Paradise Trinity Day

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Re: Paradise Trinity Day

Post by Amigoo » Thu Mar 07, 2019 5:33 am

:sunflower: Re: Quadrature 360 design

On "impossible" Quadrature, when all evidence is reviewed,
geometers can only proclaim "Lattice Pi!" ;)

:idea: At the annual Squared Circles Soirée ...
serve the 16 similar wedges to honored guests ...
and the 17th wedge to the most honored guest. 8)

:scratch: "What?! There are 17?!" Wait! There's one more :!:
Freeze the 18th inner wedge for the next Soirée.

Rod :)

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Re: Paradise Trinity Day

Post by Amigoo » Thu Mar 07, 2019 7:11 pm

:sunflower: Re: Quadrature 360 design
( http://aitnaru.org/images/Khristos_Voskrese.pdf )

Speaking of the House of Pi where Pi are Square ...
Pi are squared-circles, hosted by sqrt(2):

:idea: If Diameter = 4(sqrt(2))
= 5.6568542494923801952067548968388..

:arrow: Then Circumference = Pi x D
= 17.771531752633464988063523960243..
/ 16 = 1.1107207345395915617539702475152..
x sqrt(2) = 1.5707963267948966192313216916398..
= Pi/2

Rod :D (Pi for two is very social, 8)
but ask for two Pi Forks) ;)

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Re: Paradise Trinity Day

Post by Amigoo » Fri Mar 08, 2019 10:33 am

:sunflower: Re: Familiar Rarity design
(aka "Moons of Pi" and "MoonzaPi") :roll:

Familial inscribed squares, hosted by sqrt(2). ;)
Quadrature, short, sweet, and squared ...
with a characteristic parallelogram. 8)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Fri Mar 08, 2019 6:28 pm

:sunflower: Re: Blue MoonzaPi design
(renamed from Familiar Rarity)

Riddle of Cartesian juxtaposition ...

:scratch: When occurs "The stars at night are big and bright ..."?
When not the "impossible" season of Blue MoonzaPi. :roll:

:farao: Does the design appear as a scarab beetle entering the moon :?:
"Scarabs were popular amulets and impression seals in Ancient Egypt."

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Fri Mar 08, 2019 8:38 pm

:sunflower: Re: Blue MoonzaPi design
(from "Blue Moons of Pi")

:geek: Geometers' secret:
Legs of "scarab beetle" define perimeter (arc) of largest moon. 8)

Rod ... :bike: ... (lookin' for MoonzaPi constellation
out where "The stars at night are big and bright ...")

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Re: Paradise Trinity Day

Post by Amigoo » Mon Mar 11, 2019 6:33 pm

:sunflower: Re: Blue MoonzaPi design

Evidence of neuronal neuroticism in a Cartesian Neighborhood?
Re: https://www.psychologytoday.com/us/basics/neuroticism

"Since all personality traits, including emotional instability, exist on a spectrum,
everyone is a little neurotic to some degree—some people are just much more"

:cheers: Nevertheless ...
All these lines still do not show how to square the circle,
yet confirm that 2/sqrt(Pi) propagates in the Neighborhood.

:idea: Translation: While we do not know how to square circles,
we know what they will look like when we get there. :roll:

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Tue Mar 12, 2019 12:48 pm

:sunflower: Re: Quadrature Unfoldable design (Q90904X)

Who knew :?: :!: Pythagoras :?:

The four Pythagorean right triangles in the inner circle
unfold (pivot 90 degrees on their 90-degree angle) to create
the points required for this Cartesian composition. 8)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Tue Mar 12, 2019 8:55 pm

:sunflower: Re: Q90904X design
"Quadrature Unfoldable"

:geek: Geometers' secret:
The red straight lines have equal length,
helping identify the 4 unfolding hypotenuses,
from within the inner circle; 8) 8) 8) 8)
a circle squared by 2 right triangles.

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Wed Mar 13, 2019 5:22 am

:sunflower: Re: Q90904X design
"Quadrature Unfoldable"

:farao: Who knew :?: :!: On the E-R Bridge, "Quadrature Unfolding"
is oft a Can Of Wormholes, as in "How Now? Brown COW.
Doth the barrel contain lofty enlightenment or holy draft?"

:scratch: Who can tell? (whence the late night inspiration)
"Bottle of wine, fruit of the vine" or ...
"Barrel of draft, Quadrature's laff".

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Wed Mar 13, 2019 3:14 pm

:sunflower: Re: Q90904X design
"Quadrature Unfoldable"

:farao: Who knew :?: :!:
Since "sacred" geometry prefers symmetry, this update
pushes Q90904X into the "newly sacred" category.
:idea: Symbolism: Heaven and Earth unite. :kiss:

:geek: Geometers' secret: The straight lines of the perimeter
and three straight red lines all have equal length. HCIT 8)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Wed Mar 13, 2019 6:33 pm

:sunflower: Re: Q90904Xbox design

If you knew the box was there (in this Cartesian Neighborhood),
then you've already crossed the E-R Bridge ... in your mind. ;)

Don't linger :!: There's more journey ahead, even "extraterrestrial"
... but you originated from Earth. :roll:

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Thu Mar 14, 2019 11:33 am

:bana: On Pi Day 2019 ... (re: Q90904X design)
Don't ask "What's inside the box?" - ask "What's outside the box?"

A geometric box in a Cartesian Neighborhood, hosted by sqrt(2) ;)
and circle-squaring Pythagorean right and scalene triangles
where 2/sqrt(Pi) effectively defines the limit of Pi (IMO).

:geek: Geometers' secret:
In both circle-squaring Pythagorean right and circle-squaring scalene triangles,
the line having length between the shortest and longest lines (of the three)
is the SoCS (has length equal to Side of Circle's Square). 8)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Thu Mar 14, 2019 6:45 pm

:bana: On Pi Day 2019 ... (re: Q90904X design)
Don't ask "What's inside the box?" - ask "What's outside the box?"
:idea: But what's inside the box is the historical Pi concept (an infinite calculation by design, IMO).
Pythagorean Pi rules "outside the box" (a circle and its square share only 8 symmetric points). ;)

Re: https://www.cnn.com/2019/03/14/tech/pi- ... index.html

"Emma Haruka Iwao spent four months working on the project in which she calculated pi to 31.4 trillion digits ... a world record."

"Iwao did her number crunching primarily from Google's office in Seattle, where she works at as a developer and advocate for Google Cloud. Fittingly, she used 25 Google Cloud virtual machines to generate the enormously long number. It's the first pi record calculated on the cloud."

Rod :)

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Re: Paradise Trinity Day

Post by Amigoo » Fri Mar 15, 2019 1:59 pm

:sunflower: Re: Q90904Xbox design

Proportional Pi leavened by sqrt(2)
sqrt(2) : 2 ~ sqrt(Pi) : sqrt(2)(sqrt(Pi)

Means divided by Extremes
2( sqrt(Pi) ) = sqrt(2) ( sqrt(2)(sqrt(Pi) )

Equality of Means and Extremes
2( sqrt(Pi) ) = 2( sqrt(Pi) )

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Fri Mar 15, 2019 2:53 pm

:sunflower: Re: Q90904Xbox design
Proportional Pi leavened by sqrt(2):
sqrt(2) : 2 ~ sqrt(Pi) : sqrt(2)(sqrt(Pi)
This correspondence hints that the isosceles right triangle
and circle-squaring Pythagorean right triangle are better models
for a more finite Pi calculation (whatever that means), ...
a calculation that accepts the constraint of sqrt(2) ...
(whatever that means). :roll:

Rod :)

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Re: Paradise Trinity Day

Post by Amigoo » Fri Mar 15, 2019 3:55 pm

:sunflower: Re: Quadraturial Juxtaposition design
(moving from concept to the Neighborhood)

Juxtaposition of the Pythagorean right and isosceles right triangles,
adjoined on their equal-length hypotenuses, best illustrates :finger:
the apparent geometric "constraint of sqrt(2)". ;)

Rod ... :bike: ... (still cruisin' the Neignborhood)

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Re: Paradise Trinity Day

Post by Amigoo » Fri Mar 15, 2019 8:01 pm

:sunflower: Re: Quadraturial Juxtaposition design

:idea: Obviously :roll: (to advanced mathematicians), the new Pi calculation
follows the sqrt(2) "yellow brick road" decrements to the "end of tunnel"
with those decrements directly associated with the short line(?) of the
circle-squaring right triangle and segment of circle's circumference. :o

:bana: A sight to see (mathematically), that end of the tunnel :!:

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Fri Mar 15, 2019 10:44 pm

:sunflower: Re: Quadraturial Juxtaposition design

Obviously :roll: (to advanced geometers) ...
sqrt(2) is all over this Neighborhood :!:

:geek: In each of the two circles having sqrt(2) relationship,
adjoined isosceles right and circle-squaring right triangles
overlay a circle-squaring scalene triangle. And each line
of these 3 triangles in a circle have sqrt(2) relationship
to a similar line in the other circle. HCIT :!:

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Sat Mar 16, 2019 3:45 pm

:sunflower: Re: Juxta14159 design
Obviously :roll: (to advanced geometers) ...
sqrt(2) is all over this Neighborhood :!:
:scratch: Thinkin' "outside the box" about Area = Pi x radius^2 ...

This design, continuing exploration of Quadraturial Juxtaposition,
identifies fourth quadrant (of a square) representing the .14159..
portion of the total Pi area (3.1459..). ;)

That this .14159 is focused on sqrt2 decrement of the large circle
hints that this conceptual geometry may provide more insight. :finger:

Rod :)

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Re: Paradise Trinity Day

Post by Amigoo » Sat Mar 16, 2019 7:27 pm

:sunflower: Re: Juxta14159 design

"My point ... and I do have one" ... :roll:
Relation of perimeters of a circle's enclosing and inscribed squares
to Circumference of this circle where D = 4.0 (calculate up):

= 16.0
x 1.2732395447351626861510701069801.. (2/sqrt(Pi))^2
= 12.566370614359172953850573533118.. Circumference = Pi(D)
x 1.5707963267948966192313216916398.. Pi/2
8.0

Obviously (to advanced geometers ; - ), mathematical relationships
of inscribed square, circumference, and enclosing square are finite
... but have a few decimal points. ;)

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Sat Mar 16, 2019 10:55 pm

:sunflower: Re: Juxta14159 design
"My point ... and I do have one" ... :roll:
Relation of perimeters of a circle's enclosing and inscribed squares
to circumference of this circle where D = 4.0 (calculate up):

= 16.0
x 1.2732395447351626861510701069801.. (2/sqrt(Pi))^2
= 12.566370614359172953850573533118.. circumference = Pi(D)
x 1.5707963267948966192313216916398.. Pi/2
8.0

And further ... :duh

1.2732395447351626861510701069801.. (2/sqrt(Pi))^2 = 4/Pi
x 1.5707963267948966192313216916398.. Pi/2
= 2.0 suggesting that 8.0 x 2.0 = 16.0 ;)
Who knew :?: :!:

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sun Mar 17, 2019 5:11 pm

:sunflower: Re: Juxta14159 design
"My point ... and I do have one" ... :roll:
Relation of perimeters of circle's enclosing and inscribed squares
to circumference of this circle where D = 4.0 (calculate up):

= 16.0 enclosing square
x 1.2732395447351626861510701069801.. (2/sqrt(Pi))^2 = 4/Pi
= 12.566370614359172953850573533118.. circumference = Pi(D)
x 1.5707963267948966192313216916398.. Pi/2
8.0 inscribed square

And further ... :duh

1.2732395447351626861510701069801.. (2/sqrt(Pi))^2 = 4/Pi
x 1.5707963267948966192313216916398.. Pi/2
= 2.0 suggesting that 8.0 x 2.0 = 16.0 ;)
Who knew :?: :!:

Still further ... :duh

2 / 1.5707963267948966192313216916398.. Pi/2
= 1.2732395447351626861510701069801.. 4/Pi

2 / 1.2732395447351626861510701069801.. 4/Pi
= 1.5707963267948966192313216916398.. Pi/2

Rod ... :bike: ... (need mental gas to go further) :roll:

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Re: Paradise Trinity Day

Post by Amigoo » Mon Mar 18, 2019 3:22 am

:sunflower: Re: Juxta14159 design
"My point ... and I do have one" ... :roll:

Relation of perimeters of circle's enclosing and inscribed squares
to circumference of this circle where D = 4.0 (calculate up):

= 16.0 enclosing square
x 1.2732395447351626861510701069801.. (2/sqrt(Pi))^2 = 4/Pi
= 12.566370614359172953850573533118.. circumference = Pi(D)
x 1.5707963267948966192313216916398.. Pi/2
8.0 inscribed square

And further ... :duh

1.2732395447351626861510701069801.. (2/sqrt(Pi))^2 = 4/Pi
x 1.5707963267948966192313216916398.. Pi/2
= 2.0 suggesting that 8.0 x 2.0 = 16.0 ;)
Who knew :?: :!:

Still further ... :duh

2 / 1.5707963267948966192313216916398.. Pi/2
= 1.2732395447351626861510701069801.. 4/Pi

2 / 1.2732395447351626861510701069801.. 4/Pi
= 1.5707963267948966192313216916398.. Pi/2

Even further ... :duh

1.5707963267948966192313216916398.. Pi/2
/ 1.2732395447351626861510701069801.. 4/Pi
= 1.2337005501361698273543113749845..

3.1415926535897932384626433832795.. Pi
/ 1.2337005501361698273543113749845..
= 2.5464790894703253723021402139603.. 8/Pi
/ 2 = 1.2732395447351626861510701069801.. 4/P1

Rod ... :bike: ... (obviously, enough gas for one day) :roll:

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Re: Paradise Trinity Day

Post by Amigoo » Mon Mar 18, 2019 3:39 pm

:sunflower: Re: Spiraling Squares design
"My point ... and I do have one" ... :roll:
A quick assembly of Pi leftovers, geometrically speaking,
and suggested for Leftover Pi Day (Julian 314, Nov. 10). :roll

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Mon Mar 18, 2019 4:33 pm

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

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

Rod ... :bike: ...

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