A plane contains points $A$ and $B$ with $AB = 1$. Let $S$ be the union of all disks of radius $1$ in the plane that cover $\overline{AB}$. What is the area of $S$? $\text {(A) } 2\pi + \sqrt3 \qquad$ $ \text {(B) } \frac {8\pi}{3} \qquad$ $ \text {(C) } 3\pi - \frac {\sqrt3}{2} \qquad$ $ \text {(D) } \frac {10\pi}{3} - \sqrt3 \qquad$ $ \text {(E) }4\pi - 2\sqrt3$
Solution
The sector with radius 2 and angle $120^\circ$ (green part) has the area $\frac{4\pi}3$.
The sector with radius 1 and angle $60^\circ$( blue part) has the area $\frac{\pi}6$.
The equilateral triangle with side $AB$, which is covered twice by the green parts, has area $\frac{\sqrt3}4$.
So the answer is $2(\frac{4\pi}3+\frac{\pi}6)-2\cdot \frac{\sqrt3}4=3\pi-\frac{\sqrt3}2$.
Let $ABC$ be an equilateral triangle with side length $2$. Let $S$ be the union of all disks of radius $1$ whose centers lie inside or on $\triangle ABC$. What is the area of $S$?
