Paradise Trinity Day

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Re: Paradise Trinity Day

Post by Amigoo » Sat Sep 12, 2020 3:14 pm

:sunflower: Re: PdR Lite design*
"sans pattern begot" ;)

:geek: Quadrature, proof that Pi is divisible by sqrt(2)
when Pi is represented as Side of Circle's Square (SoCS)
... displayed in three nested circles, all squared. 8)

* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Sat Sep 12, 2020 5:33 pm

:sunflower: Re: Flutterbye design

Our transitions to Paradise require many planetary flutterbyes,
not necessarily beginning on Diversey in Chicago. ;)

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sat Sep 12, 2020 11:01 pm

:sunflower: Re: Pi Are Squares design

:scratch: Who knew?! Pi are squares!
... when they're not circles. ;)

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

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Mon Sep 14, 2020 9:46 am

:sunflower: Re: Pi Are ABC design

Quadrature with Pythagorean je ne sais quoi 8)

Rod ... :bike: ....

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Re: Paradise Trinity Day

Post by Amigoo » Mon Sep 14, 2020 10:06 am

:sunflower: Re: Pi Are ABC design
"Quadrature with Pythagorean je ne sais quoi"

:hithere Long story short:

Several nights ago, just after 2:00 AM, I was awakened by a knock on my front door. It sounded like a knock on my wood door - not on the exterior storm door (but I didn't hear the storm door open). I decided not to answer the door unless the knock occurred again - it didn't.
Best guess: a dream.

This morning at 2:22 AM (I looked at the clock), I was again awakened by a similar knock. Best guess: A Midwayer prompt ...
so I got out of bed and created the Pi Are ABC design. :roll

Conjecture: This was a good time for brain energies to focus on the task ... and to receive any proffered assistance. ;)
Pythagoras said "Hello!" ... apparently. :roll:

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Tue Sep 15, 2020 4:37 am

:sunflower: Re: Pi Are ABC design
"Quadrature with Pythagorean je ne sais quoi"

:geek: About this algebra (also in design subtitle) ...
sqrt(4-Pi) x sqrt(Pi)/sqrt(4-Pi) x 2/sqrt(Pi) = 2.0

0.92650275035220848584275966758914.. sqrt(4-Pi)
x 1.9130583802711007947403078280205.. sqrt(Pi)/sqrt(4-Pi)
x 1.1283791670955125738961589031216.. 2/sqrt(Pi)
= 2.0

Tip: hypotenuse of circle-squaring right triangle
= circle's diameter = 2.0; short side = sqrt(4-Pi)

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Wed Sep 16, 2020 2:22 pm

:sunflower: Re: Pi Are ABC design*
"Quadrature with Pythagorean je ne sais quoi"

:geek: Geometry nerds of Quadraturial persuasion know that
Quadrature is beyond comprehension unless it's esoteric. ;)

* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Thu Sep 17, 2020 1:22 pm

:sunflower: Re: Pi Are Quadrilaterals design
"Quadrature with Pythagorean je ne sais quoi"

Similar quadrilaterals with Pythagorean perspective. 8)

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sat Sep 19, 2020 1:11 pm

:sunflower: Re: Pi Are Quadrilaterals design
"Quadrature with Pythagorean je ne sais quoi"

Now, with better geometric balance (esoterically speaking). ;)
:scratch: "What?! sqrt(2) is esoteric? Who knew ?!"

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Sun Sep 20, 2020 12:48 am

:sunflower: Re: Pi Are Quadrilaterals design
"Quadrature with Pythagorean je ne sais quoi"

:roll Voted best subtitle for this revised design:

Come explore the worlds of geometry!
"Lines and triangles and squares! Oh my!"


Rod :)

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Re: Paradise Trinity Day

Post by Amigoo » Tue Sep 22, 2020 4:07 am

:sunflower: Re: Pi Are Quadrilaterals design*
"Quadrature with Pythagorean je ne sais quoi"

:geek: Geometer's secret ...
Lines of the outer '7' have twice the length
of the lines of the inner '7', via sqrt(2). ;)

* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sat Sep 26, 2020 3:57 am

:sunflower: Re: Pi Are Quadrilaterals design
"Quadrature with Pythagorean je ne sais quoi"

:geek: Line length relationships in circle-squaring right triangle
sides = sqrt(Pi), sqrt(4-Pi), 2.0 (hypotenuse and diameter)

sqrt(Pi) = 1.7724538509055160272981674833411..
sqrt(4-Pi) = 0.92650275035220848584275966758914..
sqrt(Pi) / sqrt(4-Pi) = 1.9130583802711007947403078280204..

2/sqrt(4-Pi) = 2.1586552217353950788554161024245..
2/sqrt(Pi) = 1.1283791670955125738961589031215..
(2/sqrt(4-Pi)) / (2/sqrt(Pi))
= 1.9130583802711007947403078280204.. ;)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Sat Sep 26, 2020 1:32 pm

:sunflower: Re: Pi Are Quadrilaterals design
"Quadrature with Pythagorean je ne sais quoi"

:geek: "Line length relationships in circle-squaring right triangle"
are not revelatory, but this is all about "transcendental" Pi. ;)

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sat Sep 26, 2020 2:19 pm

:sunflower: Re: Pi Are Quadrilaterals design
"Quadrature with Pythagorean je ne sais quoi"

:geek: "Line length relationships in circle-squaring right triangle"
are not revelatory, but this is all about "transcendental" Pi
... as nuanced by sqrt(2). :roll:

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Sat Sep 26, 2020 9:49 pm

:sunflower: Re: Trapezoidal Bifurcation design

:geek: A circle having a diameter = 2.0, has a circumference = 2(Pi).
The two shorter sides (chords) of the circle-squaring right triangle
associate with similar circumference/2 as the two sides (chords)
of the isosceles right triangle, adjacent to the right triangle.

:scratch: Either sqrt(2) is transcendental or Pi is not ... apparently. ;)

Lines and triangles and squares! Oh my!

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Sun Sep 27, 2020 2:22 pm

:sunflower: Re: Trapezoidal Bifurcation design,
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf

"Triangular sharing of a 2(Pi) circumference,
a confluence of transcendental Pi hosted by sqrt(2)
with the subtle ambience of a Pythagorean ABC"

:geek: Also known as a Cartesian "Whatsit?"
(of Quadraturial persuasion) :roll:

:scratch: "So, whatsit ?!" Go figure!

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Tue Sep 29, 2020 1:59 am

:sunflower: Re: Trapezoidal Bifurcation design

"Triangular sharing of a 2(Pi) circumference,
confluence of transcendental Pi hosted by sqrt(2)
with subtle ambience of Pythagorean ABC"

:geek: Geometer's secret about lower right quadrant ...

Pythagorean perspective is best represented
by a square adjoined to that circle's hypotenuse,
but the isosceles right triangle is sufficient clue
and permits enlargement of the geometry. 8)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Tue Sep 29, 2020 12:21 pm

:sunflower: Re: Trapezoidal Bifurcation design
(four sqrt(2)-nested circles all squared) 8)

:geek: The equations represented by this design
(for incorrigible geometry nerds of Quadraturial persuasion): :roll:

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

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Wed Sep 30, 2020 3:33 pm

:sunflower: Re: Trapezoidal Bifurcation design
(four sqrt(2)-nested circles all squared) 8)

:geek: Lines and triangles and squares!
... and trapezoids and parallelograms! Oh my!


Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Sat Oct 03, 2020 9:11 pm

:sunflower: Re: Trapezoidal Bifurcation design
(five sqrt(2)-nested circles, all effectively squared) 8)

:geek: Better identification of the concentric parallelograms
(concentric relative to sqrt(2) integration). ;)

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sun Oct 04, 2020 3:33 pm

:sunflower: Re: Trapezoidal Bifurcation design*
(seven sqrt(2)-nested circles, all effectively squared) 8)

:geek: Who knew :?: :!:
Quadrature is all about pattern :!:

:scratch: "What pattern? circle? square? triangle?
trapezoid? parallelogram? ..." Yes! Go figure! ;)

* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Tue Oct 06, 2020 4:57 am

:sunflower: Re: Trapezoidal Bifurcation design
(eight sqrt(2)-nested circles, all effectively squared) 8)

Now, less color noise on the perimeter of this composition. :roll

:geek: Who knew :?: :!:
This composition includes a Center of Quadrature :farao:
that's obvious when you connect the points. ;)

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Tue Oct 06, 2020 1:11 pm

:sunflower: Re: Trapezoidal Bifurcation design
(eight sqrt(2)-nested circles, all effectively squared)

Tweaked daily until it squeaks (computer mouse, that is). :roll:

:geek: Who knew :?: :!:
This composition includes a Center of Quadrature (CoQ)
that's increasingly obvious as those points connect. ;)

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Thu Oct 08, 2020 2:57 am

:sunflower: Re: Trapezoidal Bifurcation design
(eight sqrt(2)-nested circles, all effectively squared)

:geek: Geometer's secret about red and yellow lines extending inward from perimeter of their respective circles ...

While this seems esoteric, these lines (when lengthened) help define the composition's Center of Quadrature
as well as being perpendicular to a line of similar color that represents one side of that circle's inscribed square
... and having the same length as the side of that square. HCIT ?! 8)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Thu Oct 08, 2020 3:57 pm

:sunflower: Re: Trapezoidal Bifurcation design*
Quadrature is all about pattern :!:

:geek: Who knew :?: :!:
The more spidery-webby this Cartesian composition gets,
the more it's convincing that sqrt(2) hosts Pi ... 8)
and not vicey-versey. :roll:

* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf

Rod ... :bike: ...

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