2024 AMC 8 第 18 题

先试着解答 2024 AMC 8 第 18 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2024 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

18.

以 OO 为圆心的三个同心圆半径分别为 11、22 和 33。点 BB 和 CC 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
小提示:
先算半径 11 和 22 之间阴影圆环的面积
Compute the area of the shaded annulus between radii 11 and 22
大提示:
令阴影面积等于非阴影面积,解圆心角
Set shaded area equal to unshaded area and solve for the sector angle
解答:
设 θ\theta 为 ∠BOC\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 4π−π=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(9π−4π)=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+360−θ360(5π)\pi + \dfrac{360-\theta}{360}(5\pi)。
令两部分面积相等并解 θ\theta:3π+θ360(5π)=π+360−θ360(5π) \begin{gathered} 3\pi + \dfrac{\theta}{360}(5\pi) \\ = \pi + \dfrac{360-\theta}{360}(5\pi) \end{gathered} 2π=360−θ−θ360(5π) 2\pi = \dfrac{360-\theta-\theta}{360}(5\pi) 25=1−2θ360 \dfrac{2}{5} = 1 - \dfrac{2\theta}{360} 2θ=35(360) 2\theta = \dfrac{3}{5}(360) θ=108。 \theta = 108\text{。}
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 4π−π=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(9π−4π)=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+360−θ360(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+360−θ360(5π) \begin{gathered} 3\pi + \dfrac{\theta}{360}(5\pi) \\ = \pi + \dfrac{360-\theta}{360}(5\pi) \end{gathered} 2π=360−θ−θ360(5π) 2\pi = \dfrac{360-\theta-\theta}{360}(5\pi) 25=1−2θ360 \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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