Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (minor edit)
(→‎Video Solution)
 
(17 intermediate revisions by 8 users not shown)
Line 1: Line 1:
−
== Problem ==
+
== 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?
 
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?
  
Line 34: Line 34:
  
 
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>.
 +
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/abSgjn4Qs34?t=2763
 +
 +
==Video Solution==
 +
 +
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)
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_8_Problems/Problem_25&oldid=220653"