Difference between revisions of "2012 AIME I Problems/Problem 6"

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==Problem 6==
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==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>
  
==Solutions==
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==Solution==
  
===Solution 1===
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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> because 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>
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{2k\pi}{142}} = \sin{\frac{k\pi}{71}}</math> for some <math>k</math> with <math>0 \le k < 142.</math> But 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> and thus <math>n = \boxed{071.}</math>
 
  
===Solution 2===
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==Video Solution==
Note that <math>w^{143}=w</math> and similar for <math>z</math>, and they are not equal to <math>0</math> because the question implies the imaginary part is positive. Thus <math>w^{142}=z^{142}=1</math>, so the imaginary part of each is of the form <math>\sin\left(\frac{2 \pi k}{142}\right)</math> where <math>k</math> is a positive integer between <math>1</math> and <math>141</math> inclusive. This simplifies to <math>\sin\left(\frac{\pi k}{71}\right)</math>. Therefore, the imaginary part of <math>z</math> is <math>\sin\left(\frac{\pi m}{71}\right)</math>, where <math>0 \le m < 71</math>. Since <math>071</math> is prime, it is the only possible denominator, so <math>\boxed{n = 71}</math>.
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https://www.youtube.com/watch?v=cQmmkfZvPgU&t=30s
  
 
== 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}}
 
{{MAA Notice}}

Latest revision as of 19:18, 24 January 2021

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 $z \neq 0,$ because 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 Solution

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

See also

2012 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AIME Problems and Solutions

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