2023 AMC 12A 第 18 题

Question

圆

C_1

和

C_2

的半径都为

1

，两圆圆心之间的距离为

\tfrac12

。圆

C_3

是同时内切于

C_1

和

C_2

的最大圆。圆

C_4

同时内切于

C_1

和

C_2

，并且外切于

C_3

。圆

C_4

的半径是多少？

Circle

C_1

and

C_2

each have radius

1,

and the distance between their centers is

\tfrac12.

Circle

C_3

is the largest circle internally tangent to both

C_1

and

C_2.

Circle

C_4

is internally tangent to both

C_1

and

C_2

and externally tangent to

C_3.

What is the radius of

C_4?

\dfrac{1}{14}

\dfrac{1}{12}

\dfrac{1}{10}

\dfrac{3}{28}

\dfrac{1}{9}

答案：D

知识点：相切圆坐标几何

难度评级：1990

小提示：

由对称性，

C_3

的圆心在

C_1C_2

的中点；其半径为

1-\tfrac14=\tfrac34

By symmetry

C_3

is centered at the midpoint of

C_1C_2;

its radius is

1-\tfrac14=\tfrac34

大提示：

把

C_4

放在垂直平分线上；其圆心到单位圆圆心距离为

1-r

，到

C_3

圆心距离为

\tfrac34+r

。

Place

C_4

on the perpendicular bisector; its center is

1-r

from a unit circle’s center and

\tfrac34+r

from the center of

C_3

解答：

把圆心放在

O_1=\left(-\tfrac14,0\right)

和

O_2=\left(\tfrac14,0\right)

。由对称性，

C_3

以原点为圆心；与

C_1

内切给出其半径为

1-\tfrac14=\tfrac34

。

设

C_4

的半径为

r

，圆心为对称轴上的

(0,k)

。与

C_3

外切可得

k=\tfrac34+r

，与

C_1

内切可得

\sqrt{\tfrac{1}{16}+k^2}=1-r

。

代入得

\tfrac{1}{16}+\left(\tfrac34+r\right)^2=(1-r)^2

，化简为

\tfrac72 r=\tfrac38

，所以

r=\dfrac{3}{28}

。

所以正确答案是 D。

Put the centers at

O_1=\left(-\tfrac14,0\right)

and

O_2=\left(\tfrac14,0\right).

By symmetry

C_3

is centered at the origin, and internal tangency to

C_1

gives radius

1-\tfrac14=\tfrac34.

Let

C_4

have radius

r,

centered at

(0,k)

on the axis of symmetry. External tangency to

C_3

gives

k=\tfrac34+r,

and internal tangency to

C_1

gives

\sqrt{\tfrac{1}{16}+k^2}=1-r.

Substituting,

\tfrac{1}{16}+\left(\tfrac34+r\right)^2=(1-r)^2,

which simplifies to

\tfrac72 r=\tfrac38,

so

r=\dfrac{3}{28}.

Thus, the correct answer is D.

2023 AMC 12A 第 18 题

答案：D

难度评级：1990

小提示：

大提示：

解答：

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