Question

Given a triangle, its midpoint triangle is obtained by joining the midpoints of its sides. A sequence of polyhedra $\mathcal{P}_i$ is defined recursively as follows: $\mathcal{P}_0$ is a regular tetrahedron whose volume is $1.$ To obtain $\mathcal{P}_{i+1},$ replace the midpoint triangle of every face of $\mathcal{P}_i$ by an outward-pointing regular tetrahedron that has the midpoint triangle as a face. The volume of $\mathcal{P}_3$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Answer: 101

Concepts:volume power scaling of length, area, and volume recursion geometric sequence

Difficulty rating: 2990

Solution:

Attaching a tetrahedron over the midpoint triangle of a face replaces that face by $6$ equilateral triangles of half the side length: the $3$ corner triangles plus $3$ exposed faces of the new tetrahedron. So all faces of $\mathcal{P}_i$ are congruent, with side $\left(\frac{1}{2}\right)^i$ times the original, and $\mathcal{P}_i$ has $4 \cdot 6^i$ faces.

Passing from $\mathcal{P}_i$ to $\mathcal{P}_{i+1}$ glues one regular tetrahedron onto each face; each is similar to $\mathcal{P}_0$ with ratio $\left(\frac{1}{2}\right)^{i+1},$ hence has volume $\left(\frac{1}{8}\right)^{i+1}.$ The volume added is $4 \cdot 6^i \left(\frac{1}{8}\right)^{i+1} = \frac{1}{2}\left(\frac{3}{4}\right)^i.$

Therefore the volume of $\mathcal{P}_3$ is $1 + \frac{1}{2} + \frac{3}{8} + \frac{9}{32} = \frac{69}{32},$ and $m + n = 69 + 32 = 101.$

2001 AIME II Problem 12

Answer: 101

Difficulty rating: 2990

Solution:

Problem 12 in Other Years