Re: How?Prove polygon->circle, area->max

On Sun, 15 Jul 2007 08:01:07 -0700, Steve Ro
<junkmailjunk@xxxxxxxxxxx> wrote:

If the number of sides (N) of a regular polygon increases the area
approaches that of a circle (within which it is inscribed) - I want to
be able to show what value the area of the polygon approaches as N-
infinity for a given value of the circumference (C).

The base of the triangles (created by lines drawn from the centre to
each corner) has a length:-
Lb = C/N

The angle at the apex of each triangle is:-
theta=2pi(rad)/N

The adjacent length for each triangle (line from centre of polygon to
mid-point of a side) is:-
La = (Lb/2) / (tan (theta/2))

Area of each of the N triangles is:
An = (Lb/2) x La
i.e. (half x base) x height

Area of polygon is:-
Ap = N x An

Substituting for An and then La and Lb above and then simplifying
gives me:-

Ap = N.(X/Y)

where:
X=(C/2N)^2
Y=tan(pi/N)

How do I then show what value Ap approaches as N increases?
Inspecting the terms X and Y show that Y->0 as tan(0)=0 and X->0 and N-
infinity so I end up proving nothing - how should I go about this? If
N=1000 Ap aproaches (C^2)/4pi (from A(circle)=pi(R^2), and
C(circle)=2.pi.R)

C=2.pi.R, I thought I could differentiate Ap wrt N but I'm having to
scratch around my maths from years ago to prove this - I think I
should be able to show how

Ap = N.(X/Y) -> (C^2)/4pi as N->infinity

I may get there in the end but maybe I'm tackling it the wrong way or
I've missed something. A bit of help would be appreciated.

cheers

See, for example, formula (9) at
<http://mathworld.wolfram.com/RegularPolygon.html>

A = 1/2 n R^2 sin(2 pi/n)

We want to find

lim[n -> oo] sin(2 pi/n) / (1/n)

After applying L'Hospital once, this becomes

lim[n -> oo] 2 pi cos(2 pi/n)

which equals 2 pi. Hence

A = 1/2 R^2 2 pi = pi R^2

.

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