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?

(In reply to re: solution by fwaff)

Here's one additional solution, where the LV bird is slow enough so that the R bird on its return from Las Vegas passes the LV bird still on its first leg. The other possibility would be if the LV bird were so fast it was the one to pass the R bird. That case I think would require the LV bird to be much much faster than the R bird in order that the first distance from LasVegas be twice the second distance from Reno. Perhaps someone else can figure that case.

But in the case of a slower LV bird, my equations become
2y/s=1-2y
(1-y)/s=2-y

2y/(1-y)=(1-2y)/(2-y)
1-3y+2y²=4y-2y²
4y²-7y+1=0
y=(7±√33)/8
Using the + gives a solution above 1, contrary to the situation, so y=(7-√33)/8 = 0.156930 and s = 2y/(1-2y)=0.457427.

The next time they meet, will be shortly thereafter as the R bird bounces back from Reno, starting out his 3rd one-way trip while LV is still completing his 2nd. Let this distance be called z. Then
(1-z)/s = 2+z
z=(1-2s)/(s+1) = 0.058422
but this is the fraction of the way back to Las Vegas. As y is 0.156930 (which is x miles), z converts to 0.372281 y, in our units, or 0.372281 x miles.

Posted by Charlie on 2003-04-14 05:56:00