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.

Source
Suggested Reading

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.

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