You're Trapped!

It helps for visualization problems such as this one to consider "extreme cases" and see how they are related. In my solution, I will focus on positions where the circle is touching two vertices of the equilateral triangle, call these extreme circles, and then consider how we can go from one of these positions to another.

It is easy to see that there are three such possibilities. To get from one to another, we simply rotate the circle about one of the vertices of the triangle until the last vertex touches the circle. Consider the following diagram:

It depicts the scenario described, where points $D$ and $F$ are the centers of the two extreme circles, and have diameters $AG$ and $AH$ respectively. More importantly, since both diameters connect at the pivot point, the arc $GH$ represent the path taken by point $G$ as the circle is rotated.

It follows that since point $G$ is the point furthest on the first extreme circle from point $A$ , this path swept out lies on the outermost bound of points that lie inside any possible circle enclosing the equilateral triangle.

Let's analyze arc $GH$ further: Using coordinate geometry we can see that point $D$ lies at $(6,6)$ , and so $\angle DAB = 45^{\circ}$ . Since $\Delta ABC$ is equilateral, we know that $\angle CAB = 60^{\circ}$ . It then follows that $\angle CAD = 15^{\circ}$ which, by symmetry, gives us that $\angle BAF = 15^{\circ}$ . We can then conclude that $\angle DAF = 30^{\circ}$ .

Now, let's extend the diagram to show all three possible rotations at once:

This is a depiction of the entire possible area of points enclosed by the circle. The outermost red arcs represent the section where the "swept" region is the outermost bound, and the purple arcs represent the section where the "extreme circles" are the outermost bound.

Overall, the region is made up of three purple sectors, the equilateral triangle $\Delta DEF$ , and the three remaining regions, bounded by the red outermost arcs and the remaining red segments, which we will call the blobs.

Purple Sector: By the Vertical Angle Theorem, since $ADG$ and $BDL$ are line segments, we know that $\angle ADB = \angle GDL$ . Since $AD$ and $BD$ are both radii of the same extreme circle centered at $D$ , we know $AD=BD=6\sqrt{2}$ . We also know that $AB=12$ , since it is a side of the original equilateral triangle. This shows that $\Delta ADB$ is a right isosceles triangle, which leads to $\angle GDL = 90^{\circ}$ .

Each purple arc is therefore $\frac{1}{4}$ of a full extreme circle, meaning that each of their areas is $A_{arc}=18\pi$ .

Triangle DEF: Consider triangle $\Delta ADF$ . Two of its sides $AD$ and $AF$ are radii of extreme circles, and we established before that $\angle DAF=30^{\circ}$ . Therefore, by law of cosines $DF^{2}=(2)(6\sqrt{2})^{2}(1-\cos(30))=144\left(1-\frac{\sqrt{3}}{2}\right)$ .

$DF$ is a side length of equilateral triangle $DEF$ , and the area of an equilateral triangle is $\frac{s^{2}\sqrt{3}}{4}$ , so we can conclude that the area of $\Delta DEF$ is $A_{\Delta}= 36\sqrt{3}-54$

The Blob: Each blob is composed of a sector, minus the area of a triangle. Explicitly, blob $DFGH$ is composed of sector $AGH$ , minus triangle $\Delta ADF$ . Numerically, $AG$ and $AH$ are both diameters of extreme circles, so they have length $12\sqrt{2}$ . We already established that $AD=AF=6\sqrt{2}$ and $\angle DAF=30^{\circ}$ . By area of a sector and SAS area formula, the area of the blob is then $A_{blob}=\pi (12\sqrt{2})^{2} \left(\frac{30}{360} \right)-\frac{(6\sqrt{2})^{2}\sin(30)}{2}=24\pi-18$

Total Area: The total region is then comprised of three congruent purple sectors, three congruent blobs, and $\Delta DEF$ . This makes the area of the region

$A=3A_{arc}+3A_{blob}+A_{\Delta}=126\pi+36\sqrt{3}-108 \approx 350.19 \Rightarrow \lfloor A \rfloor = \boxed{350}$