Difference between revisions of "2013 AMC 10A Problems/Problem 20"

(Image Upload)
(Image Formatting)
Line 28: Line 28:
  
 
==Solution 2==
 
==Solution 2==
[[Image:AMC_10A_2013_-20.png]]
+
[[Image:AMC_10A_2013_-20.png|200px|thumb|left|alt text]]
  
 
==See Also==
 
==See Also==

Revision as of 00:47, 18 August 2014

Problem

A unit square is rotated $45^\circ$ about its center. What is the area of the region swept out by the interior of the square?


$\textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8}$

Solution 1

First, we need to see what this looks like. Below is a diagram.

[asy] size(200); defaultpen(linewidth(0.8)); path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square; fill(square^^square2,grey); for(int i=0;i<=3;i=i+1) { path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1)); draw(arcrot); fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey); draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow); } draw(square^^square2);[/asy]

For this square with side length 1, the distance from center to vertex is $r = \frac{1}{\sqrt{2}}$, hence the area is composed of a semicircle of radius $r$, plus $4$ times a parallelogram with height $\frac{1}{2}$ and base $\frac{\sqrt{2}}{2(1+\sqrt{2})}$. That is to say, the total area is $\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}$.

Solution 2

alt text

See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS