AMC10 2014 B

AMC10 2014 B · Q19

AMC10 2014 B · Q19. It mainly tests Probability (basic), Circle theorems.

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?

两个同心圆，半径分别为1和2。在外圆上独立均匀随机选择两点。连接两点的弦与内圆相交的概率是多少？

(A) $\frac{1}{6}$ $\frac{1}{6}$

(B) $\frac{1}{4}$ $\frac{1}{4}$

(D) $\frac{1}{3}$ $\frac{1}{3}$

(E) $\frac{1}{2}$ $\frac{1}{2}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $A$ be the first point chosen on the outer circle, let chords $\overline{AB}$ and $\overline{AC}$ on the outer circle be tangent to the inner circle at $D$ and $E$, respectively, and let $O$ be the common center of the two circles. Triangle $ADO$ has a right angle at $D$, $OA=2$, and $OD=1$, so $\angle OAD=30^\circ$. Similarly, $\angle OAE=30^\circ$, so $\angle BAC=\angle DAE=60^\circ$, and minor arc $BC=120^\circ$. If $X$ is the second point chosen on the outer circle, then chord $\overline{AX}$ intersects the inner circle if and only if $X$ is on minor arc $BC$. Therefore the requested probability is $\frac{120^\circ}{360^\circ}=\frac{1}{3}$.

答案（D）：设 $A$ 为外圆上选取的第一个点，外圆上的弦段 $\overline{AB}$ 与 $\overline{AC}$ 分别在 $D$ 与 $E$ 处与内圆相切，设 $O$ 为两圆的公共圆心。三角形 $ADO$ 在 $D$ 处为直角，且 $OA=2$、$OD=1$，因此 $\angle OAD=30^\circ$。同理，$\angle OAE=30^\circ$，所以 $\angle BAC=\angle DAE=60^\circ$，从而小弧 $BC=120^\circ$。若 $X$ 为外圆上选取的第二个点，则弦段 $\overline{AX}$ 与内圆相交当且仅当 $X$ 在小弧 $BC$ 上。因此所求概率为 $\frac{120^\circ}{360^\circ}=\frac{1}{3}$。

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