2013 AMC 10A Problems/Problem 20

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

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

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);
fil
draw(square^^square2); (Error making remote request. Unknown error_msg)

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 $\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } \frac{\pi}{4} + 2 - \sqrt{2}}$ .

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2013 AMC 10A Problems/Problem 20

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