Difference between revisions of "2013 AMC 10A Problems/Problem 20"
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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 | + | 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>. |
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==Solution 2== | ==Solution 2== | ||
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<asy> | <asy> | ||
size(200); | size(200); | ||
defaultpen(linewidth(0.8)); | defaultpen(linewidth(0.8)); | ||
path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square; | path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square; | ||
− | for(int i=0;i<= | + | for(int i=0;i<=6;i=i+1) |
{ | { | ||
− | path arcrot=arc(origin,sqrt(2)/2,45+ | + | path arcrot=arc(origin,sqrt(2)/2,45+270*i,270*(i+1)); |
draw(arcrot); | draw(arcrot); | ||
− | |||
} | } | ||
draw(square^^square2);</asy> | draw(square^^square2);</asy> | ||
<center><math>\textbf{(high res image; no labels)}</math></center> | <center><math>\textbf{(high res image; no labels)}</math></center> | ||
+ | [[Image:AMC 10A 2013 20.jpg]] | ||
− | + | Let <math>O</math> be the center of the square and <math>C</math> be the intersection of <math>OB</math> and <math>AD</math>. The desired area consists of the unit square, plus <math>4</math> regions congruent to the region bounded by arc <math>AB</math>, <math>\overline{AC}</math>, and <math>\overline{BC}</math>, plus <math>4</math> triangular regions congruent to right triangle <math>BCD</math>. The area of the region bounded by arc <math>AB</math>, <math>\overline{AC}</math>, and <math>\overline{BC}</math> is <math>\frac{\text{Area of Circle}-\text{Area of Square}}{8}</math>. Since the circle has radius <math>\dfrac{1}{\sqrt {2}}</math>, the area of the region is <math>\dfrac{\dfrac{\pi}{2}-1}{8}</math>, so 4 times the area of that region is <math>\dfrac{\pi}{4}-\dfrac{1}{2}</math>. Now we find the area of <math>\triangle BCD</math>. <math>BC=BO-OC=\dfrac{\sqrt {2}}{2}-\dfrac{1}{2}</math>. Since <math>\triangle BCD</math> is a <math>45-45-90</math> right triangle, the area of <math>\triangle BCD</math> is <math>\dfrac{BC^2}{2}=\dfrac {\left (\dfrac {\sqrt {2}}{2}-\dfrac{1}{2} \right)^2}{2}</math>, so <math>4</math> times the area of <math>\triangle BCD</math> is <math>\dfrac{3}{2}-\sqrt {2}</math>. Finally, the area of the whole region is <math>1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=\dfrac{\pi}{4}+2-\sqrt {2}</math>, which we can rewrite as <math>\boxed{\textbf{(C) } 2 - \sqrt{2} + \frac{\pi}{4}}</math>. | |
− | Let <math>O</math> be the center of the square and <math>C</math> be the intersection of <math>OB</math> and <math>AD</math>. The desired area consists of the unit square, plus <math>4</math> regions congruent to the region bounded by arc <math>AB</math>, <math>\overline{AC}</math>, and <math>\overline{BC}</math>, plus <math>4</math> triangular regions congruent to right triangle <math>BCD</math>. The area of the region bounded by arc <math>AB</math>, <math>\overline{AC}</math>, and <math>\overline{BC}</math> is <math>\frac{\text{Area of Circle}-\text{Area of Square}}{8}</math>. Since the circle has radius <math>\dfrac{1}{\sqrt {2}}</math>, the area of the region is <math>\dfrac{\dfrac{\pi}{2}-1}{8}</math>, so 4 times the area of that region is <math>\dfrac{\pi}{4}-\dfrac{1}{2}</math>. Now we find the area of <math>\triangle BCD</math>. <math>BC=BO-OC=\dfrac{\sqrt {2}}{2}-\dfrac{1}{2}</math>. Since <math>\triangle BCD</math> is a <math>45-45-90</math> right triangle, the area of <math>\triangle BCD</math> is <math>\dfrac{BC^2}{2}=\dfrac {\left (\dfrac {\sqrt {2}}{2}-\dfrac{1}{2} \right)^2}{2}</math>, so 4 times the area of <math>\triangle BCD</math> is <math>\dfrac{3}{2}-\sqrt {2}</math>. Finally, the area of the whole region is <math>1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=\dfrac{\pi}{4}+2-\sqrt {2}</math>, which we can rewrite as <math>\boxed{\textbf{(C) } 2 - \sqrt{2} + \frac{\pi}{4}}</math>. | ||
==See Also== | ==See Also== |
Latest revision as of 17:55, 23 July 2021
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
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.