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:

Four roads on a plane, each a straight line, are in general position. Along each road walks a traveler at a constant speed. Their speeds, however, may not be the same. It's known that traveler #1 met with Travelers #2, #3, and #4. #2, in turn, met #3 and #4 and, of course, #1. Show that #3 and #4 have also met.

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

O = D1 × D2, A = D2 × D3, B = D2 × D4, C = D1 × D3, D = D1 × D4, E = D3 × D4.

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

OA = t1v2, OB = t2v2, AC = t3v3, OE = t5v3, BD = t4v4, BE = t6v4.

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

t5 + (t2 - t1) = t6, or t5 - t6 = t2 - t1.

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

OA.CE.DB = OB.CA.DE,

or

[t1v2]·[(t5 - t3)v3]·[t4v4] = [t2v2]·[t3v3]·[(t6 - t4)v4],

which simplifies to

(1)	t1·(t5 - t3)·t4 = t2·t3·(t6 - t4).

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

OD.AC.BE = OC.AE.BD

or

(2)	(t2 + t4).t3.t6 = (t1 + t3).t5.t4

(1) and (2) can be manipulated into

t1.t4.t5 – t2.t3.t6 = t3.t4.(t1 – t2) and t1.t4.t5 – t2.t3.t6 = t3.t4.(t6 – t5),

which obviously yields the desired result.

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