Difference between revisions of "Mock AIME 3 2006-2007 Problems/Problem 8"
(Created page with "== Problem == A cube has vertices <math>(0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0)</math>, and <math>(1,1,1)</math>. At <math>t=0</math>, a particle is at <ma...") |
(→Solution) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 2: | Line 2: | ||
A cube has vertices <math>(0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0)</math>, and <math>(1,1,1)</math>. At <math>t=0</math>, a particle is at <math>(12,12,0)</math>. After one second, the particle travels to <math>(\frac{9}{16},\frac{5}{8},\frac{1}{13})</math>. 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. | A cube has vertices <math>(0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0)</math>, and <math>(1,1,1)</math>. At <math>t=0</math>, a particle is at <math>(12,12,0)</math>. After one second, the particle travels to <math>(\frac{9}{16},\frac{5}{8},\frac{1}{13})</math>. 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 <math>(\frac{1}{2},\frac{1}{2},0)</math> is the new origin. Now the point <math>(\frac{9}{16},\frac{5}{8},\frac{1}{13})</math> becomes | ||
| + | <math>(\frac{1}{16},\frac{1}{8},\frac{1}{13})</math>. Let <math> \overrightarrow{v} </math> be a vector with head <math>(\frac{1}{16},\frac{1}{8},\frac{1}{13})</math>. 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 <math>(\frac{t}{16},\frac{t}{8},\frac{t}{13})</math> after <math>t</math> seconds. The particle reaches an edge (stops) if the coordinates satisfy 2 of the 3 following conditions: | ||
| + | |||
| + | 1) <math>x</math> is an integer multiple of <math>\frac{1}{2}</math> | ||
| + | |||
| + | 2) <math>y</math> is an integer multiple of <math>\frac{1}{2}</math> | ||
| + | |||
| + | 3) <math>z</math> is an integer | ||
| + | |||
| + | From inspection, the first time that this occurs is when <math>t=\boxed{052}</math> | ||
Latest revision as of 19:48, 15 February 2015
Problem
A cube has vertices 
, and 
. At 
, a particle is at 
. After one second, the particle travels to 
. 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 
is the new origin. Now the point becomes
. Let 
be a vector with head . 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 
after 
seconds. The particle reaches an edge (stops) if the coordinates satisfy 2 of the 3 following conditions:
1) 
is an integer multiple of 
2) 
is an integer multiple of
3) 
is an integer
From inspection, the first time that this occurs is when 
