Question

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?

Answer

Answer: 840

Concepts:polyhedron double counting combinations

Difficulty rating: 2170

Small Hint:

Double-count face-vertex and face-edge incidences to find $V$ and $E$

Big Hint:

Count all vertex pairs, then subtract pairs lying together on a face, correcting the double count of edges

Solution:

The total number of face-vertex incidences is $12(4)+8(6)+6(8)=144.$ Three faces meet at each vertex, so $V=48.$ The same sum counts each edge twice, so $E=72.$

There are $\binom{48}{2}=1128$ vertex pairs. Summing pairs on faces gives $12\binom42+8\binom62+6\binom82=360.$ Every edge was counted twice in this sum and every other same-face pair once, so the number of distinct boundary pairs is $360-E=288.$ Hence $1128-288=840$ joining segments lie in the interior.

1988 AIME Problem 10

Answer: 840

Difficulty rating: 2170

Small Hint:

Big Hint:

Solution:

Problem 10 in Other Years