我发现注册的名字后的圈不是蓝的(男)就是粉(女),偏偏你这个是灰的。是何道理? |
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consider three points A, B, C such that d(A,B), d(A,C), and d(B,C) are all integers. we know -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. |
One down, 5 or 6 more to go? |
your questions are all challanging. high quality ones. |
sounds so familiar. search the web and got this: BTW, Chvatal is a professor of my deparment. |
thanks |
Answer copied from WXC: 对平面上的N个点,可以两两连接得C(N,2)条直线,(包括互相重合的在内),从每个点向这些直线引垂线,共得到有限条垂线。若这些垂线的长度都为0,则所有点都在一条直线上,命题成立。否则,在所有有限条非零长度的垂线中,必有一条最短的,其长度为d。设它是由点A向直线L所引的垂线,垂足为H。因在直线L上至少有三点,故必有两点在H的同侧(包括H点),记离H点较近的点为B,离H点较远的点为C,连接AC,从B和H点分别向AC作垂线,垂足分别为P和Q,则有0<BP≤HQ<AH=d,与d的定义矛盾。 This "simple" problem went unsolved for 50 years (1893 - 1943). Probably not noticed by first class mathematicians. I think it is easier than many IMO problems. |
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