2000 AMC 10 Problem 18

P16-20~4 minPrint

Quadrilaterals & Polygon Areas Circles Cut the figure into known shapes (triangles, rectangles, sectors)

Charlyn walks completely around the boundary of a square whose sides are each $5$ km long. From any point on her path she can see exactly $1$ km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?

Show solution without answering

Key insight

She sees everything within 1 km of the square's edge: a frame inside the square, four 5 by 1 strips outside, and four quarter-circles at the corners.

Solution

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

Why this works

"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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2, §4.3)
Introduction to Geometry, Chapter 9: Polygons (§9.3)