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2009 AMC 12B 第 18 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

18.

Rachel 和 Robert 在圆形跑道上跑步。Rachel 逆时针跑，每 9090 秒跑完一圈；Robert 顺时针跑，每 8080 秒跑完一圈。两人同时从起点出发。开始跑步后 1010 分钟到 1111 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 9090 seconds, and Robert runs clockwise and completes a lap every 8080 seconds. Both start from the start line at the same time. At some random time between 1010 minutes and 1111 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?

116\dfrac{1}{16}

18\dfrac{1}{8}

316\dfrac{3}{16}

14\dfrac{1}{4}

516\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 18\tfrac{1}{8} 圈的弧；找出在第 1010 分钟内每位跑者位于这段弧内的时间。

The picture spans 18\tfrac{1}{8} lap on each side of the start; find when each runner is within that arc during the 1010th minute

大提示：

Rachel 在 18.75-41.2518.75\text{-}41.25 秒内入镜，Robert 在 30-5030\text{-}50 秒内入镜；取重叠长度再除以 6060 秒。

Rachel is in view for 18.75-41.2518.75\text{-}41.25 s and Robert for 30-5030\text{-}50 s; take the overlap over 6060 s

解答：

照片覆盖距离起点不超过 18\dfrac{1}{8} 圈的弧。在 600600 秒时，Rachel 已跑 6236\tfrac{2}{3} 圈，距离起点还差 3030 秒；四分之一圈对应 22.522.5 秒，所以她在 30−11.25=18.7530 - 11.25 = 18.75 秒到 30+11.25=41.2530 + 11.25 = 41.25 秒之间入镜，这是第 1010 分钟内的时间。

在 600600 秒时，Robert 距离起跑线 4040 秒；四分之一圈需要 2020 秒，所以他在 3030 到 5050 秒之间入镜。两人都入镜的时间是 3030 到 41.2541.25 秒，长度为 11.2511.25 秒，占 6060 秒的比例为 11.2560=316\dfrac{11.25}{60} = \dfrac{3}{16}。

所以正确答案是 C。

The picture covers the arc within 18\dfrac{1}{8} lap of the start on each side. At 600600 s Rachel has run 6236\tfrac{2}{3} laps, 3030 s short of the line; a quarter lap takes her 22.522.5 s, so she is in view between 30−11.25=18.7530 - 11.25 = 18.75 s and 30+11.25=41.2530 + 11.25 = 41.25 s of the 1010th minute.

At 600600 s Robert is 4040 s from the line; a quarter lap takes 2020 s, so he is in view between 3030 and 5050 s. Both appear between 3030 and 41.2541.25 s, a window of 11.2511.25 s out of 60,60, giving probability 11.2560=316.\dfrac{11.25}{60} = \dfrac{3}{16}.

Thus, the correct answer is C.

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