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

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==Problem 8==
 
==Problem 8==
 
==Solution(BASH)==
 
==Solution(BASH)==
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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)
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NOT ALLOWED!
  
 
==Solution 2 (Another BASH)==
 
==Solution 2 (Another BASH)==

Revision as of 19:25, 14 March 2019

The 2019 AIME I takes place on March 13, 2019.

Problem 8

Solution(BASH)

NOT ALLOWED!

Solution 2 (Another BASH)

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

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

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