Resistive Wheel

Suppose that current $I$ enters the circuit at vertex $1$ , and leaves at vertex $n$ . Let the potentials at vertices $1$ , $2$ , ..., $n$ be $V_1$ , $V_2$ , ..., $V_n$ . Let the current from vertex $k$ to vertex $k+1$ be $I_k$ for $1 \le k \le n-1$ , and let the current from vertex $n$ to vertex $1$ be $I_n$ . Let the potential at the central vertex be $0$ , and let the current from the central vertex to vertex $k$ be $J_k$ for $1 \le k \le n$ .

Since current is neither created nor destroyed we deduce that $\begin{aligned} I_1 - I_n - J_1 & = \; I \\ I_k - I_{k-1} - J_k & = \; 0 \hspace{2cm} 2 \le k \le n-1 \\ I_n - I_{n-1} - J_n & = \; -I \end{aligned}$ so that $J_k \; = \; \left\{ \begin{array}{lll} I_1 - I_n - I & \hspace{1cm} & k = 1 \\ I_k - I_{k-1} & & 2 \le k \le n-1 \\ I_n - I_{n-1} + I & & k = n \end{array}\right.$ and we also have $J_kR \; = \; -V_k \hspace{0.5cm} 1 \le k \le n \hspace{2cm} I_kR \; =\; \left\{ \begin{array}{lll} V_k - V_{k+1} & \hspace{1cm} & 1 \le k \le n-1 \\ V_n - V_1 & & k =n \end{array}\right.$ Combining these equations we deduce that $V_{k+1} - 3V_k + V_{k-1} \; = \; 0 \hspace{2cm} 2 \le k \le n-1$ and hence we deduce that $V_k \; = \; A\alpha^k + B\alpha^{-k} \hspace{2cm} 1 \le k \le n$ where $\alpha = \tfrac12(3 + \sqrt{5})$ . Satisfying the boundary conditions gives us that $A \; =\; \frac{IR(1 - \alpha^{-1})}{(\alpha - \alpha^{-1})(1 - \alpha^n)} \hspace{2cm} B \; =\; \frac{IR(\alpha - 1)}{(\alpha-\alpha^{-1})(1 - \alpha^{-n})}$ and so the effective resistance of the circuit is $R_n \; = \; \frac{V_1 - V_n}{I} \; = \; R\left[\frac{(1 - \alpha^{-1})(\alpha - \alpha^n)}{(\alpha - \alpha^{-1})(1 - \alpha^n)} + \frac{(\alpha-1)(\alpha^{-1} - \alpha^{-n})}{(\alpha-\alpha^{-1})(1 -\alpha^{-n})}\right]$ and hence $\begin{aligned} \lim_{n \to \infty}R_n & = \; R\left[ \frac{1-\alpha^{-1}}{\alpha - \alpha^{-1}} + \frac{(\alpha-1)\alpha^{-1}}{\alpha - \alpha^{-1}}\right] \\ & = \; \frac{2(1 - \alpha^{-1})}{\alpha - \alpha^{-1}} \; = \; 1 - \frac{1}{\sqrt{5}} \end{aligned}$ making the answer $1 + (-1) + 5 = \boxed{5}$ .