A recycled rhombicosidodecahedron

My friend Cody drinks a lot of Red Juice. When bought from Costco, two 96 fluid ounce bottles of Red Juice come paired like this:

That red connector between the bottles creates a handle for easy transportation. Most people dispose of these connectors and consume the Red Juice with the help of friends. Cody manages to drink this quantity of juice in a few days and sets the connectors aside, initially out of laziness but soon building a collection of them.

One day Cody asked me what the coolest kind of polyhedron was that one could build with 121 edges. Well, the rhombicosidodecahedron can be built with 120 same-sized edges. I had no idea what Cody had in mind — I expected he had obtained some kind of polyhedral construction kit — but instead he responded with a photo of his large collection of Red Juice connectors. After seeing the photo, I was genuinely worried for Cody’s well-being and what his dentist might have to say. But my concern for Cody’s health was soon outweighed by my excitement to begin the construction of a new rhombicosidodecahedron made from these edges!

What is a rhombicosidodecahedron?

A rhombicosidodecahedron is made of triangular (20), square (30), and pentagonal (12) faces. This animated gif (from Wikipedia) demonstrates the construction of one. First imagine placing a small explosive charge at the center of an icosahedron made by gluing triangles together — when the charge explodes, the faces blow outward. Then freeze time at the moment the shared edge of two neighboring triangles separates to a distance equal to one side of these triangles. Now connect these edges up again to form squares joining the triangles. At the same time, you will create 12 pentagons coming from where the vertices of the triangles used to be. The resulting solid is a rhombicosidodecahedron, and it is therefore no coincidence that it has the same number of triangles as the icosahedron, the same number of pentagons as the dodecahedron (the icosahedron’s evil arch-enemy), and the same number of squares as the number of edges on either of these polyhedra.

How many edges do we need?

Let’s find the number of connectors required to build such a structure, with the connectors acting as edges in the design.

20 triangular faces (from icosahedron)
12 pentagonal faces (from dodecahedron)
E squares, where E is the number of edges on either the dodecahedron or the icosahedron. Sometimes I forget this, but it can be obtained by Euler’s formula:

\( \chi=V-E+F=2,\)

where \(\chi\) is the “Euler characteristic”, which is 2 for all convex polyhedra. This means that the number of edges for either the icosahedron or the dodecahedron is

\(E=12+20-2=30.\)

So both the dodecahedron and the icosahedron have 30 edges, so the rhombicosidodecahedron also has 30 squares. Then the number of edges on the rhombicosidodecahedron can be computed by adding the total number of edges belonging to each face:

\(12\times 5+20\times 3+30\times 4=240,\)

but we have actually counted the edges exactly twice, since each edge is counted from both of the neighboring faces. This tells us that the rhombicosidodecahedron has 120 edges. Another way to compute the number of faces without needing Euler’s formula at all is to notice that every edge is a neighbor to either a pentagon and a square or a triangle and a square. So right away, the number of edges must be equal to

\(12\times 5+20\times 3=120,\)

which gives the same answer. In order to have a collection of 120 connectors, Cody must have consumed 240 Red Juices, since each connector holds two bottles. This comes down to about 180 gallons of juice. Granted, he has been collecting for quite some time now.

An interesting property of the rhombicosidodecahedron is that each vertex connects exactly four edges. I’ll talk about this later.

How to build the rhombicosidodecahedron?

While discussing our plan of attack, Cody brought up a concern: because these connectors would be stacked at the circular part that holds the neck of the bottle, some Red Juice connectors would stack in asymmetrical ways. That is, a particular connector might be stacked third-out from the center on one side, but only the first-out on the other side. Aside from the poor aesthetics of angled connections, this would lead to asymmetries in the final product, as some connectors would be skewed and therefore not extending their true distance. A mathematical question comes to mind: is it possible to connect these in such a way that there isn’t any warping? Above is an illustration of a single vertex, showing which layer each connector occupies. If one of these connectors is on a different layer on a neighboring vertex, there will be warping.

My solution to this problem was to reduce it to another common mathematical problem: edge coloring. Since each vertex has exactly four edges radiating from it, we can ask if there is a way to color the edges of the rhombicosidodecahedron with exactly four colors in such a way that no two colors are adjacent. In this way, each of the four colors represents one of the four layers. It is tempting at this point to exclaim that it can certainly be done according to the four color theorem, but unfortunately we cannot apply that immediately here, because we are coloring edges and not vertices of a graph. We could also try to use Vizing’s Theorem, which states that the chromatic index of a graph is either \(\Delta\) or \(\Delta+1\), where \(\Delta\) is the largest vertex degree of the graph. Unfortunately, this does not help us here, since our maximum vertex degree is already four, so we cannot say right away whether we can color our graph with only four edges.

I decided that since I would end up needing a constructive solution anyway, I might as well find one. I began with a planar projection of the rhombicosidodecahedron (as seen on the right), beginning from the center and working my way outward.

Occasionally, I would come to an edge where I simply could not give it any color; all of my four colors were already taken by neighboring edges. At this point, I would just decide on a color for the new edge, and fix whichever edge was now conflicting. After fixing this edge, there would be exactly one other edge with a problem. This created a cascading effect that was relatively managable because the number of areas did not branch off exponentially. I continued following the chain reaction, edge by edge, until the problem got pushed off to the frontier of the area on which I had been working, at which point there were no more problems remaining. It was then safe to continue. After a bit of tedious work, I was happy to find that the entire graph was properly colored, and no edges remained. I was left with this: