A [plane](https://artofproblemsolving.com/wiki/index.php/Plane) contains points $A$ and $B$ with $AB = 1$. Let $S$ be the [union](https://artofproblemsolving.com/wiki/index.php/Union) of all disks of radius $1$ in the plane that cover $\overline{AB}$. What is the area of $S$?
$\textbf {(A) } 2\pi + \sqrt3 \qquad \textbf {(B) } \frac {8\pi}{3} \qquad \textbf {(C) } 3\pi - \frac {\sqrt3}{2} \qquad \textbf {(D) } \frac {10\pi}{3} - \sqrt3 \qquad \textbf {(E) }4\pi - 2\sqrt3$
As the red circles move about segment $AB$, they cover the area we are looking for.
On the left side, the circle must move around pivoted on $B$.
On the right side, the circle must move pivoted on $A$
However, at the top and bottom, the circle must lie on both A and B, giving us our upper and lower bounds.
This egg-like shape is $S$.


The area of the region can be found by dividing it into several sectors, namely
\begin{align*} A &= 2(\mathrm{Blue\ Sector}) + 2(\mathrm{Red\ Sector}) - 2(\mathrm{Equilateral\ Triangle}) \\ A &= 2\left(\frac{120^\circ}{360^\circ} \cdot \pi (2)^2\right) + 2\left(\frac{60^\circ}{360^\circ} \cdot \pi (1)^2\right) - 2\left(\frac{(1)^2\sqrt{3}}{4}\right) \\ A &= \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} \\ A &= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*}