# Mock AIME 3 2006-2007 Problems/Problem 8

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## Problem

A cube has vertices $(0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0)$, and $(1,1,1)$. At $t=0$, a particle is at $(12,12,0)$. After one second, the particle travels to $(\frac{9}{16},\frac{5}{8},\frac{1}{13})$. The particle bounces until it hits an edge or a corner at which point it stops. If the particle travels in straight lines with uniform speed, find the time (in seconds) when the particle stops motion.

## Solution

Shift the cube so that $(\frac{1}{2},\frac{1}{2},0)$ is the new origin. Now the point $(\frac{9}{16},\frac{5}{8},\frac{1}{13})$ becomes $(\frac{1}{16},\frac{1}{8},\frac{1}{13})$. Let $\overrightarrow{v}$ be a vector with head $(\frac{1}{16},\frac{1}{8},\frac{1}{13})$. If you imagine the particle traveling through infinite unit cubes in space instead of the particle bouncing around in the unit cube (a common method for reflection problems), the particle will be at $(\frac{t}{16},\frac{t}{8},\frac{t}{13})$ after $t$ seconds. The particle reaches an edge (stops) if the coordinates satisfy 2 of the 3 following conditions:

1) $x$ is an integer multiple of $\frac{1}{2}$

2) $y$ is an integer multiple of $\frac{1}{2}$

3) $z$ is an integer

From inspection, the first time that this occurs is when $t=\boxed{052}$