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

Revision as of 19:42, 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.

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

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

@@ Line 10: / Line 10: @@
 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);
-fil
 draw(square^^square2);</asy>

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

Revision as of 19:42, 8 February 2013

Problem

Solution

See Also