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>. | ||
− | ==Solution== | + | ==Solution 1== |
Using the logarithmic property <math>\log_{a^n}b^n = \log_{a}b</math>, we note that <math>(2x+y)^2 = 4x^2+4xy+y^2</math>. | Using the logarithmic property <math>\log_{a^n}b^n = \log_{a}b</math>, we 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>. | ||
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The cases <math>x=y</math> and <math>x=-2y</math> can be found by SFFT ([https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick Simon's Favorite Factoring Trick]) from <math>x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0</math>. | The cases <math>x=y</math> and <math>x=-2y</math> can be found by SFFT ([https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick Simon's Favorite Factoring Trick]) from <math>x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0</math>. | ||
+ | ==Solution 2== | ||
+ | Do as done in Solution 1 to get <math>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</math>. Do as done in Solution 1 to get <math>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}</math>. If <math>\frac{x}{y}=1</math>, then <math>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</math>. If <math>\frac{x}{y}=-2</math>, then <math>-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</math>. Hence our final answer is <math>21\cdot 6=\boxed{126}</math> | ||
+ | -vsamc | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2018|n=I|num-b=4|num-a=6}} | {{AIME box|year=2018|n=I|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:51, 12 April 2020
Contents
Problem 5
For each ordered pair of real numbers satisfying there is a real number such that Find the product of all possible values of .
Solution 1
Using the logarithmic property , we note that . That gives upon simplification and division by . Then, or . From the second equation, . If we take , we see that . If we take , we see that . The product is .
-expiLnCalc
Note
The cases and can be found by SFFT (Simon's Favorite Factoring Trick) from .
Solution 2
Do as done in Solution 1 to get . Do as done in Solution 1 to get . If , then . If , then . Hence our final answer is -vsamc
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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