Question

在最近的一次竞选活动中，一名候选人在一个假定为平面的国家中巡回旅行。第一天他向东走，第二天向北走，第三天向西走，第四天向南走，第五天又向东走，如此循环。若他每天行进 $\frac{n^2}{2}$ 英里，其中 $n$ 是当天的编号，那么第 $40$ 天结束时，他与出发点相距多少英里？

During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went $\frac{n^2}{2}$ miles on the $n$ th day of this tour, how many miles was he from his starting point at the end of the $40$ th day?

答案

答案：580

知识点：等差数列距离公式向量

难度评级：2070

小提示：

将 $40$ 天分成十个四天周期，分别求水平位移与竖直位移之和

Group the $40$ days into ten four-day cycles and sum horizontal and vertical displacements separately

大提示：

对周期编号 $k$ ，比较 $(4k+1)^2$ 与 $(4k+3)^2$ ，并用同样的方法比较另一对

For cycle index $k,$ compare $(4k+1)^2$ with $(4k+3)^2$ , and similarly compare the other pair

解答：

用 $k=0,1,\ldots,9$ 给十个周期编号。水平位移为 $\begin{aligned}&\frac12\sum_{k=0}^9\left((4k+1)^2-(4k+3)^2\right)\\&\quad=\sum_{k=0}^9(-8k-4)\\&\quad=-400\end{aligned}\text{。}$ 同理，竖直位移为 $\begin{aligned}&\frac12\sum_{k=0}^9\left((4k+2)^2-(4k+4)^2\right)\\&\quad=\sum_{k=0}^9(-8k-6)\\&\quad=-420\end{aligned}\text{。}$ 因此，他与出发点的距离为 $\sqrt{400^2+420^2}=580$ 。

Index the ten cycles by $k=0,1,\ldots,9.$ The horizontal displacement is $\begin{aligned}&\frac12\sum_{k=0}^9\left((4k+1)^2-(4k+3)^2\right)\\&\quad=\sum_{k=0}^9(-8k-4)\\&\quad=-400.\end{aligned}$ Similarly, the vertical displacement is $\begin{aligned}&\frac12\sum_{k=0}^9\left((4k+2)^2-(4k+4)^2\right)\\&\quad=\sum_{k=0}^9(-8k-6)\\&\quad=-420.\end{aligned}$ Therefore the distance from the start is $\sqrt{400^2+420^2}=580.$

1993 AIME 第 2 题

答案：580

难度评级：2070

小提示：

大提示：

解答：

其他年份的第 2 题