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?

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

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

Because the two figures are similar, the second figure has quadruple the area of the first figure since a fourth of the same region on the right figure is equal to the entire region of the left figure. This means the side lengths are $\sqrt{4} = 2$ times longer, meaning the radius $R$ is $1 \cdot 2 = \boxed{\textbf{(B)} 2}$ .

~alwaysgonnagiveyouup

Vide Solution 1 by SpreadTheMathLove

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

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