From: David G Radcliffe Subject: Cutting equilateral triangles Date: 4 Apr 1999 00:36:57 GMT Newsgroups: sci.math Keywords: No dissections into incongruent equilateral triangles In this note I will prove the following theorem. Theorem 1. If an equilateral triangle is dissected into a finite number of smaller equilateral triangles, then two of the smaller triangles are congruent. This is a consequence of a more general theorem: Theorem 2. If a parallelogram is dissected into finitely many equilateral triangles, then two of the triangles are congruent. If an equilateral triangle is dissected into equilateral triangles, then we can obtain a dissection of a parallelogram into equilateral triangles by attaching another (undivided) equilateral triangle along one of the sides of the original triangle. According to theorem 2, two of the triangles in this dissection are congruent. But the triangle that we attached is larger than all of the constituent triangles of the original dissection. Therefore, two of the triangles of the original dissection must be congruent. This shows that theorem 2 implies theorem 1. We now turn to the proof of theorem 2. Suppose we have are given a dissection of a parallelogram into equilateral triangles. The parallelogram is drawn so that two sides are horizontal. There is a graph associated to this dissection. Each vertex of the graph corresponds to a maximal horizontal segment, formed from the sides of the triangles of the dissected parallelogram. The vertices corresponding to the top and bottom sides of the parallelogram are called "polar vertices". Each triangle lies between two maximal horizontal segments, and thus corresponds to an edge of the graph. There may be multiple edges joining two vertices. _ __________ O \ \ /\ | \ \ / \ | | \ /----\ O | \ / \ / \ /|\ | \/___\/___\ \|/ / O_/ dissected associated parallelogram graph Now it is "obvious" that this graph is planar. Therefore, Euler's polyhedron formula V-E+F=2 holds, where V is the number of vertices, E is the number of edges, and F is the number of faces (including the unbounded face). If the graph contains two edges with the same endpoints, then the corresponding triangles are congruent, and we are done. Suppose to the contrary that the graph has no multiple edges. Then each face must have at least 3 sides, and so 3F <= 2E. Combining this with V-E+F=2, we conclude that 6 = 3V-3E+3F <= 3V-3E+2E = 3V-E, or E <= 3V-6. The degree of each non-polar vertex is even and is at least 4. If a non-polar vertex has degree 4, then the dissected parallelogram must contain two congruent triangles. If this does not hold, then each non-polar vertex has degree at least 6. Since each polar vertex has degree at least 2, it follows that the sum of the degrees is at least 4+6(V-2) = 6V-8. But the sum of the degrees is equal to twice the number of edges. Therefore E >= 3V-4, and this contradicts E <= 3V-6. Thus in any case we must have two congruent triangles, and the theorem is proved. The foregoing argument is an elaboration of the proof given in the article "Dissections Into Equilateral Triangles" by W.T. Tutte. That article appears in "Mathematical Recreations: A Collection in Honor of Martin Gardner", edited by David A. Klarner. Any errors are my own. -- David Radcliffe radcliff@uwm.edu