Difference between revisions of "2018 AIME I Problems/Problem 5"
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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>. | ||
From the second equation, <math>9x^2+6xy+y^2=3x^2+4xy+Ky^2</math>. 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>. | From the second equation, <math>9x^2+6xy+y^2=3x^2+4xy+Ky^2</math>. 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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==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 21:06, 8 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 .
Solutions
Straightforward 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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