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Difference between revisions of "2013 AMC 10A Problems/Problem 20"
Line 7: Line 7:
  
 
==Solution==
 
==Solution==
 +
 +
First, we need to see what this looks like.  Below is a diagram.
 +
 +
<asy>
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size(200);
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defaultpen(linewidth(0.8));
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path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;
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fill(square^^square2,grey);
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for(int i=0;i<=3;i=i+1)
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{
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path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));
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draw(arcrot);
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fill(arcrot--((sqrt(2)-1)/(2*sqrt(2)),0)--cycle,grey);
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draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);
 +
}
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draw(square^^square2);</asy>
 +
 +
We see that we have four quarter circles with radius <math>\frac{1}{2}</math> and four triangles (split the triangles in the above picture in half).  The former each have area <math>\frac{\pi}{16}</math>.  Thus, their total area is <math>\frac{\pi}{4}</math>.
 +
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}
 
{{AMC10 box|year=2013|ab=A|num-b=19|num-a=21}}

Revision as of 12:41, 8 February 2013

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 $\frac{\pi}{4}$.


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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