Difference between revisions of "2005 AMC 8 Problems/Problem 25"
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| − | Let the region within the circle and square be <math>a</math>. Let <math>r</math> be the radius. We know that the area of the circle minus <math>a</math> is equal to the area of the square, minus <math>a</math> . | + | Let the region within the circle and square be <math>a</math>. In other words, it is the intersection of the area of circle and square. Let <math>r</math> be the radius. We know that the area of the circle minus <math>a</math> is equal to the area of the square, minus <math>a</math> . |
We get: | We get: | ||
Revision as of 20:43, 13 November 2014
Problem
A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?
![[asy]pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(a--d--b--c--cycle); draw(circle(o, 2.5));[/asy]](http://latex.artofproblemsolving.com/4/0/b/40bd7bb55e07ffdab8d71c28dee061143a893015.png)
$\textbf{(A)}\ \frac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \frac{1\plus{}\sqrt{2}}{2} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}$ (Error compiling LaTeX. Unknown error_msg)
Solution
Let the region within the circle and square be 
. In other words, it is the intersection of the area of circle and square. Let 
be the radius. We know that the area of the circle minus is equal to the area of the square, minus .
We get:
So the answer is 
.
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Question | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
