Difference between revisions of "2019 AIME I Problems/Problem 8"
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+ | ==Solution 6== | ||
+ | Let <math>m=\sin^2 x</math> and <math>n=\cos^2 x</math>, then <math>m+n=1</math> and <math>m^5+n^5=\frac{11}{36}</math> | ||
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+ | <math>m^6+n^6=(m^5+n^5)(m+n)-mn(m^4+n^4)=(m^5+n^5)-mn(m^4+n^4)</math> | ||
+ | |||
+ | Now factoring <math>m^5+n^5</math> as solution 4 yields <math>m^5+n^5=(m+n)(m^4-m^3n+m^2n^2-mn^3+n^4)</math> | ||
+ | <math>=m^4+n^4-mn(m^2-mn+n^2)=m^4+n^4-mn[(m+n)^2-3mn]=m^4+n^4-mn(1-3mn)</math>. | ||
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+ | Since <math>(m+n)^4=m^4+4m^3n+6m^2n^2+4mn^3+n^4</math>, <math>m^4+n^4=(m+n)^4-2mn(2m^2+3mn+2n^2)=1-2mn(2m^2+3mn+2n^2)</math>. | ||
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+ | Notice that <math>2m^2+3mn+2n^2</math> can be rewritten as <math>[\sqrt{2}(a+b)]^2-mn=2-mn</math>. Thus,<math>m^4+n^4=1-2mn(2-mn)</math> and <math>m^5+n^5=1-2mn(2-mn)-mn(1-3mn)=1-5mn+5(mn)^2=\frac{11}{36}</math>. As in solution 4, we get <math>mn=\frac{1}{6}</math> and <math>m^4+n^4=1-2*\frac{1}{6}(2-\frac{1}{6})=\frac{7}{18}</math> | ||
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+ | Substitute <math>m^4+n^4=\frac{7}{18}</math> and <math>mn=\frac{1}{6}</math>, then | ||
+ | <math>m^6+n^6=\frac{11}{36}-\frac{1}{6}*\frac{7}{18}=\frac{13}{54}</math>, and the desired answer is <math>\boxed{067}</math> | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=7|num-a=9}} | {{AIME box|year=2019|n=I|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 04:41, 13 August 2020
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, 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
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.