Resistance

Aleksander from Gdynia Bilingual High School No 3, Poland, David from Guilford County (is that UK, US or elsewhere?) Andrei from Romania and Chris from CSN used different methods to solve this problem, all of them leading to solving a quadratic equation. Chris used the observation that, removing three resistors leaves the resistance of the remaining network unchanged, and quickly arrived at the quadratic equation which gives the solution. Aleksander visualised the network in a different, but equivalent, configuration and used continued fractions and David's method was similar. Andrei arrived at the same quadratic equation by first proving that the sequence of resistances $R_0, R_1, R_2, ...$ from A to B, corresponding to different number of resistance 'squares', is a steadily decreasing sequence and, as $R_n > 0$, the sequence tends to a limit.

This is Chris's solution:

Following the hint, keep in mind that the equations for series resistances and parallel resistances are: $R = R_1 + R_2$ for series and $R = (R_1 \times R_2)/(R_1 + R_2)$ for parallel.



















From the diagram I observe that: $${1\over R_n} = {1\over R} + {1\over 2R +R_{n-1}}.$$ where $R = 1$. I observe that $R_1< R_0$, and I shall prove by induction that $R_n< R_{n-1}$. This, by the recurrence relation, is equivalent to: $$R_n< R_{n-1}\Leftrightarrow {1\over R_n}> {1\over R_{n-1}} \Leftrightarrow {1\over R} + {1\over 2R + R_{n-1}} > {1\over R} + {1\over 2R + R_{n-2}}\Leftrightarrow R_{n-1} < R_{n-2}.$$ So, the sequence $(Rn)$ is strictly decreasing. Evidently $R_n > 0$ so the sequence $(R_n)$ is convergent. By considering the limit in the recurrence relation (putting $R=1$), $$\lim _{n\to \infty} R_n = {1\over {1+ {1\over 2 + \lim_{n\to \infty}R_{n-1}}}}$$ where $$\lim _{n\to \infty} R_n = \lim _{n\to \infty} R_{n-1} = R_{L}.$$ This gives $R_{L}= {2+ R_{L}\over 3 + R_{L}}$ and hence the quadratic equation $R_{L}^2 + 2R_{L} - 2= 0$. The positive solution of this equation gives the resistance of the network: $\sqrt 3 - 1$ ohms.