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

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==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>.
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==Solution 1==
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Using the logarithmic property <math>\log_{a^n}b^n = \log_{a}b</math>, we note that <cmath>(2x+y)^2 = x^2+xy+7y^2</cmath>
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That gives <cmath>x^2+xy-2y^2=0</cmath> upon simplification and division by <math>3</math>. Factoring <math>x^2+xy-2y^2=0</math> gives <cmath>(x+2y)(x-y)=0</cmath> Then, <cmath>x=y \text{ or }x=-2y</cmath>
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From the second equation, <cmath>9x^2+6xy+y^2=3x^2+4xy+Ky^2</cmath> 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>.
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-expiLnCalc
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==Solution 2==
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Do as done in Solution 1 to get <cmath>x^2+xy-2y^2=0</cmath> <cmath>\implies (\frac{x}{y})^2+\frac{x}{y}-2=0</cmath> <cmath>\implies \frac{x}{y}=\frac{-1\pm\sqrt{1+8}}{2}=1,-2</cmath> Do as done in Solution 1 to get <cmath>9x^2+6xy+y^2=3x^2+4xy+Ky^2</cmath> <cmath>\implies 6x^2+2xy+(1-K)y^2=0</cmath> <cmath>\implies 6(\frac{x}{y})^2+2\frac{x}{y}+(1-K)=0</cmath> <cmath>\implies \frac{x}{y}=\frac{-2\pm \sqrt{4-24(1-K)}}{12}</cmath> <cmath>\implies \frac{x}{y}=\frac{-2\pm 2\sqrt{6K-5}}{12}=\frac{-1\pm \sqrt{6K-5}}{6}</cmath>If <math>\frac{x}{y}=1</math> then <cmath>1=\frac{-1\pm \sqrt{6K-5}}{6}</cmath> <cmath>\implies 6=-1\pm \sqrt{6K-5}</cmath> <cmath>\implies 7=\pm \sqrt{6K-5}</cmath> <cmath>\implies 49=6K-5</cmath> <cmath>\implies K=9</cmath>If <math>\frac{x}{y}=-2</math> then <cmath>-2=\frac{-1\pm \sqrt{6K-5}}{6}</cmath> <cmath>\implies -12=-1\pm \sqrt{6K-5}</cmath> <cmath>\implies -11=\sqrt{6K-5}</cmath> <cmath>\implies 121=6K-5</cmath> <cmath>\implies 126=6K</cmath> <cmath>\implies K=21</cmath>Hence our final answer is <math>21\cdot 9=\boxed{189}</math>
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-vsamc<math>\newline</math>
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==Solution 3 (Official MAA)==
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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>
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==See Also==
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{{AIME box|year=2018|n=I|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 12:58, 14 August 2024

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 \[(2x+y)^2 = x^2+xy+7y^2\] That gives \[x^2+xy-2y^2=0\] upon simplification and division by $3$. Factoring $x^2+xy-2y^2=0$ gives \[(x+2y)(x-y)=0\] Then, \[x=y \text{ or }x=-2y\] 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$\newline$

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 $21\cdot9=189.$

See Also

2018 AIME I (Problems • Answer Key • Resources)
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Problem 6
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