Question

按如下方式构造两个半径为 $1$ 的圆。圆 $A$ 的圆心从连接 $(0, 0)$ 和 $(2, 0)$ 的线段上均匀随机选取。圆 $B$ 的圆心从连接 $(0, 1)$ 和 $(2, 1)$ 的线段上均匀随机选取，并且与第一次选择相互独立。圆 $A$ 和圆 $B$ 相交的概率是多少？

Two circles of radius $1$ are to be constructed as follows. The center of circle $A$ is chosen uniformly and at random from the line segment joining $(0, 0)$ to $(2, 0).$ The center of circle $B$ is chosen uniformly and at random, and independently of the first choice, from the line segment joining $(0, 1)$ to $(2, 1).$ What is the probability that circles $A$ and $B$ intersect?

$\dfrac{2 + \sqrt{2}}{4}$

$\dfrac{3\sqrt{3} + 2}{8}$

$\dfrac{2\sqrt{2} - 1}{2}$

$\dfrac{2 + \sqrt{3}}{4}$

$\dfrac{4\sqrt{3} - 3}{4}$

答案：E

知识点：几何概率距离公式

难度评级：2040

解答：

设两个圆心为 $(a, 0)$ 和 $(b, 1)$ ，其中 $a, b \in [0, 2]$ 。两个半径均为 $1$ 的圆相交，当且仅当圆心距至多为 $2$ ： $\begin{aligned} &\sqrt{(a - b)^2 + 1} \le 2 \\ &\iff |a - b| \le \sqrt{3}. \end{aligned}$

所有点对 $(a, b)$ 填满面积为 $4$ 的正方形 $[0, 2]^2$ 。失败区域 $|a - b| \gt \sqrt3$ 是两个直角三角形，每个的两条直角边长为 $2 - \sqrt3$ ，总面积为 $(2 - \sqrt3)^2 = 7 - 4\sqrt3$ 。

所以有利面积为 $4 - (7 - 4\sqrt3) = 4\sqrt3 - 3$ ，概率为 $\frac{4\sqrt3 - 3}{4}.$

因此，正确答案是 E。

Let the centers be $(a, 0)$ and $(b, 1)$ with $a, b \in [0, 2].$ The circles (radius $1$ each) intersect iff the distance between centers is at most $2:$ $\begin{aligned} &\sqrt{(a - b)^2 + 1} \le 2 \\ &\iff |a - b| \le \sqrt{3}. \end{aligned}$

The pairs $(a, b)$ fill the square $[0, 2]^2$ of area $4.$ The failing region $|a - b| \gt \sqrt3$ is two right triangles, each with legs $2 - \sqrt3,$ of total area $(2 - \sqrt3)^2 = 7 - 4\sqrt3.$

So the favorable area is $4 - (7 - 4\sqrt3) = 4\sqrt3 - 3,$ and the probability is $\frac{4\sqrt3 - 3}{4}.$

Thus, the correct answer is E.

2008 AMC 12B 第 21 题

答案：E

难度评级：2040

解答：

其他年份的第 21 题