## Colouring dodecahedra

This week our photos are from the blackboard of Toby Bailey, where he and I were trying to figure out how to colour the edges of a dodecahedron.

First I asked whether it was possible to colour the edges with 3 colours so that at every vertex 3 different colours met. Initially we thought it would be impossible and set out to find a proof of this. However, our investigations showed that it was actually possible! The colours around a vertex can be ordered in two ways: blue-red-green or blue-green-red. It turns out that out of the 20 vertices, 16 of them will have the same colour ordering and 4 of them with be the other one. And these 4 vertices (circled in the top picture) form the vertices of a regular tetrahedron! Magic.

After this challenge, the next obvious question was whether we could colour the edges with 5 colours so that around every face of the dodecahedron there are 5 different colours. This was a much easier task but the results were, in many ways, much more beautiful.

Since there are 30 edges to be coloured in 5 colours, there must be 6 edges of each colour. Toby noticed that there are 5 different cubes that can be embedded in a dodecahedron, and wondered if there would be a relation between the cubes and the colouring. The photo below shows where the 5 differently coloured cubes sit inside the dodecahedron. A colouring of the faces is therefore obtained by looking at the pentagram formed inside a pentagonal face, and colouring the opposite edges in the colours given.

Here is a picture of our final 5-colouring:

See if you can spot all the symmetries in this picture! How many different 5-colourings are there in total?

### 4 responses

1. This takes me back to my OU graph theory course. We were just touching on this kind of thing by the end of it and there was no follow on course. That was in 1997 so I don’t think I’ll be getting back into this soon!

July 27, 2011 at 10:22 am

• I’m pretty sure your pretty 5-coloring uses each of the (6 choose 3) = 20 combinations of three colors at the 20 vertices.

November 17, 2013 at 1:08 am

2. Dante Regiani Freitas

I´m trying really hard to solve this, but aparently you really can´t use each of the 12 possible combinations, the dodecahedron seems to always mirror the first 6 you put in it, otherwise the combinations won’t fit. Though I think it’s impossible, I can’t find a proper prove to say that for sure.

December 29, 2016 at 10:04 pm