Difference between revisions of "2013 AMC 10A Problems/Problem 20"
m (→Solution 2) |
m (→Solution 1) |
||
Line 25: | Line 25: | ||
draw(square^^square2);</asy> | draw(square^^square2);</asy> | ||
− | For this square with side length 1, the distance from center to vertex is <math>r = \frac{\sqrt{2}}{2}</math>, hence the area is composed of a semicircle of radius <math>r</math>, plus <math>4</math> times a parallelogram (or a kite with diagonals of <math> | + | For this square with side length 1, the distance from center to vertex is <math>r = \frac{\sqrt{2}}{2}</math>, hence the area is composed of a semicircle of radius <math>r</math>, plus <math>4</math> times a parallelogram (or a kite with diagonals of <math>(\sqrt{2}-1)</math> and <math>r \text{ or} \frac{\sqrt{2}}{2}</math>) with height <math>\frac{1}{2}</math> and base <math>\frac{\sqrt{2}}{2(1+\sqrt{2})}</math>. That is to say, the total area is <math>\frac{1}{2} \pi \left(\frac{\sqrt{2}}{2}\right)^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } 2 - \sqrt{2} + \frac{\pi}{4}}</math>. |
<asy> | <asy> |
Revision as of 00:54, 2 January 2021
Contents
[hide]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.