Soccer Ball Resistor Puzzle

[bob012345]

TL;DR

A truncated icosahedron (soccer ball) is made from 90 identical 1 Ω resistors, one on each edge. The exercise is to find the exact equivalent resistances between adjacent vertices.

I worked on this problem during the World Cup. A truncated icosahedron (soccer ball) is made from 90 identical 1 Ω resistors, one on each edge.

Part (a) Determine the exact equivalent resistance between two adjacent vertices across a hexagon–hexagon edge.

Part (b)Without solving a second resistor network from scratch, determine the exact equivalent resistance between two adjacent vertices across a hexagon–pentagon edge.

 

[Baluncore]

I have noticed that many of these problems reduce to a branching ladder network, a continued fraction.

 

[Baluncore]

(a) hex-hex edge R = 0.66897929
(b) hex-pent edge R = 0.6488431
Can you confirm, or what did you get ?

 

[bob012345]

(a) hex-hex edge R = 0.66897929
(b) hex-pent edge R = 0.6488431
Can you confirm, or what did you get ?

Those numbers look numerically close but can you give me the exact values? How did you get these?

 

[bob012345]

Here are my exact values.

$R_{hex-hex}=\frac{16778}{25080}$

$R_{hex-pent}=\frac{16273}{25080}$

 

[Baluncore]

I cheated by avoiding van Steenwijk’s method.

Found the graph of the TI on Wikipedia, numbered all 60 nodes, then numbered all 90 edges.
I wrote out the 90x 1 ohm resistors in SPICE format; R90 node1 node2 1

I pasted the 90 element table into LTspice, then grounded one node, while injecting a current of 1 amp into the other node. The voltage on the injection node gave the resistance of the network to the grounded node.

I checked my results by comparing two similar edges on the network. They were the same, so the graph entered was symmetrical.

 

[bob012345]

I started in a similar vein but a bit different. I drew the 2D projection of the TI modified to fit a flat x-y format. Then I drew it in LTspice also. It looks like this;

The loops numbered 1-12 are the pentagons and the others are the hexagons. This gave me the numerical numbers that matched yours but what I really wanted are the exact fractions. The next step was to use symmetry to reduce this to a much simpler circuit of which I solved the system of equations exactly to get $R_{hex-hex}$.

Once $R_{hex-hex}$ is known, there is a trivial way to get the other nearest neighbor value $R_{hex-pent}$ without having to solve another network. That is part b of this puzzle.