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Try this beautiful problem from the Pre-RMO, 2019 based on Covex Cyclic Quadrilateral.

## Covex Cyclic Quadrilateral – PRMO 2019

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

• is 107
• is 55
• is 840
• cannot be determined from the given information

### Key Concepts

Largest Area

Distance

But try the problem first…

Source

PRMO, 2019, Question 23

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

Let $\angle APB= \theta$

area of $\Delta PAB$=$\frac{1}{2}(3)(4)sin\theta$

Second Hint

area of $\Delta PAD$=$\frac{1}{2}(3)(6)sin(\pi-\theta)$

area of $\Delta PDC$=$\frac{1}{2}(8)(6)sin{\theta}$

area of $\Delta PCD$=$\frac{1}{2}(8)(4)sin(\pi-\theta)$

Final Step

or, area of quadrilateral ABCD is $\frac{1}{2}(12+18+48+32)sin{\theta}$

or, maximum area of ABCD=6+9+24+16=55.