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Published by the Canadian Mathematical Society. http://crux.math.ca/ The Back Files The CMS is pleased to offer free access to its back file of all issues of Crux as a service for the greater mathematical community in Canada and beyond. Journal title history: ➢ The first 32 issues, from Vol. 1, No. 1 (March 1975) to Vol. 4, No.2 (February 1978) were published under the name EUREKA. ➢ Issues from Vol. 4, No. 3 (March 1978) to Vol. 22, No. 8 (December 1996) were published under the name Crux Mathematicorum. ➢ Issues from Vol 23., No. 1 (February 1997) to Vol. 37, No. 8 (December 2011) were published under the name Crux Mathematicorum with Mathematical Mayhem. ➢ Issues since Vol. 38, No. 1 (January 2012) are published under the name Crux Mathematicorum. Mathematicorum Crux CRUX MATHEMATICORUM Vol. 12, No. 9 November 1986 Published by the Canadian Mathematical Society/ Publie par la Societe Mathematique du Canada The support of the University of Calgary Department Statistics is gratefully acknowledged. * of Mathematics and t * CRUX MATHEMATICORUM is a problem-solving journal at the senior secondary and university undergraduate levels for those who practise or teach mathematics. Its purpose is primarily educational, but it serves also those who read it for professional, cultural, or recreational reasons. It is published monthly (except July and August). The yearly subscription rate for ten issues is $22,50 for members of the Canadian Mathematical Society and $25 for nonmembers. Back issues: $2,75 each. Bound volumes with index: Vols. 1 & 2 (combined) and each of Vols. 3-10: $20. All prices quoted are in Canadian dollars. Cheques and money orders, payable to CRUX MATHEMATICORUM, should be sent to the Managing Editor. All communications about the content of the journal should be sent to the Editor. All changes of address and inquiries about subscriptions and back issues should be sent to the Managing Editor. Founding Editors: Leo Sauve, Frederick G.B. Maskell. Kditor: G.W. Sands, Department of Mathematics and Statistics, University of Calgary, 2500 University Drive N.W., Calgary, Alberta, Canada, T2N 1N4. Managing Editor: Dr. Kenneth S. Williams, Canadian Mathematical Society, 577 King Edward Avenue, Ottawa, Ontario, Canada, KIN 6N5• ISSN 0705 - 0348. Second Class Mail Registration No. 5432. * t Return Postage Guaranteed. t CONTENTS The Olympiad Corner: Problems: Solutions: 79 . M.S. Klamkin 1181-1190 1010, 1048-1053 229 241 . 243 ~ 228 - - 229 - THE OLYM>IAD CORNER: 79 M.S. KLAMKIN All communications Department of about Mathematics, this column University should be of Alberta, sent to Edmonton, M.S. Klamkin, Alberta, Canada, T6G 2G1. This month I give three sets of problems. The first set is a number of solutions appear at "Quickies" proposed by me and whose the end of this corner. Mathematical Quickies were initiated by C.W. Trigg in 1950 as then editor of the Problems and Questions Department of Mathematics Magazine* These are problems which can be solved laboriously, but with proper insight and Imowledge can be disposed of quickly. Knowing a problem is a Quickie is a decided hint toward solving it. Q-l. Determine a two parameter integral (x,y9z,w) solution of the Diophantine equation w2 = 2(x4 + yA + z 4 ) . Q-2. Determine a three parameter integral solution (x,y,z,w) of the Diophantine equation (xyz + yzw + zwx * wxy)2 = (yz - xw)(xz - yw)(xy Q-3. Determine the number of real solutions (x,y,z,w) - zw). of the simultaneous set of equations x 5 + x = 2y2z, ys + y = 2z2w, zs •¥ z ~ 2w2x, ws + w = 2x2y, Q-4. For af b positive determine the maximum value of y = (a - x)(x + Jx 2 + b2) over all real x (no calculus please!), Q-5. It is known that if A, B9 C are the angles of a triangle, then cot A/2 + cot B/2 + cot C/2 > 373" and cot A + cot B + cot C > J5. Prove that in fact 3(cot A + cot B + cot C) > cot A/2 + cot B/2 + cot C/2. Q~8. Determine the minimum value of = cot(?T - A)/4 -f cot (IT - B)/4 -f cot(n - C)/4 . sin A + sin B + sin C ~— where A, B, C are the angles of a triangle. - 230 - Q-7. Prove that the diagonals of a quadrilateral are perpendicular if and only if the sum of the squares of one pair of opposite sides equals the sum of the squares of the other pair of opposite sides. Q-8, If a, b, {abc - az(b c and a - x, b - yf - y) - bx(c - z) e - z are non-negative, prove that - x)}z - cy(a > 4xyz(a - x)(b - y)(c - z). Q-9. By factoring (a2 + b2 + t2)(a + b + c)(5 + c - a)(c + a - 6)(a + 5 - or otherwisej show that a triangle is acute, right, or a 2 + 2 b + c 2 is > f = 2 f Here a9 bf or < 8J? . 8a2b2c2 ) - obtuse, according as c are the sides and R is the circumradius of the triangle, Q-10. Find the shortest distance between two non-intersecting face diagonals lying in adjacent faces of a unit cube« * The next two sets of problems are by courtesy of Walther Jan^s, and as usua] we invite elegant solutions for them* 17th Austrian Mathematical Olympiad 1986 1st dayf June 4, 1986 (time allowed 4 1/2 hours) _1. Show that a square can be inscribed in any regular n-gon® 2* For s9t € M 9 let M = {(x,y) | 1 < x < s, 1 < y < t9 xfy be a given set of points in a plane• vertices belong to e m] Determine the number of rhombuses whose M and whose diagonals are parallel to the x,y coordinate axes* _3. Determine the set of all values of x 0 ,x x € IR such that the sequence defined by x X n+1 1 x = — — ,«— , 3x . - 2x n-1 n li > 1 contains infinitely many natural numbers, 2nd day (4 1/2 hours) 4_* Determine the with base 10 maximum value of n such that there exists a number N representation t/1t/2...c/ having different digits and such that n! |tf. Furthermore, for this maximum n9 determine all numbers N* the possible - 231 - Show that for every convex n-gon (n > 4 ) , the arithmetic mean of the 5. lengths of all the lengths of all the diagonals. 6. n For a sides is less than the arithmetic mean of the given positive integer, determine all functions F: M —• R such that F(x + y) = F(xy ~~ n) for all x,y € IN with xy > n. * 9th Austrian-Polish Mathematical Competition 1986 (Warsaw) 1st day ? June 25 f 1986 (time allowed 4 1/2 hours) AtA2A3 _l. is a given non-right-angled triangle* Three circles Cit C29 C 3 are mutually tangent and they pass through the vertices such that A2$A3 € Ct; A3$At € C 2 ; and Ai9A2 AtA2A3 such that the triangle determined by the centers of the e r3. Determine the possible angles for three circles is similar to the given triangle, 2* For 12 > 1, let 13-1 be a a P(l) polynomial > 3, having n distinct 0 negative real roots. Prove that that there 2 2n aQ* If every point in space is either blue or reds show exists at least one unit square whose number of blue vertices is 0, 1, or 4, 2nd day (4 1/2 hours) 4. Determine all triplets (x,y,z) of natural numbers such that z+1 z+1 o 100 x - y = 2 5* Determine all quadruples (x,y,u,v) of real numbers satisfying the simultaneous equations x2 + y 2 + u2 + v 2 = 4, xii + yv = -xv - yu, xyu + yov + uvx 4- vxy = - 2 , xyov = - 1 . 69 Determine the extreme values of the ratio of the circumradius to the inradius of a tetrahedron whose circumcenter and incenter coincide* - 232 - 3rd day (4 1/2 hours) - Team Competition Show that nl 7. is divisible by k whenever n and k are natural z such that n 8. > 41 > 0 and i has no prime factor > n. m x n matrix of distinct real numbers is given. An numbers The elements of each row are rearranged (in the same row) such that the elements are increasing from left to right. Next the elements of each column are rearranged (in the same column) such that the elements are increasing from top to bottom. Show that now the elements in each row are still increasing from left to right* Determine all continuous and monotone functions Fl IR —* IR such 9* 2 F(l) = 1 and F(F(x)) = (F(x)) that for all x e R 0 t I now give solutions to some previous corner problems as s*\?ll as the "Quickies" at the beginning of this corner• 5. Competition [1981: 43] Britain, Luxembourg, Ten gamblers Each in in Mersch, The Netherlands, started playing turn threw five dice* Luxembourg and each with that the same after payment the beginning., amount of money* At each stage the gambler who had thrown paid moment, where u is the total shown by the dice. one after the other. Great Yugoslavia). to each of his nine opponents 1/n times the amount which that at (Belgium, opponent owned They threw and p>aid At the tenth throw the dice showed a total of 12, and it turned out that every gambler had the same sum as he had at Determine if possible the totals shown by the dice at each of the other throws. Solution by Andy Liu, University of Alberta, Edmonton, Alberta. Suppose fl was one of the gamblers and N was the next gambler • Before M's play, both M and N had received exactly the same amounts from each previous Suppose M3s die-roll was 7* - x m; then M's total payment to the other gamblers amounted to — — where T was gambler and so still had the same amount x of money. the total wealth of gamblers other than fl. all the gamblers and T - x was the wealth of all the After M's payment, fl had x - . _ _ _ = x(l + —) - — m m m - 233 - Now suppose Nfs die-roll was n. while N had x(l + — ) . Then* in the same manner, after N's payment M had W 1 + b ~ &<J + b - x<* + b*1 + ^> - & 1 + ^ m m n m n m n and N h a d r - x(i + -) 1 1 1 rr x(l + i ) - _ ^ ™ ^ ~ i ^ = x(l + i ) ( l + i ) - ~ . a « a n n These two amounts must be equal, since M and N had equal amounts at the end of the game. Thus Id m + 1) =1, n » that is, m = n + 1. Since in our problem* we have 10 pleyers with the last die-roll being 12, the previous die-roll must have been 12+1 and then the die-roll before that 12+1+1, etc. Hence the ten die-rolls are 21, 20, 19, 18, 17, 16, 15, 14, 13, and 12 in that order* It is to be noted that the information about the five dice being used, though consistent with the answers, is in fact redundant. 19. [1985: 4] From the 1984 All-Union Olympiad (proposed by Balotov). Find xy + Zyz + 3zx if x, y, z are positive numbers for which x 2 + xy + y2/3 = 25, Solution by George y2/3 + z 2 = 9, Evagelopoulos, z 2 + zx + x2 = 16. and Law student, Athens, Greece, Let A = x 2 + xy + y2/3, B = y2/3 + z 2 , C = z 2 + zx + x 2 , D - xy + 2yz + 3zx. Since A = B + C, we get xy = z(x + 2z>* Also, 17 = y(x + 2z) + 3zx = y(xy)/z + 3xz = (y2/3 + z2)3x/z = 27x/z. Furthermore, D = xy + 2yz + 3zx = 2z2 + zx + 2yz + 3zx = 2z(2x + y + z), so that 2x + y + z = Z?/2z = 27x/2z2, Next, A-B+C= Since x,y > 0, x/z = 875/9* 32 = x ( 2 x + y + z ) = 27x2/2z2. Finally, D = 27x/z = 2473"* A.B. - 234 - This problem appears in KVANTf 23. From [1985: 5] the November 1984, p*61 9 1984 All-Union Olympiad (proposed by Yu. Mikheyev)• For what integers m and n do we have (5 + 3J2)m = (3 + 5jZf and (a + bM)m = (b + a ^ ) 1 5 , where a and Z> are relatively prime integers, while d is a uatural number not divisible by the square of a prime? solution We bj by George Evagelopoulos, Law student, Athens, Greece. will show that, with the assumption (implicit in the problem) that a, and d are all greater than one f the only solution to either equation is the obvious one m = n = 0. Suppose that = (5 4- avO")12 (a + bMf and that one of z» and J? is nonzero; then both are, and in fact they both have Without loss of generality! w e may take m and n positive the same sign. and a > b* must Then since 1 < a + bJ3 < b + av^"? it follows that m > n, By expanding (a + &v£7) and (b + aJ7) by the binomial theorem and equating rational and irrational parts, it also follows that (a - bj&f - (b - av6T)J (a - £>vEF)a '5 - ajcf] and thus [iTOTJ We will now show that (1) cannot hold* u = a - 5^ a + 57d and v = Let B -I- avfc/ Then since a > 5, b - a,/7 < 0, so v = a7d + 5 We will use the fact that for positive p and q, 2g q . i = 1 TpTgFTT p + q ~ p + g i n c r e a s e s with p/g* Now i f a - 5v3" > 0 , then a - bvET" u = a + bjd and (1! - 235 - a and thus u < v. . aJJ On the other hand if a - hJU < 0, then b<JI - a Kjd + a and &J7 and again it implies that u 24 a follows that u < a J7 < v. In either case 0 < uf v < 1 and m > n < v $ contradicting (1). From [1985: 5] the 1984 All-Union Olympiad (proposed by O.R. Musin)• If 72 > 3 natural numbers are written around a circle so that the ratio of the sum of the two neighbours of any number to itself is a natural number, prove that the sum of all such ratios is (a) not less than 2n f (b) not more than 3n. Solution by George Evagelopoulos, Law student, Athens, Let the 72 natural numbers be denoted by ai tan, a l + a n 2 a A • • . ,a A a l + a3 A Greece. and let n . + a. 12-1 1 12 (a) Since x/y + y/x > 2 for x,y > 0, It follows that s 12 = \a2 a Q-i 2^ L (b) l] a 1 + S 0 [ 2 J We will show that S [a3 a 2] a., a > 2n. Q- a l % 3J < 3i2 for 12 > 3 by mathematical induction. 12 For 72 = 3, we consider two cases. (I) All the numbers a, are equal* (li) At least two of at, a29 In this case S3 a 3 are not equal* 6 < 3*3, Suppose at > a2 and at > a3» Then at > £iJL£i so 2 > £» + a* . Since the latter ratio equals a natural number, at a a 1+ » + J + + a. + = a2 + a 3 . 2 ^ + 2a, Consequently§ - 236 - where k = 2a 3 /a 2 and € = 2a a /a 3 are natural numbers. Since k€ = 4* we can only have the following possibilities: * = 1, € = 4, then S 3 = 8 < 9; 1 = 2, e = 2, then S 3 = 7 < 9; A = 4, * = 1, then S 3 = 8 < 9. We assume now that the inequality is valid for S S 1 and we will show that = 2n < 3n* Otherwise? we choose < Zn. n If all the numbers a are equal, then 5 largest of the n numbers. the If there is more than one such number, we take one of them which is greater than at least one of its nearest neighbors. Let this number be a . Then n a a 1 12-1 > + a1 n so 2 > 13-1 1 Since the latter ratio equals a natural number, a = a ., 4- a., so that ^ n 73-1 1 a + a0 a 0 + a 1 + r, 13 2 = 1 + s, n-1 where a 0 + a1 13-Z a i i?-l 1 and a are natural numbers. -5 + a 0 13-1 2 Consequently the sequence a.?a2,•*.9a inductive hypothesis and so S _, < 3 (12 - ! ) • a S = -1 + n a0 a l + a 3 0 + a , + 3 < 3/2, 13-1 i Finally, - F a i a + a 0 13-2 13 , 13-1 1 ^ 13 2 + — . — _ — _ _ + — — _ — — — . -i- — — - — _ a ^ a 8... 13-1 13 1 a ! 13-Z 1 + 1 + 1 + 1 + 13-1 n-1 = S a . satisfies the + a0 Z - 237 - Editorial comment. By letting a. _ 1 + 3 Q pi - 2) + n . (n - 1) + 1 . 12 + 2 2 12 = i for all i, we obtain 12-1 1 12 = 2 + . . . + 2 + l + ( j * + 2) = 2(n - 2) + n + 3 = 3l2 - 1, which is the least upper bound* t Solutions to the Quickies Q-l. Letting z = x + y, we find that w = ±2(x2 4- xy -I- y 2 ) • Q-2, Expanding out the left and right sides, we obtain 2 x2y2z2 + 2xyzw 2 xy = 2 x 2 y 2 z 2 - xyzw 2 x 2 or xyzw(x + y + z - f w ) 2 = 0 . Thus all solutions are given by xyzw = O o r x r + y + z-l-w = Os Q-3. Multiplying the equations together, we obtain + x 4 ) - 16x 2 y 2 z 2 w 2 } = 0, xyzw{Jl(l Since 1 + x4 > 2x 2 with equality if and only if x = ±1, etc., all the solutions are given by (x,y,z,w) provided = (0,0,0,0) or (±1,±1,±1,±1) x and z have the same sign and the same for y and w. Thus there are five real solutions* Q-4» Letting x = b sinh B, sinh B + &Jsinh20 + 1) 2y = 2(a - b sinh B)(b = (2a + 5(e @ = 5 2 + be9(2a ~ e 0 ))&e @ - &e0). Thus y takes on its maximum value for 5e 2 = ^raax "" b = a, Finally, 2 + a 2""~™°™ and is taken on for _ b(a/b x _ — ^ - 6/a) _ a 2 - b-z - - 2 a ' Q-5. Let a, 5, c be the sides of the triangle, s the semiperimeter, and r the inradius. Since - 233 - cot A/2 = etc. and 1 2 cot A = cot A/2 etc. the inequality is equivalent to a r ^-HM + | „ (s - b Is - c P r a ,s - b _ . 4- a s - c 4 — or M- a ,s - 5 ,s - c 4 ___—™—. 4 , s - 6 ^ s - c or s = 3s - (a 4 5 + c) > 3r 2 s - a s~b <1] s - c Since it is known that r2s = (s - a)(s - 6)(s - c ) , (1) is equivalent successively to s2 s 2 > 3[(s - 6)(s - c) + (s - c)(s - a) 4 (s - a)(s - 6)], > 3[3s2 - 2s(a 4 I? 4 c) 4 be 4 ca 4 a&], (a 4 5 4 c ) 2 > 3(6c 4 ca 4 afc), and finally (b - c ) 2 + (c - a)2 4 (a - 6 ) 2 > 0, which is true, with equality if and only if the triangle is equilateral. Q-So For any triangle ABC$ there is a non-obtuse triangle A'B'C with angles A# = (u - A)/2, B' = (IT - B)/2, C" = (u - C)/2. So equivalently, we want the minimum of S for non-obtuse , _ cot A 7 2 + cot 5 7 2 + cot C / 2 = sin W + sinTET 4 sin IT7"" triangles,- It will turn out that we can drop the primes in this expression and consider all triangles? since the minimum will occur only for equilateral triangles• Since the area F of a triangle ABC is given by F = ft2{sin 2A 4 sin 2B 4 sin 2C}/2 = r2{cot A/2 4 cot B/2 + cot C/2} where R and r are the circumradius and inradiusf respectively, S' will equal (tf/r)2/2. equilateral It is well known that (R/r) triangles. . 2 and is taken on only for mm Finally! the minimum values of S and S' are then both - 239 - equal to 2, Q~7, This problem appeared in the 1912 Botvos Competition, are given in the Hungarian Problem Book II, Library, MAA, Washington, D.C., 1963, pp*50-54 where it is figure is coplanar* Two solutions New Mathematical assumed that the We give a vectorial solution which is simpler and which does not assume the figure is coplanar* Denoting the vertices of the quadrilateral by vectors A, B f C, D from a common origin, the4 result follows immediately from the identity (A - B ) 2 + (B - C ) 2 - (A - B ) 2 - (D - C ) 2 = 2(A - C)•(B - D ) . Q-8» The result follows immediately from the equivalent inequality since this inequality implies [xyz + (a - x)(b - y)(c - z ) ] 2 > ixyz(a - x)(b - y)(c - z) and the left side reduces to [abc - az(b - y) - bx(c - z) - cy(a - x)] 2 * The given inequality is related to a result of J*F, Rigby, "Inequalities concerning the areas obtained when one Math. triangle is inscribed in another", Mag. 45 (1972) 113-116. Let a, bf c denote the sides of a triangle AM? with an inscribed triangle DEF where BD = xf CE = vi AF = z, and let A, A , zL , A , J a b c and A~ denote the areas of ABC, AFE, BFD, CDE? and DEF, respectively. a-x Then (a - so dividing the given inequality by (abc)2 zl 2L y)y gives A A A^ A > 4 -£. A JE. - J ^1 4 or zlid > 4z! zL zl 0 abc with equality if and only if the three cevians AD, BE, Thus A® + Zt^Zt 0 0 a + 21*21. 4- A%A 05 > 421 21, A 0 c - abc and CF are concurrent, - 240 - find it. follows quickly that a 0 - b c with equality if and only if the three cevians are medians* Q-9, Letting S = a2 4- b2 + c 2 , we have (d2 -f b2 + c2)(a 4- & 4- c)(b 4- c - a)(c -ha- b)(a 8a2b2c2 4- 6 - c) - = S[2(k 2 c 2 4- c 2 d 2 4- a 2 5 2 ) - (<2^4- fc« 4- c 4 )] - 8a 2 & 2 c 2 = S'[4(&2c2 4- c2a2 4- d2£>2) - S 2 ] - 8a 2 5 2 c 2 = 4$(5 2 c 2 4- c 2 a 2 4- a 2 6 2 ) - S 3 - 8a 2 5 2 c 2 = S'3 - 2 S 2 U 2 + 5 2 4- c 2 ) 4- 4S(5 2 c 2 4- c2a2 4- d 2 5 2 ) - 8a 2 5 2 c 2 = (S - 2a2)(S - 252)(S - 2r a ) = (52 4- c 2 - a2)(c2 •¥ a2 ~ b2)(a2 + 52 - c2), (1) Also, since a6c = 4/?F and (d 4- b 4- c)(5 4- c - a)(c 4- a - b) (a 4- b - c) = 16F2 where # and F are the circumradius and area, respectively, (1) is also equal to 16F2(a2 4- b 2 of the triangle,, 4- c 2 - 8i?2) • Finally, since (1) is positivef 0, or negative according to whether the triangle is acute, right, or obtuse, the desired result follows, Q-1Q» In the following figure we have two adjacent unit cubes face to face and we will determine the shortest distance EB and FC. h between D Since* FJ\\EB, h is the distance from B to the plane of FJ and FC. of tetrahedron BFCJ = i = (|.) -Area(FCJ). * touching * The volume V Since Area(FCJ) = -g-, h = - ^ * - 241 - P R O B L E M S Tiollem proposals and solulloas should le sent to the editor, whose uild^vess appeals on the giant pafe o§ this Issue. Tioposuls should, wA>enewei possllle, le accompanied ty, a solution} legeiences, and othei InslqAts which aie llfcelf to le of help to the edltoi. An a^te^il^h (*) agtei a numlei Indicates a piollem sulmltted without a solution. Vnltylfial pioltems me paitlculaily, sought. %ut ethei lateiesttng, pioltems may also le acceptable pnowlded thef aie not too well known mid legeiences aie g,lven as to theln provenance. Gidlnailly, t l§ the oilplnatoi og a piollem can le located. 6t should not le sulmltted Ip somelodp else without his o% her peimlsalon. To facilitate the It consideration, 'four solutions, typewritten or neatly, handwritten, on signed, separate sheets, should piegeially, le mailed to the edit01 legoie fane 1, 1987, although solutions received agtei that date will also le considered until the time when a solution 6<^ punished. t * 1181. Proposed by D.S. Belgrade, Belgrade, t Mitrinovic and J.E. Yugoslavia. Pecaric, University of (Dedicated to Leo Sauve.) Let x, y, z be real numbers such that xyz(x + y + z) > 0, and let af b, c be the sides, m , m^ . w the medians a triangle• (a) (b) b and F the area of a c Prove that jyz-a2 + zxb2 + xyc2 j > AFJxyz(x 2 2 + y + z) m jyzffl -f- zxn?^ + xym 1 > 3FvSyzTx~T y°'™4" z) 1 1182. a D c* "* . Proposed by Peter Andrews and Edward T.H. Wang, Wilfrid University, Waterloo, Ontario. Let a^ ^a^, > « » fa Laurier (Dedicated to Leo Sauve*) denote positive reals where n > 2. Prove that " < tan-»£i- + tan-i£2. + .. . + tan-* Jl < ( " " 1 } " Z a2 a3 at ~ 2 and for each inequality determine when equality holds* 1183. Proposed by Roger Izard, Let Dallas, Texas. ABCD be a convex quadrilateral and let points Et G lie on BD and F, H lie on AC such that AE, BF, CG, DH bisect angles CDA respectively. parallelogram. Suppose DAB, ABC, BCD, that AE = CG and BF = DH, Prove that ABCD is a - 242 - Proposer! by J.T. Groenman, Arnhem, The Netherlands. 1184. Let ABC be a nonequilateral triangle and let 0, J, H, F denote the circumcenter? incenter* orthocenter* and the center of the nine-point Can either of the triangles OIF or IFH be equilateral? circle, respectively. t Proposed Austria. 1185. Determine by Walther Janous, (Jrsulinengywnasium, Innsbruck, the set M of all integers k > 2 such that there exists a positive real number u satisfying [u ] = n (mod k) for all natural A A numbers n$ where [x] denotes the greatest integer < x. (Problem A-5 in the 1983 Putnam Competition is equivalent to showing that 2 € M. ) 1186. Proposed by Svetoslav Bilchev, Velikova, Mathematical gymnasium, Russe, If a, £>, «: are the sides Technical University and Emilia Bulgaria. of a triangle aind s, R, r the Equipment Corp., semiperimeter, circumradius* and inradius? respectively? prove that 2(6 + c - <*)vST > 4r(4fl + r) ^R + F URs" where the sum is cyclic over a, 6, c. Proposed 1187. by Stanley Rabmowitz, Digital Nashua, New Hampshire. Find 2 +2 a polynomial with integer coefficients that has as a root, Proposed by Dan Sokolowsky, Williamsburg, 1188. Given a circle Virginia. A' and distinct points Af B in the plane of ft*, o construct a chord PQ offtsuch that B lies on the line PQ and ZPAO = 90 . t 1189. Proposed by Kee-wai Lau, HOIK/ Kong. Find the surface area and volume of the solid intersection formed by of eight unit spheres whose centres are located at the vertices of a unit cube. t 1190. the Proposed by Richard I. Hess, Rancho Palos Verdest Prove or disprove that California. - 243 - n-1 lim 2 - 2n 7 *=0 S O L U T I O N S A'# piaC-teai l& evei peimanefitty, cto-^ett. The e<&Lt&% mltl atmai/A, &e p-tea^e<$, to cofibLdet tfoi p-u&tLcatLon a&w §,otutLoa& oi new- LnA,LgAt& on p-nt^t psio&temb. 1010? [1985: 17; 1986: Ursulinengymnasium, Innsbruck, by Walther i7 an integer, contains infinitely determine all such k. many Janous, Austria. there integers k t 1 such that the sequence {3n2k Are squares? + ink2 If the answer + i3}3, is yes3, (The case k = 1 is dealt with in Crux 873 ,[1984: 335].) Comment by Richard 1. Proposed 113] K. Guy, University of Calgary, Calgary, Alberta. show that k must be of the form p^ *. .p €2 or 3D.,,,p .€2 where ro > 0 l r l m *i '©+1 and each p . is a prime s 1 mod 6; that is? the condition conjectured by Hess We [1986: 115] Is at least necessary for k to be a solution of the problem* (In the comment following this one? not however, G. Walsh shows that It is sufficient.) If we put k = kxk\ where ki z 2 2 = 3n k is squarefree? and write + 3nk2 = 3n2ktk2 then 2 2 k1k i\z t + k3 + 3nklk* that ktk2\z and it follows + *}*| , since kt is squarefree* (1) Putting z = ir1A"2z1, (1) becomes klklzl = Sn 2 !^ 2 -f 3n*2*2 + AfAf JtjL^f = 3n2 + taJ^A* + Af**. Hence l i j3n first 3iki9 2 ? f so (as in Kierstead s argument) either kt\n (2) or 3 |A"4 • Assuming put n = ^ii^i • Then (2) becomes Axzf = Zk\n\ + 3iflii2i + Af*« zf = S l ^ f + 3Jt1Jt|/?1 + ktk* so we can put zt = k1z29 and then SO 4* t z| = 3(2/2! + JSr|)2 + *«. (3) - 244 - 3)1!, we put kt If = 3k3$ and with similar calculations (2) can be written in the form 4Jr3zf = (2nt + 3 1 2 ) 2 + 3k*» (4) Thus we need to show that if (3) (resp, (4)) has infinitely many solutions for a given kt (resp, k3) then kk (resp, k3) is a product of primes all s 1 mod 6 (and k3 t 1). we show that 2fii in (3) and that 2fjt3 in (4), oince kt First thus both kt and Ar3 = 3i 3 and are squarefreef we need to show that both 3(2i2s + I f ) 2 + k* and (2nt + 3 £ 2 ) 2 + 3Jr^ have even powers of 2 in their prime factorizations. this for expressions 2 the form 3X of 2 X if and Y is enough to show But this is obvious if X and F + r . have opposite parityf and follows by induction if X Finally, It 2 2 are both odd, then 3,Y + Y and Y are both even. = 4 mod 8, so 3,Y + Y2 is 2 divisible by 4 but not 8. Next we show that k3 £ 1 in (4). Otherwise, (4) becomes 4z| = (2nt + 3 1 2 ) 2 + 3*J, or 3*5 = (2z2 + 2nt + 3*|)(2z2 - 2nt - 3i 2 ) which obviously has only a finite number of solutions for each k2« It is clear since kt p = 2 mod is square free that 3fjr3, Thus we suppose that 3 is a divisor of ^ this is impossible. Since kt some in (3) or Jf3 in (4), and it remains to show and k3 are squarefree, this means that either 3(21^ + k2)2 + k\ or (2nt -f 3ir|)2 -4- 3*J has an odd power of p in its prime factorization. 2 the form 3X + Y 2 can have this property. We show that no integer This is clear if p divides just one of X and Yf and follows by induction if p divides both of X and F, that plY. Then 3A'2 + Y2 E 0 mod p may be written 9X2Y~2 + 3 E 0 mod p, or (3XY-1)2 of = -3 mod p. So suppose - 245 - However it is known that -3 is a quadratic non-residue of such primes p. This contradiction completes the proof* 11. Comment by Gary Walsh, student, University of Calgary, Calgary, Alberta. was conjectured by Hess [1986: 115] that for any number k of the it 2 form p<.««*p £ there are or 3D. • * .p B*£29 infinitely integer z. many We give a where m > 0 and each p. is , prime E 1 mod integers n such that (n *¥ k)3 constructive proof that there 6, - n3 = z 2 for some are infinitely many counterexamples to this conjecture. use the following facts, where m and n denote squarefree odd integers We greater than 1 (for the first, see D.T\ Walker, The equation mXz - nY2 Amer. Math. Monthly = ±1, 74 (1967) 504-513; the second follows from the first by a simple induction): (i) if a positive solution (X9Y) smallest positive solution (AfB)t a to mX2 - nY2 = 1 exists, then there is and all positive solutions (A,,B,) are A & obtained from A, JH + B.Jfi = (Ajfi + Bjn) A 2#+l K with ir = 0,1,2, • • • . is the smallest positive solution to mX2 - nY2 (ii) If (A,B) B and are of opposite parityf and if A (resp« B) is even, then A B ) is even for every kf and B K p. E 1 mod 6, the A (resp* (resp. A ) is odd for every i. K K As well, we need the result proved in [1986: 113], that for k = each = 1, then problem of p^«»*p with finding infinitely many n such that (n + & ) 3 - n* is a square is equivalent to the solvability of the equation Uaz - 3& 2 = 1 for some integers a, 5» Suppose that x, y, A" are positive integers with k squarefree and k and suppose 12x solvable. (XfY) We = <y,2x). z 2 2 + 1 = ky . have that We claim that the equation Aka the equation 2 kX - 2 3Y = 1 > 1, 2 - 3i> = 1 is not is solvable by Thus y^F + 2x75" = A.JE + B.V5" = (A^F + BV5")2i+1 for kX z some - 3Y 2 integer = 1. /, where (AfB) is the smallest positive solution to Since B. = 2x, B . is even and A . is odd for every /. Now if - 246 - a, b are integers such that 41a2 - 3b2 = 1, then (2a,3) = (A ,B ) for some €t contradiction• We now have that any squarefree integer k > 1 for which 12x2 + 1 = ky2 is solvable is not a solution to the original problem. Next we integer k show that any of this form is a product of primes each congruent to 1 mod 6* If pj k, then 12x2 + 1 E 0 mod p so clearly p ^ 2 or 3. Also (6x)2 = -3 mod p so p as 2 mod 3 since -3 is a quadratic nonresidue modulo such primes. Thus p = 1 mod 6* integer k > 1 for which 12x2 + 1 = ky2 holds is a Thus any squarefree counterexample of Messrs conjecture; for instance, x = 1, k = 13, x = 3, * = 109, x = 4, 1 = 193, x = 5, k = 301, and so on. For # = 1,2,... put 12n2 + 1 = k V2 where each k nJn j- > 1 is n . squarefree. It remains to show that the set K = {k |T? = 1,2,...} is infinite. We suppose for a contradiction that K is finite, say \K\ = s. For each k € Kf the equation 1 = ky2 can be written ky2 - 3(2n)2 12/22 + positive solutions are all of the form (yt2n) = 1, and thus its = (A.,B.) where A.JE + B.V3" = (A^f 4- B7S")2i+1 and (A,B) is the smallest positive solution. Since A.^JF + B.^JS 2+1 2+1 = (A^F + B ^ " ) 2 i + 3 = {A.JF + B.^")(AvT + Bv^")2 = {A„(1A2 + 3B2) + 6ABB.}JF + {B.(kA2 + 3B2) + 21ABAi}75", we have = B.(iA2 + 3B2) + 2MBA i , B and in particular B . „. > 4B.. Thus \B. 7+I B.9B. i > s* j - I f 1 - B.I /8 Now consider the equations 12s2 + 1 = k y2 > 3s whenever i * j and - 247 - 12(s + l) 2 + 1 = k ^ y * 12<2S)2 Then from \K\ = s, two of {k ,k o with i £ j t 1, ST X + 1 = * 2s y| s . . . . ,k0 } must be equal, say k =k 1 S TJ bS ST =k and so *yL,- - 3 ^ s + *>J2 = x and *y*+J But 2(s + i), 2(s + j) 3 2 [ <s + i)i 2 = !• > 2s > s and |2(s + i) - 2(s + j)j = 2|i - j| < 2s < 3s, which is a contradiction. * 1048. * Proposed [1985: 147] * by J.T. Groenman, Arnhem, The Netherlands. In base ten, 361 = 192. Find at least three other bases in which 361 is a perfect square. Solution It by Hayo Ah1 burg, Benidorm, Alicante, Spain. is well known that Euler [1] solved the following problem: find all triangular numbers N = _ - _ _ which are also pentagonal numbers N = _ _ _ . This leads to 1 ± JTZF2~~T-JW~TT i -— E~ the expression R under the square root should be a square to make £ an Y - 1 integer. With k = —-r^— we obtain the Fermat-Pell equation where /? = 3x2 - 2 = y2. Euler gives recursion formulas, in our notation x x o0 = JYo ° = 1» . 1 = 2x y , ^n+l y .1 = 3xn + J2y n+1 n +^n n , Following methods developed by Lagrange [2], we can also give x and y in closed form: x 13 perhaps by few people? T$T^ (2 + ^ + l, (l) - 248 - [H4 (2 + 73") •]• where n = 0,1,2,.«. and [a] is the greatest integer < a. The triangular- pentagonal numbers N are now found from 3e2 - e P H M = with k = x -1 _ i ± y that is, "2T "ff" The first values are 1? 0 i 2 3 4 5 6 7 X 1 3 11 41 153 571 2131 7953 y 1 5 19 71 265 989 3691 13775 N 0 1 15 210 2926 40755 567645 7906276 Now? 1 hope the patient reader who has stayed with me thus far will be as enchanted as I am to discover cross-connections problems. In Crux between seemingly unrelated 1048 we are looking for those positive integers b (base) and y which are solutions of 3b2 + 6£ + 1 = y 2 . Substituting b = x - 1, this becomes 3x2 - 2 = y2» Voila! b = 2, 10, 40, 152, 570, 2130, 7952, etc. (The base b in closed form i s, according to (1), b = [T73^ ( 2 + JS)n ] f n= x 2 3 '''- ) References: [1] Leonhard Euler, Vollstandige Anleitung zur Algebra? St„ Petersburg, 1770, Part II, Section 2, Chapter 6, %91, Problem V. [2] Joseph Louis Lagrange, Additions to the French translation of Euler?s Algebra, Lyon, 1774, SVII, #75. Also Rruxelles, solved by SAM BAETHGE, Belgique; Antonio, Texas; C. FESTRAETS-HAMOIR, RICHARD I. HESS, Rancho Palos Vercles, California; MLTHER JANGUS, Ursulinengywnasinm, Cuyahoga Falls, Ohio; San J.A, Innsbruck, Austria; MCCALWM, Medicine Hat, FRIEND Alberta; H. KIERSTEAD JR. , STEWART METCHETTE, - 249 - Culver City, Ohio; California; BOB PRIELIPP, RABINOWITZ, Federal University Digital Republic proposer. LEROY F. MEYERS, The Ohio State Equipment of There of Wisconsin, Corp., Germany; was one partial solution * Oshkosh, Nashua, KENNETH M. New Hampshire; WILKE, Proposed [1985: 148] ABC Let A > B > C and A' and > B' > C*. Topeka, J. STANLEY SUCK, Kansas; and Essen, the received. t by Jack Garfunkel, Flushing, N.Y. A^B'C* be two nonequilateral triangles such that Prove that r r where s, r and s'9 Columbus, Wisconsin; t 1049, A'B'C University, r* are the semiperimeter and inradius of triangles ABC and , respectively. Comment by U.S. Since Klamkin, University of Alberta, Edmonton, Alberta. the perimeter of a triangle circumscribed about a given circle can o be unbounded if either two angles of the triangle are close close to o 0 , the result stated is clearly f le, to For 90 a or else specific counterexample, just let A = 120°, f A = 89°, Then A - C > A' - C , but s ^ A — = C O t -rr while r B = 30°, B* = 89°, + COt TJ C = 30°, C = 2° + COt rr ** 8 Z — T = C O t yr— + C O t rr— 4" C O t yj— ?« 59 . r Z Z Z One can also find counterexamples involving only acute triangles, for example A = 89°, B = 86,5°, C = 4.5°, A' = 88°, B' = 88°, C" = 4°. Nevertheless? since cot(x/2) is convex in (0,ir), we can obtain a positive result by the Majorization Inequality (e.g. see p.30 of E*F. Beckenbach and R, Bellman, Inequalities, Springer-Verlag, Heidelberg, 1965). A > B > C, A' > B' > C , and also that A > A' A 4- B > A' + B' ; and then - = Z cot A/2 > 2 cot A V 2 = ^ r - r* . Suppose that - 250 - * * * [1985: 148] Proposed by Stanley 1050. Corp., In Descartes. Digital Equipment Nashua, New Hampshii-e. the Show Rabinowitz, plane, how to you are given the curve known as the folium of construct the asymptote to this curve using straightedge and compasses only. Solution by J.A. McCailum, Medicine Hat, Alberta. We assume that the folium is in general position* Then the only definite point in the plane is the point where the folium intersects itself. C* C With as centre and any convenient radius draw a circle. Call this This circle will cut the two tails of the folium in points P and Q. (It may also cut the loop of the folium but we ignore these points if they occur,) Join PQ and find Join DC and produce it Its midpoint D« to cut the loop of the folium at A. then the apex of the folium. A is Trisect AC9 and produce AC to E so that CE = ^AC. Finally, at E erect a line perpendicular to ECAf and extend this line Indefinitely far in both directions. construction works This is the required asymptote to the is The because of the known symmetry of the folium and the known fact that the apex of the folium is three times as far from asymptote folium. (easy to prove point 3 C as the + y 3 = 3axy of the from the usual equation x folium)• Also solved Ohio State University, assumed that solutions by ELWN ADAMS, Gainesville, Columbus, the coordinate assuming as Ohio; Florida; and the proposer. axes were given. given not only LEROY F. Two other the axes, but also MEYERS, The Adams, readers however, submitted the equation of the folium. The proposer wonders if asymptotes to other y = tan x or y = log x, can be constructed * 1051. familiar curves, with straightedge * [1985: 187] Proposed and for example compasses. * by George Tsintsif&s, Thessaioniki , - 251 - Greece. Let a? b9 c be the side lengths of a triangle of area K, and let uf v9 w be positive real numbers. Prove that ua< +_vb^__^J^_> y + w w + u u + v BK2> When does equality occur? Some interesting triangle inequalities may result if we assign specific values to u, vf w. Find a few. I. Generalization by D.S. Mitrinovic Bel grade, Bel grade, Yugoslavia. and J.B, Pe6aricf University of Let r., , • • •, r be real numbers such that 1 n n Rk - 2 r > 0, * = 1,. ..,73, i=l and let x 1 t .. M x be positive numbers. We will prove that n r. ."1 1=1 73 1 xz > /?. i - I? - r 1 2 X 1=1 J (1) J i=l Indeed, 13 r. i=l i 2 73 1=1 2 (.1=1 1 r . - R. " x IT r . 2 1=1 J i _ 1 73 X2 1=1 " n 2 ff . l 1=1 73 2 - x. n 2 X2 J i =l 73 F^T 2 x - 2 x2 i=l * where we used Cauchy*s inequality in the last step. Equality is valid in (1) if and only if *T Putting 7? = 3, rt becomes R = w , r 2 = vf r3 = w and xl = a2, x 2 = b2, x3 = c2, (1) - 252 - _,, -r — r — V 4- W T W 4- U J ^ L _ > i(a 2 + 5 2 + c 2 ) 2 - (a4 4- 5 4 4- c 4 ) U 4- V ~ 2 = ^[2a 2 5 2 + 25 2 c 2 4- 2c 2 a 2 - (a4 4- 5 4 4- c 4 )] 1 •(a 4- 6 + c)(a 4- 6 - c)(6 4- c - a)(c 4- a - 6) with equality if and only if y 4- w Generalization II. by w4-o M.S. u 4- y ® Klamkin, University of Alberta, Edmonton, Alberta, More generally? we show that for 0 < n < 1, 4/2 4/2 ua vb U + W ¥4-11 .2/2 4/2 0 3 4& wc U 4- V (2) 75" The proposed inequality corresponds to the case n = 1, Letting S be the left side of (2), we have o r o 2[S fU 4- V 4- W , 4J3, ,4/3, 4/2*, s 4- (a + 5 4- c ) ] [(V > (a by Cauchy*s inequality. , U 4- V 4- W , 4 / 2 , U 4- V 4- W 4/21 44- _ U — _4- — V— - c W 4-, U — - 6 4/2 ,4/2 4/2 4- (w 4- if) 4- (u 4- y ) ] \V + W W + U U 4- V F 4- W 4- w) 4-5 4/2 a 4- c ) (3 Equality holds in (3) if and only if 2/? a V-fW 2/? 2/2 W4-H 1/4-F (3) is equivalent to 2S > (a 4-5 4-c ) - 2(a 4-5 + c ) , 4/2 ^ ,4/2 ^ 4/2, 2n 2n 2/2 2J2 2/2.2/2 0h A 0 A 0 25 c 4- 2c a 4- 2a 5 (a +5 4- c ) , /2 , , n 8 n w , i ) , // /? w // , // , 12 w n , , // n. = (a 4 - 5 4 - c ) ( 5 4 - c - a ) ( c 4- a - 5 ) ( a 4- 5 - c ) . Since af b9 c are the edges of a triangle, {a ,5 ,c } also form a t Ax (4) triangle whose area we denote by F . Then (4) becomes 72 S > 8F 2 72 by ? Heron s formula. We now obtain (2) by applying Oppenheimfs generalization of the Finsler-Hadwiger inequality! i.e.. - 253 - _£> 4F ' ^ AF JZ JJ 0 < n < 1 with equality (for n & 1) if and only if a = Inequalities Piihlications involving elements de la Faculte 5 = c. (See A* Oppenheim, of triangles, quadrilaterals or tetrahedra, d'Electrotechnique de 1'Universite a Belgrade, No.461 - No.497, (1974) 257-263.) Note American also that the case n = y gives the proposer's problem E3150 in the Mathematical Monthly (May 1986, p.400). It is of interest to compare the given inequality with the not so easily proven, and incomparable, inequality (See M.S. Klamkin, ub2c2 vc2a2 wa2b2 V + W W + U Tl * V - Two non-negative ft2 forms, El em, quadratic der Math., 28 (1973) 141-146.) 3 Finally, note that if we put a = 5 = c, we then have Kz = y^ a 4 , and so the given inequality reduces to the known inequality u v w > 3 + + V + W W4-U for any positive real numbers ti, v9 w. III. Further special cases by Innsbruck, £! + V ~ "2" Walther Janous, Ursulinengymnasium, Austria. (i) Putting u - v = w into the given inequality, we obtain a 4 + 5 4 + c 4 > 16IT2 which i s ( [ 1 ] , 4 , 1 0 ) . (ii) Putting u = a , y = 6 , w = c , we obtain 2s - a + 2F^~5" * 2S~=~F - 8l4'2> where s is the semiperimeter. (iii) Putting u = a" 4 , v = 6"4, w = c"4, we obtain ft«c< 5 4 -!» c + 4 c<a< c 4 + a + 4 a<ft« a 4 + 6 ^ 4 which, because of 54c4 5 4 + c4 &2c2 x /t , e x.c. , Z is stronger than ([1], 4.12). (iv) Putting u = a"2, v = 6~2, w = c"2, we obtain - 254 - 1 a2b2c2 1 > 8K'' a2 * h2 which, using 4K7? = abc where R is the circumradius, can be written 1 1 1 2R2 + b2 Because of < b2 * c2 etc. , w' this is stronger than the first inequality of ([l] f 5*24), (v) Putting u = b + c, v = c + a , w = a + 6 , we obtain '2s - a' 4 (2s - b' A '2s - c" c« a + b + > 8ff2 2s 4- c_ [2s + a [2s + j b (vi) Since 4^2 - 2i}2 = Z>aA? = c 2 A 2 , b c i r e t h e a l t i t u d e s , we o b t a i n a a a b c u v + Putting u - h2, a v h2 5 a2 W 2 V 4 A a w = h2, c " w * U A» b we obtain C + ° V 2, h* c > 2, h2 + i?2 c a A? + i32 6 _.*! + _JL_.il > i? 2 4- /> 2 a b which is ([1], 6,7). Next we apply Klamkin's duality 3a 35 3c 3/T I (a ,b ,c ,m ,m, ,m ,K) > 0 •=» I a fc c > 0 " where m t m, * m are the medians. a b c (This follows immediately from the fact the medians of a triangle with area K are themselves sides of a triangle 3a 36 3c , 3K with medians j — , -j—, -*— and area —-. See Klamkin's solution to Aufgabe 677 9 that ff.^m. <fcr Mat/?, 28 (1973) 130.) Tsintsifas 1 inequality then reads — — ~ ID- + v + w a > "A2, r - 2 1 m? + -^—m« u -f v w + tz h We may now use (5) to obtain more or Jess interesting duals of (i) - (v) v =ro."4f w = m instance, putting n - m~ 4 into (5) yields M /flM 4 b c D 4 c c 4 Z c a Q > 77 A ? — a a which we show is stronger than ([!]* 8.6)* i.e. 6 (5) For - 255 - m2 + mf +ffl2> 37SK. a ft c ~ SqIJHr i ri£, Lh i s beec >mes (M* + < -I- «/") + 2(m2mf + i » > 2 + ra2wi2) > 27tf2f a A> c a b be c a and t h u s we h a v e t o show t h a t m*m* o c 6 < 2 /s^ + 2 2 izifffl '6 c + 2ZT ffl, But this follows because of 6 c a! 1 < 2/»>2 5 c 4 -I- i?i 5 etc. c find < + m4 b c m*m* 6 c etc. m£ + b Reference: [1] 0. Bottema et al, Geometric InequalHies, Wolters-Noordhoff f Groningenf 1969. Also solved by C. FESTRAETS-HAMOIR, Bruxelles, Arnhem, The Netherlands; Middle town, Pennsylvania; author unknown, sent Bulgaria, VEDULA to who received and this N. the MURTY, proposer. office by it from the editor Pennsylvania Svetoslav (2), From a [1985: 187] Trinity State Bilchev University, solution, of Russe, Sosinskij. using the same method. by other solvers, (i) by Klamkin and Murty, and (iv) by Murty and the 1052. J. of KVANT, A.B. cases were noticed J .T. GROENMAN, There was a further Janous also obtained Klamkin's generalization A couple of Janous' special Belgique; in particular proposer. College, Cambridge, examination p&fter dated June 5, 1901. Prove that 1 2 _ 3 l -3 -5 Solution 2 1 2 3 1 3 -5 -7 2 by George Szekeres, 2 3 5 -7 *9 _ 2 University = I1^_ 9 ?L IT" e 2s 2 of New South Wales, Kensington AustraIia. How would a Trinity student of 1901 evaluate this sum? He may choose - 256 - from a number of different methods, Partial I. fractions. First, 1 2 3 (n-2) n (n±2)' J_ + 1 32n ISn 1 3 1 64 n-2 n+2 1 128 (n-2) (//+2)2J 2 (not very hard to determine the coefficients9 the answer in a way gives away the expected form of the partial fraction decomposition). From here the required sum is Y ( ^ n=3 ;/ odd -i)<"+1>/2 JL - _i (ji-2)2n3(n+2)2 32 L+ L 1 32 _i _1 16 16 1 3" Hi -1) - 1\ 4 3] 32" [ BT 1 1 1 32" + TB" + 3ZT + TBrgr + + l-i_ 33 1 F + I_- 53 1 T2"8~ T E J 1 3- + 5- 1 TB" _ 1 1 ' S" TB" 1 3 1 +X 3" TB" 1-1-+I The answer follows now from >-W and i - i _3 3 The + i 53 32 first one is trivial from the arctan series, the second is Euler's. method is perhaps the most remembers the second straightforward but assumes that This the student sum. Of course he can still bluff his way through the problem by working backwards from the given answer* TT. Theory of lesidues. This is no doubt the most direct method (my own favourite) and best suiled for examination conditions memorize any filing. Note that the required sum is as it does not require the student to - 257 - (-D } n-i ("D (2fi-l)2(2/3+l)3(2n+3)2 73-1 2 L , 12=1 (2l3-l) (213-1 (2/H-l)3(2l2+3)2 (-D (2/2~l)2(2/2+l)3(2/2-f3): n=-2 I 7 (-D D=-oo 1 1 7* S" 12-1 (-D-1 1 (2/7-1 )2(2//-fl)3(2«-f-3): n-1 2" (-D- (-1)2(1)3(3)2 (-3)2(-l)3(l)2 where (~D J3=-oo 13-1 (2i3-l)2(2ri-l-l)3(2i3+3)2 To evaluate S, we consider the function IT f(z) 2 (z - l) (z + l) 2 z 3 cos nz Let iV be a positive integer and let C be having corners (±N9±N)* the square in the complex plane From the fact that jcos H Z | > 1 for z on C we see that lim [ f(z)dz = 0, Hence the sum of all the residues of f(z) is zero. On the other handf f(z) has simple poles at z = k + ~r (& and integer), poles of order 2 at z = ±1, and a pole of order 3 at z = 0, lim z-tk+y TT z - k 2 y 2 The residue of f(z) at & + ^ is lim 71 3 (z - l ) ( z + l ) z c o s xrz 3 z-^ir+rr hrt-rhy 128w 2 3 (2k - I ) ( 2 * + I ) (21 + 3 ) (-1) 2 &-1 k 1 c o s rrz lim 2 -128 (2k - 1 ) ( 2 1 + 1 ) 3 ( 2 * + 3 ) 2 The residue of f(z) at z = 1 is Z z-#*+£ - u s i n TFZ - 258 - H lim JZ z^l the residue of f(z) IT (z + l)2z3cos nz at z = -1 is similarly d u lim Uz (Z - l) 2 Z 3 COS HZ z-r-1 and the residue of f(z) at z = 0 is Id* lim : z-»0 2 dz2\ (z - 1)2(Z 4- l)2COS HZ J: = — 2 + 2lt Thus 128S -f- u + u + _ + 2ir = 0, S = and so the required sum is 1 c x 1 - 1 7^ II1» Fourier This + « ** ^ - g- ST " "512" # series. is more elaborate than the first two, but gives quite a bit more (definitely not recommended to the run-of-the-mill student). Let S be as before and define F( sin(2/3 4- l)x „=l n=-oo (1) (2D - l)2(2n + l)3(2n 4- 3 ) 2 then clearly - = •*[?] • The following two Fourier expansions are well known e.g. Knopp, 501, #15-8): Unendliche } Reihen, (any standard 386-388, or Apostol, Mathematical S^~Tssii n ( 2 n - l)x = textbook, Analysis, (2) n=-oo i J2=-o© cos(2/? - l ) x = (2n - 1) : IT* 7fX (3) - 259 - both valid for 0 < x < IT. There are corresponding formulae for (n 9 2TI ) but we need not bother about them. First, F'"(x) = ~G(x)9 the interval Here we go. where 00 cos(2i? + l ) x G?(x) (2n - l)2(2n + 3)2 Then oo (2n + l ) 2 c o s ( 2 n <r- x) = - + l)x 2 (2n - l ) ( 2 n + 3 ) a ^ /l=-oo = + - i 5 fcr=-r sTr-Trf003*2" + 1)x n=-oo oo 1 oo X eos(2n + l ) x + cos(2i7 - 3 ) x = -T2 1 i2n . / 0 7 c L V ) os(2x) _ 1)2 I V — " ^ 2 Tsr^TTTsr+^r COs(2l2 - 1)X 1 „y . . 1— _ L _ - v H(x) Z (2n ~ I ) 2 L 72 = - o o where Hfv> H{x) - \ " 2 cos(2ri + l ) x 12n - I ) ( 2 n + 31 77=-oo and we used the identity o o A - £ A + B c o s A 4- c o s B = 2 c o s — ^ cos -—«— Thus oo H'/V\ tf m {2n - _ ^ " Z + D(sin(2n + l)x (2n - l)(JJn + 3) n=-«o oo 1 ~ V ? L 1 In - 1 + 1 ' sin(2n + l)x iin + 3 n~-oo 1 V 1 I /} = - » cos(2j2 + l ) x sin(2n + l ) x + sin(2n - 3)x In - 1 //8. (4) - 260 - oo v \ -cos(2x) > /0 sin(2n - l)x 2n - t J2 = -oo = - -n- COS by (2) and the 2x, identity A + B A - B sin A + sin B = 2 sin — n — - cos — J J — Integrating? and observing that H\^\ = 0, we get ff(x) = ~ x sin 2x, hence from (3) and (4) flT2 1 TTXl IT G"(x) = - -^ oos(2x)« U- - •£-- + g- sin 2x u IT2 nx •g sin 2x - ^- cos 2x + -j— cos 2x* Using , x sin 2x cos 2x 0 x cos 2x dx = ^— + — j — — and ~ _, -x cos 2x . sin 2x x sin 2x ax = + —* n Z 4 and noting G'(0) = 0, G|£ = 0 we get G'(x) = - yyr sin 2x + —^ sin 2x and 7T 7TX IT G(x) = -^ cos 2x - j7T cos 2x + _ sin 2x« Thus F"'(x) = - Tp^ cos 2x -I- jrr cos 2x - m sin 2x. Finally, three times and observing that F#/(0) = 0, F'LJ integrating F(0) = 0, we obtain successively «#// 7T 2 v . 0 , UX r/ F' (x) = - _: sin 2x + ^ ^r/ i l* 2 FMx) = ^ o cos 2x - . W X M 0 s IT 0 7T sin 2x 4- •— cos 2x - *«• , o . 3lT cos 2x + m o TTX ^ IT 2 sin 2x - ^ +^ , and JT F(x) = ^ Thus 2 TT V sin 2x - J 2 ? sin 2x - IT M TTX cos 2x - ^ 7T V TF + E5~ + CT - 261 - 7X3 IT vfa] IT 3 H IT3 IT as required. Admittedly the Fourier argument is heavier than the other two, but as a bonus, formula (5) evaluates the series (1) for all 0 < x < n« Also solved by C. KIERSTEAD JR., Cuyahoga Southwold, Suffolk, Townsville, Austral FESTRAFFS-EAMOIR, Falls, Ohio; England; and KEE-WAI BASIL Bruxelles, Ik^lgique; LAU, Kong; C. Hong R.C. LYNESS, RENNIE, James Cook University, la. % % 1053. FRIEND H. [1985: 187] Proposed Corp., Nashua, Exhibit % by Stanley New Rabinowitz, Digital Equipment Hampshire. a bisection between the points in the plane and the lines in the plane. Solution by Leroy F Meyers, The Ohio State University, Columbus, Ohio. Every nonvertical line in the (Euclidean) plane has a unique equation of the form y = mx + 5, where m and b are arbitrary real numbers, and conversely every pair (m,b) of real numbers determines a nonvertical line with y = mx + b. Since the vertical $ haven t lines correspondence, we must modify the correspondence. line with equation x been Thus, = b we associate the point (09b); integer and the point (mfb) included to each in the vertical to each nonvertical line with equation y = mx + b we associate the point (m + nonnegative equation 1,5) if m is a if HI is otherwise. This is the required one-to-one correspondence. Also solved by J. PARMENTER, Memorlal There omit University was one incorrect projective plane SUCK, Essen, (nn Federal of solution. the surface, Republic Newfound1 andf Two readers an easier of St. Germany; J John s, Newfound1 * M M. and. submitted solutions for the problem). I flatly decided to these. * and *