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

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So the answer is <math>\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}</math>.
 
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==Video Solution by OmegaLearn==
 
==Video Solution by OmegaLearn==
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==Video Solution==
 
==Video Solution==
  
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https://www.youtube.com/watch?v=xdeWCR666q8
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https://www.youtube.com/watch?v=xdeWCR666q8   ~David
  
 
== 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}}

Latest revision as of 08:53, 17 June 2024

Problem 25

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 $\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}$.

Video Solution by OmegaLearn

https://youtu.be/abSgjn4Qs34?t=2763

Video Solution

https://www.youtube.com/watch?v=xdeWCR666q8 ~David

See Also

2005 AMC 8 (Problems • Answer Key • Resources)
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Problem 24 		Followed by
Last Problem
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All AJHSME/AMC 8 Problems and Solutions

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