We have created a convex solid from pentagons and hexagons, where three faces meet at every vertex, each pentagon shares an edge with 5 hexagons, and each hexagon shares an edge with 3 pentagons. How many faces does the solid have?
Solution. Let the number of pentagonal faces bounding the convex body be denoted by $\ell_{5}$ and the number of hexagonal faces by $\ell_{6}$. The body has $\ell=\ell_{5}+\ell_{6}$ faces.

Since each pentagon is bounded by five edges and each hexagon by six edges, the number of edges of the body is $e=\frac{5 \ell_{5}+6 \ell_{6}}{2}$, because due to convexity, exactly two faces meet at every edge, so we counted each edge twice in $\left(5 \ell_{5}+6 \ell_{6}\right)$.

Since each pentagon has five vertices and each hexagon has six vertices, the number of vertices of the body is $c=\frac{5 \ell_{5}+6 \ell_{6}}{3}$, because according to the problem's condition, exactly three faces meet at every vertex; thus, we counted each vertex three times in $\left(5 \ell_{5}+6 \ell_{6}\right)$.

Euler's theorem is valid for every convex polyhedron:

$$
\ell+c=e+2
$$

Substituting the above into this:

$$
\ell_{5}+\ell_{6}+\frac{5 \ell_{5}+6 \ell_{6}}{3}=\frac{5 \ell_{5}+6 \ell_{6}}{2}+2
$$

From this, we get that $\ell_{5}=12$, so the polyhedron has 12 pentagonal faces.

Since every pentagon shares five edges with hexagons and every hexagon shares three edges with pentagons, the number of hexagonal faces is $\frac{5 \ell_{5}}{3}$, because in $5 \ell_{5}$, we counted each hexagonal face three times.

$$
\ell_{6}=\frac{5 \ell_{5}}{3}=\frac{5 \cdot 12}{3}=20
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_1ce5e4d2588e5a5671b1g-1.jpg?height=355&width=341&top_left_y=1107&top_left_x=881)

Thus, the body described in the problem has $\ell=\ell_{5}+\ell_{6}=12+20=32$ faces. (Traditional soccer balls are also sewn together from this many and this type of leather patches.)

Note. A construction of such a body can be found in the article "Soccer Ball $(5,6,6)$" by Gyula Nagy (KöMaL, 1996/5., p. 268).