Problem 1141

Solution:

Let us denote $c = 2020$. First consider the case when there exists a face $\mathcal{F}$ with an even number of vertices. Then it suffices to assign to the vertices of the face $\mathcal{F}$ alternately the numbers $c$ and $\frac{1}{c}$, and to all other vertices of the polyhedron the number $1$. Indeed, the product of the numbers on the face $\mathcal{F}$ is equal to $1$, and every other face either shares an edge with the face $\mathcal{F}$ (and on it the product is $1$) or has

no common vertex with $\mathcal{F}$ (in that case all the numbers on it are ones).

From now on we assume that all faces have an odd number of vertices. Consider two disjoint faces of the polyhedron $\mathcal{F}_A$ and $\mathcal{F}_B$ and vertices $A$ and $B$ on them, respectively, such that $AB$ is an edge of the polyhedron. To the vertices of these two faces, except $A$ and $B$, we will assign alternately the numbers $c$ and $\frac{1}{c}$ as in the figure, and to all other vertices of the polyhedron the number $1$. In this case too it is easy to see that the condition of the problem is satisfied.

Finally, let us verify that such faces $\mathcal{F}_A$ and $\mathcal{F}_B$ can always be found. There exists a face $\mathcal{F}$ that is not a triangle (otherwise the given polyhedron would be a tetrahedron). Consider four

consecutive faces adjacent to it $\mathcal{F}_1, \mathcal{F}_2, \mathcal{F}_3$ and $\mathcal{F}_4$. If the faces $\mathcal{F}_1$ and $\mathcal{F}_3$ are disjoint, take these two, and if they are not, then they have a whole common edge, so the three faces $\mathcal{F}_1, \mathcal{F}_3$ and $\mathcal{F}$ form a belt that separates $\mathcal{F}_2$ from $\mathcal{F}_4$, and in that case we can take $\mathcal{F}_2$ and $\mathcal{F}_4$.