Question

两个同心圆半径分别为 $1$ 和 $2$ 。在外圆上独立且均匀随机选取两点。连接这两点的弦与内圆相交的概率是多少？

Two concentric circles have radii $1$ and $2.$ Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?

$\ \dfrac{1}{6}$

$\ \dfrac{1}{4}$

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

$\ \dfrac{1}{3}$

$\ \dfrac{1}{2}$

答案：D

知识点：几何概率弦切线

难度评级：1600

解答：

在外圆上固定第一个端点 $A$ 。从 $A$ 作两条与内圆相切的弦，并把它们的另一个端点记为 $B$ 和 $C$ 。从 $A$ 出发的弦与内圆相交，当且仅当它的另一个端点位于小弧 $BC$ 上。

设 $O$ 为两圆的共同圆心， $D$ 为一个切点，则 $\triangle AOD$ 是直角三角形，其中 $OA=2$ ， $OD=1$ 。所以 $\angle OAD=30^\circ$ 。两条切线弦在 $A$ 处所成角为 $60^\circ$ ，因此所截小弧 $BC$ 的度数为 $120^\circ$ 。

所以所求概率为 $\dfrac {120^\circ}{360^\circ} = \dfrac 13 .$

所以正确答案是 D。

Fix the first endpoint $A$ on the outer circle. Draw the two chords from $A$ that are tangent to the inner circle, and call their other endpoints $B$ and $C$ . A chord from $A$ meets the inner circle exactly when its second endpoint lies on the minor arc $BC$ .

If $O$ is the common center and $D$ is a tangency point, then $\triangle AOD$ is right, with $OA=2$ and $OD=1$ . Hence $\angle OAD=30^\circ$ . The two tangent chords therefore make a $60^\circ$ angle at $A$ , so the intercepted minor arc $BC$ measures $120^\circ$ .

Therefore, the probability is $\dfrac {120^\circ}{360^\circ} = \dfrac 13 .$

Thus, the correct answer is D .

2014 AMC 10B 第 19 题

答案：D

难度评级：1600

解答：

其他年份的第 19 题