MathproblemsISSN: 2217-446X, url: http://www.mathproblems-ks.com Volume 2, Issue 1 (2012), Pages 48–68 Editors: Valmir Krasniqi, José Luis Dı́az-Barrero, Valmir Bucaj, Mihály Bencze, Ovidiu Furdui, Enkel Hysnelaj, Paolo Perfetti, József Sándor, Armend Sh. Shabani, David R. Stone, Roberto Tauraso, Cristinel Mortici. PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets, each indicating the name of the sender. Drawings must be suitable for reproduction. Proposals should be accompanied by solutions. An asterisk (*) indicates that nei- ther the proposer nor the editors have supplied a solution. The editors encourage undergraduate and pre-college students to submit solutions. Teachers can help by assisting their students in submitting solutions. Student solutions should include the class and school name. Solutions will be evaluated for publication by a com- mittee of professors according to a combination of criteria. Questions concerning proposals and/or solutions can be sent by e-mail to:
[email protected] Solutions to the problems stated in this issue should arrive before 2 April 2012 Problems 29. Proposed by Enkel Hysnelaj, University of Technology, Sydney, Australia. 1) Let f : (0, ∞) → R be a function that satisfies the property f (x2 ) + f (y 2 ) = f (xy) 2 for any (x, y) ∈ (0, ∞). Show that f (x3 ) + f (y 3 ) + f (z 3 ) = f (xyz) 3 for any (x, y, z) ∈ (0, ∞). 2) Generalize the above statement, so show that if f (x21 ) + f (x22 ) = f (x1 x2 ) 2 for any (x1 , x2 ) ∈ (0, ∞), then f (xn1 ) + f (xn2 ) + . . . + f (xnn ) = f (x1 x2 . . . xn ) n 2010 c Mathproblems, Universiteti i Prishtinës, Prishtinë, Kosovë. 48 49 for any (x1 , x2 , . . . , xn ) ∈ (0, ∞) and n a positive integer greater than 1. 30. Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buzau, Romania. Evaluate: Z 2012 x sinn t lim dt, x→0 2011 x tm where n, m ∈ N. 31. Proposed by Valmir Bucaj, Texas Lutheran University, Seguin, TX. If the vertices of a polygon, in counterclockwise order, are: 1 1 2 1 3 1 4 1 √ ,√ , √ ,√ , √ ,√ , √ ,√ ,..., a1 an+1 a2 a2n a3 a2n−1 a4 a2n−2 n−1 1 n 1 √ ,√ , √ ,√ , an−1 an+3 an an+2 where (an )n≥1 is a decreasing geometric progression, show that the area of this polygon is 3 1 1 A= √ √ −√ 2 a1 a2n an 32. Proposed by Mihály Bencze, Braşov, Romania. Let f : [a, b] → R be a two times differentiable function such that f 00 and f 0 are continuous. If m = min f 00 (x) and M = max f 00 (x), then prove that x∈[a,b] x∈[a,b] m (b2 − a2 ) M (b2 − a2 ) ≤ bf 0 (b) − af 0 (a) − f (b) + f (a) ≤ 2 2 33. Proposed by Ovidiu Furdui, Cluj, Romania. Find the value of Z π/2 √ sinn x + cosn x dx n lim n→∞ 0 34. Proposed by Mihály Bencze, Braşov, Romania. Solve the following equation r r q √ q √ √ 2 + 2 + . . . + 2 + x + 2 − 2 + . . . + 2 + x = x 2, | {z } | {z } n-times n-times where n ≥ 3. 35. Proposed by Florin Stanescu, School Cioculescu Serban, Gaesti, jud. Dambovita, Romania. Let P (x) = an xn + an−1 xn−1 + ... + a1 x + a0 be a polynomial with strictly positive real coefficients of degree n ≥ 3, such that P 0 has only real zeros. If 0 ≤ a < b show that Rb 1 dx 0 a P 0 (x) 1 P (b) P 0 (b) − P 0 (a) Rb 1 ≥ ln 0 ≥ dx b−a P (a) P (b) − P (a) a P 00 (x) Rome. Switzerland.50 Solutions No problem is ever permanently closed. Tor Vergata Univer- sity. Department of Mathematics. Let x. Let f be a real. z be positive real numbers. and g(x) = x− . 1107 + 1580x − 512x2 − 384x3 > 0. So X (1 − 3x)(3y + 3z + 4x) X X 2 = f (x) ≥ g(x) = 0 cyc (2y + 2z + 3x) cyc cyc Also solved by the proposer 23. Italy. integrable function defined on [0. (3 − 2x)2 (2 + x)2 343 3 We observe that g(x) is the tangent to the graph of f (x) at x = 1/3. we have to prove that X 4x(2 − x) (1 − x)(3 + x) X (1 − 3x)(3y + 3z + 4x) − = ≥ 0. Prove that X 4x(x + 2y + 2z) X (x + y)(3x + 3y + 4z) ≥ cyc (x + 3y + 3z)2 cyc (2x + 2y + 3z)2 Solution by Albert Stadler. Proposed by Paolo Perfetti. Italy. Proposed by Paolo Perfetti. because −3 4 1 2 + 5 1 − 27 1 3 3 864 f0 = = 3 1 2 1 2 343 3−2 2+3 3 We claim that f (x) ≥ g(x). We will be very pleased considering for publication new solutions or comments on the past problems. department of Mathematics. y. for 0 ≤ x ≤ 1. Obviously X (x + y)(3x + 3y + 3z) X (y + z)(3y + 3z + 4x) = cyc (2x + 2y + 3z)2 cyc (2y + 2z + 3x)2 The inequality then reads as X 4x(x + 2y + 2z) X (y + z)(3y + 3z + 4x) ≥ cyc (x + 3y + 3z)2 cyc (2y + 2z + 3x)2 By homogeneity. for 0 ≤ x ≤ 1. So. 1] such that . Tor Vergata Univer- sity. (3 − 2x) (2 + x) 343 3 343(3 − 2x)2 (2 + x)2 and clearly. 22. we can assume that x + y + z = 1. Indeed (1 − 3x)(4x2 + 5x − 27) 864 (1 − 3x)2 (1107 + 1580x − 512x2 − 384x3 ) 1 2 2 − x − = . cyc (3 − 2x)2 (2 + x)2 cyc (2y + 2z + 3x)2 Let (1 − 3x)(4x2 + 5x − 27) 864 1 f (x) = . Rome. Let us consider two cases : (a) F (x) > 0 for x ∈ In . F is continuous on [0. m(x − bn )). Moreover. . 1 Proof. and let F (x) = Z 1 Rx 2 m2 M 2 0 f (y)dy. (i. integrable function defined on [0.e. 1) : F (x) 6= 0} is an open set.e. The continuity of F shows that the set O = {x ∈ (0. M−m −m m if x ∈ 0. 1] such R1 that 0 f (x)dx = 0 and m = min0≤x≤1 f (x). Higher Institute for Applied Sciences and Technology. So. while F keeps a constant sign on In . we conclude that F (an ) = F (bn ) = 0. 51 R1 0 f (x)dx = 0 and m = min f (x). 1] \ O. The open set O is the union of at most denumerable family of disjoint open intervals. Syria. Since f is integrable. Let f be a nonzero real. ∀ x ∈ In . First. Since an and bn belong to [0. Thus there exist N ⊂ N and a family (In )n∈N of non-empty disjoint open sub- intervals of (0. since F (0) = F (1) = 0.) there is nothing to be proved. From m ≤ f ≤ M we conclude that. 0 < F (x) ≤ min(M (x − an ). M = max f (x). in what follows we will suppose that F 6= 0. M = max0≤x≤1 f (x). bn ). 1 M if x ∈ M −m . Prove that 1 −mM Z F 2 (x)dx ≤ (3M 2 − 8mM + 3m2 ) 0 6(M − m)2 Solution by Omran Kouba. Damascus. 1) such that O = ∪n∈N In . then F (x)dx ≤ . 1] \ O. 1]. we have Z x F (x) = F (x) − F (an ) = f (t) dt ≤ M (x − an ) an and Z bn F (x) = −(F (bn ) − F (x)) = (−f )(t) dt ≤ −m(bn − x) = m(x − bn ) x So. we observe that −mM mM m2 M 2 m2 M 2 2 (3M 2 − 8mM + 3m2 ) = − + 2 ≥ 6(M − m) 2 3(M − m) 3(M − m)2 and we will prove the following Proposition. Let us define F (x) = 0≤x≤1 0≤x≤1 Rx 0 f (y)dy. we see that F (t) = 0 for every t ∈ [0. MM −m f0 (x) = h i f1 (x) = h i −m M m if x ∈ M −m . with equality if and only if f coincides 0 3(M − m)2 for almost every x in [0. f = 0 a. for x ∈ In . 1] with one of the functions f0 or f1 defined by h h M if x ∈ 0. Suppose that In = (an . If F = 0. if and only if. and consequently Z Z bn 2 F 2 (x) dx ≤ (min(m(an − x). 3(M − m)2 3(M − m)2 with equality if and only if F (x) = max(m(x−an ). and f (x) = M h i for almost every x ∈ M bMn −ma −m . and f (x) = m h i −mbn for almost every x ∈ M aMn−m . M h (x−bn )) for every x ∈ In . That is. n So. M bMn −ma −m n . ∀ x ∈ In . That n )) for every −mbn is. 0 < −F (x) ≤ min(−m(x − an ). in both cases we have m2 M 2 Z F 2 (x) dx ≤ |In |3 In 3(M − m)2 . M (bn − x)). f (x) = M for almost every x ∈ an . f (x) = m for almost every x ∈ an . M (bn − x))) dx In an Z an +M (bn −an )/(M −m) Z bn = m2 (x − an )2 dx + M 2 (bn − x)2 dx an bn +m(bn −an )/(M −m) Z M (bn −an )/(M −m) Z −m(bn −an )/(M −m) 2 2 2 =m x dx + M x2 dx 0 0 m M2 2 3 m2 M 2 = (bn − an ) = |In |3 . we have Z x F (x) = F (x) − F (an ) = f (t) dt ≥ m(x − an ) an and Z bn F (x) = −(F (bn ) − F (x)) = (−f )(t) dt ≥ −M (bn − x) x So. if and only if. m(x − bn ))) dx In an Z an −m(bn −an )/(M −m) Z bn 2 2 = M (x − an ) dx + m2 (bn − x)2 dx an bn −M (bn −an )/(M −m) Z m(an −bn )/(M −m) Z M (bn −an )/(M −m) = M2 x2 dx + m2 x2 dx 0 0 m2 M 2 m2 M 2 = 2 (bn − an )3 = |In |3 3(M − m) 3(M − m)2 with equality if and only if F (x) = min(M (x−an ). m(x−b h x ∈ In .52 and consequently Z Z bn 2 F 2 (x) dx ≤ (min(M (x − an ). M aMn−m . (b) F (x) < 0 for x ∈ In . From m ≤ f ≤ M we conclude that. for x ∈ In . bn . bn . e. where f0 and f1 are the functions defined in the statement of the proposition.M. (bn )n≥1 be sequences of positive real numbers such that an+1 bn+1 lim 2 = lim 3 = a > 0. Tor Vergata University. Italy. Also solved by the proposer 24. Compute n→∞ n · an n→∞ n · bn s r ! bn+1 bn lim n+1 − n n→∞ an+1 an Solution 1 by Paolo Perfetti. This completes the proof. thus. Bucharest and Neculai Stanciu. Buzau. 1) and f (x) = f0 (x) a. The desired inequality is. Romania. or f (x) = f2 (x) a. 53 and consequently Z 1 XZ m2 M 2 X F 2 (x) dx = F 2 (x) dx ≤ 2 |In |3 0 In 3(M − m) n∈N n∈N !3 m2 M 2 X m2 M 2 ≤ |In | = |O|3 3(M − m)2 3(M − m)2 n∈N 2 2 m M ≤ . Department of Mathematics. Proposed by D. 3(M − m)2 3 where we used the well-known inequality n∈N λ3n ≤ P P n∈N λn . Now.. we get ((n0 + k − 1)!)3 ((n0 + k − 1)!)3 3 (a − ε)k bn0 ≤ bn0 +k ≤ (a + ε)k bn0 ((n − 1)!) ((n − 1)!)3 ((n0 + k − 1)!)2 ((n0 + k − 1)!)2 2 (a − ε)k an0 ≤ an0 +k ≤ (a + ε)k an0 ((n − 1)!) ((n − 1)!)2 The computations are straightforward. Moreover. Let (an )n≥1 . we have s bn0 +k+1 1 bn +k+1 n0 +k+1 = exp ln 0 an0 +k+1 n0 + k + 1 an0 +k+1 and k+1 k+1 (n0 + k)! a−ε bn 0 bn +k+1 (n0 + k)! a+ε bn 0 ≤ 0 ≤ (n0 − 1)! a+ε an0 an0 +k+1 (n0 − 1)! a−ε an0 . Batinetu-Giurgiu. proved. the equality case can occur if and only if O = (0. Rome. From the definition of the limit of a sequence immedia- tely follows that an+1 ∀ ε ∃ n0 : n > n0 =⇒ a − ε < 2 <a+ε n an bn+1 ∀ ε ∃ n0 : n > n0 =⇒ a − ε < 3 < a + ε n bn Thus for any k ≥ 1.e. using the fact that ex = 1 + o(1) when x → 0). namely n √ n0 n! = (n/e) 2πn(1 + o(1)). Athens. It is well known (see[1] p.46 for example) that zn being a sequence of positive numbers. we get bn +k+1 ln 0 = (k + 1) ln A + ln B + an0 +k+1 1 1 + (n0 + k) ln(n0 + k) − (n0 + k) + ln(2π) + ln(n0 + k) + ln(1 + o(1)) 2 2 Thus. n −1 (n + 1)(zn+1 )1/(n+1) bn+1 an+1 −1/(n+1) n = 3 (zn+1 ) →e n(zn )1/n n bn n2 an n+1 Now (n+1)(zn+1 )1/(n+1) s −1 1/(n+1) n r bn+1 n bn n(z ) 1/n (n + 1)(z n+1 ) → e−1 . limn→+∞ zn+1 zn = 1/n bn ` ∈ R ⇒ limn→+∞ (zn ) = `. so −1 −n zn+1 bn+1 an+1 1 n = 3 1+ → e−1 zn n bn n2 an n n+1 and limn→+∞ (zn )1/n = e−1 on account of the preceding. We set zn = nn an . 1 bn +k+1 (k + 1) ln A + (n0 + k) ln(n0 + k) − (n0 + k) ln 0 = + o(1) = n0 + k + 1 an0 +k+1 n0 + k + 1 = ln A + ln k − 1 + o(1) and then. then the limit is 1/e.54 a+ε n0 b Let us define A = a−ε and B = (n0 −1)!a . Therefore. Greece. Solution 2 by Anastasios Kotronis. n+1 − = (zn )1/n n ln an+1 an ln (n+1)(zn+1 ) 1/(n+1) n(zn )1/n n(zn )1/n since 1/(n+1) (n+1)(zn+1 )1/(n+1) n(zn )1/n −1 exp ln (n+1)(z n+1 ) n(zn )1/n −1 lim 1/(n+1) = lim 1/(n+1) (n+1)(zn+1 ) n→+∞ ln n(zn )1/n n→+∞ ln (n+1)(z n+1 ) n(zn )1/n ex − 1 = lim =1 x→0 x . Using Stirling’s formulae. yields 1 bn +k+1 Ak exp ln 0 = (1 + o(1)) n0 + k + 1 an0 +k+1 e Subtracting we obtain 1 bn +k+1 1 bn +k exp ln 0 − exp ln 0 = n0 + k + 1 an0 +k+1 n0 + k an0 +k A a+ε 1 = (1 + o(1)) = + o(1) e a−ε e Since ε is arbitrarily small. Kaczor. If E. Paris. M.J.M. M. we write the inequality claimed as −1 1 BC · M D EA FA ≥2 + 2 MA DC · EB BD · F C On account of AM-HM inequality it will be suffice to prove that if E. yz(x + 1) x+1 x+1 Last inequality trivially holds. Let M be a point lying on cevian AD.S. Moubinool Omarjee. Math. 25. Real Numbers. F be three points lying on the sides BC. and the proof is complete. = . Comanesti and Neculai Stanciu. x+1 x+1 and by the relation (R2 ) from Rec. Romania. Nowak Problems in Mathematical Analysis I. F are collinear then show that BC · M D EA FA + ≥4 MA DC · EB BD · F C Solution 1 by Titu Zvonaru. We denote BD AF AE a = BC. 4CF C 0 ∼ 4AF A0 . First. 2000. F are collinear. Barcelona. 2/2011. F are collinear then holds MD EB FC BC · = DC · + BD · MA EA FA Indeed.. 108. M. = z. Solution 2 by the proposer. E. France. we obtain that AM ayz AM yz(x + 1) = ax a ⇔ = MD x+1 · z + x+1 ·y MD xz + y Hence. and the proposer. Also solved by Albert Stadler. 55 References [1] W. the given inequality becomes ! a(xz + y) z y a + ax ≥ 4 ⇔ (xz + y)2 ≥ 4xyz ⇔ (xz − y)2 ≥ 0. assume that points E. = x. BARCELONA TECH. Spain. and 4DM D0 ∼ 4AM A0 . = y.T. Buzau. A. DC FC EB We have ax a BD = . Switzerland. Then. pp. DC = . George Emil Palade Secondary School. Sequences and Series . CA of ∆ABC. M. It is easy to check that 4EBB 0 ∼ 4AEA0 . Proposed by José Luis Dı́az-Barrero. = EA AA0 FA AA0 MA AA0 and the statement becomes BB 0 CC 0 DD0 DC · 0 + BD · 0 = BC · AA AA AA0 or DC · BB 0 + BD · CC 0 = DC · DD0 . AB. we have EB BB 0 FC CC 0 MD DD0 = . Let D. Likewise. and substituting for x in the original equation we get 2 1 y−1 y−1 f + f (y) = a +b +c (1) 1−y y y x−1 Similarly. Since 4BDB 00 ∼ 4BCC 00 . De- termine all functions f : R − {0. Proposed by Enkel Hysnelaj. We draw the parallel to B 0 C 0 that cuts line DD0 at D00 and line DD00 BD CC 0 at C 00 . b. Let- 1 ting y = 1−x . Also solved by Codreanu Ioan-Viorel. Finally. Australia. Damascus. 1} → R. c ∈ R. Higher Institute for Applied Sciences and Technology. then 00 = from which follows CC BC BD BD DD00 = CC 00 = (CC 0 − BB 0 ) BC BC Thus. Satulung. Syria. Omran Kouba. Maramures. which satisfy the relation x−1 1 f +f = ax2 + bx + c. Sydney. Romania. the same result is obtained when BB 0 > CC 0 . Seguin. x 1−x where a. F are the vertices if its medial triangle. E. and the proposer. Solution by Valmir Bucaj. TX. letting y = x . and substituting for x we get 2 y−1 1 1 f (y) + f =a +b +c (2) y 1−y 1−y . 26.56 A C' F M C'' E A' D' B' D'' B D C Figure 1. BD · CC 0 0 00 00 0 BD 0 0 0 0 BD DD = DD + D D = (CC − BB ) + BB = + BB 1 − BC BC BC BD · CC 0 BB 0 BD · CC 0 CD · BB 0 = + (BC − BD) = + BC BC BC BC and the statement follows. Texas Lutheran University. observe that equality holds when 4ABC is equilateral and D. Suppose that BB 0 < CC 0 . University of Technology. Problem 25 Now we distinguish two cases according to BB 0 < CC 0 or BB 0 > CC 0 . . 89. y−1 1 f +f = ay 2 + by + c. Thus x x xa lnx < xπ(x) < xb lnx So ax 1 bx 1 x lnx < xπ(x) < x lnx Therefore eax < xπ(x) < ebx . " 2 2 # a y−1 1 2 b y−1 1 c f (y) = + −y + + −y + . (which is reproduced on the analytic number theory books like Tom M.126-129. Journal de Math. Solution by the proposer. 2 y 1−y 2 y 1−y 2 and the result follows by setting y = x. Stone. (which is reproduced in the book Introduction to analytic and probabilistic number theory by Gerald Tenenbaum). p. apart from its first solution given by P. y 1−y after substituting in the preceding. and the proposer 27. Chandrasekharan’s book with the same title). France. Anastasios Kotronis. Albania. lnx lnx for x sufficiently large. Greece. 366-390. GA. 17 (1852). With π(x) = the number of primes ≤ x. 2. Paris. Omran Kouba. Tchebychef proved (1850) that there exist constants a and b such that x x a < π(x) < b . Switzerland. let us know that the problem solved above. Apostol’s Introduction to analytic number theory and K. Higher Institute for Ap- plied Sciences and Technology. Pures et Appl. Chebyshev in Memoire sur les nombres premiers. Monthly. Also solved by Albert Stadler. L. Amer. Comment by the Editors. Moubinnol Omarjee. 57 Adding (1) and (2) gives 2 2 y−1 1 y−1 y−1 1 1 2f (y)+f +f =a +b +a +b +2c y 1−y y y 1−y 1−y Since. Math. Athens. Proposed by David R. Damascus. Statesboro. Syria. which concludes the proof. by M. show that there exist constants a and b such that eax < xπ(x) < ebx for x sufficiently large. Nair in the article On Chebyshev-type inequalities for primes. USA. no.also it has been given an elementary solution using different methods. we get 2 2 y−1 y−1 1 1 2f (y) = a +b +a +b − ay 2 − by + c y y 1−y 1−y Finally. Adrian Naco. Georgia Southern University. and the proposer. [AC] = b. Also solved by Albert Stadler. Viveiro. 1) → R. Buzau. [BC] = a. y =1− . Syria. y and z by a b c x=1− . Switzerland. f (t) = √ = t−1/2 − t1/2 t The function f is convex since it is the sum of two convex functions.58 Also solved by Albert Stadler. Comanesti and Neculai Stanciu. Switzerland 28 Proposed by Florin Stanescu. Titu Zvonaru. Gaesti. Ro- mania. so √ x+y+z 1 f (x) + f (y) + f (z) ≥ 3f = 3f = 2 3. Bruno Salgueiro Fanego. (4) 3 3 and the desired inequality follows from (3) and (4). Prove that a b c p √ +√ +√ ≥ 2 3p. jud. Dambovita. (3) p p−a p p−b p p−c where 1−t f : (0. p−a p−b p−c where [AB] = c. Damascus. z =1− p p p Clearly we have x + y + z = 1 and a b c √ √ +√ √ +√ √ = f (x) + f (y) + f (z). School Cioculescu Serban. . Solution by Omran Kouba. George Emil Palade Secondary School. Higher Institute for Applied Sciences and Technology. Spain. Let us define the positive real numbers x. Let ABC be a triangle with semi-perimeter p. Romania. a4 be nonzero real numbers defined by ak = . sin kβ (1 ≤ k ≤ 4). 59 MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and ele- gant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. Let a > −3/4 be a real number. Calculate . a2 . Show that s r s r 3 a + 1 a + 3 4a + 3 3 a + 1 a + 3 4a + 3 + + − 2 6 3 2 6 3 is an integer and determine its value. The source of the proposals will appear when the solutions be published. α. sin(kβ + α) 22. a3 . Proposals are always wel- comed. Proposals 21. β ∈ R. Let a1 . . . 1 + a21 + a22 a1 + a2 + 1/a4 1 + a2 (a1 + a3 ) . . . . a1 + a2 + 1/a4 . 2 + 1/a24 a2 + a3 + 1/a4 . . . 1 + a2 (a1 + a3 ) a2 + a3 + 1/a4 1 + a22 + a23 . . 20z − 16y 2 = 9. . y. b. z) of real numbers such that 12x − 4z 2 = 25. c be the lengths of the sides of a triangle ABC with inradius r and cicumradius R. p2 . Let a. 24. pn and |A| > 3 · 2n + 1. 24y − 36x2 = 1. Find all triples (x. then show that it contains one subset of four distinct elements whose product is the fourth power of an integer. Let A be a set of positive integers. . 25. Prove that s r 3 a b c 1 ≤ ≤ 2r + R b + c + 2a c + a + 2b a + b + 2c 4 . . If the prime divisors of elements in A are among the prime numbers p1 . 23. A regular convex polygon of L + M + N sides must be colored using three colors: red. L + M ≥ N.60 Solutions 16. then y = 0 and 3x = 5y. Indeed. Viveiro. We claim that these conditions that are necessary are also sufficient. L + N > M > 0 and M + N > L > 0. 2 Also solved by Bruno Salgueiro Fanego. 1. A number of three digits is written as xyz in base 7 and as zxy in base 9. We distinguish two cases: (a) K is even. We begin coloring in red sides first. On the other hand. M≤ . 5. In the first case. 3. L + M > N > 0. (XI Spanish Math Olympiad (1973-1974)) Solution by José Luis Dı́az-Barrero. 2. we obtain the number 000 which does not have three digits. From the preceding. Spain. y. then it must be K K K L≤ . third. x · 72 + y · 7 + z = z · 92 + y · 9 + x from which follows 8(3x − 5z) = y. etc.. Barcelona. without painting two consecutive sides of the same color. then M + N − L + 1 ≤ N + 1. (III Spanish Math Olympiad (1965-1966)) Solution by José Luis Dı́az-Barrero. then it must be K −1 K −1 K −1 0<L≤ . Finally. Therefore. N≤ 2 2 2 That is. the number zxy in base 9 represents the number z · 92 + y · 9 + x in base 10. we have x = z = 0 or x = 5 and z = 3. The number xyz in base 7 represents the number x · 72 + y · 7 + z in base 10. 17. BARCELONA TECH. 2. yellow and blue. therefore this set of consecutive sides can not be colored alternatively yellow-blue-yellow. the non consecutive L − 1 sides are colored yellow or blue until to complete the M yellow sides and the N blue sides. and our claim is proven. Give the necessary and sufficient conditions. 3. BARCELONA TECH. WLOG we can assume that L ≥ M ≥ N. 4. independently of the parity of K. and to write it in base 9 only the integers form 0 to 8 are used. Find the number in base 10. in such a way that L sides must be red. 0<M ≤ . 6. Barcelona. M yellow and N blue. . using inequalities. Then. Both numbers in base 10 are the number 248 and this is the answer. 0<N ≤ 2 2 2 That is. x. Since x. 5.. Consequently.. 6}. L + N ≥ M and M + N ≥ L. 1. from the starting point in a circular sense until complete L red sides. y. To write a number in the system of base 7 the only integers used are 0. and from the second. Spain. Since L ≥ M. we get the number 503 in base 7 and 305 in base 9. to obtain a colored polygon with no two consecutive sides of the same color. Spain. fifth. (b) If K is odd.. z ∈ {0. Let K = L + M + N. remains to be colored L − 1 non consecutive sides and K − (2L − 1) = M + N − L + 1 ≥ 1 consecutive sides. z are integers ranging form 0 to 6. 4. then we have CN = N D = x. we have AD · CN + AC · N D = AN · CD Since N is the midpoint of the arc CD. Let M and N be the midpoints of the arcs AB and CD which do not contain C and A respectively. Spain. Applying Ptolemy’s theorem to the inscribed quadrilateral ACN D. Problem 18 AN · CD = (AC + AD)x. considering the inscribed quadrilateral CN DB we have BN · CD = (BD + BC)x. 18. we have AN AP = BN BP This completes the proof. and C N D A P B M Figure 2. Barcelona. 61 2 Also solved by José Gibergans Báguena. BARCELONA TECH. Likewise. yields AN AC + AD = BN BD + BC Applying the bisector angle theorem. Spain. Barcelona. Dividing the preceding expressions. If M N meets side AB at P. 2 . Let ABDC be a cyclic quadrilateral inscribed in a circle C. Technical University of Catalonia (BARCELONA TECH). then show that AP AC + AD = BP BC + BD (IMAC 2011) Solution by Ivan Geffner Fuenmayor. then P = 0. BARCELONA TECH. . we argue by Mathematical Induction on L. there are (k −1)(n−k) points of itersection of the chord A0 Ak with the other chords. Consider a collection C of K + 1 lines each crossing R (not just touching). 4 2 2 Solution 2 by Omran Kouba. Syria. Hence. on account of A(K). . Spain. Since one draws a (K + 1)−st line `. then there are P = intersection points and L = 4 2 n n chords and the number of regions is R = + + 1. Place n points on a circle and draw in all possible chord joining these points. and let us add the n + 1st point A0 on the arc An A1 that does not contain any other point. choose some line ` ∈ C. . and so. (IMAC-2011) Solution 1 by José Gibergans Báguena. . Syria. Let us denote by an the number of disjoint regions created. Hence the number of new regions is S + 1. R0 = 0 + 0 + 1 = 1 and A(0) holds. a2 = 2 and a3 = 4. find (with proof ) the number of disjoint regions created. The chords A0 A1 and A0 An add two regions. For each L ≥ 0. 19. Let R be an arbitrary convexn region in the plane. Spain. and whenever ` crosses a line inside of R. . BARCE- LONA TECH. Clearly a1 = 1. When no lines intersect R. First. if L lines that cross R. If no three chord are concurrent. Consequently. the number of regions determined by the K + 1 lines is. . Damascus. since the circle is convex and any intersection point is determined by a n n unique 4−tuple of points. then the number of disjoint regions created is RL = L+P +1. the chord A0 Ak passes through (k − 1)(n − k) + 1 regions. RK+1 = RK + S + 1 = (K + P + 1) + S + 1 = (K + 1) + (P + S) + 1. . with P intersection points inside R. An in this order. Damascus. Barcelona. where P +S is the total number of intersection points inside R.62 Also solved by Omran Kouba. Therefore. Higher Institute for Applied Sciences and Technology. Suppose that we have an regions obtained on placing the n points A1 . drawing the chord . Hence. let A(L) be the statement L o that for each P ∈ 1. . 2. then the number of disjoint regions created inside R is RL = L + P + 1. a new region is created when ` first crosses the border of R. Finally. Barcelona. 2 . Higher Institute for Applied Sciences and Technology. To prove the preceding claim. starting outside R. Fix some K ≥ 0 and suppose that A(K) holds for K lines and some P ≥ 0 with RK = K + P + 1 regions. Let S be the number of lines intersecting ` inside R. and apply A(K) to C\{`} with some P intersection points inside R and RK = K + P + 1 regions. And for 1 < k < n. and José Gibergans Báguena. we prove that if a convex region crossed by L lines with P interior points of intersection. A(K +1) holds and by the PMI the claim is proven. 1) → R defined by f (x) = ln . BARCELONA TECH. and f is increasing and convex. n−1 X n−1 X k k+2 an = 1+ −2 3 2 k=1 k=1 n−1 n−1 X k + 1 k X k+3 k+2 = 1+ − −2 − 4 4 3 3 k=1 k=1 n+2 n n n = 1+ −2 = + +1 4 3 4 2 which is the required number of regions. Therefore. b. Barcelona. then f 0 (x) = 4 (|x| < 1) and 2k + 1 2k + 1 k=0 k=0 ∞ 00 X k(2k − 1)x2k−2 f (x) = 4 (|x| < 1). Thus n X an+1 = an + ((k − 1)(n − k) + 1) k=1 But n X n X n X ((k − 1)(n − k) + 1) = − k 2 + (n + 1) k + (1 − n)n k=1 k=1 k=1 n(n + 1)(2n + 1) n(n + 1)2 = − + − n(n − 1) 6 2 n+2 n = −2 3 2 So. x 1−x ∞ ∞ X x2k X kx2k−1 Since f (x) = 2 for |x| < 1. Spain. Prove that s 1−a 1−b 1−c 3 1+a bc 1+b ca 1+c ab ≥ 64 b+c c+a a+b (József Wildt Competition 2009) Solution by José Luis Dı́az-Barrero. 20. BARCELONA TECH. 1). Barcelona. 63 A0 Ak adds (k − 1)(n − k) + 1 new regions. Let a. 2 Also solved by José Luis Dı́az-Barrero. c be positive real numbers such that a + b + c = 1. Consider the function f : (0. f 0 (x) > 0 and f 00 (x) > 0 for all 2k + 1 k=0 x ∈ (0. . 1 1+x Spain. we have f ≤ or equiva- 3 3 lently. we get s 1/a 1/b 1/c 3 1+a 1+b 1+c ≥8 (5) 1−a 1−b 1−c 1 1 1 WLOG we can assume that a ≥ b ≥ c. We have. " 1/a 1/b 1/c # 3 3 + (a + b + c) 1 1+a 1+b 1+c ln ≤ ln + ln + ln a+b+c 3 − (a + b + c) 3 1−a 1−b 1−c Taking into account that a + b + c = 1 and the properties of logarithms. we get 1 1 1 1 1 1 g(a) + g(b) + g(c) ≥ g(a) + g(b) + g(c) b c a a b c or 1/b 1/c 1/a 1/a 1/b 1/c 1+a 1+b 1+c 1+a 1+b 1+c ≥ 1−a 1−b 1−c 1−a 1−b 1−c From the preceding and (5) we obtain s s 1/b 1/c 1/a 1/b 1/c 1/a 3 1+a 1+b 1+c 3 1+a 1+b 1+c = b+c c+a a+b 1−a 1−b 1−c s 1/a 1/b 1/c 3 1+a 1+b 1+c ≥ ≥8 1−a 1−b 1−c Likiwise.64 a+b+c f (a) + f (b) + f (c) Applying Jensen’s inequality. ≤ ≤ and g(a) ≥ g(b) ≥ a b c 1+x g(c). we get 1 1 1 1 1 1 g(a) + g(b) + g(c) ≥ g(a) + g(b) + g(c) c a b a b c and s s 1/c 1/a 1/b 1/c 1/a 1/b 3 1+a 1+b 1+c 3 1+a 1+b 1+c = b+c c+a a+b 1−a 1−b 1−c s 1/a 1/b 1/c 3 1+a 1+b 1+c ≥ ≥8 1−a 1−b 1−c Multiplying up the preceding inequalities yields. Equality holds when a = b = c = 1/3. where g is the increasing function defined by g(x) = ln . Spain. Applying 1−x rearrangement’s inequality. 2 Also solved by José Gibergans Báguena. Barcelona. BARCELONA TECH. s 1+1 1+1 1+1 3 1+a b c 1+b c a 1+c a b ≥ 64 b+c c+a a+b from which the statement follows. applying rearrangement’s inequality again. . and we are done. a2 . . . n !2 X X ≥ ak ak (x + bk ) k=1 k=1 √ √ √ √ √ √ a1 a2 an Proof. an into CBS inequality. In this note a constrained inequality is generalized and some refine- ments and applications of it are also given. A constrained inequality that can be derived immediately from the preceding result is given in the following . . (1 ≤ k ≤ n) be positive real numbers. Introduction In [1] the following problem was posed: Let a. Setting ~u = x+b1 . a2 (x + b2 ). . . b. Let x and ak . . . bk . 1. Main Results In the sequel some generalizations and refinements of (6) are given. . x+bn and ~v = a1 . c be positive real numbers such that a + b + c = 1. . . 2. . Then n ! n ! n !2 X X ak X ak ak ≥ (x + bk )2 x + bk k=1 k=1 k=1 n !4 . an (x + bn ) into CBS inequality again and we get n !2 n ! n ! X X ak X ak ≤ ak (x + bk ) x + bk k=1 k=1 k=1 from which the statement immediately follows. . . We begin with Theroem 1. 65 MATHNOTES SECTION On a Discrete Constrained Inequality Mihály Bencze and José Luis Dı́az-Barrero Abstract. x+b2 . Our aim in this short paper is to generalize it and to give some of its applications. To proveRHS inequality we set a1 a2 an p p p ~u = x+b1 . x+bn and ~ v= a1 (x + b1 ). we have n !2 n √ !2 n ! n ! X ak X ak √ X X ak = ak ≤ ak x + bk x + bk (x + bk )2 k=1 k=1 k=1 k=1 and the q LHS inequality q isqproven. x+b2 . . . . Prove that a b c 3 (ab + bc + ca) 2 + + ≥ (6) b + b c2 + c a2 + a 4 A solution to the preceding proposal and some related results appeared in [2]. " n n ! n n !# X X X X X ≥ ak y ak + ak bk z ak + ak bk k=1 k=1 k=1 k=1 k=1 Proof. bk .66 Pn Corllary 1. Then exists c ∈ (y. Let 0 ≤ y < z and ak . (1 ≤ k ≤ n) be positive real numbers. bk . y k=1 k=1 After a little straightforward algebra the statement follows and the proof is com- plete. Let ak . Corllary 2. Then n ! n ! X X ak ak (y + bk )(z + bk ) k=1 k=1 n 2 b2a−b i aj X ak Y (y + bj )(z + bi ) j i ≥ + log (y + bk )(z + bk ) (y + bi )(z + bj ) k=1 1≤i<j≤n n !4 . Let y < z and ak . From Theorem 1. n !2 Z z n X X ≥ ak ak (x + bk ) dx. Then holds: n ! n n !2 X X ak X ak ak (1 + bk )2 + ≥2 (1 + bk )2 1 + bk k=1 k=1 k=1 Proof. we have Z z X n ! n ! Z z X n !2 X ak ak ak dx ≥ dx y (x + bk )2 y x + bk k=1 k=1 k=1 !4 . z) such that n ! n ! n !2 X X ak X ak ak ≥ (y + bk )(z + bk ) c + bk k=1 k=1 k=1 n !4 . Theroem 2. (1 ≤ k ≤ n) be positive real numbers such that k=1 ak = 1. we get n ! n ! X 2 X ak ak (1 + bk ) ≥1 (1 + bk )2 k=1 k=1 and ! !2 n n X X ak ak (1 + bk )2 ≥1 1 + bk k=1 k=1 Adding up the preceding inequalities the statement follows. bk . (1 ≤ k ≤ n) be strictly positive real numbers." n n ! n n !# X X X X X ≥ ak y ak + ak bk z ak + ak bk k=1 k=1 k=1 k=1 k=1 . Setting x = 1 in Theorem 1. Applying again Theorem 2 with y = 0 and z = 1. 67 Proof. c positive numbers with sum one is ab + bc + ca ≤ 13 (a + b + c)2 ≤ 13 and corollary 4. c be positive numbers of sum one. b. Let ak (1 ≤ k ≤ n) be positive real numbers such that k=1 ak = 1. bk . Notice that this result is a generalization and refinement of the inequality posed in [1]. Prove that a b c a2 b2 c2 + + ≥ + + b(1 + b) c(1 + c) a(1 + a) b(1 + b) c(1 + c) a(1 + a) 2bc 2ca 2ab ! a(1 + c) a−c b(1 + a) b−a c(1 + b) c−b 9 + log ≥ . Then 2ai aj X a1 X a21 Y aj+1 (1 + ai+1 aj+1 −ai+1 ≥ + log a2 (1 + a2 ) a2 (1 + a2 ) ai+1 (1 + aj+1 cyclic cyclic 1≤i<j≤n 1 ≥ X X a1 a2 1 + a1 a2 cyclic cyclic Proof. Taking into account that for all a. (1 ≤ k ≤ n) be positive real numbers such that k=1 ak = 1. c(1 + a) a(1 + b) b(1 + c) 4 Proof. dt = (z −y) 0 t + bk y t + bk c + bk k=1 k=1 k=1 Putting this in Theorem 2 the inequality claimed follows and this completes the proof. (1 ≤ k ≤ n) and an+1 = a1 into the preceding corollary the statement follows. Let a. Setting bk = ak+1 . we get Pn Corllary 3. we get a b c a2 b2 c2 + + ≥ + + b(1 + b) c(1 + c) a(1 + a) b(1 + b) c(1 + c) a(1 + a) 2bc a−c 2ca 2ab ! a(1 + c) b(1 + a) b−a c(1 + b) c−b + log c(1 + a) a(1 + b) b(1 + c) . Let ak . for n = 3 we have Corllary 5. Indeed. b. Applying Lagrange’s Mean Value Theorem to the function Z x X n !2 Z z Xn !2 n !2 ak ak X ak f (x) = dt yields. Then 2ai aj n n X ak X a2k Y bj (1 + bi ) bj −bi ≥ + log bk (1 + bk ) bk (1 + bk ) bi (1 + bj ) k=1 k=1 1≤i<j≤n 1 ≥ n ! n ! X X ak bk 1+ ak bk k=1 k=1 Pn Corllary 4. Putting x = a+b+c . CRUX. References [1] G. No. . Solutions: Problem 3062. Prove that X a 3s X a (1) ≥ ≥ . c and the radii of ex-circles ra . Vol. rb (4R + r + rb ) 4r(4R + r) rb2 + (4R + r) cyclic cyclic where the notations are usual. L. b. Problem 3062. namely x y z 3 (xy + yz + zx) + + ≤ (x + y + z = 1). we have the following inequalities similar to the ones appeared in [3]. 1982. 31. 2 Setting in the expressions of x. 5 (2005) 335. b(2s + b) 2(s2 + r2 + 4rR 4s2 + b2 cyclic cyclic X ra 3s X ra (2) ≥ ≥ . Inequalities (manuscript). rb . Dospinescu. Problem 2. using the sides a. rc . ACKNOWLEDGEMENTS The authors would thank to the Ministry of Education of Spain that has partially supported this research by grant MTM2009-13272. CRUX. 32. Let ABC be a triangle. Problem 1. z the elements of a triangle ABC and applying the previous procedure new inequalities for the triangle can be derived. y = and x = a+b+c a+b+c into X X x 3 ≥ ≥ X X x xy xy y(1 + y) 4 1 + y2 cyclic cyclic cyclic cyclic the statement follows. 6 (2006) 403-404. Prove that X a 3(a + b + c) X a ≥ ≥ b(a + 2b + c) 4(ab + bc + ca) b + (a + b + c)2 2 cyclic cyclic a b c Solution. No. b. y. For instance.68 1 9 ≥ ≥ (ab + bc + ca)(1 + ab + bc + ca) 4 Finally. 337. Vol. combining the inequality posed in [1] by Dospinescu and the inequality presented in [2] by Janous. Dı́az-Barrero and G. [3] M. 1 + y2 1 + z2 1 + x2 4 two applications are given. c be positive real numbers. Brasov. Dospinescu. [2] J. Bencze. Let a.