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.

(First solution) Draw a circle with center P.  Draw a diameter of the circle meeting its circumference in A and C.  Choose a point Q on the circumference of the circle so that angle APQ is around 45 degrees.  Mark a point B on QC so that QB has length equal to the radius of the circle.  Then P and Q is a solution for Kereki's problem for triangle ABC.

(Second solution) Now draw a line parallel to AB through Q.  It meets the circle at another point Q'.  Draw a line through Q' parallel to BC meeting AB at B' and AC at C'.  Then AB'C' is similar to ABC and P and Q' is a solution for Kereki's problem for AB'C' not similar to that for triangle ABC.

(Third solution) Mark a point N' on BC, so that B is between Q and N', and BN' has length equal to the radius of the circle.  Draw a line parallel to AB meeting the circle at Q''.  Draw a line through Q'' parallel to BC meeting AB at B'' and meeting AC at C''.  Then triangle AB''C'' is similar to ABC and P and Q'' is a solution for Kereki's problem for triangle AB''C'' not similar to those for the first two solutions.

Since the line parallel to AB in the third solution meets the circle in two places, there is actually a fourth solution.  But by now, who cares?

Posted by McWorter on 2005-05-11 04:31:06