1991 AMC 12 第 22 题

先试着解答 1991 AMC 12 第 22 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 1991 AMC 12 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。

22.

两个圆外切。直线 PABPAB 和 PA′B′PA'B' 是公切线,其中 AA 和 A′A' 在小圆上,BB 和 B′B' 在大圆上。若 PA=AB=4PA=AB=4,则小圆的面积为
Two circles are externally tangent. Lines PABPAB and PA′B′PA'B' are common tangents with AA and A′A' on the smaller circle and BB and B′B' on the larger circle. If PA=AB=4,PA=AB=4, then the area of the smaller circle is
1.44π1.44\pi
2π2\pi
2.56π2.56\pi
8π\sqrt8\pi
4π4\pi
答案:B
知识点:相切圆位似圆幂
小提示:
两个圆通过以 PP 为中心的位似相互对应
The two circles are related by a dilation centered at PP
大提示:
利用 PA=4PA=4、PB=8PB=8 求位似比,再对小圆应用切线长度关系
Use PA=4,PA=4, PB=8PB=8 to find the dilation ratio, then apply the tangent-length relation to the smaller circle
解答:
由于 PA=4PA=4 且 AB=4AB=4,有 PB=8PB=8。因此,以 PP 为中心、把小圆映到大圆的位似比为 22。若小圆半径为 rr,则 3r3r 是其圆心到 PP 的距离:两个圆心位于同一条射线上,它们到 PP 的距离之比为 22,且两圆心间的距离为 r+2r=3rr+2r=3r。对由 PP、小圆圆心和 AA 构成的直角三角形应用勾股定理,PA2=(3r)2−r2=8r2。 PA^2=(3r)^2-r^2=8r^2\text{。}所以 16=8r216=8r^2,从而 r2=2r^2=2,面积为 2π2\pi。
因此正确答案为 B。
Since PA=4PA=4 and AB=4,AB=4, we have PB=8.PB=8. The homothety centered at PP taking the smaller circle to the larger therefore has ratio 2.2. If the smaller radius is r,r, its center is 3r3r from P:P: the centers are on the same ray, their distances from PP have ratio 2,2, and their difference is r+2r=3r.r+2r=3r. Applying the right triangle formed by P,P, the small center, and AA, PA2=(3r)2−r2=8r2. PA^2=(3r)^2-r^2=8r^2. Thus 16=8r2,16=8r^2, so r2=2r^2=2 and the area is 2π.2\pi.
Thus the correct answer is B.
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