Solution to the 4 Travelers problem

The following is based on the proof to the Four Travelers problem posted by Rob Fatland at the CTK Exchange. 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.

The solution is simple and beautiful.

Assuming that no meetings have taken place yet, in the rest frame of traveller #1, travellers #2, #3, and #4 all approach #1 along straight lines that can be defined in terms of angles (say a2, a3, and a4). Of course once they meet #1 they depart along the same lines with the new headings (ak + 180°) etcetera.

We can determine a2 by observing #2 at just about any time (since a2 is constant). So, let's choose the moment that #2 meets #3. This gives us #3's angle a3 as well, the same: a2 = a3. Similarly since #2 meets #4 we have a2 = a4.

Thus from #1's perspective, #2, #3, and #4 are all travelling along the same line that goes towards #1 which exactly means that the four are collinear.

Note that the argument breaks down if the equality of angles can't be obtained. This only happens if the meeting point of, say, #2 and #3 coincides with the position of #1, i.e. when #1, #2, and #3 meet simultaneously, i.e., when their roads are concurrent. But this case is excluded by the problem's premises.

(The latter observation is due to S. Anderson. There is a dynamic illustration for Rob's argument.)

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