Can we distribute the 12 musical notes in the chromatic scale on the faces of a dodecahedron such that the set of musical intervals at each vertex is the same?
(Assumes treating inverted invervals as equivalent, i.e. perfect 5th = perfect 4th, minor second = major 7th etc)
Notes can be represented by the numbers 0-11
Interval is the difference between the numbers. Inverted interval means (12 - original interval).
Notes 0 and 2 have interval 2
Notes 1 and 11 have interval 10
Intervals 2 and 10 are equivalent inversions because 12 - 2 = 10 and 12 - 10 = 2
Context: I am interested in mapping musical notes onto polyhedra and it seemts to make sense to start with dodecahedra because they have 12 faces and there are 12 musical notes in western music. I have considered some sort of combinatorial search to solve for this, but maybe the concept is DOA for some mathematical reason.
Suppose such a labelling exists. Clearly at most three different intervals appear on edges. Also every note meets five others, only two of which are at any given interval, so at least three different intervals must be used. So there are exactly three. Let's colour these red, green and blue.
Look at the five vertices around the face labelled with $0$. There is only one way to colour their edges (up to rotation, and permuting colours) with each vertex meeting all colours.
Now the faces to the top left and bottom right have the same (blue) difference from $0$, so they are $k$ and $12-k$ for some choice of $k$. But they also have the same (green) difference from the face to the top right, so that face must be labelled $6$. Now the same argument with right and left swapped shows the face to the top left must also be labelled $6$, a contradiction.
Therefore such a labelling is not possible.