Iberoamerican Mathematical Olympiad

Solution:

Let the circle have center $O$ on $AB$ and radius $r$. Let $\angle OAD = \theta$, $\angle OBC = \varphi$. Since $ABCD$ is cyclic, $\angle ADC = 180^\circ - \varphi$, so $\angle ODA = 90^\circ - \varphi / 2$.

If $AD$ touches the circle at $X$, then
$$AD = AX + XD = r \cot \theta + r \tan (\varphi / 2).$$
Similarly,
$$BC = r \cot \varphi + r \tan (\theta / 2).$$
Put $t = \tan (\theta / 2)$. Then $\cot \theta = (1 - t^2) / 2t$, so
$$\cot \theta + \tan (\theta / 2) = \frac{1 + t^2}{2t} = \frac{1}{\sin \theta}.$$
Similarly for $\varphi$, so
$$AD + BC = \frac{r}{\sin \theta} + \frac{r}{\sin \varphi} = AO + OB = AB.$$

Suppose $AD$ and $BC$ meet at $H$ (we deal below with the case where they are parallel). Then $HCD$ and $HAB$ are similar, so
$$\text{area } HCD = \left(\frac{CD^2}{AB^2}\right) \text{area } HAB$$
and
$$\text{area } ABCD = \left(1 - \frac{CD^2}{AB^2}\right) \text{area } HAB.$$
Also
$$\frac{AB}{CD} = \frac{HA}{HC} = \frac{HB}{HD} = \frac{HA + HB}{HC + HD} = \frac{HA + HB}{HB - BC + HA - DA} = \frac{HA + HB}{HA + HB - AB}.$$
Hence
$$HA + HB = \frac{AB^2}{AB - CD},$$
which is fixed. Now for fixed $HA + HB$ we maximise the area of $HAB$ by taking $HA = HB$ and hence $AD = BC$.

Put $h = CD$, $k = AB$. So $k \cos \theta + h = k$. Hence $\cos \theta = 1 - h / k$. Hence
$$\sin \theta = \sqrt{2h / k - h^2 / k^2}.$$
So
$$\text{area } ABCD = \frac{1}{2}(h + k) \cdot \frac{1}{2} k \sin \theta = (h / 2 + k / 2) \sqrt{hk / 2 - h^2 / 4} \quad (*).$$

If $AD$ and $BC$ are parallel then $A$ and $B$ must lie on the circle, so that $\angle DAB = \angle ABC = 90^\circ$. But $ABCD$ is cyclic, so it must be a rectangle. Hence $AB = CD$ and
$$\text{area } ABCD = k^2 / 2.$$
In this case $(*)$ still gives the correct answer.