Difference between revisions of "2013 AMC 10A Problems/Problem 20"
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path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square; | path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square; | ||
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Revision as of 21:08, 1 January 2019
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
![$\textbf{(high res image; no labels)}$](http://latex.artofproblemsolving.com/d/0/c/d0cad456843171c2445debebc39ca51772eae46a.png)
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 4 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.