Difference between revisions of "2019 AIME I Problems/Problem 8"
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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. | 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. | ||
− | 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=\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/ | + | 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=\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 <math>\frac{13}{54}</math>. Therefore, the answer is <math>\boxed{067}</math>. |
eric2020, inspired by Tommy2002 | eric2020, inspired by Tommy2002 |
Revision as of 13:52, 16 March 2019
Contents
Problem 8
Let be a real number such that . Then where and are relatively prime positive integers. Find .
Solution 1
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 . 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
Solution 3 (Newton Sums)
Newton sums is basically constructing the powers of the roots of the polynomials instead of deconstructing them which was done in solution 2. Let and be the roots of some polynomial . Then, for some .
Let . We want to find . Clearly and . Newton sums tells us that where for our polynomial .
Bashing, we have
Thus . Clearly, so .
Note . Solving for , we get . Finally, .
Solution 4 (Fun)
We let and , so we have and . Noticing that might be a useful value to find, we let . Then we can work our way up to find . using quadform you get or . Since , and since can't exceed 1, can't exceed . Clearly, . And finally, looking back to previous results, we see that (it's easier not to simplify the fraction). which yields the answer .
~PCampbell
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.