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

Quadrilateral

Distance

## Check the Answer

But try the problem first…

Answer: is 55.

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.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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