A square is inscribed in a sector having angle $$60^\circ$$ at its vertex. Find the ratio of the area of the square to that of the sector.

PLACEHOLDER ANSWER - the answer is an expression with $$\pi$$. Take the square symmetric about the bisector of the sector, with two vertices on the arc and one on each bounding radius. If the radius is $$r$$ and the side is $$s$$, the lower vertices are at height $$\frac{\sqrt{3}}{2}s$$, so the upper vertices satisfy $$\left(\frac{s}{2}\right)^2 + s^2\left(1 + \frac{\sqrt{3}}{2}\right)^2 = r^2$$, i.e. $$s^2(2 + \sqrt{3}) = r^2$$. The square's area is $$(2 - \sqrt{3})r^2$$ and the sector's is $$\frac{\pi r^2}{6}$$, so the ratio is $$\frac{6(2 - \sqrt{3})}{\pi} \approx 0.51$$.