Question

Let $ABCD$ be a quadrilateral inscribed in a circle. Suppose $AB=\sqrt{2+\sqrt{2}}$ and $AB$ subtends ${135}^{\circ }$ at the centre of the circle. Find the maximum possible area of $ABCD$.

Answer 1

Let $O$ be the centre of the circle in which $ABCD$ is inscribed and let $R$ be its radius. Using cosine rule in triangle $AOB$,

we get, $2+\sqrt{2}=2R^2(1-cos{135}^{\circ })$ $[\because$ In $△ABC,\cos A=\frac{b^2+c^2-a^2}{2bc}$ where $a,b$ and $c$ are the sides of given triangle $]$

$⟹2+\sqrt{2}=R^2(2+\sqrt{2})$.

$\therefore R=1$.

Consider quadrilateral $ABCD$ as in the second figure above. Join $AC$.

For $\left[ADC\right]$ to be maximum, it is clear that $D$ should be the mid-point of the arc $AC$ so that its distance from the chord $AC$ is maximum

$\therefore AD=DC$ for $\left[ABCD\right]$ to be maximum

Similarly, we conclude that $BC=CD$

$\therefore BC,CD,DA$ subtends equal angles at the centre $O$.

Let $\angle BOC=\alpha ,\angle COD=\beta and\angle DOA=\gamma$.

Observe that $\left[ABCD\right]=\left[AOB\right]+\left[BOC\right]+\left[COD\right]+\left[DOA\right]=\frac{1}{2}sin{135}^{\circ }+\frac{1}{2}(sin\alpha +sin\beta +sin\gamma )$.

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

$\therefore \left[ABCD\right]=\frac{1}{2}sin{135}^{\circ }+\frac{3}{2}sin{75}^{\circ }=\frac{1}{2}\left(\begin{array}{c}\frac{1}{\sqrt{2}}+3\frac{\sqrt{3}+1}{2\sqrt{2}}\end{array}\right)=\frac{5+3\sqrt{3}}{4\sqrt{2}}$.

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

$\frac{sin\alpha +sin\beta +sin\gamma }{3}\le sin\left(\begin{array}{c}\frac{\alpha +\beta +\gamma }{3}\end{array}\right)=sin{75}^{\circ }$.

Hence $\left[ABCD\right]\le \frac{1}{2\sqrt{2}}+\frac{3}{2}sin{75}^{\circ }=\frac{5+3\sqrt{3}}{4\sqrt{2}}$,

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