Paradise Trinity Day

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Re: Paradise Trinity Day

Post by Amigoo » Tue Mar 19, 2019 6:46 am

:sunflower: Re: Spiraling Squares design
"In pursuit of the Big Bang of Quadrature"

As Paradise has a precise location in the Grand Universe,
Pi has a precise location in this Cartesian Neighborhood:

:geek: Location of Pi between D=2 and D=4
(calculate up and down from sqrt(Pi)):

= 4.0
x 2.2567583341910251477923178062431.. 2(sqrt(4/Pi))
1.7724538509055160272981674833411.. sqrt(Pi)
x 1.1283791670955125738961589031215.. sqrt(4/Pi)
= 2.0

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

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Tue Mar 19, 2019 4:33 pm

:sunflower: Re: EIO Squares design (in progress)

About EIO Squares ... (Enclosing, Inscribed, 'O' = circle)

:idea: Since "Old MacDonald had a farm ..."
"moo moo, baa baa, oink oink, quack quack ... EIO"
(re: https://www.youtube.com/watch?v=EGymN-Lc87M )

:geek: Calculate up and down from perimeter of circle's square ...

Perimeter of Enclosing square for D = 4.0
= 16.0
x 1.1283791670955125738961589031216.. sqrt(4/Pi)

<> 14.179630807244128218385339866729..
Perimeter of circle's square; SoCS = 2(sqrt(Pi))

/ 1.2533141373155002512078826424055.. ((sqrt(Pi)(sqrt(2))/2
= 11.313708498984760390413509793678.. 4(2(sqrt(2)))
Perimeter of Inscribed square for D = 4.0

:geek: Confirmation that "Old MacDonald had a farm" ...

Relation of Enclosing/Inscribed perimeters:
16.0 / 11.313708498984760390413509793678..
= 1.4142135623730950488016887242097.. sqrt(2)

1.1283791670955125738961589031216.. sqrt(4/Pi)
x 1.2533141373155002512078826424055.. ((sqrt(Pi)(sqrt(2))/2
= 1.4142135623730950488016887242097.. sqrt(2)

Rod ... :bike: ...
"moo moo, baa baa, oink oink, quack quack ... EIO"

:bana: Did you know :?: :!: ...
Like "impossible" Quadrature, it's impossible to say fast
"moo moo, baa baa, oink oink, quack quack ... EIO"
but possible with lotsa practice (video tape this ... :lol:
sounds like a barnyard with a group of quackers)

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Re: Paradise Trinity Day

Post by Amigoo » Tue Mar 19, 2019 7:07 pm

:sunflower: Re: EIO Squares design*

“moo moo, baa baa, oink, oink, quack quack … EIO” :roll:

* EOP: http://aitnaru.org/images/Khristos_Voskrese.pdf

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Wed Mar 20, 2019 2:18 am

:sunflower: Re: EIO Squares design

When two triangles adjoin on the squared circles farm,
the hypotenuses transform into a "hypotenusii". :lol:

“moo moo, baa baa, oink, oink, quack quack
and a loud yeller hypotenusii ... EIO” :roll:

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sat Mar 23, 2019 8:22 pm

:sunflower: Re: Rights of Pythagoras design

"Convincing concept of Quadrature
with ad infinitum of adjoined triangles
via sqrt(2) in a Cartesian Neighborhood."

:geek: Acknowledges triangles created by the 8 points
of a circle upon which rests the circle's square. 8)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Sun Mar 24, 2019 4:44 am

:sunflower: Re: Rights of Pythagoras design
Convincing concept of Quadrature
:scratch: Who knew?! Believers (of Quadraturial persuasion)
see a boxa Pi in this Cartesian Neighborhood. :roll

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sun Mar 24, 2019 5:53 am

:sunflower: Re: Rights of Pythagoras design
Convincing concept of Quadrature
:farao: What's remarkable - even mysterious - about the boxa Pi
is that it was drawn last ... using available Cartesian points.
But such is the essence of "impossible" Quadrature. ;)

:geek: Some lines were then redrawn to "push" box into background.
But such is the "impossible" patience of a curious geometer. :roll:

Rod :)

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Re: Paradise Trinity Day

Post by Amigoo » Mon Mar 25, 2019 7:57 am

:sunflower: Re: Rights of Pythagoras design
Convincing concept of Quadrature
:geek: Geometers' secret:

Wedge at bottom left, containing 'i' (for 'impossible' and 'information'),
identifies the three lines representing Pi/2, sqrt(Pi), and 2.0, since ...
2.0/sqrt(Pi) = sqrt(Pi)/(Pi/2) = 1.1283791670955125738961589031215..,
the circle-squaring constant nicknamed "iPi" for "impossible Pi". :roll:

Rod :)

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Re: Paradise Trinity Day

Post by Amigoo » Mon Mar 25, 2019 7:22 pm

:sunflower: Re: Rights of Pythagoras design
Convincing concept of Quadrature
:geek: Geometers' secret (another):

Quadrature is best comprehended by visualing movement of triangles on an axis:
Golden vertical line, representing sqrt(Pi), is both long side of adjoined right triangle
on the right and hypotenuse of adjoined right triangle on the left. 8)

Either triangle can pivot in the opposite direction, complementing this Cartesian composition
that clearly ;) shows the geometric relationship of Pi/2, sqrt(Pi), and 2.0. :roll

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Tue Mar 26, 2019 2:44 am

:sunflower: Re: Rights of Pythagoras design
best comprehended by visualizing movement of triangles on an axis
:idea: After more "visualization", a new moon appears (on the left of the design),
giving evidence of 2/sqrt(Pi) propagation of circle-squaring right triangles. 8)

:geek: The distance between three of these triangles* is (2(sqrt(Pi))^2, ;)
hinting that 2/sqrt(Pi) is a Constant Of Interest (COI, pronounced "coy") :roll:
for development of a more precise Pi value, relative to Circumference.

* and relation of their hypotenuse to long side

Rod ... :bike: ... (cruisin' in my Conveyance Of Interest)

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Re: Paradise Trinity Day

Post by Amigoo » Tue Mar 26, 2019 10:10 pm

:sunflower: Re: Rights of Pythagoras design
About sqrt(2) spiral with tandem Pi spiral ...

:geek: Ratio of hypotenuse to long side of circle-squaring
right triangle maintains at 2/sqrt(Pi) during growth of spirals
(SoCS = Side of Circle's Square):

When D = 1
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 0.88622692545275801364908374167057.. sqrt(Pi)/2, SoCS

When D = sqrt(2)
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.2533141373155002512078826424055.. (sqrt(Pi)(sqrt(2)))/2, SoCS

When D = 2.0
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.7724538509055160272981674833411.. sqrt(Pi), SoCS

:geek: Evidence of tandem Pi spiral ...

1.2533141373155002512078826424055.. (sqrt(Pi)(sqrt(2)))/2, SoCS
/ 0.88622692545275801364908374167057.. sqrt(Pi)/2, SoCS
= 1.4142135623730950488016887242097.. sqrt(2)

1.7724538509055160272981674833411.. sqrt(Pi), SoCS
/ 1.2533141373155002512078826424055.. sqrt(Pi)(sqrt(2)), SoCS
= 1.4142135623730950488016887242097.. sqrt(2)

:scratch: Analysis/conjecture:

Because Diameter is a whole number every second
sqrt(2) increment, and because both sqrt(2) and Pi spirals
grow in tandem, then the Pi value must be influenced
by sqrt(2) "wholeness" every second increment.

"Either sqrt(2) is transcendental or Pi is not"
(that is, the Pi of geometry - not of advanced math) :roll:

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Wed Mar 27, 2019 2:22 pm

:sunflower: Re: Spi_rally design
(rally of spiraling circle-squaring objects)

:idea: While trying to give isosceles right triangles more visibility,
I noticed a missing spiral: the circle-squaring scalene triangle :!:

:roll Now, circle-squaring right triangles are most visible, followed by
scalene triangles, with isosceles right triangles lurking nearby. 8)

Obviously, 2/sqrt(Pi) is the common spiral growth factor. ;)

Rod ... :bike: ... (cruisin' back to the rally)

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Re: Paradise Trinity Day

Post by Amigoo » Thu Mar 28, 2019 2:55 am

:sunflower: Re: Spi_rally design
(rally of spiraling circle-squaring objects)
2/sqrt(Pi) is the common spiral growth factor. ;)
:idea: New "out of the box" insight ...

With 2/sqrt(Pi) relationship, both Diameter and SoCS (hypotenuse and long side)
maintain that relationship no matter how many times they're divided by 2 ...
and computers will run out of known Pi digits long before division by 2
gets lost in sub-quantum values. :roll:

Then what happens :?: :!: ;)

Rod :)

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Re: Paradise Trinity Day

Post by Amigoo » Thu Mar 28, 2019 5:11 am

:sunflower: Re: Spi_rally Similar design

All lines of the two similar scalene and two similar right triangles
have sqrt(2) relationship, making this Cartesian Neighborhood
the convincing essence of "impossible" Quadrature. 8)

Rod :)

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Re: Paradise Trinity Day

Post by Amigoo » Thu Mar 28, 2019 11:11 am

:sunflower: Re: Spi_rally Similar design

Redrawn to show that every line of the circle-squaring objects in both circles
(the smaller circle a sqrt(2) sibling of the larger) has geometrically-provable
sqrt(2) relationship with a similar line. HCIT :!:

Rod ... :bike: ... (off for similar respite)

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Re: Paradise Trinity Day

Post by Amigoo » Thu Mar 28, 2019 12:03 pm

:sunflower: Re: Spi_rally Similar design
(added to: http://aitnaru.org/images/Khristos_Voskrese.pdf )

Geometers' secret ... :roll

The dark blue (circle-squaring) scalene triangle can be inscribed
in a circle having the same diameter as the small light blue circle
... and both circles host identical scalene triangles. 8)
(a conjunction of circles known as Vesica Piscis)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Fri Mar 29, 2019 3:11 pm

:sunflower: Re: Spi_rally Similar design
a conjunction of circles known as Vesica Piscis
:scratch: Who knew :?: :!: that the Vesica Piscis was so "ancient history",
relative to Quadrature. More exploration of Spi_rally Similar did not reveal
any significant geometric contribution by Vesica to this Neighborhood. :(

:idea: The geometric objects, angles, and lines of "impossible" Quadrature
are unique - recently "Out of the Box" and "New Millennium"! :roll:

;) Maybe, a circle-squaring scalene triangle is this millennium's V.P.
and incorporating 2/sqrt(Pi), 45-degree angle, et al. 8)

Rod :)

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Re: Paradise Trinity Day

Post by Amigoo » Fri Mar 29, 2019 8:33 pm

:sunflower: Re: Spi_rally Similar design
did not reveal any significant geometric contribution by Vesica
:duh Of course not :!:
Vesica is inherent in the sqrt(2) CSC* foundation of this Quadrature,
a foundation constructed again and again for several years! :roll:

* Circle inscribed in Square inscribed in Circle

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Sat Mar 30, 2019 5:55 am

:sunflower: Re: Spi_rally Similar design

Now, a bit more esoteric but ...
convincing geometric proof that Pi is divisible by 2. :shock:
Who knew :?: :!: (and note the theta-like symbol)

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sat Mar 30, 2019 12:12 pm

:sunflower: Re: Spi_rally Similar design
note the theta-like symbol
:geek: Skip the 'theta" - Quadrature prefers its own symbol
(that scalene triangle inscribed in a circle).

;) The short line of the scalene is one side of an inscribed square
(not displayed to limit busyness of "impossible" esotericism). :roll:

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Sat Mar 30, 2019 12:57 pm

:sunflower: Re: Spi_rally Similar design
Quadrature prefers its own symbol
... especially since the scalene unifies two overlapping trapezoids
composed of 3 overlapping scalene triangles. HCIT :!: :!: :!:

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Sun Mar 31, 2019 8:57 pm

:sunflower: Re: Spi_rally Similar design

Who knew :?: :!: that Pi was always divisible by 2 ...
since a circle's area square could have been any square from
an inscribed square to a square enclosing the circle, all of which
are divisible by two in a CSC* Cartesian Neighborhood. ;)

* Circle inscribed in Square inscribed in Circle

:geek: Geometers' master secret:

The 8 points (only) of a circle upon which rests the square
was the beginning of this research years ago since area square
must have sides less than the enclosing square and sides
greater than the inscribed square :!: A revelation! 8)

:cheers: This 8-point "revelation" is effective Quadraturial rebuttal
to those who insist (still) that Pi must account for the infinite
number of points on a circle. "On a plate, there's 8 ... only."
(grammarians prefer "there are") :roll:

Rod :)

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Re: Paradise Trinity Day

Post by Amigoo » Mon Apr 01, 2019 1:22 pm

:sunflower: Re: Spi_rally Similar design
Pi was always divisible by 2
:geek: Geometers' secret:
The dark blue lines show that the diagonal of the smaller trapezoid
has same length as mid-point to mid-point of the larger trapezoid
... when the circles geometrically associate by sqrt(2).

Rod :D ("Pi with two forks, please")

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Re: Paradise Trinity Day

Post by Amigoo » Tue Apr 02, 2019 9:47 am

:sunflower: Re: Two Similar design*
(further exploration of Spi_rally Similar)
Pi was always divisible by 2
Whereas Spi_rally Similar focused on the associated sqrt(2) circle of the largest circle,
Two Similar contrasts the circle-squaring objects of the largest and smallest circles,
a contrast that includes the Happy Family of Quadrature's objects: trapezoid,
scalene-, right-, and isosceles right triangles. :roll

This juxtaposition creates visual confirmation that the largest circle
(and its objects) have twice the dimensions of the smallest circle. 8)

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

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Wed Apr 03, 2019 11:55 am

:sunflower: Re: Two Similar design
Pi was always divisible by 2
:geek: Obviously, every Pi calculation to collect more digits
must end in an even number ... according to this CSC
precision (re: sqrt(2)^2 = 2). What is every odd number?
An interim calculation?

If every Pi division (or multiplication) by sqrt(2) must produce
an even ending digit (maintains CSC precision), what sequences
of Pi's digits remain as "authentic" Pi? :roll:

Maybe sqrt(2)'s ending digit is the chain's weak link. :o

Rod :stars:

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