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Archimedes Calculates Pi

In Archimedes’ first section, he calculates Pi, π, to be slightly less than 3 and one seventh.

Archimedes does not work out all the math on paper (actually parchment or sheepskin.)  Most of the portions in brackets [ ] were added by Heath in 19th century.  Notice how the equations end with 153. 

 
See below,
Archimedes  Math  Made  Simple  and  Easy
For a detailed explanation of all the mathematical steps that prove the 10 conclusions below, click on the number for that equation.

1 2 3 4 5 6 7 8 9 10 Conclusion

 

Words below in brackets [ ] and parathesis ( )  are added for clarity.
Archimedes text begins :

 

 

 

I.     Let AB be the diameter of any circle, O its center, AC the tangent at A; and let the angle AOC be one-third of a right angle. 
(See diagram on right.)

Then

 

And

First, draw OD bisecting the angle AOC and meeting AC in D.
Now     CO : AO = CD : DA     

so that

 (CO + OA) : CA = OA : AD

Therefore

Hence

so that

 

Secondly, let OE bisect the angle AOD, meeting AD in E.    
Therefore      

    

 

Thus

 

Thirdly, let OF bisect the angle AOE and meet AE in F.
We thus obtain the result [corresponding to (3) and (5) above] that

         

Thus

Fourthly, let OG bisect the angle AOF, meeting AF in G.
We have then
 

        

Now the angle AOC, which is one-third of a right angle, has been bisected four times, and it follows that angle

Make the angle AOH on the other side of OA equal to the angle AOG, and
let GA  produced (extended to)  meet OH in H.
Then angle

Thus GH is one side of a regular polygon of 96 sides circumscribed to the given circle.
And since,

       

    

while
AB = 2 OA,      GH = 2 AG.

it follows that

But

 

 Therefore the circumference of the circle (being less than the perimeter of the polygon)

 


End Quote:

So,  Conclusion :

 

And notice above that almost all the equations end with the number “153.”   
And even equation 10 could be expressed as :

 

So, in equation 10 Archimedes multiplies his denominator 153 times 96 to get 14688.

Is this by pure chance that these equations end with 153 ?  Rather, it is an indication of the brilliance and the simplicity of the design of Archimedes’ solution for solving for Pi, π. Archimedes had brilliantly laid a 30° - 60° - 90°  Δ Triangle with the longer leg along the diameter and the other leg of the triangle as a side of a polygon going closely around the circumference of the circle. This set up a ratio of

By using this denominator of 153 in each equation he was able to calculate π.

 

For a detailed explanation of all the mathematical steps that prove Archimedes 10 conclusions above, click on the number for that equation to see Archimedes Math Made Simple.
1 2 3 4 5 6 7 8 9 10 Conclusion
 

 

 

Archimedes  Math  Made  Simple

This section below details all the mathematical steps that lead to the ten conclusions Archimedes makes above in calculating Pi,  π.

Previously when mathematicians wanted to calculate the value of  Pi,  π, they measured the circumference of a circle and divided it by the measurement of its diameter.  However measurements are inherently inaccurate when a person is trying to arrive at three or more significant digits.  In an ingenious way Archimedes developed a new method that did not involve measurements at all.  It is based on pure mathematical calculations.  His method allowed a person to calculate the value of Pi as accurately as one desired.  Because the method was pure mathematics the answers were indisputable.

A 30° - 60° - 90°  triangle has unique characteristics such that the ratio of the longer leg to the shorter leg is  3 :1.  By constructing such a triangle such that the longer leg is along the radius (or diameter) of any give circle and the shorter leg is along a tangent outside of the circle Archimedes sets up this ratio of  3:1  for his method of determining the value of  Pi,  π.  This outside tangent is used to calculate the measure of a polygon constructed just outside of the circumference of the circle.  Constructing the triangle in this way places the value of 1 in the denominator which makes his calculations much easier.

 


 

 


 

Proof for  (1)

 

Properties of a 30° - 60° - 90°  Δ Triangle 
A 30° - 60° - 90°  triangle has unique characteristics that were especially useful to Archimedes.  Begin with an equilateral triangle Δ  ACB (above) that has a measure of 2 on each side. Each angle will be 60°.  Drop a line from Point A that bisects the angle  CAB.  It will bisect the opposite side CB, so that DB is a measure of 1.  The right triangle ADB has a hypotenuse, AB, with the measurement of 2. One leg, DB has a measure of 1.  Using the Pythagorean theorem, C2 = A2 + B2,  or  22 = 12 + x2 , with x being the vertical leg, or altitude of the equilateral triangle,  X can be solved to equal  3. 

In order to make the mathematical computations, Archimedes needed to substitute a rational number for 3, so he used this very close approximation for  3.

I.     Let AB be the diameter of any circle, O its center, AC the tangent at A; and let the angle  AOC be one-third of a right angle.  (See diagram on right.)

Then

(1)    AO : CA =[ √3 : 1  ]       
         AO : CA > 265 : 153   

   
Proof for  (2)

Conclusion 2, below, as also in conclusion 1, is deduced from the Properties of a 30° - 60° - 90°  Triangle.  See above.

    and

(2)    CO : CA = [  2:1 ]
                         = 306 : 153.   

.
Proof for  (3)

First, draw OD bisecting the angle  AOC and meeting AC in D. 
Now     CO : AO = CD : DA     

See :  Euclid’s Elements : Triangles Book 6, Proposition 3


 

Add 1 to each side.

Since the measure of lines  CD + DA = CA

Multiply each side by
                                      

Reverse equation.

Substitute using statements 2 and 1 above.

 

 


Therefore,

 

 

Therefore


Proof for  (4)

Using the Pythagorean Theorem, 

OD2 = OA2 + AD2

Divide each side by  AD2

Substituting using line  (3.)

  

 

 

Hence


Beginning with the previous conclusion :

The next step is to simplify and take the square root of each side.

Again, in order to make mathematical computations Archimedes needs to substitute a rational approximation for the square root because it is an irrational number:

 

So,

so that


Proof for  (5)

Secondly, let OE bisect the angle AOD, meeting AD in E.
    Therefore

Again using :Euclid’s Bk 6, Prop. 3, as when we made the first bisection of angle AOC

So,    OD : OA = DE : EA

And again adding 1 to each side gives

Since   DE + EA = DA

Multiply each side by
                                    

Reverse equation

Substituting using lines 4 and 3

 

    

Therefore,

       

    

 


Proof for  (6)

As before, use the Pythagorean Theorem, 

OE2 = OA2 + EA2

Divide each side by EA2

Substituting using line
    

Taking the square root of each side,
and again Archimedes needs to substitute a rational approximation for the irrational square root so he uses:

 

This gives us line (6)

Thus

 


Proof for  (7)

Thirdly, let OF bisect the angle AOE and meet AE in F.
We thus obtain the result [corresponding to (3) and (5) above] that

As in the previous bisections, we again use :Euclid’s Bk 6, Prop. 3, as when we made the first bisection of angle AOC

Reverse Equation

Add  +1 to each side of the equation

Since EF + FA = EA

  Multiply each side by 
                                       

Substitute using lines (6) and (5).

 

 

Therefore,

         

 

 

Proof for  (8)

Using the Pythagorean Theorem, 

OF2 = OA2 + AF2

Divide each side by  AF2

Substitute using line  (7)

 

Therefore,

Taking the square root of each side,
and again Archimedes needs to substitute a rational approximation for the irrational square root so he uses:

 

 

So,


Proof for (9)

Fourthly, let OG bisect the angle AOF, meeting AF in G.
We have then

Again using :Euclid’s Bk 6, Prop. 3, as when we made the first bisection of angle AOC

 

Reverse equation

Add +1 to each side.

Since  FG  + AG  = FA

 

Substitute using lines (8)  and  (7) 

 

  

Therefore,

or

       


Proof for (10)

Now the angle AOC, which is one-third of a right angle, has been bisected four times, and it follows that angle

 

Make the angle AOH on the other side of OA equal to the angle AOG, and let GA produced (extended to)  meet OH in H.
Then angle

Thus GH is one side of a regular polygon of 96 sides circumscribed to the given circle.

Because,

And since,

 

And since the diameter, AB is two time the radius, AO

AB = 2 AO,  

And side GH = AG + AH.   And, since AG and AH are mirror images, AG = AH .  
So,

GH = 2 AG,

it follows that

Substituting line
                    

And since,   96 x 153 = 14688

we get

 
Or   

 

 

Proof for Conclusion:

Inverting the ratio above give us

And since,

And since,

Therefore,

 Therefore the circumference of the circle (being less than the perimeter of the polygon)

 

So,

In the next section, Archimedes goes onto inscribe a polygon within the circle and finds the lower limit of Pi.  He concludes with the calculation that Pi is between the following limits.

Final Conclusion:

    


 
 
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