AMC12 2003 B

AMC12 2003 B · Q25

AMC12 2003 B · Q25. It mainly tests Probability (basic), Circle theorems.

Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distances between the points are less than the radius of the circle?

在圆上随机且相互独立地选取三个点。三点之间所有两两距离都小于圆的半径的概率是多少？

(A) \dfrac{1}{36} \dfrac{1}{36}

(B) \dfrac{1}{24} \dfrac{1}{24}

(D) \dfrac{1}{12} \dfrac{1}{12}

(E) \dfrac{1}{9} \dfrac{1}{9}

Answer

Correct choice: (D)

正确答案：（D）

Solution

The first point is placed anywhere on the circle, because it doesn't matter where it is chosen. The next point must lie within $60$ degrees of arc on either side, a total of $120$ degrees possible, giving a total $\frac{1}{3}$ chance. The last point must lie within $60$ degrees of both points. The minimum area of freedom we have to place the third point is a $60$ degrees arc(if the first two are $60$ degrees apart), with a $\frac{1}{6}$ probability. The maximum amount of freedom we have to place the third point is a $120$ degree arc(if the first two are the same point), with a $\frac{1}{3}$ probability. As the second point moves farther away from the first point, up to a maximum of $60$ degrees, the probability changes linearly (every degree it moves, adds one degree to where the third could be). Therefore, we can average probabilities at each end to find $\frac{1}{4}$, the average probability we can place the third point based on a varying second point. Therefore the total probability is $1\times\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$ or $\boxed{\text{(D)}}$

第一个点可以放在圆上的任意位置，因为选在哪里并不影响结果。第二个点必须落在第一个点两侧各 $60$ 度的弧段内，总共 $120$ 度，因此概率为 $\frac{1}{3}$。第三个点必须同时落在距前两个点各不超过 $60$ 度的弧段内。放置第三个点的自由弧长最小为 $60$ 度（当前两点相距 $60$ 度时），对应概率 $\frac{1}{6}$。放置第三个点的自由弧长最大为 $120$ 度（当前两点重合时），对应概率 $\frac{1}{3}$。当第二个点从与第一个点重合逐渐移动到最多相距 $60$ 度时，这个概率线性变化（每移动 $1$ 度，第三个点可选的范围增加 $1$ 度）。因此可取两端概率的平均得到 $\frac{1}{4}$，即在第二个点变化时第三个点可放置的平均概率。所以总概率为 $1\times\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$，即 $\boxed{\text{(D)}}$。

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