Correct solutions were received from Ivan Lisac, Burkart Venzke, Philippe Fondanaiche.  A number of incorrect solutions were submitted.

Here's an interesting solution from Burkart Venzke.

All nine points lie in the interior of the square, so pick any one and call it A.  If any two of the other eight points are colinear with A, they form a traingle of area zero.  So you can suppose that no pair is colinear; you next pick any further point, call it B, and label the other points, C, D, E,...  clockwise and in order around Astarting from B.  Now look at the eight distinct and disjoint triangles with vertices ABC, ACD, ADE,...  The union of the eight triangles lies in the interior of the square, so the sum of the areas must be less than 1.  One of the triangles must therefore have area less than 1/8.
 

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Page last updated 17 May 1999.