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 (if we realize it is, we can always backtrack) 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 a possible solution.

2014 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2014 AIME I Problems/Problem 14

Problem 14

Solution

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