Resistors and means

Relevant wiki: Simple Circuits

Take two copies of this network - left and right copy. Connect $R_x$ to the node A of the left copy, and $R_y$ to the node A of the right copy. Now, connect these two copies in parallel. You'll get exactly the same circuit you started with! This means that the total resistance of this new network is equal to the total resistance of our starting network: Where $R_{tot}$ is the total resistance of the starting network. So: $\begin{aligned} \dfrac{1}{R_x + R_{tot}}\ + \dfrac{1}{R_y + R_{tot}}\ &= \dfrac{1}{R_{tot}}\ \\ \dfrac{R_x + R_y + 2R_{tot}}{(R_x + R_{tot})(R_y + R_{tot})}\ &= \dfrac{1}{R_{tot}}\ \\ (R_x + R_{tot})(R_y + R_{tot}) &= (R_x + R_y + 2R_{tot})R_{tot} \\ R_{tot}^2 + R_{tot}(R_x + R_y) + R_xR_y &= 2R_{tot}^2 + R_{tot}(R_x + R_y) \\ R_{tot}^2 &= R_xR_y \\ R_{tot} &= \sqrt{R_xR_y} \end{aligned}$

Bonus problem: using only resistances of $R_x$ and $R_y$ , create resistive networks so that their total resistances are equal to each of the wrong answers...

Set the equivalent resistance to $R$ . Now, as the tree is infinite, the two trees under the first split have a equivalent resistance of $R$ as well.

Setting the left resistor's resistance to $x$ and the right one to $y$ , we have the following:

$\frac{1}{R} = \frac{1}{x+R} +\frac{1}{y+R}$

Multiplying everything by $R$ , $x+R$ and $y+R$ , we get $(x+R)(y+R)=R(x+y+2R)$

Distributing and cancelling: $xy=R^{2} \implies \boxed{R=\sqrt{xy}}$

Take out the initial $R_{x}$ and $R_{y}$ resistances. Since the circuit is so large it doesn't matter if we take out $2$ .

(try solving the equation infinity - 2 = ?)

now exactly as novak has described we get a circuit consisting of $2$ arms the resistance is $R + R_{x}$ in one arm and $R +R_{y}$ in another. These are in parallel with each other so,

$1 / (R+ R_{x}) + 1 / (R + R_{y}) = 1 / R$ . This gives quadratic in $R$ . Solving we get $R^2 = R_{x}R_{y}$ showing that $R$ is geometric mean of $R_{x}$ and $R_{y}$ .

Edit : Sep 7, I have not done electricity in detail yet but Rohit Gupta asked what the result would be if the resistors were all capacitors,And I worked it out and the same result follows : $C_{equivalent}^2$ $=$ $C_x$ $C_y$

The given network is an example of an infinite geometric series since there is a constant ratio between the parent nodes and the child nodes. So of course you have to use the geometric mean.