Hello Bractals and Vladimir.

I put this problem on the site sometime in October of 2002.

The solution in case the point is on the side of triangle is trivial.

There must be 3 solution lines when the point is "deep" inside the triangle.

When the point is outside or on the perimeter of the triangle where is one solution.

So there must be a locus of points with 2 solutions.

And the construction has to include an angle bisectors.

Bractal, if you use the Sketchpad you can move around vertices

A,B,C and the given point P and see if the arears of the resulting

halves remain equal ( i doubt it is the case for the solution you

provided ).

Anyway, the proof still remains to be seen.

Golland.