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

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

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

解答：

其他年份的第 18 题