AMC12 2008 B

AMC12 2008 B · Q21

AMC12 2008 B · Q21. It mainly tests Probability (basic), Coordinate geometry.

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 $(0,0)$ and $(2,0)$. 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?

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

(A) $\dfrac{2 + \sqrt{2}}{4}$ $\dfrac{2 + \sqrt{2}}{4}$

(B) $\dfrac{3\sqrt{3} + 2}{8}$ $\dfrac{3\sqrt{3} + 2}{8}$

(D) $\dfrac{2 + \sqrt{3}}{4}$ $\dfrac{2 + \sqrt{3}}{4}$

(E) $\dfrac{4\sqrt{3} - 3}{4}$ $\dfrac{4\sqrt{3} - 3}{4}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Circles centered at $A$ and $B$ will overlap if $A$ and $B$ are closer to each other than if the circles were tangent. The circles are tangent when the distance between their centers is equal to the sum of their radii. Thus, the distance from $A$ to $B$ will be $2$. Since $A$ and $B$ are separated by $1$ vertically, they must be separated by $\sqrt{3}$ horizontally. Thus, if $|A_x-B_x|<\sqrt{3}$, the circles intersect. Now, plot the two random variables $A_x$ and $B_x$ on the coordinate plane. Each variable ranges from $0$ to $2$. The circles intersect if the variables are within $\sqrt{3}$ of each other. Thus, the area in which the circles don't intersect is equal to the total area of two small triangles on opposite corners, each of area $\frac{(2-\sqrt{3})^2}{2}$. So, the total area of the 2 triangles sums to $(2-\sqrt{3})^2$. Since the total $2 \times 2$ square has an area of $4$, the probability of the circles not intersecting is $\frac{(2-\sqrt{3})^2}{4}$. But remember, we want the probability that they do intersect. We conclude the probability the circles intersect is:\[1-\frac{(2-\sqrt{3})^2}{4}=\boxed{\textbf{(E)}\frac{4\sqrt{3}-3}{4}}.\]

以 $A$ 和 $B$ 为圆心的两圆在它们的圆心距离小于两圆相切时的圆心距离时会相交。两圆相切时圆心距离等于半径之和，因此 $A$ 到 $B$ 的距离为 $2$。由于 $A$ 与 $B$ 在竖直方向相差 $1$，它们在水平方向必须相差 $\sqrt{3}$。因此当 $|A_x-B_x|<\sqrt{3}$ 时，两圆相交。现在在坐标平面上作出两个随机变量 $A_x$ 与 $B_x$。每个变量的取值范围都是从 $0$ 到 $2$。当两变量相差不超过 $\sqrt{3}$ 时两圆相交。因此两圆不相交的区域面积等于正方形两对角处的两个小三角形面积之和，每个三角形面积为 $\frac{(2-\sqrt{3})^2}{2}$。所以两个三角形总面积为 $(2-\sqrt{3})^2$。由于整个 $2\times 2$ 正方形面积为 $4$，两圆不相交的概率为 $\frac{(2-\sqrt{3})^2}{4}$。但我们要求的是相交的概率，因此 \[1-\frac{(2-\sqrt{3})^2}{4}=\boxed{\textbf{(E)}\frac{4\sqrt{3}-3}{4}}.\]

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