A courier pigeon departs Las Vegas for Reno at the same time as another courier pigeon departs Reno for Las Vegas. Both pigeons fly at constant speeds, although different from each other. They cross paths 2x miles from Las Vegas. After each arrives at their destination they immediately turn around, going back and forth without breaks. They cross paths the second time x miles from Reno.
Where will they cross paths the third time?
Let the distance between Reno and Las Vegas be the unit of distance, and the time it takes the bird first leaving Reno (the R bird) to get to Las Vegas be the unit of time. Call the speed of the bird originally starting at Las Vegas (the LV bird) s. Let x miles be y units of our distance measure.
When the two birds meed for the first time, the LV bird has traveled a distance of 2y (we can convert to miles later if necessary, but right now we're using Reno-LasVegas units), and has been traveling at rate s, and so has taken 2y/s units of time. Meanwhile the R bird has traveled 1-2y units traveling as speed 1. Therefore (2y/s)=(1-2y).
When they meet the second time, the LV bird has traveled the full distance plus y, that is 1+y units at rate s, totalling (1+y)/s units of time. At that same time the R bird has traveled the whole distance plus 1-y, or a total distance of 2-y, at speed 1, so the time is 2-y, and therefore (1+y)/s = 2-y.
So we have
2y/s = 1-2y
Divide the former by the latter to get
2y/(1+y) = (1-2y)/(2-y)
This leads to
Then as 2y/s=1-2y,
Therefore the LV bird is traveling at 2/3 the speed of the R bird. A diagram with the appropriate slopes bouncing off the two sides, shows that the next time they meet, the R bird will have made 3 full one-way trips while the LV bird will have made 2, so they will both be at Las Vegas when this happens.
Posted by Charlie
on 2003-04-14 03:28:56