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2014年英国数学奥林匹克第二轮试题及其解答

2014年英国数学奥林匹克第二轮试题及其解答
2014年英国数学奥林匹克第二轮试题及其解答

2014年英国数学奥林匹克第二轮试题及其解答

解答人: 文武光华数学工作室 田开斌

一、将一个正2014边形的每一条对角线染上n 种颜色之一,使得对于任何两条在多边形内有交点的对角线,它们的颜色不同,求n 能取到的最小可能知。

解答:首先,由于正2014边形一共有1007条对角线交于中心,所以这1007条对角线一定颜色不同,故至少需要1007种颜色。

下面构造一种1007种颜色的染法。显然,所有的对角线可以分为2014个集合,每个集

合中的对角线互相平行,且相邻两个集合间的直线的夹角为 。显然,1007条过中心的

对角线在1007个不同的集合中。我们将有过中心的对角线所在的集合及其下一个集合染为同一种颜色,这样恰好用了1007种颜色,因为颜色相同的对角线要么平行,要么交于一个顶点,所以在2014边形内部有交点的对角线的颜色一定不同。

综上所述,最少需要1007种颜色才能满足题设要求。

二、证明:不存在一个长方体,其体积、表面积和周长数值上相等(这里的周长指的是12条棱的长度之和)。

证明:若存在一个长方体,其体积、表面积和周长数值都相等。设其长、宽、高分别为a 、b 、c ,则

abc =2(ab +bc +ca )=4(a +b +c ) (*)

根据(*)知:[2(ab +bc +ca )] =4(a +b +c )abc ,即(ab +bc +ca ) =

(a +b +c )abc 。然而(ab +bc +ca ) ≥3(a +b +c )abc ,这便得到矛盾。

三、数列{a }中,a =4,a =a ?a 。

(1)证明:存在无穷多个素数,它们至少是该数列中某一项的约数;

(2)是否存在无穷多个素数,每个素数均不是该数列中任何一项的约数。

解答:(1)根据条件知:a =a (a ?1)。故对于任意m ∈N ,若正整数n >m ,都有a =a (a ?1)(a ?1)…(a ?1),从而a ?1=a ·∏(a ?1) ?1。于是知 a ?1,a =1,这说明a ?1存在一个素因子,它不能整除a 、a 、…、a 中的任一项。而a ?1|a ,所以每个a 都会多出一个前面所有项都没有的素因子。命题得证。

(2)对于一个素数p >5,若p 是该数列中某一项的因子,我们不妨设k 是使得p|a 的最小正整数。因为p|a (a ?1),且p ?a ,所以p|a ?1,从而p|a ?a ?1。因为素数p >5,所以

p|a ?a ?1

?p|4(a ?a ?1)

?(2a ?1) ≡5(mod p ) (*)

这说明,若素数p >5,使得p 是该数列中某一项的素因子,则 =1。

而根据二次互反律知 · =(?1)

· =1,且 = =?1,故当p ≡2、3(mod 5)时, =?1,从而p 不是该数列中任一项的素因子。又根据狄利克雷定理知,

这样的素数p 有无穷多个,所以存在无穷多个素数,使得每个素数均不是该数列中任何一项的约数。

四、点P 为△ABC 内一点,△ABC 内接于⊙O,延长AP 交⊙O 于A ,同样地定义B 、C 。记O 为△BCP 外心,类似地定义O 、O 。记O 为△B C P 的外心,类似地定义O 、O ,求证:O O 、O O 、O O 三线共点。

证明:如图,过B 作BB 垂线,过C 作CC 垂线,交于点M ;过B 作BB 垂线,过C 作CC 垂线,交于点N 。则O 为PM 中点,O 为PN 中点。根据圆的对称性,直线BM 、B M 关于点O 中心对称,直线CM 、C N 关于点O 中心对称,故M 、N 关于点O 中心对称,所以O 为线

段MN中点。注意到O O ∥MN,故O O 过OP中点K。同理可知O O 、O O 也都过OP 中点K,所以O O 、O O 、O O 三线共点。

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