The sum of all integers $n$ satisfying $\frac{1}{4} < \sin\frac{\pi}{n} < \frac{1}{3}$ Asked in China Junior Math Olympiad. (No calculators or tables allowed.)

Find the sum of all integers $n$ satisfying the following inequality:
  $$\frac{1}{4} < \sin\frac{\pi}{n} < \frac{1}{3}$$


$\dfrac{1}{4} < \sin\dfrac{\pi}{n}
\implies
\sin\dfrac{\pi}{n} > 0
$
$
-\pi < \dfrac{\pi}{n} < \pi
\implies
n > 0
$
$
n=1
$ does not work
$
\implies n \ge 2
$
$
-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}
\implies
\sin(x) 
$ increasing
$
\sin\dfrac{\pi}{6} = \dfrac{1}{2} > \dfrac{1}{3}
\implies
n \ge 7
$
$\dfrac{1}{4} < \sin x < x
\implies
n < 4\pi
\implies
n \le 12
$
$
\dfrac{1}{3} > \sin x \ge x - \dfrac{x^3}{6}
\implies
n \ge 10
$
But we still need to prove that $n=10,11,12$ work.
It is enough to prove that $n=10$ and $n=12$ work. There are formulas for $\sin\dfrac{\pi}{n}$ in these two cases:
$$
\sin\dfrac{\pi}{10} = \dfrac{\sqrt5 - 1}{4},
\qquad
\sin\dfrac{\pi}{12} = \dfrac{\sqrt3 - 1}{2 \sqrt2}
$$
I don't know whether this whole approach is feasible in a test...
