Covex Cyclic Quadrilateral | PRMO 2019 | Question 23

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.

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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