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

(→‎Solution 2)
(→‎Solution 3 (Newton Sums))
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==Solution 3 (Newton Sums)==
 
==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 <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 2. 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>.
 +
 +
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} 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>.
  
 
==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 13:50, 15 March 2019

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 $(1/2+z^5)+(1/2-z^5)^5=\frac{11}{36}$. 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 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 $y=1$ or $y=\frac{1}{6}$. Clearly $y=1$ 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 \[5b^2-5b+1=\frac{11}{36}\Rightarrow 5b^2-5b+\frac{25}{36}=0, 36b^2-36b+5=0, (6b-1)(6b-5)=0\] $b=\frac{1}{6} or \frac{5}{6}$. 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, $13+54=\boxed{067}$.

See Also

2019 AIME I (Problems • Answer Key • Resources)
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Problem 9
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