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>. | ||
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Note that <math>(2x+y)^2 = x^2+xy+7y^2</math>. DO NOT SQUARE THE WRONG SIDE! | Note that <math>(2x+y)^2 = x^2+xy+7y^2</math>. DO NOT SQUARE THE WRONG SIDE! | ||
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>. |
Revision as of 13:48, 14 March 2018
For each ordered pair of real numbers satisfying there is a real number such that Find the product of all possible values of .
Solution
Note that . DO NOT SQUARE THE WRONG SIDE! 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
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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