The octagon P₁P₂P₃P₄P₅P₆P₇P₈ is inscribed in a circle, with the vertices around the circumference in the given order. given that the polygon P₁P₃P₅P₇ is a square of area 5 and the polygon P₂P₄P₆P₈ is a rectangle of area 4, find the maximum possible area of the octagon.

Solution:

Let O be the center of the given circle. Let ß be the angle P₂OP₃ and let Þ be the angle P₂OP_4.

Since the polygon P₁P₃P₅P₇ is a square of area 5, we have | P₁P₃|²=5/2.

The area of the polygon P₂P₄P₆P₈ is 5 sin(Þ). Therefore Þ=arc sin (5/4).

A simple calculation shows that the area of the octagonis

P(ß)=5/2(sin(Þ-ß)+sin(ß)+cos(Þ-ß)+cos(ß)),

where 0<=ß<=Þ.

Clearly P(0)=P(Þ)=6, and the maximum is obtained at the point a

where P'(a)=0.

This yields a=/2

Maximum(P)=5(sin(Þ/2)+cos(Þ/2))=3c, where c^2=5.