APMO 2000
I've just joined in the 13^th APMO (8/3/2000 in Vietnam). I think I had a half correct. Here're the 5 problems, must be done in 4 hours:

Problem 1: Calculate the sum (x_i) / (1 - 3x_i + 3x_i²), i = 0 to 101, with x_i = i/101


Problem 2: Find all ways to fill 9 number {1,2,3,4,5,6,7,8,9} in 9 circles so that: sums of 4 numbers in each side are the same sums of 4 squares of number in each side are the same

Problem 3: Give M is the midpoint of side BC of triangle ABC. AN is the bisector of BAC. The line pass N and perpendicular to AN meets AB, AM at P, Q. The line perpendicular to AP at P cut AN at O. Prove that OQ ^ BC.
Problem 4: Prove the compound inequality with n > k:
(1/(n+1)).(nⁿ)/(k^k.(n-k)^n-k) � (n!)/(k!(n-k)!) � (nⁿ)/(k^k.(n-k)^n-k)
(The origin problem was find n so that (1² + 2² +...+ n²)/n is a perfect square. But someone in APMO assembly found it had happened in a international contest not long time ago. So, APMO replaced it with the problem above).
Problem 5: (a₀, a₁,... a_n) is a permutation of (0, 1, 2,..., n). An alteration is allowable if it swap a_i and a_j with a_i = 0 and a_j = a_i-1+1. Find all n so that we can obtained (1, 2,...,n, 0) from (1, n, n-1,..., 2, 0) with finite alterations.

VMO 2000
The 38^th VMO 1999-2000 was held in March, 13 and 14. There're 3 hours for 3 problems each day. And this is the 6 problems:

The first day
Problem 1: c is a real positive number, the sequence (x_n) determined by x_n+1 = �(c + � (c + x_n)) (n = 0,1,2... 0) if expressions in roots are not negative. Find all c so that with any x₀ � (0 ; c), this sequence is defined with all n and has finite limitation.
Problem 2: Give two circle (O₁ ; r₁) and (O₂ ; r₂). Two points M, N move on two circle with the same angle velocity. Let O₁M and O₂N meet at Q.
1. Find the locus of midpoints of MN.
2. Prove that the circumcircle of MNQ go through a fixed point.
Problem 3: Give P(x) = x³ + 153x² - 111x + 38
1. Prove that there're at least 9 numbers a in [1 ; 3²⁰⁰⁰] that 3²⁰⁰⁰ is a divisor of P(a).
2. How many a in [1 ; 3²⁰⁰⁰] that 3²⁰⁰⁰ is a divisor of P(a)?
The second day
Problem 4: Give 0 � a � p and P_n(x) = xⁿsina - xsin(na) + sin(n-1)a.
1. Show that there's exactly one quadratic trinomial has form f(x) = x² + ax + b so that P_n(x) divisible f(x) for all n > 3.
2. There's not a g(x) = x + c so that P_n(x) divisible g(x) for all n > 3.
Problem 5: Find all n > 3 so that there're n points in space sastify all of three conditions:
(1) not 3 points in the same line
(2) not 4 points in the same circle
(3) all circumcircles of three points have the same radius.
Problem 6: P(x) is a polynomial that P(x² - 1) = P(x) . P(-x)
Give A_P is the set of all x so that P(x) = 0. Find the maximum size A_P