Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 2)
Line 7: Line 7:
  
 
==Solution 2==
 
==Solution 2==
 +
 +
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}</math>.
 +
 +
-Emathmaster
  
 
==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 19:35, 14 March 2019

The 2019 AIME I takes place on March 13, 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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_8&oldid=104390"