Difference between revisions of "2019 AIME I Problems/Problem 8"
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==Solution== | ==Solution== | ||
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+ | We can substitute <math>y = \sin^2{x}</math>. Since we know that <math>\cos^2{x}=1-\sin^2{x}</math>, we can do some simplification. | ||
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+ | This yields <math>y^5+(1-y)^5=\frac{11}{36}</math>. From this, we can substitute again to get some cancellation through binomials. If we let <math>z=1/2-y</math>, we can simplify the equation to <math>(1/2+z^5)+(1/2-z^5)^5=\frac{11}{36}</math>. After using binomial theorem, this simplifies to <math>\frac{1}{16}(80z^4+40z^2+1)=11/36</math>. If we use the quadratic theorem, we obtain that <math>z^2=\pm\frac{1}{12}</math>, so <math>z=\pm\frac{1}{2\sqrt{3}}</math>. By plugging z into <math>(1/2-z)^6+(1/2+z)^6</math> (which is equal to <math>sin^{12}{x}+cos^{12}{x}</math>, we can either use binomial theorem or sum of cubes to simplify, and we end up with 13/54. Therefore, the answer is <math>\boxed{67}</math>. | ||
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+ | eric2020, inspired by Tommy2002 | ||
==Solution 2== | ==Solution 2== |
Revision as of 21:16, 14 March 2019
The 2019 AIME I takes place on March 13, 2019.
Contents
Problem 8
Let be a real number such that . Then where and are relatively prime positive integers. Find .
Solution
We can substitute . Since we know that , we can do some simplification.
This yields . From this, we can substitute again to get some cancellation through binomials. If we let , we can simplify the equation to . After using binomial theorem, this simplifies to . If we use the quadratic theorem, we obtain that , so . By plugging z into (which is equal to , we can either use binomial theorem or sum of cubes to simplify, and we end up with 13/54. Therefore, the answer is .
eric2020, inspired by Tommy2002
Solution 2
First, for simplicity, let and . Note that . We then bash the rest of the problem out. Take the tenth power of this expression and get . Note that we also have . So, it suffices to compute . Let . We have from cubing that or . Next, using , we get or . Solving gives or . Clearly is extraneous, so . Now note that , and . Thus we finally get , giving .
-Emathmaster
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.