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2013 AMC 10A Problems/Problem 20

Revision as of 18:17, 18 August 2014 by Claudeaops (talk | contribs) (Fixing Latex)

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

AMC 10A 2013 20.jpg

Let $O$ be the center of the square and $C$ be the intersection of $OB$ and $AD$. The desired area consists of the unit square, plus 4 regions congruent to the region bounded by arc $AB$, $\overline{AC}$, and $\overline{BC}$, plus 4 triangular regions congruent to right triangle $BCD$. The area of the region bounded by arc $AB$, $\overline{AC}$, and $\overline{BC}$ is (Area of Circle-Area of Square)/8. Since the circle has radius $\dfrac{1}{\sqrt {2}}$, the area of the region is $\dfrac{\dfrac{\pi}{2}-1}{8}$, so 4 times the area of that region is $\dfrac{\pi}{4}-\dfrac{1}{2}$. Now we find the area of $\triangle BCD$. $BC=BO-OC=\dfrac{\sqrt {2}}{2}-\dfrac{1}{2}$. Since $\triangle BCD$ is a 45-45-90 right triangle, the area of $\triangle BCD$ is $\dfrac{BC^2}{2}=\dfrac {\left (\dfrac {\sqrt {2}}{2}-\dfrac{1}{2} \right)^2}{2}$, so 4 times the area of $\triangle BCD$ is $\dfrac{3}{2}-\sqrt {2}$. Finally, the area of the whole region is $1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=\dfrac{\pi}{4}+2-\sqrt {2}$.

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

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