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 "2012 AIME I Problems/Problem 6"
m (→‎Solution)
 
(27 intermediate revisions by 15 users not shown)
Line 1: Line 1:
−
==Problem 6==
+
==Problem ==
 
The complex numbers <math>z</math> and <math>w</math> satisfy <math>z^{13} = w,</math> <math>w^{11} = z,</math> and the imaginary part of <math>z</math> is <math>\sin{\frac{m\pi}{n}}</math>, for relatively prime positive integers <math>m</math> and <math>n</math> with <math>m<n.</math> Find <math>n.</math>
 
The complex numbers <math>z</math> and <math>w</math> satisfy <math>z^{13} = w,</math> <math>w^{11} = z,</math> and the imaginary part of <math>z</math> is <math>\sin{\frac{m\pi}{n}}</math>, for relatively prime positive integers <math>m</math> and <math>n</math> with <math>m<n.</math> Find <math>n.</math>
  
 
==Solution==
 
==Solution==
−
Substituting the first equation into the second, we find that <math>(z^{13})^{11} = z</math> and thus <math>z^{142} = 1.</math> So <math>z</math> must be a <math>142</math>nd root of unity, and thus the imaginary part of <math>z</math> will be <math>\sin{\frac{2m\pi}{142}} = \sin{\frac{m\pi}{71}}</math> for some <math>m</math> with <math>0 \le m < 142.</math> But note that since <math>71</math> is prime and <math>m<71</math> by the conditions of the problem, the denominator in the argument of this value will always be <math>71</math> and thus <math>n = \boxed{071.}</math>
+
 
 +
Substituting the first equation into the second, we find that <math>(z^{13})^{11} = z</math> and thus <math>z^{143} = z.</math> We know that <math>z \neq 0,</math> since we are given the imaginary part of <math>z</math>, so we can divide by <math>z</math> to get <math>z^{142} = 1.</math> So, <math>z</math> must be a <math>142</math>nd root of unity, and thus, by De Moivre's theorem, the imaginary part of <math>z</math> will be of the form <math>\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}},</math> where <math>k \in \{1, 2, \ldots, 70\}.</math> Note that <math>71</math> is prime and <math>k<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71.</math> Thus, <math>n = \boxed{071}.</math>
 +
 
 +
==Video Solutions==
 +
 
 +
https://www.youtube.com/watch?v=cQmmkfZvPgU&t=30s
 +
 
 +
https://www.youtube.com/watch?v=DMka35X-3WI&list=PLyhPcpM8aMvIo_foUDwmXnQClMHngjGto&index=6 (Solution by Richard Rusczyk) - AMBRIGGS
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=5|num-a=7}}
 
{{AIME box|year=2012|n=I|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Latest revision as of 15:04, 14 May 2025

Problem

The complex numbers $z$ and $w$ satisfy $z^{13} = w,$ $w^{11} = z,$ and the imaginary part of $z$ is $\sin{\frac{m\pi}{n}}$, for relatively prime positive integers $m$ and $n$ with $m<n.$ Find $n.$

Solution

Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{143} = z.$ We know that $z \neq 0,$ since we are given the imaginary part of $z$, so we can divide by $z$ to get $z^{142} = 1.$ So, $z$ must be a $142$nd root of unity, and thus, by De Moivre's theorem, the imaginary part of $z$ will be of the form $\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}},$ where $k \in \{1, 2, \ldots, 70\}.$ Note that $71$ is prime and $k<71$ by the conditions of the problem, so the denominator in the argument of this value will always be $71.$ Thus, $n = \boxed{071}.$

Video Solutions

https://www.youtube.com/watch?v=cQmmkfZvPgU&t=30s

https://www.youtube.com/watch?v=DMka35X-3WI&list=PLyhPcpM8aMvIo_foUDwmXnQClMHngjGto&index=6 (Solution by Richard Rusczyk) - AMBRIGGS

See also

2012 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_6&oldid=248393"