Real fun with cyclic quadrilateral

Let $O$ as center of given circle { $radius=r$ } , Clearly in triangle $OAB$ using cosine law $\displaystyle{{ a }^{ 2 }={ r }^{ 2 }+{ r }^{ 2 }-2r.r\cos { \theta } \\ { r }^{ 2 }=\cfrac { { a }^{ 2 } }{ 2(1-\cos { \theta } ) } }$

$Note$ : Since it is Standard result that area of an standard triangle $ABC$ is $\Delta =\cfrac { 1 }{ 2 } bc\sin { A }$

Let angle's as $\displaystyle{\angle BOC={ x }_{ 1 },\angle COD={ x }_{ 2 },\angle DOA={ x }_{ 3 }}$

So Area of quadrilateral is :

$\displaystyle{A=\cfrac { 1 }{ 2 } \times { r }^{ 2 }\left( \sin { \theta } +\sin { { x }_{ 1 } } +\sin { { x }_{ 2 } } +\sin { { x }_{ 3 } } \right) }$

For to maximize area A , we have to maximize $\displaystyle{\sin { { x }_{ 1 } } +\sin { { x }_{ 2 } } +\sin { { x }_{ 3 } } }$ where it has constrained that $\displaystyle{{ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }=2\pi -\theta }$ . For this purpose In my opinion Jenson's Inequality is best .

So consider an function $\displaystyle{f(x)=\sin { x } }$ and Applying Jensen's Inequality , maximum will occur when :

$\displaystyle{\sin { { x }_{ 1 } } +\sin { { x }_{ 2 } } +\sin { { x }_{ 3 } } =3\sin { \left( \cfrac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } \right) } \equiv 3\sin { \left( \cfrac { 2\pi -\theta }{ 3 } \right) } }$ . So Maximum area will be :

$\displaystyle{{ A }_{ max }=\cfrac { 1 }{ 2 } \times { r }^{ 2 }\left( \sin { \theta } +3\sin { \left( \cfrac { 2\pi -\theta }{ 3 } \right) } \right) \\ { A }_{ max }=\cfrac { { a }^{ 2 } }{ 4\left( 1-\cos { \theta } \right) } \times \left( \sin { \theta } +3\sin { \left( \cfrac { 2\pi -\theta }{ 3 } \right) } \right) }$

$P=4,q=3,r=3,p+q+r=10$