Tài liệu Đề thi toán bằng tiếng anh 2012 pdf

Hanoi Open Mathematical Olympiad • • • www.hexagon.edu.vn H E XAGON® inspiring minds always Hanoi Mathematical Olympiad 2012 Senior Section 1. Let x = √ 6+2 √ 5+ √ 6−2 √ 5 √ 20 . Find the value of (1 + x 5 − x 7 ) 2012 311 2. Arrange the numbers p = 2 √ 2 , q = 3, t = 2 1+ 1 √ 2 in increasing order. 3. Let ABCD be a trapezoid with AD parallel to BC and BC = 3 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 3 −3x, where x = 1 2 ( 3  2 + √ 3+ 3  2 − √ 3). 5. Let f(x) be a function such that f(x) + 2f  x+2010 x−1  = 4020 −x for all x = 1. Find the value of f(2012). 6. For every n = 2, 3, . . . , let A n =  1 − 1 1 + 2  ×  1 − 1 1 + 2 + 3  × ··· ×  1 − 1 1 + 2 + ··· + n  . Determine all positive integers n such that 1 A n is an integer. 7. Prove that a = 1 . . . 1   2012 5 . . . 5   2011 6 is a perfect square. 8. Determine the greatest number m such that the system x 2 + y 2 = 1, |x 3 −y 3 | + |x − y| = m 3 has a solution. 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 = 3 cm, BP = 4 cm, find the value of AM/BN . 10. Suppose that the equation x 3 + px 2 + qx + 1 = 0, with p, q being some rational numbers, has three real rooots x 1 , x 2 , x 3 , where x 3 = 2 + √ 5. Find the values of p, q. 11. Suppose that the equation x 3 + px 2 + qx + r = 0 has three real roots x 1 , x 2 , x 3 where p, q, r are integers. :et S n = x n 1 + x n 2 + x n 3 , for n = 1, 2, . . . ,. Prove that S 2012 is an integer. Copyright c  2011 H E XAGON 1 Hanoi Open Mathematical Olympiad • • • www.hexagon.edu.vn 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 3 cm is painted red and then cut into 3 × 3 × 3 = 27 cubes with sides of length 1 cm. If a denotes the number of small cubes of side-length 1 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 + 1 = (x 2 − y 2 ) 2 . 15. 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 + 5z 2 = 22. 2 Hanoi Open Mathematical Olympiad • • • www.hexagon.edu.vn Solutions 1. Let x = √ 6+2 √ 5+ √ 6−2 √ 5 √ 20 . Find the value of (1 + x 5 − x 7 ) 2012 311 Solution. Notice that 6 + 2 √ 5 = ( √ 5 + 1) 2 and 6 − 2 √ 5 = ( √ 5 − 1) 2 , √ 20 = 2 √ 5 then x = 1. That is (1 + x 5 − x 7 ) 2012 311 = 1. 2. Arrange the numbers p = 2 √ 2 , q = 3, t = 2 1+ 1 √ 2 in increasing order. We have 2 1+ 1 √ 2 ≥ 2 1+ 1 2 = 2 3 2 = 2 √ 2. Since √ 2 ≤ 3 2 , then 2 √ 2 ≤ 2 √ 2. Notice that t 2 = 2 2+ √ 2 ≤ 2 2+ 3 2 ≤ 8 √ 2. Thus q 4 −t 4 = 81 −64 × 2 < 0. It follows that p < t < q. 3. Let ABCD be a trapezoid with AD parallel to BC and BC = 3 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 1 3 + 1 6 = 4. Let M be the intersection of AC and BD. A D CB By the Thales theorem we get OE BC + OF AD = OD BD + OC AC = OD BD + OB BD = 1. From this, 1 OE = 1 BC + 1 AD . Likewise, 1 OF = 1 BC + 1 AD . Hence, OE = OF . That is, 2 EF = 1 OE = 1 BC + 1 AD = 1 3 + 1 6 = 1 2 . We get EF = 4 cm. 4. What is the largest integer less than or equal to 4x 3 −3x, where x = 1 2 ( 3  2 + √ 3+ 3  2 − √ 3). Solution. By using the identity a 3 + b 3 + 3ab(a + b) = (a + b) 3 , we get (2x) 3 =  3  2 + √ 3 + 3  2 − √ 3  3 = 4 + 6x. Thus 4x 3 − 3x = 2. That is, the largest integer desired is 2. 3 Hanoi Open Mathematical Olympiad • • • www.hexagon.edu.vn 5. Let f(x) be a function such that f(x) + 2f  x+2010 x−1  = 4020 −x for all x = 1. Find the value of f(2012). Solution. Let u = x+2010 x−1 then x = u+2010 u−1 . Thus we have f  u + 2010 u − 1  + 2f (u) = 4020 − u + 2010 u − 1 . Interchanging u with x gives f  x + 2010 x − 1  + 2f (x) = 4020 − x + 2010 x − 1 . 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) = 1 3  8040 − 4020 + 2x − 2x + 4020 x − 1  = 1 3  4020 + 2x − 4020 + 2x x − 1  . Hence, f(2012) = 1 3  8044 − 8044 2011  = 2680. 6. For every n = 2, 3, . . . , let A n =  1 − 1 1 + 2  ×  1 − 1 1 + 2 + 3  × ··· ×  1 − 1 1 + 2 + ··· + n  . Determine all positive integers n such that 1 A n is an integer. Solution. The k-th summand of the product has the form a k = 1 − 1 (k + 1)(k + 2) = k(k + 3) (k + 1)(k + 2) , k = 1, 2, ··· , n −1 from which we get A n = n + 2 3n and hence 1 A n = 3 − 6 n + 2 . It follows that 1/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 Hanoi Open Mathematical Olympiad • • • www.hexagon.edu.vn 7. Prove that a = 1 . . . 1   2012 5 . . . 5   2011 6 is a perfect square. Solution. Let p = 1 . . . 1   2012 . Then 10 2012 = 9p + 1. Hence, a = p(9p + 1) + 5p + 1 = (3p + 1) 2 , which is a perfect square. 8. Determine the greatest number m such that the system x 2 + y 2 = 1, |x 3 −y 3 | + |x − y| = m 3 has a solution. Solution. We need to find the maximum value of f(x, y) f(x, y) = |x − y| + |x 3 − y 3 | when x, y vary satisfying the restriction x 2 + y 2 = 1. Rewriting this as f(x, y) = |x − y|(1 + x 2 + xy + y 2 ) = |x − y|(2 + xy). from which we square to arrive at f 2 (x, y) = (x −y) 2 (2 + xy) 2 = (1 −2xy)(2 + xy) 2 . By the AM-GM inequality we get f 2 (x, y) = (1 − 2xy)(2 + xy) 2 = (1 −2xy)(2 + xy)(2 + xy) ≤  1 − 2xy + 2 + xy + 2 + xy 3  3 =  5 3  3 . Hence, f(x, y) ≤ 5 3 .  5 3 . Equality occurs when xy = − 1 3 , x 2 + y 2 = 1. This simultaneous equations are equivalent to xy = − 1 3 , x + y = 1 √ 3 . 5 Hanoi Open Mathematical Olympiad • • • www.hexagon.edu.vn Solving for x x 2 − x √ 3 − 1 3 = 0. ∆ = 1 3 + 4 3 = 5 3 , that is x = 1 2  1 √ 3 −  5 3  , x = 1 2  1 √ 3 +  5 3  . Therefore, the value of m 3 is 5 3  5 3 . Hence, m max =  5 3 . 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 = 3 cm, BP = 4 cm, find the value of AM/BN . Solution. Notice that ∠MP A = ∠AP C − ∠MP C =  90 ◦ + ∠ABC 2  − 90 ◦ = ∠ABC 2 = ∠P BN . Similarly, ∠N P B = ∠P AM. The triangle AP M is similar to triangle P BN. Since P M = P N, we get M A.NB = P M 2 = PN 2 . Hence M A N B = M A 2 M A.N B = M A 2 P N 2 = P A 2 P B 2 = 3 2 4 2 = 9 16 . 10. Suppose that the equation x 3 + px 2 + qx + 1 = 0, with p, q being some rational numbers, has three real rooots x 1 , x 2 , x 3 , where x 3 = 2 + √ 5. Find the values of p, q. Solution. Since x = 2 + √ 5 is one root of the equation, we get x − 2 = √ 5 from which we get a quadratic polynomial x 2 − 4x −1 = 0 by squaring. (x + α)(x 2 − 4x −1) = x 3 + px 2 + qx + 1 = 0. Expanding the left hand side and comparing the coefficients give α = −1 and hence p = −3, q = −5. 11. Suppose that the equation x 3 + px 2 + qx + r = 0 has three real roots x 1 , x 2 , x 3 where p, q, r are integers. Let S n = x n 1 + x n 2 + x n 3 , for n = 1, 2, . . . ,. Prove that S 2012 is an integer. Solution. By the Vieta theorem we get x 1 +x 2 +x 3 = −p, x 1 x 2 +x 2 x 3 +x 3 x 1 = q, x 1 x 2 x 3 = −r for p, q, r ∈ Z. We can prove the following recursive relation S n = −p.S n−1 − qS n−2 − rS n−3 . From this and mathematical induction, by virtue of S 1 = −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. 6 Hanoi Open Mathematical Olympiad • • • www.hexagon.edu.vn 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. 13. A cube with sides of length 3 cm is painted red and then cut into 3 × 3 × 3 = 27 cubes with sides of length 1 cm. If a denotes the number of small cubes of side-length 1 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 = 8. Hence, a − b −c + d = −7. 14. Sovle the equation in the set of integers 16x + 1 = (x 2 − y 2 ) 2 . Solution. Since the right hand side is non-negative we have deduce that 16x + 1 ≥ 0. That is, x take positive integers only. Therefore, (x 2 − y 2 ) 2 ≥ 1, or |x − y| 2 |x + y| 2 ≥ 1. That is, x 2 ≥ 1. 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 16x + 1 ≥ 0. Since x ∈ Z, we get x ≥ 0, which implies that 16x + 1 ≥ 1. Hence, (x 2 −y 2 ) 2 ≥ 1. Thus, (x −y) 2 ≥ 1. Now that 16x + 1 = (x 2 − y 2 ) 2 = (x −y) 2 (x + y) 2 ≥ x 2 . From this we obtain the inequality, x 2 − 16x − 1 < 0. Solving this inequality gives x ∈ {0, 1, ··· ,16}. In addition, 16x + 1 is a perfect square, we get x ∈ {0, 3, 5, 14}. Only x = 0; 5 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 x 2 + 2y 2 + 5z 2 = 22. Solution. 7 . Prove that a = 1 . . . 1   2012 5 . . . 5   2011 6 is a perfect square. Solution. Let p = 1 . . . 1   2012 . Then 10 2012 = 9p + 1. Hence, a = p(9p. x 5 − x 7 ) 2012 311 Solution. Notice that 6 + 2 √ 5 = ( √ 5 + 1) 2 and 6 − 2 √ 5 = ( √ 5 − 1) 2 , √ 20 = 2 √ 5 then x = 1. That is (1 + x 5 − x 7 ) 2012 311 =