## Problem 1.

A convex polyhedron has 12 square faces, 8 regular hexagonal faces, and 6 regular octagonal faces. At each vertex of the polyhedron, exactly one square, one hexagon, and one octagon meet. How many segments connecting pairs of vertices of the polyhedron are interior to it, that is, are neither edges nor contained in a face?
Solution: Let $V$ be the number of vertices, $A$ the number of edges, $D$ the number of diagonals on the faces, and $I$ the number of interior diagonals.

Since each vertex of the polyhedron is exactly on one square face, there must be

$$
V=4 \cdot 12=48 \text { vertices. }
$$

(We would get the same result by counting with hexagons or octagons).

Since exactly 3 edges emanate from each vertex, there are $A=3 V / 2=72$ edges.

Since each square has 2 diagonals, each hexagon has 9, and each octagon has 20, it follows that $D=12 \cdot 2+8 \cdot 9+6 \cdot 20=216$ diagonals on the faces.

Finally, the number $I$ we are looking for will be equal to the total number of pairs of segments that can be formed by joining pairs of vertices in all possible ways, minus the edges and the diagonals on the faces.

$$
I=\left(\begin{array}{c}
48 \\
2
\end{array}\right)-A-D=840
$$