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Re: Paradise Trinity Day

Posted: Wed Oct 23, 2019 12:07 pm
by Amigoo
:sunflower: Re: Quinale! 2:1(16 design
(aka "The Art of Pi")

Simplified (even the PDF):
http://aitnaru.org/images/Quinale!.pdf

Rod :D

Re: Paradise Trinity Day

Posted: Sat Oct 26, 2019 2:06 pm
by Amigoo
:sunflower: Re: Quinale! 2:1(16 design
(aka "The Art of Pi")

Simplified (even PDF): http://aitnaru.org/images/Quinale!.pdf
but still here: http://aitnaru.org/images/Pi_Fork_n_Lute.pdf

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Sun Oct 27, 2019 12:48 pm
by Amigoo
:sunflower: Re: Quinale! 2:1(16 design
(aka "The Art of Pi")
If c = 2, a = sqrt(Pi), b = sqrt(4-Pi)
:idea: aka "Pythagorean Pi" since ...
a^2 + b^2 = c^2 thus Pi + (4 - Pi) = 4

:scratch: Who knew :?: :!:

Rod (now that's "transcendental") :roll:

Re: Paradise Trinity Day

Posted: Tue Oct 29, 2019 10:34 pm
by Amigoo
:sunflower: Re: Scanale! design
(more "Art of Pi")

Two overlapping scalene triangles, associated by sqrt(2).
(of course, circle-squaring in this Neighborhood) ;)

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Thu Oct 31, 2019 1:59 pm
by Amigoo
:sunflower: Re: Scanale! design
(more "Art of Pi")

8) Updated to show 3 overlapping, circle-squaring scalene triangles,
with largest and smallest an increment (or decrement) of sqrt(2)^2. ;)

Rod :D

Re: Paradise Trinity Day

Posted: Fri Nov 01, 2019 1:06 pm
by Amigoo
:sunflower: Re: Scanale! design
(more "Art of Pi")

:geek: Geometers' secret ...

All 3 overlapping, circle-squaring scalene triangles are similar; each line
of the middle triangle has sqrt(2) relationship to the respective line
of both the smaller and larger triangles ... what we would expect
of similar geometry in sqrt(2)-nested circles. ;)

:shock: Of course, this proves that the largest and smallest triangles
differ by a factor of 2 (precisely), proving Pi/2 is "transcendental". 8)

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Sat Nov 02, 2019 12:48 am
by Amigoo
:sunflower: Re: Scanale! design
(more "Art of Pi")

:geek: Geometers' secret ...

The three similar vertices of the two scalene triangles
in "adjacent" circles are equidistant ... precisely. 8)

Rod :D

Re: Paradise Trinity Day

Posted: Sat Nov 02, 2019 3:29 pm
by Amigoo
:sunflower: Re: Scanale! design
(more "Art of Pi")

:scratch: Who knew?! :roll:
Pi was always precisely divisible by 2:

:geek: Where D = 2, C = Pi(D),
2/sqrt(2) = side of inscribed square
and C/4 = Pi/2. 8)

Rod :)

Re: Paradise Trinity Day

Posted: Sat Nov 02, 2019 5:55 pm
by Amigoo
:sunflower: Re: Scanale! design
(more "Art of Pi")

:geek: Simple message of the geometry:
Obviously, Pi cannot run off into infinity
without sqrt(2) close behind (or in front). 8)

"Either sqrt(2) is transcendental or Pi is not"

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Sat Nov 02, 2019 8:22 pm
by Amigoo
:sunflower: Re: Scanale! design
(the "Art of Perfect Pi")

:geek: All astute geometers know ...
a circle of Diameter = 2/sqrt(Pi)
and Circumference = 2(sqrt(Pi))
has an Area = 1 "A Perfect Pi"

D = 2/sqrt(Pi) = 1.1283791670955125738961589031215..

C = Pi(D) = 3.1415926535897932384626433832795..
x 1.1283791670955125738961589031215..
= 3.5449077018110320545963349666823.. = 2(sqrt(Pi))

A = Pi(r^2)
= 3.1415926535897932384626433832795..
x ((1.1283791670955125738961589031215..)/2)^2

= 3.1415926535897932384626433832795..
x 0.31830988618379067153776752674503..
= 1

Rod :D

Re: Paradise Trinity Day

Posted: Sun Nov 03, 2019 1:06 am
by Amigoo
:sunflower: Re: Entangled Pi design*
"A quantum of Pi for D=2 and A = 1"

* added to: http://aitnaru.org/images/Quinale!.pdf

That's sum Pi! ;)

Rod :D

Re: Paradise Trinity Day

Posted: Sun Nov 03, 2019 3:14 am
by Amigoo
:sunflower: Re: Entangled Pi design
"A quantum of Pi for D=2 and A = 1"

:geek: Geometers' quantum secret of Entangled Pi (a riddle): :scratch:

The circle is to the square and the square to the circle
as the circle to the circle and the square to the square
bestow the square root of Pi geometrically entangled. :roll:

Rod :stars:

Re: Paradise Trinity Day

Posted: Sun Nov 03, 2019 11:07 am
by Amigoo
:sunflower: Re: Entangled Pi design
"A quantum of Pi for D=2 and A = 1"

:geek: Geometers' quantum secret of Entangled Pi
(a previous riddle ... or conundrum): :scratch:

"To square the circle one must circle the square." ;)

Rod :D

Re: Paradise Trinity Day

Posted: Wed Nov 06, 2019 11:21 am
by Amigoo
:sunflower: Re: Entangled Pi design
To square the circle one must circle the square
:geek: Geometers' secret ...
"Entangled Pi" refers to 2/sqrt(Pi), the constant
that defines the circle-squaring right triangle
(new Pythagorian perspective). ;)

:idea: The '~Pi' symbol has good potential. 8)

2/sqrt(Pi) = 1.1283791670955125738961589031215..
/2 = 0.56418958354775628694807945156077..
^2 = 1.2732395447351626861510701069801..

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Thu Nov 07, 2019 2:53 pm
by Amigoo
:sunflower: Re: Entangled Pi design
To square the circle one must circle the square
:geek: Geometers' secret ...

"Pi Fork" identified in inner circle to authenticate "impossible" Quadrature
and featuring line length ratios of sqrt(Pi) and 2/sqrt(Pi). 8)

Rod :)

Re: Paradise Trinity Day

Posted: Thu Nov 07, 2019 6:00 pm
by Amigoo
:sunflower: Re: Entangled Pi design*
To square the circle one must circle the square
* updated in: http://aitnaru.org/images/Quinale!.pdf

:geek: Geometers' secret (about "Pi Fork" ratios) ...

Given: circle-squaring right triangle (red) with light blue "Pi Fork"

Line length ratios = 2/sqrt(Pi)
- hypotenuse to long side
- two long crossed blue lines

Line length ratios = sqrt(Pi)
- top of right triangle to short blue line
- hypotenuse to adjoined blue line
- long side to adjoined blue line

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Fri Nov 08, 2019 12:11 pm
by Amigoo
:sunflower: Re: Entangled Pi design
To square the circle one must circle the square
Nested circles all squared, simplified with salience. :roll:

:geek: Looks esoteric but it's just extrapolation of lines and objects
from the foundational 2/sqrt(Pi) circle-squaring right triangle
(with inherent "Pi Fork" identified and highlighted). 8)

"The '~Pi' symbol has good potential" :finger:

Rod :D

Re: Paradise Trinity Day

Posted: Sat Nov 09, 2019 1:06 pm
by Amigoo
:sunflower: Re: Entangled Pi design
To square the circle one must circle the square
:geek: The new Pi symbol (for constant 2/sqrt(Pi)) is displayed in PDF
and is pronounced "wiggle E" (sounds like "wiggly"). :lol:

Diner: "I'll have Wiggly Pi for dessert."
Waiter: "That's served only at the House of Pi" ;)

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Sun Nov 10, 2019 3:32 pm
by Amigoo
:sunflower: Re: Entangled Pi design
To square the circle one must circle the square
:geek: The math of "Wiggle E" (sounds like "wiggly"). :lol:
(see overlapping right and isosceles right triangles
on right side of this Entangled Pi composition,
colorfully simplified by salience)

2/sqrt(Pi) = 1.1283791670955125738961589031215..

1.7724538509055160272981674833411.. sqrt(Pi)
x 1.4142135623730950488016887242097.. sqrt(2)
= 2.506628274631000502415765284811..
/ 2 = 1.2533141373155002512078826424055..
x 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.4142135623730950488016887242097.. sqrt(2)

:scratch: "Say what?!" Go figure! ;)

Rod :D

Re: Paradise Trinity Day

Posted: Wed Nov 13, 2019 12:53 pm
by Amigoo
:sunflower: Re: Entangled Pi design
To square the circle one must circle the square
:geek: Geometers' secret ...
2/sqrt(Pi) is the length of the shared hypotenuse of the 2 overlapping triangles
(right and isosceles right) with 2/sqrt(Pi) also that circle's diameter. 8)

:scratch: "And the other lines?" Go figure! ;)

Rod :)

Re: Paradise Trinity Day

Posted: Thu Nov 14, 2019 2:06 pm
by Amigoo
:sunflower: Re: Mitosis design

:geek: Yet another 2/sqrt(Pi) construction ...
with visually confirming essence of Quadrature. 8)

:scratch: "Where's the mitosis?" Geometrically telophasic.
("ïmpossible" cell division of Quadrature ... apparently) :roll:

Rod :stars:

Re: Paradise Trinity Day

Posted: Fri Nov 15, 2019 9:37 am
by Amigoo
:sunflower: Re: Mitosis design*
"geometrically telophasic"

:geek: Geometers' secret ...
Red '7' refers to the 2 adjoined trapezoids,
inscribed in those 2 circles and share one side. 8)

:cheers: Who knew :?: :!: Mitosis math: 7 - 4 = 3

4 phases: Prophase, Metaphase, Anaphase, Telophase

7 trapezoid sides - 4 trapezoid sides = 3 circles :D
and the seven "deity" levels of "impossible" Quadrature?
(Static, Potential, Associative, Creative, Evolutional, Supreme, Ultimate)

* updated in: http://aitnaru.org/images/Quinale!.pdf

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Sat Nov 16, 2019 4:44 pm
by Amigoo
:sunflower: Re: Mitosis design
"geometrically telophasic"

:geek: Geometers' secret ...
Red '77' must refer to Mitosial Mitosis. :roll:
(smaller '7' appears where design might evolve)

:farao: Quadrature ever more esoteric, however ...
"Seventy times and seven" comes to mind. (139:2.5)

Rod :stars:

Re: Paradise Trinity Day

Posted: Sun Nov 17, 2019 7:55 pm
by Amigoo
:sunflower: Re: Mitosis design
"geometrically telophasic"

:geek: Geometers' secret ...
Red '77' must refer to Mitosial Mitosis. :roll:

:scratch: ... or maybe it's numerology:
https://trustedpsychicmediums.com/angel ... 7-meaning/
Who can tell :?: :!: "Be bold and brave" ;)

Rod ... :bike: ...

Re: Paradise Trinity Day

Posted: Tue Nov 19, 2019 3:59 am
by Amigoo
:sunflower: Re: Mitosis CQ design
"geometrically telophasic w/CQ"

:geek: Geometers' secret ...
Geometrically telophasic with a Center of Quadrature. 8)

Rod ... :bike: ... ("CQ CQ ...")