Given a triangle ABC, how can you find a point P on AC and a point Q on BC, such that AP=PQ=QB?
N.B. A construction method is sought, and only compass and straightedge are allowed.
(In reply to
May not be possible by Bryan)
When either of AC or BC is more than twice the other (so that one of then is less then half the other), there is does not exist a valid solution, with P between A and C and Q between B and C.

Posted by Charlie
on 20050427 19:49:37 