QA we IMPORTANT MATHEMATICAL DISCOVERIES UC-NRLF $B 527 TDE ^ By P. D. WOODLOCK ^ r IN MEMORIAM FLORIAN CAJO J R^faaetaci IMPORTANT DISCOVERIES IN PLANE AND SOLID GEOMETRY CONSISTING OF THE RELATION OF POLYGONS TO CIRCLES AND THE EQUALIZING OF PERIMETERS TO CIRCUMFERENCES AND DRAWING CURVED LINES EQUAL TO STRAIGHT LINES THE TRISECTION OF AN ANGLE AND THE DUPLICATION OF THE CUBE By p. D. WOODLOCK nnn PRESS OF e. w. stephens publishing company columbia, missouri Copyright 1912 By'-iP. D. Wooi/i-ooK. PREFACE. In publishing this book the author feels confident that he has added something to geometrical science, which has not heretofore been known. He especially invites geometricians and mathemati- cians to examine carefully and without bias, the several propositions and problems, and their demonstrations, con- tained in the book, and he has no doubt that they w^ill find interest in every page. Some of the problems appearing in this book have occu- pied the attention of geometricians in all ages since the introduction of geometry as a science, yet all attempts at their solution have been unsuccessful down to the pres- ent time. That the author of this little work has been rewarded with the discovery of the true solutions of these problems he confidently leaves to the consideration and candid judgment of geometricians throughout the world. iv 270590 PART I. CONSISTING OF PRELIMINARY DEMONSTRATIONS LEADING TO THE EQUALIZING OF CURVED LINES TO STRAIGHT LINES. PRELIMINARY DEMONSTRATIONS. PROPOSITION A. To form a series of polygons upon a square or equilateral triangle having the same center and equal perimeters, the number of their sides being to each other consecutively in the ratio of two. Figure A. In Figure A, let BCDE be a square, and draw the diag- onals BD and CE, bisecting each other at A, which is the 8 Important Mathematical Discoveries center of the square, and draw the Hne AP. Then AP is the radius of the inscribed, and AC is the radius of the circumscribed circles. Then bisect the hues AB and AC at R and L, and draw the lines RF and LH equal to RB and LC respectively, and parallel to AP, and draw the Figure B. lines AW, AX and Ay, bisecting the lines BE, CD, and DE, and draw RN equal to RB, and parallel to AW, and draw LI equal to LC, and parallel to AX. And in like manner bisect the lines AE and AD at G and U, and draw the lines CM and GO, equal to GE, and draw UK and Preliminary Demonstrations 9 UJ, equal to UD, and draw the lines RL, LU, UG, and GR, and draw the lines FH, HI, IJ, JK, KO, OM, MN and NF. And the figure thus formed is an octagon, whose perimeter is equal to the perimeter of the square BCDE. Now the line Cr is perpendicular to LH, and HT is per- pendicular to LC. And LH being equal to LC, therefore Figure C. the line Cr equals HT. In like manner FS = BV, and FH = Vr. Hence the lines SF + FH + HT = BV + Vr + rC = BC, which is a side of the square. Now the lines SF + FH + HT are equal to one-fourth of the perimeter of the octagon. And BC is one-fourth of the perimeter 10 Important Mathematical Discoveries of the square; therefore the perimeter of the octagon FHIJ is equal to the perimeter of the square BCDE. In Figure B, let the square BCDE and the octagon FHIJ have equal perimeters, and produce AP to n, which bisects FH, and draw the lines AF and AH. Bisect AF c H B Figure D. at R, and AH at V, and draw the lines Ra and Rb, each equal to RF, and parallel to AB and An respectively. In like manner draw the lines Vc and Vd, each equal to VH, and draw ab, be, and cd, and draw the line ae, parallel to FS, and dh parallel to HT. Then ea + ab + be + cd + dh = SF + FH + HT = BC. And as ea + ab + be + cd + dh form one-fourth of the perimeter of a polygon of 16 sides, Preliminary Demonstrations 11 and by completing the remaining sides of that polygon, its perimeter is equal to the perimeter of the octagon, and also to that of the square, their sides are to each other consecutively in the ratio of two. Proceed likewise in forming polygons of 32, 64, 128, 256, etc., sides, until they become infinite in number and minuteness. In Figure C, let BCD be an equilateral triangle, and bisect its sides at W, N and M, and draw the lines CM, BN and DW, intersecting each other at the point A, which is the center of that triangle. And bisect AC at V, AB at T, and AD at L, and draw VI and TH, parallel to AW, making VI = VC, and TH = TB. And draw VJ and LK parallel to AN, and equal to VC and LD. In like manner draw LE and TF, equal to LD and TB, and draw the lines HI, IJ, JK, KE, EF, and FH. And the figure thus formed is a hexagon whose perimeter is equal to that of the triangle BCD. Now it will be seen that PH = BS, HI = SR, and 10 = RC. Hence PH + HI + 10 = BC. In like manner OJ + JK + KX = CD, and XE + EF + FP = BD. There- fore their perimeters are equal. Figure D represents an equilateral triangle, a hexagon, and a polygon of 12 sides, whose perimeters are equal to each other, and whose sides are to each other in the ratio of two. And so on to infinity. 12 Important Mathematical Discoveries PROPOSITION B. In a square, an equilateral triangle and in all regular polygons, the sum of the least and greatest radii, divided by two, is equal to the least radius of a polygon of twice the number of sides and equal perimeter. Figure E. In Figure E, let the square BCDE, and the octagon FHIJ have equal perimeters (Proposition A), and we see that the line AP is the radius of the inscribed circle, and AC is the radius of the circumscribed circle, and they are the least and greatest lines that can be drawn from the Preliminary Demonstrations 13 center A to the perimeter of the square BCDE, and for convenience we will call them the least and greatest radii ot that square. Now the line NI bisects the line AP at d, and AC at L. Hence dL is parallel to PC, and AL = LC, and LC = LH; Figure F. therefore AL = LH, and as LH = dn, hence AL = dn. Then (AP + AC) h- 2 = Ad + AL = An, which is the least radius of the octagon FHIJ. Let Figure F represent a polygon of 16 sides drawn upon the square and octagon, their perimeters being equal. Then 14 Important Mathematical Discoveries in the octagon, An and AH are the least and greatest radii, and it will be seen that (An + AH) ^2 = At; and At is the least radius of a polygon of 16 sides, whose perimeter is equal to that of the octagon, or of the square, and so on until the sides become infinite in number and minuteness. PROPOSITION C. If a series of polygons be drawn upon each other, having the same center and equal perimeters, and the number of their sides being to each other consecutively in the ratio of two, the greatest radii of each polygon bisect the angles formed at the center, by the least and greatest radii of its next preceding polygon, and the lines joining the outer points of the greatest radii of any one of these polygons, with the outer points of the greatest radii of its next suc- ceeding polygon, are perpendicular to the latter radii. In Figure J, let the square BCDE and the octagon FHIJ, have equal perimeters, and let the line be be a side of a polygon of 16 sides, and or, a side of a polygon of 32 sides, whose perimeters are equal to that of the square, or octa- gon, and produce the line An to X, making AX equal to AB or AC, and draw the arc BXC, and it will be seen that the lines nH, tc and dr, are each half a side of a polygon of 8, 16, and 32 sides respectively; and join the points CH, He, and cr. ■ Now the triangles AnH and ATH are similar, hence the angles nAH and TAH are equal. And the line AH bisects the angle nAC, and in like manner the line AC bisects the angle nAH, and the line Ar bisects the angle tAc, and so on to infinity. And in the triangle AHC, the line Ac is bi- sected at L, and LH, LC and LA are equal to each other. Hence if a circle be described on the line AC as diameter, its circumference must pass through the point H. There- fore the angle AHC is a right angle. In like manner the angles AcH, Arc, etc., are right an- Preliminary Demonstrations 15 gles, and the lines CH, He, and er are perpendicular to the bisecting lines AH, Ac, and Ar respectively. Hence if the angles be bisected indefinitely, and polygons formed, until the last polygon that can possibly be drawn and re- Figure J. tain distinct sides is reached, the lines joining the outer points of the greatest radii of any one of these polygons, with the outer points of the greatest radii of its next suc- ceeding polygon, are perpendicular to the latter radii, and the rectangle formed by the greatest radii of any one of these polygons with the least rad.ii of its next succeeding 16 Important Mathematical Discoveries polygon, is equal to the square of the greatest radius of the latter polygon. It will be seen from the foregoing demonstrations that if an infinite series of polygons be thus formed, the perim- eter of any one of them intersects each side of all its preced- ing polygons, in two places, the points of intersection being equally distant from the middle of the side, no matter how minute the sides may become, and as a circle is the ultimate form of a polygon whose sides by the process of bisection become infinitely minute, the circumference of that circle will also intersect each side of all preceding polygons in like manner. PART 11. PERIMETERS AND CIRCUMFERENCES MADE EQUAL AND CURVED LINES MADE EQUAL TO STRAIGHT LINES. PERIMETERS AND CIRCUMFERENCES. PROPOSITION I. To form a circle whose circumference is equal to the perimeter of a given square. Let BCDE be a given square (the quadrature of the circle) and AR its least radius, and AC its greatest radius, FIJK L H X Figure I. as in Figure I. And with A as center and AC as radius, draw the arc BFC, and produce the line AR to F, bisect- 19 20 Important Mathematical Discoveries ing the arc BFC at F. Then bisect the arc PC at H, and FH at L, and FL at K, and FK at J, and FJ at I, and so on, bisecting to infinity, or as far as it is within our means to bisect. And draw the lines AH, AL, AK, AJ, and AI. Then from point C draw Ct perpendicular to AH, and from the point t draw tV perpendicular to AL, and draw VP, Po, and on, perpendicular to AK, AJ, and AI, respect- ively, the perpendicular always falling on each bisecting line, until the least possible bisecting line is reached, which is represented by AI. Then draw the final perpendicular nm to the line AF, and it will be seen that nm is half the side of a polygon, whose perimeter is equal to that of the square BCDE. (Prop. C, Preliminary Demonstra- tions.) Then through the point m, with radius Am, draw the circle SmWX, and the circumference of that circle is equal to the perimeter of the square BCDE. Now let the perpendicular nm be the least possible part or division of a straight line, a mere point, and we see that it is half the side of a polygon whose perimeter is equal to that of the square, and whose sides are capable of only one division or bisection, and as the circumference of a circle, which is equal to the perimeter of a polygon, both having the same center, must intersect each side of that polygon in two places, the points of intersection being equally dis- tant from the middle of the side, therefore, the circum- ference SmWX must intersect the perpendicular nm, but nm being a mere point, cannot be either bisected, inter- sected, or divided further. Therefore the circumference SmWX must pass over and coincide with the perpendic- ular or point nm, which represents half a side of a polygon, and hence it must pass over and coincide with the remain- ing half of that side, consequently it must pass over and coincide with each and every side comprising the whole perimeter of that polygon, and the point is reached where the final figure loses its identity as a polygon, and assumes and contains all the properties of a complete circle, as rep- resented by the circle SmWX. Therefore, the circumfer- Perimeters and Circumferences 21 ence of the circle SmWX is equal to the perimeter of the square BCDE, and the arc bmd equals the line BC. Now if the line BC = 2, the perimeter of the square equals 8. Hence the circumference of the circle SmWX = 8. And as Am, or the radius of a circle whose circum- ference is 8, is equal to the ratio of a circle to its circum- scribing square, which is 1.2732395 +. Therefore Am = 1.2732395 -f . The following table gives the radii of each polygon, from the square to a polygon of 524288 sides, the perimeter of each being 8, and the number of their sides being to each other consecutively in the ratio of 2. NO. OF SIDES. SHORT RADIUS. 4 1. 8 1.2071067811865475 + 16 1.25683487303146201 + 32 1.26914627894207004 + 64 1.2722167067075638 + 128 1.2729838511604253 + 256 1.2731756083078977 + 512 1.2732235458896797 + 1024 1.27323553034872029 + 2048 1.2732385264073428 + 4096 1.2732392754215578 + 8192 1.2732394626770475 + 16384 1.27323950948983831 + 32768 1.27323952119303589 + 65536 1.2732395233334371 + 131072 1.273239524045252 + 262144 1.2732395242281144 + 524288 1.27323952427383 + 262144 1.2732395242281144 + 524288 1.27323952427383 + LONG RADIUS. 1.414213562373095 + 1.3065629648763765 + 1.28145768485268807 + 1.27528713444730576 + 1.2737509956132868 + 1.2733673656153701 + 1.2732714833914617 + 1.2732475148077608 + 1.2732415224659654 + 1.2732400244357728 + 1.2732396499325372 + 1.27323955630262909 + 1.2732395328962334 + 1.2732395254738383 + 1.2722395247570668 + 1.2732395244109769 + 1.2732395243195456 + 1.2732395243195456 + From the above we find the ratio between diameter and circumference is 3.1415927 -f. 22 Important Mathematical Discoveries PROPOSITION II. To form a circle whose circumference is equal to the perimeter of a given equilateral triangle. In Figure II, let BCD be a given equilateral triangle, and draw the perpendicular CE, and let A be the center ^ Figure II. of that triangle. Then AC is its greatest radius. Then draw AF perpendicular to BC, and the line AF is its least radius. Then produce AF to L, making AL = AC. Then Perimeters and Circumferences 23 with A as center draw the arc LC, and draw the bisecting lines AH, AI, AJ, AK, etc., to infinity, or as far as it is pos- sible to bisect. Then from the point C draw Cr perpendic- ular to AH, and rP perpendicular to AI, and po to AJ, and on to AK, etc., and draw the final perpendicular to the point m, on the line AL. And with A as center and Am as radius draw the circle mS, and the circunference of that circle is equal to the perimeter of the triangle BCD. (See Demonstration of Proposition I, Quadrature of Circle.) PROPOSITION III. To determine the Quadrature of the Circle independ- ently of all infinite or unlimited bisections. In Figure III, let BCDE be a given square and A its center, and draw the lines AF and AD, and draw DH form- ing a right angle with AD, and produce AF to H. Then on the line AH, as diameter, draw the circle HDAC. And bisect the line HD at P, and draw FJ, which intersects HD at P, and draw PA. And on the line AH, make AO = AP, and draw PL at right angles with AH. Then on the line AO draw the semicircle OnA, and draw the lines On and nA. Then from the point D draw the line Dm perpendicular to AP, and draw mR parallel to FD. And on the line On make nX = FR, and draw AX. Then with A as center and AX as radius, draw the circle XWYS, and the circumference of that circle is equal to the perimeter of the square BCDE. Now let the Hne CD = 2. Then FD = 1. And AF = 1, and FH = 1, and AD and DH are equal to each other, and each is equal to i/^ 2 , and Dp = half the i/y = .70710678+ and AD^ +JDP = AP- = 2.5. Hence A02 = 2.5. Therefore AO = i/^2l = 1.58113883 + . Now AO X AF = An2. But AO X AF = AO X 1 = AO. There- fore, An- = AO = 1.581 13883 + . Now in the right- angled triangle ADP, PA X Am = AD^ = 2. Hence PA 24 Important Mathematical Discoveries X Am^ = 4. And as PA- = 2.5, then 4 -- 2.5 = Am^. Hence Am^ = 1.6, and AR^ = 1.44, and AR = 1.2. Then FR = .2, and hence nX = .2. Now An^ = 1.58113883 + , H y Figure III. and nX2 = .04. Hence A X^ = 1.58113 883 + , + .04 = 1.62113883 + , and Ax = i/-" 1.62 11 3883+ = 1. 2732395 + , which is the ratio of a circle to its circumscribing square. Perimeters and Circumferences 25 and is the radius of tlie circle XWYS. Hence the circum- ference of that circle is equal to the perimeter of the square BCDE. And 4 -- 1.2732395+ = 3.141592 + , which is the ratio of diameter to circumference. From the foregoing demonstration (Figure III) the fol- lowing rule is obtained, viz.: Multiply the square of the radius of the inscribed circle by the i^ 2^ , and add to the product the one-hundredth part of the area of the square, and the square root of the sum is the radius of the circle whose circumference is equal to the perimeter of the given square. It will also be seen that to draw a circle whose circum- ference is equal to the perimeter of a given polygon, bring that perimeter into the form of a square, and proceed as in Figure III or Figure I, Part II. 26 Important Mathematical Discoveries PROPOSITION IV. To find the number of degrees in any angle of a given triangle. In Figure IV, let the angle BAC be a given angle, and it is required to determine the number of degrees in it. Draw the lines AB and AC equal to each other and draw BC, and draw the arc niHn equal to the line BC. Then bisect R C angle BAC by the line AH, which is the radius of the cir- cle of which the arc mHn is a part. Then find the circum- ference of that circle, and divide by the arc mHn or the line BC, and then divide the result into 360 for the degrees in said angle. PART III. THE TRISECTION OF AN ANGLE. THE TRISECTION OF AN ANGLE PROPOSITION I. In any triangle, the sum of any two of its sides, is to the third side as either one of those two sides is to that corres- ponding part of the third, cut off by a line bisecting the angle which the said two sides contain. o Figure A. Let BAC (Figure A) be a triangle, and produce BA to E, making AE = AC, and draw EC, and draw AP par- allel to BC. Then draw AD parallel to EC, and AD bi- sects the angle BAC, as will be seen by the following dem- onstration : 29 30 Important Mathematical Discoveries AE = AC, then BE = BA + AC, and the angle AEC = ACE, and as the angle BAC = angles AEC + ACE, there- fore the angle BAC = 2 ACE, and as AD is parallel to EC, the angles DAC and ACE are equal. Hence the Figure B. the angle BAC = 2 DAC, and the line AD bisects the angle BAC. Now BE = BA + AC, and in triangles BEC and BAD, BE : BC : : BA : BD, and BE : BC : : AE : AP. Hence BE : BC : : AE : DC. Therefore, BA + AC : BC : : BA : BD, and BA + AC : BC : : AC : CD. The Trisection of an Angle 31 PROPOSITION II.* In Figure C, let ABC be a right-angled isoceles triangle and it is required to trisect the right angle ABC. Bisect the angle ABC by the line BD, and bisect and angle BAC by the line AP, and draw BH = BP, and with A as center and AH as radius, draw the arc HS, meeting the line DB J yn D ^ Figure C. K C produced, at the point S, and draw AS and CS. Then draw the line SJ bisecting the angle DSA, and draw SK bisecting the angle DSC, and draw Bm and Bn parallel to SJ and SK respectively, and the lines Bm and Bn tri- sect the angle ABC. *The trisection of acute and obtuse angles will appear in the next issue of book. 32 Important Mathematical Discoveries Now in triangle ABC, let the line Ac = 2. Hence DB, DA and DC each equals 1, and AB and CB each equals i/T. And as the line AP bisects the angle BAC, it will be seen that the line BP = .58578 + , and as BP = BH, there- fore the line AH = (i/T+ .585^8 + ) = 2. Hence the line AS = 2, and the line SD = i/ 3 , and the triangle ASC is equilateral, and the angle ASC is 60 degrees, the angle ASD is 30 degrees, and the angle JSD = 15 degrees. And as the dine Bm is parallel to SJ, hence the angle mBD is 15 degrees. And as the angle ABD = 45 degrees, there- fore the angle niBD = one-third of angle ABD. In like manner the angle nBD = one-third of angle CBD. Hence the angle niBN is one-third of the angle ABC. In all cases an isoceles triangle must be formed, the angle to be trisected being the vertical angle, as in Figure C. PART IV. THE QUADRATURE AND DUPLICATION OF THE CUBE. QUADRATURE AND DUPLICATION OF THE CUBE. PROPOSITION I. In Figure I, let ABCD be a square side of a given cube, and on the line AB draw the semicircle BOA, and let BE = 1, and draw the lines EFP, FB and FA. Then AB X T 5 D P / / / X 1 X 1 X / / / / "H vA R c A Figure I, L BE = FB- = AB. Then make F^ = BF, and draw LJ. Then AB X BL = BJ^ = BH^, and draw HS at right angles with BH, and draw the square SBRT, and the square .^5 36 Important Mathematical Discoveries SBRT = AB^ of which the square ABCD is one of the square sides. F P D Figure II. Now BE = 1, then AB X BE^BF^ = AB. And BF = j/AB, and therefore BL = i/AB. Then AB X BL - i/AB^. But AB X BL = BJ-' = BH^ = BS. Hence BS = i/AB' , consequently BS^ = AB^. Hence the square SBRT = AB\ Quadrature and Duplication of the Cube 37 PROPOSITION II. To bring a slab one inch thick into the form of a cube, whose soHd contents are equal to the solid contents of said cube. In Figure II, let ABYX be a square slab one inch thick, whose solid contents = 27. Then the line AB = i.^27 = H F ^ Figure III. ^ E 5.196 + . Now let BD = 1, then AB X BD = BC^ (the semicircle BSA being drawn). Then make BE = BC, and draw the lines EL, LB and CA. Then make BP = i/ BE, 38 Important Mathematical Discoveries and it is also the 4th root of AB. Then AB X BP = BR^. And make the angle LBS = two-thirds of the angle LBR, and draw SF, then the line BS is the cube root of the slab ABYX, and is therefore the cube root of 27, which is 3. A ^ H E F P D Figure IV. Then make BW = BS, and on BW draw the semi-cir- cle BmW, and draw the lines Bm and niW. Now AB X BF = BS2 = BW2 = 9. Hence 9 ^ 5.196+ = i 3",+ then BF = 1.732+ and = i/BW. Hence BW is the re- ciuired line representing the edge of the cube which is equal to the given slab; and BF^ = AB. Hence any rectangu- lar slab may be brought into a cube. Quadrature and Duplication of the Cube 39 Or, let Figure III be a reproduction of Figure II, then BA multiplied by its square root AF = AL-. Then make An = the 8th root of BA. Then BA X An = AP. And the difference between the angles thus formed at A, which is the angle LAI, bisect by line AR, and draw the line Rm. Then Am is the cube root of AB. This method is different from that of Figure II, and has the advantage of dispensing with trisections. PROPOSITION III. To form a cube whose solid contents are twice the solid contents of a given cube, and vice versa. In Figure IV, let the square BHVK be a side of a given cube, and let BJTO represent a square slab one inch thick, and equal to the contents of the given cube. Then pro- duce BJ to A, making BA = to the diagonal of the square BJTO, and form the square ABYX, and it will be seen that the square ABYX is twice the square BJTO. Then bring the square ABYX into a cube as represented by the square side BHtb. Now the square ABYX is twice the square BJTO. It is therefore twice the cube BHVK, and the square ABYX being equal to the cube represented by the square side BHtb, hence the cube BHtb is twice the cube BHVK. It will be seen that a cube may be formed whose solid contents are any given number of times greater than a given cube. UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. 27Jul'50JP /I0V171953LU 6Dec'53Wp WVg 8 1953 LU JJ\N 6 1954LU &0e^ DECS 1954 LD 21-100wi-ll,'49(B7146sl6)476 LIBRARY USE APR 12 1960 /WK i2'S6Q LIIRARY IJISE AUG 2^ 1965 Alie24'65-10PM 11^70590 THE UNIVERSITY OF CALIFORNIA LIBRARY