proof that pi = 3.141

Estimating the value of pi by a geometric method

The circumference of the circle is smaller than the perimeter of the outer hexagon, and bigger than the perimeter of the inner hexagon.

R

R

RR

RR

The circumference of the circle is smaller than the perimeter of the outer hexagon, and bigger than the perimeter of the inner hexagon.

The perimeter of the inner hexagon is exactly 6 radii.

The perimeter of the inner hexagon is exactly 6 radii.

The perimeter of the outer hexagon is harder to work out.

R

R

RR

RR

The circumference of the circle is smaller than the perimeter of the outer hexagon, and bigger than the perimeter of the inner hexagon.

R30O

2R Tan 30 O

The length of each side is 2R Tan 30O

The length of each side is 2R Tan 30O So, the perimeter of the outer hexagon is

6 x 2R Tan 30O = 12R Tan 30O

2R Tan 30 O

2R Tan 30O

2R Tan 30O

2R Ta

n 30

O

2R Ta

n 30

O

2R Tan 30 O

The circle is sandwiched between the inner hexagon and the outer hexagon.

2R Tan 30 O

2R Tan 30O

2R Tan 30O

2R Ta

n 30

O

2R Ta

n 30

O

2R Tan 30 O

R

R

RR

RR

The length of each side is 2R Tan 30O So, the perimeter of the outer hexagon is

6 x 2R Tan 30O = 12R Tan 30O

The circle is sandwiched between the inner hexagon and the outer hexagon.

6R < Circumference < 12RTan 30O

2R Tan 30 O

2R Tan 30O

2R Tan 30O

2R Ta

n 30

O

2R Ta

n 30

O

2R Tan 30 O

R

R

RR

RR

The length of each side is 2R Tan 30O So, the perimeter of the outer hexagon is

6 x 2R Tan 30O = 12R Tan 30O

The circle is sandwiched between the inner hexagon and the outer hexagon.

6R < Circumference < 12RTan 30O

C = 2 p R

3 < p < 3.42R Tan 30 O

2R Tan 30O

2R Tan 30O

2R Ta

n 30

O

2R Ta

n 30

O

2R Tan 30 O

R

R

RR

RR

The length of each side is 2R Tan 30O So, the perimeter of the outer hexagon is

6 x 2R Tan 30O = 12R Tan 30O

I can improve on the estimate by doubling the number of sides.

I can improve on the estimate by doubling the number of sides.

Inside wedge

Outside wedge

R

R

2R T

an 1

5O2R

Sin

15O

I can improve on the estimate by doubling the number of sides.

Inside wedge

Outside wedge

R

R

2R T

an 1

5O2R

Sin

15O

12 Sin 15O < p < 12 Tan 15O

3.106 < < 3.21 p

In general, for an n-sided polygon inside the circle and another one outside of the circle,

n x Sin (180/n) < p < n x Tan (180/n)

In general, for an n-sided polygon inside the circle and another one outside of the circle,

n x Sin (180/n) < p < n x Tan (180/n)

As the number of sides increase, the value of p is squeezed more tightly by the upper and lower limits.

sides p min p max 6 3.000000 3.464102 12 3.105829 3.215390 24 3.132629 3.159660 48 3.139350 3.146086 96 3.141032 3.142715 192 3.141452 3.141873 384 3.141558 3.141663 768 3.141584 3.141610 1,536 3.141590 3.141597 3,072 3.141592 3.141594 6,144 3.141593 3.141593

In general, for an n-sided polygon inside the circle and another one outside of the circle,

n x Sin (180/n) < p < n x Tan (180/n)

This may seem like the end of the story but there is a problem with this process:

sides p min p max 6 3.000000 3.464102 12 3.105829 3.215390 24 3.132629 3.159660 48 3.139350 3.146086 96 3.141032 3.142715 192 3.141452 3.141873 384 3.141558 3.141663 768 3.141584 3.141610 1,536 3.141590 3.141597 3,072 3.141592 3.141594 6,144 3.141593 3.141593

When a calculator or spreadsheet generates the SIN or TAN of an angle, it is using a formula that depends on the value of pi.

This may seem like the end of the story but there is a problem with this process:

When a calculator or spreadsheet generates the SIN or TAN of an angle, it is using a formula that depends on the value of pi.

You can’t use a formula that relies on pi to estimate pi!That’s circular reasoning.

This may seem like the end of the story but there is a problem with this process:

You have to have a way of generating SIN and TAN values that doesn’t rely on pi. Here’s a way:You can use geometry to get the COS of 60 degrees.

...and then you can use the half angle formulas to generate all the SINs and TANs you need after that.

So starting from 60 degrees, you can calculate the Sin, Cos and Tan of increasingly small angles (as long as you have a good way of calculating square roots).

http://www.slideshare.net/yaherglanite/a-formula-for-pi

Presentations on related subjects by the same author:

A formula for pi

http://www.slideshare.net/yaherglanite/area-of-circle-proof-1789707

Deriving the circle area formula

http://www.slideshare.net/yaherglanite/area-of-a-circle-simplification

Calculating the area of a circle without using Pi (and without using a calculator)

David C, Canterbury NZ, 2009

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