2007/8 British Mathematical Olympiad
Round 1: Friday, 30 November 2007--------------------------------------------------------------------------------
1) Find the value of
(1^4+2007^4+2008^4)/(1^2+2007^2+2008^2)
2) Find all solutions in positive integers x,y,z to the simultaneous equations.
x+y-z=12
x^2+y^2-z^2=12
3) Let ABC be a triangle with an obtuse angle at A. Let Q be a point (other than A,B, or C) on the circumcircle of the triangle, on the same side chord BC as A, and let P be the other end of the diameter through Q. Let V and W be the feet of the perpendiculars from Q onto CA and AB respectively. Prove that the triangle PBC and AWV are similar. (Note: the circumcircle of the triangle ABC is the circle which passes through the vertices A,B and C.)
4) Let S be a subset of the set of numbers {1,2,3,...,2008} which consists of 756 distinct numbers. Show that there are two distinct elements a,b of S such that a+b is divisible by 8.
5)Let P be an internal point of the triangle ABC. The line through P parallel to AB meets BC at L, the line through P parallel to BC meets CA at M, and the line through P parallel to CA meets AB at N. Prove that
BL/LC X CM/MA X AN/NB (is equal or smaller than) 1/8
and locate the position of P in the triangle ABC when equality holds.
6) The function f is defined on the set of positive integers by f(1)=1, f(2n)=2f(n) and nf(2n+1)=(2n+1)(f(n)+n) for all values where n is equal or bigger than 1.
(i) Prove that f(n) is always an integer.
(ii) For how many positive integers less than 2007 is f(n)=2n 最后一道题目这么简单呀
假设
f(n)=n*g(n)
得到
g(1)=1,g(2n)=g(n),g(2n+1)=g(n)+1=g(2n)+1
得到，g(n)是n二进制表示中1的数目。
现在问有多少个小于2007的整数中二进制表示正好两位是1，
2007二进制表示为11111010111，后面就不用算了，太简单了 我来做个最简单的，余下的楼下继续:)
令 a=2007，则
原式 = [ 14 + a4 + ( 1 + a )4 ] / [ 12 + a2 + ( 1 + a )2 ]
= 1 + a + a2 = 4030057 我也来做一个，好久不做奥数题了，不足指出请指出:)
2) 已知x,y,z都是正整数，
z=x+y-12
而x^2+y^2-z^2=12
则x^2+y^2-(x+y-12)^2=12
所以xy-12x-12y+78=0
(x-12)(y-12)=66=1*66=(-1)*(-66)=6*11=(-6)*(-11)=2*33=(-2)*(-33)=3*22=(-3)*(-22)
方程中x,y等价：
i)x-12=1, y-12=66, 此时x=13,y=78,z=13+78-12=79
ii) x-12=-1,y-12=-66,此时y=-54<0,不符合要求,舍去
iii)x-12=6, y-12=11, 此时x=18,y=23,z=18+23-12=29
iv) x-12=-6,y-12=-11 此时x=6, y=1 ,z=6+1-12=-5<0,不符合要求,舍去
v)x-12=2, y-12=33, 此时x=14,y=45,z=14+45-12=47
vi) x-12=3, y-12=22, 此时x=15,y=34,z=15+34-12=37
vii)另外2种情况下，不符合要求，舍去
综上所述,有8组解,(x,y,z)为(13,78,79),(78,13,79),(18,23,29),(23,18,29),
(14,45,47),(45,14,47),(15,34,37),(34,15,37)
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