Problem of the Week

This problem is based on a suggestion from Philippe Fondanaiche.

The two diagonals, AC and BD, of the convex quadrilateral ABCD intersect in the point O.   See the figure on the left.  The areas of the triangles AOB, BOC, and COD are, respectively, 1, 2, and 3.  Among all such quadrilaterals, what is the smallest possible area for the triangle DOA?

For those in need of heartier nutrition, here is (a slightly modified version of) Philippe Fondanaiche's original poser:  A convex 2n-gon, ABC..., has the property that the diagonals of opposite vertices all meet at the common point O. , The areas of the 2n-1 triangles AOB, BOC, COD, etc...are 1,2,3,...,(2n-1).  Among all such possible polygons, what is the smallest possible area for the last triangle?  (First try to figure out if this is even possible.)

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ã2003 Alberto L. Delgado