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

Revision as of 23:53, 11 January 2020

Problem 8

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 $z=1/2-y$ , we can simplify the equation to $(1/2+z)^5+(1/2-z)^5=\frac{11}{36}$ . After using binomial theorem, this simplifies to $\frac{1}{16}(80z^4+40z^2+1)=11/36$ . If we use the quadratic formula, we obtain the that $z^2=\frac{1}{12}$ , so $z=\pm\frac{1}{2\sqrt{3}}$ . By plugging z into $(1/2-z)^6+(1/2+z)^6$ (which is equal to $\sin^{12}{x}+\cos^{12}{x}$ , we can either use binomial theorem or sum of cubes to simplify, and we end up with $\frac{13}{54}$ . Therefore, the answer is $\boxed{067}$ .

eric2020, inspired by Tommy2002

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 tenth 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 $a^{12}+b^{12}=\frac{11}{36}-\frac{7}{18}*\frac{1}{6}=\frac{13}{54}$ , 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: $\sin^8+\sin^4x\cos^4x+\cos^8x = (\sin^4x+\cos^4x)^2-\sin^4x\cos^4x)= (1 - 2\sin^2x\cos^2x)^2-\sin^4x\cos^4x) =1+4\sin^4x\cos^4x-4\sin^2x\cos^2x$ The second group: $-\sin^6x\cos^2x-\sin^2x\cos^6x = -\sin^2x\cos^2x(\sin^4x+\cos^4x)=-\sin^2x\cos^2x(1-2\sin^2x\cos^2x) = -\sin^2x\cos^2x+2\sin^4x\cos^4x$ 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, $\sin^{12}x + \cos^{12}x = (\sin^4x+\cos^4x)(\sin^8x-\sin^4x\cos^4x+\cos^8x)=(1-2\sin^2x\cos^2x)((\sin^4x+\cos^4x)^2-3\sin^4x\cos^4x)=(1-2\sin^2x\cos^2x)((1-2\sin^2x\cos^2x)^2-3\sin^4x\cos^4x)$ 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

@@ Line 17: / Line 17: @@
 ==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, <math>F(a)=a^2-a+b</math> for some <math>b=\sin^2x\cdot\cos^2x</math>.
+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>.

2019 AIME I (Problems • Answer Key • Resources)
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