Difference between revisions of "2018 AIME I Problems/Problem 5"
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==Solution== | ==Solution== | ||
| − | Note that <math>(2x+y)^2 = x^2+xy+7y^2</math>. | + | Note that <math>(2x+y)^2 = x^2+xy+7y^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>. | ||
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>. | ||
Revision as of 21:47, 20 March 2018
For each ordered pair of real numbers 
satisfying ![\[\log_2(2x+y) = \log_4(x^2+xy+7y^2)\]](http://latex.artofproblemsolving.com/a/f/b/afbbbda1a46f1a5797bb4174fb377e1ec8497dd4.png)
there is a real number 
such that ![\[\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).\]](http://latex.artofproblemsolving.com/3/6/d/36dbe8d0f02bf52e0f19883df3d2014ef023bbad.png)
Find the product of all possible values of .
Solution
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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
