Difference between revisions of "2019 AIME I Problems/Problem 8"
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Factor the first equation. <cmath>\sin^{10}x + \cos^{10}x = (\sin^2x+\cos^2x)(\sin^8x-\sin^6x\cos^2x+\sin^4x\cos^4x-\sin^2x\cos^6x+\cos^8x)</cmath> | Factor the first equation. <cmath>\sin^{10}x + \cos^{10}x = (\sin^2x+\cos^2x)(\sin^8x-\sin^6x\cos^2x+\sin^4x\cos^4x-\sin^2x\cos^6x+\cos^8x)</cmath> | ||
First of all, <math>\sin^4x+\cos^4x = 1-2\sin^2x\cos^2x</math> because <math>\sin^4x+\cos^4x=(\sin^2x + \cos^2x)^2 -2\sin^2x\cos^2x = 1 - 2\sin^2x\cos^2x</math> | First of all, <math>\sin^4x+\cos^4x = 1-2\sin^2x\cos^2x</math> because <math>\sin^4x+\cos^4x=(\sin^2x + \cos^2x)^2 -2\sin^2x\cos^2x = 1 - 2\sin^2x\cos^2x</math> | ||
− | We group the | + | We group the first, third, and fifth term and second and fourth term. The first group: <cmath> \sin^8+\sin^4x\cos^4x+\cos^8x = (\sin^4x+\cos^4x)^2-\sin^4x\cos^4x)= (1 - 2\sin^2x\cos^2x)^2-\sin^4x\cos^4x) |
− | =1+4\sin^4x\cos^4x-4\sin^2x\cos^2x</cmath> The | + | =1+4\sin^4x\cos^4x-4\sin^2x\cos^2x</cmath> The second group: <cmath>-\sin^6x\cos^2x-\sin^2x\cos^6x = -\sin^2x\cos^2x(\sin^4x+\cos^4x)=-\sin^2x\cos^2x(1-2\sin^2x\cos^2x) = -\sin^2x\cos^2x+2\sin^4x\cos^4x</cmath> Add the two together to make <cmath>1+4\sin^4x\cos^4x-4\sin^2x\cos^2x-\sin^2x\cos^2x+2\sin^4x\cos^4x = 1 - 5\sin^2x\cos^2x+5\sin^4x\cos^4x</cmath> Because this equals <math>\frac{11}{36}</math>, we have <cmath>5\sin^4x\cos^4x- 5\sin^2x\cos^2x+\frac{25}{36}=0</cmath> Let <math>\sin^2x\cos^2x = a</math> so we get <cmath>5a^2- 5a+\frac{25}{36}=0 \Rightarrow a^2-a+\frac{5}{36}</cmath> Solving the quadratic gives us <cmath>a = \frac{1 \pm \frac{2}{3}}{2}</cmath> Because <math>\sin^2x\cos^2x \le \frac{1}{4}</math>, we finally get <math>a = \frac{1 - \frac{2}{3}}{2} = \frac{1}{6}</math>. |
Now from the second equation, <cmath>\sin^{12}x + \cos^{12}x = (\sin^4x+\cos^4x)(\sin^8x-\sin^4x\cos^4x+\cos^8x)=(1-2\sin^2x\cos^2x)((\sin^4x+\cos^4x)^2-3\sin^4x\cos^4x)=(1-2\sin^2x\cos^2x)((1-2\sin^2x\cos^2x)^2-3\sin^4x\cos^4x)</cmath> Plug in <math>\sin^2x\cos^2x = \frac{1}{6}</math> to get <cmath>(1-2(\frac{1}{6}))(1-2(\frac{1}{6})^2-3(\frac{1}{6})^2) = \frac{13}{54}</cmath> | Now from the second equation, <cmath>\sin^{12}x + \cos^{12}x = (\sin^4x+\cos^4x)(\sin^8x-\sin^4x\cos^4x+\cos^8x)=(1-2\sin^2x\cos^2x)((\sin^4x+\cos^4x)^2-3\sin^4x\cos^4x)=(1-2\sin^2x\cos^2x)((1-2\sin^2x\cos^2x)^2-3\sin^4x\cos^4x)</cmath> Plug in <math>\sin^2x\cos^2x = \frac{1}{6}</math> to get <cmath>(1-2(\frac{1}{6}))(1-2(\frac{1}{6})^2-3(\frac{1}{6})^2) = \frac{13}{54}</cmath> |
Revision as of 12:00, 26 December 2019
Contents
[hide]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 formula, we obtain the 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 . 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
Factor the first equation. First of all, because We group the first, third, and fifth term and second and fourth term. The first group: The second group: Add the two together to make Because this equals , we have Let so we get Solving the quadratic gives us Because , we finally get .
Now from the second equation, Plug in to get which yields the answer
~ZericHang
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.