Approximating Pi using Geometry

Date: 08/12/98 at 00:33:38
From: Rimi Ghosh
Subject: Value of pi

I need to know a simple method to find the approximate value of pi 
(3 < pi < 4) using elementary geometry.

Date: 08/12/98 at 13:48:02
From: Doctor Rick
Subject: Re: Value of pi

Hi, Rimi. 

I had fun with this problem a while back in connection with a
Geometry Problem of the Week on the Math Forum. I'll share a partial 
answer from that work. Of course, you realize that "simple" and "fast" 
are not the same thing - the methods that are really used to 
approximate pi converge much faster than what I will show you, but 
they require some higher knowledge than geometry, like Taylor series.

The geometrical method works like this. We will try to approximate the 
area of a circle of radius 1 -- the area is pi. We do this by noting 
that the area of an inscribed polygon must be less than the area of 
the circle, and the area of a circumscribed polygon must be greater 
than the area of the circle. Here is what they look like when the 
polygons are squares:

   +----------------******----------------+
   |          *       /\       *          |
   |       *        /    \        *       |
   |     *       /          \       *     |
   |   *       /              \       *   |
   |  *     /                    \     *  |
   | *    /                        \    * |
   |*   /                            \   *|
   *  /                                \  *
   */                                    \*
   *\                                    /*
   *  \                                /  *
   |*   \                            /   *|
   | *    \                        /    * |
   |  *     \                    /     *  |
   |   *       \              /       *   |
   |     *       \          /       *     |
   |       *        \    /        *       |
   |          *       \/       *          |
   +----------------******----------------+

First take the circle and inscribe a square in it. The inscribed square 
has side sqrt(2) [that's the square root of 2], because its diagonal is 
a diameter of the circle. 

Next inscribe a regular octagon. Its area turns out to be 2*sqrt(2) = 
2.82843. It isn't very close to pi yet, but we can keep going. For this 
purpose I derived a formula for the area of a regular (2^n)-gon (a 
polygon with (2^n) sides) inscribed in a circle of radius 1. It turns 
out that this area is:

   A(n) = 2^(n-2) * s(n-1)

where s(n-1) is the edge length of the inscribed (2^(n-1))-gon. For 
example, the area of the octagon (n = 3) is 2^1 * s(2) where s(2) is 
the edge length of the square: s(2) = sqrt(2) so A(3) = 2*sqrt(2), as 
I just said. I won't try to prove the formula here, just leave it as 
an exercise for the reader.

I still need a formula for the edge length s(n). I derived a formula 
for a(n) in terms of s(n-1), so we can compute, for example, the edge 
length of the inscribed octagon from the edge length of the square. 
The formula is:

   s(n) = sqrt(2 - sqrt(4 - s(n-1)^2))

(I have discovered a truly remarkable proof which this e-mail is too 
small to contain.) For example, knowing that the edge length of the 
inscribed square is sqrt(2), we find that the edge length of the 
inscribed octagon is:

   s(3) = sqrt(2 - sqrt(4 - s(2)^2))
        = sqrt(2 - sqrt(4 - sqrt(2)^2))
        = sqrt(2 - sqrt(2)) = 0.76536

Using these two formulas, here are the areas of some regular polygons
inscribed in a circle of radius 1:

   Square                                                 2
   Octagon     2*sqrt(2)                                = 2.82843
   Hexadecagon 4*sqrt(2 - sqrt(2))                      = 3.06147
   32-gon      8*sqrt(2 - sqrt(2 + sqrt(2)))            = 3.12144
   64-gon      16*sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2)))) = 3.13655

I'm on my way to computing a very good approximation to pi. I leave it 
to you to derive formulas for the area of circumscribed polygons in a 
similar manner. The difference between the areas of the inscribed and 
circumscribed polygons will indicate how good the approximation is, 
and it will keep getting smaller.

This may be more than you wanted, Rimi, but it makes an interesting 
project. It's nice to know that we don't have to take pi on faith - 
it can be computed using geometry alone.

- Doctor Rick, The Math Forum
Check out our web site! http://mathforum.org/dr.math/