Question

如图，四个半径为 $1$ 的圆各自与正方形的两条边相切，并且都与一个半径为 $2$ 的圆外切。正方形的面积是多少？

Four circles of radius $1$ are each tangent to two sides of a square and externally tangent to a circle of radius $2,$ as shown. What is the area of the square?

$32$

$22 + 12\sqrt{2}$

$16 + 16\sqrt{3}$

$48$

$36 + 16\sqrt{2}$

答案：B

知识点：相切圆特殊直角三角形正方形（几何）

难度评级：1540

解答：

考虑连接半径 $2$ 的圆心和两个相邻小圆圆心形成的等腰直角三角形。它的两条直角边长为 $2 + 1 = 3$ ，所以斜边为 $3\sqrt2$ 。

正方形边长比这条斜边多 $2$ （两端各一个小圆半径），所以 $s = 2 + 3\sqrt2$ 。

面积为 $\begin{aligned} (2 + 3\sqrt2)^2 &= 4 + 12\sqrt2 + 18 \\ &= 22 + 12\sqrt2. \end{aligned}$

所以正确答案是 B。

Consider the isosceles right triangle joining the center of the radius- $2$ circle to the centers of two adjacent small circles. Its legs have length $2 + 1 = 3,$ so its hypotenuse is $3\sqrt2.$

The side of the square exceeds this hypotenuse by $2$ (one radius on each end), so $s = 2 + 3\sqrt2.$

The area is $\begin{aligned} (2 + 3\sqrt2)^2 &= 4 + 12\sqrt2 + 18 \\ &= 22 + 12\sqrt2. \end{aligned}$

Thus, the correct answer is B.

2007 AMC 10A 第 15 题

答案：B

难度评级：1540

解答：

其他年份的第 15 题