Solution to the 4 Travelers problem

The following is based on the solution to the Four Travelers problem sent to me by Michel Cabart. To remind, here is the problem:

Michel's solution refers to the following diagram:

The four roads are denoted D₁, D₂, D₃, and D₄. The intersections O, A, B, C, D, E are defined by

Accordingly, we assume that the first traveler moves in the direction from O to C to D, the second from O to A to B, the third from A to C to E, and the forth from B to D to E. For every traveler, i.e., for every traveler, we set up a private time frame. We assume that O serves as t = 0 on roads D₁ and D₂, on road D₃ t = 0 at A, and, for D₄, t = 0 at B.

Denoting their speeds v₁, ..., v₄, we time the progress of each of the travelers in their private time frames so that

Now, after travelers 1 and 2 part after a meeting at O, 1 continues towards C where he is going to meet traveler 3. Traveler 2 proceeds towards A where he greets traveler 3 after t₁ (time units). Traveler 3 starts his time count from this point. It takes him t₃ to get to C where he meets traveler 1. Thus the latter spent t₁ + t₃ on the road between intersections O and A. In a similar vein, the first traveler spends t₂ + t₄ between O and D.

More generally, for every triangle formed by the roads, the condition that all its vertices serve as a meeting place for respective travelers is equivalent to an identity between time intervals. In particular, in order to prove that travelers 3 and 4 meet in E, we only need to show that

Applying Menelaus' Theorem to triangle ABE and line OCD yields:

or

which simplifies to

Similarly, applying Menelaus' Theorem to ΔCED and line OAB we get

or

(1) and (2) can be manipulated into

which obviously yields the desired result.

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Copyright © 1996-2012 Alexander Bogomolny