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Difference between revisions of "2014 AIME I Problems/Problem 14"

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== Solution ==
 
== Solution ==
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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 <math>frac{3}{x-3}</math>, then the fraction becomes of the form <math>frac{x}{x - 3}</math>. 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.
+
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 <math>\frac{3}{x-3}</math>, then the fraction becomes of the form <math>\frac{x}{x - 3}</math>. 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.
  
 
<math>\frac{1}{x - 3} + \frac{1}{x - 5} + \frac{1}{x - 17} + \frac{1}{x - 19} = x - 11</math>
 
<math>\frac{1}{x - 3} + \frac{1}{x - 5} + \frac{1}{x - 17} + \frac{1}{x - 19} = x - 11</math>
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Then, we can cancel out a y from both sides, knowing that <math>x = 11</math> is a possible solution.
 
Then, we can cancel out a y from both sides, knowing that <math>x = 11</math> is a possible solution.
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After we do that, we can make the final substitution z = y^2.
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 +
<math>\frac{2}{z - 64} + \frac{2}{z - 36} = 1</math>
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<math>2z - 128 + 2z - 72 = (z - 64)(z - 36)</math>
 +
<math>4z -  200 = z^2 - 100z + 64(36)</math>
 +
<math>z^2 - 104z + 2504 = 0</math>
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 +
Using the quadratic formula, we get that the largest solution for z is <math>z = 52 + 10\sqrt{2}</math>. Then, repeatedly substituting backwards, we find that the largest value of x is <math>11 + \sqrt{52 + \sqrt{200}}</math>. The answer is 11 + 52 + 200 = 263.
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== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=13|num-a=15}}
 
{{AIME box|year=2014|n=I|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:50, 14 March 2014

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. 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 11 + 52 + 200 = 263.

See also

2014 AIME I (Problems • Answer Key • Resources)
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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