Problem:
Given a triangle, its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra Pi​ is defined recursively as follows: P0​ is a regular tetrahedron whose volume is 1. To obtain Pi+1​, replace the midpoint triangle of every face of Pi​ by an outward-pointing regular tetrahedron that has the midpoint triangle as a face. The volume of P3​ is m/n, where m and n are relatively prime positive integers. Find m+n.
Solution:
The diagram shows P1​. Notice that P0​ has 4 triangular faces, P1​ has 24 , and, inductively, Pi​ has 4⋅6i. This expression therefore counts the small tetrahedra that are attached to Pi​ to form Pi+1​. The volume of each of these small tetrahedra is (81​)i+1, and hence the volume of Pi+1​ is 4⋅6i(81​)i+1=(21​)(43​)i more than the volume of Pi​. In particular, the volume of P3​ is
1+(21​)+(21​)(43​)+(21​)(43​)2=3269​
Thus m+n=101​.
The problems on this page are the property of the MAA's American Mathematics Competitions
