Question

一只虫子白天一直走路，夜里睡觉。第一天，它从点 $O$ ，出发，面向东方，向正东走了 $5$ 个单位。每天夜里，虫子逆时针转 $60^\circ$ 。每天白天，它沿新的方向走前一天一半的距离。虫子会任意接近点 $P$ 。则 $OP^2 = \frac{m}{n}$ ，其中 $m$ 与 $n$ 是互质正整数。求 $m + n$ 。

A bug walks all day and sleeps all night. On the first day, it starts at point $O,$ faces east, and walks a distance of $5$ units due east. Each night the bug rotates $60^\circ$ counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to point $P.$ Then $OP^2 = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：103

知识点：复数等比数列

难度评级：2560

解答：

在复平面中工作，令 $O$ 为原点，正实轴表示东方。每天的位移都由前一天乘以 $z = \frac{1}{2}e^{i\pi/3}$ ，得到，所以 $\begin{aligned} P &= 5\left(1 + z + z^2 + \cdots\right) \\ &= \frac{5}{1 - z}. \end{aligned}$

由于 $z = \frac{1}{4} + \frac{\sqrt{3}}{4}i$ 有 $1 - z = \frac{3}{4} - \frac{\sqrt{3}}{4}i$ ，其模长平方为 $\frac{9}{16} + \frac{3}{16} = \frac{3}{4}$ 。因此 $OP^2 = \frac{25}{3/4} = \frac{100}{3},$ 且 $m + n = 100 + 3 = 103$ 。

Work in the complex plane with $O$ at the origin and east along the positive real axis. Each day's displacement is the previous one multiplied by $z = \frac{1}{2}e^{i\pi/3},$ so $\begin{aligned} P &= 5\left(1 + z + z^2 + \cdots\right) \\ &= \frac{5}{1 - z}. \end{aligned}$

Since $z = \frac{1}{4} + \frac{\sqrt{3}}{4}i,$ we get $1 - z = \frac{3}{4} - \frac{\sqrt{3}}{4}i,$ whose squared magnitude is $\frac{9}{16} + \frac{3}{16} = \frac{3}{4}.$ Therefore $OP^2 = \frac{25}{3/4} = \frac{100}{3},$ and $m + n = 100 + 3 = 103.$

2020 AIME I 第 8 题

答案：103

难度评级：2560

解答：

其他年份的第 8 题