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

Revision as of 11:50, 3 March 2021

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 1

Using the logarithmic property $\log_{a^n}b^n = \log_{a}b$ , we note that $(2x+y)^2 = 4x^2+4xy+y^2$ . That gives $x^2+xy-2y^2=0$ upon simplification and division by $3$ . Factoring $x^2+xy-2y^2=0$ by Simon's Favorite Factoring Trick gives $(x+2y)(x-y)=0$ 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

Solution 2

Do as done in Solution 1 to get $x^2+xy-2y^2=0\implies (\frac{x}{y})^2+\frac{x}{y}-2=0\implies \frac{x}{y}=\frac{-1\pm\sqrt{1+8}}{2}=1,-2$ . Do as done in Solution 1 to get $9x^2+6xy+y^2=3x^2+4xy+Ky^2\implies 6x^2+2xy+(1-K)y^2=0\implies 6(\frac{x}{y})^2+2\frac{x}{y}+(1-K)=0\implies \frac{x}{y}=$ $\frac{-2\pm \sqrt{4-24(1-K)}}{12}$ $\implies \frac{x}{y}=\frac{-2\pm 2\sqrt{6K-5}}{12}=\frac{-1\pm \sqrt{6K-5}}{6}$ . If $\frac{x}{y}=1$ , then $1=\frac{-1\pm \sqrt{6K-5}}{6}\implies 6=-1\pm \sqrt{6K-5}\implies 7=\pm \sqrt{6K-5}\implies 49=6K-5\implies K=9$ . If $\frac{x}{y}=-2$ , then $-2=\frac{-1\pm \sqrt{6K-5}}{6}\implies -12=-1\pm \sqrt{6K-5}\implies -11=\sqrt{6K-5}\implies 121=6K-5\implies 126=6K\implies K=21$ . Hence our final answer is $21\cdot 9=\boxed{189}$ -vsamc $\newline$ -minor edit:einsteinstudent

Solution 3 (Official MAA)

Because $x^2+xy+7y^2=\left(x+\tfrac{y}{2}\right)^2+\tfrac{27}{4}y^2>0,$ the right side of the first equation is real. It follows that the left side of the equation is also real, so $2x+y>0$ and $\log_2(2x+y)=\log_{2^2}(2x+y)^2=\log_4(4x^2+4xy+y^2).$ Thus $4x^2+4xy+y^2=x^2+xy+7y^2,$ which implies that $0=x^2+xy-2y^2=(x+2y)(x-y).$ Therefore either $x=-2y$ or $x=y,$ and because $2x+y>0,$ $x$ must be positive and $3x+y=x+(2x+y)>0.$ Similarly, $\log_3(3x+y)=\log_{3^2}(3x+y)^2=\log_9(9x^2+6xy+y^2).$ If $x=-2y\ne 0,$ then $9x^2+6xy+y^2=36y^2-12y^2+y^2=25y^2=3x^2+4xy+Ky^2$ when $K=21.$ If $x=y\ne 0,$ then $9x^2+6xy+y^2=16y^2=3x^2+4xy+Ky^2$ when $K=9.$ The requested product is $21\cdot9=189.$

Video Solution

https://www.youtube.com/watch?v=iE8paW_ICxw

@@ Line 15: / Line 15: @@
 -minor edit:einsteinstudent
+==Solution 3 (Official MAA)==
+Because <math>x^2+xy+7y^2=\left(x+\tfrac{y}{2}\right)^2+\tfrac{27}{4}y^2>0,</math> the right side of the first equation is real. It follows that the left side of the equation is also real, so <math>2x+y>0</math> and <cmath>\log_2(2x+y)=\log_{2^2}(2x+y)^2=\log_4(4x^2+4xy+y^2).</cmath> Thus <math>4x^2+4xy+y^2=x^2+xy+7y^2,</math> which implies that <math>0=x^2+xy-2y^2=(x+2y)(x-y).</math> Therefore either <math>x=-2y</math> or <math>x=y,</math> and because <math>2x+y>0,</math> <math>x</math> must be positive and <math>3x+y=x+(2x+y)>0.</math> Similarly, <cmath>\log_3(3x+y)=\log_{3^2}(3x+y)^2=\log_9(9x^2+6xy+y^2).</cmath> If <math>x=-2y\ne 0,</math> then <math>9x^2+6xy+y^2=36y^2-12y^2+y^2=25y^2=3x^2+4xy+Ky^2</math> when <math>K=21.</math> If <math>x=y\ne 0,</math> then <math>9x^2+6xy+y^2=16y^2=3x^2+4xy+Ky^2</math> when <math>K=9.</math> The requested product is <math>21\cdot9=189.</math>
 ==Video Solution==

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