TITLE: Triangle inside a square
QUESTION [6 upvotes]: 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} $?

REPLY [8 votes]: 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 
$\le l^2/2$.