A convex polyhedron has faces that are either pentagons or hexagons, and exactly three edges extend from each vertex. How many pentagonal faces does the polyhedron have? Give an example of such a polyhedron.
Let the number of pentagonal faces of the polyhedron be $n$, and the number of hexagonal faces be $k$. According to Euler's well-known theorem, for every convex polyhedron, $l + c = e + 2$, where $l$ is the number of faces, $c$ is the number of vertices, and $e$ is the number of edges. Based on this theorem, we try to estimate $n$. Since three edges emanate from each vertex, the number of vertices is $\frac{5n + 6k}{3}$. Every edge belongs to two faces, so the number of edges is $\frac{5n + 6k}{2}$. Applying Euler's theorem:

$$
n + k + \frac{5n + 6k}{3} = \frac{5n + 6k}{2} + 2, \quad \text{from which} \quad n = 12
$$

An example of such a body is the dodecahedron, in which case the number of hexagons is zero, or the polyhedron obtained by "cutting off" the vertices of a regular icosahedron. If this cutting off occurs in such a way that a regular pentagon is formed at each vertex cut, and the faces become regular hexagons, then this body is "inflated" to a soccer ball, about which we also read in the KöMaL recently. (See the back cover of the 1996/4 issue.) The spatial structure of the third allotrope of carbon, recently discovered, the $\mathrm{C}_{60}$, is also imagined to be of this type. The other fullerenes also have a spatial structure corresponding to the polyhedron in the problem.

Lajos Deli (Hajdúszoboszló, Hógyes E. Gymn., 8th grade)

Note. András Juhász (Fazekas M. Fóv. Gymn., 2nd grade) also examined the problem for $m$-sided faces instead of pentagons, and found that the sought polyhedron can only exist for $m < 6$. For the case $m = 4$, examples can be a cube or a hexagonal prism, and for $m = 3$, a tetrahedron. He also found that if every vertex of a polyhedron has three edges emanating from it, then not all of its faces can be hexagonal or larger.