Difference between revisions of "2014 AIME I Problems/Problem 14"
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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. | + | 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>. | ||
| Line 32: | Line 32: | ||
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> | ||
| + | |||
| + | ==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 == | == 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}} | ||
Latest revision as of 14:09, 13 August 2022
Contents
Problem 14
Let 
be the largest real solution to the equation
There are positive integers 
, 
, and 
such that 
. Find 
.
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 
, then the fraction becomes of the form 
. A similar cancellation happens with the other four terms. If we assume 
is not the highest solution (which is true given the answer format) we can cancel the common factor of 
from both sides of the equation.
Then, if we make the substitution 
, we can further simplify.
If we group and combine the terms of the form 
and 
, we get this equation:
Then, we can cancel out a 
from both sides, knowing that 
is not a possible solution given the answer format.
After we do that, we can make the final substitution 
.
Using the quadratic formula, we get that the largest solution for 
is 
. Then, repeatedly substituting backwards, we find that the largest value of is 
. The answer is thus 
Video Solution by Punxsutawney Phil
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 (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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
