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

Revision as of 01:31, 13 May 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 $(2x+y)^2 = x^2+xy+7y^2$ . 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 from $x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0$ .

-RootThreeOverTwo

@@ Line 11: / Line 11: @@
 ==Note==
-The cases <math>x=y</math> or <math>x=-2y</math> can be found by SFFT from <math>x^2+xy-2y^2=0</math>.
+The cases <math>x=y</math> and <math>x=-2y</math> can be found by SFFT from <math>x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0</math>.
 -RootThreeOverTwo
 ==See Also==
 {{AIME box|year=2018|n=I|num-b=4|num-a=6}}
 {{MAA Notice}}

2018 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2018 AIME I Problems/Problem 5"

Revision as of 01:31, 13 May 2018

Solution

Note

See Also