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3 points colinear? No way! (Posted on 2005-05-19) Difficulty: 5 of 5
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: So at least the first 32 are reachable... | Comment 9 of 14 |
(In reply to So at least the first 32 are reachable... by Jer)

Only 1.87n?  Depressing.  Then again, maybe a more worthy challenge.

I once made up the problem:

Consider the triangular grid with points au+bv, where u=(1,0), v=(.5,sqr(3)/2), and a and b integers in the interval [0,n).  What is the greatest number of points that can be put inside this grid so that no three points are the vertices of an equilateral triangle?

The right answer appeared to be 2n-2 until n=12, when I found 24 points no three of which are the vertices of an equilateral triangle.  This problem appears to be quite a challenge also, except the maximum seems to be a quadratic function of n rather than a linear one.

(Thanks for the posting help, Jer.
It works!)


  Posted by McWorter on 2005-05-20 17:55:29
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