Triangle Circuit

A simpler way to visualise the circuit is as follows (R=1ohm)

If the Resistance between A and B is say P as the number of 'triangles' tends to infinity then the circuit simplifies to the following

From this we get

$P=\dfrac{(P+R)(R)}{P+2R}$

putting R=1 and solving we get $P=0.618ohms$

Suppose the circuit is built with $n$ triangles, and therefore $k = 2n+1$ wires. We number the wires as follows:

The arrows show how the current will flow is point $A$ is connected to the higher voltage.

Now we add the next wire:

Applying Kirchhoff's current/junction rule to the red arrows shown, we find $I_{k+1} = I_{k-1} + I_k.$

Now close the triangle by yet another wire:

Apply Kirchhoff's voltage/loop rule to the triangle marked with red arrows: $V_{k+2} = V_k + V_{k+1}.$ Since all wires have the same resistance, we can divide ( $V/R = I$ ) and find $I_{k+2} = I_k + I_{k+1}.$

This shows that the currents in the wires satisfy the recurrence relation $I_{k+2} = I_k + I_{k+1}\ \ \ \ \ k \geq 1. \\ I_1 = I_2.$ The solution is the Fibonacci sequence, up to a factor that depends on the applied voltage: $I_k = f_k\:I_1.$

The total current flowing from A to B is $I_{AB} = I_{k+1} + I_{k+2} = f_{k+3}\:I_1$ and the voltage across A and B is $V_{AB} = R\:I_{k+2} = f_{k+2}\:I_1\:R$ so that $R_{AB} = \frac{V_{AB}}{I_{AB}} = \frac{f_{k+2}}{f_{k+1}}\:R,$ where $R = 1\ \Omega$ is the resistance of a single wire. It is well-known that the ratio between successive terms in the Fibonacci sequence is the golden ratio $\phi$ , so $R_{AB} = \frac 1\phi\:R = \frac{\sqrt 5 - 1}2\:R = \boxed{0.618033989\ \Omega}.$

If the effective resistance between $A$ and $B$ in a circuit with $n$ triangles is $R_n\,\Omega$ , then the circuit with $n+1$ triangles is equivalent to a resistance of $1\,\Omega$ and a resistance of $1+R_n\,\Omega$ in parallel, and hence $R_{n+1}^{-1} \; = \; 1 + (1 + R_n)^{-1}$ Since $R_1 = \tfrac23$ , it is a simple induction that $R_n \; = \; \frac{F_{2n+1}}{F_{2n+2}} \;,$ where $F_n$ are the Fibonacci numbers, with $F_0=0$ , $F_1=1$ , ... Thus $\lim_{n \to \infty} R_n \; =\; \phi^{-1}$ where $\phi$ is the golden section, making the answer $\boxed{0.6180339887...}$