# 1988 AIME Problems/Problem 10

## Problem

A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?

## Solution

By the Euler characteristic, we have that $V - E + F = 2$. The number of faces, $F$, is $12 + 8 + 6 = 26$. Since every point lies on exactly one vertex of a square/hexagon/octagon, we have that $V = 12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48$. Substituting gives us $E = 72$.

The number of segments joining the vertices of the polyhedron is ${48\choose2} = 1128$. Of these segments, $72$ are edges. The number of diagonals of a square is $\frac{n(n-3)}{2} = 2$, of a hexagon is $9$, and of an octagon $20$. Hence the number of face diagonals is $2 \cdot 12 + 9 \cdot 8 + 20 \cdot 6 = 216$.

Subtracting, we get that the number of space diagonals is $1128 - 72 - 216 = 840$.