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.

Forget my previous question, and assume we are dealing with a triangle ABC such that there is a solution where AP=PQ=QB.

On triangle ABC, draw point M some distance from A to C, and draw point N this same distance from B to C, such that AM=BN. As AM and BN approach zero, MN approaches AB in a linear relation. What we want to do is increase or decrease the lengths of AM and BN until we find the magic points P and Q where AM=MN=NB (AP=PQ=QB).

Start by constructing two perpendicular rays from a common point O, for instance OX projecting to the right from O and OY projecting up from O. Transpose the length of AM onto ray OX starting at O. Transpose the length of AB onto ray OY starting at O. Construct a perpendicular to OX at M, and transpose the length of MN onto this line from M upward.

Now we construct a line from the top of the transposed AB to the top of the transposed MN. This line represents the linear relationship between the lengths of AM and MN. Again, as AM approaches zero, MN approaches AB.

Construct a bisector of XOY. Label the point where this bisector intersects the sloped line drawn above as Q. Draw a perpendicular to OX through Q, and label the point of intersection P. Transpose AP and PQ onto triangle ABC.

Edited on April 27, 2005, 8:20 pm

Edited on April 28, 2005, 3:29 pm

Posted by Bryan on 2005-04-27 20:20:07