## An amazingly symmetric icosahedron edge coloring

In my last post I showed off the stellated icosahedron I made out of folded paper:

I also claimed that it uses a very particular and elegant color scheme. When I set out to make it, I did some googling to try to figure out what sort of color scheme I could use. The Sonobe units essentially act as the edges of the icosahedron, so I was looking for an edge coloring. I quickly found this page on mathpages.com, which enumerates all 780 possible (!) ways to color the edges of an icosahedron with five colors such that five different colors meet at every vertex.

Buried in the middle of that page I found this paragraph:

Incidentally, the coloring in class #6 with just one representative out of the total 780 is the unique pattern such that all five colors have an identical arrangement (up to rotation), and not only does each color adjoin each vertex only once, but each color occurs only once on the perimeter of the pentagon surrounding each vertex. With this pattern, the pentagon edge with a given color is always opposite the “spoke” with that color. It follows that, since no two vertices have the same color cycle, no two pentagon perimeters have the same color cycle.

As soon as I saw that, I knew I had found my edge coloring! The paragraph above explains that if we look at any pentagon of five triangles, the edges with any one particular color will look like this:

And if we put that all together with five colors, each pentagon will look something like this:

…except that each of the twelve pentagons has a different permutation of colors around it. There are $(5-1)! = 24$ different ways to arrange five colors in a cycle, so only half of the possible permutations actually show up. This means that given any particular choice of five colors, there are two essentially different ways to use them to color the edges of an icosahedron using this scheme; you can get from one to the other by swapping any two colors. (Incidentally, this situation seems very reminiscent of the straws thingy — I hope to spend some time trying to elucidate the precise connection between the two.) You might wonder how this squares1 with the claim in the paragraph above that this pattern is unique. Well, as a pattern it is unique—two patterns are equivalent if we can change one into the other by consistently changing certain colors into other colors. We are just observing that given a particular set of colors, there are two distinct ways to instantiate this pattern with the given colors.

Given just this description, I was able to work out the coloring for the whole icosahedron and fit the origami pieces together in the right way (it was quite a challenge keeping track of where I was, involving a heavily annotated net drawn on a piece of paper!). But it wasn’t until I had finished building the model that I started to get a better sense of what the set of edges of a given color looks like. And it wasn’t until I started writing this post that it suddenly dawned on me what this set really looks like (click for a higher-resolution version2):

As you watch the rotating icosahedron above, try focusing on any pentagon: you will see that exactly one edge of the pentagon and the opposite “spoke” are red. Here is another way to describe where the red edges go. The vertices of an icosahedron can be described by three interlocking golden rectangles, like this:

The set of edges of a given color are precisely the short edges of one of these sets of golden rectangles! And if we take this set of edges and rotate it around any one of the icosahedron’s vertices, we get four other isomorphic, disjoint sets of edges. Coloring each set of edges a different color yields the final edge coloring of the entire icosahedron (again, click for a higher-resolution version):

Just to reiterate some of the amazing properties of this edge coloring (and mention a few more):

• Each vertex has all five colors touching it.
• Each pentagon has one edge of each color.
• Each vertex/pentagon pair has a different cyclic permutation of colors.
• For each color, the set of edges having that color is completely symmetric, i.e. the edges all play the same role. Take any, say, red edge and turn the icosahedron so that edge is at the top, and the other red edges will be in exactly the same positions as they would with any other red edge being at the top.
• The set of edges for a given color correspond to the ends of a set of three golden rectangles.
• The sets of edges of each color are symmetric with respect to each other as well: for any two colors, there is some rotation of the icosahedron which sends every edge of the first color to a position that used to be occupied by an edge of the second color.
• Each of the $2 \cdot \binom{5}{3} = 20$ possible edge colorings of a triangle using three out of five colors shows up exactly once.

Amazing!

1. pentagons?

2. Also, if you want to see the code used to generate the animations, you can find it here.