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More Than 22! (Posted on 2005-08-18) Difficulty: 3 of 5
Let T(n) be a triangular portion of the triangular grid with n points on a side. It is an unsolved problem what the maximum number of points is that can be selected from T(n) so that no three selected points are the vertices of an equilateral triangle. For small values of n the maximum appears to be 2n-2. However, for T(12), shown below, I found more than 2*12-2=22 points, no three of which are the vertices of an equilateral triangle! How many can you find?
o
o o
o o o
o o o o
o o o o o
o o o o o o
o o o o o o o
o o o o o o o o
o o o o o o o o o
o o o o o o o o o o
o o o o o o o o o o o
o o o o o o o o o o o o
Examples:

Here's an example of 5 points selected from T(4), no three of which are the vertices of an equilateral triangle.
s
o o
o s s
o s o s
It turns out that this selection of 5 cannot be increased to 6 without three of the selected points being the vertices of an equilateral triangle. If we add the first point in the second row, we get
s
s o
o s s
o s o s
Notice that three of the 6 s's are the vertices of an equilateral triangle.

There is a better selection of 6 points in T(4) no three of which are the vertices of an equilateral triangle.

See The Solution Submitted by McWorter    
Rating: 4.1667 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): And this ? - 30 points...right again...I think I´m gonna rest | Comment 38 of 42 |
(In reply to re(2): And this ? - 30 points...right again...I think I´m gonna rest by pcbouhid)

Absolutely no problem, pcbouhid, this is math at its best! Hammering away and watching the problem slowly chip apart. love it.
  Posted by owl on 2005-08-19 21:04:04

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