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the following question beat me. How from given any 9 points inside a square of side 1 we can always find 3 which form a triangle with area less than $1/8$ .

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I guess you allow flat triangles. –  1015 Jan 14 '13 at 0:50

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up vote 6 down vote accepted

Draw a horizontal line to split the square into two rectangles of area $\frac{1}{2}$. One of the rectangles must contain at least 5 of the points.

Now draw a vertical line to split that rectangle into two squares of area $\frac{1}{4}$. One must contain at least 3 points.

Now you need to show that any triangle inside a square of area $\frac{1}{4}$ has area at most $1/8$, which is discussed here: Maximum area of a triangle in a square

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