This has Vietnamese version. Please go to mo2000.html

APMO 2000
I've just joined in the 13th 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 (xi) / (1 - 3xi + 3xi2), i = 0 to 101, with xi = 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)/(kk.(n-k)n-k) á (n!)/(k!(n-k)!) á (nn)/(kk.(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,... an) is a permutation of (0, 1, 2,..., n). An alteration is allowable if it swap ai and aj with ai = 0 and aj = ai-1+1. Find all n so that we can obtained (1, 2,...,n, 0) from (1, n, n-1,..., 2, 0) with finite alterations.
Now I think there's no matter with the first 4 problems, each of them can be explained in a few lines. But the most difficult is problem 5, and the solution I know is long.

VMO 2000
The 38th 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 (xn) determined by xn+1 = (c + (c + xn)) (n = 0,1,2... 0) if expressions in roots are not negative. Find all c so that with any x0 Î (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 á a á p and Pn(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 Pn(x) divisible f(x) for all n > 3.
2. There's not a g(x) = x + c so that Pn(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 AP is the set of all x so that P(x) = 0. Find the maximum size AP

Now I still don't know how to solve P2.2 and P3. If anyone know, please send it so me. Thank you!