When Wires Turn Useless

The total wire was of $15 \space \Omega$ , divided into $30$ branches. That means each branch is of $0.5 \space \Omega$ .

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 $R = \underbrace{\frac{\frac{1}{2}}{5}}_{(A,B), (A, C),(A, D),(A, E), (A, F)}+\underbrace{\frac{\frac{1}{2}}{10}}_{10 \space parallel \space resistance}+\underbrace{\frac{\frac{1}{2}}{5}}_{(L,G),(L,H),(L, I), (L, J),(L, K)}=\frac{1}{4}=0.25$

$100R=\boxed{25}$

Hm... I dont know why, but this formula works for most of such figures,
The resistance between two opposite nodes = $\dfrac{\text{Total resistance}}{\text{No. of edges} + \text{No. of parts}}$
In this figure it is $\dfrac{15}{30 + 30} = \dfrac{15}{60} = 0.25$
So, $100R = \color{#3D99F6}{\boxed{25}}$