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Geometry Level 4

∏ n = 1 15 sin ⁡ ( n π 31 ) = α 2 β \large \prod _{ n=1 }^{ 15 }{ \sin { \left( \frac { n\pi }{ 31 } \right) } } =\frac { \sqrt { \alpha } }{ { 2 }^{ \beta } }

If the above equation holds true for coprime positive integers α \alpha and β \beta , find α + β \alpha +\beta .

This problem was created in celebration of the Rio 2016 Summer Olympics (Games of the XXXI Olympiad).
Image Credit: Olympic.org .


The answer is 46.

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Jonas Katona by Jonas Katona , 22, USA

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1 solution

Chew-Seong Cheong
Jul 10, 2016

The following formula (T. Drane, pers. comm., Apr. 19, 2006) [ Eqn. 24 ] states that:

∏ n = 1 m − 1 sin ⁡ ( n π m ) = m 2 m − 1 \prod_{n=1}^{m-1} \sin \left(\frac {n \pi}m \right) = \frac m{2^{m-1}}

⟹ ∏ n = 1 30 sin ⁡ ( n π 31 ) = 31 2 30 ∏ n = 1 15 sin ⁡ ( n π 31 ) ∏ n = 16 30 sin ⁡ ( n π 31 ) = 31 2 30 ∏ n = 1 15 sin ⁡ ( n π 31 ) ∏ n = 16 30 sin ⁡ ( π − n π 31 ) = 31 2 30 ∏ n = 1 15 sin ⁡ ( n π 31 ) ∏ n = 16 30 sin ⁡ ( ( 31 − n ) π 31 ) = 31 2 30 ∏ n = 1 15 sin ⁡ ( n π 31 ) ∏ n = 1 15 sin ⁡ ( n π 31 ) = 31 2 30 ( ∏ n = 1 15 sin ⁡ ( n π 31 ) ) 2 = 31 2 30 ⟹ ∏ n = 1 15 sin ⁡ ( n π 31 ) = 31 2 15 \begin{aligned} \implies \prod_{n=1}^{30} \sin \left(\frac {n \pi}{31} \right) & = \frac {31}{2^{30}} \\ \prod_{n=1}^{\color{#D61F06}{15}} \sin \left(\frac {n \pi}{31} \right) \prod_{n=\color{#D61F06}{16}}^{30} \sin \left(\frac {n \pi}{31} \right) & = \frac {31}{2^{30}} \\ \prod_{n=1}^{\color{#D61F06}{15}} \sin \left(\frac {n \pi}{31} \right) \prod_{n=\color{#D61F06}{16}}^{30} \sin \left(\color{#D61F06}{\pi} - \frac {n \pi}{31} \right) & = \frac {31}{2^{30}} \\ \prod_{n=1}^{\color{#D61F06}{15}} \sin \left(\frac {n \pi}{31} \right) \prod_{n=\color{#D61F06}{16}}^{30} \sin \left(\frac {\color{#D61F06}{(31-n)}\pi}{31} \right) & = \frac {31}{2^{30}} \\ \prod_{n=1}^{\color{#D61F06}{15}} \sin \left(\frac {n \pi}{31} \right) \prod_{n=\color{#D61F06}{1}}^{\color{#D61F06}{15}} \sin \left(\frac {\color{#D61F06}{n} \pi}{31} \right) & = \frac {31}{2^{30}} \\ \left( \prod_{n=1}^{15} \sin \left(\frac {n \pi}{31} \right) \right)^2 & = \frac {31}{2^{30}} \\ \implies \prod_{n=1}^{15} \sin \left(\frac {n \pi}{31} \right) & = \frac {\sqrt{31}}{2^{15}} \end{aligned}

⟹ α + β = 31 + 15 = 46 \implies \alpha + \beta = 31 + 15 = \boxed{46}

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Bonus:-

Prove ∏ n = 1 m − 1 sin ⁡ ( n π m ) = m 2 m − 1 \large\prod_{n=1}^{m-1} \sin \left(\frac {n \pi}m \right) = \frac m{2^{m-1}}

Rishabh Jain - 4 years, 11 months ago

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