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 GEOMETRICAL SOLUTIONS 
 
 or THK LEtfOTHS AND DIVISION OF 
 
 CIRCULAR ARCS, 
 
 THE 
 
 * v.-!, ■ . • 
 
 QUADRATURE OF THE CIRCLE. TRISEQ 
 THE ANGLE, DUPLICATION OP TH 
 
 AND THE 
 
 QDIDRATDRE OF THB H¥PfiRBOLi. 
 
 B7 
 
 PETER FLEMING, CIVIL ENGINEER, 
 
 AUTHOR OF " A SYSTEM OF LAND SURVEYING," " A METHOD OF 
 
 MEASimiNG A BASE LINE BY ANGULAR OBSERVATION," 
 
 AND << GEOMETRICAL SOLUTIONS OF THE 
 
 QUADRATURE OF THE CIRCLE." 
 
 MONTREAL: 
 
 PRINTED FOR THE AUTHOR, 
 
 1851. 
 
PREFACE. 
 
 In treating of Geometrical Solutions, of the Quadra- 
 ture of the Circle, the perimeters of regular polygons of a 
 given circle are made the chords of another given arc, having 
 one common point on that arc— that is, the first is the peri- 
 meter of the least polygon or inscribed square (triangle) and 
 by doubling the number of sides of this polygon, the perime- 
 ter of which becomes the second chord from the same 
 point ; and by thus successively doubling the number of 
 sides, it is evident the perimeter of the ultimate polygon will 
 be equal to the circumference of the given circumscribing 
 circle. In the same manner, taking the circumscribing 
 and corresponding polygonal perimeters, that the ultimate 
 polygonal perimeter will also come to be equal to the same 
 circumference. Or let p be equal to the perimeter of the 
 inscribed square, (triangle) and be the first term of an in- 
 finite series— and p' the perimeter of the polygon double of 
 the number of sides of the first — and p" that of the polygon 
 of double the number of sides of the second &c. ; then p' — p 
 will be the second term, p" — p' the third term, and p" — p" 
 fourth term, &c. ad infinitum. In the same manner, let q 
 be equal to the perimeter of the circumscribing square 
 (triangle) of the same circle, be the first term of an infinite 
 series, q' that of the polygon of double the number of sides of 
 which q is the perimeter ; then q' — q will be the second term, 
 q"— q' the third term, and q** — q* the fourth term, &c. ad 
 infinitum ; and the first will be a converging infinite series of 
 the form p+(p'— p)+(p''_p')+(p«'— p'')+ &c., and the lar- 
 ter will also be a converging infinite series of the form q — 
 (q'— q)— (q"— q')— (q*^— q")— &c., and each series will be 
 
H. 
 
 equal to the circumference of the given circle ; but as the 
 first series is an increasing one, and the other a diminishing 
 one, it is evident that the ultimate chord of chords p, p', p" 
 &c. ad infinitum, and the ultimate chord of q, q", q", 6ic. ad 
 infinitum, must meet in a common point n on the arc, and this 
 point joined to the common point A of meeting of all the 
 chords p, p', p', and q, q', q" ; the line An, (fig. 8, Geo- 
 metrical Solutions of the Quadrature of the Circle) must be 
 the ultimate chord of both, and equal to the circumference 
 of the given circle ; by which it is resolved into a straight 
 line. Such is the result presumed to be obtained by each of 
 the Nine Geometrical Solutions of the Quadrature of the 
 Circle, which are the Geometrical Summations of the 
 infinite series, derived from the differences of polygonal 
 perimeters equal to the circumference. 
 
 In the present work we find the lengtlis of circular 
 arcs, by the same geometrical summation of the series deri- 
 ved from their perimeters — that is, by regular polygonal di- 
 vision of the arc, giving both inscribed and circumscribed 
 perimeters, by which the length of the arc in the same man- 
 ner is found ; and if the arc is made a known part of the 
 circumference, the length of the whole circumference, con- 
 sequently is a multiple of the arc given in a straight line, as 
 exemplifiei by propositions, 1 and 2. 
 
 The division of circular arcs, or of the whole circle, 
 into any number of equal parts , are resolved in the same 
 manner, by infinite series derived from the number of divi- 
 sions required, as exemplified in propositions, 3, 4, and 5, 
 and by which any equal division of circular arcs, or of the 
 whole circumference may be obtained. Consequently 
 the Trisection of the Angle comes only to be one example, 
 of an indefinite number, of the same general application. 
 
 The Duplication of the Cube and the Quadrature of 
 the Hyperbola, appear to be no more than examples of the 
 same application ; or the geometrical solutions of summation 
 of infinite converging series, of which the form is alternately 
 
111. 
 
 plus and minus in its terms — for the duplication is the sum- 
 
 3 
 
 mation of the Binomial Theorem (a4.x)|=^ 2, and is 
 only an example of an indefinite number, of which a 
 power or root may be in the same manner represented by 
 a straight line. Also, the Quadrature of the Hyperbola 
 is only the geometrical solution of summation of a series of 
 the same form, as shown by propositions, 6, 7, and 8. 
 
 It may now appear, by both the Geometrical Solution 
 of the Quadrature of the Circle, and those of this work, that 
 the whole of the hitherto unresolved problems of the an- 
 cients, namely, the Quadrature of the Circle, Trisection of 
 the Angle, Duplication of the Cube, Quadrature of the Hy- 
 perbola, and Lengths and Division of Circular Arcs, which 
 have vainly been attempted to be solved by finite lines or 
 expressions sought for upon the supposition of distinct or un- 
 connected principles for each, both by geometry and analy- 
 sis, belong only to one of general application to all, — which 
 is that of the Geometrical summation of infinite converging 
 series. 
 
 It may farther appear, that the finite solutions of these 
 ancient problems, if thus depending upon such common ba- 
 sis — the geometrical summation of infinite series — uone of 
 them could ever be found singly, or by a different construc- 
 tion for each ; and consequently, without this common basis, 
 or common construction heit r; discovered, they all might still 
 and forever remain unsolved. I ut even under this view of the 
 basis known, the duplication of the cube may have remained 
 unsolved, without the discovery of the Binomial Theorem 
 due to Sir Isaac Newton ; although in all probability, from 
 the advancements made in analysis since his time, 
 the discovery of this series would be an ordinary result. 
 All which may well account for the mysterious obstacles 
 which have so long lain in the way of all past attempts, 
 either by geometry or analysis, to solve them ; and by which 
 these problems remained to be an open challenge to the 
 

 IV. 
 
 intellectual capacity and ingenuity of past ages to remove. 
 The pretension of solving geometrically the Summa- 
 tion of Infinite Series, as given in the Geometrical Solutions 
 of the Circle and the present work, is wholly based upon the 
 variation or movement of the points representing the differ- 
 ences of the chords, or perimeters on a given arc, which 
 differences give the terms of two infinite series, each of 
 which evidently roust vanish at a common point n ; and 
 that there must be made a point of variation of each term, 
 of the points of one of the series through which 
 
 or 
 
 arcs of intersection are made, with arcs described through 
 points of the opposite series, that those intersections con- 
 tinued ad infinitum must be on a straight line, and this 
 line must meet the arc in that vanishing point, (" Geome- 
 trical Solutions of the Quadrature of the Circle,', Lemma 
 3, 4, 5, &c.) and Lemma ; by which it is evident, the 
 chord of this point will be common to both series, and be 
 without any approximation ; therefore, this construction 
 will become of general application, to find in a straight line 
 the sum of any two converging series whatever, if connected 
 by a common law, or of that in which its terms are alter- 
 nately plus and minus. 
 
 PETER FLEMING. 
 Montreal, December, 1850. 
 
GEOMETRICAL SOLUTIONS 
 
 .• ;4, IM ■ or THE LCN0TH8 AMD DIVISION OF 
 
 CIRCULAR ARCS, &c. 
 
 LEMMA 1. 
 
 FIG. 1. With any radius CA, describe the 
 semicircle ABDE, and make the chords 
 AB and BD, each equal to the radius 
 AC. Also describe from the point A, 
 through C, the arc BC, and from B, the 
 arc ACD, and bisect the arc CD in the 
 point D', and join A and D. Then from 
 any number of points a, b, c, d, e, f, &c,. 
 of the arc BD, nnake on the arc BC, 
 the arc Ba' equal to the arc B a, — the 
 arc B b' equal to the arc B b, — the arc 
 B C equal to the arc B c, &c, — and from 
 the point A, as a centre, with the dis- 
 tance A a, describe the arc a m ; and 
 from the point C through a', describe 
 the arc a' m, intersecting the arc a m in 
 the point m. In the same manner, from 
 the points A and C as centres, describe 
 the intersections n, o, p, q, r, &c., and 
 through these intersections, which let it 
 be granted are of indefinite number, draw 
 the line B, m, n, o, p, q, r, &c, cutting 
 
8 
 
 the arc CD; — This line shall be a curv- 
 ed line passing through the point I>. 
 
 For make the arc BI equal to the 
 arc BB' ; the arc CD' is by construction, 
 equal to the arc CB', and BB' is equal to 
 CD' — hence the point D', must be on the 
 intersection of the arcs B'D' and ID' — 
 consequently the point D', must be on 
 the line Bmnopqr, &c. — But the radii of 
 each intersection are unequal, and each 
 intersection of different radii — therefore 
 the line Bmnopqr, &c, must be a curv- 
 ed line, cutting the arc CD in the point 
 D'. 
 
 LEMMA 2. 
 FIG. 2. From the point A as a centre, de- 
 scribe through B the arc BCA", and from 
 D' as a centre describe through B the arc 
 BA'A" meeting the arc BCA" in the 
 point A" — and from A" as a centre de- 
 scribe the arc AD', cutting the arc BA' 
 in the point C, and the arc BC in the 
 point B , and from C through D' des- 
 cribe the arc A"D'H ; and join A'H, 
 which by construction must pass through 
 the point B'. Next make the arcs Ba 
 and Ba' equal to each other (Lemma 1 ) 
 — and Bb' equal to Bb &c, and through 
 the points a and a', b and b', c and c', &c, 
 describe from C and A, the intersections 
 
^y 
 
 / 
 
 in,n,o,p, &c. Also make D'a' equal to D' 
 a, D'b' equal toD'b &c., and through the 
 points a and a', describe from C and A'' 
 as centres the intersection m' — In the 
 same manner describe the intersections 
 n', o', p', &c, — It is evident by construc- 
 tion that a line, draun through the points 
 B, m, n, o, p; &c,and a line drawn through 
 the points D',m',n',o',p', &c. must meet 
 in a common point G, on the straight 
 line A'B'H ; for the arcs BH and BB', 
 are symmetvical with the arcs D'H and 
 D'B', to the straight line A B'H. The 
 points B, m, n, o, p, &c, aul the points 
 D\ mS n', o\ p\ &c., shall be on the arc 
 of a circle described through the points 
 BGD'. 
 
 From the points C and C', through 
 O, describe the arcs G e and G e', and 
 from the points A and A", describe the 
 arcs Ge" and Ge'", — by construction Be 
 is equal to Be", and D'e' is equal to D'e". 
 
 Descri])e through the points BGD' 
 (constructed as fig. 2) the arc D'GBb, FIG. 3. 
 and from G as a centre, with the dis- 
 tance D'B (equal AC) describe the arc 
 b a" r, intersecting the arc D'GBb in 
 the points b and r, — and from bVith the 
 same radius (AC), through G describe 
 the arc Ga, intersecting the arc b a" in 
 the point a'. Next on the arc b a" r, 
 

 "7 
 
 i(/^ 
 
 )^y 
 
 ^■ 
 
 % 
 
 make the arcs b c' and r q, each equal 
 to the arc BC or DC, and with the ra- 
 dius AC, describe through G and c', the 
 arc Gc'a, meeting the arc Ga'a in the 
 point a, and intersecting the arc BB'C 
 in the point d, and through G and q 
 describe the arc Gq intersecting the 
 arc AB'D' in the point d'. Also on the 
 arc Ga'a, make the arc Gc equal to the 
 arc D'C, and through the points b and 
 c, describe the arc be, intersecting the 
 arc Gc'a in the point b', and make 
 pq' equal to BC, and through q' and B 
 describe the arc Bq', intersecting the arc 
 bb'c in the point b" ; then from c, through 
 a and b, describe the semi-circular arc 
 a b e, and draw through the points a 
 and c the straight line ace, meeting 
 the semi-circular arc in the point e. 
 Also from q describe the arc Gh — from 
 c describe the arc Gh'H', — and from q' 
 the arc Bh". 
 
 By construction the curvilinear 
 angle d'Gg', is equal to the curvilinear 
 angle d'D'g. The curvilinear angle 
 dBg, is equal to the curvilinear angle 
 dGg, and the curvilinear angle b" b g'' 
 is equal to the curvilinear angle b"Bg'' 
 — also by the same the angle g" b 
 h" is equal to the angle g"Bh" — 
 gBh'to gGh'— g'Ghto g'D'h. Hence 
 
11 
 
 it is evident that the straight line Lh, 
 
 must pass through the intersection d', 
 
 and bisect the arc D'G, in the point g^^^ gTH^ ^ 
 
 and Lh' bisect the arc GB, in the poij 
 
 g, and Lh" bisect the arc bB in the po j 
 
 g" (the point L, the centre of the 
 
 bBGD.) 
 
 Also by construction, the 
 BGD', is related to the semi-circle 
 ABDE, the same as the arc bBG is re- 
 lated to the semi-circle abH'e ; and the 
 corresponding centres of construction 
 in each, are in position symmetrical and 
 equal — that is, the point a, corresponds 
 to the point A — the centre c, to the 
 centre C— and c' to C— b' to B' &c. 
 Also the curvilinear triangle bb'G, is 
 similar and equal to the triangle BB'D' 
 — the triangles bb"B — BdG — Gd'D', 
 are similar and equal to each other, and 
 each similar to the triangle bb'G or to 
 BB'D'. 
 
 Again, from the centre C, through 
 the point g, describe the arc gi', and 
 from A de«!cribe the arc g i ; also from c' 
 describe the arc gP and from q the arc 
 gf. Now, by construction the quadrila- 
 teral curvilinear figure BdGh', is simi- 
 lar to the quadrilateral curvilinear 
 figure BB'D'H of which D'e' is equal to 
 De", and Be" is equal to Be — (fig. 2,) 
 
12 
 
 -.1 
 
 Therefore, Gf ' must be eqnal to Gf^ 
 and Bi' equal to Bi. Also Gf must 
 be equal to IVi, and Gf equal to 
 Di' — Bi equal to Gl, and Bi equal to 
 Gr — and the same of bm equal to bm', 
 Bn equal to Bn', and the whole of the 
 corresponding arcs of the quadrilateral 
 figures bh"Bb"— Bh'Gd — and GhD'd' 
 must be equal to each other ; but the 
 point g' by construction is a point of 
 the curve D'G (fig 2.) and the point g 
 is a point of the curve BG. Also G is 
 a point of the curve Bgg'D (fig. 2.> but 
 by construction the line BgGgTK is the 
 arc of a circle described from the centre 
 L — hence the points D',g', G, g, B, must 
 be on a circular arc. 
 
 LEMMA 3. 
 
 FIG. 4. I^®* ^'^y ^^^ ^^ ^6 equal to the 
 sum of any converging series — that is 
 the arcs AlH-12-]-23-|- &c. the first, se- 
 cond, third, &c. terms. Also the 
 arc An, equal to, or the sum of another 
 series, and the arcs Al'— 1'2'— 2'3'— &c. 
 the first, second, third, &c. terms ; but 
 the one series varying as the other va- 
 ries or dependant upon each other by 
 the same law, and consequently the 
 terms of each equally vanishing at the 
 point n.* 
 
 • Iiiritance siich series miiy be formed,by the chord 
 A 1 made equal to the perimeter of the inscribed 
 
13 
 
 From any point H with the radius 
 AC, describe the arc FCA', and bisect 
 the arc CF in the point !>', and with 
 the radius AC through H and D", des- 
 cribe the arc HD", and bisect the arc 
 HD" in the point G', and from the 
 middle of the arc of the two last terms, 
 as 33', 44', 55', &c. with the distance 
 CG', cut the arc nE in the point D, 
 and from D with the distance AC or 
 radius, cut the arc nA in the point H', 
 and from H', with the same radius des- 
 cribe the arc DCA". Next bisect the 
 arc CD in the point D', and with the 
 radius AC, describe the arc HD', and 
 bisect IfD' in the point G. 
 
 Again from the point A', describe 
 through the points H' 1, 2, 3, &c. the 
 arcs H'C, It, 2s, 3r, &c. Next make the 
 arcsH'l',H'2",H'3",&c, equal to HI, 
 H'2, H'3, &c, and from the centre C des 
 cribe through the points 1 ", 2", 3", &c., 
 the arcs l"t', 2"s', 3"r , &c., making the 
 intersections d,e,f, &c., and through H', 
 d,e,f, &c., describe the circular arc H'G 
 D'. The points of intersections, d,e,f, 
 &c., shall all be only on the circular arc 
 
 square or polygon of four side^ ^f " given circle — the 
 chord A2, equal to that of eight sides, &c. — and the 
 chord Al' equal to the perimeter of the circumscribing 
 polygon of four sides, and the chord A2', equal to that 
 of eight sides, &c. (See Geom. Sol. of the Circle.) 
 
14 
 
 (Lemma 2) H'G, described from the 
 centre L or on half of the arc H'D'. 
 
 For from the centre C, with the dis- 
 tance CG, describe the arc Gg, meeting 
 the arc H'C in the point g, and let H'h 
 equal Hn, and because (always suppos- 
 ed) I'n is greater than In, the arc 3'n is 
 greater than 3n — hence Hg must be 
 greater than Hn' — therefore all the 
 intersections d, e, f...n must be on the 
 circular arc H'G (Lemma 2, fig. 2) or 
 between the points H' and G. 
 
 LEMMA 4. 
 
 FIG. 4. Through the intersections d,e,f, &c., 
 describe the circular arc H'GD' from the 
 centre L (Lemma 2.) and from the point 
 D' through the points of intersection e" 
 f' &c., the arcs eV, f f' &c. And 
 from th6 point A', describe through the 
 pointee f g, &c., the arcs ex, f'y, g'z, 
 &c., meeting the arc Hn, in the points 
 x,y,z, &c. 
 
 The arc Ix shall be the variation of 
 the point 1 towards the point 2, and the 
 arc 2y shall be the variation of the 
 point 2 towards the point 3, &c. (Lem- 
 ma 9, Solutions Quadrature of the Cir- 
 cle). Then from the point D, through 
 the point x, y, z, &c, and from the point 
 H', through the points 2', 3', 4', &c. des- 
 
15 
 
 I 
 
 cribe the intersections a, b, c, &c. The 
 points a, b, c, shall be on one straight 
 line a b c.n. 
 
 For from the point H' through the 
 points 1', 2', 3', &c. and the point D 
 through the points 1, 2, 3, &c. describe 
 the intersections /, a, b^ c, &c. and draw 
 the curve line I ab c.n. Also, con- 
 tinue 2' a to meet / 1 in the point a' — 
 and 3' b to meet 2 a in the point bf &c., 
 and draw the curve line abc,...n. — 
 Now by construction the intersection w 
 has moved to be the point a — the inter- 
 section^ has moved to be the point by 
 and c' has moved to be the point c, &c. 
 on the arcs 2'a', 3'i', 4'c', &c. and the 
 ultimate intersection of each arc a'b'c', 
 a,b,Cf a,b,c, must be in the point n, 
 (Lemma 9, Quad.) But if the points 
 X, y, z, &c., be not the points of varia- 
 tion by which the intersections a, b, c, 
 &c. shall be on one straight line with 
 the point n — \etp, 9, r, ^c. be the true 
 points — that is 1 ^ be the variation of 1 
 towards 2 — 2 q that of 2 towards 3, &c., 
 and on the arc H'C, make the arc H'p' 
 equal to the arc Kp—Hq equal to H'q — 
 H'r' equal to H'r, <^c., and through the 
 points p and p', from the points A and 
 C, describe the arcs p m' and p' ml, 
 which by construction must meet in the 
 
16 
 
 point rn' on the circular arc H' G D' 
 (Lemma 2.) In the same manner des- 
 cribe the arcs q n and 9' n' — r and r' 0' 
 &c. — and from the point D' describe 
 through the points m', w', 0/ &c — the 
 arcs m m", w' n", o' o", &c' — and from 
 C through the points m", n",o'', &c. des- 
 cribe through the point w!' &c.* the arcs 
 p" m" u &c. — hence 1" ji should be the 
 variation of 1" towards 2" — and in the 
 same manner 2" q towards 3" &c. But 
 the arc 1 'p is made the variation of 1 
 towards 2, &c., but now Yp is the varia- 
 tion p" to V and not the same ratio of 1 ^ 
 to 12, SpG.y which by constiruction should 
 be equal ; for H'p" should be equal to H 2, 
 — for H 2" is equal to H 2 — therefore 
 1 X is the true variation of 1 towards 2, 
 &c., and through which the intersec- 
 tions a,b,c, <^c., must be in one straight 
 line with the ultimate intersection n on 
 the arc B D (Lemma 9. Sol. Quad.) 
 
 CorroUary, — It is evident if the 
 points 1, 2, 3, 4, are laid off from the 
 point A by the chords Al, A2, A3, &c. 
 that the chord An will be equal to the 
 sum of infinite terms, the differences 
 of the chords — and if supposed on the 
 arc A H' n, the length of the arc An will 
 
 • The point n'', o", &c. are not shown on the 
 fijiure on account of the smalhiess of the scale. 
 
17 
 
 be equal to the sum of the infinite series, 
 orarcs Al + 1 2 +23 + 34, &c. 
 
 SOLUTIONS. 
 Case First. — To Jind the length of 
 any circvdar arc in a itraight line, 
 
 PROPOSITION 1. 
 Problem. 
 
 Let Almnl be a Quadrantal arc. 
 Draw the chord A 1, and divide the ^'^' ^' 
 arc Almnl by continued bisections in- 
 to regular inscribed and circumscribing 
 sides — first into two sides Am, ml, 
 and c d, d e — second into four sides A 1, 
 I m, m n, n 1 and f g, gh, h i, i k ^c. which 
 will be the inscribed and circumscribed 
 perimeters of the arc A 1 m n 1. Then 
 from the point A make A 1 the inscribed 
 perimeter of one side, and the chord 
 A 2 equal to the inscribed perimeter of 
 two sides — A 3 equal to the inscribed 
 perimeter of four sides, &c. Also A 1' 
 the diameter of the semi-circle equal 
 to ab the circumscribing perimeter of 
 one side — ^and A 2' equal to the circum- 
 scribing perimeter of two sides, &c. 
 
 Then from the point I (Lemma 3) 
 through the points B, 1,2,3, &c. describe 
 the arcs BC, lt,2s,3r, &c. — and on the arc 
 B C, make the arc Bl" equal to the arc 
 Bl,— B2" equal to B2— B8'' equal to 
 
18 
 
 B3, &;c. — and from the point C, des- 
 cribe through the points 1", 2", 3", «&c. 
 the arcs l"t', 2"s', 3"r', &c. making the 
 intersections d,e,f, &c. Also through 
 the intersections d,e,f, &c. from the 
 centre L (Fig. 2) describe the arc B d e 
 f...G D', bisecting the arc B C in the 
 point D', (Lemma 1 and 2.) Again from 
 the point D', through the intersections 
 e", f ", &c., describe the arcs e"e', f "f ', 
 «&c. — and from the point I, through the 
 points e, f ', &c. describe the arcs ex, 
 f y, &c. meeting the arc BD, in the 
 points X, and y, &c. 
 
 From the point B, through the points 
 2' ,3', &c., and from the point D, through 
 the points x, y, &c. describe the inter- 
 sections a,b,c, &c. — and through a,b,Cy 
 &c. draw the straight line a b c &c» 
 meeting the arc B D in the point n, and 
 join A n. For that a b c.n is a straight 
 line (Lemma 4) — let Al = x, — A2 = x', 
 — A3 = x" &c — then A n, must be equal 
 to the sum of the series x -f (x' — x) + 
 (x"— X') + (x "— x") + &c. Therefore 
 the distance A n must be equal to the 
 perimeter of infinity of sides on the arc 
 Almnl, or equal to the Quadrantal 
 arc. 
 
 CorroUary,— Hence the Quadrature 
 of the Circle ; for four times A n, must 
 
19 
 
 be equal to the whole circumference of 
 the circle of the diameter Al'. 
 
 Note. — In the case of the given arc 
 being greater than the Quadrantal arc, 
 or greater than the semi-circular arc — 
 then make the first perimeter, be from 
 bisection of the arc into two sides, and 
 the second from bisection of the first, 
 and the third from bisection of the se- 
 cond, &c — and make the diameter of 
 the semi-circle or Al', equal to the first 
 or greatest circumscribing perimeter, 
 and construct by this proposition. Fig. 
 5, and (Lemma 3.) 
 
 PROPOSITION 2. 
 
 Problem. 
 
 Of the length of the circumference w 
 Quadrature of the Circle. 
 
 Let the circle ABC be given, FIG. 6. 
 and inscribe and circumscribfi regular 
 polygons of four, eight, sixteen, &c., 
 sides, and with half of the perimeter of 
 the circumscribing square, from the 
 centre C, describe the semi-circle A—l' FIG. 7 
 — then from the point A, make the chord 
 Al, equal to the inscribed perimeter of 
 four sides, A2 equal to the perimeter of 
 eight sides, A3 equal to the perimeter 
 of sixteen sides, &c. Also make the 
 chord A2' equal to the circumscribing 
 
20 
 
 perimeter of eight sides — A'3' eqoal to 
 the perimeter of sixteen sides, &c., and 
 describe the arcs F C F and D C I, and 
 and from I describe through B the arc 
 B C, (Lemma 3.) Next from I describe 
 the arcs It, 2s, 3r, &C', and from the 
 center C, describe the arcs l"t', 2''s» 
 S'Y, &c. making the intersections d,e,f, 
 6cC; and describe the points of variation 
 X, y, z, &c. (Lemma 3 and 4> on the arc 
 B D' Then from the points B and D, 
 through the points 2^, 3', &c>, and 
 through the points x, j, z, &c- describe 
 the intersections a, b, c, &c- and draw 
 the straight line abc.. &c- meeting the 
 arc B D, in the point n, and join A and 
 n. The distance A n shall be the 
 determinate length of the circumference 
 of the given circle ABC (Fig. 6.) 
 The demonstration the same as Prop. 
 1. 
 
 CorroUary, — It is evident that the 
 distance A n may be equally derived 
 from the perimeters by division or 
 bisection of the radius of the given cir- 
 cle, only the semi-circle A...1', would 
 be ofdiameter equal to the circumscrib- 
 ing triangle of the given circle ABC 
 
 Case First. — The Division of Cir» 
 cular Arcs and the Circle into equal 
 Arc- 
 
21 
 
 LEMMA A 
 
 Let the aro H F be divided into n FIG. 8. 
 equal arcs, and the arc H n be the unity 
 of division, consequently the arc F n 
 will be equal to n — 1 parts. Now let 
 the arc H n, be equal to the Geometrical 
 
 X H X M 
 
 series x -f + + 1 
 
 n— 1 (n— l)a (0-1)3 (n— 1)4 
 
 &C' ; then the arc F n, will be equal to 
 
 <x + 
 
 + 
 
 + 
 
 + 
 
 n-l (n-I)8 (n — 1)3 fn — 1J4 
 
 &c.) n — 1; for the whole arc F n must 
 be equal to H n [n — 1.] Then let the 
 arc H 1 be equal to x, the arc 1 2, equal 
 
 X X 
 
 to , the arc 2 3, equal to , the 
 
 u-l [n — l]a 
 
 s 
 
 arc 3 4, equal to , the arc 4 5 equal 
 
 (n-l)3 
 
 to 
 
 and &c. Also the aro F 1' 
 
 [n-l]4 
 
 must be equal to x [n — 1] — the arc 1' 
 
 X (n - 1) 
 
 2' equal to ■ = x, — the arc 2' 3' 
 
 n-l 
 
 X (n-l) 
 
 (n-l) 2 
 X (n-l) 
 
 n— 1 
 
 X 
 
 -, the arc 3' 4 
 
 / e' 
 
 -, the arc 4' 5 
 
 equal to 
 
 equal to 
 
 x(n— 1)3 (n— 1)2 
 
 X (n — 1) X 
 
 equal to ■ = , &c., hence 
 
 (n— I) 4 (n— 1) 3 
 
 the arc F n, will be equal to the series 
 
 X [n— iJ + x-l- -f ^ 
 
 n-l (n-l) 2 (n-l) 3 
 
 &C, 
 
22 
 
 From the points B and D, f Leir'Tia 
 3.] as centers, through the points l' iS' 
 3' &c., and through the points 1.2-3. &c. 
 describe the intersections ab.c, dec and 
 draw the line a bc^c, and also through 
 the intersections a,b\ct &c. draw the 
 line db'c <Scc, The lines a b Cf and a'b' 
 c &c. shall meet in the point n, on the 
 arc H F. 
 
 For by construction the arc U n is 
 
 Z X 
 
 equal to the series x ^ • + 
 
 n-l (n— 1) 2 
 
 X X 
 
 H f- Ac. — and Fn is 
 
 + 
 
 (n— 1) 3 (u-1) 4 
 equal to the series x [n — 1] + x 4- 
 
 n-l 
 
 + 
 
 + 
 
 + 
 
 + &c. — 
 
 (n-l) a (n— 1) 3 (n— 1) 4 
 
 Hence by leaving out the first term x 
 [n — 1] of the latter series, equal to the 
 arc F 1', we have the arc H n equal to 
 the arc I'n, consequently the term or 
 arc 1'2' is equal to H 1, the arc 2' 3' 
 equal to the arcs 1 2, the arc 3' 4' equal 
 to the arc 2 3, the arc 4'6', equal to tha 
 arc 3 4, &c., therefore the •-i.- j of Hn 
 is the same as that of In, and each must 
 approach by equal terms or arcs to the 
 common vanishing point in n ; there- 
 fore the curve lines ab c &c. and ab'c 
 raust meet the arc H F, in the point n. 
 
88 
 
 Corrollary, — Let the series x' 
 
 X' x' X' X' 
 
 + -t 4 + 
 
 n-1 (n— 1)2 (n-l)3 (n-1) 4 
 
 + &c. be greater than the arc H n, then 
 
 x' 
 
 the series x' [n — 1] 4- x' + f- 
 
 n— 1 
 
 X' X' 
 
 -f + &c. will be greater 
 
 (n— 1)2 (n— 1)3 
 
 than F n. It is evident that there 
 cannot be an ultimate intersection in 
 the arc or point n, for n cannot be the 
 vanishing point of either series nor can 
 there be a common vanishing point ; but 
 
 let the series X" + 
 
 -f- 
 
 + 
 
 + 
 
 u— 1 [n— 1] 2 
 
 + &c. be less than 
 
 [n-1] 3 [n-1] 4 
 
 the arc H n or equal to H t. and conse- 
 
 X X" 
 
 quently x" [n — 1] + x" + + 
 
 n-1 [n-1] 2 
 
 H 1 h &c. less than 
 
 [n— 1] 3 [n-1] 4 
 
 Fn, or equal to Ft'. It is also evident 
 that the ultimate intersection described 
 from B and D, through the points t 
 and t' must be without the arc H F, and 
 be in the point t". 
 
 LEMMA B. 
 
 Let the arc H F, to be divided into FIG. 8. 
 n equal parts. Divide the length of the 
 
I ^ 
 
 94 
 
 arc in a straight line into n equal parts 
 [Case 1.] Then through any point R 
 of the same arc, draw the tangent P Q 
 and make R P and P Q each equal to 
 the half of the nth part of the straight 
 line which is equal to the arc H F, and 
 join PC and QC, intersecting the arc in 
 the points U and V, and join U and V. 
 The distance U V, shall be less than the 
 chord of the nth part of the arc H F. 
 For the arc U V is less than P Q, 
 and the chord U V is less than the arc 
 U F, therefore the chord U V must be 
 less than the chord of the nth part of 
 the arc H F. 
 
 3. 
 
 PROPOSITION 
 
 Problem. 
 
 Division of an Arc into three equal parts^ 
 or the Trisection of an Angle, 
 
 FIG. 8. Let H F be the given arc to be 
 
 trisected. Describe the distance P Q 
 
 [Lemma B.] and make the chords HI 
 
 and It, each equal to the half of U V — 
 
 and let it be supposed H n is equal to 
 
 the third part of the arc H F — then H n, 
 
 will be equal to the series x + f- 
 
 n— 1 
 
 V 
 
 &c. and the arc Fn, will be equal 
 
 [n-l] 2 
 
 to X [n — 1] + X 4- 
 
 + 
 
 a— 1 
 
 [n-l] 2 
 
 + 
 
25 
 
 &c. fLemma A.] Now on both arcs 
 make x = 1, and n = 3, we have for the 
 vahie of II n = 1 + J + J + I + &c., 
 andFn=2+l + J+i+| + &c., 
 but as the value of x is unknown in re- 
 lation to H n or F n, make Ht equal to 
 the infinite series 1' + i' +1' + |' + &c. 
 = 2*, and bisect the arc Ht, in the point 
 1, and make the arc 1 2 equal to half 
 of the arc HI = J', make the arc 2 3 
 equal to half of the arc 12= |', &c. 
 In the same manner Fl' equal to twice 
 the arc HI = 2', — the arc 1' 2' equal to 
 HI = 1', —the arc 2' 3' equal ^',— the 
 arc 3' 4' equal 5 , &c. Hence the terms, 
 or arcs on Ft' are double the corres- 
 ponding arcs on Ht, and the remain- 
 ing arc t'n must be double the arc tn. 
 
 From the point B and D, as centers 
 [Lemma 3] describe through the points 
 2' 3' 4'...t', and through the points x, y, 
 z...t, the intersections a, b, c.t", and 
 draw through these intersections the 
 straight line a b c.t" meeting the arc 
 H F in the point n. The arc H n shall 
 be the third part of the arc H F [Lemma 
 
 4.] 
 
 For Ht is equal to the series 1' -f- 
 
 J' 4. J' + g' 4- &c., the points of variation 
 
 X y z &c. become evanescent in the 
 
 Bonnycastle's Algebra. Vol. 1, page 329. 
 
26 
 
 poiut t ; and t is the ultimate point of 
 variation, — consequently the intersection 
 t" is the ultimate intersection of the 
 series Ht and Ft', and therefore must be 
 on the line a b c &c. (Len ^na 4.) But 
 the points 1. 2. 3...t and the points 1', 2', 
 3',...t' are in position proportional to the 
 arcs Hn and Fn. Therefore the straight 
 line a b c.t", must cut the arcH F in the 
 same proportion in the point n, and the 
 arc Hn must be equal to the third part 
 of the arc H F, or the arc Fn be equal 
 to twice the arc Hn. 
 
 PROPOSITION 4. 
 
 Problem. 
 
 Division of the Semi-circle into three 
 equal parts. 
 
 ^^^ ^- Let A B D E be the semi-circle— 
 with the distance At [Lemma B] from 
 A cut the semi-circle in the point t, and 
 make Et' equal to twice At. Next 
 bisect At in the point o, and make E o' 
 equal to twice A o — the arc o 1 equal to 
 the half of A o, and o' 1' equal to twice 
 ol, &c. Then from the points B and 
 D [Lemma 3] describe through che 
 point 3', 1', 5'...t', and through the points 
 of variation x, y, z...t [Lemma 4] the 
 intersections a,b, 3...t", and draw the 
 straight line abc.t", meeting the semi- 
 
circular arc in the point n. The arc 
 An shall be equal to the third part of 
 the semi-circular arc A E, [Prop. 3 
 Prob.] or the chord An shall be equal 
 to the radius A C. The demonstration 
 the same as Prop. 3. 
 
 PROPOSITION 5. 
 
 Problem. 
 
 Division of the circumference of the 
 
 Circle into seven equal parts. 
 
 Let A B C be the given circle. FIG. 10. 
 
 From any point A of the circumference 
 
 make At equal to less than the seventh 
 
 part of the circumference ABC 
 
 [Lemma B.] — and At equal to the series 
 1' 1' 1' 1' 
 
 1'+ + + + 
 
 n-l [n— 1J2 [n— 1)3 [n— 1]4 
 
 + &C., then n will be equal to 7 andn— 1 
 
 V 
 
 = 6 ; hence At is equal to 1' H + 
 
 6 
 1' 1' 1' 
 
 1 + — [Lemma A.] The sum 
 
 6» 63 6, 
 
 1' 
 
 of which series, or At is equal to * 
 
 1'— 1' 
 
 6 
 
 6 , r 
 
 = — = l' 4- — . Therefore make the 
 
 5 6 
 
 6 
 arc Al equal to — of the arc At, [Prop 3] ; 
 
 6 
 
 • Wood's Ali,n^bi:i, page 113. Bonnycastle's 
 Alg'ebra, vol. 1, pa^e 92. 
 
nextmake the arc 1 2, equal to the second 
 
 1 
 term of the series or to — of Al, the arc 
 
 6 
 
 1 1 
 
 2 3,equal to the third tenn = — = — &c. 
 
 6« 36 
 
 Also make the arc A C 1' equal to six 
 times of Al — the arc 1' 2' equal to six 
 times the arc 1 2, — the arc 2' 3' equal 
 to six times the arc 2 3^ &c. Then 
 from the point B and D, (Lemma 31 
 through the points 2' 3' 4'...t', and 
 through the points of variation x,y,z.,.t, 
 describe the intersections a,b,c...t", and 
 draw the straight line a b c.t", meeting 
 the circumference in the point n — the 
 arc An shall be the seventh part of the 
 circle ABC. 
 
 Note.- -In the same manner may be 
 described the nth part of any circle 
 whatever,! observing as n increases the 
 scale of the figure will require to 
 be enlarged accordingly, if it is desired 
 to have more intersections upon the 
 straight line a.b...t"n than a and t". 
 
 tin Legendre's Geometry, page 112, he gives the 
 following note: — "We have believed a long time 
 that these were the only polygons which we were 
 able to inscribe by the procedure of Pllementary 
 Geometry, or that which is the same, by the resolu- 
 tion of equal ions of the first and second degree ; but 
 M. Gauss has proven, in a work entitled Dixquisi- 
 tinnes Aril/tmelicfi, Lepitirr, 1801, that we nuiy in- 
 scril. J by the same nioai's the regular polyiion of 
 secenlecn sides, and in general those of •> -|- 1 sides, 
 jMovided that '2" -|- 1, .vliall be a prime nundjer. M. 
 
I 
 
 29 
 
 Case Third. — Of the Geometncal 
 Summation of the Infinite Convetging 
 Series, of which the terms are alternately 
 plus ana minus. 
 
 PROPOSITION 6. 
 
 Problem. 
 
 The Duplication of the Cube* 
 
 Let the side of the cube be required, 
 
 of which the solid content is 2, or the 
 
 double of that of which the side is 1. 
 
 Then the cube root of 2, will be the 
 
 length of the side of the cube, double 
 
 of that whose side is 1. 
 
 By the Binomial Theorem we 
 
 m X 
 
 have [a + X] ^ = a S^ (1 H . [— ] + 
 
 n a 
 m m — n x 2 m m— n m — ^2n x 3 
 
 — . .(— ) +— . . .t— ] + 
 
 n2na n2n 3n a 
 
 &c. Now make a =1, and x = 1, and 
 m 1 
 
 — = — , we shall have the numerical 
 
 n 3 
 
 1 2 10 20 
 
 series 1 -f h • + 
 
 3 
 
 22 
 
 41 
 
 18 
 71 
 
 162 
 
 486 
 
 + 
 
 &c, = -• 2, 
 
 729 17496 39366 
 
 and by reducing the first and second 
 
 Legouche Jidds, page 419 — We shall terminate these 
 applications of Trigonometry by giving, after the ex- 
 cellent work of Gauss cited page li2, the manner of 
 describing the regular polygon of 17 sides by the 
 simple resolution of equations of the second degrne." 
 
30 
 
 terms, third and fourth terms, &c., we 
 
 4 12 
 
 shall have the series — — — 
 
 3 848 
 
 72 26345038 3 
 
 &c.= a/ 2. Also 
 
 6641 12244097 
 
 by reducing the second and third term, 
 the fourth and fifth terms, the sixth and 
 seventh terms &c. we have the serids 
 
 128302 
 + + &c. = 
 
 2 46 
 
 1 + — + 
 
 9 2187 
 
 3 
 
 ^2. 
 
 19131113 
 
 FIG. 11. Draw the straight line a b and 
 
 make ac =: 1, then make a 1 equal to 
 
 4 12 
 
 — of a c, — 12 equal , — 23 equal to 
 
 3 243 
 
 72 26345038 
 
 -, and 3 4 equal to &c. Also 
 
 6661 
 
 122444097 
 
 make Al =ac= 1, 1'2' equal to — ofac, 
 
 9 
 45 128302 
 
 — 2' 3' equal to , 3'4'equal to , 
 
 2187 19131113 
 
 &c. Next describe the semi-circle A B D 
 E,with a radius,not less than the distance 
 a c, and make the chord Al equal to al, 
 the chord A2 equal to a2, the chord A3 
 eqnal to a3, and A4 equal to a4 &c. 
 Also make the chord Al' equal toal' = 
 1, A2' equal to a2', A3' equal to a3', 
 and A4 equal to a4', "^c, and describe 
 
 
31 
 
 the arcs It, 2s, 3r, &c. [Lemma S^ — 
 and the arcs l"t', 2"s', 3"r', &c., and also 
 describe the points of variation x, y, z 
 &c. tLemma 4.] Then from the points 
 B and D [Lemma 3.1 describe through 
 the points 2', 3', 4', &c., and through 
 the points x, y, z, &c., the intersections 
 a, b, c, &c., and d' :m through a, b, c, 
 &c., the straight line a b c.n, meeting 
 the semi-circular arc A B D E in the 
 point n — and join A and n. The dis- 
 tance An shall be equal to the side of 
 the cube, whose solid content is 2, or 
 equal to the sum of the infinite series 1 
 
 1 2 10 20 
 
 + — 1-— + &C. = v^2. 
 
 3 18 162 486 
 
 Note. — To obtain a number of 
 intersections a, b, c,...&c. more than 
 two or three, would require that the 
 distance a c = I, to be of a much 
 greater scale than that of this figure ; 
 for it is evident that the parts of this 
 line expressed by the terms of the series 
 become rapidly to decrease. 
 
 PROPOSITION 7. 
 
 Problem. 
 
 The Quadrature of the Hyperbola- 
 
 1 
 
 2 
 
 + i- 
 
 i + 
 
 The series 1 
 
 1 1 
 +&C. found by Lord Bounker* 
 
 •tti 
 
 * Bonnycastle's Algebra, vol. 1, page 334, Note a. 
 
FIG. 12. 
 
 equal to the Appolloniaii Hjrperbolio 
 space, or axea l)etween its aibj^mtotes, 
 the sum of which accordiug to Lorgna 
 18 .6931471.1 
 
 This series by reduction [Prop 6] 
 
 I 1 1 1 1 
 
 is found to be 1 h — -\ -\ 
 
 2 12 30 «« 90 
 
 1 1 
 + &c. = .6931471 and 1 
 
 6 » 
 11 
 &c. = .6931471. 
 
 42 72 
 
 On the Hne a biinake a l'=l, and 
 
 1 1 
 
 1'2' equal to — of al',— 2*3' equal to — ^ 
 
 6 20 
 
 1 1 
 
 -'3'4' equal to — , and 4'5' equal to — , 
 
 42 72 
 
 &0' Also make al equal to J of al', 
 
 1 1 
 
 and 1 2 equal to — , — 2 3 equal to — , 
 
 12 30 
 
 1 ^ 1 
 
 3' 4' equal to — , and 45' equal to — , &c. 
 
 56 90 
 
 Then in the same manner done in the 
 preceding [Prop. 5, 6] describe the 
 right line a b c -n, meeting the arc B D 
 in the point n, and join An, — the dis- 
 tance An shall be to 1 or al' — as the 
 area of the Hyperbola is to 1' or 
 .6931471 :1'. 
 
 tSee a Desertation on the Summation of infinite con- 
 verging series by A. M. Lorgna, Professor of Mathe- 
 matics in the Military College of Veroua. Translated 
 from the Latin by H. Claire. Table page 122. 
 
33 
 
 Corrollary, — It \,4il appear the fol- 
 lowing eleven series to the above in 
 Lorgna's Table, that the summation or 
 areas of each may be found in the same 
 manner, for they are all the same form 
 of this. 
 
 ii 
 
 PROPOSITION 8. 
 
 Paoblem. 
 
 The QuadrcUure of the Circle. 
 
 1111 
 Theseries 1 + 1 
 
 3 5 7 9 
 
 — &c. = J of the circumference of the 
 circle whose radius is 1 — first given by 
 Liebuitz**. 
 
 By reduction [Prop. 6] this series 
 is resolved into the two following : — 
 
 2 2 2 2 2 
 
 1 1 + 1 h&c. = 
 
 8 35 99 195 328 
 
 I Circumference. 
 
 2 2 2 2 
 
 IS 68 143 155 
 
 I Circumference. 
 
 Upon a straight line a b, make a 1' 
 
 2 
 
 =1, — l'2equalto — ofal', — 2'3' equal 
 
 15 FIG, 13. 
 
 2 ^ 2 
 
 to — , — 3' 4' equal to — , &c. Also 
 
 63 143 
 
 make al equal to § of al', — 1 2 equal to 
 
 * Introduction a L'analyse Infinitesimale par 
 Leonard Euler. Tom. 1, page 104. 
 
84 
 
 t 3 a 
 
 ,23 equal to — ,34 eq^nal to — , 
 
 86 
 
 99 
 
 199 
 
 4 5 equal to — , &c. Then from the 
 
 328 
 
 point A on the semi-circle A B D E, 
 make the chord Al' equal to al', A2' 
 equal to a2', A3' equal to a3', &c., and 
 describe the points of variations x, y, z, 
 &c. [Lemma 41 — and from the points B 
 and D [Lemma 3 ] describe the intersec- 
 tions a, b, c, &C-, and draw the straight 
 line a b c ..n, meeting the arc B D in 
 the point n, and join An. The straight 
 line An, shall be equal to | of the cir- 
 cumference of the circle whose radius 
 is equal to al*, or the sum of the parts 
 
 2 2 2 
 
 represented by — H 1 h &c. of 
 
 3 35 99 
 
 2 3 
 
 the circumference, or I — — — -^ 
 
 15 63 
 
 — &c.5= J of the circumference — and 
 consequently eight times An, must be 
 equal to the whole circumference of the 
 circle whose radius is equal to a 1' =1 
 
 • Newton has remarked with regard to this series, 
 that to exhibit its value to twenty places of decimals, 
 would require the computation of no less than five 
 thousand million of its terras, the performance of 
 which would occupy more than a thousand years. — 
 See Newton's second letter to Oldenburgh on the 
 Commeroium Epistolicum, page 159." (Bonnycaatle's 
 Algebra, vol. 1, page 336.) 
 
 
36 
 
 Cask Fourth — -The penmctrr ven 
 of any Polygon — to find theinsenStdvr 
 circumscribed circle of the Polygon. 
 
 " Problem* 
 
 Let the diameter A E of the semi- FIG. 14. 
 circle A B E, be equal or greater than 
 the perimeter of the circumscribing 
 triangle ot a given circle, and the chord 
 An equal to its circumference — the 
 chord A^ equal to the perimeter of the 
 inscribed triangle — As equal to that of 
 the inscribed square — Ao equal to that 
 of the inscribed Octagon, &c. Also 
 At' equal to the perimeter of the cir- 
 cumscribing triangle — As' equal to 
 the perimeter of the circumscribing 
 square, — Ao' equal to that of the 
 circumscribing Octagon, &c. [Prop. 
 1, &c., Geo- Sol. Quad, of Circle! and 
 [Prop. 2.]. Now let Aa, be the perimeter 
 of any polygon of which is required the 
 circumscribing circle — ^which in this 
 case let the polygon be the inscribed 
 square, and its perimeter to be equal to 
 ka — then with tho distance A a, from A 
 intersect the chord As in 6, and through b 
 draw b b' parallel to s C, and from b 
 with the distance hb describe the semi- 
 circle A 5 E' intersecting the chord At 
 in the point t". Next divide M" into 
 three equal parts, from which construct 
 
the equilateral triangle ABC [Fig 15.] 
 and inscribe the circle a h t — the circle 
 a be shall be the circumscribing circle 
 of the square d ef g, whose perimeter 
 . i ir( shall be equal ka or kb, . ... t 
 
 - For by intersection the triangle kbb' 
 is similar to the triangle kaC, and the 
 segment xhf is similar to the segment 
 A B ^. Therefore A« is to kb, so is kf 
 to At", equal to the perimeter of the 
 circumscribing triangle of a circle a be, 
 and d efg the inscribed square of the 
 inscribed circle a be in the equilateral, 
 triangle ABC [Fig. 15.] the side of 
 which is equal to the third part of Af . 
 Therefore the circle a b e must be 
 the circumscribing circle of a square 
 whose perimeter defg is equal to Ab 
 or Aa' 
 
 CorroUary 1, — It is evident that An' 
 must be equal to the circumference of 
 the circle a b e — kp must be equal to the 
 perimeter of the inscribedtriangle— A^ 
 must be equal to the perimeter of the 
 inscribed octagon, &c. Also kh must 
 be equal to the perimeter of the circum- 
 scribing square, and kd equal to the 
 perimeter of the circumscribing octagon 
 of the same circle. 
 
 CorroUary 2, — In the same manner 
 from the given perimeter of any polygon 
 
37 
 
 * 
 
 the inscribed circle is found— for let A«" 
 be the girenperimeter of a circumscrib- 
 ing square— then A/' must be the perime- 
 ter of the circumscribing triangle, the 
 inscribed circle of which must be the 
 circle required, — and the perimeter 
 of the circumscribing square must be 
 equal to A*", &c. 
 
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 Copies, 
 Anderson, T. B. Esq, 1 
 
 Andrew, William Esq., Math'l Pro- 
 fessor M'Gill College. 
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 Edmonstone, Willliam Esq. 
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 Garth, George Esq. 
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 Kinnear, David Esq. 
 Law, James Esq. 
 Moffat, Hon'ble George. 
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 M'Gill, Hon'ble Peter. 
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COPIES 
 
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 M'Donald, Joseph Esq. Civ. Eng. 
 
 Nelson Wolford, M. D. 
 
 Ostell, John Esq, Architect. 
 
 Palsgrave, C. T. Esq. 
 
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 Wells, John Architect. 
 
 Viger, Hon'ble D. B. 
 
 Viger J. Esq. 
 
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