In isosceles triangle $ABC$, $AB=AC$ and $\angle BAC=120^\circ$. A circle is tangent to $\overline{AB}$ at $B$ and to $\overline{AC}$ at $C$. What fraction of the area of $\triangle ABC$ lies inside the circle?

1. Set $AB=AC=1$. The circle's center $O$ lies on the angle bisector of $\angle A$, so the right triangle $ABO$ has $\angle BAO=60^\circ$ and $OB=\sqrt3$.
2. The central angle $\angle BOC=60^\circ$. The circular segment inside $\triangle ABC$ therefore has area $\frac12(\sqrt3)^2\cdot\frac{\pi}{3}-\frac12(\sqrt3)^2\sin60^\circ=\frac{\pi}{2}-\frac{3\sqrt3}{4}$.
3. Since $[ABC]=\frac12\sin120^\circ=\frac{\sqrt3}{4}$, the requested fraction is $\boxed{\frac{2\sqrt3\pi}{3}-3}$.