Baltic Way 1992 · Problem 16

Solution:

No, it cannot. Let us call a series of faces $F_{1}, F_{2}, \ldots, F_{k}$ a ring if the pairs $(F_{1}, F_{2}),(F_{2}, F_{3}), \ldots, (F_{k-1}, F_{k}),(F_{k}, F_{1})$ each have a common edge and all these common edges are parallel. It is not difficult to see that any two rings have exactly two common faces and, conversely, each face belongs to exactly two rings. Therefore, if there are $n$ rings then the total number of faces must be $2\binom{n}{2} = n(n-1)$. But there is no positive integer $n$ such that $n(n-1) = 1992$.