 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Leaving Las Vegas (Posted on 2003-04-14) 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?

 See The Solution Submitted by Ravi Raja Rating: 3.6667 (3 votes) Comments: ( Back to comment list | You must be logged in to post comments.) The 3 solutions | Comment 7 of 8 | As previously stated by others, there could be 3 solutions to this. All 3 solutions use the same general method, setting up bird-speed equations and cancelling. I'll use Cory Taylor's notation rather than my own (I like his better!)

The first is where the Reno and LV birds cross, reach their destination, turn around and cross again. The bird-speed equations are:

v(r) = (d-2x)/t(1) = (2d-x)/t(2) and
v(lv) = 2x/t(1) = (d+x)/t(2)

Get rid of the v's and simplify to get (d-2x)(d+x) = (2d-x)(2x). Solve and you get d=5x. So when the LV goes 2x, the Reno bird goes 3x. It's probably easier to just draw it out from here to get an answer, but one math route would be the following: At the third meeting point, the LV bird will have travelled a distance I'll call 'a' from the second meeting point. So we get more equations:

v(r) = (2d+x+a)/t(3) and v(lv) = (d+x+a)/t(3)

Divide the two and use 3/2 for v(r)/v(lv) and you get (after simplification) d=x+a. Because d=5x, you get a=4x, which puts the two birds exactly in Las Vegas.

The second case is where the LV bird is a total slowpoke, and the Reno bird gets to LV, turns and catches the LV bird before it even gets to Reno. You now get the equations:

v(r) = (d-2x)/t(1) = (2d-x)/t(2) and
v(lv) = 2x/t(1) = (d-x)/t(2)

Get rid of the v's and simplify to get (d-2x)(d-x) = (2d-x)(2x), which expanded out gives d�-7xd+4x�=0 (Here's where Cory Taylor went wrong, he missed a d in his equation). Solve with quadratic and get d=6.37x (rounded). So the new velocity ratio of the two birds is 4.37/2. 'a' is defined the same way, and we get a new second set of equations:

v(r) = (2d+x-a)/t(3) and v(lv) = (d-(x-a))/t(3)

Divide the two and use the new ratio of 4.37/2 and you get .37d = 6.37x-6.37a. Use d=6.37x and you end up with x-a=0.37x which is the distance from Reno where they meet for the third time.

The final case is where the Reno bird is sucking wind and the LV bird gets to Reno and turns around and catches the Reno bird before it even gets to LV. New equations:

v(r) = (d-2x)/t(1) = x/t(2) and
v(lv) = 2x/t(1) = (d+x)/t(2)

Get rid of the v's and simplify to get d�-xd-4x�=0. Use quadratic to get d=2.56x (rounded). Use 'a' again and get two new equations:

v(r) = (x+a)/t(3) and v(lv) = (d+x+a)/t(3)

Divide the two and use the new ratio of .56/2 and you get 2.56(x+a)=1.68d. Put in d=2.56x and you get x+a=1.68x which is the distance from Reno where they meet for the third time (0.88x from LV)

Summary:
1) they meet in LV
2) they meet 0.37x from Reno
3) they meet 0.88x from LV
Edited on January 15, 2004, 3:26 pm
 Posted by Jack Squat on 2004-01-15 15:24:02 Please log in:

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