1986 AMC 8 第 23 题

先试着解答 1986 AMC 8 第 23 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 1986 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。

23.

大圆的直径是 ACAC。两个小圆的圆心在 ACAC 上,并且在大圆圆心 OO 处相切。如果每个小圆半径为 11,阴影区域面积与一个小圆面积之比是多少?
The large circle has diameter AC.AC. The two small circles have their centers on ACAC and just touch at O,O, the center of the large circle. If each small circle has radius 1,1, what is the value of the ratio of the area of the shaded region to the area of one of the small circles?
在 12\dfrac12 和 11 之间
between 12\dfrac12 and 11
11
在 11 和 32\dfrac32 之间
between 11 and 32\dfrac32
在 32\dfrac32 和 22 之间
between 32\dfrac32 and 22
无法由给定信息确定
cannot be determined from the information given
答案:B
知识点:圆面积面积分割
小提示:
大圆半径是 22,因为 ACAC 等于两个小圆的直径之和
The large circle has radius 2,2, since ACAC equals two small diameters
大提示:
由对称性,阴影区域是大圆面积减去两个小圆面积后的一半
By symmetry the shaded region is half the difference between the large circle’s area and the two small circles’ area
解答:
大圆半径为 22,所以面积是 π(2)2=4π\pi(2)^2 = 4\pi,每个小圆面积是 π\pi。
由对称性,阴影区域是面积差的一半:12(4π−2π)=π\dfrac12(4\pi - 2\pi) = \pi。它与一个小圆面积 π\pi 的比值是 11。
所以正确答案是 B。
The large circle has radius 2,2, so its area is π(2)2=4π,\pi(2)^2 = 4\pi, and each small circle has area π.\pi.
By symmetry the shaded region is half the difference of the areas: 12(4π−2π)=π.\dfrac12(4\pi - 2\pi) = \pi. The ratio of this to one small circle’s area π\pi is 1.1.
Thus, the correct answer is B .
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