Difference between revisions of "2019 AIME I Problems/Problem 8"

Latest revision as of 12:19, 1 February 2024

Problem

Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$ . Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution 1

We can substitute $y = \sin^2{x}$ . Since we know that $\cos^2{x}=1-\sin^2{x}$ , we can do some simplification.

This yields $y^5+(1-y)^5=\frac{11}{36}$ . From this, we can substitute again to get some cancellation through binomials. If we let $z=\frac{1}{2}-y$ , we can simplify the equation to: $\left(\frac{1}{2}+z\right)^5+\left(\frac{1}{2}-z\right)^5=\frac{11}{36}.$ After using binomial theorem, this simplifies to: $\frac{1}{16}(80z^4+40z^2+1)=\frac{11}{36}.$ If we use the quadratic formula, we obtain $z^2=\frac{1}{12}$ , so $z=\pm\frac{1}{2\sqrt{3}}$ (observe that either choice of $z$ doesn't matter). Substituting $z,$ we get:

$\sin^{12}{x}+\cos^{12}{x}=\left(\frac{1}{2}-z\right)^6+\left(\frac{1}{2}+z\right)^6=2z^6 + \frac{15z^4}{2} + \frac{15z^2}{8} + \frac{1}{32}=\frac{13}{54}.$

Therefore, the answer is $13+54=\boxed{067}$ .

-eric2020, inspired by Tommy2002

Motivation

The motivation to substitute $z=\frac{1}{2}-y$ comes so that after applying the binomial theorem to $y^5+(1-y)^5=\left(\frac{1}{2}+z\right)^5+\left(\frac{1}{2}-z\right)^5,$ a lot of terms will cancel out. Note that all the terms with odd exponents in $\left(\frac{1}{2}+z\right)^5$ will cancel out, while the terms with even exponents will be doubled. mathboy282

Solution 2

First, for simplicity, let $a=\sin{x}$ and $b=\cos{x}$ . Note that $a^2+b^2=1$ . We then bash the rest of the problem out. Take the fifth power of this expression and get $a^{10}+b^{10}+5a^2b^2(a^6+b^6)+10a^4b^4(a^2+b^2)=\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1$ . Note that we also have $\frac{11}{36}=a^{10}+b^{10}=(a^{10}+b^{10})(a^2+b^2)=a^{12}+b^{12}+a^2b^2(a^8+b^8)$ . So, it suffices to compute $a^2b^2(a^8+b^8)$ . Let $y=a^2b^2$ . We have from cubing $a^2+b^2=1$ that $a^6+b^6+3a^2b^2(a^2+b^2)=1$ or $a^6+b^6=1-3y$ . Next, using $\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1$ , we get $a^2b^2(a^6+b^6)+2a^4b^4=\frac{5}{36}$ or $y(1-3y)+2y^2=y-y^2=\frac{5}{36}$ . Solving gives $y=\frac{5}{6}$ or $y=\frac{1}{6}$ . Clearly $y=\frac{5}{6}$ is extraneous, so $y=\frac{1}{6}$ . Now note that $a^4+b^4=(a^2+b^2)^2-2a^2b^2=\frac{2}{3}$ , and $a^8+b^8=(a^4+b^4)^2-2a^4b^4=\frac{4}{9}-\frac{1}{18}=\frac{7}{18}$ . Thus we finally get $a^{12}+b^{12}=\frac{11}{36}-\frac{7}{18}\cdot\frac{1}{6}=\frac{13}{54}$ , giving $\boxed{067}$ .

- 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 $\sin^2x$ and $\cos^2x$ be the roots of some polynomial $F(a)$ . Then, by Vieta, $F(a)=a^2-a+b$ for some $b=\sin^2x\cdot\cos^2x$ .

Let $S_k=\left(\sin^2x\right)^k+\left(\cos^2x\right)^k$ . We want to find $S_6$ . Clearly $S_1=1$ and $S_2=1-2b$ . Newton sums tells us that $S_k-S_{k-1}+bS_{k-2}=0\Rightarrow S_k=S_{k-1}-bS_{k-2}$ where $k\ge 3$ for our polynomial $F(a)$ .

Bashing, we have $S_3=S_2-bS_1\Rightarrow S_3=(1-2b)-b(1)=1-3b$ $S_4=S_3-bS_2\Rightarrow S_4=(1-3b)-b(1-2b)=2b^2-4b+1$ $S_5=S_4-bS_3\Rightarrow S_5=(2b^2-4b+1)-b(1-3b)=5b^2-5b+1=\frac{11}{36}$

Thus $5b^2-5b+1=\frac{11}{36}\Rightarrow 5b^2-5b+\frac{25}{36}=0, 36b^2-36b+5=0, (6b-1)(6b-5)=0$ $b=\frac{1}{6} \text{ or } \frac{5}{6}$ . Clearly, $\sin^2x\cdot\cos^2x\not=\frac{5}{6}$ so $\sin^2x\cdot\cos^2x=b=\frac{1}{6}$ .

Note $S_4=\frac{7}{18}$ . Solving for $S_6$ , we get $S_6=S_5-\frac{1}{6}S_4=\frac{13}{54}$ . Finally, $13+54=\boxed{067}$ .

Solution 4

Factor the first equation. $\sin^{10}x + \cos^{10}x = (\sin^2x+\cos^2x)(\sin^8x-\sin^6x\cos^2x+\sin^4x\cos^4x-\sin^2x\cos^6x+\cos^8x)$ First of all, $\sin^4x+\cos^4x = 1-2\sin^2x\cos^2x$ because $\sin^4x+\cos^4x=(\sin^2x + \cos^2x)^2 -2\sin^2x\cos^2x = 1 - 2\sin^2x\cos^2x$ We group the first, third, and fifth term and second and fourth term. The first group: $\begin{align*} \sin^8+\sin^4x\cos^4x+\cos^8x &= (\sin^4x+\cos^4x)^2-\sin^4x\cos^4x)\\ &= (1 - 2\sin^2x\cos^2x)^2-\sin^4x\cos^4x)\\ &= 1+4\sin^4x\cos^4x-4\sin^2x\cos^2x \end{align*}$ The second group: $\begin{align*} -\sin^6x\cos^2x-\sin^2x\cos^6x &= -\sin^2x\cos^2x(\sin^4x+\cos^4x)\\ &= -\sin^2x\cos^2x(1-2\sin^2x\cos^2x)\\ &= -\sin^2x\cos^2x+2\sin^4x\cos^4x \end{align*}$ Add the two together to make $1+4\sin^4x\cos^4x-4\sin^2x\cos^2x-\sin^2x\cos^2x+2\sin^4x\cos^4x = 1 - 5\sin^2x\cos^2x+5\sin^4x\cos^4x$ Because this equals $\frac{11}{36}$ , we have $5\sin^4x\cos^4x- 5\sin^2x\cos^2x+\frac{25}{36}=0$ Let $\sin^2x\cos^2x = a$ so we get $5a^2- 5a+\frac{25}{36}=0 \Rightarrow a^2-a+\frac{5}{36}$ Solving the quadratic gives us $a = \frac{1 \pm \frac{2}{3}}{2}$ Because $\sin^2x\cos^2x \le \frac{1}{4}$ , we finally get $a = \frac{1 - \frac{2}{3}}{2} = \frac{1}{6}$ .

Now from the second equation, $\begin{align*} \sin^{12}x + \cos^{12}x &= (\sin^4x+\cos^4x)(\sin^8x-\sin^4x\cos^4x+\cos^8x)\\ &= (1-2\sin^2x\cos^2x)((\sin^4x+\cos^4x)^2-3\sin^4x\cos^4x)\\ &= (1-2\sin^2x\cos^2x)((1-2\sin^2x\cos^2x)^2-3\sin^4x\cos^4x) \end{align*}$ Plug in $\sin^2x\cos^2x = \frac{1}{6}$ to get $(1-2(\frac{1}{6}))(1-2(\frac{1}{6})^2-3(\frac{1}{6})^2) = \frac{13}{54}$ which yields the answer $\boxed{067}$

~ZericHang

Solution 5

Define the recursion $a_n=(\sin^2 x)^n+(\cos^2 x)^n$ We know that the characteristic equation of $a_n$ must have 2 roots, so we can recursively define $a_n$ as $a_n=p*a_{n-1}+q*a_{n-2}$ . $p$ is simply the sum of the roots of the characteristic equation, which is $\sin^2 x+\cos^2 x=1$ . $q$ is the product of the roots, which is $-(\sin^2 x)(\cos^2 x)$ . This value is not trivial and we have to solve for it. We know that $a_0=2$ , $a_1=1$ , $a_5=\frac{11}{36}$ . Solving the rest of the recursion gives

$a_2=1+2q$ $a_3=1+3q$ $a_4=1+4q+2q^2$ $a_5=1+5q+5q^2=\frac{11}{36}$ $a_6=1+6q+9q^2+2q^3$

Solving for $q$ in the expression for $a_5$ gives us $q^2+q+\frac{5}{36}=0$ , so $q=-\frac{5}{6}, -\frac{1}{6}$ . Since $q=-(\sin^2 x)(\cos^2 x)$ , we know that the minimum value it can attain is $-\frac{1}{4}$ by AM-GM, so $q$ cannot be $-\frac{5}{6}$ . Plugging in the value of $q$ into the expression for $a_6$ , we get $a_6=1-1+\frac{1}{4}-\frac{1}{108}=\frac{26}{108}=\frac{13}{54}$ . Our final answer is then $13+54=\boxed{067}$

-Natmath

Solution 6

Let $m=\sin^2 x$ and $n=\cos^2 x$ , then $m+n=1$ and $m^5+n^5=\frac{11}{36}$

$m^6+n^6=(m^5+n^5)(m+n)-mn(m^4+n^4)=(m^5+n^5)-mn(m^4+n^4)$

Now factoring $m^5+n^5$ as solution 4 yields $m^5+n^5=(m+n)(m^4-m^3n+m^2n^2-mn^3+n^4)$ $=m^4+n^4-mn(m^2-mn+n^2)=m^4+n^4-mn[(m+n)^2-3mn]=m^4+n^4-mn(1-3mn)$ .

Since $(m+n)^4=m^4+4m^3n+6m^2n^2+4mn^3+n^4$ , $m^4+n^4=(m+n)^4-2mn(2m^2+3mn+2n^2)=1-2mn(2m^2+3mn+2n^2)$ .

Notice that $2m^2+3mn+2n^2$ can be rewritten as $[\sqrt{2}(a+b)]^2-mn=2-mn$ . Thus, $m^4+n^4=1-2mn(2-mn)$ and $m^5+n^5=1-2mn(2-mn)-mn(1-3mn)=1-5mn+5(mn)^2=\frac{11}{36}$ . As in solution 4, we get $mn=\frac{1}{6}$ and $m^4+n^4=1-2*\frac{1}{6}(2-\frac{1}{6})=\frac{7}{18}$

Substitute $m^4+n^4=\frac{7}{18}$ and $mn=\frac{1}{6}$ , then $m^6+n^6=\frac{11}{36}-\frac{1}{6}*\frac{7}{18}=\frac{13}{54}$ , and the desired answer is $\boxed{067}$

Solution 7 (Official MAA)

Let $c=\sin^2x\cdot\cos^2x,$ and let $S(n)=\sin^{2n}x+\cos^{2n}x.$ Then for $n\ge 1$ $\begin{align*} S(n)&=(\sin^{2n}x+\cos^{2n}x)\cdot(\sin^2x+\cos^2x)\\ &=\sin^{2n+2}x+\cos^{2n+2}x+\sin^2x\cdot\cos^2x(\sin^{2n-2}x+\cos^{2n-2}x)\\ &=S(n+1)+cS(n-1). \end{align*}$ Because $S(0)=2$ and $S(1)=1,$ it follows that $S(2)=1-2c, S(3)=1-3c,S(4)=2c^2-4c+1,$ and $\tfrac{11}{36}=S(5)=5c^2-5c+1.$ Hence $c=\tfrac16$ or $\tfrac56,$ and because $4c=\sin^2{2x},$ the only possible value of $c$ is $\tfrac16.$ Therefore $S(6)=S(5)-cS(4)=\frac{11}{36}-\frac16\left(2\left(\frac16\right)^2-4\left(\frac16\right)+1\right)=\frac{13}{54}.$ The requested sum is $13+54=67.$

Solution 8 (Recursion)

Let $a_n=\sin^nx+\cos^nx$ for non-negative integers $n$ . Then $a_0=2$ and $a_2=1$ . In addition, $a_n=\sin^nx+\cos^nx=\left(\sin^{n-2}x+\cos^{n-2}x\right)\left(\sin^2x+\cos^2x\right)-\sin^2x\cos^2x\left(\sin^{n-4}x+\cos^{n-4}x\right)=a_{n-2}-Xa_{n-4},$ where $X=\sin^2x\cos^2x$ . So we can compute \begin{align*} a_4&=1-2X\\ a_6&=1-3X\\ a_8&=1-4X+2X^2\\ a_{10}&=1-5X+5X^2=\frac{11}{36} \end{align*}so $X=\frac{1}{6},\frac{5}{6}$ . But by the sin double angle formula, $\sin^2x\cos^2x=\frac{1}{4}\sin^22x\leq\frac{1}{4}$ , so $X=\frac{1}{6}$ . Then $a_{12}=a_{10}-Xa_8=\frac{11}{36}-\frac{1}{6}\cdot\frac{7}{18}=\frac{13}{54}$ so the answer is $\boxed{067}$ as desired.

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 The Power Of Logic

https://youtu.be/TWQn4DvBATc

~ Hayabusa1

@@ Line 1: / Line 1: @@
-The 2019 AIME I takes place on March 13, 2019.
+==Problem==
+Let <math>x</math> be a real number such that <math>\sin^{10}x+\cos^{10} x = \tfrac{11}{36}</math>. Then <math>\sin^{12}x+\cos^{12} x = \tfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
-==Problem 8==
+==Solution 1==
-==Solution(BASH)==
-Remember <math>sin^2(x)+cos^2(x)=1</math>. This means <math>sin^{10}(x)+cos^{10}(x)=(sin^2(x)+cos^2(x))sin^{10}(x)+(sin^2(x)+cos^2(x))cos^{10})(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{10}(x)+sin^{10}(x)cos^{2}(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{2}(x)[sin^8(x)+cos^8(x)]</math>. Let us look at $sin^8(x)+cos^8(x)=(sin^2(x)+cos^2(x))sin^{8}(x)+(sin^2(x)+cos^2(x))cos^{8})(x)=sin^{10}(x)+cos^{10})(x)
+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:
+<cmath>\left(\frac{1}{2}+z\right)^5+\left(\frac{1}{2}-z\right)^5=\frac{11}{36}.</cmath>
+After using binomial theorem, this simplifies to:
+<cmath>\frac{1}{16}(80z^4+40z^2+1)=\frac{11}{36}.</cmath>
+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> (observe that either choice of <math>z</math> doesn't matter). Substituting <math>z,</math> we get:
+<cmath>\sin^{12}{x}+\cos^{12}{x}=\left(\frac{1}{2}-z\right)^6+\left(\frac{1}{2}+z\right)^6=2z^6 + \frac{15z^4}{2} + \frac{15z^2}{8} + \frac{1}{32}=\frac{13}{54}.</cmath>
+Therefore, the answer is <math>13+54=\boxed{067}</math>.
+-eric2020, inspired by Tommy2002
+===Motivation===
+The motivation to substitute <math>z=\frac{1}{2}-y</math> comes so that after applying the binomial theorem to <math>y^5+(1-y)^5=\left(\frac{1}{2}+z\right)^5+\left(\frac{1}{2}-z\right)^5,</math> a lot of terms will cancel out. Note that all the terms with odd exponents in <math>\left(\frac{1}{2}+z\right)^5</math> will cancel out, while the terms with even exponents will be doubled.
+'''mathboy282'''
+==Solution 2==
+First, for simplicity, let <math>a=\sin{x}</math> and <math>b=\cos{x}</math>. Note that <math>a^2+b^2=1</math>. We then bash the rest of the problem out. Take the fifth power of this expression and get <math>a^{10}+b^{10}+5a^2b^2(a^6+b^6)+10a^4b^4(a^2+b^2)=\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1</math>. Note that we also have <math>\frac{11}{36}=a^{10}+b^{10}=(a^{10}+b^{10})(a^2+b^2)=a^{12}+b^{12}+a^2b^2(a^8+b^8)</math>. So, it suffices to compute <math>a^2b^2(a^8+b^8)</math>. Let <math>y=a^2b^2</math>. We have from cubing <math>a^2+b^2=1</math> that <math>a^6+b^6+3a^2b^2(a^2+b^2)=1</math> or <math>a^6+b^6=1-3y</math>. Next, using <math>\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1</math>, we get <math>a^2b^2(a^6+b^6)+2a^4b^4=\frac{5}{36}</math> or <math>y(1-3y)+2y^2=y-y^2=\frac{5}{36}</math>. Solving gives <math>y=\frac{5}{6}</math> or <math>y=\frac{1}{6}</math>. Clearly <math>y=\frac{5}{6}</math> is extraneous, so <math>y=\frac{1}{6}</math>. Now note that <math>a^4+b^4=(a^2+b^2)^2-2a^2b^2=\frac{2}{3}</math>, and <math>a^8+b^8=(a^4+b^4)^2-2a^4b^4=\frac{4}{9}-\frac{1}{18}=\frac{7}{18}</math>. Thus we finally get <math>a^{12}+b^{12}=\frac{11}{36}-\frac{7}{18}\cdot\frac{1}{6}=\frac{13}{54}</math>, giving <math>\boxed{067}</math>.
+'''- 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 <math>2</math>. Let <math>\sin^2x</math> and <math>\cos^2x</math> be the roots of some polynomial <math>F(a)</math>. Then, by Vieta, <math>F(a)=a^2-a+b</math> for some <math>b=\sin^2x\cdot\cos^2x</math>.
+Let <math>S_k=\left(\sin^2x\right)^k+\left(\cos^2x\right)^k</math>. We want to find <math>S_6</math>. Clearly <math>S_1=1</math> and <math>S_2=1-2b</math>. Newton sums tells us that <math>S_k-S_{k-1}+bS_{k-2}=0\Rightarrow S_k=S_{k-1}-bS_{k-2}</math> where <math>k\ge 3</math> for our polynomial <math>F(a)</math>.
+Bashing, we have
+<cmath>S_3=S_2-bS_1\Rightarrow S_3=(1-2b)-b(1)=1-3b</cmath>
+<cmath>S_4=S_3-bS_2\Rightarrow S_4=(1-3b)-b(1-2b)=2b^2-4b+1</cmath>
+<cmath>S_5=S_4-bS_3\Rightarrow S_5=(2b^2-4b+1)-b(1-3b)=5b^2-5b+1=\frac{11}{36}</cmath>
+Thus <cmath>5b^2-5b+1=\frac{11}{36}\Rightarrow 5b^2-5b+\frac{25}{36}=0, 36b^2-36b+5=0, (6b-1)(6b-5)=0</cmath>
+<math>b=\frac{1}{6} \text{ or } \frac{5}{6}</math>. Clearly, <math>\sin^2x\cdot\cos^2x\not=\frac{5}{6}</math> so <math>\sin^2x\cdot\cos^2x=b=\frac{1}{6}</math>.
+Note <math>S_4=\frac{7}{18}</math>. Solving for <math>S_6</math>, we get <math>S_6=S_5-\frac{1}{6}S_4=\frac{13}{54}</math>. Finally, <math>13+54=\boxed{067}</math>.
+==Solution 4==
+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>
+We group the first, third, and fifth term and second and fourth term. The first group:
+<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,
+<cmath>
+\begin{align*}
+\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)
+\end{align*}
+</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>
+which yields the answer <math>\boxed{067}</math>
+~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>
+==Solution 7 (Official MAA)==
+Let <math>c=\sin^2x\cdot\cos^2x,</math> and let <math>S(n)=\sin^{2n}x+\cos^{2n}x.</math> Then for <math>n\ge 1</math>
+<cmath>\begin{align*}
+S(n)&=(\sin^{2n}x+\cos^{2n}x)\cdot(\sin^2x+\cos^2x)\\
+&=\sin^{2n+2}x+\cos^{2n+2}x+\sin^2x\cdot\cos^2x(\sin^{2n-2}x+\cos^{2n-2}x)\\
+&=S(n+1)+cS(n-1).
+\end{align*}</cmath>
+Because <math>S(0)=2</math> and <math>S(1)=1,</math> it follows that <math>S(2)=1-2c, S(3)=1-3c,S(4)=2c^2-4c+1,</math> and <math>\tfrac{11}{36}=S(5)=5c^2-5c+1.</math> Hence <math>c=\tfrac16</math> or <math>\tfrac56,</math> and because <math>4c=\sin^2{2x},</math> the only possible value of <math>c</math> is <math>\tfrac16.</math> Therefore <cmath>S(6)=S(5)-cS(4)=\frac{11}{36}-\frac16\left(2\left(\frac16\right)^2-4\left(\frac16\right)+1\right)=\frac{13}{54}.</cmath> The requested sum is <math>13+54=67.</math>
+==Solution 8 (Recursion)==
+Let <math>a_n=\sin^nx+\cos^nx</math> for non-negative integers <math>n</math>. Then <math>a_0=2</math> and <math>a_2=1</math>. In addition,<cmath>a_n=\sin^nx+\cos^nx=\left(\sin^{n-2}x+\cos^{n-2}x\right)\left(\sin^2x+\cos^2x\right)-\sin^2x\cos^2x\left(\sin^{n-4}x+\cos^{n-4}x\right)=a_{n-2}-Xa_{n-4},</cmath>where <math>X=\sin^2x\cos^2x</math>. So we can compute
+\begin{align*}
+a_4&=1-2X\\
+a_6&=1-3X\\
+a_8&=1-4X+2X^2\\
+a_{10}&=1-5X+5X^2=\frac{11}{36}
+\end{align*}so <math>X=\frac{1}{6},\frac{5}{6}</math>. But by the sin double angle formula, <math>\sin^2x\cos^2x=\frac{1}{4}\sin^22x\leq\frac{1}{4}</math>, so <math>X=\frac{1}{6}</math>. Then<cmath>a_{12}=a_{10}-Xa_8=\frac{11}{36}-\frac{1}{6}\cdot\frac{7}{18}=\frac{13}{54}</cmath>so the answer is <math>\boxed{067}</math> as desired.
+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 The Power Of Logic==
+https://youtu.be/TWQn4DvBATc
+~ Hayabusa1
 ==See Also==
 {{AIME box|year=2019|n=I|num-b=7|num-a=9}}
 {{MAA Notice}}
+[[Category:Intermediate Algebra Problems]]

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