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File này là đề thi Olympiad Toán Hà Nội Mở rộng năm 2014 dành cho các em học sinh lớp 8 và lớp 10. Nếu cảm thấy tự tin về vốn kiến thức toán học cũng như anh ngữ của mình, kì thi Hanoi Opening Mathematical Olympiad, hay còn dc viết tắt là HOMC, sẽ là 1 cơ hội thuận lợi cho các bạn cọ xát, thử thách bản thân và đồng thời cũng có thể thấy dc vốn kiến thức của mình đang ở ngưỡng nào mà phấn đấu thêm. Không những thế, kì thi HOMC còn giúp cho các bạn làm quen với cách học và giải toán bằng tiếng anh, từ đó các bạn sẽ cảm thấy tự tin hơn khi đối mặt với các kì thi toán quốc tế, hoặc đơn giản hơn là khi đọc các tài liệu nước ngoài Chúc các bạn sẽ vững tin và bay cao trong sự nghiệp của mình
Hanoi Open Mathematical Competitions 2014 Junior Section Answers and solutions Sunday, 23 March 2014 08h30-11h30 Q1. Let the numbers x and y satisfy the conditions x 2 + y 2 − xy = 2 x 4 + y 4 + x 2 y 2 = 8 The value of P = x 8 + y 8 + x 2014 y 2014 is (A): 46; (B): 48; (C): 50; (D): 52; (E) None of the above. Answer. (B) We have x 2 + y 2 = 2 + xy. It follows x 4 + y 4 + 2x 2 y 2 = x 2 y 2 + 4xy + 4. Equivalently, x 4 + y 4 + x 2 y 2 = 4(xy + 1), 8 = 4(xy + 1)) ⇔ xy + 1 = 2 ⇔ xy = 1 and then x 4 + y 4 = 7, xy = 1. Hence x 8 + y 8 = (x 4 + y 4 ) 2 − 2x 2 y 2 = 49 − 2 = 47 and we find P = 47 + 1 = 48. Q2. How many diagonals does 11-sided convex polygon have? (A): 43; (B): 44; (C): 45; (D): 46; (E) None of the above. Answer. (B) The number of diagonals of 11-sided convex polynom is difined by 11(11 − 1) 2 − 11 = 44. Q3. How many zeros are there in the last digits of the following number P = 11 × 12 × · · · × 88 × 89 (A): 16; (B): 17; (C): 18; (D): 19; (E) None of the above. 1 Answer. (C) We have P = 5 18 · A, where A is the integer containing sufficiently many even factors. Thus, the answer is 18. Q4. If p is a prime number such that there exist positive integers a and b such that 1 p = 1 a 2 + 1 b 2 , then p is (A): 3; (B): 5; (C): 11; (D): 7; (E) None of the above. Answer. (E) Write m := (a, b). It implies a = mr b = ms. We then have 1 p = 1 m 2 r 2 + 1 m 2 s 2 . It follows p(r 2 + s 2 ) = mr 2 s 2 . Since r 2 + s 2 is not divisable by neither r 2 nor s 2 , we have p . . .r 2 and p . . .s 2 . As p is a prime number, r 2 = s 2 = 1. We find m = 2 and p = 2. So, the answer is (E). Q5. The first two terms of a sequence are 2 and 3. Each next term thereafter is the sum of the nearestly previous two terms if their sum is not greather than 10, 0 otherwise. The 2014th term is (A): 0; (B): 8; (C): 6; (D): 4; (E) None of the above. Answer. (B) We have the sequence as follows 2, 3, 5, 8, 0, 8, 8, 0, 8, 8, . . . Starting from the fifth term, the sequence is periodic with the period 3. We then have a n = 0, if n − 4 = 1 (mod 3) 8, otherwise. Note that 2014 − 4 = 2010 = 0 (mod 3). Thus, the answer is 8. 2 Q6. Let a, b, c be the length sides of a given triangle and x, y, z be the sides length of bisectrices, respectively. Prove the following inequality 1 x + 1 y + 1 z > 1 a + 1 b + 1 c . Solution. Let AD be bisector of angle BAC and D belongs to BC. Draw DE AB, E belongs to AC. Then triangle EAD is iscoceles and DE = EA = d. We have ED AB = EC CA and then AE AC + ED AB = EC CA + AE CA = 1. It follows d b + d c = 1 ⇔ 1 b + 1 c = 1 d . Hence 1 2 1 b + 1 c = 1 2d < 1 x (1) since 2d > AD. Similarly, 1 2 1 c + 1 a < 1 y , (2) 1 2 1 c + 1 a < 1 y . (3) From (1), (2) and (3) we find 1 x + 1 y + 1 z > 1 a + 1 b + 1 c , which was to be proved. Q7. Determine the integral part of A, where A = 1 672 + 1 673 + · · · + 1 2014 . Solution. Consider the sum S = 3n+1 k=n+1 1 k . Note that there are 2n + 1 terms in the sum and the middle term is 1 2n + 1 . So we can write the sum in the form S = 1 2n + 1 + n k=1 1 2n + 1 + k + 1 2n + 1 − k 3 = 1 2n + 1 + 2 2n + 1 n k=1 1 1 − k 2n + 1 2 . On the other hand, using the inequalities 1 + a < 1 1 − a < 1 + 2a for 0 < a < 1 2 , we get 1 1 − k 2n + 1 2 > 1 + k 2n + 1 2 and 1 1 − k 2n + 1 2 < 1 + 2 k 2n + 1 2 . Hence 1 2n + 1 + 2 2n + 1 n k=1 1 + k 2n + 1 2 < S < 1 2n + 1 + 2 2n + 1 n k=1 1 + 2 k 2n + 1 2 ⇔ 1 + 2 (2n + 1) 3 n k=1 k 2 < S < 1 + 4 (2n + 1) 3 n k=1 k 2 ⇔ 1 + n(n + 1) 3(2n + 1) 2 < S < 1 + 2 3 n(n + 1) (2n + 1) 2 . It is easy to check that 2 9 ≤ n(n + 1) (2n + 1) 2 < 1 4 , ∀n ≥ 1. This leads to 29 27 < S < 7 6 and then [S] = 1 and [A] = 1. Q8. Let ABC be a triangle. Let D, E be the points out side of the triangle so that AD = AB, AC = AE and ∠DAB = ∠EAC = 90 ◦ . Let F be at the same side of the line BC as A so that F B = F C and ∠BF C = 90 ◦ . Prove that triangle DEF is a right-isoceles triangle. Solution. Note that CD ⊥ BE and ∆F EB = ∆F DC. Hence, F E = F D. (1). On the other hand, ∠DF C = ∠EF B then ∠DF E = 90 0 . (1) (1) and (2) together imply |deltaDEF | is right-iscoceles, q.e.d. Q9. Determine all real numbers a, b, c such that the polynomial f(x) = ax 2 + bx + c satisfies simultaneously the folloving conditions |f(x)| ≤ 1 for |x| ≤ 1 f(x) ≥ 7 for x ≥ 2 4 Figure 1: Solution. Let g(x) = 2x 2 − 1 − f(x), then deg g ≤ 2. Note that g(−1) = 1 − f(−1) ≥ 0, g(0) = −1 − f(0) ≤ 0, g(1) = 1 − f (1) ≥ 0 and g(2) = 7 − f (2) ≤ 7 − 7 = 0. Hence the equation g(x) = 0 has at least 3 roots. It follows g(x) ≡ 0 and then f (x) = 2x 2 − 1. Thus (a, b, c) = (2, 0, −1). Q10. Let S be area of the given parallelogram ABCD and the points E, F belong to BC and AD, respectively, such that BC = 3CE, 3AD = 2AF. Let O be the intersection of AE and BF. The straightlines AE and BF meet CD at points M and N, respectively. Determine area of triangle MON. Solution. Let a = AB and h be the height of the parallelogram. Since ∆EAB ∼ ∆EMC, it follows CM AB = EC EB = 1 2 and CM = a 2 . Similarly, we have N D = a, then M N = a + a + a 2 = 5a 2 . Let h, h 1 , h 2 be heights of the given parallelogram, ∆OM N and ∆OAB, respectively, then h 1 h 2 = MN AB = 5 2 . 5 It follows h 1 h = h 1 h 1 + h 2 = 5 7 . Hence h 1 = 5h 7 . These follow that S M ON = 1 2 × MN × h 1 = 1 2 × 5a 2 × 5h 7 = 25ah 28 = 25 28 S. Q11. Find all pairs of integers (x, y) satisfying the following equality 8x 2 y 2 + x 2 + y 2 = 10xy. Solution. We have 8x 2 y 2 + x 2 + y 2 = 10xy ⇔ 8xy (xy − 1) + (x − y) 2 = 0. (1) Consequently, if x; y is an integral root of the equation then xy (xy − 1) ≤ 0 ⇒ 0 ≤ xy ≤ 1. Because x; y are integral, there are two possibilities: - If xy = 0 then from (1) we have a root x = y = 0. - If xy = 1 then from (1) we have other roots x = y = 1; x = y = −1. Thus the equation has three solutions (x, y) = (0, 0), (x, y) = (1, 1), (x, y) = (−1, −1). Q12. Find a polynomial Q(x) such that (2x 2 − 6x + 5)Q(x) is a polynomial with all positive coefficients. Solution. Note that (2x 2 − 6x + 5)(2x 2 + 6x + 5)(4x 4 + 6x 2 + 25) = 16x 8 + 164x 4 + 625. Hence (2x 2 −6x+5)(2x 2 +6x+5)(4x 4 +6x 2 +25)(x 3 +x 2 +x+1) = (16x 8 +164x 4 +625)(x 3 +x 2 +x+1) = 16x 11 +16x 10 +16x 9 +16x 8 +164x 7 +164x 6 +164x 5 +164x 4 +625x 3 +625x 2 +625x+625. So we can choose Q(x) = (2x 2 + 6x + 5)(4x 4 + 6x 2 + 25)(x 3 + x 2 + x + 1). Q13. Let a, b, c > 0 and abc = 1. Prove that a − 1 c + c − 1 b + b − 1 a ≥ 0. Solution. We prove the following inequality a c + c b + b a ≥ ab + bc + ca. (1) 6 Indeed, from inequality (c − 1) 2 (2c + 1) ≥ 0 for c > 0, it follows 2c 3 + 1 ≥ 3c 2 ⇔ 1 c + 2c 2 ≥ 3c ⇔ ab + 2c 2 ≥ 3abc 2 ⇔ a c + 2c b ≥ 3ac. Similarly, c b + 2b a ≥ 3bc, b a + 2a c ≥ 3ab. These three inequalities imply a c + c b + b a ≥ ab + bc + ca = 1 c + 1 a + 1 b . Hence a c + c b + b a ≥ 1 c + 1 a + 1 b , which is a − 1 c + c − 1 b + b − 1 a ≥ 0. The equality holds iff a = b = c = 1. Q14. Let be given a < b < c and f (x) = c(x − a)(x − b) (c − a)(c − b) + a(x − b)(x − c) (a − b)(a − c) + b(x − c)(x − a) (b − c)(b − a) . Determine f(2014)? Solution. Let g(x) = f(x) − x. Then deg g ≤ 2 and g(a) = g(b) = g(c) = 0. Hence g(x) ≡ 0 and f(x) = x for all x ∈ R. It follows f(2014) = 2014. Q15. Let a 1 , a 2 , . . . , a 9 ≥ −1 and a 3 1 + a 3 2 + · · · + a 3 9 = 0. Determine the maximal value of M = a 1 + a 2 + · · · + a 9 . Solution. For a ≥ −1, we find (a + 1) a − 1 2 2 ≥ 0. It follows a 3 − 3 4 a + 1 4 ≥ 0 ⇔ 3a ≤ 4a 3 + 1, ∀a ≥ −1. Hence 3(a 1 + a 2 + · · · + a 9 ) ≤ 4(a 3 1 + a 3 2 + · · · + a 3 9 ) + 9 = 9. So M ≤ 3. The equality holds for a 1 , a 2 , . . . , a 9 ∈ − 1, 1 2 . For example, we can choose a 1 = −1, a 2 = a 3 = · · · = a 9 = 1 2 , then a 3 1 + a 3 2 + · · · + a 3 9 = 0 and a 1 + a 2 + · · · + a 9 = 3. Thus, max M = 3. —————————————– 7