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=\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 the that <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>. | + | 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 the that <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 | ||
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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 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) | + | We group the first, third, and fifth term and second and fourth term. The first group: |
− | =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>. | + | <cmath> |
+ | \begin{align*} | ||
+ | \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 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | The second group: | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | -\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 | ||
+ | \end{align*} | ||
+ | </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, | Now from the second equation, | ||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
− | \sin^{12}x + \cos^{12}x | + | \sin^{12}x + \cos^{12}x &= (\sin^4x+\cos^4x)(\sin^8x-\sin^4x\cos^4x+\cos^8x)\\ |
− | &= (\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)((\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) | &= (1-2\sin^2x\cos^2x)((1-2\sin^2x\cos^2x)^2-3\sin^4x\cos^4x) | ||
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~ZericHang | ~ZericHang | ||
+ | |||
+ | ==Solution 5== | ||
+ | Define the recursion <math>a_n=(\sin^2 x)^n+(\cos^2 x)^n</math> | ||
+ | We know that the characteristic equation of <math>a_n</math> must have 2 roots, so we can recursively define <math>a_n</math> as <math>a_n=p*a_{n-1}+q*a_{n-2}</math>. <math>p</math> is simply the sum of the roots of the characteristic equation, which is <math>\sin^2 x+\cos^2 x=1</math>. <math>q</math> is the product of the roots, which is <math>-(\sin^2 x)(\cos^2 x)</math>. This value is not trivial and we have to solve for it. | ||
+ | We know that <math>a_0=2</math>, <math>a_1=1</math>, <math>a_5=\frac{11}{36}</math>. | ||
+ | Solving the rest of the recursion gives | ||
+ | |||
+ | <cmath>a_2=1+2q</cmath> | ||
+ | <cmath>a_3=1+3q</cmath> | ||
+ | <cmath>a_4=1+4q+2q^2</cmath> | ||
+ | <cmath>a_5=1+5q+5q^2=\frac{11}{36}</cmath> | ||
+ | <cmath>a_6=1+6q+9q^2+2q^3</cmath> | ||
+ | |||
+ | |||
+ | Solving for <math>q</math> in the expression for <math>a_5</math> gives us <math>q^2+q+\frac{5}{36}=0</math>, so <math>q=-\frac{5}{6}, -\frac{1}{6}</math>. Since <math>q=-(\sin^2 x)(\cos^2 x)</math>, we know that the minimum value it can attain is <math>-\frac{1}{4}</math> by AM-GM, so <math>q</math> cannot be <math>-\frac{5}{6}</math>. | ||
+ | Plugging in the value of <math>q</math> into the expression for <math>a_6</math>, we get <math>a_6=1-1+\frac{1}{4}-\frac{1}{108}=\frac{26}{108}=\frac{13}{54}</math>. Our final answer is then <math>13+54=\boxed{067}</math> | ||
+ | |||
+ | -Natmath | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | |||
==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 05:41, 13 August 2020
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 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.