2008 AIME II 第 9 题

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9.

一个粒子位于坐标平面上的 (5,0)(5, 0)。定义粒子的一次移动为:先绕原点逆时针旋转 π/4\pi/4 弧度,再沿 xx 轴正方向平移 1010 个单位。已知粒子移动 150150 次后的位置为 (p,q)(p, q),求小于或等于 p+q|p| + |q| 的最大整数。

A particle is located on the coordinate plane at (5,0).(5, 0). Define a move for the particle as a counterclockwise rotation of π/4\pi/4 radians about the origin followed by a translation of 1010 units in the positive xx-direction. Given that the particle's position after 150150 moves is (p,q),(p, q), find the greatest integer less than or equal to p+q.|p| + |q|.

答案:19
知识点:复数单位根等比数列
难度评级:2840
解答:

将平面看作复平面,一次移动把 zz 变为 ωz+10\omega z + 10,其中 ω=eiπ/4\omega = e^{i\pi/4}。从 z0=5z_0 = 5 开始迭代, z150=5ω150+10(ω149+ω148++ω+1). \begin{aligned} z_{150} &= 5\omega^{150} \\ &\quad {}+ 10 \scriptsize\left(\omega^{149} + \omega^{148} + \cdots + \omega + 1\right). \end{aligned}

因为 ω8=1\omega^8 = 1150=818+6150 = 8 \cdot 18 + 6,所以 ω150=ω6=i\omega^{150} = \omega^6 = -i。 在等比和中,每 88 个连续幂之和为 00,因此这 150150 项化为 1+ω++ω51 + \omega + \cdots + \omega^5 =ω6ω7= -\omega^6 - \omega^7 =i(2222i)= i - \left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\right)。因此 z150=5i+10(22+(1+22)i)=52+(5+52)i. \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}

所以 p+q=52+5+52|p| + |q| = 5\sqrt{2} + 5 + 5\sqrt{2} =5+10219.14= 5 + 10\sqrt{2} \approx 19.14, 小于或等于它的最大整数是 1919

Identify the plane with the complex plane, so a move sends zz to ωz+10\omega z + 10 with ω=eiπ/4.\omega = e^{i\pi/4}. Starting from z0=5z_0 = 5 and iterating, z150=5ω150+10(ω149+ω148++ω+1). \begin{aligned} z_{150} &= 5\omega^{150} \\ &\quad {}+ 10 \scriptsize\left(\omega^{149} + \omega^{148} + \cdots + \omega + 1\right). \end{aligned}

Since ω8=1\omega^8 = 1 and 150=818+6,150 = 8 \cdot 18 + 6, we get ω150=ω6=i.\omega^{150} = \omega^6 = -i. In the geometric sum, every block of 88 consecutive powers adds to 0,0, so the 150150 terms reduce to 1+ω++ω51 + \omega + \cdots + \omega^5 =ω6ω7= -\omega^6 - \omega^7 =i(2222i).= i - \left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\right). Therefore z150=5i+10(22+(1+22)i)=52+(5+52)i. \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 p+q=52+5+52|p| + |q| = 5\sqrt{2} + 5 + 5\sqrt{2} =5+10219.14,= 5 + 10\sqrt{2} \approx 19.14, and the greatest integer less than or equal to this is 19.19.

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