Buckyball a.k.a. Fullerene-60 2 ohm resistor maze: compute average effective resistance

There are two types of edge. There are $60$ edges of type $1$ , which connect a hexagon to a pentagon, and there are $30$ edges of type $2$ , which connect two hexagons. There are two edges of type $1$ and one edge of type $2$ at each vertex.

Consider the situation where a current $I$ enters the network at one vertex $V$ , while $\tfrac{1}{60}I$ leaves the network at each vertex. By symmetry, the amount of current flowing along the type $1$ edges from the vertex $V$ will be equal. Thus the current flowing on the type $1$ edges away from $V$ will be $\tfrac{59}{60}uI$ , and the current flowing on the type $2$ edge away from $V$ will be $\tfrac{59}{60}(1-2u)I$ for some $u$ .

Superimpose this current flow with a similar one, where $-I$ enters the network at a vertex $W$ that is adjacent to $V$ , with $-\tfrac{1}{60}I$ leaving the network at each vertex. The total effect of this is to have a current flow with $I$ entering at $V$ , and $I$ leaving at $W$ . The current flowing along the edge $VW$ will be $2 \times \tfrac{59}{60}uI = \tfrac{59}{30}uI$ if $VW$ is of type $1$ , and $2 \times \tfrac{59}{60}(1-2u)I = \tfrac{59}{30}(1-2u)I$ if $VW$ is of type $2$ . Thus the potential difference between vertices $V$ and $W$ is $\tfrac{59}{15}uI$ if $VW$ is of type $1$ , and $\tfrac{59}{15}(1-2u)I$ if $VW$ is of type $2$ . Thus the effective resistance of the network between neighbouring vertices $V,W$ is $\tfrac{59}{15}u$ of $VW$ is of type $1$ , and $\tfrac{59}{15}(1-2u)$ if $VW$ is of type $2$ . Thus the average effective resistance is $\tfrac{1}{90}\big[60 \times \tfrac{59}{15}u + 30 \times \tfrac{59}{15}(1-2u)\big] \; = \; \tfrac{59}{45}$

We still need to determine one of these two resistances. It turns out that $u < \tfrac13$ , and so the answer is $\boxed{59004560}$ . We can either perform a brute force calculation to determine $u$ , or we can make a $50:50$ guess as the whether $n = 60$ or $30$ , and get the answer in at most two gos. Solving for one resistance by solving the simultaneous equations obtained from Kirchoff's laws, particularly if we use symmetry to reduce the network somewhat, should not be too hard...

The least effort method is to look at this Mathworld page , which gives the effective resistances for a whole array of Platonic and Archimedena polyhedra, but without specifying which vertex pairs correspond to which effective resistances. There are 23 different possible values of effective resistance for the truncated icosahedron, and a simple check shows that there are only two of these effective resistances $R_1,R_2$ for which $2R_1 + R_2 \; = \; \tfrac{59}{15}$ and these are (note that the Mathworld calculations assume that each portion of the network has resistance $1\,\Omega$ instead of $2\,\Omega$ ) $R_1 \; = \; \tfrac{16273}{12540} \hspace{2cm} R_2 \; = \; \tfrac{8389}{6270}$ Then $R_1$ must be the Type 1 resistance, and $R_2$ the type 2 resistance. Since $R_1 < R_2$ , we deduce that $n=60$ .

$\text{name}=\text{"Buckyball"};$

$\text{nvertices}=\text{PolyhedronData}[\text{name},\text{VertexCount}];$

$\text{edges}=\text{Table}[e[[1]]\unicode{f3d4}e[[2]],\{e,\text{PolyhedronData}[\text{name},\text{Edges}]\}];$

The resistors are entered as conductances; hence, the $\frac12$ .

$g=\text{PlanarGraph}\left[\text{Range}[\text{nvertices}],\text{edges}, \\ \ \ \text{VertexLabels}\to \text{Name},\text{EdgeWeight}\to \text{ConstantArray}\left[\frac{1}{2},\text{Length}[\text{edges}]\right]\right];$

$\text{resistance}=\text{With}\left[\left\{\Gamma =\text{PseudoInverse}\left[\text{With}\left[\{\text{wam}=\text{WeightedAdjacencyMatrix}[\text{\$\#\$1}]\},\text{DiagonalMatrix}\left[\text{Tr}\text{/@}\text{wam}^T\right]-\text{wam}\right]\right]\right\}, \\ \ \ \text{Outer}[\text{Plus},\text{Diagonal}[\Gamma ],\text{Diagonal}[\Gamma ]]-\Gamma -\Gamma ^T\right]\&;$

$r=\text{resistance}(g);$

$\text{result}=\text{Tally}[\text{Table}[r[[e[[1]],e[[2]]]],\{e,\text{edges}\}]] \Longrightarrow \left( \begin{array}{cc} \frac{16273}{12540} & 60 \\ \frac{8389}{6270} & 30 \\ \end{array} \right)$

$f=\frac{\text{Plus}\text{@@}\text{Table}[\text{Times}\text{@@}k,\{k,\text{result}\}]}{\text{Plus}\text{@@}\text{Table}[k[[2]],\{k,\text{result}\}]} \Longrightarrow \frac{59}{45}$

$1\kern 0.125em 000\kern 0.125em 000\ \text{Numerator}[f]+100\ \text{Denominator}[f]+60 \Longrightarrow 59\kern 0.125em 0045\kern 0.125em 60$