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If the area of rectangle $ABCD$ with integral sides is $48$ If $E$ be any point on AD. How to find the sum of maximum and minimum area ( of $\triangle BCE$?

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Are $A$, $B$, $C$, $D$ in counterclockwise (or clockwise) order? Then the question is very easy. – André Nicolas Oct 21 '11 at 20:32
Hint: draw a diagram. If necessary remember how to compute the area of a triangle and a rectangle. Note the answer will be different if AD is a diagonal, rather than a side as is indicated by the way that the question is posed. – Mark Bennet Oct 21 '11 at 20:32
@André Nicolas:I am not sure. – Quixotic Oct 21 '11 at 20:37
@André: I’d be willing to make a small bet that $\overline{AD}$ is a side of the rectangle. That’s a cute little question. – Brian M. Scott Oct 21 '11 at 20:43
It is usual for the vertices to be named in order. If you draw the diagram as Mark Bennet suggests it will help. There is also a big hint in the fact that you are not given the dimensions of $ABCD$. Is it $6 \times 8, 1 \times 48$ or what? – Ross Millikan Oct 21 '11 at 20:51
up vote 2 down vote accepted

Assuming the usual clockwise or anticlockwise notations.

Since we know that the areas of two or more triangle having the same base and lying between the same parallel lines will be equal, so here we would have equal maximum and minimum area for $\triangle BCE$.

Again, the area of a triangle is given by $\frac{1}{2}\times x \times y$ where $x$,$y$ are the sides of rectangle. Here the constraint is $x \times y = 48 $ So choosing any value of $x,y$ satisfying this constraint would give the same area for any $\triangle BCE$.

Hence, in this case, the maximum area + minimum area = $48$

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I think you can even post it as a non-CW answer. The site explicitly permits (and encourages?) that. – Srivatsan Oct 21 '11 at 22:36

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