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i found this in a puzzle book. Number the edges of a cube from 1 to 12 so that the sum of the edges of each face is the same. I could find a solution by trial and error. But it raises the general question. Is the solution unique? How many 'different' solutions are there? (different meaning one solution is not just a symmetry operation of another). Can this be done for polyhedra other than a cube? For instance can the edges of an dodecahedron (12 pentagon faces) or an icosahedron be numbered so that the sum of the edges is the same for each face??

a previous question on MathStack provided specific solution for the cube problem but did not address the general questions that flow from this specific example.

Ethan Bolker
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steve newman
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  • The same sort of numbering of edges for the tetrahedron will not allow all four faces to have equal edge-sums. The total of all four faces' edge-sums would be... – hardmath Jul 04 '17 at 01:10

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One quick test is that the sum of each face has to be the average number times the number of edges on the face. If you add the face sum over all the faces it gives the sum of all the numbers from $1$ to the maximum times the number of edges that meet at a vertex because that is the number of times you count each vertex. An icosahedron has $30$ edges so the average edge has value $15.5$ and the face must total $46.5$. That is impossible, so you can't do the icosahedron. This argument will extend to any Platonic solid with an even number of edges and faces that have an odd number of sides. The cube is the only Platonic solid with faces that have an even number of edges. The next test, not needed here, is that you must be able to achieve the proper sum in enough ways. For the cube, the average edge has value $\frac {13}2$ so the face sum is $26$. You have to have at least $6$ ways to sum to $26$. The cube clearly passes because there is a solution.

Ross Millikan
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  • this answer (Ross Millikan) seems to deal with the related, but somewhat different problem of numbering the vertices. My original question was about numbering the edges. – steve newman Jul 04 '17 at 18:02
  • @stevenewman: the argument is identical and I have corrected it. As each face has the same number of edges as vertices the conclusion that you can't do any Platonic solid whose faces have an odd number of edges still applies. That shows the cube is the only one that can be solved. – Ross Millikan Jul 05 '17 at 00:59