An extremely fair die

Question

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 $ABC, ACD, ADE, AEB, FBC, FCD, FDE, FEB$.

Answer 1

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.