2022 AMC 10B 第 22 题

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22.

SS 为坐标平面中所有同时与下面三个圆相切的圆的集合:x2+y2=4,x2+y2=64x^{2}+y^{2}=4,\qquad x^{2}+y^{2}=64\text{,}以及 (x5)2+y2=3(x-5)^{2}+y^{2}=3\text{。}SS 中所有圆的面积之和。

Let SS be the set of circles in the coordinate plane that are tangent to each of the three circles with equations x2+y2=4,x2+y2=64,x^{2}+y^{2}=4,\qquad x^{2}+y^{2}=64, and (x5)2+y2=3.(x-5)^{2}+y^{2}=3. What is the sum of the areas of all circles in S?S?

 48π\ 48 \pi

 68π\ 68 \pi

 96π\ 96 \pi

 102π\ 102 \pi

 136π\ 136 \pi

答案:E
知识点:相切圆圆面积分类讨论
难度评级:2390
小提示:

根据所求圆与较小同心圆内切还是外切来配对讨论

Pair cases according to whether it is internally or externally tangent to the smaller concentric circle

大提示:

每个所求圆都与最大的同心圆内切

Every desired circle is internally tangent to the largest concentric circle

解答:

将半径为 2288 的两个同心圆分别称为内圆和外圆。设所求圆的半径为 rr,圆心到原点的距离为 dd。所求圆必与外圆内切,因此 d+r=8d+r=8

若所求圆与内圆外切,则 dr=2d-r=2,从而 (r,d)=(3,5)(r,d)=(3,5)。若所求圆包含内圆,则 rd=2r-d=2,从而 (r,d)=(5,3)(r,d)=(5,3)。因此,所求圆的半径必为 3355

第三个已知圆的半径为 3\sqrt3,圆心为 (5,0)(5,0)。对 rr 的两个可能值,所求圆都可以与第三个圆内切或外切,所以两圆心之间的距离为 r3r-\sqrt3r+3r+\sqrt3。在这四种情形中,若该距离为 qq,则都有 dq<5<d+q|d-q|<5<d+q。因此,以原点为圆心、dd 为半径的圆与所求圆圆心的轨迹交于关于 xx 轴对称的两点,如图所示。于是,每一种半径都有 44 个所求圆。

所以总面积为 4(52π+32π)=136π4(5^2\pi+3^2\pi)=136\pi\text{。}因此,正确答案是 E

Call the concentric circles of radii 22 and 88 the inner and outer circles. Let a desired circle have radius rr and let its center be distance dd from the origin. It must be internally tangent to the outer circle, so d+r=8.d+r=8.

If it is externally tangent to the inner circle, then dr=2,d-r=2, giving (r,d)=(3,5).(r,d)=(3,5). If it contains the inner circle, then rd=2,r-d=2, giving (r,d)=(5,3).(r,d)=(5,3). Thus every desired circle has radius 33 or 5.5.

The third given circle has radius 3\sqrt3 and center (5,0).(5,0). For either value of r,r, a desired circle may be internally or externally tangent to it, so the distance between their centers is r3r-\sqrt3 or r+3.r+\sqrt3. In all four cases, if this distance is q,q, then dq<5<d+q,|d-q|<5<d+q, so the circle of possible centers intersects the circle of radius dd about the origin in two points, symmetric across the xx-axis, as shown. Hence there are 44 desired circles of each radius.

The total area is therefore 4(52π+32π)=136π.4(5^2\pi+3^2\pi)=136\pi. Thus, E is the correct answer.

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