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2012 AIME I Problems/Problem 6

Revision as of 00:52, 26 February 2014 by MSTang (talk | contribs) (Solution 1)

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 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 $\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}}$ for some $k$ with $0 \le k < 142.$ 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 $n = \boxed{071.}$

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 (ProblemsAnswer KeyResources)
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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