2008 AMC 10A Problem 17

Picture a band, $3$ units wide, wrapped around the outside of the triangle. Split it into straight parts and corner parts.

1. Straight parts. Next to each side, the points within $3$ units of that side form a rectangle $6$ long and $3$ wide. Three sides give
$$3 \times (6 \times 3) = 54.$$
2. Corner parts. At each corner of the triangle, the two rectangles do not meet; there is a gap. Points in the gap are within $3$ units of the corner point itself, so the gap is a slice of a circle of radius $3$.
3. How big is each slice? Around the corner point there are $360^\circ$ in total. The triangle takes $60^\circ$, and the two rectangles each take $90^\circ$. The slice gets what is left:
$$360^\circ - 60^\circ - 90^\circ - 90^\circ = 120^\circ.$$
4. Three slices of $120^\circ$ make $360^\circ$, a full circle of radius $3$. Its area is
$$\pi \times 3^2 = 9\pi.$$
5. The whole band has area $54 + 9\pi$.

The answer is $\boxed{\textbf{(B)}\ 54+9\pi}$.

A band of fixed width around any shape with straight sides and no dents is always rectangles on the sides plus circle slices at the corners. The slices always add up to one full circle, because walking once around the shape you turn a full $360^\circ$. So the band's area is (perimeter) × (width) + $\pi$ × (width)$^2$; here $18 \times 3 + 9\pi$.