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

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draw((2,0)--(b+1,1),dashed);
 
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==Solution 2==
 
==Solution 2==
 
[[Image:AMC 10A 2013 20.jpg]]
 
[[Image:AMC 10A 2013 20.jpg]]

Revision as of 14:37, 12 February 2017

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--(0,0)--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 $\frac{1}{2} \pi (1/\sqrt{2})^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } 2 - \sqrt{2} + \frac{\pi}{4}}$.

[asy] size(150);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--(0,0)--cycle,grey);} //draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);} draw(square^^square2); //draw((-.5,.5)--(.5,-.5)^^(0,sqrt(.5))--(0,-sqrt(.5)),dotted);draw((.5,.5)--(-.5,-.5),dotted); [/asy] (To turn each dart-shaped piece into a parallelogram, cut along the dashed line and flip over one half.) [asy] size(150,Aspect);real r=sqrt(2);real b=2-2/r; draw((0,0)--(-1,1)--(b-1,1)--(0,r)--cycle);draw((0,1)--(b-1,1)--(b/2-1,1-b/2));draw((0,0)--(b-1,1),dashed); fill((2,0)--(b+1,1)--(b+2,0)--cycle,lightgray);draw((.5,.5)--(1,.5),EndArrow); draw((2,0)--(1,1)--(b+1,1)--(b+2,0)--(2,0)^^(b+1,1)--(b/2+1,1-b/2)^^(2,0)--(2+b/2,b/2)); draw((2,0)--(b+1,1),dashed); [/asy]

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 $\frac{\text{Area of Circle}-\text{Area of Square}}{8}$. 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 $1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=\dfrac{\pi}{4}+2-\sqrt {2}$, which we can rewrite as $\boxed{\textbf{(C) }    2 - \sqrt{2} + \frac{\pi}{4}}$.

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

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