A geometry problem by Áron Bán-Szabó

Suppose that $ABCD$ is a quadrilateral with maximum area, given that it has perimeter $4$ . Suppose that the points $A,B,C$ have been chosen. Then the length $CD + DA$ is fixed, and so $D$ lies on an ellipse with $A,C$ at the foci. The area of $ABCD$ will be greatest when the height of $D$ above the $AC$ is greatest, and this will happen when $CD = AD$ . Applying the same argument to the other vertices tells us that $ABCD$ must be a rhombus of side $1$ , and the rhombus of side $1$ with greatest area is a square. The largest possible area for a quadrilateral of perimeter $4$ is $1$ , achieved by a square of side $1$ .

Thus a quadrilateral of perimeter $4$ and area $1$ is a square of side $1$ . Both of its diagonals have length $\sqrt{2}$ , so the product of their diagonals is $\boxed{2}$ .

We will show that if a quadrilateral's area is $n$ , and the quadrilateral's perimeter is $k$ , then the

$n=(\frac{k}{4})^2$

equation will only be true, if the quadrilateral is a square.

First, it is clear that we only need to test the convex quadrilaterals, because with a given perimeter we can test a convex quadrilateral instead of a concave quadrilateral.

Using the relationship between arithmetic mean and geometric mean:

$(\frac{k}{4})^2=(\frac{\frac{a+c}{2}+\frac{b+d}{2}}{2})^2\geq \frac{a+c}{2}*\frac{b+d}{2}=\frac{ad+bc+ab+cd}{4}$

where $a$ , $b$ , $c$ , $d$ are the sides of the quadrilateral. The equality will only be true if $a+c=b+d$ .

From the figure above it is easy to see that $ad+bc\geq ax+cy = 2n$ , and $ad+bc= ax+cy = 2n$ will only be true if $DAB\angle =DCB\angle =90°$ .

Similarly $ab+cd\geq 2n$ , and $ab+cd=2n$ will only be true if $ABC\angle =ADC\angle =90°$ .

From these we get $n=(\frac{k}{4})^2$ will only be true, if $ABCD$ is a square. So by using the pythagoras theorem we get the only possible value of $AC*BD=\sqrt{2}^2=2$ .

[This is not a complete solution]

Show that if a quadrilateral has area $A$ , then the maximum value of the product of the diagonals is $2A$ . Hint: Area.

Find the equality conditions (where the product $= 2A$ ), and show that the perimeter is between $2A$ and $4A$ . (I am not fully certain about the range.)

Hence, we get an answer of 2 (corresponding to $A = 1$ ).

Let the quadrilateral be ABCD as in the figure. Let AO=x, BO=y, CO=w and DO=z. Let $\angle AOD=\theta$ and $\angle AOB=\alpha$ . Now, we have

$[AOD]=$ $\frac{1}{2}$ $xwsin \theta$

$[AOB]=$ $\frac{1}{2}$ $xysin \alpha$

$[BOC]=$ $\frac{1}{2}$ $yzsin \theta$

$[COD]=$ $\frac{1}{2}$ $zwsin \alpha$

Adding these results, we get

$[ABCD]=1=\frac{1}{2} [(xw+yz)sin\theta + (xy+zw)sin\alpha]$

$=>2=(xw+yz)sin\theta + (xy+zw)sin\theta$
[ $sin\alpha=sin(π-\theta)=sin\theta$ ]

$=>2=(xw+yz+xy+zw)sin\theta$

$=>2=(x+z)(y+w)sin\theta$

$=>AC.BD=2cosec\theta$

$Now, cosec\theta ≥1$

$Thus, AC.BD ≥2$

Now, we will show that equality holds since the perimeter is 4.

For extreme values, there is likely to be some sort of symmetry. So the quadrilateral is likely to be rectangle or ||gram. However ||gram diagonals are not equal, and having the long side very long needs a large perimeter. Thus we try a rectangle that has longer diagonals than a square for same area.
Let a,b be the sides of the rectangle.
$Area=a*b=1, ~P=2(a+b)=4.~\implies~ a+b=2. \\ (a+b)^2=4, ~~ (a-b)^2=(a+b)^2-4ab=4 - 4=0.\\ \therefore~ a=b. \\ \implies~It~ is~ a ~square ~side~ 1.\ ~This~is ~the~only~solution. \\ Product ~of~ the~ diagonal~=\sqrt2*\sqrt2=\Large 2.$

a quadrilateral with perimeter= 4 and area =1 is square, then product of its diagonal = 2.

By pure intuition, optimization problems in geometry most usually end up in symmetry. In this case, 4a = 4 & a^2 = 1 fit a square of side 1. Hence sqrt(2)^2=2. Notice that Ptolemy's theorem is exactly what the problem is asking for, i.e., product of diagonals = sum of products of two opposite sides; sqrt(2) sqrt(2) = 1×1+1×1.