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 BOUGHT WITH THE INCOME OF THE 
 
 SAGE ENDOWMENT FUND 
 
 THE GIFT OF 
 
 HENRY W. SAGE 
 
 1891 
 
 MATHEMATICS 
 
Cornell University Library 
 QA 31.A67G3 1909 
 
 Geometrical solutions derived from mecha 
 
 3 1924 001 507 627 
 
Cornell University 
 Library 
 
 The original of this book is in 
 the Cornell University Library. 
 
 There are no known copyright restrictions in 
 the United States on the use of the text. 
 
 http://www.archive.org/details/cu31924001507627 
 
GEOMETRICAL SOLUTIONS 
 
 DERIVED FROM 
 
 MECHANICS 
 
 A TREATISE BY ARCHIMEDES 
 
 RECENTLY DISCOVERED AND TRANSLATED FROM THE GREEK BY 
 
 DR. J. L. HEIBERG 
 
 CHICAGO 
 THE OPEN COURT PUBLISHING COMPANY 
 
 LONPON AGENTS 
 KBGAN PAUL, TRENCH, *ft©BNER & QQ., Ltd. 
 
GEOMETRICAL SOLUTIONS 
 
 DERIVED FROM 
 
 MECHANICS 
 
 A TREATISE OF ARCHIMEDES 
 
 RECENTLY DISCOVERED AND TRANSLATED FROM THE GREEK BY 
 
 DR. J. L. HEIBERG 
 
 PROFESSOR OF CLASSICAL PHILOLOGY AT THE UNIVERSITY OF COPENHAGEN 
 
 WITH AN INTRODUCTION BY 
 
 DAVID EUGENE SMITH 
 
 PRESIDENT OF TEACHER'S COLLEGE, COLUMBIA UNIVERSITY, NEW YORK 
 
 ENGLISH VERSION TRANSLATED FROM THE GERMAN BY LYDIA G. ROBINSON 
 AND REPRINTED FROM "THE MONIST," APRIL, 19OQ. 
 
 CHICAGO 
 THE OPEN COURT PUBLISHING COMPANY 
 
 LONDON AGENTS 
 
 KEGAN PAUL, TRENCH, TRUBNER & CO., LTD. 
 
 1909 
 
 7 
 
A. "B-K^L 
 
 Copyright by 
 
 The Open Court Publishing Co. 
 
 1909 
 
INTRODUCTION. 
 
 IF there ever was a case of appropriateness in discovery, 
 the rinding of this manuscript in the summer of 1906 
 was one. In the first place it was appropriate that the dis- 
 covery should be made in Constantinople, since it was here 
 that the West received its first manuscripts of the other ex- 
 tant works, nine in number, of the great Syracusan. It was 
 furthermore appropriate that the discovery should be made 
 by Professor Heiberg, facilis princeps among all workers 
 in the field of editing the classics of Greek mathematics, 
 and an indefatigable searcher of the libraries of Europe 
 for manuscripts to aid him in perfecting his labors. And 
 finally it was most appropriate that this work should ap- 
 pear at a time when the affiliation of pure and applied 
 mathematics is becoming - so generally recognized all over 
 the world. We are sometimes led to feel, in considering 
 isolated cases, that the great contributors of the past have 
 worked in the field of pure mathematics alone, and the 
 saying of Plutarch that Archimedes felt that "every kind 
 of art connected with daily needs was ignoble and vulgar" 1 
 may have strengthened this feeling. It therefore assists 
 us in properly orientating ourselves to read another treat- 
 ise from the greatest mathematician of antiquity that sets 
 clearly before us his indebtedness to the mechanical appli- 
 cations of his subject. 
 
 Not the least interesting of the passages in the manu- 
 
 1 Marcellus, 17. 
 
2 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 script is the first line, the greeting to Eratosthenes. It is 
 well known, on the testimony of Diodoros his countryman, 
 that Archimedes studied in Alexandria, and the latter fre- 
 quently makes mention of Konon of Samos whom he knew 
 there, probably as a teacher, and to whom he was indebted 
 for the suggestion of the spiral that bears his name. It is 
 also related, this time by Proclos, that Eratosthenes was a 
 contemporary of Archimedes, and if the testimony of so 
 late a writer" as Tzetzes, who lived in the twelfth century, 
 may be taken as valid, the former was eleven years the 
 junior of the great Sicilian. Until now, however, we have 
 had nothing definite to show that the two were ever ac- 
 quainted. The great Alexandrian savant, — poet, geog- 
 rapher, arithmetician, — affectionately called by the stu- 
 dents Pentathlos, the champion in five sports, 2 selected by 
 Ptolemy Euergetes to succeed his master, Kallimachos the 
 poet, as head of the great Library, — this man, the most 
 renowned of his time in Alexandria, could hardly have 
 been .a teacher of Archimedes, nor yet the fellow student of 
 one who was so much his senior. It is more probable that 
 they were friends in the later days when Archimedes was 
 received as a savant rather than as a learner, and this is 
 borne out by the statement at the close of proposition I 
 which refers to one of his earlier works, showing that this 
 particular treatise was a late one. This reference being 
 to one of the two works dedicated to Dositheos of Kolonos, 3 
 and one of these (De lineis spiralibus) referring to an 
 earlier treatise sent to Konon, 4 we are led to believe that 
 this was one of the latest works of Archimedes and that 
 Eratosthenes was a friend of his mature years, although 
 
 2 His nickname of Beta is well known, possibly because his lecture room 
 was number 2. 
 
 'We know little of his works, none of which are extant. Geminos and 
 Ptolemy refer to certain observations made by him in 200 B. C, twelve years 
 after the death of Archimedes. Pliny also mentions him. 
 
 4 T(3i' tvotI ~K6vo3va awvGToXevTwv 6e(ap7]/xaTwv. 
 
INTRODUCTION. 3 
 
 one of long standing. The statement that the preliminary 
 propositions were sent "some time ago" bears out this idea 
 of a considerable duration of friendship, and the idea that 
 more or less correspondence had resulted from this com- 
 munication may be inferred by the statement that he saw, 
 as he had previously said, that Eratosthenes was "a capable 
 scholar and a prominent teacher of philosophy," and also 
 that he understood "how to value a mathematical method 
 of investigation when the opportunity offered." We have, 
 then, new light upon the relations between these two men. 
 the leaders among the learned of their day. 
 
 A second feature of much interest in the treatise is the 
 intimate view that we have into the workings of the mind 
 of the author. It must always be remembered that Archi- 
 medes was primarily a discoverer, and not primarily a com- 
 piler as were Euclid, Apollonios, and Nicomachos. There- 
 fore to have him follow up his first communication of theo- 
 rems to Eratosthenes by a statement of his mental proces- 
 ses in reaching his conclusions is not merely a contribution 
 to mathematics but one to education as well. Particularly 
 is this true in the following statement, which may well be 
 kept in mind in the present day: "I have thought it well 
 to analyse and lay down for you in this same book a pecu- 
 liar method by means of which it will be possible for you 
 to derive instruction as to how certain mathematical ques- 
 tions may be investigated by means of mechanics. And I 
 am convinced that this is equally profitable in demonstrat- 
 ing a proposition itself; for much that was made evident 
 to me through the medium of mechanics was later proved 
 by means of geometry, because the treatment by the former 
 method had not yet been established by way of a demonstra- 
 tion. For of course it is easier to establish a proof if one 
 has in this way previously obtained a conception of the 
 questions, than for him to seek it without such a prelim- 
 inary notion .... Indeed I assume that some one among the 
 
4 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 investigators of to-day or in the future will discover by the 
 method here set forth still other propositions which have 
 not yet occurred to us." Perhaps in all the history of 
 mathematics no such prophetic truth was ever put into 
 words. It would almost seem as if Archimedes must have 
 seen as in a vision the methods of Galileo, Cavalieri, Pascal, 
 Newton, and many of the other great makers of the mathe- 
 matics of the Renaissance and the present time. 
 
 The first proposition concerns the quadrature of the 
 parabola, a subject treated at length in one of his earlier 
 communications to Dositheos. 5 Pie gives a digest of the 
 treatment, but with the warning that the proof is not com- 
 plete, as it is in his special work upon the subject. He has, 
 in fact, summarized propositions VII-XVII of his com- 
 munication to Dositheos, omitting the geometric treat- 
 ment of propositions XVIII-XXIV. One thing that he 
 does not state, here or in any of his works, is where the 
 idea of center of gravity 6 started. It was certainly a com- 
 mon notion in his day, for he often uses it without defining 
 it. It appears in Euclid's 7 time, but how much earlier we 
 cannot as yet say. 
 
 Proposition II states no new fact. Essentially it means 
 that if a sphere, cylinder, and cone (always circular) have 
 the same radius, r, and the altitude of the cone is r and that 
 of the cylinder 2r, then the volumes will be as 4 : i : 6, 
 which is true, since they are respectively %7rr 3 , %l7r 3 , and 
 2vr s . The interesting thing, however, is the method pur- 
 sued, the derivation of geometric truths from principles 
 of mechanics. There is, too, in every sentence, a little 
 suggestion of Cavalieri, an anticipation by nearly two thou- 
 sand years of the work of the greatest immediate precursor 
 of Newton. And the geometric imagination that Archi- 
 
 B TeTpaywi/itrfibs wapafioXijs. 
 
 6 Kirrpa jSapwc, for "barycentric" is a very old term. 
 
 7 At any rate in the anonymous fragment De levi et ponderoso, sometimes 
 attributed to him. 
 
INTRODUCTION. 5 
 
 medes shows in the last sentence is also noteworthy as one 
 of the interesting features of this work : "After I had thus 
 perceived that a sphere is four times as large as the cone . . . 
 it occurred to me that the surface of a sphere is four times 
 as great as its largest circle, in which I proceeded from the 
 idea that just as a circle is equal to a triangle whose base is 
 the periphery of the circle, and whose altitude is equal to 
 its radius, so a sphere is equal to a cone whose base is the 
 same as the surface of the sphere and whose altitude is 
 equal to the radius of the sphere." As a bit of generaliza- 
 tion this throws a good deal of light on the workings of 
 Archimedes's mind. 
 
 In proposition III he considers the volume of a sphe- 
 roid, which he had already treated more fully in one of his 
 letters to Dositheos, 8 and which contains nothing new from 
 a mathematical standpoint. Indeed it is the method rather 
 than the conclusion that is interesting in such of the sub- 
 sequent propositions as relate to mensuration. Proposition V 
 deals with the center of gravity of a segment of a conoid, and 
 proposition VI with the center of gravity of a hemisphere, 
 thus carrying into solid geometry the work of Archimedes 
 on the equilibrium of planes and on their centers of grav- 
 ity. 9 The general method is that already known in the 
 treatise mentioned, and this is followed through propo- 
 sition X. 
 
 Proposition XI is the interesting case of a segment of 
 a right cylinder cut off by a plane through the center of 
 the lower base and tangent to the upper one. He shows 
 this to equal one-sixth of the square prism that circum- 
 scribes the cylinder. This is well known to us through the 
 formula v — 2r 2 h/3, the volume of the prism being 4r 2 h, 
 and requires a knowledge of the center of gravity of the 
 cylindric section in question. Archimedes is, so far as we 
 
 8 IIep2 KavoeiStSv /ecu <T<fiaipoei8eup. 
 
 "'Emir^Swv looppoiriCbv y Kivrpa fiapuv iirnr&uv. 
 
6 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 know, the first to state this result, and he obtains it by his 
 usual method of the skilful balancing of sections. There 
 are several lacunae in the demonstration, but enough of 
 it remains to show the ingenuity of the general plan. The 
 culminating interest from the mathematical standpoint lies 
 in proposition XIII, where Archimedes reduces the whole 
 question to that of the quadrature of the parabola. He 
 shows that a fourth of the circumscribed prism is to the 
 segment of the cylinder as the semi-base of the prism is to 
 the parabola inscribed in the semi-base ; that is, that Yip : 
 v = x /ib: {%■ /4&) , whence v = V&p. Proposition XIV is in- 
 complete, but it is the conclusion of the two preceding prop- 
 ositions. 
 
 In general, therefore, the greatest value of the work 
 lies in the following: 
 
 i. It throws light upon the hitherto only suspected re- 
 lations of Archimedes and Eratosthenes. 
 
 2. It shows the working of the mind of Archimedes in 
 the discovery of mathematical truths, showing that he often 
 obtained his results by intuition or even by measurement, 
 rather than by an analytic form of reasoning, verifying 
 these results later by strict analysis. 
 
 3. It expresses definitely the fact that Archimedes was 
 the discoverer of those properties relating to the sphere 
 and cylinder that have been attributed to him and that are 
 given in his other works without a definite statement of 
 their authorship. 
 
 4. It shows that Archimedes was the first to state the 
 volume of the cylinder segment mentioned, and it gives 
 an interesting description of the mechanical method by 
 which he arrived at his result. 
 
 David Eugene Smith. 
 Teachers College, Columbia University. 
 
GEOMETRICAL SOLUTIONS DERIVED FROM 
 
 MECHANICS. 
 
 Archimedes to Eratosthenes, Greeting: 
 
 Some time ago I sent you some theorems I had discovered, 
 writing down only the propositions because I wished you to find 
 their demonstrations which had not been given. The propositions 
 of the theorems which I sent you were the following: 
 
 1. If in a perpendicular prism with a parallelogram 2 for base 
 a cylinder is inscribed which has its bases in the opposite paral- 
 lelograms 2 and its surface touching the other planes of the prism, 
 and if a plane is passed through the center of the circle that is the 
 base of the cylinder and one side of the square lying in the opposite 
 plane, then that plane will cut off from the cylinder a section which 
 is bounded by two planes, the intersecting plane and the one in 
 which the base of the cylinder lies, and also by as much of the 
 surface of the cylinder as lies between these same planes; and the 
 detached section of the cylinder is % of the whole prism. 
 
 2. If in a cube a cylinder is inscribed whose bases lie in oppo- 
 site parallelograms 2 and whose surface touches the other four planes, 
 and if in the same cube a second cylinder is inscribed whose bases 
 lie in two other parallelograms 2 and whose surface touches the 
 four other planes, then the body enclosed by the surface of the 
 cylinder and comprehended within both cylinders will be equal to 
 % of the whole cube. 
 
 These propositions differ essentially from those formerly dis- 
 covered ; for then we compared those bodies (conoids, spheroids 
 and their segments) with the volume of cones and cylinders but none 
 of them was found to be equal to a body enclosed by planes. Each 
 of these bodies, on the other hand, which are enclosed by two planes 
 and cylindrical surfaces is found to be equal to a body enclosed 
 
 "This must mean a square. 
 
8 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 by planes. The demonstration of these propositions I am accordingly 
 sending to you in this book. 
 
 Since I see, however, as I have previously said, that you are 
 a capable scholar and a prominent teacher of philosophy, and also 
 that you understand how to value a mathematical method of in- 
 vestigation when the opportunity is offered, 1 have thought it well 
 to analyze and lay down for you in this same book a peculiar method 
 by means of which it will be possible for you to derive instruction 
 as to how certain mathematical questions may be investigated by 
 means of mechanics. And I am convinced that this is equally profit- 
 able in demonstrating a proposition itself ; for much that was made 
 evident to me through the medium of mechanics was later proved 
 by means of geometry because the treatment by the former method 
 had not yet been established by way of a demonstration. For of 
 course it is easier to establish a proof. if one has in this way pre- 
 viously obtained a conception of the questions, than for him to seek it 
 without such a preliminary notion. Thus in the familiar propositions 
 the demonstrations of which Eudoxos was the first to discover, 
 namely that a cone and a pyramid are one third the size of that 
 cylinder and prism respectively that have the same base and alti- 
 tude, no little credit is due to Democritos who was the first to make 
 that statement about these bodies without any demonstration. But 
 we are in a position to have found the present proposition in the 
 same way as the earlier one ; and I have decided to write down and 
 make known the method partly because we have already talked 
 about it heretofore and so no one would think that we were spread- 
 ing abroad idle talk, and partly in the conviction that by this means 
 we are obtaining no slight advantage for mathematics, for indeed 
 I assume that some one among the investigators of to-day or in the 
 future will discover by the method here set forth still other propo- 
 sitions which have not yet occurred to us. 
 
 In the first place we will now explain what was also first made 
 clear to us through mechanics, namely that a segment of a parabola 
 is Ys of the triangle possessing the same base and equal altitude; 
 following which' we will explain in order the particular propositions 
 discovered by the above mentioned method; and in the last part 
 of the book we will present the geometrical demonstrations of the 
 propositions. 4 
 
 4 In his "Commentar," Professor Zeuthen calls attention to the fact that 
 it was already known from Heron's recently discovered Metrica that these 
 propositions were contained in this treatise, and Professor Heiberg made the 
 same comment in Hermes. — Tr. 
 
GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 9 
 
 1. If one magnitude is taken away from another magnitude and 
 the same point is the center of gravity both of the whole and of the 
 part removed, then the same point is the center of gravity of the 
 remaining portion. 
 
 2. If one magnitude is taken away from another magnitude and 
 the center of gravity of the whole and of the part removed is not 
 the same point, the center of gravity of the remaining portion may 
 be found by prolonging the straight line which connects the centers 
 of gravity of the whole and of the part removed, and setting off 
 upon it another straight line which bears the same ratio to the 
 straight line between the aforesaid centers of gravity, as the weight 
 of the magnitude which has been taken away bears to the weight 
 of the one remaining [De plan, aequil. I, 8]. 
 
 3. If the centers of gravity of any number of magnitudes lie 
 upon the same straight line, then will the center of gravity of all the 
 magnitudes combined lie also upon the same straight line [Cf. ibid. 
 
 I, 5]- 
 
 4. The center of gravity of a straight line is the center of that 
 
 line [Cf. ibid. I, 4]. 
 
 5. The center of gravity of a triangle is the point in which the 
 straight lines drawn from the angles of a triangle to the centers of 
 the opposite sides intersect [Ibid. I, 14]. 
 
 6. The center of gravity of a parallelogram is the point where 
 its diagonals meet [Ibid. I, 10]. 
 
 7. The center of gravity [of a circle] is the center [of that 
 circle] . 
 
 8. The center of gravity of a cylinder [is the center of its axis]. 
 
 9. The center of gravity of a prism is the center of its axis. 
 
 10. The center of gravity of a cone so divides its axis that the 
 section at the vertex is three times as great as the remainder. 
 
 11. Moreover together with the exercise here laid down I will 
 make use of the following proposition: 
 
 If any number of magnitudes stand in the same ratio to the 
 same number of other magnitudes which correspond pair by pair, 
 and if either all or some of the former magnitudes stand in any 
 ratio whatever to other magnitudes, and the latter in the same ratio 
 to the corresponding ones, then the sum of the magnitudes of the 
 first series will bear the same ratio to the sum of those taken from 
 the third series as the sum of those of the second series bears to 
 the sum of those taken from the fourth series [De Conoid. I]. 
 
10 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 Let a/8y [Fig. i] be the segment of a parabola bounded by the 
 straight line ay and the parabola a/?y. Let ay be bisected at 8, 8/3e 
 being parallel to the diameter, and draw a/?, and /Jy. Then the 
 segment a/3y will be % as great as the triangle a/?y. 
 
 From the points a and y draw a£ 1 1 S/?«, and the tangent y£ ; 
 
 produce [y/J to *, and 
 make k6 = y*] . Think of 
 yd as a scale-beam with 
 the center at k and let /*£ 
 be any straight line what- 
 ever 1 1 eS- Now since y£a 
 is a parabola, y£ a tan- 
 gent and y8 an ordinate, 
 then e/3 = /38; for this in- 
 deed has been proved in 
 the Elements [i. e., of 
 conic sections, cf. Quadr. 
 parab. 2]. For this rea- 
 y son and because fa and 
 /*£||e8, iiv-vi, and £* = 
 Ka. And because ya:a£ 
 = ix£ : $0 ( for this is shown 
 in a corollary, [cf. Quadr. parab. 5]), ya:a£ = y/c: K v; and y K = K 0, 
 therefore 0« : «v = n£ : &>• And because v is the center of gravity of 
 the straight line /x|, since iw = v£, then if we make 717 = £0 and 6 as 
 its center of gravity so that 1-0 = 0,;, the straight line rB-q will be in 
 equilibrium with /*£ in its present position because dv is divided in 
 inverse proportion to the weights tt, and & and Ok : KV = /x| : V t ; there- 
 fore k is the center of gravity of the combined weight of the two. 
 In the same way all straight lines drawn in the triangle £ay||eS are 
 in their present positions in equilibrium with their parts cut off by 
 the parabola, when these are transferred to 0, so that * is the center 
 of gravity of the combined weight of the two. And because the 
 triangle yfa consists of the straight lines in the triangle yfa and the 
 segment a/?y consists of those straight lines within the segment of 
 the parabola corresponding to the straight line $0, therefore the 
 triangle Cay in its present position will be in equilibrium at the 
 point k with the parabola-segment when this is transferred to 9 as 
 its center of gravity, so that k is the center of gravity of the combined 
 
GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. II 
 
 weights of the two. Now let y/c be so divided at x that yK = 3«x; 
 then x will be the center of gravity of the triangle afy, for this 
 has been shown in the Statics [cf. De plan, aequil. I, 15, p. 186, 
 3 with Eutokios, S. 320, 5ff.]. Now the triangle fay in its present 
 position is in equilibrium at the point k with the segment /Jay when 
 this is transferred to 6 as its center of gravity, and the center of 
 gravity of the triangle fay is x > hence triangle afy : segm. a/Jy when 
 transferred to 6 as its center of gravity =6k:kx. But 0k = 3«x; 
 hence also triangle afy = 3 segm. a/Jy. But it is also true that triangle 
 fay = 4Aa/8y because £k = k<x, and aS = 8y; hence segm. a/?y = % the 
 triangle a/3y. This is of course clear. 
 
 It is true that this is not proved by what we have said here; 
 but it indicates that the result is correct. And so, as we have just 
 seen that it has not been proved but rather conjectured that the 
 result is correct we have devised a geometrical demonstration which 
 we made known some time ago and will again bring forward 
 farther on. 
 
 n. 
 
 That a sphere is four times as large as a cone whose base is 
 equal to the largest circle of the sphere and whose altitude is equal 
 to the radius of the sphere, and that a cylinder whose base is equal 
 to the largest circle of the sphere and whose altitude is equal to the 
 diameter of the circle is one and a half times as large as the sphere, 
 may be seen by the present method in the following way: 
 
 Let aj8y8 [Fig. 2] be the largest circle of a sphere and ay and (88 
 its diameters perpendicular to each other ; let there be in the sphere 
 a circle on the diameter (38 perpendicular to the circle a/?y8, and 
 on this perpendicular circle let there be a cone erected with its 
 vertex at a; producing the convex surface of the cone, let it be 
 cut through y by a plane parallel to its base ; the result will be the 
 circle perpendicular to ay whose diameter will be ef. On this 
 circle erect a cylinder whose axis =ay and whose vertical bound- 
 aries are eX. and fi/. Produce ya making aO = ya and think of y0 as 
 a scale-beam with its center at a. Then let juv be any straight line 
 whatever drawn ||/38 intersecting the circle ajSyS in £ and o, the 
 diameter ay in a, the straight line ae in ir and af in p, and on the 
 straight line pv construct a plane perpendicular to ay ; it will inter- 
 sect the cylinder in a circle on the diameter juv ; the sphere a/3y8, in 
 a circle on the diameter |o ; the cone o«f in a circle on the diameter 
 
12 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 X 
 
 
 V a. 
 
 
 Tt 
 
 
 f? 
 
 \x 
 
 
 
 A 
 
 / 
 
 {/" 
 
 a ?\°\ 
 
 S 
 
 
 
 H 
 
 
 s 
 
 t 
 
 
 f ; 
 
 ' a 
 
 t 
 
 Fig. 2. 
 
 Ti-p. Now because yo X a(T^= uer X ott (for ay = ou, a<r = ir<r), and yaX 
 o<r = a£* = fr 2 + air 2 then /«r x cttt = £<j 2 + <"r 2 . Moreover, because ya : aa 
 
 = pa: air and ya = a$, there- 
 fore 6a : aa = juo- : air — /jut 2 : 
 fur x air- But it has been 
 proved that to 2 + air 2 — /no-x 
 air ; hence ad : a<r=[W 2 :£ct 2 + 
 air 2 - But it is true that 
 pa 2 : £o- 2 + cttt 2 = /iv 2 :|o i + 
 irp 2 = the circle in the cyl- 
 inder whose diameter is 
 /wrthe circle in the cone 
 whose diameter is 7rp + the 
 circle in the sphere whose 
 diameter is £o; hence 8a: 
 a<r = the circle in the cyl- 
 inder : the circle in the 
 sphere + the circle in the 
 cone. Therefore the circle in the cylinder in its present position 
 will be in equilibrium at the point ^ with the two circles whose 
 diameters are £o and vp, if they are so transferred to that 8 is the 
 center of gravity of both. In the same way it can be shown that 
 when another straight line is drawn in the parallelogram £\ 1 1 ef , 
 and upon it a plane is erected perpendicular to ay, the circle pro- 
 duced in the cylinder in its present position will be in equilibrium 
 at the point a with the two circles produced in the sphere and the 
 cone when they are transferred and so arranged on the scale-beam 
 at the point 9 that 6 is the center of gravity of both. Therefore 
 if cylinder, sphere and cone are filled up with such circles then the 
 cylinder in its present position will be in equilibrium at the point a 
 with the sphere and the cone together, if they are transferred and 
 so arranged on the scale-beam at the point 8 that 9 is the center of 
 gravity of both. Now since the bodies we have mentioned are in 
 equilibrium, the cylinder with k as its center of gravity, the sphere 
 and the cone transferred as we have said so that they have 8 as 
 center of gravity, then 8a : ok = cylinder : sphere + cone. But 8a- 
 2ok, and hence also the cylinder =2X (sphere + cone) . But it is also 
 true that the cylinder =3 cones [Euclid, Elem. XII, 10] , hence 3 
 cones = 2 cones + 2 spheres. If 2 cones be subtracted from both 
 sides, then the cone whose axes form the triangle ae£, = 2 spheres. 
 But the cone whose axes form the triangle oe£ = 8 cones whose axes 
 
GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 1 3 
 
 form the triangle a/38 because e£ = 2/3S, hence the aforesaid 8 cones 
 = 2 spheres. Consequently the sphere whose greatest circle is a/3y8 
 is four times as large as the cone with its vertex at a, and whose 
 base is the circle on the diameter /38 perpendicular to ay. 
 
 Draw the straight lines <j>/3x and i^8o> 1 1 ay through j8 and 8 in 
 the parallelogram A£ and imagine a cylinder whose bases are the 
 circles on the diameters <j>$ and x M and whose axis is ay. Now 
 since the cylinder whose axes form the parallelogram <£u is twice 
 as large as the cylinder whose axes form the parallelogram <^8 and 
 the latter is three times as large as the cone the triangle of whose 
 axes is a/38, as is shown in the Elements [Euclid, Elem. XII, 10], the 
 cylinder whose axes form the parallelogram <j><i> is six times as large 
 as the cone whose axes form the triangle a/88. But it was shown 
 that the sphere whose largest circle is a/3y8 is four times as large 
 as the same cone, consequently the cylinder is one and one half 
 times as large as the sphere, Q. E. D. 
 
 After I had thus perceived that a sphere is four times as large 
 as the cone whose base is the largest circle of the sphere and whose 
 altitude is equal to its radius, it occurred to me that the surface of 
 a sphere is four times as great as its largest circle, in which I pro- 
 ceeded from the idea that just as a circle is equal to a triangle whose 
 base is the periphery of the circle and whose altitude is equal to 
 its radius, so a sphere is equal to a cone whose base is the same as 
 the surface of the sphere and whose altitude is equal to the radius 
 of the sphere. 
 
 in. 
 
 By this method it may also be seen that a cylinder whose base 
 is equal to the largest circle of a spheroid and whose altitude is 
 equal to .the axis of the spheroid, is one and one half times as large 
 as the spheroid, and when this is recognized it becomes clear that 
 if a spheroid is cut through its center by a plane perpendicular to 
 its axis, one-half of the spheroid is twice as great as the cone whose 
 base is that of the segment and its axis the same. 
 
 For let a spheroid be cut by a plane through its axis and let 
 there be in its surface an ellipse a/3y8 [Fig. 3] whose diameters are 
 ay and /38 and whose center is k and let there be a circle in the 
 spheroid on the diameter j6S perpendicular to ay; then imagine a 
 cone whose base is the same circle but whose vertex is at a, and 
 producing its surface, let the cone be cut by a plane through y 
 
14 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 parallel to the base; the intersection will be a circle perpendicular 
 to ay with e£ as its diameter. Now imagine a cylinder whose base 
 is the same circle with the diameter «£ and whose axis is ay ; let ya 
 be produced so that aO-ya; think of By as a scale-beam with its 
 center at a and in the parallelogram A.£ draw a straight line fiv 1 1 ef , 
 and on fiv construct a plane perpendicular to ay ; this will intersect 
 the cylinder in a circle whose diameter is fiv, the spheroid in a circle 
 whose diameter is £o and the cone in a circle whose diameter is 
 irp. Because ya : aa = ea : air — fur : air, and ya = ad, therefore da : acr = 
 fia : air. But fia \ air — per 2 : fia X air and fia X air = ira 2 + <j£ 2 > for atr X cry : 
 <r| 2 = aK x Ky : k,8 2 = ok 2 : up 2 (for both ratios are equal to the ratio 
 
 between the diameter and the 
 parameter [Apollonius, Con. 
 I, 21]) = aa 2 : air 2 therefore 
 aa 2 laaXay = ira 2 '. a£ 2 = air 2 '. 
 airXir/i., consequently pirxira 
 = a$ 2 - If ttct 2 is added to both 
 sides then fMT X air = ira' + a£ 2 . 
 Therefore 6a: aa = pa 2 lira 2 + 
 o£ 2 - But fia 2 : a£? + ai& = the 
 circle in the cylinder whose 
 diameter is iw : the circle with 
 the diameter |o + the circle 
 with the diameter irp; hence 
 the circle whose diameter is 
 pv will in its present position 
 be in equilibrium at the point 
 a with the two circles whose 
 diameters are £o and wp when they are transferred and so arranged 
 on the scale-beam at the point a that 6 is the center of gravity of 
 both; and 6 is the center of gravity of the two circles combined 
 whose diameters are £o and up when their position is changed, 
 hence a : aa- the circle with the diameter fiv : the two circles whose 
 diameters are £o and irp. In the same way it can be shown that 
 if another straight line is drawn in the parallelogram Af II ef and on 
 this line last drawn a plane is constructed perpendicular to ay, then 
 likewise the circle produced in the cylinder will in its present posi- 
 tion be in equilibrium at the point a with the two circles combined 
 which have been produced in the spheroid and in the cone respec- 
 tively when they are so transferred to the point 6 on the scale-beam 
 that 6 is the center of gravity of both. Then if cylinder, spheroid 
 
GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 1 5 
 
 and cone are filled with such circles, the cylinder in its present posi- 
 tion will be in equilibrium at the point u. with the spheroid + the 
 cone if they are transferred and so arranged on the scale-beam at 
 the point a that 6 is the center of gravity of both. Now k is the 
 center of gravity of the cylinder, but 6, as has been said, is the 
 center of gravity of the spheroid and cone together. Therefore 
 6a : cue = cylinder : spheroid + cone. But a6 = 2a.K, hence also the 
 cylinder = 2 x (spheroid + cone) = 2 x spheroid + 2 x cone. But the 
 cylinder = 3 x cone, hence 3 x cone = 2 x cone + 2 x spheroid. Subtract 
 2 x cone from both sides ; then a cone whose axes form the triangle 
 aef = 2 x spheroid. But the same cone = 8 cones whose axes form 
 the A a/88; hence 8 such cones = 2 x spheroid, 4X cone = spheroid; 
 whence it follows that a spheroid is four times as great as a cone 
 whose vertex is at a, and whose base is the circle on the diameter 
 /?8 perpendicular to Ae, and one-half the spheroid is twice as great 
 as the same cone. 
 
 In the parallelogram A.f draw the straight lines <f>x and xj/wWay 
 through the points /8 and 8 and imagine a cylinder whose bases 
 are the circles on the diameters <$>$ and x«>) and whose axis is ay. 
 Now since the cylinder whose axes form the parallelogram <j><a is 
 twice as great as the cylinder whose axes form the parallelogram 
 <j>8 because their bases are equal but the axis of the first is twice as 
 great as the axis of the second, and since the cylinder whose axes 
 form the parallelogram <£8 is three times as great as the cone whose 
 vertex is at a and whose base is the circle on the diameter /?S per- 
 pendicular to ay, then the cylinder whose axes form the parallelo- 
 gram <j>m is six times as great as the aforesaid cone. But it has 
 been shown that the spheroid is four times as great as the same 
 cone, hence the cylinder is one and one half times as great as the 
 spheroid. Q. E. D. 
 
 IV. 
 
 That a segment of a right conoid cut by a plane perpendicular 
 to its axis is one and one half times as great as the cone having 
 the same base and axis as the segment, can be proved by the same 
 method in the following way : 
 
 Let a right conoid be cut through its axis by a plane inter- 
 secting the surface in a parabola apy [Fig. 4] ; let it be also cut 
 by another plane perpendicular to the axis, and let their common 
 line of intersection be /3y. Let the axis of the segment be 8a and 
 
l6 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 let it be produced to 6 so that 0a = aS. Now imagine 80 to be a 
 scale-beam with its center at a; let the base of the segment be the 
 circle on the diameter py perpendicular to a8; imagine a cone whose 
 base is the circle on the diameter py, and whose vertex is at a. 
 Imagine also a cylinder whose base is the circle on the diameter py 
 and its axis aS, and in the parallelogram let a straight line jiv be 
 drawn ll/8y and on pv construct a plane perpendicular to a8; it will 
 intersect the cylinder in a circle whose diameter is /iv, and the seg- 
 ment of the right conoid in a circle whose diameter is £o. Now 
 since pay is a parabola, 08 its diameter and |<r and /?8 its ordinates, 
 then [Quadr. parab. 3] 8a : aa = /88 2 : %<j*. But 8a = a6, therefore 
 Oa: a <T = t*a 2 :<r£ 2 . But /x<r 2 :o-| 2 = the circle in the cylinder whose 
 diameter is pv : the circle in the segment of the right conoid whose 
 
 diameter is £0, hence Oa :acr — the 
 circle with the diameter fiv : the 
 circle with the diameter £0 ; there- 
 fore the circle in the cylinder 
 whose diameter is /tv * s m its 
 present position, in equilibrium 
 at the point u. with the circle 
 whose diameter is £0 if this be 
 transferred and so arranged on 
 the scale-beam at 6 that 6 is its 
 center of gravity. And the cen- 
 ter of gravity of the circle whose 
 diameter is /iv is at a, that of the 
 circle whose diameter is £0 when 
 its position is changed, is 0, and we have the inverse proportion, 
 6a:aa = the circle with the diameter fiv : the circle with the diameter 
 £0. In the same way it can be shown that if another straight line 
 be drawn in the parallelogram ey 1 1 /3y the circle formed in the 
 cylinder, will in its present position be in equilibrium at the point a 
 with that formed in the segment of the right conoid if the latter 
 is so transferred to 6 on the scale-beam that 8 is its center of grav- 
 ity. Therefore if the cylinder and the segment of the right conoid 
 are filled up then the cylinder in its present position will be in 
 equilibrium at the point a with the segment of the right conoid if 
 the latter is transferred and so arranged on the scale-beam at 6 that 
 6 is its center of gravity. And since these magnitudes are in equi- 
 librium at a, and k is the center of gravity of the cylinder, if a8 is 
 bisected at k and 6 is the center of gravity of the segment trans- 
 
 Fig. 4. 
 
GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. I 1 / 
 
 ferred to that point, then we have the inverse proportion 9a:a.K = 
 cylinder : segment. But 6a = 2ok and also the cylinder = 2 x segment. 
 But the same cylinder is 3 times as great as the cone whose base is 
 the circle on the diameter /?y and whose vertex is at a ; therefore it 
 is clear that the segment is one and one half times as great as the 
 same cone. 
 
 v. 
 
 That the center of gravity of a segment of a right conoid which 
 is cut off by a plane perpendicular to the axis, lies on the straight 
 line which is the axis of the segment divided in such a way that 
 the portion at the vertex is twice as great as the remainder, may 
 be perceived by our method in 
 the following way: 
 
 Let a segment of a right 
 conoid cut off by a plane per- 
 pendicular to the axis be cut by 
 another plane through the axis, 
 and let the intersection in its sur- 
 face be the parabola a/3y [Fig. 5] 
 and let the common line of inter- 
 section of the plane which cut off 
 the segment and of the intersect- 
 ing plane be /3y ; let the axis of 
 the segment and the diameter of 
 the parabola a/?y be a.8 ; produce 
 8a so that a6 = aS and imagine 80 
 to be a scale-beam with its center 
 at a ; then inscribe a cone in the segment with the lateral boundaries 
 /80 and ay and in the parabola draw a straight line £0 1 1 /3y and let 
 it cut the parabola in | and o and the lateral boundaries of the cone 
 in ir and p. Now because £o- and /?S are drawn perpendicular to the 
 diameter of the parabola, 8a : a<r = /3S 2 : £&* [Quadr. parab. 3]. But 
 Sa:a(T = /?8:™- = £8 2 :/38x7ro-, therefore also /?3 2 : £<r 2 = /?S 2 : £8 x ircr. 
 Consequently £&* = /?S x ira and /?S : £<r = £r : tto-, therefore f3&:ircr = 
 go 2 : cttt 2 . But /?8 : ira = 8a : a<r = 0a : acr, therefore also 0a : ao- = fcr 2 : air'. 
 On |o construct a plane perpendicular to a8; this will intersect the 
 segment of the right conoid in a circle whose diameter is £0 and the 
 cone in a circle whose diameter is irp. Now because 8a: acr = |<r z : air 2 
 and go 2 : aw 2 = the circle with the diameter $0 : the circle with the 
 
l8 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 diameter -up, therefore 6a : ao-=the circle whose diameter is £o : the circle 
 whose diameter is vp. Therefore the circle whose diameter is £o 
 will in its present position be in equilibrium at the point a with the 
 circle whose diameter is irp when this is so transferred to 6 on the 
 scale-beam that 6 is its center of gravity. Now since a is the center 
 of gravity of the circle whose diameter is £o in its present position, 
 and is the center of gravity of the circle whose diameter is -up 
 if its position is changed as we have said, and inversely 0a:acr = the 
 circle with the diameter £o : the circle with the diameter -np, then 
 the circles are in equilibrium at the point a. In the same way it 
 can be shown that if another straight line is drawn in the parabola 
 1 1 /?y and on this line last drawn a plane is constructed perpendicular 
 to a8, the circle formed in the segment of the right conoid will in 
 its present position be in equilibrium at the point a with the circle 
 formed in the cone, if the latter is transferred and so arranged on 
 the scale-beam at 6 that 6 is its center of gravity. Therefore if the 
 segment and the cone are filled up with circles, all circles in the 
 segment will be in their present positions in equilibrium at the point 
 u. with all circles of the cone if the latter are transferred and so ar- 
 ranged on the scale-beam at the point that 6 is their center of 
 gravity. Therefore also the segment of the right conoid in its 
 present position will be in equilibrium at the point a with the cone if 
 it is transferred and so arranged on the scale-beam at 6 that 6 is its 
 center of gravity. Now because the center of gravity of both mag- 
 nitudes taken together is a, but that of the cone alone when its 
 position is changed is 6, then the center of gravity of the remaining 
 magnitude lies on ad extended towards a if a* is cut off in such a 
 way that a$-.aK = segment : cone. But the segment is one and one 
 half the size of the cone, consequently ad = %an and k, the center of 
 gravity of the right conoid, so divides a8 that the portion at the 
 vertex of the segment is twice as large as the remainder. 
 
 VI. 
 
 [Th'e center of gravity of a hemisphere is so divided on its 
 axis] that the portion near the surface of the hemisphere is in the 
 ratio of 5 : 3 to the remaining portion. 
 
 Let a sphere be cut by a plane through its center intersecting 
 the surface in the circle a/fy8 [Fig. 6], ay and /38 being two diameters 
 of the circle perpendicular to each other. Let a plane be con- 
 
GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. IO, 
 
 structed on /?8 perpendicular to ay. Then imagine a cone whose base 
 is the circle with the diameter /38, whose vertex is at a and its 
 lateral boundaries are fta and a8 ; let ya be produced so that ad = ya, 
 imagine the straight line By to be a scale-beam with its center at a 
 and in the semi-circle fia.8 draw a straight line £0 1 1 (38 ; let it cut 
 the circumference of the semicircle in | and o, the lateral boundaries 
 of the cone in ir and p, and ay in c . On |o construct a plane perpen- 
 dicular to ae; it will intersect the hemisphere in a circle with the 
 diameter £0, and the cone in a circle with the diameter m. Now 
 because ay.ae = & 2 : ae 2 and £a 2 = ae 2 + e£ 2 and ae = ew, therefore ay : ae 
 = £e 2 + en 2 : eTr 2 . But $e 2 + €ir 2 :€Tr 2 -the circle with the diameter £0 + 
 the circle with the diameter irp : the circle with the diameter wp, and 
 ya = a0, hence 6a : ae = the circle with the diameter |o + the circle with 
 the diameter np : circle with the diameter irp. 
 Therefore the two circles whose diameters 
 are £0 and irp in their present position are in 
 equilibrium at the point a with the circle 
 whose diameter is wp if it is transferred and 
 so arranged at 6 that 6 is its center of gravity. 
 Now since the center of gravity of the two 
 circles whose diameters are £0 and irp in their 
 present position [is the point e, but of the 
 circle whose diameter is irp when its position 
 is changed is the point 6, then 6a : ae = the 
 circles whose diameters are] $0 [, irp: the 
 circle whose diameter is irp. In the same 
 way if another straight line in the] hemi- 
 sphere /?a8 [is drawn 11/38 and a plane is 
 constructed] perpendicular to [ay the] two 
 [circles produced in the cone and in the hemi- 
 sphere are in their position] in equilibrium at a [with the circle 
 which is produced in the cone] if it is transferred and arranged on 
 the scale at 6. [Now if] the hemisphere and the cone [are filled 
 up with circles then all circles in the] hemisphere and those [in the 
 cone] will in their present position be in equilibrium [with all 
 circles] in the cone, if these are transferred and so arranged on the 
 ( scale-beam at 6 that 6 is their center of gravity; [therefore the 
 hemisphere and cone also] are in their position [in equilibrium at 
 the point a] with the cone if it is transferred and so arranged [on 
 the scale-beam at 6] that $ is its center of gravity. 
 
20 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 VII. 
 
 By [this method] it may also be perceived that [any segment 
 whatever] of a sphere bears the same ratio to a cone having the 
 same [base] and axis [that the radius of the sphere + the axis of the 
 
 opposite segment : the axis of the opposite segment] 
 
 and [Fig. 7] on /iv construct a plane perpendicular to ay; it will 
 intersect the cylinder in a circle whose diameter is par, the segment 
 of the sphere in a circle whose diameter is $0 and the cone whose 
 base is the circle on the diameter e£ and whose vertex is at o in 
 
 a circle whose diameter 
 & is irp. In the same way 
 as before it may be 
 shown that a circle whose 
 diameter is pv is in its 
 present position in equi- 
 a librium at a with the two 
 circles [whose diameters 
 are fo and np if they are 
 so arranged on the scale- 
 beam that 6 is their cen- 
 ter of gravity. [And the 
 same can be proved of 
 all corresponding cir- 
 cles.] Now since cylin- 
 der, cone, and spherical 
 segment are filled up 
 with such circles, the 
 cylinder in its present 
 position [will be in equilibrium at a] with the cone + the spherical 
 segment if they are transferred and attached to the scale-beam at 0. 
 Divide arj at <f> and x so that a X = xv and v <p = y 3 a<f> ; then x will be the 
 center of gravity of the cylinder because it is the center of the axis 
 arj. Now because the above mentioned bodies are in equilibrium 
 at o, cylinder : cone with the diameter of its base e£ + the spherical 
 segment /3a8 = 6a.:a X . And because ^^3^ then [ yv x ,<£]= ^ x^y. 
 Therefore also yr)Xr^>- %/?)f 
 
 t/x/f 
 
 \\e 
 
 \o 
 
 
 p- 
 
 
 1 ff 
 
 ■;e\ 
 
 
 
 
 
 ■f \ 
 
 
 /e 
 
 6 V 
 
 
 V 
 
 )p 
 
 \ 
 
 Fig 7. 
 
 vna. 
 
 In the same way it may be perceived that any segment of an 
 ellipsoid cut off by a perpendicular plane, bears the same ratio to 
 
GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 21 
 
 a cone having the same base and the same axis, as half of the axis 
 of the ellipsoid + the axis of the opposite segment bears to the axis 
 of the opposite segment 
 
 VIII. 
 
 produce ay [Fig. 8] making a6 - ay and y£ = the radius of the sphere ; 
 imagine y& to be a scale-beam with a center at a, and in the plane 
 cutting off the segment inscribe a circle with its center at 77 and its 
 radius = arj; on this circle construct a cone with its vertex at a and 
 its lateral boundaries ae and a£. Then draw a straight line kX 1 1 «£ ; 
 let it cut the circumference of the 
 segment at k and \, the lateral bound- 
 aries of the cone ae£ at p and o and ay 
 at ir. Now because ay : air = ok 2 : air 2 
 and Ka 2 = atr 2 + irK 2 and air 2 = iro 2 (since 
 also arj 2 = er) 2 ) , then ya : air = kit 2 + iro 2 : 
 oir 2 . But kit 2 + iro 2 : iro 2 = the circle 
 with the diameter k\ + the circle with 
 the diameter op: the circle with the 
 diameter op and ya = a8; therefore 
 6a: air = the circle with the diameter 
 k\+ the circle with the diameter op: 
 the circle with the diameter op. Now 
 since the circle with the diameter k\ + 
 the circle with the diameter op : the 
 circle with the diameter op = a6:ira, 
 let the circle with the diameter op be 
 transferred and so arranged on the 
 scale-beam at 6 that 8 is its center of 
 gravity; then 8a : air = the circle with 
 
 the diameter k\ + the circle with the diameter op in their present 
 positions : the circle with the diameter op if it is transferred and 
 so arranged on the scale-beam at 8 that 8 is its center of gravity. 
 Therefore the circles in the segment /?aS and in the cone ae£ are in 
 equilibrium at a with that in the cone acf. And in the same way 
 all circles in the segment f3a8 and in the cone ait, in their present 
 positions are in equilibrium at the point a with all circles in the 
 cone ae£ if they are transferred and so arranged on the scale-beam 
 at 6 that 8 is their center of gravity ; then also the spherical segment 
 
22 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 a/38 and the cone aef in their present positions are in equilibrium 
 at the point a with the cone eat if it is transferred and so arranged 
 on the scale-beam at 6 that is its center of gravity. Let the cyl- 
 inder pv equal the cone whose base is the circle with the diameter 
 e£ and whose vertex is at a and let wq be so divided at <f> that arj = 4<j>rj ; 
 then </> is the center of gravity of the cone eat, as has been previously 
 proved. Moreover let the cylinder pv be so cut by a perpendicularly 
 intersecting plane that the cylinder a is in equilibrium with the 
 cone eat. Now since the segment a/38 + the cone ea£ in their present 
 positions are in equilibrium at u. with the cone eat if it is trans- 
 ferred and so arranged on the scale-beam at that 8 is its center 
 of gravity, and cylinder /w = cone eat, and the two cylinders /u + v 
 are moved to 6 and jw is in equilibrium with both bodies, then will 
 also the cylinder v be in equilibrium with the segment of the sphere 
 at the point a. And since the spherical segment j8a8 : the cone whose 
 base is the circle with the diameter /?8, and whose vertex is at a = 
 irj\ryy (for this has previously been proved [De sph. et cyl. II, 2 
 Coroll.]) and cone /?a8 : cone ca£ = the circle with the diameter 
 y88 : the circle with the diameter e£ = /3»f : 77c 2 , and fi-q 2 = yq x rja, 
 qe 2 = rja 2 , and y-qXr/a: rja 2 = yq : -qa, therefore cone /3a8 : cone eat — 
 y-q-.-qa. But we have shown that cone /3a8 : segment /3a8 = yij : i/£, 
 hence 8t' laov segment /3a8 : cone eat-^q-.-qa. And because ax'-xv = 
 qa + 4rjy : a-q + 2-qy so inversely -qx : x a = 2 yv + v a '■ 4yv + v a an d by addi- 
 tion rja : ax "= 6yq + 2-qa : 17a + 4-qy. But tj£ = %(6rjy + 2-qa) and y<j> F= 
 
 Vi (4w + v a ) > f° r that is evident. Hence -qa : ax = £17 : y<£> consequently 
 also ^rj-.rja = y<j>:xa- But it was also demonstrated that £q:-qa= the 
 segment whose vertex is at a and whose base is the circle with the 
 diameter /?8 : the cone whose vertex is at a and whose base is the 
 circle with the diameter et; hence segment j3aZ : cone eat = y<j>: X a.- 
 And since the cylinder /t is in equilibrium with the cone eat at a, and 9 
 is the center of gravity of the cylinder while is that of the cone 
 eat,, then cone eat, : cylinder p = 6a:a<f> = ya:a<l>. But cylinder fiv = 
 cone eat ; hence by subtraction, cylinder jx : cylinder v = a<f> : y<f>. And 
 cylinder a V = cone eat ; hence cone eat ■ cylinder v = ya : y<f> = 6a : y<£. 
 But it was also demonstrated that segment /3aS : cone eaf = y<£: X a; 
 hence Si' tcrou segment /8a8 : cylinder v = £a:a X . And it was demon- 
 strated that segment /Ja8 is in equilibrium at a with the cylinder v 
 and is the center of gravity of the cylinder v, consequently the 
 point x is also the center of gravity of the segment /3a8. 
 
GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 23 
 
 IX. 
 
 In a similar way it can also be perceived that the center of grav- 
 ity of any segment of an ellipsoid lies on the straight line which is 
 the axis of the segment so divided that the portion at the vertex 
 of the segment bears the same ratio to the remaining portion as the 
 axis of the segment +4 times the axis of the opposite segment 
 bears to the axis of the segment + twice the axis of the opposite 
 segment. 
 
 x. 
 
 It can also be seen by this method that [a segment of a byper- 
 boloid] bears the same ratio to a cone having the same base and axis 
 as the segment, that the axis of the segment + 3 times the addition 
 to the axis bears to the axis of the segment of the hyperboloid + twice 
 its addition [De Conoid. 25] ; and that the center of gravity of the 
 hyperboloid so divides the axis that the part at the vertex bears the 
 same ratio to the rest that three times the axis + eight times the 
 addition to the axis bears to the axis of the hyperboloid + 4 times 
 the addition to the axis, and many other points which I will leave 
 aside since the method has been made clear by the examples already 
 given and only the demonstrations of the above given theorems re- 
 main to be stated. 
 
 XI. 
 
 When in a perpendicular prism with square bases a cylinder is 
 inscribed whose bases lie in opposite squares and whose curved 
 surface touches the four other parallelograms, and when a plane is 
 passed through the center of the circle which is the base of the 
 cylinder and one side of the opposite square, then the body which 
 is cut off by this plane [from the cylinder] will be % of the entire 
 prism. This can be perceived through the present method and 
 when it is so warranted we will pass over to the geometrical proof 
 of it. 
 
 Imagine a perpendicular prism with square bases and a cyl- 
 inder inscribed in the prism in the way we have described. Let the 
 prism be cut through the axis by a plane perpendicular to the plane 
 which cuts off the section of the cylinder; this will intersect the 
 prism containing the cylinder in the parallelogram a/? [Fig. 9] and 
 the common intersecting line of the plane which cuts off the section 
 of the cylinder and the plane lying through the axis perpendicular 
 
24 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 to the one cutting off the section of the cylinder will be j8y ; let the 
 axis of the cylinder and the prism be y8 which is bisected at right 
 angles by e£ and on d, let a plane be constructed perpendicular to 
 yS. This will intersect the prism in a square and the cylinder in a 
 circle. 
 
 /^ X 
 
 
 N. X, 
 
 ye 
 
 70 
 
 o 
 Fig. 10. 
 
 Now let the intersection of the prism be the square p.v [Fig. 10], 
 that of the cylinder, the circle £owp and let the circle touch the sides 
 of the square at the points £, o, it and p ; let the common line of 
 intersection of the plane cutting off the cylinder-section and that 
 passing through e£ perpendicular to the axis of the cylinder, be k\; 
 this line is bisected by tt8£. In the semicircle oirp draw a straight 
 line err perpendicular to w^, on <rr construct a plane perpendicular 
 to £ir and produce it to both sides of the plane enclosing the circle 
 io-ap ; this will intersect the half-cylinder whose base is the semi- 
 circle oirp and whose altitude is the axis of the prism, in a parallelo- 
 gram one side of which = or and the other = the vertical boundary 
 of the cylinder, and it will intersect the cylinder-section likewise 
 in a parallelogram of which one side is or and the other w [Fig. 9] ; 
 and accordingly w will be drawn in the parallelogram 8e 1 1 fi<a and 
 will cut off ei = irx- Now because ey is a parallelogram and vi 1 1 By, 
 and ed and /?y cut the parallels, therefore dd : 81 = toy : yv = $u> : w. Eut 
 j8aj : w - parallelogram in the half-cylinder : parallelogram in the 
 cylinder-section, therefore both parallelograms have the same side 
 or; and £0 = 0*-, i0 = x#; and since tt0 = 0£ therefore 0£:0x= paral- 
 lelogram in half-cylinder : parallelogram in the cylinder-section. 
 Imagine the parallelogram in the cylinder-section transferred and 
 so brought to f that £ is its center of gravity, and further imagine 
 
GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 25 
 
 ir£ to be a scale-beam with its center at 8 ; then the parallelogram in 
 the half-cylinder in its present position is in equilibrium at the 
 point 8 with the parallelogram in the cylinder-section when it is trans- 
 ferred and so arranged on the scale-beam at £ that £ is its center of 
 gravity. And since % is the center of gravity in the parallelogram 
 in the half-cylinder, and | that of the parallelogram in the cylinder- 
 section when its position is changed, and £8 : $x - the parallelogram 
 whose center of gravity is x '■ tne parallelogram whose center of 
 gravity is £, then the parallelogram whose center of gravity is x 
 will be in equilibrium at with the parallelogram whose center of 
 gravity is £. In this way it can be proved that if another straight 
 line is drawn in the semicircle oirp perpendicular to ir8 and on this 
 straight line a plane is constructed perpendicular to ird and is pro- 
 duced towards both sides of the plane in which the circle £oirp lies, 
 then the parallelogram formed in the half-cylinder in its present 
 position will be in equilibrium at the point 6 with the parallelogram 
 formed in the cylinder-section if this is transferred and so arranged 
 on the scale-beam at £ that £ is its center of gravity ; therefore also 
 all parallelograms in the half-cylinder in their present positions will 
 be in equilibrium at the point 6 with all parallelograms of the 
 cylinder-section if they are transferred and attached to the scale-beam 
 at the point |; consequently also the half-cylinder in its present 
 position will be in equilibrium at the point 8 with the cylinder- 
 section if it is transferred and so arranged on the scale-beam at £ 
 that I is its center of gravity. 
 
 XII. 
 
 Let the parallelogram pv be perpendicular to the axis [of the 
 circle] £0 [np] [Fig. 11]. Draw 8 p. and 
 8t) and erect upon them two planes per- 
 pendicular to the plane in which the 
 semicircle oirp lies and extend these 
 planes on both sides. The result is a 
 prism whose base is a triangle similar 
 to 8prj and whose altitude is equal to 
 the axis of the cylinder, and this prism 
 is % of the entire prism which contains 
 the cylinder. In the semicircle oirp and 
 in the square p.v draw two straight lines 
 k\ and tv at equal distances from «•£; 
 these will cut the circumference of the semicircle oirp at the points 
 
 Fig. 11. 
 
26 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 k and t, the diameter op at <r and £ and the straight lines ftj and 6/1 
 at <£ and x- Upon k\ and ™ construct two planes perpendicular 
 to op and extend them towards both sides of the plane in which lies 
 the circle £oirp; they will intersect the half-cylinder whose base is 
 the semicircle oirp and whose altitude is that of the cylinder, in a 
 parallelogram one side of which =k<t and the other = the axis of 
 the cylinder; and they will intersect the prism (hjp. likewise in a 
 parallelogram one side of which is equal to Ax and the other equal 
 to the axis, and in the same way the half-cylinder in a parallelogram 
 one side of which = t£ and the other = the axis of the cylinder, and 
 the prism in a parallelogram one side of which =v<j> and the other 
 = the axis of the cylinder 
 
 XIII. 
 
 Let the square a/J-yS [Fig. 12] be the base of a perpendicular 
 prism with square bases and let a cylinder be inscribed in the prism 
 
 whose base is the circle e£-q0 which 
 touches the sides of the parallelogram 
 a/?y8 at e, f, i] and 6. Pass a plane 
 through its center and the side in the 
 square opposite the square a^yS corre- 
 sponding to the side yS ; this will cut 
 off from the whole prism a second prism 
 which is Yi the size of the whole prism 
 and which will be bounded by three 
 parallelograms and two opposite tri- 
 angles. In the semicircle z£,-q describe 
 a parabola whose origin is ^e and whose 
 axis is £k, and in the parallelogram S v draw par 1 1 k£ ; this will cut 
 the circumference of the semicircle at £, the parabola at A, and 
 p.vxv\ = v£ 2 (for this is evident [Apollonios, Con. I, 11]). Therefore 
 liv.vk^Kyf-.Xo 2 . Upon p. v construct a plane parallel to &,; this will 
 intersect the prism cut off from the whole prism in a right-angled 
 triangle one side of which is p, v and the other a straight line in the 
 plane upon y8 perpendicular to y 8 at v and equal to the axis of the 
 cylinder, but whose hypotenuse is in the intersecting plane. It will 
 intersect the portion which is cut off from the cylinder by the plane 
 passed through 07 and the side of the square opposite the side yS 
 in a right-angled triangle one side of which is tf and the other 
 a straight line drawn in the surface of the cylinder perpendicular 
 
 Fig. 12. 
 
GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 27 
 
 to the plane kv, and the hypotenuse 
 
 and all the triangles in the prism : all the triangles in the cylinder- 
 section = all the straight lines in the parallelogram 817 : all the straight 
 lines between the parabola and the straight line 07. And the prism 
 consists of the triangles in the prism, the cylinder-section of those 
 in the cylinder-section, the parallelogram fy of the straight lines 
 in the parallelogram 8r)\\ k£ and the segment of the parabola of the 
 straight lines cut off by the parabola and the straight line eq ; hence 
 prism : cylinder-section = parallelogram -qS : segment e£-q that is 
 bounded by the parabola and the straight line eq. But the parallelo- 
 gram S17 = % the segment bounded by the parabola and the straight 
 line E77 as indeed has been shown in the previously published work, 
 hence also the prism is equal to one and one half times the cylinder- 
 section. Therefore when the cylinder-section ■= 2, the prism = 3 and 
 the whole prism containing the cylinder equals 12, because it is four 
 times the size of the other prism ; hence the cylinder-section is equal 
 to % of the prism, Q. E. D. 
 
 XIV. 
 
 [Inscribe a cylinder in] a perpendicular prism with square 
 bases [and let it be cut by a plane passed through the center of the 
 base of the cylinder and one side of the opposite square.] Then this 
 plane will cut off a prism from the whole prism and a portion of 
 the cylinder from the cylinder. It may be proved that the portion 
 cut off from the cylinder by the plane is one-sixth of the whole 
 prism. But first we will prove that it is possible to inscribe a solid 
 figure in the cylinder-section and to circumscribe another composed 
 of prisms of equal altitude and with similar triangles as bases, so 
 that the circumscribed figure exceeds the inscribed less than any 
 given magnitude 
 
 But it has been shown that the prism cut off by the inclined plane 
 <% the body inscribed in the cylinder-section. Now the prism 
 cut off by the inclined plane : the body inscribed in the cylinder- 
 section = parallelogram 8*7 : the parallelograms which are inscribed 
 in the segment bounded by the parabola and the straight line vq. 
 Hence the parallelogram 817 <% the parallelograms in the segment 
 bounded by the parabola and the straight line eq. But this is im- 
 possible because we have shown elsewhere that the parallelogram 
 877 is one and one half times the segment bounded by the parabola 
 
 and the straight line 07, consequently is 
 
 not greater 
 
28 GEOMETRICAL SOLUTIONS DERIVED FROM MECHANICS. 
 
 And all prisms in the prism cut off by the inclined plane: all 
 prisms in the figure described around the cylinder-section = all 
 parallelograms in the parallelogram $77 : all parallelograms in the 
 figure which is described around the segment bounded by the 
 parabola and the straight line eq, i. e., the prism cut off by the in- 
 clined plane : the figure described around the cylinder-section = 
 parallelogram 817 : the figure bounded by the parabola and the 
 straight line ey. But the prism cut off by the inclined plane is 
 greater than one and one half times the solid figure circumscribed 
 around the cylinder-section 
 

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