The problem can be generalized to an arbitrary number of roads, N>3, which
makes it even more striking: Assume the first two Travelers met and have also
met all the remaining N-2 fellows. Prove that the remaining ones all have met
each other.
- One of the visitors noticed that under the conditions of the
problem, all the Travelers stay on a moving straight line. To cite:
Perhaps another solution to consider may be more naturally phrased and reveal more information about the motions involved. Draw a line connecting Travelers 1 and 2, and see how it varies through time. Because of general position of the travelers paths and the constancy of their speed, we see that Travelers 3 and 4 must also be on the line as it travels given that 1 and 2 meet them. Now general position gives that 3 and 4 must also meet, since the line intersects their paths.
Perhaps this argument needs to be made more precise, but on first investigation, I see no flaw. What do you think?
- John Mason made the following remark:
If two travellers set out from a common point along straight lines, travelling at constant speeds, then the ratios of the distances travelled in a given time is the ratio of their speeds. So lines drawn at various times joining their positions will be parallel. But is this not Thales theorem?
It is well known amongst sailors that if two ships are travelling such that the angle between them is constant (that is, the angle between the line of travel of one, and the line of sight of the bow of the other ship), then the two ships are on a collision course. This is really useful if you are a little sailboat and there is a big lake freighter in lake Ontario, for instance. I see this 'rule of thumb'as Thales theorem in another form.
