Paradise Trinity Day

[Amigoo]

Re: PdR Lite design*
"sans pattern begot"

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.

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

Rod

[Amigoo]

Re: Flutterbye design

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

Rod ... ...

[Amigoo]

Re: Pi Are Squares design

Who knew?! Pi are squares!
... when they're not circles.

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

Rod

[Amigoo]

Re: Pi Are ABC design

Quadrature with Pythagorean je ne sais quoi

Rod ... ....

[Amigoo]

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

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.

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

Rod

[Amigoo]

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

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 ... ...

[Amigoo]

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

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

[Amigoo]

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

Similar quadrilaterals with Pythagorean perspective.

Rod ... ...

[Amigoo]

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

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

Rod

[Amigoo]

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

Voted best subtitle for this revised design:

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

Rod

[Amigoo]

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

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 ... ...

[Amigoo]

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

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

[Amigoo]

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

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

Rod ... ...

[Amigoo]

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

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

Rod

[Amigoo]

Re: Trapezoidal Bifurcation design

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.

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

Lines and triangles and squares! Oh my!

Rod

[Amigoo]

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"

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

"So, whatsit ?!" Go figure!

Rod ... ...

[Amigoo]

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"

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.

Rod

[Amigoo]

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

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

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

Rod

[Amigoo]

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

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

Rod

[Amigoo]

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

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

Rod ... ...

[Amigoo]

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

Who knew
Quadrature is all about pattern

"What pattern? circle? square? triangle?
trapezoid? parallelogram? ..." Yes! Go figure!

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

Rod

[Amigoo]

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

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

Who knew
This composition includes a Center of Quadrature
that's obvious when you connect the points.

Rod ... ...

[Amigoo]

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

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

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

Rod

[Amigoo]

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

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 ?!

Rod

[Amigoo]

Re: Trapezoidal Bifurcation design*
Quadrature is all about pattern

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

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

Rod ... ...