1960 AMC 12 第 34 题

先试着解答 1960 AMC 12 第 34 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1960 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

34.

两名游泳者分别位于一个长 9090 英尺的泳池两端,同时开始沿泳池长度方向游泳,一人的速度为每秒 33 英尺,另一人的速度为每秒 22 英尺。他们来回游了 1212 分钟。假设转身不耗费时间,求他们相互经过的次数。
Two swimmers, at opposite ends of a 9090-foot pool, start to swim the length of the pool, one at the rate of 33 feet per second, the other at 22 feet per second. They swim back and forth for 1212 minutes. Allowing no loss of time at the turns, find the number of times they pass each other.
2424
2121
2020
1919
1818
答案:C
知识点:相对速度反射(几何)最小公倍数
小提示:
每当游泳者转身时,将泳池作镜像延拓,使两条路径都变成直线
Reflect the pool each time a swimmer turns so both paths become straight
大提示:
两人的位置变化共同以 180180 秒为周期;检查一个完整周期,并计入端点处的位置重合
The joint position pattern repeats every 180180 seconds; inspect one full cycle, including endpoint coincidences
解答:
速度较快者的位置每 6060 秒重复一次,速度较慢者的位置每 9090 秒重复一次。因此,两人的位置变化共同以 180180 秒为周期。在一个周期内,他们的位置在 t=18,54,90,126,162 t=18,54,90,126,162 秒时重合。原题答案采用的计数方式把 t=90t=90 秒时两人在端点同时转身也计作一次相遇。因此每个周期计 55 次,1212 分钟内共有 44 个周期,得到 5⋅4=205\cdot4=20。
因此,正确答案是 C。
The faster swimmer’s position repeats every 6060 seconds, and the slower swimmer’s position repeats every 9090 seconds. Thus their joint position pattern repeats every 180180 seconds. During one such cycle, their positions coincide at t=18,54,90,126,162 t=18,54,90,126,162 seconds. The keyed interpretation counts the simultaneous turn at the endpoint when t=90t=90 as one of these encounters. Thus there are 55 counted encounters per cycle and 44 cycles in 1212 minutes, giving 5⋅4=20.5\cdot4=20.
Therefore, the correct answer is C.
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