AMC8 2024

AMC8 2024 · Q18

AMC8 2024 · Q18. It mainly tests Circle theorems, Area & perimeter.

Three concentric circles centered at $O$ have radii of $1$, $2$, and $3$. Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angles $BOC$, as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of $\angle{BOC}$ in degrees?

三个以 $O$ 为中心的同心圆，半径分别为 $1$、$2$ 和 $3$。点 $B$ 和 $C$ 在最大圆上。两个较小圆之间的区域被涂黑，两个较大圆之间由中心角 $BOC$ 限定的部分也被涂黑，如图所示。假设涂黑和未涂黑区域面积相等。$\angle{BOC}$ 的度数是多少？

(A) 108 108

(B) 120 120

(D) 144 144

(E) 150 150

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $x=\angle{BOC}$. We see that the shaded region is the inner ring plus a sector $x^\circ$ of the outer ring. Using the formula for the area of a circle ($A = \pi r^2$), we find that the area of $x$ is $\left( 4 \pi - \pi \right)+\frac{x}{360} \left( 9 \pi - 4 \pi \right)$. This simplifies to $3 \pi + \frac{x}{360}(5 \pi)$. The unshaded portion is comprised of the smallest circle plus the sector $(360-x)^\circ$ of the outer ring, which evaluates to $\pi + \frac{360-x}{360}(5 \pi)$. We are told these are equal. Therefore, $3 \pi + \frac{x}{360}(5 \pi) = \pi + \frac{360-x}{360}(5 \pi)$. Solving for $x$ reveals $x=\boxed{\textbf{(A) } 108}$.

设 $x=\angle{BOC}$。我们看到涂黑区域是内环加上外环的 $x^\circ$ 扇形。使用圆面积公式 ($A = \pi r^2$)，$x$ 的面积是 $\left( 4 \pi - \pi \right)+\frac{x}{360} \left( 9 \pi - 4 \pi \right)$。这简化为 $3 \pi + \frac{x}{360}(5 \pi)$。未涂黑部分包括最小圆加上外环的 $(360-x)^\circ$ 扇形，计算为 $\pi + \frac{360-x}{360}(5 \pi)$。已知两者相等。因此，$3 \pi + \frac{x}{360}(5 \pi) = \pi + \frac{360-x}{360}(5 \pi)$。解得 $x=\boxed{\textbf{(A) } 108}$。

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