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

**Topic**

Geometry

**Contest**

Pre Regional Math Olympiad

2019

**Reading**

Mathematical circle .

Sequential Hints

The first step is very clear, you have to some how locate the point P .

claim : The intersection of the diagonal is the required point .

Proof : suppose that there is a point P' other than P (the intersection of diagonal) .

now try to apply triangular inequality in \(\triangle\)PBD & \(\triangle\)APC.

(do the rest yourself , and complete the proof)

so , the area of [ABCD]=\(\frac{1}{2}\)\(\sin\theta\)(PA*PB+PB*PC+PC*PD+PD*PA)

now, the area will maximum when \(\sin\theta\)=1

so , just put the value of PA, PB, PC,PD

You get the answer 55

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