Difference between revisions of "2019 AIME I Problems/Problem 8"
(→See Also) |
Adam zheng (talk | contribs) m (→Solution 1) |
||
Line 6: | Line 6: | ||
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=\frac{1}{2}-y</math>, we can simplify the equation to <math>(\frac{1}{2}+z)^5+(\frac{1}{2}-z)^5=\frac{11}{36}</math>. After using binomial theorem, this simplifies to <math>\frac{1}{16}(80z^4+40z^2+1)=\frac{11}{36}</math>. If we use the quadratic formula, we obtain | + | 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=\frac{1}{2}-y</math>, we can simplify the equation to <math>(\frac{1}{2}+z)^5+(\frac{1}{2}-z)^5=\frac{11}{36}</math>. After using binomial theorem, this simplifies to <math>\frac{1}{16}(80z^4+40z^2+1)=\frac{11}{36}</math>. If we use the quadratic formula, we obtain <math>z^2=\frac{1}{12}</math>, so <math>z=\pm\frac{1}{2\sqrt{3}}</math>. By plugging z into <math>(\frac{1}{2}-z)^6+(\frac{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 23:14, 1 January 2024
Contents
[hide]Problem
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 , 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, by Vieta, 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
Solution 5
Define the recursion We know that the characteristic equation of must have 2 roots, so we can recursively define as . is simply the sum of the roots of the characteristic equation, which is . is the product of the roots, which is . This value is not trivial and we have to solve for it. We know that , , . Solving the rest of the recursion gives
Solving for in the expression for gives us , so . Since , we know that the minimum value it can attain is by AM-GM, so cannot be .
Plugging in the value of into the expression for , we get . Our final answer is then
-Natmath
Solution 6
Let and , then and
Now factoring as solution 4 yields .
Since , .
Notice that can be rewritten as . Thus, and . As in solution 4, we get and
Substitute and , then , and the desired answer is
Solution 7 (Official MAA)
Let and let Then for Because and it follows that and Hence or and because the only possible value of is Therefore The requested sum is
Solution 8 (Recursion)
Let for non-negative integers . Then and . In addition,where . So we can compute
A quick note: this solution uses recursion and is similar to the solution 7 above. It was from trumpeter, posted in the AoPS Forums, Contest Discussion.
Video Solution By North America Math Contest Go Go Go
https://www.youtube.com/watch?v=a3Ps425bYUM
~ Please sub.!
Video Solution By The Power Of Logic
~ Hayabusa1
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.