www.VNMATH.com Hanoi Mathematical Olympiad 2012 Senior Section Let x = √ √ √ 6+2 √ 5+ 6−2 20 √ √ Arrange the numbers p = 311 Find the value of (1 + x5 − x7 )2012 2, 1+ √1 q = 3, t = 2 in increasing order Let ABCD be a trapezoid with AD parallel to BC and BC = cm, DA = cm Find the length of the line segment EF parallel to the two bases and passing through the intersection of the two diagonals AC, BD, E is on CD, F on AB √ √ 3 What is the largest integer less than or equal to 4x3 −3x, where x = 21 ( + 3+ − 3) Let f (x) be a function such that f (x) + 2f of f (2012) x+2010 x−1 = 4020 − x for all x = Find the value For every n = 2, 3, , let An = 1− 1+2 × 1− 1+2+3 Determine all positive integers n such that An × ··· × − 1 + + ··· + n is an integer Prove that a = is a perfect square 2012 2011 Determine the greatest number m such that the system x2 + y = 1, |x3 − y | + |x − y| = m3 has a solution Let P be the intersection of the three internal angle bisectors of a triangle ABC The line passing through P and perpendicular to CP intersects AC and BC at M, N respectively If AP = cm, BP = cm, find the value of AM/BN 10 Suppose that the equation x3 + px2 + qx + 1√= 0, with p, q being some rational numbers, has three real rooots x1 , x2 , x3 , where x3 = + Find the values of p, q 11 Suppose that the equation x3 + px2 + qx + r = has three real roots x1 , x2 , x3 where p, q, r are integers :et Sn = xn1 + xn2 + xn3 , for n = 1, 2, , Prove that S2012 is an integer Copyright c 2011 HEXAGON www.VNMATH.com 12 Let M be a point on the side BC of an isosceles triangle ABC with BC = BA Let O be the circumcenter of the triangle and S be its incenter Suppose that SM is parallel to AC Prove that OM ⊥ BS 13 A cube with sides of length cm is painted red and then cut into × × = 27 cubes with sides of length cm If a denotes the number of small cubes of side-length cm that are not painted at all, b the number of cubes painted on one side, c the number of cubes painted on two sides, and d the number of cubes painted on three sides, determine the value of a − b − c + d 14 Sovle the equation in the set of integers 16x + = (x2 − y )2 15 Determine the smallest value of the expression s = xy−yz−zx, where x, y, z are real numbers satisfying the condition x2 + 2y + 5z = 22 www.VNMATH.com Solutions Let x = √ √ √ 6+2 √ 5+ 6−2 20 √ 311 Find the value of (1 + x5 − x7 )2012 √ √ √ √ √ √ Solution Notice that + = ( + 1)2 and − = ( − 1)2 , 20 = then x = That is 311 (1 + x5 − x7 )2012 = √ 1+ √1 Arrange the numbers p = 2 , q = 3, t = √ 1+ 12 = 2 = 2 √ √ √ Since ≤ 23 , then 2 ≤ 2 Notice that t2 = 22+ √ 2 1+ √1 in increasing order We have 2 ≥ √ ≤ 22+ ≤ Thus q − t4 = 81 − 64 × < It follows that p < t < q Let ABCD be a trapezoid with AD parallel to BC and BC = cm, DA = cm Find the length of the line segment EF parallel to the two bases and passing through the intersection of the two diagonals AC, BD, E is on CD, F on AB Hint Making use of the similarity of triangles The line segment is the harmonic means of the two bases, = +2 = Let M be the intersection of AC and BD B C A D OE BC BC OF + AD + AD OD BD OD OB By the Thales theorem we get = + OC AC = BD + BD = From this, OE = 1 1 = Hence, OE = OF That is, + Likewise, OF = = + BC AD EF OE BC 1 1 = + = We get EF = cm AD What is the largest integer less than or equal to 4x3 −3x, where x = 21 ( 2+ Solution By using the identity a3 + b3 + 3ab(a + b) = (a + b)3 , we get (2x)3 = 2+ √ 3+ 2− √ 3 Thus 4x3 − 3x = That is, the largest integer desired is = + 6x √ 3+ 2− √ 3) www.VNMATH.com x+2010 x−1 Let f (x) be a function such that f (x) + 2f of f (2012) Solution Let u = x+2010 x−1 then x = u+2010 u−1 u + 2010 u−1 f = 4020 − x for all x = Find the value Thus we have + 2f (u) = 4020 − u + 2010 u−1 + 2f (x) = 4020 − x + 2010 x−1 Interchanging u with x gives x + 2010 x−1 f Let a = f (x), b = f x+2010 x−1 Solving the system a + 2b = 4020 − x, b + 2a = 4020 − x + 2010 x−1 for a in terms of x gives = a = f (x) = 2x + 4020 x−1 4020 + 2x 4020 + 2x − x−1 8040 − 4020 + 2x − Hence, f (2012) = 8044 − 8044 2011 = 2680 For every n = 2, 3, , let An = 1− 1+2 × 1− 1+2+3 Determine all positive integers n such that An × ··· × − 1 + + ··· + n is an integer Solution The k-th summand of the product has the form ak = − k(k + 3) = , k = 1, 2, · · · , n − (k + 1)(k + 2) (k + 1)(k + 2) from which we get An = n+2 3n It follows that 1/An is an integer if and only if n + is positive = 3− An n+2 factor of Notice that n ≥ 2, we get n = and hence www.VNMATH.com Prove that a = is a perfect square 2012 2011 Solution Let p = Then 102012 = 9p + Hence, 2012 a = p(9p + 1) + 5p + = (3p + 1)2 , which is a perfect square Determine the greatest number m such that the system x2 + y = 1, |x3 − y | + |x − y| = m3 has a solution Solution We need to find the maximum value of f (x, y) f (x, y) = |x − y| + |x3 − y | when x, y vary satisfying the restriction x2 + y = Rewriting this as f (x, y) = |x − y|(1 + x2 + xy + y ) = |x − y|(2 + xy) from which we square to arrive at f (x, y) = (x − y)2 (2 + xy)2 = (1 − 2xy)(2 + xy)2 By the AM-GM inequality we get f (x, y) = (1 − 2xy)(2 + xy)2 = (1 − 2xy)(2 + xy)(2 + xy) ≤ − 2xy + + xy + + xy = 3 Hence, f (x, y) ≤ Equality occurs when xy = − , x2 + y = This simultaneous equations are equivalent to 1 xy = − , x + y = √ 3 www.VNMATH.com Solving for x ∆= + x x2 − √ − = 3 = 53 , that is x= √ − Therefore, the value of m3 is 5 , x= Hence, mmax = √ + Let P be the intersection of the three internal angle bisectors of a triangle ABC The line passing through P and perpendicular to CP intersects AC and BC at M, N respectively If AP = cm, BP = cm, find the value of AM/BN = − 90◦ = ∠ABC Solution Notice that ∠M P A = ∠AP C − ∠M P C = 90◦ + ∠ABC 2 ∠P BN Similarly, ∠N P B = ∠P AM The triangle AP M is similar to triangle P BN Since M A2 M A2 A P A2 P M = P N , we get M A.N B = P M = P N Hence M N B = M A.N B = P N = P B = 32 = 16 42 10 Suppose that the equation x3 + px2 + qx + 1√= 0, with p, q being some rational numbers, has three real rooots x1 , x2 , x3 , where x3 = + Find the values of p, q √ √ Solution Since x = + is one root of the equation, we get x − = from which we get a quadratic polynomial x2 − 4x − = by squaring (x + α)(x2 − 4x − 1) = x3 + px2 + qx + = Expanding the left hand side and comparing the coefficients give α = −1 and hence p = −3, q = −5 11 Suppose that the equation x3 + px2 + qx + r = has three real roots x1 , x2 , x3 where p, q, r are integers Let Sn = xn1 + xn2 + xn3 , for n = 1, 2, , Prove that S2012 is an integer Solution By the Vieta theorem we get x1 +x2 +x3 = −p, x1 x2 +x2 x3 +x3 x1 = q, x1 x2 x3 = −r for p, q, r ∈ Z We can prove the following recursive relation Sn = −p.Sn−1 − qSn−2 − rSn−3 From this and mathematical induction, by virtue of S1 = −p ∈ Z, we get the desired result 12 Let M be a point on the side BC of an isosceles triangle ABC with AC = BC Let O be the circumcenter of the triangle and S be its incenter Suppose that SM is parallel to AC Prove that OM ⊥ BS www.VNMATH.com Solution Let OM meet SB at H N is the midpoint of AB Since O is the circumcenter of triangle OBC which is isosceles with CA = CB and SM AC we have ∠SOB = 2∠OCB = ∠ACB = ∠SM B It follows that quadrilateral OM BS is concyclic Hence, ∠HOS = ∠SBM = ∠SBN which implies the concyclicity of HOBN Hence, ∠OHB = ∠ON B = 90◦ , as desired 13 A cube with sides of length cm is painted red and then cut into × × = 27 cubes with sides of length cm If a denotes the number of small cubes of side-length cm that are not painted at all, b the number of cubes painted on one side, c the number of cubes painted on two sides, and d the number of cubes painted on three sides, determine the value of a − b − c + d Solution Just count from the diagram of the problem, we get a = 1, b = 4, c = 12, d = Hence, a − b − c + d = −7 14 Sovle the equation in the set of integers 16x + = (x2 − y )2 Solution Since the right hand side is non-negative we have deduce that 16x + ≥ That is, x take positive integers only Therefore, (x2 − y )2 ≥ 1, or |x − y|2 |x + y|2 ≥ That is, x2 ≥ It is evident that if (x, y) is a solution of the equation, then (x, −y) is also its solution Hence, it is sufficient to consider y ≥ From the right hand side of the equation, we deduce that 16x + ≥ Since x ∈ Z, we get x ≥ 0, which implies that 16x + ≥ Hence, (x2 − y )2 ≥ Thus, (x − y)2 ≥ Now that 16x + = (x2 − y )2 = (x − y)2 (x + y)2 ≥ x2 From this we obtain the inequality, x2 − 16x − < Solving this inequality gives x ∈ {0, 1, · · · , 16} In addition, 16x + is a perfect square, we get x ∈ {0, 3, 5, 14} Only x = 0; give integer value of y The equation has solutions (0; 1), (0; −1), (5; 4), (5; −4) 15 Determine the smallest value of the expression s = xy−yz−zx, where x, y, z are real numbers satisfying the condition x2 + 2y + 5z = 22 Solution