Question

八个半圆如图沿边长为 $2$ 的正方形内侧排列。与所有这些半圆相切的圆的半径是多少？

Eight semicircles line the inside of a square with side length $2$ as shown. What is the radius of the circle tangent to all of these semicircles?

$\dfrac{1+\sqrt2}4$

$\dfrac{\sqrt5-1}2$

$\dfrac{\sqrt3+1}4$

$\dfrac{2\sqrt3}5$

$\dfrac{\sqrt5}3$

答案：B

知识点：相切圆勾股定理

难度评级：1660

解答：

从正方形中心到一个半圆圆心的距离可由直角三角形的斜边求出。

一条直角边是从正方形中心到边中点的距离，长度为 $1$ 。

另一条直角边是从边中点到半圆圆心的距离，长度为 $\dfrac 12$ 。这也说明半圆半径为 $\dfrac 12$ 。

因此正方形中心到半圆圆心的距离为再减去半圆半径 $\dfrac 12$ ，得到小圆半径 $\sqrt{1^2 + \left(\dfrac 12\right)^2} = \dfrac {\sqrt 5}2.$ $\dfrac{\sqrt 5 -1}2 .$

所以正确答案是 B。

The distance from the center of the square to the center of the semicircles can be found as a hypotenuse of a right triangle.

One of the legs is from the center of the square to the center of one of the sides which is of distance $1.$

The other leg is from the center of the side to the center of one of the semicircles which is of distance $\dfrac 12.$ This also shows that the radius of the semicircles is $\dfrac 12.$

Therefore, the distance from the center of the square to the center of the semicircle is $\sqrt{1^2 + \left(\dfrac 12\right)^2} = \dfrac {\sqrt 5}2.$ Then we subtract $\dfrac 12$ for the radius of the semicircle. This makes the radius of the circle $\dfrac{\sqrt 5 -1}2 .$

Thus, the correct answer is B .

2014 AMC 10B 第 22 题

答案：B

难度评级：1660

解答：

其他年份的第 22 题