2005 AMC 8 Problems/Problem 25

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?

$[asy] pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(a--d--b--c--cycle); draw(circle(o, 2.5)); [/asy]$

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

Solutions

Solution 1

Let the region within the circle and square be $a$ . In other words, it is the area inside the circle $\textbf{and}$ the 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 $\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}$ .

Solution 2

We realize that since the areas of the regions outside of the circle and the square are equal to each other, the area of the circle must be equal to the area of the square.

$\pi r^2=4$

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

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

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

Solution 3

If you want to do complicated for no reason to waste time:

We know the side length of the square is 2. Let the part of the line that is in the circle be $x$ , and the other lengths be $1- \frac{x}{2}$ . Now the radius of the circle is $\sqrt{1+\frac{x^2}{4}}$

Now we find the areas of the square corners and circle corners which the problem says are equal, and after eliminating the $2(1-\frac{x}{2})^2$ on both sides from $2(1-\frac{x}{2})^2$ = $(1+\frac{x^2}{4}\pi} - 4 + 2(1-\frac{x}{2})^2$ (Error compiling LaTeX. Unknown error_msg) Now after some algebra we get the same answer as others, A.

2005 AMC 8 (Problems • Answer Key • Resources)
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