2024 AMC 8 第 18 题

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18.

OO 为圆心的三个同心圆半径分别为 112233。点 BBCC 在最大圆上。两个较小圆之间的区域被涂阴影,两个较大圆之间由圆心角 BOCBOC 截出的部分也被涂阴影,如下图所示。若阴影区域和非阴影区域面积相等,BOC\angle BOC 的度数是多少?

Three concentric circles centered at OO have radii of 1,1, 2,2, and 3.3. Points BB and CC 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 angle BOC,BOC, as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of BOC\angle BOC in degrees?

108108

120120

135135

144144

150150

答案:A
知识点:圆环扇形一次方程
难度评级:1540
解答:

θ\thetaBOC\angle BOC 的度数。

阴影区域的一部分是半径 22 圆与半径 11 圆之间的圆环,面积为 4ππ=3π4\pi-\pi=3\pi。另一部分是最大圆的一个扇形减去半径 22 的圆,面积为 θ360(9π4π)=θ360(5π)\dfrac{\theta}{360}(9\pi-4\pi) = \dfrac{\theta}{360}(5\pi)。所以阴影面积为 3π+θ360(5π)3\pi + \dfrac{\theta}{360}(5\pi)

非阴影区域由最小圆和外圆环未阴影部分组成,面积为 π+360θ360(5π)\pi + \dfrac{360-\theta}{360}(5\pi)

令两部分面积相等并解 θ\theta3π+θ360(5π)=π+360θ360(5π) \begin{gathered} 3\pi + \dfrac{\theta}{360}(5\pi) \\ = \pi + \dfrac{360-\theta}{360}(5\pi) \end{gathered} 2π=360θθ360(5π) 2\pi = \dfrac{360-\theta-\theta}{360}(5\pi) 25=12θ360 \dfrac{2}{5} = 1 - \dfrac{2\theta}{360} 2θ=35(360) 2\theta = \dfrac{3}{5}(360) θ=108. \theta = 108.

所以正确答案是 A

Let θ\theta be the measure of BOC.\angle BOC.

One component of the shaded region is the area of the circle with radius 22 minus the area of the circle with radius 1.1. This part has area 4ππ=3π.4\pi-\pi=3\pi. The remaining area is a sector of the biggest circle minus the area of the circle with radius 22. This has area θ360(9π4π)=θ360(5π).\dfrac{\theta}{360}(9\pi-4\pi) = \dfrac{\theta}{360}(5\pi). Hence, the total area of the shaded region is 3π+θ360(5π).3\pi + \dfrac{\theta}{360}(5\pi).

Next, we note that the unshaded region is composed of the smallest circle and the unshaded portion of the outer ring. This will have a total area of π+360θ360(5π)\pi + \dfrac{360-\theta}{360}(5\pi)

Lastly, we equate the area of both regions and solve for θ:\theta: 3π+θ360(5π)=π+360θ360(5π) \begin{gathered} 3\pi + \dfrac{\theta}{360}(5\pi) \\ = \pi + \dfrac{360-\theta}{360}(5\pi) \end{gathered} 2π=360θθ360(5π) 2\pi = \dfrac{360-\theta-\theta}{360}(5\pi) 25=12θ360 \dfrac{2}{5} = 1 - \dfrac{2\theta}{360} 2θ=35(360) 2\theta = \dfrac{3}{5}(360) θ=108. \theta = 108.

Thus, A is the correct answer.

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