2000 AMC 10 Problem 18

Charlyn sees every point within $1$ km of some point of the square's boundary. Split that region into the part inside the square and the part outside.

Inside: a point is visible unless it is more than $1$ km from all four sides, i.e. unless it lies in the central $3 \times 3$ square. Visible inside area: $5^2 - 3^2 = 16$.

Outside: along each side she sees a $5 \times 1$ rectangle, for $4 \cdot 5 = 20$. At each corner, the points within $1$ km of the corner vertex fill in a quarter-disc of radius $1$; the four quarter-discs form one full circle of area $\pi$.

Total: $16 + 20 + \pi = 36 + \pi \approx 39.14$, which rounds to $39$.

The answer is $\boxed{\textbf{(C)}\ 39}$.

"All points within distance $r$ of a path" is a band of width $2r$ around the path. Along straight stretches the band is rectangular; at a convex corner the outside of the band is rounded by a circular sector, while the inside of the band has a sharp corner that the two adjacent strips already cover. Decompose into rectangles, a hollow square, and sectors, then add.

• Forgetting that the 3 by 3 center is out of sight, or treating the outer corners as full unit squares instead of quarter-circles (which gives 40).
• Modeling the visible region as four $5 \times 2$ strips centered on the sides ($40$), which double-counts the inner corners and squares off the outer ones.
• Rounding $36 + \pi$ with a crude $\pi \approx 4$ to get $40$.