2008 AMC 10A Problem 17

Picture the band of width $3$ wrapped around the outside of the triangle. It splits into two kinds of pieces.

Along each side, the points within $3$ units of that side (measured perpendicularly, outside the triangle) form a $6 \times 3$ rectangle. Three sides give
$$3 \cdot (6 \cdot 3) = 54.$$

At each vertex the two rectangles leave a wedge-shaped gap. Points there are within $3$ of the vertex itself, so the gap is a sector of radius $3$. Its angle is $360^\circ$ minus the triangle's $60^\circ$ interior angle minus the two $90^\circ$ rectangle corners, which is $120^\circ$. Three such sectors total $360^\circ$, a full disk of radius $3$:
$$\pi \cdot 3^2 = 9\pi.$$

The region has area $54 + 9\pi$, so the answer is $\boxed{\textbf{(B)}\ 54+9\pi}$.

The set of points within distance $d$ of a convex polygon is the polygon plus rectangles of height $d$ on each side plus sectors at each vertex, and the sector angles are the exterior angles, which always sum to $360^\circ$. So for any convex polygon the outer band has area (perimeter)$\cdot d + \pi d^2$. Here that is $18 \cdot 3 + 9\pi$.

The answer must contain a $\pi$ term from the rounded corners and a non-$\pi$ term from the straight parts, and the region is clearly smaller than a full disk of radius $3 + 2\sqrt3$ (choice D). Only (B) and (C) fit that shape, and (C) has too much area.

• Using 60-degree sectors at the vertices (the interior angle) instead of 120-degree sectors, or leaving out the sectors entirely.
• Including the triangle's own area $9\sqrt3$, which the problem excludes ("outside the triangle").