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.

Hi!  This is my first post, so don't laugh too hard at how mathematically flawed my submission probably is.  Be gentle with me!  I've probably even made a mistake that's already been discussed -- I get confused easily :)

Please look at the PDF file I've created while checking out my explanation below.

1) Select a random radius and strike arcs from centerpoints A and B - where the arcs intersect AC and BC will be points P' and Q'.

2) From centerpoints P' and Q', strike arcs back along the lines created by segments AC and BC.  These will be points A' and B'. 

3) We now have a solution that works for angle ACB, but comprised of points A', B', C, P' and Q' instead of the desired points A, B, C, P, and Q.

4) OK, here's where I'm probably screwing up.  It seemed logical for me to think that for any polyhedron abqp where ap=pq=bq, that the ratio ab:pq should remain constant, as long as the angle acb remains constant.  (I'm sure, however, that this assumption will be quickly and easily disproven by someone who knows some actual geometry and/or calculus, unlike myself :) 

Anyhoo, we can now determine the length of the desired segment PQ, by transposing angle theta (created by the right triangle formed by segments A'B' and P'Q') onto the endpoint of segment AB, and seeing where along the perpendicular of AB the angle intersects.  Is this maybe not the correct length of PQ?

5) Strike point P along AC by setting the compass to a radius of PQ.

6) Strike point Q along BC with the same compass settings.

I think that  AP=PQ=BQ!  But I also smoke a lot of stuff that I shouldn't  :)

Thanks for reading -

Dan Schmiedeler

Posted by Dan on 2005-05-05 06:37:59