Coloring Two Faces of an Icosahedron

Question

Suppose that you have a solid regular icosahedron (a polytope with 20 sides all of which are equilateral triangles), and all the sides are white.

(a) In how many ways can exactly two sides be painted red?

(b) In how many ways can one side be painted red and another one side painted blue?

Two colorings considered the same if one can be converted into the other by rotating the icosahedra in space.

I tried doing it by mapping which sides connected to others, but it didn't seem to help. Any help would be greatly appreciated!

Answer 1

Following @JMoravitz's idea, consider what happens if you color one triangle red and you color a neighboring triangle that shares an edge blue. All such colorings are equivalent. The same is true for red-red colorings. On the other hand, if you color a triangle that only shares one vertex blue, then there are two such inequivalent colorings. However, there there is only one red-red coloring.

To complete these arguments, you to first explain why any given triangle can always be rotated to the one initially colored red. Then consider rotations about the centroid of the red triangle. In the case of red-red colorings, you need to consider a rotation about the shared vertex---this interchanges the two red triangles.

This takes care of the 9 triangles that are distance one (measured in terms of moves from one triangle to a neighboring one that shares an edges) or two away from the initial triangle. There are 6 more triangles that are distance three away, 3 distance four away, and only one that is distance five away. You can visualize the distance as the distance between the vertices of the dual graph, the dodecahedral graph.

Answer 2

As Christian Blatter said in the comments, coloring faces of the regular icosahedron is the same as coloring vertices of its dual, the regular dodecahedron.

Now, the regular dodecahedron $D$ is a so-called distance-transitive polytope (as all the other Platonic solids too). This means, that for any two pairs $(v,w),(v',w')$ of vertice with pair-wise identical graph-theoretical distance along the edge-graph (that is, $\mathrm{dist}(v,w)=\mathrm{dist}(v',w')$), there is a symmetry of $D$ that maps $v$ to $v'$ and $w$ to $w'$.

In other words, for any $\delta\in\Bbb N$, there is, up to symmetry, at most one pair $(v,w)$ of vertices (same or different color, does not matter) with distance $\delta$. And since the edge-graph of the dodecahedron has diameter five, $\delta$ can take on the values $1,2,3,4$ and $5$.

So the answer to (a) and (b) is five.

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As a bonus, you got an answer for all the Platonic solids: the answer to (a) and (b) is always the diameter of the edge-graph of its dual. And the same holds for all regular polytopes in all dimensions, excluding the 4-dimensional exceptions (which are not distance-transitive).