HEXAGON inspiring minds always Let x = 6+2 + Hanoi Mathematical Olympiad 202 6 2 Senior Section 20 Find the value of + x x 7 202 2 Arrange the numbers p = 2 2, q =, t = 2 + 2 in increasing order Let ABCD be a trapezoid with AD parallel to BC and BC = cm, DA = 6 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 4 What is the largest integer less than or equal to 4x x, where x = 2 2 + + 2 Let fx be a function such that fx+2f x+200 x = 4020 x for all x Find the value of f202 6 For every n = 2,,, let A n = + 2 + 2 + Determine all positive integers n such that A n is an integer 7 Prove that a = }{{ } }{{} 6 is a perfect square 202 20 8 Determine the greatest number m such that the system has a solution x 2 + y 2 =, x y + x y = m + 2 + + n 9 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 = 4 cm, find the value of AM/BN 0 Suppose that the equation x + px 2 + qx + = 0, with p,q being some rational numbers, has three real rooots x,x 2,x, where x = 2 + Find the values of p,q Suppose that the equation x + px 2 + qx + r = 0 has three real roots x,x 2,x where p,q,r are integers :et S n = x n + xn 2 + xn, for n =,2,, Prove that S 202 is an integer Copyright c 20 HEXAGON
2 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 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 4 Sovle the equation in the set of integers 6x + = x 2 y 2 2 Determine the smallest value of the expression s = xy yz zx, where x,y,z are real numbers satisfying the condition x 2 + 2y 2 + z 2 = 22 2
Solutions Let x = 6+2 + 6 2 20 Find the value of + x x 7 202 Solution Notice that 6 + 2 = + 2 and 6 2 = 2, 20 = 2 then x = That is + x x 7 202 = 2 Arrange the numbers p = 2 2, q =, t = 2 + 2 in increasing order We have 2 + 2 2 + 2 = 2 2 = 2 2 Since 2 2, then 22 2 2 Notice that t 2 = 2 2+2 2 2+ 2 8 2 Thus q 4 t 4 = 8 64 2 < 0 It follows that p < t < q Let ABCD be a trapezoid with AD parallel to BC and BC = cm, DA = 6 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 = 4 Let M be the intersection of AC and BD + 6 B C A D By the Thales theorem we get OE BC + OF AD = OD BD + OC AC = OD BD + OB BD = From this, OE = BC + AD Likewise, OF = BC + AD Hence, OE = OF That is, 2 EF = OE = BC + AD = + 6 = We get EF = 4 cm 2 4 What is the largest integer less than or equal to 4x x, where x = 2 2 + + 2 Solution By using the identity a + b + aba + b = a + b, we get 2x = 2 + + 2 = 4 + 6x Thus 4x x = 2 That is, the largest integer desired is 2
Let fx be a function such that fx+2f x+200 x = 4020 x for all x Find the value of f202 Solution Let u = x+200 x then x = u+200 u Thus we have u + 200 f + 2fu = 4020 u + 200 u u Interchanging u with x gives x + 200 f + 2fx = 4020 x + 200 x x Let a = fx, b = f x+200 x Solving the system a + 2b = 4020 x, b + 2a = 4020 x + 200 x for a in terms of x gives a = fx = 2x + 4020 8040 4020 + 2x x = 4020 + 2x 4020 + 2x x Hence, 6 For every n = 2,,, let A n = + 2 f202 = 8044 8044 = 2680 20 + 2 + Determine all positive integers n such that A n is an integer Solution The k-th summand of the product has the form a k = + 2 + + n k + k + 2 = kk +, k =,2,,n k + k + 2 from which we get A n = n + 2 n and hence A n = 6 n + 2 It follows that /A n is an integer if and only if n + 2 is positive factor of 6 Notice that n 2, we get n = 4 4
7 Prove that a = }{{ } }{{} 6 is a perfect square 202 20 Solution Let p = } {{ } Then 0 202 = 9p + Hence, 202 which is a perfect square a = p9p + + p + = p + 2, 8 Determine the greatest number m such that the system has a solution x 2 + y 2 =, x y + x y = m Solution We need to find the maximum value of fx, y fx,y = x y + x y when x,y vary satisfying the restriction x 2 + y 2 = Rewriting this as from which we square to arrive at By the AM-GM inequality we get Hence, fx,y = x y + x 2 + xy + y 2 = x y 2 + xy f 2 x,y = x y 2 2 + xy 2 = 2xy2 + xy 2 f 2 x,y = 2xy2 + xy 2 = 2xy2 + xy2 + xy 2xy + 2 + xy + 2 + xy = fx,y Equality occurs when xy =, x2 + y 2 = This simultaneous equations are equivalent to xy =, x + y =
Solving for x x 2 x = 0 = + 4 =, that is x = 2 Therefore, the value of m is, x = 2 Hence, m max = + 9 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 = 4 cm, find the value of AM/BN Solution Notice that MPA = APC MPC = 90 + ABC 2 90 = ABC 2 = PBN Similarly, NPB = PAM The triangle APM is similar to triangle PBN Since PM = PN, we get MANB = PM 2 = PN 2 Hence MA 2 4 2 = 9 6 NB = MA2 MANB = MA2 = PA2 = PN 2 PB 2 0 Suppose that the equation x + px 2 + qx + = 0, with p,q being some rational numbers, has three real rooots x,x 2,x, where x = 2 + Find the values of p,q Solution Since x = 2 + is one root of the equation, we get x 2 = from which we get a quadratic polynomial x 2 4x = 0 by squaring x + αx 2 4x = x + px 2 + qx + = 0 Expanding the left hand side and comparing the coefficients give α = and hence p =, q = Suppose that the equation x + px 2 + qx + r = 0 has three real roots x,x 2,x where p,q,r are integers Let S n = x n + xn 2 + xn, for n =,2,, Prove that S 202 is an integer Solution By the Vieta theorem we get x +x 2 +x = p, x x 2 +x 2 x +x x = q, x x 2 x = r for p, q, r Z We can prove the following recursive relation S n = ps n qs n 2 rs n From this and mathematical induction, by virtue of S = p Z, we get the desired result 2 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 6
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 = SMB It follows that quadrilateral OMBS is concyclic Hence, HOS = SBM = SBN which implies the concyclicity of HOBN Hence, OHB = ONB = 90, as desired 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 =, b = 4, c = 2, d = 8 Hence, a b c + d = 7 4 Sovle the equation in the set of integers 6x + = x 2 y 2 2 Solution Since the right hand side is non-negative we have deduce that 6x + 0 That is, x take positive integers only Therefore, x 2 y 2 2, or x y 2 x + y 2 That is, x 2 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 0 From the right hand side of the equation, we deduce that 6x + 0 Since x Z, we get x 0, which implies that 6x + Hence, x 2 y 2 2 Thus, x y 2 Now that 6x + = x 2 y 2 2 = x y 2 x + y 2 x 2 From this we obtain the inequality, x 2 6x < 0 Solving this inequality gives x {0,,,6} In addition, 6x+ is a perfect square, we get x {0,,,4} Only x = 0; give integer value of y The equation has solutions 0;,0;,; 4,; 4 Determine the smallest value of the expression s = xy yz zx, where x,y,z are real numbers satisfying the condition x 2 + 2y 2 + z 2 = 22 Solution 7