consider three points A, B, C such that d(A,B), d(A,C), and d(B,C) are all integers. we knowwww.ddhw.com -d(A,B) <= d(A,C) - d(B,C) <= d(A,B). that means when A, B are fixed, the choice on C are on a family of hyperbolas (finite number) generated by A and B. now if D is not colinear with A and B, but both d(A,D) and d(C,D) are integers. so C is also on the family of hyperbolas generated by A, D. as we can see, with A, B, D three points fixed, there can be only finite number of points left to choose, which are those points lie on the intersection of two families of hyperbolas. The actual candidates are even less because the distances between them have to be integer too. www.ddhw.com
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哈哈哈哈 O(∩_∩)O哈哈~ 好久不来了，不小心看了一眼自己主页，发现自动关联了自己以往发过的帖，点进这个看了一眼，笑个半死。。。敢情我当初还发过这么搞笑的东西啊，我自己完全不记得了 
雷达卡
发表于 2005-3-9 23:42:46