SOLUTIONS A 2 + B 2 = AB + BA = 2 4

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1 SOLUTIONS /4 SOLUTIONS No problem is ever permnently closed The editor is lwys plesed to consider for publiction new solutions or new insights on pst problems Sttements of the problems in this section originlly pper in : 4(9), p Proposed by Dniel Sitru Determine ll A, B M (R) such tht: Ç å 4 A + B =, 4 8 Ç å AB + BA = We received 7 correct solutions nd will feture the solution by Joseph DiMuro Summing the two equtions, we obtin: (A + B) = A + AB + BA + B = Å ã We cn digonlize this mtrix in order to find its squre roots: Å ã Å ã Å ã (A + B) = P DP 64 /4 / =, /4 / A + B = P D / P = Å ã Å ã Å ã ±8 /4 / /4 / We cn lso subtrct the originl two equtions to obtin: (A B) = A AB BA + B = Å ã 4 4 = ± As before, we digonlize this mtrix: Å ã Å ã Å ã (A B) = P DP 36 /3 /3 =, /3 /3 Å ã 4 8 Å ã Å ã Å ã Å ã A B = P D / P ±6 /3 /3 = = ± /3 /3 Copyright c Cndin Mthemticl Society, 7

2 46/ SOLUTIONS Now we hve the two equtions A + B = ± Å ã Å ã 4 8, A B = ±, which cn esily be solved to produce four possible pirs of mtrices for A nd B One solution is A = Å ã 3 6, B = Å ã The other solutions my be obtined by interchnging A nd B, nd/or replcing A nd B with their negtives 48 Proposed by D M Bătineţu-Giurgiu nd Neculi Stnciu Let ABC be right-ngle tringle with A = 9 nd BC =, AC = b nd AB = c Consider the Fiboncci sequence F n with F = F = nd F n+ = F n+ + F n for ll non-negtive integers n Prove tht F m (bf n + cf p ) + for ll non-negtive integers m, n, p F n (bf p + cf m ) + F p (bf m + cf n ) 3 We received 8 correct solutions nd present the solution by Adnn Ali From the Cuchy-Schwrz Inequlity, (b + c )(F k + F l ) = (F k + F l ) (bf k + cf l ), for ll k, l Thus, F m F n (bf n + cf p ) + (bf p + cf m ) + (bf m + cf n ) Fm (Fn + Fp ) + Fn (Fp + Fm) + Fp (F m + Fn) 3, where the lst inequlity follows from Nesbitt s Inequlity Equlity holds iff F m = F n = F p nd b = c Editor s Comments As solvers pointed out, the fct tht the F n s were Fiboncci numbers ws irrelevnt; it ws only necessry tht they were nonnegtive F p 483 Proposed by Ovidiu Furdui Clculte lim n n n n x + n cos x dx We received solutions, of which 6 were correct nd complete We present the solution by Michel Btille Crux Mthemticorum, Vol 4(9), November 6

3 SOLUTIONS /47 We show tht the required limit is Let f n (x) = +n cos x The π-periodicity of f n nd the chnge of vribles x = tn (t), dx = dt +t esily yield (k+)π/ (k )π/ for ny k, n N f n (x) dx = π/ π/ f n (x) dx = This sid, for every n N with n, let p n = n Then, (p n ) π n < (p n + ) π nd I n = Clerly, π/ xf n (x) dx + π/ nd for k {,,, p n }, Similrly, Thus, (k ) π n π p n n + k= p n k= (k+)π/ (k )π/ dt n + + t = π n + π + nd In = n xf n(x) dx n xf n (x) dx + xf n (x) dx (p n ) π xf n (x) dx π π/ π f n (x) dx = π/ 4 n + π n + (p n ) π (k ) I n (k+)π/ (k )π/ xf n (x) dx (k + ) π π n + n xf n (x) dx n f n (x) dx nπ (p n ) π n + π 4 n + + π p n (k + ) + nπ n + n + k= so tht π (p n ) n + I n π ( π n π ) Å p n + n = πp n π n + + π 4p n + n p n ã Since p n n π s n, we obtin I n π p n n + n n s n The result follows Copyright c Cndin Mthemticl Society, 7

4 48/ SOLUTIONS 484 Proposed by Michel Btille In the plne, let Γ be circle nd A, B be two points not on Γ Determine when MA MB is not independent of M on Γ nd, in these cses, construct with ruler nd compss I nd S on Γ such tht ß ß IA MA IB = inf MB : M Γ SA MA nd SB = sup MB : M Γ We feture the proposer s solution; we received no others Becuse A is not on Γ, inversion in the circle with centre A nd rdius AB tkes Γ to nother circle, cll it Γ For ny point M on Γ, this inversion tkes the pir of points M, B to nother pir M, B, whose distnces stisfy consequently, MB = AB M B AM AB = AM AM M B AM = MA M B AB AB ; MA MB = AB BM () MA From (), MB is independent of M on Γ if nd only if BM is constnt This occurs if nd only if B is the centre of Γ ; tht is, if nd only if A nd B re n inverse pir with respect to Γ Otherwise, let the dimeter of Γ through B intersect Γ t C nd D with BC > BD Then MA MB is miniml when BM is mximl; tht is, when M = C; MA MB is mximl when BM is miniml, in which cse M = D Thus, I coincides with C (the imge of C under our inversion), nd S coincides with D The construction of I nd S is immedite once Γ hs been drwn The circle Γ cn be redily constructed from the inverses of three points of Γ (s in the figure): Editor s Comments For ny two fixed points A nd B, the locus of points M for which MA MB is constnt is clled the circle of Apollonius; inversion in tht circle interchnges A nd B See, for exmple HSM Coxeter nd SL Greitzer, Geometry Revisited (Mthemticl Assocition of Americ, 967), exercise 4, pges 4 nd 7 Also, Theorem 4 there provides the distnce formul used bove to obtin () Crux Mthemticorum, Vol 4(9), November 6

5 SOLUTIONS /49 48 Proposed by José Luis Díz-Brrero Correction Let ABC be n cute tringle Prove tht 4»sin(cos A) cos B +»sin(cos 4 B) cos C +»sin(cos 4 C) cos A < 3 We received eight submissions, six of which re correct We present the solution by Titu Zvonru It is well known tht cos A + cos B + cos C 3 [Item 6 on p of the book Geometric Inequlities by O Bottem et l; Groningen, 969] Using this, together with the fcts tht sin x < x for < x < π, xy + yz + zx x + y + z, nd (x + y + z) 3(x + y + z ) we then hve» 4 sin(cos A) cos B < 4 cos A cos B cos A cyc cyc cyc» 3(cos A + cos B + cos C) 3( 3 ) = 3 3 Editor s comments Arkdy Alt» proved the stronger result tht the given upper bound could be replced by 3 4 sin which is less thn 3 3 since sin < This new upper bound is ttined if nd only if the tringle is equilterl His proof used the Cuchy-Schwrz Inequlity, concvity of the functions sin x nd cos x, Jensen s Inequlity s well s the fct tht cos A = + r R nd the Euler s Inequlity r R 486 Proposed by Dniel Sitru Let be f : [, ] R; f twice differentible on [, ] nd f (x) < for ll x [, ] Prove tht f(x)dx 6 f(x)dx + 4f() We received seven solutions nd present two of them Solution, by AN-nduud Problem Solving Group From the given conditions, f is concve on [, ] inequlity we get 6 f(x)dx + 9 On the other hnd, we hve Å ã f = f f(x)dx 6 = 6 Å f(x)dx + 9 f(x)dx + 8 f() + 8 f Using Hermite-Hdmrd s Å f() + f ã 9 f() + 8 Å ã 9 f, Å ã Å ãã Copyright c Cndin Mthemticl Society, 7

6 4/ SOLUTIONS so 8 f From here, we get f(x)dx 6 Å ã f() + 6 Å ã f f(x)dx + 4f() + 6 Å ã f Using Hermite-Hdmrd s inequlity, we get Å ã Ç f = f + å f(x)dx Å ã f Hence, we get f(x)dx 6 f(x)dx + 4f() f(x)dx Solution, by Leonrd Giugiuc In f(x) dx, we mke the substitution x x 4 We need to prove f(x) dx = Å ã 4x + f dx 6 Å ã 4x + f dx 4 Å ã 4x + f dx Å ã 4x + f dx nd cler frctions to get f(x) dx + 4f() f(x) dx + f() ò ï 4 f(x) + f() But f (x) < x [, ], so f is concve on [, ] nd from here Å ã 4x + f 4 f(x) + f() Integrting, we conclude tht Å ã 4x + f dx ï 4 f(x) + ò f() Editor s Comments Henry Ricrdo observed tht this problem ppers s problem MA (with solution) in the Dniel Sitru s book Mth Phenomenon, published in English by the Romnin publisher Editur Prlel 4 in 6 dx dx Crux Mthemticorum, Vol 4(9), November 6

7 SOLUTIONS /4 487 Proposed by Lorin Scenu If S is the re of tringle ABC, prove tht m (b + c) + m 4S sin A, where b nd c re the lengths of sides tht meet in vertex A, nd m is the length of the medin from tht vertex; furthermore, equlity holds if nd only if b = c nd A = We received seven correct submissions nd present the solution by Leonrd Giugiuc Let A be the reflection of A in the midpoint M of BC Becuse AMC = A MB, we hve AA = m, A B = b, A BA = π A, nd [A AB] = [ABC] = S (where the squre brckets denote re) Let m = m nd denote by r, R, nd s (= m+b+c ) the inrdius, circumrdius, nd semiperimeter, respectively, of A AB We need to prove tht which is equivlent, in turn, to m(b + c) + m 8S sin A BA, m(m + b + c) 8S sin(π A) ms r 8S sin A m sin A 4r R r But the finl line is Euler s inequlity pplied to A AB, which completes the proof Equlity holds for Euler s inequlity if nd only if A AB is equilterl, which implies tht b = c nd A =, s desired 488 Proposed by Ardk Mirzkhmedov Let, b nd c be positive rel numbers such tht b + b c + c + b c = 4 Prove tht + b + c + bc( + b + c) (b + bc + c) We received four submissions ll of which re correct We present the solution by the proposer, expnded by the editor with some detils We first show tht the given condition implies or + bc + b b + c + c c + b = () (b + c )(c + b ) + b(c + b )( + bc ) + c( + bc )(b + c ) = ( + bc )(b + c )(c + b ) () Copyright c Cndin Mthemticl Society, 7

8 4/ SOLUTIONS Let S nd P denote the left side nd the right side of (), respectively Then by strightforwrd computtions, we find S = cyc (4bc + b 3 + c + 3 b c) nd = bc + 4( b 3 + b c 3 + c 3 ) + bc( 3 b + b 3 c + c 3 ) = bc + 4( b 3 + b c 3 + c 3 ) + bc(4 b c ) = 6bc + 4( b 3 + b c 3 + c 3 ) 3 b 3 c 3 P = (4b + c 3 + b c + bc 3 )(c + b ) = 8bc + 4(c 3 + b 3 + b c 3 ) + bc( 3 b + b 3 c + c 3 ) + 3 b 3 c 3 = 8bc + 4( b 3 + b c 3 + c 3 ) + bc(4 b c ) + 3 b 3 c 3 = 6bc + 4( b 3 + b c 3 + c 3 ) 3 b 3 c 3 Hence, S = P which estblishes () Now, for ll u, v, w >, we hve by the Cuchy-Schwrz Inequlity tht Setting (u + v + w)( u + b v + c w ) ( + b + c) (3) u = + bc, v = b + c b nd w = c + b c, we then hve by () nd (3) tht so = + bc + b b + c b + c ( + b + c) c + b c u + v + w, ( + bc ) + (b + c b) + (c + b c) = u + v + w ( + b + c) from which it follows tht + b + c + bc( + b + c) (b + bc + c) 489 Proposed by Dniel Sitru nd Leonrd Giugiuc Let, b, c nd d be rel numbers with < < b < c < d Prove tht b + c b + d c > 3 + ln d There were 4 correct solutions We present four of them here Most of the solvers pproched the problem long the lines of one of the first two solutions Crux Mthemticorum, Vol 4(9), November 6

9 SOLUTIONS /43 Solution Since x > + ln x for x, s desired Solution b + c b + d c > Å + ln b ã ( + + ln c ) Å + + ln d ã = 3 + ln d b c Applying the AM-GM Inequlity, we find tht the left side is not less thn Ç å d d 3 3 > 3 + ln 3 = 3 + ln d Solution 3, by Kee-Wi Lu For < < b < c < d, let f(, b, c, d) = b + c b + d c ln d, g(, b, c) = b + c b + ln c, h(, b) = b + ln b An nlysis of the prtil derivtives revels tht ech of its functions strictly increses in its finl vrible, so tht f(, b, c, d) > f(, b, c, c) = g(, b, c) > g(, b, b) = h(, b) > h(, ) = 3, which yields the desired result Solution 4, by the proposers Let f(x) = /x A digrm shows tht whence s desired (b )f() + (c b)f(b) + (d c)f(c) > b + c b + d > ln d ln, c d dx x, Editor s Comments Two solvers provided strightforwrd generliztion for n incresing sequence { k } of n + positive rels: n k= k+ k > n + ln n+ Copyright c Cndin Mthemticl Society, 7

10 44/ SOLUTIONS 49 Proposed by Nermin Hod zić nd Slem Mlikić Let, b nd c be non-negtive rel numbers such tht + b + c = 3 Prove tht 3b + 6c bc + b 3c + 6 c + c 3 + 6b b 3 8 We received two correct solutions We present the solution of the proposers, slightly modified by the editor Using Jensen s inequlity for f(x) = x + b + c 3b + 6c bc + Å (3b + 6c bc) + b + c which we cn rerrnge to b + b + c + b(3c + 6 c) + b + c (which is convex on (, )), we hve 3c + 6 c + + c(3 + 6b b) + b + c c + b + c ã, 3 + 6b b 3b + 6c bc + b 3c + 6 c + c 3 + 6b b ( + b + c) 3(b + bc + c ) + 6(b + bc + c) 3bc In order to prove the inequlity given in the question, it thus suffices to show ( + b + c) 3(b + bc + c ) + 6(b + bc + c) 3bc 3 8, which holds (by cross multiplying nd rerrnging) if nd only if 8( + b + c) 9(b + bc + c ) + 8(b + bc + c) 9bc 8( + b + c ) 9(b + bc + c ) + (b + bc + c) 9bc By the Cuchy-Schwrz inequlity, b + bc + c + b + c Note for lter tht equlity holds if nd only if = b = c = Hence, it suffices to show tht 6( + b + c ) 9(b + bc + c ) 9bc Finlly, since + b + c = 3, this reduces to (b + bc + c ) bc () Assume tht b c Then ( b)(b c), equivlent to b + bc b + c Multiply both sides by > nd rerrnge to get bc b + c b Note tht the cubic g(b) = 3b b 3 hs locl mximum t b =, nd in fct for ll b we hve 3b b 3 g() = Hence bc + b + c b + 3b b 3, which is equivlent to bc + b + c + c b b( + b + c ) + 3b Since +b +c = 3, this shows tht bc+ b + c+c b, which is equivlent to (), concluding the proof Crux Mthemticorum, Vol 4(9), November 6