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.
Josh's solution points directly at a 22-point configuration for T[12].
Choose all the points along the bottom and left side except for the
bottom left point. In general, this approach shows that the 2n-2 is at
least a lower bound for T[n].
Nice eye, Josh.
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Posted by owl
on 2005-08-18 19:34:49 |
Extending Josh's Sol'n
