Problem:
Ten points in the plane are given, with no three collinear. Four distinct segments joining pairs of these points are chosen at random, all such segments being equally likely. The probability that some three of the segments form a triangle whose vertices are among the ten given points is , where and are relatively prime positive integers. Find .
Solution:
Begin generally with points in the plane, no three of which are collinear. There are segments joining the points, and ways to choose four segments. Because two triangles can share at most one side, four segments cannot form two triangles. Therefore, it suffices to count the ways of choosing a triangle and one additional segment. There are ways to choose the vertices of a triangle, and then ways to choose an additional segment. Hence the probability of obtaining a triangle when is
so that .
The problems on this page are the property of the MAA's American Mathematics Competitions