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IN MEMORIAM 
 FLOR1AN CAJOR1 
 
 C 
 
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 ^/VX_ 
 
 CURIOSITIES of MATHEMATICS 
 
 FOR THE INSTRUCTION 
 
 MATHEMATICIANS 
 
 AND THE BENEFIT OF 
 
 THE BRITISH ASSOCIATION FOR THE 
 
 . ADVANCEMENT OF SCIENCE 
 
 JAMES SMITH, ESQ. 
 
 MEMBER OF THE MERSEY DOCKS AND HARBOUR BOARD, AND 
 EX-CHAIRMAN OF THE LIVERPOOL LOCAL MARINE BOARD. 
 
 LIVERPOOL 
 
 EDWARD HOWELL, CHURCH STREET, 
 
 LONDON 
 
 SIMPKIN, MARSHALL & CO., STATIONERS' HALL COURT 
 
 1870 
 
 
 All Rights Reserved. 
 
CURIOSITIES of MATHEMATICS 
 
 FOR THE INSTRUCTION 
 
 OK 
 
 MATHEMATICIANS 
 
 AND THE BENEFIT OF 
 
 THE BRITISH ASSOCIATION FOR THE 
 ADVANCEMENT OF SCIENCE 
 
 BY 
 
 JAMES\SMITH, ESQ. 
 
 MEMBER OF THE MERSEY DOCKS AND HARBOUR BOARD, AND 
 EX-CHAIRMAN OF THE LIVERPOOL LOCAL MARINE BOARD. 
 
 LIVERPOOL 
 
 EDWARD HOWELL, CHURCH STREET, 
 
 LONDON 
 
 SIMPKIN, MARSHALL & CO., STATIONERS' HALL COURT. 
 
 1870 
 
 All Rights Reserved. 
 
CAJORI 
 
Of 
 
 TO THE READER. 
 
 For many years I have been a regular attender at 
 the Annual Meetings of " The British Association for 
 the Advancement of Science," and it is well known to the 
 Mathematical Members of the Association, that I repudiate 
 the assertion of Mathematicians, that a square exactly 
 equal in superficial area to a given circle cannot be 
 found. On the contrary, I maintain, that by practical 
 Geometry a square equal to a given circle can be constructed, 
 isolated, and exhibited, standing on its equivalent circle ; 
 and this I have proved in a variety of ways, in my published 
 works. I have also proved : — First : That the area of a 
 circle is equal to the sum of the areas of squares on the 
 sides of a right-angled triangle, of which the sides that 
 contain the right angle are in the ratio of 3 to 4, and 
 the middle side the radius of the circle. Second : That 
 the area of a circle is equal to the area of a square on the 
 hypothenuse of a right-angled triangle, of which the sides 
 that contain the right angle are in the ratio of 7 to 1, and 
 the sum of these two sides equal to the diameter of the 
 circle. 
 
 The symbol 7r is adopted to denote the circumfer- 
 ence of a circle of diameter unity : and ir times the 
 square of the radius = circumference X semi-radius, in 
 every circle. No honest Geometer or Mathematician 
 will dispute, that the equation, tt (r 4 ) = (c X s r) in 
 every circle, whatever be the value of -k . 
 
 ™ •*.?«* ■low*? 
 
Hence : If p denote the perimeter of a regular 
 hexagon : c denote the circumference of its circumscrib- 
 ing circle : a denote the area of the circle : and x denote 
 the area of a regular dodecagon inscribed within the 
 
 circle : — 
 
 p : c : : x : a 
 
 • ■-(!)• = -(f) 
 
 But this equation cannot be proved arithmetically with 
 an indeterminate value of 7r : neither can we get this 
 equation from a given arithmetical value of c, with a 
 false value of 71 . 
 
 Let c be represented by any arithmetical quantity, 
 say 360, and it will not be disputed by any honest 
 Mathematician, that the circumference of a circle is 
 divided into 360 equal parts, for all practical purposes. 
 
 Then : 
 
 — — -3— = lie "2 = diameter of the circle: 
 
 3l 3-125 2 
 
 IK*2 ^ t r ^.l • 1 Radius 57*6 
 
 — b = 57-6 = radius of the circle: = D ' 
 
 2 D/ 22 
 
 = 28'8 = semi-radius of the circle: 6 (r) = (6x 57"6) 
 
 = 345-6 = p: and 6 (r x s r) — 6(57-6 x 28'S) = 
 
 6(1658-88) = 9953* 2 8 = x. 
 
 ••,3*(f)'=3i(f) 
 
 thatis,3i( 34 f) 2 = 3H 995 f S ) 
 Again : 
 
 Divide the diameter of the circle into two parts A B 
 and B C, so that A B shall be to B C in the ratio of 
 7 to 1. 
 
 Then : 
 Diameter = lip = ^ = fi Q . ? (R Q = (; x ^.^ 
 
5 
 ioo-8 = A B : therefore, (AB'+B C 2 ) = (ioo"8 2 + i4'4 2 ) 
 = (10160-64 + 207-36) = 10368 = 3£ ( ), and 
 
 this equation = a = area, of the circle. 
 Hence : 
 
 H(<0 = !!(36o) = 345-6 =p 
 And, U («) = H (10368) = 9953'28 = x 
 
 .•.31 \£\ = 31 f- j as in the previous example. 
 
 No other value of tt but that which makes 8 circum- 
 ferences of a circle exactly equal to 25 diameters can 
 produce these results, and it follows, not as an assumption, 
 but as a logical deduction, that V = 3' I2 S ls tne true 
 arithmetical value of the circumference of a circle of 
 diameter unity. It also follows, that §4 (3*1 25) = 
 
 24_x_32 2 5 75 _ - _ t} e £ ; ^. Z£/A! value of the perimeter 
 
 25 25 J r 
 
 of an inscribed regular hexagon to a circle of diameter 
 unity. Hence : From the known relations that exist be- 
 tween a regular polygon of six sides and a regular polygon 
 of twelve sides inscribed in the same circle, we can 
 ascertain the area of the circle from a given radius, but 
 from no other polygons, whether inscribed or circumscribed 
 to a circle. The reason is obvious. The perimeter of all 
 polygons of more than 6 sides, and the area of all 
 polygons of more than 12 sides are incommensurable. 
 
 The Rev. W. Allen Whitworth, Fellow of St. John's 
 College, Cambridge, and formerly Professor of Mathema- 
 tics in Queen's College, Liverpool, charged me in a Letter 
 he addressed to me, dated November 9th. 1363, with 
 bringing a false accusation against " recognised Mathe- 
 maticians" and in his Letter made the following assertion : 
 " I kjioiv and ahvays teach that the value of it is a finite and 
 
determinate quantity ; " and yet, the Reverend gentleman 
 also teaches, that the arithmetical value of 7r is 314159 
 with a never ending string of decimals. To give his 
 views something like an air of consistency he asserts, that 
 the series 1 + i + i + \ + T V + £2 + &c stands for 2. 
 
 In the Liverpool Leader of August 21, 23, and 
 September 11, 1869, there appeared three Articles from 
 the pen of Mr. Whitworth under the title of " Curiosities 
 of Mathematics. In one of them, in which he introduced 
 my name, he professes to prove that the arithmetical 
 value of 7r is greater than 3' 125. These articles were to 
 be continued, but after waiting several months without 
 seeing any signs of it, I wrote a series of Letters in the 
 Leader in which I exposed the fallacy in the premisses, 
 reasoning, and conclusion of Mr. Whitworth. (Thes e 
 Letters appear in my last published work.) To these 
 Letters he never replied, but in April, 1870, I received a 
 curious epistle from him, in which he said : — The series of 
 Articles on " Curiosities of Mathematics" will probably be 
 continued when you (I) have finished writing in the Leader 
 under the same title." This led me to make another 
 considerable pause. Still Mr. Whitworth remained 
 silent, and I concluded my series with Letters which will 
 be found in the following pages, and which must speak 
 for themselves. 
 
 James Smith. 
 Barkeley House, Seaforth, 
 1st August, 1870. 
 
From the "LIVERPOOL LEADER," June iith, 1870. 
 CURIOSITIES OF MATHEMATICS. 
 
 TO THE EDITOR OF THE LIVERPOOL LEADER. 
 
 Sir, 
 
 The last letter I " received from your contributor, the 
 Rev. W. Allen Whitworth, he commenced by saying " the copyright 
 of the articles in the Leader, entitled ' Curiosities of Mathematics,' 
 is my property," and he valued each of these articles at twenty 
 guineas. They appeared in the Leader of August 21, 28, and 
 September nth, 1869, without any signature, but with an intimation 
 that they were to be continued. He next observed — " The series 
 will probably be continued when you (I) have ceased inserting 
 advertisements under the same title." It is more that two months 
 since a letter of mine appeared in your Journal, and as yet there is 
 no sign of Mr. Whitworth continuing the series. Has he abandoned 
 his intention ? Nous verrons. I notice from the announcement in 
 the public papers that he has recently been appointed to the 
 incumbency of Christ Church, Liverpool. 
 Diagram. 
 
8 
 
 The geometrical figure represented by the diagram may be 
 constructed in the following way : — 
 
 Draw the straight line A B and bisect it in O. With O as 
 centre and O A or O B as radius, describe the circle. With A B as 
 perpendicular, describe the right-angled triangle ABC, making B 
 the right angle, and A B to B C in the ratio of 8 to 6. From D the 
 point of intersection between A C the hypothenuse of the triangle 
 ABC and the circumference of the circle, draw the straight line 
 D M perpendicular to A B, and, therefore, parallel to B C. From 
 D draw the straight line D N perpendicular to B C, and, therefore, 
 parallel to A B. Bisect B C in P, and join D P, D B, and D O. 
 
 Other methods of construction will suggest themselves to any 
 " reasoning geometrical investigator." 
 
 By Euclid: Prop. 31 ; Book 3. 
 
 A D B is a right-angled triangle. 
 By Euclid : Prop. 4 and Prop. 12 ; Book 2. 
 
 D O 2 + O A 2 + 2 (O A x O M) = A D 2 
 
 DP 2 + P B 2 + 2 (P B x P N) = D B 2 
 
 By Euclid : Prop. 47 ; Book I. 
 
 D M 2 + M A 2 = A D 2 
 
 D M 2 + M B 2 = D B 2 
 
 D O 2 — O M 2 = D M 2 
 
 D P 2 — P N 2 = D N 2 
 
 D N 2 + N B 2 = D B 2 
 
 D N 2 + N C 2 = D C 2 
 
 A D 2 + D B 2 = A B 2 
 
 B D 2 + D C 2 = B C 2 
 
 A B 2 + B C 2 = A C 2 
 
 By Euclid : Prop. 8 ; Book 6. 
 
 The triangles DMA and D M B on each side of D M are 
 similar triangles, and similar to the whole triangle A D B. 
 
 The triangles D N B and D N C on each side of D N are similar 
 triangles, and similar to the whole triangle B D C. 
 ABxAM = AD 2 
 ABx MB = DB 2 
 By Euclid : Prop. 2, Book 2 ; and Prop. 35, Book 3. 
 (A B x A M) + (A B x M B) = A B 2 
 
(A B x A O) + (A B x O B) = A B a 
 (B C x B N) + (B C x N C) = B C 
 (B C x B P) + (B C x P C) - B C- 
 M B N D is a rectangular parallelogram, and it follows that the 
 side N B = the side D M ; and the side D N = the side M B. 
 
 Now, B C the hypothenuse of the right-angled triangle B D C, 
 which is also the base of the right-angled triangle A B C, is divided 
 into two equal parts, B P and P C, and into two unequal parts, B N 
 and N C, by construction. Will your contributors, the Rev. W. Allen 
 Whitworth and the Rev. Geo. B. Gibbons— both " recognised 
 mathematicians "—favour your mathematical readers with a solution 
 of the following problem ? 
 
 Problem. 
 Find the exact arithmetical ratio, expressed in whole numbers, 
 between the lengths of B N and N C, the two unequal parts into 
 which B C the hypothenuse of the right-angled triangle B D C, is 
 divided. 
 
 I hope Mr. Whitworth and Mr. Gibbons will not think I am 
 asking too great a favour at their hands. 
 I am, Sir, 
 
 Yours obediently, 
 
 James Smith. 
 Barkeley House, Seaforth, 
 June 6th, 1870. 
 
 From the "LIVERPOOL LEADER," June i8th, 1870. 
 CURIOSITIES OF MATHEMATICS. 
 
 to the editor of the liverpool leader. 
 
 Sir, 
 
 I sent copies of your last number to upwards of eighty 
 members of the British Association for the Advancement of Science, 
 and have written and posted the following letters : — 
 2 
 
IO 
 
 J. S. to the Editor of the Athenaeum. 
 Sir, 
 
 In this days number of the Liverpool Leader, of which 
 I send you a copy by this post, you will find a letter of mine, in 
 which I give an interesting and important problem for solution. I 
 might have given the following method of constructing the geometri- 
 cal figure on which the problem is founded. 
 
 Let A B be a'straight line divided into two unequal parts, A M 
 and M B, in the ratio of 5 to 3. Bisect A B in O, and with O as 
 centre and O A or O B as radius, describe the circle. With A B as 
 perpendicular describe the right-angled triangle ABC, making A B 
 and B C the sides that contain the right angle in the ratio of 4 to 3. 
 A C the hypothenuse of the triangle ABC intersects the circum- 
 ference of the circle at the point D. Join D O, D M, and D B. 
 From D draw the straight line D N perpendicular to B C, and, 
 therefore, parallel to A B, and join D P. 
 
 The problem can be solved without the aid of the differential 
 calculus or any other of the higher branches of Mathematics. 
 Will you kindly give the solution of the problem in your next 
 number, which will greatly oblige several of your Mathematical 
 readers, and cannot fail to interest all your scientific readers ? 
 I am, Sir, 
 
 Yours respectfully, 
 
 J.S. 
 nth June, 1870. 
 
 J. S. to PROFESSOR DE MORGAN. 
 
 Sir, 
 
 In to-day's number of the Liverpool Leader, of which I 
 send you a copy by this post, you will find a Letter of mine, in 
 which I give a problem for solution. 
 
 The problem can be solved without the aid of the differential 
 calculus or any other of the higher branches of Mathematics. Will 
 you kindly give the solution in the next number of the Athenaumi 
 Or, if you prove that the solution is impossible, and thus make my 
 
II 
 
 statement to be an untruth, you will, by doing so, prove that " James 
 Smith, of Liverpool, is nailed by himself to the barn-door as the 
 delegate of miscalculated and disorganised failure." 
 Your dutiful tutelary, 
 
 J.S. 
 nth June, 1870. 
 
 J. S. to The Rev. W. ALLEN WHITWORTH. 
 
 Sir, 
 
 In the Liverpool Leader of to-day, a copy of which 
 accompanies this communication, you will find a letter of mine, in 
 which I give an interesting and novel problem for solution. 
 
 As the Editor of the Leader has decided not to insert any 
 further letters on " Curiosities of Mathematics " unless paid for as 
 advertisements, in consideration of your writing a Letter to the 
 Leader giving the solution of the problem, to appear in that Journal 
 either of the 18th or 25th of this month, I hereby engage to pay for 
 its insertion as an advertisement. 
 
 Yours respectfully, 
 
 J.S. 
 nth June, 1870. 
 
 J. S. to The Rev. GEORGE B. GIBBONS. 
 
 My Dear Sir, 
 
 I send you by this post a copy of to-day's Liverpool 
 Leader, in which you will find a Letter of mine, giving a remarkable 
 and novel problem for solution. 
 
 If, in the course of next week, you send me a Letter giving the 
 solution of the problem, I will take care that it is inserted in the 
 Leader of the 25 th of this month. 
 
 I am, dear Sir, 
 
 Faithfully yours, 
 
 J.S. 
 \\th June, 1870. 
 
12 
 
 J. S. TO E. L. GARBETT. 
 Sir, 
 
 On Saturday last I posted to your address a copy of the 
 Liverpool Leader of that day's date. In it you would see a Letter 
 of mine, in which I give a problem. I am prepared to pay you your 
 price (3 Id) for a solution of it, and can assure you that I should 
 think I never spent threepence and one-eighth of a penny better. 
 
 Yours truly, 
 
 J. S. 
 
 13M Jtme, 1870. 
 
 Whether your contributors, Mr. Whitworth and Mr. Gibbons, 
 do, or do not, favour us with a solution of my problem. I do hope 
 to see some notice taken of it in the next number of the Athenceum. 
 I am, Sir, 
 
 Yours obediently, 
 
 James Smith. 
 Barkeley House, Seaforth, 
 June 14th, 1870. 
 
 From the "LIVERPOOL LEADER," June 25TH, 1870. 
 CURIOSITIES OF MATHEMATICS 
 
 to the editor of the liverpool leader. 
 
 Sir, 
 
 I have received from the Rev. Geo. B. Gibbons, for 
 insertion in your valuable journal, the following communication, in 
 which he gives what he thinks a solution of the problem, founded 
 on the geometrical figure represented by the diagram in my Letter 
 which appeared in the Leader of the nth inst. : — 
 
13 
 
 
 
 
 G. B. G 
 
 . TO J. S. 
 
 
 My 
 
 Dear 
 
 Sir, 
 
 
 
 
 
 
 
 Your Letter reached me 
 
 yesterday, but the 
 
 spap 
 
 er only 
 
 arrived this day. 
 
 Using the figure there given : 
 
 
 
 
 By similar triangles, 
 
 
 
 
 
 AM 
 
 _ DN 
 
 
 
 
 
 M D 
 
 — NC 
 
 
 
 
 Or, 
 
 AM 
 BN 
 
 MB 
 
 — NC 
 
 
 
 
 Tha 
 
 .is: 
 
 
 
 BN AM , ■ , 
 
 NC = MB the ratl ° reqmred ' 
 Again : By the numerical values given 
 AJVI _ 4 
 M D _ 3 
 BM _ 3 
 M D ~ 4 
 .-. Dividing one by the other : 
 
 & = £ = §»*■—*** 
 
 We can easily find the values of all the lines employed. 
 Put: 
 
 Then: 
 
 But, 
 
 Substituting in the first equation the value of/, deduced from 
 the second. 
 
 If U-) = 8 
 25 (x) = 128 
 
 x - 5'i2 = AM 
 ; = AW = 2-88 = M B 
 MD = BN = fAM 
 
 = |(5-i2) = 3-84 
 C N =|BM 
 
 = |(2-88) = 2- 16 
 
 AM = 
 
 - X 
 
 MB = 
 
 --y 
 
 x+y 
 
 _ 
 
 X 
 
 16 
 9 
 
14 
 
 A D 2 = (5-I2) 9 + (3'84) 2 = 40-96 
 .-. AD = 6-4 
 
 DC 2 =(2-i6) ! + (2-S8 2 ) - 12-96 
 .-.DC -3-6 
 
 Yours very truly, 
 
 G. B. Gibbons. 
 Werrington, Launceston, 
 
 \\th June, 1870 {at night). 
 
 Some of the absurdities involved in Mr. Gibbons's supposed 
 solution of the problem will, I think, be self-evident to many of your 
 readers ; but I will deal with these absurdities in a future communi- 
 cation. 
 
 I am, Sir, 
 
 Yours obediently, 
 
 James Smith. 
 Barkeley House, Seaforth, 
 June 14th, 1870. 
 
 CURIOSITIES OF MATHEMATICS (OR OF WHAT?) 
 
 Sir, 
 
 Having been honoured by Mr. Smith with a special 
 challenge to solve his problem of June nth, allow me to point out 
 that, by Euclid, B. 6, Prop. 8, the triangle B D C is similar to A B C, 
 and its parts B N D, D N C, are similar to the same. 
 
 Wherefore, by Prop. 4, A B : B C : 
 
 : BN 
 
 ND 
 
 And . . : 
 
 : ND 
 
 NC 
 
 And as Mr. Smith makes A B : B C : 
 
 : 4 
 
 : 3 
 
 By B. 5, def. 10, B N : N C in the duplicate ratio of 4 : 3 — that 
 is, 16 : 9. 
 
 Now, it is really too bad to tax the time of two clergymen — 
 " recognised Mathematicians " or not — with this as a " curiosity ; " 
 and having said so, and answered it for them, I will beg to substitute 
 one worthy their calculation. 
 
i5 
 
 7, MORNINGTON ROAD, N.W., 
 
 1 6th June, 1870. 
 Dear Sir, 
 
 I am not surprised at your flood of replies to Mr. 
 Smith's mere schoolboy " problem." Having received both it and a 
 Letter from him, without knowing what the Rev. Mr. Whitworth's 
 " curiosity" of last year might have been, I thought well to suggest 
 one, but should not feel called on to spend anything on publishing 
 it. If, after Mr. Smith is settled, you like to introduce this as an 
 independent " curiosity," with my name, or (preferably) without, this 
 amended version will work. 
 
 I am, Sir, 
 
 Yours truly, 
 
 E. L. GARBETT. 
 
 A "CURIOSITY OF MATHEMATICS," SUGGESTED TO 
 THE REV. W. ALLEN YVHITWORTH. 
 A testator left a property to be thus employed, after being turned 
 into money : — His executors were to advertise in a certain number 
 of newspapers of the largest circulation, requesting the Archbishop 
 of the province to answer these three questions. The last chapter of 
 Scripture that contains these words, " If any man have an ear let him 
 hear," ends, he said, with a riddle concerning a " name," whose 
 Letters, taken as numerals in the arithmetic of that day, were to 
 notate a given number. The Archbishop to be asked to state (1) 
 whether the New Testament contained any name — that is, any 
 noun in the nominative case (whether substantive, adjective, or 
 participle) — that answered the said riddle ; (2) whether it contained 
 more than one ; and (3) what the name, if any, was, cr the names, 
 if more than one, were. On his Grace answering publicly all three 
 questions, the advertising was to cease, the remaining money to be 
 laid up six months, and then applied to such purpose as the said 
 Archbishop might direct, unless any of his answers had meanwhile 
 been proved false, but in that case paid to the person proving it so. 
 And unless or until his Grace might answer all three questions, the 
 advertising was to be repeated daily till a certain sum named was 
 
[6 
 
 expended. On this occurring the same questions were thenceforth 
 to be put to the Roman Catholic Archbishop alone, in the same 
 manner, and with all the same conditions ; and unless or until he 
 might answer, the advertisements were to continue till the whole 
 money were exhausted therein. 
 
 On execution of this will commencing, the Protestant Archbishop 
 set a clerk on Bruder's " Concordantiae Noir Testamenti " (Leipsic, 
 1842), to examine every noun occurring in the nominative, beginning 
 from A, working 8 hours per diem. The average time to find and 
 examine each was 3 minutes. Meanwhile, the daily cost of repeat- 
 ing the question was ^5. Presently he answered question 1 ; but 
 before he could answer 2 or 3, and when it appeared the next day 
 would have completed the work, the day came that, in accordance 
 with the will, the enqu;ry had to be transferred to the Romish 
 prelate. He, guessing that the Protestant had probably began 
 alphabetically and used the above Concordance, set two clerks on 
 the same book, but beginning from the end, or rather the last 
 possible noun ^as no word containing Y or W could answer, the 
 numeral meaning of either of these Letters being too high). As 
 these clerks checked each other, and were paid by job instead of 
 time, their joint progress was just thrice that of the Protestant ; and 
 presently, without their going through the book, their master was 
 enabled both to confirm the answer given to question 1, and answer 
 2 and 3. After six months, none of these answers being disproved, 
 he obtained the disposal of the remaining legacy. 
 
 PROBLEM. 
 Without knowing the sum left, or that ultimately thus applied, 
 what would have been saved from the advertising expense had the 
 Protestant got through his work a day sooner ? 
 
 E. L. Garbett. 
 
 Sir, 
 
 Mr. James Smith, under this heading, propounds in 
 your columns of Saturday last (p. 245) a " problem,'" addressed to 
 Mr. Whitworth and Mr. Gibbons ; and as he has done me the 
 honour to send me a copy of your paper, I presume he wishes for 
 
17 
 
 an answer from me. The " problem " is not a very formidable one, 
 any tyro can see that from the similarity of the triangles, 
 BN:NC::AD:DC::AB 2 :: BD 
 i.e., by the data : : 64 : 36 : : 16:9. 
 
 Your obedient Servant, 
 
 Alex. Ed. Miller. 
 24, Old Buildings, Lincoln's Inn, 
 June \3tJ1, 1870. 
 
 From the "LIVERPOOL LEADER," July 2nd, 1870. 
 CURIOSITIES OF MATHEMATICS. 
 
 to the editor of the liverpool leader. 
 
 Sir, 
 
 The past week has been very prolific of correspondence 
 on the above subject. I give the following, which I think will be 
 both amusing and instructive to your scientific readers : — 
 
 E. L. GARBETT to J. S. 
 
 Dear Sir, 
 
 As I pointed out to the Editor of the Liverpool 
 Leader, on receiving his paper of June 11, nothing more is necessary 
 to prove the ratio of B N to N C, which you demand on page 245, 
 than to observe that the parts A B D, B D C, into which you divide 
 ABC, are (by Euclid, Book 6, Prop. 8) similar thereto ; and again, 
 those of B D C— viz., B N I). D N C— are, for the same reason, 
 similar to the same. As you made A B : B C : : 4 : 3, this is also 
 (by Prop. 4 of same book) the ratio of the legs containing the right 
 angle in any of these 5 triangles. As B N : N D : : 4 : 3, and also 
 D N : N C : : 4 : 3 ; therefore, B N : N C in the duplicate ratio of 
 4 to 3, — that is, as 16 to 9. 
 
 This is also the ratio of A M to M B, and of A D to D C, in 
 s 
 
18 
 
 the figure constructed, as directed on page 245. But you now, on 
 page 262, describe a different figure. Instead of finding M, as 
 before, by dropping D M perpendicular to A B, you tell us this line 
 is divided into AM, MBin the ratio of 5 to 3 ; in which case M D 
 will not be a perpendicular— though this shifting of the point M 
 nowise affects your problem, for which O, M, and P, are wholly 
 superfluous. It makes your figure a new one, containing only six 
 right-angled triangles, while that of last week contained nine. 
 
 I cannot see why anyone should be at expense to advertise 
 proofs of mistakes like these, in which any schoolboy advanced 
 beyond Book I. or II would correct you ; and how could Professor 
 De Morgan ask the Athentzum to cover therewith paper due to its 
 readers' information or amusement ? 
 
 As it appears a clergyman began, and another attends to the 
 Liverpool " Curiosities of Mathematics," I gave the Editor, af er 
 answering you, a new " Curiosity founded on your late Catholic will 
 case," and the subject of your late 666-problemed friend. He can 
 bring that forward whenever he may think fit. Meanwhile, I expect 
 you will advertise whatever those you publicly called on may have 
 had to say to your problem. 
 
 I am, Sir, 
 
 Faithfully yours, 
 
 E. L. G. 
 
 7, MOKNINGTON ROAD, N.W., 
 
 20/^ June, 1870. 
 
 J. S. TO E. L. GARBETT. 
 
 Dear Sir, 
 
 I am in receipt of your favour ot yesterday, for 
 which I thank you, as it has led me to detect an omission in my 
 letter on page 262 of the Liverpool Leader. 
 
 In giving the method of constructing the geometrical figure on 
 which the problem is founded in the letter to the Athenceum, the 
 concluding sentence should have run thus -.—Bisect B C in P, and 
 from D draw the straight line D N perpendicular to B C, and, 
 
*9 
 
 therefore, parallel to A B, and join D P. It is hardly conceivable 
 that you could be ignorant of the fact, that the omission of the words 
 " bis est B C in P " was a mere lapsus. 
 
 Assuming you to have made this correction, may I ask you to 
 construct two diagrams— one by the method given on page 245 of the 
 Liverpool Leader, and the other by the method given on page 262 
 of that Journal. The geometrical figures represented by the two 
 diagrams will be similar in all respects, and I defy you to prove the 
 contrary, unless you are prepared to prove " Euclid at Fault" in his 
 theorem : Prop. 31, Book 3. It is sheer nonsense to say that by the 
 method of construction given on page 262 we get a figure in which 
 M D is not perpendicular to A B ; and worse than nonsense to say : — 
 " Though this shifting of the point M nowise affects your problem, 
 for which O, M, and P are wholly superfluous." 
 
 Referring to the diagram in my letter in the Liverpool Leader 
 of the nth inst., the following things are self-evident : — 
 
 BP+PN = BN; and, PC — PN = NC;andBN+NC = 
 2(B P) = B C, by construction. 
 
 M D = B N, and M B = D N, for they are opposite sides of the 
 parallelogram MBND; and it follows, that M D + N C = B C ; 
 but it does not follow that AMD and A B C are similar triangles, 
 or that A M D and D N C are similar triangles. 
 
 I have received a letter from the Rev. Geo. B. Gibbons, giving 
 a supposed solution of my problem. He, of course, intended it for 
 insertion in the Liverpool Leader, and it will appear in next Satur- 
 day's number. I may tell you that he makes the ratio between B N 
 and N C the same as you do— that is, as 16 to 9 ; but you are both 
 wrong. 
 
 I know that 666 is the number of the Apocalyptic beast, but I 
 do not know what you mean when you speak of " the subject of 
 your (my) late 666-problemed friend." 
 
 Now, 24 + 7 = 31 ; and 24 — 7=17 ; and B N : N C : : 31 : 17. 
 
 Proof: Let BP = 666. 
 
 Then : Because B C is bisected in P, 2(B P) = 2(666)= 1332 = BC. 
 
 Hence: B P + ^B P) = 666+ 194-25 = 860-25 = 6 N 
 
 And: PC— /^P C) = 666— 194-25 = 47175 = N C 
 
 But : 860-25 : 47 r 75 : : 3 1 : 17 : 
 
 .-. BN : N C : : 31 : 17. 
 
20 
 
 In the following way you might have made B N : N C : : 30 : 18, 
 and with a much greater show of plausibility than by your 16 to 9 
 ratio. 
 
 Let A B = 8 
 Then: j(A B) = f (8) = B C, by construction = 6. 
 KA Q = |(8) = M B, by construction = 3. 
 ABC and D N C are apparently similar triangles, and D N = 
 MB, by construction. 
 
 .-. f(D N) = f( 3 ) = 2-25 = NC, and B C — N = 6 — 2-25 = 375 . 
 
 375 
 225 
 
 Multiply the two terms of this ratio by 8, and we get the equiva- 
 lent ratio ||. But, B N is not to N C in the ratio of 30 to 18. 
 
 The concluding sentence of the last paragraph of your letter 
 shall have my attention. 
 
 I am, Sir, 
 
 Faithfully yours, 
 
 J.S. 
 Barkeley House, Seaforth, 
 June 21st, 1870. 
 
 E. L. GARBETT to J. S. 
 Dear Sir, 
 
 I have yours of yesterday, and the only thing it 
 calls on me to answer is, that by " your 666-problemed friend " I meant 
 him whose book you sent to Professor De Morgan, and of whom the 
 latter said that " if there were anything in Christianity he would 
 have been no fool.''* I forget his name, he who published two vol- 
 umes, therefore more than 666 puzzlements about the beast's name, 
 and never settled more than a bishopwhat that name was after all. 
 
 * This refers to the late Dr. Thorn of Liverpool, author of a work 
 entitled: — "The number and names of the Apocalyptic Beasts," but this 
 was not the book I sent to Professor De Morgan. Dr. Thorn was a 
 personal and intimate friend of mine, and was well known and highly 
 respected in my native town. Professor De Morgan made Dr. Thorn into 
 a paradoxer as well as myself, and coupled and devoted a whole chapter to 
 us, in his Budget of Paradoxes. " J. S. 
 
21 
 
 You can construct your two diagrams on the front and back of 
 
 tracing paper as large as the docks if you can get it (I advise that 
 
 the circle should be only 3^ diameters round), and, when you make 
 
 M correspond in both, I will procure and pay you, in 3^ days, 3^ 
 
 times the gold in Solomon's revenue. But, till then, I cannot pay 
 
 one more postage on the matter. — not one., observe,— and you owe 
 
 me 3ld. for showing the ratio of B N to N C. 
 
 Yours faithfully, 
 
 E. L. G. 
 
 7, MORNINGTON ROAD, 
 12nd June, 1870. 
 
 J. S. to E. L. GARBETT. 
 
 Sir, 
 
 Referring to the geometrical figure represented by the 
 diagram in my Letter in the Liverpool Leader of the nth instant, 
 let c denote the circumference of the circle ; let p denote the 
 perimeter of an inscribed regular hexagon ; let a denote the area 
 of the circle ; and let x denote the area of a regular dodecagon 
 inscribed within the circle. 
 
 Now, Sir, if you be " a reasoning geometrical investigator," a 
 
 Christian, and "no fool," you will— with the information I have 
 
 given you in my last communication— perform the following feats: — 
 
 First : Find the exact arithmetical value of the circumference of 
 
 a circle of diameter unity. 
 
 Second : Prove that DN:DP::/:c 
 Third : Prove that DN:DP::*:a. 
 
 Faithfully yours, 
 
 J. S. 
 Barkeley House, Seaforth, 
 i^rdjuney 1870. 
 
The Rev. GEO. B. GIBBONS to J. S. 
 
 My Dear Sir, 
 
 I duly received yesterday the Liverpool 
 
 Leader of 18th June. I notice a difference between the presented 
 
 "constructions " in the Papers of nth and 18th June. In the 
 
 former, M is fixed by your giving D M perpendicular to A B. In 
 
 A M 5 
 the latter, M is fixed by the given ratio „ __ = -■ The rest of 
 
 the construction being the same in both, these fixtures will not place 
 M in the same point. D M is no longer perpendicular to A B. 
 For, if D M be perpendicular to A B, making A M = 5 and B M = 
 3, we get AM x BM = DM 2 , or 15 = DM 2 , — that is, D M= ^15. 
 „ CB DM vT; 
 But 'AB = AM- "f notf. 
 
 I sent you the solution you asked, according to conditions of 
 
 nth June, in Liverpool Leader, and then £-££ = -. 
 
 B J\l 9 
 
 Werrington, Launceston, 
 22nd June, 1870. 
 
 Yours truly, 
 
 G. B. G. 
 
 J. S. TO the Rev. GEO. B. GIBBONS. 
 
 My Dear Sir, 
 
 I am in receipt of your favour of the 22nd 
 instant, which has had my careful attention. In giving a method of 
 constructing the geometrical figure on which my problem is founded 
 in the Letter to the Editor of the Athensum, as given in the Leadet 
 of the 1 8th instant, the last sentence should have run thus:— 
 Bisect B C in P, and from D draw a straight line D N perpendicular 
 to B C, and therefore, parallel to A B, and join D P. I can hardly 
 conceive it possible that you can have a doubt in your mind as to 
 the fact, that the omission of the words, bisect B C in P, was a mere 
 lapsus. 
 
23 
 
 I admit that there is " a difference between the presented ' con- 
 struction ' in the papers (Leader) of nth and 18th inst. ; " but, 
 whether the construction be made by one method or the other, the 
 result will be the same— that is to say, if diagrams be drawn by both 
 methods, the geometrical figures so constructed will be similar in all 
 respects. You say :— " In the former, M is fixed by your giving 
 D M perpendicular to A B. In the latter, M is fixed by the given 
 
 AM K 
 ratio p.— -^ = -• The rest of the construction being the same in both, 
 
 these fixtures will not place M in the same point. D M is no longer 
 perpendicular to A B." In this you are decidedly wrong. " If 'you 
 ca?it see it I can't help it, but the fact remains notwithstanding." 
 You next observe :— " For, if D M be perpendicular to A B, making 
 AM = 5 and B M = 3, we get A M x B M = D M* or 15 = D M 2 — 
 that is, D M = J 15" Now, I admit that, by Euclid (Prop. 13, 
 Book 6), D M is a mean proportional between A M and B M, and 
 = ^15 when AM+BM=AB= 8. But your conclusion, 
 
 -X~5 = x4* = - - not 4'" is a & ross absurdity, for, C B : A B 
 : 6 : 8, by construction. 
 
 Referring to the geometrical ligure represented by the diagram 
 in my Letter in the Liverpool Leader of the nth instant, it is self- 
 evident that BN + NC = BC;and because B C is bisected in P, 
 B P + P N = B N. and P C — P N = N C. You may tell me that 
 these are simply truisms, whatever be the value of P N. Granted ! 
 But can you find the exact arithmetical value of P N ? Can you 
 prove that B N + N C = 6, when A B = 8, by your analogy or pro- 
 portion, B N : N C : : 16 : 9? Pray, Sir, let there be no shirking of 
 these questions ! 
 
 Let B C = 48. 
 Then : V - 24 = B P 
 
 A (B ?) = 1 -^ = 7 = P N 
 
 24 
 
 BP + PN = 24 + 7 = 3i = BN 
 
 P C— PN = 24—7 = 17 = NC 
 .-. BN :N C ::3i : 17 
 .-. B N + N C = 31 + 17 = 48 = the given value of B C. 
 
2 4 
 
 Again : Let B C = 50. 
 
 Then : » T ° = 25 = B P 
 
 v T (B P) = 7 -^ S = 7-291666... = P N 
 24 
 
 BP+PN = 25+ 7'29i666. . = 32-291666...;= B N 
 
 And, B N : N C: : 31 : 17 
 
 But, 31:17:. 32-291666... : 17-708332. . . 
 
 .-. NC= 17708332... 
 
 .". BN + NC = 32-291666... + 17-708332... =49-999998... 
 
 and is less than 50, the given value of B C ; but it does not follow, 
 
 that B N is NOT to N C in the ratio of 31 to 17. 
 
 Now, if, in the right-angled triangle ABC the sides A B and 
 B C be 8 and 6, a straight line drawn from the right angle B 
 perpendicular to its opposite side A C will divide the triangle ABC 
 into two similar triangles A D B and BDC, and these triangles will 
 be similar to the whole triangle ABC, and B D will be a mean propor- 
 tional between AD and DC ; but it does not follow, because the two 
 sides that contain the right angle in a right-angled triangle are in 
 the ratio of 8 to 6, that under all circumstances a straight line drawn 
 from the right angle perpendicular to its opposite side will divide the 
 triangle into two similar triangles. 
 
 For example : Let A B = 8. Then : +(8) =64 = A D : |(A B) = 
 f(AD) = 4-8 = BD : |(BD) = 3-6 = DC (in your letter of the 14th inst. 
 you make A D = 6-4 and D C=3"6), and B D C is a right angled 
 triangle, of which the sides B D and D C, which contain the right 
 angle, are in the ratio of 8 to 6 : therefore, by Euclid, Prop. 47, 
 Book i,(BD 2 + DC 2 j=(4'8 2 + 3"6 2 ) = 23-04 + 1296) = 36 = B C* ; 
 therefore, */36 = 6 = BC. Now, DN = MB = 3 ; and BN = M B = N /i5 
 when A B = 8 ; and by Euclid, Prop. 13, Book 6, D N is a mean 
 proportional between BN and N C— that isBN:ND::ND:NC; 
 or > \f i 5 '■ 3 : : 3 : N C. But, ^15 : 3 : : 3 : ^54, and you will find 
 that this would make B C greater than 6, which is impossible. 
 Surely it cannot be necessary to proceed further. Well, then, from 
 what precedes, the following questions arise:— Is it Euclid that is at 
 fault ? Or, is it mathematics that are at fault ? Or, is geometry 
 " a mockery, a delusion, and a snare f" 
 
 The fact is, Euclid and mathematics arc both at fault. Euclid's 
 Propositions 8 and 13, Book 6, are not of general and universal 
 
25 
 
 application in practical geometry. These Propositions are incon- 
 sistent with each other, and both are inconsistent with many other 
 Propositions of Euclid when tested by "that indispensable instrument 
 of science, Arithmetic." 
 
 I presented you with a copy of my work, " The Geometry of the 
 Circle ;" and, in conclusion, may refer you to Diagram 8, and my 
 Letter to the Rev. Professor Whitworth, page 140. I may also refer 
 you to the Letters to J. M. Wilson, Esq., pages 334, 335, &c. 
 
 Faithfully yours, 
 
 J. S. 
 June 25th, 1870. 
 
 I think I have put your contributor, the Rev. Geo. B. Gibbons, 
 
 in a fix, out of which he will rind it difficult to wriggle. 
 
 I am, Sir, 
 
 Yours obediently, 
 
 James Smith. 
 Barkeley House, Seaforth, 
 
 June 27th, 1870. 
 
 From the " LIVERPOOL LEADER," July c/th, 1870. 
 CURIOSITIES OF MATHEMATICS. 
 
 to the editor of the liverpool leader. 
 
 Sir, 
 
 The Editor of the Athenceum and Professor De Morgan 
 declined to solve the problem given in my Letter in the Leader of 
 the 1 ith June, as requested in my communications to them of that 
 date. This led me to write them again. 
 
 J. S. to the Editor of the Athenaeum. 
 
 Sir, 
 
 In your publication of this day I find no reference to the 
 problem I gave in my Letter in the Liverpool Leader of the nth 
 
26 
 
 inst., a copy of which I sent you. The following is the solution of the 
 problem, and proves the necessity of including " arithmetical 
 considerations " in the study of Geometry. 
 Let A B = 8. 
 Then : B C = | (A B), by contruction = 6. 
 But : B C is bisected in P. 
 
 .-. BP = PC = 3. 
 Hence : B P + ^ (B P) = B N, and, PC-^(PC) = N C. 
 .-. B N = 3-875, and, N C = 2-125. 
 .-. BN + NC = 3-875 + 2-125 = BC = 6. 
 
 . " . -. ^ expresses the ratio between B N and N C. 
 
 Multiply the two terms of this ratio by 8, and we obtain the 
 equivalent ratio §} ; and it follows, that B N is to N C in the ratio 
 of 31 to 17. The ratio is expressed in whole numbers, and with 
 arithmetical exactness. Q. E. D. 
 
 Now, Sir, conceive a dozen of the Members of the Royal Society 
 of England, a dozen Members of the Royal Institute of France, and 
 a dozen of the leading Members of the British Association for the 
 Advancement of Science, with Professor De Morgan (the greatest 
 Mathematician in Europe, according to my correspondent R. F. 
 Glaister, Esq.) as their secretary, assembled in solemn conclave to 
 solve my problem. They would no doubt decide that the solution 
 of the problem is impossible, aye, and would probably say, just as 
 impossible as to square the circle, or find the exact ratio of diameter 
 to circumference in a circle. 
 
 But, Sir, I have given you the solution of the problem, and no 
 Mathematician in the world can controvert it; and if you fail to 
 take the earliest opportunity of giving the problem and its solution 
 in the columns of the Atheticeui/i, will you play the part of an honest 
 scientific journalist ? If you find my solution to be irrefragable and 
 decline to admit the fact, what will you be ? Do you suppose that 
 the AthencBum can for any length of time stifle scientific truth, and 
 "bolster up" false systems, by maintaining a dogged silence with 
 regard to unpalatable truths, because they happen to jar with long- 
 cherished prejudices ? Magna est Veritas, et prcevalebit. 
 Yours respectfully, 
 
 i<S/// J ion', 1X70. J. S. 
 
2 7 
 J. S. to PROFESSOR DE MORGAN. 
 
 Sir, 
 
 In the Liverpool Leader of last week I gave an 
 
 interesting and novel problem for solution, and I sent you a copy of 
 
 the paper. As I have no doubt you found yourself incompetent to 
 
 comply with the request I made in my Letter to you of the nthinst., 
 
 I now give you the solution of the problem. (See the foregoing 
 
 Letter.) 
 
 Now, Sir, if you can " upset " my solution of the problem and 
 won't, what are you ? If you find it to be irrefragable and hesitate 
 to admit the fact, what will you be ? 
 
 I should like to see you grapple with this Letter in a final 
 supplementary chapter to your " Budget of Paradoxes." 
 Your dutiful tutelary, 
 
 J. s. 
 i&kjune, 1870. 
 
 The Rev. Professor Whitworth and the learned Professor De 
 Morgan were too cunning to be caught in the trap I set for them in 
 my Letter in the Leader of the nth June last. Not so Gibbons, 
 Garbett, and Miller, and I may tell you that another gentlemen well 
 known as the recognised "Mathematical authority" of Liverpool 
 swallowed the bait, and was caught in the trap. All these gentlemen 
 are agreed in making B N to N C (see diagram) in the ratio of 16 to 
 9 ; but they all fail to perceive, that if orthodox theories were true, 
 the solution of my problem would be an utter impossibility. Pro- 
 fessor De Morgan's Mathematical knowledge will not enable him to 
 solve the problem on orthodox theories, but his temper will not 
 permit him to take the humiliating step of admitting it. 
 
 If De Morgan have any self-respect, he will either prove my 
 solution of the problem to be fallacious, or withdraw, and apologise 
 for, the scurrilous attacks he has made upon me from time to time 
 in the columns of the A thence ■um. 
 
 If I were to attempt to give my problem and its solution in 
 Section A of the British Association at its forthcoming Meeting, I 
 have no doubt I should be grossly insulted, as I have been over and 
 over again in the last ten years, when I have submitted Papers 
 which 1 was wishful to read in that Section. 
 
28 
 
 In the Athencsum of September 9, 1865, or in the Transactions 
 of the British Association for that year, you will find the President's 
 (Professor Phillips) inaugural address, which concludes as follows :— 
 
 " Here, indeed, is the stronghold of the British Association. 
 Wherever and by whatever means sound learning and useful know- 
 ledge are advanced, there to us are friends. Whoever is privileged 
 to step beyond his fellows on the road of scientific discovery will 
 receive our applause, and, if need be, our help. Welcoming and 
 joining in the labour of all, we shall keep our place among those 
 who clear the roads, and remove the obstacles from the paths of 
 science ; and whatever be our own success in the rich fields that lie 
 before us, however little we may now know, we shall prove that, in 
 this our day, we knew at least the value of knowledge, and joined 
 hands and hearts in the endeavour to promote it." 
 
 I shall leave it to your readers to reconcile the fulsome and 
 extravagant professions of Professor Phillips, with the common 
 practice of the Association. I know from personal experience, and 
 from the treatment I have seen others meet with in Section A of the 
 British Association, that, if a gentlemen offer to read a Paper, and 
 his views are known to jar with the long-cherished prejudices of 
 Mathematicians, his Paper will certainly be rejected ; and if he 
 attempt to take part in a discussion on a Paper, he will as certainly 
 be insulted. 
 
 I had written so far when this day's Leader came into my 
 hands. On reading the erratum of Mr. E. L. Garbett, I was led to 
 address the following short letter to that gentleman :— 
 
 Sir, A 
 
 ABC represents a right-angled 
 triangle, of which B is the right angle. 
 If A B = it + a-' + & 
 And AC = s' + £ + x' 
 How many times is A B 2 contained in the sum 
 of the squares of the three sides of the triangle ? 
 
 This cannot be a " puzzlement " to you, or 
 to a bishop, and I shall not object to pay 2d. 
 postage for an answer to the question. 
 
 Faithfully yours, 
 
 J. S. 
 
2 9 
 
 Surely Mr. E. L. Garbett is sufficient of '• a Christian and a 
 Gentleman" to answer my question ! Will he expound my " riddle ?" 
 Nous verrons ! 
 
 I am, Sir, 
 
 Faithfully yours, 
 
 James Smith. 
 Barkeley House, Seaforth, 
 2nd July, 1870. 
 
 From the "LIVERPOOL LEADER," July i6th, 1870. 
 CURIOSITIES OF MATHEMATICS 
 
 to the editor of the liverpool leader. 
 
 Sir, 
 
 For several weeks past, Letters of mine have appeared 
 in your valuable journal under the above title, and with this com- 
 munication I shall conclude the series. 
 
 Again, referring to the geometrical figure represented by the 
 diagram in my Letter in the Leader of the 1 ith June : 
 Let A B the diameter of the circle = 8. 
 Then : B C = f (A B), by construction = 6. 
 
 A O + O M = A M = ! (A B), by construction 5 
 M B = |(0 B), by construction = 3. 
 But B C is bisected in P. 
 .\BP = PC= MB = 3. 
 
 By Euclid : Prop. 8, Book 6. 
 A B x A M = 8 x 5 = 40 = A D- 
 .'.AD= J40. 
 
 ABxMB = 8x3 = 24 = DB 2 
 .-. D B = J24 
 
 AMxMB = 5X3=ij = MD» 
 .-. M D = JI$. 
 
30 
 
 By Euclid : Prop. 47, Book 1 . 
 
 A B 2 + B C 2 = 8 2 + 6- = 64 + 36 = 100 = A C 2 
 .-. J100 = 10 = A C ; that is = A D + D C 
 2 2 
 
 A D 2 + D B 2 - ^40 + J24 = 40 + 24 = 64 - A B 2 
 
 .-. AB = 76 4 "= 8 
 
 2 
 
 B C 2 — D B 2 = 6 2 — ^24"= 36 - 24 = 12 = D C 2 
 
 .-. D C = J 12. 
 
 Now, D N = M B for they are opposite sides of the parallelogram 
 
 MBND; and for the same reason M D = B N. 
 
 Hence : *(BP) = £(3) = 7 *^ 3 - = ^ = -§75 = NP. 
 
 .-. N P + P B = -875 + 3 = 3-875 = N B . 
 By Euclid : Prop. 47, Book 1 . 
 DN 2 + NB 2 =3 2 +3-875 2 = 9+ 15-015625 -24-015625 =D B 2 , 
 and is greater than 24. 
 
 But, D N 2 + N P 2 = 3 2 + -875 2 = 9 + 765625 = 9765625 = 
 DP 2 . 
 
 .-. V9765625 =3-125 = DP. 
 
 By Euclid : Prop. 12, Book 2. 
 D P 2 + P B 2 + 2 (P B x P N) = B D 2 
 .-. 3 -i25 2 + 3 2 + 2(3 x -875) 
 
 - 9-765625 + 9 + 5-25 = 24-015625 
 = D B 2 , and is greater than 24. 
 By Euclid, Prop. 8, Book 6, M D = J 15 when- A B the diameter 
 of the circle = 8 ; and by Euclid, Prop. 13, Book 6, M D is a mean 
 proportional between A M and M B ; therefore, A M : M D : : M D 
 : M B— that is, 5 •' ^15 : : J15 : 3 ; therefore, M B = 3. By Euclid, 
 Prop. 13, Book 6, N D is a mean proportional between B N and N C 
 therefore, B N : N D : : N D : N C— that is, ^15 : 3 : : 3 : Js m 4 
 therefore, NC= Js'4- But, ^15 = 3-872..., and Jy4 = 2-323... 
 therefore, B N + N C = 3-872... + 2-323... »= 6-195... and is greater 
 than BN + NC —that is, greater than 6, the known value of the 
 line B C when A B - 8. 
 
 Again : By Euclid, Prop. 13, Book 6, B D is a mean proportional 
 between A D and D C: and by Euclid, Prop. 8, Book 6, A D = ^40, 
 
3i 
 
 and B D = V 2 4 when A B - 8 ; therefore, AD:BD ::BD:DC 
 — that is, ^40 : J 24 : ^24 : ^144; therefore, D C = N / I 4'4- But, 
 ^40 = 6-325. .. and N / I 4"4 = 3794--- ! therefore, AD + DC r 
 6'325... + 3794... = io - H9..., and is greater than AD + DC — 
 that is, greater than 10, the known value of the line A C when A B 
 = 8, and B C = 6. 
 
 The Rev. Geo. B. Gibbons, in his Letter of the 14th June, 
 which appeared in the Leader of the 25th June, makes A D = 6'4 
 and D C = y6, when AB = 8; and, by Euclid, Prop. 13, B D is a 
 mean proportional between A D and D C ; therefore, A D x D C 
 = 6-4 x 3 - 6 = 23*04 = B D 2 , and is less than 24, the value of B D 2 
 as ascertained by Euclid, Prop. 8, Book 6. 
 
 According to Gibbons, Garbett, Miller, and my Liverpool 
 friend, B N is to N C in the ratio of 16 to 9. 
 
 Let B C the hypothenuse of the right-angled triangle B D C 
 = 25. 
 
 If these gentlemen be right in their ratio, then, according to 
 common sense, B N = 16 and N C = 9. By Euclid, Prop. 2, 
 Book 2, (B C x B N) + (B C x N C) = (25 x 16) + (25 x 9) = 
 (400 + 225) = 625 = B C 2 ; therefore, ^625 = 25 = B C. But, 
 when the whole line B C = 6, I have proved that B N is to N C in 
 the ratio of 31 to 17, making BN= 3-875 and N C = 2-125 ; and 
 by Euclid, Prop. 2, Book 2, (B C x B N) + (B C x N C) = 
 (6 x 3-875) + (6 x 2-125) = (23-25 + 12-75) = 36 = B C 2 ; there- 
 fore, J 36 = 6 = B C. 
 
 Out of these facts the following questions arise:— Is B N to N C 
 in the ratio of 16 to 9 ? Is B N to N C in the ratio of 31 to 17 ? 
 Or, are both these ratios fallacious ? It is self-evident that both 
 cannot be true. These matters are within the scope, and ought 
 to command the attention of, the British Association ; but if any 
 gentleman were to attempt to deal with them in Section A of the 
 Association, he would most assuredly be insulted. 
 
 I had written so far when I received the following communi- 
 tion : — 
 
 Sir, 
 
 On the other side is a solution of your problem, without 
 calling in the aid of the 6th Book, or the doctrine of similar triangles. 
 I think there is no flaw in the process. 
 
32 
 
 Solution. 
 
 By hypothesis, A B = 4 a and B C = 3 a. 
 
 . • . By Euclid, Prop. 47, Book 1, A C = 5 a. 
 
 , . ' AB x BC . AC x DB 
 
 Area of triangle A B C = also = — — 
 
 .-. AB x BC = AC x DB 
 
 4. a x 3 a = 50 DB 
 . • . D B = V a ' 
 
 But B D C is a right-angled triangle. 
 .•.BC s ~BD a =DC ! 
 
 9« 2 — Vt « 2 = dC2 
 
 BC x DN 
 also = 
 
 i*« a 
 
 -DC 
 
 
 
 ** 
 
 = DC 
 
 
 
 Area of triangle 
 
 BDC= BD 
 
 x DC 
 
 BD 
 
 x DC = BC x DN 
 
 
 V w 
 
 1. x 4 (fl) = 
 
 3a x DN 
 
 
 DN 
 
 = «(*) 
 
 
 
 BN 
 
 - V(BD ! 
 
 — DN 2 ) = V 
 
 (W « 2 ) 
 
 And, 
 
 NC =30 
 
 — If* = H* 
 
 
 BN : 
 
 N C : : 48 
 
 : 27 : : 16 : 9. 
 
 
 
 
 Yours, 
 
 &c, 
 
 X. Y. Z. 
 
 4/A 7?//y, 1870. 
 
 X. Y. Z. arrives at the same conclusion as Gibbons, Garbett, and 
 Miller as to the ratio of B N to N C, but by a somewhat different 
 " process." X. Y. Z. and Gibbons concur, in making A D = 6-4 and 
 D C = 3-6, when A B = 8. But, D N = 3 by construction, and by 
 computation BN = 3-875, and N C = 2-125, when BC = 6. Now, 
 the "process" of X. Y. Z. would make DN = 2-88, B N = 3*84, and 
 N C = 2- 16. But, B N = M D, and D N = MB ; and, by Euclid, 
 Prop. 8, Book 6, M D = Ji^ when MB = 3. Can B N and M D, 
 which are opposite sides of the parallelogram M B N D, have 
 different arithmetical values ? Can D N and M B, which are also 
 opposite sides of the parallelogram M B N D, have different arith- 
 metical values. Gibbons fancies he gets over this difficulty, by 
 boldly asserting that MD is not perpendicular to AB, and so upsets 
 
33 
 
 M 
 
 the fixed value of A M, making AM = 5*12, and consequently 
 M B = 2-88. X. Y. Z. carefully omits all reference to the parallelo- 
 gram M B N D, and this parallelogram is essential to the proof of the 
 correctness of my solution of the problem. 
 
 The foregoing remarks will be plain enough to any honest 
 mathematician, if he give what follows his careful consideration. 
 
 Diagram. 
 With A B as perpendicular, 
 describe the right-angled triangle 
 ABC, making the sides A B and 
 B C in the ratio of 8 to 6. 
 Divide A B into two parts, A M 
 and M B, so that A M shall be to 
 M B in the ratio of 5 to 3. From 
 the point M draw the straight 
 line M D, perpendicular to A B, 
 and, therefore, parallel to B C, 
 to meet A C the hypothenuse of 
 the triangle A B C in the point D, 
 and join D B. From the point 
 D draw the straight line D N 
 perpendicular to B C, and, there- 
 fore, parallel to A B, to meet B C the hypothenuse of the triangle 
 B D C in the point N. Bisect B C in P, and join D P. 
 
 In the construction of this geometrical figure we avoid the 
 introduction of a circle and all reference to a circle. 
 
 Let B C the hypothenuse of the right-angled triangle B D C = 
 X 1 s ' — that is, the number of the Apocalyptic beast = 666. 
 
 It is self-evident that the whole line B C is divided into two 
 equal parts B P and P C, and two unequal parts B N and N C. It 
 is also self-evident that B N is divided into two unequal parts B P 
 
 and PN. 
 
 Problem. 
 
 Find the exact ratio expressed in whole numbers between B P 
 and P N, and between B N and N C. 
 
 Solution. 
 BC _ 666 
 2 ~ 2 
 
 333 
 
 BP 
 
34 
 
 A(BP) = 3 ^ = 13-875, and, 7(i3-87S) = 97-125 = PN 
 
 .-. BP + PN = 333 + 97-125 = 430-125 = BN 
 But, B C is bisected in P. 
 
 .-. PC — PN = 333 — 97-125 = 235-875 = NC 
 . ■ . BN + NC = 43o-i25 + 235-875 = 666 =the given value of BC. 
 Hence : B P : P N : : 24 : 7 
 that is, 333 : 97-125 :: 24 : 7 
 
 and, BN : NC 1:31: 17 
 
 that is, 430-125:235-875 1:31 : 17 
 And it follows, that D N : D P : : ft : c, where ^ denotes the perimeter 
 of a regular hexagon, and c denotes the circumference of its circum- 
 scribing circle. I think X. Y. Z. will find no flaw in the process. 
 
 Saturday afternoon's post brought me the following extraordinary 
 communication : — 
 
 185, Islington, Liverpool, 9th July, 1870. 
 
 Dear Sir, 
 
 Before you republished my articles on the 
 " Curiosities of Mathematics," I informed you that 1 valued the 
 copyright at twenty guineas for each article, and that (unless you 
 immediately accepted some lower terms which I then proposed) I 
 should expect payment at that rate. 
 
 You did not accept the terms which I proposed, but you did 
 deliberately use my property with full cognizanceof the price which 
 1 put upon it, (as you have since admitted in one of your Letters to 
 the Leader.) 
 
 I am happy to be assured that no question can arise as to the 
 intrinsic value of my papers. I had a right to put what price I chose 
 upon them : just as you might put any fancy price on any article of 
 yours which I desired to make use of. As you were cognizant of 
 my price before publication, I have no hesitation in asking you to 
 remit to me the full amount. 
 
 I may add, that if you desire any more articles of the same 
 character at the same price, I shall be happy to supply them, as I 
 
35 
 
 am anxious to raise funds to pay off debts on my church and 
 schools. 
 
 I am, dear Sir, 
 
 Yours faithfully, 
 
 W. Allen Whitworth. 
 
 This bark seems very like the bark of a cur. Mr. Whitworth is 
 quite at liberty to bite, by instructing his solicitor to take proceed- 
 ings against me to enforce payment of his demand. I have given 
 another problem in this Letter, and I can assure Mr. Whitworth that 
 certain leading members of "The British Association for the 
 Advancement of Science " would be delighted to find an article of 
 his in the Liverpool Leader, proving, by means of orthodox and 
 recognised systems, that my solution of the problem is " a. mocke?y, 
 a delusion, and a snare!' If Mr. Whitworth will insert such an 
 article, I shall be happy to subscribe twenty guineas to a fund, to 
 enable him " to pay off the debts on his church and schools." 
 
 Mr. Gibbons has not replied to my Letter of the 25th June, which 
 appeared in the Leader of the 2nd inst., neither have I a reply from 
 Mr. Garbett to my letter to him which appeared in your last week's 
 number. It would seem as if both had become " dumb dogs" that 
 " cannot bark." Both are silenced,, but neither has the mathematical 
 honesty to throw up the sponge, and admit that he is beaten in a fair 
 battle. In this Letter I have given the British Association something 
 worthy of attention at its forthcoming meeting ; but I may tell you 
 that, as a confraternity, I have found the members of that body to 
 be men possessed of heads so crammed with uncommon sense, 
 " that there is not a cranny left for reasoning to get in at" and I 
 shall not again lay myself open to insult by offering to read a Paper 
 in Section A. 
 
 In conclusion : It will be self-evident to any reader, if only a 
 school-boy, that Gibbons, Garbett, Miller, and X. Y. Z. (why had the 
 latter not the moral courage to give his name?) make the ratio 
 between the lengths of two unequal lines the same as the ratio 
 between the areas of squares on two other unequal lines in the 
 diagram. Nothing can be more absurd, and yet such is the 
 conclusion arrived at by the Mathematician's application of Mathe 
 matics to Geometry. It is sufficient that I have caught in my trap 
 
36 
 
 five " blind mice," and I shall not notice any further Letters, whether 
 public, private, or anonymous, unless and until your contributors 
 Whitworth and Gibbons have either admitted the correctness of my 
 solution of the problem given in the Leader of the nth June, or 
 proved it to be fallacious. 
 
 I am, Sir, 
 
 Yours obediently, 
 
 James Smith. 
 Barkeley House, Seaforth, 
 nth July, 1870. 
 
 From the "LIVERPOOL LEADER," July 23RD, 1870. 
 CURIOSITIES OF MATHEMATICS 
 
 to the editor of the liverpool leader. 
 
 Sir, 
 
 The Letter under the above title, which appeared in 
 the Leader of Saturday, I intended to be the last of the series, but 
 my correspondence of last week is of such importance, that I must 
 ask you to give it a place in your next number. 
 
 . ALEX. ED. MILLER TO J. S. 
 
 Sir, 
 
 As you have mentioned my name in your Letter to the 
 Leader of Saturday, perhaps you will be good enough to insert the 
 following reply to your " solution" : — 
 
 P N is not = & (B P), but = & (B P). 
 For, BN x BC= BD 2 
 
 . BD AB .„ 
 and 'DC= BC = * 
 
 .-. DC 2 = A(BD 2 ): BC* = BD*(i * A) = fMBD 2 ) 
 .\BN = H(BC)=$!(BP) 
 
 3N = (Jf — i)BP = &(0P) 
 Now, if B C = 6 
 
37 
 
 BN = & x 6 = 3-84 
 NC = 5 »3 x 6 = 2-16. 
 
 Q. E. D. 
 
 In the question given in Saturday's Leader in Greek numerals, 
 I suspect that s' is a misprint for ?'. If s' is right, the answer is 
 3-8763 very nearly. If <?' is right, the answer is = 3*125. 
 
 I cannot guess what " orthodox theory " would render the 
 solution of the problem given in June 11 an utter impossibility ; to 
 me it seemed a very simple deduction from Euclid, Prop. 47, Book 
 1, and Prop. 8, Book 6. 
 
 Your obedient servant, 
 
 Alex. Ed. Miller. 
 nth July, 1870. 
 
 J. S. TO ALEX. ED. MILLER. 
 
 Sir, 
 
 In reply to your favour of yesterday, I beg to inform 
 you that the only Letter of yours to the Editor of the Liverpool 
 Leader that I have seen is the one that appeared in that Journal of 
 the 25th June. You may have sent him the supposed solution of my 
 problem that you now give me, but it is the first time I have seen it. 
 
 Let A B be a straight 
 line divided into two unequal 
 parts A M and M B in the 
 ratio of 5 to 3. With A B as 
 perpendicular describe the 
 right-angled triangle ABC, 
 making A B to B C in the 
 ratio of 8 to 6. From the 
 point M draw a straight line 
 perpendicular to A B, and, 
 therefore, parallel to B C, to 
 meet A C the hypothenuse of 
 the triangle ABC in the 
 point D, and join D B. From 
 the angle D draw a straight 
 
38 
 
 line parallel to A B, and therefore, perpendicular to B C, to meet 
 B C the hypothenuse of the right-angled triangle B D C in the point 
 N. Bisect B C in P, and join D P. 
 
 If you bisect A B in O, and with O as centre and O A or O B 
 as radius describe a circle, you will find that the circumference of 
 the circle will not cut A C the hypothenuse of the triangle A B C in 
 the point D, but in a point nearer C* 
 
 Let AB =8. 
 Then : By construction : 
 
 f(AB) = |(8)= 6 = BC 
 f(AB) = f (8) = 5 = AM 
 and,|(AB) = f (8)= 3 = MB 
 But, B C is bisected in P. 
 .-. BP= PC = MB = 3. 
 
 By Euclid, Prop. 8, Book;6. 
 A M x M B = 5 x 3 = 15 = M D 2 
 But B N = M D, for they are opposite sides of the parallelogram 
 MBN D. 
 
 .-. B N = x /i5 = 3-872. &c, and is greater than 3-84, which you 
 make the value of B N in the figure in my Letter to the Leader of 
 June 11. 
 
 Hence : BN=||x6= 3-875 
 
 N C = H x 6 = 2-125 
 .-. BN : NC : .-3-875 : 2-125 
 .-. 31 : 17 : : 3-875 : 2-125 
 And the product of the means is equal to the product of the 
 extremes. 
 
 You will find that AMD and D N C are not similar triangles. 
 The symbol to which you refer is somewhat unshapely, but 
 stands for the Greek numeral 6, and your answer 3-125 is correct. 
 
 I yesterday gave the Editor of the Leader the MSS. of a Letter 
 which will appear in next Saturday's number, of which I will send 
 you a copy. With that Letter I intended to conclude the series on 
 "Curiosities of Mathematics;" but if, after perusing it, you still 
 wish your Letter of yesterday to appear in the Leader, please say 
 so, and I will have it inserted with my reply in the following week's 
 number. 
 
 Your obedient Servant, 
 12//* July, 1870. James Smith. 
 
 * This paragraph contains an absurd assertion, "without any attempt 
 at proof." J. S. 
 
39 
 
 P.S. — I think I sent you a copy of the Liverpool Leader of the 
 2nd inst, but I suspect that you have not read a Letter of mine that 
 appeared in it. I send you another copy by this post. 
 
 ALEX. ED. MILLER to J. S. 
 
 Sir, 
 
 I am obliged by yours of yesterday. I only sent a 
 solution of your problem of June n, which is the one you allude to. 
 My Letter of Monday was called forth by yours which appeared in 
 the Leader of July 9, in which you say that " Gibbons, Garbett, and 
 Miller fell into the trap " you had set by answering that problem by 
 the proportion 16:9. I still maintain that 16 : 9 is the true answer 
 to the problem of June 11, and I should wish my Letter of Monday 
 to appear, unless you formally and publicly withdraw the expression 
 above quoted. To-day you send me a totally new construction, in 
 which B D C is no lo nger a right-angled triangle, and therefore your 
 solution is wrong, because AMxMB is not = MD 2 (BDA not being 
 a right angle), and therefore BN, which is = M D, is not ^15. If 
 it were, you would have B N : N C : : *JT$ : ( J 5 — J 3) J3> which 
 cannot be expressed in whole numbers. You yourself practically 
 admit that B D C is not a right angle when you say, the circle 
 described with A B as diameter will not meet A C in D, but a point 
 somewhat nearer C. That is true. Let that point be E, and join 
 B E. Now, B E A is the angle in a semi-circle. 
 
 .-. BEDisa right-angled triangle. 
 .-. BDCis less than a right-angle. 
 .-. M D 2 is greater than AM x M B. 
 The true answer to your new diagram, which is even simpler than 
 the original one, is : — 
 
 B N : N C : : A D : D C : : A M : M B : 5 : 3. Q. E. D. 
 
 If you publish your Letter of yesterday, please add this to it ; if 
 not, you can do as you please about it. 
 
 Your obedient Servant, 
 
 Alex. Ed. Miller. 
 13th July, 1870. 
 
40 
 J. S. TO ALEX. ED. MILLER. 
 
 Sir, 
 
 I am obliged by yours of the 13 th inst., but not more 
 obliged than surprised. The truth is, you attempt to reconcile 
 Geometry with your notion of Mathematics, losing sight of the fact 
 that Mathematics must necessarily harmonise with the exact science 
 of Geometry ; and when rightly applied, can never be inconsistent 
 with practical Geometry. 
 
 Take the figure in my Letter to you of the 12th inst. From the 
 angle A draw a straight line parallel to D B to meet another straight 
 line drawn from the angle B parallel to A D in a point F, and so 
 construct a rectangular parallelogram AFBD. AB will be a 
 diagonal of the parallelogram and divide it into two similar and 
 equal right-angled triangles. Again : From the angle B draw a 
 straight line parallel to D C to meet another straight line drawn from 
 the angle C parallel to B D in a point G, and so construct a 
 rectangular parallelogram D B G C. B C will be a diagonal of the 
 parallelogram and divide it into two similar and equal right-angled 
 triangles. Surely these facts will convince you of the absurdity of 
 your assertion, that " B D C is less than a right angle ; " and I 
 cannot help thinking you will see how you and others have been 
 catiglit in a trap. 
 
 It is assumed by Mathematicians that straight lines drawn to 
 points in A C, from any point in A B parallel to B C, and from any 
 point in B C parallel to A B, will produce two right-angled triangles 
 similar to the triangle ABC. This is not true, and you will find 
 that A M D, ABC, and DNC are not similar triangles in the 
 figure in my Letter of the 12th inst. 
 
 I have never said that any Proposition of Euclid is not true 
 under any circumstances. What I have said, and what I maintain, 
 is, that certain Propositions of Euclid are inconsistent with certain 
 other Propositions, and are, therefore, not of general and universal 
 application in practical Geometry, and, consequently, are not true, 
 under all circumstances. 
 
 Now, with reference to the figure in my Letter in the Liverpool 
 Leader of June 11, you might have made out, by a certain method of 
 computation, that AM = 5-12 and BM = 2'SS when AB the 
 
4i 
 
 diameter of the circle = 8 ; making A D = 6%, D C = 3-6, BN = 
 3-84, and N C - 2*16 ; and, if this were true, it would follow, that 
 AMD, ABC, and D N C would be similar right-angled triangles, 
 having the sides that contain the right angle in the ratio of 3 to 4. 
 Therefore, A B 2 xBD = (AD + DC) 2 
 and, D B 2 x D C 2 = (B N + N C) 2 
 
 .-.DP 2 x P B 2 = 2 (P B x P N) =DB ! 
 From these facts other results follow, which I may leave you to 
 trace out. 
 
 In conclusion : Your Letters and my answers shall be inserted 
 in the next number of the Liverpool Leader. 
 
 Your obedient Servant, 
 
 J.S. 
 16th July, 1870. 
 
 We shall see what Mr. Miller makes of this communication. 
 If he answer it, I shall have another Letter for you, with which I 
 shall certainly conclude my series on " Curiosities of Mathematics." 
 I am, Sir, yours obediently, 
 
 James Smith. 
 Barkeley House, Seaforth, 
 1 %th July, 1870. 
 
 From the "LIVERPOOL LEADER," July 30TH, 1870. 
 CURIOSITIES OF MATHEMATICS 
 
 to the editor of the liverpool leader. 
 
 Sir, 
 
 I have received several communications during the past 
 week, but it is not necessary to trouble you with the whole of them. 
 It is sufficient to give you the following correspondence :— 
 
 6 
 
4 2 
 X. Y. Z. to JAMES SMITH. 
 
 X. Y. Z. takes exception to two points in your communication 
 appearing in the Leader of the 16th instant. Referring to your 
 figure, you say that the triangle B D C is a right-angled triangle. 
 You do not prove that it is so, and it can easily be proved (to my 
 satisfaction at least) that it is not. 
 
 Again: — -In your solution you say (without a shadow of proof) 
 that B N = ^ (B P) ; and assuming this, you have of course no 
 difficulty in proving that B N : N C : : (24 + 7) : (24 — 7) : : 31 : 17. 
 
 Without some common basis of argument it is of course useless 
 attempting to argue ; but if you admit (and I should think of all men 
 you would be the last to deny) the universality of the simple rule 
 that the area of a plane rectilinear triangle is one-half the product 
 of the base and the perpendicular upon the base, you may easily 
 satisfy yourself that the proportion you gave as the supposed solu- 
 tion of your problem is not correct. 
 
 Referring again to your figure — 
 
 r.- 1 a n /- AxB 8x6 
 Area of triangle A B C = — - — = = 24. 
 
 But, triangle A B C = triangle A D B + triangle B D C. 
 
 r * • it>t^ BCxDN 6x3 
 Now, area of triangle B D C = = - J = 9. 
 
 .-. Area of triangle A D B 
 
 But, area of triangle ADB = = 4 M D. 
 
 .-. 4 M D = 15, and M D = if- 
 
 But, B N = M D, by construction. 
 .-. B N = V 6 - 
 
 And, since B P = 3, P N will = f and N C = \. 
 •BN : NC: : 15 : 9. 
 I need hardly say that the figure so constructed is materially 
 different from that formerly constructed with the aid of the circle ; 
 and if in your figure of Saturday you bisect A B in O, and with O B as 
 a radius describe a circle, such circle will not pass through the point 
 D ; for, if D O be joined, the line D O will not be equal to O B, 
 which it would be if D were a point in the circumference. 
 \Wi July, 1870. 
 
 24 — 9= 15. 
 
 
 AB x MD 
 
 8M D 
 
 2 
 
 2 
 
43 
 
 X. Y. Z. fails to perceive that ABC and D N C are not similar 
 triangles. Mr. Miller, in his Letter of the nth inst, assumes ABC 
 and D N C to be similar triangles, and I threw him off the scent in 
 my Letter of the 12th inst. (these Letters appear in your last 
 number) by suggesting the argument of X Y Z. 
 
 By hypothesis, let AM =5-12 and MB= 2-88. 
 Then: AM + M B = 5-12 + 2'88 = 8 = A B 
 
 ABxAM = 8x 5 -i2 = 40-96 = AD S 
 ••• A D = J&yb = 6-4. 
 
 ABxMB = 8x 2-88 = 23-04 = B D 2 
 .-. BD = J 22, -04 = 4-8. 
 
 B C 2 — B D 2 =6 2 — 4'8 2 =36 - 23-04= i2-o6 = D C 2 
 .. DC= ^7^6 = 3-6. 
 Hence : A B x B C = (A D x D B) + (B D x D C), and this 
 would make ABC and D N C similar right-angled triangles. 
 
 Mathematicians have never discovered that if the length of a 
 straight line be represented by J is, its true length is 3 + ^ (3) = 
 3-875, and not 3-872, &c, the extracted root of ^'15. 
 
 ALEX. ED. MILLER to J. S. 
 
 Dear Sir, 
 
 I do not see that our correspondence can be 
 usefully continued, but I owe it to your courtesy to answer your last. 
 I admit that Mathematics can never " be inconsistent with practical 
 Geometry," which, indeed, is one of the phases of Mathematics, and 
 it is on that very account that I maintain, that a proposition shown, 
 as yours has been shown, to err against one or more of the funda- 
 mental principles of Geometry, can never be set on its legs by any 
 amount of Arithmetic. In the construction you suggest in your 
 last, the parallelograms will not be rectangles, and therefore nothing 
 to the point. In opposition to you, I maintain, and till you show 
 me specific instances in Euclid to the contrary, will continue to 
 believe, that there is no Proposition of Euclid inconsistent, under 
 any circumstances, with any other, and that no plane rectilinear 
 
44 
 
 figure can be drawn inconsistent in any particular with any Proposi- 
 tion in, or necessary deduction from, elementary Geometry {i.e., 
 Euclid). 
 
 It is not assumed by Mathematicians that straight lines drawn 
 from any point in A B parallel to B C, &c, cut the triangle ABC 
 into similar triangles ; this is distinctly proved in Euclid, Prop. 2, 
 Book 6, and is universally true. 
 
 All your work in reference to the figures in the Liverpaol 
 Leader require the concurrent existence of two inconsistent postulates : 
 (i) B D C = a right-angled triangle, and (2) A M : M B : : 5 : 3. I 
 have shown that these cannot co-exist (when A B : B C : : 8 : 6) 
 unless you can have either (1) a plane triangle the sum of whose 
 angles is greater than two right angles, or (2) a semi-circle con- 
 taining an angle less than a right angle. These are both impossi- 
 bilities, therefore your postulates are mutually destructive. Utrum 
 harum maris accipe, when accepted it proves the other false. 
 Your obedient servant, 
 
 Alex. Ed. Miller. 
 24, Old Buildings, Lincoln's Inn, 
 \Wi July, 1870. 
 
 P.S. — What you say I "might have shown," is precisely what I 
 did show you. 
 
 A. E. M. 
 
 J. S. TO ALEX. ED. MILLER. 
 
 Dear Sir, 
 
 I am in receipt of yours of the 18th, which has 
 surprised me more than did your favour of the 13th inst. 
 
 You admit that " Mathematics can never be inconsistent with 
 practical Geometry," and I as freely admit (1) that the three angles 
 of a plane triangle are together equal to two right angles ; and (2) 
 that any two straight lines drawn from the extremities of a diameter 
 of a circle to a point in the circumference, will produce a right-angled 
 triangle. Like you, " I do not see that our correspondence can be 
 usefully continued." 
 
45 
 
 Yesterday morning's post brought me a diagram from Mr. E. 
 L. Garbett. I acknowledged its receipt, and send you the diagram 
 herewith. Take ample time to carefully examine it, and then please 
 return it. I shall direct your attention to some facts with reference 
 to it, and with this our controversy must terminate. 
 
 A B C is a right-angled triangle, with the base B C bisected in 
 P, and Mr. Garbett makes the sides B C and A B, that contain the 
 right angle, 600 and 800, making AC the hypothenuse 1000. The 
 sides of the triangle A B C he divides into parts, and gives the 
 arithmetical value of the separate parts. In the figure there are two 
 parallelograms, each of which is divided by a diagonal into two 
 similar and equal right-angled triangles. The sides of the small 
 right-angled triangle of which D'D is the hypothenuse, are 9, 12, and 
 15, and it follows, that this triangle and the triangle ABC are 
 similar, but not equal, right-angled triangles ; and whether you and 
 Mr. Garbett can, or can not, see it, it also follows, that the angles 
 D'and D in the triangles B D' C and B D C are right angles .* The 
 triangle ABC and the triangle A M' D' are similar right-angled 
 triangles, having the sides that contain the right angle in the ratio 
 of 3 to 4. The triangles AMD and A M' D' are both within the 
 circle, but the angle D of only one of them touches it. 
 
 Mr. Garbett is obviously not satisfied, or he would not have 
 been at the trouble of constructing and sending me the diagram. 
 
 Faithfully yours, 
 
 Barkeley House, Seaforth, 
 2.0th July, 1870. 
 
 J. S. 
 
 J. S. TO ALEX. ED. MILLER. 
 Dear Sir, 
 
 Your favour of the 19th inst. came to hand last 
 evening. If I were to answer it and similar communications, our 
 correspondence might never terminate. 
 
 * Mr. Miller was quick to detect the absurdity of the assertion : — " The 
 angle D' and D in the triangles BD'C and BDC are right angles," but 
 failed to discover the glaring absurdities involved in the arithmetical values 
 Mr. Garbett puts on the lines in the diagram. 
 
4 6 
 
 I may tell you, that in my published works I have proved many 
 things by the Sines of angles, and also by Logarithms* 
 
 Having dropped you and Mr. Garbett into a quagmire of 
 geometrical inconsistency, contradiction, and confusion, I must leave 
 you there. My last series of letters in the Liverpool Leader on 
 " Curiosities of Mathematics " will shortly appear in the form of a 
 
 pamphlet. 
 
 Faithfully yours, 
 
 J. S. 
 Barkeley House, Seaforth, 
 21st July., 1870. 
 
 One of the greatest blunders of Mathematicians is, that 01 
 making the geometrical and trigonometrical sines of angles the same. 
 Mr. Alex. Ed. Miller fell into this blunder in a Letter of his which 
 appeared in the Correspondent more than four years ago. Unequal 
 angles may have a common sine geometrically, but never trigono- 
 metrically. Unequal triangles may trigonometrically have the same 
 sines and cosines ; indeed, if in unequal right-angled triangles the 
 angles are equal, which is not an impossibility, the trigonometrical 
 sines and cosines of the angles will necessarily be equal. One of 
 Mr. Miller's assumptions is, that triangles are similar if the angles 
 are equal, although the ratios of side to side are unequal. 
 
 I shall have no more Letters for you on " Curiosities of Mathe- 
 matics," unless and until your contributor, the Rev. W. Allen 
 
 Whitworth, renews his. 
 
 I am, Sir, 
 
 Yours obediently, 
 
 James Smith. 
 Barkeley House, Seaforth, 
 z^th July, 1870. 
 
 When Mr. Miller returned Mr. Garbett's diagram, he 
 sent with it the following communication : — 
 
 * In his Letter Mr. Miller introduced a fallacious argument, founded on 
 the trigonometrical Sines of Angles. 
 
47 
 
 24, Old Buildings, Lincoln's Inn, 
 26th July, 1870. 
 
 Dear Sir, 
 
 I return Mr. Garbett's diagram, which seems all 
 right. 
 
 If you do not see that your construction, coupled with your 
 admissions : (1) that " any two straight lines drawn from the ex- 
 tremities of a diameter of a circle to a point in the circumference, 
 will produce a right-angled triangle," and (2) that " if you bisect A B 
 in O, and with O A or O B as radius describe a circle, the circum- 
 ference will not cut A C in the point D, but in a point nearer C," 
 amounts to saying, that from the point B two distinct perpendiculars 
 can be drawn to A C, one to the point D, and the other to the plane 
 where A C meets the circumference. In other words, that in Mr. 
 Garbett's diagram, A D B is a right-angled triangle, whether the D 
 is red or black. (In his diagram Mr. Garbett put some of the lines 
 and letters in red ink, and others in black.) If you can shew by any 
 abstract reasoning in the world, geometrical or otherwise, that two 
 distinct perpendiculars can be drawn from the same point to any 
 straight line whatever, you may earn the right to talk of having left 
 your opponents in a quagmire of inconsistencies ; but until some 
 better proof of this is given than the reiterated assertion that " it 
 follows, whether Mr. Garbett can see it or not," you may possibly 
 believe such a thing to be possible, but you will be the only man in 
 England, who knows what an angle is, who will do so. 
 
 I should not have troubled you with this but for your note in 
 Saturday's Leader :—" We shall see what Mr. Miller makes of this 
 communication." And you have seen it, and may make what you 
 can of it. 
 
 Your obedient servant, 
 
 Alex. Ed. Miller. 
 
 P.S.— I do not understand the distinction you seem to draw 
 between Geometry and Mathematics; at any rate, I have throughout, 
 except when I consented to follow you into Arithmetic — been dealing 
 with Geometry only, and that ot the simplest kind. 
 
 A. E. M. 
 
48 
 
 For an answer to Mr. Miller's P.S. I may refer him to 
 the concluding part of my Letter in the Liverpool Leader of 
 July 30th, (see pages 41-6). I admit, that unequal straight 
 lines drawn from the same point ; can never be perpen- 
 dicular to the same straight line. By suggesting a false 
 hypothesis, and by making an absurd assertion, I seem 
 to have disturbed Mr. Miller's equanimity, and thrown 
 him altogether off the right scent. Mathematician like, 
 he is quick to detect a fallacy, but apparently incom- 
 petent to comprehend a new geometrical truth, however 
 stringently demonstrated. 
 
 The Reader will observe from dates, that Mr. Miller 
 kept Mr. Garbett's diagram a full week, and I may fairly 
 infer, that he gave it a careful examination. It is obvious 
 that it did not suggest anything new to him. 
 
 The annexed diagram is much smaller, but in all 
 other respects similar to Mr. Garbett's. He did not give 
 me his method of constructing the diagram, and I can 
 only conjecture it from his values of the various lines. 
 He may have constructed it in the following way: — 
 
 Draw the straight line B Pand produce it to N', making 
 PN' = { (B P). Produce P N' to C, making N'C = 
 3 (P N'). From the point B draw B A perpendicular to 
 B C, making B A = \ (B C), and join A C. Bisect A B 
 in O, and with O as centre and O A or O B as radius, 
 describe the semi-circle A D B. From the point D where 
 the semi-circle cuts A C, draw D M perpendicular to AB, 
 and therefore parallel to B C. Join DO and D P, and 
 from the point D draw D N perpendicular to B C, and 
 therefore parallel to A B. From N' draw a straight line 
 perpendicular to B C to meet A C the hypothenuse of 
 the right-angled triangle A B C in the point D', and from 
 
€ 
 
• 
 
49 
 
 D' draw the straight line D' M' parallel to D M. Join 
 B D and B D/ 
 
 My method of constructing the diagram is as follows : 
 
 I first draw two straight lines of indefinite length at 
 right angles to each other, making B the right angle. I 
 then open my compasses at random, and from the angle 
 B mark off six equal parts together equal to B C, and 
 eight of such parts together equal to A B, join A C, ana 
 bisect B C in P. The rest of the construction must be 
 the same as Mr. Garbett's, whatever we make our starting 
 point. 
 
 Mr. Garbett makes A B = goo, B P = 300, = B C 
 600, and A C = 1000 : by construction, and so far we are 
 agreed : that is to say, we are agreed, that A B is to 
 B C in the ratio of g to 6, and A C to B C in the ratio of 10 
 to 6, by construction ; and no amount of Mathematics 
 can upset these ratios. 
 
 Now, any person may take a pair of compasses, and 
 with B as centre and B P as radius, describe a circle. The 
 circumference of such circle will cut the line A B in the 
 point M, and it follows, that B M must equal B P, for 
 they are radii of the circle. But, Mr. Garbett makes 
 B P = 300, by construction ; and by computation makes 
 B M only = 288. Again : Mr. Garbett makes D' M' = 
 D' C = 375. With D' as centre and D' M' as radius, 
 describe a circle. The circumference of such circle will 
 not pass through the point C, but through a point in the 
 line D' C nearer D'. Here again, computation is not in 
 harmony with construction. Is it conceivable that any- 
 thing could be more absurd ? 
 
 Mr. Miller says of Mr. Garbett's diagram, "it seems 
 all right ; " which surely means, that he agrees with 
 
50 
 
 Mr. Garbett, as to the arithmetical values he has put on the 
 various lines of which his geometrical figure is composed. 
 De Morgan, Whitworth, and all other " recognised 
 Mathematicians, attempt to make Geometry fit their 
 Mathematics, instead of endeavouring to make Mathe- 
 matics harmonize with Geometry. Mr. Garbett's diagram 
 is the best illustration of this fact that has ever come 
 under my observation. 
 
 I might point out absurdity upon absurdity with 
 reference to this remarkable geometrical figure. Why 
 should I? In Geometry, one reductio ad absurdum 
 demonstration is as good as a thousand. I have given 
 two. X. Y. Z. (can X. Y. Z be J. T. T. who wrote on 
 this subject some years ago, and whose Letters appeared 
 in the Daily Post under the sobriquet of Pi) Miller, 
 Garbett, and Gibbons, might readily furnish many more, 
 if they could but get rid of prejudice, and go honestly 
 to work. I have done my duty ! Let Mathematicians 
 and the British Association do theirs. 
 
 James Smith. 
 
 Barkeley House, Seaforth, 
 4th August, 1870. 
 
ERRATA. 
 Page 24 : nth line from bottom— For B N = M B : read B N = M D. 
 Page 29: nth line from bottom — For by construction 5 : read by con- 
 struction = 5. 
 
Andrew and David Russell, Printers, Liverpool 
 
UNIVERSITY OF CALIFORNIA LIBRARY 
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