Try this beautiful problem from the Pre-RMO, 2019 based on Covex Cyclic Quadrilateral.
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.
Largest Area
Quadrilateral
Distance
Answer: is 55.
PRMO, 2019, Question 23
Geometry Vol I to IV by Hall and Stevens
Let \(\angle APB= \theta\)
area of \(\Delta PAB\)=\(\frac{1}{2}(3)(4)sin\theta\)
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 \(\Delta PCD\)=\(\frac{1}{2}(8)(4)sin(\pi-\theta)\)
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.
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