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Algebra Difficulty 4.8 AIME Prove it China

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

Solution

As sinx\sin x is a convex function for x(0,π6)x \in (0, \frac{\pi}{6}), we have 3πx<sinx<x\frac{3}{\pi}x < \sin x < x. Then
sinπ13<π13<14,sinπ12>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},
sinπ10<π10<13,sinπ9>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
sinπ13<14<sinπ12<sinπ11<sinπ10<13<sinπ9. \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.

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