AMC8 2025

AMC8 2025 · Q18

AMC8 2025 · Q18. It mainly tests Quadratic equations, Area & perimeter.

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?

左边的圆半径为1单位。圆与内接正方形之间的区域被涂影。右边的圆中，圆与内接正方形之间的区域的四分之一被涂影。两个圆的涂影区域面积相等。右边圆的半径$R$是多少单位？

(A) \ \sqrt2 \ \sqrt2

(B) \ 2 \ 2

(D) \ 4 \ 4

(E) \ 4\sqrt2 \ 4\sqrt2

Answer

Correct choice: (B)

正确答案：（B）

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 $R=\boxed{\text{(B) 2}}$

左边圆涂影区域的面积是圆的面积减去正方形的面积，即$\pi-2$。右边圆的涂影面积是圆与正方形之间区域的$\frac{1}{4}$，即$\dfrac{\pi R^2-2R^2}{4}$，可因式分解为$\dfrac{R^2(\pi-2)}{4}$。由于涂影面积相等，有$\pi-2=\dfrac{R^2(\pi-2)}{4}$，简化得$R^2=4$。取平方根，$R=\boxed{\text{(B) 2}}$

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