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 re(3): Two solutions for one triangle! by Charlie)

Ah, the mystery deepens, perhaps into "The Kereki Code"!  I think there is a unique solution to Kereki's problem if and only if triangle ABC is equilateral!

Here's one example of a triangle with a 60 degree angle at C having two solutions of Kereki's problem.  Let the vertices of triangle ABC be A=(-sqr(3),0), B=(sqr(3)/2,sqr(3)-3/2), and C=(1,0).  Then the angle at C is 60 degrees.  One choice for P and Q is P=(0,0), Q=(0,sqr(3)).  I believe there is another choice with P in the third or fourth quadrant and Q in the fourth quadrant (please forgive my laziness in not computing the second solution).

Your formula is great when the angle at C is not 60 degrees, but I think you need a different argument to find two solutions when the angle at C is 60 degrees.

Posted by McWorter on 2005-05-08 19:22:45