Resistive Möbius Network

Reorganising the circuit, we consider a loop of $2n$ vertices, labelled $1$ to $2n$ , which are connected in a ring. In addition, there are wires connecting vertex $j$ with vertex $n+j$ for $1 \le j \le n$ . Note that these wires do not meet at the centre of the diagram, but pass by each other without interaction. All wires represent a resistor of size $R\,\Omega$ . Suppose that a current of $I$ Amps enters the circuit at vertex $1$ , and leaves at vertex $2n$ . Suppose that the current in the wire from vertex $u$ to vertex $u+1$ is $I_u$ for $1 \le u \le 2n-1$ , and that the current in the wire from vertex $2n$ to vertex $1$ is $I_{2n}$ . Suppose also that the current in the wire from vertex $u$ to vertex $n+u$ is $J_u$ for $1 \le u \le n$ . Conservation of current tells us that $\begin{array}{rclclcrclcl} I_{2n} + I & = & I_1 + J_1 & & & & I_{n+j-1} + J_j & = & I_{n+j} & \hspace{0.25cm} & 1 \le j \le n-1 \\ I_{j-1} & = & I_j + J_j & \hspace{0.25cm} & 2 \le j \le n & \hspace{0.5cm} & I_{2n-1} + J_n & = &I + I_{2n} \end{array}$ Thus $\begin{aligned} I_{j-1} - I_j & = \; J_j \; = \; I_{n+j} - I_{n+j-1} \\ I_{j-1} + I_{n+j-1} & = \; I_j + I_{n-j} \end{aligned}$ for all $2 \le j \le n-1$ , so that $I_j + I_{n+j} = K$ for $1 \le j \le n-1$ . Since $I_{2n} + I - I_1 \; =\; J_1 \; = \; I_{n+1} - I_n$ we deduce that $I_n + I_{2n} = I_1 + I_{n+1} - I$ , and hence $I_j + I_{n+j} \; = \; \left\{ \begin{array}{lll} K & \hspace{1cm} & 1 \le j \le n-1 \\ K-I & & j=n \end{array} \right.$

Suppose now that the voltage at the vertex $j$ is $V_j$ , for $1 \le j \le 2n$ . Then we have $\begin{array}{rclll} V_{2n} - V_1 & = \; RI_{2n} \\ V_j - V_{j+1} & = \; RI_j & \hspace{1cm} & 1 \le j \le 2n-1 \\ V_j - V_{n+j} & = \; RJ_j & & 1 \le j \le n \end{array}$ and so $\begin{aligned} RK \; = \; RI_j + RI_{n+j} & = \; V_j - V_{j+1} + V_{n+j} - V_{n+j+1} \\ V_{j+1} + V_{n+j+1} & = \; V_j + V_{n+j} - RK \end{aligned}$ for $1 \le j\le n-1$ , and hence $V_j + V_{n+j} \; = \; A - jRK \hspace{2cm} 1 \le j \le n$ for some constant $A$ . Since $RK - RI \; = \; RI_n + RI_{2n} \; = \; V_n - V_{n+1} + V_{2n} - V_1 \; =\; (A - nRK) - (A - rK) \; = \; -(n-1)RK$ we deduce that $K= \tfrac{I}{n}$ , so that (without loss of generality, we can choose $A=0$ ) $I_j + I_{n+j} \; = \; \left\{ \begin{array}{lll} \frac{1}{n}I & \hspace{0.5cm} & 1 \le j \le n-1 \\ -\tfrac{n-1}{n}I & & j = n \end{array}\right. \hspace{2cm} V_j + V_{n-j} \; = \; -\tfrac{j}{n}IR \hspace{1cm} 1 \le j\le n$ Solving the equations $V_j - V_{n+j} = RJ_j$ and $V_j + V_{n+j} = -\tfrac{j}{n}IR$ for $1 \le j \le n$ simultaneously gives $V_j \; = \; \tfrac12R\left(J_j - \tfrac{j}{n}I\right) \hspace{1cm} V_{n+j} \; = \; -\tfrac12R\left(J_j + \tfrac{j}{n}I\right) \hspace{2cm} 1 \le j \le n$ Another round of substitution into the given equations yields $V_{j+1} - 4V_j + V_{j-1} \; = \; -\tfrac{j}{n}IR \hspace{2cm} 2 \le j \le n-1$ Solving this recurrence relation, there must be constants $P$ and $Q$ such that $\begin{aligned} V_j & = \; P\alpha^j + Q\alpha^{-j} - \tfrac{j}{2n}IR \\ V_{j+n} &= \; -P\alpha^j - Q\alpha^{-j} - \tfrac{j}{2n}IR \\ RJ_j & = \; 2\big(P\alpha^j + Q\alpha^{-j}\big) \end{aligned}$ for $1 \le j \le n$ , where $\alpha = 2 + \sqrt{3}$ . Since $\begin{aligned} RJ_1 & = \; IR + V_{2n} - 2V_1 + V_2 \\ RJ_n & = \; V_{n-1} - 2V_n + V_{n+1} \end{aligned}$ we can solve a pair of simultaneous equations to obtain $P,Q$ . If we do this. we obtain that the effective resistance of this circuit is $R_n \; =\; \frac{V_1 - V_{2n}}{I} \; =\; \frac16\left(6 - \sqrt{3} + \frac{2\sqrt{3}}{1 + (2+\sqrt{3})^n} - \frac{3}{n}\right)R$ Putting $R = 1$ , we deduce that $\lim_{n \to \infty} R_n \; =\; \tfrac16(6 - \sqrt{3}) \; = \; \frac{11}{2(6 + \sqrt{3})}$ making the answer $11 +2 + 6 + 3 = \boxed{22}$ .