Fifteen pennies lie on the table in the shape of a triangle Fifteen pennies lie on the table in the shape of a triangle, with ﬁve pennies
on each side. For some reason, the pennies are painted either black or white. 
Prove that there exist three pennies of the same color whose centers are the vertices of an equilateral triangle.
Any other approach other than brute force? So due to symmetry we can fix a base. 
 A: We proceed by assuming that such a configuration is possible, and attempt to construct it by avoiding the structure. We should eventually reach a contradiction.
Label the coins as such:
$\begin{array} { l l l l l l l l l l l l l l l l l }
&&&&&1\\
&&&&2&&3\\
&&&4&&5&&6\\
&&7&&8&&9&&10\\
&11&&12&&13&&14&&15\\
\end{array}
$
There are $10 + 6 + 3 + 1 = 20$ upward pointing equilateral triangles. There are $6+1 = 7$ downward pointing equilateral triangles. There are 2 slanted equilateral triangles (2-10-12, 3-7-14).
Since 1-11-15 is not monochromatic, WLOG 11 and 15 are black and 1 is white. Consider 10.
If 10 is black, then from 10-14-15, 14 is white.
If 10 is white, then from 1-7-10 so 7 is black. Then from 7-11-12 so 12 is white. Then from 2-10-12 so 2 is black. Then from 2-11-14 so 14 is white.
Hence, in either case, 14 must be white. A similar argument shows that 12 is white.
At least one of 4 and 6 is black. By considering 4-7-13 or 6-13-15, we get that 13 is white.
Since 12, 13, 14 are white, hence 5, 8, 9 are black. But this is an equilateral triangle.
We thus reach a contradiction.
A: This is a bit of a late hit, but my solution is, I think, substantially different and of possible interest.
Assume a configuration without a monochromatic equilateral triangle exists.
5-8-9 is not monochromatic, so by symmetry we can assume without loss of generality that 5 is white and 8 and 9 are black.
That means 13 must be white.
So 10 must be black to avoid 5-13-10.
That makes 6 white to avoid 9-10-6.
3 must be white to avoid 3-8-10 and black to avoid 3-5-6.
Contradiction.
