Honeycomb Resistance

Let's first rearrange the circuit a bit: Then we use this: In the rearranged circuit to get this: Now, since the circuit is symmetric about line $l$ , using the symmetry point-splitting rule, we can split points C and D to get this circuit: The circuit is still symmetric, but this time it is divided into two equal parts - top and bottom part. Let's take a look at only the top part: Top part and bottom part have the same equivalent resistance, so the starting circuit can be represented as: The equivalent resistance of the honeycomb is $\dfrac{R_{eq}}{2}$ , so, all we need to do is find $R_{eq}$ . Let's rearrange this top part of the ciruit: By applying the triangle-to-star transformation rule, we get that:

$R_A = \dfrac{3R\times 3R}{3R + 3R + R}\ = \dfrac{9R}{7}\ \\ R_B = R_C = \dfrac{3R\times R}{3R + 3R + R}\ = \dfrac{3R}{7}\$

After transformation and another rearrangment, we get this final circuit: Which is very easy to solve: $R_{eq} = \dfrac{9R}{7}\ + \dfrac{( \dfrac{3R}{7}\ + 3R + \dfrac{3R}{7}\ ) \times ( \dfrac{3R}{7}\ + R + \dfrac{3R}{7}\ )}{( \dfrac{3R}{7}\ + 3R + \dfrac{3R}{7}\ ) + ( \dfrac{3R}{7}\ + R + \dfrac{3R}{7}\ )}\ + \dfrac{9R}{7}\ = \dfrac{18R}{7}\ + \dfrac{ \dfrac{27R}{7}\ \times \dfrac{13R}{7}\ }{ \dfrac{27R}{7}\ + \dfrac{13R}{7}\ }\ = \dfrac{ (13 \times 27 + 18 \times 40)R}{7 \times 40}\ = 3.825$

So, the solutions is $100000 \times \dfrac{R_{eq}}{2}\ = \boxed{191250}$

First step: Reduce the solution to an easy figure, which can help us analyze the circuit easily

In the above figure, the blue line represents the equivalent resistance of the outer half of each hexagon through which it is drawn.

For instance each side $= 1 \space \Omega$ and therefore $R_{ACDE}=R_{AE}=3 \space \Omega$ .

After the reduction the figure looks something like this:

The $\color{#D61F06}{red}$ line shows the dead resistance by the principal of equipotential points in the circuit.

In the outer Hexagon $A,C,D,B,F, E$ the $\color{#3D99F6}{blue}$ arrow shows the flow of current, and while flowing through the two branches the current never divides as the nodes which are in the path are connected by dead resistances.

For the inner hexagon the current first enters through the branch $(AG)$ and then divides equally at node $G$ , but after that it never divides and rather joins at node $J$ .

So, total $R=\large \underbrace{\underbrace{(3+3+3)}_{ACDB} \space parallel \space to \space \underbrace{(3+3+3)}_{AFEB}}_{outer hexagon} \space parallel \space to \space \underbrace{(1+ \underbrace{\frac{3}{2}}_{inner hexagon} + 1)}_{inner circuit}$

$R= \large \underbrace{(9) \space parallel \space to \space (9)}_{outer hexagon} \space parallel \space to \space \underbrace{(\frac{7}{2})}_{inner hexagon}$

$R= \frac{63}{32}=1.96875$

$100000R=\color{magenta} {\boxed{196875}}$

I don't know to write a good solution using latex but here I can provide a ' hint ' .

"Consider the middle hexagon, here , by symmetry, no current will flow in the branches connecting its top two vertices & bottom two vertices to the other Hexagons ! " Hence we disconnect those branches & our circuit simplifies to three resistances of $9$ , $9$ & $\frac{7}{2}$ ohms connected in parallel ! Solving this we get, $\text {R=}\frac {63}{32}$ ; Hence, final answer is $\boxed {100000R=196875}$ .