Difference between revisions of "2005 AMC 8 Problems/Problem 25"
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| − | ==Problem== | + | == 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? | 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); | + | <asy> |
| + | pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); | ||
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+\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 1== | + | === Solution 1 === |
| − | |||
Let the region within the circle and square be <math>a</math>. In other words, it is the area inside the circle <math>\textbf{and}</math> the 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> . | Let the region within the circle and square be <math>a</math>. In other words, it is the area inside the circle <math>\textbf{and}</math> the 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> . | ||
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So the answer is <math>\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}</math>. | So the answer is <math>\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}</math>. | ||
| − | ==Solution 2== | + | === 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. | 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. | ||
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So the answer is <math>\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}</math>. | So the answer is <math>\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}</math>. | ||
| − | ==See Also== | + | == See Also == |
{{AMC8 box|year=2005|num-b=24|after=Last problem}} | {{AMC8 box|year=2005|num-b=24|after=Last problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:35, 19 October 2020
Contents
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/c/8/c/c8c667941886cc09d8e6a51330698d67e111d5d3.png)
Solution 1
Let the region within the circle and square be 
. In other words, it is the area inside the circle 
the 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 
.
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.
So the answer is .
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last problem | |
| 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. 
