MATHEMATICAL RECREATIONS: CONTAINING SOLUTIONS OF MANY VERY DIFFICULT AND IMPORTANT EQUATIONS, AND OF SEVERAL USEFUL PROBLEIMIS GEOMETRY, SURVEYING AND ASTRONOMY, TOGETHER WITH A METHOD OF FINDING THE ROOTS OF EQUATIONS BY PROJECTION. BY I. N. ROBINSON, A. M. Author of a Course of Mathematics. ALBANY: ERASTUS H. PEASE & CO., 82 STATE STREET; JACOB ERNST, CINCINNATI. 1851. Entered according to Act of Congress, in the year 1851, by HORATIO N. ROBINSON, in the Clerk's office of the Northern District of New-York. ALBANY' J. MUNSELL, PRINTER. PREFACE. When an author comes before the public with a new book, it is customary for him to make some excuse, or to explain some of his reasons for so doing; but in this instance we shall depart from this custom, for we prefer that the leading motive should not be generally known. It was not, however, the promptings of vanity, nor the hope of gain: the real motive was far more modest and practical. Some few will perceive the original object designed to be carried out in the work, and to some few I have explained it; and to these it will appear far more perfect and consistent than to others. Without this hint, some would be surprised at the simplicity of several of the problems, and consider them wholly unworthy of a place in a work like this. This book is not designed for mathematicians already made. Such persons require no aid from me. It is for those who have a desire to become mathematicians, and for those who would render their knowledge a little more practical than can be deduced firom the ordinary class-books in this country. The lovers of the mathematics will perhaps be pleased with the manner of finding the roots of equations of the second and third degrees, which is here carried out more fully and practically than in any work that has come under the author's inspection. The subject is touched upon in Hutton's Mathematics as formerly published; but 1 claim to have elucidated, modernized, and rendered it practically useful, so far as respects real roots; and there is reason to hope iv PREFACE. that certain abstract principles which are to be found in a recent work by Mr. John Paterson of Albany, will ere long enable us satisfactorily to geometrise imaginary roots. In discussing this subject, I arrived at equations of the fourth degree prior to those of the third. This course is doubtless open to criticism, and may perhaps be condemned by some, as deviating too widely from the usual mode of communicating instruction by proceeding from the simpler to the more complicated cases; but we would caution the censorious, and remind them that different roads, from different points of the compass, may yet lead to the same citadel. I have not referred to the differential and integral calculus; fearing, if such were the case, that many would lay aside the book, to whom it would otherwise be highly useful. Some useful and practical problems from these sciences may nevertheless be added at a future day, if circumstances should warrant such a course. MATHEMATICAL RECREATIO(NS. EQUATI ONS "1 12 1. Given (x — ) +(1 —) =: to find the values of x. x; Put (x —)= P, and (1 — Q; ----------— ]' X then the given equation becomes P+ Q = -------- [2] Multiply [2] by P- Q; then we have P2-Q2 (P-Q)X. ------- - --- --— [3] From [1], we find 2- Q2 (x- 1); therefore x- = (P- Q)x, or 1- = P-Q. - -----—... [4 x By adding [2] and [4], we find 1 —+x = 2P; x that is, 1+P2 = 2 P. By transposition, 1-2 p+P2 _ 0, or 1 —P = 0, or P 1. Therefore, by the first of' equations [1], we perceive that x — or x21 - i Hence= ( / (Math. Recreations.) I MATHEMATICAL RECREkATIONS,. 2. Given (2x'- +Cj2 (a2- a -: to find the values of x. We observe that this equation is in the same form as the preceding, and would be identical if we changed x2 to x, a4 to 1. Therefore the value of x2 in this equation will be of the same form as the value of x in the first example, except it will cortairn the factor a2, because the square root has been once extracted That is, z _ (li:E/), or x = Ta( 2. But this conclusion is too summary to satisfy the young algebraist; therefore it is proper to take some of the intermediate steps. Put (2 ) _ P, (a2 = ---------- - then the given equation becomes x2 P+Q =- -- -- Muitiply [21 by (P-Q); then we have P-2_Q2 P-= Q). But the value of (P'2 -Q') drawn from [1], is (x'2-a2); X2 therefore (P- Q) 2_a2 a3 or P-Q —-, - --- --- -- - - [3By adding equations [2: and [3], we find a 3X2 2P = a-k4 2 a Multiply this equation by a; then 2aP a2_-+ z2 that is, 2 aP = a'+P'2 or 0 = a2 —2aP+.P2. Square root, 0 = a.-P, or P = a,.tP2 ~ (i2 P a. EQUATIONS. 3 From the first of equations [1], we find a4 a 2 14 I 3. Given x4(1+ -)2 (3x2+x) = 70 to find the values of x, Observe that (3x2+x) - 3x2(1+3). Put (1+ )- y; then the given equation becomes x4y2 -3x2y = 70. 9 289 Completing the square, X4y2-_32y+- = 3 17 2 2 x2y- 10, or -7; that is, x2+- 10, or -7. 3 Whence x = 3, or - or 6(1-v1 -iL). 4. Given (x+)+Vx2 — y2 - 9(x+Y) (x+y) — /x2 —y 8 y and (x2-+y)2+(x-y) 2x(x2~+y)+ 506, J to find the values of x and y. iMultiply the fraction, numerator and denominator, which forms the first member of the first equation, by (x+y)+ V/x2_ — (( X+y)+ V;2_y2)2 9(X+?) then we shall have = (y 2xy+ 2y2 8y Divide both denominators by 2y; then (7x+y+ V2+_y2)2 9(x+y) x+-Y 4 4 MATHEMATICAL RECR EATIONS. Multiply by (x+y), and then extract the square root of both members; then xy+V/2 —y2 = -(Xq -y) ---------- [a] Hence 2Vx2-y2 = (X+y): Squaring, 4(x2-y2) (x+y)2. Divide by (x+y); then we have 3x 4(x-y) x+y, or y 5 In the second equation, put x2-t-y = P; then p2+ — y = 2xP+506, or P2 —2xP = -x+y+506. Add x2 to both members of this equation; then PP2-2xP+x2 = x2+y-x+ 506, or (P-x)2= x2+y-x+506, or (x29+y —)2 = (x2+y-x)+506; that is, Q2 —Q = 506, or Q = 23 or -22; that is, x2+y — = 23 or -22, or x'2 x-x = 23 or -22, or x2- 2=x- 23 or -22: 23 69 Hence x= 5 or -- y= 3 or -65 To obtain all the roots in these compound equations, we must notice the double sign, and use all the results that arise from extracting square roots; and we shall find a greater or less number of roots, according to the number of times that we extract the square root. 5. Find the values of x from the following equation. 2x2(x3~+a)2 = 22(x+~2a)+a2(x-a). Put a = nx; then 2x3(1+n,3) = 2X2(x+2nx)+nn22(x-nX). Divide both members by X3; then we have 2(1+n3) = 2(1 +2n)+n2( 1-n) = 2+4n+n2 —n3. Add 1 to both members, and transpose -n3; then we have (l+n3)+2(1+na3)1 = 3+4n+n2. EQUATIONS. 5 Add 1 to both members, to complete the squares; then extract the square root, and we shall have (1+n3)+= — = (2-n). Taking the plus sign, dropping unity from both sides, and squaring, gives l+n3 ( 1+n)2. -------- - [A] By division, 1 —n+n2 l+n, or n = 2.. Eq. [A] can be put in this form: (1- n+n')-( 1+n)X(1+n) = 0. This equation can be verified by making either factor equal zero the first factor gives n = 2; the second, n = - -i Whence, x: la or -a. 6G n 8 7 x 8 7 Put Vx = y; then y- =y y-2; y4 —2y3 —8y+ 16= 7y2; y — 2y3 —7y2-8y+ 16 = 0. Extract the square root, or rather attempt the extraction ~ y4-2y' —7y2-8y+ 16 = 0 ( y2_-y y4 2y2 —y) -2y3 —'7y2 -2y_+ y2 -8y2-8y+ 16 Now the equation may be put in the following form ~ (y2_Y)2 )_8y2 — y+ 16 = 0. If -8y2 were -Sy2, the equation would be in a quadratic form; but we can cause +8$y2 to appear in the first member of the equation, by adding 16y2 to both members: then we shall have (y2 —y)2+Sy2 —8yS+ 1 6 =16y2, or (y2~_y)2+8(y2//_y)+ 16 = 16y2. By evolution, y2 — y+4 = ~-4y; therefore y2-5y = -4, and y2+3y = -4; or y = 4 or 1, and y/= (I= -7-3) But V: =; x:- = 4 or 1, x = 16 or 1. 6 MATHEMATICAL RECEATIONS. 7. Given xy - 125x+-30Dy, and y2-x-2 -90000, to find x and y. A direct solution of this problem, and without taking especial care, will produce a resulting equation containing very high numbers: to avoid this inconvenience, we must resort to artifice. Observe that 5x25=125, 12x25 = 300, (300)2 = 90000. Put a - 25, b=- 12; then xy = b5a+aby, y2 —x2 a2b2; -----—. [1] aby aby= -. C. — --—.. O --------—. [23 2 a2b2y2 y'- 10ay~25a2 Substitute this value of x2 in [1]; then we have 2 a2b2y2 y__- I 0ay-+-25a y4_- 10ay3+25a2y2-a2b2y2 = a2b2y2_ lOa3b2y+25a4b2; y4- 10ay3 +(25a2-2a2b2)y2 + 10a3b2y = 25a4b2. Decompose the first member of this last equation into factors, by attempting to extract the square root, thus: y4- 10ay' + 2 5ct2y2 -2a2b2y2 + 10a3b2y(y2- 5ay y4 2y2-5ay) - 10ay3 + 25a2y2 - 1Oay3 + 25a2y2 -2a2b2(y2- 5ay) This result shows that the last equation may be written in the following form: (yV2 5ay)22a2b2(y2 -bay) 25a 4b2. Put y2 —5ay - Q; then we have the quadratic Q2-2a(b2Q = 25a4b62; Q2 -2a2b2Q + a,4b4 25a4b2 + a4b4 a4b2(25+b2). = a4b2( 169). By evolution, Q —a2b2 13a2b; Q = 13a2b+ a2b2 = a=b( 13 b) - 25a2b - a3b. - Observe that b2' 144; therefore (25+b)= 169: Also a = 25; therefore 295ab = a'b. EQUATIONS. Hence y2 —5ay = a3b; 4y2 —20ay+25a2 =- 4a3b+25a 4a3b+a3 a3(4b+ l) -- 49a3 = 49a2.25' 2y-5ca = 7.a.5; 2y = 40a; y = 20a = 500. This value of y, substituted in equation [2], gives ab.20a 25.12.4 - -= - -16.25. 15a 3 8. dGiven ( x-+xy = 99, to find the values of 8. Given y+xy2+ xy3+y4 = 0 x and y. Divide one equation by the other, and we have +y __ 11 y+ —y y' +y 10" Reducing the first member by dividing numerator and denominator by (1+y), and l-y+y2-y+y _ 11 Clearing of fractions, 10 —10y+ 110y- i0y — 10y3+ y4 = lly+- ly; 0O( +y2+y4) = 21(y+y'). Divide both members by y2; then we have lo( +l+yj2) = )1( +y) y y 1 1 Now put (-+y) = P; then y-+2+y2 = P2 Whence 10(P2 —1) = 2 ].P, or -p21-P.P 1, Hence / —y= -1 ~ whence y = 2. 8 MATHEMATICAL RECREATIONS. Put this value of y in the first equation, x+xy5 - 33x = 99; x = 3. These equations obviously result fr-om the following problem: " There are six numbers in georiietrical progression - the sum of the first and sixth is 99, andr the sum of the four means is 90; what are the numbers?" Ans. 3, 6, 12, 24, 48, 96. X3 X? / 3 The numbers may be denoted by-9 y, Y and V. (See ROBINSON'S lIgebra, page 198, art. 123, and art. 112.) We call attention to this, to show that equations may sometimes be changed to others having no visible connection. 9. Find two numbers, whose product shall be equal to the difference of their squares, and the sum of their squares equal to.the difference of their cubes. In other words, find the values of x and y from the following equations: xy - x -y2, - - - - [1 ] y = y3 -y3 -- [2] Put x -vy; then [1] and [2] become Vy2 -_ V2y2-_y2, or v = v __-, ----- [3] v2y2+y2 = V3y 3 y3, or 2+ 1 = (V3-1)y..... [4] From [3], 1-v = v2; whence 2+v = v+1. ------ [5] Multiply the first of equations [5] by v, and we have V. V2 = VI; that is, 2v+ 1 = v3 or 2v = v3-1.........[6] By the aid of [51 and [6], we perceive that [4] reduces to this:f:or:m: 2 + v = 2Py, 2+v r = ------------------ [7] Equation [3], resolved as a quadratic, gives 2v = 14=t-/ and 2+v = 2(=:/+5). EQUATIONS, 9 These values, put in the second member of equation [7], give 2 (_,/s+) G- - I::5 by actual division. But x vy = (l V)x-V = ( i5). 10. Given 4V' — -48 xy = Y xy-18' to find the rational values of x and y. Ans. x-8, or — 12; y = 2, or - -. Transpose — 2y and -x in the first equation, and we have xy+x-V-(y2- 1) = 2y2+2y. Divide both members by (y+ 1); then x-V- (y- 1) = 2y. - ------------ [a] This is as far as we can reduce this equation, without the aid of the other equation; therefore clear the second equation of fractions, and we have x2y2_ 18xy = 4V/" —48. Add 49 to both members; then x2y2- 18xy+49 = 4/ -+ 1. Add 4xy to both members, to make squares of them; then x2y2 —14xy+49 = 4xy+4V' 7+ 1. By evolution, xy-7 - 2Vxj+ 1, n] or xy-2v'/ = 8; xy —2/,y+ = 9; V/' — = +3;, = 4 or -2. - ---------- [m] Put these values of JV in equation [a], and we have — 4(y — ) = -, - --- [b] or x+2(y-1) = 2. ------------------ [] From [b], x-6y +4 = O; ---------------- [d] f~rom [C], x = 2. (aliath. Recreations.) 2 10 IVIATHEMATICAAL RECREATIONS. From equations [d] and [ml, we find y = 2 or -4. These values of y, put in equation [mln], give x = 8 or — 12. By giving the double sign to the second member of [n], other values can be foundl which may or may not be imaginary. Observe that we divided one of the equations by (y~+ 1); therefore (y+ 1) may be put equal to zero, and — 1 must be one root of that equation. 11. Find some of the values of' x and y ifrom the following equations: 3x-xJ 4 —2y+S = 2-y, — - [1] V/x+y 3 x 2x —3 3y.5x2 Double equation [1], then add ( 2 +4) to both members; then we have 5 +4+6x —2x 4 —2y+8 = (8-2y+ 41 Put 5x — 2y+8 = P, and transpose; then 5x'1 +6x+4 -- P2+2xP. Now add x2 to both members, and 9x2 +6x+4 =- P2+2xP+x2. By evolution, 2X+2 P+x. or 2-+2 =P. By squaring, -+2x+4 = p - — 2+8, 4 4-2y or 2x x2 —2y+4.-..[a] EQUATIONS. i Multiply eq. [2] by 4', and transpose -3xz; then we have.,/x+y From [a], we find 3x2 — 6y == 6x- 12; therefore 2 (2cx-3)4x2.yrt~j ~~:t~ —12. v'/x+y Dividing by 2, and afterwards multiplying by V'x+y, gives x+y = (2x —3)2x+(3x —6)V/ y. From eq. [a], we find x-y = (2 +2) and v+-y = -+23; therefbre (2 = x2-6x+(3x-6) +2. 1 @2 Put J +2 = Q; thern we have 02-(3x-6)Q = 4zx —6z. Add ( 2 —3)2 to both members, to complete the squares; then Q2_(3x6) x3) 25x-(3x-6)=-Q-( 3)2= — 15-x+9. 2 ~4 By evolution, Q~;bx-3 --— 3 -- b whence Q =x; that is, -+2 = x, or g-+2 = x2, or x 2. This value of x, put in'[a], gives y = 2. By taking the double sign to the second member of equation [b], we can find other values of x and y. 12 MATHEMATICAL RECREATIONS. 12. Find some of the values of x and y from the following equations: 5- 2VY+ X (V 3V 9-)2 - - -- -V —- [ I] 16J —-- x —16. ----------- [2] Y Y 7- 1OV/ - xy- 16y. By transposition, 7+16y - xy+ 10,/V. Add 25, 32+16y = xy+10V/-y +25. By evolution, 4V2y- = V/+5. -- --------- [a] By transposing (V —3v/y)' in eq.[1], and expanding, we have 9X2 5-2/y+2 +x+ w-6V,+9y - -- [b] Equation [a] shows us that 6,/,j = 24V2-+ —30. Substituting this value of 6V/y in [b], we have 9X2 5-2-V/y — 2+x-24V/~-2+30+ 9y =-9 Uniting and transposing, gives 9x2 35+9y —26v/y +2 = 6~ - Vx. Subtract 17 from both members; then we may write the equation 9x2 thus: 9(2+y)-26V/ 2 - x- 17. Put V+y —- = P, and complete the square of the first member by Art. 99 in Robinson's Algebra; then we shall have 169 9xw 16 9P2 —26P+- =9 - 9XBy evolution, 3P13 3 4 3 whence P = 8+ 1 -- [c] or 4P - X+4. -- EQUATIONS. 13 Comparing equations [a] and [d], we perceive that X2 I +4 or -_I = Vx-. 2 x2 By squaring, — x+ 1 = xy or -4 I+- y[n] Equation [c] is += By squaring, 64+ 2+y, 64 4Equations [n] and [m] give x2 x — 1 +, 64 4 4 x x2 1 or -—, orx=4, y=-. 64 x GEOMETRICAL PROBLEMS. 1. From two given points on the same side of a line given in position, it is required to draw two straight lines which shall contain a given angle, and be terminated in that line. Let A, B be the two given points, - and CD the line given in position. Join AB; and from the point A, and on the opposite side of AB from CD, make the angle ABV equal to c - half the given angle; and from the point B, make the angle BAV also equal to half the given angle, thus forming the isosceles triangle AVB. About the triangle AVB describe a circle, cutting the line given in position in the two points C and D. Join AC and CB, or AD and BD, and they will be the lines required. DEMONSTRATION, Join CVa The angle VCB = BAV, because each is measured by half the same arc VB. Also the angle ACV = ABV, because each is measured by half the same are AV. Now by addition the angle ACB = BAV+ABV = the given angle. Q. E D. Renmark. The given angle can not be an angle of any magnitude: it must be sufficiently small to permit the circle to cut the line which is given in position. When the greatest possible angle is,required, the circle will pass through the two points A and B, and touch the line CD. This observation suggests the following problem - GCEOMETRICAL PROBLEMS. 15 2. From two given points on the same side of a line given in position, it is required to draw two lines that shall meet in that line, at that point where they would formn the GREATEST POSSIBLE angle. Let A, B be the two given points, A and CD the line given in position. Produce AB and CD until they meet at F. Take FH of such a magnitude as to make (FPH)2 -A xFB.':-.D -.-o Through the three points A, B and H, describe a circle; the line FH will touch that circle at the point H. Join AH, BH, and the angle AHB is the angle required; and it is the greatest angle that can be made by lines drawn from A. and B, to terminate in the line CD. DEMONSTRATION. Take any other point than H, as C on the line CD, and join BC and AC; these lines will cut the circumference of the circle, because CD touches the circle at H.. BC cuts the circle in 0. Join AO. Now the angle AOB - AHB, because each is measured by half the same are AB; but AOB is greater than ACB, because it is the exterior angle of the triangle AOC; that is, AHB is greater than ACB. In the same manner we could prove that the angle AHB would be greater than ACB, if the point C were in any other position on the line CD, except H; therefore AHB is the greatest possible angle required. Remark. in case AB and CD would not ineet, AB would be parallel to CD; and then we would find the point I b13y making the triangle AHB isosceles, having AB for its base, and its vertex terminated in the line CD. 16 MATHEMATICAL RECREATIONS. 3. From two given points on the same side of a line given in position, it is required to draw two straight lines that shall meet in that line, and mnake the LEAST POSSIBLE sum. Let A, B be the two given points, and A CD the line given in position. From either point as B, let fall the perpendicular BD on the line CD, and produce it to F, making BD = DF. Join C. j AF, and from 0 the intersection of CD "E;: join OB. AOf OB is less than the sum of any other two lines drawn from the points A and B, and terminated in the line CD. DEMONSTRATION. Because BD = DiF; OD common to the two rightangled triangles BDO, ODF; OB = OF: therefore AO+ OB AF. Join AC, CB, CF. AC+CB = AC+CF, because CB=CF; but (AC+CF) are together greater than the straight line AF; that is, (AC+CB) are together greater than (AO+OB). Now let C be at any other point in the line CD, except 0, and still we shall have (AC+ CB) greater than (AO+ OB): hence (AO+OB) is the least possible. Remark. We may observe that the angle BOD- AOC. Consider CD a plane mirror, and BO a ray of light passing from B to 0: it would then be reflected in the line OA, as we well know by observation; therefore when light passes from one point, and is reflected to another, it always passes through the shortest distance in space. GEOMETRICAL PR OBLEMS, 17 4. If from any two points in the circuziference of a circle, there be drawn two straight lines to a point in a tangent to that circle, they will gmake the greatest angle when drawn to the point of contact. Let ATB be the circle; Tt, a tangent /0-~ at the point T; and A, B, any two points in the circumference. Draw AT, BT, At, Bt, and join AO.'A \\1 The angle ATB = AOB, because they stand on the same are AB. But the angle AOB is greater than the angle AtB, because it is the exterior angle of the triangle AtO. Therefore the angle ATB, which is equal to AOB, is greater than the angle AtB; and the same would be true if t were at any other point on the tangent except T. 5. Fromt any point wn'ithin? ct given cicle, to draw a straight line, which shall make with the circumference an angle less than any angle made by any other' line drawn from that point. Let P be any point in the circle. Draw PC to the centre; and through ) Pw triangle, On the o/,, shorter side CB as // radius, describe the circle cutting the I base AB in F, and the longer side AC in E. Join EF; AD:3produce AC to D, and join DB & BE. AE is the difference of the sides, and AF is the difference of the segments of the base made by a perpendicular from C. The triangle AEF gives the following proportion: AE: AF = sinAFE: sinAEF.. [1] 62 MATHEMATICAL RECREATIONS. This proportion is the theorem, as will appear when we show the value of the angles. Because EFBD is a quadrilateral inscribed in a circle, EFB+ D = 180% But EFB+AFE = 180~. By subtraction, D-AFE = 0, or D = AFE. In the same manner we can prove that the angle AEF = ABD; hence sinAEF = sinABD. But sinABD = cosEBF, because the- sine of an angle greater than 900 is equal to the cosine of the excess over 90~; therefore proportion [1] becomes AE:' AF = sinD:, cosABE. -[2] The angle D is half the vertical angle ACB, and the: angle ABE is the half difference of the angles CBA and CAB, for BCD is the sum of those angles; therefore BEC = EBC = the half sum. Consequently EBA or ABE is the half difference; for half the sum of two numbers, added to half their difference, is equal to the greater number. Hence proportion [2] or [1] becomes AE: AF - sin -1-ACB: cos ~-(CBA-CAB). Q. E. D. 6. AJs the greater of two sides of a triangle is to the less, so is radius to the tangent of a certain angle. Demonstrate that radius is to the tangent of the difference between this angle and half a right angle, as the tangent of half the sum of the angles at the base of the triangle, is to the tangent of half their di]erence. To obtain that certain angle, we must put the sides at right ~. angles with each other. On the longer side as radius, describe a circle. Divide the: E diameter into two parts AE, AI EG, so that AE shall be the /, sum of the sides; then EG will / / be the difference of the sides. / From C, draw CD at right an- - gles to AG. Join AD, DG, and TRIGONOMETRICAL THEOREMS AND PROBLEMS. 63 draw EH parallel to AD. The angle ADG = 90~, because it is in a semicircle; hence EHG = 90~, AD = DG, EH -- HG. Let a be the greater side of any triangle, represented in magnitude (but not in position) by CD; and c the lesser side, represented in magnitude by CE: then it is plain that a: c =: tanCDE. This angle, taken out of the tables, and subtracted from 45~, will give the angle EDG. By proportional triangles, we have AE:EG-DH:.HG or EH; that is, a+c a-c D= H: EH. But DH: EH- = R: tanEDG; whence ac: a-c = R tanEDG. From scholium to theorem 3, we have a+c: a-c = tan. sum angles at base: tan - diff.; whence R: tanEDG - tan a sum angles base: tan -2 diff. Q. E. D. APPLICATION. The application of this theorem is to find the angles of triangles, when the logarithms of two sides, and the angle included between them, are given. It avoids the necessity of finding the sides from the logarithms; and is very useful in astronomy in finding the longitude of a planet as seen from the earth, when the position of the earth, and the position of the planet as seen from the sun, are both given. The astronomical tables give the position of the earth from the sun, and the logarithm of its distance. They also give the position of each and every one of the planets as seen from the sun, and the logarithms of their distances. From these elements, this theorem will give the position of a planet, at any definite time, as seen from the earth. We give the following EXAMPLE. JSt noon (Greenwich time),.March 1, 1846, the earth, as seen from the sun, was in longitude 1600 33' 53", and the logarithm 64 MATHEMATICAL RECREATIONS. of its distance from the sun was - 1,996293.* At the same time the longitude of Jupiter, as seen from the sun, was 46~ 54' 54", and the logarithm of its distance was 0,698498. What was the longitude of Jupiter as seen from the earth? (As we are not teaching astronomy, we take no account of the planet's latitude, which would affect its geocentric longitude in some slight degree.) Let S be the position of the sun. Describe a circle around it with any radius, - - and take any point as T for the first point, or / zero point of lon- - 01;I1O gitude. Let SE s represent the dis- " o tance between the earth and sun I- i20 at this particular time; and to cor- /respond with -our 1,example, the position of E the earth must be as represented in the figure. Place Jupiter at the point I to correspond with its given longitude and distance. The object is to find the position of the line El, in respect to c? —; or, what is the same thing, the angle'YSH (in case SH is p,.rallel to EI). COMPUTATION. From 1600 33' 53" Take 46 54 54 The angle ESI = 113~ 38' 59" 180 0 0 2) 66021' 1" }(I+SEI) = 330~ 10' 30,5". * The unit is the mean distance between the earth and sun, and its logarithm is 0,00000. At this time the distance was less than the mean distance, and of course the index to the logarithm is negative. TRIGONOMETRICAL THEOREMS AND PROBLEMS. 65 The half difference of these angles is as yet an unknown quantity, which is our immediate object to find; and we represent it by x. By the theorem: As 0,698498 Is to - 1,996293 So is 10,000000 To tan of angle 110 13' 40", whose log is 9,297795 45 0 0 330 46' 20" As 10,000000: tan 33~ 46' 20". 9,825257 tan 33 10 30,5 9 9,815417: tanx23 36 50 - 9,640674 Diff. 9~ 33/ 40"= SIE = HSI. Hence from Ion. 46~ 54' 54" take 9 33 40 Lon. of Jupiter from the Earth, 37~ 21' 14". If desirable, the logarithm of El can be readily found. A Nautical Almanac will give any number of similar examples. 6. Two ships in the same north latitude, but 126 miles distant from each other, sail 248 miles south, and their distance from each other is then 200 miles. Required the latitude left, and the latitude arrived at. Ans. Lat. left, 830 2'; arrived at, 78~ 54. This problem corresponds to the supposition that the earth is a perfect sphere, and 60 miles to a degree. The ships were at A and B, and U" ^v~csailed to C and D. AC = 248 miles, or 4~ 8'; -D " _4 AB = 126 miles, and CD = 200 miles. The two plane triangles ABH, "1,: CDE are similar, and give the _ D proportion AH: CE = AB; CD. (Math. Recreations.) 9 66 MATHEMATICAL RECREATIONS. But AH is the sine of the arc PA, and CE is the sine of the arc PC. Let x -= the arc from P to the midway point between A and C, and put a = half the arc AC; then AH = sin (x-a), and CE = sin (x+a); whence sin(x-a): sin(x+a) = 126: 200, or sinxcosa —cosxsina: sinxcosa+cosxsina - 63: 100. By adding and subtracting antecedents and consequents, 2 sinx cosa: 2 cosx sina = 163: 37, sinx or — cosa: sina 163 37, GOSX or tanx: tana 163: 37; 163 whence tanx = - tana. Put a - 20 4'; then, log tana = 8,557336 log 163 = 2,212188 10,769524 log 37 = 1,568202 tanx = 90 2' log = 9,201322 PA=(9~ 2')-(2~ 4')=( 6~ 58') = polar distance, or lat 830 2'. PC =(9 2') + (2~ 4')=( 11~ 6') = polar distance, or lat 78~ 54'. 7. ~. ship at sea came in sight of two islands, one bearing north, the other west. J/lfter sailing N 250 W 8 miles, the ship was at equidistance from the islands; and after sailing 3 miles further on the same course, she was in a line between the islands. What was her distancefrom the islands when they were first seen?' We adduce this example to show the power of algebraical analysis as applied to geometry: a solution purely geometrical, we could never attain. TrIGONOMETRICAL THEOREMS AND PROBLEMS. 67 Let S be the position of the ship, N the northern island, W the western, one. Sn is the direction the ship sails: So = 8 miles; on = 3 miles. Put SN=y, SW=x, So=a, Sn=b,,, and the angle nSm = 25~- A; then Sn = a cosA, /o_ _lt r or = Sq = a sinA; Smn = np = b cosA, =nm = Sp - bsinA;,Nr = y-acosA, x H. - s Wq = x-a sinA. Because o is at equal distances from N and W, we have (x-a sinA)2+ a2os2A = (y —a cosA)2+ a2sin2A... [1] By reduction, we find x2 —2 ax sinA = y2 —2 ay cosA.. [2] By similar triangles, we have Wp: np = WS: SN; that is, x-bsinA: bcosA = x: y; [3] b cosA.x whence Y x-b sinA' x - b sinA This value of y, put in equation [2], gives b2cos2A.x2 2 ab cos2A.x _ —2 ax sinA --. (x - b sinA)' x-b sinA' b2Cos2A.x or (x-2a sinA)(x-b sinA) = - - 2abcos2A. x- sinA Here we have an equation of the third degree, and its solution will give SW; but as we care nothing for the numerical result, we here leave it. 68 ASTRONOMICAL PROBLEMS. 1. In latitude 40~ 48' north, the sun bore south 79~ 16' west at 3h 38m P. M. apparent time. Required his altitude and declination at that time. In this example there is nothing difficult; and any one who can construct the figure, can solve the problem. The time from noon z gives the angle / /. zPS = 54~ 30': EZ = 40~ 48'; / hence ZP = 49~ 12'. / The object of the / problem is to find Sm l -... and Sn. In the rightangled spherical triangle PQZ, we have the side PZ and the angles P and Q; and therefore we can find ZQ, PQ, and the angle PZQ. Because we have the bearing of the sun, we have the angle PZS = 100~ 44'; from which subtract PZQ, and we shall have QZS; and having the side ZQ and the right angle Q, we can obtain all the parts of the rightangled triangle QZS. Then the complement of ZS is the sun's altitude, and PQ+ QS is the sun's polar distance, which is the complement of the declination. In this case, altitude = 36~ 46'; sun's declination, 15~ 32' N. ASTRONOMICAL PROBLEMS. 69 2. In latitude 16~ 4' north, when the sun's declination was 23~ 4' north; required the time in the afternoon, and the sun's altitude and bearing, when his azimuth neither increases nor decreases. Ans. Apparent time 3h 9m 26s P. M.; altitude, 45~ 1'; bearing north 73~ 16' west. 1; X~:~,. Let dd' be the small circle of the sun's \r I \ [ \"declination. / \The problem requires the point where I/ /p......\.1 }> "!r the great circle ZSN _____...\ =....\ will touch the small -circle dd'. Let it / touch at S; and then \! \ / \ / we have the rightangled spherical triangle PSZ, in which PS = 66~ 58', and PZ = 73~ 56'. Spherical trigonometry gives the proportion R: cosPS = cosZS: cosPZ, or R: sin(23~ 2') = cosZS: sin(160 4'); whence ZS = 44~ 59', or the altitude = 450 1'; PZS = 73~ 16', and ZPS = 470 21' 30" = 3h 9m 26. 3. In latitude 16~ 40' north, when the sun's declination wats 23~ 18' north, I observed it twice in the forenoon bearing north 68~ 30' east. Required the times of observation, and the sun's altitude at each time. Ans. Times, 6h 15m 40s, and 10h 32m 48s A. M.; altitudes, 9~ 59' 36", and 680 29' 42". The last figure will serve for the problem, with a little aid of' imagination. In this example the circle ZSN must cut the small circle dd' in two points, and the angle PZS is given 68~ 30', No further remarks are necessary. 70 MATHEMATICAL RECREATIONS. 4. Given the latitude, the sun's declination, or the declination of any planet or star, to find the semidiurnal arc for the sun, planet or star. Let Hh be the horizon, EQ the equator, Pp the six o'clock E meridian, S or S' the < position of the sun or star, according as its declination is north or south. The portions of the equator represented by Cq, Cq', changed into time, and added to six Q hours if the latitude and declination are both north, or subtracted from six hours if one is north and the other south; the sum or difference, as the case may be, will be the interval between the meridian position of the sun, planet or star, and its rise from or fall to the horizon. We find the arcs Cq, Cq' by means of the triangles CqS, Cq'S'. Sq is the declination, and the angle SCq is the complement of the latitude. By trigonometry, R. sinCq = tanSq. cotSCq. When radius is unity, this equation becomes sinCq = tanD.tanL; where D represents the declination whatever it is, and L the latitude. We give the preceding as a lemma to the following problem: ASTRONOMICAL PROBLEMS. 71 4. In 1850, the right ascension of Spica was 13h 17m 18s, its south declination 10~ 22' 37"; the right ascension of Alrcturus was 14h 8m 49s, and north declination 19~ 57' 56"..1 traveller to Caltfornia observed that Spica set 2h 26m 14s before Jsrcturus (allowance made for refraction). What was his north latitude? Let L = the latitude; d = 10~ 22' 37", D = 19~ 57' 56". The star Spica came upon the observer's meridian at some definite time that we denote by JM. Then MJl + (6 —q5) _ the time Spica set; JM + Oh 51m 31s + (6+ -) = the time Arcturus set. Observe that Oh 51m 31s is the difference of the right ascensions of the two stars. Cq and Cq' are right ascension arcs, and an arc is changed into time by dividing it by 15; of course, making one hour the unit. By subtracting the time Spica set, from the time Arcturus set, we shall obtain an expression equal to 2h 26m 14s; Cq Cq' that is, Oh 51m 31s + 15+ -q 2h 26m 14s, - [1] 15 -5 or Cq+Cq' = 15(lh 34rn 43s)..[2] Equation [1] expresses time: equation [2] expresses arc. When we divide arc by 15, we obtain time: when we multiply time by 15, we obtain arc; hence 1 hour multiplied by 15, gives 15~. Therefore the sum of the arcs Cq+ Cq' = 23~ 41'... -.....[3] By the preceding problem, sinCq' = tand. tanL,.-. — -. —-.[4] and sinCq - tanD.tanL. [5] Put a = 23~ 41', and observe that Cq' = a-Cq; whence sinCq' =- sin(a-Cq).. sina cosCq - cosa sinCq. From [5], 1-sin2Cq = 1-tan2D. tanL; that is, cos2Cq- l —tan2D.tan2L. 72 MATHEMATICAL RECREATIONS. The values of sinCq', sinCq, and cosCq=V/1-tan2D.tan2L, put in [6], gives tand. tanL = sina/1-tan2D.tan2L - cosa. tanD. tanL.. [7] By transposition and division, tand+ cosa. tanDtanL 1tan2D.tanL. ("-"~"sina -tanL = VI-tan2D.tan2L. Let this coefficient be represented by mn; then nm2tan2L = 1-tan2D.tan2L. By division, rn2 = tn —ttan2D, By tan2LtanD, or m2 +tan2D = cot2L, because t cot2L. We are now left to apply this equation. First find m, which is (n- +tn). We take the logarithms (sina tana from the table of logarithmic sines and tangents, with 10 dropped from the index, because our equation refers to radius unity, thus: tan d.. - 1,262978 tan D. — 1,560270 sin a. - 1,603882 tan a... -- 1,642090 0,3623.X. - 1,559096 0,8283 -.- — 1,918180 Whence m = 0,3623+0,8283 = 1,1906; m2= 1,4175; tan2D = 0,1320. Therefore cot2L = 1,5495; cotL = 1,2447; log + 10 = 10,095000, corresponding to latitude 38~ 46' north. ASTRONOMICAL PROBLEMS. 73 6. On the 14th of JNovember 1829, the star.Menkar (whose right ascension was 2h 53m 21s, and declination 30 24' 52" north) rose 48m 3s before Alldebaran (whose right ascension was 4h 26m 7s, and declination 160 19' 31" north): From what latitude was the observation made? Ans. Lat. 39~ 33' north. Let il be the time Menkar was on the observer's meridian; then M- (6+ ar) = time Menkar rose, and. M+r- (6 + Ar- c) = time Aldebaran rose Arc are Diff. rr+ = 48m 3s. The difference of the right ascensions of the two stars is represented by r. Hence r =- h 32m 46s, and Arc_arc = Oh 44m 43s = 0,74527; 15 15 whence Arc —arc - (0,74527)15 = 110 10' 45". For more distinctness, put a = 11~ 10' 45", x=Arc, y=arc; then x = a+y, and sinx = sin(a+y) - sina cosy+ cosa siny. Let d = 3~ 24' 52", D = 160 19' 31", and L = the required latitude; then sin x = tan D. tan L, and sin y = tan d. tan L. By proceeding as in the previous problem, we find cot2L m2qn+tan2d, tanD in which m =. —cota tand. sina (Math. Recreations.) 10 74 DIVISION OF LANDS. The two following problems are from GUMMERE'S Surveying, and are considered very difficult. I. There is a piece of land bounded as fbllows: Beginning at the southwest corner; thence, 1. N 14~ 00' W, Distance 15,20 chains - a; 2. N 70 30E, E 20,43 - b; 3. S 6 00 E, E " 22,79 " -c; 4. N 86 30 WA, " 18,00 " =d. Within this lot there is a spring; the course to itfrom the second corner is S 75~ E, distance 7,90 chains. It is required to cut off ten acres from the west side of this lot, by a line running through the spring. Where will this line meet the fourth side, that is, how far from the first corner? Ans. 4,6357 chains. First make a draft of the field. It is as here represented., \.V l'.-peace k A l& Produce the sides b and d, the second and fourth, until they meet at G. Let S be the position of the spring, and join SG. We may or may not find the contents of the field*: it is not necessary for the location of the line LSH. When we have four sides olnly, and all the angles, as in this field, the best method of finding the contents is by conceiving it to be two triangles. Thus, in this case, the area is represented by 1 ab.sin ABC + A dc.sin CDA, DIVISION OF LANDS. 75 It is necessary to find the area of the triangle ABG. Conceive a meridian line run through B; then we perceive that the angle ABG- =14 + 70~ 30'= 84~ 30'. Conceive also a meridian line to be run through A, and then we perceive that the angle BAG = 86~ 30' - 140 = 720 30'; whence AGB = 23~. With the angles and the side AB= 15,20, we readily find AG:38,72, BG = 37,10, and the area AGB = 280,65 square chains. It is necessary to find the line GS and the angle BGS. From the given direction of the lines BC and BS, we find the angle GBS = 145~ 30'; and then from the triangle GBS, we find BGS = 5~ 51' 30", and GS = 43,83. Also we have the angle SGA -- 17 8' 30". To the area of the triangle AGB - 280,65, add 10 acres, or 100 square chains: then the area of the triangle GLH must equal 380,65 square chains; but GL and GH are both unknown. Put GL = y, GH - x: then we shall have the equation xy sin 23~ = 2(380,65).... [11 It is obvious that the sum of the two triangles LGS, SGH is equal to the triangle GLH. Put GS = n = 43,83, sin23 - P, sin( 170 8' 30") = Q, sin(50 51' 30") = R, and 2(380,65) or 761,3 = a: then we have Pxy = a, [1 and Rmy+ Qmx - a......[2] a From [1], y = p-. This value put in [2], gives aRm +Qmx = a;...[3] whence x — x= --.. [4] VQ PQ' a all We now find the numerical values of - and a- by logarithms, as follows: As our radius is unity, we diminish the indices of the logarithmic sines by 10. 76 MATHEMATICAL RECREATIONS. log a, 2,881556 log a, 2,881556 log m, 1,641771 log R, - 1,008880 log Q, -1,469437 1,890436 1,111208 a 1,111.208 58,932. 1,770348 logP, -1,591878 log Q, - 1,469437 - 1,061315 - 1,061315 674,72 —........ 2,829121 Equation [4) now becomes x2 —58,932x = -674,7; whence x = GH = 43,366: from which subtract GA - 38,72, and we have AH the distance required = 4,646, which differs from the given answer about one link of the chain. LEMMA, Find the point in any trapezoid, through which any straight line which meets the parallel sides will divide the trapezoid into two equal parts. Let ABCD be the trapezoid. Bisect the parallel sides AB and C F L E CD in the points n and m. Join mn, and bisect mn in 0, and O is I the point required. Any line meeting the parallel A q: n1B sides, and passing through 0, will divide the trapezoid into two equal trapezoids. It is obvious that the line mn divides the figure into two equal parts, because the sum of the parallel sides is the same in each. Now draw any other line through O, as pOq: the trapezoid pqBD = mnBD; because the triangle Omp = qOn, and one triangle is cut off and the other is put on at the same time. The triangles are equal, because mO= On, the angle pmO = Onq, and the opposite angles at 0 are equal: therefore mp = nq; and whatever more DIVISION OE LANDSd 77 than Cm is taken on one side, the equal quantity qn less than An is taken on the other side. Another method of finding the point 0, is to bisect AC in HI and draw HO parallel to AB or CD, and equal to one-fourth the sum of AB and CD. IJ. PROBLEM. There is a piece of land bounded as follows: Beginning at the westernmost point of the field; thence, 1. N 35~ 15' E, 23,00 chains; 2. N 75 30 E, 30,50 c; 3. S 3 15 E, 46,49 "; 4. N 66 15 W, 49,64 cc It is required to divide thisfield into four equal parts, by two lines, one running parallel to the third side, the other cutting the first and third sides..Find the distance of the parallel line from the first corner measured on the fourth side, and the bearing of the other line. Gummere's answer ~ Distance to the parallel, 32,50 chains; Bearing of the other side, S 880 22' E. Construct the figure as here represented, according, to the directions and dis- -, -' -d - -tances; AB being the first side, &c. (; <-' is t\ Produce the sides AD, BC'" \ \ j to meet in G. The angle GAB = 660 15' + 35~ 15' = 1010 30', and GBA = 750 30' - 35~ 15' = 40~ 15'; hence the angle G -- 38~ 15'. With the side AB and the angles, the sides and area of the triangle GAB can be readily found. 78 MATHEMATICAL RECELA'TIONS. The area of the field is equal to (11,5)(30,5)sin( 139~ 45') + (24,82)(46,49)sin 63~. This expression gives 1254,86 square chains for the whole area, and of course one of the four equal divisions is 313,715 square chains. The area of the triangle GAB = 270,5 square chains; GA = 24,05, GB = 36,40: hence GD = 73,69, GC=66,90. Let PQ be drawn parallel to DC, and let it divide the field into two equal parts; then the two triangles GDC and GQP are equiangular. The area of the triangle GDC = 270,5+1254,86 = 1525,36; the area of the triangle GQP - 270,5+ 627,43- 897,93. Put GQ = x; then (73,69)2: 2 1525,36: 897,93 [x=56,54.1 whence GQ-GA AQ = 56,54-24,05 = 32,49. We have now all the parts of the trapezoid QDCP, and can find the point O within it, through which any line will divide it into two equal parts (See lemma). But a line must be drawn from O in such a direction as to divide the area ABPQ into two equal parts. Let On be that line. Join OH. OH and the angle OHm can be computed; also BH and the angle BHm, and the area of the triangle BHP; to which add 313,71, and we shall have the whole area Hnm. Put Hn x, Hm- = y, HO = a, area Hnm = i; then -x6y sinBHP -= J,.[11 and -ax sinBHO —2ay sinOHm H. - --- -- [23 From [1] and [2] we can compute x and y, which give the points n and m, and therefore the direction nm. The angle AHQ or BHP is found thus: GAB = 1010 30)'; therefore HAQ = 78~ 30'. Because PQ is parallel to CD, the angle AQH = D = 63~; whence the third angle of' the triangle AQH, that is, AHQ = 38~ 30'. DIVISION OF LANDS. 79 Having the angles and the side AQ = 32,49 of this triangle, we find QH = 51,16 and AH = 46,30. But AB = 23; therefore BH = 23,30. By the aid of the triangle GQP, we find PQ = 35,89; hence QH-QP = HP = 15,27. Now the area of the triangle BHP is expressed thus: -(23)( 15,27)sin(380 30') -- ( 1,5)( 15,27)sin(38~ 30') = 109,32 sq. chains. To this add BnmP = 313,71 "; Area Hnm, designated by As = 423,03 sq. chains. By means of the two triangles HBP and Hnm, we have the following proportion, which is in fact but a modification, or rather a simplification of equation [1] xy: (23,3)(15,27) = 423,03: 109,32; whence xy =- 1390,6. a a [3] With the value of s9, equation [2] becomes xsinBHO -- ysinOH- 846,06 [4] WVe must now find the value of the angles BHO and OHm, and the numerical value of HO, designated by a. The area of the trapezoid PQDC is 627,42 square chains, and its parallel sides are DC - 46,49 and PQ = 35,89. Half their sum is 41,19: therefore divide 627,42 by 41,19, and we have 15,22 for the perpendicular distance between DC and PQ; and from 0 to PQ is 7,61. The distance from O to the middle point of QD is one half of 41,19, or 20,59; but this is greater by 3,89 than the distance from Q to the perpendicular from 0; that is, from Q to the perpendicular is 20,59 less 3,89, or 16,70. From OH- 51,16, take 16,70, and we have 34,49 for the longer side of a rightangled triangle whose hypothenuse is HO. Now we have 34,46: 7,61 = rad: tanOHm. This proportion gives OHmn = 12~ 27', and HO - 35,29 = a. 80 MATHEMATICAL RECREATIONS. To BHP = 380~30', add 120 27', and we have BHO=-50 57.o Now equation [4] becomes sin(50~ 57')x - y sin( 12~ 27') = 846,06 35,29 Substituting the value of y from equation [3], we have sin(500 57')x 1390,6. sin( 12 27) 23,97; x2- 23,97 x 1390,6. sin( 120 27') X2 — 23,97sin(50o 57') log 1390,6 -— _ 3,143020 sin( 12~ 27') —. -1,333624 2,476644 sin(500 57').. - 1,884254 log 391,2 2,592390 2 — 23,97x = 391,2; x - 35,1. From HA = 46,30, take 35,1, and we have An - 11,29. The triangle HnO is nearly isosceles: HO = 35,29; Ha - 35,1. I make the angle HnO = 64~ 57', which gives the direction of nO south 79~ 48' east. I have made an error in numerical computation, or this part of Gummere's answer is erroneous. PROBLEMS IN MENSURATION. 81 I received the two following problems from Mr. D. M. KIMBALL, Of Warren, Massachusetts. PROBLEM I. Suppose five men own a grindstone in equal shares. The stone is 7 feet 9 inches in diameter, 4 inches thick at the eye, and the eye-holefour inches square, and tapers to a point at the outer edge. JVow i/ each man were to grind off his proportional share separately, how many inches will each one diminish its diameter? In the first place we must compute the solidity of the stone, and divide it by 5. Let AB represent the diameter of the stone, and CD its axis; then ACBD will represent a diamnetral, 1 / _ F section. If the plane ACBD revolve on its axis CD, it will generate two __ _ v1 \ I equal cones; and if from these we e / G'O -I take the cylinder QR and the two _-, i_ _small cones on its ends, we shall have the solidity of the stone. AB -= 93 inches; AV = 44,5; VP = 2, and we find OD by the following proportion: 44,5: 2: 46,5 OD = 93 ~~~~~1B 44,5~ The expression for the solidity of one cone is 0,7854(93)2 1; for two cones, 7854(93)262 9464,3 cubic inches. 44,5 Although the eye-hole is square, there is a cylindrical waste space about the axis, whose diameter is the diagonal of that square. The square of this diameter is 32; hence the cylinder RQ is expressed by 0,7854(32)4. To this add the two small cones at its ends. The altitudes of these cones are 93 89 4 OD —PV - - 44,5 44,5 - 4,5 (JMlath. Recreations. 11 82 MATHEMATICAL RECREATIONS. For two cones, the expression is 0,7854(32)8 = 1,506. 133,5 The cylinder RQ, plus the cones = 102,037 cubic inches. From the whole mass --—.... --- --- 9464,300, Subtract. 102,037; The available part of the stone = -- 9362,263. The cubic inches to each therefore must be 1872,4526. After the first man has ground off his share from the whole mass, there must remain - 7591,848 cubic inches; After two have ground, there remains 5716,396 " After three have ground, _ 3846,944 " After four have ground, - 1974,492; After the fifth, there remains 102,050' which is the waste space about the axis. The main part of the problem is now to come; that is, to discover how much each one will diminish the diameter. One general expression, however, will answer for the whole. Suppose the first man grinds down from A to EF. FH will then be the semidiameter. Conceive the plane CDEF to revolve on the axis CD: it will generate a solid -— 7591,848 cubic inches, as we have just determined. But this solid will consist of a cylinder and two equal cones. The length of the cylinder is GH, and the altitude of one of the cones is HD. Let FH = y, and HD = x. Then AO OD — y: x; 93 93 that; is, - - x 2 44,5 4 hence x -Y. I'he expression for the solidity of the cylinder is 22Q93 -x) F h(4n,5 t For the two cones, the expression is 2~y2px. PROBLEMS IN MENSURATION. 83 Therefore the expression for the whole solid must be 1862 y2x+ 2x 186+y2 4~y2o 44,5 s X, or 44,5 -~yx. Substituting the value of x in terms of y, we have the following general expression, which, being put equal to the cubic inches which remain after the first, second, third, or fourth person has ground off his share, will give an equation which, being solved, will give y, and of course 2y, the diameter of the stone at each of these points, thus ~ G86^yi 16-y3 = 7591,848. 44,5 267 This equation will give the diameter of the stone after the first man has ground off his share. The above equation reduces to 93,267 2 =267 or y3- 69,75y2 -(7591,848)50 26 --- 50,26[1] The following equation will give the diameter of the stone, after two shares are ground off: 267 y — 69,75y2 = — (5719,396)0 26 -- [21 the following, when three shares are ground off: 267 3 -- 69,75y2 = -(3846,944)5-. —.; [3] and the following, when four shares are taken: 267 y3 - 69,75y9 = -(1974,492)50 265. [4] These equations, being of the third degree, may have three real roots; and the mere mechanical operator might happen on one that would not apply to this problem, and from thence condemn the equations; but these circumstances will never trouble the initiated& 84 MATHEMATICAL RECREATIONS. Solving equation [1], we find three roots; one near 56, another 33,19 nearly, and a third near the locality of minus 21. As neither the first nor the third can apply to our problem, we take no pains to determine them more accurately. The second root, 33,19 inches nearly, shows that the first share taken off would diminish the radius of the stone 13,3 inches, or the diameter 26,6 inches. Equation [2] has three roots; one near 62, the second 26,48 nearly, and the third near minus 19: the second applies to our problem. Equation [3] gives three roots; the first near 63,5, the second 22,77, and the third near minus 16,5. Equation [4] also gives three roots; one near 65, the second 13,67, and the third near minus 12. The fifth man will grind down to the eye. The radius of the circle, or y, will then be 2,828, and should be found in the following equation' y3 69,75y2 (102,037) 26 = -541,8. 50,265 A solution of this equation gives y = 2,845; showing a slight discrepancy, probably arising from the neglect of decimals. Whence the answer to the question is as follows: The number of cubic inches in the stone is 9362,263. The share for each of the five men is.... 1872,452o The first share takes 26,6 inches from the diameter; the second, 13,42 inches; the third, 7,42 " the fourth, 18,20 6C "; the fifth, 21,666 6" PROiLEMS IN MENSuRATION. 85 PROBLEM II. There is a grindstone 6 feet in diameter, with a sixainch circular eye-hole at the centre. The inner rim of the stone at the eye is but 2 inches in thickness, and outer rim 6 inches. What is the solid content of this stone, and what will be its diameter when one-third worn away? A L Let the plane ABab revolve on the axis Apd ___ CD, and conceive it to generate the stone., ABCD, revolving on CD, will generate a cylinder, which will just enclose the stone, 1. n o, \_/; two conical frustums, and the cylindrical 1 _ ~ t eye-space. Therefore if from the cylinder generated by the plane ACDB, we subtract the two frustums and the small cylinder whose section is abcd, we shall have the solidity of the stone. The solidity of the cylinder is found by the following product: 0,7854(72)26 - 24429,08 cub. inches. The frustums 4(2~+72.6+62)0,7854- 5918,774 c" 18510,30 cc The small cylinder 0,7854(36)2.. - - 56,55 C" Cubical content of the stone in inches - 18453,75 "C One-third of this. - 6151,25 C" When the stone is worn down one-third, let its outer rim be at PQ; then PR will represent its diamneter. Now it is obvious that the cylinder generated by the revolution of the plane PQRS on the axis, together with the two frustums at its ends, must equal the whole original solid, diminished by 6151,25 cubic inches; that is, equal ~. 24429,08 less -_ 6151,25 which is equal to 18277,83 Draw ao perpendicular to PR and AC. Put ao = x, and 86 1MATHEMATICAL RECREATIONS. PR = 2y; then Po = y-3, and by proportional triangles we have x y-3= 2: 33; whence x — = ~.. [1] 33 The cylinder generated by the revolution of the plane PQRS on the axis CD, is expressed by -y2(2x+2). The expression for the two frustums is (+S+3y)(4- 2x) 3 Whence -y2(2x+2)+q(3 2 36y +y2)(4-2x) -18277,83. 3 Dividing by 2gr or 6,2832, and putting 36 = a, we have y2(q+ 1)+(a2+ay+y2)(2-x) 18277,83 3 6,2832 18277,83 or 3y2x+ 3y2+-(a2+ay+y2)(2-x) - 182 83; that is, 2,0944 3y2x+ 3y-2+2a2 +2ay+2y2-(a2+- ay)x-y2x - 8727 nearly, By uniting terms, 2y2x-+5y2+2ay-(a2+ay)x = 6135. Substituting the value of x as found in equation [1], in the preceding, gives 4y- 12+ y + 2ay-(a2 +ay)( ) - 6135. Put b = 6135, and multiply by 33: we then have 4y3 — 12y2+ 165y2+66ay-(a2+ay)(2y-6) = 33b, or 4y3+ 153y2+66ay-2a2y-2ay2+6a2+6ay -- 33b. Observe that 2a2 = 72a, 2a = 72, and 6a2 = 7776, because a = 36. These substitutions being made, we have 4y3+81y2 = 33b-7776, or 4y3+81y2 = 194679, or y3+20,25y2 _ 48669,75. This equation solved, gives one root = 30,73, and two imaginary roots. This result shows that the diameter of the stone is now 61inches, and the original diameter is diminished 10~ inches. 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