2008 AIME II Problem 9

Attempt Problem 9 of the 2008 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AIME II solutions, or check the answer key.

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9.

A particle is located on the coordinate plane at (5,0).(5, 0). Define a move for the particle as a counterclockwise rotation of π4\frac{\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|.

Answer: 19
Concepts:complex numberroots of unitygeometric sequence
Difficulty rating: 2840
Small Hint:

In the complex plane a move sends zz to ωz+10,\omega z + 10, where ω=eiπ4\omega = e^{\frac{i\pi}{4}}

Big Hint:

After iterating, group the resulting geometric sum into blocks of 88 using ω8=1\omega^8 = 1

Solution:

Identify the plane with the complex plane, so a move sends zz to ωz+10\omega z + 10 with ω=eiπ4.\omega = e^{\frac{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++ω5=ω6ω7=i22+22i. \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 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=5+10219.14, \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 19.19.

Problem 8#8
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