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 3 points colinear? No way! (Posted on 2005-05-19)
For any grid, x by x, figure out a formula for the greatest number of points that can be put on the inside of the grid such that no three points are colinear.

 No Solution Yet Submitted by Corey Rating: 3.0000 (1 votes)

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 re(3): So at least the first 32 are reachable... | Comment 11 of 14 |
(In reply to re(2): So at least the first 32 are reachable... by Jer)

Thanks.  I already have 14 problems pending.  Assuming the first one is accepted, I estimate it will be at least two weeks before it appears (and maybe 3 or 4 additional months before the fourteenth appears).  This site has many talented and prolific contributors.

An easier version used for a mathematics competition at Ohio State University states:

Let T be the set {au+bv | a, b in [0,n-1), u=(1,0), v=(.5,sqr(3)/2)}.  Show that, given any subset A of T, with |A|>n(n+1)/3, some three elements of A are the vertices of an equilateral triangle.

I'll consider posting the variation as you suggest, jer.  It bothers me, however, to post problems for which a solution is unknown.

 Posted by McWorter on 2005-05-20 18:29:45

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