2001 AIME II Problem 12

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

Small hint:

Each face of

\mathcal{P}_i

is replaced by

6

half-size faces, so

\mathcal{P}_i

has

4 \cdot 6^i

faces

Big hint:

The tetrahedra added at step

i + 1

each have volume

\left(\frac{1}{8}\right)^{i+1},

so the total volume added is

\frac{1}{2}\left(\frac{3}{4}\right)^i

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.