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Old 07-24-2007   #4
Scarlet Manuka
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If there are n sides, each vertex has (n-3) diagonals leading from it (there are n-1 other vertices, two of which are its neighbours; a line to any other is a diagonal). So the number of diagonals is n (n-3) / 2 (each diagonal can be started from either end). So we have to solven - n (n-3) / 2 = 0 n (n-3) - 2n = 0 n (n - 5) = 0 n = 0 or 5, obviously 0 doesn't count. So the answer is 5.