Problem:
The probability that a set of three distinct vertices chosen at random from among the vertices of a regular -gon determine an obtuse triangle is . Find the sum of all possible values of .
Solution:
Consider the cases of even and odd values of separately. If is even, let , and label the vertices of the polygon consecutively from through . Assume without loss of generality that one of the chosen vertices is . If the diametrically opposite vertex is also chosen, the resulting triangle is a right triangle. Among all other choices for the two remaining vertices, the resulting triangle is obtuse if and only if the difference between the indices of the vertices is at most . This will always be the case if both vertices have indices of the same sign. There are ways to choose two vertices with positive indices and the same number of ways to choose two vertices with negative indices. If exactly one of the chosen vertices has positive index with , then exactly choices of a vertex with a negative index result in an obtuse triangle. Therefore the number of obtuse triangles with exactly one vertex of positive index is , and the total number of obtuse triangles is . The total number of possible triangles with one vertex at is , so the probability of an obtuse triangle is
Setting the probability equal to gives , and so . If is odd, let , and label the vertices of the polygon consecutively from through . The total number of triangles with one vertex at is . An argument similar to the previous argument shows that the number of obtuse triangles with remaining vertices of positive index is for each of the three cases , and . The probability of an obtuse triangle is therefore
Setting the probability equal to gives , and so . The sum of all possible values of is .
The problems on this page are the property of the MAA's American Mathematics Competitions