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.
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.