2012 AIME I Problems/Problem 6

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 $n.$

Solution

Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{143} = z.$ $z \neq 0,$ because $\text{Im}(z)$ is given as $\sin(\text{something ugly}),$ so we can divide by $z$ to get $z^{142} = 1.$ So, $z$ must be a $142$ nd root of unity, and thus the imaginary part of $z$ will be of the form $\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}},$ where $k \in \{1, 2, \ldots, 141\}.$ But 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 $71$ and thus $n = \boxed{071.}$

2012 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2012 AIME I Problems/Problem 6

Problem 6

Solution

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