Golden Number linked to 666
Formal Proof needed

Hello,

While working on my websites on 666 myth ( http://www.666myth.co.nr/ French= http://www.666mythe.co.nr/ ) and on Isomorphous Triplets ( http://www.chez.com/cosmos2000/Numbers/IsomorphousTriplets.html ) I have found recently direct links between the Beast Number 666 and the famous Golden Number Phi, well-known by Pythagoras, Leonardo da Vinci ... and in Sacred Geometry. This unexpected and incredible relation 666 versus Φ may constitute a way for the Rehabilitation of 666 !!! ...

First we have this curious relations:

666 = 7^3 pi (Phi – 1) = 7^3 pi phi or 666 = (6/5) 7^3 Phi
with pi = 3.141593 Phi = 1.618034 phi = Phi -1 = 0.618034 7^3 = cubic of 7

Besides, after computing some sinus and cosines , expressed in Degrees and absolute values, we obtain:

Phi /2 = sin 666º = cos 324º = cos 216º = cos 144º = cos 36º = 0.80901699...
Phi = 2sin 666º = 2cos 324º = 2cos 216º = 2cos 144º = 2cos 36º = 1,61803399...
Phi = sin 666º + cos 216º = sin 666º + cos (6x6x6)º
Phi = cos 144º + cos 36º = cos º + cos (6 x 6)º
Phi = sin 666º + cos 144º = sin 666º + cos º

But, I am wondering if there could exist formal mathematical proofs for this 666 and Golden Number links.

Thanks for advice and help.

Titus (cosmos2000 (@) iquebec.com)

the # of the beast is not 666, but 616. it was in the news recently

www.ConcurrentInverse.com

http://www.cut-the-knot.org/htdocs/dcforum/DCForumID4/606.shtml

There will always be a use for large prime numbers, eg 111, when
expressing approximations to irrational numbers. There are 11*6 feet
in a chain, 13 weeks in a quarter, and 2*11/7 = 3.1428 = Pi + 0.0013.

666 - 720 = 54 = 270/5 deg = 3 Pi / 10 radians
pi/5 is related to the golden ratio
616 - 630 = 14 is not likely to be related to the golden ratio

5**1/2 = 2.2360679774997896964091736687313
(5/7)**1/2 = 0.84515425472851657750961832736595
(5/7)**1/3 = 0.89390353509656765128791642511372

1/ Phi = Phi -1
Phi**2 - Phi -1 = 0
2*Phi = 1 +-(1+4)**0.5 = 1 +- (5)**0.5 = 2* 1.6180339887498948482045868343656

Phi and pentagonal symmetry are discussed in depth at
http://www.mcs.surrey.ac.uk/Personal/r.Knott/Fibonacci/phi2DGeomTrig.html#trigmore

1/(5)**1/2 - 0.44721359549995793928183473374626
pentagonal symmetry pi/5 = 36deg
sin36 = 0.58778525229247312916870595463907
cos36 = 0.80901699437494742410229341718282
= Phi /2 = (1+(5)**0.5)/4
tan36 = 0.72654252800536088589546675748062 = (5-2*5**0.5)**0.5

cos72 = ((5**0.5-1)**0.5/2

Thus in pentagonal symmetry the is a correspondence or relationship of
Phi, 5**0.5 and Pi/5 . Why or how can this be derived?

36-54 triangle hypotenuse = 4
side opposite 36 = 2 Phi = 2.35114100 = 2*1.17557050
side opposite 54 = = 3.236067977

ref: http://www.mcs.surrey.ac.uk/Personal/r.Knott/Fibonacci/simpleTrig.html#xactrig
exact trig values
ref: http://www.mcs.surrey.ac.uk/Personal/r.Knott/Fibonacci/phi2DGeomTrig.html#trigmore

I have observed the phenomenon when I calculated the dodecahedral and
icosahedral angles. Consider a pentagonal jello mold with each if
the five sides a cylinder centered at the center of the pentagon base
with unit vectors going from the center of the pentagon and scribing
an arc to the vertical. At some angle the unit vector points are
unity apart; this defines the dodecahedral symmetry. The sides of
the base are then 2*sin(Pi/5) = 1.176 . If the five unit vectors are
raised by an angle alphathen the sides of the elivated pentagon are
2*cos(alpha)*sin(Pi/5) = 1 .

dodecahedral, icosahedral symmetry
In calculating the unit vectors to the vertices I observed
Phi and/or the square root of 5 cropping up.

cos(pi - dihedral) = 1/5**0.5

We have levels of complexity
Pi/2 2**0.5, Pythagoras, square root
Pi/6 3**0.5
Pi/5 Phi, golden, quadratic equation root
Pi/7 cubic equation root
...

My discussion of possible and impossible things is at
http://www.issi1.com/corwin/possible.txt

ref: http://www.issi1.com/corwin/platonic.txt
http://www.issi1.com/corwin/unit_v.txt
http://www.issi1.com/corwin/vertices.txt
ref: www.vb-helper.com/tutorial_platonic_solids.html

cc:golden numerology
cosmos2000@iquebec.com
www.666myth.co.nr/
cosmos2000@iquebec.com?subject=666 Bivalent Myth
www.chez.com/cosmos2000/Numbers/IsomorphousTriplets.html

Wm.C. Corwin.info, Ph.D.,P.E.
www.ConcurrentInverse.com
billc@issi1.com
(c) copyright 2005 Wm.C.Corwin

TITLE: Proof of simple exact formulas for dimensions of the
dodecahedron and icosahedron.

cc:
webmaster@www.geom.uiuc.edu
www.geom.uiuc.edu/docs/reference/CRC-formulas/handbook.htm

mathworld@wolfram.com EricWWeisstein
mathworld.wolfram.com/TrigonometricAdditionFormulas.html

cosmos2000@iquebec.com
www.666myth.co.nr/

george@georgehart.com
www.georgehart.com/virtual-polyhedra/platonic-index.html

RobertW@kagi.com
www.software3d.com/Stella.html

enquiry@ronknott.com
http://www.mcs.surrey.ac.uk/Personal/r.Knott/Fibonacci/simpleTrig.html#xactrig

efr@st-andrews.ac.uk
www-groups.dcs.st-andrews.ac.uk/~history/HistTopics/Golden_ratio.html

feedback@vb-helper.com
www.vb-helper.com/tutorial_platonic_solids.html

The proof of the exact formulas in closed form for the dodecahedron
and icosahedron is at http://www.issi1.com/corwin/calculator/proof.txt .
I have not yet read the formulas or their derivation elsewhere but
George Hart has informed me that they have been published. I don't
think the derivation is on the internet so I am putting it there so
references and links to them can be made. Interest in proofs
of exact formulas for pentagonal symmetry was expressed at
http://www.cut-the-knot.org/htdocs/dcforum/DCForumID4/606.shtml .
I have used the numerical results to make ray tracing applications at
http://www.issi1.com/corwin/icosahedron.jpg .

RESULTS:

dodecahedron
face normal angle arccos(1/5**0.5) 63.4349488 deg

edge length (50 - 22*5**0.5)**0.5
= 0.89805595
inscribed radius 1
center of edge radius ((5 - 5**0.5)/2)**0.5
= 1.17557
superscribed radius 3**0.5 * (5 - 2*5**0.5)**0.5
= 1.25840857
pentagon height (5/2)**0.5 * ( 3 - 5**0.5)**0.5
= 1.381966

surface area 30 * 2**0.5 * (65 - 29*5**0.5)**0.5
volume 10 * 2**0.5 * (65 - 29*5**0.5)**0.5
= 5.55029
4pi/3 = 4.1887
icosahedron
face normal angle arccos(5**0.5/3)

edge length 6**0.5 * (7 - 3*5**0.5)**0.5
inscribed radius 1
center of edge radius (3/2)**0.5 * (3 - 5**0.5)**0.5
superscribed radius 3**0.5 * (5 - 2*5**0.5)**0.5
triangle heigth 3/2**0.5 * (7 - 3*5**0.5)**0.5

surface area 30 * 3**0.5 * (7 - 3*5**0.5)
volume 10 * 3**0.5 * (7 - 3*5**0.5)
= 5.05406

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My discussion of possible and impossible things is at
http://www.issi1.com/corwin/calculator/possible.txt
Other relevant things are at
http://www.issi1.com/corwin/platonic.txt
http://www.issi1.com/corwin/unit_v.txt
Ray tracing example
http://www.issi1.com/corwin/icosahedron.jpg

After doing these derivations I told whoever I thought may be interested
and George Hart replied to me that it was well known and in the books:

Regular Polytopes by H.S.M.Coxeter Prof. Mathematics University of Toronto
Dover isbn 0486614808
Zome Geometry by George Hart and Henri Picciotto KeyCurriculumPress 2001

At
http://www.georgehart.com/virtual-polyhedra/references.html
13 of the 187 references may be particularly useful in calculations.

Please visit my website.

Best Regards,

Vincent

www.theosopher.com