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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 $sin^8(x)+cos^8(x)=(sin^2(x)+cos^2(x))sin^{8}(x)+(sin^2(x)+cos^2(x))cos^{8})(x)=sin^{10}(x)+cos^{10})(x)
+
Remember <math>sin^2(x)+cos^2(x)=1</math>. This means <math>sin^{10}(x)+cos^{10}(x)=(sin^2(x)+cos^2(x))sin^{10}(x)+(sin^2(x)+cos^2(x))cos^{10})(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{10}(x)+sin^{10}(x)cos^{2}(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{2}(x)[sin^8(x)+cos^8(x)]</math>. Let us look at <math>sin^8(x)+cos^8(x)=(sin^2(x)+cos^2(x))sin^{8}(x)+(sin^2(x)+cos^2(x))cos^{8})(x)=sin^{10}(x)+cos^{10})(x)
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 +
==Solution 2 (Another BASH)==
 +
First, for simplicity, let </math>a=\sin{x}<math> and </math>b=\cos{x}<math>. Note that </math>a^2+b^2=1<math>. We then bash the rest of the problem out. Take the tenth power of this expression and get </math>a^{10}+b^{10}+5a^2b^2(a^6+b^6)+10a^4b^4(a^2+b^2)=\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1<math>. Note that we also have </math>\frac{11}/{36}=a^{10}+b^{10}=(a^{10}+b^{10})(a^2+b^2)=a^{12}+b^{12}+a^2b^2(a^8+b^8)<math>. So, it suffices to compute </math>a^2b^2(a^8+b^8)<math>. Let </math>y=a^2b^2<math>. We have from cubing </math>a^2+b^2=1<math> that </math>a^6+b^6+3a^2b^2(a^2+b^2)=1<math> or </math>a^6+b^6=1-3y<math>. Next, using </math>\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1<math>, we get </math>a^2b^2(a^6+b^6)+2a^4b^4=\frac{5}{36}<math> or </math>y(1-3y)+2y^2=y-y^2=\frac{5}{36}<math>. Solving gives </math>y=1<math> or </math>y=\frac{1}{6}<math>. Clearly </math>y=1<math> is extraneous, so </math>y=\frac{1}{6}<math>. Now note that </math>a^4+b^4=(a^2+b^2)-2a^2b^2=\frac{2}{3}<math>, and </math>a^8+b^8=(a^4+b^4)^2-2a^4b^4=\frac{4}{9}-\frac{1}{18}=\frac{7}{18}<math>. Thus we finally get </math>a^{12}+b^{12}=\frac{11}{36}-\frac{7}{18}*\frac{1}{6}=\frac{13}{54}<math>, giving </math>\boxed{067}$.
  
 
==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 19:16, 14 March 2019

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

Problem 8

Solution(BASH)

Remember $sin^2(x)+cos^2(x)=1$. This means $sin^{10}(x)+cos^{10}(x)=(sin^2(x)+cos^2(x))sin^{10}(x)+(sin^2(x)+cos^2(x))cos^{10})(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{10}(x)+sin^{10}(x)cos^{2}(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{2}(x)[sin^8(x)+cos^8(x)]$. 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)

==Solution 2 (Another BASH)== First, for simplicity, let$ (Error compiling LaTeX. Unknown error_msg)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}$.

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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