Solution to the 4 Travelers problem
Miguel Branchtein, Brasil
July 27, 2003
Here's another solution to the 4 travelers' problem.
We draw the four straight lines where travelers 1, 2, 3 and 4 respectively travel. Their meeting points are A, B, C, D, E and F.
Given that they meet in A, B, C, D and E, we must show that they also meet in F.
The straight line FG is just an auxiliary line, drawn from F paralel to line 1. The point where it crosses line 2, we'll call it G.
We call the distance between points A and B, and so forth. We call V1 the speed of traveler 1, and so forth.
From triangle ABC, we see that, in order that travelers 2 and 3 meet in point C, in the same time traveler 2 goes from A to C, traveler 1 must go from A to B, and then traveler 3 goes from B to C. That is,
Similarly, from triangle ADE, we have
At last, from triangle CEF, we get
Given (1) and (2), we have to prove (3).
We substitute 1/V3, from (1), and 1/V4 , from (2), into (3).
Putting together all terms in 1/V1 and 1/V2
In order to evaluate these, we observe that triangle GFC is similar to triangle ABC:
And triangle GFE is similar to triangle ADE:
Substituting these relations in (4), we get
proving eq. (4), and consequently eq. (3).
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