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标题: 平面共线点集 [打印本页]

作者: fzy    时间: 2005-3-8 02:06
标题: 平面共线点集

A) 平面上有一个有限点集S,满足下列条件:S中任意两点P1,P2的联线或延长线上都有S中另外一点P3。证明S的所有点都在一条直线上.www.ddhw.com

B) 平面上有一个无穷点集S,其中任意两点的距离是整数.证明S的所有点都在一条直线上.
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作者: 我很高兴    时间: 2005-3-8 03:56
标题: 回复:平面共线点集

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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作者: husonghu    时间: 2005-3-8 04:11
标题: 嗨,这是给不想申明性别的网友的自由啊!这样很好:世界更丰富多彩。[:)]

  嗨,这是给不想申明性别的网友的自由啊!这样很好:世界更丰富多彩。





作者: 新用户    时间: 2005-3-8 05:31
标题: 回复:回复:平面共线点集

【华闻解答】为什么有些发贴人的笔名旁边有一个®,而且有不同的颜色,而我的却没有?那是因为...
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作者: fzy    时间: 2005-3-9 18:26
标题: Nobody wants to do my problems. Feel Bad. [:((]

  Nobody wants to do my problems. Feel Bad.





作者: sean9991    时间: 2005-3-9 23:42
标题: question B

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.
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作者: fzy    时间: 2005-3-10 00:13
标题: 回复:question B

One down, 5 or 6 more to go?
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作者: sean9991    时间: 2005-3-10 00:54
标题: try my best

your questions are all challanging.  high quality ones.
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作者: sean9991    时间: 2005-3-10 02:02
标题: question A

sounds so familiar.  search the web and got this:
http://mathworld.wolfram.com/SylvestersLineProblem.html
BTW, Chvatal is a professor of my deparment.www.ddhw.com
 
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作者: fzy    时间: 2005-3-10 02:43
标题: 回复:try my best

thanks
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作者: fzy    时间: 2005-3-14 20:01
标题: Answer to question A

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。www.ddhw.com

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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