Re: Polygons of large area

On 2008-04-21, riderofgiraffes <mathforum.org_am@xxxxxxxxxxxxxx> wrote:
I guess we can hand-wave and show that the area is bounded above

Showing that it's bounded above isn't that difficult: you've already
showed that for any polygon, there is a convex polygon with larger
area, and the area of a convex polygon is just the sum of areas of
triangles. That sum is bounded by 1/2*max(height)*sum(bases), the sum
of bases is the perimeter, and max(height) is no more than half the
perimeter. So for perimeter P, the area is bounded by P^2/4.


If we could show that the LUB is L^2 cot(pi/n)/n then we're done.
I suspect that's too much to ask.

I suspect so too. The most elementary proof I can think of that the
area actually has a *maximum* (not just a LUB) involves at least some
real analysis.


- Tim
.

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