Problem:
Let be a convex polygon with sides, . Any set of diagonals of that do not intersect in the interior of the polygon determine a triangulation of into triangles. If is regular and there is a triangulation of consisting of only isosceles triangles, find all the possible values of .
Solution:
The answer is , where and are nonnegative integers. In other words, is either a power of 2 (when ) or the sum of two nonequal powers of 2 (with being considered as a power of 2 ).
We start with the following observation.
Lemma. Let be a convex polygon with . Suppose that is cyclic and its circumcenter does not lie in its interior. If there is a triangulation of consisting only of isosceles triangles, then , where is a positive integer.
Proof. We call an arc minor if its arc measure is less than or equal to . By the given conditions, points lie on the minor arc of the circumcircle, so none of the angles is acute. (See the left-hand side diagram shown below.) It is not difficult to see that is longer than each other side or diagonal of . Thus must be the base of an isosceles triangle in the triangulation of . Therefore, must be even. We write . Then is an isosceles triangle in the triangulation. We can apply the same process to polygon and show that is even. Repeating this process leads to the conclusion that for some positive integer .
The results of the lemma can be generalized by allowing if we consider the degenerate case .
We are ready to prove our main result. Let denote the regular polygon. There is an isosceles triangle in the triangulation such that the center of lies within the boundary of the triangle. Without loss of generality, we may assume that , with (that is, ), is this triangle. Applying the Lemma to the polygons , and , we conclude that there are (where and are nonnegative integers) vertices in the interiors of the minor arcs , , respectively. (In other words, .) Hence
where and are nonnegative integers. The above discussion can easily lead to a triangulation consisting of only isosceles triangles for . (The middle diagram shown above illustrates the case . The right-hand side diagram shown above illustrates the case .)
This problem was suggested by Gregory Galperin.
The problems on this page are the property of the MAA's American Mathematics Competitions