Consider a convex quadrilateral whose diagonals have lengths p,q and whose area is A.

What is the maximum area of a rectangle that circumscribes the given quadrilateral?

I'm just assuming now that the rectangle does not have to fully contain the quadrilateral. In cases where the diagonal's lengths are not too different, this is no restriction.

As noted, diagonals can be moved with a parallel motion without changing the rectangle's area. Therefore we can assume diagonals meet at their midpoints and the quadrilateral is actually a parallelogram. We do not really need this for the following, but it helps to picture/draw the situation. Now it is easy to see that

a = p cos α

and

b = q cos β

where α and β are the angles between a and p resp. b and q. We have α + β + θ= pi/2, where θ is the angle between p and q.

Now the rectangle's area R is

R = ab = p q cos α cos β = p q cos α sin (α+θ)

If we see R as a function of α and set R'(α) = 0, we can work out that R maxes out at

α= π/4 - θ/2

Inserting this into the equation for R gives

R_max = 1/2 (sin θ+ 1 ) p q

And thanks to Larry's area formula for quadrilaterals this gives

R_max = A + p q / 2

which - as asked for - expresses the maximum area as a function of A, p and q and which is indeed a nice formula :)

Edited on January 8, 2020, 4:56 pm

Posted by JLo on 2020-01-08 16:53:22