Maths Olympiad Prep

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Problem 962

AMC 12 late, AIME early
Algebra Difficulty 4.8 Prove it China Mathematical Competition · China

The sum of all the positive integers nn satisfying 14<sin⁡πn<13\frac{1}{4} < \sin \frac{\pi}{n} < \frac{1}{3} is ______.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

As sin⁡x\sin x is a convex function for x∈(0,π6)x \in (0, \frac{\pi}{6}), we have 3πx<sin⁡x<x\frac{3}{\pi}x < \sin x < x. Then
sin⁡π13<π13<14,sin⁡π12>3π×π12=14, \sin \frac{\pi}{13} < \frac{\pi}{13} < \frac{1}{4}, \sin \frac{\pi}{12} > \frac{3}{\pi} \times \frac{\pi}{12} = \frac{1}{4},
sin⁡π10<π10<13,sin⁡π9>3π×π9=13, \sin \frac{\pi}{10} < \frac{\pi}{10} < \frac{1}{3}, \sin \frac{\pi}{9} > \frac{3}{\pi} \times \frac{\pi}{9} = \frac{1}{3},
that is
sin⁡π13<14<sin⁡π12<sin⁡π11<sin⁡π10<13<sin⁡π9. \sin \frac{\pi}{13} < \frac{1}{4} < \sin \frac{\pi}{12} < \sin \frac{\pi}{11} < \sin \frac{\pi}{10} < \frac{1}{3} < \sin \frac{\pi}{9}.
Therefore, all the possible values of positive integers nn are
10,11,1210, 11, 12, and their sum is 33.
The answer is 33.

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