# An extremely fair die

I have in mind a special kind of die, one that is even more fair than uniform density dice. If the die has $$n$$ faces with labels going from $$1$$ through $$n$$, then some faces will be larger than others (of course). I want to engineer the die so that faces with large numbers are surrounded by faces with lower numbers, and vice versa. Quantitatively I want the sum of the faces proximate to each vertex to be the same for all vertices.

Out of the Platonic solids it seems this is only possible for a d8, an octohedron. Can you find a way to label the faces of the die from $$1$$ to $$8$$ so that the sum of numbers around each vertex is the same for each of the $$6$$ vertices? For consistency let's label the faces like so:

Start with square $$BCDE$$, each letter corresponding to a vertex in the $$x-y$$ plane. Vertex $$A$$ is above this square in the $$z$$ axis and vertex $$F$$ is below. The eight faces are $$ABC, ACD, ADE, AEB, FBC, FCD, FDE, FEB$$.

## 1 Answer

The following arrangement works:

   6
1
4 7 8 3
2
5

where you should imagine the four numbers in the middle on four triangular faces meeting at a vertex, and then the ones around the outside being on the remaining faces adjacent to those.