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

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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 theorem, we obtain that $z^2=\pm\frac{1}{12}$ , 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 $a^4+b^4=(a^2+b^2)-2a^2b^2=\frac{2}{3}$ , 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, $F(a)=a^2-a+b$ 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 (Not Fun)

We let $a = \sin^2(x)$ and $b = \cos^2(x)$ , so we have $a+b=1$ and $a^5 + b^5 = \frac{11}{36}$ . Noticing that $ab$ might be a useful value to find, we let $c = ab$ . Then we can work our way up to find $c$ . $a + b = 1$ $(a+b)(a+b) = 1(a+b)$ $a^2 + 2ab + b^2 = 1$ $a^2 + b^2 = -2c + 1$ $(a+b)(a^2 + b^2) = -2c + 1$ $a^3 + ab^2 + a^2b + b^3 = -2c + 1$ $a^3 + b^3 + ab(a + b) = -2c + 1$ $a^3 + b^3 + c = -2c + 1$ $a^3 + b^3 = -3c + 1$ $(a + b)(a^3 + b^3) = -3c + 1$ $a^4 + ab^3 + a^3b + b^4 = -3c + 1$ $a^4 + b^4 + ab(a^2 + b^2) = -3c + 1$ $a^4 + b^4 + c(-2c + 1) = -3c + 1$ $a^4 + b^4 - 2c^2 + c = -3c + 1$ $a^4 + b^4 = 2c^2 - 4c + 1$ $a^5 + ab^4 + a^4b + b^5 = 2c^2 - 4c + 1$ $a^5 + b^5 + ab(a^3 + b^3) = 2c^2 - 4c + 1$ $\frac{11}{36} + c(-3c + 1) = 2c^2 - 4c + 1$ $\frac{11}{36} = 5c^2 - 5c + 1$ $5c^2 - 5c + \frac{25}{36} = 0$ using quadform you get $c = \frac{1}{6}$ or $c = \frac{5}{6}$ . Since $c = \sin^2(x)\cos^2(x) = (\sin(x)\cos(x))^2 = (\frac{\sin(2x)}{2})^2$ , and since $\sin(2x)$ can't exceed 1, $c$ can't exceed $(\frac{1}{2})^2 = \frac{1}{4}$ . Clearly, $c = \frac{1}{6}$ . And finally, $a^5 + b^5 = (a + b)(a^5 + b^5)$ $a^5 + b^5 = a^6 + ab^5 + a^5b + b^6$ $\frac{11}{36} = a^6 + b^6 + c(a^4 + b^4)$ looking back to previous results, we see that $a^4 + b^4 = 2c^2 - 4c + 1 = \frac{14}{36}$ (it's easier not to simplify the fraction). $\frac{11}{36} = a^6 + b^6 + c(\frac{14}{36}) = a^6 + b^6 + \frac{14}{216}$ $a^6 + b^6 = \frac{11}{36} - \frac{14}{216} = \frac{13}{54}$ which yields the answer $\boxed{067}$ .

~PCampbell

@@ Line 31: / Line 31: @@
 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 (Fun)==
+==Solution 4 (Not Fun)==
 We let <math>a = \sin^2(x)</math> and <math>b = \cos^2(x)</math>, so we have <math>a+b=1</math> and <math>a^5 + b^5 = \frac{11}{36}</math>. Noticing that <math>ab</math> might be a useful value to find, we let <math>c = ab</math>. Then we can work our way up to find <math>c</math>.
 <cmath>a + b = 1</cmath>

2019 AIME I (Problems • Answer Key • Resources)
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