
1. Let $A B C D$ be a quadrilateral inscribed in a circle. Suppose $A B=$ $\sqrt{2+\sqrt{2}}$ and $A B$ subtends $135^{\circ}$ at the centre of the circle. Find the maximum possible area of $A B C D$.
![](https://cdn.mathpix.com/cropped/2024_06_05_177a0e3dd76b8909557ag-1.jpg?height=360&width=1090&top_left_y=478&top_left_x=515)


Solution: Let $O$ be the centre of the circle in which $A B C D$ is inscribed and let $R$ be its radius. Using cosine rule in triangle $A O B$, we have

$$
2+\sqrt{2}=2 R^{2}\left(1-\cos 135^{\circ}\right)=R^{2}(2+\sqrt{2})
$$

Hence $R=1$.

Consider quadrilateral $A B C D$ as in the second figure above. Join $A C$. For $[A D C]$ to be maximum, it is clear that $D$ should be the mid-point of the arc $A C$ so that its distance from the segment $A C$ is maximum. Hence $A D=D C$ for $[A B C D]$ to be maximum. Similarly, we conclude that $B C=C D$. Thus $B C=C D=D A$ which fixes the quadrilateral $A B C D$. Therefore each of the sides $B C, C D, D A$ subtends equal angles at the centre $O$.

Let $\angle B O C=\alpha, \angle C O D=\beta$ and $\angle D O A=\gamma$. Observe that

$$
[A B C D]=[A O B]+[B O C]+[C O D]+[D O A]=\frac{1}{2} \sin 135^{\circ}+\frac{1}{2}(\sin \alpha+\sin \beta+\sin \gamma)
$$

Now $[A B C D]$ has maximum area if and only if $\alpha=\beta=\gamma=\left(360^{\circ}-\right.$ $\left.135^{\circ}\right) / 3=75^{\circ}$. Thus

$$
[A B C D]=\frac{1}{2} \sin 135^{\circ}+\frac{3}{2} \sin 75^{\circ}=\frac{1}{2}\left(\frac{1}{\sqrt{2}}+3 \frac{\sqrt{3}+1}{2 \sqrt{2}}\right)=\frac{5+3 \sqrt{3}}{4 \sqrt{2}}
$$

Alternatively, we can use Jensen's inequality. Observe that $\alpha, \beta, \gamma$ are all less than $180^{\circ}$. Since $\sin x$ is concave on $(0, \pi)$, Jensen's inequality gives

$$
\frac{\sin \alpha+\sin \beta+\sin \gamma}{3} \leq \sin \left(\frac{\alpha+\beta+\gamma}{3}\right)=\sin 75^{\circ}
$$

Hence

$$
[A B C D] \leq \frac{1}{2 \sqrt{2}}+\frac{3}{2} \sin 75^{\circ}=\frac{5+3 \sqrt{3}}{4 \sqrt{2}}
$$

with equality if and only if $\alpha=\beta=\gamma=75^{\circ}$.
