2014 AIME I 第 14 题

Add $4$ to both sides, splitting one unit to each fraction: each becomes $\frac{x}{x-k}.$ Then substitute $t = x - 11.$

Symmetric pairs combine to $\frac{2}{t^2 - 64} + \frac{2}{t^2 - 36} = 1,$ a quadratic in $t^2$

Add $4$ to both sides, giving one unit to each fraction: since $\frac{k}{x-k} + 1 = \frac{x}{x-k},$ the equation becomes $$\begin{aligned} &\scriptsize x\left(\frac{1}{x-3} + \frac{1}{x-5} + \frac{1}{x-17} + \frac{1}{x-19}\right) \\ &= x^2 - 11x = x(x - 11). \end{aligned}$$ Besides $x = 0,$ we can divide by $x$ and substitute $t = x - 11,$ which pairs the fractions symmetrically: $$\frac{2t}{t^2 - 64} + \frac{2t}{t^2 - 36} = t.$$

Besides $t = 0,$ dividing by $t$ gives $\frac{2}{t^2 - 64} + \frac{2}{t^2 - 36} = 1.$ With $u = t^2,$ clearing denominators gives $2(u - 36) + 2(u - 64)$ $= (u - 36)(u - 64),$ i.e. $u^2 - 104u + 2504 = 0,$ so $u = 52 \pm \sqrt{200}.$

The largest solution is $m = 11 + \sqrt{52 + \sqrt{200}} \approx 19.1,$ which exceeds the other candidates $0,$ $11,$ and $11 \pm \sqrt{52 \pm \sqrt{200}}.$ Therefore $a + b + c$ $= 11 + 52 + 200 = 263.$