MathproblemsISSN: 2217-446X, url: http://www.mathproblems-ks.com Volume 4, Issue 1 (2014), Pages 231-262 Editors: Valmir Krasniqi, Jos´ e Luis D´ıaz-Barrero, Armend Sh. Shabani, Paolo Perfetti, Mohammed Aassila, Mih´aly Bencze, Valmir Bucaj, Emanuele Callegari, Ovidiu Furdui, Enkel Hysnelaj, Anastasios Kotronis, Omran Kouba, Cristinel Mortici, Jozsef S´ andor, Ercole Suppa, David R. Stone, Roberto Tauraso, Francisco Javier Garc´ıa Capit´an. 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 neither 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 committee 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 June 15, 2014 Problems 88. Proposed by Hun Min Park, Korea Advanced Institute of Science and Technology, Daejeon, South Korea. Suppose that three real numbers a, b, c(0 ≤ a, b, c, ≤ 1) satisfy the following equality; X a − b a · =0 1 − ab 1 − a2 cyc Prove that X a = b = c. X (Note that means ’cyclic sum’ f (x, y, z) = f (x, y, z) + f (y, z, x) + f (z, x, y)) cyc cyc 89. Proposed by Mohammed Aassila, Strasbourg, France. Let S be the set of positive integers that does not contain the digit 7 in their decimal representation. Prove that X1 < +∞. n n∈S c 2010 Mathproblems, Universiteti i Prishtin¨ es, Prishtin¨ e, Kosov¨ e. 231 232 90. Proposed by Omran Kouba, Higher Institute for Applied sciences and Technology, Damascus, Syria. Let n be a positive integer, prove that n X √ √ . b kc . n . . (−1) . ≤ . . (i. “George Emil Palade” School. 256603.and let the sequence {bn }n≥0 be defined by bn = a2n L2n . 1). k=0 and determine the cases of equality. then n 1 ln n Wn ∼ √ 1 − 2n π where Wn := (2n−1)!! (2n)!! is Wallis product. such that the sum 2 2 2 |BP | |CP | |AP | + + b2 c2 a2 is minimal. n ∈ N.M. Romania. 95. and Neculai Stanciu. . Bulgaria. An approximation formula of Wallis product. Binzhou City. Buzˇ au. Athens. 93. B˘ atinet¸u-Giurgiu. Romania. Shandong Province. Bucharest. Calculate Z 1Z 1 ln(1 − x) ln(1 − xy)dxdy. Technical University of Cluj-Napoca. Prove that bn is squarefree. Greece. 3 2n + 1 94. for every n ∈ N \ {1}. Proposed by Anastasios Kotronis. for n ∈ N where {Ln }n≥0 is the sequence of Lucas numbers. evaluate +∞ X n+1 (−1) n=1 2n+1 x3 −1 n+1 x n tan x − x + − · · · + (−1) . 91. For x ∈ (−1. Proposed by D. “Matei Basarab” National College. bn is not a perfect square). Proposed by Ovidiu Furdui. Proposed by Li Yin. Department of Mathematics. Let {an }n≥0 be a sequence of positive integer numbers such that 5 does not divide an for all. Binzhou University. Find a point P in the plane of a given triangle ABC. Romania. where a = BC. Proposed by Sava Grozdev and Deko Dekov (Jointly). b = CA and c = AB.e. China. For all n ∈ N. Cluj-Napoca. 0 0 92. . xy + yz + zx = 3v 2 . Syria. Proof Let x + y + z = 3u. the minimum of a(x2 + y 2 + z 2 ) + 9xyz xy + yz + zx when x. Rome. y and z are nonnegative real numbers such that x + y + z = 1. z ∈ R if and only if a ≥ 1. Answer: The minimum is 2a if a ≤ 1 and a + 1 if a ≥ 1. We homogenize by writing a(x2 + y 2 + z 2 )(x + y + z) + 9xyz (xy + yz + zx)(x + y + z) which in terms of the variables (u. Tor Vergata University. Italy. As for the second case we set x = y and get (x − z)2 (2ax − 2x + az) a(x2 + y 2 + z 2 )(x + y + z) + 9xyz −a−1= (xy + yz + zx)(x + y + z) x(x + 2z)(2x + z) which is positive for any x. x+y+z . a) Let a ≥ 1. Proposed by Omran Kouba. xyz = w3 . The minimum is thus the maximum between 2a and a + 1 which is 2a if a ≤ 1 while a + 1 if a ≥ 1. We will be very pleased considering for publication new solutions or comments on the past problems. Solution 2 by AN-anduud Problem Solving Group. The standard theory says that. v. Mongolia. Ulaanbaatar. Find. once fixed the valued of (u. Solution 1 by Paolo Perfetti. Damascus. In the first case we set z = 0 and get a(x2 + y 2 ) ≥m xy and it is evident that m = 2a. v) ≥ 0 This is a linear function in w3 and then it holds if and only if it holds for the minimum value of w3 . Department of Mathematics. 81. Let m be the searched minimum. the minimum value of w3 occurs when at least one among the variables is zero or when two of the are equals. Higher Institute for Applied sciences and Technology. w) becomes a(9u2 − 6v 2 )3u + 9w3 −m≥0 9uv 2 that is 9w3 + R(u. in terms of a > 0.233 Solutions No problem is ever permanently closed. Schur’s inequality is equivalent to 9xyz x2 + y 2 + z 2 + ≥ 2(xy + yz + zx). v). we have x2 + y 2 + z 2 ≥ xy + yz + zx. From this we find that (a − 1)(x2 + y 2 + z 2 ) ≥ (a − 1)(xy + yz + zx). Determine the sequence {cm }m≥0 such that m−1 X ck lim nm F (n) − = cm n→+∞ nk k=0 where. Greece. for m = 0. Actually. Hence a(x2 + y 2 + z 2 ) + 9xyz min = a + 1. Solution 1 by Moti Levy. xy + yz + zx This inequality becomes equality when x = y = z = 31 . Moti Levy. Athens. Rehovot.234 Using the above inequality and the given condition x + y + z = 1. RIsrael. Hence a(x2 + y 2 + z 2 ) + 9xyz = 2a. the sum is considered to be 0. San Jose. Rehovot. xy + yz + zx with equality when x = y = 21 . and z = 0. (3) On the other hand we have a(x2 + y 2 + z 2 ) + 9xyz ≥ a(x2 + y 2 + z 2 ) + 9axyz (4) From (3) and (4) we have a(x2 + y 2 + z 2 ) + 9xyz ≥ 2a(xy + yz + zx) or equivalently a(x2 + y 2 + z 2 ) + 9xyz ≥ 2a. California. From (1) we have. ∞ j−1 X (−1) 1 1 1 = = 1 kn + 1 kn 1 + kn kj j=1 1 nj . the problem deals with 1 xn finding asymptotic expansion of F (n) = 0 e − 1 dx. 82. min xy + yz + zx Also solved by Arkady Alt. a(x2 + y 2 + z 2 ) + 9axyz ≥ 2a(xy + yz + zx). and the proposer. 1 Let F (n) := k≥1 (kn+1)k! . Israel. (2) Add (1) and (2) we get a(x2 + y 2 + z 2 ) + 9xyz ≥ (a + 1)(xy + yz + zx) or equivalently a(x2 + y 2 + z 2 ) + 9xyz ≥ a + 1. xy + yz + zx b) Let 0 < a < 1. USA. we have x2 + y 2 + z 2 + 9xyz ≥ 2(xy + yz + zx) (1) Also clearly.Proposed P by Anastasios Kotronis. 1) . Italy. Tor Vergata University. . 2. We have ! r ∞ r j X X X X cm (−1) 1 c m = nr+1 − F (n) − m j+1 j+1 m n n k!k n m=0 m=1 j=0 k≥1 r ∞ r p−1 X p−1 X X X X 1 (−1) 1 c (−1) m = nr+1 + − p p p p m n k!k n k!k n p=1 p=r+1 m=1 k≥1 and the limit as n → ∞ yields (−1)r k≥1 ∞ X k=1 1 . 1. k!k r+1 . 1. Department of Mathematics. 2. c1 = 2 F2 (1. . . . (kn + 1)k! n n n The sequence {cm } can be expressed by the generalized hypergeometric function. cm = (−1) m−1 m+1 Fm+1 (1.235 ∞ X k=1 ∞ ∞ j−1 X X (−1) 1 = (kn + 1)k! k j k! j=1 = k=1 ∞ X j=1 1 nj ∞ j−1 1 X (−1) .3179 Solution 2 by Paolo Perfetti. . nj k j k! k=1 Let ( {cm }m≥0 = (−1) m−1 ∞ m−1 X (−1) k m k! k=1 then ∞ X k=1 ) m≥0 1 c3 c1 c2 = + 2 + 3 + ··· . . 2. . For m = 1. By induction we suppose that k!k for any 0 ≤ m ≤ r. n m F (n) − m−1 X k=0 ck nk ! = nF (n) = k≥1 whose limit n → +∞ is evidently c1 = cm = ∞ X (−1)m−1 k=1 nr+1 k!k m ∞ X 1 X (−1)j k!k j=0 (kn)j ∞ X k=1 1 . in particular. 1. Rome. 2. . 2. (c0 = 0) n→+∞ Let m = 1. X F (n) = k≥1 ∞ X 1 X (−1)j 1 = (kn + 1)k! k!kn j=0 (kn)j k≥1 The absolute convergence of the series allows us to take the limit n → +∞ under the sum so for m = 0 we have lim F (n) = 0. . 1) = −γ + Ei (1) ∼ = 1. (k − 1)! p=1 p! 0 (k − 1)! 0 Also solved by Moubinool Omarjee. For z ∈ Ω we consider G(z) defined by the formula G(z) = ∞ X 1 p! (p + z) p=1 Clearly. and for every nonnegative integer m. we have m−1 X ck 1 + O . F (n) = nk nm k=0 Moreover. but they can. and for every m ≥ 0 (with the same convention as in the statement of the problem. This proves that G. Editor’s Comment: The proposer of this problem indicated that it is a generalization of problem U278 of Mathematical Reflections. be expressed as integrals: Z Z ∞ (−1)k−1 X 1 ∞ k−1 −pt (−1)k−1 ∞ k−1 e−t ck = t e dt = t e − 1 dt.Daejeon.) we have zG(z) = m−1 X ck z k + O(z m ) k=0 with c0 = 0 and ck = G(k−1) (0)/(k − 1)! when k is a positive integer. . Let {an } strictly increasing sequence of positive integers such that gcd(ai . we have G (m) (z) = ∞ X (−1)m m! p! (p + z)m+1 p=1 In particular. France. Syria. But F (n) = 1 1 n G( n ).236 Solution 3 by Omran Kouba. for |z| < 1 we have G(z) = ∞ X G(m) (0) m z m! m=0 Thus for z in the neighborhood of 0. Let bn = an+1 − an . j(i < j). Higher Institute for Applied sciences and Technology. is analytic in Ω and that. Prove that the sequence {bn } is unbounded(has no upper bound). alternatively. Let Ω = {z ∈ C : <z > −1}. Korea Advanced Institute of Science and Technology. itself. according to (1) we have ck = (−1)k−1 ∞ X 1 . this is a series of analytic functions on Ω (namely: z 7→ 1/(k! (z + k)). Note that the ck ’s do not seem to have a closed form. Damascus.Proposed by Hun Min Park. 83.) that converges to G uniformly on every compact subset of Ω. Paris. South Korea. p! pk p=1 This yields the desired conclusion. for every m ≥ 0 and every z ∈ Ω. so for large n. aj ) = 1 for any i. and the proposer. n ln n (a − a ) ≥ 1 = n. 1 π (X) ≥ ln X. Syria. Let X be a positive integer. Let us define the sequence {pn }. By definition of A (x) .) Thus. which are less or equal to x. which leads us to the conclusion that the sequence {bn } is unbounded. Since gcd(ai . Israel (Jointly). . n=1 Suppose. Now. aA(X)+1 − a1 ≥ X − 2 2 Clearly. Then aA(X)+1 − a1 ≤ A (X) M. . Damascus. A(X) X bn = aA(X)+1 − a1 . which are less or equal to x. X/ ln X 2M π(X) On one hand. . By the definitions above. Since ai and aj are relatively prime (for i 6= j) then pi 6= pj . a ≥ .we have aA(X)+1 > X. there exists an absolute positive constant A such that π(x) ≤ A lnxx hence an an n ≤ π(an ) ≤ A ≤A ln an ln n where with a0 = 0. where pn is the largest prime number that divides an . A (x) ≤ P (x) ≤ π (x) . i. 2 By equations ((1)) and (2). hence X X = . according to the weak form of the prime number theorem. (2) π (X) A (X) 1 ≥ ≥ X X 2M Multiplying both sides by ln X > 0. . an ] and consequently. (3) n→∞ n . Higher Institute for Applied Sciences and Technology. where π(x) represents the number of primes p that are smaller or equal to x.e. Solution 2 by Omran Kouba. Rehovot. pn are distinct primes from the interval [1. (1) where π (x) is the number of prime numbers less or equal to x. we have an = Pn we used the trivial Pn inequality n ≤ an (since. X ≤ aA(X)+1 − a1 ≤ A (X) M..237 Solution 1 by Shmuel Isaac and Moti Levy. that the sequence {bn } is bounded by the positive constant M > 0. X ≥ 2a1 . π(an ) ≥ n. on the contrary. Let A (x) be the number of all {an } sequence terms. the Prime Number Theorem states that limX→∞ X/ ln X = 1 but on 1 the other hand limX→∞ 2M ln X = ∞. This is a contradiction. aj ) = 1 for every distinct i and j we conclude that p1 . and aA(x)+1 > x. aA(x) ≤ x. Let P (x) be the number of {pn } sequence terms. p2 . Let pn be the smallest prime number that divides an . and in particular k k−1 n k=1 k=1 A an lim = +∞. and the proposer. ln n Solution 3 by Paolo Perfetti. The generalized hyperbolic cosine function is defined as coshp (x) = sinh0p (x). so the problem remains open. . Binzhou City. The sequence {aj } with the smallest entries aj for any j is the sequence of the prime numbers. Also solved by Arthur Handle. Any other sequence {a0j } is such that a0j ≥ aj and then asymptotically a0j ≥ c0 j ln j for any j large enough. n ln n and lim sup n→∞ bn > 1. = arcsinp (1) = p 1/p 2 0 (1 − t ) πp The inverse of arcsinp (x) on 0. China. the generalized inverse hyperbolic sine function Z x 1 dt . Let 1 < p < ∞. This implies that n X aj+1 − aj = an+1 − a1 ≤ nc1 k=1 but this contradicts a0j ≥ c0 j ln j if n is large enough. 0) p generalizes the classical inverse hyperbolic sine function. The generalized cosine function is defined as cosp (x) = dx sinp (x). prove that ln x sinp x − x cosp x < sinp x p sinp x and ln sinhp x x coshp x − sinhp x > . x ∈ [0. Tor Vergata University. Similarly.Proposed by Li Yin. We argue by contradiction by assuming that there exists a constant c1 such that bn ≤ c1 . For p > 2 and π x ∈ (0. Refining upon the proof presented above we see that we have lim inf n→∞ an ≥ 1. This contradiction proves that {bn } cannot be bounded. Department of Mathematics. 2 is called the generalized sine function and d denoted by sinp . 2p ). ∞). 0≤x≤1 arcsinp (x) = (1 − tp )1/p 0 and Z 1 πp 1 dt. The inverse of arcsinhp is called the generalized hyperbolic sine function and denoted by sinhp . Shandong Province. Binzhou University. suppose that the sequence {bn } is bounded by some constant M then it follows immediately that n−1 an − a1 1X = bk ≤ M n n k=1 and consequently lim supn→∞ ann ≤ M which is a absurd according to (3). Department of Mathematics and Information Science. x ∈ (−∞. Rome.238 Now. Remark. (1 + tp )1/p arcsinhp (x) = 0 −arcsinh (−x) . Italy. we can generalize the inverse of arcsin as follows: Z x 1 dt. x p sinhp x One incomplete solution was received. 84. Rehovot. p > 0. Using the following well known indentities L1 + L2 + · · · + Ln = Ln+2 − 3 and F1 + F2 + · · · + Fn = Fn+2 − 1 we have !3 n X n X F2 · Lk ! n X F3 k k=1 k=1 k L2k k=1 Lk ! ≥ n X !5 Fk . for 1 ≤ k ≤ n. respectively Ln represents the nth Fibonacci number respectively the nth Lucas number.239 85. we use the inequality (3) due to J. Romania. Lm Lpk (Ln+2 − 3) k k=1 k=1 To prove (2). “George Emil Palade” School. yk > 0. B˘ atinet¸u-Giurgiu. n X xp+1 k k=1 ykp Pn p+1 ( k=1 xk ) ≥ Pn p . A more general version of this problem appeared in ”The Fibonacci Quarterly”. Bucharest. and Neculai Stanciu. we get n X !3/5 n X F2 k · Lk k=1 = k=1 n X 3/5 Lk 53 n X k=1 · n X F3 !1/5 k k=1 L2k 2 1/5 !5 1/5 3 1/5 !5 1/5 n n X X F F k k · · Lk L2k k=1 k=1 ! n 3/5 X F Fk .Proposed by D. Buzˇ au. ( k=1 yk ) (3) By Radon’s inequality. and xk ≥ 0.M. p > 0. November 2013. ! n ! n m+p+2 X X F p+1 Fkm+1 (Fn+2 − 1) k ≥ m > 0. Solution 1 by AN-anduud Problem Solving Group. Romania. Mongolia. Applying the H¨older’s inequality. (1) k=1 It is enough to prove the above inequality. Radon. Solution 2 by Moti Levy. · k2/5 = Lk k=1 !3/5 2/5 3/5 Lk · Lk k=1 ≥ !1/5 Fk 1/5 Lk Hence (1) is proved. Ulaanbaatar. Prove that n n X Fk2 X Fk3 (Fn+2 − 1)5 · ≥ Lk L2k (Ln+2 − 3)2 k=1 k=1 where Fn . Let n be a positive integer. Volume 51. Israel. (2) m+p . n X F m+1 k=1 ! k n X F p+1 Lm k k=1 k Lpk ! Pn p+m+2 ( k=1 Fk ) ≥ Pn p+m ( k=1 Lk ) (4) . Number 4. “Matei Basarab” National College. bn we have: n n 2(1−λ)p X X ak a2λp (a1 + · · · + an )2p k ≤ · . Then for every positive numbers a1 . . 23 . the desired inequality is obtained. Let x1 . . . . 2µ(p−1) 2(1−µ)(p−1) (b1 + · · · + bn )2p−2 k=1 bk k=1 bk Proof. Let p. . an and b1 . µ < 1. . . with xk = a/ a ˜ and yk /˜b with a ˜ = a1 + · · · + an and ˜b = b1 + · · · + bn . we obtain ! ! n n X X (a1 + · · · + an )5 a3k a2k ≤ · . . n X k=1 n X Fk = Fn+2 − 1 (5) Lk = Ln+2 − 3 (6) k=1 The required result is obtained by substituting (5) and (6) in (3). . Applying this. λ. (b1 + · · · + bn )3 b2k bk k=1 k=1 Finally. . . k=1 n X k=1 Lk = k=1 n X k=1 (Lk+2 − Lk+1 ) = Ln+2 − L2 = Ln+2 − 3. bk = Lk and noting that n n X X Fk = (Fk+2 − Fk+1 ) = Fn+2 − F2 = Fn+2 − 1. by setting ak = Fk . Higher Institute for Applied Sciences and Technology. Damascus. µ be real numbers with p > 1. λ. . . xn and y1 . µ) = 52 . . and 0 < λ. Syria. . We will use the following lemma: Lemma. then by H¨older’s inequality we have ! p1 ! r1 ! p1 n n n X X X xpk xpk 1/r · yk ≤ yk = p/r y p−1 k=1 k=1 k k=1 yk That is. . µ < 1. yn be positive numbers such that n n X X xk = yk = 1 k=1 1 p Let r > 1 be defined by 1= n X xk = n X xk 1/r k=1 k=1 + yk k=1 1 r = 1. 2µ(p−1) 2(1−µ)(p−1) (b1 + · · · + bn )2p−2 k=1 bk k=1 bk Taking. 35 . we obtain n n 2(1−λ)p X X ak a2λp (a1 + · · · + an )2p k ≤ · . by the Cauchy-Schwarz inequality: 1≤ n X xpk y p−1 k=1 k = n X xλp k µ(p−1) y k=1 k · (1−λ)p xk (1−µ)(p−1) yk v u n uX ≤t k=1 x2λp k 2µ(p−1) · yk n X 2(1−λ)p xk 2(1−µ)(p−1) k=1 yk for 0 < λ. Solution 3 by Omran Kouba. .240 The following sums are well known. (p. 241 Remark. VA. 86.Proposed by Ovidiu Furdui. Technical University of Cluj-Napoca. Department of Mathematics. Tor Vergata University. we have ! ! n n X X Fkm−p Fkp (Fn+2 − 1)m ≥ . ´ Also solved by G. Cluj. and the proposer. Let x = t2 . The integral reads as ! r Z 1 Z 1 Z 1 p 1 2 ln tdt + 2 − 1 dt ln 1 + 2 ln(t + 1 − t )dt = 2 t2 0 0 0 Z 1 . University of De Las Palmas. Angel Plaza. Romania. Newport News. Grain Canaria. · m−q q−2 (Ln+2 − 3)m−2 L L k=1 k k=1 k with the same proof. C. x 0 Solution 1 by Paolo Perfetti. Rome. Similarly. Italy. for m > 2 and 0 < q. USA. Calculate √ √ Z 1 ln( x + 1 − x) √ dx. Greubel. p < m. Spain. 1 . ln t dt = (t ln t − t). = −1 0 0 p 2 In the second integral we change t = 1/ 1 + y and it becomes Z ∞ Z ∞ y 1 ln(1 + y) . . ∞ 1 p p ln(1 + y) dy = − dy . z2 = (−1 − 2)/2. 0 (t ← π − t) 2 . The change of variables x = sin2 t shows that x Z π/2 I=2 ln (sin t + cos t) cos t dt. where z1 = (−1 + √ Solution 2 by √ Arkady San Jose. R 1 ln x + 1 − x √ Let I := 0 dx. USA. We get evidently √ √ √ −1 1 − z1 −1 2 − 2 −1 ln(3 + 2 2) √ √ ln √ √ √ = ln = ln(3 − 2 2) = 2 1 − z2 2 2+ 2 2 2 and the integral finally is √ √ 2 ln(3 + 2 2) − 2. California. √ Alt. 0 Z =2 π/2 ln (sin t + cos t) sin t dt. + 2 )3/2 2 1 + y 0 (1 + y 1+y 1 + y2 0 0 Z ∞ Z ∞ 1 dt = 2 dz = 2 |{z} |{z} 1 + sinh t z + 2z − 1 1 0 y=sinh t z=et Z ∞ 1 1 1 − dz z − z z − z z − z2 1 2 1 1 √ 2)/2. I= 0 √ Z = 2 ln √ 2 cos θ cos θ dθ (t ← −π/4 √ = π/4 2 (ln 2) 2 Z π/4 √ π/4 {z I1 } | Z iπ/4 I2 = sin θ ln(cos θ) + √ ln(cos θ) cos θ dθ {z I2 } 2. b. Syria. z be positive real number such that x2 + y 2 + z 2 = 3. The proposed problem is equivalent to the following problem. Omran Kouba. Greece. Newport News. I1 = √ Z cos θ dθ + 2 π 4 (u = sin θ) Also solved by Albert Stadler. Higher Institute for Applied Sciences and Technology. Let a. Rehovot. University of Prishtina. and π/4 Z −π/4 √ 1/ 2 −π/4 sin2 θ dθ cos θ 2 1 u 2 ln √ + 2 du 1 − u2 2 0 √ √ Z 1/ 2 √ 1 1 2 =− ln 2 − 2 + + du 2 1+u 1−u 0 √ √ √ 2 2+1 =− ln 2 − 2 + ln √ 2 2−1 √ √ √ 2 =− ln 2 − 2 + 2 ln( 2 + 1) 2 √ √ Finally I = −2 + 2 2 ln( 2 + 1). and the proposer. = + θ) −π/4 −π/4 | Clearly. c be positive real numbers such that a + b + c = 3. Moti Levy. France. Greubel. y. Israel. (2) . G. Damascus. Ulaanbaatar. Mongolia. 87. AN-anduud Problem Solving Group. Prove that x+y y+z z+x + + ≥ 3. Athens. VA. 1 + ca 1 + ab 1 + bc Solution 1 by AN-anduud Problem Solving Group. USA. x. we have 2 x+y y+z z+x + + ((x + y)(1 + xy)2 + (y + z)(1 + yz)2 + (z + x)(1 + zx)2 ) 1 + xy 1 + yz 1 + zx ≥ ((x + y) + (y + z) + (z + x))3 = 8(x + y + z)3 . Paris. (1) 1 + xy 1 + yz 1 + zx Applying H¨ older’s inequality. Republic of Kosova. C. Anastasios Kotronis.242 Taking the half sum we obtain Z π/2 ln (sin t + cos t) (cos t + sin t)dt.Proposed by Dorlir Ahmeti. Switzerland. Moubinool Omarjee. Prove that √ √ √ √ √ √ a+ b b+ c c+ a √ + √ + √ ≥ 3. Department of Mathematics.243 Thus (1) would follow from (2) If we prove the next inequality 8(x + y + z)3 ≥ 9((x + y)(1 + xy)2 + (y + z)(1 + yz)2 + (z + x)(1 + zx)2 ) or equivalntly 9(x5 + y 5 + z 5 ) + 48xyz ≥ 25(x3 + y 3 + z 3 ). Italy. Rome. Tor Vergata University. . To prove (3) we note that X X X 9 x5 + 48xyz − 25 x3 = (x + y)(x − y)2 (x + y − z)2 cyc cyc (3) cyc + X z(x − y)2 ((x + y − 6z)2 + 35xy) cyc + 25xyz X (x − y)2 ≥ 0 cyc Also solved by Paolo Perfetti. and the proposer. a1 . Proposals 61. 62. b1 . Let A(x) be a polynomial with integer coefficients such that for 1 ≤ k ≤ n + 1. . 65. · · · . P Y and P Z to the sides BC. Draw perpendiculars P X. Proposals are always welcomed. Compute the value of BX + CY + AZ PX + PY + PZ 63. respectively. and 6 of five grams. then prove that ! ! n n n X 1 n+1 X n2 2n − 2 X 2 2 Re a k bk ≤ n |ak | + |bk | 2 n n−1 n k=0 k=0 k=0 . an and b0 . Let a0 . . Let P be an interior point to an equilateral triangle ABC. 5 of two grams. The source of the proposals will appear when the solutions be published. . CA and AB. 9y 2 − 4x2 = 60.244 MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. How many ways are there to weigh of 31 grams with a balance if we have 7 weighs of one gram. . If n ≥ 2. holds: A(k) = 5k Find the value of A(n + 2). respectively? 64. bn be complex numbers. Find all real solutions of the following system of equations: p p x2 + y 2 + 6x + 9 + x2 + y 2 − 8y + 16 = 5. we have n m+1 n −(m+1) 1 1 1 1 1+ 1− <1⇔ 1+ 1+ <1 n m+1 n m a1 a2 . 50 ≡ 1 (mod 13). 51 ≡ 5 (mod 13). for k = 0 is 1. n where a1 . 31 ≡ 3 (mod 13). What are the smallest and the biggest? How many are there in total? (50th Catalonian Mathematical Olympiad) Solution by Eloi Torrent Juste. Find all positive integers n smaller that 201314 such that 3n ≡ 3 (mod 13) and 5n ≡ 5 (mod 13). . statement 1 ≤ 12k + 1 ≤ 201314 and therefore 0 ≤ k ≤ 12 The smallest positive integer. 32 ≡ 9 (mod 13). Therefore. . Spain. BARCELONA TECH. . AULA Escola Europea. . 2 Also solved by Jos´ e Luis D´ıaz-Barrero. Thus. an are positive numbers not all the same. So. According to the 201313 or 0 ≤ k ≤ 16776. a multiple of 12 and n = 12k + 1. etc. powers of three when divided by 13 present a cycle of order 3. and so forth. Spain. Barcelona. The biggest. BARCELONA TECH. Prove that n m+1 1 1 1+ < 1+ n m (Training Sessions of Catalonian Team for OME 2013) Solution by Jos´ e Luis D´ıaz-Barrero. n − 1 must be multiple of 3 and 4 at the same time. . + an . 57. Barcelona. That is. namely √ n a1 + a2 + . 53 ≡ 8 (mod 13). for k = 16776 is 201313 and the total number is 16776 + 1 = 16777. . . Applying AM-GM inequality. Spain. powers of five when divided by 13 present a cycle of order 4. an < and the statement immediately follows. a2 .245 Solutions 56. we have s n m+1 1 1 n+m+1 1+ 1− n m+1 1 1 n 1+ + (m + 1) 1 − n m+1 < =1 n+m+1 From the preceding. The integers n searched are of the form n = 3k + 1 and at the same time of the form n = 4h + 1. Barcelona. 54 ≡ 1 (mod 13). 33 ≡ 1 (mod 13). . We have that 30 ≡ 1 (mod 13). 52 ≡ 12 (mod 13). Let n. Likewise. m be positive integers. . Italy. Also solved by Arkady Alt. 58. and let a be a positive real number. we have a < ln(1 + a) < a. cos C) in identity cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = 1 with x y z . Find the minimum value of x4 + y 4 + z 4 + x2 y 2 z 2 (Training Catalonian Team for OME 2014) Solution 1 by Arkady Alt.246 Solution 2 by Omran Kouba. there is a real number ξ ∈ (0. California. C are the measures of the angles of an acute triangle ABC. Paolo Perfetti. By replacing (cos A. Let f be the function defined for x > −1 by f (x) = ln(1 + x). B. USA. Damascus.1 . USA. Rome. a a 1+ξ 1+a that is. . BARCELONA TECH. Using the Mean Value Theorem. that is x. for a > 0. . we obtain 2 2 2 x2 y2 z2 x y z + + +2· · · =1 4 4 4 2 2 2 Thus x2 + y 2 + z 2 + xyz = 4 Since triangle ABC is acute. Spain. a) such that ln(1 + a) f (a) − f (0) 1 1 0 = = f (ξ) = ∈ . California. and Jos´ e Gibergans-B´ aguena. y. Let x = 2 cos A. San Jose. San Jose. Syria. Since the lower bound 4 can be attained if x = y = z ⇐⇒ A = B = C. Department of Mathematics. Tor Vergata University. y = 2 cos B and z = 2 cos C. where A. the desired minimum is 4. cos B. 1+a Applying the upper inequality with a = 1/n and the lower one with a = 1/m. Barcelona. z > 0. then by QM-AM inequality we have 2 2 x4 + y 4 + z 4 + x2 y 2 z 2 x + y 2 + z 2 + xyz ≥ = 1 =⇒ x4 + y 4 + z 4 + x2 y 2 z 2 ≥ 4 4 4. Higher Institute for Applied Sciences and Technology. we get 1 1 < 1 < (m + 1) ln 1 + n ln 1 + n m Taking exponentials yields the desired inequality. Damascus. Solution 4 by Paolo Perfetti. We employ the so called ”uvw” theory which can be found at The art of problem solving forum. Italy. Remark. the latter occurs when x = y (or cyclic). It is a known standard result that 1 ≤ cos A + cos B + cos C ≤ 3/2 and then 2 ≤ 2 cos A + 2 cos B + 2 cos C ≤ 3. The answer is 4 and it is attained only when the triangle is equilateral. Define three new quantities x + y + z = 3u. Tor Vergata University. Syria. Higher Institute for Applied Sciences and Technology. the minimum value of the expression claimed is 4 and it is attained when 4ABC is equilateral. 4 4 At x + y fixed. xyz = w3 We have x4 +y 4 +z 4 +x2 y 2 z 2 = (w3 )2 +12uw3 +81u4 −108u2 v 2 +18v 4 = (w3 )2 +12uw3 +R(u. x4 + y 4 + z 4 + x2 y 2 z 2 − 4 = x4 + y 4 + z 4 + x2 y 2 z 2 − 2(x2 + y 2 + z 2 + xyz) + 4 = (x2 − 1)2 + (y 2 − 1)2 + (z 2 − 1)2 + (xyz − 1)2 ≥ 0 with equality if and only if x2 = y 2 = z 2 = xyz = 1. note as in the preceding solutions we have x2 + y 2 + z 2 + xyz = 4 Thus.247 Solution 2 by Jos´ e Luis D´ıaz-Barrero BARCELONA TECH. xy + yz + zx = 3v 2 . Clearly we have 2 ≤ x + y + z ≤ 3. It follows that the minimum of the parabola occurs when w = 0 or when w is minimum once fixed the values of u and v. then we have x2 + y 2 + z 2 + xyz = 4 cos2 A + 4 cos2 B + 4 cos2 (A + B) − 8 cos A cos B cos(A + B) = 4(cos2 A + cos2 A − cos2 A cos2 A + sin2 A sin2 B) h i = 4 sin2 B(cos2 A + sin2 A) + cos2 B = 4 Taking into account AM-QM inequality yields r x4 + y 4 + z 4 + x2 y 2 z 2 x2 + y 2 + z 2 + xyz 1= ≤ 4 4 from which follows x4 + y 4 + z 4 + x2 y 2 z 2 ≥ 4. v) This is a convex increasing parabola if w3 ≥ 0 whose minimum has negative abscissa. The minimum 1 corresponds to a degenerate isosceles triangle while the maximum to an equilateral triangle. If w = 0 we have for instance z = 0 that is C = π/2. Department of Mathematics. According to the theory. or equivalently A = B = C = 60◦ . First. So. Spain. Barcelona. √ the minimum of x + y . Note that the condition that ABC is acute is unnecessary. Solution 3 by Omran Kouba. This yields x4 + y 4 = 2x4 = 8 . occurs when x = y that is A = B = π/4 or z = y = 2. Rome. Since A + B + C = π. BARCELONA TECH. Barcelona. More generally. and let µ = n1 (z1 + . Clearly we have n X |zk − µ|2 = k=1 = n X k=1 n X |zk |2 − 2 n X <(zk µ) + k=1 n X |µ|2 k=1 n X |zk |2 − 2n|µ|2 + n|µ|2 = k=1 |zk |2 − n|µ|2 k=1 That is n X k=1 |zk |2 = n|µ|2 + n X k=1 |zk − µ|2 (1) . Prove that −1 1 1 1 81 + + ≤ 3|x+y +z|2 +|2x−y −z|2 +|2y −x−z|2 +|2z −x−y|2 |x|2 |y|2 |z|2 (Training Catalonian Team for OME 2014) Solution 1 by Omran Kouba. . 6a3 y = 6a2 ay ≤ 54ay 1011y 2 + 693a2 > 1674ay = (1620 + 54)ay so we get 324y 4 + 442y 2 ≥ 108ay 3 + 27a2 y 2 and this is implied by 324y 4 + 442y 2 ≥ 108 · 3y 3 + 27 · 9y 2 ⇐⇒ 324y 4 + 199y 2 ≥ 324y 3 and this finally follows by the AGM 324y 4 + 199y 2 ≥ 507y 3 . consider n nonzero complex numbers z1 . .248 If z = y and x = a − 2y. Spain. x. Also solved by Jos´ e Gibergans-B´ aguena. a) 3 3 729 (3y − a)2 = (324y 4 − 108ay 3 + 1458y 2 − 1620ay + 702a2 − 27a2 y 2 − 6a3 y − a4 ) 729 Now we prove that P (y. a) > 0. Let x. + zn ). y. Syria. Damascus. . . a ≤ 2 ≤ 3 we get a 4 a 6 (3y − a)2 x4 + y 4 + z 4 + (xyz)2 − 3 − = P (y. 4} = 4. z be nonzero complex numbers. . Higher Institute for Applied Sciences and Technology. z can be equal if and only if A = B = C = π/6 whence x = y = z = 1 and x4 + y 4 + z 4 + (xyz)2 − 4 ≥ 0 The searched minimum is thus min{8. . We have showed that a 4 a 6 − ≥0 x4 + y 4 + z 4 + (xyz)2 − 3 3 3 and the difference equals zero if x = y = z = a/3. y. 59. zn . Indeed 702a2 − a4 = a2 (702 − a2 ) ≥ a2 (702 − 9) = 693a2 . On account of their definition. the HM -AM inequality shows that !−1 n n X X 1 ≤ n2 |zk |2 |zk |2 (2) k=1 k=1 Combining (1).249 Now. and rearranging. we obtain . (2). 2 . . . 2 !−1 . n n n . . X . X . X X . 1 . . 4 . zj . . n ≤ n. . zk . + . (n − 1)zk − . . |zk |2 . . BARCELONA TECH. Barcelona. Spain. Solution 2 by Jos´ e Luis D´ıaz-Barrero. z2 . We have . k=1 k=1 k=1 j6=k The proposed inequality corresponds to n = 3 and (z1 . z). z3 ) = (x. y. . . . . . . . ! . x + y + z . 2 1 . 2x − y − z . 2 . 2y − x − z . 2 . 2z − x − y . 2 . . + . . +. . +. . . . . . . . . 3 3 . yields 1 3 |x|2 + |y|2 + |z|2 ≥ 1 1 1 3 + 2+ 2 2 |x| |y| |z| and taking into account that the given inequality is equivalent to 3 1 ≤ (|x + y + z|2 ) 1 1 1 9 + 2+ 2 |x|2 |y| |z| 1 + |2x − y − z|2 + |2y − x − z|2 + |2z − x − y|2 27 from which the statement follows. 2 |2x − y − z| = (2x − y − z) (2x − y − z) 2 2 2 = 4 |x| + |y| + |z| − 2 (xy + xy) + (yz + yz) − (zx + zx) . San Jose. Notice that equality holds when |x| = |y| = |z| and we are done. 2 |2z − x − y| = (2z − x − y) (2z − x − y) 2 2 2 = |x| + |y| + 4 |z| − 2 (zx + zx) + (xy + xy) − (zx + zx) . California. = Solution 3 by Arkady Alt. . USA. 2 |2y − x − z| = (2y − z − x) (2y − z − x) 2 2 2 = |x| + 4 |y| + |z| − 2 (yz + yz) + (zx + zx) − (xy + xy) . 3 3 3 (x + y + z)(x + y + z) 9 1 (2x − y − z)(2x − y − z) (2y − x − z)(2y − x − z) (2z − x − y)(2z − x − y) + + + 3 9 9 9 1 2 2 2 = |x| + |y| + |z| 3 Applying AM-HM inequality. Since 2 |x + y + z| = (x + y + z) (x + y + z) 2 2 2 = |x| + |y| + |z| + (xy + xy) + (yz + yz) + (zx + zx) . + i1 i2 1 1 · 2···n + 1 X 1≤i1 <. + i1 i2 X 1≤i1 <.<in ≤n+1 =n+1− 1 i1 i2 . + i1 i1 i2 1 · 2···n + 1 1≤i1 ≤n+1 = n+1 Y 1+ k=1 1 k 1≤i1 <i2 ≤n+1 = n+1 Y k=1 k+1 (n + 2)! = =n+2 k (n + 1)! Then X 1≤i1 ≤n+1 1 + i1 X 1≤i1 <i2 ≤n+1 =S− 1+ 1 + ...250 then 2 2 2 2 3 |x + y + z| + |2x − y − z| + |2y − z − x| + |2z − x − y| 2 2 2 2 2 2 2 2 2 = 3 |x| + |y| + |z| + 6 |x| + |y| + |z| = 9 |x| + |y| + |z| and the original inequality becomes 1 81 2 |x| + 1 |y| 2 + !−1 1 |z| 2 or 1 9 2 |x| + 1 2 |y| + !−1 1 2 |z| 2 2 2 ≤ 9 |x| + |y| + |z| 2 2 2 ≤ |x| + |y| + |z| . Compute the following sum: X X 1 + i1 1≤i1 <i2 ≤n+1 1≤i1 ≤n+1 1 + . BARCELONA TECH..<in ≤n+1 1 i1 i2 . in (Training Spanish Team for VJIMC 2014) Solution 1 by Jos´ e Gibergans-B´ aguena. . in 1 (n + 1)! 2 and we are done. . . . 60.. Barcelona..... Spain. USA.. Spain. Let us denote by S the following sum: X X 1 1 1 S =1+ + + . Solution 2 by Arkady Alt.. Barcelona. where latter inequality holds because by Cauchy’s inequality ! 1 1 1 2 2 2 |x| + |y| + |z| ≥9 2 + 2 + 2 |x| |y| |z| Also solved by Jos´ e Gibergans-B´ aguena. BARCELONA TECH. California. Consider the polynomial n+1 n Y X 1 P (x) = x+ = xn+1 + an+1 + ak xn+1−k k k=1 k=1 . San Jose. . + an = P (1) − 1 − . independently. Rome. . Barcelona. Let Sn be the proposed sum and let Nn+1 = {1. and Angel Plaza (jointly). (n + 1)! (n + 1)! From the other hand n+1 n+1 Y Y k+1 1 P (1) = = =n+2 1+ k k Since an+1 = k=1 k=1 Hence. University of Las Palmas. . Gran Canaria. . .. n + 1}. . One from Angel Plaza. Syria. Valencia. k=1 k=0 X⊂Nn+1 |X|=k j∈X that can be proved by induction. and another from Yiri D. 1≤k ≤n+1 i1 i2 . Himar A. Spain. . Fabelo (students). . + an = (n + 2) − 1 − 1 1 =n+1− .<ik ≤n+1 1 . Editor’s Comment: On the last issue we forgot to acknowledge two more solutions of Problem 55 of the MathContest section we received. (n + 1)! (n + 1)! Solution 3 by Omran Kouba. We will use the following formula: n+1 n+1 Y X X Y (1 + bk ) = bj . a1 + a2 + . . Tor Vergata University. . . BARCELONA TECH. Italy and Jos´ e Luis D´ıaz-Barrero. 2. Higher Institute for Applied Sciences and Technology. Department of Mathematics. ik 1 1 then original sum is a1 + a2 + . . Spain. Taking bj = 1/j and noting that terms corresponding to X = ∅ and X = Nn+1 are missing in the proposed sum. Damascus. we see that n+1 Y 1 1 1 −1− =n+1− 1+ Sn = k (n + 1)! (n + 1)! k=1 Also solved by Paolo Perfetti.251 Then by general Vieta’s Theorem X ak = 1≤i1 <i2 <. on each interval [an . Anastasios Kotronis Abstract. n→+∞ which is the desired result. we can deduce the following results: Proposition 1.M. so there exists a real number ξn between an and an+1 . n→+∞ (2) f : R∗+ → R∗+ is a monotone nondecreasing function with a continuous derivative f 0 . For n ∈ N∗ . because limn→∞ an = a. an+1 ]. on account of 1 and 2. +∞) with a continuous derivative. Main Results For this section we assume that: (1) {an }n≥1 is a positive sequence such that lim an = a ∈ R∗+ and n→+∞ lim n (an+1 − an ) = b ∈ R. if an+1 6= an the function f satisfies the assumptions of Lagrange’s theorem. Now. such that f (an+1 ) − f (an ) = (an+1 − an ) f 0 (ξn ). lim n (f (an+1 ) − f (an )) = bf 0 (a). we do not need the full strength of 2. . This remark will help us in the next proposition. (and this remains valid with ξn = an if an = an+1 . In this note we present new methods to calculate the limits of some particular sequences and their generalizations that appeared in some problem solving journals. and that (3) {xn }n≥1 is a positive sequence for which there exists a number t ∈ R such that xn+1 lim = x ∈ R∗+ . 1.Ba ¸ u-Giurgiu.) thus n (f (an+1 ) − f (an )) = n (an+1 − an ) f 0 (ξn ). Neculai Stanciu. we have limn→∞ ξn = a. Note that. lim n (f (an+1 ) − f (an )) = lim n (an+1 − an ) · lim f 0 (ξn ) = bf 0 (a). n→+∞ Proof.252 MATHNOTES SECTION Calculating the limits of some real sequences ˘ tinet D. we only need that f be a real valued function defined on (0. n→+∞ n→+∞ Remark. and f 0 is continuous. therefore. n→+∞ xn nt+1 With the above assumptions. to the function ln(f ) to conclude that f 0 (a) . n→+∞ lim unn = e. then (n + 1)t nt n+1 lim Cn = xe−(t+1) n→+∞ . thus So. n vn 1 √ · · n+1 vn+1 unn = 1 + n vn+1 and limn→∞ unn = e follows also from Proposition 3. using 3. n→+∞ Proof. n→+∞ and lim n(un − 1) = 1. √ xn+1 nt · for n ≥ 2. Proof. Cn = √ √ n x xn+1 n − . According to the preceding remark. lim n (ln f (an+1 ) − ln f (an )) = b Proposition 3. n→+∞ Proposition 5. limn→+∞ f (an+1 ) f (an ) n =e bf 0 (a) f (a) . lim n→∞ n→∞ vn for every positive sequence {vn }n≥1 . If for n ≥ 2. √ n x (n + 1)t n n+1 For the following propositions we set un = Proposition 4. 46]: √ vn+1 = ` =⇒ lim n vn = `. we have lim n→∞ vn+1 vn = e1+t x . n→∞ f (a) Now. p. Finally. lim un = 1. Consider vn = nn(t+1) /xn . lim n→+∞ nt+1 = √ n x n et+1 x . we can Apply Proposition 1.253 Proposition 2. and the proposition follows immediately from the well-known property [17. since un − 1 n(un − 1) = ln(unn ) · . Keeping the notation of the previous proof we have √ n v 1 n un = 1 + √ n+1 n vn+1 √ n √ vn = nt+1 / n xn . we have limn→∞ un = 1. Proof. Also. the Proposition follows using the continuity of the exponential function. We have vn+1 (1 + n)(n+1)(t+1) xn xn (1 + n)(n+1)(t+1) = = vn xn+1 nn(t+1) xn+1 nn(t+1) 1+t 1+t n xn nt+1 1 1 = 1+ 1+ xn+1 n n Thus. from Proposition 3. ln un we conclude immediately that lim n(un − 1) = 1. (n + 1)t nt nt+1 n+1 so on account of Propositions 3 and 4: √ n x n lim Cn = lim · lim n(un − 1) = xe−(t+1) . (n + 1)t n √ n+1 x (f (a) + bf 0 (a)) . √ n Bn = f (an ) xn t n − 1 · ln tnn nt+1 ln tn we get lim Bn = f (a) n→+∞ x et+1 · 1 · ln e f (a)+bf 0 (a) f (a) = x (f (a) + bf 0 (a)) . Proposition 6. n→+∞ n→+∞ nt+1 n→+∞ This concludes the proof of the Proposition. we set Bn then lim Bn = n→+∞ = f (an+1 ) √ n x xn+1 n − f (an ) t . et+1 Remark.254 Proof. we obtain limn→+∞ account of Propositions 2. If for n ≥ 2. and on f (a)+bf 0 (a) f (a) . et+1 e . 4: lim n→+∞ tnn = lim n→+∞ f (an+1 ) f (an ) n Writing · lim unn = e bf 0 (a) f (a) n→+∞ tn −1 ln tn e=e = 1. We have √ √ √ n x n x xn+1 n n Cn = − = · n(un − 1). Proposition 6 may also be proved as follows: We have √ √ n x n+1 x n+1 n Bn = f (an+1 ) − f (a ) n (n + 1)t nt √ √ √ √ n x n x n x n+1 x n+1 n n n = f (an+1 ) − f (a ) + f (a ) − f (a ) n+1 n+1 n (n + 1)t nt nt nt n+1 √ √ √ n x n x xn+1 n n = f (an+1 ) − + t+1 · n (f (an+1 ) − f (an )) t t (n + 1) n n and Propositions 1. We have √ n Bn = f (an ) where tn := xn nt f (an+1 ) un − 1 f (an ) √ n = f (an ) xn (tn − 1). et+1 Proof. 4 (a) Now. nt f (an+1 ) f (an ) un . 3 and 5 yield lim Bn = f (a) n→+∞ x x + t+1 bf 0 (a) = x (f (a) + bf 0 (a)) e−(t+1) . since limn→+∞ tn == ff (a) · 1 = 1. xn = uvnn . t = 0 n→+∞ n→+∞ nxn v . Evaluate limn→+∞ n+1 (n + 1)! − n n! . f (x) = 1 we have a = lim an . p √ √ √ n Bn = n+1 n + 1 n+1 (n + 1)! − n n n!. from Proposition 6. n+1 A3. Calculate limn→+∞ n+1 vn+1 − n vnn . Evaluate limn→+∞ and from Proposition 1: √ √ n+1 b = lim (an+1 − an ) n = lim n+1− nn n n→+∞ n→+∞ √ √ √ (n + 1) n+1 n + 1 − n n n − n+1 n + 1 = lim n→+∞ = 1 − 1 = 0. Applications On the following we preserve the notation of the previous section.255 2. n ∈ N∗ . we get xn+1 u a = lim an . x = lim = ∈ R∗+ . x = lim n→+∞ n→+∞ xn+1 (n + 1)! = lim = 1. f (x) = 1. and f : R∗+ −→R∗+ . √ Solution: With xn = n!.) Solution: We have un+1 vn un+1 n c vn u · lim lim = lim = ∈ R∗+ n→+∞ nvn+1 un n→+∞ nc+1 un n→+∞ vn+1 v and taking f : R∗+ −→R∗+ . Now. n ∈ N∗ . an = n n. Solution: With xn = n!. so lim n→+∞ √ n+1 p √ √ n n + 1 n+1 (n + 1)! − n n n! = e−1 . {vn }n≥1 be real positive sequences. an a sequence satisfying the assumptions in 1. n+1 so lim p n+1 n→+∞ (n + 1)! − √ n n! = e−1 . Bn is known as Traian Lalescu’s sequence (see [5]). p √ A1. and f : R∗+ −→R∗+ . Let {un }n≥1 . Bn = p √ n (n + 1)! − n!. x = lim = 1. and c ∈ R∗ with limn→+∞ nuc+1 = q q un v u u n+1 n+1 u ∈ R∗+ and limn→+∞ nc vn = v ∈ R∗+ ... f (x) = x we have xn+1 (n + 1)! a = 1. n ∈ N∗ and {an }n≥1 any sequence satisfying the assumptions in 1. t = 0 = lim n→+∞ nxn n→+∞ n!n A2. √ p √ √ n+1 n + 1 n+1 (n + 1)! − n n n n! . n→+∞ nxn n!n t=0 and from Proposition 6. (This application is a generalization of [1]. g) = nf (x) Calculate limn→+∞ Bn (f. Let f. √ g(x) n+1 (n+1)! √ Solution: Setting wn (x) = . with the notation in [1]. g) = n = f (x) √ g(x) n n! (wn (x) − 1) = n !g(x) √ n n! (wn (x) − 1) n !g(x) √ n n! wn (x) − 1 n · ln (wn (x)) n ln wn (x) so g(x) 1 lim Bn (f. g : R → R such that f (x) + g(x) = 1. we have n n! p n+1 lim wn (x) = lim n→+∞ n→+∞ = and limn→+∞ (n + 1)! n+1 ! n+1 n · · √ n n n! !g(x) g(x) 1 ·e·1 =1 e wn (x)−1 ln wn (x) = 1. e ve For c = 2. so 2a 3b n n s r ! a bn+1 bn n+1 = lim − n = e−1 n→+∞ an+1 an ae lim n+1 n→+∞ n = and we have solved [1]. x ∈ R and g(x) √ g(x) p n n+1 (n + 1)! n! − . bn ∈ R∗+ and limn→+∞ nan+1 = limn→+∞ nbn+1 = a ∈ R∗+ . an . n→+∞ e . n ∈ N∗ . g) = · 1 · ln eg(x) = g(x)e−g(x) . g). vn = an . Furthermore. p n+1 n lim (wn (x)) = n→+∞ = = lim n→+∞ (n + 1)! √ n n! (n + 1)! lim · n→+∞ n! lim n→+∞ !n !g(x) !g(x) 1 p (n + 1)! !g(x) n+1 n+1 p (n + 1)! = eg(x) . n+1 Now we note that Bn (f. A4. Bn (f. it is un = bn .256 and r un+1 − vn+1 r un vn x x (f (a) + bf 0 (a)) = (1 + b · 0) e e x u = = . Problem 3764. Problem 234. Greece. Problem 704. No. School Science and Mathematics journal.M. http://www.M. Revista Matematic˘ a duin Timi¸soara. Problem 67. N.B˘ atinet¸u-Giurgiu. [16] D. 2013. so lim Bn (f.pentagon. N. p. Problem 24. http://www. [11] D. p. g) = Ln = n (n + 1)! n! − . akotronis@gmail. [5] Trian Lalescu.B˘ atinet¸u-Giurgiu.M. 1900-1901.Stanciu. Problem 715. 2013. Vol.hu/~matsefi/Octogon/volumes/Wildt_2013_1. Octogon Mathematical magazine. p. April 2005. http://www.com. On Lalescu Sequences. No.M.Stanciu.16.M.38.2.org/pentagon/Vol_71_Num_1_Fall_2011. N.asymmetry.B˘ atinet¸u-Giurgiu. p. Problem W3. Issue 2. Issue 3. N. p.
[email protected]. Vol.B˘ atinet¸u-Giurgiu. Neculai Stanciu: Department of Mathematics.54. References [1] D.B˘ atinet¸u-Giurgiu.71.oei. Vol.B˘ atinet¸u-Giurgiu.pdf [7] D.B˘ atinet¸u-Giurgiu.kappamuepsilon. Crux Mathematicorum. http://www.M.1.1. Problem 579. Problem 3713.pdf [15] D.Stanciu.49. p. Problem 5208. p. http://mathproblems-ks. x ∈ R.Stanciu.M.B˘ atinet¸u-Giurgiu.pentagon.M.org/Websites/ssma/images/Problems%20Section/ April-2012.801. S ¸ iruri Lalescu.33-38. Vol.S. [4] D.1. No. Bucharest.org/pentagon/Vol_71_Num_2_Spring_2012.2. http://mathproblems-ks. Nr. Geogre Emil Palade General School. A.1A.13.M. 2012. [6] D. Revista Escolar de la Olimpiada Iberoamericana de Matem´ atica. sin2 x √ sin2 x p n n+1 cos2 x Bn (f.php?id=92 [14] D. Vol.3. Vol. http://www.M. N. The Pentagon.257 If f (x) = cos2 x.Stanciu. Romania. Problem 11676.1. 2000.Stanciu. g(x) = sin2 x.Stanciu. Vol.2.rsme.M. [12] D.J.kappamuepsilon. 2013. Edition XXIII.gr.es/gacetadigital/english/vernumero.38. N.Stanciu.63. Mathproblems Mathematical journal. pdf [17] W.Stanciu. 2012. Octogon Mathematical magazine. 2011. p.91.1-2. 2012. No. N.285.B˘ atinet¸u-Giurgiu.33. one can solve [1] to [16] and many other problems from various math problem solving journals. No. p. 2013.T.B˘ atinet¸u-Giurgiu. N.pdf [13] D.. M. 1985.com.1. http://www. p.Stanciu. N. No.71.B˘ atinet¸u-Giurgiu.7.es/oim/revistaoim/numero49/ Probs241_245.M. No. Sequences and Series.M. April. Anastasios Kotronis: Athens.com/?wpfb_dl=4 [2] D. N. The Pentagon.3. November 2012. La Gazeta de la RSME. Romania. http://www. September 2012. www. Problem 692. Real Numbers.B˘ atinet¸u-Giurgiu: Department of Mathematics. April 2013. p. pp. J´ ozsef Wildt International Mathematical Competition. p. Mathproblems Mathematical journal.21. Gazeta Matematic˘ a. Problem 2414. Issue 4. 2011. pp. No. g) = n→+∞ 2 lim Ln = sin2 x · e− sin n→+∞ x and the solution of [2] follows. . Vol.72.Stanciu. Buz˘ au.Stanciu. Vol.M.com/?wpfb_dl=7 [9] D. pp.pentagon. Nowak Problems in Mathematical Analysis I.33-38. February 2012. Remark: With the methods presented above.B˘ atinet¸u-Giurgiu.44. p.42. N. 2012. Kaczor. N. N. D. The Pentagon.M.com/?wpfb_dl=10 [3] Maria B˘ atinet¸u-Giurgiu.M. Matei Basarab National College. Mathproblems Mathematical journal.502.pdf [10] D.285.VI.ssma. Crux Mathematicorum.uni-miskolc. Vol. then with the notation in [2]. Vol.502.9. Problem 43.198-202.pdf [8] D.B˘ atinet¸u-Giurgiu.kappamuepsilon. p. Vol. Vol.119.Stanciu. No. http://mathproblems-ks. No.229. The American Mathematical Monthly.org/pentagon/Vol_72_Num_1_Fall_2012.B˘ atinet¸u-Giurgiu. y) such that p p p p x2 + 2x + 1 + x2 − 4x + 4 + x2 − 2xy + y 2 + y 2 − 6x + 9 = 4. “Tor Vergata” University. c > 0 and b+c b c a + + ≥2 b+c c+a a+b 22. Romania. Proposed by Stanescu Florin. Dept. Math.258 JUNIOR PROBLEMS Solutions to the problems stated in this issue should arrive before June 15. We have a rectangular chessboard 3 × 20 made of square boxes 1 × 1. Italy. “George Emil Palade” School. Republic of Kosova If a ≥ 3/2 then prove that a. University of Prishtina. The color of the balls in the fourth and fifth baskets is white. rc are the lengths of the rays of the excircles. rb . Proposed by Dorlir Ahmedi. B˘ atinet¸u-Giurgiu. Bucharest. How many different configurations can occur to Mary? 24. Gaesti. Math. How many are the different ways to do that? 23.M. 2014 Proposals 21. Rome. jud. Mary picks up a ball from each basket in such a way that the sum of the five numbers is 96.Dambovita. in the second is green and in the third is red. Find all pairs of real numbers (x. We also have 30 dominoes of size 1 × 2 or 2 × 1 that we want to use to cover the chessboard. b. Mary has 5 baskets each containing 97 colored balls numbered from 0 to 96. . and p the semiperimeter of the triangle 25. “Matei Basarab” National College. Buzˇ au. Dept. Proposed by Callegari Emanuele. Romania and D. Romania Prove that in a triangle ABC the following inequality holds: 27r ra rb rc p ≤ + + ≤ 2p a b c 2r where ra . Proposed by Proposed by Neculai Stanciu. Proposed by Callegari Emanuele. Rome. The color of the balls in the first basket is blue. r is the radius of the circle inscribed in the triangle. Serban Cioculescu school. “Tor Vergata” University. Italy. “Matei Basarab” National College. Maramures. Proposed by D. of the equation becomes √ √ 1 x+ x−2 1 √ √ √ +√ √ =√ √ = ( x − 2)( x − 1) x( x − 1) x( x − 1)( x − 2) √ 2( x − 1) 2 2 √ √ =√ √ =√ √ = x( x − 1)( x − 2) x( x − 2) x−2 x which equals the r.h. Pite¸sti.s. Neculai Stanciu. Damascus. Daniel V˘ acaru. B˘ atinet¸u–Giurgiu.h.s.. x − 2 x 6= 0 ⇐⇒ x 6= 4 Thus if x ∈ (0. 4}. If A = B = C the triangle is equilateral and the equality trivially holds. +∞)\{1. “ Matei Basarab. Bucharest. x 6= 1 √ √ x > 0 ∧ x − x 6= 0 ⇐⇒ x 6= 1. B˘ atinet¸u–Giurgiu.M. Romania. Prove that the acute triangle ABC is equilateral if and only if tan2 A tan2 B tan2 C + + =9 sin2 B + cos2 C sin2 C + cos2 A sin2 A + cos2 B Solution by the authors.“ George Emil Palade” School. √ x − 3 x + 2 6= 0. Bucharest. x− √ x 6= 0. Suppose now that the equality holds. Buz˘ au. Proposed by D. Omran Kouba. Romania. Two incorrect solutions have been submitted. Romania. √ x − 2 x 6= 0 √ √ √ x > 0 ∧ x − 3 x + 2 = ( x − 2)( x − 1) 6= 0 ⇐⇒ x 6= 4.259 Solutions 16. the l. Syria and the proposers (independently). The existence conditions are x > 0. 17. “ George Emil Palade” School. Buzˇ au. National College. Neculai Stanciu. tan2 A sin B + cos2 C cyc X X tan2 A 1 2 2 =P (sin B + cos C) 2 2 2 sin B + cos2 C cyc (sin B + cos C) cyc cyc X 1 X tan2 A = (sin2 B + cos2 C) 2 3 cyc sin B + cos2 C cyc 9=U = X 2 . Determine all real number x satisfying 1 √ x−3 x+2 + 1 2 √ = √ x− x x−2 x Solutions by Codreanu Ioan Viorel .M. Romania. Romania. Romania. Romania and Titu Zvonaru. and the proposer. so BP BD · BC = BA BA2 Similar from the power of the point C with respect to the circumcircle of triangle ABD we obtain CD · BC CQ = CA CA2 Since AD is the symmedian from A we have AB 2 BD = DC AC 2 CQ From last three equation we have BP = BA CA .M. through Jensens’inequality gives !2 2 X A+B+C π ≥ 3 tan = 3 tan2 = 27 27 = 3U ≥ tan A 3 3 cyc Equality holds if and only if A = B = C. The injectivity of the tangent in the given π domain imposes A = B = C = that is the triangle is equilateral. x ∈ (0. Using the power of the point B with respect to the circumcircle of triangle ADC we obtain BP · BA = BD · BC. π/2). Show that P Q is parallel to BC. Neculai Stanciu. teramo. Syria. . Solution by D. 3 Also solved by Titu Zvonaru. Higher Institute for Applied Sciences and Technology. so P Q is parallel to BC.e. Romania. Comˇ ane¸sti. the tangent is defined on cyc (0.260 and Cauchy–Schwartz’s inequality yields 1 9 = U ≥ (tan A + tan B + tan C)2 3 X 2 that is 27 = 3U ≥ tan A . Romania and Omran Kouba. Bucharest. Damascus. Buzˇ au.Proposed by Ercole Suppa. Com˘ anesti¸. Remark The equivalence BP CQ BD AB 2 = ⇐⇒ = BA CA DC AC 2 shows that the reverse is true. Italy Let K be the symmedian point of 4ABC. Denote by P and Q the intersection points (different from A) of AB and AC with the circumcircles of triangles 4ADC and 4ABD respectively. Also solved by Omran Kouba. π/2) and then the convexity of tan x. let D = AK ∩ BC. 18. Syria. Damascus. if P Q is parallel to BC then AD is the symmedian from A. Romania. Since ABC is acute. Bˇ atinet¸u-Giurghiu. i. Albania. Find all integer solutions of the equation 3x + x4 = 5x . 2} are the only solutions to this equation. Buzˇ au and Titu Zvonaru. if x is a negative integer then clearly 3x + x4 > 3x > 5x and x is not a solution. Mathematical Group Galaktika Shqiptare. Solution by Omran Kouba. Republic of Kosova. Higher Institute for Applied Sciences and Technology. and the proposer. if x is a nonnegative integer solution. Hence x2n − x0 = x2n+1 − x1 = n X (x2k − x2k−2 ) = n k=1 n X (x2k+1 − x2k−1 ) = n k=1 . Clearly. Math. Rome. Syria The answer is x ∈ {0. a4 > 1 and. It is clear that x = 0 and x = 2 are solutions.M. Neculai Stanciu. ≥5 = an 1+n 1+4 125 Thus an > 1 for n ≥ 4. so that x 5 − 7 × 3x 6 6 > 5x . Romania. for n ≥ 4. Bˇ atinet¸u-Giurghiu. 20.Proposed by Armend Shabani. Syria. which is equivalent to (b) Also solved by Arber Igrishta. Let us now prove (a) and (b): x 4 (a) We have 35 ≥ 53 > 7. University of Prishtina. Higher Institute for Applied Sciences and Technology. Let a0 = a1 = a2 = 1 and for n ≥ 1 an an+1 an + an−1 an+2 = Find an for any n. Thus {0. with this notation we have xn+1 = 1 + xn−1 . 2}. D. Bucharest. Solution by Omran Kouba. Damascus. Ioan Viorel Codreanu. if x ≥ 4 then 6 (a) 5x − 3x > 5x 7 6 x · 5 > x4 (b) 7 Combining (a) and (b) we see that for x ≥ 4 we have 5x − 3x > x4 and x is not a solution to the proposed equation. Maramures. Romania. (b) Let an = 6 7 · 5n n4 . we have 4 4 n an+1 4 256 =5 > 1. Proposed by Paolo Perfetti. Damascus. Let xn = an /an+1 . Now. then x must be even since both 3x and 5x are odd integers in this case. Moreover. Romania. Dept. “ Tor Vergata” University. 5x − 3x = 5x + 7 7 7 Which is (a). Satulung. Italy.261 19. Indeed. Comˇ ane¸sti. for n ≥ 1.262 Thus. xn n! · (n + 1)! Finally 1 . So. an = . bn/2!c · dn/2!e where b·c and d·e are the floor and ceiling functions respectively. (n!)2 and a2n+1 = a2n 1 = . we have a2n−2 = x2n−2 x2n−1 = n2 a2n Multiplying. x2n = x2n+1 = n + 1 for n ≥ 0. we obtain a2n = 1 . Pages 263-302 Editors: Valmir Krasniqi. Shabani. Given points U and P in the plane of 4ABC. Mih´aly Bencze. Issue 2 (2014). 263 . Armend Sh. Emanuele Callegari. Enkel Hysnelaj. we say that the Prasolov product of U and P is defined. Prove that Z x ∞ X x x2 xn np ex − 1 − − − ··· − = ex Qp (t)dt. Ercole Suppa. url: http://www. Valmir Bucaj. The editors encourage undergraduate and pre-college students to submit solutions.com Volume 4. Note that if U is the orthocenter of 4ABC and P is the nine-point center of 4ABC. Let Ua Ub Uc be the cevian triangle of U . In this case the intersection point of the lines is the Prasolov product of U and P. Proposals should be accompanied by solutions. Prishtin¨ e. Universiteti i Prishtin¨ es. Technical University of Cluj-Napoca.mathproblems-ks. Ub and Uc in P . Let p ≥ 1 be an integer and let x ∈ R. Mohammed Aassila. Rb and Rc the reflections of Ua . Romania. Proposed by Ovidiu Furdui. Cluj-Napoca. Anastasios Kotronis. Student solutions should include the class and school name. Francisco Javier Garc´ıa Capit´an. Drawings must be suitable for reproduction. 1! 2! n! 0 n=1 where Qp is a polynomial of degree p which satisfies the equation Qp+1 (x) = xQ0p (x) + xQp (x) with Q1 (x) = x 97. Denote by Ra . Roberto Tauraso. Bulgaria. If the lines ARa .com Solutions to the problems stated in this issue should arrive before October 15. then the Prasolov product is known as the Prasolov point. Kosov¨ e. respectively. An asterisk (*) indicates that neither the proposer nor the editors have supplied a solution. Omran Kouba. Teachers can help by assisting their students in submitting solutions. Cristinel Mortici. Questions concerning proposals and/or solutions can be sent by e-mail to: mathproblems-ks@hotmail. PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets. Prove c 2010 Mathproblems. Proposed by Sava Grozdev and Deko Dekov (Jointly). BRb and CRc concur in a point. Ovidiu Furdui. 2014 Problems 96. Solutions will be evaluated for publication by a committee of professors according to a combination of criteria. David R. Jos´ e Luis D´ıaz-Barrero. Jozsef S´ andor. Stone. Paolo Perfetti.Mathproblems ISSN: 2217-446X. each indicating the name of the sender. Let U a be the point at which the A-excircle meets the side BC of 4ABC. Binzhou City. China. Proposed by Li Yin. 1]. Bucharest. Consider a real function f : [a.264 that the Prasolov product is defined. city Gaesti. Calculate " # 2n+1 ∞ Y 4 1 (n − 1)(n + 1) 1+ e2 n (2n − 1)(2n + 1) n=2 100. x→∞ exists for some α ∈ [0. +∞) → (0. 98. +∞) → R be a bounded continuous function. Find n→∞ Z γn p n lim (2n − 1)!! f (x)dx. Romania. f (x)dx ≤ n−1 bn − an n a 102. Show that ! n +∞ X 4 n! X nk nk − = + O(n−1 ). Consider a continuous function f : (0. Romania. Show that for every positive integer n with n ≥ 2 the following inequality holds Z b (b − a)(bn f (b) − an f (a)) bf (b) − af (a) n − . Athens. Let f : (0. Proposed by Florin Stanescu. Find the value(s) of a. Let (γn )n≥1 be the sequence defined by γn = − ln n + k=1 k1 and let γ = lim γn . n Romania. n→∞ γ 101. Prove that the lines ARa . Dambovita. Greece. Buzˇ au. . nn k! k! 3 k=0 k=n+1 99. provided U is the Nagel point of 4ABC and P is the Spieker center of 4ABC. Denote by Ra . Bucharest. respectively. Ub and Uc in P . Suppose that the limit lim xα |f (x + 2) − 2f (x + 1) + f (x)| = a ∈ R. B˘ atinet¸u-Giurgiu. ∞). jud. Department of Mathematics. b] → R (with a > 0. Romania. and define U b and U c similarly. “Matei Basarab” National College. The problem could be re-formulated as follow. Proposed by D. 256603. Shandong Province. and Neculai Stanciu. BRb and CRc concur in a point. Serban Cioculescu school. Proposed by Marcel Chirit¸ˇ a. Binzhou University. “George Emil Palade”P School.) having a positive and increasing derivative. Proposed by Anastasios Kotronis. Let P be incenter of the medial triangle of 4ABC. Rb and Rc the reflections of Ua .M. which implies a = b = c. c 6= 1. Solution 2 by Moshe Goldstein and Moti Levy. Suppose that three real numbers a. Rome. cyc cyc Solution 1 by Paolo Perfetti. c < 1. 2 ∞ XX cyc k=0 2 2 (a2 )k = ∞ XX (ak )2 = cyc k=0 We know that a + b + c ≥ ab + bc + ca. South Korea. (1 − ac) (1 − a2 ) (1 − c2 ) It follows that a = c. X (Note that means ’cyclic sum’ f (x. Italy. b. ≤ 1) satisfies the following equality. Korea Advanced Institute of Science and Technology. then X a−b a a c a−c c−a = + 0= 2 2 1 − ab 1 − a 1 − ac 1 − a 1 − ca 1 − c2 cyc (1 + ac) . y)). c(0 ≤ a. x. 1 − a2 1 − b2 1 1 > ≥ 0. Department of Mathematics. a 6= b. b.Proposed by Hun Min Park. b. then ∞ XX (ab)k cyc k=0 (ak )2 + (bk )2 + (ck )2 ≥ (ab)k + (bc)k + (ca)k with the equality if and only if a = b = c. we may assume that a ≥ b ≥ c. It must be a. Rehovot. Daejeon. 1 − ab 1 − bc 1 1 > ≥ 0. x) + f (z. c. 1 − ca 1 − bc = (a − c) (1) (2) (3) . by virtue 0 ≤ a. Suppose that a = b. Now suppose. The result follows. b. X a − b a =0 · 1 − ab 1 − a2 cyc Prove that X a = b = c. that is a > b ≥ c. Without loss of generality. 88. y. The equality is X cyc X 1 1 = 1 − a2 1 − ab cyc which is. We will be very pleased considering for publication new solutions or comments on the past problems. Israel. z) + f (y. Then a b > ≥ 0.265 Solutions No problem is ever permanently closed. z. z) = f (x. y. Tor Vergata University. > Solution 3 by Omran Kouba. all different from 7. c) = 0 then a = b = c. and the proposer. n n∈S Solution 1 by Henry Ricardo. France. excluding 0 and 7. D. c) = = ∞ X (a2n + b2n + c2n − an bn − bn cn − cn an ) n=1 ∞ X 1 (an − bn )2 + (bn − cn )2 + (cn − an )2 2 n=1 1 (a − b)2 + (b − c)2 + (c − a)2 2 So. USA. c ∈ [0. For a. 89. Bucharest. Romania and Neculai Stanciu. b. b. Higher Institute for Applied sciences and Technology. c) = Noting that a−b a b−c b c−a c · + · + · 1 − ab 1 − a2 1 − bc 1 − b2 1 − ca 1 − c2 x−y x (1 − xy) − (1 − x2) 1 1 · = = − 2 2 2 1 − xy 1 − x (1 − xy)(1 − x ) 1−x 1 − xy ∞ X = (x2n − xn y n ) n=1 we conclude that F (a. b. (5) 1 − ab 1 − bc cyc cyc P a−b a Inequalities (4) and (5) imply that cyc 1−ab 1−a2 > 0. San Jose. “George Emil Palade” School. Romania(Jointly). hence a = b and we are done. Strasbourg. and 9 choices for the remaining i − 1 digits. . California. We can write S = S1 + S2 + S3 + · · · . b. where Si is the sum of all terms of the harmonic series whose denominators contain exactly i digits. 1) we consider F (a. X a−b (4) 1 X (a − b) = 0. USA.M. Syria. Mohammed Aassila. Now the number of i-digit numbers that do not contain the digit 7 is 8 · 9i−1 : There are 8 choices for the first digit. if F (a. New York. Bˇ atinet¸uGiurgiu. Prove that X1 < +∞. Damascus. New York Math Circle. Let S be the set of positive integers that does not contain the digit 7 in their decimal representation. Buzˇ au. “Matei Basarab” National College. which contradicts the assumption in the problem statement.266 It follows from (1) that X X a−b c a−b a > . ≥ Also solved by Arkady Alt. 2 2 1 − ab 1 − a 1 − c 1 − ab cyc cyc Using (2) and (3) we obtain. n0 = 8 and nk+1 = 9nk . . S= ∞ [ Sk . Therefore S= ∞ X i=1 Si ≤ ∞ X 8 · 9i−1 i=1 10i−1 i−1 ∞ X 9 =8 = 80. the number of terms in Sk+1 is less than 8 9 + 92 + · · · + 9k < 9k+1 . By the induction hypothesis. k=1 where Sk := S ∩ 10k−1 . 2. α10k + 1. So m ≥ 10i−1 .267 Furthermore. . it is less than 9 + 9 2 + · · · + 9k . 10k . Solution 3 by AN-anduud Problem Solving Group. n n∈Sk P∞ P Clearly. J. Suppose that it is true that the number of terms in Sk is less than 9k . 10 i=1 so S converges by comparison. which implies that 1/m ≤ 1/10i−1 . 1. (α + 1)10k . . Remark: This method can be used to show convergence of the sum of reciprocals of integers that do not contain the digit k ∈ {0. therefore ak+1 = n∈Sk+1 n1 < 910k . The other intervals have the same number of terms. 21(2). Math. Denote the number of terms of the given series between 101k and 10k+1 by k nk for k ≥ 0. Hence nk = 8 · 9 . Now we show by mathematical induction that the number of terms in Sk is less than 9k . then we have to show that the number of terms in Sk+1 is less than 9k+1 . 9}. . which is equal to the number of terms of the interval S ∩ 1. Hence. where m is an i-digit number. The number of terms in S1 is 8 < 91 . (Feb. each number in Si is of the form 1/m. . ∞ ∞ X1 X X 9k = ak < = 90. . 48-50. k ≥ 0. . we split the set Sk+1 into nine intervals Sk+1 = 9 [ α=1 S ∩ α10k . 1914) pp. of course. . Rehovot. The seventh intersection S ∩ 7 · 10k . 10k . Ulaanbaatar. . Solution 2 by Moti Levy. k=1 ak = n∈S n1 . Amer. Israel. A Curious Convergent Series. 8 · 10k is empty. Kempner. P k+1 Each term in Sk+1 is not less than 10k . We partition the set S into disjoint subsets {Sk }. Monthly. Define the sequence {ak }k≥1 by ak = X 1 . To show this. Mon1 golia. n 10k−1 n∈S k=1 k=1 Reference: A. Haroun Meghaichi. Higher Institute for Applied sciences and Technology. Let n be a positive integer. Barcelona. Syria. Center of Recreational Mathematics. Algeria. Proposed by Omran Kouba. and the proposer. University of Science and Techonology. Damascus. Campus of Bellaterra. Also solved by Roberto de la Cruz Moreno. Houari Boumediene. Spain.268 Then X 1 < n n∈S. prove that . n<10m+1 1+ 1 1 + ··· + 2 9 + 1 1 + ··· + 10 99 + ··· + 1 1 + ··· + 10m 99 · · · 9 1 1 · n1 + · · · + m · nm 10 10 2 m ! 9 9 9 1 =8 1+ + ··· + <8· 9 = 80. 90. Syria. Damascus. 10 10 10 1 − 10 < 1 · n0 + Therefore P 1 n∈S n < +∞. Higher Institute for Applied sciences and Technology. Omran Kouba. . n . X √ . . . √ n . (−1)b kc . ≤ . . etc. If (p − 1) ≤ n ≤ p2 −1 then d ne = p and the sequence n p be√a positive on k (−1)b c is composed of p runs. Solution 1 by Moti Levy. . 1}. p. 1. 1. . The sequence n √ o k (−1)b c k≥0 is composed of consecutive runs of +1 or −1. for m = 1. 2 If (p − 1) ≤ n < p2 − 1 then . −1. √ 2 Let integer. . k=0 2 The length of the m-th run is m2 − (m − 1) = 2m − 1. Israel. k=0 and determine the cases of equality. −1} and the third run is {1. Rehovot. The first run is {1}. and the m−1 sign of the units in the m-th run is (−1) . . the second run is {−1. 1. n . . p−1 . ! . X . X . √ . . . kc . m−1 p−1 2 . b (−1) (−1) (2m − 1) + (−1) n + 1 − (p − 1) . . . ≤. . . . . m=1 k=0 . p p−1 2 = (−1) (p − 1) + (−1) n + 1 − (p − 1) . . . . p 2 = . (−1) (p − 1) − n + 1 − (p − 1) . . . . . . 2 . = . (p − 1) − n + 1 − (p − 1) . = . p2 − 1 − n − p. If n = p2 − 1. . < p. . p . . n . X . . X √ . √ . . . kc . m−1 b n . (−1) (2m − 1). = p = (−1) . =. . . . . . . k=0 m=1 and these are the cases of equality. Tor Vergata √ University. Department of Mathematics.n} X X q=0 √ b n+1−1c X (−1)q k=q 2 (−1) q (q+1)2 −1 X q=0 k=q 2 √ b n+1−1c (q+1)2 −1 X (−1) q X q=0 √ b nc 1+ 1+ k=q 2 √ X (−1)q q=b n+1−1c+1 √ b nc X √ q=b n+1c (−1) n X 1 k=q 2 q n X 1 k=q 2 √ √ If n = p2 − 1 it is b n + 1c > b nc and the second contribution is absent so for n = p2 − 1 we have √ b n+1−1c X (−1)q q=0 (q+1)2 −1 X √ b n+1−1c X 1= (−1)q (2q + 1) q=0 √ b n+1−1c k=q 2 = (−1) √ (b n + 1 − 1c + 1) = (−1)bp−1c (bpc) = (−1)bp−1c p Doing the modulus we get . Thus n X (−1)b √ kc = k=0 = = √ b nc min{(q+1)2 −1. Rome. Italy. Let q ≤ k < q + 1.269 Solution 2 by Paolo Perfetti. . √ . b n+1−1c (q+1)2 −1 X X √ (−1)q 1= p = dpe = d ne . Indeed it yields p2 − r + 1 < (1 + p − 1)2 ⇐⇒ r > 1 √ √ which evidently holds true. We have that √ n + 1 < 1 + b nc. we have (−1) √ q=b n+1−1c+1 q n X k=q 2 1= .n} X X q (−1) = √ b n+1−1c q=0 k=q 2 √ b n+1−1c = X X q=0 (−1)q (2q + 1) + (−1)b q=0 √ b n+1−1c √ (−1) q (q+1)2 −1 X k=q 2 nc 1+ √ b nc X √ (n − (b nc)2 + 1) = √ √ √ (b n + 1 − 1c + 1) + (−1)b nc (n − (b nc)2 + 1) = √ √ √ √ = (−1)b n+1−1c b n + 1c + (−1)b nc (n − (b nc)2 + 1) = √ √ √ √ = −(−1)b n+1c b n + 1c + (−1)b nc (n − (b nc)2 + 1) = (−1) √ √ Since b n + 1c = b nc. This implies that b n + 1c = b nc. Thus we have √ b nc min{(q+1)2 −1. q=0 k=q 2 Now let n be such that n = p2 − r where 2 ≤ r ≤ 2p − 1 in such a way that (p − 1)2 ≤ n ≤ √ p2 − 2. 270 . . √ √ √ √ . . . −(−1)b n+1c b n + 1c + (−1)b nc (n − (b nc)2 + 1). ≤ . √ . √ √ √ ≤ . −b n + 1c + (n − (b nc)2 + 1). Cluj-Napoca. 91. Romania. Syria. Higher Institute for Applied sciences and Technology. since Z 0 1 n 2 x ln xdx = Z 0 ∞ 0 = ln2 x − ln x ln2 x − ln x dx = 3 t2 e−(n+1)t dt = Γ(3) 2 = (n + 1)3 (n + 1)3 . Damascus. Also solved by Arkady Alt. Since Z 1 h (1 − xy) iy=1 (1 − x) ln(1 − xy)dy = − ln(1 − xy) − y ln(1 − x) − 1 =− x x y=0 0 we see that ln2 (1 − x) 2 I= ln (1 − x) − ln(1 − x) − dx x 0 Z 1 ln2 x = ln2 x − ln x − dx 1−x 0 Z 1 Now.Proposed by Ovidiu Furdui. University of Science and Techonology. noting that 3x − 3x ln x + x ln2 x we see that Z 1 0 Also. Haroun Meghaichi. = n + 1 − (n − (b nc)2 + 1) ≤ √ ≤ n+1−1 and the last step is to show √ √ n + 1 − 1 ≤ d ne Recall that n = p2 − r thus we need to show p √ √ p2 − r + 1 ≤ d ne = 1 + b nc = 1 + p − 1 which clearly holds. Algeria. 0 0 Solution 1 by Omran Kouba. and the proposer. USA. San Jose. Let the considered integral be denoted by I. California. Calculate Z 1Z 1 ln(1 − x) ln(1 − xy)dxdy. Technical University of Cluj-Napoca. Houari Boumediene. The answer is 3 − 2ζ(3). It is easy to see that for k a positive integer: X1 n≥1 1 − n n+k =1+ 1 1 1 + + · · · + = Hk 2 3 k (1) the k-th Harmonic Number.271 we see that Z 1 0 ∞ Z 1 X ln2 x dx = xn ln2 xdx 1−x 0 n=0 = ∞ X 2 = 2ζ(3) (n + 1)3 n=0 Finally I = 3 − 2ζ(3). is justified by the constant sign of the summands-integrands. the change of the way of summation and of summationintegration order. in what follows. I:= Z 1 0 =− = Z Z 1 ln(1 − x) ln(1 − xy) dx dy = − 0 Z 1 X yk Z 0 k≥1 1 k 1 0 X yk X 1 k n 0 k≥1 n≥1 Z 1 xn+k dx dy = 0 Z Z 1 0 Z 1 Z 1 k 0 k≥1 X yk Z 0 k≥1 1 X (xy)k XX 0 k≥1 n≥1 k 1 xk 0 ln(1 − x) dx dy X xn dx dy n n≥1 k y dy nk(n + k + 1) XX 1 1 y k dy = nk(n + k + 1) 0 nk(k + 1)(n + k + 1) k≥1 n≥1 k≥1 n≥1 X XX X1 1 1 1 1 = − = − k(k + 1)2 n k(k + 1)2 (n + k + 1) k(k + 1)2 n (n + k + 1) n≥1 k≥1 n≥1 k≥1 X X 1 1 1 1 1 − + − =− 2 k + 1 k (k + 1) n (n + k + 1) n≥1 k≥1 X X 1 X 1 X1 1 1 1 1 =− − − +1− − k+1 k n (n + k + 1) k2 n n+k = XX xk ln(1 − x) dx dy = Z 1 n≥1 k≥1 k≥1 n≥1 :=A+1−B For A. Greece. whenever it takes place. Athens. Solution 2 by Anastasios Kotronis. Now. from (1) and summing by parts we have: 1 1 Hk+1 . . +∞ X 1 − Hk+1 = − + (Hk+2 − Hk+1 ) . k+1 k k 1 k+1 k≥1 k≥1 3 X 1 1 = + − =2 2 k+1 k+2 A=− X k≥1 . Algeria. University of Science and Techonology. we can integrate by parts to get : n+1 1 Z 1 Z 1 n+1 (t − 1) ln(1 − t) 1 t −1 tn ln(1 − t) dt = − dt n + 1 n + 1 t − 1 0 0 0 n Z −1 X 1 k −Hn+1 = (1) t dt = n+1 n+1 0 k=0 Where Hn is the n-th harmonic number. 471–472”. See also page 6 of the Collected Contributions of M. Math. So B = 2B − 2ζ(3) and hence B = 2ζ(3). Monthly. k k(k + 1)2 0 0 0 0 k=1 k=1 Now.595886. from (1). k k+1 k(k + 1) k k+1 n+1 n+1 k=1 k=1 . B= X Hk k≥1 and furthermore B= X 1 X1 1 − k2 n n+k n≥1 k≥1 = XX k≥1 n≥1 =2 XX 1 = nk(n + k) k≥1 n≥1 XX k≥1 n≥1 =2 =2 X N ≥1 1 1 + n(n + k)2 k(n + k)2 X 1 n+k−1=N ====== 2 2 n(n + k) N ≥1 N XX N ≥1 k2 X n. Klamkin. Houari Boumediene. Solver’s Note: A reference to the details presented here is “ M.272 For B. therefore Z 1Z 1 ∞ Z 1Z 1 k k ∞ X X y x Hk+1 I= ln(1−x) ln(1−xy) dxdy = − ln(1−x) dxdy = . Monthly. S. Klamkin to the Amer.k≥1 n+k−1=N 1 n(n + k)2 X X HN +1 − N1+1 HN 1 = 2 = 2 n(N + 1)2 (N + 1)2 (N + 1)2 n=1 N ≥1 N ≥1 X HN +1 1 −2 = 2(B − 1) − 2(ζ(3) − 1) 2 (N + 1) (N + 1)3 N ≥1 where ζ is the Riemann zeta function. Math. For n ∈ N>0 . S. Solution 3 by Haroun Meghaichi. The first sum is easy to calculate : n n X X Hk+1 Hk+1 1 Hk Hk+1 1 Hn+1 − = + − =2− − . 59 (1952) pp.Amer. Finally I = 3 − 2ζ(3) ≈ 0. we can simplify the latter series to be : A= ∞ X k=1 Hk+1 = k(k + 1)2 ∞ X Hk+1 k=1 k Hk+1 − k+1 ! − ∞ X Hk k=2 k2 ! . Romania. Algeria. E. Rome. B˘ atinet¸u-Giurgiu. 4} This is true because of the periodicity of the mod sequence : mod(n + 5k. 1. 5)|n ∈ N} = {0. “George Emil Palade” School. for n ∈ N where {Ln }n≥0 is the sequence of Lucas numbers. k + 1 n→+∞ n+1 n+1 To calculate the other sum we use the following result for m > 0: Z ∞ Z 1 2 1 x=e−t/(m+1) 2 m t2 e−t dt = x ln x dx = 3 (m + 1) (m + 1)3 0 0 (2) From (1) we get : B= ∞ X Hk k=1 k2 =− ∞ Z X k=1 1 k−1 t k 0 ln(1 − t) dt = Z 0 1 ln2 (1 − t) dt = t Z 0 1 ln2 t dt. It follows that bn is not a perfect square for every n ∈ N. 5) = mod(n. Buzˇ au. Department of Mathematics. . Solution 2 by Haroun Meghaichi. 109-113. 1. By its definition. and Neculai Stanciu. and let the sequence {bn }n≥0 be defined by bn = a2n L2n . Houari Boumediene. Also solved by Arkady Alt.Proposed by D. Bucharest. pp. k ∈ N. Paolo Perfetti. Etc. (k + 1)3 I = A − (B − 1) = 3 − 2ζ(3). 1.. Mongolia. for every n ∈ N \ {1}. Let {an }n≥0 be a sequence of positive integer numbers such that 5 does not divide an for all. VA.. Cohn. Rehovot. 1.M. Israel. B. 92. Israel. Rehovot. Ulaanbaatar. Italy. n) be the remainder on division of m by n.C. it is clear that : tk ln2 t dt = ∞ X k=0 2 = 2ζ(3). Prove that bn is not a perfect square. Newport News. . bn is a perfect square if and only if L2n is a perfect square. It was proved in the article. 1−t Using the result (2) we get B= ∞ X k=0 Now. J. California. University of Science and Techonology. AN-anduud Problem Solving Group. USA. 0. Moti Levy. San Jose. H. and the proposer. Let mod(m. that if Lk is perfect square then k = 1 or k = 3. 5) for any n. Solution 1 by Moti Levy. With a quick evaluation we have {mod(n2 . “Matei Basarab” National College. n ∈ N. 4. Greubel. Romania. 4. ”Square Fibonacci Numbers. USA. Tor Vergata University.} = {0.273 Since Hn ∼ ln(n) and ln(n) = o(n) we get : A= ∞ X Hk+1 k=1 k − Hk+1 1 Hn+1 = lim 2 − − = 2.” Fibonacci Quarterly 2 1964. 3}. 5) = 1 mod(L4k+6 . 3} Which means that bn ∈ / {0. 5) = mod(2a2n . then b0 is not a perfect square. mod(n. Syria. 4) = 1. we have 2n = 4k where k ∈ N>0 . 1. 4k + 2. 5) = 3 4 lemma : if if if if mod(n. 4) = 2. Let x3 x2n+1 fn (x) = tan−1 x − x + − · · · + (−1)n+1 3 2n + 1 Clearly we have 2n fn0 (x) = X 1 1 1 − (−x2 )n+1 (−1)n+1 x2n+2 2 k − (−x ) = − = 1 + x2 1 + x2 1 + x2 1 + x2 k=0 Thus n+1 fn (x) = (−1) Z 0 x t2n+2 dt 1 + t2 . 4} for all n ∈ N. therefore : mod(L2n . 5) = 2. 5) = mod(1 + 2. 4k + 1. Conclusion : bn is not a perfect square for any n ∈ N\{1}. mod(n. Also solved by Omran Kouba. suppose it is true for some integers 4k. 4}. Athens. Damascus. 5). Let’s take a look on mod(Ln . Damascus. For x ∈ (−1. Here’s a 2 1 mod(Ln . 4 + 3. (−1)n+1 n tan−1 x − x + 3 2n + 1 n=1 +∞ X Solution 1 by Omran Kouba. Greece. 5) = mod(L4k+5 + L4k+4 . For n = 0. 93. then bn is not a perfect square for n > 1. 5) ∈ {2. we have mod(b0 .Proposed by Anastasios Kotronis. 1). 5) = 3 mod(L4k+7 . 3} it is clearly true. Syria. 5) = 4 Which means that (lemma 1) is true for any integer n by strong induction. which means that b0 ∈ / {0. 5) ∈ {2. For n > 1. 5) = mod(L4k+6 + L4k+5 . 1. 1. evaluate x2n+1 x3 − · · · + (−1)n+1 . 4) = 0. 5) = mod(4 + 3. 2. (lemma 1) Proof : for n ∈ {0. which implies that mod(bn . Higher Institute for Applied sciences and Technology. 5) ∈ {1. 4) = 3. Higher Institute for Applied Sciences and Technology. 4}. 5) = mod(2 + 4. 5) = 2 mod(L4k+5 . then mod(L4k+4 . 5) = mod(L4k+3 + L4k+2 . 5) = mod(3 + 1. 5) = mod(L4k+4 + L4k+3 . mod(n.274 Hence mod(a2n . and the proposer. 275 It follows that ∞ X n+1 (−1) n=1 1 nfn (x) = 2 1 = 2 Z x 0 Z x 0 t3 1 + t2 t3 1 + t2 ∞ X (2n)t n=1 ∞ X n=0 t2n 2n−1 !0 ! dt dt 0 t3 1 dt 2 1 − t2 0 1+t x t4 dt 2 2 2 0 (1 + t )(1 − t ) 1 = 2 Z = Finally. 4 4k + 1 4 4 4k − 1 k=1 k=2 k=2 . we can rearrange the series as ∞ X n=1 k−1 X n=1 ∞ X (−1)n+1 n (−1)k+1 k=n+1 ∞ k−1 X x2k+1 x2k+1 X = (−1)k (−1)n n 2k + 1 2k + 1 n=1 k=2 (−1)n n is equal to −k/2 if k is even and (k − 1)/2 if k is odd. Department of Mathematics. Clearly ∞ X (−1) n=1 ∞ X = n+1 n tan (−1)n+1 n n=1 −1 ∞ X x3 x2n+1 − . Rome. . noting that t4 1 = (1 + t2 )(1 − t2 ) 8 Z x 1 1 2 2 2 + − − + (1 + t)2 (1 − t)2 1 − t 1 + t 1 + t2 we conclude that ∞ X 1 1−x x n+1 (−1) nfn (x) = + ln + arctan x 4 1 − x2 1+x n=1 which is the desired conclusion. Italy. It follows ∞ X (−1)k k=2 =− =− k−1 x2k+1 X (−1)n n = 2k + 1 n=1 ∞ X kx4k+1 k=1 ∞ X k=1 4k + 1 − ∞ ∞ k=2 k=2 1X x4k−1 1 X x4k−1 (2k − 1) + = 2 4k − 1 2 4k − 1 ∞ ∞ ∞ 1 X x4k+1 1 X 4k−1 3 X x4k−1 x4k+1 + − x + . + (−1)n+1 x−x+ 3 2n + 1 (−1)k+1 k=n+1 = x2k+1 2k + 1 Since |x| < 1. Tor Vergata University. . Solution 2 by Paolo Perfetti. =x 4 y4 4 1 − x 1 − y 4 1−x 2 0 0 ∞ X ∞ X Moreover ∞ X x4k−1 = k=2 ∞ X Z 1 Z 1 x7 . + (−1) . therefore. 3 2n+1 +∞ P x n x n+1 −1 + . 1 − x4 Z 1 x4k+1 4k+1 = y 4k dy = x 4k + 1 0 k=1 k=1 Z x Z 1 y4 1 1+x 1 x4 y 4 dy = dy = −x + ln + arctan x.. 4 1 − x2 4 4 1+x − Solution 3 by Arkady Alt.California. + −x2 2 1+x n=1 ! n+1 +∞ +∞ n−1 P P x2(n+1) P 1 − −x2 1 x4 +∞ n+1 (−1) n − = n· = n x2 = 2 2 2 2 1+x 1+x 1+x 1 + x n=1 n=1 n=1 x4 1 1 1 1 1 1 · = − + + 2 + 4 (x2 + 1) 1 + x2 (1 − x2 )2 4 (x − 1) 4 (x + 1) 8 (x − 1)2 8 (x + 1) x Rx t4 dt 1 1−t t −1 and. USA. 2 4 4 3 By summing up the three contributions we obtain x4k−1 y 4k−2 dy = 1 x 1 x5 x 1 1+x 1 1 x7 − + ln + arctan x − + 4 1 − x4 4 16 1 − x 8 4 1 − x4 3 1 1 1 x3 + − arctan x + ln(1 + x) − ln(1 − x) − = 4 2 4 4 3 1 x 1 1 1−x =− − arctan x − ln . 1 − x4 x8 y 6 dy = 1 − (xy)4 Z x t6 dt = 4 0 0 0 1−t k=2 Z x 1 t2 1 1 t+1 1 1 1 2 − + + − − t dt = = 2 2 1 + t2 41+t 41−t 4 0 x 1 x2 x 1 1 x2 x x3 = − arctan x + − + ln(1 + x) − ln(1 − x) − − − = 2 2 8 4 4 4 8 4 3 1 1 1 x3 − arctan x + ln(1 + x) − ln(1 − x) − ... San Jose.276 ∞ X x4k+1 = k=1 x5 . 4 1 + x 1 − x2 . S (x) = 0 = ln + + tan (t) = 2 4 1 + t 1 − t2 (1 + t2 )(1 − t2 ) 0 1 1−x x ln + + tan−1 (x) .. (−1) n tan (x) − x − Let S (x) := 3 2n + 1 n=1 +∞ n P 1 n+1 = Then S 0 (x) = (−1) n − 1 − x2 + . Israel. Find a point P in the plane of a given triangle ABC. b = CA and c = AB. → r 2.277 Also solved by AN-anduud Problem Solving Group. Mongolia. Haroun Meghaichi. Algeria. 2 2 2 − − − − − − − − − − wm |→ z −→ r m | = wm (→ z −→ r m ) · (→ z −→ r m ) = wm |→ z | + |→ r m | − 2→ z ·→ rm 2 2 −c − → − −c − → − −c − → − −c |2 + |→ − −c · → − wm |→ r m | = wm (→ r m ) · (→ r m ) = wm |→ r m | − 2→ rm 2 2 2 − − −c − → − − −c |2 + 2w (→ − → − → − wm |→ z −→ r m | − wm |→ r m | = wm |→ z | − |→ m c − z )· r m Summing. w2 . Bulgaria. Rehovot. . Now we apply the lemma to our problem: . Solution 1 by Moti Levy. m=1 Then the following holds: n X m=1 2 − − wm |→ z −→ r m| = n X m=1 2 2 −c − → − −c − → − wm |→ r m | + |→ z| Proof. . . such that k=1 wk = 1. Ulaanbaatar. . n X m=1 2 − − wm |→ z −→ r m| − n X m=1 2 −c − → − wm |→ r m| n n X X 2 − −c |2 −c − → − − = |→ z | − |→ wm + 2 (→ z )· wm → rm m=1 m=1 2 − −c |2 + 2 (→ −c − → − −c = |→ z | − |→ z )·→ 2 − −c |2 + 2 |→ −c |2 − 2→ − −c = |→ z | − |→ z ·→ 2 2 − − −c + |→ −c |2 = |→ −c − → − = |→ z | − 2→ z ·→ z| . such that the sum 2 2 2 |BP | |CP | |AP | + + b2 c2 a2 is minimal. Let → r 1. wn n −c denote the weighted average be n positive weights. Let w1 . Israel. This proves the lemma. 94. and the proposer. University of Science and Techonology. It follows from the lemma that the vector which minimizes the weighted sum Pn → − → − 2 → − m=1 wm | z − r m | is c . Moti Levy. . where a = BC. → r nP be n arbitrary points on a plane. . Rehovot. . Let → vector defined by n X → −c := − wm → r m. − − − Lemma 1. Houari Boumediene.Proposed by Sava Grozdev and Deko Dekov (Jointly). . This shows that f (M ) ≥ f (G) with equality if and only if M = G. λn be n positive numbers.278 − − − Let → r A . . . and let the weights be: wA = 1 a2 + 1 b2 1 b2 1 c2 2 2 + = c2 a2 a2 b2 + a2 c2 + b2 c2 a b a2 b2 + a2 c2 + b2 c2 b2 c 2 wC = 2 2 . . We consider. Damascus. f (M ) − f (G) = = = = = n X k=1 n X k=1 n X −−−→ −−→ λk Ak M 2 − Ak G2 −−−→ −−→ −−−→ −−→ λk Ak M − Ak G · Ak M + Ak G −−→ −−→ −−→ λk GM · GM + 2Ak G k=1 n X k=1 n X k=1 λk λk ! ! n −−→ X −−→ |GM |2 + 2GM · λ k Ak G k=1 |GM |2 . and the lemma follows.→ r B and → r C be vectors from the origin to the triangle vertices. a b + a2 c2 + b2 c2 then the original problem becomes: − Find a vector → r P which minimizes the weighted sum 2 2 2 − − − − − − wA |→ rP −→ r A | + wB |→ rP −→ r B | + wC |→ rP −→ r C| . Consider n points A1 . λk ))1≤k≤n . If the triangle is equilateral then the point P is the centroid. that is G is the unique point G defined by n X −−→ λk GAk = ~0. . and let λ1 . wB = The answer is → − − − − r P = wA → r A + wB → r B + wC → r C. k=1 then G is the unique point in the plane P where f attains its minimum. Indeed. . the real function f : P → R defined by n X f (M ) = λk |Ak M |2 k=1 Let also G be the barycenter of the wighted points ((Ak . . . Syria. Solution 2 by Omran Kouba. Higher Institute for Applied Sciences and Technology. . Proof. . We will use the following Lemma: Lemma 2. An in the plane P. Proposed by Li Yin. and (C. Thus the desired minimum is attained at the unique point P which is the barycenter of the weighted points (A. 256603. Italy. Binzhou University. Shandong Province. China. Rome. b12 ). using Stirling’s formula. This point is the first Brocard point Ω.279 In the proposed problem we have n = 3. On one hand. Damascus. An approximation formula of Wallis product. 2 ex = 1 + o(1). Tor Vergata University. We use 1 ln(1 − x) = −x − x2 + o(x2 ). A2 = B. a12 ). (when the triangle ABC is labeled in counterclockwise order. Department of Mathematics. For all n ∈ N. Binzhou City. Solution 1 by Paolo Perfetti. then n ln n 1 Wn ∼ √ 1 − 2n π where Wn := (2n−1)!! (2n)!! is Wallis product. Higher Institute for Applied Sciences and Technology. √ n! ∼ (n/e)n 2πn(1 + o(1)) We get 1 √ π ln n 1− 2n n 1 1 1 ln n ln n = √ exp n ln 1 − = √ exp n − + o( ) 2n 2n n π π 1 ln n 1 = √ exp − + o(1) = √ √ (1 + o(1)) 2 π π n √ √ 1 (2n)! (2n)2n e−2n 2π 2n(1 + o(1)) = √ √ (1 + o(1)) Wn = 2n = 2 (n!)2 22n n2n e−2n 2πn π n It follows lim n→∞ √1 π Wn 1− ln n n 2n =1 Solution 2 by Omran Kouba. Department of Mathematics. (B. λ1 = 1/c2 and λ1 = 1/a2 . A1 = A.) Also solved by the proposer. Wn = 2n (1) 2 (n!)2 2 (2πn)n2n e−2n πn . it is the unique point inside ABC such that ∠ΩAB = ∠ΩBC = ∠ΩCA. Syria. 95. we have √ (2n)! 2 πn(2n)2n e−2n 1 ∼ 2n =√ . c12 ). A3 = C and λ1 = 1/b2 . Algeria. 2n π πn and the desired conclusion follows from (1) and (2).280 and on the other hand n ln n ln n 1− = exp n ln 1 − 2n 2n 2 ln n ln n = exp n − +O 2n n2 2 ln n ln n = exp − +O 2 n 2 1 ln n = √ exp O n n 2 1 ln n =√ 1+O n n Thus n ln n 1 1 √ 1− ∼√ . Haroun Meghaichi. . Mongolia. Ulaanbaatar. Houari Boumediene. (2) Also solved by AN-anduud Problem Solving Group. University of Science and Techonology. How many perfect deltas are there in T20 ? . The source of the proposals will appear when the solutions be published. such that segment AE intersects segment CD. b. c. b. Denote by I the point of intersection of CD and EF . A delta is said to be perfect if the length of its side is even. A triangle is called a delta if its vertex is at the top. For all positive real numbers a. d be digits such that d > c > b > a ≥ 0. Proposals 65. How many numbers of the form 1a1b1c1d1 are multiples of 33? 66. Consider a sequence of equilateral triangles Tn as represented below: T1 T2 T3 T4 T5 The length of the side of the smallest triangles is 1. and assume that K is different from I and J. c. let E be a point on line BC outside segment BC. Given trapezoid ABCD with parallel sides AB and CD. d prove the inequality p p p p √ a4 + c4 + a4 + d4 + b4 + c4 + b4 + d4 ≥ 2 2(ad + bc) 68. for example. there are 10 deltas in T3 . Let a. Assume that there exists a point F inside segment AD such that ∠EAD = ∠CBF . Let K be the midpoint of segment EF . Proposals are always welcomed. Prove that K belongs to the circumcircle of 4ABI if and only if K belongs to the circumcircle of 4CDJ 67.281 MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. and by J the point of intersection of AB and EF . We place knights in the chess board squares such that no two knights attack one another and compute the sum of the numbers of the cells on which the knights are placed. the knight can attack the square in the opposite corner. For any 2 × 3 or 3 × 2 rectangle that has the knight in its corner square. with the numbers 1 through 64 placed in the squares as in the diagram below. 1 9 17 25 33 41 49 57 2 10 18 26 34 42 50 58 3 11 19 27 35 43 51 59 4 12 20 28 36 44 52 60 5 13 21 29 37 45 53 61 6 14 22 30 38 46 54 62 7 15 23 31 39 47 55 63 8 16 24 32 40 48 56 64 Assume we have an infinite supply of knights. What is the maximum sum that we can attain? Note. Consider a chess board.282 69. . satisfies the given 3 system and it is the desired solution. Find all real solutions of the following system of equations: p p x2 + y 2 + 6x + 9 + x2 + y 2 − 8y + 16 = 5. 8/3). 4) is equal to 5. y) = −1. Moreover. Spain. then the unique solution of the given system is (−1. from 4x − 3y + 12 = 0 9y 2 − 4x2 = 60 we obtain ⇔ 3y = 4x + 12 2 (4x + 12) − 4x2 = 60 ⇔ 3y = 4x + 12 12 (x + 7) (x + 1) = 0 8 3 16 (x. Solution 2 by p Arkady Alt. p California. 62. 9y 2 − 4x2 = 60. 0) and B(0. Let P be an interior point to an equilateral triangle ABC. 8 By substitution immediately follows that only (x. First we observe that points (x. Spain. y) = −7. y) solution of the system must lie on AB. y) = −1. BARCELONA TECH. Barcelona. yields 2 4 x + 4 − 4x2 = 60 ⇔ x2 + 8x + 7 = 0 9 3 with roots x = −7 and x = −1. P Y and P Z to the sides BC. Since only the second lie in [−3. This let us to conclude that points (x. y) that satisfy the first equation are those that the sum of their distances to A(−3. AULA Escola Europea. Draw perpendiculars P X. Substituting these values in 3 3 the second equation. (50th Catalonian Mathematical Olympiad) Solution 1 by Eloi Torrent Juste. Barcelona. USA. 3 (x. 0]. The equation of 4 4 AB is y = x + 4 or x. x + 4 with −3 ≤ x ≤ 0. Compute the value of BX + CY + AZ PX + PY + PZ (First BARCELONATECH MATHCONTEST 2014) .283 Solutions 61. Also solved by Jos´ e Luis D´ıaz-Barrero. if a point P lies out of the segment AB then AP + P B > AB = 5. CA and AB. Squaring both sides of the equation x2 + y 2 + 6x + 9 + x2 + y 2 − 8y + 16 = 5 we have 2 p p x2 + y 2 + 6x + 9 + x2 + y 2 − 8y + 16 = 25 ⇔ 4x − 3y + 12 = 0 Then. respectively. San Jose. we have AZ 2 + ZP 2 = (a − CY )2 + P Y 2 BX 2 + P X 2 = (a − AZ)2 + P Z 2 CY 2 + P Y 2 = (a − BX)2 + P X 2 Developing and adding up. AY P . aAZ = AB · AP hence −−→ −−→ −→ −−→ −−→ −→ a(BX + CY + AZ) = BC · BP + CA · CP + AB · AP −−→ −→ −−→ −−→ −→ −−→ −−→ −−→ = (BC + CA + AB) ·BP + CA · CB + AB · AB | {z } ~ 0 −→ −−→ −−→ −−→ = CA · CB + AB · AB 1 = a2 + a2 2 so 3 a 2 (2) √ BX + CY + AZ = 3. First let us denote the side length of the triangle by a. aCY = CA · CP . BARCELONA TECH. BXP and CXP. BX + CY + AZ = . Spain. −−→ −−→ −→ −−→ −−→ −→ aBX = BC · BP .284 Solution 1 by Omran Kouba. B. If a is the length of the side of 4ABC. That is. from (1) and (2) we get Solution 2 by Jos´ e Luis D´ıaz-Barrero. then on account of Pithagoras theorem. Damascus. The area of the triangle can be calculated in two ways and we get √ 3 2 a = a(P X + P Y + P Z). Joining A. C with P we obtain three pairs of right triangles: AZP. CY P. Higher Institute for Applied Sciences and Technology. hence 2 √ 3 PX + PY + PZ = a (1) 2 On the other hand. BZP. √ √ a(P X + P Y + P Z) a2 3 a 3 = ⇔ PX + PY + PZ = 2 4 2 From the preceding immediately follows that √ BX + CY + AZ = 3 PX + PY + PZ and we are done. yields 3a 2 On the other hand the sum of the areas of triangles AP B. PX + PY + PZ BX + CY + AZ = Thus. BP C. CAP is the area of 4ABC. Syria. Barcelona. . How many ways are there to weigh of 31 grams with a balance if we have 7 weighs of one gram. . . respectively? (Training Catalonian Team for OME 2014) Solution 1 by Jos´ e Luis D´ıaz-Barrero BARCELONA TECH. 4. . z = AZ. . x = BX. USA. a term with x31 is obtained by taking some term x from the first parentheses. 5 of two grams. 4P AY. . l ∈ {0. 63. 6. Higher Institute for Applied Sciences and Technology. + x46 + x47 . 3. Spain. . k + 2l + 5m = 31. + x10 ) (1 + x5 + x10 + . 4P ZA. . therefore. . . and we are done. + x10 ) (1 + x5 + x10 + . 7}. and z from the third. 2. BARCELONA TECH. . w = P Z and h = be height 2 of the equilateral triangle ABC. (1 + x + x2 + . Barcelona. . San Jose. Spain. 2. = 3. m ∈ {0. 30} We claim that the number of solutions of this equation equals the coefficient of x31 in the product (1 + x + x2 + . 10}. 10. 4P Y C. 8. 4P CX. + x30 ) = 1 + x + . y = CY. 4P BZ. . 20. Since. . 7}. . v = P Y. . + x7 ) (1 + x2 + x4 + . 4. . and 6 of five grams. Each such possible selection of x. Solution 2 by Omran Kouba. Syria. 1. 5. California. . . . 4P XB. u = P X. Damascus. . 2 PX + PY + PZ Also solved by Jos´ e Gibergans-B´ aguena. . then ah au av az [ABC] = [P BC] + [P CA] + [P AB] ⇔ = + + ⇐⇒ u + v + w = h 2 2 2 2 and x+y+z x+y+z 2 (x + y + z) BX + CY + AZ √ = = = PX + PY + PZ u+v+w h a 3 Applying Pythagorean theorem to chain of right triangles 4P XB. x + y + z = cyc √ 3a BX + CY + AZ and. z ∈ {0. 15. We are looking for the number of triplets (k. y ∈ {0. 6. . . 6}. y and z contributes 1 to the considered coefficient of x31 in the product. + 10x30 + 10x31 + 10x32 + . The required number is the number of solutions of x + y + z = 31 with x ∈ {0. + x7 ) (1 + x2 + x4 + . . Let a =√BC = CA = a 3 AB. we obtain 2 2 u + x2 = w2 + (a − z) 2 w2 + z 2 = v 2 + (a − y) 2 2 v + y 2 = u2 + (a − x) Adding all equations we get X X 2 2 u2 + x2 = w + (a − z) ⇔ 3a2 = 2a (x + y + z) cyc S0. . then the number of ways to obtain 31 grams is 10.285 Solution 3 by Arkady Alt. in such a way that x+y+z = 31. . some term y from the second. Barcelona. 5}. l. 5. . m) such that k ∈ {0. 25. . + x30 ) Indeed. Barcelona. 6). 4. California. 3). 5 − t. 5. 4)} • m = 5. >0 5 5 5 31 − 2y 21 + 2t 2t + 1 then. 0 ≤ z ≤ 6} . y. 3. |S| = 2+ − = 12 + − . • m = 4. 64. (1. USA. BARCELONA TECH. . l. Syria. 3} yields a solution. 4). (0. Solution 3 by Arkady Alt. 5} yields a solution. 3. 5 5 5 Also solved by Jos´ e Gibergans-B´ aguena. 5. z) | x. 0 ≤ y ≤ 5. 0 ≤ x ≤ 7. So every l ∈ {2. and we get four solutions: (k. 0. holds: A(k) = 5k Find the value of A(n + 2). San Jose.286 Since k + 2l ≤ 17 we conclude that 5m ≥ 14. z) | t. z ∈ Z and 24 ≤ 2y + 5z ≤ 31. y. 4. l. m) ∈ {(7. 2. and this yields the unique solution (k. 1. 2. Let A(x) be a polynomial with integer coefficients such that for 1 ≤ k ≤ n + 1. 4). 5). 5 5 5 18 + 2t 2t + 3 28 − 2y = =3+ and 5 5 5 2t + 1 2t + 3 ≤z ≤4+ . So. m) = (1. denoting t := 5 − y. m) ∈ {(6. 1. 5)} • m = 6. then k + 2l = 16 or k = 2(8 − l) ≥ 2(8 − 5) = 6 this gives the unique solution (k. 5). 0. 28 − 2y ≤ Since in integers 24 ≤ 2y+5z ≤ 31 ⇐⇒ 24−2y ≤ 5z ≤ 31−2y ⇐⇒ 5 31 − 2y 31 − 2y 28 − 2y z≤ and for any 0 ≤ y ≤ 5 holds ≤ 6. 2. So every l ∈ {0. S = (31 − 2y − 5z. we obtain = = 4+ . z) | y. Damascus. 5. (3. m) = (6. 6}. 0 ≤ z ≤ 6} = {(31 − 2y − 5z. 5 5 5 5 t=0 t=0 X X 5 5 6 X 2t + 1 2t + 3 2t + 1 2t + 1 Noting that − = − = 5 5 5 5 t=0 t=1 t=0 2·0+1 2·6+1 13 − =− = −2 we get |S| = 12 − 2 = 10. then k + 2l = 11 or k − 1 = 2(5 − l) ≤ 6. then k + 2l = 6 or k = 2(3 − l). then k + 2l = 1. y. 5). 3 + 5 5 5 5 X X 2t + 3 2t + 1 2t + 1 2t + 3 Hence. so m ∈ {3. Spain. 0 ≤ y ≤ 5. z ∈ Z and 0 ≤ t ≤ 5. 3. Higher Institute for Applied Sciences and Technology. 4. and we get four solutions: (k. l. (2. l. (4. z ∈ Z and x + 2y + 5z = 31. the total number of ways to weight 31 grams is 10. (5. • m = 3. (Training UPC Team for IMC 2014) Solution 1 by Omran Kouba. We have to compute the number of elements of the set S := {(x. 4). Indeed. Solution 2 by Jos´ e Gibergans-B´ aguena. We observe that for all k ≥ 1. holds: k X k j k k k k 5k = (1 + 4)k = 4 = + 4 + . This allows us to conclude that An (X + 1) − An (X) = 4An−1 (X). + 4 j 0 1 k j=0 . So. (which are the only possible values left for n. In view of the above. 2}.) satisfies A(1) = 5 and A(2) = λ. then 3 must divide A(4) − A(1) = 620 which is absurd. Similarly. (for an arbitrary λ ∈ Z. when n ≥ 3. A similar conclusion also holds when n = 2.. the polynomial A(X) = 20X − 15 + λ(X − 1)(X − 2) satisfies A(1) = 5. which is the desired conclusion. BARCELONA TECH. we have (b−a) | (P (b)− P (a)). and Qn(k) = 5k for 1 ≤ k ≤ n.) the polynomial A is not uniquely determined by the conditions that it has integer coefficients and that it satisfies A(k) = 5k for 1 ≤ k ≤ n + 1. if a polynomial P has integer coefficients then for every distinct integers a and b. given An (X) for some n > 0 we consider 1 Qn (X) = (An (X + 1) − An (X)) 4 Clearly deg Qn = n − 1. Indeed. of An is guaranteed. • When n ∈ {0. 1. Spain. Let An (X) be a polynomial of degree n such that A(k) = 5k for 1 ≤ k ≤ n + 1. A(2) = 25 and takes an arbitrary odd value at X = 3..287 Remarks: • If n ≥ 3 no such polynomial exists. note that the existence and uniqueness. from the above formula we see that bn = 5n+1 + 4bn−1 This is equivalent to n+1 bn−1 5 = . bn = 4n+1 n k+1 X 5 k=0 4 = 5(5n+1 − 4n+1 ). Let bn = An (n + 2). First. 4n+1 4n 4 Adding these equalities and noting that b0 = 5 we see that n k+1 bn 5 X 5 − = 4n+1 4 4 bn − k=1 Thus. Now. Hence Qn (X) = An−1 (X). when n = 0 the polynomial A(X) = λ(X − 1) + 5(2 − X). when n = 1. Find the value of A(n + 2). Barcelona. Solution. I propose a modified statement of the problem as follows: Generalization of Proposal 64. if there is a polynomial A with integer coefficients such that A(1) = 5 and A(4) = 54 . from general theorems about Lagrange interpolation. (a − n + 1) = n n! for any real a and n ≥ 1. Barcelona. then prove that ! ! n n n X n2 2n − 2 X 1 n+1 X 2 2 |bk | |ak | + Re a k bk ≤ n 2 n n−1 n k=0 k=0 k=0 (Training UPC Team for IMC 2014) Solution by Jos´ e Luis D´ıaz-Barrero. a1 . deg(B(x)) = n and it is easy to see that A(k) = B(k) for 1 ≤ k ≤ n + 1. an and b0 . San Jose. we prove that for all n ≥ 2.. . Spain. If n ≥ 2. n3 2n−2 n−1 n d holds. 65. + 4 1 2 n n+1 n+1 2 n+1 n n + 1 n+1 n + 1 n+1 =5 1+ 4+ 4 + . . the following identity Pn k=0 k=0 k2 n 2 k = . We begin with the following claim: Let α. Clearly. Spain. 1 2 n where a a(a − 1)(a − 2) . California. . · · · ... . . b1 . · · · . + 4 .. we conclude that A(x) = B(x) for all x ∈ R. + 4 + 4 − 4 1 2 n n+1 n+1 = 5(5n+1 − 4n+1 ) Also solved by by Arkady Alt. USA and Jos´ e Luis D´ıaz-Barrero. a1 . n X k=0 2 |ak | + |α| 2 n X k=0 2 |bk | − 2Re α = n X k=0 n X k=0 ak bk k=0 ! = n X (ak − αbk )(ak − αbk ) k=0 |ak − αbk |2 ≥ 0 and the inequality claimed follows. Thus we have n+1 n+1 2 n+1 n A(n + 2) = B(n + 2) = 5 1 + 4+ 4 + . we consider the polynomial x−1 x−1 2 x−1 n B(x) = 5 1 + 4+ 4 + . So. BARCELONA TECH. . . . . Barcelona. then it holds ! ! n n n X X 1 X Re α |ak |2 + |α|2 |bk |2 ak bk ≤ 2 k=0 k=0 Indeed. Let a0 .. an and b0 . bn be complex numbers. . applying two times the operator x dx to the binomial identity Pn Pn (1 + x)n = k=0 nk xk we get nx(1 + x)n−1 = k=0 k nk xk and n X n−1 2 n−2 2 n nx(1 + x) + n(n − 1)x (1 + x) = k xk k Now. bn be complex numbers. b1 ..288 Now. Indeed. a0 . BARCELONA TECH. From the preceding the inequality k k=0 claimed immediately follows and the proof is complete. n To prove the statement. we get ! ! n n n X X n3 2n − 2 X 1 2 2 n−1 |bk | |ak | + (n + 1) n2 Re ak bk ≤ n 2 n−1 k=0 k=0 k=0 n X n on account that k = n 2n−1 . Damascus. The last identity can be easily obtained k k=0 n X n k setting x = 1 in nx(1 + x)n−1 = k x . Now let us come to our problem. Higher Institute for Applied Sciences and Technology. we put into the claim α = k for 0 ≤ k ≤ n. Syria.289 Multiplying the first and the third and equating terms of degree n. Thus un ≥ 1 for every √ n ≥ 2. Using the simple inequality X+Y ≥ 4XY we conclude that v v ! u n u n X n n 2 X u uX √ 1 n+1 X n 2n − 2 2 2 2 t t |a | + |b | ≥ u |a | |bk |2 k k n k 2n n n−1 n k=0 k=0 k=0 k=0 . and we k obtain ! ! 2 X n n n X n 1 X 2 2 n 2 k Re ak bk ≤ |ak | + k |bk | k 2 k k=0 k=0 k=0 After adding up the above expressions. the sequence {un }n≥2 is increasing with u2 = 23 ≥ 1. Solution 2 by Omran Kouba. Let un be defined by n+1 n2 2n − 2 n + 1 2n − 2 un = 4 n · n = n−1 2 n 2 (n − 1) n 4 n−1 It is easy to check that un+1 (n + 2)(2n − 1) n−2 −1= −1= ≥1 un 2n(n + 1) 2n(n + 1) So. we obtain 2 n X 2n − 1 2n − 2 2 n k =n + n(n − 1) n−1 n−2 k k=0 (2n − 1)(2n − 2)! (2n − 2)! n3 2n − 2 =n + n(n − 1) = n!(n − 1)! n!(n − 2)! n−1 n as claimed. . ! n n n . X . X X . . ≥ |ak ||bk | ≥ . ak bk . ≥ Re ak bk . . . k=0 k=0 where used the Cauchy-Schwarz inequality. k=0 Also solved by Jos´ e Gibergans-B´ aguena. Barcelona. . BARCELONA TECH. Spain. The only even solution of equation (2) is k = 2. We study the solutions of the equation σ(22k − 1) = 3 · 22k−1 . and for prime powers p one has σ(pa ) = pa + . p−1 (1) We define also σ(1) = 1. p (3) There is equality only for n = p. Then remark that n = 22k − 1 = 24m − 1 = 16m − 1. p Let now be k an even solution to (2). Let k = 2m (m ≥ 1). Main results Theorem. Indeed. + p + 1 = pa+1 − 1 . p+1 k Now. 2. . First remark that the inequality pk+1 − 1 p+1 k ≥ ·p p−1 p (4) is true. with equality only for k = 1. AMS Subject Classification. By studying certain composition function of σ with other arithmetical functions (see [1]) we have encountered the equation σ(22k − 1) = 3 · 22k−1 . It is well-known that this function is multiplicative. a conjecture is stated. Proof. which is divisible by 5. 1. inequality (3) follows. Introduction Let σ(n) denote the sum of divisors of the positive integer n. (2) The aim of this note is to completely determine all solutions of (2) when k is even. for any m ≥ 1. Then σ(n) ≥ p+1 · n. Let p be a prime divisor of n. Let n = pk · N . For k odd. Keywords. inequalities. Sum-of-divisors function. where p is prime and (p. The following auxiliary result will be used: Lemma. . Proof. 11A25. . N ) = 1. since σ(n) = σ(pk ) · σ(N ) ≥ · p · N . which is true.290 MATHNOTES SECTION On an equation for the sum-of-divisors function ´ zsef Sa ´ ndor Jo Abstract. a simple computation shows that (4) is equivalent to pk−1 ≥ 1. as k ≥ 1 and p ≥ 2. 291 Clearly n is always divisible by 3, as n = 4k − 1. Let n = 3a · 5b · N be the prime factorization of N , where N is not divisible by 15. By applying the Lemma, we can write: 4 a 6 b 24 σ(n) ≥ ·3 · 5 · σ(N ) ≥ n, as σ(N ) ≥ N. 3 5 15 Therefore, the inequality 24 2k 3 · 22k−1 ≥ (2 − 1) (5) 15 must hold true. Written equivalently: 15 2k · 2 ≥ 8(22k − 1), or 22k ≤ 24 so k ≤ 2. 2 As k is even, we get k = 2. Remark. Unfortunately, the above method cannot be applied for k = odd. A such solution of equation (2) is k = 5. Conjecture. The only odd solution k to equation (2) is k = 5. References [1] J. S´ andor, On the composition of some arithmetic functions, III (in preparation) [For parts I and II, see: Studia Univ. Babe¸s-Bolyai, Math., 34(1989), 7-14; resp. J. Ineq. Appl. Pure Math., 6(2005), no. 2, art. 73 (electronic)]. J´ ozsef S´ andor: Babe¸s-Bolyai University Department of Mathematics, Str. Kog˘ alniceanu nr. 1 400084 Cluj-Napoca, Romania E-mail:
[email protected] 292 JUNIOR PROBLEMS Solutions to the problems stated in this issue should arrive before October 15, 2014 Proposals ´ 26. Proposed by Francisco Javier Garc´ıa Capit´ an, I.E.S. Alvarez Cubero de C´ ordoba, Spain. Let ABC be a rectangle triangle, CD the altitude corresponding to the right angle, and I, Ja , Jb the incenters of the triangles ABC, CAD and CDB, respectively. Find the area of the triangle IJa Jb in terms of the hypothenuse c and the inradius r of the triangle ABC. 27. Proposed by Ercole Suppa, Teramo, Italy. Let 4ABC a triangle with incenter I and circumcenter O. Let D, E, F be the touch points of the incircle (I) with BC, CA, AB respectively. Let V be the center of circle passing through F which touches BC at B. Let W be the center of circle passing through E which touches BC at C. Prove that if AC + AB = 2 · BC then O, I, V , W are collinear. 28. Proposed by Paolo Perfetti, “Tor Vergata” Univ., Rome, Italy. Let a, b, c be real positive numbers. Prove that r r r r 3a2 + 2b2 + 3c2 3b2 + 2c2 + 3a2 3c2 + 2a2 + 3b2 3 (a + b)(b + c)(c + a) + + ≥3 2ac 2ba 2cb abc 29. Proposed by Armend Sh. Shabani, University of Prishtina, Republic of Kosova. Let p be a fixed positive integer, greater than 2. Show that the mapping f : N → N ∪ {0} defined by h n i h 2n i h (p − 1)n i f (n) = + + ... + 2 3 p is neither injective nor surjective. 30. Proposed by D.M. Bˇ atinet¸u-Giurgiu, Bucharest, Romania and Neculai Stanciu, Buzˇ au, Romania. Determine all positive integers numbers a, b, c, d which satisfy the equation a2 + b3 + c4 = d5 . 293 Solutions 21. Proposed by Dorlir Ahmedi, University of Prishtina, Republic of Kosova a a, b, c > 0 and b+c ≥ 32 then prove that If a b c + + ≥ 2. b+c c+a a+b Solution by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria The proposed inequality is equivalent to ∆ ≥ 0 where ∆ = a(c + a)(a + b) + b(b + c)(a + b) + c(b + c)(c + a) − 2(b + c)(a + c)(a + b)171 Now, withwe s= b + c prove and p that = bc, we have Therefore should 2 3 ∆ = a3 −c a+2 sa + a(p b +−a s ) + s − 4ps + ≥8 c By assumption a = xs for some x b≥ 3/2, so or equivalently x ∆ = s3c x3b− xa2 − a(3 + u) + u + + + 4 ≥ 8. 2 b c b c 4p b−c wherecu =b1 − 2 = ∈ [0, 1]. aNowa if P (x) = x3 − x2 − x4 (3 + u) + u then Since + ≥ 2sone hasb + tocprove that + ≥ 6. b c b c 3 5u 3 12 − u 3 3 0 00 (3) From Cauchy-Schwarz inequality we have P = , P = , P = 7, P =6 2 8 2 4 2 2 a a b c + · + ≥4 Thus, b 3c a a k X P (k) (3/2) 3 Therefore P (x) = x− . ak! 3 2 a a k=0 + ≥4· ≥ 4 · ≥ 6. c b + c ∆ ≥ 20 which is the desired inequality. So, P (x) ≥ 0 for every x ≥b 32 . Consequently Note solved that equality holds if and only if ua= 0sand x = 3/2 that (a, b, c) =Stanciu, (2λ, λ, λ) Also by Titu Zvonaru, Comˇ ne¸ ti, Romania andis Neculai for some λ > 0. ”George Emil Palade” School, Buzˇ au, Romania, and the propser Also solved by Titu Zvonaru, Comˇ ane¸sti, Romania and Neculai Stanciu, 22. Proposed by Emanuele Callegari, Math. Dept., “Tor Vergata” University, “George Emil Palade” School, Buzˇ au, Romania; and the proposer Rome, Italy We have a rectangular chessboard 3 × 20 made of square boxes by Emanuele Callegari, “Tor University, 122. × 1. Proposed We also have 30 dominoes of size 1Math. × 2 or 2Dept., × 1 that we Vergata” want to use to cover Rome, Italy. We have a rectangular chessboard × 20 the chessboard. How many are the different ways3to do made that? of square boxes 1 × 1. We also have 30 dominoes of size 1 × 2 or 2 × 1 that we want to use to cover the Solution by the proposer chessboard. How many are the different ways to do that? The answer is 413403. Solution by the proposer. Let Xn be the number of different coverings of a chessboard of the following type coverings of a chessboard of the following type Let Xn be the number of different 3 2n Figure 1 with with little little rectangles rectangles of of dimensions dimensions 22× ×11 or or 11× ×2. 2. Observe that, if the chessboard is of type (2n + 1) × 3, the coverings are trivially zero because the chessboard has an odd number of cells. In order to solve our problem we must compute X10 . A possible strategy could be to look for a recursive formula for Xn but, after a few attempts, we probably realize that it is hard to do it. In a certain sense, it is better to be more ”ambitious” and to search non only Xn , but also the number Yn of different coverings of a chessboard of the following type: The answer is 413403. Let Xn be the number of different coverings of a chessboard of the following type 3 294 2n Figure 1 with little rectangles of dimensions 2 × 1 or 1 × 2. Observe that, if the chessboard Observe that, if the chessboard is of type (2n+1)×3, the coverings are trivially zero is of type (2n + 1) × 3, the coverings are trivially zero because the chessboard has because the chessboard has an odd number of cells. In order to solve our problem an odd number of cells. we must compute X10 . In order to solve our problem we must compute X10 . A possible strategy could be to look for a recursive formula for X but, after a A possible strategy could be to look for a recursive formula for Xnn but, after a few attempts, we probably realize that it is hard to do it. In a certain sense, it is few attempts, we probably realize that it is hard to do it. In a certain sense, it is better to be more ”ambitious” and to search non only Xn , but also the number Yn better to be more ”ambitious” and to search non only Xn , but also the number Yn of different coverings of a chessboard of the following type: of different coverings of a chessboard of the following type: 3 172 172 2n Figure 2 In fact, In172 fact, in in spite spite of of the the fact fact that that itit isis hard hard to to find find aa recursive recursive formula formula for for X Xn (and (and In fact, in),spite of the fact that it rather is hardsimple to findtoa write recursive formulaformula for Xnn (and also for Y we discover that it is a recursive n also for Yn ), we discover that it is rather simple to write a recursive formula for for alsofact, for in Ynspite ), we of discover simple write a recursive formula for both, simultaneously. In the factthat thatititisisrather hard to find atorecursive formula for Xn (and both, simultaneously. both, simultaneously. To this aim, let us start to place a little rectangle to cover the left upper cell of the also for Y ), we discover that it is rather simple to write a recursive formula for n To this aim, let us start to place a little rectangle to cover the left upper cell of the To thissimultaneously. aim, letchessboard us start to place a little can rectangle to cover the left upper cell of the non-rectangular both, non-rectangular chessboard of of fig.2. fig.2. We We can do do itit in in two two ways: ways: non-rectangular chessboard of fig.2. We can do it in two ways: To this aim, let us start to place a little rectangle to cover the left upper cell of the non-rectangular chessboard of fig.2. We can do it in two ways: 3 3 3 3 3 3 2n 2n 2n 3 2n 4 Figure Figure 2n 3 2n 4 Figure Figure When we start in the way of fig.3 the part of the chessboard that remains uncovered Figure 3 Figure 4remains start the way of fig.3 theXpart of the chessboard thatthe is When of thewe kind of in fig.1, so that it has coverings. On other uncovered hand, if n different When we start in the way of fig.3 the part of the chessboard that remains uncovered is start of the ofinfig.1, so of that it has Xn to different coverings. Onremains the other hand, if weWhen inkind the way of way fig.4 we are forced in the following way we start the fig.3 the part ofcontinue the chessboard that uncovered iswe of the kind of way fig.1,ofso thatwe it are has Xn different coverings. the other if in the fig.4 to continue in theOn following wayhand, is ofstart the kind of fig.1, so that it hasforced Xn different coverings. On the other hand, if we start in the way of fig.4 we are forced to continue in the following way we start in the way of fig.4 we are forced to continue in the following way 3 3 3 2n 2n 5 Figure 2n 5 Figure so that the part of chessboard still uncovered has Yn−1 different coverings. As a Figure 5 has Yn−1 different coverings. As a so that partthat of chessboard still uncovered result we the obtain result obtain so thatwethe part that of chessboard Ystill uncovered has n−1 different coverings. As a Xn + Yn−1 . YYn−1 so that the part of chessboard still has different coverings. As a n =uncovered result we obtain that n =. X n+Y n−1 result we obtain that Yn = Xn + YYn−1 Now, let us. consider the rectangle in fig.1. Now, let us consider the rectangle in fig.1. We can cover the left upper cell in two =X We canlet cover the left the upper cell inYntwo ways: n + Yn−1 . Now, us consider rectangle in fig.1. We can cover the left upper cell in two ways: ways: let us consider the rectangle in fig.1. We can cover the left upper cell in two Now, ways: 3 3 3 3 3 3 2n 2n 2n 6 2n 7 Figure Figure 2n 6 2n 7 Figure Figure If we cover it in the way of fig.6 we are forced to continue in the following way: IfIfwe cover it in the way of fig.6 we are forced to continue in Figure 6 Figure 7following we cover it in the way of fig.6 we are forced to continue inthe the followingway: way: If we cover it in the way of fig.6 we are forced to continue in the following way: 3 3 3 2n 2n Figure 8 2n 8 Figure so that the part of chessboard still uncovered has Yn−1 different coverings. On the 8 so that the part of chessboard stillFigure uncovered has Y different coverings. On the Now, let consider us consider rectangle in fig.1. cover upper in two Now, let us the the rectangle in fig.1. We We cancan cover the the left left upper cell cell in two ways: ways: 3 3 3 3 2n 2n 2n 2n Figure 6 Figure Figure 6 Figure 7 7 295 If cover we cover in the of fig.6 we are forced to continue in the following If we it initthe wayway of fig.6 we are forced to continue in the following way:way: 3 3 2n 2n Figure Figure 8 8 so that part of chessboard uncovered Y different different coverings. so that that thethe part of chessboard chessboard stillstill uncovered hashas different coverings. OnOn thethe n−1n−1 so the part of still uncovered has YYn−1 coverings. On the other hand, if we start in the way of fig.7, we can continue in two ways: other hand, if we start in the way of fig.7, we can continue in two ways: other hand, if we start in the way of fig.7, we can continue in two ways: 3 3 2n 2n Figure Figure 9 9 3 3 2n 2n Figure Figure 10 10 The region still uncovered in fig.9 has Yn−1 different coverings while the one of fig.10 has Xn−1 . Thus we obtain: Xn = 2Yn−1 + Xn−1 . It is now easy to verify by direct inspection that X1 = 3 and Y1 = 4. Therefore we obtain the recursive formula: ( Xn = 2Yn−1 + Xn−1 (1) Yn = Xn + Yn−1 with initial data: ( X1 = 3 Y1 = 4. After 10 iterations of (1), we obtain X10 = 413403, which is the answer to the problem. 23.Proposed by Emanuele Callegari, Math. Dept., “Tor Vergata” University, Rome, Italy. Mary has 5 baskets each containing 97 colored balls numbered from 0 to 96. The color of the balls in the first basket is blue, in the second is green and in the third is red. The color of the balls in the fourth and fifth baskets is white. Mary picks up a ball from each basket in such a way that the sum of the five numbers is 96. How many different configurations can occur to Mary? Solution by the proposer The answer is 200425. We need the Lemma: m+k−1 Lemma If k and m are strictly positive integers, then there exist exactly k−1 k−uples of non negative integers (n1 , n2 , . . . , nk ) such that n1 + n2 + · · · + nk = m. This is a standard combinatorial result whose proof is omitted. Let (n1 , n2 , n3 , n4 , n5 ) be any choice of balls, i.e for any i = 1, . . . , 5 let ni be the number that is written on the ball that has been picked up from the i-th . are . . a). 2 Therefore the number N2 of all (x. n2 . 2 2 2 2 (3) Thanks to the elementary properties of the binomial coefficients. Let us find N2 .. thanks to the Lemma.. Clearly. . from (2) and (3) it follows: 2 3 4 5 97 98 99 N2 + S = + + + + ··· + + = (4) 2 2 2 2 2 2 3 and 2 4 3 6 5 N2 − S = + − + − + . . n3 . Let’s call N1 the number of solutions of type I and N2 the number solutions of type II. thanks to the Lemma. (2) 2 2 2 2 2 a=0 To compute (2). n5 ) such that n1 + n2 + n3 + n4 + n5 = 96. (1) Note that. The sum N1 + N2 is our number. the number of 5-tuples of non negative integer of the form (x. 48. n4 . In the first one (solution of type I) n4 6= n5 while in the second one (solution of type II) n4 = n5 . z) satisfying (1). 2 2 2 2 2 98 97 . 1 + · 2 99 3 99 3 (5) + 492 . y. basket. such that x + y + z + a + a = 96. a. 100 4 Two incorrect solutions were received. . for every fixed a = 0. if we find N2 .e. z. the 3−tuples of 98 − 2a non negative integer (x. a. y. the relation above will allow us to find also N1 and therefore. Two different situations may occur. a) satisfying (1) is 48 X 98 − 2a 2 4 6 98 N2 = = + + + ··· + . 2 2 As a consequence of (4) and (5) we have: N2 = 1 1 · ((N2 + S) + (N2 − S)) = · 2 2 Thus. Clearly 100 2N1 + N2 = total number of 5−tuples = (0) 4 Thus. define the quantity 3 5 7 97 S= + + + ··· + .296 100 is the number of different 5−tuples 4 (n1 . + 49 2 (6) = 2000425. from (0) and (6) it follows: 1 1 N1 +N2 = ·((2N1 + N2 ) + N2 ) = · 2 2 which is the result. + − = 1 + 3 + 5 + · · · + 97 = 492 . y. i. z.. . 1. also N1 + N2 . f (x) + f (y) + f (z) ≥ 3f = 3 2 Proof. Bˇ atinet¸u-Giurgiu. Buz˘ au.Dambovita. we have 1 1 1 1 27 ≤ + + ≤ 2 x(1 − x) y(1 − y) z(1 − z) 2xyz 1 x(1−x) 1 + 1−x is convex on (0.297 24. Next. Romania and D. and p the semi-perimeter of the triangle. Serban Cioculescu school. Comˇ ane¸sti. Since ra (p − a) = rb (p − b) = rc (p − c) = rp = Area(ABC) we conclude.Proposed by Stanescu Florin. On the other hand. Solution by Omran Kouba. that ra rb rc rp rp rp + + = + + a b c a(p − a) b(p − b) c(p − c) r 1 1 1 = + + p x(1 − x) y(1 − y) z(1 − z) So. Gaesti. and the proposer 25. the function f (x) = = 1 x uv which is the first inequality. jud. Titu Zvonaru. r is the radius of the circle inscribed in the triangle. y) such that p p p p x2 + 2x + 1 + x2 − 4x + 4 + x2 − 2xy + y 2 + y 2 − 6y + 9 = 4. Prove that in a triangle ABC the following inequality holds: 27r ra rb rc p ≤ + + ≤ 2p a b c 2r where ra . Bucharest. z with x + y + z = 1. since u+v ≤ u+v 4 for every positive u and v. so x+y+z 27 . we come to the proposed problem. Romania.M. z = 1 − pc . “George Emil Palade” School. Higher Institute for Applied Sciences and Technology. y. Proposed by Neculai Stanciu. by the lemma we conclude that ra rb rc r 27r ≤ + + ≤ p a b c 2pxyz But xyz = r2 by Heron’s formula. Romania. . Romania. Romania Find all pairs of real numbers (x. Matei Basarab National College. Damascus. p2 Also solved by Neculai Stanciu. rb . and the desired inequality follows. 1). Syria We will use the following lemma: Lemma: for positive numbers x. y = 1 − pb . First. rc are the lengths of the rays of the excircles. by setting x = 1 − ap . we have zx xy yz + + xyz(f (x) + f (y) + f (z)) = y+z z+x x+y y+z+z+x+x+y 1 ≤ = 4 2 and the upper inequality follows. Buzˇ au. y) ∈ {(2. we conclude that |x + 1| + |x − y| + |y − 3| ≥ |x + 1 + y − x| + |3 − y| ≥ |1 + y + 3 − y| = 4. For x = 2 equation (1) becomes |2 − y| + |y − 3| = 1. a)|2 ≤ a ≤ 3}.298 Solution by the proposer. hence |x − 2| = 0 ⇔ x = 2. (1) Since |a| + |b| ≥ |a + b| and | − a| = |a| for every real numbers a and b. (2) . We note first that p p p p x2 + 2x + 1 + x2 − 4x + 4 + x2 − 2xy + y 2 + y 2 − 6y + 9 = 4 is equivalent to |x + 1| + |x − 2| + |x − y| + |y − 3| = 4. After solving last equation one obtains solutions (x. From (1) and (2) we have: 4 = |x + 1| + |x − 2| + |x − y| + |y − 3| ≥ |x − 2|. ) Solution: Let T (2n) be the number of ways to tile a 3 × 2n rectangular chessboard using dominoes. .) This problem seeks a number for T (20). (To motivate this convention. Sullivan Emmanuel College. Dominos on a rectangular board Solution to Junior Problem 22 MathProblems (4)1 (2014) Brendan W. In this note we address the question of counting the number of different ways to do that? (Problem proposed by Callegari Emanuele. observe that T (2) = 3: Now. “Tor Vergata” University. Math. USA Abstract: We have a rectangular chessboard 3 × 20 made of square boxes 1 × 1. consider a 3 × 2n board. The articles. We will describe all of the cases wherein one can construct a domino tiling of this board. proposed at MathProblems or elsewhere. We also have 30 dominoes of size 1 × 2 or 2 × 1 that we want to use to cover the chessboard. We solve this problem by proving the more general recursive formula T (2n) = 4 · T (2n − 2) − T (2n − 4) for n ≥ 2. Rome. Dept. more specifically.299 SOLUTION OF A PROBLEM ANOTHER PERSPECTIVE! This section of the Journal presents a solution of a problem. First. with the additional convention that T (0) = 1. These disjoint cases are based on the dominoes in the (2n − 2)-th and (2n − 1)-th columns of the board. (Note: there are trivially no domino tilings of a 3×k board when k is odd. consider: How many ways can we tile an empty board? One way: I just did it!) This will lead us to the answer: T (20) = 413403. should contain ideas and information that readers may find suitable to use in other similar problems. The presentation of such solution should be done in the style of an article. Italy. appended to a proper tiling of the first 2n − 2 columns. if the one and only crossing occurs on the top row. We know that there are 3 ways to tile the final two columns (since T (1) = 3). the situations where the one and only crossing occurs on the bottom row yield no tilings. Finally. By exhibiting three cases. ··· 3 × 2n − 2 portion boundary (i) 0 crossings. Firstly. this case contributes the term 3 · T (2n − 2) to the total number of tilings. when the one and only crossing occurs in the middle row. This means we have a proper tiling of the final two columns. we can attempt to tile the surrounding squares in two ways: ··· ? ? ··· Secondly. we have one possibility: ··· ? . we can show that this case actually contributes no proper tilings: each possibility yields an irreconcilable situation in the 2n-th column. (ii) 1 crossing. as well.300 these cases depend on how many domino tiles cross the boundary between those two columns. and there are T (2n−2) ways to tile the remaining portion. Thus. by obvious symmetry. We can extend this to the left by converting the leftmost vertical domino into two horizontal ones. this will be analogous to the cases where the two crossings occur on the bottom two rows.301 Thus. which can be appended to any proper tiling of the first 2n − 4 columns. (By symmetry. then this yields no tilings: ? ··· ? Next. the terms we derive presently will then be doubled to account for all the tilings. Furthermore. the number of such tilings is T (2n − 4). which can be appended to any proper tiling of the first 2n − 6 columns. in blocks of two columns. Thus. the number of such tilings is T (2n − 6). . This is the interesting case! First. (iii) 2 crossings. Thus. each case can be extended arbitrarily far to the left. we consider the more fruitful cases where the two crossings occur on the top two rows. yield proper tilings. like so: ··· 3 × 2n − 6 portion This amounts to a particular proper tiling of the final 6 columns. One possibility is as follows: ··· 3 × 2n − 4 portion This amounts to a particular proper tiling of the final 4 columns. thus. indeed. Thus. we will generate tilings that correspond to this summation of terms: T (2n − 4) + T (2n − 6) + T (2n − 8) + · · · T (2) + T (0). the case of exactly 1 crossing yields no tilings. we observe that if the two crossings occur on the top and bottom row.) What we will observe is that such cases do. “appended” to the only tiling of the empty board. . or even just counting to observe that T (2) = 3 and T (4) = 11. The next step is to simplify this recurrence to two terms. Consider the difference T (2n) − T (2n − 2). we include the terms of the summation mentioned above to account for all of these tilings.e. originally. In total. Plugging in n = 10 (i. then. Observe that this case yields no proper tilings: ··· ? Overall. That final case amounts to a particular proper tiling of the entire board. T (2n) = 2 3 2 3 3+ 3 This can be used somewhat more efficiently for larger n. 2n = 20) yields the same answer as above.e. as mentioned above. generating functions). Finally. Therefore. so one can solve for a closed form using standard techniques (i. the cases where the two crossings of the boundary column occur on the bottom two rows are symmetrically analogous.302 Observe that this extension process can be continued until we reach the leftmost side of the board. This yields & ' √ ! √ √ ! √ n √ n −1 + 3 (2 + 3)n+1 1+ 3 √ √ √ · (2 + 3) + · (2 − 3) = . we can plug numbers into this recurrence and deduce that n 0 1 2 3 T (2n) 1 3 11 41 4 5 153 571 6 7 2131 7953 8 9 29681 110771 10 413403 Thus. note that the above recurrence is linear. Also. (iv) 3 crossings. and observe that we can cancel terms: T (2n) − T (2n − 2) = 3T (2n − 2) + 2T (2n − 4) + 2T (2n − 6) + · · · + 2T (0) − 3T (2n − 4) + 2T (2n − 6) + · · · + 2T (0) = 3T (2n − 2) − T (2n − 4) Adding T (2n − 2) to both sides. Knowing that T (0) = 1 (by convention). the desired answer is that there are T (20) = 413403 ways to properly tile a 3 × 20 rectangular chessboard using dominoes. we have now deduced that T (2n) = 3 · T (2n − 2) + 2 · n−2 X T (2k) k=0 provided n > 1 (otherwise the first part of the case involving 2 crossings above is invalid). we include the terms Pn−2 2 · k=0 T (2k). we conclude that T (2n) = 4T (2n − 2) − T (2n − 4) as claimed. for n ≥ 1. Ercole Suppa.com Solutions to the problems stated in this issue should arrive before June 15. Issue 1 (2013). each indicating the name of the sender. Questions concerning proposals and/or solutions can be sent by e-mail to: mathproblems-ks@hotmail. url: http://www.Mathproblems ISSN: 2217-446X. Prove that ∞ X ∞ X (i − 1)!(j − 1)! i+j a a a = 2Li2 − 2Li2 − − 2Li2 (a). Roberto Tauraso. c 2010 Mathproblems. Francisco Javier Garca Capit´ an. Proposals should be accompanied by solutions. Ovidiu Furdui. Enkel Hysnelaj.com Volume 3. where ζ(z) represents the Riemann zeta function. Technical University of Cluj-Napoca. J´ ozsef S´andor. Solutions will be evaluated for publication by a committee of professors according to a combination of criteria. Stone. Anastasios Kotronis. An asterisk (*) indicates that neither the proposer nor the editors have supplied a solution. Armend Sh. Kosov¨ e. Cluj-Napoca. 2) Show that ∞ X ∞ X (i − 1)!(j − 1)! i=1 j=1 (i + j)! = ζ(2). Universiteti i Prishtin¨ es. Proposed by Ovidiu Furdui. Romania. Paolo Perfetti. Cristinel Mortici. PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets. 118 R z ln(1 − t) dt for 0 t . Teachers can help by assisting their students in submitting solutions. 2013 Problems 58. Emanuele Callegari. (i + j)! 2−a 2−a i=1 j=1 where Li2 (z) is the Dilogarithm function defined by Li2 (z) = − all z ∈ {z ∈ C : |z| ≤ 1}. 1) Let a be a real number such that −1 ≤ a ≤ 1. Student solutions should include the class and school name. Pages 118–139 Editors: Valmir Krasniqi. Jos´ e Luis D´ıaz-Barrero. The editors encourage undergraduate and pre-college students to submit solutions.mathproblems-ks. Drawings must be suitable for reproduction. Valmir Bucaj. David R. Mih´aly Bencze. Prishtin¨ e. Shabani. If a ∈ R+ . Prove that √ p Ln L2n+2 Ln+1 L2n+3 2 + + (Ln + Ln+2 ) ≥ 2 6 · Ln Ln+1 · Ln+2 . c1 x2 + 2b2 x + a1 = 0 has real roots. Bucharest. Romania and Neculai Stanciu. Romania and Neculai Stanciu. Proposed by Mih´ aly Bencze. Shandong Province. ck ∈ R∗ (k = 2 1. 1]. Romania (Jointly). 0. x ∈ [0. Buzˇ au. Department of Mathematics and Information Science. Proposed by Anastasios Kotronis. China. Matei Basarab National College. B˘ atinet¸u-Giurgiu. c.119 59. Show that X n≥0 1 = (2n + 1)(3n + 2) · · · (kn + k − 1) k 1 X m−1 k (−1) (m − 1)mk−2 k! m=2 m ! m−1 X π π 2`π `π cot − ln m + cos · ln sin . b ∈ R. Buzˇ au. d. George Emil PaladeP Secondary School. 63. Gaesti. Sydney. 0 0 65.M. Matei Basarab National College. Bucharest. b. xk ∈ R∗+ . which satisfy the relation 7+x f (x) + f = ax + b 2−x where a. bk . 1≤k≤n then prove that n X aXn + bxk k=1 cXn − dxk ≥ (an + b)n . 256603. Binzhou University. −1. Romania Show that if ak . Romania. 2 m m m `=1 64. University of Technology. Athens. George Emil Palade Secondary School. Binzhou City. Let n be a positive integer. 1. 61. Ln = Ln−1 + Ln−2 . Proposed by Florin Stanescu. Proposed by D. then at least one of the equations a1 x2 + 2c2 x + b1 = 0. Let k ≥ 3 be an integer number. Proposed by D. 60. B˘ atinet¸u-Giurgiu. Let f : [0. such that a1 + b21 + c21 = a22 + b22 + c22 . Determine all functions f : R \ {−2. 1] → R be a continuous function. Ln+3 Ln + Ln+2 where Ln represents the nth Lucas number defined by L0 = 2. Bra¸sov. Ron mania (Jointly). then prove that Z 1 2 Z 1 f 2 (x)dx ≥ 24 f (x)dx . Proposed by Enkel Hysnelaj. 0 where F (x) = Rx 0 1/2 f (t)dt. 2} → R. If Z 1 Z 1 2 x f (x)dx = −2 F (x)dx. and for all n ≥ 2. Greece. L1 = 1.∗ Proposed by Li Yin. Australia. 2). Xn = k=1 xk and cXn > d max xk .M. b1 x2 + 2a2 x + c1 = 0. Let Ωn . cn − d 62. Serban School Cioculescu. 3818 · · · . . show that r r 1 (Ωn+1 ) n+1 n+2 α n + 2 6 β 6 1 n+3 n+3 n (Ωn ) with the best possible constant factors α = 2 and β = ln( 34 ) √ ln( π 2 ) = 2. . that is Ωn = . . For all natural numbers n. 2 .120 π n/2 represents the volume of the unit ball in Rn . Γ(n/2 + 1) R ∞ x−1 −t where Γ(x) = 0 t e dt. x > 0. n = 1. 121 Solutions No problem is ever permanently closed. Let f : (a. +∞) the sequence {f (k + x)}k≥0 is decreasing to zero. Syria. We will be very pleased considering for publication new solutions or comments on the past problems. and its sum satisfy the following inequality: ∞ X 1 1 (−1)k f (x + k) ≤ − f 0 (x). First. So. letting m tend to +∞. If ` < 0. since f 0 is increasing and negative. there exists a real number ` ≤ 0 such that limx→∞ f 0 (x) = `. Athens. 50. for every x > a. we must have ` = 0. Now. Then. Proposed by Anastasios Kotronis. f 0 ≤ 0 and f must be nonnegative and decreasing. then the convexity of f implies that f (x) ≥ f (b) + (x − b)f 0 (b) and this leads to the contradiction limx→∞ f (x) = +∞. More generally we have : Proposition. ∞) → R be a continuously differentiable P∞ convex function such that limx→∞ f (x) = 0. then the convexity of f implies that for y > b we have f (y) − f (b) ≤ f 0 (y)(y − b) ≤ `(y − b). +∞). Show that +∞ X k=0 (−1)k 1 = + O(n−1 ln−2 n) ln(n + k) 2 ln n n → +∞. for each x ∈ (a. note that the conditions on f imply that f is nonnegative decreasing and that limx→∞ f 0 (x) = 0. if for some b ∈ (a. let S(x) be defined. Taking the limit as y tend to +∞ we obtain the contradiction −f (b) ≤ −∞. Indeed. 0 ≤ − f (x) + 2 2 k=0 Proof. On the other hand. and P the alternating series test proves the convergence of the series (−1)k f (x + k). Damascus. +∞) we have f 0 (b) > 0. So. by S(x) = ∞ X (−1)k f (x + k) = k=0 ∞ X f (2k + x) − f (2k + 1 + x) k=0 On the other hand f (x) = f (x + 2m + 2) + m X f (2k + x) − f (2k + 2 + x) k=0 So. for x ∈ (a. So. Higher Institute for Applied Sciences and Technology. the series k=0 (−1)k f (x + k) does converge. Greece. we obtain f (x) = ∞ X k=0 f (2k + x) − f (2k + 2 + x) . Solution by Omran Kouba. again using the convexity of f . Indeed. Calculate: n X (n + k)a−1 lim n 2 − exp n→∞ (n + k)a + b !! k=1 Solution 1 by Moti Levy. The first step is to show that the limit does not dependent of a and b. Romania (Jointly). B˘ atinet¸u-Giurgiu. George Emil Palade Secondary School. Proposed by D. Rome. Tor Vergata University. Matei Basarab National College. 51. Moubinool Omarjee. X (−1)k 1 1 1 ≤ ≤ + 2 ln x ln(x + k) 2 ln x 2x ln2 x k=0 from which the statement follows. and the proposer. Also solved by Paolo Perfetti. Let a ∈ (1. f (x) = 1/ ln(x) we obtain +∞ ∀ x > 1. France. we have f (2k + 1 + x) ≤ 1 (f (2k + x) + f (2k + 2 + x)) 2 It follows that 0≤ m X f (2k + x) − 2f (2k + 1 + x) + f (2k + 2 + x) ≤ f 0 (2m + 2 + x) − f 0 (x) k=0 So letting m tend to +∞ we obtain using (3) that 0 ≤ 2S(x) − f (x) ≤ −f 0 (x). Buzˇ au. we get 2S(x) − f (x) = ∞ X f (2k + x) − 2f (2k + 1 + x) + f (2k + 2 + x) k=0 But using the convexity of f we have f (2k+1+x)−f (2k+x) ≥ f 0 (2k+x) and f (2k+2+x)−f (2k+1+x) ≤ f 0 (2k+2+x) Thus f (2k + x) − 2f (2k + 1 + x) + f (2k + 2 + x) ≤ f 0 (2k + 2 + x) − f 0 (2k + x) On the other hand. ∞) and b ∈ (0. Bucharest. ∞).122 Combining the preceding. Paris. +∞) → R. Applying the Lemma to the function f : (1. Department of Mathematics.M. we have n n a−1 X (n + k) 1X = a n (n + k) + b 1+ k=1 k=1 1 k n + b n(n+k)a−1 . Romania and Neculai Stanciu. Italy. 123 If we put en. 0 ≤ δ ≤ .k = b . n 2 na 2 2 na 2 na 1 + nk k=1 !! n n a−1 X 1X 1 (n + k) − n 2 − exp n 2 − exp a n (n + k) + b 1+ k=1 k=1 ! !! n n a−1 X (n + k) 1X 1 = n exp − exp a k n (n + k) + b 1+ n k=1 k=1 ! ! 2 n 1 b 1X 1 1 1 b ≤ n exp + exp (δ) n 2 na 2 2 na 1 + nk !! k n k=1 Z1 n 1X 1 n 1+ k=1 k n ≤ 1 dx = ln 2 1+x 0 !! !! n n a−1 X 1X 1 (n + k) n 2 − exp − n 2 − exp a n (n + k) + b 1 + nk k=1 k=1 ! ! 2 2 1 b 1 1 b 1 b 1 1 b ≤ n exp (ln 2) exp (δ) = 2n exp (δ) + + 2 na 2 2 na 2 na 2 2 na 1 b2 exp (δ) 4 n2a−1 It follows from the preceding that = b na−1 + n a−1 X (n + k) lim n 2 − exp a n→∞ (n + k) + b k=1 !! !! n = lim n 2 − exp n→∞ 1X 1 n 1+ k=1 k n . . n(n+k)a−1 n 1X 1 n 1+ k=1 n 1X = n ≤ 1 n k=1 n X k=1 1 ≤ k ≤ n.k − = k n 1 + nk + en. 0 ≤ δ ≤ h 2 ! ! n n 1X 1 1X 1 − exp exp b n n 1 + nk 1 + nk + n(n+k) a−1 k=1 k=1 ! ! n 2 1 b 1 1 b 1 b 1X 1 ≤ exp + exp (δ) . h2 exp (x + h) − exp (x) = exp (x) h + exp (δ) .k ≤ n k n − 1X n 1+ k=1 b na and 1 k n + b n(n+k)a−1 ! ! n 1 1X en.k n k=1 ! Z 1 n 1 en.k 1 + nk 1 + nk + en. then 0 < en.k 1 1 b b 1X b ≤ a ≤ a 2 dx = 2 na . k 2 k 2 n n n 0 (1 + x) 1+ 1+ 1 1+ k=1 n 0≤ 1 n n X 1 1+ k=1 k n − 1 n n n X k=1 1 1+ k n + b n(n+k)a−1 ≤ 1 b 2 na By Taylor’s theorem. Since the function 1+x is convex then the area of the triangle formed by the secant is greater than ∆n . Namely. 2 1 So now we estimate ∆n .124 P n The second step is evaluation of limn→∞ n 2 − exp n1 k=1 Let ∆n := ln 2 − 1 n 1 k 1+ n . ! ! n−1 n−1 1 1 X 1 1 1 X n = − ∆n ≤ k k k+1 2n 2n 1 + 1 + k+1 1 + 1 + n n n n k=0 k=0 ! ! ! n−1 n X 1 X 1 3 1 1 ≤ 2 2 = 2 2 + 2n 2n 4 1 + nk 1 + nk k=0 k=1 ! Z 1 n 3 1 1X 1 1 dx 1 3 3 + + 2 = = ≤ + 2. ! ! n n n−1 1 1 3 1 X 1 1X 1 1X 1 ∆n ≥ 2 = = − k 2 k 2 k 2 2n 2n n 2n n 4n k=1 1 + n k=1 1 + n k=0 1 + n ! Z 1 1 dx 3 1 3 ≥ = − 2. 2 2 2 2n n 8n 2n 0 (1 + x) 8n 4n 8n 1+ k k=1 n 1 1+x Since the function is convex then the area of the triangle formed by the tangent is less than ∆n . 2 − 4n 2n 4n 8n 0 (1 + x) . That is. Pn 1 k=1 1+ k n then ! n 1X 1 = exp n 1 + nk k=1 exp (ln 2) − exp ! n 1X 1 n 1+ k=1 k n 1 ∆n + ∆2n exp δ 2 and !! n lim n 2 − exp n→∞ 1X 1 n 1+ k=1 k n ! n = lim exp n→∞ 1X 1 n 1+ k=1 k n 1 2 n∆n + n∆n exp δ . . . . . ∆n − 1 . ≤ 3 . . 4n . 4 2 Solution 2 by Anastasios Kotronis. Athens. We use the fact that Hn = ln n + γ + 1 + O(n−2 ) 2n n → +∞. Greece. 8n2 !! n 1X 1 lim n 2 − exp n→∞ n 1 + nk k=1 ! n 1X 1 1 2 = lim exp n∆n + n∆n exp δ n→∞ n 2 1 + nk k=1 ! n 1X 1 1 2 = lim exp lim n∆ + lim n∆ exp δ n n n→∞ n→∞ n→∞ 2 n 1 + nk k=1 1 1 = exp (ln 2) +0 = . (1) . Paolo Perfetti. Adrian Naco. Polytechnic University. Italy. Damascus. Sa.b. km (2) Applying Ces` aro–Stolz criterium. 52. Albania.1} ) −→ . Tor Vergata University. Moubinool Omarjee. France.2} + O(n ) = +O(n− min{a−1. b c 3 a 2 2 2 cos + cos + cos > b+c a+c a+b 2 . Department of Mathematics. c > 0.n ) = n 2 − 2 1 − 4n 2 2 Also solved by Omran Kouba.2} . 4n = ln 2 − 1 −b 4n On account of the above 1 1 a>1 1 − min{a.125 Note that for m > 1. Higher Institute for Applied Sciences and Technology. and the proposer.n n n X X 1 b (n + k)a−1 −2a = 1− + O(n ) := (n + k)a + b n+k (n + k)a k=1 k=1 = H2n − Hn − b n X k=1 1 + O(n−2a ) (n + k)a+1 n X 1 + O(n−2 ) (n + k)a+1 k=1 ! 2n n X X 1 1 1 = ln 2 − −b − + O(n−2 ) 4n k a+1 k a+1 k=1 k=1 X X 1 1 1 − b − + O(n−2 ) = ln 2 − 4n k a+1 k a+1 k≥n+1 k≥2n+1 1 = ln 2 − + O n− min{a. Syria.b. m−1 Now for a > 1 and b ∈ R. Tirana. it is X k≥n+1 1 = O(n1−m ). we have lim nm−1 n→+∞ −(n + 1)−m 1 = lim m n→+∞ (n + 1)1−m − n1−m k k≥n+1 −1 = lim m − 1 + O(n−1 ) X n→+∞ → 1 . Let a. prove that. Rome. Proposed by Yuanzhe Zhou. n (2 − exp Sa. b. Paris. The School of Physics and Technology (SPT) at Wuhan University. Case 1) z < 1 2 : f (x) + f (y) + f (z) ≥ (−2 (1 − cos 2) x + 1) + (−2 (1 − cos 2) y + 1) + (−2 (1 − cos 2) z + 1) = −2 (1 − cos 2) (x + y + z) + 3 = −2 (1 − cos 2) + 3 > 0 Case 2) z ≥ 1 2 : x+y < 1 2 f (x) + f (y) + f (z) ≥ −2 (1 − cos 2) x + 1 − 2 (1 − cos 2) y + 1 − 1 ≥ −2 (1 − cos 2) (x + y) + 2 − 1 = 1 − (1 − cos 2) (x + y) ≥ 1 − 1 (1 − cos 2) > 0. By repeated integration by parts one can easily show that for any non negative integer k and any positive integer n. and the proposer.. 2 ≤x≤1 then f (x) ≥ g (x) for 0 < x ≤ 1. Greece. Technical University of Cluj-Napoca. 53. Setting x = a b c . Then it will be suffice to consider the following two cases: 1) z < 21 and 2) z ≥ 21 . lnn x Pm−1 dx = n! n+m−1 i m (−1) · (m−1)! · i=1 (−1) · s(m − 1. 1 2 x + y + z = 1. 2 Also solved by Omran Kouba. k)z . + 1 2 cos 2α . 1−x 1−y 1−z 2 Using the well-known trigonometric identity cos2 α = 2x . 0 < x < 12 g (x) = 1 −1. Prove that ( Z 1 (−1)n · n! · ζ(n + 1). k). m ≥ 2. the inequality becomes f (x) = cos 1−x f (x) + f (y) + f (z) > 0. Solution 1 by Anastasios Kotronis. i) · ζ(n + 1 − i). + +c k+1 (k + 1)2 (k + 1)3 (k + 1)n+1 . Let n and m be nonnegative integers with n > m − 1. Romania. we have Z k n x ln x dx = x k+1 lnn x n lnn−1 x n(n − 1) lnn−2 x (−1)n n! − + − . Damascus. Cluj-Napoca. 0 (1 − x) where ζ denotes the Riemann zeta function.126 Solution by Moti Levy. Athens. The Stirling numbers of the first kind denoted by s(n. m = 1. and setting x + y + z = 1. Proposed by Ovidiu Furdui. y= . are the special by the generating function z(z − 1)(z − 2) · · · (z − n + 1) = Pn numbers defined k k=0 s(n. z= . 1 2. Let g (x) be defined by −2 (1 − cos 2) x + 1. The function f (x) is concave for 0 < x < so we can find a piecewise linear lower bound for f (x).. a+b+c a+b+c a+b+c then the inequality becomes x y z 3 cos2 + cos2 + cos2 > . Higher Institute for Applied Sciences and Technology. we may assume that 0 < x ≤ y ≤ z ≤ 1. Syria. WLOG. Z 1 n ln x dx = (−1)n n!ζ(n + 1) 1 −x 0 = and for m ≥ 2: Z 0 1 lnn x dx (1 − x)m = (−1)n+m−1 X m−1 X (−1)i s(m − 1. Damascus. 1−x k=0 . we have Z 1 Z 1X X k + m − 1 Z 1 lnn x k+m−1 k n dx= x xk lnn x dx ln x dx = m k k 0 (1 − x) 0 k≥0 0 k≥0 X k + m − 1 (−1)n n! = k (k + 1)n+1 k≥0 −k−1 X k+m−1 X m−1 m−1 n n+m−1 = (−1) n! = (−1) n! (k + 1)n+1 (k + 1)n+1 k≥0 k≥0 But for m ≥ 2. 1) Now. 1). −k−1 m−1 = 1. β+1 (α + 1) (α + 1)β+1 0 0 0 Now.127 Furthermore. First let us calculate the integral 0 xα lnβ (1/x) dx for α > −1 and β > 0. so. for x ∈ [0. i)ζ(n − i + 1) i=1 and we get what we wanted. we will use that (1 − x)−m = X k + m − 1 k≥0 k xk x ∈ (−1. Syria. i)(−k − 1)i . Solution 2 by Omran Kouba. Higher Institute for AppliedR Sciences and 1 Technology. (m − 1)! i=0 and for m = 1. Using the change of variables x → e−t we have Z 1 Z ∞ Z ∞ 1 Γ(β + 1) β α β −(α+1)t x ln (1/x) dx = t e dt = uβ e−u dt = . we have −k − 1 (−k − 1)(−k − 1 − 1)(−k − 1 − 2) · · · (−k − 1 − (m − 1 − 1)) = m−1 (m − 1)! m−1 X 1 s(m − 1. i) n! (m − 1)! (k + 1)n−i+1 i=0 k≥0 = (−1)n+m−1 m−1 X X n! 1 (−1)i s(m − 1. i) (m − 1)! i=0 (k + 1)n−i+1 k≥0 = (−1)n+m−1 n! (m − 1)! m−1 X (−1)i s(m − 1. we have ∞ X (− ln x)n = xk (− ln x)n . from the definition of Stirling numbers. 1] and exchange the order of summation Z 1 n ∞ Z 1 ∞ X X ln x 1 = (−1)n n!ζ(n + 1) dx = xk lnn x dx = (−1)n n! (k + 1)n+1 0 1−x 0 k=0 k=0 This gives the case m = 1. i)z i = z(z−1)(z−2) · · · (z− m + 2). 0) = 0 for m ≥ 2. i)(−1)i (j + 1)i = (−1)m−1 (j + 1)(z + 2) · · · (j + m − 1) i=0 or m−1 m−1+j (−1)m−1 X s(m − 1. by definition we have i=0 s(m−1. and the proposer. i)ζ(β + 1 − i) m (m − 1)! 0 (1 − x) i=0 Also solved by Moti Levi. (In fact.128 Since all the terms are positive we can integrate on [0. we obtain m−1 X s(m − 1. for a real β and an integer m such that 1 ≤ m ≤ β we have Z 1 β m−1 ln (1/x) (−1)m−1 Γ(β + 1) X dx = (−1)i s(m − 1. 1) we have ∞ ∞ X X −m 1 m−1+j j j = (−x) = x (1 − x)m j m−1 j=0 j=0 (1) Pm−1 On the other hand. i) (j + 1)i xj (1 − x)m (m − 1)! i=0 j=0 It follows that Z 0 1 Z 1 m−1 ∞ X lnn x (−1)m−1 X n dx = (−1)i s(m − 1. i) n+1−i (m − 1)! (j + 1) i=0 j=0 = m−1 (−1)n+m−1 n! X (−1)i s(m − 1. Replacing z by −(j + 1). Now suppose that m ≥ 2. For x ∈ [0. i)ζ(n + 1 − i) (m − 1)! i=0 which is the desired result. more generally. i)(−1)i (j + 1)i = (m − 1)! i=0 m−1 Replacing in (1) we obtain m−1 ∞ X (−1)m−1 X 1 i = (−1) s(m − 1. the above formula includes the case m = 1 since s(0. 0) = 1. The same proof shows that. since s(m − 1. i) (j + 1)i xj ln xdx (1 − x)m (m − 1)! i=0 0 j=0 m−1 ∞ X n! (−1)m−1 X i i n (−1) s(m − 1. i) (j + 1) (−1) = (m − 1)! i=0 (j + 1)n+1 j=0 m−1 ∞ n+m−1 X X (−1) n! 1 = (−1)i s(m − 1.) Remark. . Moti Levi. Let x. The dominated convergence theorem allows us to take the limit under integral getting 1 nt + 1 nt − 1 n lim f (t) 4 sinh sinh dt = n→+∞ R 2 2 + 1 Z Z nt − 1 n nt + 1 sinh dt = f (t)et dt f (t) lim 4 sinh n→+∞ 2 2 R+ R+ Z Also solved by Omran Kouba. the space of Lebesgue integrable functions. α be real positive numbers. and the proposer. . France. Syria. Proposed by Enkel Hysnelaj. z. University of Technology. Greece. Paris. Tor Vergata University. Let f : [0. putting etn from which follows e2 − x(e2 + 1) ≤ 0 This holds true since x ≥ 1. Athens. Anastasios Kotronis. Proposed by Moubinool OMARJEE. +∞[→ R be a measurable function such that g(t) = et f (t) ∈ L1 (R+ ) . Rome. Damascus.129 54. we have √ √ √ √ 1 x e √ −√ e x− √ √ ≤x e x e x sinh Indeed. Italy. Now define nt − 1 nt + 1 sinh fn (t) = f (t) 4 sinh 2 2 n1 |fn | ≤ et |f (t)| ∈ L1 (R+ ) by the Lebesgue integrability of et f (t). Show that if X nx3 + (n + 1)x =α x2 + 1 cycl then X1 2α 27n3 > + 2 x 3 9n α + α3 cycl where n is a natural number. y. Department of Mathematics. Australia. Higher Institute for Applied Sciences and Technology. We prove that nt − 1 nt + 1 sinh ≤ etn 2 2 = x. Find n1 Z nt + 1 nt − 1 lim f (t) 4 sinh sinh dt n→+∞ R 2 2 + where sinh(x) = ex −e−x 2 Solution by Paolo Perfetti. 55. Sydney. that k = An + Bn .53 = 201. 50 is false. Let x = y = z = 1 and α = 301. Syria. 3 1 2α 27n3 2×301. BARCELONA TECH. Spain.5+301. Damascus. Barcelona. cycl Clearly 3 > 201. . and the proposer. Higher Institute for Applied Sciences and Technology. 56.130 Solution by Moti Levi. Fn−1 Fn Fn+1 Now 3 2 2 2 Fn3 − Fn+1 + Fn+2 Fn−1 = −Fn−1 (Fn2 + Fn+1 + Fn Fn+1 ) + Fn+2 Fn−1 2 2 2 = Fn−1 Fn+2 − Fn − Fn+1 − Fn Fn+1 2 = Fn−1 (Fn+1 + Fn )2 − Fn2 − Fn+1 − Fn Fn+1 = Fn−1 (2Fn Fn+1 − Fn Fn+1 ) = Fn−1 Fn Fn+1 Thus An = 1 and consequently k = 1. which implies that the claim in Problem 55 is not true.5. Let n ≥ 2 be a positive integer.5 + 9×100227×100 x = 3 and 3 + 9n2 α+α3 = 3 ×301. An = Bn = 2 2 Fn+2 Fn2 Fn+1 (Fn + Fn−1 − Fn+1 ) = 0 Fn−1 Fn Fn+1 On the other hand An = 3 2 + Fn+2 Fn−1 Fn3 − Fn+1 . Proposed byJos´e Luis D´ıaz–Barrero. Let 2 2 Fn+2 Fn+1 Fn2 + − Bn Fn−1 Fn+1 Fn Fn+1 Fn−1 Fn 1 1 1 2 2 = Fn2 Fn+1 Fn+2 + − Fn−1 Fn+1 Fn Fn+1 Fn−1 Fn so. then n = 100 satisfies the X condition. Author’s Comment: I assume that either I did not understand the problem statement or there are typo errors. Find all possible values of the number k such that 2 2 2 2 2 Fn+2 ) Fn+2 (1 + Fn2 Fn+1 ) Fn2 ) Fn2 (1 + Fn+1 F 2 (1 + Fn+2 + = k + n+1 Fn−1 Fn+1 Fn Fn+1 Fn−1 Fn Solution by Omran Kouba. Also solved by Moti Levi. 50. Clearly. Proposals are always welcomed. F themselves form an equilateral triangle. Compute sin 1◦ + sin 2◦ + . 42. A knight is placed at random in one of the n2 squares of a chessboard. + sin 133◦ + sin 134◦ cos 1◦ + cos 2◦ + . Let n be an odd positive integer. Find the value of the following expression n+m 1 X k X n m 7 23n i j k=0 i+j=k 45. Proposals 41. .131 MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. . The source of the proposals will appear when the solutions be published. It is possible that knight back to its initial position after passing once through each square? 43. E. m be positive integers. Let n. + cos 133◦ + cos 134◦ 44. Let AGB. Prove that the centers of the circumcircles of those equilateral triangles D. Find a cubic polynomial which zeros are the square of the zeros of p(x) = x3 + 2x2 + 3x + 4. . . . BHC and CKA be equilateral triangles constructed on the sides of any triangle ABC. CFIS. 2 Also solved by Jos´ e Gibergnas-B´ aguena. Jos´ e Gibergnas-B´ aguena. We have that 17(17n−1 + 19n−1 ) < 17n + 19n < 19(17n−1 + 19n−1 ) as can be easily checked. Im(αβ + βγ + γα) = 0. Rumania . or Therefore. b). BARCELONA TECH. Higher Institute for Applied Sciences and Technology. Spain . Im β−γ Im αβ − γβ − αγ + |γ|2 (α − γ)(β − γ) Im = =0 |β − γ|2 |β − γ|2 which is equivalent to Im αβ − γβ − αγ = 0 Since for any complex number z = x + iy is Im(−z) = −y = Im z. (Training Sessions of Spanish Team for First Stage OME 2013) Solution by Jos´ e Luis D´ıaz-Barrero. and only if. Damascus. y) is x−c y−d z−γ = ⇐⇒ Im =0 a−c b−d β−γ α−γ = 0.132 Solutions 36. and we are done. Ioan Viorel Codeanu. Satulung. . Viveiro. Find all positive integers n such that 17n−1 + 19n−1 divides 17n + 19n . (Training Sessions of Spanish Team for First Stage OME 2013) Solution by Francesc Gispert S´ anchez (student). then 17n +19n = 18(17n−1 +19n−1 ) = (17+1)17n−1 +(19−1)19n−1 = 17n +17n−1 + 19n − 19n−1 from which follows 17n−1 = 19n−1 . Let α. BARCELONA TECH. Jos´ e Luis D´ıaz-Barrero. Spain .The preceding is only possible for n = 1 for which 17n−1 + 19n−1 = 2 that divides 17n + 19n = 36. Barcelona.α. Spain. β. Pitesti. β = (a. Daniel V´ acaru. Bruno Salguerico. z = (x. Spain and Omran Kouba. Barcelona. β and γ be three distinct complex numbers. Barcelona. The equation of the line through γ = (c. Higher Institute for Applied Sciences and Technology. Spain and Omran Kouba. BARCELONA TECH. Spain. d). Show that they are collinear if. 2 Also solved by Bruno Salguerico. Rumania. Damascus. γ are collinear if. Since 17n−1 + 19n−1 divides 17n + 19n . 37. Spain. Syria. BARCELONA TECH. BARCELONA TECH. Syria. and only if. then Im αβ − γβ − αγ = Im(αβ) + Im(−γβ + Im(−αγ) = Im(αβ + βγ + γα) = 0 and we are done. Viveiro. Barcelona. Barcelona. 133 38. as we wanted to prove. (Training Sessions of Spanish Team for First Stage OME 2013) Solution by Jos´ e Gibergnas-B´ aguena. 39. chords AD and BC meet P Q at points X and Y respectively. 77% passed Physics and 89% passed Geometry. Viveiro. Barcelona. Through the midpoint M of a chord P Q of a circle. Damascus. Algebra. 4M XX 0 ∼ 4M Y Y 0 . = y2 CY x2 XD = y1 YB from which follows x2 x1 x2 x1 x2 AX · XD P X · XQ = = · = = y2 y1 y2 y2 y1 CY · Y B PY · Y Q (a − x)(a + x) a 2 − x2 (a2 − x2 ) + x2 a2 = 2 = 2 =1 = 2 2 2 2 (a + y)(a − y) a −y (a − y ) + y a and x = y. Spain and Jos´ e Luis D´ıazBarrero. Spain. 82% passed Algebra. yields x x1 = . The students of a University Course in Mathematics take their exams in Calculus. Prove that M is the midpoint of XY. We begin by dropping perpendiculars x1 = XX 00 and y1 = Y Y 00 from X and Y to AB. Spain and Omran Kouba. Higher Institute for Applied Sciences and Technology. It is known that 73% passed Calculus exam. x2 = XX 0 and y2 = Y Y 0 from X and Y to CD. x = XM. Syria. At least. how many students have passed the exam of all four subjects? . y y2 x1 AX . y = M Y. any other chords AB and CD are drawn. Butterfly’s Theorem Letting a = P M = M Q. Physics and Geometry. BARCELONA TECH. 4AXX 00 ∼ 4CY Y 0 .4DXX 0 ∼ 4BY Y 00 . = 2 Also solved by Bruno Salguerico. BARCELONA TECH. and observing the pairs of similar triangles: 4M XX 00 ∼ 4M Y Y 00 . Barcelona. y y1 x2 x = . C A x1 P X Y' x'' M y2 Q Y y1 x2 x' Y'' B D Figure 1. Prove that 2/3 tan2 α 1 1 X +3 ≥2 3 tan β tan γ tan α + tan β + tan γ cyclic (Mediterranean Mathematical Olympiad 2012) . Barcelona. Viveiro. So. N (C ∪ A) = N (C) + N (A) − N (C ∩ A) and N (C ∩ A) = N (C) + N (A) − N (C ∪ A) Since N (C) = 73. Let us denote by B ∩ P = C ∩ A ∩ P = D. N (B ∩ P ) = N (B) + N (N ) − N (B ∪ P ) = 55 + 77 − 100 = 32 Finally. Spain. Let us denote by N (B) = N (C ∩ A) the number of student that passed the exam of Calculus and Algebra under the hypothesis that N (C∪A) = 100. So. N (D ∩ G) = N (D) + N (G) − N (B ∪ P ) = 32 + 89 − 100 = 21 2 and we are done. So. Barcelona. N (C ∩ A) = 73 + 82 − 100 = 55 Now. we calculate the number of students that passed Calculus and Algebra. BARCELONA TECH.134 (Training Sessions of Spanish Team for First Stage OME 2013) Solution by Francesc Gispert S´ anchez (student). Also solved by Bruno Salguerico. Spain. 40. N (A) = 82. Then. CFIS. Let α. then N (C∩A) will be minimum when N (C∪A) = 100 that it is the maximum value that it may attain. Then N (B ∪ P ) = N (B) + N (P ) − N (B ∩ P ) and N (B ∩ P ) = N (B) + N (N ) − N (B ∪ P ) The minimum value of N (B ∩ P ) will be got when N (B ∪ P ) = 100. β. N (D ∪ G) = N (D) + N (G) − N (D ∩ G) and N (D ∩ G) = N (D) + N (G) − N (B ∪ P ) The minimum value of N (D ∩G) corresponds to the maximum value of N (B ∪P ) = 100. we compute the number of students that passed the four exams. Algebra and Physics. γ be the angles of an acute triangle ABC. Spain and Jos´ e Luis D´ıazBarrero BARCELONA TECH. That is. First. we calculate the minimum number of students that pass the exam of Calculus. we have √ 3 3 tan α tan β tan γ tan3 α + tan3 β + tan3 γ + 3 tan α tan β tan γ tan α + tan β + tan γ √ √ 3 3 3 3 tan α + tan β + tan γ 3 tan α tan β tan γ 3 3 tan α tan β tan γ = +3 −2 3 tan α tan β tan γ tan α + tan β + tan γ tan α + tan β + tan γ s 9(tan3 α + tan3 β + tan3 γ) ≥4 4 −2 (tan α + tan β + tan γ)3 So. BARCELONA TECH. Using the well-known identity tan α tan β tan γ = tan α + tan β + tan γ and rearranging terms. applying two times AM-GM inequality. Barcelona. we can write the inequality claimed as √ tan3 α + tan3 β + tan3 γ 3 3 tan α tan β tan γ + ≥2 3 tan α tan β tan γ tan α + tan β + tan γ Now. Viveiro. BARCELONA TECH. Spain and Jos´ e GibergnasB´ aguena. That is when 4ABC is equilateral and we are done. 2 Also solved by Bruno Salguerico. Spain. Spain. it will be suffice to prove that s 3 3 3 4 9(tan α + tan β + tan γ) ≥1 (tan α + tan β + tan γ)3 The preceding is equivalent to see that 9(tan3 α + tan3 β + tan3 γ) ≥ (tan α + tan β + tan γ)3 It trivially holds from r 3 3 3 tan α + tan β + tan γ 3 tan α + tan β + tan γ ≤ 3 3 Equality holds when tan α = tan β = tan γ.135 Solution by Jos´ e Luis D´ıaz-Barrero. . Barcelona. 2 (3) . on page 52. Ba In Memory of Ion Ionescu Abstract. No. III (15 September 1897 – 15 August 1898). and applying well–known theorems we deduce that: d2 + d02 = 2 · AM 2 + 2 · A00 M 2 (1) d2 − d02 = 4 · A00 M 2 · M H. The solution of the problem 273. Ion Ionescu. and we construct around the side BC the equilateral triangles BCA0 and BCA00 . Muzicescu (Student. Prove that there is no triangle for which the inequality √ 4S 3 > a2 + b2 + c2 is satisfied. 15 October 1897. Ia¸si). In triangle AA0 A00 . the founder of Mathematical Gazette. 1. Vol. M. 2 . The aim of this note is to prove that the Weitzenb¨ oc’s inequality must be named the Ionescu-Weitzenb¨ oc’s inequality. III (15 September 1897–15 August 1898). Introduction In Romanian Mathematical Gazette. Let ABC be a triangle. appeared in Gazeta Matematicˇa. so M = AA0 ∩ BC will be the middle of these two lines. Prove that there is no triangle for which the inequality √ 4S 3 > a2 + b2 + c2 is satisfied.136 MATHNOTES SECTION The Inequality Ionescu . Let AA0 = d and AA00 = d0 . 12. Vol.G. (2) and from the triangle ABC.Weitzenb¨ ock ˇ tinetu-Giurgiu and Neculai Stanciu D. as follows: Problema 273. we have b2 + c2 = 2 · AM 2 + a2 . published the following problem: *273. Solution by N. No. on pages 281-283. 15 August 1898. bc sin A · sin 30◦ which yields 2bc sin(A + 30◦ ) > b2 + c2 . On the other hand we have (b − c)2 ≥ 0. (9) which is impossible. G. a2 = b2 + c2 − 2bc cos A. VI Lic. making the substitutions given by (5) and (6) we get √ 2d02 = a2 + b2 + c2 − 4S 3. i. Assuming this condition is satisfied we can divide (7) by (8). because otherwise the first member would be negative and could not be greater like the second member which is positive. Solution by I. 3 = cot 30◦ = 2 sin 30◦ Replacing. Therefore sin(A + 30◦ ) > 1. Maria Rugesu (Student Ia¸si) and by Mrs.137 On the other hand. since the sine function cannot be greater than the unity. So we must have A + 30◦ = 90◦ . so the given inequality is impossible. Sichitiu (Sc. 2 By the relations (1). because M H is evidently equal to the height of this triangle. the triangle must be equilateral. simplifying with 2 and letting the second member by only b2 + c2 we have cos 30◦ + cos A > b2 + c2 .e. G. so 2bc ≤ b2 + c2 . (5) √ 2 02 d − d = 2a · M H 3.e. Penescu (Bucharest). If (8) and (9) become equalities then so does (7). Urechilˇ a and I. De Art. (2). Sp. which in turn gives 2 2 2 √ A A H B M C A a + b + c ≥ 4S 3. Th. because A00 M is the height of the equilateral triangle with length of the side a. Ionescu (Student Cl. i. (3). (7) (8) ◦ The inequality (7) is satisfied for A < 150 . Hence the given inequality becomes equality for any equilateral triangle. ¸si Geniu) and Corneliu P. (6) but 2a · M H represents 4 times the area of the triangle ABC. and (4) we deduce: d2 + d02 = a2 + b2 + c2 . Vladimirescu (Rˆımnicul Vˆılcei). We have √ 1 cos 30◦ S = bc sin A. . Moscuna (Bucharest) and I. So the inequality (7) and therefore the given inequality are absurd for any triangle. Galat¸i). M. we have: 1 √ (4) A00 M = a 3. Solution by Miss. A = 60◦ and b = c. i. Emil G. Department of Mathematics. I. George Emil Palade School. 137-146 the article Uber eine Ungleichung in der Dreiecksgeometrie. 3 2bc + c2 + b2 − a2 2bca2 − b2 − c2 > (a2 + b2 + c2 )2 . the given inequality is impossible for all triangles. because all the terms on the left-hand side are positive. But the last inequality is impossible. The last inequality becomes equality if and only if a = b = c. Nicolaescu. where he proved that: In any triangle ABC. Eleven proofs of the Ionescu–Weitzenb¨ ock ’s inequality were presented by Arthur Engel in his book Problem solving strategies. 1998. with usual notations.e. 5. 4a2 b2 + 4a2 c2 + 4b2 c2 − 4a4 − 4b4 − 4c4 > 0. Springer Verlag. 3 4b2 c2 − (a2 − b2 − c2 )2 > (a2 + b2 + c2 )2 . 2 (a2 − b2 )2 + (a2 − c2 )2 + (b2 − c2 )2 < 0. The inequality Ionescu-Weitzenb¨ ock was given as a problem at the third IMO. 8–15. Matei Basarab National College. pp. Roland Weitzenb¨ ock published in Mathematische Zeitschrift. 2 2a4 + 2b4 + 2c4 − 2a2 b2 − 2a2 c2 − 2b2 c2 < 0. Vintilˇ a. Bucharest. 3 (b + c)2 − a2 a2 − (b − c)2 > (a2 + b2 + c2 )2 . by Mrs. and therefore from this moment the inequality of Weitzenb¨ ock must be named the inequality Ionescu–Weitzenb¨ ock. Hence.138 The given inequality becomes successively 4 p √ p(p − a)(p − b)(p − c) · 3 > a2 + b2 + c2 . 3 2a2 b2 + 2a2 c2 + 2b2 c2 − a4 − b4 − c4 > a4 + b4 + c4 + 2a2 b2 + 2a2 c2 + 2b2 c2 . 1-2. Moreover. we have 23 demonstrations and 10 generalizations other than those published of the Ionescu–Weitzenb¨ ock ’s inequality. July. 1961. Iliovici. holds the inequality √ a2 + b2 + c2 ≥ 4 3S We observe that the inequality of Ion Ionescu is the same as the inequality of Weitzenb¨ ock. RoDepartment of Mathematics. Nit¸escu. A. 3(a + b + c)(b + c − a)(a + c − b)(a + b − c) > (a2 + b2 + c2 )2 . V. 6p(2p − 2a)(2p − 2b)(2p − 2c) > (a2 + b2 + c2 )2 . No. 16p(p − a)(p − b)(p − c)3 > (a2 + b2 + c2 )2 . In the year 1919. We discovered the above while we were working on an article about Weitzenb¨ ock ’s inequality. Vol. when the triangle is equilateral. V. Also solved. in different ways. Hungary. Romania ˇ u. Cambureanu and C. Vespr´em. Buza mania . respectively. respectively. Bucharest. Romania. c are real positive numbers. 2013 Proposals 1. B˘ atinet¸u–Giurgiu.E. circumradius R and inradius r.. b. Let us consider the intersection points P = A0 C ∩ BC 0 .M B˘ atinet¸u–Giurgiu. Q = AC 0 ∩ B 0 C and T = P Q ∩ AB. Proposed by Paolo Perfetti. Let AA . BB . area S.139 JUNIOR PROBLEMS Solutions to the problems stated in this issue should arrive before June 15. Math.S. Proposed by Ercole Suppa. (Dedicated to Juan Bosco Romero M´arquez) 4. Neculai Stanciu. c.M. Proposed by D. Bucharest. find with proofs the coordinates of all the extremum points of x2 x4 + x3 + 4x + 16 5. b. . “Matei Basarab” National College. Alvarez Cubero de Priego 0 0 0 de C´ ordoba. Italy. Without using calculus. I. Prove that √ a3 b3 c3 4 3 + + ≥ S b·R+c·r c·R+a·r a·R+b·r R+r 2. let E = BP ∩ AC. “Tor Vergata” Univ. Spain. CC be any three diameters of the same circle. let P be an arbitrary point on AD. Let ABC be a triangle with side lengths a. Buz˘ au. “Matei Basarab” National College. Dept. E1 . Italy. Romania. Neculai Stanciu. F1 be the second intersection of BP. Romania. Romania. Proposed by D. Buz˘ au. Proposed by Francisco Javier Garc´ıa Capit´ an. Show that E. “George Emil Palade” School. In ∆ABC let D be the midpoint of BC. F1 . Rome. F are concyclic. F = CP ∩ AB and let E1 . Show that T C is tangent to the circle. Teramo. then a 1 1 1 ≤ + a2 + bc 4 b c ´ 3. “George Emil Palade” School. Prove that if a. CP with the circumcircle. Let x ∈ R and (Ln (x))n≥2 be the sequence defined by sin2 x √ sin2 x p 2 n n+1 Ln (x) = ncos x (n + 1)! − n! Calculate lim Ln (x). Prishtin¨ e. Ovidiu Furdui. Solutions will be evaluated for publication by a committee of professors according to a combination of criteria. Buzˇ au. Pages 140–170 Editors: Valmir Krasniqi. PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets. Proposed by D. Kosov¨ e. Emanuele Callegari. 1 ≤ k ≤ n.com Volume 3.Mathproblems ISSN: 2217-446X.M. Department of Mathematics and Information Science. url: http://www. then show that v u n n n n X X X u X x2k x4k 6 x3k tn x6k . Francisco Javier Garca Capit´ an. Paolo Perfetti. Stone. 2013 Problems 66. n→∞ c 2010 Mathproblems. George Emil Palade Secondary School. If xk > 0. Mih´aly Bencze. Issue 2 (2013). Cristinel Mortici. An asterisk (*) indicates that neither the proposer nor the editors have supplied a solution. Drawings must be suitable for reproduction. Ercole Suppa. Proposals should be accompanied by solutions. B˘ atinet¸u-Giurgiu.com Solutions to the problems stated in this issue should arrive before September 15. Mohammed Aassila. Questions concerning proposals and/or solutions can be sent by e-mail to: mathproblems-ks@hotmail. Binzhou University. Shandong Province. Teachers can help by assisting their students in submitting solutions. Student solutions should include the class and school name. Valmir Bucaj.mathproblems-ks. Romania and Neculai Stanciu. Jos´ e Luis D´ıaz-Barrero. k=1 k=1 k=1 k=1 67. The editors encourage undergraduate and pre-college students to submit solutions. Shabani. 140 . Roberto Tauraso. China. Armend Sh. Bucharest. Jozsef S´andor. Li Yin. Anastasios Kotronis. Universiteti i Prishtin¨ es. Matei Basarab National Colege. Romania (Jointly). each indicating the name of the sender. Binzhou City. 256603. David R. Enkel Hysnelaj. 141 68. Department of Mathematics and Information Science. University of Technology. Proposed by Mih´ aly Bencze. Ploiesti. Solve the following system of equations 1 1 1 x + y+z = a 1 1 1 y + z+x = b 1 1 1 z + x+y = c where a. we can generalize the inverse of arcsin as follows: Z x 1 0≤x≤1 arcsinp (x) = dt (1 − tp )1/p 0 and Z 1 πp 1 = arcsinp (1) = dt. Let 1 < p < ∞. b. Binzhou University. Australia. log xk + 2 k=1 1≤i<j≤m k=1 where a > 0. 70. Proposed by Anastasios Kotronis. Bra¸sov. 72∗ . p 1/p 2 0 (1 − t ) πp The inverse of arcsinp (x) on 0. 69. Greece. cosp (x) sinp (x) 0 . Proposed by Li Yin. Solve the following equation: m m Y X X m(m + 1) 2 2 log a = (m + 1) log a log log xi log xj + xk . Let an = Show that r r 1 π −3/2 π −1/2 2 −1 n + n + n + O(n−2 ). Binzhou City. 71. 256603. If k ≥ . Shandong Province. an = 2 3 12 2 Pn k=0 n k k!n−k . The generalized hyperbolic cosine function is defined as cosp (x) = d dx sinp (x). Calculate Z π2p ln cosp (x) ln sinp (x) Ip = dx. Let a and b be positive real 1 numbers such that a + b = 2. 2 is called the generalized sine function and denoted by sinp . Proposed by Enkel Hysnelaj. c ∈ < − {0}. Proposed by Vasile Cirtoaje. Romania. China. then prove that 2 kb aa bb ka ≥ 1. Athens. Sydney. Romania. We will be very pleased considering for publication new solutions or comments on the past problems.142 Solutions No problem is ever permanently closed. Solution by Paolo Perfetti. It is well-known that (i − 1)!(j − 1)! Γ(i)Γ(j) Γ(i + j) Γ(i)Γ(j) 1 = = (i + j)! Γ(i + j) Γ(i + j + 1) Γ(i + j) i + j Moreover Γ(i)Γ(j) = Γ(i + j) Z 1 xi−1 (1 − x)j−1 dx 0 and then ∞ X ∞ X (i − 1)!(j − 1)! i=1 j=1 (i + j)! a i+j = ∞ X ∞ Z X 1 xi−1 (1 − x)j−1 dx i=1 j=1 0 ∞ 1 ai+j i+j and ∞ X ∞ Z X 1 xi−1 (1 − x)j−1 dx 0 i=1 j=1 ∞ XX ai+j = i+j i=1 j=1 Z xi−1 (1 − x)j−1 dx 0 Z a y i+j−1 dy 0 The geometric series yields Z 1 Z dx 0 or a dy 0 xy 1 y(1 − x) 1 = x(1 − xy) 1 − x 1 − y(1 − x) y Z 1 Z dx 0 a dy 0 y 1 1 − xy 1 − y(1 − x) . Tor Vergata University. Italy. Romania. Department of Mathematics. (i + j)! i=1 j=1 where ζ(z) represents the Riemann zeta function. Rome. Technical University of Cluj-Napoca. 1) Let a be a real number such that −1 ≤ a ≤ 1. i+j a 2−a 2) Show that ∞ X ∞ X (i − 1)!(j − 1)! = ζ(2). Proposed by Ovidiu Furdui. Prove that ∞ X ∞ X (i − 1)!(j − 1)! a a = 2Li2 − 2Li2 − − 2Li2 (a) (i + j)! 2−a i=1 j=1 Rz dt for all where Li2 (z) is the Dilogarithm function defined by Li2 (z) = − 0 ln(1−t) t z ∈ {z ∈ C : |z| ≤ 1}. 58. Cluj-Napoca. Integrating by parts Z 1 . and F (1) = ζ(2) and this proves both 1) and 2). we get F (a) = 2Li2 F 0 (a) = −2 ln 1 − a 2−a a 2−a 2 +2 (2 − a)2 ln 1 + a 2−a −a 2−a −2 ln(1 − a) ln(1 − a) +2 = −2 2 (2 − a) a 2−a which is G0 (a).143 Z 1 Z 1 Z a y 1 1 y 1 dy dx = dy + dx 1 − xy 1 − y(1 − x) 2−y 0 1 − xy 1 − y + xy 0 0 0 Z a ln(1 − y) dy dy = −2 2−y 0 Z a Thus we have ∞ X ∞ X (i − 1)!(j − 1)! (i + j)! i=1 j=1 ai+j = −2 Z a dy 0 ln(1 − y) dy = G(a) 2−y Differentiating a a − 2Li2 − − 2Li2 (a) 2−a 2−a and keeping in mind the minus sign. Now we show G(1) = ζ(2). We assume that Li2 (1) = ζ(2) and Li2 (−1) = −ζ(2)/2. 1 Z 1 ln(2 − y) ln(1 − y) . G(1) = −2 dy = 2 ln(2 − y) ln(1 − y). Advanced Institute of Science and Technology. Matei Basarab National Colege. Hun Min Park. and for all n ≥ 2. Proposed by D. Dajeon. Romania and Neculai Stanciu. +2 dy |{z} = 2−y 1−y 0 0 0 1−y=−x Z −1 ln(1 − x) 2 dx = −Li2 (−1) = ζ(2) x 0 On the other hand F (1) = 2Li2 (1) − 2Li2 (−1) − 2Li2 (1) = ζ(2) Also solved by Omran Kouba. Syria. Romania (Jointly). Ln+3 Ln + Ln+2 where Ln represents the nth Lucas number defined by L0 = 2. Moti Levy. Buzˇ au. Prove that √ p Ln L2n+2 Ln+1 L2n+3 2 + + (Ln + Ln+2 ) ≥ 2 6 · Ln Ln+1 · Ln+2 . Higher Institute for Applied Sciences and Technology. Damascus. Let n be a positive integer. Ln = Ln−1 + Ln−2 . South Korea. and the proposer 59. Bucharest. B˘ atinet¸u-Giurgiu. George Emil Palade Secondary School. . Israel . L1 = 1.M. −1. Solution by AN-anduud Problem Solving Group. Ulaanbaatar. x+1 7+x 2 · 2−x −7 2g(x) − 7 = 7+x =x g(x) + 1 2−x + 1 7+x f (x) + f = ax + b 2−x In (1) if we do the change of variables 2x − 7 7+x . and 1 + 14 = Ln + Ln+1 4 25 16 > 5 2 −1 > √ 6 − 1. b ∈ R. Determine all functions f : R \ {−2. University of Las Palmas de Gran Canaria. Also solved by Omran Kouba. Spain. Higher Institute for Applied Sciences and Technology. 2} → R. Consider the function g(x) = 2−x (g ◦ g)(x) = g(g(x)) = 7+ 7 + g(x) = 2 − g(x) 2− 7+x 2−x 7+x 2−x = 2x − 7 . So. 1. by the AM-GM inequality. ≥ 1+ Ln + Ln+1 Since 2 Ln 1 ≥ . s Ln L2n+2 Ln+1 L2n+3 Ln+1 L2n+3 1 Ln L2n+2 + ≥ · 2 Ln+3 Ln + Ln+2 Ln+3 Ln + Ln+2 s p Ln+3 = · Ln Ln+1 · Ln+2 Ln + Ln+2 p > Ln Ln+1 · Ln+2 . Ln Ln+1 < Ln L2n+1 = Ln+2 2 . Then lia. By the AM-GM inequality. Sydney. the conclusion follows. x→ 2−x x+1 (g ◦ g ◦ g)(x) = g(g(g(x))) = (1) . Mongolia. which satisfy the relation 7+x f (x) + f = ax + b 2−x where a. p √ (Ln + Ln+2 )2 6 − 1 · Ln Ln+1 · Ln+2 . So it is enough to prove that ≥ 2 p Again. Ulaanbaatar. and the proposer. Australia. 60. Mongo7+x . 0. University of Technology. Proposed by Enkel Hysnelaj. Damascus. AN-anduud Problem Solving Group.x → . it is sufficient to prove that √ 6 − 1 L2n+2 (Ln + Ln+2 )2 ≥ 2 √ 2Ln + Ln+1 ≥ 6−1 Ln + Ln+1 2 √ Ln 6 − 1. Syria. Moti Levy.144 Solution by Angel Plaza. Israel . Polytechnic University. Thus f is convex. d. By assumption tk ∈ [0. we find that n n X X aXn + bxk t1 + · · · + tn (an + b)n = f (tk ) ≥ nf = nf (1) = cXn − dxk n cn − d k=1 k=1 which is the desired inequality. Damascus. Buzˇ au. cn/d) we have f 00 (x) = 2dn(bc+ad) (cn−dx)3 > 0. Now. Let ρk = 1. University of Prishtina. 1≤k≤n then n X aXn + bxk k=1 cXn − dxk ≥ (an + b)n cn − d Solution 1 by Omran Kouba. Let f be the function defined on the interval [0. 61. Matei Basarab National Colege. xk ∈ R+ . Republic of Kosova. . Bucharest. . Damascus. . for x ∈ [0. France. let tk = nxk /Xn for k = 1. Syria. Tirana. .145 then 7+x 2x − 7 7+x f +f =a· +b 2−x x+1 2−x 2x − 7 2x − 7 f + f (x) = a · +b x+1 x+1 Now adding up (1) and (3) and subtracting (2). Israel . c. If a ∈ R+ . Damascus. Proposed by D. Syria. Moubinool OMARJEE. Dannan. Syria. Romania and Neculai Stanciu. therefore. Higher Institute for Applied Sciences and Technology. Higher Institute for Applied Sciences and Technology. b. cn/d) for every k. Paris. Fozi M. n. Israel . cn/d) by an + bx f (x) = cn − dx Clearly. George Emil Palade Secondary School. Dorlir Ahmeti. 2. n a b c k=1 d bX d + ρk b ≥n − ρk d Let f (x) = b d a b c d a b c d xk Xn + − +x −x then 0 < ρk < 1 and Pn k=1 Pn n k=1 n ρk ρk Pn k=1 ρk = . We re-write the inequality using ρk . using the convexity of f . Albania. Solution 2 by Moti Levy. yields 2x − 7 7 + x − +b 2f (x) = a · x + x+1 2−x from which follows ax3 + (2a + b)x2 − (5a + b)x + 21a − 2b f (x) = 2(x + 1)(x − 2) (2) (3) Also solved by Moti Levy. and the proposer. Ron o Pn ∗ mania (Jointly). Omran Kouba.M. Adrian Naco. Xn = k=1 xk and cXn > d max xk . B˘ atinet¸u-Giurgiu. Mathematics Research Center. Ulaanbaatar. Let k ≥ 3 be an integer. 63. Mongolia. 62. Greece. Maramure¸s. then at least one of the equations a1 x2 + 2c2 x + b1 = 0. We argue by contradiction. ck ∈ R∗ (k = 2 1. AN-anduud Problem Solving Group. Then. Athens. Bra¸sov. Maramure¸s. Satulung. and Angel Plaza.O. Omran Kouba. n 1X f (ρk ) ≥ f n Also solved by Ioan Viorel Codreanu. 4a22 − 4b1 c1 < 0. all the discriminants are negative. Damascus. student. University of Las Palmas de Gran Canaria. Yogyakarta. Romania. Proposed by Anastasios Kotronis. Paolo Perfetti. So. Tor Vergata University. Higher Institute for Applied Sciences and Technology. Satulung. Solution by Juan Gabriel Alonso (E. Syria. Spain. (a1 − b1 )2 + (b1 − c1 )2 + (c1 − a1 )2 < 0. Department of Mathematics. Indonesia. Bellaterra Campus. 4c22 − 4a1 b1 < 0. Tor Vergata University. 2 a21 + b21 + c21 − 2 (a1 b1 + b1 c1 + c1 a1 ) < 0. Spain (Jointly). 2). bk . Roberto De la Cruz Moreno. Republic of Kosova. Dorlir Ahmeti. Garo´ e ´ Atlantic School). Show that if ak . University of Prishtina. Israel . The inequality follows (c−dx)2 from Jensen’s inequality for convex functions. Proposed by Mih´ aly Bencze. Also solved by Moti Levy.S. Muhammad Thoriq. Rome. Let us suppose that none of the given equations has any real root. 4b22 − 4c1 a1 < 0 from where 2 a22 + b22 + c22 − 2 (a1 b1 + b1 c1 + c1 a1 ) < 0. First Course. Ioan Viorel Codreanu. Romania. such that a1 + b21 + c21 = a22 + b22 + c22 . Rome. and the proposer. n k=1 k=1 0 The function f (x) is convex in the interval 0.146 then the inequality in terms of the function f is ! n 1X ρk . Show that X n≥0 1 (2n + 1)(3n + 2) · · · (kn + k − 1) k 1 X m−1 k = (−1) (m−1)mk−2 k! m=2 m ! m−1 X π π 2`π `π cot − ln m + cos · ln sin 2 m m m `=1 . Gadjah Mada University. Ulaanbaatar. c1 x2 + 2b2 x + a1 = 0 has real roots. Paolo Perfetti. Italy. Department of Mathematics. Italy. ANanduud Problem Solving Group. b1 x2 + 2a2 x + c1 = 0. dc since its derivative f (x) = ad+bc is monotonically non-decreasing in the interval. Romania. Mongolia and the proposer. which indeed it is a contradiction. Barcelona. Department of Mathematics. j6=m j−1 m−1 − j m k 1 Y 1 m j j=2. ψ q 2 q q q l=1 ψ (bm ) = ψ m−1 m m π π (m − 1) X = − (γ + ln 2) − ln m − cot + cos 2 m l=1 Since cot (m − 1) π π = − cot m m 2πl (m − 1) m ln sin πl m . The partial fractions of un are m=2 am n+bm . j6=m j=2. Israel . 0 < p < q. j6=m j−m mj = k Y j=2. j6=m 1 mk−2 k m Thus m am = (−1) mk−2 (m − 1) k m (4) Pk m=2 am = 0 follows from the convergence of our series that limn→∞ nun = 0. m=2 Y Pk Let Q (n) . m n X The Digamma function ψ (z) can be used to evaluate the infinite sum un . where the m coefficients {am } are given by am = k X d (Q (n)) = dn m=2 k Y 1 d dn (Q(n))|n=−bm . j6=m k Y = k Y (bj − bm ) = j=2.147 Solution by Moti Levy. Pk Pk Pk nam Hence 0 = limn→∞ nun = limn→∞ m=2 n+b = limn→∞ m=2 1+ambm = m=2 am . (n + bm ) . q π πp X 2πlp πl p = −γ − ln 2q − cot + cos ln sin . j6=m k Y m m (−1) (m − 2)! (k − m)! k! 1 1 m! (k − m)! 1 m m = (−1) = (−1) k−2 k−2 m k! m−1 m (m − 1) = (j − m) j=2. k Y k! m−1 1 Let un = (2n+1)(3n+2)···(kn+k−1) = n+bm . n≥0 It is known that X un = − k X am ψ (bm ) (5) m=2 n≥0 The Digamma function for rational arguments is given by Gauss’s Digamma Theorem. (n + bj ) j=2. where bm = m . j6=m d (Q (n))|n=−bm = dn k Y j=2. or otherwise the series will be greater than harmonic series and thus diverges. then prove that Z 1 2 Z 1 2 f (x)dx ≥ 24 f (x)dx 0 0 Solution 1 by Omran Kouba. 1]. Syria and the proposer. the required result follows ! k m−1 X X X k π 2lπ π lπ m−1 k−2 un = (−1) m (m − 1) cos cot − ln m + . Note that 1 Z 1 Z 1 −2 F (x) dx = (1 − 2x)F (x) − (1 − 2x)f (x) dx 1/2 x=1/2 so Z 1 Z 2 1 x f (x) dx = −F (1) − 0 1/2 (1 − 2x)f (x) dx 1/2 or equivalently Z 1/2 Z 2 (1 + x )f (x) dx + 0 1 2 Z (1 + (1 − x) )f (x) dx = 1/2 g(x)f (x) dx = 0 0 where ( 1 + x2 g(x) = 1 + (1 − x)2 if x ∈ [0. 0 where F (x) = Rx 0 1/2 f (t)dt. If Z 1 Z 1 2 x f (x)dx = −2 F (x)dx. Higher Institute for Applied Sciences and Technology. we have Z 1 Z 1 f (x)dx = (1 + λg(x))f (x)dx 0 0 1 . m (γ + ln 2) am = 0. Damascus. x ∈ [0. Gaesti. then m ψ (bm ) = − (γ + ln 2) − ln m + X π π cot + cos 2 m l=1 Pk Using the fact that X un = − n≥0 2lπ m πl ln sin . for an arbitrary λ ∈ R.148 and cos 2l (m − 1) π m = cos 2lπ m . Syria. Damascus. Higher Institute for Applied Sciences and Technology. 1] → R be a continuous function. ln sin m 2 m m m m=2 n≥0 l=1 Also solved by Omran Kouba. Romania. 1] Now. Let f : [0. Proposed by Florin Stanescu. 64. Serban School Cioculescu. 12 ) if x ∈ [ 12 . we obtain ! m−1 X π π lπ 2lπ cot − ln m + ln sin cos 2 m m m m=2 k X am m=2 (6) l=1 and by substituting (4) in (6). λ+ 283 4 Thus. Moreover. using the Cauch-Schwarz inequality. this is the best 260 possible inequality since we have equality if f (x) = 1 − 283 g(x).149 So. Rome. we obtain 2 Z 1 Z 1 Z 2 (1 + λg(x)) dx · f (x)dx ≤ f 2 (x) dx 0 0 0 1 But 1 Z 13 283 2 283 4 λ+ λ = + 6 240 849 240 (1 + λg(x))2 (x) dx = 1 + 0 2 260 . 2 ≤ 0 1 2 0 = Z 2 1 2 0 Z 1 Z Z + ! 21 2 1 f 2 (x)dx = 0 1 2 ! 21 2 4 x dx 0 Z Z = 0 1 f 2 (x)dx 1 40 . the minimum value that this integral may take is 849 and it corresponds to 260 λ = − 283 . Z 1Z x Z 1 Z 21 Z 1 Z 1 Z 1 F (x)dx = f (y)dy = dx dyf (y) f (y)dy + dx = 1 2 1 2 1 2 1 2 Z Z 1 2 0 1 2 0 y 1 f (x)(1 − x)dx f (x)dx + 1 2 0 This yields 1 Z 1 2 Z 2 x f (x)dx = − 0 1 Z f (x)dx − 2 f (x)(1 − x)dx. First we note that. 1 2 0 and then Z 1 Z 1 f (x)(x − 1)2 dx − f (x)dx = − 1 2 0 1 2 Z f (x)x2 . Choosing this value for λ we conclude that Z 1 2 Z 1 4 f (x)dx ≤ f 2 (x) dx 849 0 0 which is much stronger than the proposed inequality. 0 We get by Cauchy–Schwarz and the fact that the integrands in the second line are positive 1 Z 2 f (x)dx = Z ≤ f (x)(x − 1) dx + 1 f 2 (x)dx Z 1 2 f (x)dx 0 Z !2 1 2 f (x)x ! 21 1 (x − 1)4 dx Z 1 2 + x4 dx 0 ! 21 1 4 (x − 1) dx 1 2 which implies our inequality. Tor Vergata University. Solution 2 by Paolo Perfetti. Italy. Department of Mathematics. Shandong Province. See Theorem 1 in ”Feng Qi and Bai-Ni Guo.844 60 and is monotonic v Proof. where R ∞ x−1 −t Γ(x) = 0 t e dt . 256603. Lemma 2. Israel .150 Also solved by Moti Levy. that is Ωn = Γ(n/2+1) . China. x 0 fv (x) = − ln (Γ (x + 1)) + xψ (x + 1) − 0 dfv (x) v (x + 2) = x ψ (x + 1) − 2 dx (x + 1) vx2 . Prove that r r 1 (Ωn+1 ) n+1 n+2 α n + 2 6 β 6 1 n+3 n+3 (Ωn ) n with the best possible constant factors α = 2 and β = ln( 34 ) √ ln( π 2 ) = 2. We will use the following results: Lemma 1: For x > 0. Sharp Inequalities for polygamma functions. Paris. France and the proposer. 2 (x + 1) x + 1 + π62 ! where 6 Pv (x) = (x + 1) + (1 − v (x + 2)) x + 1 + 2 π 2 . May 1. For all positive integers n. Let Ωn π n/2 the volume of the unit ball in Rn . Moubinool Omarjee. . Proposed by Li Yin.1984v1. arXiv:0903. The function 1 (Γ (x + 1)) x hv (x) = v (x + 1) is strictly increasing for x ≥ 1 and v ≤ vc = decreasing for x ≥ 0 and v ≥ 1. π 2 +1 π 2 +3 ∼ = 0. Binzhou City. . . 1 (x + 1) 2 + 1 x+ 1 0 > ψ (x + 1) > 3 2 (x + 1) 2 + 1 x+1+ 6 π2 (7) Proof.3818 · · · . 65∗ . x > 0. gv (x) = 1 ln Γ (x + 1) − v ln (x + 1) . Binzhou University. 2 . 2009”. x+1 ! (8) We will show that dfv (x) > 0 for x ≥ 1 and v ≤ vc dx Substitute the right hand of Lemma 1 in (8) . Israel . Let gv (x) = ln hv (x) and fv (x) = x2 dg dx . n = 1. Solution by Moti Levy. Department of Mathematics and Information Science. dfv (x) >x dx 1 (9) 1 v (x + 2) 2 + 6 − 2 x + 1 + π2 (x + 1) (x + 1) x = Pv (x) . The fact f1 (0) = 0 and (10) imply that gv (x) is monotonic decreasing for x ≥ 0 and v ≥ 1. which is equivalent to 1 1 n+1 Γ n2 + 1 n Γ n+1 2 +1 0< β1 ≤ β1 n n+1 2 +1 2 +1 Squaring both sides. so the inequality in the problem The ratio (Ωn+1 )1 1 (Ω) n (Γ( n2 + 32 )) n+1 statement is equivalent to s s n1 n n n Γ + 1 + 1 +1 2 2 α ≤ β n2 3 (11) 1 n 3 ≤ n 3 n+1 + + 2 2 2 2 Γ + 2 2 Let us begin with the right hand side of inequality (11). 2 −4 ∼ The fact fvc (1) = 1 − 12 ππ2 +1 > 0 and (8) imply that gv (x) is +3 − γ = 4. 2 (x + 1) x + 32 dfv (x) <x dx 1 (10) ! where 3 Qv (x) = (x + 1) + (1 − v (x + 2)) x + 2 7 5 2 = (1 − v) x + 3 − v x + − 3v 2 2 2 The polynomial Qv (x) is negative for x ≥ 0 and v ≥ 1. 1 1 n+1 (Γ( n2 +1)) n is equal to . Γ n 2 n 2 n1 +1 2 Γ ≤ β2 +1 n+1 2 n+1 2 +1 1 n+1 +1 2 β2 (12) .151 It is clear that f or v < vc and x ≥ 1 Pv (x) > Pvc (x) dPvc (x) 4 6 = 2 x− 2 2 dx π +3 π (π + 3) Pvc (1) = 0 dPvc (1) dx > 0 and Pvc (1) = 0. It follows that hv (x) is strictly increasing for x ≥ 1 and v ≤ vc . It follows that hv (x) is also monotonic decreasing for x ≥ 0 and v ≥ 1.868 × 10 strictly increasing for x ≥ 1 and v ≤ vc . We substitute the left hand of Lemma 1 in (8) to show that Since dfv (x) dx dfv (x) <0 dx for x ≥ 0 and v ≥ 1 1 v (x + 2) 2 + 2 3 − x+ 2 (x + 1) (x + 1) x = Qv (x) . We return now to the original problem. then Pvc (x) ≥ 0 for x ≥ 1 and consequently > 0 for x ≥ 1 and v ≤ vc . 1 n+1 n1 2 Γ n+1 Γ n2 + 1 2 + 1 2 ≤ α2 α2 n+1 n + 1 2 2 +1 (13) By Lemma 2.152 By Lemma 2. √ 1 The asymptotic expansion of the gamma function leads to Γ (x) → 2πxx− 2 e−x as x → ∞. (Γ(x+1)) x (x+1)v tends to infinity and this fact implies that α cannot be . inequality (13) holds for α2 ≥ 1 or α ≤ 2. = e ∞ if v < 1. 1 If v < 1 then greater than 2.368 0. ≤ 2 ( 21 +1) β (Γ(1+1)) 2 implies (1+1) β ∼ = 2.381 8. ln( 34 ) We conclude that inequality (12) holds for n ≥ 1 when β ≥ 2 ln ( π4 ) ∼ = 2. ( π4 ) Let us continue with the left hand side of the inequality (10): s 1 n Γ n2 + 1 n 2 +1 α 1 n 3 ≤ 2 + 2 Γ n2 + 32 n+1 1 1 Γ n2 + 1 n Γ n2 + 21 + 1 n+1 ≤ α1 α1 n n 1 2 + 2 +1 2 +1 Squaring both sides. inequality (12) holds for n ≥ 2 and 2 β ≤ Now we check inequality (12) for n = 1.381 8 and ln( 43 ) 2 ln is the minimal constant value. If n = 1 then that 2 β ≤ 4 ln( π ) ln( 34 ) ln( 43 ) or β ≥ 2 ln ( π4 ) π 2 +1 π 2 +3 or β ≥ 2 (Γ( 21 +1)) 2 ∼ 2 ππ2 +3 +1 = 2. 1 1 x+1 √ x1 (x + 1)1+ 2x (Γ (x + 1)) x e− x lim 2π = lim v v x→∞ x→∞ (x + 1) (x + 1) 1 1 1 = lim x 2x x1−v = lim x1−v x→∞ x→∞ e e if v > 1. 0 1 if v = 1. Prove that for all x with 0 ≤ x ≤ 1/2. Let n be a positive integer. Proposals are always welcomed. n 49. n be positive integers such that p is prime and p < n. Let p. 47. .153 MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. . (Here [x] p p p p represents the integer part of the real number x. Let S = {1. The source of the proposals will appear when the solutions be published. 50. then find the value of ∠BAC. and find all values of n for which an − 207 is the cube of a positive integer. then prove that p · divides − . Prove that an is divisible by 2013 for all n ≥ 1. (p − 1)! = 1. Let Γ be the circumcircle of a triangle ABC and let E and F be the intersections of the bisectors of ∠ABC and ∠ACB with Γ. If EF is tangent to the incircle γ of 4ABC. 2743}. For all positive integer n we consider the numbers an = 46 + 1943. it holds sin2n x + cos2n x ≥ (1 − 2x2 )n + (2x2 )n . .) 48. If p divides n + 1 2 n n n n and . Proposals 46. 3. Find the smallest positive integer n such that the product of any n distinct elements in S is divisible by 2743. . 2. 42. q(x2 ) = (x2 − r2 )(x2 − s2 )(x2 − t2 ) = (x − r)(x + r)(x − s)(x + s)(x − t)(x + t) Since p(−x) = (−x − r)(−x − s)(−x − t) = −(x + r)(x + s)(x + t). Viveiro. We have that x + y + z = a2 + b2 + c2 = (a + b + c)2 − 2(ab + bc + ca) = (−2)2 − 2 · 3 = −2 xy + yz + zx = (ab)2 + (bc)2 + (ca)2 = (ab + bc + ca)2 − 2(abbc + bcca + caab) = = 32 − 2abc(b + c + a) = 9 − 2(−4)(−2) = −7 xyz = (abc)2 = (−4)2 = 16 Finally the required polynomial is q(x) = x3 + 2x2 − 7x − 16 Solution 2 by Jos´ e Luis D´ıaz-Barrero. Spain. (First Stage OME 2013) Solution 1 by Adrian Naco. then q(x2 ) = (x − r)(x + r)(x − s)(x + s)(x − t)(x + t) = p(x)[−p(−x)] = (x3 + 2x2 + 3x + 4)(x3 − 2x2 + 3x − 4) = x6 + 2x4 − 7x2 − 16 Hence. It is possible that knight back to its initial position after passing once through each square? (Training Sessions of Spanish Team for Second Stage OME 2013) . q(x) = x3 + 2x2 − 7x − 16 is a solution and also all polynomials obtained by it after multiplying it by a constant. Barcelona. BARCELONA TECH. Barcelona. BARCELONA TECH. Let a. p(x) = (x − r)(x − s)(x − t). Omran Kouba. z the roots of the required polynomial. c the roots of the given polynomial and x. Let n be an odd positive integer. Damascus. Then. Polytechnic University. A knight is placed at random in one of the n2 squares of a chessboard. We are searching a polynomial of the form q(x) = (x − r2 )(x − s2 )(x − t2 ). Syria. b. Let r. Spain. 2 Also solved by Bruno Salgueiro Fanego. Find a cubic polynomial which zeros are the square of the zeros of p(x) = x3 + 2x2 + 3x + 4. Albania. y. Tirana. We have. Jos´ e GibergansB´ aguena. s and t be the zeros of p(x).154 Solutions 41. Spain. Higher Institute for Applied Sciences and Technology. Barcelona. Suppose that chess knight passes once through each square of the chess board and go back to its initial position. On the other hand. + sin 133◦ + sin 134◦ cos 1◦ + cos 2◦ + . we have that a + d + e + h and b + g + c + f are even on account that the sum of two positive integers whose difference is even is also even. Y -2+i -1+2i h 1+2i a g 2+i b X -2-i f c e 2-i d -1-2i 1-2i Let us denote by a. Omran Kouba. e. Then. Viveiro. respectively. c. . 2 Also solved by Bruno Salgueiro Fanego. Barcelona. since the polygonal is closed it sum is zero. Taking as origin in the plane the initial position of the knight the 8 possible new positions after jumping once are ±1 ± 2i and ±2 ± i. When the polygonal describe by the knight be closed the sum of the vectors drawn from the origin representing its movements will be zero. When knight runs any polygonal on the chessboard to a point P we can get the coordinates of P drawing from the origin (initial position of the knight) vectors equivalents to the vectors that represent its movements and adding up their components. g. . respectively. yields a + 2b + 2c + d − e − 2f − 2g − h = 0 and 2a + b − c − 2d − 2c − f + g + 2h = 0 and (a + d) − (e + h) = 2(f + g − b − c) and (b + g) − (c + f ) = 2(d + c − a − h) Then. h the number of times that knight moves in the directions of the positions ±1 ± 2i and ±2 ± i. Spain. It is not possible that knight back to its initial position after passing once through each square of a chessboard with an odd number os squares. . BARCELONA TECH. Spain. Syria 43. Finally. b. BARCELONA TECH. That is. Barcelona. f. d. Higher Institute for Applied Sciences and Technology. Spain.155 Solution by Jos´ e Gibergans-B´ aguena. + cos 133◦ + cos 134◦ . Damascus. So. Spain and Jos´ e Luis D´ıaz-Barrero. . Francesc Gispert S´ anchez. Contradiction. we get that a + b + c + d + e + f + g + h is even. we have a + b + c + d + f + g + h = n2 (odd). Compute sin 1◦ + sin 2◦ + . (1+2i)a+(2+i)b+(2−i)c+(1−2i)d+(−1−2i)e+(−2−i)f +(−2+i)g+(−1+2i)h = 0 from which follows (a + 2b + 2c + d − e − 2f − 2g − h) + (2a + b − c − 2d − 2c − f + g + 2h)i = 0 Identifying real and imaginary parts. BARCELONA TECH. Barcelona. .. Spain. Barcelona. Bruno Salgueiro Fanego. BARCELONA TECH. Damascus. . + sin 133◦ + sin 134◦ =1+ 2 ◦ ◦ ◦ ◦ cos 1 + cos 2 + . Find the value of the following expression n+m 1 X k X n m 7 23n i j k=0 i+j=k (Training Sessions of Spanish Team for Second Stage OME 2013) Solution by Jos´ e Luis D´ıaz-Barrero BARCELONA TECH. . Spain. BARCELONA TECH.+sin 133◦ +sin 134◦ = (1+ 2)(cos 1◦ +cos 2◦ +. and after rearranging terms. Barcelona. Viveiro. We have n+m 1 X k X n m 7 23n i j k=0 i+j=k . sin 1◦ − cos 1◦ = sin 2◦ − cos 2◦ = . . From the identity cos(135◦ − n◦ ) = cos 135◦ cos n◦ + sin 135◦ sin n◦ we get sin n◦ − cos n◦ = √ 2 cos(135◦ − n◦ ) Then. . Barcelona.. .. . Romania and Jos´ e GibergansB´ aguena. + cos 133◦ + cos 134◦ ) √ = 2 (cos 1◦ + cos 2◦ + ... yields (sin 1◦ + sin 2◦ + .. Let n. Higher Institute for Applied Sciences and Technology. + cos 133◦ + cos 134◦ ) From the preceding immediately follows √ sin 1◦ +sin 2◦ +.. . Omran Kouba. Pitesti. . Spain. m be positive integers. Spain and Jos´ e Luis D´ıazBarrero.. BARCELONA TECH.156 (Training Sessions of Spanish Team for First Stage OME 2013) Solution by Francesc Gispert S´ anchez. .+cos 133◦ +cos 134◦ ) and √ sin 1◦ + sin 2◦ + . + sin 133◦ + sin 134◦ ) − (cos 1◦ + cos 2◦ + . . . Spain. . + cos 133 + cos 134 2 Also solved by Daniel Vacaru. . ◦ ◦ ◦ ◦ sin 133 − cos 133 = sin 134 − cos 134 = √ √ √ √ 2 cos 134◦ 2 cos 133◦ 2 cos 2◦ 2 cos 1◦ Adding up the preceding equalities. Syria 44. Barcelona. So. Hence ∠AP G = ∠ABG = 60◦ (both have the common arc AG). Spain. Viveiro. since ∠AP G + ∠AP C = 180◦ . We begin with the following well-known result. Lines KB. Indeed. + = 3n 7 + 2 0 k 1 k−1 k 0 k=0 Now we observe that ! ! n m + 0 k ! ! n m + . From the preceding immediately follows that quadrilaterals KAP C and HBP C are concyclic.157 n+m 1 X k n m n m n m + . Syria. then points G. HA and GC in the following figure meet at point P . + 1 k−1 ! ! n m + . Triangles KCB and ACH are congruent because the second is a 60◦ rotation of the first about point C.. n+m 1 X k X n m 7 = 23n i j n+m 1 X k n+m 7 23n k = 1 (1 + 7)n+m = 8m 23n k=0 i+j=k k=0 2 and we are done. ∠CKP = ∠CAP and ∠CHP = ∠CBP . Let AGB. Spai. Damascus. (Training Sessions of Spanish Team for the Second Stage OME 2013) Solution 1 by Francesc Gispert S´ anchez.. Spain. H C K P B A G . Also solved by Bruno Salgueiro Fanego. Proof.. To obtain the total number of selections we have added the number of possibilities for 0 ≤ r ≤ k. Barcelona. (This point is called Fermat’s point). Thus ∠AP C = ∠AP B = ∠BP C = 120◦ . Higher Institute for Applied Sciences and Technology.. On the other hand. + r k−r ! ! n m = k 0 ! n+m k as it is well-known. each term nr k−r on the LHS represents the number of ways of choosing r women and k − r men. BARCELONA TECH. Omran Kouba. Finally. BARCELONA TECH. 45. BHC and CKA be equilateral triangles constructed on the sides of any triangle ABC. then AP BG is also concyclic. the RHS is the number of ways of choosing k people m from a set of n women and m men.. C are collinear. Since ∠AP B and ∠AGC add up to 180◦ . Prove that the centers of the circumcircles of those equilateral triangles D. P. E. F themselves form an equilateral triangle. Francesc Gispert S´ anchez. Lemma [Fermat’s point]. Therefore.. P C are respectively radical axis of two circles. we get ∠DBE = ∠ABH and triangles DBE and ABH are similar. then we get DE = DF. P B. 2 2 BD = BL and BE = BM as it is well-known. Likewise. by Spanning (capacius) arc. then 4BKC and 4AHC are congruent and AH = BK. Then. K C F P A R Q E H S D B G Scheme for Napoleon’s theorem solution 1. . E are the centers of the equilateral triangles ABG and CBH.158 Let P be the intersection point of the circumcircles of the equilateral triangles ABG. Moreover. P R⊥DF and P S⊥DE as it is well-known. BARCELONA TECH. then ∠ABL = ∠CBM = 30◦ . In the following figure we have that ∠ACK = ∠BCH = 60◦ . Solution 2 by Jos´ e Gibergans-B´ aguena. So. BCH and CAK. triangles AKB and ADF are similar too. Triangles BCM and ABL are 3 3 similar. That is ∠AP C = ∠BP C = ∠AP B = 120◦ Since P A. A similar procedure leads us to obtain that DF = F E. Produce BD and DE to L and M respectively. Adding to each ∠ACB. Spain. Hence. Likewise we obtain that ∠RDS = ∠QSD = 60◦ and the proof is complete. Now considering the quadrilateral F RP Q we have ∠RF Q + ∠AP C = 180◦ or equivalently ∠RF Q + 120◦ = 180◦ from which follows ∠RF Q = 60◦ . So. then P Q⊥EF. Barcelona. we have ∠BCK = ∠ACH. Then since D. AB BD = AH DE and AB AD = BK DF Since we have already seen that BK = AH and AD = BD. Since sides AC = KC and CH = CB. ∠AP C = 180◦ −∠AKC (cyclic). then adding ∠ABC to each. AB BL 3/2 BD BD = = = BC BM 3/2 BE BE Since ∠CBE + ∠ABD = ∠CBH. 4DEF is equilateral and the proof is complete. Spain. Then holds 4A1 A2 A2 is equilateral ⇔ z1 + ei2π/3 z2 + ei4π/3 z3 = 0 Proof. AH DE Solution 3 by Jos´ e Luis D´ıaz-Barrero and Jos´ e Gibergans-B´ aguena. z2 . A3 of a positively oriented triangle. It is well-known that 4A1 A2 A2 is equilateral and positively oriented if and only if A3 is obtained from A2 by rotation about A1 through an angle of π/3. Namely. AB BD Remark. Let z1 . z3 − z1 = eiπ/3 (z2 − z1 ) from which follows z3 = z1 + √ ! 1 3 +i (z2 − z1 ) = 2 2 √ ! 1 3 −i z1 + 2 2 √ ! 1 3 +i z2 2 2 . Barcelona. Indeed. The relation = can also be obtained from the right triangles AH DE BM H and ALB. To prove our result we need the following Lemma. BARCELONA TECH. A2 . z3 be the coordinates of the vertices A1 .159 K C F M H E A B D L G Scheme for Napoleon’s theorem solution 2. in 4BM H we have cos 30◦ = BM 3/2 BE = . BH BH cos 30◦ = BL 3/2 BD = AB AB and in 4ALB holds from which follows AB BD = . Let zA . On account of the Lemma. zD = 3 1 zF = zA + zC + zK . zB + zG + 2 zA = 0. Omran Kouba. zB .160 Therefore. B. Spain. . 3 1 zB + zC + zH . 2 Also Solved by Bruno Salgueiro Fanego. Viveiro. HBC and KAC are respectively 1 zA + zB + zG . C. Damascus. Higher Institute for Applied Sciences and Technology. zE = 3 Finally. The centers of gravity of triangles GAB. zC + zH + 2 zB = 0. where = ei2π/3 . Syria. zC be the affixes of vertices A. √ ! 1 3 z1 + e z2 + e z3 = z1 + − + i z2 2 2 √ !" √ ! √ ! # 1 3 3 1 3 1 + − −i z1 + z2 −i +i 2 2 2 2 2 2 √ ! √ ! 1 3 1 3 = z1 + − + i z2 − z1 + z2 = 0 −i 2 2 2 2 i2π/3 i4π/3 and the proof is complete. we have 3 zD + zF + 2 zE = zA + zB + zG + zA + zC + zK + 2 zB + zC + zH = (zA + zK + 2 zC ) + (zC + zH + 2 zB ) + (zB + zG + 2 zA ) = 0 and 4DEF is equilateral. respectively. we have zA + zK + 2 zC = 0. 2. ∀x. such that x > y. The generalization of problem Pn ki Let f : R → R be monotone such that i=1 f (x) is everywhere continuous. Analogously. . So. ∀n ∈ {1. such that x > y. . 2. . ◦ f )(x) ).161 MATHNOTES SECTION Generalization of Continuous Functions Adrian Naco Abstract. f is a decreasing monotone function. {z } | | {z } ki times ki times Prove that the function f ki . is continuous. y ∈ R. f is an increasing monotone function. if for i = k f k (x) > f k (y) then for i = k + 1 ⇒ f k (x) > f k (y) ⇒ f (f k (x)) > f (f k (y)) ⇒ f k+1 (x) > f k+1 (x) Case 2. }. . where ki is an odd (respectively even) positive integer (let us pose f ki (x) = f (f . . we apply the induction method on i. ) = (f ◦ f ◦ . f (x) > f (y). . . f (x) . for i = 1. we have that for i = 1 f (x) < f (y) ⇔ f 2−1 (x) < f 2−1 (y) . n}. . Let us prove that the function f 2k−1 is also a decreasing monotone function. reasoning as above. Introduction The last generalisation will come from Problem 42 (see [1]). . . Proof. In these notes we present a generalization of Problem 42. 3. . Applying the induction method on i we have that. ∀x. If f is a monotone function then 1) the function f ki (x) is of the same nature as f (related to monotonicity) if ki is an odd positive number. 2) the function f ki (x) is an increasing monotone one if ki is an even positive number. for i ∈ {1. . 1. . y ∈ R. Lema 1. . Case 1. f 2k is an increasing monotone function if f is a decreasing monotone one. ∀x. then . So. Let us define as. such that x > y. Lema 2. we have that for i = 1 f (x) < f (y) ⇒ f (f (x)) > f (f (y)) ⇔ f 2 (x) > f 2 (y) f 2k (x) > f 2k (y) then for i = k + 1 if for i = k ⇒ f 2k (x) > f 2k (y) ⇒ f (f 2k )(x) < f (f 2k )(y) ⇒ f (f (f 2k ))(x) > f (f (f 2k ))(y) ⇒ f 2+2k (x) > f 2+2k (y) ⇒ f 2(k+1) (x) > f 2(k+1) (y) Thus. If f is a monotone function and if ki is an odd (respectively even) positive number. y ∈ R.162 f 2k−1 (x) < f 2k−1 (y) then for i = k + 1 if for i = k ⇒ f 2k−1 (x) < f 2k−1 (y) ⇒ f (f 2k−1 )(x) > f (f 2k−1 )(y) ⇒ f (f (f 2k−1 ))(x) < f (f (f 2k−1 ))(y) ⇒ f 2+2k−1 (x) < f 2+2k−1 (y) ⇒ f 2(k+1)−1 (x) < f 2(k+1)−1 (y) Analogously. ∀x ∈ R. ∆f ki (x) = f ki (x + ∆x) − f ki (x). ∀i ∈ N . we prove the Lemma 1. . X . X n . . n . . k . ki . . = . ∆f i (x). ∆f (x) . . . . the function f ki is of the same nature as the function f . Based on Lema 1. ki is an odd positive number. Case 1. i=1 i=1 Proof. .. ∀i ∈ N . if f is a decreasing (increasing) monotone function then for ∆x > 0 (∆x < 0) we have that ∆f ki (x) = f ki (x + ∆x) − f ki (x) < 0 ⇒ ⇒ ⇒ −∆f ki (x) > 0 .163 Thus. . . . −∆f ki (x) = . . ∆f ki (x). . . X n n . X . . k ki i . . −∆f (x) = . ∆f (x). i=1 i=1 ⇒ . . n . X n . . . . k . X ki i . . . . −∆f (x) . = . ∆f (x). . ⇒ . . . n n . . . X . k . X ki . = . ∆f i (x). . − (x) ∆f . . . . ⇒ . . . n n . . . k . X . X ki i . . . . ∆f (x). = . ∆f (x). . i=1 i=1 i=1 i=1 i=1 i=1 and for ∆x < 0 (∆x > 0) we have that ∆f ki (x) = f ki (x + ∆x) − f ki (x) > 0 ⇒ ⇒ ⇒ ∆f ki (x) > 0 . . . . k ki i . ∆f (x) = . ∆f (x). . . n n . X X . . k . ∆f i (x). ∆f ki (x) = . . i=1 ⇒ i=1 . . . n n . . . k . X . X ki i . . . . ∆f (x). = . ∆f (x). . i=1 i=1 Case 2. f ki is an increasing monotone function as f is simply a monotone function.. Thus for ∆x < 0 we have that ∆f ki (x) = f ki (x + ∆x) − f ki (x) < 0 ⇒ ⇒ ⇒ −∆f ki (x) > 0 . ∀i ∈ N . . ki is an even positive number. [10pt] Based on Lema 1. . . . −∆f ki (x) = . . ∆f ki (x). . . X n n . X . k . ki i . . −∆f (x) = . ∆f (x). i=1 i=1 ⇒ . X . . X n . . n . . k . ki . . = . ∆f i (x). −∆f (x) . . . . ⇒ . X . X . n . . n . . k . ki . − . = . ∆f i (x). ∆f (x) . . . . ⇒ . X . X . n . . n . . k . ki i . . . . ∆f (x). = . . ∆f (x). i=1 i=1 i=1 i=1 i=1 i=1 . 164 and for ∆x > 0 we have that ∆f ki (x) = f ki (x + ∆x) − f ki (x) > 0 ⇒ ⇒ ⇒ ∆f ki (x) > 0 . . . . ∆f ki (x) = . . ∆f ki (x). . . n n . X X . . k ki i . . ∆f (x) = . ∆f (x). i=1 ⇒ i=1 . . X . X n . . . . k . n ki i . . . . ∆f (x). = . ∆f (x). . . is continuous. Pn Since i=1 f ki (x) is everywhere continuous then ∀a ∈ R lim x→a n X f ki (x) = i=1 n X f ki (a) i=1 or equivalently ∀ > 0 ∃δ() > 0 ∀x ∈ (a − δ. . . we prove Lemma 2. ) = (f ◦ f ◦ . . . Pn Theorem 1. where ki is an odd (respectively even) positive integer (let us pose f ki (x) = f (f . . . Let f : R → R be monotone such that i=1 f ki (x) is everywhere continuous. f (x) . n}. for i ∈ {1. 2. a + δ) ⇒ . ∀n ∈ N . Proof. i=1 i=1 Thus. ◦ f )(x) ). . . . | {z } | {z } ki times ki times Prove that the function f ki . . n n X . . X k ki i . < . f (x) − f (a) . . ⇔ . . n . . X k . [f i (x) − f ki (a)]. < . . ⇔ . . n . . X ki . ∆f (a). . < . i=1 i=1 i=1 i=1 ⇔ ⇒ . n . X . . k . ∆f i (a). < . . . i=1 . . k . . ∆f i (a). < . . . Albania E-mail address: nacoad@yahoo. Department of Mathematics. References [1] C. The last inequality confirms that the function f ki is everywhere continuous. Faculty of Mathematical Engineering and Physical Engineering. MORTICI. . Proposed Problem 42. MathProblems. Polytechnic University of Tirana. page 70.com . 2. 2. n} and i 6= i0 . . The problem proposed to solve is a special case where ki0 = 1 and ki is an odd number for i ∈ {1. Note. . 2 (2012). + ≥ ra2 + 2rb rc rb2 + 2rc ra rc2 + 2ra rb 4R + r . Spain. Proposed by Armend Sh. b = CA and let ra . rc be the ex–radii respectively.165 JUNIOR PROBLEMS Questions concerning proposals and/or solutions can be sent by e-mail to: juniorproblems-ks@hotmail. If a + b = 3c show that ra + rc ≥ 4rc . Romania. Republic of Kosova. Proposed by Ercole Suppa. “Matei Basarab” National College. Shabani. I. University of Prishtina. y) = 1 and x. Romania Prove that in any triangle ABC with the usual notations. University of Prishtina. Teramo Italy. Proposed by Dorlir Ahmeti. Buz˘ au.S. Let (x. Neculai Stanciu. Prove or disprove: for every positive integer k holds (kx − y. Bucharest. ´ 9. Describe the region of the plane occupied by the vertices of triangles ABC having circumcenter O and orthocenter H. holds 2 1 1 1 3 + . Alvarez Cubero de Priego de C´ ordoba.E. 10. Republic of Kosova.com Proposals 6. rb . D. 8. Two points O and H are given in position on the plane. k 3 x + k 2 x − k 2 y + kx − ky + x − y) = 1. 7. Find the first digit of the number 999999 .M B˘ atinet¸u–Giurgiu. “George Emil Palade” School. In ∆ABC let a = BC. y > 0. Francisco Javier Garc´ıa Capit´ an. Satulung. b. Maramure¸s. Romania. then a 1 1 1 ≤ + a2 + bc 4 b c Solution by Daniel Vacaru.166 Solutions No problem is ever permanently closed. Romania. Prove that √ a3 b3 c3 4 3 + + ≥ S b·R+c·r c·R+a·r a·R+b·r R+r Solution by Ioan Viorel P Codreanu. Let ABC be a triangle with side lengths a. Neculai Stanciu. Ioan Viorel Codreanu. Italy. c are real positive numbers. Also solved by the proposers 2. “George Emil Palade” School. Bucharest. BB . We will be very pleased considering for publication new solutions or comments on the past problems. Romania. Romania. 20 Problem Solving Group. Roma. . “Matei Basarab” National College. circumradius R and inradius r. Neculai Stanciu. Spain.M. Using the Bergstrom Inequality we have P 2 2 X X a3 a a4 = ≥P bR + cr abR + acr (abR + acr) P 2 P The inequality a ≥ ab yields P 2 2 P 2 2 P 2 2 P 2 a a a a P P P 2 = = ≥ (abR + acr) (R + r) ab (R + r) a (R + r) The Ionescu–Weitzenb¨ ock inequality X √ a2 ≥ 4 3S completes the proof. respectively. c. b.A. “George Emil Palade” School. G. Proposed by Francisco Javier Garc´ıa Capit´ an. Romania. Prove that if a. Romania (with P few changes by the editor) stands for cyc .S. One has equivalently a2 b + a2 c + b2 c + bc2 ≥ 4abc By the AGM one has √ p 4 a2 b + a2 c + b2 c + bc2 ≥ 4 4 (a2 b)(a2 c)(b2 c)(bc2 ) = 4 a4 · b4 · c4 = 4abc as desired Also solved by Arb¨ er Avdullahu. Proposed by D.M B˘ atinet¸u–Giurgiu. Buz˘ au. I. Maramure¸s. Proposed by D. Romania.E. CC be any three diameters of the same circle. and the proposers. “Matei Basarab” National College. Buz˘ au. area S. ´ 3.R. Bucharest. 1. Alvarez Cubero de Priego 0 0 0 de C´ ordoba. Pitesti. Satulung. Let AA . B˘ atinet¸u–Giurgiu. Italy. Remark. By applying Pascal theorem to the degenerated hexagon CCC 0 ABB 0 we get that the points T 0 = CC ∩ AB. Proposed by Paolo Perfetti. Dept. (Dedicated to Juan Bosco Romero M´arquez) Solution by proposer. Italy. Call T 0 the intersection of AB and the line through C tangent to the circle. Therefore T 0 lies on P Q. sin(α − β) TB = 2sin2 β .. Show that T C is tangent to the circle. Rome.A. find with proofs the coordinates of all the extremum points of x2 x4 + x3 + 4x + 16 Solution by G. sin(α − β) TC = 2 sin α sin β . It is enough to prove that T 0 = T . sin(α − β) that lead to T C 2 = T A · T B. Jubilado. but it also lies on AB from its definition. Without using calculus. “Tor Vergata” Univ. and therefore T C is perpendicular to the circle.20 Problem Solving Group. Math. Another solution consists of considering the angles ∠AOC = 2α and ∠COB = 2β and arriving to the relations TA = 2sin2 α . Roma. Universidad de Sevilla 4. Q = AC 0 ∩ B 0 C and T = P Q ∩ AB.167 Let us consider the intersection points P = A0 C ∩ BC 0 . O = CC 0 ∩ BB 0 are Q = C 0 A ∩ B 0 C collinear.R. Hece T 0 = P Q ∩ AB = T . Also solved by Ricardo Barroso Campos. . Ceva’s theorem yields AF · BD · CE = F B · DC · EA which. let P be an arbitrary point on AD. F are concyclic. since BD = DC. Solution by proposer. Teramo.168 We note that f (x) := x2 4x2 . CP with the circumcircle. Considering the power of P with respect to the circumcircle of 4ABC we obtain P B · P E1 = P C · P F1 ⇒ PC P E1 = PB P F1 (1) Since the lines AD. f (0) = 0 ≤ f (x) ≤ 1 = f (−2). respectively. rewrites as AF AE = FB EC (2) . CF are concurrent. it follows immediately that for all x ∈ R. 4 Moreover for x > 0 f (x) := 1 (x2 + 16/x2 ) 1 1 p = ≤ p = f (2). 2 2 2 2 + (x + 4/x) 12 2 x · 16/x + 2 x · 4/x Also solved by the proposer 5. = x4 + x3 + 4x + 16 (x + 2)2 ((2x − 3)2 + 7) + 16x2 then. E1 . let E = BP ∩ AC. Italy. F = CP ∩ AB and let E1 . Show that E. F1 . Proposed by Ercole Suppa. BE. F1 be the second intersection of BP. In ∆ABC let D be the midpoint of BC. F1 . E1 . hence 4P EF ∼ 4P BC so PF PC = (3) PE PB From (1) and (3) it follows that P E1 PF = ⇐⇒ P E · P E1 = P F · P F1 PE P F1 proving that E. Also solved by Ricardo Barroso Campos. CF are three concurrent cevians and E. as desired. Jubilado. Reverting the steps of the above proof it is easy to show that also the converse is true i. BE. Universidad de Sevilla. . Remark. if AD. F1 .169 Therefore by Thale’s theorem F E k BC.e. F are concyclic then D is the midpoint of BC. F are concyclic. E1 . Cristinel Mortici. Department of Mathematics and Information Science. 170 . 2013 Problems 73. University of Technology. Proposals should be accompanied by solutions.Proposed by Enkel Hysnelaj. Jos´ e Luis D´ıaz-Barrero. David R. Jozsef S´andor.Mathproblems ISSN: 2217-446X. Australia. Armend Sh. Prishtin¨ e. Universiteti i Prishtin¨ es. China. The editors encourage undergraduate and pre-college students to submit solutions. Ercole Suppa. Determine all the real valued functions f such that ( ) ax + 1 f (x) + 3f =x bx where a and b are real numbers that satisfy b ̸= 0 and a2 + 2b = 0 or a = 0. Shandong Province. Paolo Perfetti. Solutions will be evaluated for publication by a committee of professors according to a combination of criteria. each indicating the name of the sender. 256603. Mih´aly Bencze.com Volume 3. Mohammed Aassila. PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets. Valmir Bucaj. For 0 < x < 1. An asterisk (*) indicates that neither the proposer nor the editors have supplied a solution. Anastasios Kotronis. Student solutions should include the class and school name. Stone. Francisco Javier Garca Capit´an. Roberto Tauraso. Drawings must be suitable for reproduction.mathproblems-ks. x n(1 − x) c ⃝2010 Mathproblems. Emanuele Callegari.com Solutions to the problems stated in this issue should arrive before December 15. Kosov¨ e. Enkel Hysnelaj. Questions concerning proposals and/or solutions can be sent by e-mail to: mathproblems-ks@hotmail. n ∈ N show that 2x n+1 + > n + 3. Binzhou City. Shabani. Ovidiu Furdui. Sydney. Issue 3 (2013). Binzhou University. url: http://www. Li Yin. Omran Kouba. Pages 170–197 Editors: Valmir Krasniqi. Teachers can help by assisting their students in submitting solutions. 74. 4 (mod 5) . Matei Basarab National Colege. George Emil Palade Secondary School. Ln and Tn are the n−th Fibonacci. Athens. Proposed by Mohammed Aassila. b. show that 0 (mod 5) . n = 15 (mod 20) 2 (mod 5) . Proposed by Anastasios Kotronis. 80∗ . Let R+ = (0. Prove that : ( ) (1) 2 | cos x| + | cos y| + | cos z| + | cos(y + z)| + | cos(x + z)| + | cos(y + x)| ≥ 3 (2) | cos x|+| cos y|+| cos z|+| cos(y +z)|+| cos(y +x)|+| cos(x+z)|+3| cos(x+ y + z)| ≥ 3. Bucharest. If A. n ≥ 1.M. For t ∈ R calculate the following limits: ) ( sin2 (t) sin2 (t) cos2 (t) . Strasbourg. (1) Prove that if there exists a positive integer k such that b2 − 4ac = k 2 a2 then +∞ ∑ n=1 an2 1 ∈ Q. Serban School Cioculescu. Republic of Kosova. z ∈ R. ∞) and f. Let a. France. n = 10 (mod 20) . Greece. Lucas and Triangular number respectively. y. g : R∗+ → R∗+ be functions such that: g(x + 1) f (x) lim (f (x + 1) − f (x)) = a ∈ R∗+ . Proposed by D. B ∈ M2 (C) are two matrices. Romania and Neculai Stanciu.171 75. show that the following statements are equivalent: 2 (1) (AB − BA) = O2 . lim = b ∈ R∗+ and the limits lim . (g(x + 1)) x+1 − (g(x)) x and limx→∞ x→∞ 76. Romania. Find all positive integers n such that the product n ∏ (9k 2 + 1) k=1 is a perfect square. B˘ atinet¸u-Giurgiu. University of Prishtina. c be integers such that a ̸= 0 and an2 + bn + c ̸= 0 for all n ∈ N \ {0}. If x. 78. If Fn . n ̸= 0 (mod 5) 1 (mod 5) . Romania. Gaesti. Buzˇ au. (1) lim (f (x)) (g(x + 1)) x+1 − (g(x)) x x→∞ ) ( cos2 (t) cos2 (t) sin2 (t) (2) lim (f (x)) . Proposed by Mih´ aly Bencze. Proposed by Alban Kryeziu. . n = 0 (mod 20) 77. + bn + c (2) Is the converse true ? 79. n = 5 (mod 20) 2Fn5 + n2 Fn (Tn + n) − Ln − Ln+1 = 3 (mod 5) . Romania (Jointly). Proposed by Florin Stanescu. x→∞ x→∞ xg(x) x→∞ x (g(x))1/x x exist. Brasov. (2) det (AB (A + I2 ) − BA (B + I2 )) = det (ABA − BAB) . 66. k=1 then we will prove n+1 ∑ k=1 xk n+1 ∑ k=1 x5k ≤ (n + 1) n+1 ∑ x6k (3) k=1 Again we have ∀k ∈ {1. We will be very pleased considering for publication new solutions or comments on the past problems. Department of Mathematics and Information Science. then v u n n n n ∑ ∑ ∑ u ∑ x2k x6k . k = 1. 2 · · · n. · · · . n} : (x5n+1 − x5k )(xn+1 − xk ) ≥ 0 ⇔ x6n+1 + x6k ≥ xn+1 x5k + x5n+1 xk . Now we will assume that our inequality is holds for n and then we will prove for n + 1. n ∑ x2k k=1 · n ∑ k=1 ( x4k = n √ ∑ √ x3 · xk ) ( · k=1 n √ √ ∑ x3k · x5k ) k=1 v v u n u n n n ∑ ∑ u∑ u∑ 3 t ≤ xk · xk · t x3k · x5k k=1 k=1 k=1 v u n n n ∑ ∑ u∑ 3 t = xk · xk · x5k k=1 k=1 k=1 k=1 If we prove following inequality then our proposed problem well be solved n ∑ xk · k=1 n ∑ x5k ≤ n · k=1 n ∑ x6k (1) k=1 In order to prove (1) we use induction. x4k 6 x3k tn k=1 k=1 k=1 k=1 Solution 1 by AN-anduud Problem Solving Group. Applying the Cauchy-Schwartz inequality we have. Proposed by Li Yin. Mongolia. In other words we assume that n ∑ xk · k=1 n ∑ k=1 x5k ≤ n n ∑ x6k (2). Show that if xk > 0. For n = 1 is trivial. Binzhou University. China. 2. Binzhou City. 256603. Shandong Province.172 Solutions No problem is ever permanently closed. Since (x51 − x52 )(x1 − x2 ) ≥ 0 ⇔ x61 + x62 ≥ x51 x2 + x52 x1 ⇔ 2(x61 + x62 ) ≥ (x1 + x2 )(x51 + x52 ) the base of induction is true. Ulaanbaatar. Let n = 2. 1. Rome. We use the well known asymptotic expansion of the Gamma function (See ”Handbook of Mathematical Functions” by Abramowitz and Stegun. Bucharest. Calculate lim Ln (x). Equation 6. Rehovot. c and d we have. Switzerland. Let x ∈ R.37 on page 257). k. Israel. Γ (n) = 2πe n n . Also solved by Albert Stadler. Damascus. Syria. Proposed by D. nx6n+1 + n ∑ x6k ≥ xn+1 · k=1 n ∑ x5k + x5n+1 · k=1 n ∑ xk (4). Higher Institute for Applied Sciences and Technology. Herrliberg. 6x2i x2j x4k x4l ≤ x3i x3j x6k + x3i x3j x6l + x3i x3k x6l + x3i x6k x3l + x3j x3k x6l + x3j x6k x3l Adding these inequalities yields ) )2 ( n )2 ( n ( n )2 ( n ∑ ∑ ∑ ∑ 6 3 4 2 xk xi xk ≤ 6n 6 xi i=1 k=1 i=1 k=1 and the desired conlusion follows by taking square roots. Romania (Jointly). given by (( )sin2 x ( √ )sin2 x ) √ 2 n n+1 Ln (x) = ncos x (n + 1)! − n! . George Emil Palade Secondary School. and (Ln (x))n≥2 . Paolo Perfetti. n→∞ Solution 1 by Moti Levy. Buzˇ au. according to the AM-GM inequality: 6a2 b2 c4 d4 ≤ a3 b3 c6 + a3 b3 d6 + a3 c3 d6 + a3 c6 d3 + b3 c3 d6 + b3 c6 d3 So. Solution 2 by Omran Kouba. Moti Levy. ( ( )) √ 1 −n n− 12 1+O . l ≤ n. j. for all indices 1 ≤ i. Israel. Italy.M.173 so summing all together we have. For positive real numbers a. Department of Mathematics. b. 67. Tor Vergata University. B˘ atinet¸u-Giurgiu. Buchenrain. Matei Basarab National Colege. and the proposer . Rehovot. Romania and Neculai Stanciu. k=1 From (2) and (4) we get ( n ∑ xk + xn+1 ) · k=1 ⇔ n+1 ∑ k=1 n ∑ x5k + x5n+1 · k=1 xk · n ∑ k=1 ⇔ xk ≤ (n + 1) k=1 x5k + x5n+1 · (n+1 ∑ n ∑ n+1 ∑ n ∑ x6k + nx6n+1 k=1 xk ≤ (n + 1) k=1 n ∑ x6k + (n + 1)x6n+1 k=1 ) (n+1 ) n+1 ∑ ∑ 5 xk · xk ≤ (n + 1) x6k k=1 k=1 k=1 Finally (3) is proved. 1+n n n2 α α . Damascus. Solution 2 by Omran Kouba. we have ( ( )) α 1 (n + 1) = n 1 + + O . ( ) √ 1 1 log(n!) = n log n − n + log n + log 2π + O . for α ̸= 0.174 Ln (x) = ncos lim Ln (x) n→∞ = lim ncos ( 2 x n→∞ − (√ 2 2π x ( ) sin2 x sin2 x (Γ (n + 2)) n+1 − (Γ (n + 1)) n ( 2x (√ ) sin sin2 x n+2 2 n+1 2π e− sin x n+1 (n + 2) ) sinn2 x ( e − sin2 x n+1 n (n + 1) sin2 x n+ 1 2 n+1 n+ 3 2 n+1 ) )) ) sin2 x sin2 x (n + 2) − (n + 1) n→∞ ( ) 2 sin2 x sin2 x − sin2 x =e lim n1−sin x (n + 2) − (n + 1) n→∞ (( )sin2 x ( )sin2 x ) n+2 n+1 − sin2 x =e lim n − n→∞ n n (( )sin2 x ( )sin2 x ) 2 2 1 = e− sin x lim n 1+ − 1+ n→∞ n n ( ( )) 2 sin2 x sin2 x − sin2 x =e lim n 1 + − 1+ n→∞ n n = e− sin x = e− sin x 2 2 lim ncos 2 x ( sin2 x. n n2 ( ) 1 1 1 = +O . for large n. Syria. Higher Institute for Applied Sciences and Technology. Our starting point is the well-known asymptotic expansion. ( √ ( ) ( )) α log(n!) α log n α log 2π 1 α/n (n!) = exp = exp α log n − α + + +O n 2n n n2 ( ) √ ( ) α log n α log 2π 1 = e−α nα exp + +O 2n n n2 ( √ ( 2 )) α log n α log 2π log n −α α =e n 1+ + +O 2n n n2 Now. 2 n In particular. 2 The answer is (sin2 x)e− sin x . 1+n n n2 ( ) log(1 + n) log n log n = +O . France. Rome. Tor Vergata University. 68. Let a and b be positive real 1 numbers such that a + b = 2. α/(n+1) √ ( 2 )) α log(n + 1) α log 2π log n 1+ = e (n + 1) + +O 2(n + 1) n+1 n2 √ ( ( )) ( ( 2 )) α 1 α log n α log 2π log n −α α =e n 1+ +O 1+ + +O 2 n n 2n n n2 ) ( √ ( 2 ) α log n α(1 + log 2π) log n = e−α nα 1 + + +O 2n n n2 −α α ( 2 ) ( ) log n n1−α ((n + 1)!)α/(n+1) − (n!)α/n = αe−α + O . Department of Mathematics. Ploiesti. Italy. ( ((n + 1)!) Thus. Also solved by AN-anduud Problem Solving Group. Romania. Solution by Paolo Perfetti. Lyc´ ee henri IV. Department of Mathematics. Paris. Rome. then prove that 2 kb aa bb ka ≥ 1. we get f (x) = (1 + x)k(1−x) ln(1 + x) + (1 − x)k(1+x) ln(1 − x) ≥ 0 Since f (x) = f (−x) we can consider 0 ≤ x ≤ 1. Mongolia. Moubinool Omarjee. Tor Vergata University. Ulaanbaatar. Moreover f (x) ≥ 0 is equivalent to 1 (1 + x)k(1−x) ln(1 + x) ≥ (1 − x)k(1+x) ln 1−x and doing a second logarithm we come to 1 . Proposed by Vasile Cirtoaje. Italy. The logarithm and b = 2 − a yield ak(2−a) ln a + (2 − a)ka ln(2 − a) ≥ 0 Setting a = 1 + x. −1 ≤ x ≤ 1. h(x) = k(1 − x) ln(1 + x) + ln ln(1 + x) − k(1 + x) ln(1 − k) + ln ln ≥0 1−x h′ (x) = −k ln(1 + x) + + (1 − x)k 1 (1 + x)k + − k ln(1 − x) + + 1+x (1 + x) ln(1 + x) 1−x 1 (1 − x) ln(1 − x) or h′ (x) = 2k 1 1 1 + x2 − k ln(1 − x2 ) + + 1 − x2 (1 + x) ln(1 + x) (1 − x) ln(1 − x) . Paolo Perfetti. n and the desired conclusion follows by taking α = sin2 x. If k ≥ . and the proposer.175 So. Also solved by the proposer. Link: http://www.php?f=52&t=351014&p=1891000#p1891000. −4104kx2 + 792x2 < 0. 2 3 1 1 ln(1 − x) ≥ −(x + x2 + x3 ) 2 3 we have 1 + x2 1 ( )+ − k ln(1 − x2 ) + 1 − x2 (1 + x) x − 12 x2 + 13 x3 1 ( ) . x + 1620 − 6480k < 0 2 Thus we have obtained R′ (x) ≥ 0.com/Forum/viewtopic. . Since ) ( 3 ′ x2 + O(x3 ) h (x) = (2k − 1) + 5k − 4 it follows h′ (x) ≥ 0 if k ≥ 1/2 holds.artofproblemsolving. Moreover by 5 1 h(x) = (2k − 1)x + ( k − )x3 + O(x5 ) 3 4 it follows h(0) = 0 and then h(x) > 0 for any x. where P (x) = − 2(16kx10 + 40kx8 − 87kx6 + 192x6 + 696x4 − 1485kx4 − 4104kx2 + + 792x2 + 1620 − 6480k)x and Q(x) = (x + 1)2 (2x2 − 3x + 6)2 (x − 1)2 (2x2 + 3x + 6)2 We want R (x) ≥ 0 so are reduced to show ′ 16kx10 +40kx8 −87kx6 +192x6 +696x4 −1485kx4 −4104kx2 +792x2 +1620−6480k ≤ 0 Indeed by 0 ≤ x ≤ 1 and k ≥ 1/2 we have 16kx10 + 40kx8 − 87kx6 + 192x6 ≤ −21kx6 + 192x6 = 363 6 x 2 363 6 x + 696x4 − 1485kx4 − 4104kx2 + 792x2 + 1620 − 6480k ≤ 0 2 Clearly we have 363 6 696x4 − 1485kx4 < 0. + 1 2 1 3 = R(x) (1 − x) −x − 2 x − 3 x h′ (x) ≥ 2k ) ( 5 x2 + O(x4 ) R(x) = (2k − 1) + 5k − 4 and R′ (x) = P (x) Q(x) .176 Since 1 1 ln(1 + x) ≥ x − x2 + x3 . Editorial Comment: The problem in question has also been posed in the forum The Art of Problem Solving in the past. South Korea. Korea Advanced Institute of Science and Technology(KAIST). from ∆ = 0 we conclude that log xk = s = log a for every k and the desired conclusion follows. So the determinant has to be non-negative. Romania. Higher Institute for Applied Sciences and Technology. k=1 W check easily that ∆= m ∑ log2 xk − ms2 + m(m + 1)s2 − 2m(m + 1)s log a + m(m + 1) log2 a k=1 = m ∑ ( 2 log xk + k=1 m ∑ =2 m ∑ )2 log xk − 2m(m + 1)s log a + m(m + 1) log2 a k=1 ∑ log2 xk + 2 log xi log xj + m(m + 1) log2 a 1≤i<j≤m m ∏ k=1 − 2(m + 1) log a log xk = 0. then the equation becomes ∑ ∑ ∑ m(m + 1) 2 ui 2 + (ui uj ) + b = (m + 1)b ui 2 i≤m i<j≤m i≤m We can rewrite the formula as a quadratic equation with respect to b. Proposed by Mih´ aly Bencze.177 69. uk = log xk . Solve the following equation: m ∑ ∑ log2 xk + 1≤i<j≤m k=1 ∏ m(m + 1) log2 a = (m + 1) log a log xk . ∑ ∑ ∑ m(m + 1) 2 b − (m + 1) ui b + u2i + (ui uj ) = 0 2 i≤m i≤m i<j≤m This equation must have at least one real root. Solution 2 by Hun Min Park. Let b := log a. k=1 But. ∑ ∑ ∑ ∑ D = (m+1)2 u2i + 2 (ui uj ) −2m(m+1) ui + (ui uj ) ≥ 0 i≤m i<j≤m i≤m i<j≤m 2 ∑ ∑ ∑ ∑ 1 ∴ D = (−m + 1) u2i + 2 (ui uj ) = ui − m u2i ≥ 0 m+1 i≤m i<j≤m i≤m i≤m . and define ∆= m ∑ (log xk − s)2 + m(m + 1)(s − log a)2 . 2 m log xi log xj + k=1 where a > 0. Daejeon. Bra¸sov. let s = m log(x1 x2 · · · xm ). Solution 1 by Omran Kouba. 1 Indeed. The answer is x1 = x2 = · · · = xm = a. Syria. Damascus. Also solved by AN-anduud Problem Solving Group. Ulaanbaatar. Show that √ √ πn 2 1 π 4 + + − + O(n−3/2 ). an = 2 3 12 2n 135n . .178 Besides. Syria. Greece. that is t = log a. Let ξk = log xk − log a. Show that : √ √ π −1/2 2 −1 1 π −3/2 an = n + n + n + O(n−2 ). Higher Institute for Applied Sciences and Technology. Israel. Solution 3 by Moti Levy. Rehovot. we obtain (t − log a)2 = 0. ξ T = [ξ1 . 1 >)2 = 0 where. Damascus. . This implies that ∥ξ∥ = 0 or ξ = 0. and < ·. Thus the solution of the equation is x1 = x2 = · · · = xm = a. ∑n ( ) 70. We conclude that the solution of the equation is x1 = x2 = · · · = xm = a. 2 ∑∑ ∑∑ ∑ ∑ 1∑∑ 1 1 (ui −uj )2 = (u2i +u2j )− ·2 (ui uj ) = m u2i − ui 2 2 2 i≤m j≤m and m ∑ i≤m i≤m j≤m u2i − ∑ i≤m j≤m i≤m i≤m 2 ui ≥ 0 by Cauchy-Schwarz inequality. Proposed by Anastasios Kotronis. · > is the familiar Euclidien norm. Let an := k=0 nk k!n−k . The original is wrong. So i≤m ∑∑ (ui − uj )2 = − i≤m j≤m 2 D=0 m+1 This can be true only when u1 = u2 = · · · = um . Calling this common value t and replacing back in the equation. Mongolia and the proposer. i=1 j=1 k=1 In vector notation. then the equation is equivalent to the quadratic form m ∑ ξk2 + m ∑ m ∑ ξi ξj = 0. 2 3 12 2 Solution by Omran Kouba. 2 ∥ξ∥ + (< ξ. ξk ]. 1 is the m dimensional vector whose all entries are equal to 1. Athens. . it should be corrected as follows: ( ) ∑nstatement Let an = k=0 nk k!n−k . . k+1 2 4 2 k+1 k−1 ∑ αk = µj µk+1−j . one checks easily that P is a solution to the differential equation y ′ −y = xn that satisfies P (0) = −n!. The function φ defines a strictly increasing bijection from (−1. whose inverse is the known Lambert function W . by integration on [0. (see [1] and the references therein. from (P (x)e−x )′ = xn e−x we conclude. the function u 7→ 1e W ′ ( u−1 e ) has the following expansion 1 ′ W e ( u−1 e ) = ∞ ∑ kµk (2u)k/2−1 k=1 where (µk )k≥0 is defined by the following recurrence relations: k − 1 ( µk−2 αk−2 ) αk µk−1 + − − .179 Consider the simple differential equation y ′ − y = xn . n]. Considering only the first few terms. the change of variables u ← W (t) transforms the integral expression for an into the following one ∫ 0 n!en an = n − n (−et)n W ′ (t) dt n −1/e ( ) ∫ 1 n!en u−1 1 = n −n du (1 − u)n W ′ n e e 0 But. we have ( ) √ 2 11 √ 1 86 1 ′ u−1 W =√ − + 2u − u + O(u u) e e 3 24 135 2u . In particular P (n) = −nn an Conversely. If P is a polynomial solution to this equation. α0 = 2 and α1 = −1 (see [1]). +∞). +∞) onto (−1/e.µ1 = 1. −1 where φ(x) = xex . for 0 < u < 1. then clearly deg P = n and for 0 ≤ k ≤ n and ( ) n! n (k+1) (k) n−k n−k P (x) − P (x) = n(n − 1) · · · (n − k + 1)x = x = k!xn−k (n − k)! k ∑n ( ) adding these equalities we see that P (x) = − k=0 nk k!xn−k . On the other hand. µk = j=2 with the initial values µ0 = −1.) So. that ∫ n −an nn e−n + n! = tn e−t dt 0 or equivalently n!en 1 an = n − n n n ∫ n n n−t t e 0 n!en = n − n(−e)n n ∫ n!en dt = n − n(−e)n n ∫ 1 (−u)n e−nu du 0 0 (φ(u))n du. Mongolia and the proposer. Gonnet and D. Knuth& On the Lambert W Function. Buchenrain. J. pp. Ulaanbaatar. Proposed by Enkel Hysnelaj. for instance we have: √ √ √ ( ) πn 2 1 π 4 1 π 8 1 √ an = + + − + + +O . University of Technology. c ∈ ℜ − {0}. Rome. Herrliberg. Tor Vergata University. E. 5. 2 3 12 2n 135n 288 2n3 2835n2 n2 n [1]&R. Jeffrey and D. Hare and D. In the given system of equations if we do the following substitution . Italy. 71. E. This asymptotic expansion can be pursued further.& ADVANCES IN COMPUTATIONAL MATHEMATICS. Sydney. Department of Mathematics. Also solved by Albert Stadler. Australia. G. Corless and G. and the proposer. which is the announced result.180 and consequently √ ( ) √ √ n!en n n! π 2n 11 2n n! π 86 n 1 √ √ − + − + + O nn 24 2Γ(n + 5/2) 135 n2 + 3n + 2 n n 2 Γ(n + 3/2) 3(n + 1) √ ( ( ) ) n 1 1 11 4 n!e n! 2π 2 √ + +O = n −n + − n Γ(n + 3/2) 2 24(2n + 3) 3 135n n n ) ( ( )) ( n!en √ (n!)2 22n 1 1 1 2 4 1 √ = n − 2 − +O + − +O n (2n)! 2 48n n2 3 135n n n √ ( ) 1 Using the well-known Stirling expansion: n! = nn e−n 2πn 1 + 12n + O(n−2 ) we find that )2 ( )−1 ( ) ( √ (n!)2 22n 1 1 1 √ = πn 1 + 1+ +O (2n)! 12n 24n n n √ ( ) √ 1 π 1 √ = πn + +O 8 n n n √ ( ) √ n!en 1 2π 1 √ = 2πn + +O nn 12 n n n an = Thus √ an = πn 2 1 + + 2 3 12 √ π 4 − +O 2n 135n ( 1 √ n n ) . (1996). Solution 1 by AN-anduud Problem Solving Group. Switzerland. Paolo Perfetti. M.329–359. b. Solve the following system of equations 1 1 1 x + y+z = a 1 1 1 y + z+x = b 1 1 1 z + x+y = c where a. H.& Vol. Remark. 181 1 x = u. Hence } 1 (u + v) + 2w = (a + b)t ⇒ w = (a + b − c)t u + v = ct 2 } 1 (u + w) + 2v = (a + c)t ⇒ v = (a + c − b)t u + w = bt 2 } 1 (w + v) + 2u = (c + b)t ⇒ u = (c + b − a)t w + v = at 2 Hence now we using the last identities in (∗). c t= 4 . we get 1 1 1 1 1 1 xy + zx = yz + xy = zx + yz a a b b c c 1 1 0 xy 1 a a ∴ 1b 1b 0 yz = 1 t (t ∈ R) · · · (5) zx 1 0 1c 1c which is a linear system in xy. (2) × y(z + x). 2(a + b − c) x= Solution 2 by Hun Min Park. Daejeon. yz. South Korea. y1 = v. 1 1 1 x + y+z = a · · · (1) 1 1 1 · · · (2) y + z+x = b 1 1 1 + = · · · (3) z x+y c (1) × x(y + z). b. zx. If u + w = bt then v + w = at. zx) = (a + b − c. a − b + c) 2 . Solving this system. (3) × z(x + y) 1 1 1 ⇒ x + y + z = x(y + z) = y(z + x) = z(x + y) · · · (4) a b c From (4). yz. then finding t in terms of a. Korea Advanced Institute of Science and Technology(KAIST). we obtain t (xy. 2ab + 2bc + 2ca − a2 − b2 − c2 Hence 2ab + 2bc + 2ca − a2 − b2 − c2 2(b + c − a) 2ab + 2bc + 2ca − a2 − b2 − c2 y= 2(a + c − b) 2ab + 2bc + 2ca − a2 − b2 − c2 z= . z1 = w then we have vw u + v+w = uw v + u+w = uv w + u+v = 1 a 1 b 1 c (∗) ⇔ b a c b a c = = = u+w v+w u+v u+w v+w u+v . u + v = ct. −a + b + c. 1 a2 + b2 + c2 − 2(ab + bc + ca) . Syria. Department of Mathematics. China.182 i. a + b − c ̸= 0) Let s s s x= . y= . Damascus. and the proposer. 256603. 2 −a + b + c 1 a2 + b2 + c2 − 2(ab + bc + ca) y=− . −a + b + c. Proposed by Li Yin. y. Binzhou University. Shandong Province. Italy. a ( ) 1 1 1 x+y+z = + + s −a + b + c a − b + c a + b − c −((a2 + b2 + c2 ) − 2(ab + bc + ca)) = s (−a + b + c)(a − b + c)(a + b − c) ( ) 1 1 1 1 1 x(y + z) = · + s2 a a −a + b + c a − b + c a + b − c 1 2a = s2 a (−a + b + c)(a − b + c)(a + b − c) 2 = s2 (−a + b + c)(a − b + c)(a + b − c) 1 As x + y + z = x(y + z) by (4). 2 a−b+c 1 a2 + b2 + c2 − 2(ab + bc + ca) z=− 2 a+b−c x=− Also solved by Omran Kouba. Tor Vergata University. a 2 −((a2 + b2 + c2 ) − 2(ab + bc + ca)) s= s2 (−a + b + c)(a − b + c)(a + b − c) (−a + b + c)(a − b + c)(a + b − c) x:y:z= 1 ∴ s = − (a2 + b2 + c2 − 2(ab + bc + ca)) 2 Thus. Department of Mathematics and Information Science. Let 1 < p < ∞. we can generalize the inverse of arcsin as follows: ∫ x 1 0≤x≤1 dt arcsinp (x) = (1 − tp )1/p 0 . z= (s ∈ R) −a + b + c a−b+c a+b−c 1 and calculate two real numbers x + y + z. Paolo Perfetti.e xy : yz : zx = a + b − c : −a + b + c : a − b + c This result gives xyz xyz xyz 1 1 1 : : = : : yz xz xy yz xz xy 1 1 1 = : : −a + b + c a − b + c a + b − c (As x. Rome. 72∗ . a − b + c. Binzhou City. z = ̸ 0. x(y + z). Higher Institute for Applied Sciences and Technology. p 0 (1 − u)2/p u ∫ 1 1 ln x ln(1 − x) = 3 dx with the change of variables x = 1 − u. Syria. (This is just an integration by parts.) Finally. 1) we have (∞ )( ∞ ) ∞ ∑ xn ∑ ∑ ln(1 − x) n = = x Hn xn − 1−x n n=1 n=1 n=1 ∑n where Hn = 1/k is the nth harmonic number. for x ∈ [0. p 0 x2/p 1 − x Now. using the k=1 positivity of the functions involved we can interchange the signs of summation and integral: (∞ ) ∫ 1 ∫ 1 ∞ 1 ln(1/x) ∑ 1 ∑ n Ip = 3 Hn x dx = 3 Hn xn−2/p ln(1/x) dx p 0 x2/p p n=1 0 n=1 ∞ Hn 1 ∑ 3 p n=1 (n + 1 − 2/p)2 ∫1 1 where we used the simple result 0 tα ln(1/t) dt = (1+α) 2 which is valid for α > −1.183 and ∫ 1 πp 1 = arcsinp (1) = dt. Higher Institute for Applied Sciences and Technology. The generalized hyperbolic cosine function is defined as cosp (x) = d dx sinp (x). Now. p n=1 (p(n + 1) − 2)2 = Also solved by Moti Levy. Israel. Rehovot. Calculate: ∫ π2p ln cosp (x) ln sinp (x) Ip = dx. 2 is called the generalized sine function and denoted by sinp . . Taking the derivative of the identity x = sinp (arcsinp (x)) we see that √ cosp (arcsinp (x)) = p 1 − xp . Consequently. making the change of variables x ← arcsinp (t) in the considered integral we see that ∫ 1 √ ln p 1 − tp ln t dt √ √ Ip = p p p t 1 − t 1 − tp 0 ∫ 1 p 1 ln(1 − t ) ln t = dt p 0 (1 − tp )2/p t ∫ 1 1 ln(1 − u) ln u = 3 du with the change of variables u = tp . 1]. for x ∈ [0. p 1/p 2 0 (1 − t ) [ πp ] The inverse of arcsinp (x) on 0. cosp (x) sinp (x) 0 Solution by Omran Kouba. ∞ 1∑ Hn Ip = . Damascus. n such that m − n = p with p prime and 2017 ≤ p ≤ 3011. Let P = {A. 52. B} be a partition of the set of positive integers N. Show that ( )2 ( ) n n 2n − 1 1 ∑ k = n k n−1 k=0 . Let n be a positive integer. The partition P is called admissible if for any positive integers m.184 MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. 53. Proposals 51. Show that the orthocenter of an acute triangle coincides with the center of the circle inscribed in the triangle formed by the feet of its altitudes. then n ∈ A and m ∈ B or viceversa. Let n be a nonnegative integer. Compute 2n+1 ∑ j=1 (2n + 1)2 j3 − 3(2n + 1)j + 3j 2 55. Proposals are always welcomed. 54. The source of the proposals will appear when the solutions be published. Find all admissible partitions of N. Without using the series expansion of the hyperbolic functions. show that the function f : R → R defined by f (x) = sinh2 x is not a polynomial. That would be 2743 minus 1 (for 2743) minus 12 (for numbers with factor 211) minus 210 (for numbers with factor 13). since the number of those elements of S that are prime to 211 is 2730. and the number of those elements of S that are prime to 13 is 2532. Spain. note that the prime factorization of 2743 is 13×211. If p divides n + 1 ([ ] ) [ ]2 ( ) [ ] n n n n and . So. Spain. . That is. BARCELONA TECH. Spain. contains a multiple of 211 and a multiple of 13. Thus the product of the elements in A must be divisible by 2743. The answer is 2731. 3. the wanted minimum n is greater than 2730. 2. any subset A of S with 2731 elements. Let S = {1. 2 Also solved by Jos´ e Gibergans-B´ aguena. So. then there are 210 numbers with a factor of 13 before 27437 and 12 numbers with a factor of 211 before 2743. So. This essentially means that you have all the 13’s in the list but no number of factor 211 to pair them up with 13 resulting in some multiple of 2743. Solution 2 by Jos´ e Luis D´ıaz-Barrero.185 Solutions 46. On the other hand. (Training Sessions of Spanish Team for IMO 2013) Solution 1 by Omran Kouba. Syria. BARCELONA TECH. there exists k ∈ N such that n = kp + p − 1 and = k. Romania.) (Diana Alexandrescu. the product of the 2730 elements that are not divisible by 211. IMAC 2013) Solution 1 by Diana Alexandrescu. then prove that p · . Since 2743 = 13 × 211. 47. Barcelona. (these are the numbers in S \ {211k : 1 ≤ k ≤ 13}. then [ ] n p | n + 1 − p. . Now the worst possible condition you can have is that you have chosen all the numbers that have a factor of 13 but no number with a factor of 211 which are 210 in total. Barcelona. Damascus. Let p. Since p | n+1. Bucharest. the number we wanted to find is 2730 + 1 = 2731. N = 2743 − 1 − 12 − 210 = 2520. Find the smallest positive integer n such that the product of any n distinct elements in S is divisible by 2743. So now you have N + 210 = 2520 + 210 = 2730. 2743}. . (p − 1)! = 1. (Here [x] divides − p p p p represents the integer part of the real number x. Bucharest. n be positive integers such that p is prime and p < n. Higher Institute for Applied Sciences and Technology. Now. Romania and Jos´ e Luis D´ıaz-Barrero. Now. p . Indeed. and we are done. .) is not divisible by 2743. Now you need any one of the remaining 13 numbers to get a product divisible by 2743. BARCELONA TECH. Assume you have chosen all the numbers in the set except those that have a factor of either 13 or 211 and call this number N . Barcelona. 2. Barcelona. Spain. . . IMAC 2013) Solution by Jos´ e Luis D´ıaz-Barrero. So. because m divides (p−1)! np an (m. 48. (p−1)!) = ( ) 1. Spain. Let m = ⌊n/p⌋. Damascus.186 we have ( ) [ ] ( ) n n kp + p − 1 − = −k p p p (kp + p − 1)(kp + p − 2) . Higher Institute for Applied Sciences and Technology. BARCELONA TECH. Barcelona. (p−1)!) = 1 and (p. (Iv´an Gueffner. let p be the largest power of p that divides m. (p−1)!) = 1. Now. or n = mp + p − 1. according to Gauss’ ( ) ( ) Lemma. we conclude that m | np . Now. . Let us denote by I the incentre of △ABC. p 2 Solution 2 by Omran Kouba. .m′ . BARCELONA TECH. hence n + 1 − mp = p. (α = 0 if p . . △IBC ≃ △IF E. . (kp + p − 1) − k(p − 1)! = (p − 1)! k(k · p · r + (p − 1)!) − k(p − 1)! = (p − 1)! k2 · p · r = ∈N (p − 1)! ([ ] ) ( ) [ ] n n n Since . p} and p divides n + 1 − mp. Spain. . . Note that q(p − 1)! = (m′ pα+1 + 1)(m′ pα+1 + 2) · · · (m′ pα+1 + p − 1) ≡ (p − 1)! (mod m′ pα+1 ) But since (m. (p − 1)! = 1. That is pm2 divides m(q−1) = np −⌊ np ⌋.m. we have n + 1 − mp ∈ {1. BARCELONA TECH. then (p − 1)! divides r and therefore − is p p p [ ]2 n divisible by p · as we wanted to prove. Barcelona.) We have m = pα m′ with p . which is the desired conclusion. then find the value of ∠BAC. (p−1)!) = 1 we conclude that (m′ pα+1 . Since ( ) n mp(mp + 1) · · · (mp + p − 1) = p p! (n) we see that (p − 1)! p = m(mp + 1) · · · (mp + p − 1). From the figure immediately follows ∠IBC = ∠IF E and ∠ICB = ∠IEF . Since both have the same height (the radii of the incircle) because BC and EF are both tangent . Also solved by Jos´ e Gibergans-B´ aguena. ( ) 1 n α So. Syria. let q be the integer defined by q = m p . Let Γ be the circumcircle of a triangle ABC and let E and F be the intersections of the bisectors of ∠ABC and ∠ACB with Γ. If EF is tangent to the incircle γ of △ABC. and consequently q ≡ 1 (mod m′ pα+1 ). (kp + 1)(kp) = −k p! k(kp + 1)(kp + 2) . IMAC 2013) Solution 1 by Nicolae Papapcu. Spain and Omran Kouba.187 A F E I C B to γ. n 49. if we denote ∠ABC = 2α and ∠ACB = 2β. Since 95 = 81 · 81 · 9 = (61+20) (61+20)·9 and 20·20·9 = 3600 = 61·59+1. we begin observing that 2013 = 3 · 11 · 61 = 33 · 61. Prove that an is divisible by 2013 for all n ≥ 1. Now. 2 Also solved by Daniel Vacaru. (Nicolae Papapcu. Therefore. Syria. then we have 95 ≡ 1 (mod 61). Spain. Barcelona. BARCELONA TECH. Romania. Jos´ e Gibergans-B´ aguena. Romania. then for all n ≥ 1. Pitesti. Romania and Jos´ e Luis D´ıazBarrero BARCELONA TECH. Now. Slobozia. an = 46 + 1943 ≡ 1952 (mod 61) ≡ 0 (mod 61) and 61|an . Slobozia. m ∈ N. For all positive integer n we consider the numbers an = 46 + 1943. from isosceles △IF B follows ∠IBF = ∠IF B = ∠CF B = ∠BAC = 180◦ − 2(α + β) Adding up the angles of △IF B yields (α + β) + 180◦ − 2(α + β) + 180◦ − 2(α + β) = 180◦ from which follows α + β = 60◦ and ∠BAC = 60◦ . Since 46 = 4096 = 61·67+9 ≡ 9 (mod 61) and 6n−1 = (5+1)n−1 = 5m+1. and find all values of n for which an − 207 is the cube of a positive integer. then △IBC = △IF E. holds ( )6n−1 ( 6 )5m+1 n ≡ 95m · 9 (mod 61) ≡ 9 (mod 61) = 4 46 = 46 n So. Higher Institute for Applied Sciences and Technology. Damascus. To prove the first part. IB = IF. Barcelona. . for all n ≥ 1. then ∠BIF = α + β. Barcelona. we observe that an −207 = 46 +1736 is an n n even integer. Romania. −9) and the solutions of the second n−1 one (6. BARCELONA TECH. For n > 1. −6). + 1) = 1023 · q = 33 · 31 · q and 33|an the jointly with the preceding yields 2013 = 33 · 61|an for all n ≥ 1. (−8. or x2 + xy + y 2 = 217. From 46 +1736 = (2x)3 follows 22·6 −3 +217 = x3 n−1 n−1 or 23(4·6 −1) + 217 = x3 . The solutions of the first system are (9. = (1024 − 1)(1024p−1 + . Spain and Omran Kouba. x2 + xy + y 2 = 31. Syria. x − y = 1. it holds sin2n x + cos2n x ≥ (1 − 2x2 )n + (2x2 )n (Marius Dragan. and we are done. Higher Institute for Applied Sciences and Technology. we have ( )p n 46 −1 − 1 = 45p − 1 = 45 − 1 = 1024p − 1. Spain. we obtain )n ( 2 )n ( sin2 2x sin 2x 2n 2n sin x + cos x ≥ 1 − + 2 2 . Prove that for all x with 0 ≤ x ≤ 1/2. Finally. putting u = cos2 x and v = sin2 x we get 2 sin x cos 2n 2 2n x + cos x sin x≥ ( sin2 2x 2 )n Adding up the above expressions. Putting 24·6 −1 = y in the last equation yields x3 − y 3 = 217 ⇔ (x − y)(x2 + xy + y 2 ) = 217 = 7 · 31 Since x − y < x2 + xy + y 2 . then we have two possibilities { { x − y = 7. p ∈ N. we have ( )n sin2 x un + cos2 x v n ≥ sin2 x u + cos2 x v Putting u = sin2 x and v = cos2 x in the preceding. then n−1 y = 24·6 −1 = 8 = 23 from which follows n = 1. For n = 0 and n = 1 the statement trivially holds. . say 2x with x ∈ N. −1) and (1. 50. . n To solve the second part of the statement. yields ( )n sin2 2x sin2 x sin2n x + cos2 x cos2n x ≥ 1 − 2 Likewise.188 ( n ) n On the other hand. Barcelona. since y = 24·6 −1 is a positive integer. Damascus. Since 6n − 1 ≡ 0 (mod 5). Then. we will apply Jensen’s inequality to the convex function f : R → R defined by f (t) = tn . Bucharest. Romania and Jos´ e GibergansB´ aguena. Bucharest. Then. BARCELONA TECH. 8). an = 46 − 4 + 1947 = 4 46 −1 − 1 + 33 · 59. 2 Also solved by Jos´ e Gibergans-B´ aguena. Let n be a positive integer. IMAC Shortlist 2013) Solution 1 by Marius Dragan. then 6n − 1 = 5p. 2 2 2 ( ) sin 2x 4x 1 sin 2x From 0 ≤ ≤ ≤ follows g ≥ g 2x2 . Syria. Combining the 2 2 2 2 preceding result. Solution 2 by Omran Kouba. Roma. since 2 ≤ 4 we conclude that cos x ≥ sin x and consequently (cos x)2j ≥ (sin x)2j . for 0 ≤ j ≤ n − 1 On the other hand. Italy and Jos´ e Luis D´ıazBarrero. j=0 Now. Universit` a degli studi di Tor Vergata Roma. g is decreasing. Note that (2x2 )n − sin2n x = (2x2 − sin2 x) n−1 ∑ sin2j x(2x2 )n−1−j . we obtain sin2n x + cos2n x ≥ (1 − 2x2 )n + (2x2 )n 2 and we are done. 2 Also solved by Perfetti Paolo. 1/2] → R defined by g(t) = (1 − t)n + tn . Consider x ∈ [0. First we have the well-known inequality: 0 ≤ sin x ≤ x so that 2x2 − sin2 x ≥ 0 1 π Also. BARCELONA TECH. 1/2]. . Higher Institute for Applied Sciences and Technology. from x ∈ [0. j=0 cos2n x − (1 − 2x2 )n = (2x2 − sin2 x) n−1 ∑ cos2j x(1 − 2x2 )n−1−j . ( then )g (t) ≤ 0. Dipartimento di Matematica. Damascus. Spain. Since ′ g ′ (t) = n(tn−1 − (1 − t)n−1 ) and 0 ≤ t ≤ 1/2.189 Now we consider the function g : [0. So. for 0 ≤ j ≤ n − 1 Using the preceding we conclude that cos2n x − (1 − 2x2 )n ≥ (2x2 )n − sin2n x which is equivalent to the desired inequality. Barcelona. 1/2] we conclude that 1 − 2x2 ≥ 2x2 and consequently (1 − 2x2 )n−1−j ≥ (2x2 )n−1−j . 2]. we need the following lemmas. [2] Let A ∈ Mn be nonsingular. Then σ1 (A)σ2 (A) · · · σn (A) = | det A| In what follows . Introduction Let Mn be the space of n × n complex matrices.1 in [3]. In this paper. we give the simple proof of Theorem 2. 2. Results and Proofs To proof of Theorem 2.1 in [3]. [1] Let A ∈ Mn be nonsingular. The purpose of this paper is to give a simple proof of Theorem 2. Theorem 1.j=1 It is well known that lower bounds for the smallest singular value σn of a nonsingular matrix A ∈ Mn have many potential theoretical and practical applications [1. the Frobenius norm of A is defined by n ∑ ∥A∥F = 21 |aij | i. Then ( σn ≥ | det A| · n−1 ∥A∥2F − ℓ2 ) (n−1) 2 (1) . 1.1. we give a simple proof of a lower bound for the smallest singular value of nonsingular matrices which is much simpler than the proof of result of LIMIN ZOU [3]. Lemma 1. For A = [aij ] ∈ Mn . [3] Let A ∈ Mn be nonsingular. Then ∥A∥2F = σ12 + σ22 + · · · + σn2 Lemma 2.190 MATHNOTES SECTION A Simple proof of lower bound for the smallest singular value Adiyasuren Vandanjav and Batzorig Undrakh Abstract. Let σi 1 ≤ i ≤ n be the singular values of A ∈ Mn and suppose that σ1 ≥ σ2 ≥ · · · ≥ σn ≥ 0. . · σn )2 · by the AM-GM inequality 1 ≥ | det A|2 · ( 2 2 σ +σ +···+σ 2 1 ( = | det A| 2 2 )n−1 n−1 n−1 n−1 ∥A∥2F − σn2 )n−1 ( ≥ | det A| · 2 n−1 ∥A∥2F Hence we have the following two inequalities. From Theorem 2. The sequence {ℓm : m ≥ 1} converges and lim ℓm ≤ σn m→∞ Proof.. ( )n−1 n−1 2 2 σn ≥ | det A| ∥A∥2F − σn2 and ( σn ≥ | det A| · n−1 ∥A∥2F )n−1 (2) ) (n−1) 2 =ℓ (3) From (3) we have σn ≥ ℓ.. · σn−1 = (σ1 σ2 · . · σn−1 )2 1 = | det A|2 · 2 2 2 σ1 · σ2 · . σn2 1 (σ1 σ2 · . So this limit point is less than or equal to σn . Let us define a sequence {ℓm : m ≥ 1} as follows ℓ1 = | det A| · ( ℓm = | det A| · n−1 ∥A∥2F − ℓ2m−1 ( n−1 ∥A∥2F ) n−1 2 and ) (n−1) 2 Corollary 1. By the Weierstrass theorem the sequence {ℓm } has a limit point.. . So ∥A∥2F − σn2 ≤ ∥A∥2F − ℓ2 (4) Thus by (2) and (4) one can have ( σn2 ≥ | det A|2 · Hence ( σn ≥ | det A| · n−1 ∥A∥2F − ℓ2 n−1 ∥A∥2F − ℓ2 )n−1 ) (n−1) 2 This completes the proof..191 Proof.1 we can see that the sequence {ℓm : m ≥ 1} is increasing for all m ≥ 1 and bounded above by σn .. A. Matrix analysis. Horn.com. [3] Limin Zou. Vol. Department of Mathematical Analysis National University of Mongolia Ulaanbaatar.A.R. Cambridge University Press. 4 (2012). Cambridge University Press. batzorig u@yahoo. Horn.com . 1991. No. 625-629. Cambridge. C. A lower bound for the smallest singular value.R. 6. 1985. C. Journal of Mathematical Inequalities. Topics in Matrix Analysis. [2] R. Mongolia E-mail address: v adiyasuren@yahoo. Jonhson. Jonhson.192 References [1] R. Let (ma . student. x. Find at least one positive integer n such that n is perfect square and n20 + n13 has 2013 digits . y. Republic of Kosova. Romania. m (z + t)m (zwa2 + twb2 ) cyclic 15. z. 13). B˘ atinet¸u–Giurgiu. Romania. 4. then the following inequality holds ( ) ∑ xm2a + ym2b m+1 √ (x + y)m+1 ≥ 3 3 S. 84. MA 02325. Bridgewater. z are the integer solutions of equation x2 + y 2 = z 2 then x3 + y 3 + z 3 is a composite number. 14. 12. University of Prishtina. ∞). University of Prishtina. 2013 Proposals 11. Provide explicit constructions of such triples to show that there are infinitely many such odd numbers N. t ∈ (0. mb . “Matei Basarab” National College. 13. . Proposed by Tom Moore.193 JUNIOR PROBLEMS Solutions to the problems stated in this issue should arrive before December 15. Buz˘ au. Proposed by Dorlir Ahmeti. 5). Neculai Stanciu. Proposed by Valmir Krasniqi. ∞). Proposed by Mihaly Bencze. Find all functions f : R → R such that f (x + 2y − f (y)) = f (x + y) and f (x3 ) = (f (x))3 . Brasov. (wa . 12.M. The examples (3. Mathematics Department. “George Emil Palade” School. Republic of Kosova. (5. Bucharest. wc ) and S be respectively the medians. Prove that if x. wb . the bisectors and the area of a triangle ABC. (13. Conant Science and Mathematics Center Bridgewater State University. USA. mc ). Show that if m ∈ [0. y. Proposed by D. Romania. 85) show that the same odd number may occur as the ”hypothenuse” and the ”odd leg” of a primitive Pythagorean triple (PPT). Our inequality becomes A A A + ≥ s−a s−b s−c that is 1 1 1 + ≥ s−a s−b s−c or 2 2 2 + ≥ b+c−a a+c−b a+b−c From the given statement a + b = 3c and from simple algebra we get 1 1 4 + ≥ 2b − a 2a − b a+b From the AM-HM inequality we get ( ) 1 1 1 2 + ≥ 2 2b − a 2a − b 2b − a + 2a − b We know that ra = and then the inequality Also solved by Alban Ismaili. b = CA and let ra . rc = where A is the area and s the s−a s−b s−c semiperimeter. y > 0. Let (x. prove or disprove: For every positive integer k holds (kx − y. rc be the ex–radii respectively. k 3 x + k 2 x − k 2 y + kx − ky + x − y). If a + b = 3c show that ra + rc ≥ 4rc Solution by Arb¨ er Avdullahu. Macedonia. Proposed by Ercole Suppa. Maramure¸s. Kosovo. Satulung.194 Solutions 6. namely d = 1. Romania and the proposer . From d|x and d|y we get d|(x. Let d = (kx − y. A A A . Daniel V˘ acaru. Kumanovo. Proposed by Armend Shabani. From d|kx − y and d|x we get d|(kx − (kx − y)) namely d|y. In ∆ABC let a = BC. Ferizaj. Pi¸testi. k 3 x + k 2 x − k 2 y + kx − ky + x − y) = 1 Solution by Ioan Viorel Codreanu. Ferizaj. Kosovo. Department of Mathematics and Computer Sciences Faculty of Nature Sciences University of Prishtina. y) = 1 and x. y). ra = . Mathematical Group Galaktika Shqiptare. namely d|x. Albania. From d|(kx − y) and d|(k 3 x + k 2 x − k 2 y + kx − ky + x − y) we deduce that d|(k 3 x − k 2 y − k 2 (kx − y) + k 2 x − ky − k(kx − y) + kx − y − (kx − y) + x). Ioan Viorel Codreanu. Satulung. Also solved by Arb¨ er Avdullahu. Maramure¸s. Romania. Teramo Italy. rb . Romania and the proposer 7. We know that ∀a. student University of Prishtina.672] = 3.3. where b = [log 999999 ].195 The editors in chief apologize for the typo ra + rc ≥ 4rc instead of ra + rb ≥ 4rc in the issue 2 vol. for for every positive integer n. Let a be the first digit of 999999 . Let log 999999 − b = B. Since log 999999 − [log 999999 ] = 2996. b if a > b ⇒ f (a) > f (b). 8. We conclude that the last digit of 999999 is 3.565 = B. Republic of Kosova. we have ( )n ( )999 1 1 1+ <e<3⇒ 1+ <3 n 999 (5) (6) From (1) and (2) we have ( )999 5 1 5 1000999 5 102997 102997 2 · 102997 < 1+ <3⇒ < <3⇒ < <3⇒ < 999999 < 999 999 2 999 2 999 2 999 3 5 From and 10 102997 = · 102996 > 3 · 102996 3 3 2 · 102997 = 4 · 102996 5 we have 3 · 102996 < 999999 < 4 · 102996 So the first digit of number 999999 is 3 . We log both sides and we get b + log a ≤ log 999999 < b + log(a + 1). Now ( )6 76 5 1 5 6 6 2 · 7 − 5 · 6 = 2018 > 0 ⇒ 6 > ⇒ 1 + > 6 2 6 2 Since ( )999 ( )6 1 1 5 f (999) > f (6) ⇒ 1 + > 1+ > 999 6 2 Next. Second solution by proposer ( )n Let f be a function given by f (n) = 1 + n1 .565 − 2996 = 0. Then a = [10B ]. it follows that a = [100. So 10b a ≤ 999999 < 10b (a + 1). . Ferizaj. Solution by Arb¨ er Avdullahu. Proposed by Dorlir Ahmeti. Kosovo. Find the first digit of the number 999999 .565 ] = [3. Neculai Stanciu. Bucharest. y) then ( −x − h 2 )2 + y2 < x2 + y 2 ⇔x2 + h2 + 2hx + y 2 < 4x2 + 4y 2 4 ⇔3x2 + 3y 2 − 2hx − h2 > 0. Now. Pite¸sti. the point A must be exterior to a circle centered at L with radius 2 · OH/3. if H = (−h. Solution by the proposer A O L H B M C Without loss of generality.E. C such that ABC has circumcenter O and orthocenter H is equivalent to that the midpoint of HA′ is interior to the circumcircle. ( )2 ( )2 h 2h ⇔ x− + y2 > .S. Romania. Prove that in any triangle ABC with the usual notations. 10. If G is the centroid. Romania. I. ”Matei Basarab” National College.196 ´ 9. Bˇ atinetu Giurgiu. M. if A′ is the reflection of A on O. 2 ra2 + 2rb rc rb + 2rc ra rc2 + 2ra rb 4R + r Solution by Daniel Vˇ acaru. BC and HA′ share the same midpoint M . Hence. Consequently. holds ( 3 )2 1 1 1 + + ≥ . one has x y z x+y+z . Buzˇ au. ”George Emil Palade” School. Romania One knows that b2 c2 (a + b + c)2 a2 + + ≥ . Two points O and H are given in position on the plane. Describe the region of the plane occupied by the vertices of triangles ABC having circumcenter O and orthocenter H. Proposed by D. let O be the origin. Alvarez Cubero de Priego de C´ ordoba. from HG : GO = 2 : 1 and 3G = A + B + C we have 3G = H + 2O = H ⇒ A + B + C = H ⇒ B + C = H − A. 3 3 If L divides the segment HO in the ratio HL : LO = 4 : −1. the existence of B. For any point A. Spain. 0) and A = (x. Proposed by Francisco Javier Garc´ıa Capit´ an. (s − a)(s − b)(s − c) S2 Finally ( )2 ( 3 )2 1 1 1 3 + + ≥ = . ra2 + 2rb rc rb2 + 2rc ra rc2 + 2ra rb ra + rb + rc 4R + r as desired. Also solved by Ioan Viorel Codreanu. One has ra +rb +rc = r2 +s2 +4Rr = ( a ) ( bc sin A ) S2 s(s − a)(s − b)(s − c) +s2 +2R·2r = +s2 + · . = 2 = ra + rb2 + rc2 + 2ra rb + 2rb rc + 2rc ra ra + rb + rc Also ( 1 ( ab + bc + ca − s2 ) S S 1 1 ) S + + =S + + =S . s s Therefore one obtains = ( ra + rb + rc = S ) ( r2 + 4Rr ) r2 + 4Rr = Ss = 4R + r.197 1 1 1 (1 + 1 + 1)2 + + ≥ ra2 + 2rb rc rb2 + 2rc ra rc2 + 2ra rb (ra2 + 2rb rc ) + (rb2 + 2rc ra ) + (rc2 + 2ra rb ) )2 ( 32 3 . s−a s−b s−c s−a s−b s−c (s − a)(s − b)(s − c) where S is the triangle’s area. Maramure¸s. 2 2 s s sin A s (s − a)(s − b)(s − c) abc + s2 + = ab + bc + ca. Satulung. Romania and the proposer. . Pages 197–230 Editors: Valmir Krasniqi. Higher Institute for Applied sciences and Technology. PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets. Greece. Ercole Suppa. Determine the sequence {cm }m≥0 such that m−1 X ck lim nm F (n) − = cm n→+∞ nk k=0 where. Find. Universiteti i Prishtin¨ es. Solutions will be evaluated for publication by a committee of professors according to a combination of criteria. y and z are nonnegative real numbers such that x + y + z = 1. Francisco Javier Garc´ıa Capit´an. in terms of a > 0. Student solutions should include the class and school name. Roberto Tauraso.com Volume 3. Cristinel Mortici. Issue 4 (2013). 82. An asterisk (*) indicates that neither the proposer nor the editors have supplied a solution. Stone. c 2010 Mathproblems. Proposed by Omran Kouba. Syria. for m = 0. The editors encourage undergraduate and pre-college students to submit solutions. Kosov¨ e.Mathproblems ISSN: 2217-446X. ProposedPby Anastasios Kotronis. Prishtin¨ e. Damascus. Teachers can help by assisting their students in submitting solutions. Anastasios Kotronis. 2014 Problems 81. Enkel Hysnelaj. Shabani. Ovidiu Furdui. Athens. Jos´ e Luis D´ıaz-Barrero. 197 .mathproblems-ks. Drawings must be suitable for reproduction. Jozsef S´ andor. Mih´aly Bencze. the minimum of a(x2 + y 2 + z 2 ) + 9xyz xy + yz + zx when x. Mohammed Aassila. David R. each indicating the name of the sender. Armend Sh. Proposals should be accompanied by solutions. url: http://www. Omran Kouba. Paolo Perfetti. the sum is considered to be 0. 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 March 15. 1 Let F (n) := k≥1 (kn+1)k! . Emanuele Callegari. Valmir Bucaj. x 0 . Prove that the sequence {bn } is unbounded(has no upper bound).Daejeon. j(i < j). Calculate √ √ Z 1 ln( x + 1 − x) √ dx. The inverse of arcsin hp is called the generalized hyperbolic sine function and denoted by sinhp . Department of Mathematics and Information Science. 2 is called the generalized sine function and d denoted by sinp . and Neculai Stanciu. Binzhou City. Proposed by Ovidiu Furdui. Shandong Province. Proposed by Li Yin. Proposed by D. Romania. Bucharest. (1 + tp )1/p arcsin hp (x) = 0 − arcsin h (−x) . “Matei Basarab” National College. Binzhou University. Korea Advanced Institute of Science and Technology. 2 (1 − tp )1/p 0 πp The inverse of arcsinp (x) on 0. Let n be a positive integer. B˘ atinet¸u-Giurgiu. 0) p generalizes the classical inverse hyperbolic sine function. 86. South Korea . Cluj. πp 2 ). aj ) = 1 for any i. Romania. ∞). The generalized hyperbolic cosine function is defined as coshp (x) = For p > 2 and x ∈ (0. Let bn = an+1 − an .198 83. Prove that n n X Fk2 X Fk3 (Fn+2 − 1)5 · ≥ Lk L2k (Ln+2 − 3)3 k=1 k=1 where Fn . Romania. respectively Ln represents the nth Fibonacci number respectively the nth Lucas number. dx prove that ln and ln x sinp x − x cosp x < sinp x p sinp x x coshp x − sinhp x sinhp x > . x ∈ [0. x p sinhp x 85. x ∈ (−∞. Technical University of Cluj-Napoca. 84. “George Emil Palade” School. Let 1 < p < ∞. Buzˇ au. the generalized inverse hyperbolic sine function Z x 1 dt . Proposed by Hun Min Park.M. China. Let {an } strictly increasing sequence of positive integers such that gcd(ai . we can generalize the inverse of arcsin as follows: Z x 1 06x61 dt arcsinp (x) = (1 − tp )1/p 0 and Z 1 πp 1 = arcsinp (1) = dt. The generalized cosine function is defined as cosp (x) = dx sinp (x). Similarly. d sinhp (x). University of Prishtina. If 1 . 73. we obtain y 1 f + 3f = y b y 1 = f + 3f b y of Antioquia. We consider the function n g(x) = ax+1 g ◦ g ◦ . y b y Therefore 1 f (x) = 8 3 −x . Sydney. Proposed by Dorlir Ahmeti. b. Prove that √ √ √ √ √ a+ b b+ c c+ a √ + √ + √ ≥ 3. Let a. Determine all the real valued functions f such that ax + 1 f (x) + 3f =x bx where a and b are real numbers that satisfy b 6= 0 and a2 + 2b = 0 or a = 0. Note that bx and denote by g (x) = | {z } n−times ax + 1 a + x(a2 + b) a − xb = = .199 87. y y . b (1) (2) If we subtract (1) from three times (2). bx b(1 + ax) b(1 + ax) a − xb a2 + b −b −1 g = = = . . Australia. Republic of Kosova. 1 + ca 1 + ab 1 + bc √ Solutions No problem is ever permanently closed. yields 8f 1 3y 1 = − . We will be very pleased considering for publication new solutions or comments on the past problems. ◦ g (x). Colombia. a 6= 0 and b 6= 0. . c be positive real numbers such that a + b + c = 3. b(1 + ax) b(a − bx) b(a − bx) a − bx −1 = x.Proposed by Enkel Hysnelaj. then g . g a − bx g 2 (x) = g 3 (x) = g 4 (x) = Since g 4 (x) = x. University we take a = 0. bx We now suppose that a2 + 2b = 0. Solution 1 by Robinson Higuita. x = y1 and x = yb . University of Technology. and f (ϕ(x)) + 3f (x) = ϕ(x) eliminating f (ϕ(x)) we find that 8f (x) = 3ϕ(x) − x. Damascus. according to the functional equation: f (x) + 3f (ϕ(x)) = x. Adding and simplifying we obtain −80f (x) = −27g 3 (x) + 9g 2 (x) − 3g 1 (x) + x. USA. France. if a = 0 then ϕ (x) = ϕ ◦ ϕ(x) = x. after simplification. Let ϕ(x) = ax+1 bx . Roberto de la Cruz Moreno. Mathematical Research Center. Barcelona. Ulaanbaatar. Spain. On the other hand. Moubinool Omarjee. . 1. and the proposer. in the case a 6= 0 we have (−3)k f (ϕk (x)) − (−3)k+1 f (ϕk+1 (x)) = (−3)k ϕk (x). clearly. 40bx(1 + ax)(2 + ax) Also solved by Arkady Alt. Syria. San Jose. 8bx And. Paris. −3f (g(x)) − 9f (g (x)) = −3g(x). f (x) = 3 + 16ax − 22bx2 + b2 x4 . Adding these equalities we obtain f (x) − (−3)4 f (x) = 3 X (−3)k ϕk (x) k=0 and. −27f (g 3 (x)) − 81f (x) = −27g 3 (x). if a 6= 0 then a − bx b(ax + 1) 1 ϕ3 (x) = ϕ2 ◦ ϕ(x) = bx − a ϕ4 (x) = ϕ3 ◦ ϕ(x) = x ϕ2 (x) = ϕ ◦ ϕ(x) = So. AN-anduud Problem Solving Group. for k = 0. Higher Institute for Applied Sciences and 2 Technology. or equivalently f (x) = 3 − bx2 .200 f (x) + 3f (g(x)) 2 = x. 80 Solution 2 by Omran Kouba. Campus de Bellaterra. Lyc´ ee Henri IV. Mongolia. in the case a = 0 we have. California. 9f (g 2 (x)) + 27f (g 3 (x)) = 9g 2 (x). from which follows f (x) = 27g 3 (x) − 9g 2 (x) + 3g 1 (x) − x . 2. 3. Rehovot. University of Las Palmas.Proposed by D. Buzˇ au. Romania and Neculai Stanciu. B˘ atinet¸u-Giurgiu. Daniel Vˇ acaru. . Gran Canaria. Lyc´ ee Henri IV. University of Antioquia. Angel Plaza.Li Yin. Colombia. China. the inequality claimed is equivalent to f (x) = (n + 2) (n + 1) x2 − 2n (n + 2) x + n (n + 1) > 0. Damascus. Binzhou City. n ∈ N show that n+1 2x + > n + 3. Italy. “Matei Basarab” National Colege. California. Also solved by Robinson Higuita. Shandong Province. Mongolia. we get the desired inequality. 75. Paolo Perfetti. Korea Advanced Institute of Science and Technology ´ (KAIST). Paris. Dividing by nx(1 − x). Tor Vergata University. Department of Mathematics and Information Science. Ulaanbaatar. Minh Can. Higher Institute for Applied Sciences and Technology. Solution 2 by Omran Kouba. We check easily that n n+2 ((n + 1)x − n)2 + . x n(1 − x) Solution 1 by Moti Levy. f (x) = Solution 3 by AN-anduud Problem Solving Group. Irvine Valley College. Spain. First we carry out the following computation n+1 2x + >n+3 x n(1 − x) ⇔ n(n + 1)(1 − x) + 2x2 > n(n + 3)x(1 − x) ⇔ (n2 + 3n + 2)x2 + (n2 + n) > (2n2 + 4n)x Hence we need to prove the last inequality. Bucharest. By AM-GM inequality we have. Irvine. Department of Mathematics. Romania. Syria. p (n2 + 3n + 2)x2 + (n2 + n) ≥ 2 (n2 + 3n + 2)(n2 + n)x2 > (2n2 + 4n)x So the desired inequality is proved. Pite¸sti. Romania (Jointly). 1). France and the proposer. Hence f (x) has no real roots. USA. Moubinool Omarjee. which implies that f (x) > 0 for all real x. South Korea. Germany. n+1 1+n This proves that f (x) > 0 for every x ∈ R. Rome. Reiner Martin. which is positive positive for x ∈ (0.201 74. The parabola f (x) is straight with negative discriminant 2 2 ∆ = n2 (n + 2) − n (n + 1) (n + 2) = −n (n + 2) < 0. Israel. For 0 < x < 1. Hun Min Park. Re-arranging terms.M. consider the real function f : R → R defined by f (x) = n(n + 1)(1 − x) + 2x2 − n(n + 3)x(1 − x). Daejeon. “George Emil Palade” Secondary School. Soden-Neuenhain. 256603. For a fixed positive integer n. Binzhou University. We begin by showing that. x→∞ and lim x→∞ g(x + 1) = b ∈ R∗+ . if limx→∞ (g(x)) x x does exist and limx→∞ g(x+1) xg(x) = b ∈ R∗+ . x→∞ xg (x) x→∞ x g (x + 1) (p (x + 1)) ≈ x xg (x) x (px) x+1 1Editorial Comment: This argument is not complete. x→∞ x lim (g(x)) x→∞ x 1/x lim . x→∞ cos2 (t) cos2 (t) sin2 (t) . . calculate the following limits: sin2 (t) sin2 (t) cos2 (t) x+1 x (1) lim (f (x)) (g(x + 1)) − (g(x)) . then 1 (g (x)) x b lim = x→∞ x e 1 by the following argument 2. Israel. Rehovot. a counter example is provided by f (x) = x + sin(πx). xg(x) Suppose also the existence of the limits f (x) . x via the following argument1: Suppose limx→∞ f (x) x = c ∈ R∗+ .202 Let R+ = (0. (2) lim (f (x)) (g(x + 1)) x+1 − (g(x)) x x→∞ Solution 1 by Moti Levy. it would prove that the existence of limx→∞ f (x)/x imply the existence of limx→∞ (f (x + 1) − f (x)). Suppose limx→∞ g (x) ≈ (px) (g(x)) x x = p ∈ R∗+ . then a = lim (f (x + 1) − f (x)) x→∞ f (x + 1) f (x) = lim (x + 1) − x x→∞ x+1 x f (x + 1) f (x) = lim (x + 1) − x x→∞ x+1 x = lim (c (x + 1) − cx) = c. For t ∈ R. then x x+1 x+1 (x + 1) 1 =p =p 1+ xx+1 x x+1 g (x + 1) 1 b = lim = lim p 1 + = pe. if ∗ limx→∞ f (x) x = c does exist and limx→∞ (f (x + 1) − f (x)) = a ∈ R+ then lim x→∞ f (x) = a. g : R∗+ → R∗+ be functions such that lim (f (x + 1) − f (x)) = a ∈ R∗+ . x→∞ 1 Similarly. 2Editorial Comment: The previous remark applies here also. if it were correct. ∞) and f. g(x)g(x)1/x ! α/(x+1) α b(x) 1−α = (h(x)) k(x) · x −1 . xg(x) h(x) = f (x) . x→∞ e x e lim (f (x)) cos2 t Similarly. it can be shown that cos2 t cos2 t cos2 t b sin2 t sin2 t x+1 x − (g (x)) =a cos2 t. or n n→∞ n→∞ . lim (f (x)) (g (x + 1)) x→∞ e Solution 2 by Omran Kouba. Higher Institute for Applied Sciences and Technology. x and Fα (x) = (f (x))1−α k(x) = (g(x))1/x . and by Cen→∞ Pn saro’s lemma we conclude that lim n1 ( k=1 ak ) = a. if an = f (n + 1) − f (n) we have by assumption lim an = a. we will use the following notation: b(x) = g(x + 1) . α g(x) x ! α/(x+1) α g(x + 1) 1−α = (h(x)) k(x) · x −1 .203 sin2 t sin2 t 1 1 (g (x + 1)) x+1 − (g (x)) x x→∞ !sin2 t !sin2 t 1 1 x+1 x 2 2 2 (g (x + 1)) (g (x)) cos t sin t = lim (f (x)) (x + 1) − xsin t x→∞ x+1 x ! sin2 t sin2 t b b sin2 t cos2 t sin2 t (x + 1) − x = lim (ax) x→∞ e e ! sin2 t sin2 t b b 1−sin2 t sin2 t cos2 t sin2 t = lim a (x) (x + 1) − x x→∞ e e ! sin2 t sin2 t 2 b 1 = acos t lim x 1+ −1 x→∞ e x sin2 t 2 sin2 t b lim x 1+ −1 = acos t x→∞ e x sin2 t 2 sin2 t 2 2 b sin t b = acos t lim x = acos t sin2 t. hence lim f (n) = a. Syria. k(x) αl(x) α x b(x) e −1 1−α = (h(x)) k(x) ln . For a real α. Damascus. x+1 k(x) l(x) α k(x) · x Now. and positive real x. x g(x + 1) α x+1 l(x) = 1 b(x) ln x + 1 k(x) α − g(x) x Let us rewrite the Fα (x) as follows: ! α x+1 g(x + 1) Fα (x) = (h(x))1−α −1 . n = 0 (mod 20) Solution 1 by Omran Kouba. so (α. β) = (0. Tor Vergata University. but they differ by the initial conditions. This proves that Xn = (αn + β)3n for some constants α and β. this proves that lim (b(x)/k(x)) = e. The characteristic equation of this linear recurrence sequence is λ2 − λ − 1 = (λ − 3)2 = 0. and the proposer. Athens. n 6= 0 (mod 5) 1 (mod 5). n = 15 (mod 20) 2 (mod 5). x→∞ Now. 2) and Ln = 2 · 3n (mod 5) for every n. − g(x) x→∞ Also solved by Paolo Perfetti. Lucas and Triangular number respectively. If Fn . We will work in the finite field F5 = Z/5Z.Proposed by Anastasios Kotronis. Greece. Similarly. Now. Therefore. Damascus. x→∞ x→∞ In particular. for (Fn )n≥0 . we have F0 = 0 and F1 = 1. we have L0 = 2 and L1 = 1. n→∞ x→∞ lim h(x) = a. Syria. Rome.204 lim h(n) = a. . and by Ces´aro’s lemma we conclude that n→∞ n 1X bk = ln b − 1. n→∞ n lim k=1 hence lim n→∞ g(n)1/n n = b/e. and lim l(x) = 0 So. But. Both sequences (Fn )n≥0 and (Ln )n≥0 satisfy the the same recurrence linear relation Xn+2 = Xn+1 + Xn . so (α. n = 10 (mod 20) . if x→∞ bn = ln g(n + 1) (n + 1)n+1 − ln g(n) nn = ln g(n + 1) ng(n) 1 − (n + 1) ln 1 + n then lim bn = ln b − 1. Department of Mathematics. Therefore. Higher Institute for Applied Sciences and Technology. 0) and Fn = 2n · 3n (mod 5) for every n. (1) lim (f (x)) g(x + 1) x→∞ 2t cos cosx2 t 2 2 2 sin2 t x+1 (2) lim (f (x)) g(x + 1) = cos2 te− cos t asin t bcos t . 76. by assumption the limit n→∞ lim k(x) does exist. taking x→∞ x→∞ the limits in the final expression of Fα we see that α lim Fα (x) = αa1−α be−1 = αe−α a1−α bα . n = 5 (mod 20) 5 2 n ≥ 1. But. For (Ln )n≥0 . by assumption the limit lim h(x) does exist. or lim k(n) = b/e. Ln and Tn are the n−th Fibonacci. show that 0 (mod 5). Italy. 2Fn + n Fn (Tn + n) − Ln − Ln+1 = 3 (mod 5). lim k(x) = b/e. β) = (2. (1) corresponds to α = sin2 t and (2) corresponds to α = cos2 t : 2t sinx2 t sin 2 2 2 cos2 t x+1 − g(x) = sin2 te− sin t acos t bsin t . 4 (mod 5). 205 Now. This proves the Lemma and shows that 20 is a period of the sequence Gn modulo 5. Tn+20 = Fn+5 ≡ 3Fn (1) mod 5. Fibonacci and Lucas Numbers with Applications. If Gn ≡ i mod 5. Solution 3 by AN-anduud Problem Solving Group. we have proved that the statement holds. And the desired conclusion follows. Therefore Fn+20 ≡ 3Fn+15 ≡ 81Fn ≡ Fn 5 mod 5. Let Gn = 2Fn5 + n2 Fn (Tn + n) − Ln − Ln+1 . and the induction follows from the identity : Fn − 2n · 3n = (Fn−1 − 2(n − 1) · 3n−1 ) + (Fn−2 − 2(n − 2) · 3n−2 ) − 10(n + 1) · 3n−2 .page 414]). 2001. we have 5 Gn+20 ≡ 2Fn+20 + (n + 20)2 Fn+20 (Tn+20 + (n + 20)) − Ln+20 − Ln+21 ≡ 2Fn5 2 + n Fn (Tn + n) − Ln − Ln+1 mod 5 ≡ Gn mod 5 mod 5. . Ln+20 ≡ 3Ln+19 ≡ (81) Ln ≡ Ln (2) mod 5. Therefore. Mongolia. if n 6= 0 (mod 5) then Un = 0 (mod 5). the reader can find the first twenty values of Gn mod (5) and note that the problem is true for these values. and if n = 5m then Un = 2 · 35m = 2 · 3m (mod 5). Solution 2 by Robinson Higuita. So. n(n+1) . we use mathematical induction to prove Fn ≡ 2n · 3n (mod 5). n ≥ 1 (1) We can easily see that (1) holds for n = 1. then Gn ≡ Gr mod (5).com. Koshy. First. John Wiley. New York. Thus. because 34 = 1 (mod 5). and consequently n3 (n + 3) Fn − Ln+2 2 = 4n3n + n4 (n + 3)3n − (18)3n = (4n + n5 + 3n4 − 3)3n def Un ≡ 2Fn5 + n2 (Tn + n)Fn − Ln − Ln+1 = 2Fn + = (4n + n + 3n4 − 3)3n = (n4 − 1)3n+1 But n4 − 1 = 0 if 5 . 2. since 5 is prime we know that Fn5 = Fn (mod 5). Ulaanbaatar.n and n4 − 1 = −1 if 5 | n. First of all. University of Antioquia. Ln+1 ≡ 3Ln mod 5. if n = 20k + r with k nonnegative integer and 0 ≤ r < 20. we suppose Gn ≡ i mod 5. (3) Using (1). In wolframalpha. It is well know that Tn = therefore (n + 20)(n + 20 + 1) n(n + 1) ≡ ≡ Tn mod 5. Colombia. we claim that Lemma 1. 2 2 On the other hand. References [1] T. then Gn+20 ≡ i mod 5. (2) and (3). 2 Indeed. it is known that (see for example [1. We prove that (1) ⇒ (2). If E. (∗) To simplify the calculations we let X = AB − BA and Y = ABA − BAB. We need the following result whose proof is left to the reader. combining (3) and 34 ≡ 1 (mod 5) the desired result follows. Gaesti. ClujNapoca. Technical University of Cluj-Napoca. F ∈ M2 (C) and x ∈ C the following equality holds: det(E + xF ) = det(E) + (T r(E) · T r(F ) − T r(EF )) · x + (det(F )) · x2 . Solution by Ovidiu Furdui. Romania. First we note that T r(X) = 0 and since X 2 = O2 one has that det(X) = 0. Also solved by Moti Levy. show that the following statements are equivalent: 2 (1) (AB − BA) = O2 . . n ≥ 1 and (1) we have Ln ≡ 2(n + 1) · 3n+1 + 2(n − 1) · 3n−1 = 3n−1 (18(n + 1) + 2n − 2) = 3n−1 (20n + 16) ≡ 3n−1 (mod 5) (2) By Fermat’s little theorem. We have based on formula (∗) combined with T r(X) = 0 and T r(XY ) = 0 that det(X + Y ) = det(X) + (T r(X)T r(Y ) − T r(XY )) + det(Y ) = det(X) + det(Y ). If A. F = Y and x = 1 and we get that det(AB(A + I2 ) − BA(B + I2 )) = det(X + Y ) = det(X) + (T r(X)T r(Y ) − T r(XY )) + det(Y ) = det(Y ) = det(ABA − BAB). Romania.206 Using Ln = Fn+1 + Fn−1 . A calculation shows that AB(A + I2 ) − BA(B + I2 ) = X + Y and XY = (AB − BA)(ABA − BAB) = ABABA − AB 2 AB − BA2 BA + BABAB. using (1) and (2) we get n(n + 3) · 2n · 3n − 3n−1 − 3n 2 = 4n · 3n + (n5 + 3n4 ) · 3n − 3n−1 − 3n 2Fn5 + n2 Fn (Tn + n) − Ln − Ln+1 ≡ 2Fn + n2 · ≡ 3n−1 (9n4 + 12n + 3n − 1 − 3) ≡ 3n−1 (4n4 + 1) 0. if U and V are two matrices. then T r(U V ) = T r(V U ) and this implies that T r(XY ) = T r(ABABA − AB 2 AB − BA2 BA + BABAB) = T r(ABABA − BA2 BA) + T r(BABAB − AB 2 AB) = 0. B ∈ M2 (C) are two matrices. Rehovot. Now we prove the implication (2) ⇒ (1). Serban School Cioculescu. Recall that. n ≡ 0 (mod 5) (3) Consequently. and the proposer. 77. n 6≡ 0 (mod 5) ≡ 3n−1 .Proposed by Florin Stanescu. Now we apply formula (∗) with E = X. (2) det (AB (A + I2 ) − BA (B + I2 )) = det (ABA − BAB) . Israel. x + m m=1 ψ (1) = −γ. b. with x = b 2a − k 2 and N = k + 1 we obtain ∞ X k X 1 2a = ∈ Q. arXiv:math-ph/9804010v1 April 12. However. 2 + bn + c an b − ka + 2am n=1 m=1 Solution 2 by Hun Min Park. Korea Advanced Institute of Science and Technology(KAIST). South Korea. Paris. Part (1). Rehovot. France. Finding Exact Values For Infinite Sums. 1998”. Applying (1) to our problem. 78. (n + α) (n + β) β−α n=1 (1) See. an2 + bn + c ak 2a 2 2a 2 n=1 Using the recurrence formula ψ (N + x) = ψ (1 + x) + N −1 X 1 .Proposed by Mohammed Aassila. (1) Prove that if there exists a positive integer k such that b2 − 4ac = k 2 a2 then +∞ X n=1 an2 1 ∈ Q. Rehovot. the Cayley-Hamilton Theorem implies that O2 = X 2 − T r(X) · X + det(X) · I2 = X 2 . Israel. Syria. Moubinool Omarjee. P∞ 1 The sum n=1 (n+α)(n+β) can be expressed by ψ. the Digamma function ∞ X ψ (β + 1) − ψ (α + 1) 1 = . Israel. for example. and the proposer. + bn + c (2) Is the converse true ? Solution 1 by Moti Levy. c be integers such that a 6= 0 and an2 + bn + c 6= 0 for all n ∈ N \ {0}. we get ∞ X 1 1 b k b k = ψ + +1 −ψ − +1 .207 Since (2) holds the preceding equality implies that det(X) = 0. Lyc´ ee Henri IV. . We have ∞ X ∞ 1 1X = an2 + bn + c a n=1 n + n=1 b 2a + k 2 1 n+ b 2a − k 2 . Moti Levy. Damascus. Let a. Higher Institute for Applied Sciences and Technology. b Since an2 + bn + c 6= 0 for all n ∈ N \{0} then 2a ± k2 are not negative integers. Also solved by Omran Kouba. Strasbourg. Daejeon. France. and the problem is solved. “Costas Efthimioua. 1−α 2−α k−α N −α+1 N +k−α 1 S= ak 1 1 + ··· + 1−α k−α ∈ Q. then: = b a 1 + (2(i + 1) − 1)k + 2j + b a S(n) = n=1 b a 1 − (2i + 1)k + 2j + X Therefore S= ! S = 1 1 − n−α n−α+k 1 ka lim UN N →+∞ 1 1 1 1 1 + + ··· + − − ··· − . an2 + bn + c a(α − β) n=1 n − α n − β n=1 where. suppose that a > 0. Then roots of an equation an2 + bn + c = 0 are given as √ −b ± b2 − 4ac b k x= =− ± 2a 2a 2 This gives X 1 S(n) := 2 an + bn + c n≥1 ! 1X 1 12X 1 1 = = − b b a ak 2n + ab − k 2n + ab + k n + 2a + k2 n + 2a − k2 n≥1 n≥1 ! 1 1 2 X X − = ak 2(ki + j) + ab − k 2(ki + j) + ab + k 1≤j≤k i≥0 ! 1 2 X X 1 = − ak (2i − 1)k + 2j + ab (2i + 1)k + 2j + ab 1≤j≤k i≥0 The inner sum can be simplified: 1 (2i − 1)k + 2j + X i≥0 1 = −k + 2j + = 1 −k + 2j + b a + b a 1 − (2i + 1)k + 2j + i≥0 2 X 1 ak −k + 2j + 1≤j≤k Solution 3 by Proposer. for N ≥ k: N X UN = Hence −b+ka 2a b a +∞ X Since α − β = k. Without loss of generality. . (1) Let α = aX 2 + bX + c. then ! b a ∈Q and β = −b−ka 2a be the roots of +∞ X 1 1 1 1 = − .208 Part (1). Proposed by Mih´ aly Bencze. which appeared in RMS Revue Math´ematiques Sp´eciales Q79: pp. z ∈ R. then tan πq p = πλ with λ ∈ Q. If x. 79. Brasov. California. by contradiction. In By the expansion of the Fourier series of the function x 7→ cos qx p over [−π. First we will prove that for any x. San Jose. 1987-1988. y. Romania. France. this is impossible. 2k + 1 k=0 Since λ ∈ Q and π is a transcendental number. 203-204. that S ∈ Q. Also solved by Moubinool Omarjee. Paris. By De Moivre’s formula we have p−1 bX 2 c p (−1)k (πλ)2k+1 = 0. Lyc´ ee Henri IV. Andr´e-Jeannin. Editor’s Comment: Moubinool Omarjee pointed out that this problem had been previously proposed by R. USA. y ∈ Rthe following inequality holds |cos x| + |cos y| + |cos (x + y)| ≥ 1. Solution 1 by Arkady Alt. (1) Since |cos (x + y)| = |cos x cos y − sin x sin y| ≥ ||cos x| |cos y| − |sin x| |sin y|| it suffices to prove the inequality ||cos x| |cos y| − |sin x| |sin y|| ≥ 1 − |cos x| − |cos y| or equivalently . ∈ 2 2 2 p n −q p n=1 p − ac ∈ Q \ N. π] we have π 1 S= 2− . 2q 2pq tan πq q Assume.209 (2) I have only a partial solution for this question: if b = 0 and this case: +∞ X q 1 with S= 6 N. Prove that: (1) 2 | cos x| + | cos y| + | cos z| + | cos(y + z)| + | cos(x + z)| + | cos(y + x)| ≥ 3 (2) | cos x|+| cos y|+| cos z|+| cos(y +z)|+| cos(y +x)|+| cos(x+z)|+3| cos(x+ y + z)| ≥ 3. . p p . . . uv − 1 − u2 · 1 − v 2 . 1] . ≥ 1 − u − v. v ∈ [0. where u. (2) Inequality (2) holds since . . p p p p . . . uv − 1 − u2 · 1 − v 2 . ≥ 1 − u2 · 1 − v 2 − uv and p 1 − u2 · p p p 1 − v 2 − uv ≥ 1 − u − v ⇐⇒ 1 − u2 · 1 − v 2 ≥ (1 − u) (1 − v) √ √ √ √ √ √ ⇐⇒ 1 − u 1 − v 1 + u 1 + v − 1 − u 1 − v ≥ 0. . z ≤ π2 . y + z ≤ π2 It follows that x + y + z ≤ 3π 4 . y. z) ≥ 2 1 − π π π 2 2 2 + 1 − (x + y) + 1 − (y + z) + 1 − (x + z) π π π 8 1 π = 9 − (x + y + z) ≥ 9 − 8 × 3 × = 3. which is equivalent to the second inequality. hence we may consider only 0 ≤ x. π π 2 G(x. y. y. 1 − π2 x f (x) = 2 if π2 ≤ x ≤ π. Rehovot. X cyclic (|cos x| + |cos (y + z)| + |cos (x + y + z)|) ≥ 3. (b) Let α = x. z) and G(x. 2 | cos x| + | cos y| + | cos z| + | cos(y + z)| + | cos(x + z)| + | cos(y + x)| and | cos x| + | cos y| + | cos z| + | cos(y + z)| + | cos(y + x)| + | cos(x + z)| + 3| cos(x + y + z)| Since the inequalities are symmetric in the variables x. y.210 (a) Applying inequality (1) we obtain X cyclic (|cos x| + |cos y| + |cos (x + y)|) ≥ X 1 = 3. x + z ≤ π2 . The function x 7→ |cos x| is periodic with period π. it is sufficient to consider only the following cases: Case 1: x. y. z ≤ π. πx − 1 By definition. 2y 2z 2x +1− +1− F (x. x + y ≤ π2 . π π 4 If x + y + z ≤ π2 then 2x 2y 2z +1− +1− π π π 2 2 2 2 + 1 − (x + y) + 1 − (y + z) + 1 − (x + z) + 3 1 − (x + y + z) π π π π 12 12 π =9− (x + y + z) ≥ 9 − × = 3. z) ≥ 1 − . respectively. z and the function |cos x| is symmetric with respect to π2 . Solution 2 by Moti Levy. z) to denote. y. cyclic which is equivalent to first inequality. We will write F (x. f (x) satisfies f (x) ≤ |cos x| . β = y + z then by (1) we have |cos x| + |cos (y + z)| + |cos (x + y + z)| = |cos α| + |cos β| + |cos (α + β)| ≥ 1. y. and therefore. Israel. Define the piecewise linear π-periodic function f . by if 0 ≤ x ≤ π2 . x + z ≤ π2 . y + z ≥ π2 and x + y + 3z ≥ 3π 2 . z) ≥ 1 − Case 2: x. π π 2 Case 4: x. y. y. x + y ≤ π2 . y + z ≥ π2 . π π 2 G(x. y. y. z) ≥ 2 1 − +1− +1− π π π 2 2 2 + 1 − (x + y) + (y + z) − 1 + 1 − (x + z) π π π 1 4 4 = 7 − (8x + 4y + 4z) ≥ 7 − (2x + y + z) ≥ 7 − × π = 3. It follows that x + z ≥ π2 . z) ≥ 2 1 − x + 1 − y + 1 − z π π π 2 2 2 + 1 − (x + y) + (y + z) − 1 + (x + z) − 1 π π π 4 4 π = 5 − (y + z) ≥ 5 − × = 3. z) ≥ 1 − 2 2 2 x+1− y+ z−1 π π π 2 2 2 2 + 1 − (x + y) + (y + z) − 1 + (x + z) − 1 + 3 (x + y + z) − 1 π π π π 4 4 3π = −3 + (x + y + 3z) ≥ −3 + × = 3. z ≤ π2 . y. z) ≥ 2 1 − x + 1 − y + z − 1 π π π 2 2 2 + 1 − (x + y) + (y + z) − 1 + (x + z) − 1 π π π 8 4 8 4 π 8 8 π = 1 + z − (x + y) ≥ 1 + z − × = z − 1 ≥ × − 1 = 3. y. It follows that 2x + y + z ≤ π 2y 2z 2x F (x. y.211 If x + y + z ≥ π 2 then 2x 2y 2z +1− +1− π π π 2 2 2 2 + 1 − (x + y) + 1 − (y + z) + 1 − (x + z) + 3 (x + y + z) − 1 π π π π = 3. y ≤ π2 . π π π π 2 π π 2 G(x. π Case 3: x. z ≥ π2 . x + y ≤ π2 . hence 2 2 2 x+1− y+1− z π π π 2 2 2 2 + 1 − (x + y) + (y + z) − 1 + 1 − (x + z) + 3 (x + y + z) − 1 π π π π 4 = 1 + (y + z) ≥ 3. π π π Moreover y + z ≥ π2 implies that x + y + z ≥ π2 . z) ≥ 1 − . G(x. z ≤ π2 . x + y ≤ π2 . y. y + z ≥ π2 2 2 2 F (x. x + z ≥ π2 . 2 2 2 F (x. University of Prishtina. y. In what follows the set of integers r that satisfy a ≤ r ≤ b. Notation. q − 1) will be denoted by Rq (n) = n − qbn/qc.Proposed by Alban Kryeziu. b). Editorial Comment: For a positive integer n let an = n Q (9k 2 + 1). 80∗ . 2 2 2 F (x. z) ≥ 1 − Also solved by AN-anduud Problem Solving Group. and it will solve “half” the problem. The next lemma deals with the case p = 2. For every n ≥ 1 we have ν2 (an ) = b n+1 2 c. x + y ≥ π2 . Also. z ≥ π2 .212 Case 5: x. and νp (an ) = 0 for p ∈ [3]. x + 2z ≥ π and 2x + 2y + 3z ≥ 5π 2 . Lemma 2. Thus ν2 (9k 2 + 1) = 0 if k is even and ν2 (9k 2 + 1) = 1 if k is odd. This proves that p ∈ [1]. Indeed. then p must divide one of the factors of an and consequently −1 must be a square modulo p. and the proposer. Proof. and the unique residue class of n modulo q that belongs to the interval ∇(0. if k is odd. If m is a positive integer. It follows that x + z ≥ π2 . z) ≥ 2 x−1+1− y+ z−1 π π π 2 2 2 + 1 − (x + y) + (y + z) − 1 + (x + z) − 1 π π π 4 4 4 π 4 = −3 + (x + 2z) − y ≥ −3 + × π + × = 3. and p is a prime. Hence . We will try k=1 to prove that an is not a perfect square for any positive integer n. note that ( 1 9k 2 + 1 ≡ k 2 + 1 (mod 4) ≡ 2 (mod 4) (mod 4) if k is even. y + z ≥ π2 . we will write νp (m) to denote the largest nonnegative integer k such that pk divides m. Find all positive integers n such that the product n Y (9k 2 + 1) k=1 is a perfect square. Ulaanbaatar. y ≤ π2 . π π π π 2 2 2 2 x + 1 − y + z − 1+ π π π 2 2 2 2 (x + y) − 1 + (y + z) − 1 + (x + z) − 1 + 3 (x + y + z) − 1 π π π π 4 5π 4 =5 = −5 + (2x + 2y + 3z) ≥ −5 + × π π 2 G(x. Republic of Kosova. y. and for l ∈ {1. Mongolia. 3} we will write [l] to denote the set of primes p such that p ≡ l (mod 4). Let us start by the following simple observation : If an ≡ 0 (mod p) for some odd prime p. will be denoted by ∇(a. the set of prime numbers is denoted by P. n o. n+1 . . ν2 (an ) = . and k is odd . k : 1 ≤ k ≤ n. . = b c 2 which is the desired conclusion. exactly one of the two numbers Rp (a) and Rp (−a) belongs to ∇(0. Setting n = 2m + 1 we see that 36 m(m+1) + 5 = kp2 . } and the above condition is equivalent to 2k − 1 ≡ 1 (mod 36) or 25(2k − 1) ≡ 1 (mod 36). the equation 9x2 + 1 ≡ 0 (mod p) has exactly two solutions modulo p which are a and −a. 7}. } and the condition p2 k ≡ 5 (mod 36) is equivalent to k ≡ 5 (mod 36) or 5k ≡ 1 (mod 36).) From the above discussion we see that p2 | (9n2 + 1) implies that and consequently p < n. an is not a perfect square in this case. Both −1 and (3−1 )2 are quadratic residues modulo p. Finally. let p = 4m + 1 and note that 9(2m + j)2 + 1 = 9j 2 − 9j + 3 − m + 9m + 9j − 2 p So. we may suppose that p ∈ [1]. If n = 1. To see the second inequality. The case p = 2 is trivially true. − 3 2 6 Proof. we have the following cases: • 9n2 + 1 = 2kp2 for some positive integer k. 25. But this condition is equivalent to 2 1 m 1 j− < − 2 9 12 . Proof. It will be denoted by tp . Now. since p ∈ [1] we know that (p2 mod 36) ∈ {1. 2m+j is not a solution to (E) if −1−3m < 9j 2 −9j +3 < m.213 Corollary 1. But. 25. and if the least positive solution is denoted by tp . We conclude that 9n2 + 1 = 2kp2 ⇒ (k 2 mod 36) ∈ {5. since p ∈ [1] we know that (p2 mod 36) ∈ {1. 29}. we consider the equation 9x2 + 1 ≡ 0 (mod p) (E) Then. that is (k mod 36) ∈ {5. If p is a prime number such that p2 | (9n2 + 1) then p < n. 9np2+1 ≥ 10. 7} ≥ 13. For a prime p ∈ [1]. So. (since k cannot be equal to 1. and so is their product −9−1 . 9n2 + 1 ≥ 10p2 > 9p2 + 1 Lemma 4. 29} In particular. Consequently. (E) has exactly two solutions modulo p. if k is an integer satisfying 0 ≤ k < p − 1/3 then 1 ≤ 9k 2 + 1 < p and consequently k is not a solution to (E). (p −√1)/2). This implies that n is odd. and consequently. Setting n = 2m we 36m2 + 1 = (2k − 1)p2 This implies that (2k − 1)p2 ≡ 1 (mod 36). But. that is (k mod 18) ∈ {1. then √ √ p−1 p p−4 tp ∈ . 2 (mod 4) then ν2 (an ) is odd. The following lemma will be useful for our discussion later. This implies that n is odd. Thus. there is an integer a such that a2 ≡ −9−1 (mod p). • 9n2 + 1 = (2k − 1)p2 for some positive integer k. that is k ≡ 1 (mod 18) or k ≡ 7 (mod 18). This implies that 2 2 kp ≡ 5 (mod 36). Lemma 3. We conclude that 9n2 + 1 = (2k − 1)p2 ⇒ (k In particular 2 9n +1 p2 mod 18) ∈ {1. There is an integer n0 ≤ 1000 such that 2p pk − 4 (C) ∀k ≥ n0 . Now since pk | (9t2pk + 1) and (9t2pk + 1) | an we see that pk | an . . That took around one hour of search and it confirmed the conjecture for all the primes p ∈ [1] with p ≤ 109 . so. Proof. For every integer n the number an is not a perfect square. and an cannot by the square of an integer. it seems that Conjecture (C) is plausible. But what is the right value of n0 ? It seems that n0 = 961 works. We may suppose that n ≥ 7. if p2k | an then p2k | (9t2pk + 1). so that p1 = 5. . But note that Y an ≡ 9r2 − 9t2pk (mod pk ) 2 9tpk + 1 1≤r≤n r6=tpk ≡ 9n Thus an 9t2p +1 k Y 1≤r≤n r6=tpk (r − tpk ) Y (r + tpk ) (mod pk ) 1≤r≤n r6=tpk 6≡ 0 (mod pk ). with pn0 = 16829. Remark 1. With this notation we have the following property: Lemma 5. In view of the known asymptotic results in Number Theory.214 or √ √ p p−4 p p−4 − < 2m + j < + 2 6 2 6√ √ p−4 p . p2 = 13. Then √ √ √ pk − 4 pk − 4 pk+1 − 4 2pk − 2tpk ≥ pk + > pk+1 − > pk+1 − ≥ 2tpk+1 3 3 3 and the lemma follows for k ≥ n0 . + So. Therefore νpk (an ) = 1. We did verify this conjecture using Mathematica on a personal computer. . pk+1 < pk + 3 Suppose that k ≥ n0 . . Now. Let k be the largest positive integer such that tpk ≤ n. This ends the proof of the lemma.. According to the previous lemma. using computer. which is absurd. We will use the following conjecture : Conjecture 1. let us arrange the primes in [1] in an increasing sequence (pk )k≥1 . For every k ≥ 3 we have tpk+1 + tpk < pk . but we could not find an explicit reference or statement that confirms it. Proof. we verify the correctness of the lemma when 3 ≤ k < n0 . as usual. Now. we must have n < pk − tpk . non of the integers in the interval ] p2 − p−4 6 2 6 [ is a solution to (E). Corollary 2. but then we conclude from Lemma 3 that pk < tpk . p3 = 17. where tpk was defined in the previous Lemma 4. . Prove that n m+1 1 1 1+ < 1+ n m 58.. C are the measures of the angles of an acute triangle ABC. in . B. . Find all positive integers n smaller that 201314 such that 3n ≡ 3 (mod 13) and 5n ≡ 5 (mod 13). y. Let x. Prove that −1 1 1 1 81 + + ≤ 3|x+y +z|2 +|2x−y −z|2 +|2y −x−z|2 +|2z −x−y|2 |x|2 |y|2 |z|2 60. Let x = 2 cos A. where A. What are the smallest and the biggest? How many are there in total? 57. Proposals are always welcomed. z be nonzero complex numbers. .<in ≤n+1 1 i1 i2 . The source of the proposals will appear when the solutions be published.. + i1 i2 X 1≤i1 <. m be positive integers. Proposals 56.. y = 2 cos B and z = 2 cos C. Let n. Find the minimum value of x4 + y 4 + z 4 + x2 y 2 z 2 59.215 MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. Compute the following sum: X X 1 + i1 1≤i1 ≤n+1 1≤i1 <i2 ≤n+1 1 + . We begin observing that(Training the partition N = A ∪ B = {1. a quadrilateral DHKC is cyclic. we= note the angles ∠HDK and ∠ECK are congruent (they sub∠HDC + ∠HKC 180that . ∠EDH = ∠HDK. 5. 3. . . ◦ ing this circle. aThus. 4. Thus. Solution by the Jos´ ecenter Luis D´ BARCELONA TECH. BARCELONA TECH. of positive integers N. Spain. and. } ∪ {2. a circle inscribed in a triangle lies Barcelona. P is called admissible if for any positive integers m. BARCELONA TECH. . The partition Spain. and we are done. consider ◦ Forthe this purpose. Findpartition all 52. Likewise. thenofn the ∈ Aset and ∈ B orintegers vice-versa. HE. B}Salgueiro. Show that the orthocenter of an acute triangle coincides with the center of the circle inscribed in the triangle formed by the feet of its (Training Sessions of altitudes. . Spain. Let = {A. 6. ◦ ◦ + ∠HKCDHKC. proving that HD. . . circumscribing circle∠EAH about=the cyclicand quadri∠HDK = ∠HCK. So. n such that m − n = p with 2017B} ≤ be p ≤ 3011. we conclude that ∠EAH = ∠EDH. So. therefore. the problem is reduced to proving that HD. Sessions of Spanish Team for IMO 2014) Spain.p prime Let Pand = {A. HE. 2 ∠HDK = ∠HCK. Spain. ∠EDH tend the same arcweHK). ∠HDC = 90 DHKC and ∠HKC = Observ90◦ . it is sufficient to prove that ∠EDH = ∠HDK. the∠HDC quadrilateral Wetherefore. Since the center of a circle inscribed in a triangle lies at the is point of intersection of the bisectors of the angles of the triangle. consider quadrilateral We have. be a partition of the setSpain.216 216 Solutions Solutions 51. Find all of m Spanish for IMO 2014) admissible partitions of N. 6. . and HK are bisectors of the angles of the triangle DEK. BARCELONA TECH. 3. . at the point of Spain. Barcelona. circumscribing a circle about the cyclic ing this circle. and HK are bisectors of the angles of the triangle DEK. . and we are done. We begin observing that the partition N = A ∪ B = {1. conclude that ∠EAH = ∠EDH. Barcelona. n such that m − n = p with p prime and 2017 ≤ p ≤ 3011. . } . Likewise. BARCELONA TECH. andP Bruno Viveiro. . Barcelona. But the angles ∠EAH and ∠HCK are congruent as angles with mutually sides. Barcelona. Since ofıaz-Barrero. But the angles ∠EAH and ∠HCK are congruent as angles lateral AEHD. a partition of mpositive N. Show that the orthocenter of an acute triangle coincides with the center of the circle inscribed in the triangle formed by the feet of its altitudes. Solution 1 by Jos´ e Gibergans-B´ aguena. Also solved by Jos´ e Gibergans-B´ aguena. 51. 52. have. Observtend the same arc HK). . Barcelona. . Hence..(Training then n Sessions ∈ A and ∈ B Team or vice-versa. 2 Also solved perpendicular by Jos´ e Gibergans-B´ aguena. . is sufficient to prove that=∠EDH ∠HDK. Spanish Team for IMO 2014) (Training Sessions of Spanish Team for IMO 2014) Solution by Jos´ e Luis D´ıaz-Barrero. a quadrilateral is cyclic. BARCELONA TECH. 5. ∠EAH = ∠EDH and with mutually perpendicular sides. ∠EDH = ∠HDK. ∠HDC 90◦ and=∠HKC = 90 . . } Solution 1 by Jos´ e Gibergans-B´ aguena. B K E H A D C figure for for the 51 51 figure thesolution solutionofofproblem problem For this purpose. the problem reduced to intersection of the bisectors of the angles of the triangle. . Indeed. we note that the angles ∠HDK and ∠ECK are congruentquadri(they sublateral AEHD. Indeed. = 180 . The admissible partitions of N. P is called admissible if for any positive integers m. Hence. and. } ∪ {2. 4. itDHKC. . . } ∪ B Next. Then n1 = n − p ≥ 1 and n1 ∈ A. Barcelona. then n1 ∈ B. 7.217 is admissible because for all n ∈ A (odd) we have that n − p = m (even) and therefore m ∈ B. . . } ⊂ B Now we have to see that the even numbers 2. . q + 4. 2099. . 3. For any ki ∈ Z. That is. = (q + p − 3) − (p − 1). q ≤ 3011. 2 ∈ B. then A = N and there is no partition. 2. .} and the unique admissible partition is N = A ∪ B = {1. Let n2 = n + 2. . = (q + p − 3) − (q − 1) = p Therefore. q + p − 3 ∈ A and q − 2 = (q + 2) − 4 = (q + 4) − 6 = . We claim that if n ∈ A. if n ∈ B with n > p. 4. Let n2 = n+2. If 2 ∈ A. . . Then n1 −n2 = n+q−(n+2) = q−2 = p. q ≤ 3011 (there are several options like 2027. . i = 1. n2 = n1 + p + 2 = n1 + q ∈ B. . . and we conclude that ap + bq + k1 p + k2 q ∈ A ⇔ 2|(k1 + k2 ) To determine the positive integers belonging to A. 5. q be prime numbers such that 2017 ≤ p. we determine the other elements of A and the elements of B. . p + 5. 9. . the odd numbers q. . . WLOG we can call A the set containing 1 and B the other set of P. we multiply by k both terms of Bezout’s identity and yields k = kap + kbq ∈ A ⇔ ap + bq + (k − 1)ap + (k − 1)bq ∈ A ⇔ 2|(k − 1)a + (k − 1)b = (k − 1)(a + b) ⇔ 2|(k − 1) . 2029. p−1 also belong to B. . . The same occurs for ap + bq + k2 q. Spain. . . So. Now if we see that for all n ∈ B then n + 2 ∈ B we are done. To prove it. BARCELONA TECH. . p − 1. Then there exist a. 4. are in B as we wanted to prove. 4. . . So. . p + 1 ∈ B because 1 ∈ A and we conclude that the set of {p + 1. Indeed. . p + 1. . 6. . . B = {2. Let p. Furthermore. . p + 3. Now we chose the number 1 it will be in A or B. then n + 2 ∈ A. If n1 = n+q. } ∪ {2. . = . 3001). b ∈ Z such that ap + bq = 1 (Bezout) as it is well-known. . } 2 Solution 2 by Jos´ e Luis D´ıaz-Barrero. we consider two twin primes p and q = p + 2 such that 2017 ≤ p. . then n + 2 ∈ B. (q + 2) − p. and therefore n2 ∈ A as we wanted to prove. 3. . . Then. Since we have assumed that 1 ∈ A. . q + 2. the numbers 2 4 p−1 = q − p.. } ⊆ A Now we consider the number 2. we have ap + bq + k1 p ∈ A ⇔ k1 is even. . So. . . 5. p + 3. we can assume that n ∈ B and n > p. N = A ∪ B = {1. and ap + bq + k1 p ∈ B ⇔ k1 is odd. . then {1. . Indeed. 6. . we conclude that all odd positive integers belong to A and all the even positive integers belong to B. B(x) = ax + b.. Damascus. BARCELONA TECH. if we assume that f (x) is some polynomial of degree m then clearly we must have m > 0. + an−1 xn−1 + an xn . So. A(x) = an xn + a0 and B(x) = 2n x + (1 − 2n )a0 . . Assume that A(x) = a0 + a1 x + . where B(x) is a polynomial are of the form A(x) = an xn + a0 and B(x) = 2n x + (1 − 2n )a0 . B} is the only admissible partition of N. a0 = aa0 + b. Syria. So. ak = 0 for 1 ≤ k ≤ n − 1 and b = (1 − 2n )a0 ... show that the function f : R → R defined by f (x) = sinh2 x is not a polynomial. Indeed. then deg(A(x)) = deg(A(2x)) = deg(B(x)) · deg(A(x)) Since the degree of a polynomial is a nonnegative integer. k − 1 is even and k is odd. yields 2n an = aan . First. Suppose that f is a polynomial.. a0 . . or else B(x) is linear. P = {A. an 6= 0. Spain.. . .. Namely. 2a1 = aa1 . Without using the series expansion of the hyperbolic functions.. 2 53.. Then. (Training Sessions of Spanish Team for IMC 2013) Solution 1 by Omran Kouba. Finally. if A(2x) = B(A(x)). then we have either A(x) is a constant k and B(k) = k. Higher Institute for Applied Sciences and Technology. b ∈ R. an ∈ R as claimed. we have A(2x) = a0 + 2a1 x + . 2 Solution 2 by Jos´ e Luis D´ıaz-Barrero. . This fact contradicts Bezout’s identity. a. from which follows a = 2n . . a0 . an ∈ R. 2n−1 an−1 = aan−1 ..218 because if 2|(a+b) then a and b have the same parity. and we are done. Therefore. Barcelona. + 2n−1 an−1 xn−1 + 2n an xn and B(A(x)) = (aa0 + b) + aa1 x + . and the equality f (2x) = 4f (x)(1+f (x)) leads to the contradiction m = deg f (2x) = deg(f (x)) + deg(1 + f (x)) = 2m Thus f cannot be a polynomial.. To prove that f (x) = sinh2 x is not a polynomial we argue by contradiction. So. f (2x) = sinh2 2x = 4 sinh2 x(1 + sinh2 x) = . Note that f (2x) = sinh2 (2x) = 4 sinh2 x cosh2 x = 4f (x)(1 + f (x)) We argue by contradiction. . + aan−1 xn−1 + aan xn Equating the preceding polynomials.. Then. we will see that the only polynomials A(x) such that A(2x) can be written as A(2x) = B(A(x)). Bruno Salgueiro. Spain. since 1 − aj = a2n+1 − j then after pairing terms. Italy. we have 2S1 = 2n+1 Xh j=0 i 2n+1 X f (aj ) + f (a2n+1 − j) = 1 = 2n + 2 j=0 from which follows S = (2n + 1)S1 = (2n + 1)(n + 1). Let us denote by S 2n+1 X = j=1 2n+1 X = j=0 j3 (2n + 1)2 − 3(2n + 1)j + 3j 2 j3 (2n + 1)2 − 3(2n + 1)j + 3j 2 Dividing both terms by 2n + 1 we get S1 = 2n+1 X S j3 = 2n + 1 (2n + 1)3 − 3(2n + 1)2 j + 3j 2 (2n + 1) j=0 2n+1 X = j=0 Putting aj = (j/(2n + 1))3 1 − 3(j/(2n + 1))j + 3(j/(2n + 1))2 j . 54. and we are done. Barcelona. Furthermore. BARCELONA TECH. Barcelona. 2 Also solved by Paolo Perfetti. Jos´ e Gibergans-B´ aguena. Department of Mathematics. 2 . 0 ≤ j ≤ 2n + 1 we get 2n + 1 S1 = 2n+1 X j=0 a3j 1 − 3aj + 3a2j Since 1 − 3x + 3x2 = (1 − x)3 + x3 . Hence. Viveiro.219 B(sinh2 x) = B(f (x)). where B(x) = 4x(1 + x). Tor Vergata University. Spain. Rome. f (x) = sinh2 x is not a polynomial and we are done. Compute 2n+1 X j=1 j3 (2n + 1)2 − 3(2n + 1)j + 3j 2 (Training Sessions of Catalonian Team for OME 2013) Solution 1 by Jos´ e Luis D´ıaz-Barrero BARCELONA TECH. Let n be a nonnegative integer. Spain. then the function f (x) = x3 x3 = 2 1 − 3x + 3x (1 − x)3 + x3 satisfies that f (x) + f (1 − x) = 1. But B(x) is a quadratic polynomial that jointly with f (2x) = B(f (x)) lead us to a contradiction on account of the previous claim. Since = . then n−k k k n−k k=0 k=0 2 n X n n − k in the right term. we obtain 2 n X n n 2n 2n − 1 k = =n k 2 n n−1 k=0 and the statement follows. Show that 2 n n 2n − 1 1 X k = n k n−1 k=0 (Training Sessions of Catalonian Team for OME 2014) Solution 1 by Jos´ e Gibergans-B´ aguena. BARCELONA TECH. Athens. Syria. 2 2 n n X X n n n n k . Barcelona. Damascus. Damascus.220 Solution 2 by Omran Kouba. Sn = . Then. 2 Also solved by Jos´ e Gibergans-B´ aguena. BARCELONA TECH. we have 2n+1 2n+1 X X (2n + 1 − j)3 Sn j3 = = 2n + 1 (2n + 1 − j)3 + j 3 |{z} (2n + 1 − j)3 + j 3 j=0 j=0 j←2n+1−j = 1 2 2n+1 X j=0 2n + 2 (2n + 1 − j)3 + j 3 = =n+1 (2n + 1 − j)3 + j 3 2 from which follows Sn = (n + 1)(2n + 1). Syria. Putting t = k = Spain. 55. Barcelona. Since n X n−1 n Sn = k−1 n−k k=1 then we see that Sn is the coefficient of X n−1 of polynomial n n X X n − 1 n X k = (1 + X)n−1 (1 + X)n = (1 + X)2n−1 X j−1 j − 1 k j=1 j=0 2n − 1 That is. Spain. Bruno Salgueiro. 2 Solution 2 by Omran Kouba. we get (n − t) . Spain. Let Sn = n1 k=0 k nk . we denote the desired sum by Sn . t t=0 2 2 X 2 n n n X X n n n (n − k) k + k = 2 k k k k=0 k=0 k=0 n 2 X n 2n = n =n k n k=0 Finally. Greece. First. Viveiro. and we are done. Anastasios Kotronis. Higher Institute for Applied Sciences and Technology. 2 n−1 . Then. Higher Institute for Applied Sciences and 2 Pn Technology. Let n be a non negative integer. 0} X r r r s xm (1 + x)s = (1 + x)r+s = m m t−m m=0 The coefficient of order t of the above function is min{t.r} X r s m t−m m=max{t−s. Moreover 2 X 2 X 2 n n n X n n n = − (n − p) n p n − p p p p=0 p=0 p=0 and then we come to prove that 2 n n 2 X n nX n 2n − 1 p = =n p 2 p=0 p n−1 p=0 or n 2 X n p=0 p (2n)! = = n!n! 2n n This is a well know result.S.r} X r s r+s = m t−m t m=max{t−s. “generalfunctionolgy”. Following the hint there.0} and taking r = s = t = n it must be equal to the coefficient of order n of (1+x)r+s = (1 + x)2n which is 2n 2 n so proving the result. Greece. Pn n 2 = n 2n−1 Changing n − k = p. Rome. . Spain. p=0 (n − p) n−p n−1 .r} xt X m=max{t−s. BARCELONA TECH. we prove the more general result min{t.Wilf. Athens. Spain. Tor Vergata University. Italy. for instance on web there is the book of H.0} and the result follows by taking r = s = t = n. To achieve the sum we write m+s s X s X m+p s t = xm (1 + x)s x x = p t − m t=m p=0 and then m+s X t=m min{t. Barcelona. Also solved by Jos´ e Luis D´ıaz-Barrero. Viveiro. Anastasios Kotronis. Department of Mathematics.221 Solution 3 by Paolo Perfetti. Bruno Salgueiro. In recent years. ϕ1 . [7] and references therein). unifications. we have 9 abc s2 d2 + e2 + f 2 ≤ ≤ 8s 3 where a = BC. Introduction If x1 . in particular we give a stronger version of Janous’s inequality [8] and we give a short and simple proof to a geometric inequality proposed by Mongolia to 1988 IMO [9. 223]. x2 . 1. 425. First. It is important to point out that the Wolstenholme inequality provided an effective approach to combine both geometric and arithmetic inequalities. Janous proposed the following problem: Let d. Based on the Wolstenholme inequality. Main Results In [8. 2. p. x3 . In 1867. and Mitrinovi´c et al. [2] p.222 MATHNOTES SECTION Some applications of the Wolstenholme inequality ˘ Mohammed Aassila and Marian Dinca Abstract. Murty et al. Let d. [5]. refinements and applications (see Bottema et al. the following celebrated cyclic inequality x21 + x22 + x23 ≥ 2x1 x2 cos ϕ1 + 2x2 x3 cos ϕ2 + 2x3 x1 cos ϕ3 (1) is known as Wolstenholme’s inequality in the literature (see Mitrinovi´c et al. b = CA and c = AB. a large number of new inequalities have been discovered with applications. e and f be the sides of the triangle determined by the three points at which the internal angle-bisectors of given ∆ ABC meet the opposite sides. [4]. In this note we give two applications of Wolstenholme’s inequality. Then. We will prove the stronger inequality: d2 + e2 + f 2 ≤ Theorem 1. we give a stronger version of Janous’s inequality and second we give a simple and short solution of a problem proposed to IMO 1988. Wolstenholme first introduced inequality (1) with a historical account in his book [3]. ϕ2 . p. 336]. e and f be the sides of the triangle determined by the three points at which the internal angle-bisectors of given ∆ ABC meet the opposite sides. [1] p. 421). The purpose of this note is to give some applications to the investigation of geometric inequalities. considerable attention has been given to study this inequality in many different directions including its various generalizations. Prove that s2 3 where s is the semi-perimeter of ∆ ABC. [6] and Milovanovi´c et al. . ϕ3 are real numbers and ϕ1 + ϕ2 + ϕ3 = π. Mitrinovi´c et al. Since wa ≤ √ bc cos A . s IB 0 = 2wb . ≤ 3abc(a + b + c) ≤ 3abc a cos2 2 3 cyclic . = . Hence. C 0 . y = bwb . we deduce that X X X 2 (awa )2 + (awa )2 3 (awa )2 d2 + e2 + f 2 ≤ cyclic cyclic = 4s2 cyclic 4s2 . BB 0 = wb and CC 0 = wc . 2 wc ≤ √ ab cos C 2 we deduce that 3 X (awa )2 cyclic X A 2 cos 2 X cyclic A . X X A + B 1 (awa )2 + (awa )(bwb ) cos d2 + e2 + f 2 = 2 2s 2 cyclic cyclic By the Wolstenholme inequality xy cos α + yz cos β + zx cos γ ≤ x2 + y 2 + z 2 2 (2) B+C C+A applied with x = awa . s In conclusion: d2 = = IA02 + IB 02 − 2IA0 · IB 0 cos ∠A0 IB 0 (awa )2 + (bwb )2 + 2(awa )(bwb ) cos 4s2 A+B 2 and similarly for e2 and f 2 . 0 +AA a + b + c wa a+b+c Consequently IA0 = 2wa .223 Proof. B 0 . Let AA0 = wa . β = 2 . Let the internal angle bisectors meet the opposite sides at the points A0 . s IC 0 = 2wc . b+c Thus ab A0 C IA0 a = = b+c = 0 AA AC b b+c and IA0 IA0 a IA0 a = . We have A0 B c = 0 AC b and hence A0 C = ab . 2 wb ≤ √ ac cos B . γ = 2 . z = cwc and α = A+B 2 . and AA0 ∩ BB 0 ∩ CC 0 = {I}. 3 3 X X cos βi ≤ cos αi . α3 we deduce that sin α1 sin α2 cos β3 + sin α2 sin α3 cos β1 + sin α3 sin α1 cos β2 sin2 α1 + sin2 α2 + sin2 α3 2 sin α1 sin α2 cos α3 + sin α2 sin α3 cos α1 + sin α3 sin α1 cos α2 . βi > 0 for 1 ≤ i ≤ n (n > 1) and that n X αi = i=1 Then we have n X cos βi i=1 sin αi n X βi = π.224 Because of the Chebyshev inequality X A 1 X A 2 2 a cos ≤ (a + b + c) · cos 2 3 2 cyclic cyclic 1 X 1 + cos A = (a + b + c) · 3 2 cyclic 3 + cos A + cos B + cos C ≤ (a + b + c) 6 ≤ (a + b + c) · 3+ 6 3 2 Hence 3 X (awa )2 ≤ abc(a + b + c) · cyclic = 3 (a + b + c). z = sin α3 . Suppose αi > 0. i. α = β1 . β = β2 and γ = β3 . p. Since α3 = π − α1 − α2 and β3 = π − β1 − β2 then the inequality (3) is equivalent to cos β1 cos β2 cos(β1 + β2 ) cos α1 + cot α2 − − ≥ cot(α1 + α2 ) − (4) sin α1 sin α2 sin(α1 + α2 ) for α1 + α2 < π and β1 + β2 < π. ≤ = When dividing by sin α1 sin α2 sin α3 we obtain the inequality (3). We prove it for n = 3. y = sin α2 . α2 . i=1 Proof. . For n = 3 and the angles of the triangle being α1 . The result is clear for n = 2. (3) sin αi i=1 i=1 We apply the inequality of Wolstenholme (2) with : x = sin α1 .e. i=1 ≤ n X cot αi . 4 9s 9 = abc · 4 2 and using the AM-GM inequality we deduce the desired inequality. 223]. We give a proof by induction. Our second result is a geometric inequality proposed by Mongolia to 1988 IMO [9. Theorem 2. 47 (1984) 413–417. Mitrinovi´c. ˘ Mohammed Aassila Marian Dinca 11 rue de l’anneau Street Aleea Buhusi N. 1989. Mati´c.com . Crux Math. Pe˘cari´c. N. Classical and New Inequalities in Analysis. J. J. by induction. Mitrinovi´c. Pe˘cari´c. Wolters-Noordhoff. France Bucuresti. [9] D.V.Z. Springer. [8] W. M. Problem 715. J. Ningbo Univ. Jani´c. Addenda to the monograph: Recent advances in geometric inequalities (I). and i=3 − i=3 n X sin ! βi !. A Book of Mathematical Problems. 1993. Volenec. Cambridge. Petrovi´c.S. Problem 2652. Murty. [4] O. The IMO Compendium. ˘ Milovanovi´c. Mitrinovi´c. Romania E-mail : aassilam@yahoo. P. [7] G. Satyanarayana. 4 (2) (1991) 108-120. K. Djordjevi´c. J. Kluwer Academic Publishers. R. D.S. Crux Math. V. Dordrecht. [2001 : 336]. 1 67200 Strasbourg. 2.E. Bottema. Geometric Inequalities. A. Acta Sci. 2006. Djuki´c. Wolstenholme. References [1] D. Fink.S. Mitrinovi´c. Vasi´c. αi i=1 The proof. Kluwer Academic Publishers. Z. V. London.E. J. Chen. Janous. 1867.N. Some discrete inequality of Opial’s type. [3] J. is now finished. Recent Advances in Geometric Inequalities.fr E-mail :
[email protected]. V. I. I. Milovanovi´c. 9 (1983) 58–62.R.S. · · · . R. [6] V. [2] D. Klamkin. Groningen. Dordrecht. Pe˘cari´c. Jankovi´c.M. Math. Volenec. (5) αi i=3 By adding (4) and (5) we deduce that n X i=1 cot αi − n X cos βi i=1 sin αi ≥ cot(α1 + α2 ) + cot cos(β1 + β2 ) − − sin(α1 + α2 ) n X ! αi i=3 cos sin n X i=3 n X ! βi ! αi i=3 ≥ cot n X i=1 cos ! αi − sin n X i=1 n X ! αi !. n − 2 n X i=3 cot αi − n X cos βi i=3 sin αi ≥ cot n X αi n X cos ! i=3 βi < π.225 Suppose n X αi < π and i=1 n X βi < π imply that n X i=1 n X αi < π and i=3 assume the inequality holds for n − k arguments k = 1. 1969 [5] D.M. Dept.M. B˘ atinet¸u–Giurgiu. Rome. Show that P Q is parallel to BC. Republic of Kosova. D. “George Emil Palade” School. Neculai Stanciu. University of Proishtina. “Matei Basarab” National College. 2014 Proposals 16. Italy Let a0 = a1 = a2 = 1 and for n ≥ 1 an+2 = Find an for any n. D. Buz˘ au. an an+1 an + an−1 . 19. teramo.226 JUNIOR PROBLEMS Solutions to the problems stated in this issue should arrive before March 15. Buz˘ au. Romania. Romania. 20. B˘ atinet¸u–Giurgiu.M. Paolo Perfetti. Italy. Let K be the symmedian point of 4ABC. Math. Proposed Armend Shabani. Neculai Stanciu. Bucharest. Find all integer solutions of the equation 3x + x4 = 5x . Determine all real number x satisfying 1 1 2 √ √ = √ + x−3 x+2 x− x x−2 x 17. Romania. Proposed by Ercole Suppa. “Tor Vergata” University. Bucharest. Denote by P and Q the intersection points (different from A) of AB and AC with the circumcircles of triangles 4ADC and 4ABD respectively. “ Matei Basarab” National College. let D = AK ∩ BC. Prove that the acute triangle ABC is equilateral if and only if tan2 A tan2 B tan2 C + + =9 2 2 sin B + cos2 C sin C + cos2 A sin A + cos2 B 2 18. Romania. “ George Emil Palade” School. k k=1 4−k k P4 4 a is the polynomial with odd degree. Romania. Romania. From second condition we have: 4 X 4 4−k k f x4 + x a = f (x + a)4 = (f (x))4 = f (x4 ). Writing m2 + n2 = x2 − y 2 we can consider 2 2 2 2 +1 −1 the triplet (x2 − y 2 . (5. i. Proposed by Tom Moore. Correction. f (x) = 1. Hence we obtained a PPT with odd cathetus equal with (x2 − y 2 = (x − y)(x + y) = m2 + n2 . Comˇ anesti. the functions are f (x) = x. Proposed by Valmir Krasniqi. USA. x2 + y 2 ) where x = m +n and y = m +n . 2mn. Mathematics Department. Damascus. University of Prishtina.227 Solutions 11. therefore f is periodic with period a. Neculai Stanciu. it takes every value on Since k=1 k x the set of real number it follows that f is constant. If a 6= 0. then we obtain f (x + a) = f (x). 12. Solution by Proposer For y = 0. Provide explicit constructions of such triples to show that there are infinitely many such odd numbers N. 13). where m > n and m and n are coprime and with different parities. Let (m2 − n2 . Conant Science and Mathematics Center Bridgewater State University. Also solved by Omran Kouba. It is clear that f (0) = −a. “George Emil Palade” School Buzˇ au. 84. x > y and x − y = 1 yields that they are consecutive. so coprime and with different parities. Higher Institute for Applied Sciences and Technology. We will differ two cases: a = 0 and a 6= 0.M. “Matei Basarab” National College. Syria. We are looking for another PPT such that odd length cathetus to be equal with m2 + n2 . The examples (3. y are positive integers. The proof is complete. Solution by D. the hypotenuse of another PPT. Republic of Kosova. Romania and Titu Zvonaru. Find all functions f : R −→ R such that f (x + 2y − f (y)) = f (x + y) and f (x4 ) = (f (x))4 . 5). f (x) = 0. 85) show that the same odd number may occur as the ”hypothenuse” and the ”odd leg” of a primitive Pythagorean triple (PPT). If f (0) = 0. Bucharest. 12. (13. MA 02325. Bˇ atinetu-Giurgiu. then we obtain x = −y.e. . 2xy. Bridgewater. Finally after substitutions. m2 + n2 ) be a PPT. 4. we will obtain f (x − f (0)) = f (x). and from initial equation we will have f (y) = y. and proposer. Because 2 2 m and n have different parities and x. In conclusion. Taking k = 2 we get n = 4 · 10100 .Proposed by Dorlir Ahmeti. Solution by proposer. Find at least one positive integer n such that n is perfect square and n20 + n13 has 2013 digits . Let x = 10101 . we have 102000 < f (10100 ) = 102000 + 101300 < 2 · 102000 (1) Since 102000 + 101300 has 2000 digits. and the largest one is m02 where m0 = b102013/40 c. Now 10100 < n < 10101 ⇒ 10100 < k 2 · 10100 < 10 · 10100 ⇒ 1 < k 2 < 10 So k = 2 or k = 3. we see that n > 10100 . Let x = 10100 . we have Since 10 2020 + 10 102020 < f (10101 ) = 102020 + 101313 < 2 · 102020 (2) 10100 < n < 10101 . Let f be a function that f (x) = x20 + x13 . student. where m = d102012/40 e. is m2 . f is increasing on positive integers. University of Prishtina. Editorial Comment: The smallest such integer n. From (1) and (2) we have 100 Because 10 is a perfect square the best way to find a positive integer n satisfying the proposed condition is to look for n of the form k 2 · 10100 . (3) 1313 has 2020 digits. we note that 20 13 420 · 102000 < 4 · 10100 + 4 · 10100 < 2 · 420 · 102000 Clearly 125 < 128 ⇒ 53 < 27 ⇒ 512 < 228 ⇒ 212 · 512 < 240 ⇒ 1012 < 420 . So 102 012 < 420 · 102000 and 2 · 420 · 102000 < 102013 . but 20 13 420 · 102000 < 4 · 10100 + 4 · 10100 < 2 · 420 · 102000 . We have to prove that this yields an answer. Republic of Kosova. n = 4 · 10100 is a perfect square and n20 + n13 has exactly 2013 digits.228 13. and 625 > 512 ⇒ 54 > 29 ⇒ 12 4 5 16 5 >2⇒ >8> ⇒ 5 · 512 > 1228 4 4 5 ⇒ 5 · 512 · 512 > 1240 ⇒ 5 · 1012 > 420 . we get n < 10 101 . Clearly. 20 13 20 13 Thus 102012 < 4 · 10100 + 4 · 10100 < 102013 and 4 · 10100 + 4 · 10100 has 2013 digits. . Let (ma . San Jose. California. B˘ atinet¸u–Giurgiu. (wa . t ∈ (0. b. then the following inequality holds X xm2a + ym2b m+1 √ (x + y)m+1 S. β. Buz˘ au. wc ) and S be respectively the medians. Show that if m ∈ [0. Also solved by Arkady Alt. wb . Bucharest. the bisectors and the area of a triangle ABC. Romania. 3 ≥ 3 m (z + t)m (zwa2 + twb2 ) cyclic Solution by Titu Zvonaru (slightly modified by the editors) First we shall prove that wa ≤ ma Indeed. α.M. m ≥ 0 m α β γ (α + β + γ)m X 3 and taking into account that m2a = (a2 + b2 + c2 ) we deduce 4 cyc P m+1 2 2 xm + ym X xm2a + ym2b m+1 a b cyc m = ≥ P 2 + tw 2 )m (zw a b zw2 + tw2 cyc cyc m+1 a b m+1 (x + y) cyc m2a (x + y) cyc m2a m ≥ m = = P P (z + t) cyc wa2 (z + t) cyc m2a P √ (x + y)m+1 (x + y)m+1 cyc m2a 3 (x + y)m+1 X 2 = a ≥3 3 S = m m (z + t) 4 (z + t) (z + t)m cyc P P The last inequality follows by the Ion Ionescu–Weitzenb¨ock inequality. x. c. cyc cyc Then. √ a2 + b2 + c2 ≥ 4 3S Remark of the solver The authors of this problem discovered after 117 years that the inequality was presented first by Ion Ionescu in Romanian Mathematical Gazette (1897) and afterwards by Roland Weitzenb¨ ock (1919). “ Matei Basarab” National College. Proposed by D. from the inequality of Radon am+1 bm+1 cm+1 (a + b + c)m+1 + m + m ≥ . Romania . by the AM-GM inequality we have r √ p 2 bc p 1p 2b2 + 2c2 − a2 2 2 = ma wa = s(s − a) ≤ s(s − a) = (b + c) − a ≤ b+c 2 4 X X Adding up we obtain wa2 ≤ m2a . a. ∞). mb . Ioan Viorel Codreanu. Romania. mc ). y. z. “George Emil Palade” School. i.229 14. Maramures. Satulung. Neculai Stanciu. Usa. γ > 0.e. ∞). Editorial Comment: We must exclude trivial cases like (x. y = 2ab and z = a2 + b2 . which implies that x3 + y 3 + z 3 is a composite number. Brasov. where x. Omran Kouba. Pitesti. . Higher Institute for Applied Sciences and Technology. Proposed by Mihaly Bencze. Romania. Syria. The solutions of the equation are: x = a2 − b2 . Damascus. Prove that if x. y. Then one easily finds that x3 +y 3 +z 3 = 2a2 (a+b)(a(a2 −ab+b2 )+6b3 ). Also solved by Daniel Vacaru. z ∈ Z. z) = (1.230 15. y. y. z are the integer solutions of equation x2 +y 2 = z 2 then x3 +y 3 +z 3 is a composite number. 1) for which x3 + y 3 + z 3 = 2. 0. Romania. Solution by Arber Igrishta. y) ∈ (0. . Proposals should be accompanied by solutions. 1) Let f : (0. so show that if f (x21 ) + f (x22 ) = f (x1 x2 ) 2 for any (x1 . Prishtin¨ e. Mih´aly Bencze.Mathproblems ISSN: 2217-446X. ∞).com Volume 2. David R. Jos´ e Luis D´ıaz-Barrero. Teachers can help by assisting their students in submitting solutions. then f (xn1 ) + f (xn2 ) + . . Questions concerning proposals and/or solutions can be sent by e-mail to: mathproblems-ks@hotmail. each indicating the name of the sender. Proposed by Enkel Hysnelaj. Roberto Tauraso. ∞). z) ∈ (0. The editors encourage undergraduate and pre-college students to submit solutions. Drawings must be suitable for reproduction. . Ovidiu Furdui. ∞). An asterisk (*) indicates that neither the proposer nor the editors have supplied a solution. Cristinel Mortici. Australia. Shabani. Solutions will be evaluated for publication by a committee of professors according to a combination of criteria. J´ozsef S´andor. ∞) → R be a function that satisfies the property f (x2 ) + f (y 2 ) = f (xy) 2 for any (x. Sydney. Paolo Perfetti. xn ) n c 2010 Mathproblems. y. . Kosov¨ e. Issue 1 (2012). url: http://www. Enkel Hysnelaj. Student solutions should include the class and school name. 2) Generalize the above statement. x2 ) ∈ (0. Show that f (x3 ) + f (y 3 ) + f (z 3 ) = f (xyz) 3 for any (x. 48 . University of Technology. + f (xnn ) = f (x1 x2 .mathproblems-ks. Armend Sh. PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets. Pages 48–68 Editors: Valmir Krasniqi. Universiteti i Prishtin¨ es. Stone. Valmir Bucaj.com Solutions to the problems stated in this issue should arrive before 2 April 2012 Problems 29. then prove that x∈[a... Evaluate: Z 2012 x sinn t dt... √ .b] M = max f 00 (x). . If 0 ≤ a < b show that Rb 1 0 dx 1 P (b) P 0 (b) − P 0 (a) a P 0 (x) ≥ ln ≥ Rb 1 0 b−a P (a) P (b) − P (a) dx a P 00 (x) . are: 1 1 2 1 3 1 4 1 . √ . such that P 0 has only real zeros. Proposed by Ovidiu Furdui. Bra¸sov. Find the value of Z π/2 √ n lim sinn x + cosn x dx n→∞ 0 34.. √ . Romania. Proposed by Mih´ aly Bencze. . + a1 x + a0 be a polynomial with strictly positive real coefficients of degree n ≥ 3. Proposed by Valmir Bucaj. 31. George Emil Palade Secondary School.√ a1 an+1 a2 a2n a3 a2n−1 a4 a2n−2 1 n 1 n−1 . Seguin.√ . Let P (x) = an xn + an−1 xn−1 + . Romania. √ an−1 an+3 an an+2 where (an )n≥1 is a decreasing geometric progression. Texas Lutheran University. b] → R be a two times differentiable function such that f 00 and f 0 are continuous. {z } | {z } | n-times n-times where n ≥ 3. √ .√ . m ∈ N. .49 for any (x1 .√ . Proposed by Florin Stanescu. .√ . √ . Proposed by Neculai Stanciu. + 2 + x + 2 − 2 + . 35. . School Cioculescu Serban.b] m (b2 − a2 ) M (b2 − a2 ) ≤ bf 0 (b) − af 0 (a) − f (b) + f (a) ≤ 2 2 33. x2 . . If m = min f 00 (x) and x∈[a. TX. Buzau. in counterclockwise order. Dambovita. If the vertices of a polygon. lim x→0 2011 x tm where n. 30. Solve the following equation r r q q √ √ √ 2 + 2 + . xn ) ∈ (0.. jud. Romania. Let f : [a. Romania. . Cluj. Bra¸sov. Proposed by Mih´ aly Bencze. . + 2 + x = x 2. show that the area of this polygon is 3 1 1 A= √ −√ √ 2 a1 a2n an 32.√ . Gaesti. ∞) and n a positive integer greater than 1. Romania. integrable function defined on [0. Let x.50 Solutions No problem is ever permanently closed. Indeed (1 − 3x)(4x2 + 5x − 27) 864 1 (1 − 3x)2 (1107 + 1580x − 512x2 − 384x3 ) − x − = . Proposed by Paolo Perfetti. Rome. department of Mathematics. Switzerland. 3 We observe that g(x) is the tangent to the graph of f (x) at x = 1/3. 1] such that . we have to prove that X 4x(2 − x) (1 − x)(3 + x) X (1 − 3x)(4x2 + 5x − 27) − ≥ 0. Obviously X (x + y)(3x + 3y + 4z) X (y + z)(3y + 3z + 4x) = (2x + 2y + 3z)2 (2y + 2z + 3x)2 cyc cyc The inequality then reads as X 4x(x + 2y + 2z) cyc (x + 3y + 3z)2 ≥ X (y + z)(3y + 3z + 4x) cyc (2y + 2z + 3x)2 By homogeneity. = (3 − 2x)2 (2 + x)2 (3 − 2x)2 (2 + x)2 cyc cyc Let f (x) = (1 − 3x)(4x2 + 5x − 27) . Italy. We will be very pleased considering for publication new solutions or comments on the past problems. Tor Vergata University. So X (1 − 3x)(4x2 + 5x − 27) X X = f (x) ≥ g(x) = 0 2 2 (3 − 2x) (2 + x) cyc cyc cyc Also solved by the proposer 23. for 0 ≤ x ≤ 1. Proposed by Paolo Perfetti. Italy. z be positive real numbers. y. (3 − 2x)2 (2 + x)2 and g(x) = 864 343 1 x− . Let f be a real. Tor Vergata University. So. 2 2 (3 − 2x) (2 + x) 343 3 343(3 − 2x)2 (2 + x)2 and clearly. Department of Mathematics. we can assume that x + y + z = 1. for 0 ≤ x ≤ 1. 22. because −3 4 1 2 + 5 1 − 27 3 3 1 864 f0 = = 1 2 1 2 3 343 3−2 2+ 3 3 We claim that f (x) ≥ g(x). Prove that X 4x(x + 2y + 2z) X (x + y)(3x + 3y + 4z) ≥ (x + 3y + 3z)2 (2x + 2y + 3z)2 cyc cyc Solution by Albert Stadler. Rome. 1107 + 1580x − 512x2 − 384x3 > 0. 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. So. m(x − bn )). Since f is integrable. Since an and bn belong to [0.51 R1 0 f (x)dx = 0 and m = min f (x). we observe that m2 M 2 mM −mM m2 M 2 2 2 (3M ≥ − 8mM + 3m ) = − + 6(M − m)2 2 3(M − m)2 3(M − m)2 and we will prove the following Proposition. (i. 1] such R1 that 0 f (x)dx = 0 and m = min0≤x≤1 f (x). Prove that Z 1 F 2 (x)dx ≤ 0 −mM (3M 2 − 8mM + 3m2 ) 6(M − m)2 Solution by Omran Kouba. then . with equality if and only if f coincides F (x)dx ≤ 0 3(M − m)2 0 for almost every x in [0. m if x ∈ 0. in what follows we will suppose that F 6= 0. Moreover. 1] \ O. Thus there exist N ⊂ N and a family (In )n∈N of non-empty disjoint open subintervals of (0. Damascus. 1] \ O. f = 0 a. M = max0≤x≤1 f (x). ∀ x ∈ In . since F (0) = F (1) = 0. while F keeps a constant sign on In . . Let f be a nonzero real. If F = 0. 1) : F (x) 6= 0} is an open set. The continuity of F shows that the set O = {x ∈ (0. 1]. Suppose that In = (an . and let F (x) = Z 1 Rx m2 M 2 2 f (y)dy.e. for x ∈ In . Higher Institute for Applied Sciences and Technology. 1) such that O = ∪n∈N In . First. 1] with one of the functions f0 or f1 defined by h h −m M M if x ∈ 0. and consequently 0 < F (x) ≤ min(M (x − an ). F is continuous on [0.) there is nothing to be proved. From m ≤ f ≤ M we conclude that. we see that F (t) = 0 for every t ∈ [0. integrable function defined on [0.1 M −m M −m Proof.e. The open set O is the union of at most denumerable family of disjoint open intervals. M = max f (x). Let us define F (x) = Rx 0 0≤x≤1 0≤x≤1 f (y)dy. we conclude that F (an ) = F (bn ) = 0. M −m M −m f0 (x) = f1 (x) = h i h i M m if x ∈ −m . 1 M if x ∈ . Syria. bn ). Let us consider two cases : (a) F (x) > 0 for x ∈ In . f (x) = M for almost every x ∈ an . M (bn − x)). M aMn−m h i M an −mbn for almost every x ∈ M −m . m(x−b n )) for every x ∈ In . 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. = 2 3(M − m) 3(M − m)2 with equality if and only if F (x) = max(m(x−an ). M x ∈ In . if and only if. ∀ x ∈ In . (b) F (x) < 0 for x ∈ In . and consequently Z Z F 2 (x) dx ≤ In 0 < −F (x) ≤ min(−m(x − an ). From m ≤ f ≤ M we conclude that. and f (x) = m is. if and only if. M bMn −ma . and f (x) = M −m h i M bn −man for almost every x ∈ M −m . M (bn − x))) dx an an +M (bn −an )/(M −m) Z m2 (x − an )2 dx = an Z bn M 2 (bn − x)2 dx + bn +m(bn −an )/(M −m) = m2 Z 0 M (bn −an )/(M −m) x2 dx + M 2 Z −m(bn −an )/(M −m) x2 dx 0 m2 M 2 m2 M 2 (bn − an )3 = |In |3 . That h −mbn . for x ∈ In . That h (x−bn )) for every n is. f (x) = m for almost every x ∈ an . bn 2 (min(m(an − x). m(x − bn ))) dx an an −m(bn −an )/(M −m) In Z M 2 (x − an )2 dx = an Z bn m2 (bn − x)2 dx + bn −M (bn −an )/(M −m) = M2 m(an −bn )/(M −m) Z x2 dx + m2 M (bn −an )/(M −m) Z 0 x2 dx 0 m2 M 2 m2 M 2 (bn − an )3 = |In |3 = 2 3(M − m) 3(M − m)2 with equality if and only if F (x) = min(M (x−an ). bn .52 Z F 2 (x) dx ≤ bn Z 2 (min(M (x − an ). . bn . in both cases we have Z F 2 (x) dx ≤ In and consequently Z 1 XZ F 2 (x) dx = 0 ≤ m2 M 2 3(M − m)2 2 ≤ X m2 M 2 |In |3 2 3(M − m) n∈N !3 X m2 M 2 |O|3 |In | = 3(M − m)2 F 2 (x) dx ≤ In n∈N m2 M 2 |In |3 3(M − m)2 n∈N 2 m M . where f0 and f1 are the functions defined in the statement of the proposition. Now.M. 3(M − m)2 3 P P where we used the well-known inequality n∈N λ3n ≤ n∈N λn . Italy. we have s bn0 +k+1 1 bn0 +k+1 n0 +k+1 ln = exp an0 +k+1 n0 + k + 1 an0 +k+1 . thus. Batinetu-Giurgiu. Let (an )n≥1 .e.. we get ((n0 + k − 1)!)3 ((n0 + k − 1)!)3 k (a − ε) b ≤ b ≤ (a + ε)k bn0 n n +k 0 0 ((n0 − 1)!)3 ((n0 − 1)!)3 ((n0 + k − 1)!)2 ((n0 + k − 1)!)2 k (a − ε) a ≤ a ≤ (a + ε)k an0 n n +k 0 0 ((n0 − 1)!)2 ((n0 − 1)!)2 The computations are straightforward. Bucharest and Neculai Stanciu.53 So. From the definition of the limit of a sequence immediately 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. Moreover. the equality case can occur if and only if O = (0. Department of Mathematics. or f (x) = f1 (x) a. (bn )n≥1 be sequences of positive real numbers such that bn+1 an+1 = lim 3 = a > 0. Compute lim n→∞ n · bn n→∞ n2 · an s r ! bn+1 bn n+1 lim − n n→∞ an+1 an Solution 1 by Paolo Perfetti. Romania. The desired inequality is. 1) and f (x) = f0 (x) a. This completes the proof. Also solved by the proposer 24. proved. Buzau. Tor Vergata University. Rome. Proposed by D.e. It is well known (see[1] p. n −1 (n + 1)(zn+1 )1/(n+1) bn+1 an+1 n −1/(n+1) = 3 (zn+1 ) →e n bn n2 an n+1 n(zn )1/n Now s n+1 r (n+1)(zn+1 )1/(n+1) 1/(n+1) n − 1 1/n bn+1 n bn (n + 1)(z ) n(z ) n+1 n → e−1 . then the limit is 1/e. Therefore. using the fact that ex = 1 + o(1) when x → 0).46 for example) that zn being a sequence of positive numbers. We set zn = nnbnan . yields 1 bn0 +k+1 Ak exp ln (1 + o(1)) = n0 + k + 1 an0 +k+1 e Subtracting we obtain 1 bn0 +k+1 bn0 +k 1 exp ln ln − exp = n0 + k + 1 an0 +k+1 n0 + k an0 +k a+ε 1 A + o(1) = (1 + o(1)) = e a−ε e Since ε is arbitrarily small. Solution 2 by Anastasios Kotronis. Greece. limn→+∞ zn+1 zn = ` ∈ R ⇒ limn→+∞ (zn )1/n = `. namely bn0 +k+1 = (k + 1) ln A + ln B + an +k+1 0 1 1 + (n0 + k) ln(n0 + k) − (n0 + k) + ln(2π) + ln(n0 + k) + ln(1 + o(1)) 2 2 ln Thus. Athens.54 and (n0 + k)! (n0 − 1)! a−ε a+ε k+1 bn +k+1 (n0 + k)! bn 0 ≤ 0 ≤ an0 an0 +k+1 (n0 − 1)! a+ε Let us define A = a−ε and B = √ n n! = (n/e) 2πn(1 + o(1)). 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. ln − = (zn )1/n 1/(n+1) an+1 an n(zn )1/n ln (n+1)(zn+1 ) n(zn )1/n . Using Stirling’s formulae. 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. we get bn0 (n0 −1)!an0 a+ε a−ε k+1 bn 0 an0 . Let D. Math. Romania. x+1 x+1 and by the relation (R2 ) from Rec. M. 4CF C 0 ∼ 4AF A0 . = x.J. 2000. Switzerland. 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.55 since lim n→+∞ (n+1)(zn+1 )1/(n+1) n(zn )1/n ln −1 1/(n+1) (n+1)(zn+1 ) n(zn )1/n = 1/(n+1) n+1 ) exp ln (n+1)(z −1 n(zn )1/n lim 1/(n+1) n→+∞ n+1 ) ln (n+1)(z n(zn )1/n = ex − 1 =1 x→0 x lim References [1] W. a + ax yz(x + 1) x+1 x+1 Last inequality trivially holds. CA of ∆ABC. France. Buzau. Moubinool Omarjee. Proposed by Jos´e Luis D´ıaz-Barrero. pp.S.M. M. F be three points lying on the sides BC. It is easy to check that 4EBB 0 ∼ 4AEA0 . assume that points E. and 4DM D0 ∼ 4AM A0 . Real Numbers. the given inequality becomes ! a(xz + y) z y ≥ 4 ⇔ (xz + y)2 ≥ 4xyz ⇔ (xz − y)2 ≥ 0. 25.. we obtain that AM = MD ax x+1 ayz AM yz(x + 1) ⇔ = a · z + x+1 · y MD xz + y Hence.T. = . First. A. BARCELONA TECH. Barcelona. AB. Nowak Problems in Mathematical Analysis I. F are collinear. Sequences and Series . Let M be a point lying on cevian AD. Also solved by Albert Stadler. and the proposer. E. = y. F are collinear then show that BC · M D EA FA + ≥4 MA DC · EB BD · F C Solution 1 by Titu Zvonaru. we have EB BB 0 FC CC 0 MD DD0 = . Paris. Spain. We denote AF AE BD a = BC. Then. 2/2011. = 0 0 EA AA FA AA MA AA0 . M. M. = z. Comanesti and Neculai Stanciu. 108. and the proof is complete. Solution 2 by the proposer. F are collinear then holds MD EB FC BC · = DC · + BD · MA EA FA Indeed. DC = . George Emil Palade Secondary School. Kaczor. DC FC EB We have ax a BD = . If E. E. Satulung. We draw the parallel to B 0 C 0 that cuts line DD0 at D00 and line BD DD00 = from which follows CC 0 at C 00 . the same result is obtained when BB 0 > CC 0 . c ∈ R. observe that equality holds when 4ABC is equilateral and D. 1} → R. Sydney. Higher Institute for Applied Sciences and Technology. DD0 = DD00 + D00 D0 = BD BD · CC 0 BD (CC 0 − BB 0 ) + BB 0 = + BB 0 1 − BC BC BC BB 0 BD · CC 0 CD · BB 0 BD · CC 0 + (BC − BD) = + BC BC BC BC and the statement follows. Suppose that BB 0 < CC 0 . Australia. Omran Kouba.56 and the statement becomes A C' F M E B' A' C'' D' D'' B D C Figure 1. = Also solved by Codreanu Ioan-Viorel. F are the vertices if its medial triangle. Damascus. 26. x 1−x where a. Proposed by Enkel Hysnelaj. . Finally. Since 4BDB 00 ∼ 4BCC 00 . Maramures. University of Technology. Syria. then CC 00 BC BD BD DD00 = CC 00 = (CC 0 − BB 0 ) BC BC Thus. which satisfy the relation x−1 1 f +f = ax2 + bx + c. Romania. Likewise. Problem 25 DC · BB 0 CC 0 DD0 + BD · = BC · AA0 AA0 AA0 or DC · BB 0 + BD · CC 0 = DC · DD0 Now we distinguish two cases according to BB 0 < CC 0 or BB 0 > CC 0 . Determine all functions f : R − {0. and the proposer. b. Adrian Naco. Tchebychef proved (1850) that there exist constants a and b such that x x a < π(x) < b . letting y = x−1 x . 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 Similarly. Omran Kouba. f (y) + f and substituting for x we get 2 y−1 1 1 =a +b +c y 1−y 1−y (2) Adding (1) and (2) gives 2 2 y−1 1 1 y−1 1 y−1 +b +b +f =a +a +2c 2f (y)+f y 1−y y y 1−y 1−y Since. USA. Seguin. Moubinnol Omarjee. Stone. TX. 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. Paris. Georgia Southern University. which concludes the proof. lnx lnx for x sufficiently large. and the proposer 27. Solution by the proposer. Thus x x xa lnx < xπ(x) < xb lnx So 1 x lnx ax 1 bx < xπ(x) < x lnx Therefore eax < xπ(x) < ebx . 2 and the result follows by setting y = x. Let1 ting y = 1−x . Proposed by David R.57 Solution by Valmir Bucaj. Texas Lutheran University. Higher Institute for Applied Sciences and Technology. Also solved by Albert Stadler. show that there exist constants a and b such that eax < xπ(x) < ebx for x sufficiently large. f y 1−y after substituting in the preceding. 1 y−1 +f = ay 2 + by + c. With π(x) = the number of primes ≤ x. Damascus. Switzerland. GA. Albania. Syria. Statesboro. . France. a f (y) = 2 " y−1 y 2 + 1 1−y # 2 −y 2 b + 2 y−1 y + 1 1−y c −y + . School Cioculescu Serban. by M. Higher Institute for Applied Sciences and Technology. Switzerland 28 Proposed by Florin Stanescu. . Romania. (3) √ √ p p−a p p−c p p−b where 1−t f : (0. 1) → R. Math. Comanesti and Neculai Stanciu. y =1− . let us know that the problem solved above. Prove that p c b a √ +√ +√ ≥ 2 3p. Romania. so √ x+y+z 1 f (x) + f (y) + f (z) ≥ 3f = 3f = 2 3. Bruno Salgueiro Fanego. [BC] = a. (which is reproduced in the book Introduction to analytic and probabilistic number theory by Gerald Tenenbaum). Apostol’s Introduction to analytic number theory and K. Viveiro. Monthly. y and z by a b c x=1− . f (t) = √ = t−1/2 − t1/2 t The function f is convex since it is the sum of two convex functions.58 Comment by the Editors. (which is reproduced on the analytic number theory books like Tom M. Athens. Chandrasekharan’s book with the same title). [AC] = b. Also solved by Albert Stadler. Chebyshev in Memoire sur les nombres premiers. 89. Pures et Appl. Solution by Omran Kouba. Damascus. Let ABC be a triangle with semi-perimeter p. Buzau.also it has been given an elementary solution using different methods. Journal de Math. and the proposer. 3 3 (4) and the desired inequality follows from (3) and (4). L. Also solved by Albert Stadler. Amer. apart from its first solution given by P. Nair in the article On Chebyshev-type inequalities for primes.126-129. Greece. jud. 17 (1852). no. 2. p−a p−c p−b where [AB] = c. Syria. Anastasios Kotronis. Spain. Dambovita. George Emil Palade Secondary School. p. Gaesti. 366-390. z =1− p p p Clearly we have x + y + z = 1 and b c a +√ √ +√ √ = f (x) + f (y) + f (z). Titu Zvonaru. Switzerland. Let us define the positive real numbers x. Show that s s r r a + 3 4a + 3 a + 3 4a + 3 3 a + 1 3 a + 1 + + − 2 6 3 2 6 3 is an integer and determine its value. Let a > −3/4 be a real number. The source of the proposals will appear when the solutions be published. α. Proposals 21.59 MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. 22. a3 . a2 . Proposals are always welcomed. Calculate . Let a1 . a4 be nonzero real numbers defined by ak = (1 ≤ k ≤ 4). β ∈ R. . 1 + a21 + a22 a1 + a2 + 1/a4 . . a1 + a2 + 1/a4 2 + 1/a24 . . 1 + a2 (a1 + a3 ) a2 + a3 + 1/a4 sin(kβ + α) . sin kβ 1 + a2 (a1 + a3 ) a2 + a3 + 1/a4 1 + a22 + a23 . . . . . . . y. then show that it contains one subset of four distinct elements whose product is the fourth power of an integer. 24y − 36x2 = 1. 25. Prove that s a b c 1 r 3 ≤ ≤ 2r + R b + c + 2a c + a + 2b a + b + 2c 4 . Find all triples (x. pn and |A| > 3 · 2n + 1. b. c be the lengths of the sides of a triangle ABC with inradius r and cicumradius R. . 20z − 16y 2 = 9. If the prime divisors of elements in A are among the prime numbers p1 . 24. Let a. . 23. z) of real numbers such that 12x − 4z 2 = 25. . p2 . Let A be a set of positive integers. we get the number 503 in base 7 and 305 in base 9. A regular convex polygon of L + M + N sides must be colored using three colors: red. and our claim is proven. then M + N − L + 1 ≤ N + 1. third.60 Solutions 16. Therefore. 4. from the starting point in a circular sense until complete L red sides. and to write it in base 9 only the integers form 0 to 8 are used. remains to be colored L − 1 non consecutive sides and K − (2L − 1) = M + N − L + 1 ≥ 1 consecutive sides. 2 Also solved by Bruno Salgueiro Fanego. L + N ≥ M and M + N ≥ L. then y = 0 and 3x = 5y. x. y. we obtain the number 000 which does not have three digits. We claim that these conditions that are necessary are also sufficient.. (XI Spanish Math Olympiad (1973-1974)) Solution by Jos´ e Luis D´ıaz-Barrero. Give the necessary and sufficient conditions. Barcelona. L + M ≥ N. A number of three digits is written as xyz in base 7 and as zxy in base 9. L + M > N > 0. The number xyz in base 7 represents the number x · 72 + y · 7 + z in base 10. Let K = L + M + N. 3. the non consecutive L − 1 sides are colored yellow or blue until to complete the M yellow sides and the N blue sides. fifth. We begin coloring in red sides first. . then it must be K −1 K −1 K −1 . BARCELONA TECH. independently of the parity of K. Barcelona. BARCELONA TECH. On the other hand. using inequalities. Consequently. without painting two consecutive sides of the same color. 2 2 2 That is.. and from the second. 6}. 6. Then. Viveiro. 5. then it must be K K K M≤ . the number zxy in base 9 represents the number z · 92 + y · 9 + x in base 10. (III Spanish Math Olympiad (1965-1966)) Solution by Jos´ e Luis D´ıaz-Barrero. Finally. Spain. Spain. x · 72 + y · 7 + z = z · 92 + y · 9 + x from which follows 8(3x − 5z) = y. in such a way that L sides must be red. In the first case. etc. 17. Indeed. Since x.. 2. We distinguish two cases: (a) K is even. L + N > M > 0 and M + N > L > 0. z are integers ranging form 0 to 6. To write a number in the system of base 7 the only integers used are 0. 1. Find the number in base 10. y. Spain. M yellow and N blue. 5. Both numbers in base 10 are the number 248 and this is the answer. yellow and blue. WLOG we can assume that L ≥ M ≥ N. 3. (b) If K is odd. From the preceding. 1. 0<M ≤ . 2. we have x = z = 0 or x = 5 and z = 3. z ∈ {0. N≤ L≤ . 4. therefore this set of consecutive sides can not be colored alternatively yellow-blue-yellow. Since L ≥ M.. to obtain a colored polygon with no two consecutive sides of the same color. 0<N ≤ 0<L≤ 2 2 2 That is. we have AD · CN + AC · N D = AN · CD Since N is the midpoint of the arc CD. Spain. yields AC + AD AN = BN BD + BC Applying the bisector angle theorem. Spain. 18. Technical University of Catalonia (BARCELONA TECH). Let M and N be the midpoints of the arcs AB and CD which do not contain C and A respectively. then show that AC + AD AP = BP BC + BD (IMAC 2011) Solution by Ivan Geffner Fuenmayor. BARCELONA TECH. Problem 18 AN · CD = (AC + AD)x. 2 . If M N meets side AB at P. then we have CN = N D = x. Applying Ptolemy’s theorem to the inscribed quadrilateral ACN D. considering the inscribed quadrilateral CN DB we have BN · CD = (BD + BC)x. and C N D A P B M Figure 2. Barcelona. Let ABDC be a cyclic quadrilateral inscribed in a circle C. Likewise. Barcelona. Dividing the preceding expressions.61 2 Also solved by Jos´ e Gibergans B´ aguena. we have AN AP = BN BP This completes the proof. First. RK+1 = RK + S + 1 = (K + P + 1) + S + 1 = (K + 1) + (P + S) + 1. BARCELONA TECH. Let R be an arbitrary convex plane.62 Also solved by Omran Kouba. find (with proof ) the number of disjoint regions created. the chord A0 Ak passes through (k − 1)(n − k) + 1 regions. The chords A0 A1 and A0 An add two regions. Finally. . then the number of disjoint regions created inside R is RL = L + P + 1. Consequently. on account of A(K). . choose some line ` ∈ C. Suppose that we have an regions obtained on placing the n points A1 . . then P = 0. Barcelona. Higher Institute for Applied Sciences and Technology. Barcelona. An in this order. the number of regions determined by the K + 1 lines is. . Hence the number of new regions is S + 1. 2 . A(K +1) holds and by the PMI the claim is proven. drawing the chord . Spain. Damascus. Fix some K ≥ 0 and suppose that A(K) holds for K lines and some P ≥ 0 with RK = K + P + 1 regions. Since one draws a (K + 1)−st line `. and let us add the n + 1st point A0 on the arc An A1 that does not contain any other point. 4 2 2 Solution 2 by Omran Kouba. (IMAC-2011) Solution 1 by Jos´ e Gibergans B´ aguena. and Jos´ e Gibergans B´ aguena. When no lines intersect R. Higher Institute for Applied Sciences and Technology. To prove the preceding claim. Spain. Therefore. For each L ≥ 0. BARCELONA TECH. since the circle is convex and any intersection point is determined by a n n unique 4−tuple of points. let A(L) be the statement n region in the o L that for each P ∈ 1. Hence. starting outside R. if L lines that cross R. Damascus. Let S be the number of lines intersecting ` inside R. And for 1 < k < n. a2 = 2 and a3 = 4. and so. R0 = 0 + 0 + 1 = 1 and A(0) holds. . . 2. with P intersection points inside R. Syria. we argue by Mathematical Induction on L. Syria. Let us denote by an the number of disjoint regions created. where P +S is the total number of intersection points inside R. If no three chord are concurrent. then there are P = intersection points and L = 4 2 n n chords and the number of regions is R = + + 1. we prove that if a convex region crossed by L lines with P interior points of intersection. there are (k −1)(n−k) points of itersection of the chord A0 Ak with the other chords. Hence. Place n points on a circle and draw in all possible chord joining these points. Consider a collection C of K + 1 lines each crossing R (not just touching). . Clearly a1 = 1. a new region is created when ` first crosses the border of R. then the number of disjoint regions created is RL = L+P +1. and apply A(K) to C\{`} with some P intersection points inside R and RK = K + P + 1 regions. and whenever ` crosses a line inside of R. 19. . Barcelona. 1). 1+x 1 . 1) → R defined by f (x) = ln x 1−x ∞ ∞ X X x2k kx2k−1 Since f (x) = 2 for |x| < 1. BARCELONA TECH.63 A0 Ak adds (k − 1)(n − k) + 1 new regions. 20. Prove that s 3 1+a b+c 1−a bc 1+b c+a 1−b ca 1+c a+b 1−c ab ≥ 64 (J´ozsef Wildt Competition 2009) Solution by Jos´ e Luis D´ıaz-Barrero. and f is increasing and convex. . then f 0 (x) = 4 (|x| < 1) and 2k + 1 2k + 1 k=0 k=0 ∞ X k(2k − 1)x2k−2 00 f (x) = 4 (|x| < 1). c be positive real numbers such that a + b + c = 1. an n−1 X k k+2 −2 3 2 k=1 k=1 n−1 n−1 X X k + 1 k 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 = 1+ n−1 X 2 which is the required number of regions. Let a. Barcelona. BARCELONA TECH. b. Spain. f 0 (x) > 0 and f 00 (x) > 0 for all 2k + 1 k=0 x ∈ (0. Therefore. Also solved by Jos´ e Luis D´ıaz-Barrero. Spain. Thus an+1 = an + n X ((k − 1)(n − k) + 1) k=1 But n X ((k − 1)(n − k) + 1) = − k=1 n X k 2 + (n + 1) k=1 = = n X k + (1 − n)n k=1 n(n + 1)(2n + 1) n(n + 1)2 + − n(n − 1) − 2 6 n+2 n −2 3 2 So. Consider the function f : (0. ≤ ≤ and g(a) ≥ g(b) ≥ a b c 1+x . 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 1+a 1+b 1+c 1+a 1+b 1+c 3 3 = b+c c+a a+b 1−a 1−b 1−c s 1/a 1/b 1/c 1+a 1+b 1+c 3 ≥ ≥8 1−a 1−b 1−c Multiplying up the preceding inequalities yields. Equality holds when a = b = c = 1/3. and we are done. applying rearrangement inequality again. 3 ln a+b+c 3 + (a + b + c) 3 − (a + b + c) " 1/a 1/b 1/c # 1 1+a 1+b 1+c ≤ ln + ln + ln 3 1−a 1−b 1−c Taking into account that a + b + c = 1 and the properties of logarithms. s 1+1 1+1 1+1 1+a b c 1+b c a 1+c a b 3 ≥ 64 b+c c+a a+b from which the statement follows. where g is the increasing function defined by g(x) = ln 1−x rearrangement’s inequality. Spain. . 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 1+a 1+b 1+c 1+a 1+b 1+c 3 3 = b+c c+a a+b 1−a 1−b 1−c s 1/a 1/b 1/c 1+a 1+b 1+c 3 ≥ ≥8 1−a 1−b 1−c Likewise. Barcelona. 2 Also solved by Jos´ e Gibergans B´ aguena. We have.64 Applying Jensen’s inequality. we get s 1/a 1/b 1/c 1+a 1+b 1+c 3 ≥8 (5) 1−a 1−b 1−c 1 1 1 WLOG we can assume that a ≥ b ≥ c. we have f a+b+c 3 ≤ f (a) + f (b) + f (c) or equiva3 lently. Applying g(c). BARCELONA TECH. . we have n X k=1 ak x + bk √ √ a1 a2 an x+b1 . . Then ! n ! !2 n n X X X ak ak ak ≥ (x + bk )2 x + bk k=1 k=1 k=1 !4 . 1. Introduction In [1] the following problem was posed: Let a. ~u = x+b1 x+b2 x+bn and ~ into CBS inequality again and we get !2 ! n ! n n X X X ak ak ≤ ak (x + bk ) x + bk k=1 k=1 from which the statement immediately follows. a2 . bk . x+bn √ !2 = k=1 and ~v = !2 √ n X ak √ ak ≤ x + bk k=1 √ n X k=1 √ √ a1 . an (x + bn ) . n !2 n X X ≥ ak ak (x + bk ) k=1 Proof. . . . In this note a constrained inequality is generalized and some refinements and applications of it are also given. . To proveRHS inequality we set q LHS inequality q p p p a1 a2 an v= a1 (x + b1 ). . (1 ≤ k ≤ n) be positive real numbers. . . . an into CBS ! ak n X k=1 ak (x + bk )2 ! and the isq proven. Main Results In the sequel some generalizations and refinements of (6) are given. Let x and ak . b. . Our aim in this short paper is to generalize it and to give some of its applications. Setting ~u = inequality. . . We begin with Theroem 1. . Prove that b c 3 a + + ≥ (6) (ab + bc + ca) 2 b + b c2 + c a2 + a 4 A solution to the preceding proposal and some related results appeared in [2]. . k=1 A constrained inequality that can be derived immediately from the preceding result is given in the following . 2. .65 MATHNOTES SECTION On a Discrete Constrained Inequality ´ ly Bencze and Jose ´ Luis D´ıaz-Barrero Miha Abstract. c be positive real numbers such that a + b + c = 1. . a2 (x + b2 ). x+b2 . 66 Pn Corllary 1. Let ak . From Theorem 1. bk . Then holds: !2 ! n n n X X X a a k k ≥2 + ak (1 + bk )2 (1 + bk )2 1 + bk k=1 k=1 k=1 Proof. bk . Let y < z and ak . Setting x = 1 in Theorem 1. (1 ≤ k ≤ n) be positive real numbers such that k=1 ak = 1. (1 ≤ k ≤ n) be strictly positive real numbers. Theroem 2. Then exists c ∈ (y. z) such that ! n ! !2 n n X X X ak ak ≥ ak (y + bk )(z + bk ) c + bk k=1 k=1 k=1 !# !4 . Corllary 2. (1 ≤ k ≤ n) be positive real numbers. Then ! ! n n X X ak ak (y + bk )(z + bk ) k=1 k=1 i aj b2a−b n 2 X Y j i ak (y + bj )(z + bi ) ≥ + log (y + bk )(z + bk ) (y + bi )(z + bj ) 1≤i<j≤n k=1 ≥ n X !4 ." ! n 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 . we get ! n n X X 2 ak (1 + bk ) k=1 k=1 ak (1 + bk )2 and n X ! n X ak (1 + bk )2 k=1 k=1 ak 1 + bk ! ≥1 !2 ≥1 Adding up the preceding inequalities the statement follows. n !2 Z z n X X ≥ ak ak (x + bk ) dx." ak y k=1 n X k=1 ak + n X ! ak bk k=1 z n X k=1 ak + n X !# ak bk k=1 Proof. y k=1 k=1 After a little straightforward algebra the statement follows and the proof is complete. Let 0 ≤ y < z and ak . bk . we have ! !2 ! n Z z X Z z X n n X ak ak dx ≥ dx ak (x + bk )2 x + bk y y k=1 k=1 k=1 !4 . Notice that this result is a generalization and refinement of the inequality posed in [1]. b. c positive numbers with sum one is ab + bc + ca ≤ 13 (a + b + c)2 ≤ 13 and corollary 4. Applying Lagrange’s Mean Value Theorem to the function !2 !2 !2 Z x X Z z X n n n X ak ak ak f (x) = dt yields. Then 2ai aj X Y aj+1 (1 + ai+1 aj+1 −ai+1 X a21 a1 ≥ + log a2 (1 + a2 ) a2 (1 + a2 ) ai+1 (1 + aj+1 cyclic 1≤i<j≤n cyclic 1 ≥ X cyclic a1 a2 1 + X a1 a2 cyclic Proof. Taking into account that for all a. 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 ! 9 a(1 + c) a−c b(1 + a) b−a c(1 + b) c−b ≥ . (1 ≤ k ≤ n) and an+1 = a1 into the preceding corollary the statement follows. c be positive numbers of sum one. bk .67 Proof. Applying again Theorem 2 with y = 0 and z = 1. Indeed. we get Pn Corllary 3. b. Let ak (1 ≤ k ≤ n) be positive real numbers such that k=1 ak = 1. (1 ≤ k ≤ n) be positive real numbers such that k=1 ak = 1. Let a. Let ak . dt = (z −y) t + bk t + bk c + bk 0 y k=1 k=1 k=1 Putting this in Theorem 2 the inequality claimed follows and this completes the proof. for n = 3 we have Corllary 5. + log c(1 + a) a(1 + b) b(1 + c) 4 Proof. Setting bk = ak+1 . Then 2ai aj n n Y bj (1 + bi ) bj −bi X X a2k ak ≥ + log bk (1 + bk ) bk (1 + bk ) bi (1 + bj ) k=1 1≤i<j≤n k=1 1 ≥ n X ! ak bk 1+ k=1 n X ! ak bk k=1 Pn Corllary 4. 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) . Vol. Problem 2.68 ≥ 1 9 ≥ (ab + bc + ca)(1 + ab + bc + ca) 4 Finally. Solutions: Problem 3062. c be positive real numbers. 1 + y2 1 + z2 1 + x2 4 two applications are given. 2 Setting in the expressions of x. Dospinescu. y Solution. Brasov. CRUX. . ACKNOWLEDGEMENTS The authors would thank to the Ministry of Education of Spain that has partially supported this research by grant MTM2009-13272. No. combining the inequality posed in [1] by Dospinescu and the inequality presented in [2] by Janous. rb . using the sides a. 32. we have the following inequalities similar to the ones appeared in [3]. Prove that X X a 3(a + b + c) a ≥ ≥ 2 b(a + 2b + c) 4(ab + bc + ca) b + (a + b + c)2 cyclic cyclic a a+b+c . 6 (2006) 403-404. 31. z the elements of a triangle ABC and applying the previous procedure new inequalities for the triangle can be derived. b. c and the radii of ex-circles ra . 337. Let ABC be a triangle. namely y z 3 x + + ≤ (xy + yz + zx) (x + y + z = 1). Problem 3062. Prove that X X a 3s a (1) ≥ ≥ . b(2s + b) 2(s2 + r2 + 4rR 4s2 + b2 cyclic cyclic X X ra 3s ra (2) ≥ ≥ . References [1] G. rb (4R + r + rb ) 4r(4R + r) rb2 + (4R + r) cyclic cyclic where the notations are usual. [2] J. Dospinescu. rc . Problem 1. No. Bencze. L. Vol. D´ıaz-Barrero and G. For instance. Let a. CRUX. [3] M. Putting x = X X xy cyclic cyclic c and x = a+b+c into X X 3 x x ≥ ≥ xy y(1 + y) 4 1 + y2 = b a+b+c cyclic cyclic the statement follows. 1982. Inequalities (manuscript). b. 5 (2005) 335. y. If A. R 1 c): f 0 (1) 2 (f (1) − f (0)) − f 0 (1) 0 (f 0dx =1 (x))2 c 2010 Mathproblems. Pages 69–90 Editors: Valmir Krasniqi. b): f 0 is increasing and strictly positive. Teachers can help by assisting their students in submitting solutions. Gaesti. 1]. ∀x ∈ [0. Paolo Perfetti. Athens. Romania. Romania. Proposed by Florin Stanescu. B ∈ M2 (R) then prove that 2 2 2 (det A) + det(AB + BA) + 2 (det B) ≥ det A2 − B 2 + 4 det AB 38. Ovidiu Furdui. 69 . Roberto Tauraso.com Solutions to the problems stated in this issue should arrive before 2 June 2012 Problems 36. Stone. Jos´ e Luis D´ıaz-Barrero. Solutions will be evaluated for publication by a committee of professors according to a combination of criteria. Issue 2 (2012). Valmir Bucaj. Questions concerning proposals and/or solutions can be sent by e-mail to: mathproblems-ks@hotmail. PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets. Proposed by Mih´ aly Bencze. Proposed by Anastasios Kotronis. The editors encourage undergraduate and pre-college students to submit solutions. Dambovita. each indicating the name of the sender. Universiteti i Prishtin¨ es.Mathproblems ISSN: 2217-446X. An asterisk (*) indicates that neither the proposer nor the editors have supplied a solution. 1] → R that have the following properties a): f is three times differentiable with f 000 (x) ≥ 0. Mih´aly Bencze. Enkel Hysnelaj. Determine all functions f : [0.com Volume 2. Armend Sh. Kosov¨ e. Prishtin¨ e. Greece. jud.mathproblems-ks. Drawings must be suitable for reproduction. Bra¸sov. Shabani. Cristinel Mortici. J´ozsef S´andor. Proposals should be accompanied by solutions. School Cioculescu Serban. David R. Student solutions should include the class and school name. url: http://www. Evaluate the sum +∞ X 1/2 1 1/2 1 1 −1/2 n+1 1/2 n 2 −1+ − + · · · + (−1) n 1 2 2 4 n 2 n=1 37. Tor Vergata University. Prove that f is continuous. then prove that: ! sink A + sink B + sink C 1 sink A sink B sink C . University of Technology. 41. Romania. Sydney. Evaluate Z π/2 2 4(cos x)2 (ln(cos x)) dx 0 42. such that x3 y3 z3 t3 + + + x−1 y−1 z−1 t−1 is an integer. 1]. Proposed by Cristinel Mortici. B and C are the angles of a triangle. If A. Rome. Let f : R → R be monotone such that f + f ◦ f ◦ f is continuous. Buzˇ au. Find all distinct positive integers x. Australia. z.70 39. t all greater than 2. Proposed by Paolo Perfetti. Proposed by Neculai Stanciu. . Valahia University of Tˆ argovi¸ste. Department of Mathematics. Proposed by Enkel Hysnelaj. Romania. 40. y. Italy. ≤ + + A+B+C 3 A B C where k ∈ (0. George Emil Palade Secondary School. We will be very pleased considering for publication new solutions or comments on the past problems. xn ) n for any (x1 . ∞). Australia. . 1) On account of the condition in the statement. Ulaanbaatar. so show that if f (x21 ) + f (x22 ) = f (x1 x2 ) 2 for any (x1 . y) ∈ (0. ∞). . 1) Let f : (0. 29. f (x3 ) + f (y 3 ) + f (z 3 ) = f (xyz) 3 2) First. University of Technology. when n = 2 it trivially holds. then f (xn1 ) + f (xn2 ) + . . Solution 1 by AN-anduud Problem Solving Group. x2 ) ∈ (0. we prove the statement for n = 2k arguing by mathematical induction. . xn ) ∈ (0. Proposed by Enkel Hysnelaj. 2) Generalize the above statement.71 Solutions No problem is ever permanently closed. . Sydney. ∞). x2 . Show that f (x3 ) + f (y 3 ) + f (z 3 ) = f (xyz) 3 for any (x. Indeed. . Now suppose that for n = 2k is true and . y. ∞) → R be a function that satisfies the property f (x2 ) + f (y 2 ) = f (xy) 2 for any (x. . Mongolia. z) ∈ (0. + f (xnn ) = f (x1 x2 . . ∞) and n a positive integer greater than 1. we have f (x3 ) + f (y 3 ) + f (z 3 ) + f (xyz) 4 3 3 2 3 3 2 )2 2 )2 2 2 )2 f (z f (x + f (y ) + f ((xyz) 1 = + 2 2 2 3 3 1 1 f (xy) 2 + f z 2 (xyz) 2 2 3 3 1 1 = f ((xy) 4 )2 + f (z 4 (xyz) 4 )2 2 = 3 3 1 = f (xy) 4 · z 4 · (xyz) 4 = f (xyz) Hence. . it is a straightforward task to show by induction on n that for every integer n > 1 and for all . y) ∈ (0. Solution 2 by Omran Kouba. . . ∞) → R be a function that satisfies f (x2 ) + f (y 2 ) = f (xy) (1) 2 We will consider two cases according to the value of f (1) : (a): f (1) = 0. So. We prove directly generalization 2). Then f (xn1 ) + · · · + f (xnn ) + f (xnn+1 ) + · · · + f (xnN ) N n n n n N N N N N ) ) f ((x1 ) ) + · · · + f ((xnN )N ) + f ((xn+1 )N ) + · · · + f ((xN = N n n = f (x1 · · · xn ) N · (xn+1 · · · xN ) N N −n n n = f (x1 · · · xn ) N · (x1 · · · xn ) n · N = f (x1 · · · xn ) Hence. for all (x1 . ∞)2 . Damascus. . . . +∞)n . choosing x1 = x2 = · · · = xn = t we get nf (t) = f (tn ) for every t > 0. we have 1 2k+1 k+1 2X k+1 f (x2i ) = 2k 1 X1 2k+1 2k+1 f (x ) + f (x ) k i 2 +i 2k i=1 2 = 2 1 X 2k f (x · x ) k i 2 +i 2k i=1 i=1 k (by the hypothesis) = f (x1 x2 · · · x2k+1 ) Thus our claim is proved. Choosing y = 1 in (1) we see that f (x2 ) = 2f (x) for every x > 0. . ∀ (x. f (xn1 ) + · · · + f (xnn ) = f (x1 · · · xn ) n as we wanted to prove. . By the given condition. ∞)2 . (b): f (1) 6= 0. that f (x1 ) + f (x2 ) + · · · + f (xn ) = f (x1 x2 · · · xn ) Also. xn+2 .72 then we prove it for n = 2k+1 . Higher Institute for Applied Sciences and Technology. (x1 . ∞) → R defined by f (x) = f (x) − f (1). . + f (xnn ) = f (x1 x2 · · · xn ) n as desired. y) ∈ (0. let f : (0. Syria. xN such that xn+1 = xn+2 = · · · = xN = (x1 x2 · · · xn ) n . In this case we obtain the conclusion by applying the preceding case to the function f : (0. . xn ) ∈ (0. . Using this from (1) we have for all (x. . Now assume that 2k < n < 2k+1 = N and we choose 1 xn+1 . +∞)n is f (xn1 ) + f (xn2 ) + . xn ) ∈ (0. f (x) + f (y) = f (xy) Now. . . . Indeed. Polytechnic University of Tirana. University of Prishtina. Adrian Naco. Proposed by Neculai Stanciu. Evaluate Z 2012 x sinn t dt. We consider the following cases: (1) For n − m ≥ −1 we have Z bx Z bx n sinn t dt = tn−m 1 + O(t2 ) m t ax ax Z bx = tn−m 1 + O(t2 ) ax bx Z = = = tn−m + O tn−m+2 dt ax . Dambovita. Romania. Athens. More generally. Albania. Buzˇ au. Department of Mathematics. Department of Mathematics. let 0 < a < b and n. George Emil Palade Secondary School. m ∈ N. Serban Cioculescu School. lim x→0 2011 x tm where n.73 Also solved by Islam Foniqi. m ∈ N. Solution by Anastasios Kotronis. Prishtin¨ e. and the proposer 30. Republic of Kosova. Greece. Romania. Gaesti. Florin Stanescu. bx . bx . . n−m+1 . n−m+3 . n−m+1 + O . t t . n−m+3 . . . ax n−m≥0 ax . . bx bx ln |t|. . + O t2 . . tm ym ax a(−x) (1) (2) For n − m ≤ −2 we distinguish two cases: • If n − m is odd. n−m≥0 x→0 −−−→ . . n − m = −1 ax ax n−m+1 n−m+1 b −a xn−m+1 + O(xn−m+3 ). we get Z bx Z b(−x) sinn t sinn y n−m+1 dt = (−1) dy. n − m ≥ 0 n−m+1 b ln a + O(x2 ). n − m = −1 n − m = −1 Carrying out the change of variable t = −y. b ln a . ( 0. then for some 0 < ε < 1 and while x → 0+ we have sin t (1 − ε)n sinn t 1 ≤ 1 ⇒ m−n ≤ m ≤ m−n and t t t t Z bx n n−m+1 n−m+1 b − a sin t bn−m+1 − an−m+1 (1 − ε)n ≤ ≤ (n − m + 1)xm−n−1 tm (n − m + 1)xm−n−1 ax (1 − ε) ≤ . Damascus. in clockwise order. Texas Lutheran University.. and the proposer 31. are: 1 1 2 1 3 1 4 1 . Syria. Let P1 = √1a1 . n−m≥0 Z bx n b = ln a . √ . show that the area of this polygon is 6 1 (n + 3)(n − 2) n(n + 1) A= √ − √ +√ √ 4 a1 a2n+2 a2n an+2 Solution 1 by Omran Kouba. Tor Vergata University.. Then the area A of the polygon (P1 . Higher Institute for Applied Sciences and Technology. . Finally. Department of Mathematics. (Correction) Proposed by Valmir Bucaj. then similarly while x → 0+ we have Z bx sinn t bn−m+1 − an−m+1 bn−m+1 − an−m+1 ≤ ≤ m−n−1 m (n − m + 1)x t (n − m + 1)xm−n−1 ax R bx n t Thus limx→0+ ax sin tm dt = +∞ and from (1) Z bx Z bx n sin t sinn t lim− dt = lim − dt = −∞ tm tm x→0 x→0+ ax ax and the limit does not exist. √ . √a2n+2−k . Paolo Perfetti.74 R bx n t Thus limx→0+ ax sin tm dt = +∞ and from (1) Z bx Z bx sinn t sinn t lim dt = lim dt = +∞ m t tm x→0− ax x→0+ ax • If n − m is even. .√ . Higher Institute for Applied Sciences and Technology..√ . Syria. Rome..√ a1 an+1 a2 a2n a3 a2n−1 a4 a2n−2 n−1 1 n 1 . √a1n+1 and for 2 ≤ k ≤ n let 1 Pk = √kak .√ . √ . . √ . TX. . P1 Pk ) 2 k=2 . Seguin. Damascus. . . P2 . n − m ≤ −2 and n − m = odd does not exist. n − m ≤ −2 and n − m = even (1 − ε)n Also solved by Omran Kouba. If the vertices of a polygon. Italy. n − m = −1 sin t lim dt x→0 ax tm = +∞. . collecting yields = 0. Pn ) is given by n−1 −−−−−→ −−−→ 1X det(P1 Pk+1 . √ √ √ √ an−1 an+3 an an+2 where (an )n≥1 is a decreasing geometric progression. n−1 k+1 1 1 X . . √ak+1 − √a1 = . √ 1 . a2n+1−k − √a1n+1 2 A= k=2 . . − √1a1 . . 1 √ √ − a2n+2−k an+1 . √k ak 1 . 75 Noting that ak a2n+1−k = a1 a2n and ak+1 a2n+2−k = a1 a2n+2 we conclude that n−1 X k+1 k 1 k+1 k A= −√ −√ −√ √ √ a1 a2n+2 a1 a2n an+1 ak+1 ak k=2 1 1 1 −√ −√ √ a1 a2n+2−k a2n+1−k 1 n 2 n(n + 1) − 6 n(n − 1) − 2 − −√ = √ √ √ −√ 2 a1 a2n+2 2 a1 a2n an+1 an a2 1 1 1 −√ −√ √ a1 a2n an+2 But an+1 an = a1 a2n and a2 an+1 = a1 an+2 . So. Bra¸sov. We have Z f (b) − f (a) = a b . Albania. 32. be a two times differentiable function such that f 00 and f 0 are continuous. Tor Vergata University. Romania. the above formula simplifies to 1 (n + 3)(n − 2) n(n + 1) 6 A= √ − + √ √ √ 4 a1 a2n+2 a2n an+2 as claimed. Also solved by Adrian Naco. and the proposer. Department of Mathematics.b] 2 m (b − a ) M (b2 − a2 ) ≤ bf 0 (b) − af 0 (a) − f (b) + f (a) ≤ 2 2 Solution by Paolo Perfetti. Department of Mathematics. Italy. Polytechnic University of Tirana. If m = min f 00 (x) and M = max f 00 (x). then prove that x∈[a. (Correction) Proposed by Mih´ aly Bencze. b] → R. Let f : [a.b] 2 x∈[a. with 0 ≤ a < b. Rome. b Z . f (x)dx = xf (x). Serban Cioculescu School. Florin Stanescu. Department of Mathematics. Also solved by Omran Kouba. Angel Plaza. . Universidad de Las Palmas de Gran Canaria. − 0 b 0 a xf 00 (x) dx a from which follows 0 Z 0 b bf (b) − af (a) − f (b) + f (a) = xf 00 (x) dx a Moreover.b] b Z xf (x)dx ≤ max f (x) a a and then b2 − a2 ≤ m 2 00 Z a x∈[a. Spain. Damascus. Dambovita. Romania.Adrian Naco.b] b xf 00 (x)dx ≤ M b xdx a b2 − a2 2 concluding the proof. Gaesti. Higher Institute for Applied Sciences and Technology. Syria. 00 Z b Z 00 xdx ≤ min f (x) x∈[a. we have √ Z π4 2 lim yn = cos x dx = n→∞ 2 0 Finally. Polytechnic University of Tirana. | {z } | {z } n-times n-times where n ≥ 3. Department of Mathematics. Syria. Z lim n→∞ π 2 √ n √ sinn x + cosn x dx = 2 · 0 2 √ = 2 2 Also solved by Albert Stadler. Then. Greece. Proposed by Ovidiu Furdui. Romania. Romania. Damascus. Z π2 Z π4 fn (x)dx = 2 fn (x) dx 0 √ n sinn x + cosn x = 0 Let us denote by π 4 Z yn = Z fn (x)dx = 0 π 4 n 1 cos x (1 + (tan x) ) n dx 0 n 1 The functions gn (x) = cos x (1 + (tan x) ) n are continuous on 0. 1 1 π n n |gn (x)| ≤ cos x 1 + tan ≤ cos x · 2 n ≤ cos x · 2 ≤ 2 4 So. 33. Mongolia. . Tor Vergata University. Albania. University of Prishtina. Cluj. Rome. π4 and the se π quence (gn ) converges to the function x 7→ cos x on 0. Switzerland. and the proposer. . + 2 + x = x 2. Ulaanbaatar. . Prishtin¨ e. Department of Mathematics. Bra¸sov. Department of Mathematics. Find the value of Z π/2 √ n sinn x + cosn x dx lim n→∞ 0 Solution by Moubinool Omarjee. Anastasios Kotronis. Republic of Kosova. Adrian Naco. Albania. Solution by Islam Foniqi. We see that x has to be in (0. AN-anduud Problem Solving Group.76 Department of Mathematics. Omran Kouba. . Furthermore. Athens. + 2 + x + 2 − 2 + . Polytechnic University of Tirana. Paolo Perfetti. 4 as can be easily checked. Proposed by Mih´ aly Bencze. Italy. Let fn (x) = fn π2 − x . so we . 2). Solve the following equation r r q q √ √ √ 2 + 2 + . and the proposer 34. Higher Institute for Applied Sciences and Technology. Paris France. by the Dominated Convergence theorem. Higher Institute for Applied Sciences and Technology. such that P 0 has only real zeros. AN-anduud Problem Solving Group.77 p can take x = 2 cos y where 0 < y < π2 . ξk < 0. The roots of P 0 (x) = 0 are evidently all negative and P 0 (x) = nan n−1 Y (x − ξk ). Mongolia. Department of Mathematics. The equation y(1− 21n ) = − π4 does not have solution because n ≥ 3 and 0 < y < π2 . Proposed by Florin Stˇ anescu. jud. x = 2 cos y = 2 cos 4(22n +1) when n ≥ 3. . Rome. If 0 ≤ a < b show that Rb 1 0 dx 1 P (b) P 0 (b) − P 0 (a) a P 0 (x) ≥ ln ≥ Rb 1 0 b−a P (a) P (b) − P (a) dx 00 a P (x) Solution by Paolo Perfetti. Also solved by Omran Kouba. 1 ≤ k ≤ n − 1 k=1 Thus we have n−1 P 00 (x) X 1 = P 0 (x) x − ξk k=1 that it is a decreasing function as well as P 01(x) and P 001(x) . Using the formula 2(1 + cos 2a) = 2 cos a. Damascus. Tor Vergata University. Let P (x) = an xn + an−1 xn−1 + . . Italy. and the proposer 35. Romania. Therefore. Dambovita. but the equation y(1 + 21n ) = π4 has the solution n n π π y = 4(22n +1) which is clearly between 0 and π2 . Now 0 Z b 00 P (b) P (x) ln = dx 0 0 P (a) a P (x) Z b 00 Z b Z b P (x) 1 1 dx · dx ≤ (b − a) dx 0 00 0 a P (x) a P (x) a P (x) on account of Chebyshev’s inequality for integrals applied to decreasing (increasing) functions. School Cioculescu Serban. we have r q √ y 2 + 2 + · · · + 2 + x = 2 cos n−1 2 {z } | (n−1)−times and the given equation becomes r r √ y y 2(1 + cos n−1 ) + 2(1 − cos n−1 ) = 2 2 cos y 2 2 or √ y y cos n + sin n = 2 cos y 2 2 which can be written as y π cos( n − ) = cos y 2 4 Now 2yn − π4 = y or 2yn − π4 = −y which is equivalent to y(1 − 21n ) = − π4 or y(1+ 21n ) = π4 . with 0 < y < π2 . Gˇ ae¸sti. + a1 x + a0 be a polynomial with positive real coefficients of degree n ≥ 3. . Ulaanbaatar. This proves the LHS inequality. Syria. AN-anduud Problem Solving Group. while PP 0 (x) Z P 0 (x)dx · Z Comment by the Editor. Gazeta Matematicˇ a. Seria B. Also solved by Omran Kouba. b b Z b P 00 (x) dx ≥ (b − a) P 00 (x) dx 0 a a P (x) a and this follows also by applying Chebyshev’s result again with P 0 (x) increasing 00 (x) is decreasing. 3 (2012) 113–121. and the proposer . This problem has appeared as part of the following paper by the same author: Aplicat¸i ale inegalitˇa¸tii lui Cebi¸sev in formˇa integralˇa. Damascus. Mongolia. Syria. Anul CXVII. nr. Ulaanbaatar.78 The RHS inequality is actually R b 00 Z b 00 P (x) dx P (x) dx ≥ (b − a) Rab 0 a P (x) P 0 (x) dx a That is. Higher Institute for Applied Sciences and Technology. Equation x3 − 2x2 − x + 1 = 0 has three real roots a > b > c. Prove that the equation √ √ √ (x + y 3)4 + (z + t 3)4 = 7 + 6 3 does not have rational solutions. Show that there exist two unique points α. Prove that is an irrational number. 29. 27. Find the value of ab2 + bc2 + ca2 . Proposals are always welcomed. The source of the proposals will appear when the solutions be published. 1] such that f (α) = α and f (β) = 1 − β. an a n=1 n .79 MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. . lim n→+∞ a1 a2 . Find all polynomials p(x) with real coefficients such that p(a + b − 2c) + p(b + c − 2a) + p(c + a − 2b) = 3p(a − b) + 3p(b − c) + 3p(c − a) for all a. β ∈ [0. Let f : [0. 1] be a differentiable function such that |f 0 (x)| = 6 1 for all x ∈ [0. 28. c ∈ R. Let {an }n≥1 be a strictly increasing sequence of positive integers such that ∞ X 1 an+1 = +∞. 1] → [0. 30. 1]. . Proposals 26. b. Syria. Alexandros Sygkelakis. Spain. we get x + y = a + 1 √ √ and xy = −a3 /27. Viveiro. 22. Let a1 . (XXVI Spanish Math Olympiad 1989-1990) Solution by Bruno Salgueiro Fanego. Omran Kouba. a4 be nonzero real numbers defined by ak = k ≤ 4). α. we call z = 3 x + 3 y and cubing. Albania. β ∈ R. and Jos´ e Luis D´ıaz-Barrero. a3 . and Paolo Perfetti. University of Prishtina. Department of Mathematics. Department of Mathematics. Kosov¨ e. yields √ √ √ z 3 = x + y + 3 3 xy 3 x + 3 y = a + 1 − az or equivalently. Putting s s r r a + 3 4a + 3 a + 3 4a + 3 3 a + 1 3 a + 1 x= + and y = − 2 6 3 2 6 3 and adding up and multiplying up the preceding expressions. Adrian Naco. respectively. Tor Vergata University. Iraklion Crete. BARCELONA TECH. Barcelona. Department of Mathematics Faculty of Mathematical Engineering and Physical Engineering Polytechnic University of Tirana. Higher Institute for Applied Sciences and Technology. So. BARCELONA TECH. Let a > −3/4 be a real number. Show that s s r r a + 3 4a + 3 a + 3 4a + 3 3 a + 1 3 a + 1 + + − 2 6 3 2 6 3 is an integer and determine its value. then it does not have real roots. z 3 + az − (a + 1) = 0 ⇔ (z − 1)(z 2 + z + a + 1) = 0 Since the discriminant of z 2 + z + a + 1 = 0 is δ = −(3 + 4q) < 0. Calculate . Prishtin¨ e. Islam Foniqi. Rome. Barcelona. Italy. Damascus. s s r r a + 3 4a + 3 a + 3 4a + 3 3 a + 1 3 a + 1 + + − =1 2 6 3 2 6 3 2 and we are done. Now.80 Solutions 21. Greece. Also solved by Jos´ e Gibergans B´ aguena. a2 . Spain. Spain. . 1 + a21 + a22 a1 + a2 + 1/a4 . . a1 + a2 + 1/a4 2 + 1/a24 . . (1 ≤ sin kβ 1 + a2 (a1 + a3 ) a2 + a3 + 1/a4 1 + a22 + a23 . 1 + a2 (a1 + a3 ) a2 + a3 + 1/a4 sin(kβ + α) . . . . . . (J´ozsef Wildt Mathematics Competition 2005) . Spain. we evaluate .81 Solution by Jos´ e Gibergans B´ aguena. Barcelona. BARCELONA TECH. First. . . . . a1 . a1 1 a2 . 1 a2 . . . . . 1 . a4 1 a4 . ∆ = . . 1 1/a4 1 . . = . . a4 . . a2 1 a3 . a2 1 a3 . 1 = [a1 (a3 − a4 ) + a2 (a4 − a2 ) + a4 (a2 − a3 )] a4 Taking into account that ak − ah sin(kβ + α) sin(hβ + α) − sin kβ sin hβ sin(kβ + α) sinh β − sin(hβ + α) sin kβ = sin kβ sin hβ 1 cos[(k − h)β + α] − cos[(k − h)β − α] = 2 sin kβ sin hβ sin(k − h)β sin α = − sin kβ sin hβ = we have − sin α sin 3β sin(2β + α) − sin 2β sin(β + α) sin β sin(4β + α) − sin(4β + α) sin 2β sin 3β sin 2β sin 3β − sin α sin(4β + α) sin β sin(4β + α) sin β = − = 0. it is easy to see by direct calculations that . sin(4β + α) sin 2β sin 3β sin 2β sin 3β Now. . . 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 . . . 2 . a1 1 a2 . . . = . . 1 1/a4 1 . . = 0 . a2 1 a3 . x2 . . pn and |A| > 3 · 2n + 1. Omran Kouba. Spain. where xi . . Higher Institute for Applied Sciences and Technology. . . Viveiro. and we are done. BARCELONA TECH. Spain 23. . If the prime divisors of elements in A are among the prime numbers p1 . To each element a in A we associate an n-tuple (x1 . and Jos´ e Luis D´ıaz-Barrero. then show that it contains one subset of four distinct elements whose product is the fourth power of an integer. . Spain. Damascus. 2 ∆= Also solved by Bruno Salgueiro Fanego. . Barcelona. Let A be a set of positive integers. Barcelona. BARCELONA TECH. . Syria. xn ). p2 . (Training Sessions of Catalonia Team for OME 2012) Solution 1 by Jos´ e Luis D´ıaz-Barrero. The product of them is a perfect square √ √ ai1 ai2 · aj1 aj2 = x2 . This means that xyzt = π(B) π(B 0 ) is the square of an integer. we can select 2n + 1 such pairs or more. We will show that |G0 | > 2n . . Thus. √ √ √ √ a11 a12 . k 7→ (p1 (k). . Similarly. aj1 aj2 with the same 0 − 1 n-tuple. and the product √ of these two elements is then a square. B 7→ p1 ( π(B)). . Now. pn ( π(B)) 0 can not be injective. every subset of 2n + 1 elements of A contains two distinct elements say a11 . Remark. from the set A. This contradiction shows that |G0 | > 2n . Hence. . So. Then the subset A0 = ∪B∈G0 B has 2|G0 | ≤ 2n+1 elements. Likewise. That is. Indeed. b} is a good subset of P (2) (A) with |G1 | = 1 + |G0 |. where x is an integer. on the contrary. and 1 otherwise. 2 . by the PHP. pn (k)) can not be injective. These n-tuples are the “pigeons”. Syria. . p y} and pB = √ 0 {z. β} belongs to G then π(B) = αβ is a perfect square. the map p p p Ψ : G0 → {0. We will also use the notation p (a) to denote νp (a) mod 2. This is the desired conclusion. since |G0 | > 2n = |{0. a21 a22 . This implies that ab is a perfect square and consequently G1 = G0 ∪ {a. p2 (k). or equivalently that xyzt is a forth power.2 Since all the previous numbers have the same divisors. and there must be two distinct elements B = {x. the previous arguments give √ √ us two of them ai1 ai2 . . 1}n . 1}n |. suppose that. This contradicts the maximality of |G0 |. which has at least 3 · 2n + 1 elements. . . . Let P (2) (A) denote the collection of subsets of A each consisting of two elements P (2) (A) = {B ⊂ A : |B| = 2} We will say that a subset G of P (2) (A) is good if it satisfies the following two properties (i): If B and C are two distinct elements of G then B ∩ C = ∅. Consider a good subset G0 of P (2) (A) of maximal cardinality among good subsets of P (2) (A). a31 a32 . t} in G0 . . a12 with the same associated n-tuple. from the remaining numbers in A we can choose a21 and √ a22 such that a21 a22 is an integer. . the map Φ : A \ A0 → {0. Solution 2 by Omran Kouba. Higher Institute for Applied Sciences and Technology. a2n +1. a11 a12 is an integer. 1}n . Note that we only assumed that |A| ≥ 3 · 2n + 1. As usual. p2 ( π(B)). |G0 | ≤ 2n . (ii): If B = {α.1 a2n +1.82 is 0 if the exponent of pi in the prime factorization of a is even. and there must be two distinct elements a and b in A \ A0 such that Φ(a) = Φ(b). and consequently |A \ A0 | ≥ 3 · 2n + 1 − 2n+1 = 2n + 1. Consider the 2n + 1 integer numbers that are the square roots of products of the two elements of each pair. The “holes” are the 2n possible choices of 00 s and 10 s. In particular. for a positive integer a and a prime p we will write νp (a) to denote the largest integer α such that pα divides a. . such that Ψ(B) = Ψ(B ). ai1 ai2 aj1 aj2 = x4 and we are done. as all exponents are even. Damascus. Spain. Solution 2 by Jos´ e Gibergans B´ aguena. 5) does not satisfy the third equation 10v − u2 = 9. Setting 6x = t. Department of Mathematics Faculty of Mathematical Engineering and Physical Engineering . 20z = 9 + 16y 2 . 2 20z − 24y = 9 + 16y − 24y. Maramure¸s. Spain. 12x to the second. 24y − 36x2 = 1. 24y − 12x = 1 + 36x2 − 12x. Rome. But. Moreover 10v = 9 + u2 ≥ 6u ≥ 25 + v 2 =⇒ (v − 5)2 ≤ 0 which implies v = 5. Switzerland. triple (1/6. 12x − 20z = 25 + 4z 2 − 20z. 10v − u2 = 9 Since 1 + t2 ≥ 2t then we have 6u ≥ 25 + v 2 . y. Satulung. Subtracting 20z to both members of the first equation. v) = (25. BARCELONA TECH. and 24y to the third. 24y = 1 + 36x2 . z) of real numbers such that 12x − 4z 2 = 25. Tor Vergata University. 313/3. u. Adding up the last three equations. Omran Kouba. yields (1 − 6x)2 + (3 − 4y)2 + (5 − 2z)2 = 0 which is possible only when 1 − 6x = 0. 2 Also solved by Bruno Salgueiro Fanego. 4y = u.83 Also solved by Jos´ e Gibergans B´ aguena. Damascus. Syria. ⇔ 24y − 12x = (1 − 6x)2 . Herrliberg. Viveiro. First. This in turn implies t = 25 from the first equation and u = 313/3 from the second but the triple (t. Higher Institute for Applied Sciences and Technology. 20z − 24y = (3 − 4y)2 . we obtain 12x − 20z = (5 − 2z)2 . 5/2) does not satisfy the system and therefore it does not have solution. 3/4. Spain 24. we write the given system in the most convenient form 12x = 25 + 4z 2 . BARCELONA TECH. Department of Mathematics. 6u − t2 = 1. Romania. Albert Stadler. 5 − 2z = 0. 3 − 4y = 0. (Training Sessions for COM-2011) Solution 1 by Paolo Perfetti. Italy. Barcelona. So. the given system has no solutions. and Ioan Viorel Codreanu. Iv´ an Geffner Fuenmayor. Find all triples (x. Barcelona. Spain. Adrian Naco. 20z − 16y 2 = 9. Barcelona. BARCELONA TECH. 2z = v we get 2t − v 2 = 25. 2(a + b + c) and using the well-known inequality 9R2 ≥ (a2 + b2 + c2 ) we get 9R2 a2 + b2 + c2 (a2 + b2 + c2 )(a + b + c) 3R = ≥ =2 abc r 3Rr 3abc 3 2(a+b+c) Now it is enough to prove that (a2 + b2 + c2 )(a + b + c) a2 (b + c) + b2 (c + a) + c2 (a + b) ≥ 3abc abc which can be written as 2 2(a2 + b2 + c2 )(a + b + c) ≥ 3(a2 (b + c) + b2 (c + a) + c2 (a + b)) This inequality is equivalent to 2(a3 + b3 + c3 ) ≥ a2 (b + c) + b2 (c + a) + c2 (a + b) Using Schur inequality a2 (b + c) + b2 (c + a) + c2 (a + b) ≤ 3abc + a3 + b3 + c3 it is enough to prove that 2(a3 + b3 + c3 ) ≥ 3abc + a3 + b3 + c3 ⇔ a3 + b3 + c3 ≥ 3abc . Spain. a) First we prove that L ≤ M . Barcelona. Let L be the left hand side. c be the lengths of the sides of a triangle ABC with inradius r and cicumradius R. BARCELONA TECH.84 Polytechnic University of Tirana. Prove that s a r b c 1 3 ≤ ≤ 2r + R b + c + 2a c + a + 2b a + b + 2c 4 (Training Sessions of Spanish Math Team for IMO 2011) Solution 1 by Islam Foniqi. Indeed. M the middle and P the right side of the inequality. Let a. University of Prishtina. Albania. r 1 2r + R 1 3 (b + c + 2a) (c + a + 2b) (a + b + 2c) ⇔ ≥ L≤M ⇔ ≥ L M r a b c From AM − GM we have r 1 b + c + 2a c + a + 2b a + b + 2c 3 (b + c + 2a) (c + a + 2b) (a + b + 2c) ≤ ( + + ) a b c 3 a b c So its enough to prove that 2r + R b + c + 2a c + a + 2b a + b + 2c 3R a+b b+c c+a 3· ≥ + + ⇔ ≥ + + r a b c r c a b Since a+b+c abc S4ABC = r = . 2 4R then we obtain abc Rr = . Prishtin¨ e. 25. b. Kosov¨ e. and Jos´ e Luis D´ıaz-Barrero. Department of Mathematics. First. Damascus. b) Now we prove that M ≤ P. if and only if a = b = c. from a) and b) we have proved that L ≤ M ≤ P as required. √ √ Using the AM-GM inequality. On account of the AM − GM inequality we have 1 a b c 1 a(b + x)(c + x) + b(c + x)(a + x) + c(a + x)(b + x) M≤ ( + + )= 3 a+x b+x c+x 3 (a + x)(b + x)(c + x) To prove that M ≤ P it is enough to see that 3(a + x)(b + x)(c + x) ≥ 4(a(b + x)(c + x) + b(c + x)(a + x) + c(a + x)(b + x)) Using the fact that x = a + b + c the last inequality becomes 3 2(a + b + c) ≥ 9abc + 5(a + b + c)(ab + bc + ca) But 3(ab + bc + ca) ≤ (a + b + c)2 so we get 3 2(a + b + c) − 9abc − 5(a + b + c)(ab + bc + ca) 1 5 ≥ 2(a + b + c)3 − 9abc − (a + b + c)3 = (a + b + c)3 − 9abc ≥ 0 3 3 Finally. Syria. if a + b + c is denoted by u then (u + a)(u + b)(u + c) = 2u3 + u(ab + bc + ca) + abc. Note that we have proved this inequality for any positive real numbers a. so (u + a)(u + b)(u + c) ≥ 54abc + 9abc + abc = 64abc. Let us denote x = a + b + c. if s denotes the semi-perimeter of ABC then we know that a + b + c = 2s. b and c.85 This clearly holds on account of the identity a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca) and the fact that a2 + b2 + c2 ≥ ab + bc + ca as it is well-known. This is equivalent to the upper inequality : s a b c 1 3 ≤ b + c + 2a c + a + 2b a + b + 2c 4 with equality. Higher Institute for Applied Sciences and Technology. using the fundamental inequality. ab + bc + ca = s2 + r2 + 4rR. if L= 2a + b + c a a + 2b + c a a + b + 2c a then (2s + a)(2s + b)(2s + c) 16s3 + 2s(ab + bc + ca) + abc = abc abc 16s2 + 2s(s2 + r2 + 4rR) 9s2 + r2 = 1+ =3+ 4rRs 2rR Now. we have u ≥ 3 3 abc and ab + bc + ca ≥ 3( 3 abc)2 . abc = 4rRs. L = . Solution 2 by Omran Kouba. The equality in both sides holds if and only if the triangle is equilateral. So. we have p s2 ≤ 2R2 + 10rR − r2 + 2(R − 2r) R2 − 2rR ≤ 4R2 + 6rR − r2 . On the other hand. we have a+b+c 1 f ≥ (f (a) + f (b) + f (c)) 3 3 That is. Spain. b = z + x and c = x + y for which r (x + y)(y + z)(z + x) xyz p R= and r = x+y+z 4 (x + y + z)xyz Then. f is concave and applying Jensen’s inequality. Peˇcari´c.86 (See for instance [1. 1989. there exist three positive numbers x. (a + b + c)/3 1 = 4 1 + (a + b + c)/3 ≥ ≥ = a 1 b c + + 3 1+a 1+b 1+c s a b c 3 1+a 1+b 1+c s b c a 3 b + c + 2a c + a + 2b a + b + 2c on account of AM-GM inequality. z such that a = y + z. BARCELONA TECH. Therefore we have proved that L−1/3 ≥ r/(2r + R) which is the desired inequality. we will see that and b c z+x x+y + = + c b x+y z+x (x + y)(y + z)(z + x) z+x x+y ≥ + 4xyz x+y z+x . On the other hand. y. Solution 3 by Jos´ e Luis D´ıaz-Barrero. Consider the function f : (0. J. R (x + y)(y + z)(z + x) = r 4xyz Now.) We conclude that L≤3+ 36R2 + 54rR − 8r2 60rR + 36R2 − 8r2 = 2rR 2rR It follows that 3 2r + R −L ≥ r R4 + 6rR3 − 6r2 R2 − 22r3 R + 4r4 r3 R (R − 2r)(R3 + 8R2 r + 9Rr2 ) + r2 (R − 2r)2 ≥ 0. using the duality principle. Volenec. Recent Advances in Geometric Inequalities. +∞) → R define by f (t) = 1+t : 0 −2 00 −3 Then. So. Kluwer Academic Publishers.S.Theorem A. Mitrovni´c. To prove the LHS inequality we can assume that a + b + c = 1 on account t of the homogeneity. [1] D. E. we have f (t) = (1 + t) and f (t) = −2(1 + t) < 0 for all t > 0. page 2]. and V. = r3 R where we used the well-known inequality R ≥ 2r. we have that b c R + ≤ (cyclic) c b r In fact. Barcelona. BARCELONA TECH. Spain and Jos´ e Gibergans B´ aguena. Spain. Tor Vergata University. Bruno Salgueiro Fanego. Italy.87 or equivalently. 2 Also solved by Ioan Viorel Codreanu. Rome. Spain. Iv´ an Geffner Fuenmayor. BARCELONA TECH. Viveiro. Paolo Perfetti. Barcelona. From the preceding and taking into account GM-HM inequalities. Maramure¸s. Department of Mathematics. Barcelona. y+z 1 1 + ≥ 2 4xyz (x + y) (z + x)2 which follows immediately from the fact that (x + y)2 ≥ 4xy and (z + x)2 ≥ 4zx. Satulung. . Romania. we have s a b c 3 b + c + 2a c + a + 2b a + b + 2c −1 b + c + 2a c + a + 2b a + b + 2c ≥3 + + a b c −1 r b b c c a a ≥ + + + + + ≥3 6+ b a c b a c 2r + R Equality holds when a = b = c because in this case R = 2r and we are done. . B= + . + an a1 an By applying H¨ older’s Inequality. + 1 + p np 1 n(n − 1) n−1 = A + B + B 2p−1 n2 + 1 n(n2 + 1) (n p−1 + 1)p−1 n (3) 1 p )p−1 n p−1 Applying Cauchy-Schwarz Inequality. + an From (3). .. an and for all positive integer p > 1. (2) ap1 an (a1 + . . (4) . . Find the largest constant cn > 0 such that. . . . (n2 +1)p Proof. Our aim in this short note is to generalize it. we have 1 1 n2 + ··· + ≥ a1 an a1 + . . 1. + + .+ p + ≥ cn (p) + .. For all positive numbers a1 .88 MATHNOTES SECTION Note on an Algebraic Inequality Vandanjav Adiyasuren and Bold Sanchir Abstract. 2 1 1 1 1 1 1 + · · · + + ≥ c + · · · + + n a21 a2n (a1 + · · · + an )2 a1 an a1 + · · · + an A solution to the preceding proposal and some related results appeared in [2]. for all positive real numbers a1 . .. the following inequality holds: p 1 1 1 1 1 1 +. . we get 1 1 1 p + ···+ p + a1 an (a1 + ... Introduction In [1] the following problem was posted: Let n be a positive integer. 2. . . (4) we get (2). In this note a constrained inequality is generalized. + an )p a1 an a1 + ... Denote A= 1 1 1 1 1 + .. + a1 an a1 + . + + . . Main Results Theorem 1.. . . + an ) (1 + . + an )p p 1 1 1 1 ≥ + ··· + + a1 an n(a1 + . . + an where cn (p) = (n3 +1)p 2p−1 n p−1 p−1 +1 . an . .. .. . .<im ≤n 1≤i1 <. + aim k ai1 + ..k (α. p > 1... . + aik 1≤i1 <.. .. β) + ai1 + .. . + aim X 1≤i1 <. . .. .<ik ≤n β .<im ≤n α + ai1 + . .m. . . . . k be positive integers. . .. . we get (5).m.<ik ≤n p X X 1 1 1 1 ≥ + m ai1 + .. + aim ) (ai1 + . . Let a1 . For all positive real numbers a1 ..<ik ≤n 1≤i1 <.<im ≤n B X = 1≤i1 <. . ai1 + .89 Corollary 1. Choosing p = 2 in (2)..k (α. Let n. . + aim ai1 + .<im ≤n 1≤i1 <···<ik ≤n p X X α β [p] ≥ cn. ... + + a21 an (a1 + .. β > 0 with kβ − mα > 0.. . . . + aik 1 .<ik ≤n (6) where [p] cn. we hve X X 1 1 + p (ai1 + .. + aim Using H¨ older’s Inequality. . + an (5) Proof. Then 2 1 1 1 n3 + 1 1 1 1 + . . .. . Denote X A = 1≤i1 <. . we get X X 1 1 L := + (8) p (ai1 + .. + aik )p 1≤i1 <.. m. .<im ≤n 1≤i1 <. ... + aim k p mp p p (m p−1 (nk ) + k p−1 (nm ))p−1 (kβ − mα)(n−m 1 α(km )(kβ − mα) k−m ) A+ B + B n−m k kβ k(βm(k−m ) + αk(km )) mβ(βm(n−m k−m ) + αk(m )) = × k p mp (m p p−1 p (nk ) + k p−1 (nm ))p−1 !p . + aim ) (ai1 + .<im <n × 1 (n m) p m p−1 + p−1 (nk ) p k p−1 kβ − mα 1 = A + kβ mβk × X 1≤i1 <.. ai1 + .. . . + aik )p 1≤i1 <. . + an )2 (n + 1)2 a1 an a1 + . an be positive numbers. ..<im ≤n p 1 ai1 + . k > m. . + 2 + ≥ 2 + . β) = k p mp p p (m p−1 (nk ) + k p−1 (nm ))p−1 (km )(kβ − αm) 1 + k kβ βm(βm(n−m k−m ) + αk(m )) !p (7) Proof. .. an and α. Theorem 2. .. + aik 1≤i1 <. Ulaanbaatar.. . .<im ≤n (9) ai1 + .com Institute of Mathematics...<ik ≤n m k X k(km ) n−m m(k−m ) 1≤i <.<ik ≤n i1 p X X α β [p] = cn.<ik ≤n 1 1 + .. Ulaanbaatar. . + aim = 1 X (n−m k−m ) 1≤i 1 <. National University of Mongolia. (3) (2011).. .(4) (2011).. + aim aik−m+1 + . . . .. (9) we get X L = 1≤i1 <. Andreescu and D. E-mail address: om sanchir@yahoo. β) + ai1 + ..<im ≤n Equality occurs when a1 = · · · = an . .k (α. (http://awesomemath. + aik | {z } (k m) ≥ = 1 k m X (k ) m (ai1 (n−m k−m ) 1≤i1 <. .org/wp-content/uploads/reflections/2011 4/MR3solutions. + aim ai1 + . .<im ≤n k ≤n 2 + . References [1] T.pdf) Department of Mathematical Analysis.<ik ≤n 1 (ai1 + ... ... ... we get X 1 1≤i1 <. . .m. Mongolia.<i 1 Using (8). .. + aik p ≥ p k m p p−1 p p−1 1≤i1 <..com . + aim )p X 1≤i1 <.. Mongolia. 13-14. E-mail address: V Adiyasuren@yahoo. + aik )p α(km )(kβ − mα) 1 A+ B n−m mβ(βm(k−m ) + αk(km )) (m (nk ) + k (nm ))p−1 kβ p k X (kβ − mα)(n−m ) k( ) 1 k−m m + k )) m(n−m ) a + .<im ≤n 1≤i1 <. Mathematical Reflections. + ai1 + .. Andrica. . [2] Solution of problem U193. Problem U193.90 Using Cauchy-Schwarz inequality.. . National University of Mongolia. Mathematical Reflections. + aik ) 1 ai1 + · · · + aik 1 + (ai1 + . + aik k(βm(n−m ) + αk( m k−m k−m 1≤i1 <... .. PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets. Cristinel Mortici. . Prishtin¨ e. each indicating the name of the sender. Pages 91–117 Editors: Valmir Krasniqi. George Emil Palade Secondary School.Mathproblems ISSN: 2217-446X. David R. J´ozsef S´andor. An asterisk (*) indicates that neither the proposer nor the editors have supplied a solution. Calculate 1 1 1 1 x+1 x lim (x + a) (Γ(x + 2)) sin − x (Γ(x + 1)) sin x→∞ x+a x 44. Kosov¨ e. Armend Sh. Teachers can help by assisting their students in submitting solutions. Valmir Bucaj. . Roberto Tauraso. B˘ atinet¸u-Giurgiu. Questions concerning proposals and/or solutions can be sent by e-mail to: mathproblems-ks@hotmail. Proposed by D. Bucharest. Universiteti i Prishtin¨ es. Mih´aly Bencze. Let A denote the Glaisher–Kinkelin constant defined by n Y 2 2 A = lim n−n /2−n/2−1/12 en /4 k k = 1. Jos´ e Luis D´ıaz-Barrero.282427130 . Shabani. Buzˇ au. Enkel Hysnelaj. Solutions will be evaluated for publication by a committee of professors according to a combination of criteria.com Volume 2. Matei Basarab National Colege. Stone. Proposals should be accompanied by solutions. 2012 Problems 43. The editors encourage undergraduate and pre-college students to submit solutions. Romania. Cluj-Napoca. Romania and Neculai Stanciu. 91 . Let a be a positive real number and Γ(x) be the Gamma function (or Euler’s second integral). Technical University of Cluj-Napoca. Drawings must be suitable for reproduction.com Solutions to the problems stated in this issue should arrive before September 3. Ovidiu Furdui. Issue 3 (2012). Proposed by Ovidiu Furdui.M. . n→∞ Prove that k=1 ∞ X ζ(2p + 1) − 1 p=1 p+2 =− γ 7 − 6 ln A + ln 2 + . url: http://www. 2 6 c 2010 Mathproblems. Student solutions should include the class and school name. Romania (Jointly). Paolo Perfetti.mathproblems-ks. c d Prove that there exist (A. USA. Prove that p 3 (a − b)2 (b − c)2 (c − a)2 a b c + + ≥3+ b c a ab + bc + ca . Let f : (0. Bra¸sov. Statesboro. Romania. jud. Romania. Paris. Proposed by Anastasios Kotronis. c be positive real numbers. Proposed by Florin Stanescu. Gaesti. Evaluate +∞ X (−1)n−1 n=1 ζ(2n) . Let a. C) ∈ M2 (Z) × M2 (Z) × M2 (Z) such that M = A2 + B2 + C 2. Proposed by Mih´ aly Bencze. Dambovita. School Cioculescu Serban. b. Georgia Southern University. Identify the graph of the equation y 4 − 10x2 y 3 + 25x4 y 2 − 50x6 y + 24x8 = 0 48. Stone. and Serafeim Tsipelis.92 P∞ 1/k s for <(s) > 1. Proposed by Moubinool Omarjee. GA. n where ζ is the Riemann zeta function. 0 < a < b such that R b f (x) f (b) f (a) − (i): a 2 dx = x b a 2 f 0 (a) = (ii): f (a) a (a) Find an example of a function that satisfies the preceding conditions (b) Show that there exists a c ∈ (a. Let M = ∈ M2 (Z). ∞) be a differentiable function for which there exist real numbers a and b. Greece. France. Proposed by David R. b) such that 0 c 2 f (c) f (c) (c−a) f (c) − 2c = e f (a) a 49. 47. Greece (Jointly). ∞) → (0. Ionnina. where ζ is the Riemann zeta function defined by ζ(s) = k=1 46. B. Athens. a b 45. Evaluate the sum +∞ X 1/2 1 1/2 1 1/2 1 n 2−1/2 − 1 + − + · · · + (−1)n+1 1 2 2 4 n 2n n=1 Solution by Omran Kouba. Syria.93 Solutions No problem is ever permanently closed. 36. ρ) and n ≥ 0 we define the rest Rn (x) by Rn (x) = f (x) − n X ∞ X ak xk = k=0 ak xk k=n+1 Then for every x ∈ (−ρ. Damascus. For x ∈ (−ρ. Athens. For a given x ∈ (−ρ. we have ∞ X nRn (x) = n=1 x2 00 f (x) 2 Proof. ρ) let r be a positive real number such that |x| < r < ρ and let Mr = sup{|ak |rk : k ≥ 0}. ρ). Higher Institute for Applied Sciences and Technology. Proposed by Anastasios Kotronis. ρ). We will be very pleased considering for publication new solutions or comments on the past problems. The value of the sum to be computed is obtained immediately applying the following P∞ Lemma. Suppose that f has a power series expansion k=0 ak xk that converges in (−ρ. Then |Rn (x)| ≤ ∞ X |ak |rk k=n+1 |x| r k ≤ Mr |x| r − |x| |x| r n This implies the absolute convergence of the following two series : S1 = ∞ X n(n + 1) Rn (x) 2 n=1 and S2 = ∞ X n(n − 1) Rn (x) 2 n=1 On one hand we have S1 − S2 = ∞ X n(n + 1) n=1 2 n(n − 1) − 2 Rn (x) = ∞ X n=1 nRn (x) . Greece. Ulaanbaatar. Consider the polynomial P (x) = det(A + Bx) = x2 det B + αx + det A. Switzerland. Albert Stadler. Department of Mathematics. Polytechnic University of Tirana. B ∈ M2 (R). √ Applying the Lemma to the function f (x) = 1 + x and the point x = −1/2 gives √ ∞ X 1/2 1 1/2 1 1 2 −1/2 n+1 1/2 n 2 −1+ − + · · · + (−1) =− n 1 2 2 4 n 2 16 n=1 = S1 − and we are done. Ioannina. Also solved by Adrian Naco. Greece. Albania. Bra¸sov. Mongolia. Romania.94 and on the other hand. since Rn−1 (x) = an xn + Rn (x). B ∈ M2 (R) then prove that 2 2 2 (det A) + det(AB + BA) + 2 (det B) ≥ det A2 − B 2 + 4 det AB Solution by AN-anduud Problem Solving Group. Let A. by the lemma we have det(A2 − B 2 ) = 2(det A2 + det B 2 ) − det(A2 + B 2 ) . Kostas Tsouvalas. Proof. then we have P (1) + P (−1) = det(A + B) + det(A − B) On the other hand P (1) + P (−1) = (det B + α + det A) + (det B − α + det A) = 2(det A + det B) From the preceding. Now we prove statement. Greece. Proposed by Mih´ aly Bencze. we have S2 = ∞ X n(n − 1) Rn (x) 2 n=2 = ∞ ∞ X X n(n − 1) n(n − 1) Rn−1 (x) − an xn 2 2 n=2 n=2 = ∞ ∞ X x2 X n(n + 1) Rn (x) − n(n − 1)an xn−2 2 2 n=2 n=1 x2 00 f (x) 2 Now the Lemma follows by combining the preceding results. First we will prove following Lemma. and the proposer 37. If A. Then det(A + B) + det(A − B) = 2(det A + det B). Lemnos. Tsipelis Serafeim. Indeed. we get det(A + B) + det(A − B) = 2(det A + det B) as desired. School Cioculescu Serban. Higher Institute for Applied Sciences and Technology. Applying Cebyshev’s inequality to the . School Cioculescu Serban. Dambovita. R 1 c): f 0 (1) 2 (f (1) − f (0)) − f 0 (1) 0 (f 0dx =1 (x))2 Solution by the proposer. ∀x ∈ [0. Since g 0 (x) = −xf 000 (x) ≤ 0. and the proposer. we obtain (det(A + B))2 + (det(A − B))2 ≥ 8 det(AB) By the wel–known inequalities ∀x. Omran Kouba. France. 1] → R. jud. Also solved by Moubinool Omarjee. Gaesti. Dambovita. Proposed by Florin Stanescu. defined by g(x) = f 0 (x) − xf 00 (x). 1] → R that have the following properties a): f is three times differentiable with f 000 (x) ≥ 0. Romania. Romania. Syria.95 So it is enough to prove following inequality det(A2 + B 2 ) + det(AB + BA) ≥ 4 det(AB) Again. 1]. 2 x2 + y 2 ≥ 2xy and 1 (det(A + B) + det(A − B))2 2 1 (2(det A + det B))2 2 2((det A)2 + (det B)2 + 2 det A det B) 2(2 det A det B + 2 det A det B) 8 det A det B = 2 det(AB) ≥ = = ≥ = and we are done. det(A2 + B 2 + AB + BA)+ det(A2 + B 2 − AB − BA) = det(A + B)2 + det(A − B)2 = (det(A + B))2 + (det(A − B))2 From the preceding. we get (det(A + B))2 + (det(A − B))2 (x+y)2 . 1]. y ∈ R : x2 + y 2 ≥ the lemma. jud. Determine all functions f : [0. by the lemma we have det(A2 + B 2 + AB + BA)+ det(A2 + B 2 − AB − BA) = 2(det(A2 + B 2 ) + det(AB + BA)) On other hand. Gaesti. From Z 0 1 x f 0 (x) 0 Z dx = 0 1 x f 0 (x) f 0 (x) − xf 00 (x) 2 (f 0 (x)) 0 = f 0 (x)−xf 00 (x) dx. we have f 0 (x) − xf 00 (x) 2 (f 0 (x)) dx Consider the function g : [0. Damascus. Switzerland. then g is decreasing. Paris. Florin Stanescu. b): f 0 is increasing and strictly positive. 38. (f 0 (x))2 dx and 1 = f 0 (1) Z 0 1 ∀x ∈ [0. Albert Stadler. ∀x ∈ [0. 1]. then f (x) = dx2 + cx + f (0) 2 . . Since P (x) = 2 ln x. If A. George Emil Palade Secondary School. Z x (f 0 (t) − tf 00 (t)) dt = cx. Buzˇ au. 1 (f 0 (x))2 or g are constant 1. then f 0 (t) − tf 0 (t) = a. 1] thenf (x) = ax + b with a. Proposed by Neculai Stanciu. b ∈ R and a > 0. 2. with solutions y = eP (x) q(x)e−P (x) dx. then we have Z f (0) 2 ln x − c e−2 ln x dx f (x) =e − x Z f (0) 1 2 =x − −c dx x x2 f (0) c + +d =x2 2x2 x f (0) =dx2 + cx + . b > 0 39. ∀x ∈ (0. 1]. 1]. 1].96 decreasing functions g(x) and 1 = 0 f (1) Z 1 0 Z 1 ≥ 0 Z 1 = 0 Z = 0 1 . 1] 2 f 0 (x) = Since f is continuous on [0. Thus. ∀t ∈ [0. (f 0 (x))2 we have f 0 (x) − xf 00 (x) dx 2 (f 0 (x)) Z 1 dx (f 0 (x) − xf 00 (x)) dx 2 · (f 0 (x)) 0 Z 1 dx 0 0 f (1) − f (0) − f (1) + f (x)dx 2 (f 0 (x)) 0 1 dx 2 (f 0 (x)) (2 (f (1) − f (0)) − f 0 (1)) From the preceding and the condition in c) it follows that functions. where P (x) is a fixed primitive function. ∀x ∈ [0. ∀x ∈ (0. If f 0 (x) = a. Taking into account the above and observing the conditions given in the statement. A+B+C 3 A B C where k ∈ (0. 1] ⇒ f (x) − f (0) − xf 0 (x) + 0 Z x f 0 (t)dt = cx 0 and 2 f (0) f (x) + − − c . Romania. ∀x ∈ [0. we obtain f (x) = ax2 + bx + c. 1] x x We observe that the last expression is a first order linear differential equation of R the form: y 0 = p(x)y + q(x). then prove that: ! sink A + sink B + sink C 1 sink A sink B sink C ≤ + + . with a ≥ 0. If g(t) = a. B and C are the angles of a triangle. Sydney. Solution by Omran Kouba. f 0 (x) = sin g 0 (x) = k cos x − kx sin x − cos x = (k − 1) cos x − kx sin x ≤ 0 We have x ∈ [0. Maramures. b. 40. WLOG we can assume A ≥ B ≥ C. University of Technology. From this result we can easily see that f (x) is decreasing function. y. Albania. Indeed. Ulaanbaatar. . he showed that If 0 < x1 ≤ x2 ≤ . Mongolia.97 Solution by AN-anduud Problem Solving Group. Proposed by Enkel Hysnelaj. such that y3 z3 t3 x3 + + + x−1 y−1 z−1 t−1 is an integer. d) ∈ M : + + + = 1 a b c d . 1]. then we will reduce the considered problem to the simpler one. Then. n n xi k=1 xi k=1 where k ∈ (0. If sink A sink B sink C ≤ ≤ A B C then by Chebyshev’s Inequality. Namely. x ∈ [0. π). z. Syria. k−1 x (kx cos x − sin x) x2 Consider function g(x) = kx cos x − sin x. and the proposer. π) ⇒ g(x) ≤ g(0) = 0. Department of Mathematics. d) of positive integers such that 1 < a < b < c < d and let 1 1 1 1 S = (a. Also solved by Adrian Naco. b. Australia. Let M denote the set of quadruples (a. . Higher Institute for Applied Sciences and Technology. c. Damascus. Our solution will consist of two steps: first we will solve an apparently simpler problem (the next lemma). c. we have X sink A ≤ X sink A 1 X A 3 A cyclic cyclic cyclic Now we need to prove that the function f (x) = sink x x is decreasing. t all greater than 2. Lemma. ≤ xn < π. then Pn n k 1 X sink xi k=1 sin xi P ≤ . Editors Comment: Using similar techniques Adrian Naco proved a generalized version of the above problem. Polytechnic University of Tirana. Satulung. Find all distinct positive integers x. Romania. Ioan Viorel Codreanu. (2. (2. t) ≤ 1 1 1 1 77 + + + = <2 2 3 4 5 60 . This yields the following two solutions: 1= (2. 5}.98 Then S consists exactly of the following six elements : (2. u2 . 7. uσ(4) ). 4. (2. for (x. y. (2. 4. 8. 3. Proof. 20) and (2. so c − 6 is a positive divisor of 36 which is smaller than 6. Let us come to our problem. c − 4 = 1 or c − 4 = 2. Now. 3. 5. (2. 42). (2. c − 6 ∈ {1. So let us consider the following cases: (i) 7 If b = 5 then 1c + d1 = 10 . 5. u4 ) is a solution to the problem then all 24 permutations (uσ(1) . (2. z. uσ(2) . 3. Let us denote by N the set of quadruples of integers (x. are also solutions. Since u3 1 = 1 + u + u2 + . f (x. 20). 10. 18) and (2. 24). 2. such a solution will be called an ordered solution. According to the remark above. d ≥ 8 and this would imply that 1= 1 1 1 1 157 1 1 1 1 + + + ≤ + + + = <1 a b c d 2 6 7 8 168 which is absurd. 4. 8. t) that satisfy 3 ≤ x < y < z < t. so c − 4 is a positive divisor of 16 which is smaller than 4. 4}. z. 3. uσ(3) . t) = (u1 . t) ∈ N . 15) This concludes the proof of the Lemma. 3. if b ≥ 6 then we would have c ≥ 7. y. u3 . 4. z. t) ∈ N . So a = 2. y. d) ∈ S. 24). 10. or equivalently (d − 4)(c − 4) = 16. 6. This yields the following four solutions: (2. 9. u−1 u−1 we see that x3 y3 z3 t3 + + + x−1 y−1 z−1 t−1 is an integer for some (x. t. 6.e. 9. if and only if. c ≥ 5 and d ≥ 6. t) = 1 1 1 1 + + + x−1 y−1 z−1 t−1 is an integer. b. z. 12) (iii) If b = 3 then 1c + d1 = 16 . i. Note that the expression given in the statement is symmetric in x. (ii) If b = 4 then 1c + d1 = 41 . which is absurd. So if (x. y. z. 42). But −2 ≤ c − 6 < d − 6. But. y. 3. 15). z. Therefore b ∈ {3. Among these solutions there is only one that satisfies x < y < z < t. c. 4. 3. or equivalently (d − 6)(c − 6) = 36.e. we have f (x. This would imply that 1 1 1 1 1 1 1 1 57 + + + ≤ + + + = <1 a b c d 3 4 5 6 60 which is impossible. 12). If a ≥ 3 then we would have b ≥ 4. But 1 ≤ c − 4 < d − 4. i. Consider (a. 18). 3. 3. 7. y. where σ ∈ S4 . but 6 ≤ c < d implies that 20d > 10(c + d) = 7cd ≥ 42d. y. it is enough to find all the solutions to the problem that belong to N . z. 4. t) ∈ N = (x. Imz = [−i. z. t − 1) ∈ S where S is identified in the lemma. t) ∈ N : f (x. 5. z. −∞). 2 2 2 2 2 2 −∞ (1 + z )(z + a) −∞ (1 + z )(z + a) where Ln(z) = ln(|z|) + iϑ. 0 ≤ t ≤ π}. Tor Vergata University. (x. t) ∈ N : f (x. Italy. 19). 6. y. Proposed by Paolo Perfetti. Setting x = arctan t we get 2 Z +∞ ln(1 + t2 ) dt (1 + t2 )2 0 We write d da Z 0 +∞ 2 Z +∞ ln(1 + at2 ) 2 ln(1 + at2 )t2 dt = dt 2 2 (1 + t ) (1 + t2 )2 (1 + at2 ) 0 Z +∞ Z +∞ √ √ ln(1 + t2 )t2 ln(1 + t2 )t2 =2 a a dt = dt 2 2 2 (1 + t2 )(a + t2 )2 0 −∞ (1 + t )(a + t ) Standard theorem on the exchange between integral and derivatives allow us to differentiate under the integral. z. z. y. 2 z Ln(z+i) Let us consider the complex function f1 (z) = (1+z 2 )(z 2 +a)2 and cut the complex plane along the set Rez = 0. y. 9. 41. 43). z. Department of Mathematics. t) ∈ N : (x − 1. (3. y. 11. We compute the counterclockwise integral over the two curves γ1 (t) = {z ∈ C : z = t. −r ≤ t ≤ r}. Mongolia. and the total number of solutions is 144. 4. 16). 8. 4. 7.99 Therefore. γ2 (t) = {z ∈ C : z = reit . (3. y. and the proposer. y − 1. Rome. 5. (3. 4. (3. Evaluate Z π/2 2 4(cos x)2 (ln(cos x)) dx 0 Solution by the proposer. z − 1. 10. Z γ1 ∪γ2 √ f1 (z)dz = 2πi(Resf1 (i) + Resf1 (i a)) . t) = 1 = (x. Also solved by AN-anduud Problem Solving Group. 13). 25). Considering the complex function z 2 Ln(z 2 + 1) z 2 Ln(z + i) z 2 Ln(z − i) = + 2 2 2 2 2 2 (1 + z )(z + a) (1 + z )(z + a) (1 + z 2 )(z 2 + a)2 we may write the two integrals Z +∞ Z +∞ √ √ z 2 Ln(z + i) z 2 Ln(z − i) a + a . Ulaanbaatar. 21). So the ordered solutions to our problem are (3. (3. √ √ − − = a7 + a8 + a9 + a10 (1 + z 2 )2 (z − i a)2 (1 + z 2 )(z − i a)3 = lim√ 1 √ 2i(1 − a)((1 + a) √ −i ln(1 + a) a4 − a8 = √ a(1 − a) √ √ −i a ln(1 + a) a5 − a9 = (1 − a)2 √ i ln(1 + a) a6 − a10 = √ a(1 − a)2 a3 − a7 = and then Z +∞ √ z 2 Ln(z + i) z 2 Ln(z − i) a + a = 2 2 2 2 2 2 −∞ (1 + z )(z + a) −∞ (1 + z )(z + a) √ √ √ √ −2π ln 2 a π a π ln(1 + a) 2πa ln(1 + a) √ + + + (a − 1)2 (1 − a) (1 − a)2 (1 − a)(1 + a) √ Z +∞ .100 2 z Ln(z−i) Moreover by defining f2 (z) = (1+z 2 )(z 2 +a)2 and cutting the complex plane along the set Rez = 0. γ2 (t) = {z ∈ C : z = re−it . −2π ≤ t ≤ −π}. Imz = [i. we calculate the clockwise integral over the two curves γ1 (t) = {z ∈ C : z = t. √ f2 (z)dz = −2πi(Resf2 (−i) + Resf2 (−i a)) Z γ1 ∪γ2 Now we compute the following residues: −2πiLn(2i) 2πiLn(−2i) −π2 ln 2 2πi (Resf1 (i) + Resf2 (−i)) = − = 2i(a − 1)2 2i(a − 1)2 (a − 1)2 √ d z 2 Ln(z + i) √ Resf1 (i a) = lim√ = z→i a dz (1 + z 2 )(z + i a)2 z2 2zLn(z + i) √ 2+ √ + = lim√ 2 (1 + z 2 )(z + i a)2 z→i a (z + i)(1 + z )(z + i a) 2z 2 Ln(z + i) . +∞). −r ≤ t ≤ r}. 2z 3 Ln(z + i) √ 2− √ − = a3 + a4 + a5 + a6 2 2 (1 + z ) (z + i a) (1 + z 2 )(z + i a)3 √ Resf2 (−i a) = lim√ z→−i z 2 Ln(z − i) d √ = a dz (1 + z 2 )(z − i a)2 z2 2zLn(z − i) √ 2+ √ + 2 (1 + z 2 )(z − i a)2 z→−i a (z − i)(1 + z )(z − i a) 2z 3 Ln(z − i) 2z 2 Ln(z − i) . +∞)}. Rew ∈ [1. w ∈ C\{w ∈ C : Imw = 0.101 Now we must integrate respect to the variable a (we assume 0 < a < 1 but a > 1 would yield of course the same result) and we get +∞ ln(1 + t2 )t2 dt = 2 2 2 −∞ (1 + t )(a + t ) √ 1 1 1 1− a √ − √ + ln √ − 2π ln 2 + 2(1 − a) 2(1 + a) 2 1 + a √ √ 1 3 1 √ − ln(1 + a) − ln(1 − a) + π − 2 1+ a 2 √ √ √ 1− a 1 2 π − ln (1 + a) − ln 2 · ln(1 − a) + Li2 + 2 2 √ √ √ √ √ 1− a 1 (1 + a) ln(1 + a) 1 √ ln(1 − a) + + ln 2 · ln(1 − a) − Li2 + + 2π 4 4 2 1− a ! √ √ 1 2 1 1 ln(1 + a) 1 √ √ + ln (1 + a) + C + + 2 1+ a 21+ a 2 Rz P∞ k where as usual Li2 (z) = k=1 zk2 = − 0 ln(1 − w) dw w . The integrations are standard so we explain only the third integral √ Z Z Z ln(1 + y) ln(1 + y) ln(1 + a) da |{z} = dy + dy = a−1 1+y y−1 2 a=y ! Z √ ln(1 − 1−y ) 1 2 ln 2 2 ln (1 + y) + − dy = (y = a) 2 y−1 1−y √ √ √ 1 2 1− a ln (1 + a) + ln 2 · ln(1 − a) − Li2 ( ) 2 2 √ Z a For a = 0 we have 1 π3 π 0 = −πLi2 ( ) + C = 0 =⇒ C = − ln2 2 12 2 Now we select the terms diverging for a → 1− and obtain √ √ √ 1 π 1 √ − 2π ln 2 ln(1 − a) + − ln(1 − a) − π ln 2 · ln(1 − a)+ 2 2 2(1 − a) √ √ √ √ 1 1 (1 + a) ln(1 + a) −π π √ + 2π ln(1 − a) + + ln 2 · ln(1 − a) → ln 2 − 4 4 2 2 1− a The terms not diverging for a → 1− contribute − and finally we get π 3 ln 2 + ln2 2 2 2 . i.e. Aplil–2012 (Omran Kouba). That problem R π/2 2 asked to evaluate 0 (ln(2 sin x)) dx. Allegany College of Maryland. Ψ0 (x) = 1 1 1 1 − lim = .102 − π 3 π π π3 π π π3 ln 2 + ln2 2 − ln 2 − + − ln2 2 = −π ln 2 + π ln2 2 − + 2 2 2 2 12 2 2 12 Comment by Paolo Perfetti: The proposed problem was inspired by num. USA and Anastasios Kotronis. 2 2 2k 2k − 1 k=1 +∞ (4) X Ψ0 (2) = k=1 2 1 π = −1 (k + 1)2 6 +∞ 3 (4) X 1 π2 Ψ0 =4 = − 4. 11639 of Americam Mathematical Monthly. x n→+∞ x + n x (3) gives +∞ X k=1 1 (x + k − 1)2 x > 0. 2 2 (2k + 1) 2 k=1 1for k ≥ 1. (4) On account of the above we have (2) Ψ(1) = −γ (3) Ψ(2) = Ψ(1) + 1 = 1 − γ +∞ X 3 (3) 1 (2) 1 1 Ψ = 2+Ψ = 2−γ+2 − = 2 − γ − 2 ln 2. Solution 2 by Cody Thompson. using two basic representations of the Digamma function Ψ. 1 k − on [a.: Γ0 (x) Γ(x) 0 Ψ(x) = (ln Γ(x)) = Ψ(x) = −γ + +∞ X 1 k=1 1 − k x+k−1 x∈R−N (1) x ∈ R − N. b] with a > 0 1 x+k−1 has a continuous derivative and P+∞ 1 k=1 (x+k−1)2 converges uniformly . (2) we get some special values for Ψ and Ψ0 . At first. Athens. Greece (Jointly). Chaneysville. differentiating (4). Pennsylvania. From (2) we have that for x ∈ R − N: Ψ(x + 1) − Ψ(x) = Furthermore. clearly we have f (x− ) ≤ f (x) ≤ f (x+ ) . Higher Institute for Applied Sciences and Technology. We argue by contradiction. Prove that f is continuous. Switzerland. and Albert Stadler. Syria. Rome. France. Suppose first that f is nondecreasing. Department of Mathematics. Proposed by Cristinel Mortici. Solution 1 by Paolo Perfetti. The result is l− + L− = lim f + f ◦ f ◦ f < lim f + f ◦ f ◦ f = l+ + L+ x→x− 0 x→x+ 0 Namely. = 12 2 2 2 Also solved by Omran Kouba. Romania. 42. Paris. Solution 2 by Omran Kouba. Let f : R → R be monotone such that f + f ◦ f ◦ f is continuous. = 2 2 2 2Γ a+1 1 − t2 0 2 and by Leibniz’s rule. Thus we have limx→x− f (x) = l− and limx→x+ f (x) = l+ and l− ≤ f (x0 ) ≤ l+ but l+ − l− > 0. differentiating twice under the integral sign and using (1) we get 1 √ Γ a2 a+1 a+1 t ln2 t π a 2 2 a √ dt = − 2Ψ + Ψ · Ψ Ψ + 8 Γ a+1 2 2 2 2 1 − t2 2 a a+1 . Syria. Let x be an arbitrary real number. Italy. Ψ0 − Ψ0 2 2 1 a−1 Z 0 For a = 3 and from the special values of Ψ and Ψ0 evaluated above we have that Z 0 π/2 √ π Γ 32 3 3 4 cos x ln (cos x) dx = · Ψ2 − 2Ψ (2) Ψ + Ψ2 (2) + 2 Γ (2) 2 2 3 Ψ0 − Ψ0 (2) 2 3 π π + π ln2 2 − π ln 2 − . 1 − t2 0 0 But √ Z 1 a−1 πΓ a2 t a 1 t2 =x 1 √ dt = B . Damascus. 0 0 Now f ◦ f ◦ f is also monotone increasing and L− = limx→x− f ◦ f ◦ f ≤ limx→x+ f ◦ 0 0 f ◦ f = L+ . Moubinool Omarjee. Valahia University of Tˆ argovi¸ste.103 Now. the discontinuity of f + f ◦ f ◦ f at x0 . Tor Vergata University. for the given integral we have Z Z π/2 cos x=t 2 2 4 cos x ln (cos x) dx = 4 t2 ln2 t √ dt. The monotonicity (increasing for instance) yields the existence of the right and left limits at any point and in particular at x0 . Higher Institute for Applied Sciences and Technology. Let’s suppose that f is discontinuous at x0 . Damascus. Also solved by AN-anduud Problem Solving Group. Romania. A similar argument shows that the conclusion holds if we assume that f is nonincreasing. and the proposer. . but f ◦ f is nondecreasing so f ◦ f ◦ f (t) ≤ f ◦ f ◦ f (x) ≤ f ◦ f ◦ f (u) or g(t) − f (t) ≤ g(x) − f (x) ≤ g(u) − f (u). and taking the limit as t approaches x by smaller values. which is the desired conclusion since x is arbitrary. Gaesti. and as u approaches x by larger values. using the continuity of g at x. so f is continuous at x. jud.104 Now. if t < x < u then f (t) ≤ f (x) ≤ f (u). Mongolia. we conclude that g(x) − f (x− ) ≤ g(x) − f (x) ≤ g(x) − f (x+ ) or f (x+ ) ≤ f (x) ≤ f (x− ) Combining the above expressions we conclude that f (x− ) = f (x+ ) = f (x). Albert Standler. School Cioculescu Serban. Ulaanbaatar. Dambovita. Florin Stanescu. Switzerland. Thus. Show that the three lines AB. We draw a line that cuts C1 at points A and B. Compute 1 lim n→∞ n3 Z 0 n n2 + x2 dx 5−x + 7 . 1+2 1 + ak k=1 where Fn represents the nth Fibonacci number defined by F1 = F2 = 1 and for n ≥ 3. Let P be an interior point of triangle ABC and let HA . AX BY and AY BX are parallel. 33. The source of the proposals will appear when the solutions be published. an+1 = (a2n + 1). 34. Let {an }n≥1 be the sequence of real numbers defined by a1 = 3. AY which cut C2 at points AX and BY and lines BX. respectively. BY which cut C2 at points BX and BY respectively. Two circles C1 and C2 have in common two points X and Y . Four dice are thrown at the same time on a table. and P AB. P AC. a2 = 5 and for 1 all n ≥ 2. Prove that 2 !2 n r X Fk < Fn+2 . 35. HB . Proposals 31. Next we draw the lines AX. HC be the orthocentres of triangles P BC. Prove that triangles HA HB HC and ABC have the same area. Fn = Fn−1 + Fn−2 . 32. Proposals are always welcomed. Find the probability that the sum of the points appeared in the upper faces lies between 14 and 18 points.105 MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. On account of Lagrange’s theorem. there exist α. b. UNIQUENESS: Suppose that exist α. Herrliberg Switzerland. EXISTENCE: Since f is derivable in [0. 1]. there exists θ ∈ (α. ( α < α0 ) such that f (α) = α and f (α0 ) = α0 . 2 Also solved by Adrian Naco. Barcelona. This completes the proof. Applying recursively the preceding. BARCELONA TECH. if we assume that there exist β. 1] and applying Bolzano’s theorem we get that exists α ∈ [0. α0 ) ⊂ [0. there exists θ0 ∈ (β. 1] → [0. ( β < β 0 ) such that f (β) = 1 − β and f (β 0 ) = 1 − β 0 . Tirana. Italy. Considering the functions g(x) = f (x) − x and h(x) = f (x) − (1 − x) that are continuous in [0. β 0 ) ⊂ [0. 1] be a differentiable function such that |f 0 (x)| = 6 1 for all x ∈ [0. Rome. Spain and Albert Stadler. Spain. We give the solution of D´ıaz-Barrero. 1] with h(β) = 0. Barcelona. d ∈ Q the expressions √ √ √ √ (a + b 3)(c + d 3) = (ac + 3bd) + (ad + bc) 3 = A + B 3 and √ √ √ √ (a − b 3)(c − d 3) = (ac + 3bd) − (ad + bc) 3 = A − B 3 are equivalents. 1] such that g(α) = 0 and β ∈ [0. NY. β ∈ [0.106 Solutions 26. First. 1]. BARCELONA TECH. Let f : [0. 1] such that α0 − α f (α0 ) − f (α) = =1 f 0 (θ) = α0 − α α0 − α contradiction. 1]. then f is continuous in [0. and Omran Kouba. 1] such that f (α) = α and f (β) = 1 − β. 1]. β 0 ∈ [0. 1] such that f 0 (θ0 ) = f (β 0 ) − f (β) (1 − β 0 ) − (1 − β) = = −1 β0 − β β0 − β contradiction. Prove that the equation √ √ √ (x + y 3)4 + (z + t 3)4 = 7 + 6 3 does not have rational solutions. (22nd Vojtˇech Jarn´ık International Mathematical Competition 2012) Solution by Jos´ e Luis D´ıaz-Barrero. Department of Mathematics. β ∈ [0. Higher Institute for Applied Sciences and Technology. 27. Likewise. That is. 1]. Paolo Perfetti. c. 1] for which f (α) = α and f (β) = 1 − β. α0 ∈ [0. Henry Ricardo. Syria. USA. Show that there exist two unique points α. Damascus. Tor Vergata University. (Longlist of Catalonian Mathematical Olympiad 2010) Solution by Jos´ e Luis D´ıaz-Barrero. On account of Lagrange’s theorem. yields √ √ √ √ (x + y 3)4 = X + Y 3 ⇔ (x − y 3)4 = X − Y 3 . Albania. we observe that for any a. 2 Also solved by Jos´ e Gibergans B´ aguena. BARCELONA TECH. 28. we have −2n = −2. Higher Institute for Applied Sciences and Technology. Furthermore. As we already know that p(0) = 0. Syria. c ∈ R holds (a + b − 2c)2 + (b + c − 2a)2 + (c + a − 2b)2 = 6(a2 + b2 + c2 ) − 6(ab + bc + ca) = 3(a − b)2 + 3(b − c)2 + 3(c − a)2 Now notice that if p(x) and q(x) are solutions to our problems then so is λ p(x) + µ q(x) for any λ. For a = b = c. we conclude that for any real numbers a1 . a2 the polynomial a2 x2 + a1 x is a solution. so.107 and √ √ √ √ (z + t 3)4 = Z + T 3 ⇔ (z − t 3)4 = Z − T 3 Therefore. and Jos´ e Luis D´ıaz-Barrero. c ∈ R. c ∈ R. Equation x3 − 2x2 − x + 1 = 0 has three real roots a > b > c. The polynomial p(x) = x is a solution to our problem because for all a. n = 2. BARCELONA TECH. 2 Also solved by Albert Stadler. Find all polynomials p(x) with real coefficients such that p(a + b − 2c) + p(b + c − 2a) + p(c + a − 2b) = 3p(a − b) + 3p(b − c) + 3p(c − a) for all a. (Benelux Mathematical Olympiad 2010) Solved by Omran Kouba. We give the solution of D´ıaz-Barrero. √ √ √ √ √ √ (x + y 3)4 + (z + t 3)4 = 7 + 6 3 ⇔ (x − y 3)4 + (z − t 3)4 = 7 − 6 3 √ √ √ The last equality is impossible because (x−y 3)4 +(z−t 3)4 ≥ 0 while 7−6 3 < 0. these are the only solutions. Spain. µ ∈ R. So we get the polynomial equation p(−2x) = p(x)+3p(−x). is a solution since for all a. Therefore. Barcelona. Now set b = c = 0. then we have p(a) + p(−2a) + p(a) = 3p(a) + 3p(−a) for all a ∈ R. 29. For n even we obtain 2n = 4. Let an 6= 0 be the coefficient of xn in p(x). so n = 1. a2 are real numbers (possibly zero). we have (a + b − 2c) + (b + c − 2a) + (c + a − 2b) = 3(a − b) + 3(b − c) + 3(c − a) Likewise. and Jos´ e Gibergans B´ aguena. b. Spain. b. since we have already shown that there can be no other solutions. Since in the LHS of the polynomial equation the coefficient of xn is (−2)n an while in the RHS the coefficient of xn is an + 3 (−1)n an . p(x) = x2 . Barcelona. . b. Spain. we must have p(x) = a2 x2 + a1 x. and for n odd. we have 3p(0) = 9p(0). Herrliberg Switzerland. Barcelona. BARCELONA TECH. hence p(0) = 0. Now suppose that p is not the zero polynomial. and let n ≥ 0 be the degree of p. then (−2)n = 1 + 3 (−1)n . Find the value of ab2 + bc2 + ca2 . Damascus. Note that the zero polynomial is a solution to this equation. where a1 . σ3 = abc = −1. = abc. v = a2 b + b2 c + c2 a Now. v = −1. Therefore. BARCELONA TECH. X < Y . Syria. Let u v w = a + b + c. Therefore. X < Y. Barcelona. then Y − X = (a − c)(c − b)(b − a) > 0. we conclude that σ1 = a + b + c = 2. we have u = 2. Let us define u and v by u = ab2 + bc2 + ca2 . c are the roots of x3 − 2x2 − x + 1 = 0. Since. b. we have XY = (a3 + b3 + c3 − 3abc)abc + (a3 b3 + b3 c3 + c3 a3 − 3a2 b2 c2 ) + 9a2 b2 c2 = u(u2 − 3v)w + v(v 2 − 3uw) + 9w2 = −12 Taking into account Cardan-Vi`ete formulae again. Spain. Since a. By Cardan-Vi`ete formulae. Higher Institute for Applied Sciences and Technology.108 (Training Sessions for COM-2012) Solution 1 by Jos´ e Gibergans B´ aguena. Since a > b > c. 2 Solution 2 by Omran Kouba. using a + b + c = 2 we have u+v = ab(a + b) + bc(b + c) + ca(c + a) = ab(2 − c) + bc(2 − a) + ca(2 − b) = 2(ab + bc + ca) − 3abc = −2 + 3 = 1 Similarly using abc = −1 we have b c a c b a uv = + + + + c a b b a c 2 2 2 a b c bc ac ab = 3+ + + + + 2 + 2 bc ac ab a2 b c 1 1 1 3 3 3 = 3−a −b −c − 3 − 3 − 3 a b c (1) . Damascus. we have that X and Y are the roots of the equation t2 − t − 12 = (t + 3)(t − 4) = 0. ab2 + bc2 + ca2 = −3 and we are done. σ2 = ab + bc + ca = −1. then X = −3 and Y = 4. = ab + bc + ca. Let us denote by X = ab2 + bc2 + ca2 and by Y = ac2 + ba2 + cb2 . w = −1 and X + Y = a(ab + ac) + b(bc + ba) + c(cb + ca) = (a + b + c)(ab + bc + ca) − 3abc = uv − 3w = 1 Taking into account the identity x3 + y 3 + z 3 − 3xyz = (x + y + z)(x2 + y 2 + z 2 − xy − yz − zx). uv = 3−11−4 = −12. b. Italy. Thus. Department of Mathematics. so u and v are the roots of the second degree equation Z 2 − Z − 12 = 0. . so σ1 σ2 1 +3 2 σ3 σ3 1+6−3=4 −3 Finally. Then we would have a q n=1 n q(a1 · a2 · · · an ) n ∞ X X 1 1 + p(a1 · a2 · · · an ) = p(a1 · a2 · · · an ) ak ak k=1 k=n+1 so the following number p (a1 · a2 · · · an ) ∞ X k=n+1 1 ak must be an integer for any positive integer n. b. Tor Vergata University. That is u = −3. c 23 + 6 − 3 = 11. This allows us to identify u as the smallest root of the second degree equation Z 2 − Z − 12 = 0. Let {an }n≥1 be a strictly increasing sequence of positive integers such that ∞ X an+1 1 lim = +∞. . we have proved that u +v = 1 and uv = −12. an+k ) > ck (a1 · · · an ) ∞ X 1 p Now suppose that = is a rational number. Rome. Barcelona. . . . an a n=1 n (Training Sessions of Spanish Math Team for VJIMC 2012) Solution 1 by Paolo Perfetti. 30. BARCELONA TECH. and then an+2 > c (a1 · · · an an+1 ) > c2 (a1 · · · an )2 > c2 (a1 · · · an . 1 1 1 a. that is u < v. we have u − v = (a − b)(b − c)(c − a) < 0. c are the roots of x3 − σ1 x2 + σ2 x − σ3 = 0 we see that a3 + b3 + c3 = σ1 (a2 + b2 + c2 ) − (a + b + c)σ2 + 3σ3 = σ1 (σ12 − 2σ2 ) − σ1 σ2 + 3σ3 = σ13 − 3σ1 σ2 + 3σ3 = Similarly. 2 Also solved by Jos´ e Luis D´ıaz-Barrero. are the roots of x3 − 1 1 1 + 3+ 3 3 a b c σ2 2 σ3 x = = + σ23 σ33 σ1 σ3 x − 1 σ3 = 0. Prove that is an irrational number. From the statement immediately follows that for any c > 1 there exists nc such that for all n > nc holds an+1 > c (a1 · a2 · · · an ). Spain.109 Since a. On the other hand. . n→+∞ a1 a2 . But if c > p + 1 > 1 then holds ∞ ∞ X X 1 1 p 0 < p(a1 · a2 · · · an ) ≤p = <1 ak ck−n c−1 k=n+1 k=n+1 contradicting what it was assumed and the proof is complete. an1 q ak ∞ X a1 a2 . then ak k=n1 +1 a1 a2 . an for every n ≥ n0 . . there is some n→∞ a1 a2 . . . an1 an1 +1 an1 +2 . . So. . . From (5) immediately follows that for all k > n1 + 1. that an+1 ≥ 2an for every n ≥ n0 . Barcelona. . 0< ∞ X k=n1 +1 a1 a2 . .110 2 Solution 2 by Jos´ e Luis D´ıaz-Barrero. . an1 q X = + ak ak k=1 k=1 ∞ X Since p(a1 a2 . . . let α denote the sum of the series. an ∞ X 1 1 1 ≤ n for all n ≥ n0 and the series converges. Since lim = +∞. . . is ≥ 2. . . . an1 q ≤ ak which is not possible because ∞ X k=n1 +1 k−n1 ∞ X 1 1 = 3 2 k=n1 +1 a1 a2 . BARCELONA TECH. 2 Solution 3 by Omran Kouba. . Higher Institute for Applied Sciences and an+1 Technology. an1 q is an integer. Therefore. an1 q a1 a2 . an n0 ≥ 3 such that an+1 ≥ a1 a2 . . . . for q(a1 a2 . Now. . . then there n→+∞ a q a a 1 2 . This proves the ∞ X 1 convergence of the series . in particular. an1 q ak is also an integer. This completes the ak proof. an ) all n ≥ n1 holds an+1 ≥ 3q(a1 a2 . where p and q are integers. . . . an1 q are both integers. . . Syria. . . . for all n ≥ n1 . an ) ≥ 3 (5) On the other hand. . . Assume that ∞ X 1 p an+1 = . an+1 2 a n=1 n see that its sum is an irrational number we argue by contradiction. an+1 Spain. an ) ≥ 2n . for all a1 a2 . an1 ) = n1 ∞ ∞ X X a1 a2 . ak k=1 R n = α − Sn = ∞ X k=1 1 an+k . and let a n=1 n Sn = n X 1 . . an an+1 that for all n ≥ n0 . To n ≥ n0 . Since lim = +∞. an+1 ≥ 2(a1 a2 . . ak−1 )q 1 0< = ≤ ak an1 +1 an1 +2 . an1 ) and k=n1 +1 k=n1 +1 a1 a2 . . . Since clearly a2 ≥ 2 we conclude. p(a1 a2 . Since lim = +∞. . From the preceding. . an n=1 n an+1 exists n1 ≥ 2 such that ≥ 3. ak−1 ak 3 Therefore. . . . holds k−n1 (a1 a2 . . an1 )q (a1 a2 . then there exists a positive integer n0 such n→+∞ a1 a2 . Damascus. k−1 an+k an+1 2 an+1 k=1 k=1 Now. ∞ X a1 a2 . . . So. aN q a1 a2 . . . am ) − q(a1 a2 . as it the finite sum on the right. . and k ≥ 1. . aN q = + a a an n n n=1 n=1 n=N +1 We note that pa1 a2 . . where p and q are integers such that (p. . aN q because of the way we chose N . am )Sm is clearly an integer. The hypothesis implies the existence of a positive integer N such that aN +1 /a1 a2 . . . am 1 2 < 0 < Rm ≤ am+1 qa1 a2 .111 For n ≥ n0 . However. . Barcelona. suppose that α is a rational number. . 2 Also solved by Jos´ e Gibergans B´ aguena. aN q > 2 and an+1 > 2an for n ≥ N + 1..57. Because lim = +∞. an am+1 that > 2q. American Mathematical Monthly. . This contradiction proves that α is an irrational. . . . . . . . . am ) − q(a1 a2 . . q) = 1. . am )Sm < 1 which is absurd since the number p(a1 a2 . whose sum is 1. that is α = p/q where p and q are an+1 positive integers. 2 P∞ Solution 4 by Henry Ricardo. am but this is equivalent to 1 p 0 < − Sm < q qa1 a2 . Paolo Perfetti acknowledged us that this problem was proposed by Paul Erd¨ os and appeared with solution as Problem 4321. . . . BARCELONA TECH. + 1/2n + . USA. . Suppose that n=1 1/an = p/q. Vol. aN = N ∞ ∞ X X X a1 a2 . we have k−1 k−1 Y an+j Y1 an+1 1 = ≤ = k−1 . . Spain. . . No. . an+k a 2 2 n+j+1 j=1 j=1 Thus. Editors Comment. This contradiction establishes the desired result.5 (1950) 347–348. there is an integer m > n0 such n→∞ a1 a2 . aN q a1 a2 . the series is term-by-term less an n=N +1 than the series 1/2 + 1/4 + . aN is an integer. a1 a2 . for n ≥ n0 we have Rn = 1 an+1 ∞ ∞ X 2 an+1 1 X 1 ≤ ≤ . Then pa1 a2 . . . . . NY. am or 0 < p(a1 a2 . If for every x ∈ (0. Introduction The starting point was when I wanted to provide my students of “Basic Calculus” class. we see that the family (fγ )γ∈[0. then α ≤ 1 and β ≥ 6/5. our results are summarized in the following two statement Proposition 2. the preceding inequality states that 3(tan x − x) is somewhere between x3 and tan3 x. Using this family. where fγ was defined previously. So. with a way to prove that tan x − x 1 lim = 3 x→0 x 3 without using any advanced topics like the L’Hˆopital’s rule.3] defined on [0. π/2) we have fα (x) ≤ 3(tan x − x) ≤ fβ (x). it was just the beginning of my investigation. it is natural to be interested in identifying the best α and β such that fα (x) < 3(tan x − x) < fβ (x) for 0 < x < π/2. Theorem 1. but where exactly? In order to describe our results. In fact.3] is increasing in the sense that fα ≤ fβ for α < β. I came up with the following Proposition 1. 1. In this note we deal with some inequalities for the tangent function that are valid for all x in (−π/2. For every x ∈ (0. π/2) holds x+ tan3 x x3 < tan x < x + 3 3 Clearly. we can reformulate the inequality writing it in the form π f0 (x) < 3(tan x − x) < f3 (x). π/2). The following two inequalities hold . for 0 < x < 2 So. We were able to completely answer this question. But this was not the end of the story. the above limit follows immediately from the Proposition 1.112 MATHNOTES SECTION An Optimal Inequality of the Tangent Function Omran Kouba Abstract. an important role is played by the family of functions (fγ )γ∈[0. These inequalities are optimal in the sense that the best values of the exponents involved are obtained. π/2) by fγ (x) = x3−γ tanγ x Taking into account the well-known inequality tan x ≥ x for 0 ≤ x < π/2. 113 (a) For every x ∈ (0. π/2). Proposition 4. If for some 0 ≤ α. π/2). . 2. Thus. For instance. π/2) where x1 ≈ 1. Results and Proofs Clearly. and that the lower bound in (6) is sharper than the lower bound in (6) for x ∈ (x1 .2332. Equivalently. we have α β 3(tan x − x) tan x tan x ≤ ≤ x x3 x for every x ∈ (0. π/2) we have 3(tan x − x) < f6/5 (x). For every x ∈ (0. π/2) the following inequality holds: x+ x3 tan3 x < tan x < x + 3 3 Proof. we have g 0 (x) = tan2 x − x2 > 0 and h0 (x) = tan4 x > 0. the Becker-Stark’s inequality [1] states π2 π2 x 8x < tan x < 2 . π/2) we have f1 (x) < 3(tan x − x). π/2). 3 Clearly. but it can be proved directly. 2. 3 tan3 x + x − tan x. among other things. 3. the next Proposition 3 follows from our main Theorem 1. So the two results are complementary but not comparable. (b) For every x ∈ (0. 2 − 4x π − 4x2 for 0 < x < π 2 Also. and the desired inequality follows since g(0) = h(0) = 0. Proposition 3. both g and h are increasing on the interval (0. it is worth mentioning that there is a lot of similar inequalities involving trigonometric functions in the literature ([1. for x ∈ (0. let g and h be the functions defined on [0. one has 4 2 x3 2 4 x3 x+ + x tan x < tan x < x + + x4 tan x (7) 3 15 3 π Numerical evidence shows that the upper bound in (6) is sharper than the upper bound in (7) for x ∈ (0. x0 ) where x0 ≈ 1.5255. that for 0 < x < π2 . π2 holds 1 1 x + x2 tan x < tan x < x + x9/5 tan6/5 x (6) 3 3 Before proving our results. β ≤ 3. π/2) by g(x) = tan x − x − h(x) = x3 . then α ≤ 1 and β ≥ 6/5. for all x ∈ 0. Indeed. 4]). in [4] the authors prove. for every real x we have ∞ ∞ ∞ X X X (−1)n x2n (−1)n 32n x2n (−1)n−1 32n−1 x2n−1 φ(x) = (9 − 24x2 ) −9 − 4x (2n)! (2n)! (2n − 1)! n=0 n=0 n=1 = ∞ X 9 + 24(2n)(2n − 1) − 9 · 32n + 4(2n) · 32n−1 n=0 ∞ X =3 n=0 32n2 − 16n + 3) + (8n − 27) · 32n−2 (−1)n x2n (2n)! (−1)n x2n (2n)! . Proof. Suppose that for x ∈ (0. he preceding and the fact that α ≤ φ(x) ≤ β for every x ∈ (0. π/2). Let φ be the function defined on R by φ(x) = (9 − 24x2 ) cos(x) − 9 cos(3x) − 4x sin(3x) Then φ(x) > 0 for 0 < x ≤ π/2. x3 x Now. π/2) we have α β tan x 3(tan x − x) tan x ≤ ≤ . since φ(x) = log(tan x) + log(1 − x/ tan x) + log 3 − 3 log(x) log(tan x) − log x we conclude that lim − φ(x) = 1 x→( π 2) On the other hand. In order to determine the sign of φ(x) for x ∈ (0. we need the following technical lemma. π/2) by tan x 3(tan x − x) log φ(x) = log . φ(x) = 6 5 6 . lim+ φ(x) = To prove our main theorem. we will use power series expansion. x 3 15 we get tan x x2 log + O(x4 ) = x 3 3(tan x − x) 2 2 4 log = log 1 + x + O(x ) x3 5 2 = x2 + O(x4 ) 5 + O(x2 ). since in the neighborhood of 0 we have tan x x2 2 =1+ + x4 + O(x6 ). π/2]. Lema 1. imply that α ≤ 1 and β ≥ 65 as desired. 5 x→0 Therefore. x x3 x that is α ≤ φ(x) ≤ β where φ is defined on (0.114 Proof. and consequently Thus. Clearly. then the preceding series is alternate. Now. Thus. (2n)! where. (2n)! (2n + 2)! √ is decreasing for any x ∈ (0. and Lemma √ 1 follows since π2 < 3. An and Bn are positive for n ≥ 4 on account √ that Un > 0 for n ≥ 4. 3 . Noting that T0 = T1 = T2 = T3 = 0 we conclude that φ(x) = 3 ∞ X n=4 (−1)n Tn 2n x (2n)! Since Tn > 0 for n ≥ 4. it is straightforward to check that Bn+1 = 128n4 + 640n3 + 1036n2 + 437n + 69(n − 1). for a real x we have φ(x) = 3 ∞ X (−1)n n=0 Tn 2n x . It follows that the sequence Tn 2n (2n)! x . Tn 2n Tn+1 x > x2n+2 . 3]. ∀ n ≥ 4. a simple computation shows that Un = (4n2 + 6n + 2) 32n2 − 16n + 3 + (8n − 27)9n−1 − 3(32n2 + 48n + 19 + (8n − 19)9n ) = 128n4 + 128n3 − 116n2 − 158n − 51 + (32n3 − 60n2 − 362n + 459)9n−1 = Bn + An · 9n−1 . and. An+4 = 32n3 + 324n2 + 694n + 99. (8) where Bn = 128n4 + 128n3 − 116n2 − 158n − 51 An = 32n3 − 60n2 − 362n + 459 Now. Tn = 2(4n − 1)2 + 1 + (8n − 27)9n−1 . as n≥4 √ we have already explained. 3] (2n + 2)(2n + 1)Tn > x2 Tn+1 or. this implies that φ(x) > 0 for x ∈ 0. π/2]. 3 . because the first term in the last series is positive. Un = (2n + 2)(2n + 1)Tn − 3Tn+1 . equivalently.115 Thus. Let Un be defined by. Using Un again. we conclude that for n ≥ 4 and x ∈ (0. √ i ∀ x ∈ 0. π/2] then this would n≥4 imply that φ(x) > 0 for x ∈ (0. if we prove Tn that the sequence (2n)! x2n is decreasing for any x ∈ (0. we are ready to prove our Main result. This implies that h is positive on the interval (0. h is increasing. recalling the expression of cos(3x) and sin(3x) in terms of cos x and sin x we see that (9 − 24x2 ) cos(x) − 9 cos(3x) − 4x sin(3x) g 0 (x) = . π/2). (b) Again. π/2). Similarly h has a derivative on [0. is that α ≤ 1 and β ≥ 6/5. since limx→0+ g(x) = 0. x x3 We have 6 9 5 sin2 x g 0 (x) = + − cos x sin x x cos x(sin x − x cos x) = (9 − 6x2 ) cos x + x(4 sin3 x − 3 sin x) − 9 cos3 x x cos x sin x (sin x − x cos x) So. Proof. Proof of Theorem 1. but it is straightforward to check that this is equivalent to the fact that 3(tan x − x) > x2 tan x for x ∈ (0. π/2). (a) Consider the function g defined on the interval (0. π/2) given by where h(x) = x − 3 + tan2 x (3 − tan2 x)(1 + tan2 x) h0 (x) = 1 − 3 (3 + tan2 x)2 2 4 tan x = (3 + tan2 x)2 So. and from the fact that the functions considered are even. Finally. 3 tan x . Using this Lemma we see that g is increasing on (0. Theorem 1.116 Now. But limx→0+ g(x) = 0. π/2) by tan x 3(tan x − x) g(x) = 6 log − 5 log . with h(0) = 0. (9) 4x cos2 x sin x (tan x − x) where φ is the function considered in Lemma 1. π/2) which is the desired inequality. for the following inequality α β x2 tan x tan x x2 tan x 1+ < <1+ 3 x x 3 x to hold for every nonzero real x from (−π/2. and this is equivalent to 3(tan x − x) < x9/5 tan6/5 x which is the desired inequality. Going back we conclude that g is also increasing on (0. we will consider an auxiliary function. 4x cos2 x sin x (tan x − x) φ(x) = . we conclude that g is positive on (0. π/2) by g(x) = 3 − x2 − 3x cot x We have g 0 (x) = x − 3 cot x + 3x cot2 x = (1 + 3 cot2 x)h(x). 2 Corollary 1. The necessary and sufficient condition. on the real numbers α and β. so g is positive on (0. π/2). π/2). . π/2). This follows from Proposition 4. Let g be the function defined on (0. W On new proofs of inequalities involving trigonometric functions. [4] Chen. Univ. Fiz. F. 602-633(1978). Syria E-mail address: omran kouba@hiast. On a hierarchy of quolynomial inequalities for tan x.1 (2003). A Double Inequality for Remainder of Power Series of Tangent Function. Mat. No. Department of Mathematics. and Li. Nova Science Publichers. E. B. Qiao. 19–22. Elektrotechn. 351–355. B. (2003).-M. Proofs of Wilkers inequalities involving trigonometric functions.. Li. 133–138. No. F. 34.. and Stark. 109–112.. Tamkang Journal of Mathematics.. Beograd. Fak. W and Qi. (2003). F. Math. B..117 References [1] Becker. Appl. 6. [3] Guo. [2] Guo... Damascus. Qi. Ch. Ser.. 2. M.-N. Inequality Theory and Applications.-N... Publ.edu. Inequal.sy . and Qi. L. No. 4..-P. Higher Institute for Applied Sciences and Technology. Let a ∈ (1. Bucharest. Matei Basarab National Colege. 51. Kosov¨ e. Buzˇ au. The editors encourage undergraduate and pre-college students to submit solutions. Ovidiu Furdui. ∞) and b ∈ (0. each indicating the name of the sender. Student solutions should include the class and school name. b. Calculate !! n X (n + k)a−1 lim n 2 − exp n→∞ (n + k)a + b k=1 52.com Volume 2. Let a. Athens. Prishtin¨ e. Proposed by D.M. An asterisk (*) indicates that neither the proposer nor the editors have supplied a solution. Armend Sh. Paolo Perfetti. J´ozsef S´andor. The School of Physics and Technology (SPT) at Wuhan University. Solutions will be evaluated for publication by a committee of professors according to a combination of criteria. Roberto Tauraso. url: http://www. PROBLEMS AND SOLUTIONS Proposals and solutions must be legible and should appear on separate sheets. Enkel Hysnelaj.Mathproblems ISSN: 2217-446X. Greece. c > 0. Romania (Jointly). George Emil Palade Secondary School. Shabani. Proposed by Yuanzhe Zhou. cos b 2 c 2 a 2 3 + cos + cos > b+c a+c a+b 2 c 2010 Mathproblems. Questions concerning proposals and/or solutions can be sent by e-mail to: mathproblems-ks@hotmail. Romania and Neculai Stanciu. Jos´ e Luis D´ıaz-Barrero. B˘ atinet¸u-Giurgiu. Show that +∞ X k=0 (−1)k 1 = + O(n−1 ln−2 n) ln(n + k) 2 ln n n → +∞. Drawings must be suitable for reproduction. Stone. Teachers can help by assisting their students in submitting solutions.com Solutions to the problems stated in this issue should arrive before February 15. ∞). Issue 4 (2012). Mih´aly Bencze. Valmir Bucaj. 2013 Problems 50. prove that. Proposed by Anastasios Kotronis. Universiteti i Prishtin¨ es. Pages 118–143 Editors: Valmir Krasniqi. David R. Cristinel Mortici. Proposals should be accompanied by solutions. 118 .mathproblems-ks. France. Australia. Show that if X nx3 + (n + 1)x =α x2 + 1 cycl then X1 2α 27n3 > + 2 x 3 9n α + α3 cycl where n is a natural number. Proposed by Enkel Hysnelaj. Republic of Kosova.∗ Proposed by Naim L. Let n and m be nonnegative integers with n > m − 1. the space of Lebesgue integrable functions. . k)z . University of Prishtina. z. The Stirling numbers of the first kind denoted by s(n. Proposed by Moubinool Omarjee. sinh n→+∞ R 2 2 + where sinh(x) = ex −e−x 2 55. are the special by the generating function z(z − 1)(z − 2) · · · (z − n + 1) = Pn numbers defined k s(n. k). Prove that k=0 1 Z 0 lnn x dx = (1 − x)m ( (−1)n · n! · ζ(n + 1). Let f : [0. BARCELONA TECH. Romania.119 53. α be real positive numbers. Braha. m = 1. Find all possible values of the number k such that 2 2 2 2 2 Fn+2 ) Fn+2 Fn2 (1 + Fn+1 ) (1 + Fn2 Fn+1 F 2 (1 + Fn+2 Fn2 ) + = k + n+1 Fn−1 Fn+1 Fn Fn+1 Fn−1 Fn 57. Cluj-Napoca. the following series converges: fn = f1 b(n−1)/2c X j=0 + f0 b(n−2)/2c X j=0 2n−1−2j 2n−2−2j X j Y 1 i1 ··· ij ≤n−2 m=1 X j Y 2 i1 ··· ij ≤n−1 m=1 where a b if and only if a + 1 < b. 56. 1 λ(im + 1)2 1 λ(im + 1)2 −1 − 1 . Department of Mathematics and Computer Sciences. University of Technology. P i n! (−1)n+m−1 · (m−1)! · m−1 i=1 (−1) · s(m − 1. y. +∞) → R be a measurable function such that g(t) = et f (t) ∈ L1 (R+ ). m ≥ 2. Let x. then prove for which values of λ. Technical University of Cluj-Napoca. Paris. where ζ denotes the Riemann zeta function. Barcelona. Sydney. Find n1 Z nt + 1 nt − 1 f (t) 4 sinh lim dt. i) · ζ(n + 1 − i). If f0 and f1 are constants. 54. Proposed by Ovidiu Furdui. Proposed byJos´e Luis D´ıaz–Barrero. Spain. Let n ≥ 2 be a positive integer. We will be very pleased considering for publication new solutions or comments on the past problems. e 2e 2e e and 1 1 1 1 1 x −x (Γ(x + 1)) sin = +O x−1 ln2 x → . we get 1 x ln x ln 2π 1 + + + O x−1 ln2 x . x+a x e e .M. (Γ(x + 2)) x+1 = + e 2e 2e e Furthermore.120 Solutions No problem is ever permanently closed. x 1 (x + a) sin = 1 + O(x−2 ) x+a 1 x and therefore 1 1 x (Γ(x + 1)) x sin = x 1 1 (x + a) (Γ(x + 2)) x+1 sin = x+a On account of the above 1 x+1 (x+a) (Γ(x + 2)) sin x ln x ln 2π + + + O x−1 ln2 x e 2e 2e x ln x ln 2π 1 + + + + O x−1 ln2 x . 43. B˘ atinet¸u-Giurgiu. Buzˇ au. we have Γ(x + 1) = so √ 2πxx+1/2 1 + O(x−1 ) ex ln (Γ(x + 1)) = x ln x − x + and x → +∞. Calculate 1 1 1 1 x+1 x sin lim (x + a) (Γ(x + 2)) − x (Γ(x + 1)) sin x→∞ x+a x Solution by Anastasios Kotronis. Matei Basarab National Colege. Athens. e 2e 2e Putting x + 1 in the preceding expression instead of x. Bucharest. George Emil Palade Secondary School. 1 x sin = 1 + O(x−2 ). Let a be a positive real number and Γ(x) be the Gamma function (or Euler’s second integral). Greece. Romania and Neculai Stanciu. ln x ln 2π + + O(x−1 ) 2 2 ln x ln 2π −2 (Γ(x + 1)) = exp ln x − 1 + + + O(x ) 2x 2x x ln x ln 2π = + + + O x−1 ln2 x . From Stirling’s formula. Proposed by D. Romania (Jointly). University of Athens. p+2 2 6 p=1 P∞ where ζ is the Riemann zeta function defined by ζ(s) = k=1 1/k s for <(s) > 1. Paolo Perfetti. Italy. First. Proposed by Ovidiu Furdui. Let A denote the Glaisher–Kinkelin constant defined by A = lim n n→∞ −n2 /2−n/2−1/12 n2 /4 e n Y k k = 1. we have ∞ X ζ(2p + 1) − 1 p+2 p=1 = Since ∞ X z 2r ∞ X z 2p+1 p=1 1 = 3 p+2 z r=1 z4 −z − r 2 2 ∞ X ∞ X k=2 1 1 2p+1 p + 2 k p=1 ! 1 =− 3 z z4 ln(1 − z ) + z + 2 2 2 then ∞ X ζ(2p + 1) − 1 p=1 p+2 = ∞ X 1 1 1 1 3 = − k + ln 1 − + k 2p+2 p + 2 k2 k2 2k 4 p=1 ∞ X ∞ X k=2 1 k=2 Now. =− Solution 1 by Paolo Perfetti. . Department of Mathematics. we have X n n n n X X X 1 3 3 3 k ln 1 − 2 = k ln(k + 1) + k ln(k − 1) − 2 k 3 ln k k k=2 = k=2 n X k=2 + + + n X k=2 n X k=2 n X k=2 k=2 (k + 1)3 ln(k + 1) + k=2 n X k=2 (k − 1)3 ln(k − 1) − 2 (3(k − 1) ln(k − 1) + 3(k + 1) ln(k + 1)) 3(k − 1)2 ln(k − 1) − 3(k + 1)2 ln(k + 1) (ln(k − 1) − ln(k + 1)) n X k=2 k 3 ln k . Romania. Damascus.121 Also solved by Omran Kouba. Tor Vergata University. Rome. . Greece. and the proposer 44.282427130 . Rome. . Cluj-Napoca. Technical University of Cluj-Napoca. Syria. k=1 Prove that ∞ X ζ(2p + 1) − 1 γ 7 − 6 ln A + ln 2 + . Konstantinos Tsouvalas. Athens. Department of Mathematics. Italy. Tor Vergata University. Higher Institute for Applied Sciences and Technology. 2 n X 1 1 γ 1 = ln n + − + o(1) 2k 2 2 2 k=2 Adding up the three preceding contributions. Athens. we have +∞ X ζ(2p + 1) − 1 p=1 p+2 = X p≥1 X X n−1 2p+1 1 X 1 − 1 = p+2 n2p+1 p+2 n≥1 p≥1 n≥2 X X n−1 2p+1 X 1 1 3 − + n + n ln 1 − 2 = p+2 2n n n≥2 p≥1 n≥2 N N X X 1 1 1 1 + n + n3 ln 1 − 2 = − lim + n + n3 ln 1 − 2 N →+∞ N →+∞ 2n n 2n n n=2 n=2 = − lim = − lim N →+∞ HN − 1 (N + 2)(N − 1) + + ln 2 2 = − lim N →+∞ n N 2 Y n −1 n=2 HN N2 N 3 + + − + ln AN 2 2 2 2 n2 3 !! . First we observe that for |x| < 1 is p X x2p+1 1 X x2 1 x4 2 2 = = − ln(1 − x ) + x + p+2 x3 p x3 2 p≥3 p≥1 = − x 1 ln(1 − x2 ) − − 2 x x3 Now. and the statement follows. Solution 2 by Anastasios Kotronis.122 and n X k=2 1 k 3 ln 1 − 2 = (n + 1)3 ln(n + 1) − n3 ln n − 8 ln 2 k +6 n X k ln k − 3n ln n + 3(n + 1) ln(n + 1) − 6 ln 2 n X k ln k − 3n2 ln n + n2 − 3n ln n − k=1 + −3n2 ln n − 3(n + 1)2 ln(n + 1) + 12 ln 2 − ln n − ln(n + 1) + ln 2 =6 k=1 1 n − ln n + 2 3 Moreover n X k= k=2 n(n + 1) − 1. Greece. we get 6 n X k=1 k ln k − 3n2 ln n − 3n ln n − whose limit when n → ∞ is 6 ln A − ln 2 + γ 2 ln n 3 2 γ 7 + n − ln 2 + − 2 2 2 6 − 76 . Mongolia. there exists x. and the proposer. 3 3N 2 /2 (N +1) 2 2 2 3 2N e then on account of the preceding. Then. Syria. B. z ∈ Z. Proposed by Moubinool Omarjee. we get +∞ X ζ(2p + 1) − 1 γ 7 = lim − − 6 ln A + ln 2 + + o(1) N →+∞ p+2 2 6 p=1 as desired. k ∈ Z. Then we have a − 1 − x2 = d − 1 − t2 and let us denote a − 1 − x2 = d − 1 − t2 = yz. Higher Institute for Applied Sciences and Technology. Konstantinos Tsouvalas. 45. Greece. k ∈ Z or a − d = 4k. Paris. c d Prove that there exist (A. a b ∈ M2 (Z). (a) Let a − d = 2k + 1. France. Ulaanbaatar. Damascus. y. University of Athens. z t c − xy − yz 0 0 1 .123 But 3 3 N N Y (n − 1)n Y (n + 1)n = n n3 n n3 n=2 n=2 AN = 3 N −1 N 1 Y (k+1)3 −k3 (N + 1)N Y (k−1)3 −k3 k · k N N3 223 k=2 k=3 N3 = = 1 (N + 1) · 2 N (N +1)3 N N Y k 6k k=1 −N 2 /2−N/2−1/12 N 2 /4 e N Y k=1 k k !6 3 · 2 (N + 1)N N 3N +3N +1/2 . Athens. we have 2 a b x + yz + 1 b = c d c t2 + yz + 1 2 x + yz xy + yt 1 b − xy − yt = + xz + zt t2 + yz c − xy − yz 1 2 x y 1 0 0 b − xy − yt = + + z t c − xy − yz 0 0 1 2 2 2 x y 1 0 0 b − xy − yt = + + . Let M = Solution by AN-anduud Problem Solving Group. Also solved by Omran Kouba. C) ∈ M2 (Z) × M2 (Z) × M2 (Z) such that M = A2 + B2 + C 2. t ∈ Z such that a − d = x2 − t2 . Since 2k + 1 = (k + 1)2 − k 2 and 4k = (k + 1)2 − (k − 1)2 . 2N (N +1)3 e3N 2 /2 Since HN = ln N + γ + o(1) and 3 ln 2 (N + 1)N N 3N +3N +1/2 N2 N ln N 1 =− − − + − ln 2 + o(1). 0 0 Also solved by Omran Kouba. Greece (Jointly). Letting t → i. sin z = eiz − e−iz . Then we also can show a b 2 + (k + 1)2 + yz b = c d c (k − 1)2 + yz (k + 1)2 + yz 2ky = 2kz (k − 1)2 + yz 2 b − 2ky + c − 2kz 0 2 2 k+1 y 1 0 = + z k−1 c − 2kz 0 2 1 b − 2ky + . Ionnina. ClujNapoca. p. n where ζ is the Riemann zeta function. Formula 11. [1. 2001. Evaluate +∞ X (−1)n−1 n=1 ζ(2n) . Athens. Choi. Proposed by Anastasios Kotronis. Syria. 160]) ∞ X ζ(2k) k=1 t2k = ln Γ(1 + t) + ln Γ(1 − t). 2i References [1] H. Solution 1 by Ovidiu Furdui. Technical University of Cluj-Napoca. k |t| < 1. Let a − 2 − (k + 1)2 = d − (k − 1)2 = yz. x.124 (b) Let a − d = 4k + 2. we get ∞ X (−1)k−1 k=1 ζ(2k) = − ln (Γ(1 + i) · Γ(1 − i)) k = − ln (i · Γ(i) · Γ(1 − i)) π·i = − ln sin(π · i) π e − e−π . Also we have a − d − 2 = 4k = (k + 1)2 − (k − 1)2 ⇔ a − 2 − (k + 1)2 = d − (k − 1)2 . in the preceding formula. and Serafeim Tsipelis. Romania. 46. M. and the proposer. = ln 2π where the last equality follows on account of Euler’s formula. Srivastava and J. Series Associated with the Zeta and Related Functions. Greece. More generally it is known that (cf. y ∈ Z. k ∈ Z.. . Higher Institute for Applied Sciences and Technology. Damascus. Kluwer Academic Publishers. Konstantinos Tsouvalas. AN-anduud Problem Solving Group. University of . Damascus.e. (nπ)2 n=1 Finally. Syria. Moreover. · · · . π n n=1 Also solved by Omran Kouba. The easiest way of determining the coefficients c0 and c1 is to observe that sinh z eg(z) = Q ∞ z2 z n=1 1 + (nπ) 2 is an even function. ∞ Y 1 sinh π = 1+ 2 . πi. Rome. Higher Institute for Applied Sciences and Technology. and hence g(z) = 0. Italy. University of Antioquia. < +∞ |nπi| n=1 nπ (nπ)2 n=1 n=1 i. Putting z → 0 in the above expression we conclude that ec0 = 1. −πi.125 Solution 2 by Robinson Higuita and Joel Restrepo(Jointly). = n n (k 2 )n n (k 2 )n n=1 n=1 k=1 But 1 ln 1 + 2 k Thus ∞ X n=1 We see that (−1) n−1 ζ(2n) n = ∞ X k=1 k=1 ∞ X (−1)n−1 1 = . We know that ∞ X 1 ζ(2n) = . Ulaanbaatar. Colombia. the convergence exponent of this sequence obviously is τ = 1. Thus. ∞ Y z2 sinh z = z 1+ . Mongolia. Paolo Perfetti. nπi. and hence g(z) = c0 + c1 z. · · · . the order of the entire function sinh z obviously is ρ = 1. Hence c1 = 0. 1− nπi nπi (nπ)2 n=1 n=1 Further. All zeros of the entire function sinh z are first order and they are the points 0. n (k 2 )n n=1 1 ln 1 + 2 = ln k ∞ Y 1 sinh π 1+ 2 = n π n=1 ! ∞ Y 1 + k2 k=1 k2 . while . Department of Mathematics. κ = 1. −nπi. Tor Vergata University. and hence ec0 +c1 z = ec0 −c1 z . ∞ ∞ ∞ X X X 1 1 1 = = +∞. Hence. by Theorem (Hadamard’s Factorization) ∞ ∞ Y Y z2 z −z/nπi z z/nπi 1+ sinh z = eg(z) z 1+ e e = eg(z) z . (k 2 )n k=1 therefore ∞ X (−1)n−1 n=1 ∞ ∞ X ∞ ∞ X (−1)n−1 1 ζ(2n) X (−1)n−1 X 1 = . ∀x ∈ (0. 0)} o y [n = 0 . Clearly. ∞). Let f : (0.126 Athens. A numeric evaluation shows that λ1 ≈ 0. Greece. Polytechnic University. 0 < a < b such that R b f (x) f (b) f (a) (i): a 2 dx = − x b a 0 f (a) 2 (ii): = f (a) a (a) Find an example of a function that satisfies the preceding conditions (b) Show that there exists a c ∈ (a. Statesboro. 48. Proposed by David R. Syria. Also solved by the proposer. 1) and λ2 ∈ (7. GA. Proposed by Florin Stanescu. Joaquin Rivero Rodriguez. School Cioculescu Serban. and the proposer. b) such that 0 c 2 f (c) f (c) (c−a) f (c) − 2c = e f (a) a Solution by Adrian Naco. Georgia Southern University. Now. Dambovita. Stone. Tirana.63270759 and λ2 ≈ 7. Cosequently. Damascus. where k is a constant number. (x. Romania. 8).49826796. ∞) x=a . y) ∈ R2 | y 4 − 10x2 y 3 + 25x4 y 2 − 50x6 y + 24x8 = 0 . a simple study of the variations of f shows that the equation f (t) = 0 has exactly two real zeros λ1 ∈ (0. G is the union of the two parabolas P1 and P2 of equations y = λ1 x2 and y = λ2 x2 respectively. ∞) be a differentiable function for which there exist real numbers a and b. USA. satisfy the condition (ii). (a) The function f (x) = kx2 . Let G = (x. jud. G = {(0. 47. x−a F (x) = 0 x ∈ (a. ∞) → (0. Athens. Higher Institute for Applied Sciences and Technology. y) ∈ R∗ × R : f x2 where f (t) = t4 − 10t3 + 25t2 − 50t + 24. Gaesti. Albania. Spain. Identify the graph of the equation y 4 − 10x2 y 3 + 25x4 y 2 − 50x6 y + 24x8 = 0 Solution by Omran Kouba. (b) The function F is defined by f (a) f (x) −ln ln x2 a2 . considering the fact that F (α) = F (γ) = 0 and based on Rolle’s Theorem. β) such that x2 x=γ Thus 0 0 f (x) f (γ) 2 = − =0 lim F (x) = ln x→γ x2 f (γ) γ x=γ The function F is continuous in [α. ∞) and even continuous in the point a. based on Darboux 0 f (x) = 0. x−a x=c 0 f (c) f (c) f (a) 2 0 (c − a) f (c) − c − ln c2 − ln a2 f (c) 2 f (c) f (a) = 0 ⇒ (c−a) − − ln 2 −ln 2 = 0 c−a f (c) c c a or (c−a) f (c) f (a) ln 2 = ln 2 + ln e c a and f (c) f (a) (c−a) = e c2 a2 0 f (c) − 2c f (c) 0 f (a) − 2c f (c) 0 f (c) − 2c f (c) (c−a) f (c) f (a) ⇔ ln 2 = ln 2 + ln e c a 2 (c−a) f (c) c ⇔ = e f (a) a 0 f (a) − 2c f (c) . β]. Indeed from the definition 0 0 f (x) f (a) 2 lim+ F (x) = ln = − = 0 = F (0) x2 f (a) a x→a x=a On account of (i) and applying the integration by part method we have 0 b Z b Z b f (x) f (x) f (x) dx = x 2 dx x − 2 x x x2 a a a Z b f (b) f (a) f (x) = dx = 0 − − b a x2 a 0 0 f (x) f (x) Case 1. considering the x2 x2 0 f (x) above result. Theorem it implies that there exists a point γ ∈ (α. we get = 0 for all x ∈ (0. ∞) ⇒ f (x) = kx2 . There are at least two points α. γ) such that F 0 (c) = 0 or = 0. we have that there f (a) ln f (x) 0 x2 − ln a2 exists c ∈ (α. Case 2. Then. thus. That is. the condition (b) holds for every c ∈ (0. ∞). If ≥ 0 or ≤ 0 for all x ∈ (0. with k ∈ R in x2 which case.127 which is everywhere continuous and differentiable in (0. ∞) such that 0 0 f (x) f (x) >0 and <0 x2 x=α x2 x=β Since the function f (x) x2 is continuous and differentiable in [α. β ∈ (a. γ] and differentiable in (α. γ). ∞). Romania. Tirana. Albania. . (n ≥ 3). 49. b. Proposed by Mih´ aly Bencze. c be positive real numbers. Based on Chauchy-Schwarz inequality we have X 2 X 2 X 2 2 X n n n n xi xi √ √ xi = xi xi+1 ≤ xi xi+1 √ √ xi xi+1 xi xi+1 i=1 i=1 i=1 i=1 = X n i=1 xi 2 xi xi+1 X n xi xi+1 i=1 ⇒ X n xi 2 X n n X xi xi 2 = ≥ ni=1 X x x x i=1 i+1 i=1 i i+1 xi xi+1 i=1 Furthermore X n 2 xi X n n X xi xi 2 i=1 −3 −3= −3≥ n X x xx i=1 i+1 i=1 i i+1 xi xi+1 i=1 = n X xi 2 + 2 i=1 n X xi xj 1≤i<j≤n n X xi xi+1 3 ≥ 2 xi + 2 i=1 n X xi xi+1 i=1 n X i=1 xi xi+1 3 − n X i=1 n X − xi xi+1 i=1 xi xi+1 xi xi+1 i=1 xi xi+1 i=1 n X i=1 n X n X ≥ n X i=1 2 xi − n X i=1 n X 1 2 xi xi+1 i=1 xi xi+1 = X n 2 (xi − xi+1 ) i=1 n X i=1 xi xi+1 . Ulaanbaatar.128 Also solved by AN-anduud Problem Solving Group. Mongolia. x2 . Prove that v uY u n n t (xi − xi+1 )2 n X n i=1 xi . Let x1 . Polytechnic University. and the proposer. be positive real numbers. . The generalization of the problem 49. Bra¸sov. ≥3+ n X x 2 i=1 i+1 xi xi+1 i=1 where xn+1 = x1 . Prove that p 3 (a − b)2 (b − c)2 (c − a)2 a b c + + ≥3+ b c a ab + bc + ca Solution 1 by Adrian Naco. . xn . Let a. ∀i ∈ {1. . . . For n ≥ 4 the equality is not true because n X xi xj > 1≤i<j≤n n X xi xi+1 i=1 Solution 2 by Eric Milesi Vidal. Barcelona. . we get ca2 ab2 bc2 + + b c a Equality holds if and only if a = b = c and we are done. First we subtract 3 from each side of the inequality p 3 (a − b)2 (b − c)2 (c − a)2 a−b b−c c−a + + ≥ b c a ab + bc + ca Now we prove that the following inequalities hold X p 1 1 1 (a − b)2 + (b − c)2 + (c − a)2 (a−b)ca( + + ) ≥ ≥ 3 (a − b)2 (b − c)2 (c − a)2 a b c 3 cyclic RHS is holds applying AM −GM inequality. . X cyclic a2 ≥ X ab and cyclic 3× X ca2 X ≥3× ab b cyclic cyclic The first inequality is well-known. Spain. . Ulaanbaatar. Department of Mathematics. . . BARCELONA TECH. Ioan Viorel Codreanu. we may assume WLOG that ca ba a ≥ b ≥ c so cb a ≤ b ≤ c Since both sequences are sorted in the opposite way by applying the rearrangement inequality. Maramure.129 ≥ 1 2 v uY u n n (xi − xi+1 )2 nt i=1 n X n 2 = xi xi+1 v uY u n n t (xi − xi+1 )2 i=1 n X i=1 xi xi+1 i=1 The last inequality is based on the well-known AM-GM inequality √ a1 + a2 + . The equality holds only for n = 3 and xi = x2 = . a2 . Tor Vergata University. Italy. and ai = (xi − xi+1 )2 . For the second. we have see that X X X ca2 (a − b)2 + (b − c)2 + (c − a)2 3ca2 a2 −ba−bc+ ≥ ⇔ a2 + ≥4 ab. Satulung. . Paolo Perfetti. ab + bc + ca ≤ Also solved by AN-anduud Problem Solving Group. = xn . b 3 b cyclic cyclic cyclic or equivalently. . n}. Equality holds if and only if a = b = c. an where a1 . Romania and the proposer. . . To prove LHS inequality. . . an ∈ R+ . Rome. 2. Mongolia. . an ≥ n n a1 a2 . β and γ be three distinct complex numbers. The students of a University Course in Mathematics take their exams in Calculus. any other chords AB and CD are drawn. Algebra. At least. Prove that M is the midpoint of XY. Prove that 2/3 1 X tan2 α 1 +3 ≥2 3 tan β tan γ tan α + tan β + tan γ cyclic . chords AD and BC meet P Q at points X and Y respectively. Physics and Geometry. 39. Proposals are always welcomed. and only if. Let α. 82% passed Algebra. 77% passed Physics and 89% passed Geometry. It is known that 73% passed Calculus exam. Through the midpoint M of a chord P Q of a circle. Show that they are collinear if. γ be the angles of an acute triangle ABC. how many students have passed the exam of all four subjects? 40. Im(αβ + βγ + γα) = 0.130 MATHCONTEST SECTION This section of the Journal offers readers an opportunity to solve interesting and elegant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. Proposals 36. 37. Find all positive integers n such that 17n−1 + 19n−1 divides 17n + 19n . Let α. β. The source of the proposals will appear when the solutions be published. 38. \ = BXY \ by spanning arc Spain. Finally. Spain. Next we draw the lines AX. AY which cut C2 at points AX and BY and lines BX. and the three lines AX BY . since AXB = AY B by spanning Y Y \ \ arc. Adrian Naco. Two circles C1 and C2 have in common two points X and Y . Omran . BARCELONA TECH. the last one is equal \ because they are angles opposite by the vertex. Show that the three lines AB. 2 Also solved by Jos´ e Luis D´ıaz-Barrero. we have \ = XA \ \ \ BAY Y BX + Y A Y X = Y A Y BX \ and Y \ Since line AAY cuts lines AB and AY BX and the angles BAY AY BX are equal. Likewise. Now. Polytechnic University. AY BX and AB are parallel. BY which cut C2 at points BX and BY respectively. Also by using again the half inscribed angle property. which is equal to AY B. Moreover. BXP is equal to QXB \ (opposite by the vertex). Barcelona. Tirana. Barcelona. AX BY and AY BX are parallel. (Short List of International Mathematical Arhimede Contest 2012) Solution by Guillem Alsina Oriol. and BXY = BXP + P XY . In the following figure we observe that BAY \ \ \ \ \X (or capacius arc). Albania. We know that \ AX\ BY BX is equal to AX XBX by spanning arc. Furthermore. we only need to see that AX BY is also parallel to these two lines. we have \ P XY = Y\ AY X.131 Solutions 32. It will be suffice to see that the angles AX\ BY BX and AY\ BX BY are equal. AY\ to AXB BX BY is \ \ \ equal to A\ Y B . then AX BY BX = AY BX BY . Q C1 BX AX X A BY B Y P C2 AY So. then lines AB and AY BX are parallel. We draw a line that cuts C1 at points A and B. BARCELONA TECH. and they are equal to XA Y BX on account of the half inscribed angle. Spain. ten ways of getting a sum score of 6 and so on. third and fourth dies represented by the same polynomial and the outcome of throwing are represented quite naturally by the polynomial (z+z 2 +z 3 +z 4 +z 5 +z 6 )(z+z 2 +z 3 +z 4 +z 5 +z 6 )(z+z 2 +z 3 +z 4 +z 5 +z 6 )(z+z 2 +z 3 +z 4 +z 5 +z 6 ) By expanding this. (Catalonian Mathematical Olympiad 2012) Solution by Jos´ e Luis D´ıaz-Barrero BARCELONA TECH. Damascus. there are four ways of getting a sum score of 5. and Oscar Rivero Salgado. respectively. Find the probability that the sum of the points appeared in the upper faces lies between 14 and 18 points.132 Kouba. Four dice are thrown at the same time on a table. We have to count how many possibilities there are to obtain sums of 15. 33. Higher Institute for Applied Sciences and Technology. we compute the distance from vertex A to the orthocenter of 4ABC. Therefore. H). Nf = 140 + 125 + 104 = 369 and the number possible sums Np = 1296. Barcelona. HB . Spain. BARCELONA TECH. z 16 . That is. then AC 0 AH AC 0 AC 0 AH = ⇔ = ⇒ AH = a CB CC 0 a CC 0 CC 0 0 0 Furthermore. 16 and 17 points. Spain. BARCELONA TECH. Spain. The second. the probability required is 41 369 Nf = = P(14 < s < 18) = Np 1296 144 2 Also solved by Jos´ e Gibergans-B´ aguena. To do it we consider the polynomial z + z2 + z3 + z4 + z5 + z6 Here. in triangle ACC we have AC = b cos A and CC 0 = b sin A. Barcelona. (Training Sessions of Spanish Team for IMO 2012) Solution by Jos´ e Luis D´ıaz-Barrero. we get z 4 + 4 z 5 + 10 z 6 + 20 z 7 + 35 z 8 + 56 z 9 + 80 z 10 + 104 z 11 + 125 z 12 + 140 z 13 + 146 z 14 +140 z 15 + 125 z 16 + 104 z 17 + 80 z 18 + 56 z 19 + 35 z 20 + 20 z 21 + 10 z 22 + 4 z 23 + z 24 and we find that there is one way of obtaining a sum score of 4. Barcelona. That is. BARCELONA TECH. HC be the orthocenters of triangles P BC. Thus. P AC. First. respectively. BARCELONA TECH. the powers of z keep track of the different faces of the dice and the coefficients of the powers of z show the number of occurrences of each face. and P AB. z 17 . 34. Spain. |AH| = dist(A. Barcelona. Barcelona. Prove that triangles HA HB HC and ABC have the same area. the number of favorable sums are the coefficients of z 15 . Eric Milesi Vidal. Since 4BCC 0 ∼ 4AHC 0 . Barcelona. respectively. BARCELONA TECH. Syria and Jos´ e Gibergans-B´ aguena. Let P be an interior point of triangle ABC and let HA . Substi- . Spain. P HB HC y P HC HA we obtain A(H H H ) = A(ABC) cot α cot β + cot β cot γ + cot γ cot α . we have P HA = −a cot α tres ´angulos. per ejemplo β = 90◦ (Figura 2). BC y AB y un es obtuso y el otro es agudo. 2) Supongamos que uno de los a´ngulos es recto.AB Fij´emonos que elisa´ngulo al principio tenemos P HA to C = −c ◦ C = 180◦ − B ya que los lados H P y HC P son. tenemos que cot(α + β) = − cot γ o. = A(ABC) cot α cot γ 2 A(P HA HC ) = 2 P H A P H C sin y ac cot α cot γ sin B = = A(ABC) cot α cot γ. Al least two of these three angles are obtuse and the other is obtuse. Let α = ∠BP C. 2 2 HA A z Figura 1 y γ β αP HC x B C HB Sumando. B. C we get the triangles P AB. Sean α = ∠BP C. (*) De aqu´ı resulta A(HA HB HC ) = A(ABC). B y C del tri´ angulo obtenemos los tres tri´ angulos P AB. AH = a cot A and in this case is cot A < 0. So we will examine these cases Cuando unimos el punto three arbitrario P conseparately. Notice that the angle y = los γ treson casos per separado. then point H lie on the exterior of 4ABC and we AH = a cot 90◦ = 0. El otro puede ser obtuso. γ = ∠AP B. α + β + γ = 360 . equivalentemente. perpendicular the sides and and one obtuse and the = −a cot α y P HBC cot γ. The area A(P−H∠AP A HC ) of triangle P HA HC is A ◦ respectivamente. but in this particular case is A = H and A b > 90◦ .133 A H C C b A c H c b C A a B A C a B tuting in the preceding expression yields AH b cos A = = | cot A| a b sin A from which follows AH = a | cot A|. So. Joining the arbitrary point with the vertices A. Evidentemente. and 1 − cot α cot β cot α + cot β cot α cot β + cot β cot γ + cot γ cot α = 1. y = ∠H AP H C = 180 other acute. Likewise. A(HA HB HAC B) =C A(P HA HB ) + A(P HB HC ) + A(P HC HA ) Como que α + β = 360 − γ. obviamente. las a´reas de los tres tri´angulos P HA HB . perpendiculares a los lados. cot γ = − cot(α + β) = o bien. right or acute. and P HEstudiaremos account of the preceding. and P CA. C = −c cot ◦ ◦ ∠HA P H1)C = 180 − ∠AP C = 180 − B because the sides HA P dicho and HC P are. P H A P H C sin y ac cot α cot γ sin B ElAa´rea A(P H HCA(P ) =HAHC ) del tri´angulo P HA=HC es. P BC y P CA. Supongamos que los tres a´ngulos son obtusos (Figura 1). los v´ertices A. recto o agudo. P BC. De estos (i) Suppose that the three angles are obtuse. β = ∠AP C. P HB HC y P HC HA obtenemos A(HA HB HC ) = A(P HA HB ) + A(P HB HC ) + A(P HC HA ) i Adding up the areas of the three triangles P HA HB . Then. γ = ∠AP C. If A 0 ◦ have AC = b cos(180 − A) and CC 0 = b sin(180◦ − A). β = ∠AP B. we can obtain the distances from H to the acute vertices (angles) of a right or obtuse triangle ABC. pues. Obviously α + β + γ = 360◦ . com m´ınimo dos son obtusos. Por lo que hemos respectively. Entonces HB = P y P H A P H C sin y ac cot α cot γ sin B . Notice that if 4ABC is a right triangle with b = 90◦ the expression is also valid. Barcelona. But in this case we have P HB = b cot β. BARCELONA TECH. and Ioan Viorel Codreanu. γ = es agudo. A(HA HB HC ) = A(P HA HC ) − A(P HA HB ) − A(P HC HB ) = cot α cot γ − (− cot α cot β) − (− cot γ cot β) A(ABC) = (cot α cot β + cot β cot γ + cot γ cot α)A(ABC) = A(ABC) 2 Also solved by Jos´ e Gibergans-B´ aguena. B C A B C HA Figura 2 HA A A Figura 3 HB P = HB P B C HC B C HC Pero en este caso tenemos P HB = b cot β. 3). P HA = −a cot α i P HC = −c cot γ y. Oscar Rivero Salgado. Barcelona. if cot β = 0 then cot α cot γ = 1. AP C= = β < 90◦ (Figura cot β + cot βH cot γH+ cot γ cot α)A(ABC) = A(ABC). Spain. P HA = −a cot α and P HC = −c cot γ. Romania. cotβ. Compute 1 lim n→∞ n3 Z 0 n n2 + x2 dx 5−x + 7 . equivalently. Barcelona. case. So. say β = 90◦ (Figura 2). 35. por lo On account of (1). El punto P es exterior al tri´angulo H H H y tenemos A(H H H ) = A A(P HA HC ) − A(P HA HB ) − A(P HC HB ). A(HA HB HC ) = A(P HA HC ) − A(P HA HB ) − A(P HC HB ) = [ = cot cot γ angles − (− cot αα. BARCELONA TECH.134 A(HA HB HC ) = lA(ABC) cot α cot β + cot β cot γ + cot γ cot α Since α + β = 360 − γ. Eric Milesi Vidal. Spain. Then HB = P and P H A P H C sin y ac cot α cot γ sin B A(HA HB H = A(HAahora P HCque )= = A(ABC) cot α cot γ 3) C )Supongamos uno de los a´ngulos α. por ejemplo. (1) (ii) Assume that one of the angles is right. Maramure. AP C= 2 2 β < 90◦ (Figura 3). β) γ − (− γ cot β)say A(ABC) (iii) Suppose that one ofαthe is cot acute. Point P is exterior to= (cot theα triangle A B HC and we have A(HA HB HC ) = A(P HA HC ) − A(P HA HB ) − A(P HC HB ). then we have cot(α + β) = − cot γ or. and the results follows in this tanto. β. Satulung. BARCELONA TECH. Spain. 1 − cot α cot β cot γ = − cot(α + β) = cot α + cot β or cot α cot β + cot β cot γ + cot γ cot α = 1 from which follows A(HA HB HC ) = A(ABC). Since the function h(x) = f (x) − ` tends to zero when x → ∞. Lemma. then |g(x)| ≤ M for some M ∈ R and . 1]. Let f : (0.135 (Training Sessions of Spanish Math Team for IMC 2012) Solution by Jos´ e Gibergans-B´ aguena. Spain. +∞] → R and g : [0. If lim f (x) = `. We begin with a Lemma. then Z Z Z x x 1 n 1 n ` n x f (x)g dx = h(x)g dx + g dx n 0 n n 0 n n 0 n Since g(x) is continuous in [0. Barcelona. 1] → R be continuous functions. BARCELONA TECH. then x→∞ Z 1 Z x 1 n g(x) dx lim f (x)g dx = ` n→∞ n 0 n 0 Proof. . Z n Z x . M n . 1 . ≤ . dx h(x)g |h(x)| dx . . Now. 1 and g(x) = 1 + x2 into the preceding Lemma and taking Setting f (x) = −x 5 +7 1 1 into account that lim −x = .n n n 0 (2) (3) 0 If H(x) is a primitive of |h(x)| then. we get x→∞ 5 +7 7 Z n 2 Z . applying L’Hospital’s rule. we have Z Z 1 ` n x g dx = ` g(t) dt n 0 n 0 and from (1) the proof immediately follows. setting x/n = t. we have H(n) = lim |h(x)| = 0 x→∞ n and the RHS of (3) tends to zero when n → ∞. 1 1 n + x2 1 1 1 . . x3 4 . lim 3 dx = + x g(x) dx = . . Barcelona. Editors Comment. Romania. Rome. University. = −x n→∞ n +7 7 0 7 3 21 0 0 5 lim n→∞ and we are done. Satulung. Maramure. 2 Also solved by Jos´ e Luis D´ıaz-Barrero. Higher Institute for Applied Sciences and Technology. Athens. BARCELONA TECH. Tor Vergata. . Italy and Omran Kouba. Ioan Viorel Codreanu. Anastasios Kotronis. Spain. Paolo Perfetti. Syria. Greece. Department of Mathematics. Damascus. Problem 31 was retracted by the proposer. The following lemmas are found respectively in Graham’s book [1]. and the ceiling function of x is denoted by dxe verifying x ≤ dxe < x + 1. Lemmas. The paper first proves several inequalities related with the floor function. and then deduces and proves a mean-value formula for the floor function with an integer variable. The function z(α). 2. z(83) = z((01010011)2 ) = 3. This paper mainly presents the way to obtain and to prove it. it holds (P1): bxc + byc ≤ bx + yc ≤ bxc + byc + 1 . which is defined in [2]. e. Preliminaries This section presents some necessary preliminaries for later sections. I called it a mean-value formula of the floor function. discrete mathematics. z(0) = z((00000000)2 ) = 1. 1.2. For arbitrary real numbers x. y and an integer n. and they are widely applied in number theory. Definition and Symbols. calculus and computer science. z(1) = z((00000001)2 ) = 2.136 MATHNOTES SECTION A mean-value formula for the floor function on integers Wang Xingbo Abstract. 2. Readers can see that the properties of the two functions are miraculous and fascinating. Due to having discrete characteristics of integers. The floor function of a real number x is denoted by bxc and it satisfies bxc ≤ x < bxc + 1. 2. A general introduction to the two functions can be seen in detail in Graham’s book [1].. The inequalities and the formulae are useful in some aspects related to analysis and computation of the floor function. I found and proved the following formula j α k δ−1 1 α+δ = + + 2 2z(α) 2z(α) 2z(α) Since the previous expression is quite similar to the formula f (x0 + ∆x) = f (x0 ) + f 0 (ξ)∆x in calculus. the two functions are still lack of analytic tools like the mean-value theorem in calculus to analyze their intermediary status. However. Hence it is worth to draw a mean-value theorem for either of the two functions. In a study on problems of binary trees.1. Introduction The floor function and the ceiling function are two special functions that have integer values. and Wang’s paper[2].g. represents the position of the first 0-bit that occurs from the least significant bit (lsb) of α’s binary representation. the fractional part of x is denoted by {x} and it satisfies x = bxc + {x}. it also can be seen that there still remain quite a lot of problems for us to study. Lemma 1. By lemma 2. x ≥ y ⇒ bxc ≥ byc The following result shows the case bxc = byc. Note that 2I 0 ≤ 2( α mod 2I α ) < 1 ⇔ 2{ I } < 1 2I 2 .if bξ + δc = bξc. m = m Lemma 2. For a real number ξ and a real number δ > 0. then bξc ≤ ξ ≤ ξ + ω ≤ ξ + δ < bξ + δc + 1 and the condition bξ + δc = bξc becomes bξc ≤ ξ + ω < bξc + 1 That is the definition of the floor function. then for an arbitrary real number ω such that 0 ≤ ω ≤ δ. z(α) is the smallest positive solution of the following modulo-inequality with unknown x. where m > 0 is an integer. Next. it holds bξ + ηc = bξc Proof. it holds bξ + ωc = bξc For a real number ρ < 0. bxc > byc ⇒ x > y. n bxc = bnxc ⇔ n{x} < 1 bnxc x + n1 +.. Theorem 2. 0 ≤ α mod 2x < 2x−1 3. Main Results In the following we present our main results. let I = z(α). Since 0 ≤ ω ≤ δ.particularly. For arbitrary real numbers x. For integer α ≥ 0. x ≤ y ⇒ bxc ≤ byc . it immediately leads to Hence the theorem holds. We begin with (see [4]) Theorem 1.bxc+ x + 12 = b2xc n n = bxc+ n−1 + 1. it holds j α k j α k 2 z(α) = z(α)−1 2 2 (4) Proof. Indeed. by the condition ρ < 0 and ρ ≤ η ≤ 0. it holds i. bxc < byc ⇒ x < y ii. we prove the second conclusion. bξc ≤ ξ + η < bξc + 1 Theorem 3. For an arbitrary integer α ≥ 0. if bξ + ρc = bξc. then for an arbitrary real number η such that ρ ≤ η ≤ 0.137 (P2): (P3): (P4): (P5): (P6): bxc − byc − 1 ≤ bx − yc ≤ bxc − byc < bxc − byc + 1 bn + xc = n + bxc n bxc ≤ bnxc . yields bξ + ρc ≤ ξ + ρ ≤ ξ + η ≤ ξ < bξc + 1 Considering bξ + δc = bξc..+ x + n−1 . I satisfies 0 ≤ α mod 2I < 2I−1 2I which is equal to 0 ≤ 2( α mod ) < 1. y. For convenience. In the case I = 1. Also use the symbol I = z(α) for convenience. j α k α + 1 α − 1 = = + 2I 2I 2I Now. it knows α−1 1 − 12 ≤ α−1 < 2αI 2I − 2 < 2αI − 12 < α+1 2I 2I α 2I < α+1 2I ≤ α−1 2I + 1 2 < α 2I + 1 2 < α+1 2I + (6) 1 2 First by using (4). it holds j j k k jαk α α+1 1 α α−1 1 − = − 1 = − 1. When I > 1. we prove it immediately obtains 1 α 1 = I + 2 2 2 j k α+1 1 α − = I 2I 2 2 In fact. 2I 2 2I 2I 2 2I j k α α+1 1 α+1 1 α−1 1 α − = + = + . it is obvious that α−1 1 < α+1 − 12 = 2αI 2I − 2 < 2αI − 12 = α−1 2I 2I (5) α α−1 1 α 1 1 α+1 α+1 = + = + + < < I I I I I 2 2 2 2 2 2 2 2 With a direct computation and by the property (P3) of lemma 1. when I > 1. it yields j k j k α−1 1 α 1 1 α 1 1 α + = I + − I ≥ I + − = I 2I 2 2 2 2 2 2 2I 2 (7) (8) .138 By the property (P4) of lemma 1. + = I +1 2I 2 2I 2I 2I 2 2 By using (4). jαk j α k α mod 2I α ) < 1 ⇔ 2{ } < 1 ⇔ 2 = I−1 2I 2I 2I 2 Theorem 4. it holds j α 1 α k α−1 1 − = − = −1 2 2 2z(α) 2z(α) 2z(α) j α+1 1 α k + = z(α) + 1 z(α) 2 2 2 j k α+1 1 α α−1 1 α+1 α 1 − = z(α) = + = = z(α) + 2 2 2 2z(α) 2 2z(α) 2z(α) 2 Proof. it yields 0 ≤ α mod 2I < 2I−1 ⇔ 0 ≤ 2( which proves the statement. it yields jαk 1 α−1 α 1 α − = = + − 1 = − 1. it yields j α 1 α k jαk jαk + = I−1 − I = I I 2 2 2 2 2 Referring to the theorem 2 and the formula (6). = = 2I 2 2I 2I 2 2I 2I 2 Hence the theorem holds in the case I = 1. For an arbitrary integer α ≥ 0. referring to the property (P1) of lemma 1. Let α ≥ 0. z(α) = 1 = z(α) + z(α) 0. −3. then it holds α+δ α +χ (11) = 2z(α) 2z(α) Proof. −2. as follows. δ) takes an identical value 2αI + χ. and it leads to jαk α+1 1 α+1 1 α+1 1 + − − = + 1 = +1= I +1 I I I 2 2 2 2 2 2 2 Furthermore. Also by the theorem 2. 0. δ) = = α+1 2I I α+δ 2I +χ− 1 2 = j α+(χ− 12 )×2I +1 2I =χ+ α+1 2I − 1 2 k =χ+ α (13) 2I (iii).. δ) = = = + χ = χ + 2I 2I 2I 2I (14) . z(α) > 1 2 2 Theorem 5. −1. 1.. on the whole interval Iχ . δ = (χ − 12 ) × 2I + 1. it holds jαk α−1 1 1 α 1 α 1 α 1 1 + + < + +1 = + +1 = − − +1 = 2I 2 2I 2 2I 2I 2 2I 2I 2 2I Thus the formula (8) holds. δ = χ × 2I . By theorem 4. it is merely necessary to prove that v(α. it holds jαk jαk α+1 1 α+1 1 − + = −1= I +1−1= I I I 2 2 2 2 2 2 (9) (10) The formulas (7) (8) (9) (10) show that the theorem 4 holds in the case I > 1.139 By the properties (P2) and (P3) of lemma 1 and referring to (7).. According to thetheorem 2. z(α) z(α) (χ − 21 ) × 2 < δ ≤ (χ + 12 ) × 2 .. this is a job to compute the values of v(α. Hence theorem is proven.. When δ = (χ − 12 ) × 2I + 1. use the symbol z(α) z(α) Iχ to denote the interval (χ − 12 ) × 2 < δ ≤ (χ + 21 ) × 2 and keep using I = z(α). δ = (χ + 21 ) × 2I and δ = (χ + 12 ) × 2I + 1. For convenience. hence the conclusions drawn in the theorem 4 can be directly adopted here. it yields v(α. which merely depends upon α and χ. it yields j α+(χ− 12 )×2I k v(α. we know that the expression 2α−1 satisfies z(α) j k α−1 α −1. 2. 3. δ) to denote 2α+δ z(α) .. Note that the condition of the theorem 5 is the same as that of the theorem 4. When δ = (χ − 12 ) × 2I . δ) at five points δ = (χ − 12 ) × 2I . When δ = χ × 2 .we use the symbol v(α. δ) = α+δ = I 2 2I (12) = 2αI + χ − 21 = χ + 2αI − 21 = χ − 1 + 2αI (ii). . δ and χ be integers such that χ = . it yields j k jαk α+δ α + χ × 2I α v(α. (i). δ) and the variable δ can be illustrated by figure 1. it yields v(α. it yields v(α. δ) does take an identical value of 2αI + χ on the whole interval Iχ . δ) does take the value of 2αI + χ on every interval Iχ . we state and prove the following Theorem 6. δ) takes an identical value of 2αI + χ − 1 on the interval j k left to Iχ . it holds . By the theorem 5. it gets to know that the length of the interval Iχ is 2I . δ) = = α 2I + α+δ 1 2 2I = j α+(χ+ 12 )×2I 2I k + χ = χ + 2αI + 12 = χ + 2αI (v). Note that the results j k(12) and (16) also imply that. the relationship between the function v(α.140 (iv). δ). When δ = (χ + 12 ) × 2I . For arbitrary integers α ≥ 0 and δ . Relationship between the variable δ and the function v(α. δ) on Iχ can be translated from the property on it. for a fixed χ . it is true that v(α. and takes an identical value 2αI + χ + 1 on the interval right to Iχ . δ) changes with the interval I0 = (−2I−1 . We call the interval I0 a principal interval since the property of the function v(α. By expressing the interval Iχ in its equivalent form Iχ = (−2I−1 + χ × 2I . This ends the proof of the theorem 5. as χ changes. δ) = α+δ =χ+1+ 2I α+1 2I = − j 1 2 α+(χ+ 12 )×2I +1 2I =χ+1+ k α (15) (16) 2I Obviously. 2I−1 ] to be a basic unit. and the value merely depends upon α and χ since I = z(α) depends on α . Figure 1. v(α.(14) and (15) show that v(α. j the k results in (13). 2I−1 + χ × 2I ] it shows that the function v(α. Finally. When δ = (χ + 12 ) × 2I + 1. Through the figure. j k Hence. 141 α+δ 2z(α) j α k δ − 1 1 = z(α) + z(α) + 2 2 2 (17) Proof. it follows δ 1 δ − 2I−1 δ − 2I−1 − 1 δ−1 1 δ−1 1 − = = + 1 = − + 1 = + 2I 2 2I 2I 2I 2 2I 2 l m j k δ I−1 I−1 I − 12 = δ−22I = δ−2 2I+2 −1 2Ij k I−1 = δ−2 2I −1 + 1 = δ−1 + 12 2I Hence the theorem 6 holds. for(i=1. \quad if(a\%\_2i$<$\_2i\_1) \{ /* Compute $2^i$ */ . then by the theorem 5. Keep using the symbol I = z(α) and suppose δ ∈ Iχ without loss of generality.i++) \{ \_2i\_1 = 1$\ll$(i-1). The C-language program is as follows.\quad /* Compute $2^{(i-1)}$*/ \_2i=\_2i\_1$\ll${1}.h" int GetI(int a) /* Find the smallest I that fits $0\leq a mod 2^i < 2^{(i-1)}$ */ \{ int i.. \#include"math.I. 4. \_2i. int \_2i\_1. it yields I I 1 1 (χ − ) × 2 < δ ≤ (χ + ) × 2 (18) 2 2 Performing a simple transformation on (18) leads to 1 1 δ δ − ≤χ< I + I 2 2 2 2 That is 1 1 δ δ − ≤χ< I − +1 2I 2 2 2 Referring to definition of the ceiling function yields δ 1 χ= I − 2 2 By the properties (P6) and (P3) of the lemma 1. Computer Test Computations related to this paper can be tested on personal computers. \} void main() \{ int I.142 I=i.j$<=$i.j. int \_2i. double delta. \} void Test(double a.X. getchar().j++)Test(i.5). for(i=10000. int I) /*Perform Test*/ \{ int U. Y=(int)floor((delta-1)/\_2i+0.U. for(int j=-i. \} \fi .i<20000.i.Y. \quad /* COmpute $2^I$*/ U=(int)floor((a+delta)/\_2i).delta.\_2i. \} \} return I. if(U!=X) printf("Err:\%d \%d \%d \%d \%d $\backslash{n}$". \_2i=1$\ll$I.a.i++) \{I=GetI(i).I). \quad\quad\quad \quad/* I is obtained */ break. X=(int)floor(delta/\_2i). \} printf("Finished!").I. X+=Y.X). com . & Tech. 4th ed. [3] D. 67-101.2.Functions Related to Binary Representation of Integers. The author sincerely present thanks to them all. Graham. ISBN 0-201-55802-5. Knuth and O. Patashnik. Foshan Bureau of Sci.528000 E-mail address: wxbmail@msn. Reading.Integer Functions. Department of Mechatronic Engineering. L. [2] Wang Xingbo. under projects 2011AA100021. PR China.8-12. Journal of Inequalities and Special Functions. Guangdong Province. pp. Ch. issue 2. New York: Chelsea. References [1] R. pp.143 Acknowledgments.(2011). under project 2012B010600018. & and Tech. Foshan University. D. 1994. 1993. vol. Solved and Unsolved Problems in Number Theory. 2nd ed. 2010C012 and Chancheng Government under projects 2011GY006. E. MA: Addison-Wesley. 2011B1023.3 in Concrete Mathematics: A Foundation for Computer Science. Shanks. The research work is supported by Department of Guangdong Sci.