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

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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}</math>
+
<math>\frac{1}{x - 3} + \frac{1}{x - 5} + \frac{1}{x - 17} + \frac{1}{x - 19} = x - 11</math>
 +
 
 +
Then, if we make the substitution y = x - 11, we can further simplify.
 +
 
 +
<math>\frac{1}{y + 8} + \frac{1}{y + 6} + \frac{1}{y - 6} + \frac{1}{y - 8} =y </math>
 +
 
 +
If we group and combine the terms of the form  <math>y - n</math> and  <math> y + n</math>, we get this equation:
 +
 
 +
<math>\frac{2y}{y^2 - 64} + \frac{2y}{y^2 - 36} = y</math>
 +
 
 +
Then, we can cancel out a y from both sides, knowing that <math>x = 11</math> is a possible solution.
  
 
== 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 20:43, 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.

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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