Question

两名游泳者分别位于一个长 $90$ 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 $3$ 英尺，另一人的速度为每秒 $2$ 英尺。他们来回游了 $12$ 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 $90$ -foot pool, start to swim the length of the pool, one at the rate of $3$ feet per second, the other at $2$ feet per second. They swim back and forth for $12$ minutes. Allowing no loss of time at the turns, find the number of times they pass each other.

$24$

$21$

$20$

$19$

$18$

答案：C

知识点：相对速度反射（几何）最小公倍数

难度评级：1870

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 $180$ 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 $180$ seconds; inspect one full cycle, including endpoint coincidences

解答：

速度较快者的位置每 $60$ 秒重复一次，速度较慢者的位置每 $90$ 秒重复一次。因此，两人的位置变化共同以 $180$ 秒为周期。在一个周期内，他们的位置在 $t=18,54,90,126,162$ 秒时重合。原题答案采用的计数方式把 $t=90$ 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 $5$ 次， $12$ 分钟内共有 $4$ 个周期，得到 $5\cdot4=20$ 。

因此，正确答案是 C。

The faster swimmer’s position repeats every $60$ seconds, and the slower swimmer’s position repeats every $90$ seconds. Thus their joint position pattern repeats every $180$ seconds. During one such cycle, their positions coincide at $t=18,54,90,126,162$ seconds. The keyed interpretation counts the simultaneous turn at the endpoint when $t=90$ as one of these encounters. Thus there are $5$ counted encounters per cycle and $4$ cycles in $12$ minutes, giving $5\cdot4=20.$

Therefore, the correct answer is C.

1960 AMC 12 第 34 题

答案：C

难度评级：1870

小提示：

大提示：

解答：

其他年份的第 34 题