Question

一个半径为 $r$ 的圆被三个圆围住，这三个圆的半径分别为 $1, 2$ 和 $3$ ，它们都与内圆外切，并且彼此也外切，如图所示。

$r$ 等于多少？

A circle of radius $r$ is surrounded by three circles, whose radii are $1, 2,$ and $3,$ all externally tangent to the inner circle and externally tangent to each other, as shown in the diagram below.

What is $r?$

$\dfrac{1}{4}$

$\dfrac{6}{23}$

$\dfrac{3}{11}$

$\dfrac{5}{17}$

$\dfrac{3}{10}$

答案：B

知识点：相切圆坐标几何

难度评级：2120

解答：

三个外圆圆心 $A, B, C$ 的两两距离分别为 $AB = 1 + 2 = 3$ ， $AC = 1 + 3 = 4$ ， $BC = 2 + 3 = 5$ ，构成 $3$ - $4$ - $5$ 直角三角形。现在对四个相互相切的圆应用笛卡尔圆定理，曲率分别为 $1, \tfrac12, \tfrac13$ ，以及 $\tfrac1r$ ： $\frac1r = 1 + \tfrac12 + \tfrac13$ $+ 2\sqrt{\tfrac12 + \tfrac16 + \tfrac13}$ $= \tfrac{11}{6} + 2\sqrt{1}$ $= \tfrac{23}{6}$ 。取倒数得 $r = \tfrac{6}{23}$ 。因此正确答案是 B。

The three outer centers $A, B, C$ are pairwise $AB = 1 + 2 = 3,$ $AC = 1 + 3 = 4,$ and $BC = 2 + 3 = 5$ apart, a $3$ - $4$ - $5$ right triangle. Now apply Descartes' Circle Theorem with curvatures $1, \tfrac12, \tfrac13,$ and $\tfrac1r,$ all mutually tangent: $\frac1r = 1 + \tfrac12 + \tfrac13$ $+ 2\sqrt{\tfrac12 + \tfrac16 + \tfrac13}$ $= \tfrac{11}{6} + 2\sqrt{1}$ $= \tfrac{23}{6}.$ Inverting, $r = \tfrac{6}{23}.$ Therefore, the answer is B.

2025 AMC 10A 第 22 题

答案：B

难度评级：2120

解答：

其他年份的第 22 题