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Difference between revisions of "2018 AIME I Problems/Problem 5"

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==Solution==
 
==Solution==
  
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Note that <math>(2x+y)^2 = x^2+xy+7y^2</math>.  
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Note that <math>(2x+y)^2 = 4x^2+4xy+y^2</math>.  
 
That gives <math>x^2+xy-2y^2=0</math> upon simplification and division by <math>3</math>. Then, <math>x=y</math> or <math>x=-2y</math>.
 
That gives <math>x^2+xy-2y^2=0</math> upon simplification and division by <math>3</math>. Then, <math>x=y</math> or <math>x=-2y</math>.
 
From the second equation, <math>9x^2+6xy+y^2=3x^2+4xy+Ky^2</math>. If we take <math>x=y</math>, we see that <math>K=9</math>. If we take <math>x=-2y</math>, we see that <math>K=21</math>. The product is <math>\boxed{189}</math>.
 
From the second equation, <math>9x^2+6xy+y^2=3x^2+4xy+Ky^2</math>. If we take <math>x=y</math>, we see that <math>K=9</math>. If we take <math>x=-2y</math>, we see that <math>K=21</math>. The product is <math>\boxed{189}</math>.

Revision as of 18:43, 21 August 2018

Problem 5

For each ordered pair of real numbers $(x,y)$ satisfying \[\log_2(2x+y) = \log_4(x^2+xy+7y^2)\]there is a real number $K$ such that \[\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).\]Find the product of all possible values of $K$.

Solution

Note that $(2x+y)^2 = 4x^2+4xy+y^2$. That gives $x^2+xy-2y^2=0$ upon simplification and division by $3$. Then, $x=y$ or $x=-2y$. From the second equation, $9x^2+6xy+y^2=3x^2+4xy+Ky^2$. If we take $x=y$, we see that $K=9$. If we take $x=-2y$, we see that $K=21$. The product is $\boxed{189}$.

-expiLnCalc

Note

The cases $x=y$ and $x=-2y$ can be found by SFFT (Simon's Favorite Factoring Trick) from $x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0$.

See Also

2018 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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