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SEA UN CIENTIFICO CON LA BIBLIA: JOHANNES KEPLER-"WHERE THERE IS MATTER, THERE IS GEOMETRY"
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Reply  Message 2 of 50 on the subject 
From: BARILOCHENSE6999 Sent: 14/02/2016 17:16

Reply  Message 3 of 50 on the subject 
From: BARILOCHENSE6999 Sent: 10/08/2017 17:52
Resultado de imagen para poliedro san marcos

Reply  Message 4 of 50 on the subject 
From: BARILOCHENSE6999 Sent: 22/12/2018 14:15
'G', the letter of 'God', the 'Great Architect', the 'Geometer', for G is the 7th letter -
the Zayin/Sword of Christ which is contrasted by the Seven Rays of Light (L.U.C.I.F.E.R./A.B.R.A.X.A.S)

    

The Freemasonic symbology of the architecture in Washington, D.C. is replete with occult concepts and references.

The Square and Compass is aligned horizontally to revere the direction in which the Sun as well as the 'Sun behind the Sun'  - the Star of Set, or Sirius which is also included in the layout as the White House is not only at the left side of the Compass, but it  is at the bottom of the averse pentagram aligned with the North.

 
The Star Sirius is a binary star, and the lesser star - Sirius B - is depicted as the smaller star to the bottom left; 'The Pentagon'. Even the hieroglyph of Sirius contains the predominant elements of the architecture; 
 the Obelisk (Washington Monument), the 5  pointed star, and the Dome (The Capital Building).
 
 
 
What's more is not only is the White House 13 blocks south  the Masonic Temple, an inverted cross is formed by the lower line of the pentagram which is also 13 blocks.
As I mentioned in Phoenix Rising, 1313 is the number of THE AEON OF MAAT (and Αποκαλυψαι, Apocalypse/Reveal) , which is the dual compnent of THE AEON OF HORUS (1212). Together they are 2525, the dual Stars.

Reply  Message 5 of 50 on the subject 
From: BARILOCHENSE6999 Sent: 08/08/2020 17:21

3.4 Phi and Right-angled Triangles - the Kepler Triangle

The German astronomer and mathematician Johannes Kepler (1571-1630) had a great interest in both Pythagorean triangles and the golden ratio which was known then mainly by the term used by Euclid "the division of a line into extreme and mean ratio":
Geometry has two great treasures; one is the Theorem of Pythagoras; the other, the division of a line into extreme and mean ratio. 
The first we may compare to a measure of gold; the second we may name a precious jewel.
Mysterium Cosmographicum 1596
Here we have both together in a single unique triangle.

Pythagorean triangles are right-angled triangles that have sides which are whole numbers in size. Since the golden section, Phi, is not a pure fraction (it is irrational), we will not be able to find a Pythagorean triangle with two sides in golden section ratio.

However, there is as a right-angled triangle that does have sides in the golden ratio. It arises if we ask the question:

Is there a right-angled triangle with sides in geometric progression, that is 
the ratio of two of its sides is also the ratio of two different sides in the same triangle?
a ar arr sidesIf there is such a triangle, let its shortest side be of length a and let's use r as the common ratio in the geometric progression so the sides of the triangle will be a, ar, ar2
Since it is right-angled, we can use Pythagoras' Theorem to get:
(ar2)2 = (ar)2 + (a)2
a2r4 = a2r2 + a2
We can divide through by a2 :
r4 = r2 + 1
and if we use R to stand for r2 we get a quadratic equation:
R2 = R + 1 or
R2 – R – 1 = 0
which we can solve to find that
R = 
1 ± √5
2
  = Phi or –phi

1-rootPhi-Phi triangleSince R is r2 we cannot have R as a negative number, so
R = r2 = Phi so 
r = √Phi
The sides of the triangle are therefore
a, a √Phi, a Phi
and any right-angled triangle with sides in Geometric Progression has two pairs of sides in the same ratio √Phi and one pair of sides in the Golden Ratio!

 

We will meet this triangle and some interesting properties of its angles later on this page in the section on Trigonometry and Phi.

3.4.1 How to construct Kepler's Triangle

FIND G on AB with MG=MT

It is very easy to construct Kepler's Triangle if we start from a golden rectnagle as described above.... 
FIND G on AB with MG=MT

Make the golden rectangle... 
FIND G on AB with MG=MT

and draw a circle centred on one corner and having the longer side of the rectangle as radius... 
FIND G on AB with MG=MT

The point where the circle crosses the other longer side marks one vertex of Kepler's triangle, the centre of the last circle is another and the right angle of the rectangle is the third. 

  •  A trigonometric intersection D Quadling, Math. Gaz. (2005), Note 89.70
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/phi2DGeomTrig.html


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