2013 AMC 10A Problems/Problem 20
Contents
Problem
A unit square is rotated about its center. What is the area of the region swept out by the interior of the square?
Solution 1
First, we need to see what this looks like. Below is a diagram.
For this square with side length 1, the distance from center to vertex is , hence the area is composed of a semicircle of radius , plus times a parallelogram (or a kite with diagonals of and ) with height and base . That is to say, the total area is .
(To turn each dart-shaped piece into a parallelogram, cut along the dashed line and flip over one half.) Alternatively, you can move the dart-shaped piece to the other side and make a kite.
Solution 2
Let be the center of the square and be the intersection of and . The desired area consists of the unit square, plus regions congruent to the region bounded by arc , , and , plus triangular regions congruent to right triangle . The area of the region bounded by arc , , and is . Since the circle has radius , the area of the region is , so 4 times the area of that region is . Now we find the area of . . Since is a right triangle, the area of is , so times the area of is . Finally, the area of the whole region is , which we can rewrite as .
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.