Four rods are hinged at their ends to form a convex quadrilateral with fixed side lengths. Show that the quadrilateral has a maximum area when it is cyclic. Sue Liu, S5, Madras College, St Andrews used two methods to solve this.

First she applied Brahmagupta's formula for the area of a quadrilateral.

${\displaystyle \Delta =\sqrt{(s-a)(s-b)(s-c)(s-d)-\mathrm{abcd}\cos {}^2\beta }}$

where $s=\frac{1}{2}(a+b+c+d)$ and $\beta =\frac{1}{2}(A+C)$ or $\frac{1}{2}(B+D)$.

Hence the area is clearly the greatest when $\mathrm{abcd}\cos {}^2\beta$ is least. Since $\cos {}^2\beta$ is always positive, this value is least when $\beta$ is 90 degrees as $\cos {90}^o=0$. Hence $\frac{1}{2}(A+C)=90$ and so $A+C=180$ showing that the opposite angles in the quadrilateral add up to ${180}^o$ and so the area of a quadrilateral with fixed lengths of sides is greatest when it is cyclic.

This method gives a proof of the required result but you have to assume Brahmagupta's formula and Sue's second method uses only the formula for the area of a triangle.

The area of the quadrilateral $\mathrm{ABCD}$ can be expressed as the sum of the areas of triangle $\mathrm{ABD}$ and $\mathrm{BCD}$. Let the area of $\mathrm{ABCD}$ be $\Delta$ then

${\displaystyle \Delta =\frac{1}{2}\mathrm{ad}\sin A+\frac{1}{2}\mathrm{bc}\sin C}$

from which

${\displaystyle 2\Delta =\mathrm{ad}\sin A+\mathrm{bc}\sin C}$

.

Squaring gives

${\displaystyle 4{\Delta }^2=a^2{d}^2\sin {}^{2}A+2\mathrm{abcd}\sin A\sin C+b^2{c}^2\sin {}^{2}C\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(1)}$

Also, we notice that (using the cosine rule for the length of the diagonal $\mathrm{DB}$)

${\displaystyle a^2+d^2-2\mathrm{ad}\cos A=b^2+c^2-2\mathrm{bc}\cos C}$

or written differently

${\displaystyle a^2+d^2-b^2-c^2=2\mathrm{ad}\cos A-2\mathrm{bc}\cos C+4b^2{c}^2\cos {}^{2}C\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(2)}$

Now we see a similarity with equation (1). If we multiply equation (1) by 4 and add equation (2) we get

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{{\Delta }^2+(a^2+d^2-b^2-c^2){}^2}& =& 4a^2{d}^2(\sin {}^{2}A+\cos {}^{2}A)\\ \multicolumn{1}{c}{}& & +4b^2{c}^2(\sin {}^{2}C+\cos {}^{2}C)\\ \multicolumn{1}{c}{}& & -8\mathrm{abcd}(\cos A\cos C-\sin A\sin C).\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(3)\end{array}}$

We know that $\sin {}^2\alpha +\cos {}^2\alpha =1$ and the compound angle formula

${\displaystyle \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta }$

for all $\alpha ,\beta$, it follows that (3) becomes

${\displaystyle 16{\Delta }^2+(a^2+d^2-b^2-c^2)=4a^2{d}^2+4b^2{c}^2-8\mathrm{abc}\cos (A+C)}$

since $a,\hspace{1em}b,\hspace{1em}c,\hspace{1em}d$ are all fixed, it follows that $\Delta$ is largest when $\cos (A+C)$ is the smallest, that is, when $\cos (A+C)=-1$. Hence the area is largest when $A+C=\pi$. So the area of the quadrilateral is largest if it is cyclic.

A variant of this method is to use the formula for the area, then

${\displaystyle \frac{d\Delta }{\mathrm{dA}}=\frac{1}{2}\mathrm{ad}\cos A+\frac{1}{2}\mathrm{bc}\cos C\frac{\mathrm{dC}}{\mathrm{dA}}}$

and from the cosine rule

${\displaystyle 2\mathrm{ad}\sin A=2\mathrm{bc}\sin C\frac{\mathrm{dC}}{\mathrm{dA}}}$

Hence

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{\frac{d\Delta }{\mathrm{dA}}}& =& \frac{1}{2}\mathrm{ad}\cos A+\mathrm{ad}\cos C\frac{\sin A}{\sin C}\\ \multicolumn{1}{c}{}& =& \frac{1}{2}\mathrm{ad}(\cos A\sin C+\sin A\cos C)\\ \multicolumn{1}{c}{}& =& \frac{1}{2}\mathrm{ad}\sin (A+C).\end{array}}$

Hence, for the maximum area, $\sin (A+C)=0$ and $A+C=\pi$ which makes the quadrilateral cyclic.