Effective resistance across 2 adjacent vertices of a dodecahedron with each edge $r$

What will be the effective resistance across 2 adjacent vertices of a regular dodecahedron (12 faces) with each edge having resistance $r$?

Here is the source for the problem, it's problem 20. on the sheet.

In the link, a typical problem is solved using symmetry (infinite square lattice, resistance between adjacent vertices) and then it is stated

It appears that such a symmetrization technique can be also applied to ﬁnite lattices.

I am looking for some ideas as to how to apply this technique to a dodecahedron. More generally, I am looking for examples where such symmetrization techniques can be used for finite lattices.

• You can certainly cut it in half and identify nodes at equal potential. The edges between those nodes can then be cut to simplify further. Alternatively you can apply a star-mesh transform to it, which should lead to the geometric dual and that may have fewer nodes (20 down to 12... maybe I am missing something)? – CuriousOne Jan 14 '16 at 20:18
• @CuriousOne I get your idea, but won't that be too hectic for a dodecahedron? Afterall 12 nodes isn't too less. – Aritra Das Jan 15 '16 at 5:37

In order to make the problem symmetric, consider this: What happens, when you take the dodecahedron and drive current $I$ into it from vertex $A$ and drive $I/20$ out from all every vertex (including $A$)? By Kirchhoff's laws and symmetry, there is a current

$I_{A-out} = \frac{(I-I/20)}{3} = \frac{19}{60}I$

going from $A$ to neighbouring vertexes.

Now suppose $B$ is a neighbour of $A$ and we do the same for $B$, but with current $-I$ (so current is flowing into $B$ from nearby vertexes). Again we find there is a current

$I_{B-in} = \frac{19}{60} I$

flowing into $B$. Now we superpose these solutions - we get a solution, where there is current $I$ going into $A$ and it all comes out from $B$. Also notice that $\pm I/20$ leaving from each vertex has also vanished. However the edge connecting $A$ and $B$ has current

$I' = I_{A-out} + I_{B-in} = \frac{19}{30} I$

So the voltage between $A$ and $B$ is $U = I'r$, thus

$U/I' = \frac{19}{30} r$

• @Floris: With dodecahedron, it is possible to do it with any two nodes, because it is possible to find the current (when all current comes from $A$) in any edge using symmetry only. Later you can find the voltage between $A$ and $B$ by adding several voltages on resistors. However notice that in an infinite lattice, this isn't so simple (or even possible) because current doesn't distribute evenly between nodes further away. – kristjan Jan 14 '16 at 21:45
• @CuriousOne: Well, it essentially proves that every circuit with 20 nodes, such that 3 edges are connected to each node and each node is symmetrical to any other node (because if that was not the case, my assumption of even distribution of currents might become false) has an effective resistance of $\frac{19}{30} r$. I suspect that this might be the only circuit that would match these constraints. However, if there are others, it proves that all these circuits have the same resistance. – kristjan Jan 16 '16 at 14:18