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 My method by Federico Kereki)

Very nice!  One little quibble though.  When you say "Through N draw a parallel to AB: it will intersect the circle at N'", it is not guaranteed that the parallel will intersect the circle.  However, one of the two choices for N is guaranteed to intersect the circle; but it requires proof.

Oh, drat!  One cannot just choose any M on AC!  It could happen that AB'C' turns out to be a single point!  No matter.  Just choose any M for which AB'C' is not degenerate.

It appears that your construction includes all examples exhibited so far, and, I suspect, all possible solutions.

However, now I am no longer convinced that there are only one or two solutions.  There are potentially two choices for N, and, for each of these choices, there are possibly two choices for N'.  Hence there could be as many as FOUR solutions!  I've been experimenting but have yet to find more than two solutions for the triangles I played with (and, no matter how I chose M and N, there was still only one solution for an equilateral triangle).

Posted by McWorter on 2005-05-11 01:06:24