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

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

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.

@@ 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?
-<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(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>
-==Solution==
+== Solutions ==
+=== Solution 1 ===
-Let the region within the circle and square be <math>a</math>. In other words, it is the intersection of the area of circle and 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> .
 We get:
@@ Line 21: / Line 24: @@
 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.
@@ Line 32: / Line 35: @@
 So the answer is <math>\boxed{\textbf{(A)}\ \frac{2}{\sqrt{\pi}}}</math>.
-==Solution 2==
+==Video Solution by OmegaLearn==
-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.
+https://youtu.be/abSgjn4Qs34?t=2763
-<math>\pi r^2=4
-</math>r^2=\frac{4}{\pi}<math>
+==Video Solution==
-</math>r=\frac{2}{\sqrt{\pi}}<math>
+https://www.youtube.com/watch?v=xdeWCR666q8   ~David
-So the answer is </math>\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.
-<math>r^2=\frac{4}{\pi}</math>
-<math>r=\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 <br /> Question}}
+{{AMC8 box|year=2005|num-b=24|after=Last Problem}}
 {{MAA Notice}}

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