curve.
Find O the middle point of BD.
Fold a number of radii of the generating circle
through O dividing the semi-circumference to the
right into equal arcs, say, four.
Divide AD into the same number of equal parts.
IN PAPER FOLDING 147
Through the ends of the diameters fold lines at
right angles to BD.
Let EFP be one of these lines, F being the end of
a radius, and let G be the corresponding point of sec-
tion of AD, commencing from D. Mark off FP equal
to GA or to the length of arc BF.
Then P is a point on the curve.
Other points corresponding to other points of sec-
tion of AD may be marked in the same way.
The curve is symmetric to the axis BD and corre-
sponding points on the other half of the curve can be
marked by folding on BD.
The length of the curve is 4 times BD and its area
3 times the area of the generating circle.
THE TROCHOID.
281. If as in the cycloid, a circle rolls along a
straight line, any point in the plane of the circle but
not on its circumference traces out the curve called a
trochoid.
THE EPICYCLOID.
282. An epicycloid is the path described by a point
on the circumference of a circle which rolls on the
circumference of another fixed circle touching it on
the outside.
THE HYPOCYCLOID.
283. If the rolling circle touches the inside of the
fixed circle, the curve traced by a point on the cir-
cumference of the former is a hypocycloid.
I 4 8 GEOMETRIC EXERCISES
When the radius of the rolling circle is a sub-
multiple of the fixed circle, the circumference of the
latter has to be divided in the same ratio.
These sections being divided into a number of
equal parts, the position of the center of the rolling
circle and of the generating point corresponding to
each point of section of the fixed circle can be found
by dividing the circumference of the rolling circle into
the same number of equal parts.
THE QUADRATRIX.*
284. Let OACB be a square. If the radius OA of
a circle rotate uniformly round the center O from the
position OA through a right angle to OB and if in the
same time a straight line drawn perpendicular to OB
move uniformly parallel to itself from the position
OA to BC ' ; the locus of their intersection will be the
quadratrix.
This curve was invented by Hippias of Elis (420
B. C.) for the multisection of an angle.
If P and P' are points on the curve, the angles
A OP and A OP' are to one another as the ordinates
of the respective points.
THE SPIRAL OF ARCHIMEDES.
285. If the line OA revolve uniformly round O as
center, while point P moves uniformly from O along
OA, then the point P will describe the spiral of Archi-
medes.
*Beman and Smith's translation of Klein' 's Famous Problems of Elemen-
tary Geometry, p. 57.
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48. The Principles of Human Knowledge. By GEORGE BERKELEY. 250 (is.6d.).
,4g. Three Dialogues Between Hylas and Philonous. By GEORGE BERKELEY.
250 (is. 6d.).
50. Public Worship: A Study in the Psychology of Religion. By JOHN P.
HYLAN. 250 (is. 6d.).
51. Descartes' Meditations, with Selections from the Principles. 350 (is. 6d.).
52. Kant's Prolegomena. 500 (2S. 6d.). (In preparation.)
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