Question

一个粒子位于坐标平面上的 $(5, 0)$ 。定义粒子的一次移动为：先绕原点逆时针旋转 $\frac{\pi}{4}$ 弧度，再平移 $10$ 个单位，方向为 $x$ 轴正方向。已知粒子移动 $150$ 次后的位置为 $(p, q)$ ，求小于或等于 $|p| + |q|$ 的最大整数。

A particle is located on the coordinate plane at $(5, 0).$ Define a move for the particle as a counterclockwise rotation of $\frac{\pi}{4}$ radians about the origin followed by a translation of $10$ units in the positive $x$ -direction. Given that the particle’s position after $150$ moves is $(p, q),$ find the greatest integer less than or equal to $|p| + |q|.$

答案

答案：19

知识点：复数单位根等比数列

难度评级：2840

小提示：

在复平面中，一次移动把 $z$ 变为 $\omega z + 10$ ，其中 $\omega = e^{\frac{i\pi}{4}}$

In the complex plane a move sends $z$ to $\omega z + 10,$ where $\omega = e^{\frac{i\pi}{4}}$

大提示：

迭代后，将所得等比和按每 $8$ 项一组，再利用 $\omega^8 = 1$ 化简

After iterating, group the resulting geometric sum into blocks of $8$ using $\omega^8 = 1$

解答：

将平面看作复平面，一次移动把 $z$ 变为 $\omega z + 10$ ，其中 $\omega = e^{\frac{i\pi}{4}}$ 。从 $z_0 = 5$ 开始迭代， $\begin{aligned} z_{150} &= 5\omega^{150} \\ &\quad {}+ 10 \scriptsize\left(\omega^{149} + \omega^{148} + \cdots + \omega + 1\right) \end{aligned}\text{。}$

因为 $\omega^8 = 1$ 且 $150 = 8 \cdot 18 + 6$ ，所以 $\omega^{150} = \omega^6 = -i$ 。在等比和中，每 $8$ 个连续幂之和为 $0$ ，因此这 $150$ 项化为 $\begin{aligned} &1 + \omega + \cdots + \omega^5 \\ &= -\omega^6 - \omega^7 \\ &= i - \frac{\sqrt{2}}{2} \\ &\quad {}+ \frac{\sqrt{2}}{2}i \end{aligned}\text{。}$ 因此 $\begin{aligned} z_{150} &= -5i \\ &\quad {}+ 10 \scriptsize\left(-\frac{\sqrt{2}}{2} + \left(1 + \frac{\sqrt{2}}{2}\right)i\right) \\ &= -5\sqrt{2} + \left(5 + 5\sqrt{2}\right)i \end{aligned}\text{。}$

所以 $\begin{aligned} |p| + |q| &= 5\sqrt{2} + 5 + 5\sqrt{2} \\ &= 5 + 10\sqrt{2} \\ &\approx 19.14 \end{aligned}\text{，}$ 小于或等于它的最大整数是 $19$ 。

Identify the plane with the complex plane, so a move sends $z$ to $\omega z + 10$ with $\omega = e^{\frac{i\pi}{4}}.$ Starting from $z_0 = 5$ and iterating, $\begin{aligned} z_{150} &= 5\omega^{150} \\ &\quad {}+ 10 \scriptsize\left(\omega^{149} + \omega^{148} + \cdots + \omega + 1\right). \end{aligned}$

Since $\omega^8 = 1$ and $150 = 8 \cdot 18 + 6,$ we get $\omega^{150} = \omega^6 = -i.$ In the geometric sum, every block of $8$ consecutive powers adds to $0,$ so the $150$ terms reduce to $\begin{aligned} &1 + \omega + \cdots + \omega^5 \\ &= -\omega^6 - \omega^7 \\ &= i - \frac{\sqrt{2}}{2} \\ &\quad {}+ \frac{\sqrt{2}}{2}i. \end{aligned}$ Therefore $\begin{aligned} z_{150} &= -5i \\ &\quad {}+ 10 \scriptsize\left(-\frac{\sqrt{2}}{2} + \left(1 + \frac{\sqrt{2}}{2}\right)i\right) \\ &= -5\sqrt{2} + \left(5 + 5\sqrt{2}\right)i. \end{aligned}$

Thus $\begin{aligned} |p| + |q| &= 5\sqrt{2} + 5 + 5\sqrt{2} \\ &= 5 + 10\sqrt{2} \\ &\approx 19.14, \end{aligned}$ and the greatest integer less than or equal to this is $19.$

2008 AIME II 第 9 题

答案：19

难度评级：2840

小提示：

大提示：

解答：

其他年份的第 9 题