Icosahedron Resistance

Call the opposite vertices $A$ and $B$ . If you inject a current, $I$ , into one vertex, then by symmetry $1/5$ of it will go through each neighboring edge of $A$ . Call this group $1$ . And $1/5$ will go through each of $B$ 's neighboring edges. Call this group $2$ . And $1/10$ of the current will go through the resistors that join group $1$ to group $2$ . So, the total voltage will be $(1/5 + 1/5 + 1/10)({50 ohms})*I$ . So , the effective resistance is $(1/5 + 1/5 + 1/10)*50$ ohms = $\boxed{25}$ ohms.

Each resistor was of $50 \space \Omega$ .

The figure above shows the sides of the regular Icosahedron, joined by the set of points $(A \space - \space L)$ . A and L are opposite to each other and hence will be followed in the solution.

Now, points $B, C, D, E, F$ are at same potential, because:

They all have equal resistance and current divides equally among them. Say the current passing through them is $\frac{I}{5}$ .

because of this , the resistances in between $(B \space and \space C), (C \space and \space D), (D \space and \space E), (E \space and \space F), (F \space and \space B)$ becomes dead, and will not be counted in the final circuit.

Now, the current again divides equally at each of the points $B, C, D, E, F$ into two parts. which means the current between the branches which can be seen in the circuit that are $(B,G), (C,G), (C, H), (D,H),(D, I),(E, I)$ and others which makes a total of 10 resistances in which the current flowing is $\frac{I}{10}$ .

Again the points $G, H, I, J, K$ are at same potential and again the resistances joining these in between these nodes become dead. At last we are left with five resistances in between $G, H, I, J, K, L$ and $L$ , which will again be in parallel.

Now, we have $5$ resistance in parallel which is in series with $10$ resistance in parallel which again is in series with $5$ resistance in parallel.

So, total resistance $R = \underbrace{\frac{50}{5}}_{(A,B), (A, C),(A, D),(A, E), (A, F)}+\underbrace{\frac{50}{10}}_{10 \space parallel \space resistance}+\underbrace{\frac{50}{5}}_{(L,G),(L,H),(L, I), (L, J),(L, K)}=\frac{100}{4}=\boxed{25}$

This is support of Geoff Pilling's solution:

The red diamond node (#1) is the entry node. The green diamond node (#2) is the exit node. The other nodes are colored circles. All the nodes at the same voltage are the same color and therefore could be shorted together without changing the answer. If the shorts were done, then you have 5 resistors in parallel, 10 resistors in parallel and then 5 resistors in parallel, with each set being in series. $\frac15+\frac1{10}+\frac15 \Longrightarrow \frac12$ . Then, multiply by the standard $50\Omega$ resistance.