Paradise Trinity Day

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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "PT 'n Q" design
"Quadrature authenticated with Flutterby Pi"

:geek: Isosceles trapezoid is best representation of "impossible" Quadrature
for it alone identifies both the circle and its square as well as comprises
the circle-squaring right triangle and circle-squaring scalene triangle,
especially when the trapezoid is nested within sqrt(2) replication. 8)

Rod :D
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "Pi 'n Phi" then "iQspiral" designs,
added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf

"Pythagorean circle-squaring right triangle
is geometric foundation of Quadrature,
where Pi + (4-Pi) = 4 (a^2 + b^2 = c^2)"

:geek: Both transcendental Pi and algebraic Phi describe spirals,
but Phi is so artistically balanced! However, Pi is inherently
associated with a circle ... therefore, sqrt(2). :roll

For D = 2, sqrt(Pi) / sqrt(4-Pi)
= 1.9130583802711007947403078280205.. Phi of Pi :roll:
D/d = D / (sqrt(Pi)/sqrt(4-Pi))^2

Phi = (1 + sqrt(5)) / 2
= 1.6180339887498948482045868343656..
D/d = D / ((1+sqrt(5))/2)^2

:idea: Rule: The circle is not squared unless ratio
of circle's diameter to Side of Circle's Square (SoCS)
= 2/sqrt(Pi) = 1.128379167095512573896158903..
= right triangle with sides sqrt(Pi), sqrt(4-Pi), 2.0
where sqrt(Pi)^2 + sqrt(4-Pi)^2 = 4 (re: PT) ;)

:geek: Regarding spirals' growth factors ...

Paired straight line segments maintain 2/sqrt(Pi) length relationship
(= 1.1283791670955125738961589031215..); linked segments within
each spiral increase/decrease by sqrt(2). Who knew?! :lol:

Both spirals show circle's center-to-spiral lengths
opposite 180 degrees to be a precise ratio of 4. :o
Outer sqrt(2) spiral uses hypotenuse of right triangle
while inner Pi spiral uses right triangle's long side. 8)

:geek: Ratio of Areas of circle-squaring right triangles

1.772453850905516027298167483341.. long side
x 0.92650275035220848584275966758914.. short side
= 1.6421833677363238765144275396003.. rectangle area
/ 2 = 0.82109168386816193825721376980015.. large triangle area

0.44311346272637900682454187083529.. long side
x 0.23162568758805212146068991689728.. short side
= 0.10263646048352024228215172122502.. rectangle area
/ 2 = 0.05131823024176012114107586061251.. small triangle area

0.82109168386816193825721376980015.. large triangle area
/ 0.05131823024176012114107586061251.. small triangle area
= 16 = Ratio of Areas 180 degrees opposite

:geek: Ultimately ...
Quadrature is about adjoined isosceles right and circle-squaring right triangles
... which create the circle-squaring scalene triangle, all overlapped
by two overlapping isosceles trapezoids (not highlighted). :roll:

So much Quadraturial GOO (Geometric Object Overlap):
- Lengths of isosceles right triangle: 2, sqrt(2), sqrt(2)
- Lengths of circle-squaring right triangle: 2, sqrt(Pi), sqrt(4-Pi)
- Lengths of circle-squaring scalene triangle: sqrt(2), sqrt(Pi), (sqrt(Pi)/sqrt(2) + sqrt(4-Pi)/sqrt(2))
- Lengths of overlapping isosceles trapezoids: Go figure! :?

:farao: A repositioned* design displays good symbolism
for suggesting the humanity-to-divinity journey
... where Heaven and Earth finally unite ...
an inner journey in the Grand Universe. 8)

* "upside down and backwards" :roll:

Rod :D
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "Cameo Pi" design*
"Pi in the Cat's Cradle"

:geek: ... and all in the Pi Corral. 8)

* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
plus "Seasonal Pi" (ornamental Pi for the season) :hithere

:farao: Is that a destiny anchor :?: :!:


Postcript: "/2" of the Collatz Conjecture (re: 3X+1 or /2)
suggests relationship to CSC** foundation of the "Pi Corral". :o

** Circle inscribed in Square inscribed in Circle ...
with sqrt(2) influence on decreasing line lengths
of the circle-squaring triangles. 8)

See also: viewtopic.php?f=15&t=27959&start=3550#p214938

:scratch: Re: "Postcript:" Spelling must be Freudian*** ...
Pi of the circle must be complemented with the square
... if heaven and earth conjunction matters. ;)

*** "very deeply hidden desires or feelings"


Rod ... :bike: ...
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "The Pythagogon" design*
More "Lines and Triangles and Squares, Oh My!" :lol:

When the Pythagorean Theorem squares the circle,
where 2/sqrt(Pi) defines both circle and its square,
= 1.1283791670955125738961589031215..

D = 2.0, 8 lines of perimeter = 1.0 each,
Side of Circle's Square (SoCS) = sqrt(Pi)
= 1.7724538509055160272981674833411..

* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Includes lines of Pi Fork, where line length ratios
= 2.0, sqrt(Pi), or 2/sqrt(Pi). :roll


Simple Sighman met the Pi Man 'vancing to the fair;
Said Simple Sighman to the Pi Man "Let me sav yor square."
Said the Pi Man to Simple Sighman "Fresh frost in symmetry."
Said Simple Sighman to the Pi Man "Oh, laud of jometry!"
;)

Rod ... :bike: ...
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "Pythagogon 444" design*

Geometry showing the square enclosed in the Pythagogon,
a square proving Pi/4 + (4-Pi)/4 = 4/4 ... and begging again
"Whence the transcendence of Pi in a squared circle?!" :roll:

Pythagorean symmetry proves sqrt(2) is transcendental or Pi is not.
:scratch: Say what?! Pi is evenly divisible by sqrt(2)? Who knew :?: :!:

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

Rod ... :bike: ...
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "Pythagogon 444" design
The Big SCOOP of Square Pi

Squared Circle Objects Of Pythagogonia (SCOOP)
which ask "Whence the transcendence of Pi?"
(aka "The Pi Corral") :roll:

Given: Circle-Squaring Right Triangle (C-SRT)
where a^2 + b^2 = c^2 (Pi + (4-Pi) = 4.0)
BS = Big Square, RT = right triangle
isrt = isosceles right triangle


:arrow: Calculate (square of C-SRT side)/4

2.0 circle's diameter, C-SRT hypotenuse
^2 = 4.0, area of its square
/ 4 = 1.0, area of isrt1

1.7724538509055160272981674833411.. sqrt(Pi), C-SRT's long side
^2 = 3.1415926535897932384626433832795.. Pi
/ 4 = 0.78539816339744830961566084581988.. area of isrt2

0.92650275035220848584275966758914.. sqrt(4-Pi), C-SRT short side
^2 = 0.8584073464102067615373566167205..
/ 4 = 0.21460183660255169038433915418012.. area of isrt3


:arrow: Calculate area of C-SRT

1.7724538509055160272981674833411.. sqrt(Pi), C-SRT long side
x 0.92650275035220848584275966758914.. sqrt(4-Pi), C-SRT short side
= 1.6421833677363238765144275396003..
/ 2 = 0.82109168386816193825721376980014.. area of C-SRT


:arrow: Calculate area of BS right side

0.92650275035220848584275966758914.. sqrt(4-Pi), C-SRT short side
/ sqrt(2) = 0.65513637756203355309393588562466.. small RT short side

1.7724538509055160272981674833411.. sqrt(Pi), C-SRT long side
/ sqrt(2) = 1.2533141373155002512078826424055.. small RT long side
x 0.65513637756203355309393588562466.. small RT short side
= 0.82109168386816193825721376980012.. area of two similar RTs
+ 1.0 area of isrt1
= 1.8210916838681619382572137698001.. area of BS right side


:arrow: Calculate area of BS left side

0.82109168386816193825721376980014.. area of C-SRT
+ 0.78539816339744830961566084581988.. area of isrt2
+ 0.21460183660255169038433915418012.. area of isrt3
= 1.8210916838681619382572137698001.. area of BS left side


:arrow: Proof area of C-SRT = total area of BS right side RTs

0.82109168386816193825721376980014.. area of C-SRT
0.82109168386816193825721376980014.. area of right side small RTs

:arrow: Proof area of BS left side isrt2 + isrt3 = area of BS right side isrt1

+ 0.78539816339744830961566084581988.. area of isrt2
+ 0.21460183660255169038433915418012.. area of isrt3
= 1.0 = area of isrt1

:arrow: Proof area of BS left side = area of BS right side 8)

1.8210916838681619382572137698001.. area of BS left side
1.8210916838681619382572137698001.. area of BS right side

:bana: Ultimately, these values can have unknown decimal places,
but be precisely balanced on both sides of the Big Square ...
proving that Pi is precisely divisible by sqrt(2), then by 2. ;)

Stated otherwise, every circle squared by the 2/sqrt(Pi) ratio
has a sqrt(2) sibling ... that has a sqrt(2) sibling ...
with all circles containing this same ratio! 8)


Rod :stars:
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "Pythagogon 444" design*
The Big SCOOP of Square Pi

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

:scratch: Pythagorean Sacred Geometry
when C-SRT area = length of diagonal?

Given: Diameter = 2.0, C-SRT hypotenuse
0.92650275035220848584275966758914.. sqrt(4-Pi), short side
1.7724538509055160272981674833411.. sqrt(Pi), long side
C-SRT = Circle-Squaring Right Triangle

0.92650275035220848584275966758914.. sqrt(4-Pi), short side
x 1.7724538509055160272981674833411.. sqrt(Pi), long side
= 1.6421833677363238765144275396003..
/ 2 = 0.82109168386816193825721376980014.. C-SRT area

0.92650275035220848584275966758914.. sqrt(4-Pi), short side
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi), sqrt(Pi)/(Pi/2)
= 0.82109168386816193825721376980014.. length of diagonal

Rod :D
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Re: Paradise Trinity Day

Post by Sandy »

It's good to see you working on this thread again, Rod. :sunflower:
xxSandy
“We measure and evaluate your Spiritual Progress on the Wall of Eternity." – Guardian of Destiny, Alverana.
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "C-SRT" design*
"Triangulation of Pi, fork'd"

* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
[ quadrature highlighting signature integration of Pi:sqrt(Pi)
(represents the circle) and 2:sqrt(2) (represents the square);
symbolic uniting of Heaven and Earth (they say) ] ;)

A study of areas and line length relationships of the C-SRT
and its sqrt(2) sibling, both containing the Pi Fork lines. 8)
If any geometric object can absorb transcendental Pi,
this unique Circle-Squaring Right Triangle (C-SRT)
has good potential to be that "Pi Corral". ;)

:scratch: Conundrum? Of all circle-squaring right triangles
identified by the 2/sqrt(Pi) ratio, only one contains
a diagonal (green line) that has length equal to
the area of the triangle: sqrt(4-Pi) / (2/sqrt(Pi))

:scratch: "What about increments x 10?" Go figure :!:


Given: C-SRT triangle sides = 2.0, hypotenuse, circle's diameter
1.7724538509055160272981674833411.. sqrt(Pi), long side
0.92650275035220848584275966758914.. sqrt(4-Pi), short side

2.0
/ 1.7724538509055160272981674833411.. sqrt(Pi)
= 1.1283791670955125738961589031215.. 2/sqrt(Pi), C-S ratio

1.7724538509055160272981674833411.. sqrt(Pi), C-SRT long side
x 0.92650275035220848584275966758914.. sqrt(4-Pi), C-SRT short side
= 1.6421833677363238765144275396003.. 2(C-SRT area)
/ 2 = 0.82109168386816193825721376980014.. C-SRT area

0.92650275035220848584275966758914.. sqrt(4-Pi), C-SRT short side
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi), C-S ratio
= 0.82109168386816193825721376980014.. diagonal length, C-SRT area


:bana: More numbas ...

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

1.7724538509055160272981674833411.. sqrt(Pi)
/ 1.1283791670955125738961589031216.. 2/sqrt(Pi)
= 1.5707963267948966192313216916397.. Pi/2
/ 1.2533141373155002512078826424055.. sqrt(Pi)/sqrt(2)
= 1.2533141373155002512078826424055.. sqrt(Pi)/sqrt(2)

1.2533141373155002512078826424055.. sqrt(Pi)/sqrt(2)
^2 = 1.5707963267948966192313216916397.. Pi/2
x 1.1283791670955125738961589031216.. 2/sqrt(Pi)
= 1.7724538509055160272981674833411.. sqrt(Pi)


:study: In comparison of three C-SRT's, only one has Area = green diagonal.
A transcendental C-SRT :?: HCIT :?: :!: This also reveals more integration
of the "Pi Fork" where all lines of this right triangle have either sqrt(Pi)
or 2/sqrt(Pi) length relationship to another line. "Impossible" quadrature. :lol:


1.0454464017541266302735942239054.. short side
x 2.0 long side
= 2.0908928035082532605471884478108..
/ 2 = 1.0454464017541266302735942239054.. area

1.0454464017541266302735942239054.. short side
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 0.92650275035220848584275966758911.. diagonal


0.92650275035220848584275966758914.. short side
x 1.7724538509055160272981674833411.. long side
= 1.6421833677363238765144275396003..
/ 2 = 0.82109168386816193825721376980014.. <<< area

0.92650275035220848584275966758914.. short side
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 0.82109168386816193825721376980014.. <<< diagonal


0.82109168386816193825721376980014.. short side
x 1.5707963267948966192313216916398.. long side
= 1.2897678009819452445732253163675..
/ 2 = 0.64488390049097262228661265818377.. area

0.82109168386816193825721376980014.. short side
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 0.72767355850930909975566083382494.. diagonal


Ro ... :bike: ...
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "C-SRT" design*
"Triangulation of Pi, fork'd"
* in http://aitnaru.org/images/Yin_Yang_Pi.pdf

:scratch: Quadrature channeled by Pi-casso?
... with this additional geometry:
If D = 2 = sqrt(Pi) x (2/sqrt(Pi)),
D = 1 = (sqrt(Pi)/2) x (2/sqrt(Pi)) 8)


:hithere Field Guide ...

This Cartesian presentation begins with a circle of D = 2.0, then circle's square constructed
with quadrature-defining right triangle ratio of 2/sqrt(Pi) (diameter/hypotenuse : long side).
Next, the circle's sqrt(2) sibling is created (D = sqrt(2)) and its square is constructed.

Then, long side (SoCS) of each triangle creates a circle with its square,
giving four circles squared, tightly integrating both sqrt(2) and sqrt(Pi). 8)
Geometry 101 (they say) :lol:

Ro ... :bike: ...
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "iamPi" design*
(integrated and maternity Pi where sqrt(Pi) and sqrt(2) of both large circles,
via sqrt(2), create smaller circle having "geometric DNA" of larger circles) 8)

Where a^2/4 + b^2/4 = c^2/4 for D = 2(2/sqrt(Pi))
and with D = 2.0, 2(2/sqrt(Pi))/sqrt(2)

Highlights a square's division into two sets of equivalent objects,
with one set representing sqrt(Pi) and the other sqrt(2).
("Pythagorean Pi" comes to mind) ;)

:idea: Geometers of quadraturial persuasion have long known that it's permissible to
"Have your cake and eat Pi too!" on Leftover Pi Day since there may be token Pi
(not "toke 'n Pi" ... but might be). :roll:

:bana: Practice now decorating a cake for Leftover Pi Day,
November 10, 2022 (Julian day 314).


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

While other Cartesian objects qualify as "circle-squaring",
this quadrature du jour features a certain quadrilateral
present in all 7 circles, effectively squared. :roll

Morbus Cyclometricus for the new millennium. :D


Ro, Ro, Ro your bote ... :viking:
Verily, verily, verily, verily ...
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "iamPi Three" design*
"A work in progress" ;)

:scratch: "Why "iamPi Three?"
It follows "iamPi Too". :roll:

:bana: Pythagorean Pi (in the sky until proven).
Start by proving sqrt(2) is a "Pi Corral". ;)


:hithere Geometer's secret ...
Circle-squaring scalene triangle in each of two concentric circles
have line length dimensions that differ by sqrt(2). Who knew :?: :!:
Now, the inspired "iamPi Three" makes sense (3 sides). 8)

* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
(highlights the two triangles' sqrt(2) juxtapositional scalenity) :roll

One side of circle-squaring scalene triangle is side of circle's inscribed square,
another side of the triangle is a side of the circle's area square. HCIT :!:


Ro ... :bike: ...
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "MC Scalenity" design*
"Pattern is the foundation of 'impossible' quadrature."

:arrow: Geometer's secret ...
Note two overlapping right triangles on left side of scalene triangles.
These pivot 90 degrees to expand/contract to the next sqrt(2) sibling. 8)

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

Ro, Ro, Ro your bote ...
Verily, verily, verily, verily ... ;)
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "MC JOI" design*
(Morbus Cyclometricus Juxtaposition Of Interest)

Despite colorful complexity, this geometry limits its claim
to the obvious Cartesian marriage of sqrt(Pi) and sqrt(2). 8)

:scratch: The design motif? "A Pi 'n Tie Affair"
(what men wear at the Squared Circles Soirée,
with "men" and "tie" currently user-defined). :roll:

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

Ro ... :bike: ...
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "Concentrizity of Pi" design,
added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
(more geometry that speaks for itself) ;)

"Sponsored by 2/sqrt(Pi)
= 1.1283791670955125738961589031215..
= (2/sqrt(Pi))^2 / (2/sqrt(Pi))
= sqrt(Pi) / (Pi/2)
= 2(1 / sqrt(Pi))"

Postscript ... Several weeks of redo was required to properly highlight Concentrizity's special letter,
finally more obvious in this Cartesian composition of concentric Pi with circles "impossibly" squared. 8)

"Pythagorean Pi with regal essence", they say :!:


Ro ... :bike: ...
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: "MC Squared" design*
When geometry reveals who's in charge of the universe. ;)

Pythagorean Pi in Cartesian space, hosted by sqrt(2).
Not how to get there, but what to expect when you arrive. 8)

* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
(still on the alluring path of Morbus Cyclometricus) :hithere

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Re: Paradise Trinity Day

Post by Amigoo »

Re: https://temporarytemples.co.uk/project/ ... 18-07-2022

:bana: How intriguing that Barbury Castle crop circle appears to show
6 spiral arms, each containing 8 right triangles, each having the
circle-squaring ratio (hypotenuse/long side) of 2/sqrt(Pi). 8)

Go figure :!: by calculating adjacent triangles. :roll
2 / sqrt(Pi) = 1.1283791670955125738961589031215..
sqrt(Pi) / sqrt(4-Pi) = 1.9130583802711007947403078280205..

:arrow: To calculate spiral growth, start with a triangle
having sides 2.0, sqrt(Pi), sqrt(4-Pi). Quadrature 101 ;)

:idea: Long side of largest triangle becomes hypotenuse of next inner triangle.
Geometry nerds of quadraturial persuasion need no calculator (they say) :roll:
... and they say a Pi spiral would not have hexagonal segmentation,
albeit a somewhat similar crop circle pattern could be created. 8)


:sunflower: iTandem design*, retrieved from last year's busyness,
suggests that sqrt(Pi) and sqrt(2) spirals are so integrated
and beyond the expertise of current crop circle makers
that the morontia machines will need reprogramming
to reflect this "impossible" Quadrature. :lol:

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


Rod :)
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: icTandem design*,
Inner circle with similar right triangle
having 1/16 area of beginning triangle
(proves Pi is precisely divisible by 16). :shock:

:geek: Long story short ...
1. Pi exists as relationship to a circle.
2. A circle has an inscribed square.
3. sqrt(2) defines inscribed square.
4. sqrt(2) spiral in 360 degrees
reveals the 16 segmentation.
5. sqrt(2) is a "Pi Corral". ;)

* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
(icTandem is inner circle portion of iTandem spirals)

:hithere Better delineation of circle-squaring right triangle
that has 1/16 the area of the spirals' starting triangle.

Pythagorean Pi will be served on Leftover Pi Day (Julian 314). :roll


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Re: Paradise Trinity Day

Post by Amigoo »

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

A few more lines were added to the Quadrature Flutterby,
further confirming sqrt(2) hosting of the tandem spirals. 8)

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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: icTandem design,
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf*

Flutterby Pi leavened by sqrt(2)

Now with overlapping, circle-squaring scalene triangles
where similar sides have sqrt(2) length association
(visual proof to good geometers). ;)

More lines added since "impossible" Quadrature
is all about pattern, balance, and sqrt(2) ... plus
"Lines and triangles and squares! Oh my!" :lol:

"What's the point?" In deed! 8)

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Re: Paradise Trinity Day

Post by Amigoo »

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

More Flutterby tweaks! :roll

Smallest circle contains 1/16 right triangle; this circle is part of the larger tandem spirals.
Completed middle circle shouts that Quadrature has a universe center. Who Knew :?: :!:

8) Now recurring mouse malaise predicts that
this cradled Pi RIPs the "impossible" of Quadrature. ;)

Ro :D
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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: iQ, Plane View design*
"impossible" Quadrature, simplified. :roll

"The propeller on the plane stays mainly in Cartesian domain." :lol:
* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf


:hithere Re: http://aitnaru.org/images/Tasty_Pi.pdf
Short 'n sweet, but "impossible" (they say), :roll:
and including Phi of Pi design, as well as
Pythagorean Objet d'art design:

Intriguing point-to-point precision in this Pythagorean Objet d'art,
juxtaposition of triangles adjacent to circle-squaring Pythagorean,
where dark blue lines of the cross have equal length. 8)

:bana: Geometer's secret ...

The blue cross portion of this Cartesian composition suggests a similar square
(blue lines of equal length), subdivided into 4 identical squares, can be created.
And this proves that Pi-associated objects have identical total area
as the non-Pi-associated objects. What does this mean?

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


:hithere Speaking of new boxa Pi ...

Phi defines two lines (one subdivided, giving three lines),
all having unique length relationship. 8)

Phi of Pi defines 3 sides of a circle-squaring right triangle,
reflecting the Pythagorean Theorem with Pi + (4-Pi) = 4
(each side** of the triangle squared where hypotenuse
has length equal to circle's diameter). 8) 8)
** sides = 2, sqrt(Pi), sqrt(4-Pi)

:arrow: The constants (line length ratios) ...

Phi of Pi = long side/short side = sqrt(Pi)/sqrt(4-Pi)
= 1.9130583802711007947403078280205..

iQuadrature = "impossible Quadratrure"
= hypotenuse/long side = 2/sqrt(Pi)
= 1.1283791670955125738961589031215..


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Re: Paradise Trinity Day

Post by Amigoo »

:sunflower: Re: Pythagorean Quadrature design,
added to: http://aitnaru.org/images/Tasty_Pi.pdf

The Pi Corral of the Pythagorean Theorem
where (a^2)/4 + (b^2)/4 = (c^2)/4, highlighting
3 sqrt(2) generations of circle-squaring triangles. 8)

Therefore area of circle-squaring right triangle
equals combined areas of two right triangles
adjoined to larger isosceles right triangle. :roll

:hithere New day, new perspective!
"Either sqrt(2) is transcendental or Pi is not" ;)

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Re: Paradise Trinity Day

Post by Amigoo »

:bana: The Circular Conundrum of Pizza Pi*

Since Circumference = Pi x Diameter,
Diameter cannot be known precisely
when Circumference has fixed value.

When D = 2, C = 6.283185307179586476925286766559..
When C = 10, D = 3.1830988618379067153776752674503..

:duh While it's intuitive that a circle's circumference
(a circle) might not have a fixed value, it's not intuitive
circle's Diameter (straight line) would not have fixed value.

:idea: In other words, a Circumference and a Diameter
can have fixed value, but not at the same time. :roll:

:bana: This explains why a pizza pie is always ordered
by the diameter and that "small, medium, large"
suggests a Pizza Secretly Purveyed Sans Pi. :shock:

* P S P S P image added to: http://aitnaru.org/images/Tasty_Pi.pdf
(red lines highlight 2/sqrt(Pi) authentication of Quadrature) 8)

:love Not to worry about sharing!
No matter your slice of Pi, it's always divisible by 2:
... 2/sqrt(Pi) = 1.1283791670955125738961589031215..
1/(sqrt(Pi)/2) = 1.1283791670955125738961589031215..
.5/(sqrt(Pi)/4) = 1.1283791670955125738961589031215..


Rod :lol:
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Re: Paradise Trinity Day

Post by Amigoo »

:study: Re: https://www.quantamagazine.org/entangle ... -20160428/
"Entanglement Made Simple" (related to entanglement of P S P S P circles?)

"Entanglement arises in situations where we have partial knowledge of the state of two systems."

:idea: Long story short (apparently) ...
"Entanglement" exists in the measurement,
but not necessarily in the reality. :roll:


:hithere On the other hand (not entangled) ...

Re: https://scienceexchange.caltech.edu/top ... tanglement
"What Is Entanglement and Why Is It Important?"

"Though scientists still debate how the seemingly bizarre phenomenon of entanglement arises,
they know it is a real principle that passes test after test ... It may be tempting to think that
the particles are somehow communicating with each other across these great distances,
but that is not the case" :o

:idea: Perhaps, an example ...

If two gyroscopes are made to spin at the same speed and angle,
they maintain relationship even when separated at great distance
... but do not communicate with each other :!:


Rod :)
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