TITLE: Is the area of a polygonal linkage maximized by having all vertices on a circle?
QUESTION [11 upvotes]: Consider a (non-stellated) polygon in the plane.  Imagine that the edges are rigid, but that the vertices consist of flexible joints.  That is, one is allowed to move the polygon around in such a way that the vertices stay a fixed distance from their adjacent neighbors.  Such a system is called a polygonal linkage.  
As the linkage varies in its embedding in the plane, the area of the interior varies.  The question is, When is the area maximized?
I have a specific answer I suspect is correct, but I am having trouble showing.  I believe it is true that every polygonal linkage has as embedding where all the vertices lie on one circle (this isn't hard to show in the case when the linkage starts non-stellated).  My claim is that the area is maximized exactly when all the vertices lie on a circle.  
I can show this for a 4-sided polygon, but with techniques that do not generalize.
Also, my requirement that the polygon be non-stellated was only so that it was clear that there was a way to flex it to have all vertices on a circle.  This question extends to the stellated case, but the question there is whether every stellated linkage can be flexed to one which is non-stellated.

REPLY [3 votes]: This is proved in I. M. Yaglom and V. G. Boltyanskii's book Convex Figures, on page 208. The essence of the argument is suppose you have two polygons $M$ and $N$ whose corresponding sides have the same lengths, but one of them, $M$, is inscribed in a circle. Then you can glue the segments of the circle outside $M$ to the corresponding sides of $N$. The resulting figure won't be a circle, but it is known that amongst all compact convex figures of given perimeter, the circle is the one with largest area (the solution to the isoperimetric problem), So $M$ plus bits has more area than $N$ plus the same bits, so $M$ has more area than $N$.
A lot of this goes back to the work of Steiner.