tan graphs in relation to the area of a circle

[morgan868]

Hi,

I hav this maths coursework that i hav to do, which involves finding the shape with the greatest area with a perimeter of 1000 metres.
Obviously its the circle, but i now hav to figure out how and why tan graphs relate to why a circle will hav a greater area than any polygon of the same perimeter, no matter how many sides it has.
Could you explain to me how the tan graph can explain this.

Thanks,

TG

[krtxmrtz]

I have used a triangle for the demo figure as it's easy enough to draw, but assume it's an n-sided polygon.

If r is the radius of the circle, then:

Area(circle) = Pi * r²
Perimeter(circle) = 2 * Pi * r

Perimeter(polygon) = n * a
Area(polygon) = (1/2) * a * h = b * h

h = (a/2) * tan(alpha)

Then:

Area(polygon) = (1/4) * n * a² * tan(alpha)

Because the perimeters of both the circle and the polygon are equal:

2 * Pi * r = n * a
so that
a = 2 * Pi * r / n

Therefore:

Area(polygon) = (1/n) * (Pi * r)² * tan(alpha)

The ratio of areas is:

Area(Circle) / Area(Polygon) = n / [Pi * tan(alpha)]

It's very easy to derive the value of alpha:

alpha = Pi/2 - Pi/n

so that

Area(Circle) / Area(Polygon) = n / [Pi * tan(Pi/2 - Pi/n)]

Obviously, as the number of sides n grows this ratio tends to infinite/infinite.

Now, it's time to use the tan graph
First make a substitution:
Pi/2 - Pi/n = x so that
n = 1/(1/2 - x/Pi)
and when the number n of sides grows, x tends to Pi/2

With this substitution:

Area(circle)/Area(polygon) = [1/(1/2-x/Pi)] / (Pi * tan x)

If you plot both 1/(1/2 - x/Pi) and tan(x) you'll see that the former function stays above tan, except at the limit when n is infinite and x=Pi/2 where both functions become infinite. So, while n is finite, the numerator is larger than the denominator and the area of the circle is always larger.

[krtxmrtz]

A further simplification using the fact that
tan(a - b) = (tan(a) - tan(b)) / (1 + tan(a)*tan(b))

tan(Pi/2 - Pi/n) = (tan(Pi/2) - tan(Pi/n)) / (1 + tan(Pi/2)*tan(Pi/n)) =
(1 - tan(Pi/n)/tan(Pi/2)) / (1/tan(Pi/2) + tan(Pi/n))

As tan(Pi/2) is infinite, this finally results in tan(Pi/2 - Pi/n) = 1/tan(Pi/n)

Area(Circle) / Area(Polygon) = tan(Pi/n) / (Pi/n)

Similarly to the previous post, you can now call x = Pi/n and plot both tan(x) and x on the same graph (see attached plot). The tan(x) curve stays above the x curve. For n -> infinite x->0 and both curves meet, i.e. equal areas when the number of sides is infinite, a circle.

[morgan868]

Hi Krtxmrtz,

Thanks very much for the help, very much appreciated. Although I havent got around to going over it in detail, i have glanced over it and it looks very promising, just what i was looking for. If i can ever help you with anything i will try, but by your apparent genius i dont think i could really help you with anything! Thanks again,

Tom

[krtxmrtz]

Hi Krtxmrtz,
Thanks very much for the help, very much appreciated.

You're very welcome, though I feel undeservedly flattered by your comment and though I get along well with geometry I am quite a newbie in many other areas... GID (glad it helped)

[morgan868]

Hi krtxmrtz,

I've got the general idea of it, but I need to be certain - what do a and h mean? I think it is a=base, and h=height. This may seem obvious but I have been using slightly different letters to represent the sides. Please could you help.

Thanks,

Tom

[krtxmrtz]

Hi krtxmrtz,
I think it is a=base, and h=height.

Yes.
a = base or, in general, side of the polygon, and h is the perpendicular from the center of the polygon to any of its sides.