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).

(In reply to Another question for Sam by logischer Verstand)

If four roads passed through the same point, then any three of them would violate the conditions of the problem. Saying that "no three pass through the same point" (as opposed to no point has three roads passing through it") makes the "or greater" implicit.

Posted by DJ on 2004-05-02 07:58:04