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.

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

We see that we have four quarter circles with radius $\frac{1}{2}$ and four triangles (split the triangles in the above picture in half). The former each have area $\frac{\pi}{16}$ . Thus, their total area is $\frac{\pi}{4}$ .

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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