Question

一条 $100$ 英尺长的自动人行道以每秒 $6$ 英尺的恒定速度移动。Al 走到人行道起点并站在上面。Bob 在两秒后走到人行道起点，并以每秒 $4$ 英尺的恒定速度沿人行道向前步行。再过两秒，Cy 到达人行道起点，并在人行道旁以每秒 $8$ 英尺的恒定速度快步向前走。在某个时刻，这三人中的一人恰好位于另外两人的正中间。在那个时刻，求中间那人与人行道起点之间的距离，单位为英尺。

A $100$ foot long moving walkway moves at a constant rate of $6$ feet per second. Al steps onto the start of the walkway and stands. Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of $4$ feet per second. Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constant rate of $8$ feet per second. At a certain time, one of these three persons is exactly halfway between the other two. At that time, find the distance in feet between the start of the walkway and the middle person.

答案

答案：52

知识点：路程、速度与时间一次方程分类讨论

难度评级：2020

解答：

从 Al 踏上人行道开始，设经过的时间为 $t$ 秒。Al 站在人行道上，所以位置是 $6t$ ； Bob 的速度为 $6 + 4 = 10$ 英尺每秒，所以位置是 $10(t - 2)$ ； Cy 在人行道旁以 $8$ 英尺每秒行走，所以位置是 $8(t - 4)$ 。当 $t \ge 4$ 时三人都已经在运动。

中间人的位置的两倍必须等于另外两人的位置之和。若 Bob 在中间， $20(t-2) = 6t + 8(t-4)$ 得 $t = \frac{4}{3} \lt 4$ ，不可能。若 Cy 在中间， $16(t-4) = 6t + 10(t-2)$ 化简为 $-64 = -20$ ，无解。若 Al 在中间， $12t = 10(t-2)$ $+ 8(t-4) = 18t - 52$ ，所以 $t = \frac{26}{3}$ 。

此时 Al 的位置为 $6 \cdot \frac{26}{3} = 52$ 英尺，而 Bob 和 Cy 的位置分别为 $\frac{200}{3}$ 和 $\frac{112}{3}$ ，二者的平均值确实是 $52$ 。中间人与起点相距 $52$ 英尺。

Measure time $t$ in seconds from when Al steps on. Al stands on the walkway, so he is at $6t;$ Bob moves at $6 + 4 = 10$ feet per second, so he is at $10(t - 2);$ Cy walks beside the walkway at $8$ feet per second, so he is at $8(t - 4).$ All three are moving once $t \ge 4.$

The middle person's position doubled must equal the sum of the other two. If Bob were in the middle, $20(t-2) = 6t + 8(t-4)$ gives $t = \frac{4}{3} \lt 4,$ impossible. If Cy were in the middle, $16(t-4) = 6t + 10(t-2)$ reduces to $-64 = -20,$ with no solution. If Al is in the middle, $12t = 10(t-2)$ $+ 8(t-4) = 18t - 52,$ so $t = \frac{26}{3}.$

At that moment Al is at $6 \cdot \frac{26}{3} = 52$ feet, while Bob and Cy are at $\frac{200}{3}$ and $\frac{112}{3},$ whose average is indeed $52.$ The middle person is $52$ feet from the start.

2007 AIME I 第 2 题

答案：52

难度评级：2020

解答：

其他年份的第 2 题