2025 AMC 8 Problems/Problem 18

Revision as of 21:15, 29 January 2025 by Mrtnvlknv (talk | contribs)

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