Prove or disprove that there exists a quadrilateral ABCD of area 1 such that all points P have the property that at least one triangle -- PAB, PBC, PCD, PDA -- has an irrational area.
I have a feeling that no ABCD exists. My method is trying to find a point in ABCD such that all triangles have area 1/4. That would mean the distance from P to the sides is inversely proportional to the sides. Or I could be completely on the wrong track.
Also if one triangle has area $a+\sqrt{b}$ another triangle should have area $k-\sqrt{b}$. Thanks!