AIME 2014 I · 第 14 题

Solution 1

The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to $\frac{3}{x-3}$, then the fraction becomes of the form $\frac{x}{x - 3}$. A similar cancellation happens with the other four terms. If we assume $x = 0$ is not the highest solution (which is true given the answer format) we can cancel the common factor of $x$ from both sides of the equation.

$$\frac{1}{x - 3} + \frac{1}{x - 5} + \frac{1}{x - 17} + \frac{1}{x - 19} = x - 11$$ Then, if we make the substitution $y = x - 11$, we can further simplify.

$$\frac{1}{y + 8} + \frac{1}{y + 6} + \frac{1}{y - 6} + \frac{1}{y - 8} =y$$ If we group and combine the terms of the form $y - n$ and $y + n$, we get this equation:

$$\frac{2y}{y^2 - 64} + \frac{2y}{y^2 - 36} = y$$ Then, we can cancel out a $y$ from both sides, knowing that $x = 11$ is not a possible solution given the answer format. After we do that, we can make the final substitution $z = y^2$.

$$\frac{2}{z - 64} + \frac{2}{z - 36} = 1$$ $$2z - 128 + 2z - 72 = (z - 64)(z - 36)$$ $$4z - 200 = z^2 - 100z + 64(36)$$ $$z^2 - 104z + 2504 = 0$$ Using the quadratic formula, we get that the largest solution for $z$ is $z = 52 + 10\sqrt{2}$. Then, repeatedly substituting backwards, we find that the largest value of $x$ is $11 + \sqrt{52 + \sqrt{200}}$. The answer is thus $11 + 52 + 200 = \boxed{263}$

Note: When $x$ is barely larger than $19$, then $\frac{19}{x-19}$ is very large, so the left side of the equation approaches infinity as $x$ approaches $19$ from the side greater than $19$. However, we also know as $x$ gets very large, the fractions get smaller as the left side approaches $0$. Since the quadratic on the right side is increasing and positive when $x=19$, the equation will be true at a certain $x>19.$ So, we don't have to assume there is an answer $x>0.$

Solution 2

Proceed as with Solution 1 until we get the following.

$\frac{2y}{y^2 - 64} + \frac{2y}{y^2 - 36} = y \implies$ $\frac{1}{y^2 - 64} + \frac{1}{y^2 - 36} = \frac{1}{2}.$

Here, we may also use a slightly different substitution, $z = y^2 - 50.$ This gives:

$\frac{1}{z - 14} + \frac{1}{z + 14} = \frac{1}{2} \implies$ $\frac{2z}{z^2 - 196} = \frac{1}{2} \implies$ $z^2 - 4z - 196 = 0.$

We now have a simpler quadratic, eliminating tedious and potentially error-prone calculations. Following through, we get $z = 2 + \sqrt{200} \implies y = \sqrt{52 + \sqrt{200}}$ as desired.

Solution 3

Stolen from forums

\begin{align} & \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}={{x}^{2}}-11x-4 \\ & x-11=u\Rightarrow x=u+11 \\ & \Rightarrow \frac{3}{u+8}+\frac{5}{u+6}+\frac{17}{u-6}+\frac{19}{u-8}=(u+11)u-4 \\ & \Rightarrow \frac{3(u-8)+19(u+8)}{{{u}^{2}}-64}+\frac{5(u-6)+17(u+6)}{{{u}^{2}}-36}={{u}^{2}}+11u-4 \\ & \Rightarrow \frac{22u+128}{{{u}^{2}}-64}+\frac{22u+72}{{{u}^{2}}-36}={{u}^{2}}+11u-4 \\ & \Rightarrow \frac{2({{u}^{2}}+11u)}{{{u}^{2}}-64}+\frac{2({{u}^{2}}+11u)}{{{u}^{2}}-36}={{u}^{2}}+11u \\ & \Rightarrow \left\{ \begin{matrix} {{u}^{2}}+11u=0\Rightarrow u=0,-11 \\ \frac{2}{{{u}^{2}}-64}+\frac{2}{{{u}^{2}}-36}=1\Rightarrow ({{u}^{2}}-36)({{u}^{2}}-64)=4{{u}^{2}}-200 \\ \end{matrix} \right. \\ & {{u}^{4}}-104{{u}^{2}}+2504=0\Rightarrow {{u}^{2}}=52\pm 10\sqrt{2} \\ & \Rightarrow u=\pm \sqrt{52\pm 10\sqrt{2}}\Rightarrow x=11\pm \sqrt{52\pm 10\sqrt{2}} \end{align}

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