RMO geometry problem
Consider a 20-sided convex polygon K, with vertices A1,A2,A3.............A20 in that order. Find the number of ways in which three sides of K can be chosen so that every pair among them has at least two sides of K between them. For example, (A1A2, A4A5,A11A12) is an admissible triple but (A1A2, A4A5 ,A19A20) is not
The answer is 520.
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1 solution
Having chosen A1A2, we cannot choose A2A3, A3A4, A20A1 nor A19A20.
Thus we have to choose two sides separated by 2 sides among 15 sides A4A5, A5A6, ..., A18A19.
If A4A5 is one of them, the choice for the remaining side is only from 12 sides A7A8, A8A9, ..., A18A19.
If we choose A5A6 after A1A2, the choice for the third side is now only from A8A9, A9A10, ..., A18A19 (11 sides).
Thus the number of choices progressively decreases and finally for the side A15A16 there is only one choice, namely, A18A19.
Hence the number of triples with A1A2 as one of the sides is 12 + 11 + 10 + ··· + 1 = 12 × 13 2 = 78.
Hence the number of triples then would be (78 × 20)/3 = 520.