15th Asian Pacific Mathematics Olympiad 2003 Problems

15th Asian Pacific Mathematics Olympiad 2003 Problems 1.  The polynomial a8x8 +a7x7 + ... + a0 has a8 = 1, a7 = -4, a6 = 7 and all its roots positive and real. Find the possible values for a0. 2.  A unit square

Read more »

Posted in: APMO

14th Asian Pacific Mathematics Olympiad 2002 ProblemsA1.  xi are non-negative integers. Prove that x1! x2! ... xn! ≥ ( [(x1 + ... + xn)/n] ! )n (where [y] denotes the largest integer not exceeding y). When do you have equality?

13th Asian Pacific Mathematics Olympiad 2001 Problems A1.  If n is a positive integer, let d be the number of digits in n (in base 10) and s be the sum of the digits. Let n(k) be the number formed by deleting the last k digits of n. Prove that n = s + 9 n(1) + 9 n(2) + ... + 9 n(d).

12th Asian Pacific Mathematics Olympiad 2000 ProblemsA1.  Find a13/(1 - 3a1 + 3a12) + a23/(1 - 3a2 + 3a22) + ... + a1013/(1 - 3a101 + 3a1012), where an = n/101. A2.  Find all permutations a1, a2, ... , a9 of 1, 2, ... , 9 such that a1 + a2 + a3 + a4 = a4 + a5 + a6 + a7 = a7 + a8 + a9 + a1 and a12 + a22 + a32 + a42 =

Read more »

Posted in: APMO

11th Asian Pacific Mathematics Olympiad 1999 ProblemsA1.  Find the smallest positive integer n such that no arithmetic progression of 1999 reals contains just n integers. A2.  The real numbers x1, x2, x3, ... satisfy xi+j ≤ xi + xj for all i, j. Prove that x1 + x2/2 + ... + xn/n ≥ xn.

10th Asian Pacific Mathematics Olympiad 1998 ProblemsA1.  S is the set of all possible n-tuples (X1, X2, ... , Xn) where each Xi is a subset of {1, 2, ... , 1998}. For each member k of S let f(k) be the number of elements in the union of its n elements. Find the sum of f(k) over all k in S.

9th Asian Pacific Mathematics Olympiad 1997 ProblemsA1.  Let Tn = 1 + 2 + ... + n = n(n+1)/2. Let Sn= 1/T1 + 1/T2 + ... + 1/Tn. Prove that 1/S1 + 1/S2 + ... + 1/S1996 > 1001. A2.  Find an n in the range 100,

Read more »

Posted in: APMO

8th Asian Pacific Mathematics Olympiad 1996 Problems

8th Asian Pacific Mathematics Olympiad 1996 ProblemsA1.  ABCD is a fixed rhombus. P lies on AB and Q on BC, so that PQ is perpendicular to BD. Similarly P' lies on AD and Q' on CD, so that P'Q' is perpendicular to BD. The distance between PQ and P'Q' is more than BD/2. Show that the perimeter of the hexagon APQCQ'P' depends only on the distance between PQ and P'Q'.

Read more »

Posted in: APMO

7th Asian Pacific Mathematics Olympiad 1995 Problems

7th Asian Pacific Mathematics Olympiad 1995 ProblemsA1.  Find all real sequences x1, x2, ... , x1995 which satisfy 2√(xn - n + 1) ≥ xn+1 - n + 1 for n = 1, 2, ... , 1994, and 2√(x1995 - 1994) ≥ x1 + 1. A2. 

Read more »

Posted in: APMO

6th Asian Pacific Mathematics Olympiad 1994 ProblemsA1.  Find all real-valued functions f on the reals such that (1) f(1) = 1, (2) f(-1) = -1, (3) f(x) ≤ f(0) for 0 < x < 1, (4) f(x + y) ≥ f(x) + f(y) for all x, y, (5) f(x + y) ≤ f(x) + f(y) + 1 for all x, y.

5th Asian Pacific Mathematics Olympiad 1993 ProblemsA1.  A, B, C is a triangle. X, Y, Z lie on the sides BC, CA, AB respectively, so that AYZ and XYZ are equilateral. BY and CZ meet at K. Prove that YZ2 = YK.YB. A2.

Read more »

Posted in: APMO

4th Asian Pacific Mathematics Olympiad 1992 Problems A1.  A triangle has sides a, b, c. Construct another triangle sides (-a + b + c)/2, (a - b + c)/2, (a + b - c)/2. For which triangles can this process be repeated arbitrarily many times?

3rd Asian Pacific Mathematics Olympiad 1991 ProblemsA1.  ABC is a triangle. G is the centroid. The line parallel to BC through G meets AB at B' and AC at C'. Let A'' be the midpoint of BC, C'' the intersection of B'C and BG, and B'' the intersection of C'B and CG. Prove that A''B''C'' is similar to ABC.

2nd Asian Pacific Mathematics Olympiad 1990 ProblemsA1.  Given θ in the range (0, π) how many (incongruent) triangles ABC have angle A = θ, BC = 1, and the following four points concyclic: A, the centroid, the midpoint of AB and the midpoint of AC?

1st Asian Pacific Mathematics Olympiad 1989 ProblemsA1.  ai are positive reals. s = a1 + ... + an. Prove that for any integer n > 1 we have (1 + a1) ... (1 + an) < 1 + s + s2/2! + ... + sn/n! .A2.  Prove that 5n2 = 36a2 + 18b2 + 6c2 has no integer solutions except a = b = c = n = 0. A3.  ABC is a triangle. X lies on the segment AB so that AX/AB = 1/

Read more »

Posted in: APMO