2009 AMC 12B 第 18 题

Question

Rachel 和 Robert 在圆形跑道上跑步。Rachel 逆时针跑，每

90

秒跑完一圈；Robert 顺时针跑，每

80

秒跑完一圈。两人同时从起点出发。开始跑步后

10

分钟到

11

分钟之间的某个随机时刻，一位站在跑道内的摄影师拍了一张照片，照片显示以起跑线为中心的四分之一圈跑道。两人都在照片中的概率是多少？

Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every

90

seconds, and Robert runs clockwise and completes a lap every

80

seconds. Both start from the start line at the same time. At some random time between

10

minutes and

11

minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?

\dfrac{1}{16}

\dfrac{1}{8}

\dfrac{3}{16}

\dfrac{1}{4}

\dfrac{5}{16}

答案：C

知识点：几何概率路程、速度与时间

难度评级：1930

小提示：

照片覆盖起点两侧各

\tfrac{1}{8}

圈的弧；找出在第

10

分钟内每位跑者位于这段弧内的时间。

The picture spans

\tfrac{1}{8}

lap on each side of the start; find when each runner is within that arc during the

10

th minute

大提示：

Rachel 在

18.75\text{-}41.25

秒内入镜，Robert 在

30\text{-}50

秒内入镜；取重叠长度再除以

60

秒。

Rachel is in view for

18.75\text{-}41.25

s and Robert for

30\text{-}50

s; take the overlap over

60

s

解答：

照片覆盖距离起点不超过

\dfrac{1}{8}

圈的弧。在

600

秒时，Rachel 已跑

6\tfrac{2}{3}

圈，距离起点还差

30

秒；四分之一圈对应

22.5

秒，所以她在

30 - 11.25 = 18.75

秒到

30 + 11.25 = 41.25

秒之间入镜，这是第

10

分钟内的时间。

在

600

秒时，Robert 距离起跑线

40

秒；四分之一圈需要

20

秒，所以他在

30

到

50

秒之间入镜。两人都入镜的时间是

30

到

41.25

秒，长度为

11.25

秒，占

60

秒的比例为

\dfrac{11.25}{60} = \dfrac{3}{16}

。

所以正确答案是 C。

The picture covers the arc within

\dfrac{1}{8}

lap of the start on each side. At

600

s Rachel has run

6\tfrac{2}{3}

laps,

30

s short of the line; a quarter lap takes her

22.5

s, so she is in view between

30 - 11.25 = 18.75

s and

30 + 11.25 = 41.25

s of the

10

th minute.

At

600

s Robert is

40

s from the line; a quarter lap takes

20

s, so he is in view between

30

and

50

s. Both appear between

30

and

41.25

s, a window of

11.25

s out of

60,

giving probability

\dfrac{11.25}{60} = \dfrac{3}{16}.

Thus, the correct answer is C.

2009 AMC 12B 第 18 题

答案：C

难度评级：1930

小提示：

大提示：

解答：

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