Paradise Trinity Day

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Amigoo
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Re: Paradise Trinity Day

Post by Amigoo » Fri Feb 19, 2021 6:46 am

:sunflower: Re: "TLO" design (aka "Three Moonsa Pi"),
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf

:scratch: Who can tell :?: :!:
'7' may allude to "scalene" of the circle-squaring triangles ...
according to geometers at the Temple of Meganta at TLO. ;)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Mon Feb 22, 2021 4:14 am

:sunflower: Re: "TLO" design (aka "Three Moonsa Pi")

:geek: A Pi divided against itself cannot stand ...
alone in a transcendental nirvana:

3.5449077018110320545963349666823.. 2(sqrt(Pi))
/ 3.1415926535897932384626433832795.. Pi
= 1.1283791670955125738961589031215.. 2/sqrt(Pi)
(defines a circle-squaring right triangle) :roll

:bana: The next Real Wheel of Pi design is not
The Last One since this foundational geometry
was created last year then rolled forward. ;)

This "Real Wheel of Pi, properly aged" design,
highlights Pi Corral with succinct concentricity: 8)

For D = 2, SoIS = sqrt(2), SoCS = sqrt(Pi)
1.2533141373155002512078826424055.. (sqrt(2)sqrt(Pi))/2
x 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.4142135623730950488016887242096.. sqrt(2)
(BTW little D = 2/Pi, C = 2)

:scratch: How did G101-Qqq get in the toy box :?:
("Geometry 101 with triple+ Quadrature") 8)

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sat Feb 27, 2021 2:22 pm

:sunflower: Re: "G101-Qqq" design,
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf

:geek: With Quadrature there is always better pattern
... when sqrt(2) manages the "impossible" Pi Corral. :roll

:idea: The Cartesian clue that originated more than a decade ago
is that Quadrature depends on 8 points of a circle whereupon rests
the circle's square. Quadrature's geometry derives from this clue
... since the infinite points of a circle are insignificant! ;)

:hithere For math nerds ...
This geometry suggests a decreasing value of Pi (not linear infinity)
as the nested sqrt(2) circles decrease in size. What progression
is this: if D = 2(sqrt(Pi)), SoCS = Pi , if D = 2, SoCS = sqrt(Pi),
if D = 1, SoCS = sqrt(Pi)/2 :?:

:hithere Regarding the '7' pattern ...
This natural pattern in sqrt(2)-nested circles was believed
to associate with a soft chime at the Temple of Meganta,
a sound heard at every time display having a '7' digit;
146 times in a 24-hour day. Perhaps but a myth. :?

:hithere Regarding '146' ... 1 x 4 x 6 = 2 x 4 = 8
= points on circle upon which rests circle's square. 8)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Fri Mar 05, 2021 11:37 am

:sunflower: Re: "Just Puzzling" design

It's just puzzling how Quadrature relates to Pythagorean squares :!:
... especially considering that this design began as exploration
of the sqrt(2) spiral, apparent foundation of the Pi Corral. 8)

:geek: Regarding large dark blue circle ...
D = 4, SoCS = 2(sqrt(Pi))

:idea: Obviously, Pi are equally square! ...
on both sides of this circle's diameter.
(transcendental / 2) / 2 .. comes to mind. ;)

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Tue Mar 09, 2021 5:55 pm

:sunflower: Re: "Just Puzzling" design,
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf

:geek: Quadrature's infinite nibbles of Pi by geometric Pi-ranhas. :lol:

:idea: Essential formulary of the Pi Corral:
2 / 1.7724538509055160272981674833411.. ( 2/sqrt(Pi) )
= 1.1283791670955125738961589031215.. (defines unique right triangle)
^2 = 1.2732395447351626861510701069801..

:stars: Lotsa numbas for circle-squaring right triangle
inscribed in a circle whose Circumference = 4.0

a = 1.1283791670955125738961589031215.. 2/sqrt(Pi)
^2 = 1.27323954473516268615107010698..

b = 0.63661977236758134307553505349006.. ((2/sqrt(Pi))^2)/2
x 0.92650275035220848584275966758914.. (sqrt(4-Pi))^2
= 0.58982997002716101129132484048001..
^2 = 0.34789939354224165695100130437553..

c = 1.27323954473516268615107010698.. (2/sqrt(Pi))^2
^2 = 1.6211389382774043431020714113555..

1.27323954473516268615107010698.. a^2
+ 0.34789939354224165695100130437553.. b^2
= 1.6211389382774043431020714113555.. c^2

:bana: Pop Quiz (from the School of Pi-ranhas)
Q: When is area of a circle equal to its diameter?
A: When circle's circumference = 4 = Pi(D)

When Circumference = 4.0
Diameter = (2/sqrt(Pi))^2
SoCS = 2/sqrt(Pi)
Area = (2/sqrt(Pi))^2
= Diameter

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Thu Mar 11, 2021 1:11 pm

:sunflower: Re: "C=4~P=4" design,
added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf

"I thought that Harey Pi would be the end of me." :roll:

:geek: Geometer's secret (a riddle) ...
"A whimsical blending of concepts for this quiescing
Quadraturial juxtaposition of Circumference and Perimeter."

Addition of the "Pi Fork" reveals a better Harey Pi
... which reveals Circumference (circle of a square)
and Perimeter (square of a circle) having length = 4. 8)

:hithere In time for Pi Day, March 14, the authentic Harey Pi
In the Pi Fork's circle-squaring right triangle,
each two colored lines have sqrt(Pi) relationship. :roll

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sat Mar 13, 2021 2:22 pm

:sunflower: Re: "C=4~P=4" design,
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf

:geek: Harey's new millennium mantra ...
"The circle cannot be squared, but squared circles exist!" :scratch:

:bana: New theme for Pi Day 2021 ...
The universe asks "Can you hear me now? Put your ears on!" :albino:

:hithere Voted best song for a New Millennium Pi ...
https://www.youtube.com/watch?v=ZgsUXUBpS8Qthem

:hithere Voted best Pi for a New Millennium song ...
2/sqrt(Pi) = 1.1283791670955125738961589031215..
(defines an inscribed Pythagorean right triangle*
that defines a circle and its square) 8)

* a = sqrt(4-Pi), b = sqrt(Pi), c = 2 [or sqrt(4) = 2] ;)

:hithere And then Pythagoras suggested
"Get Real! and ethereal! ... like me!" :roll:

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sat Mar 20, 2021 3:44 pm

:sunflower: Re: "C=4~P=4" design*

"Quadraturial Salience of Scalenity",
aka "The Three Ts of Q" [pronounced "three tease"] :lol:
(circle-squaring Trapezoid, right Triangle, scalene Triangle)

:geek: "Impossible" Quadrature is best savored with a real Pi Fork
where 2-line sets have sqrt(Pi) or 2/sqrt(Pi) relationship. 8)

:scratch: "So, is the circle finally squared?!"
Is it possible for the circle not to be squared?! :shock:
(yes, mathematically ... so they say) :roll:

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

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sun Mar 28, 2021 3:14 pm

:sunflower: Re: "C=4~P=4" design
"Salience of Scalenity" :farao:

:geek: If you comprehend this esoteric geometry,
consider yourself "impossibly" mooned by symbolism
... worthy of the next Roswell craft. :roll:

Or this might be "esoterically integrated geometry
of Quadraturial perspective" (aka "Morbus iQ"). :lol:


:duh Speaking of Morbus iQ ...

Given circle-squaring right triangle
where a = sqrt(Pi), b = sqrt(4-Pi), c = 2

"impossible" ratio of Quadrature (irQ)
a/b = ((a+b)/b)-1 :lol:

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

2.6989566012577245131409271509303.. sqrt(Pi) + sqrt(4-Pi)
/ 0.92650275035220848584275966758914.. sqrt(4-Pi)
= 2.9130583802711007947403078280204..
-1 = 1.9130583802711007947403078280204..

8) "Ps & Qs" of Pythagorean Quadrature:
Since a^2 + b^2 = c^2, Pi + (4-Pi) = 4


Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Wed Apr 07, 2021 10:10 am

:sunflower: Re: "C=4~P=4" design
"Salience of Scalenity" :farao:

:geek: Better color logic (both green squares square their respective circle).
Geometry that may have potential for a labyrinth (and might attract UFOs). 8)

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sun Apr 18, 2021 4:55 am

:sunflower: Re: "42b" design*
"... or not 2b" :roll:

:geek: Concentric salience (aka "Patterns rule!"). 8)

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

Rod :)

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Re: Paradise Trinity Day

Post by Amigoo » Mon Apr 19, 2021 9:01 am

:sunflower: Re: "Pi Shells" design
"The only game in town." ;)

:farao: A Sunday night dashed doodle
where diameters increase by 2/sqrt(Pi). :hithere
Diameters of the two small circles:
top = sqrt(Pi)/2, bottom = sqrt(Pi)/4

:geek: About reference to "dashed doodle"
... to some, a coded riddle (tune). :?

Rod :)

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