2001 AMC 12 第 18 题

Question

一个以

A

为圆心、半径为

1

的圆和一个以

B

为圆心、半径为

4

的圆外切。第三个圆与前两个圆相切，并且与它们的一条公共外切线相切，如图所示。第三个圆的半径是

A circle centered at

A

with a radius of

1

and a circle centered at

B

with a radius of

4

are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle is

\dfrac{1}{3}

\dfrac{2}{5}

\dfrac{5}{12}

\dfrac{4}{9}

\dfrac{1}{2}

答案：D

知识点：相切圆勾股定理

难度评级：1820

小提示：

对两个半径为

r

和

s

、都与同一直线相切且彼此相切的圆，它们在该直线上的切点相距

2\sqrt{rs}

For two circles of radii

r

and

s

both tangent to a line and to each other, their contact points on the line are

2\sqrt{rs}

apart

大提示：

小圆的切点把线段分成两段：

2\sqrt{x} + 2\sqrt{4x} = 2\sqrt{4}

。

The contact point of the small circle splits the segment:

2\sqrt{x} + 2\sqrt{4x} = 2\sqrt{4}

解答：

当两个半径为

r

和

s

的相切圆都贴在同一直线上时，它们在直线上的切点距离为

2\sqrt{rs}

。

两个大圆的切点相距

2\sqrt{1 \cdot 4} = 4

。设夹在它们中间的小圆半径为

x

，则它到两边的切点距离相加为

2\sqrt{1 \cdot x} + 2\sqrt{4 \cdot x} = 4\text{。}

于是

6\sqrt{x} = 4

，所以

\sqrt{x} = \dfrac{2}{3}

且

x = \dfrac{4}{9}

。

因此，正确答案是 D。

When two mutually tangent circles of radii

r

and

s

both rest on a line, the distance between their points of tangency is

2\sqrt{rs}.

The big circles’ contact points are

2\sqrt{1 \cdot 4} = 4

apart. Placing the small circle of radius

x

between them, its two tangent distances add up:

2\sqrt{1 \cdot x} + 2\sqrt{4 \cdot x} = 4.

Then

6\sqrt{x} = 4,

so

\sqrt{x} = \dfrac{2}{3}

and

x = \dfrac{4}{9}.

Thus, the correct answer is D.

2001 AMC 12 第 18 题

答案：D

难度评级：1820

小提示：

大提示：

解答：

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