Try this beautiful problem from PRMO, 2019 based on Maximum area

## Maximum area | PRMO-2019 | Problem-23

Let $\mathrm{ABCD}$ be a convex cyclic quadrilateral. Suppose $\mathrm{P}$ is a point in the plane of the quadrilateral such that the sum of its distances from the vertices of ABCD is the least. If ${\mathrm{PA}, \mathrm{PB}, \mathrm{PC}, \mathrm{PD}}={3,4,6,8} .$ What is the maximum possible area of ABCD?

- $20$
- $55$
- $13$
- \(23\)

**Key Concepts**

Geometry

Triangle

Area

## Check the Answer

Answer:\(55\)

PRMO-2019, Problem 23

Pre College Mathematics

## Try with Hints

Given that $\mathrm{PA}=\mathrm{a}, \mathrm{PB}=\mathrm{b}, \mathrm{PC}=\mathrm{c}, \mathrm{PD}=\mathrm{d}$

Now from the above picture area of quadrilateral ABCD

Area=$[\mathrm{APB}]+[\mathrm{BPC}]+[\mathrm{CPD}]+[\mathrm{DPA}]$

Therefore area $\Delta=\frac{1}{2} \mathrm{ab} \sin \mathrm{x}+\frac{1}{2} \mathrm{bc} \sin \mathrm{y}+\frac{1}{2} \mathrm{cd} \sin \mathrm{z}+\frac{1}{2}$ da $\sin \mathrm{w}$

$\Delta_{\max }$ when $x=y=z=w=90^{\circ}$

$\Delta_{\max }=\frac{1}{2}(\mathrm{a}+\mathrm{c})(\mathrm{b}+\mathrm{d})$

Now ac $=$ bd (cyclic quadrilateral) As $(a, b, c, d)=(3,4,6,8)$

$\Rightarrow{(a, c)(b, d)}={(3,8)(4,6)}$

So $\Delta_{\max }=\frac{1}{2} \times 11 \times 10=55$

## Other useful links

- https://www.cheenta.com/problem-from-inequality-prmo-2018-problem-23/
- https://www.youtube.com/watch?v=U_LztQXd12A&t=121s