1973 AMC 12 第 29 题

先试着解答 1973 AMC 12 第 29 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 1973 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。

29.

两个男孩从圆形跑道上的同一点 AA 出发,沿相反方向运动。他们的速度分别为每秒 55 英尺和每秒 99 英尺。若他们同时出发,并在首次同时回到 AA 点相遇时结束,则除去起点与终点,他们相遇的次数为
Two boys start moving from the same point AA on a circular track but in opposite directions. Their speeds are 55 ft. per sec. and 99 ft. per sec. If they start at the same time and finish when they first meet at the point AA again, then the number of times they meet, excluding the start and finish, is
1313
2525
4444
无穷多次
infinity
以上都不是
none of these
答案:A
知识点:相对速度最大公约数交点计数
小提示:
因为 55 与 99 互质,确定两人首次都完成整数圈数的时刻
Because 55 and 99 are relatively prime, determine when both boys first complete whole numbers of laps
大提示:
在终止时刻之前,每当两人合计行程增加一整圈时就会相遇
Before that finish time, meetings occur whenever their combined distance is another whole lap
解答:
设跑道长度为 LL。由于 gcd⁡(5,9)=1\gcd(5,9)=1,两人首次同时回到 AA 的正时刻为 t=Lt=L:他们分别完成了 55 圈与 99 圈。相对速度为 5+9=145+9=14,所以在时刻 LL 之前,他们在 t=kL14,k=1,2,…,13 t=\frac{kL}{14}, \qquad k=1,2,\ldots,13 时相遇。因此除去起点与终点,共相遇 1313 次。
所以正确答案是 A。
Let the track length be L.L. Since gcd⁡(5,9)=1,\gcd(5,9)=1, the first positive time when both boys are back at AA is t=Lt=L: they have completed 55 and 99 laps. Their relative speed is 5+9=14,5+9=14, so before time LL they meet at t=kL14,k=1,2,…,13. t=\frac{kL}{14}, \qquad k=1,2,\ldots,13. Thus there are 1313 meetings excluding the start and finish.
Therefore, the correct answer is A.
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