2019 AIME I Problems/Problem 8

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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 (ProblemsAnswer KeyResources)
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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