2022 AMC 12B 第 21 题

Question

设

S

为坐标平面中同时与三个圆

x^2 + y^2 = 4

、

x^2 + y^2 = 64

和

(x - 5)^2 + y^2 = 3

相切的所有圆的集合。

S

中所有圆的面积之和是多少？

Let

S

be the set of circles in the coordinate plane that are tangent to each of the three circles with equations

x^2 + y^2 = 4,

x^2 + y^2 = 64,

and

(x - 5)^2 + y^2 = 3.

What is the sum of the areas of all circles in

S?

48\pi

68\pi

96\pi

102\pi

136\pi

答案：E

知识点：相切圆圆面积分类讨论

难度评级：2170

小提示：

同时与两个同心圆（半径

2

和

8

）相切的圆，其半径为

3

或

5

。

A circle tangent to both concentric circles (radii

2

and

8

) has radius

3

or radius

5

大提示：

半径为

3

的圆的圆心在半径为

5

的圆上，半径为

5

的圆的圆心在半径为

3

的圆上。分别数出其中有多少个还与第三个圆相切。

The radius-

3

circles have centers on the circle of radius

5;

the radius-

5

circles on the circle of radius

3.

Count how many of each are also tangent to the third circle

解答：

前两个圆同心，半径分别为

2

和

8

。同时与它们相切的圆，要么半径为

3

，圆心到原点的距离为

5

；要么半径为

5

，圆心到原点的距离为

3

。

第三个圆的圆心为

(5, 0)

，半径为

\sqrt3

。对每个候选半径

s\in\{3,5\}

，与第三个圆相切要求圆心到

(5,0)

的距离为

s+\sqrt3

或

s-\sqrt3

。这两个距离圆各与相应的圆心轨迹交于两个对称的点。因此恰有四个半径为

3

的圆和四个半径为

5

的圆满足条件。

面积和为

4 \cdot \pi(3)^2 + 4 \cdot \pi(5)^2

= 36\pi + 100\pi = 136\pi

。

所以正确答案是 E。

The first two circles are concentric with radii

2

and

8.

A circle tangent to both either has radius

3

with center at distance

5

from the origin, or radius

5

with center at distance

3

from the origin.

The third circle has center

(5, 0)

and radius

\sqrt3.

For each candidate radius

s\in\{3,5\},

tangency requires the center’s distance from

(5,0)

to be

s+\sqrt3

or

s-\sqrt3.

Each of these two distance circles intersects the appropriate center-locus in two symmetric points. Hence exactly four radius-

3

circles and four radius-

5

circles work.

The sum of the areas is

4 \cdot \pi(3)^2 + 4 \cdot \pi(5)^2

= 36\pi + 100\pi = 136\pi.

Thus, the correct answer is E.

2022 AMC 12B 第 21 题

答案：E

难度评级：2170

小提示：

大提示：

解答：

其他年份的第 21 题