# 【文摘】CMO2002

1.The lengths of BC, CA and AB of triangle ABC are a, b and c respectively, with b < c. D is a point on BC such that AD bisects angle A.
(i) Find the necessary and sufficient condition for ensuring there are
points E, F on segments AB, AC, other than the vertices, which satisfy BE = CF and < BDE = < CDF. (Express the condition in terms of angles A, B and C);
(ii) Under this condition, express the length of BE in terms of a, b and c.

(1)求在线段AB,AC内分别存在点E,F(不是顶点)满足BE=CF和 角BDF=角CDF

(2)在点E,F存在的情况下,用a,b,c表示BE的长.

2. The polynomial sequence {P_n(x)} is defined as:
P_1(x) = x^2-1, P_2(x) = 2x (x^2-1)
P_n+1(x) P_n-1(x) = (P_n(x))^2 – (x^2-1)^2, n = 2, 3, …
Let S_n be the sum of the absolute values of the coefficients of P_n(x). For any positive integer n, find the non-negative integer k_n so that 2^k_n just divides S_n

2.设多项式序列{Pn(x)}满足:
P1(x)=x^2-1;P2(x)=2x(x^2-1);且Pn+1(x)Pn-1(x)=[Pn(x)]^2-(x^2-1)^2 n=2,3,……

Sn*2^(-Kn)为奇数. 【参考答案

3. 18 football teams plays a tournament. In each round, the teams are divided into 9 groups and each group plays a game. They play a total of 17 rounds so that each team can play a game with every other team. Find the maximum possible value of n, so that after n rounds, there always exists 4 teams among which only one game is played.

3.18支足球队进行单循环赛,共比赛17轮,使得每支球队都与另外17支球队各赛一场,

4. P1, P2, P3, P4 are any 4 distinct points on a plane , find the minimum value of the following:
( Summation (1<=i<j<=4) PiPj ) / ( min (1<=i<j<=4) PiPj )

4.对平面上任意4个不同的点P1,P2,P3,P4 ，

(说明,sigma代表循环求和,minPiPj是所有连线中的最小长度)

5. A rational point is a point with rational numbers as x and y coordinates.
Prove that all rational points on the plane can be divided into 3 disjoint sets satisfying:
(i) In any circle with a rational point as the center there are points from every one of the 3 sets;
(ii) On any straight line there does not exist 3 points each of which
belongs to each of the 3 sets.

5.横纵坐标均为有理数的点称为”有理点”.证明平面上的全体有理点可以分为3个两两不相交的集合， 满足条件:
(1)在以每个有理点为圆心的任意圆内一定包含这三个集合中每个集合的点
(2)在任意一条直线上不可能有三个点分别属于这个集合

6. Given c, 1/2 < c < 1, find the minimum value of constant M, so that for any positive integer n >= 2 and positive real numbers a1 <= a2 <= … <= a_n ，satisfying (1/n)Summation(k=1 to n) k a_k = c Summation(k=1 to n) a_k ,
we have Summation (k=1 to n) a_k <= M Summation(k=1 to m) a_k
where m is the greatest integer not exceeding cn.

6.给定c, 1/2 < c < 1 ,及实数0<a1<=a2<=……<=an ，