Originally posted by DeepThought
No, you've misunderstood what I'm doing. Draw a cyclic polygon on a piece of card, also draw the circumscribing circle, this circle includes each of the vertices of the polygon on its circumference. Cut out the polygon and also cut along the circumference of the circle, so you get n bits of card which are segments of the circle. The straight parts of ...[text shortened]... hen we have proved the conjecture. I haven't a clue how you'd go about proving that though.
Aha! Still confusing, but much clearer than before. As I now understand:
1) For every cyclic polygon, circumscribe a circle and link the constructed arc segments to their respective polygon sides.
2) The area of the polygon is equal to the area of the circumscribed circle minus the arc segments (which have fixed area).
3) The sum of the exterior perimeters of the arc segments is equal to the circumference of the circle.
4) As the vertices are moved, the area changes but the exterior perimeter of the figure does not.
5) Using the result that a circle encompasses the greatest area for a given perimeter, greater than any arbitrary curve with the same perimeter, the configuration of the polygon with attached segments that creates the original circumscribed circle contains the greatest area.
This proves the result for a cyclic polygon. However, it does not apply to non-cyclic polygons and it is not always possible to make a cyclic polygon out of a non-cyclic one. So the proof is still elusive.