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 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)

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 my solution | Comment 5 of 8 |
I really looked forward to this one, and finally got to it.

The way I see it, there are three possible ways that the birds can meet twice, once if the LV bird reaches its destination and meets the R bird again before the R bird has completed its first journey, one with the opposite happening, and one where both bird have completed one full journey and are on their return flight.

The formulas I used work out to be similar in all three cases which made checking each scenario fairly easy.

Here we go - I'll just give the basics of my method, if you want to chekc it out go ahead! This problem actually turned out to be much easier than I expected it to be (Don't worry though Ravi - I think the 4 was still justified!)

This problem related to the RATIO of two velocities - the actual velocities are irrelevant. This means that we can assume values for certain (combinations) of variables, which is of great value. I chose to assume that the time elapsed at the first crossing was known and labelled it t(1). Also, the distance between Reno and Las Vegas should be treated as a known, which I called D. The velocities of the birds are v(lv) and v(r), for las vegas and reno respectively. Now using the general physics formula that v=d/t, out pop expressions for the velocities in terms of d (known), x (unknown) and t(1)(known).

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

Now considering the most likely case first (that both birds complete one full trip and are returning), we can (since the birds each have a CONSTANT velocity) equate these with similar equations created when looking at t(2), namely;

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

Now equating these two sets of equations and cancelling (a lot) where possible, we get two equations both in the form t(2)=t(1)*some other stuff. Combining these and cancelling the time factors results in the following simple equation;

(d+x)/2x=(2d-x)/(d-2x)

When this equation is solved, it yields an x value of one fifth that of d (again it's simply the ratio here that is important, as we're treating d as a known value), which can then be used to solve the actual problem.

As it turns out, two of the three initial cases coincide as the birds meet for the second time at the instant that one of the two birds is completing its first journey.

Checking the third possible case (following the same steps) yields the following quadratic;

(4)(x^2)-(4d+3)(x)+(d^2)=0 While this is certainly solvable, it gives only negative values for the "known" variable d, which isn't possible under the boudaries of this question, which, finally leaves us with just the one solution.

This means that the only allowable solution is that the third meeting is at the tunaround point in Las Vegas.

 Posted by Cory Taylor on 2003-04-17 08:19:18

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