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Estimating the value of pi by a geometric method
23

Proof that pi = 3.141....

Jun 23, 2015

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David Coulson

A geometric method for localising the value of Pi
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Page 1: Proof that pi = 3.141....

Estimating the value of pi by a geometric method

Page 2: Proof that pi = 3.141....

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

Page 3: Proof that pi = 3.141....

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.

Page 4: Proof that pi = 3.141....

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.

Page 5: Proof that pi = 3.141....

R30O

2R Tan 30 O

The length of each side is 2R Tan 30O

Page 6: Proof that pi = 3.141....

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

Page 7: Proof that pi = 3.141....

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

Page 8: Proof that pi = 3.141....

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

Page 9: Proof that pi = 3.141....

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

Page 10: Proof that pi = 3.141....

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

Page 11: Proof that pi = 3.141....

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

Page 12: Proof that pi = 3.141....

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

Page 13: Proof that pi = 3.141....

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)

Page 14: Proof that pi = 3.141....

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.

Page 15: Proof that pi = 3.141....

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)

Page 16: Proof that pi = 3.141....

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

Page 17: Proof that pi = 3.141....

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:

Page 18: Proof that pi = 3.141....

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:

Page 19: Proof that pi = 3.141....

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.

Page 20: Proof that pi = 3.141....

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

Page 21: Proof that pi = 3.141....

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).

Page 22: Proof that pi = 3.141....

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)

Page 23: Proof that pi = 3.141....

David C, Canterbury NZ, 2009

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