SQUARED A SOLUTION OF THE PROBLEM OF SQUARING THE CIRCLE Showing the Exact Ratio between RADIUS AND QUADRANT BY Ebenezer Erskine Wallace E. E. WALLACE MONMOUTH, ILLINOIS I903 Copyright, 1903, BY EBENEZER ERSKINE WALLACE PREFACE Up to the present time the problem of finding the exact ratio between the diameter and the circumference of a circle has remained unsolved. For more than one hundred years the leading scientific societies of the world have declared the problem to be impossible of solution. It is, thbrefore, with some diffidence that I submit the following pages to the consideration of mathematicians and others interested in the solution of a problem heretofore seemed unsolvable. And yet, having worked on the problem for a period of time covering more than fifteen years, it is with considerable assurance of its correctness that I subinit the solution, as I believe it to be. 1 may be pardoned for such assurance when it is considered that I have worked out the problem by two separate and distinct theorems and find the answers to be the same to six decimal places. This would be impossible if either theorem were incorrect. The answer thus obtained for the ratio of the circumference to the diameter to six decimal places is 3.141941+. By calculating upon the old theory of approximation and subdividing the sides until I had over three million sides to a polygon inside of a circle, I obtained the answer 3.1418+, which tends to show that the theorems herewith submitted are correct. I am well aware that the correctness of these theorems will be called in question, and if they are not mathematically correct they must fall. I submit them, however, with the confident belief that the time will come when they`will be universally recognized as the proper solution of this problem. EBENEZER ERSKINE WALLACE. MONMOUTI, ILLINOIS, May 18, 1903. 3 TABLE A VSides Remainders or Loss 6 X 0.5 -_ -- = 3. 12 X 1.258819045102520 8460462834 =3.105828541230240 24 X 2.130526183513181 3899039747 =3.132628404316344 48 X 3.065403124924338 0515367387 = 3.139349996368224 96 X 4.032719080665409 2579935404 =3:141031743879264 192 x 5.016361730547870 1772,725885 3.141452260191040 384 X 6.008181132952953 7982770512= 3.141555053934336 768 X 7.004090603756546 7977365189= 3.141583685027328 1536 X 8.002045306156319 2273950508 =3.141590256105984 3072 X 9.001022653612915 779694336 =3.141591898874880 6144 X 10.000511326873302 4221834471 = 3.141592309567488 12288 X 11.000255663445004 = 3.141592412233728 24576 X 12.000127831723547 8350493851 = 3.141592437891072 49152 X 13.000063915861904 4732978108 =3.141592444305408 98304 X 14.000031957930970 0447026860 = 3.141592446074880 196608 X 15.000015978965486 3-191307784 3.141592446201488 393216 X 16.000007989482743 4208153432 3.141592446271488 786432 X 17.0000039947413717 ----- - 3.1415924464287744 1572864 X 18.000001997370685866 ----- 3.141592446453940224 3145728 X 19.000000998685342937 ---- - = 3.141592446466523136 TABLE B Side is to Side As Loss is to Loss 17 1 1 17.0000000000548152861088220 17 2 2 17.0000000000127399180702221 17 3 3 17.0000000000008437784870086 17 4 4 17.0000000000021131061251629 17 5 5 17.0000000000002760761228788 17 6 6 17.0000000000020444470896519 17 7 17.0000000000008168799173397 178 8 8 17.0000000000001164211782755 17 9 9 17.0000000000019960171721350 17 10 10 17.0000000000000540394788565 17 11 11 17.17 12 12 17.0000000000000267215932746 17 13 13 17.0000000000000075727649685 17 14 14 17.0000000000000003573711636 17. 15 15 17.0000000000000012765456446 17 16 16 17.0000000000000008051066557 -- -- -- --.0000000000758527031120600 4 FIRST THEOREM Analytic Study of the Relation of the Radius to the Circumference There is a true ratio existing between the radios ancl the quadrant of the circle, and it can be found as truly as the hypotenuse of a right angle triangle when the two sides are known. Fig. 1, Theorem First, and the following example worked out to four clecimal places will. fully explain. I submit this proposition: The square of the sumn of D the two sides AC and BC is C \ to the difference of their squares (which is square of radius) as the sum of the two \ sides AC and BC is to their difference. Then: À A B To find the quadrant BEC add BD square + DC square -EF square to the square of the sumn of the two sides AC and BC; for want of a better term, call it the square of the curve. Solution of the proposition to four clecimal places. Fig. 1 AC =.50 BC =.7071 1.2071 1.2071 12071 84497 24142 1.2071 1.45709041 square of the sum of the two sides AC and BC. Find EF by taking 'the square of radius AB.5 X.5 = -25- =.125 d.125 t. 3535 9 65J 350 325 703J 2500 2109 7065J 39100 35325 5 Then BD square + DC square.50 -.3535.1465 = EF.1465 7325 8790 5860 1465.02146225 BD square + DC square =.50 - EF square =.02146225.47853775 The square of the sum of the two other sides = 1.45709041 + BD square + DC square - EF square =.47853775 1.93562816 /1.93562816 11.3912 sum of the two sides. i 23) 93 69 269J 2456 2421 2781J 3528 2781 74716 The square of the sum of the two sides AC and BC-=1.45709041 + BD square + DC square - EF square =.47853775 1.93562816 Then sq. of sum of 2 sides sq. of radius: suim of 2 sides diff. of 2 sides 1.93562816:.25: 1.3912:.1796.25 69560 27824.347800000 l.1796.193562816 1542371840 1354939712 1874321280 1742065344 1322559360 1161376896 Sumn of the two sides,.. 1.3912 Difference of the two sides,...1796 2) 1.2116.6058.1796 Quadrant of circle,......7854 4 3.1416 By calculating this proposition to fifteen decinal places the result will be 3.1419413724. 6 Rule for Demonstrating First Proposition To the sqCuare of th1e sm of rCadius and diagonal ~LL ) / caddc twice t7te square of radius ns ins haf the 0,, / square of the difference between radius and diatonal z / '(or square of t/2e curve). 1 I, THEN the sum of the squares wil/l be to the dif-, ference of their squares (uhicht is square of radius) / as the sum of the two sides is to their difference; then subtractc tzhe difference from the sumt of the two sides, divide by twoo, ancd adc the difference., Tte r esult will be the quadrant of the circle. /, The circumference of the ellipsis I /,I can be calculated by the saine, I Theoreim. ~/ %,For example, take 12 for greater radius and 2 / /,' ~ for minor radius. '12 X 12 = 144 2x 2 4 /,', /148 = 12.16552506059 = the diagonal. 2. ~' ~I;)'~ 14.165525060592= /' 1 # 200.6621002422033231711481 = square of sum of /; t47.65685424961348803742 [radius and diagonal. I / 248.3189544918168112085681 if /' <248.3189544918168112085681=- 15.7581393093,~'/,Add twice the mean square of the greater /radius and the minor radius 2 X 12 X 2 = 48 - half t / /the square of the difference between the minor /'.radius and its diagonal. 2 X 2 =4 X 2 =8., / ~/8 =2.8284271246-2 =.82842712462=,/ I.68629150077302392516. 2=.34314575038651196258 (half square of difference) '~~/ /8Subtract this from 48 and result will be /~,/'; 47.65685424961348803742=sq. of the curve. j~~/,-~' Sq. of sum of 2 sides+sq. of curve: \~~~, /! ~sq. of their difference (which is ~C/ ' i _C sq. of greater radius):: sumn of 2 /,1 ^^^/ ~, -t /sides: difference of 2 sides 248.318954491857 144: '/ "'7 /''\ 15.7581393093: 9.1381345623 "/ *'/ /1~\ ~15.7581393093 X 144 /;/ /\ _____ ---- = — 9.1381345623 248.318954491857 Sum of the two sides= O1 ' -i, - 15.7581393093 Difference " = 9.1381345623 2J 6.6200047470 3.3100023735 9.1381345623 12.4481369358 \7 Aswe. 49.7925 43 7 Answer. 49.7925477432 To make this solution of the question better understood, take the same example where 12 is the greater radius and 2 is the minor radius, as in Fig. 5. AH=-12 /148 112.165525 AC= 2 1 12 X12 =144 22) 48 2x 2= 4 44 148 241J 400 241 BC-Diagonal of the smaller 2426) 15900 radius. 14556 24325J 134400 121625 EF=One-half of difference between diagonal ancl smaller radius. 243305 J 1277500 1216525 2433102) 6097500r 4866204 HO=The quadrant of the ciricumference. 2433104J 123129600 12.165525= diagonal. 2. = smaller radius. 14.165525 -=sun of two sides. 14.165525 70827625 28331050 70827625 70827625 84993150 14165525 56662100 14165525 200.662098525625 47.656855 248.318953-sum of squares of two sides.,248.318953 15.758139=-sum of the sides. 1 25j 148 125 307) 2331 2149 3145J 18289 15725 31508J 256453 252064 315161J 438900 315161 3151623J 12373900 9454869 31516269 J 291903100 8 Find diagonal BC Fig. 5. AB=2X22=4 V 8 2.828427 AC=-2x2=4 4 8 48J 400 Diagonal 2.828427 384 -Radius 2. 562J 1600.828427 1124.828427 5648J 47600 5798989 45184 1656854 -— _~- ~3313708 56564J 241600 6627416 226256 1656854 6674162 565682J 1534400 1131364 2J.686291294329 402036.343145647164 = half the square of difference of 12 X 2 -24 X 2 48. diagonal and radius. -.343145 47.656855 Then 248.318953: 144:: 15.758139: 9.138134 144 63032556 63032556 15758139 2269.172016 248.318953 J 2269.172016 9.138134 = difference of two sides. 2234870577 343014390 248318953 946954370 744956859 1142022110 993275812 148746298 15.758139 =sum of two sides. 9.138134= difference 2J 6.620005 3.310002 9.138134 12.448136= quadrant the ellipsis whose greater radius is 12 4 and -of 49.792544= circumference and whose minor radius is 2. 9 After calculating the area of the circle and its circumference according to First Theorem (Fig. 1), and also calculating the circumference of the ellipsis according to the same Theorem (Fig. 5), getting a little greater answer than is given in'geometry, I made a careful calculation by approxiB~ C Fig. 7. mation, taking a six-sided polygon inside of a circle whose diameter is 1 (Fig. 7), being divided into six segments, each and every side being equal to radius. Then DC =.5X 6=3, as per table of answers of subdivisions of sides calculated up to the sixteenth division.000007989482743X393216 (number of sides) =3.141592446271488. The remainder vanished after the sixteenth subdivision. As ail areas having like sides are in proportion to the squares of such like sides, so 0oulcd ail losses or remainders be in proportion to the squares of such sides. Look on table of answers up to the nineteenth subdivision calculated to fifteen decimal places. (Table A.) To the right I have carries out ten places more, calling them remainders or losses. Table B will show calculations of the same. By taking the seventeenth subdivision for the first term, calculating all the losses from Table A in the remainder column, adding them up I find.00000000007585270311206.00, as this is the loss in calculating one side, there being 786432 X.0000000000758527031120600-.0000586539930294522099200 Losses added. 3.141592446271488 Answer at sixteenth subdivision. 3.14165110026451745220992 Answer. If I had taken the eighteenth subdivision for the first term, the sides would have been doubled and the sides would have been diminished almost one-half, which would have increased the losses almost four times, so I will multiply.00005865399302945220992 by 4 4.00023461597211780883968 3.141592446271488 3.141827062243605 This answer is obtained on the theory of obtaining the result by approximation, and I give it as evidence of the correctness of Theorem First, Fig. 1. 10 SECOND THEOREM I have found another solution of the same question, wlich I will endeavor to explain by calling attention to Fig. 6. Taking a circle whose diameter is 1, construct a square AB touching the circle, each side of the square being equal to the diameter. Then construct a minor square within the circle marked GF Fig. 6. Then <AB x GF- =the quadrant of the circle. Multiply the quadrant of the circle CCCC by AB and the result obtained will be the area of the circle. A!! E p /F Fig 6. I find by inspection that six-sixteentls of 1 square (or outside square) =.375 square, which is a little too small by a little over.005. This difference is found by calculating the difference of the squares. 11 Now refer to Fig. 7, whose diameter is 1. By constructing a six-sided polygon inside, and subdividing it into six segments, each side will be equal to radius. In this solution use squares and difference of squares. Let AC Fig. 7 =.5 X.5 =.2.25 Then - = 4 Let AB Fig. 7 =.1875.0625 Let BC Fig.7 —.0625.1875 --- 3.0625 16 X.0625-1= area of square outside of circle. 6 X.0625 —.375 +.005673315448546 difference that will be explained..380673315448546=area of the inside square. /.380673315448546 = <.61698728953564843381 X 1.7854885846 = qadrant 4 3.1419415384- circumference, I will endeavor to explain how I get the difference. Let A - A + B = 1 If A can do one-half of a piece B = ~ of work and B can do half of a piece C + D = 1 of work in one day, both working ' C = together would do the same piece of work in half the time. " D=| Then 2 X ~ = =.25 = Square of the time it would take both. C would do I of the same piece in one day. D would do - of the same piece in one day. Then |- X - = - =.234375 = square of the time it would take both. The difference between A and B.25 Square and C and D.234375 Square.015625.0625 = BC (Fig. 7). Then.0625 - =.015625 4 If (A and B)2 capacity is.015625 greater than (C and D)2 while A and B are working out this difference (.015625), how much would D and C do? A and B is to C and D as the difference,.015625,would be to C and D capacity..234375 X.015625.25:.234375::.015625:-=..0146484375 =(CandD)2 capacity. 12 12 If A and B capacity =.015625, and D and C capacity =.0146484375, what would be the mean capacity of A and B and C and D?.0146484375 X.015625.=0002288818359375 B (A) /.0002288818359375.01512884119612272207 Square of capacity of all four being prime numbers till the square markecl A. Divide this difference by 16 and multiply by 6; the result will be the difference sought..01512884119612272207. 16=.000945552574757670129375.000945552574757670129375X 6.005673315448546020776250 add.375.380673315448546020776250 area of inside square /.380673315448546020776250=.61698728953564843381-GF (Fig. 6) equal side of inside square. 1=AB <.61698728953564843381 X =.7854853846 quadrant of circle. 4 CCCC (Fig. 6)... 3.1419415384= circumference..7854853846 X 1 (AB)=the area of the circle whose diameter is 1. Multiply the quadrant of the circle CCCC by /i X.7854853846=ED (Fig. 6) =.886276=the square of the circle. / F /3 Fig. 3. 13 That is, four times ED in a square will enclose the same area that four times the quadrant of the circle will enclose. I consider this Theorem more correct than the first because the area.375 is almost enough for the inside square. 'Then the difference is obtained from differences of squares of prime numbers until I extract the square root of the fourth power marked B, and then I find a remainder marked A. The fourth power marked B is a prime number. And the capacity of ABCD jointly= the square marked A. That is the mean between A acndB, and C and D. Feeling well satisfied that I have truly and accurately calculated the area and circumference of the circle, By First Theorem I get the following answer. 3.1419413724 By Second Theorem.....3.1419415384 By Approximation......... 3.1418270622 As further evidence to show the relation of the circumference of the circle to the diameter, look at Fig. 3, since the areas of all circles are iii proportion to the squares of their diameters. The larger circle contains twice the area of the smaller circle whose radius is.5. Then 4 of the greater circle segments B and C =- the smaller circle. B - the square of the radius. If BC -= the area of the smaller circle, AC would also = the area of the smaller circle. Then A would evidently = square of radius. Then the square of the radius is to the whole area of the circle as the diameter would be to the circumference. The area of segment C is equal to the sum of the areas of segments D and D'. TO FIND THE AREA OF ANY CIRCLE, multiply the square of its diameter by.3806733154 and the result will be the area of the inside square GF (Fig. 6). But to calculate any circle whose diameter is greater or less than 1, take -,- of AC square (Fig. 7) or -- of AB square or ~ of BC square. For example, suppose AC, the greater radius, is 12. Then 12X 12 = 144. 144 16 - 9. 9 X 2 = 4-.25 - X - = 1-=.234375. 9 now becomes the difference. Then.25:.234375:: 9: 8.4375. 9.25 J 2.109375 [ 8.4375 = ~ and c capacity. 200 109 100 93 75 187 175 125 14 Then the mean between the difference or ratio will be as follows: 2 X =.25 =9 = C 2 (Fig. 7). 8.4375 9 3 X 5 =.234375 =8.4375 —capacity of DO. 75.9375. 8.714212 64 3 167J 1193 8J 26.142636 1169 3.267829 = difference 24 cl ianeter 1741 J 2475 242 576 1741 3 17424J 73400 8 1728 69696 216 = 3 or ~ of 576. 174282 J 370400 348564 216. 1742841J 2183600 3.267829 = difference. 1742841 219.267829 = area of inside square. 17428422 J 44075900,219.267829 14.807694 = GF (Fig. 6). 1 24J 119 96 288J 2326 14.807694 2304 24 29607 J 227829 59230776 207249 29615388 296146J 2058000 355.384656 1776876 2961529J 28112400 26653761 145863900 1355.384656 18.851648 1 4 28J 255 75.406592 Circumference 224 of circle whose diameter is 24. 368J 3138 2944 3765J 19446 18825 37701J 62156 37701 377026J 2445500 2262156 3770324J 18334400 15081296 325310400 15 The question arises why the factors - and 8 are used to obtain the foregoing difference. A circle whose diameter is 1 will have a radius =.5, and the square of radius =.25. Suppose the same diameter 1 is taken, Let.375 = smaller radius and ".625 = greater radius. Then.375 X. 625 -.234375 = square of mean radius. Then.25 -.234375 =.015625 = BC square. (Fig. 7).625 4 Then the square of the radius of a perfect circle being.25, is to the square of the mean radius of the imperfect circle as the difference between the squares of the radius of the perfect circle and the imperfect circle (.234375) would be to the imperfect circle's difference, or ratio. Then.25:.234375::.015625:.0146484375.015625.25J.003662109375 L.0146484375 25 1171875 -- 468750 116 1406250 100 1171875 162 234375 150.003662109375 121 100 210 200 109 100 93 75 187 175 125 Then the mean between these two differences or ratios.015625 X.0146484375-.0002288818359375 Then i.0002288818359375=. 01512884119612272207 The mean square of the two ratios or difference.01512884119612272207- 16 =-.000945552574757670 add.0625.06344555257475767 6.38067331544854602 Area of inside square as in former example. I calculated the six-sided polygon inside of a circle whose diameter is 1 to over three million sides, and by bringing up the losses I find a result of 3.1418 -+. But the two foregoing Thleorems or Propositions are not by Approximation and are either true or false. I believe them to be true and correct. EBENEZER ERSKINE WALLACE. 16