Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle ABC$ lies outside the circle? $\textbf{(A) } \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad$ $ \textbf{(B) } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad$ $ \textbf{(C) } \frac{1}{2} \qquad$ $ \textbf{(D) } \sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad$ $ \textbf{(E) } \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}$
Solution
Let $O$ be the center. Since $\angle ABO=\angle ACO=90^\circ$ and $\angle A=60^\circ$, $\angle BOC=120^\circ$.
Assume $AB=AC=6$. $AO$ bisects $\angle A$, so $OB=AB\tan30^\circ=2\sqrt3$.
The part of the disk inside the triangle is the segment cut off by $BC$, with area equal to the sector minus $\triangle OBC$: $\frac13\cdot \pi \cdot (2\sqrt3)^2-\frac12\cdot (2\sqrt3)^2\cdot \sin 120^\circ=4\pi-3\sqrt3$.
So the area lies outside the circle = $9\sqrt3 - (4\pi-3\sqrt3)=12\sqrt3-4\pi$.
The ratio is $(12\sqrt3-4\pi) : 9\sqrt3=\frac{4}{3}-\frac{4\sqrt{3}\pi}{27}$.
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?
