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.

Accidentally found TWO choices for P and Q for a particular triangle!  Let A=(sqrt(2)/2,0), B=(0,1), C=(1,0) be the vertices of of a triangle ABC.  Then one choice for P and Q is P=(0,0) and Q=(1/2,1/2).  However, another choice for P and Q is, approximately, P=(1.7564216,0) and Q=(.741,.259).

I have yet to find a straigtedge-compass construction, but I wonder if Kereki's construction includes one or both of the above two examples.

Posted by McWorter on 2005-05-07 01:45:46