2025 AMC 8 Problems/Problem 18

Problem

The circle shown below on the left has a radius of 1 unit. The region between the circle and the inscribed square is shaded. In the circle shown on the right, one quarter of the region between the circle and the inscribed square is shaded. The shaded regions in the two circles have the same area. What is the radius $R$ , in units, of the circle on the right?

$[asy] unitsize(40); real a = 0.707; fill(circle((a,a), 1), grey); fill((0,0)--(0,1.414)--(1.414,1.414)--(1.414,0)--cycle, white); draw((0,0)--(0,1.414)--(1.414,1.414)--(1.414,0)--cycle); draw(circle((a,a), 1)); draw((0.707,0.707)--(1.414,1.414)); dot((0.707,0.707)); label("$1$", (1,1), SE); fill(circle((4+a, a), 2*a), grey); fill(shift((4+a,a)) * ((-2,-2)--(1,-2)--(1,2)--(-2,2)--cycle), white); draw(shift((4+a,a)) * ((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle)); draw(circle((4+a, a), 2*a)); draw((4+a,a)--(5+a,1+a)); dot((4+a,a)); label("$R$", (a+4.5,a+0.5), SE); [/asy]$

$\textbf{(A)}\ \sqrt2\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 2\sqrt2\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ 4\sqrt2$

Solution

The area of the shaded region in the circle on the left is the area of the circle minus the area of the square, or $\big(\pi-2)$ . The shaded area in the circle on the right is $\dfrac{1}{4}$ of the area of the circle minus the area of the square, or $\dfrac{\pi R^2-2R^2}{4}$ , which can be factored as $\dfrac{R^2(\pi-2)}{4}$ . Since the shaded areas are equal to each other, we have $\pi-2=\dfrac{R^2(\pi-2)}{4}$ , which simplifies to $R^2=4$ . Taking the square root, we have $R=\boxed{\text{(B)\ 2}}$

~mrtnvlknv

Solution 2

We start with the first area. Since the square is inscribed, its diagonal is $2\implies$ its side length is $\sqrt{2}\implies$ its area is $2$ , therefore the first area is $\pi-2$ . The second area is $\dfrac{R^2\pi-2R^2}{4}$ , found in a similar manner. Writing and solving the equation, we have $\pi-2=\dfrac{R^2\pi-2R^2}{4}\implies4(\pi-2)=R^2(\pi-2)\implies R=2.$ The answer is $\boxed{\text{(B) }2}$ ~Tacos_are_yummy_1

Solution 3 (Using similarity)

Since the two figures are given to be similar, and the 4 regions in the first circle are equal in area to only one of those regions of the second circle, we know that the second figure's area is 4 times the first figure's area. Therefore, the side lengths are multiplied by a factor of √4 = 2, so the answer is 1 * 2 = $1 \cdot 2 = \boxed{\textbf{(B)} 2}$ .

~Sigmacuber

Solution 4

(Really similar to Solution 1, but has a more larger explanation for the answer) First find the area of the smaller circle with the known radius. Since the radius is $1$ , the area of the circle is $\pi$ , and finding the side length of the square would equal to $\sqrt{2}$ . Squaring that gives us the area of the square which is $2$ , and the shaded area of the first one is equal to $\pi - 2$ . The second area can be shortened to $\frac{R^2\pi}{4} - \frac{\left(\sqrt{2R^2}\right)^2}{4}$ , since we only need $\frac{1}{4}$ th of the sections. We then make them equal, which makes this equation:

$\pi - 2 = \frac{R^2\pi}{4} - \frac{\left(\sqrt{2R^2}\right)^2}{4}$

Simplifying this equation would be $4\pi - 8 = R^2\pi - 2R^2$ . We can factor out the $R^2$ which makes $4\pi - 8 = R^2\left(\pi - 2\right)$ . Dividing out the $\pi - 2$ would give us $4$ on the left side. $4 = R^2$ which leaves us to choice $\boxed{\text{(B) }2}$ . ~Imhappy62789

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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

Page

Toolbox

Search