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Polygons in a Cube (Posted on 2011-12-11) Difficulty: 2 of 5
Given 1985 points inside a unit cube, prove that precisely 32 of these points can be chosen in such a way that every closed polygon (possibly degenerate) with these points as vertices has perimeter less than 8√3

No Solution Yet Submitted by K Sengupta    
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Solution Solution Comment 1 of 1
This looks like one of those pigeon-hole problems.  So take 1985-1=1984.  Since we are looking at a cube, factor 1984 into 31*4^3.  Divide the cube into 4^3=64 subcubes.  

At least one of the subcubes has at least 32 of the 1985 points.  1984 points can avoid this by having 31 points in each subcube, as implied the factorization of 1984.  Choose a subcube with 32 (or more) points.

The furthest apart two points can be in the subcube is sqrt(3)/4 units, occuring when the points occupy opposite corners.  32*sqrt(3)/4 = 8*sqrt(3).  

With all the points distinct, most of the individual distances will be less than sqrt(3)/4.  Therefore every closed polygon (possibly degenerate) with these 32 points as vertices has perimeter less than 8*sqrt(3).

  Posted by Brian Smith on 2016-07-17 10:43:30
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