Suppose that we have a square with side $L$. Given 3 non collinear points inside this square, can we affirm that the area of the triangle formed linking these points is less than (or equals) $\frac{L^2}{2} $?
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Since moving up from a line at the bottom of the square only reduces area and the maximum height you can get is $L$, it’s $\leq {L^2 \over 2}$. – Ry- Oct 22 '13 at 23:12
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Suppose one side of the triangle is parallel to a side of the square. Then the triangle's base is $\le L$ and the height is also $\le L$, so the area is $\le $L^2/2$.
If no sides of the triangle are parallel to any side of the square, consider a line sweeping across the square parallel to one side. Since no side of the triangle is parallel to any side of the square, the sweeping line will meet the points of the triangle three times. At the middle one of these, the triangle will be divided into two triangles with a common base of side $\le L$ and altitudes that sum to $\le L$, so their combined area is $\le l^2/2$.
marty cohen
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