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.

 A: 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
$U/I' = \frac{19}{30} r$
(which is the correct answer).

I decided to solve this problem instead of simply giving ideas, because the solution used some ideas from previous problems on the sheet, so it is probably more useful for other users this way.
