Question

如图，一个半径为 $1$ 的圆被 $4$ 个半径为 $r$ 的圆围住。 $r$ 是多少？

A circle of radius $1$ is surrounded by $4$ circles of radius $r$ as shown. What is $r?$

$\sqrt2$

$1+\sqrt2$

$\sqrt6$

$3$

$2+\sqrt2$

答案：B

知识点：相切圆正方形（几何）二次方程

难度评级：1680

解答：

连接四个外圆的圆心形成正方形。相邻外圆相切，所以每条边长为 $2r$ 。

正方形对角线经过中心圆，长度为 $1+r+r+1=2+2r$ 。边长为 $2r$ 的正方形对角线为 $2r\sqrt2$ ，所以 $2(2r)^2=(2+2r)^2$ 。

展开得 $1+2r+r^2=2r^2$ ，即 $r^2-2r-1=0$ 。正根为 $r=1+\sqrt2$ 。

所以正确答案是 B。

Connect the centers of the four outer circles to form a square. Adjacent outer circles are tangent, so each side has length $2r.$

The diagonal of the square passes through the center circle, giving length $1+r+r+1=2+2r.$ Since a square with side $2r$ has diagonal $2r\sqrt2,$ we get $2(2r)^2=(2+2r)^2.$

Expanding gives $1+2r+r^2=2r^2,$ so $r^2-2r-1=0.$ The positive root is $r=1+\sqrt2.$

Thus, the correct answer is B.

2007 AMC 10B 第 18 题

答案：B

难度评级：1680

解答：

其他年份的第 18 题