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Geometry problem
Let B be a set of more than 2^(n+1)/n distinct points with coordinates of the form (1,1, . . . ,1) in n-dimensional space, with n >= 3. Show that there are three distinct points in B which are the vertices of an equilateral triangle.
1 個解答
- 1 十年前最愛解答
Solution: Let S be the set of all points with coordinates of the form (1;1; :::;1)
in n-dimensional space, and for each p 2 S, let Ap be the set of all points in S that
dier from p in exactly one coordinate. Each Ap contains n points, so if we count all of
the points in every Ap as we let p range over all points in B, we will count more than
n 2n+1=n = 2n+1 points. Since there are 2n points in S, by the pigeonhole principle,
we must count some point x at least three times. Thus, there are three points in B
that dier from x in exactly one coordinate. These three points must each dier from
each other in two coordinates, and so they form an equilateral triangle with side length
2p2.
