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>. |
<asy> | <asy> | ||
Line 44: | Line 44: | ||
draw((2,0)--(b+1,1),dashed); | draw((2,0)--(b+1,1),dashed); | ||
</asy> | </asy> | ||
+ | Alternatively, you can move the dart-shaped piece to the other side and make a kite. | ||
+ | <asy> | ||
+ | size(75,Aspect);real r=sqrt(2);real b=2-2/r; | ||
+ | draw((r-1,1)--(b-1,1)); | ||
+ | draw((0,0)--(b-1,1)--(0,r)--(r-1,1)--cycle); | ||
+ | draw((0,r)--(0,0),dashed); | ||
+ | </asy> | ||
+ | |||
==Solution 2== | ==Solution 2== | ||
+ | <asy> | ||
+ | size(200); | ||
+ | defaultpen(linewidth(0.8)); | ||
+ | path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square; | ||
+ | for(int i=0;i<=6;i=i+1) | ||
+ | { | ||
+ | path arcrot=arc(origin,sqrt(2)/2,45+270*i,270*(i+1)); | ||
+ | draw(arcrot); | ||
+ | } | ||
+ | draw(square^^square2);</asy> | ||
+ | <center><math>\textbf{(high res image; no labels)}</math></center> | ||
[[Image:AMC 10A 2013 20.jpg]] | [[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}}{ | + | 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>. |
==See Also== | ==See Also== |
Latest revision as of 16:55, 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
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.