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?
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Examples:
Here's an example of 5 points selected from T(4), no three of which are the vertices of an equilateral triangle.
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
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.
I've got to eat supper now, but I'll make a few remarks before I eat.
In answer to owl, I made up the problem, so the only known maxima are what you guys find. It appears that there's a new maximum for T(9) discovered by Tristan!
And the new maxima you guys have found based on Tristan's idea look like they can be increased even more. Perhaps you guys will find new maxima before I get back from supper.
Smith's 26 point selection contains an equilateral triangle, the third s from the top in the 6 s's down the left side, the third s in the 6th row, and the third s in the line of s's slanting up right from the bottom.
Will check in again in about an hour.
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Posted by McWorter
on 2005-08-19 20:55:26 |
Wow! You guys are terrific!
