Difference between revisions of "2005 AMC 8 Problems/Problem 25"

(Solution)
(Problem)
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draw(a--d--b--c--cycle);
 
draw(a--d--b--c--cycle);
 
draw(circle(o, 2.5));</asy>
 
draw(circle(o, 2.5));</asy>
<math> \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}</math>
+
<math> \textbf{(A)}\ \frac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \frac{1+\sqrt{2}}{2} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}</math>
  
 
==Solution==
 
==Solution==

Revision as of 09:19, 24 August 2015

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] $\textbf{(A)}\ \frac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \frac{1+\sqrt{2}}{2} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}$

Solution

Let the region within the circle and square be $a$. In other words, it is the intersection of the area of circle and square. Let $r$ be the radius. We know that the area of the circle minus $a$ is equal to the area of the square, minus $a$ .

We get:

$\pi r^2 -a=4-a$

$r^2=\frac{4}{\pi}$

$r=\frac{2}{\sqrt{\pi}}$

So the answer is $\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}$.

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
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Problem 24
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All AJHSME/AMC 8 Problems and Solutions

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