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?
$\textbf{(A) } \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \textbf{(B) } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad \textbf{(E) } \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}$
Let the radius of the circle be $r$, and let its center be $O$. Since $\overline{AB}$ and $\overline{AC}$ are tangent to circle $O$, then $\angle OBA = \angle OCA = 90^{\circ}$, so $\angle BOC = 120^{\circ}$. Therefore, since $\overline{OB}$ and $\overline{OC}$ are equal to $r$, then (pick your favorite method) $\overline{BC} = r\sqrt{3}$. The area of the equilateral triangle is $\frac{(r\sqrt{3})^2 \sqrt{3}}4 = \frac{3r^2 \sqrt{3}}4$, and the area of the sector we are subtracting from it is $\frac 13 \pi r^2 - \frac 12 \cdot r\sqrt{3} \cdot \frac{r}2 = \frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4$. The area outside of the circle is $\frac{3r^2 \sqrt{3}}4-\left(\frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4\right) = r^2 \sqrt{3} - \frac{\pi r^2}3$. Therefore, the answer is \[\frac{r^2 \sqrt{3} - \frac{\pi r^2}3}{\frac{3r^2 \sqrt{3}}4} = \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}\]

Note
The sector angle is $120$ because $\angle B$ and $\angle C$ are both 90 degrees meaning $\angle B + \angle C = 180^\circ$, so $ABCO$ is cyclic. Thus, the angle is $180-60=120^\circ$
~mathboy282
Alternately, $\angle ABC$ is $60^\circ$ and $\angle ABO$ is $90^\circ$, making $\angle CBO$ $30^\circ$. Symmetry allows us to use the same argument to get $\angle BCO = 30^\circ$. Since the interior angles of $\triangle BCO$ must sum to $180^\circ$, that leaves $120^\circ$ for central angle $\angle BOC$.
~wescarroll

Multiple Choice Shortcut
Assuming WLoG that the equilateral triangle's side length $s$ and therefore area $A$ are algebraic ("$\pi$-free"):
The "crust" is a circle sector minus a triangle, so its area is $a \pi - b$, where $a$ and $b$ are algebraic. Thus the answer is $(A - (a \pi -b))/A = (A+b)/A - a\pi/A$.
Once you see that $s$ is $\sqrt{3} \times$ the circle's radius, and that the circle's 30°-30°-120° triangle is two halves of an equilateral triangle, infer that the smaller circle-sector triangle's area is $A/(\sqrt 3)^2 = A/3$, and so the algebraic part of the answer $(A+b)/A = (A - (-A/3)) /A = 4/3$.
The transcendental ("$\pi$") part of the answer is $-a \pi /A$, and since $a$ and $A$ are algebraic,  $\textbf{(E)}$ is the only compatible answer choice.