2014 AIME I Problems/Problem 14

Problem 14

Let $m$ be the largest real solution to the equation

$\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4$

There are positive integers $a$, $b$, and $c$ such that $m=a+\sqrt{b+\sqrt{c}}$. Find $a+b+c$.

Solution

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 $11 + 52 + 200 = \boxed{263}$

See also

2014 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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