Difference between revisions of "2019 AMC 12A Problems/Problem 12"
Sevenoptimus (talk | contribs) (Fixed formatting and added a simpler solution) |
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-WannabeCharmander | -WannabeCharmander | ||
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+ | ==Slightly simpler solution== | ||
+ | |||
+ | After obtaining <math>k + \frac{4}{k} = 6</math>, notice that the required answer is <math>(k - \frac{4}{k})^{2} = k^2 - 8 + \frac{16}{k^2} = \left(k^2 + 8 + \frac{16}{k^2}\right) - 16 = \left(k+\frac{4}{k}\right)^2 - 16 = 6^2 - 16 = 20</math> as before. | ||
==Solution 2== | ==Solution 2== | ||
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Deconstructing this expression using log rules, we get <math>(\log_2{x}-\log_2{y})^2</math>. | Deconstructing this expression using log rules, we get <math>(\log_2{x}-\log_2{y})^2</math>. | ||
− | Plugging in our | + | Plugging in our known values, we get <math>((3-\sqrt{5})-(3+\sqrt{5}))^2</math> or <math>(-2\sqrt{5})^2</math>. |
− | Our answer is 20 <math>\boxed{B}</math> | + | Our answer is 20 <math>\boxed{B}</math>. |
==Solution 3== | ==Solution 3== |
Revision as of 19:16, 10 February 2019
Contents
[hide]Problem
Positive real numbers and satisfy and . What is ?
Solution
Let , then and . Then we have .
We equate , and get . The solutions to this are .
To solve the given,
-WannabeCharmander
Slightly simpler solution
After obtaining , notice that the required answer is as before.
Solution 2
Thus or
We know that .
Thus
Thus
Thus
Thus
Solving for , we obtain .
Easy resubstitution makes
Solving for we obtain .
Looking back at the original problem, we have What is ?
Deconstructing this expression using log rules, we get .
Plugging in our known values, we get or .
Our answer is 20 .
Solution 3
Multiplying the first equation by we obtain .
From the second equation we have .
Then, .
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.