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.

I can't think of a "finite" method to solve this. But I did think of an iterative method. I will assume that the triangle given has a solution.

Because AP=PQ. P is on the perpendicular bisector of the segment AQ. Similarly, Q is on the perpendicular bisector of BP.

Pick on a point X1 on AC. Connect BX1. Then construct the perpendicular bisector BX1 and label the intersection of it with BC as Y1. It is important to pick an X1 such that it intersects. That will just have to be done by "eye-balling." And the assumption that the triangle has a solution tells us that we can do so.

Then connect AY1. Construct the perpendicular bisector of AY1. Label the intersection of that and AC as X2.

Proceed in a similar fashion to construct perpendicular bisectors and find the intersections. The intersections will eventually converge to points P and Q.

Posted by np_rt on 2005-04-27 22:40:14