Re: Maths Area Problem

"Jim Ferry" <corklebath@xxxxxxxxxxx> wrote in message
news:1191129044.029094.254450@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

On Sep 29, 11:02 am, "Philippe 92" <nos...@xxxxxxxxxxxx> wrote:

The maximum area is when the quadrilateral becomes cyclic.
Then the area is given by Brahmagupta formula, generalizing
Heron's.

A = sqrt((s-a)*(s-b)*(s-c)*(s-d))

A sci.math thread I read recently gave (essentially) this elegant
proof of the fact that the maximal area of any polygon with
prescribed
side lengths is cyclic:

If the side lengths determine a polygon at all, they determine a
cyclic one. Now affix each segment of the circumscribed circle to
the
sides, and consider the total area of the polygon plus the affixed
segments as you vary the angles between edges. The configuration of
maximal area is the circle, because the circle has maximal area of
any
shape with this fixed perimeter. Because the sum of the segment
areas
is fixed, it follows that the polygon must also attain maximal area
in
this configuration.

Yes, elegant.

I imagined the polygon to contain a (2-dimensional) gas. For each side
of the polygon, consider the total force on it due to the pressure of
the gas. This force acts at right angles to the side, at its center.
If the polygon is in equilibrium the lines of action of these forces
must be concurrent, hence the polygon is cyclic.

This argument is actually somewhat wrong (4 non-concurrent forces can
be in equilibrium), but it leads to the right answer. I don't know if
the argument can be patched up.

--
Clive Tooth
http://www.shutterstock.com/cat.mhtml?gallery_id=61771


.

Relevant Pages

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  • Re: Maths Area Problem
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  • Re: Polygons of large area
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