# 2008 AIME II Problems/Problem 9

## Problem

A particle is located on the coordinate plane at $(5,0)$. Define a move for the particle as a counterclockwise rotation of $\pi/4$ 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 $|p| + |q|$.

## Solution

### Solution 1

Let P(x, y) be the position of the particle on the xy-plane, r be the length OP where O is the origin, and $\theta$ be the inclination of OP to the x-axis. If (x', y') is the position of the particle after a move from P, then $x'=rcos(\pi/4+\theta)+10 = \sqrt{2}(x - y)/2 + 10$ and $y' = rsin(\pi/4+\theta) = \sqrt{2}(x + y)/2$. Let $(x_n, y_n)$ be the position of the particle after the nth move, where $x_0 = 5$ and $y_0 = 0$. Then $x_{n+1} + y_{n+1} = \sqrt{2}x_n+10$, $x_{n+1} - y_{n+1} = -\sqrt{2}y_n+10$. This implies $x_{n+2} = -y_n + 5\sqrt{2}+ 10$, $y_{n+2}=x_n + 5\sqrt{2}$. Substituting $x_0 = 5$ and $y_0 = 0$, we have $x_8 = 5$ and $y_8 = 0$ again for the first time. Thus, $p = x_{150} = x_6 = -5\sqrt{2}$ and $q = y_{150} = y_6 = 5 + 5\sqrt{2}$. Hence, the final answer is $5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}$

If you're curious, the points do eventually form an octagon and repeat. Seems counterintuitive, but believe it or not, it happens.

### Solution 2

Let the particle's position be represented by a complex number. Recall that multiplying a number by cis $\left( \theta \right)$ rotates the object in the complex plane by $\theta$ counterclockwise. In this case, we use $a = cis(\frac{\pi}{4})$. Therefore, applying the rotation and shifting the coordinates by 10 in the positive x direction in the complex plane results to $a_{150} = (((5a + 10)a + 10)a + 10 \ldots) = 5a^{150} + 10 a^{149} + 10a^{148}+ \ldots + 10$

where a is cis $\left( \theta \right)$. By De-Moivre's theorem, $\left(cis( \theta \right)^n )$=cis $\left(n \theta \right)$. Therefore, $10(a^{150} + \ldots + 1) = 10(1 + a + \ldots + a^6) = - 10(a^7) = - 10(\frac{ \sqrt {2} }{2} - \frac{i\sqrt {2}} {2})$

Furthermore, $5a^{150} = - 5i$. Thus, the final answer is $5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}$

### Solution 3

Each move has two parts, a rotation and a translation. But consider what happens to a translation after four rotations: it is cancelled out by another translation in the opposite direction. Thus the particle repeats position every 8 moves. So we only have to move backwards two steps from move 152 = 8(19).

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 