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Junior problems - Phần 4
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Nội dung Text: Junior problems - Phần 4
- Junior problems J181. Let a, b, c, d be positive real numbers. Prove that 3 3 3 3 a2 + d2 b2 + c2 a+b c+d ≤ + + 2 2 a+d b+c Proposed by Pedro H. O. Pantoja, Natal-RN, Brazil J182. Circles C1 (O1 , r) and C2 (O2 , R) are externally tangent. Tangent lines from O1 to C2 intersect C2 at A and B, while tangent lines from O2 to C1 intersect C1 at C and D. Let O1 A ∩ O2 C = {E } and O1 B ∩ O2 D = {F }. Prove that EF ∩ O1 O2 = AD ∩ BC. Proposed by Roberto Bosch Cabrera, Florida, USA J183. Let x, y, z be real numbers. Prove that 2 (x2 + y 2 + z 2 )2 + xyz (x + y + z ) ≥ (xy + yz + zx)2 + (x2 y 2 + y 2 z 2 + z 2 x2 ). 3 Proposed by Neculai Stanciu, George Emil Palade, Buzau, Romania J184. Find all quadruples (x, y, z, w) of integers satisfying the system of equations x + y + z + w = xy + yz + zx + w2 − w = xyz − w3 = −1. Proposed by Titu Andreescu, University of Texas at Dallas, USA 2xy J185. Let H (x, y ) = x+y be the harmonic mean of the positive real numbers x and y . For n ≥ 2, find the greatest constant C such that for any positive real numbers a1 , . . . , an , b1 , . . . , bn the following inequality holds C 1 1 ≤ + ··· + . H (a1 + · · · + an , b1 + · · · + bn ) H (a1 , b1 ) H (an , bn ) Proposed by Dorin Andrica, Babes-Bolyai University, Cluj-Napoca, Romania J186. Let ABC be a right triangle with AC = 3 and BC = 4 and let the median AA1 and the angle bisector BB1 intersect at O. A line through O intersects hypotenuse AB at M and AC at N . Prove that MB NC 4 · ≤. MA NA 9 Proposed by Valcho Milchev, Kardzhali, Bulgaria 1 Mathematical Reflections 1 (2011)
- Senior problems S181. Let a and b be positive real numbers such that 1 1 |a − 2b| ≤ √ and |2a − b| ≤ √ . a b Prove that a + b ≤ 2. Proposed by Titu Andreescu, University of Texas at Dallas, USA S182. Let a, b, c be real numbers such that a > b > c. Prove that for each real number x the following inequality holds 1 (x − a)4 (b − c) ≥ (a − b)(b − c)(a − c)[(a − b)2 + (b − c)2 + (c − a)2 ]. 6 cyc Proposed by Dorin Andrica, Babes-Bolyai University, Cluj-Napoca, Romania a2 −1 n S183. Let a0 ∈ (0, 1) and an = an−1 − n ≥ 1. Prove that for all n, 2, √ n 1 1 n+1+ n ≤ − < . 2 an a0 2 Proposed by Arkady Alt, San Jose, California, USA 1 1 1 + · · · + n , n ≥ 2. Prove that S184. Let Hn = + 2 3 √ n eHn > n! ≥ 2Hn . Proposed by Tigran Hakobyan, Vanadzor, Armenia S185. Let A1 , A2 , A3 be non-collinear points on parabola x2 = 4py, p > 0, and let B1 = l2 ∩ l3 , B2 = l3 ∩ l1 , B3 = l1 ∩ l2 where l1 , l2 , l3 are tangents to the parabola [ A1 A2 A3 ] at points A1 , A2 , A3 , respectively. Prove that is a constant and find [B1 B2 B3 ] its value. Proposed by Arkady Alt, San Jose, California, USA S186. We wish to assign probabilities pk , k = 0, 1, 2, 3, to random variables X1 , X2 , and X3 taking values in the set {0, 1, 2, 3} (some of them possibly with prob- ability 0), such that the Xi , i = 1, 2, 3, will be identically distributed with P (Xi = k ) = pk , k = 0, 1, 2, 3, and X1 + X2 + X3 = 3. Prove that this is possible if and only if p2 + p3 ≤ 1/3, p1 = 1 − 2p2 − 3p3 , and p0 = p2 + 2p3 . Proposed by Shai Covo, Kiryat-Ono, Israel 2 Mathematical Reflections 1 (2011)
- Undergraduate problems U181. Consider sequences (an )n≥0 and (bn )n≥0 , where a0 = b0 = 1, an+1 = an + bn , and bn+1 = (n2 + n + 1)an + bn , n ≥ 1. Evaluate limn→∞ Bn , where (n + 1)2 n2 −√ . Bn = √ n+1 a na n+1 n Proposed by Neculai Stanciu, George Emil Palade, Buzau, Romania 1 U182. Find all continuous functions f on [0, 1] such that f (x) = c if x ∈ 0, and 2 1 f (x) = f (2x − 1) if x ∈ , 1 , where c is a given constant. 2 Proposed by Arkady Alt, San Jose, California, USA U183. Let m and n be positive integers. Prove that n (m + 2n)m+n+1 − nm+n+1 1 n ≤ . (m + n + 1)(m + n)m+n+1 k+m+1 k k=0 Proposed by Dorin Andrica, Babes-Bolyai University, Cluj-Napoca, Romania b U184. Let f, g : [a, b] → R be differentiable functions such that f (x)dx = 0. Prove a that there is some c ∈ (a, b) satisfying b b f (c) g (x)dx + g (c) f (x)dx = 2f (c)g (c). c c Proposed by Duong Viet Thong, National Economics University, Vietnam U185. Determine if there is a non-constant complex analytic function satisfying the conditions: (i) f (f (z )) = f (z ) for all complex numbers z (ii) there is a complex number z0 , such that f (z0 ) = z0 . Proposed by Harun Immanuel, Airlangga University, Indonesia 3 Mathematical Reflections 1 (2011)
- U186. Let (A, +, ·) be a finite ring of characteristic ≥ 3 such that 1 + x ∈ U (A) {0} for each x ∈ U (A). Prove that A is a field. Proposed by Sorin Radulescu, Aurel Vlaicu College, Bucharest and Mihai Piticari, Dragos Voda College, Campulung Moldovenesc, Romania 4 Mathematical Reflections 1 (2011)
- Olympiad problems O181. Let a, b, c be the sidelengths of a triangle. Prove that abc abc abc ≥ a + b + c. + + −a + b + c a−b+c a+b−c Proposed by Titu Andreescu, Unversity of Texas at Dallas, USA and Gabriel Dospinescu, Ecole Normale Superieure, France O182. On side BC of triangle ABC consider m points, on CA n points, and on AB s points. Join the points from sides AB and AC with the points on side BC . Determine the maximum number of the points of intersection situated in the interior of triangle ABC . Proposed by Dorin Andrica, Babes-Bolyai University, Cluj-Napoca, Romania O183. Evaluate 2010 kπ tan4 . 2011 k=1 Proposed by Gabriel Dospinescu, Ecole Normale Superieure, France O184. Points A, B, C, D lie on a line in this order. Using a straight edge and a compas construct parallel lines a and b through A and B, and parallel lines c and d through C and D, such that their points of intersection are vertices of a rhombus. Proposed by Mihai Miculita, Oradea, Romania O185. Find the least integer n ≥ 2011 for which the equation x4 + y 4 + z 4 + w4 − 4xyzw = n is solvable in positive integers. Proposed by Titu Andreescu, University of Texas at Dallas, USA O186. Let n be a positive integer. Prove that each odd common divisor of 2n − 1 2n n+1 , ,..., n n n is a divisor of 2n − 1. Proposed by Dorin Andrica, Babes-Bolyai University, Cluj-Napoca, Romania 5 Mathematical Reflections 1 (2011)
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