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

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===Solution 2===
 
===Solution 2===
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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>, and <math>071</math> is prime, so it is the only possible denominator, and thus is the answer.
+
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>.
  
 
== 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}}

Revision as of 11:29, 24 December 2012

Problem 6

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 $n.$

Solutions

Solution 1

Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{142} = 1.$ So $z$ must be a $142$nd root of unity, and thus the imaginary part of $z$ will be $\sin{\frac{2m\pi}{142}} = \sin{\frac{m\pi}{71}}$ for some $m$ with $0 \le m < 142.$ But note that $71$ is prime and $m<71$ by the conditions of the problem, so the denominator in the argument of this value will always be $71$ and thus $n = \boxed{071.}$

Solution 2

Note that $w^{143}=w$ and similar for $z$, and they are not equal to $0$ because the question implies the imaginary part is positive. Thus $w^{142}=z^{142}=1$, so the imaginary part of each is of the form $\sin\left(\frac{2 \pi k}{142}\right)$ where $k$ is a positive integer between $1$ and $141$ inclusive. This simplifies to $\sin\left(\frac{\pi k}{71}\right)$. Therefore, the imaginary part of $z$ is $\sin\left(\frac{\pi m}{71}\right)$, where $0 \le m < 71$. Since $071$ is prime, it is the only possible denominator, so $\boxed{n = 71}$.

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
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