Problem:
Given a triangle, its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra Piβ is defined recursively as follows: P0β is a regular tetrahedron whose volume is 1. To obtain Pi+1β, replace the midpoint triangle of every face of Piβ by an outward-pointing regular tetrahedron that has the midpoint triangle as a face. The volume of P3β is m/n, where m and n are relatively prime positive integers. Find m+n.
Solution:
The diagram shows P1β. Notice that P0β has 4 triangular faces, P1β has 24 , and, inductively, Piβ has 4β
6i. This expression therefore counts the small tetrahedra that are attached to Piβ to form Pi+1β. The volume of each of these small tetrahedra is (81β)i+1, and hence the volume of Pi+1β is 4β
6i(81β)i+1=(21β)(43β)i more than the volume of Piβ. In particular, the volume of P3β is
1+(21β)+(21β)(43β)+(21β)(43β)2=3269β
Thus m+n=101β.
The problems on this page are the property of the MAA's American Mathematics Competitions