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.
As the length of BC approaches 0, AQ approaches AC, thus either AP or PQ or both > QB. Do we need some parameters for this puzle?
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Posted by Bryan
on 2005-04-27 19:29:18 |