- 1 - 2010 年部分省市中考数学试题分类汇编 压轴题(四 ) 23.(安徽省)如图,已知111ABCA B C△∽ △ ,相似比为 k(k>1),且ABC△的三边长分别为 a、b、c(a>b>c),1 1 1ABC△ 的三边长分别为1a 、1b 、1c . (1)若 c=a1,求证:a=kc; [证 ] (2)若 c=a1,试给出符合条件的一对111ABCA B C△和△ ,使得 a、b、c 和1a 、1b 、1c 都是正整数,并加以说明; [解 ] (3)若 b=a1,c=b1,是否存在111ABCA B C△和△ 使得 k=2?请说明理由. [解 ] 解:(1)证:111ABCA B C△∽ △ ,且相似比为11(1).ak kkakaa, , 又1.caakc, ······················································································· (3 分) (2)解:取111864432.abcabc,,,同 时 取,, ················· (8 分) 此时1111112abcABCA B Cabc,△∽ △ 且1.ca ···································(10 分) 注:本题也是开放型的,只要给出的ABC△和111A B C△ 符合要求就相应赋分. (3)解:不存在这样的ABC△和111A B C△ .理由如下: 若2k ,则111222.aabbcc, , 又1ba,1cb, 112244aabbc, 第 23 题图 - 2 - 2 .bc·······································································································(12 分) 24bcccca,而bca, 故不存在这样的ABC△和111A B C△ ,使得2.k ·········································(14 分) 注:本题不要求学生严格按反证法的证明格式推理,只要能说明在题设要求下2k 的情况不可能即可. 24.(芜湖市 本小题满分14 分)如图,在平面直角坐标系中放置一矩形ABCO,其顶点为A(0,1)、B(-3 3,1)、C(-3 3,0)、O(0,0).将此矩形沿着过E(-3,1)、F(-4 33 ,0)的直线EF 向右下方翻折,B、C 的对应点分别为B′、C′. (1)求折痕所在直线EF 的解...