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

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For each ordered pair of real numbers <math>(x,y)</math> satisfying <cmath>\log_2(2x+y) = \log_4(x^2+xy+7y^2)</cmath>there is a real number <math>K</math> such that <cmath>\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).</cmath>Find the product of all possible values of <math>K</math>.
 
For each ordered pair of real numbers <math>(x,y)</math> satisfying <cmath>\log_2(2x+y) = \log_4(x^2+xy+7y^2)</cmath>there is a real number <math>K</math> such that <cmath>\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).</cmath>Find the product of all possible values of <math>K</math>.
  
==Solutions==
+
==Solution==
  
==Straightforward Solution==
 
 
Note that <math>(2x+y)^2 = x^2+xy+7y^2</math>. DO NOT SQUARE THE WRONG SIDE!
 
Note that <math>(2x+y)^2 = x^2+xy+7y^2</math>. DO NOT SQUARE THE WRONG SIDE!
 
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>.

Revision as of 13:48, 14 March 2018

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 $(2x+y)^2 = x^2+xy+7y^2$. DO NOT SQUARE THE WRONG SIDE! 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

See Also

2018 AIME I (ProblemsAnswer KeyResources)
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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