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


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 19^{th} 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’ actual text
[translated into English ] begins :



I. Let AB be the diameter of any circle, O its center, AC the
tangent at A; and let the angle
∠ AOC be
onethird 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 onethird 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, C^{2} =
A^{2} + B^{2},
or 2^{2} = 1^{2} +
x^{2 }, 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 onethird 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,
OD^{2} = OA^{2} + AD^{2}
Divide each side by AD^{2}
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,
OE^{2} = OA^{2} + EA^{2}
Divide each side by EA^{2}
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,
OF^{2} = OA^{2} + AF^{2}
Divide each side by AF^{2}
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 onethird 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:

See the connection between Archimedes and the Gospel. Read
153
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