Icosahedron of Resistors

I prefer systematic computations of these kinds of problems. Here, we have $12$ nodes, and we can use Kirchhoff's circuit laws use the fact the net current at any node is $0$ , so that we have $12$ equations of the form ( $v$ being voltages, and resistors being $1$ Ohm)

$5{ v }_{ a }-{ v }_{ b }-{ v }_{ c }-{ v }_{ d }-{ v }_{ e }-{ v }_{ f }=0$

except that for two of the nodes (vertices) in which we're trying to determine the effective resistance of the net, the equations are of the form

$5{ v }_{ 1 }-{ v }_{ b }-{ v }_{ c }-{ v }_{ d }-{ v }_{ e }-{ v }_{ f }-\frac { 1 }{ R } ({ v }_{ 1 }-{ v }_{ 2 })$

where ${ v }_{ 1 }$ and ${ v }_{ 2 }$ are those two nodes, set at some arbitrary voltages differing by $1$ and $R$ is the equivalent resistance of the icosahedron net. Solve for $R$ and find that it's $\dfrac { 11 }{ 30 }$ . You can find the effective resistance between other pairs of nodes in this fashion, the only $2$ other possible resistance values being $\dfrac { 7 }{ 15 }$ and $\dfrac { 1 }{ 2 }$

simply use the formula R=2(v-1)r/v.n where v- total vertices; n- no. of edges meeting a vertex; r- resistance of an edge; NOTE: one can use this formula to calculate equivalent resistance between any two adjacent vertices of a regular polyhedron

Let vertices $1$ and $2$ be adjacent. Consider the setup where a current of $1A$ enters through vertex $1$ , and a current $\frac{1}{11}A$ leaves through the other $11$ vertices. Note that, due to symmetry, a current $\frac{1}{5}A$ flows through each of the $5$ edges leaving vertex $1$ . Hence, the voltage between vertices $1$ and $2$ is: $V_{1} - V_{2} = ( \frac{1}{5} A)(1Ω) = \frac{1}{5}V$ Consider a second setup, where a current of $1A$ leaves through vertex $2$ , and a current of $\frac{1}{11}A$ enters through the other $11$ vertices. Again, note that a current of $\frac{1}{5}$ A flows through each of the $5$ edges entering vertex $2$ . Hence, the voltage between vertices $1$ and $2$ is: $V_{1} - V_{2} = ( \frac{1}{5} A)(1Ω) = \frac{1}{5}V$ Using the superposition principle we have:

• A current of $\frac{12}{11}A$ enters through vertex $1$ ,
• A current of $\frac{12}{11}A$ leaves through vertex $2$ ,
• No current enters or leaves through the other $10$ vertices, and
• The voltage between vertices $1$ and $2$ is:

$V_{1} - V_{2} = \frac{1}{5}V + \frac{1}{5}V = \frac{2}{5}V$ We have constructed precisely the experimental setup that serves to determine the effective resistance between vertices $1$ and $2$ . That is, we have inserted a current at vertex $1$ , taken it out at vertex $2$ , and measured the voltage between the two points. The effective resistance between vertices $1$ and $2$ is therefore given by:

$V = RI \longrightarrow \frac{2}{5}V = (R_{eff})(\frac{12}{11}A) \longrightarrow \boxed{R_{eff} = \frac{11}{30}Ω}$ $\frac{a}{b} = \frac{11}{30} \longrightarrow a = 11 , b = 30$ $\large \boxed{a + b = 41}$