Putnam 2000, Problem A-3

The octagon $P_1P_2P_3P_4P_5P_6P_7P_8$ is inscribed in a circle, with the vertices around the circumference in the given order. Given that the polygon $P_1P_3P_5P_7$ is a square of area 5, and the polygon $P_2P_4P_6P_8$ is a rectangle of area 4, find the maximum possible area of the octagon.

Solution:
The maximum area is \(3\sqrt{5}\).  
We deduce from the area of \(P_1P_3P_5P_7\) that the radius of the circle is \(\sqrt{5}/2\). An easy calculation using the Pythagorean Theorem then shows that the rectangle \(P_2P_4P_6P_8\) has sides \(\sqrt{2}\) and \(2\sqrt{2}\). For notational ease, denote the area of a polygon by putting brackets around the name of the polygon.  

By symmetry, the area of the octagon can be expressed as  
\[
[P_2P_4P_6P_8] + 2[P_2P_3P_4] + 2[P_4P_5P_6].
\]  

Note that \([P_2P_3P_4]\) is \(\sqrt{2}\) times the distance from \(P_3\) to \(P_2P_4\), which is maximized when \(P_3\) lies on the midpoint of arc \(P_2P_4\); similarly, \([P_4P_5P_6]\) is \(\sqrt{2}/2\) times the distance from \(P_5\) to \(P_4P_6\), which is maximized when \(P_5\) lies on the midpoint of arc \(P_4P_6\). Thus the area of the octagon is maximized when \(P_3\) is the midpoint of arc \(P_2P_4\) and \(P_5\) is the midpoint of arc \(P_4P_6\). In this case, it is easy to calculate that \([P_2P_3P_4] = \sqrt{5} - 1\) and \([P_4P_5P_6] = \sqrt{5}/2 - 1\), and so the area of the octagon is \(3\sqrt{5}\).