2004 AMC 12A Problem 24

A radius-$1$ disk covers segment $\overline{AB}$ exactly when its center is within $1$ of both $A$ and $B.$ That region $R$ is the lens where the two unit circles centered at $A$ and $B$ overlap. Each unit circle passes through the other’s center, so the lens is bounded by two $120^\circ$ arcs. Two $120^\circ$ sectors of area $\tfrac{\pi}{3}$ overlap in two equilateral triangles of total area $\tfrac{\sqrt3}{2},$ giving $R$ area $\tfrac{2\pi}{3} - \tfrac{\sqrt3}{2}.$ The set $S$ consists of all points within $1$ of $R.$ Beyond $R$ itself, this adds two $60^\circ$ sectors of radius $1$ (each area $\tfrac{\pi}{6}$) and two $120^\circ$ annuli of outer radius $2$ and inner radius $1$ (each area $\pi$). Therefore the area of $S$ is $$\begin{aligned} &\left(\tfrac{2\pi}{3} - \tfrac{\sqrt3}{2}\right) + 2 \cdot \tfrac{\pi}{6} \\ &\quad {}+ 2\pi = 3\pi - \tfrac{\sqrt3}{2}. \end{aligned}$$ Thus, the correct answer is C.