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

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==Problem==
 
 
== Problem 14 ==
 
== Problem 14 ==
  
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== Solution ==
 
== 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 <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 <math>x = 0</math> is not the highest solution (if we realize it is, we can always backtrack) we can cancel the common factor of <math>x</math> from both sides of the equation.
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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 <math>x = 0</math> is not the highest solution (which is true given the answer format) we can cancel the common factor of <math>x</math> 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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<math>\frac{2y}{y^2 - 64} + \frac{2y}{y^2 - 36} = y</math>
 
<math>\frac{2y}{y^2 - 64} + \frac{2y}{y^2 - 36} = y</math>
  
Then, we can cancel out a <math>y</math> from both sides, knowing that <math>x = 11</math> is a possible solution.
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Then, we can cancel out a <math>y</math> from both sides, knowing that <math>x = 11</math> is not a possible solution given the answer format.
 
After we do that, we can make the final substitution <math>z = y^2</math>.
 
After we do that, we can make the final substitution <math>z = y^2</math>.
  
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Using the quadratic formula, we get that the largest solution for <math>z</math> is <math>z = 52 + 10\sqrt{2}</math>. Then, repeatedly substituting backwards, we find that the largest value of <math>x</math> is <math>11 + \sqrt{52 + \sqrt{200}}</math>. The answer is thus <math>11 + 52 + 200 = \boxed{263}</math>
 
Using the quadratic formula, we get that the largest solution for <math>z</math> is <math>z = 52 + 10\sqrt{2}</math>. Then, repeatedly substituting backwards, we find that the largest value of <math>x</math> is <math>11 + \sqrt{52 + \sqrt{200}}</math>. The answer is thus <math>11 + 52 + 200 = \boxed{263}</math>
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Note: When <math>x</math> is barely larger than <math>19</math>, then <math>\frac{19}{x-19}</math> is very large, so the left side of the equation approaches infinity as <math>x</math> approaches <math>19</math> from the side greater than <math>19</math>. However, we also know as <math>x</math> gets very large, the fractions get smaller as the left side approaches <math>0</math>. Since the quadratic on the right side is increasing and positive when <math>x=19</math>, the equation will be true at a certain <math>x>19.</math> So, we don't have to assume there is an answer <math>x>0.</math>
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==Video Solution by Punxsutawney Phil==
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https://youtu.be/pNsmv333SE0
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==Video Solution by Mathematical Dexterity (Pure magic!)==
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https://www.youtube.com/watch?v=7b7IPOYZbrk
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==Video Solution: Math Bear presents... A Fun Algebra Equation Problem==
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https://www.youtube.com/watch?v=1hdYnbm4Tvo
  
 
== 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 23:01, 1 December 2024

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


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 $x>0.$

Video Solution by Punxsutawney Phil

https://youtu.be/pNsmv333SE0

Video Solution by Mathematical Dexterity (Pure magic!)

https://www.youtube.com/watch?v=7b7IPOYZbrk

Video Solution: Math Bear presents... A Fun Algebra Equation Problem

https://www.youtube.com/watch?v=1hdYnbm4Tvo

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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