JC-NRLF 
 
 273 
 
 AND 
 ,TR I SECTION 
 
 OF 
 
 .ANGLES 
 

 LIBRARY 
 
 OF THE 
 
 UNIVERSITY OF CALIFORNIA. 
 
 
 
Dividing 
 Angles 
 
 
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 I 1 
 
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 | J. C, WILLMON | 
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 a 
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 * ittitoittifc* xivi/ ***** ifcifcitoVfc H 
 
 LOS ANGELES, CALIFORNIA 
 19O4 
 
 COPYRIGHTED 1904 3 
 
PREFACE 
 
 To bisect an angle and to divide it into multiples 
 of two is easy; to trisect an angle is much more 
 difficult, and a rule that will divide angles into ANY 
 number of equal parts has long been sought for, but 
 has never, so far as I know been heretofore discovered. 
 The problem was supposed to be impossible, and prob- 
 able is impossible by means of Elementary Geometry. 
 The line i e f k is a slightly curved line, therefore 
 the solution does not come within the domain of 
 Euclidian or Elementary Geometry. In "Higher Ge- 
 ometry" it is allowable to solve problems by means 
 of curved lines drawn through a series of fixed points 
 and my solution of dividing angles as herein given 
 comes within the domain of "Higher Geometry", and 
 I believe the solution to be new and correct. 
 
 J. C. WILLMON, 
 
 Los Angeles, Cal. 
 
 August 9, 1904. 
 
PROBLEM 
 
 To divide any angle into any number of equal angles: 
 Let b a c be the given angle which it is proposed 
 to divide into any number of equal parts, as for 
 example, five. 
 
 Take a point 1 on the line a c make a 1 = 1, 2 = 
 2, 3 = 3, 4 = 4, 5. With a as center and a 1 as 
 radius describe the arc 1 d. With a as center and a 
 5 as radius describe the arc 5 k j. With a as center 
 describe a number of indefinite arcs, as e e, i i &c. 
 With a radius = 1 d mark 5 points on the arcs 5 k 
 j, e e, i i, &c; then. 
 
 5 k = 5. 1 d 
 e e = 5, 1 d 
 i i = 5, 1 d 
 
 Through the points i e k draw the slightly curved 
 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 h g = I d. Then f h = 5 1 d and 
 h g = 1 d and the angle g a h = ]-5 angle b a c. 
 
 PROOF 
 
 5 k = 5 1 d by construction, 
 e e = 5 i d by construction, 
 i i = 5 1 d by construction, 
 fh=5k=ee=ii by construction. 
 .". f h = 5 1 d by construction, 
 h g = 1 d by construction. 
 fh=51d=5hgby construction, 
 h g = 1-5 f h by construction. 
 /. angle g a c = 1-5 angle b a c. 
 
THE 
 
 SECRET OF THE CIRCLE 
 
 AND 
 
 TRISECTION OF ANGLES 
 
 BY 
 
 JEREMY CARLISLE WILLMON 
 
 McBRIDE PRESS 
 316 W. SECOND STREET, Los ANGELES, CAL. 
 
4 
 
 COPYRIGHT BY 
 
 JEREMY C. 
 
 1903 
 
R A 
 
 OF TH 
 
 UNIVERSITY 
 
 PREFACE. 
 
 The most famous geometrical and mathematical 
 problem of the ages is the squaring of the circle. 
 
 The problems that have heretofore defied solu- 
 tion by fixed rule or method are : 
 
 i st. The construction of a square that shall 
 exactly equal in area any given circle. 
 
 2nd. The construction of a circle that shall 
 exactly equal in area any given square. 
 
 3rd. The construction of a straight line that 
 shall exactly equal the circumference of any given 
 circle. 
 
 4th. The construction of a circle the circumfer- 
 ence of which shall exactly equal any given straight 
 line. 
 
 The circumference of a circle has been proven by 
 mathematical calculations, and the ratio of its circum- 
 ference to its diameter has been computed to 400 
 decimal places. 
 
 5 
 
 124491 
 
To reduce a circle to a square or a square to a 
 circle, or to reduce the circumference of a circle to a 
 straight line, or a straight line to the circumference of 
 a circle, has always, in the past, required complicated 
 mathematical calculations, always with chances of 
 making mistakes in figures, and the results at best 
 only * 'approximate." There may have been many 
 "approximate" solutions of the above problems. 
 
 Mathematical calculations serve to prove the 
 length of a straight line, when properly constructed, 
 that shall equal the circumference of a circle, but the 
 calculations do not and cannot make the line. 
 
 Let it be conceded that the circumference of a 
 circle may be measured by making a wheel or disc of 
 the same diameter and circumference as the circle, 
 but to do this for every circle to be calculated would 
 be utterly impracticable. 
 
 These four problems are all solved by the right- 
 angled triangle ABC, Figure i. 
 
 By mathematical calculations the results are 
 proven "correct to infinity." 
 
 The triangle is to be constructed according to the 
 geometrical diagram a b c, Figure 2. 
 
 It is to be made of metal or other material^ and 
 
may be of any size. For the sake of simplicity I have 
 adopted the following: A t C = 10 inches; A B = 
 8.862269254-!- inches ; B C = 4.6325138 + inches. 
 Then BAG forms an angle of 27 36' . Whatever 
 the size of the triangle may be, the proportions between 
 the parts are always the same. In geometrical prob- 
 lems the proofs must be, not "approximate," but ex- 
 act. The inch, the foot, the metre and other stand- 
 ards of measurement may not be exactly the same ten 
 thousand years from now as they are today ; but math- 
 ematical and geometrical truths are the same always, 
 unchangeable and eternal. Therefore the triangle ABC, 
 Figures i and 2, as a standard of measurement between 
 circles, squares, straight lines and circumferences may 
 be made of any size, anywhere at any time, and the 
 angle being always the same and the triangle being 
 always similar, is a geometrical truth, "the same yes- 
 terday and today and forever." 
 
 Let it be conceded : 
 
 ist. That it is as easy to construct a circle as it 
 is to construct a straight line. 
 
 2nd. That it is as easy to construct a mechanical 
 wheel or disc as it is to construct a mechanical square. 
 
 3rd. That a wheel or disc may be revolved along 
 
a plane surface in a straight line, and that a straight 
 line may be constructed ^and marked equal to the 
 perimeter of the wheel or disc and equal to the cir- 
 cumference of a circle of the same diameter. 
 
 4th. That this line may be divided into 4 equal 
 parts. 
 
 5th. That the geometrical diagram a b c, Figure 
 2, may be constructed and used as a standard for con- 
 structing similar angles and triangles and the right- 
 angled Triangle A B C, Figure i, may be constructed 
 with the angle BAG similar to angle b a c in Figure 
 2. Then with the triangle a square may be con- 
 structed equal in area to any given circle ; a circle 
 may be constructed equal in area to any given square ; 
 a straight line may be constructed equal to the cir- 
 cumference of any given circle ; and a circle may be 
 constructed whose circumference equals any given 
 straight line. 
 
 Therefore the 4 problems are solved. 
 
 Another celebrated problem is the trisection of 
 angles by elementary geometry (straight lines and cir- 
 cles). It was attempted in the schools of the ancients, 
 and has probably been attempted by most students 
 and professors of geometry. It has, I am informed, 
 
 8 
 
been solved by "Higher Geometry" by means of an 
 "auxiliary curve," but I have not seen the solution. 
 I include the problem herein, a simple method by ele- 
 mentary geometry, trusting it may prove interesting 
 to the reader. 
 
 To anyone finding objections to my methods of 
 solving any of the foregoing problems, I would say 
 that I am willing to have my solutions of any or all of 
 them relegated to obscurity when better methods are 
 discovered for accomplishing the same purposes. 
 
 I have read that these problems are impossible, 
 and that none but ignorant persons would waste time 
 attempting to solve them. I will confess that my 
 schooling did not include geometry nor any of the 
 higher branches of learning ; and therefore it may be 
 well for me to explain how I came to solve the prob- 
 lems. 
 
 It all came about from making an analysis of the 
 poem "In the Distance." Noticing that it contained 
 7 verses and 7 lines to the verse led me to looking for 
 further coincidences, and through this my attention 
 was attracted to triangular numbers and triangles. 
 The idea came to me that through this poem would be 
 solved the secret of the circle, and I turned my atten- 
 
 9 
 
tion to the same, with the results as herein given. 
 The poem having led to the solution of the problems 
 under consideration, it is published in connection here- 
 with, together with a number of the coincidences ; 
 they may lead to the solution of other problems. 
 None of the coincidences were noticed or discovered 
 until the poem 'was complete, word for word and letter 
 for letter as it now is ; not a word was changed or 
 abbreviated to bring about any of these results. I do 
 not attempt to explain the coincidences. I have 
 endeavored to demonstrate the facts and truths herein, 
 only as I found them. 
 
 J. C. WILLMON. 
 Los ANGELES, CAL., July 13, 1903. 
 
 10 
 
Figure 1 
 
 Figure 2 
 
 11 
 
Figure 4 
 
 13 
 
C 1 
 
 Figure 6 
 
 15 
 
Figure T 
 
I. 
 
 Triangle ABC, Figure i , is a mechanical triangle 
 constructed similar to geometrical triangle a b c, 
 Figure 2. 
 
 II. 
 
 Construct a right triangle A B C so that A C = 
 10 ; A B = 8.862269254 + ; B C 4.6325138 +, or 
 in the same proportions. 
 
 Figure 2. Take any circle a k c f ; construct a 
 round wheel or disc of the same diameter as the circle 
 a k c f ; roll the wheel or disc one revolution on a plane 
 surface in a straight line f j ; then the line f j = the 
 circumference of the circle a k c f . Divide f j into 4 
 equal parts ; let f g = ^ f j. Draw the diameters k f 
 and a c I to each other ; make a d = f g ; erect the 
 ]_db cutting the circumference at b ; connect b and a 
 and b and c. Then triangle a b c is the triangle re- 
 quired. 
 
 THEOREM. 
 
 Figure 2. In any circle a k c f , if a d on the 
 diameter a c be made = % of the circumference of the 
 circle and the line d b be erected _L to a c, cutting the 
 circumference of the circle at b, then a b one side of 
 a square containing the same area as the circle. Con- 
 versely, if a b be made = to one side of a square con- 
 
 19 
 
taining the same area as the circle, and the line b d be 
 made _J_.to the diameter a c, cutting the diameter at d, 
 then a d = % circumference of the circle a k c f . 
 
 PROOF. 
 
 a b c is a right A being inscribed in a semi-circle. 
 
 Suppose the" radius a e = 5, then a c = 10. 
 Then f j = 31.415926535897932 + = circumference by 
 by construction ; f g = ^ : f j 7.853981633974483 + - 
 
 a d = f g by construction. 
 
 e d = a d ae = 2.853981633974483 +. 
 
 dc = ec e d = 2.146018366025516 + . 
 
 e b = radius = 5 by construction. 
 
 b d = i/ e b 2 e d 2 = 4.1054584 + Ref I. 
 
 a b j/ a d 2 + b d 2 = 8.862269254+ Ref. II. 
 
 b c = ]/dc 2 + bd 2 =4.6325138 + Ref. III. 
 
 .*. a c = 10 ; a b = 8.862269254 + ; be == 4.63- 
 25138+ Q. E. D. 
 
 Ref. I b d is a leg of a right A e b d. 
 
 Ref. II a b is the hypothenuse of right A a b d. 
 
 Ref. Ill b c is the hypothenuse of right Abed. 
 
 PROBLEM I. 
 
 To construct a square equal in area to any given 
 circle. 
 
 Construction : Take any circle a b c g, Figure 
 3 ; draw diameter a e c ; construct on the diameter 
 
 20 
 
a ec the A a b c similar to A A B C, Figure i. On 
 a b construct the square a b d f , which is the required 
 square. 
 
 PROOF. 
 
 Suppose the diameter a e c = 6 ; A abcis simi- 
 lar to A A B C, Figure i by construction. 
 
 .'. ab = ^ of AB, Figure i = 5.3173615524+. 
 
 5.3173615524 +' 2 28.2743338823+ : area 
 
 square a b d f . 
 
 Diameter a e c = 6. .'. area of circle a b c g 
 28.2743338823 +. 
 
 .'. The square constructed on a b = circle = 
 a b c g. Q. K. F. 
 
 PROBLEM II. 
 
 To construct a circle that shall equal in area any 
 given square. 
 
 Construction : Take any square a b d g, Figure 
 4 ; on a b construct A a b c and a b f , similar to A 
 ABC, Figure i ; the lines a c and b f, cutting or 
 being produced until they cut the sides of the square 
 at c and f, with e as center and e a as radius, describe 
 the circle a b c f , which is the required circle. 
 
 PROOF. 
 
 Suppose the side a b = 6 and the area = 36. 
 
 a c and b f are equal ; they bisect each other at e. 
 
 2L 
 
.'. ae eb = ef ec= radius of circle a b e f. 
 
 A a be, Figure 4, is similar to A A B C, Figure i , 
 by construction .'. the sides are proportional. 
 
 .'. If a b = 6, a c = 6.770275 +. 
 
 a e = y 2 a c 3.3851375 + radius of circle 
 a b c f . 
 
 Circumference X \ radius = area of circle. 
 
 .*. area of circle 36. 
 
 .*. circle a b c f = square a b d g. Q. E. F. 
 
 PROBLEM III. 
 
 To construct a line equal to the circumference of 
 a given circle. 
 
 Construction : Take any circle a b c h, Figure 5 ; 
 draw diameter a c ; on a c construct A a b c, similar 
 to A A B C, Figure i. From b drop the _j_ b d ; pro- 
 duce a c to g, making a g = 4 a d ; then a g = the 
 circumference of the circle a b c h. 
 
 PROOF. 
 
 Suppose a c = 6. 
 
 In Figure 2, a d = 7.8539816 +. 
 
 In Figure 5, triangle a b d is similar to triangle 
 a bd, Figure 2. 
 
 . '. Therefore a d Figure 5 = T % a d Figure 2 = 
 4.71238898 + . 
 
 4.71238898 + X 4 = 18.84955592 + := a g. 
 
 a c = 6 .'. circumference a b c h= 18.849- 
 55592+. 
 
 .-. a g circumference. Q. E. F. 
 
PROBLEM IV. 
 
 To construct a circle, the circumference of which 
 shall equal any given straight line. 
 
 Construction : Take any straight line a g, Figure 
 6 ; divide a g into 4 equal parts ; let a d = ^ a g; 
 at d erect the _[_ d b ; at a construct angle c a b 
 angle CAB, Figure i ; the side of angle cab cutting 
 or being produced until it cuts d b at b ; draw j e the 
 perpendicular bisector of a b cutting a g at e ; with e 
 as center and e a and e b as radius describe the circle 
 a b c f , which is the circle required. 
 
 The line, circle and angle of problem 4 being 
 similar to the lines,, circles and angles of problems I, 
 II and III, the reader may apply the proof. 
 
 PROBLEM V. 
 To Trisect an Angle. 
 
 Take any angle bac, Figure 7, not greater than a 
 right angle ; take any point, e on the line a b and 
 draw the line e d || a c ; with e as center and e a as 
 radius describe the arc a g f h ; draw the line a d so 
 that f d = e a ; draw f i || d e ; with a as center and 
 a d as radius describe the arc c d j b ; then angle d a c 
 
 = Y$ angle bac. 
 
 PROOF. 
 
 d f = e f by construction .'. angle e d f = angle 
 d e f. 
 
Angle c a d = angle a d e. Ref. I. 
 
 Angle a d e = angle d e f . Ref. II. 
 
 Angle e d f = angle e f i. Ref. I. 
 
 Angle a d e + angle d e f + angle d f e = a 
 straight angle. Ref. III. 
 
 But angle a f i + angle e f i H- angle d f e = a 
 straight angle. 
 
 .'. Angle a f i -f- angle e f i = angle a d e + 
 angle d e f . 
 
 But angle d e f = angle e f i = angle a d e. 
 
 .*. a f i = a d e = c a d. 
 
 .'. Angle a f e = 2 angle cad. 
 
 But angle a f e = angle d a b. Ref. II. 
 
 .*. Angle d a b = 2 angle cad. 
 
 .'. Angle c a d = ft angle b a c. 
 
 Q. E. F. 
 
 Ref. I. When two || lines are cut by a third line 
 the opposite interior angles are equal. 
 
 Ref. II. In an isosceles A the angles opposite 
 the equal sides are equal. 
 
 Ref. III. The sum of the angles of a A is equal 
 to a straight angle. 
 
 NOTE. The triangle, Figure i, may be found 
 useful in solving many additional problems, notably 
 the surfaces and contents of spheres, cylinders and 
 cones, as well as the converse of the problems. 
 
 24 
 
IN THE DISTANCE. 
 
 The countless leg-ions passed away, 
 And all the hosts on earth today, 
 L,ike vanished dreams may be forgot, 
 Their names and deeds remembered not, 
 
 Their gilded glories gone ; 
 Their works as rust and desert dust, 
 
 Fame's phantom shadows flown. 
 
 II 
 
 Or like enchanted music rung 1 , 
 Our songs attuned to cadence sung 1 , 
 Or names by nrystic fate renowned, 
 By glamoured ancient glories crowned, 
 
 With all that fame endears, 
 It naught would be to you or me, 
 
 Far down the distant years. 
 
 Ill 
 
 A few at most our troublous days ; 
 Unto the vast unknown we gaze ; 
 A glimmer of immortal dawn, 
 A star of hope still shining on, 
 
 Gleams through the darkest sky ; 
 A trust that good shall cross the flood, 
 
 And only evil die. 
 25 
 
IV 
 
 Where doubt exists a hope may live ; 
 None know the gif ts that time may give ; 
 Above our highest hopes and far 
 Beyond the dreamer's brightest star, 
 
 Have faith! for us may rise 
 The future's dawn, the shores unknown, 
 
 The fadeless lOden skies. 
 
 V 
 
 Ivet patience ever shield thy breast 
 From storm-tossed waves of wild unrest, 
 And love make all thy pathways bright, 
 Contentment make thy burdens light ; 
 
 Ivet gloomy thoughts forlorn, 
 And griefs and fears, the pains and tears, 
 
 All pass like mists of morn. 
 
 VI 
 
 Haste not to leap the fabled stream ; 
 What waits beyond we may not dream ; 
 Rejoice today, yet meekly trust, 
 That only good above our dust, 
 
 By fate, somewhere, somehow, 
 From acts of ours may grow as flowers, 
 
 In far-off years from now. 
 
 VII 
 
 Trust not in fame nor wealth to bless ; 
 Go help the poor and soothe distress ; 
 Be brave, be true and do your best ; 
 Do good until with God you rest, 
 
 In some far wondrous home, 
 And all will be as well with thee, 
 Through all the years to come. 
 26 
 
COINCIDENCES. 
 
 There are three words in the title and 300 words 
 in the 7 verses ; 13 letters in the title and 1300 letters 
 in the complete work ; there are 1287 letters in the 7 
 verses, not counting the 13 letters in the title. 
 
 There are 7 verses, 7 lines to the verse, 7 sentences 
 and 7 periods. There are 7 times 7 words in the 7th . 
 verse. There are also coincidences corresponding to 
 the number of minutes to the hour, the number of 
 hours to the day, and the number of months in the 
 year. There are as many syllables to the verse (52) 
 as there are weeks in the year, and 52 punctuation 
 marks are used in the 7 verses. 
 
 There are 365 syllables in the 7 verses. 7X52 
 364. (The extra syllable is in the line erding with 
 the word flowers.) Also, the second and fourth verses 
 combined have 365 letters, and the fourth and sixth 
 verses combined have 365 letters, corresponding to the 
 number of days in one year. 
 
 Only one exclamation point is used ; it divides the 
 words exactly ; 150 words preceding it and 150 follow- 
 ing it. 
 
 The line numbers of the third verse, 15 + 16 + 17 
 + 18 + 19+20+21 = 126-^-3 the number of the verse, 
 equals 42, the number of words in the verse. The 
 seventh line of the third verse is the shortest line in 
 the work ; there are three times as many words in the 
 
 27 
 
verse as there are letters in this line. There are 13 X 
 1 3 1 69 letters in the verse; the sum of all numbers 
 from i to 169 equals 14365. The two left-hand figures 
 will represent the number of letters in the shortest 
 line, and the three right-hand figures the number of 
 syllables in the verses and the number of days in 
 the year. 
 
 The first letter of the alphabet is used as a word 
 and for the commencement of words 33 times ; 33 
 commas are used ; there are 33 letters in the longest 
 line and 33 lines preceding it. 
 
 There are 29 letters in the first line, 29 letters in 
 the last word of each line of the first verse combined, 
 29 letters in the last word of the middle line of each 
 verse combined and 29 letters in the last word of each 
 verse combined. 
 
 There are 24 letters in the last line and 24 letters 
 in the first word of each verse combined ; the sum of 
 all numbers from i to 24 = 300, the number of words 
 in the 7 verses. 
 
 The number of letters in the alphabet 26, multi- 
 plied by the number of verses 7 = 182, the number of 
 letters in the 7th verse. Seven lines have each 26 let- 
 ters, viz: 2d, i5th, i9th, 24th, 36th, 38th and 45th; 
 these numbers added together, 2 + 15 + 19+24+36+ 
 38+45 = 179, the number of letters in the (2-6-) 2nd 
 and 6th verses. The sum of all numbers from i to 26 
 = 351, the number of letters in the 3d and 7th verses 
 combined. 
 
 28 
 
The most wonderful of all numbers is 1287. The 
 number of verses 7, multiplied by the number of let- 
 ters 1287 = 9009; the answer reads the same either 
 way backward or forward. 1287 with the right-hand 
 figure transposed and added to the left results in 828, 
 which reads the same either way backward or forward. 
 
 The sum of all numbers from i to 1287 = 828828, 
 which reads the same either way backward or forward, 
 divided in the middle either half reads the same either 
 way ; 828828 divided by 2 and the answer divided in 
 the middle, either half is the same and reads the same 
 either way. In all the series of triangular numbers, 
 the number 1287 has no parallel; no other number 
 will produce parallel or similar results. In the last 
 half (150) of the words are 625 letters; 1287 -f 625 
 1912. The sum of all numbers from i to 1912 = 
 1828828. 
 
 (To find the sum of all numbers from i up to any 
 number required, square the greater number; to the 
 product add the greater number and divide by 2. 
 Example: 10 X 10 = 100 + 10 = no ~- 2 = 55, 
 the sum of all numbers from i to 10 ) 
 
 1287 X in = 
 
 1287 X 222 = 
 
 1287 X 333 = 
 1287 X 444 = 
 1287 X 555 = 
 1287 X 666 = 
 
 29 
 
 142857 
 285714 
 428571 
 57H28 
 7H285 
 857H2 
 
Each answer contains the same figures in the 
 same order, but beginning at a different place. Multi- 
 plied by 259 the answer is six threes (333333). Mul- 
 tiplied by 518 the answer is six sixes (666666). Mul- 
 tiplied by a trinity of sevens (777) the answer is six 
 nines (999999)- 
 
 The decimal of \ is .142857. 1000000 -r- 7 
 142857.142857. 
 
 The decimal of T yy is .001287. IOOQOOO -*- 777 
 = 1287.001287. 
 
 The decimal of 1287 is .000777. 1000000 -8- 1287 
 = 777.000777. 
 
 MYSTIC NUMBERS. 
 
 (5) (3) 5th verse 3d line : 5X3 = T 5- The sum 
 of all numbers from i to 5 equals 15. "Love" is the 
 1 5th word of the 5th verse in the 3d line, and the 
 1 79th word of the work. (7) (4) 7th verse 4th line : 
 7 X 4 = 28. The sum of all numbers from i to 7 = 
 28. "God" is the 28th word of the 7th verse in the 
 4th line, and the 279th word of the work. 
 
 (10). The sum of all numbers from i to 10 = 
 55 ; "Mystic" is the 55th word of the work in the 
 loth line. 
 
 Commencing with the Sun as i, Mercury 2, 
 Venus 3, Earth 4, Mars*5, The Asteroids 6, Jupiter 7, 
 Saturn 8, Uranus 9, Neptune io,_Comets n, the Fixed 
 
Stars and Nebula 12, and i3th the Unknown ; 13 
 multiplied by the number of verses, 13 X 7 = 91. 
 ''Unknown" is the gist word of the work. 
 
 The word numbers added, 55+91 + 179 + 279 
 = 604. The figures of the answer added separately, 
 6+0 + 4=1 o. The sum of all numbers from i to 
 10 = 55. The figures in the numbers of the words 
 added separately, 5 + 5 + 9 + 1 + 1+7+9+2 
 + 7 + 9 = 55, the mystic number. 
 
 A CURIOUS GEOMETRICAL COINCIDENCE. 
 
 If an equilateral triangle be constructed of any 
 size and a circle be inscribed therein ; and towards 
 each vertex of the triangle additional circles be in- 
 scribed which are to the preceding circle as 3 to 7, 
 with -|- of their diameters and -| of their circumferences 
 cut by the larger circle, and so on in the same ratio 
 to infinity, each circle will touch the triangle, and 
 none will pass its bounds. 
 
v r~*> 
 
 YD 
 
 
 4