There are 400 points on a plane. Prove that different distances among them are not less then 15.
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Expert's answer
2010-05-31T11:30:25-0400
Let the number of different distances between the points is equal to k. Let's fix two points. Then all other points are points of intersection of two concentric circumferences, consisting of k circumferences. Consequently, the total number of points does not exceed (2k² +2). It is left to notice that (2*14²) +2=394<400.
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