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THE 
 
 SECRET OF THE CIRCLE 
 
 AND 
 
 T 
 
 HE SQUARE 
 
 BY 
 
 J. C. WILLMON 
 
 ^LIF 
 AUTHOR'S EDITION 
 
 McBRIDE PRESS 
 316 W. Second Street. Los Angeles, Cal. 
 
COPYRIGHT BY 
 
 JEREMY C. WILLMON 
 1905 
 
OF TUt 
 
 OF 
 
 PREFACE. 
 
 It is my intention to demonstrate the possibility of 
 constructing a straight line equal to any given arc of a 
 circle, and through this problem to construct a square, 
 equal in area to any circle and a circle equal in area to 
 any square, with solutions of kindred geometrical pro- 
 blems. Many other geometrical problems are possible 
 through these demonstrations. 
 
 Another problem that may prove interesting, is 
 the division of angles into any number of equal parts; 
 this problem comes under the domain of the higher 
 plane curves 
 
 Through the problem of constructing a line equal 
 to the square of any line, it becomes possible to con- 
 struct a line equal to any line multiplied by any other 
 line, and through the two problems it becomes possible 
 to construct a line equal to the cube of any line. 
 
In Figure i (which is not referred to in any of the 
 problems and proofs, but is included herein that the 
 reader may gain a clearer idea of the Secret of the Circle 
 and of the figures and problems following) A C is the 
 diameter, and A J is tangent to the circle at A; the first 
 lines bisecting the two semi-circles equal the perimeter 
 of a square inscribed in the circle; the second bisectors 
 equal a polygon of eight sides inscribed in the circle; 
 the third a polygon of sixteen sides; the fourth a poly- 
 gon of thirty-two sides; the fifth a polygon of 64 sides; 
 the sixth a polygon of 12S sides; the seventh a polygon 
 of 256 sides; the eighth a polygon of 512 sides; the 
 ninth a polygon of 1024 sides; the tenth a polygon of 
 2048 sides and so on; when the bisections and right 
 angles approach and coincide with the tangent, the line 
 J A J equals the circumference of the circle. The area 
 ot a circle equals radius X Yi circumference; then rec- 
 tangle a e f j = area of the circle. If a d be made % 
 circumference and d b be made j^ a c cutting the circle 
 at b, then a j X a e = a d X a c = a b'^; then a b = one 
 side of a square containing the same area as the circle. 
 
 J. C. WILI.MON, 
 Los Angeles, Cal., December 1904. 
 
 3 
 
Figure I 
 
Figure ii 
 
 5> 
 
Figure III 
 
 11 
 
Figure IV 
 
 13 
 
FIGURE V 
 
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 Figure vi 
 
 17 
 
) 
 
 Figure vii 
 
 19 
 
PROBLEM I 
 
 To find a line which shall equal an}^ giv^en arc. 
 Let the giv^en arc be a k f , Figure 2, draw chord 
 a f; at a erect the tangent a j _\_ to a c the diameter of 
 circle a k f b; draw a k s the bisector of < j a f; draw 
 f s j_ to a f; draw a t the bisector of < j a s; draw s t 
 _L to a s; and so on indefinitely, approaching the limit 
 a j ; then a j is the required line. 
 
 PROOF 
 
 Let ebe the center of the circle. 
 
 Bisect < a e f at k; then bisect ■< a e k at 1, etc. 
 
 (Ref I. An < at the center is measured by }4 
 the intercepted arc.) 
 
 The bisector a k s of < j a f , bisects the arc a k f 
 at k. 
 
 The bisector a t of < j as, bisects the area k at 1. 
 
 (Ref IL An < formed by a tangent and a chord 
 is measured by j4 the intercepted arc.) 
 
 The bisectors of <s formed by the bisectors of <s 
 jaf;jas;jat, etc., and the bisectors of <s a e f ; 
 
 21 
 
a ek, a el, etc., being measured by the same arcs, meet 
 in a point in the circumference of the circle at the 
 points k, 1, o, etc. 
 
 (leisj_aks, Ref. III. The _\_ erected at the mid- 
 dle of a chord passes through the center of the circle 
 and bisects the arc of the chord.) 
 
 f s is ^ to a f by construction .'. k e and f s are 
 
 parallel. 
 
 .*. ak = ks, (Ref. IV) If a line be drawn through 
 two sides of a A parallel to the third side it divides the 
 sides proportionally. 
 
 .'. a k = ^ a s. 
 
 1 e is j^ to and bisects a k at r . '. is parallel to s t. 
 .-. a r = ^ a k = ^ a s, etc. 
 .-. a 1 = ^ a t (Ref. IV.) 
 .-. By same proof a o =^ ^ a X etc. 
 .-. as = 2ak;at = 4al; a X =8ao, etc., in- 
 definitely until the limit is reached. 
 
 ( Ref. V. The difference between a variable and its 
 limit is a variable whose limit is zero.) 
 
 Then a j = arc a k f . 
 .-. a j is the required line. 
 
 22 
 
PROBLEM II 
 
 To draw a square = in area to a circle. 
 
 Take any circle a f c g Figure 3. Draw diameter 
 a c; make a d ^ J^ circumference of circle as per pro- 
 blem I. Draw d b j_ to a c cutting the circumference 
 at b; connect a b; on a b construct the square a b i h 
 which is the required □• 
 
 PROOF 
 
 Post: The circumference of a circle X ^ radius 
 = area. 
 
 Post: ^ the circumference of a circle X radius 
 = area. 
 
 .*. ^ circumference X 2 radius = area. 
 
 .*. a d X a c = area of circle a f c g. a b"-^ = a d 
 X a c. (If from a point in the circumference a j^ be 
 drawn to the diameter, and chords from the point to 
 the extremities of the diameter, the j_ is a mean propor- 
 tional between the segments of the diameter and each 
 chord is a mean proportional between its adjacent seg- 
 ments and the dian>eter. ) 
 
 .*. a d: a b :: a b: a c. 
 .-. a b- = a d X a c. 
 
 28 
 
But a d X a c = area of circle a f c g as above, and 
 a b'^ = area of square. 
 
 . '. the square a b i h := circle a f c g which was to 
 be constructed. 
 
 Note. — The angle bac Figure 3, when construct- 
 ed, may be used for solving problems 2, 3, 4 and 5, and 
 other allied problems, by constructing equal angles. 
 (The nearest approximate to angle, b a c is 27*^*36' — 
 and for convenience, for practical work, it may be 
 marked on a protractor, or a triangle containing the 
 essential angle constructed of wood metal or other ma- 
 terial may be used.) 
 
 PROBLEM III 
 
 To draw a circle = in area to a square. 
 
 Take any square a b i h, Figure 3, on a b at a con- 
 struct <Cbac=<bac, Figure 3; on b a at b construct 
 <abn = <^bac, Figure 3; with e as center, and e 
 a or e b as radius draw circle a f c g; then the area of 
 the circle a f c g = area of the square a b i h. 
 
 24 
 
PROOF 
 
 This problem is the converse of problem II, and the 
 reader may apply the proof. 
 
 PROBLEM IV 
 
 To draw a circle with circumference = a straight 
 line. 
 
 Take any line and divide in into 4 = parts; let a d 
 Figure 3, represent % line; at d erect a ^ d b; on line 
 a d at a construct <bac=:<bac; Figure 3, the 
 line a b cutting or being produced until it cuts d b at b; 
 draw u e the ]_ bisector of a b cuttino: a d at e; with 
 e as center, and e a or e b as radius, draw circle a f c g; 
 then circumference of circle afcg=^4ad. 
 
 PROOF 
 
 The proof in problem II may be applied to problem IV. 
 
 PROBLEM V 
 
 To divide any angle into any number of equal 
 angles. 
 
 25 
 
Let b a c, Figure 4, be the given angle which it 
 is proposed to divide into any number of equal parts, 
 as for example, five. 
 
 Take a point i on the line a c; make a i = 1-2 ^ 
 2-3 = 3-4 = 4-5; with a as center and a i as radius 
 describe the arc i d; with a as center and a 5 as radius 
 describe the arc 5 k j ; with a as center describe a num- 
 ber ofindefinite arcs, as e e= i i, etc.; with a radius = 
 chord I d mark 5 points on each arc 5 k j, e e, i i, etc.; 
 then 
 
 5 chords in 5 k = 5 chord i d. 
 
 5 chords in e e = 5 chord i d. 
 
 5 chords in i i = 5 chord i d. 
 
 Through the pointsi e k, etc., draw the plane curve 
 line i e f k cutting the line b a at f; with a as center 
 and a f as radius describe the arc f g h; make chord h 
 g ^ chord I d; then 5 chords in f h = 5 chord i d, 
 and chord h g = chord i d and angle g a h = ^ 
 angle b a c. 
 
 PROOF 
 
 5 chords in 5 k = 5 chord i d by construction. 
 5 " " e e = 5 "id 
 5 " «' i i = 5 •* I d 
 
 26 
 
 (( 
 
 u {« 
 
fh = 5^ = G6 = iit)y construction, (by chord 
 -= I d.) 
 
 .*. f h = 5, I d by construction, (by chord = i d.) 
 
 Chord h g = chord i d by construction. 
 
 5 chords in f h := 5 chords i d by construction. 
 
 h g = i f h by construction. 
 
 .*. angle g a c = ^- angle b a c. 
 
 PROBLEM VI 
 
 To draw a line = square of any given line. 
 
 Take any line a b, Figure 5; on a b or on a b pro- 
 duced, take a definite length a e (not less than j4 the 
 given line) with e as center and e a as radius describe 
 the semi-circle a f c; with a as center and a b as radius 
 describe the arc b f cutting the semi-circle at f; from f 
 drop the }_fd; then ad X a c = a b-; but a c = 2 a 
 e and a e is definite by construction; .'. a c being = 2 
 a e is definite by construction; .*. a b (which is indef- 
 inite) taken a c (definite by construction) times equals 
 the square of a b. 
 
 PROOF 
 
 a f ^ a b by construction. 
 
 But af^ = adXac(if from a point in the cir- 
 
cumference a j^ be drawn to the diameter, and chords 
 from the point to the extremities of the diameter, the 
 _j_ is a mean proportional between the segments of the 
 diameter and each chord is a mean proportional between 
 its adjacent segments and the diameter.) 
 
 .*. a b- = a d X a c. 
 
 .'.ad taken a c times = a b squared. 
 
 PROBLEM VII 
 
 To draw a line = any given line X by any other 
 given line. 
 
 Take an}" two lines a b and be, Figure 6, construct 
 on a c the semi circle age and erect the x & b- then g 
 b2 = a b X b c. 
 
 On the line b g or on b g produced take a definite 
 length b f (not less than ,^1; b g ) with f as center and 
 f b as radius describe the semi-circle b i h; with b as 
 center and b g as radius describe the arc g i cutting the 
 semi-circle b i h at i; draw i d j^ to b h; then i h'^ = h 
 d X h b and i b- = b d X b h; but b h = 2 b f is def- 
 inite by construction .". b h is definite by construction; 
 .•.a line may be drawn = d b taken h b times = b %^ 
 (b g = b i)= a bXb c. (See problem VI , Figure 5.) 
 
 28 
 
PROOF 
 
 b g-^ = a b X b c. 
 
 (Ref. I. It from a point in the circumference a j_ 
 be drawn to the diameter and chords from the points to 
 the extremities of the diameter, the j_ is a mean pro- 
 portional between its adjacent segment and the diam- 
 eter. ) 
 
 A line may be drawn = square of any line as per 
 problem VI. 
 
 b h = 2 b f is definite by construction. 
 
 .'. b d X b h = bg- = a b X be = required line. 
 
 PROBLEM VIII 
 
 To draw a line = cube of an}- given line. 
 
 Take any line a e, Figure 7; on the line a e or on 
 the line a e produced take a definite length a c (not 
 less than ^ the given line) with c as center and c a as 
 radius describe the semi-circle a f b; with a as center 
 and a e as radius cut the arc a f b at f; draw f d j^ a e; 
 then a d X a b = a e- as per problem VI. 
 
 Draw a line ^ a e X a e- (a line may be drawn = 
 any line X by any other line, problem VII;) then a d X 
 a b X a e = a e'^ 
 
 29 
 
PROOF 
 
 a b = 2 a c is definite by construction. 
 
 .'.a line may be drawn =^ a d X a b X a e"- as per 
 problem 6. 
 
 a e = the given line. 
 
 .'. a line = a e X a e- may be drawn as per pro- 
 blem 7. 
 
 .*. a line niav be drawn ^ a d > ab X ae = a e'\ 
 
 
 30 
 
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