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The world has long been fascinated with the proportion between the Circumference and Diameter of a Circle, known as Pi (c/d = π from the equation c=πd). Note that the ratio of the circumference and its diameter is constant, no matter the size of the circle.
The ancient Egyptians and Babylonians knew Pi was slightly larger than 3, but it took Archimides, in the 3rd century BC, to devise a method to calculate Pi theoretically. Archimides realized that a regular polygon could be used to approximate a circle, the more sides you include, the closer the approximation. Since the ratio of the Perimeter to the Diagonal in Polygons can be figured exactly, and these were basically approximations of the Circumference and Diameter of a circle, he realized Pi could be approximated by using regular polygons. The more sides used, the better the approximation.
Today we have tricks that make the job even easier, but the original principle of using inscribed polygons still is the heart of the approximation.
So lets start with a square, a regular polygon of 4 sides;
It's not hard to see that the diagonals of the red square is the same as the diameters of the circle. The ratio of perimeter to the diagonal, if each side of the square is one would be:
(4/1.414)= 2.829
Note that this number is less than pi, as we would expect, since the perimeter of the square is clearly shorter than the circumference of the circle that it inscribes. This is true of any polygon inscribed in a circle. So as we add sides, we will be approaching the value of Pi from below.
Take a Hexagon incribed in a circle:
Notice in the diagram above is that the hexagon is a combination of 6 equilateral triangles. Let the sides of the equilateral triangles equal 1. From the diagram above you can see that the perimeter is 6 and the diagonal is 2 so the ratio of the diagonal to the perimeter of the hexagon is 6/2 = 3, still less than Pi = 3.1415 but getting closer.
Next let's look at an Octagon
An Octagon can be constructed out of 8 identical isosceles triangles like the one below:
And we know that angle BAC = 45° since 8 times 45° is 360° (one rotation). Let AB and AC equal 1. This tells us two things, it says that the diagonal of our octagon is equal to 2. It also allows us to calculate length BC, which in turn will tell us the perimeter (8 x BC).
Using the Law of Cosines on our isosceles triangle:
We get BC = √((1)2 + (1)2 -2(1)(1)Cos(45º)) = √(2-1.414) = .7653
Thus the perimeter of the Octagon is .7654 x 8 = 6.1232
And the ratio of the Perimeter to the Diagonal is 6.1232 / 2 = 3.0616
We're getting closer. We should note that for ANY even sided polygon (6-sided or larger) the above procedure could be used, with only the angle BAC and the number of times the line segment BC is multiplied changing. For instance:
A Hexacontagon (60 sided polygon) could be constructed out of 60 isosceles triangles. Each triangle would have an apex angle of 6º (60 x 6 = 360). Again we let the two equal sides of the isosceles triangles equal 1 and use the law of cosines to find the other side (which corresponds to a Hexacontagon side) is equal to .1046719. Multiply by 60 and we get 6.2803, which when divided by the diagonal 2 will give you 3.1402, which gets us much closer to pi.
How about a 36000 sided polygon? That's .01º or sides of 6.28318529 which when divided by the diagonal (2) gives 3.141592649, where pi is actually 3.14159265. Clearly as we add numbers of sides, we get closer and closer to the value of pi.
There are a ton of great links out there for this:
Nova Approximating Pi
A Piece of Pi
Joy of Pi
And of course (what would I do without it) Wikipedia
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Re: How to Calculate Pi with Polygons
