Four roads on a plane, each a straight line, are in general position so that no two are parallel and no three pass through the same point. Along each road walks a traveler at a constant speed. Their speeds, however, may not be the same. It's known that two of the travelers have met each of the other three already.

Prove that the other two have also met each other.

The problem can be generalized to an arbitrary number of roads, which makes it even more striking: Assume that two of the travelers met and have each met all the remaining fellows.

Prove that, if this is the case, the remaining ones all have met each other (ie, if two travelers have met everyone, then everyone has met everyone).

I drew this out and got a picture that shows one road going one direction (say North) and the 3 other roads aiming Northwest at varying degrees. This way, the one walker (going N) is sure to intersect all other roads and *by chance* be walking at an appropriate speed so that he runs into each other walker in turn.

Since each of the other roads would be aiming NW, each of them would intersect in various places (and by some bit of good timing they bump into one another)

Another idea would be that two of the roads cross perpendicular to one another. The other two roads would then pass through both of the perp. roads at different angles, so they would then meet as well...

Posted by Kim on 2006-12-22 04:14:39