# How many ways can we color a regular tetrahedron with two colors?

A regular tetrahedron is a polyhedron with four faces that are congruent equilateral triangles. How many ways can we color it with two colors? Two colorations are the same if we can rotate a colored tetrahedron in one way, so that it looks like the second coloration. The two colors do not necessarily have to be used.

• This would be a much more interesting question if you offered some significant thoughts of your own. You can always edit the question to improve it. You might do well to consider this: math.meta.stackexchange.com/questions/1803/… – David K Nov 17 '17 at 2:37
• What are you coloring - the faces? the vertices? Something else? Also - assuming the faces - are you also stipulating that each face is colored entirely in one color? (And perhaps you don't necessarily define how the edges and the vertices are colored?) – mathguy Nov 17 '17 at 2:40
• What are you coloring? the faces – jaimeisaac Nov 17 '17 at 2:49
• This MSE link might prove useful reading. – Marko Riedel Nov 17 '17 at 22:59

Color patterns in cases where certain colorings are considered equal are counted by the so-called Polya Theory, AKA "Burnside Lemma". But we don't need this theory in the simple example at hand.

The two colors can be partitioned on the four faces of the tetrahedron in a $0+4$, $1+3$, or $2+2$ way. The $0+4$ and $1+3$ types can be realized in two ways each: $0$ red and $4$ blue, or $4$ red and $0$ blue; similarly for $1+3$. For the $2+2$ type its different: If you paint two triangles red these two triangles share an edge, and the two remaining triangles, to be painted blue, share the opposite edge. It follows that there is only one color pattern for $2+2$.

As I am understanding the question, you have 2 colors say red/blue,

As all the equilateral triangles are adjacent, they are five ways to color it using 2 colors.

Al faces red

1 face blue, 3 faces red

2 faces blue, 2 faces red

3 faces blue, 1 face red

All faces blue