Difference between revisions of "2013 AMC 10A Problems/Problem 20"
Claudeaops (talk | contribs) m (Centering the Image) |
Claudeaops (talk | contribs) (Solution 2) |
||
Line 28: | Line 28: | ||
==Solution 2== | ==Solution 2== | ||
− | [[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 4 regions congruent to the region bounded by arc <math>AB</math>, <math>\overline{AC}</math>, and <math>\overline{BC}</math>, plus 4 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 (Area of Circle-Area of Square)/8. 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=</math>\dfrac{\sqrt {2}}{2}-\dfrac{1}{2}<math>. Since </math>\triangle BCD<math> is a 45-45-90 right triangle, the area of </math>\triangle BCD<math> is </math>\dfrac{BC^2}{2}=\dfrac{({\dfrac{\sqrt {2}}{2}-\dfrac{1}{2})^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)=<math>\dfrac{\pi}{4}+2-\sqrt {2}</math>. | ||
==See Also== | ==See Also== |
Revision as of 19:01, 18 August 2014
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 with height and base . That is to say, the total area is .
Solution 2
Let be the center of the square and be the intersection of and . The desired area consists of the unit square, plus 4 regions congruent to the region bounded by arc , , and , plus 4 triangular regions congruent to right triangle . The area of the region bounded by arc , , and is (Area of Circle-Area of Square)/8. 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 . \dfrac{\sqrt {2}}{2}-\dfrac{1}{2}\triangle BCD\triangle BCD\dfrac{BC^2}{2}=\dfrac{({\dfrac{\sqrt {2}}{2}-\dfrac{1}{2})^2}{2}\triangle BCD\dfrac{3}{2}-\sqrt {2}1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=.
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.