Difference between revisions of "2018 AIME I Problems/Problem 5"
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+ | ==Solution 3 (Official MAA)== | ||
+ | Because <math>x^2+xy+7y^2=\left(x+\tfrac{y}{2}\right)^2+\tfrac{27}{4}y^2>0,</math> the right side of the first equation is real. It follows that the left side of the equation is also real, so <math>2x+y>0</math> and <cmath>\log_2(2x+y)=\log_{2^2}(2x+y)^2=\log_4(4x^2+4xy+y^2).</cmath> Thus <math>4x^2+4xy+y^2=x^2+xy+7y^2,</math> which implies that <math>0=x^2+xy-2y^2=(x+2y)(x-y).</math> Therefore either <math>x=-2y</math> or <math>x=y,</math> and because <math>2x+y>0,</math> <math>x</math> must be positive and <math>3x+y=x+(2x+y)>0.</math> Similarly, <cmath>\log_3(3x+y)=\log_{3^2}(3x+y)^2=\log_9(9x^2+6xy+y^2).</cmath> If <math>x=-2y\ne 0,</math> then <math>9x^2+6xy+y^2=36y^2-12y^2+y^2=25y^2=3x^2+4xy+Ky^2</math> when <math>K=21.</math> If <math>x=y\ne 0,</math> then <math>9x^2+6xy+y^2=16y^2=3x^2+4xy+Ky^2</math> when <math>K=9.</math> The requested product is <math>21\cdot9=189.</math> | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 12:50, 3 March 2021
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 . Factoring by Simon's Favorite Factoring Trick gives Then, or . From the second equation, . If we take , we see that . If we take , we see that . The product is .
-expiLnCalc
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
Solution 3 (Official MAA)
Because the right side of the first equation is real. It follows that the left side of the equation is also real, so and Thus which implies that Therefore either or and because must be positive and Similarly, If then when If then when The requested product is
Video Solution
https://www.youtube.com/watch?v=iE8paW_ICxw
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.