2000 AMC 10 Problem 18

1. Charlyn can see every point that is within $1$ km of some point on her path. Split what she sees into the part inside the square and the part outside.
2. Inside the square. A point inside is out of sight only if it is more than $1$ km from every side. Those points make a smaller square in the middle, $5 - 1 - 1 = 3$ km on a side. So she sees the $5 \times 5$ square minus the $3 \times 3$ middle:
$$25 - 9 = 16.$$
3. Outside, along the sides. Walking along one side, she sees a strip $5$ km long and $1$ km wide beyond it. Four sides give $4 \times 5 = 20$.
4. Outside, at the corners. Beyond each corner, the strips do not reach. The points she can see there are within $1$ km of the corner point itself, which is a quarter of a circle of radius $1$. Four quarter-circles make one full circle, with area $\pi \cdot 1^2 = \pi \approx 3.14$.
5. Add up.
$$16 + 20 + \pi \approx 39.14,$$
which rounds to $39$.

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

"Everything within $1$ km of a path" is a band around the path. Along straight parts the band is made of rectangles. At an outside corner it is rounded, because the points there are measured from the corner itself. Break the region into rectangles, the square frame and circle pieces, then add. Drawing a quick sketch of one corner makes the rounded part obvious.

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 $\boxed{\textbf{(C)}\ 39}$.