Difference between revisions of "2018 AIME I Problems/Problem 5"
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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}=</math> <math>\frac{-2\pm \sqrt{4-24(1-K)}}{12}</math> <math>\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 9=\boxed{189}</math> | 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}=</math> <math>\frac{-2\pm \sqrt{4-24(1-K)}}{12}</math> <math>\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 9=\boxed{189}</math> | ||
− | -vsamc\newline | + | -vsamc<math>\newline</math> |
-minor edit:einsteinstudent | -minor edit:einsteinstudent | ||
Revision as of 11:30, 29 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 -minor edit:einsteinstudent
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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