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Algebra Difficulty 4.8 AIME Prove it United States

Problem:
Let ω=cos2π727+isin2π727\omega = \cos \frac{2\pi}{727} + i \sin \frac{2\pi}{727}. The imaginary part of the complex number
k=813(1+ω3k1+ω23k1) \prod_{k=8}^{13}\left(1+\omega^{3^{k-1}}+\omega^{2 \cdot 3^{k-1}}\right)
is equal to sinα\sin \alpha for some angle α\alpha between π2-\frac{\pi}{2} and π2\frac{\pi}{2}, inclusive. Find α\alpha.

Solution

Solution:
727=362727 = 3^6 - 2. Our product telescopes to
1ω3131ω37=1ω121ω6=1+ω6, \frac{1-\omega^{3^{13}}}{1-\omega^{3^{7}}} = \frac{1-\omega^{12}}{1-\omega^{6}} = 1 + \omega^{6},
which has imaginary part sin12π727\sin \frac{12\pi}{727}, giving α=12π727\alpha = \frac{12\pi}{727}.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.