Triangle $ABC$ is equilateral with side length $6$. Suppose that $O$ is the center of the inscribed
circle of this triangle. What is the area of the circle passing through $A$, $O$, and $C$?
$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$
Construct the circle that passes through $A$, $O$, and $C$, centered at $X$. 
Also notice that $\overline{OA}$ and $\overline{OC}$ are the angle bisectors of angle $\angle BAC$ and $\angle BCA$ respectively. We then deduce $\angle AOC=120^\circ$.
Consider another point $M$ on Circle $X$ opposite to point $O$. 
As $AOCM$ is an inscribed quadrilateral of Circle $X$, $\angle AMC=180^\circ-120^\circ=60^\circ$.
Afterward, deduce that $\angle AXC=2·\angle AMC=120^\circ$.
By the Cosine Rule, we have the equation: (where $r$ is the radius of circle $X$)
$2r^2(1-\cos(120^\circ))=6^2$
$r^2=12$
The area is therefore $\pi r^2 = \boxed{\textbf{(B)}\ 12\pi}$.
~Wilhelm Z