Path of Less Resistance

The problem

The resistance of the original circuit and the new circuit are, respectively, $R_1 = (3 + 12) | (12 + R);\ \ \ \ \ R_2 = (3 | 12) + (12 | R).$ Here, $a|b$ is a shorthand notation for the parallel-circuit calculation: $a|b = 1/(\tfrac1a+\tfrac1b) = ab/(a+b)$ .

The solution

The equation we need to solve is $21\cdot R_1 = 25\cdot R_2$ . It is easy to prove that $\boxed{R = 8}$ is a solution: $R_1 = (3 + 12) | (12 + 8) = 15 | 20 = \frac{60}7;\ \ \ \ R_2 = (3 | 12) + (12 | 8) = \frac{12}5 + \frac{24}5 = \frac{36}5; \\ 21\cdot\frac{60}7 = 180 = 25\cdot\frac{36}5.$ The question, of course, is how this solution is found.

The method

Expanding the expressions for $R_1, R_2$ we have $R_1 = (3 + 12) | (12 + R) = \frac{15(12+R)}{27+R}; \\ R_2 = (3 | 12) + (12 | R) = \frac{12}5 + \frac{12R}{12+R}.$ Thus we obtain the equation $25\left(\frac{12}5 + \frac{12R}{12+R}\right) = 21\cdot \frac{15(12+R)}{27+R}.$ We simplify and divide the entire equation by the common factor 15: $4 + \frac{20R}{12+R} = \frac{21(12+R)}{27+R}.$ Multiply everything by the product of the denominators, $(12+R)(27+R)$ : $4(12+R)(27+R) + 20R(27+R) - 21(12+R)(12+R) = 0;$ expand and collect like terms: $3R^2 + 192R - 1728 = 0,\ \ \ \ R^2 + 64R - 576 = 0.$ Solve this, e.g. by splitting off a square: $(R + 32)^2 = 1024 + 576 = 1600,\ \ \ R = -32\pm\sqrt{1600} = -72\ \ \text{or}\ \ 8.$ The negative solution makes no physical sense, so we found that $R = 8$ .