Paradise Trinity Day

[Amigoo]

Re: Quadrature 360 design

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

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

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

Rod

[Amigoo]

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):

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

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

Rod (Pi for two is very social,
but ask for two Pi Forks)

[Amigoo]

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

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

Rod

[Amigoo]

Re: Blue MoonzaPi design
(renamed from Familiar Rarity)

Riddle of Cartesian juxtaposition ...

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

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

Rod

[Amigoo]

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

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

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

[Amigoo]

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"

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

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

Rod ... ...

[Amigoo]

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.

Rod

[Amigoo]

Re: Q90904X design
"Quadrature Unfoldable"

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

Rod ... ...

[Amigoo]

Re: Q90904X design
"Quadrature Unfoldable"

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?"

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

Rod

[Amigoo]

Re: Q90904X design
"Quadrature Unfoldable"

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

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

Rod

[Amigoo]

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.

Rod ... ...

[Amigoo]

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).

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).

Rod

[Amigoo]

On Pi Day 2019 ... (re: Q90904X design)

Don't ask "What's inside the box?" - ask "What's outside the box?"

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

[Amigoo]

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

[Amigoo]

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).

Rod

[Amigoo]

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
the apparent geometric "constraint of sqrt(2)".

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

[Amigoo]

Re: Quadraturial Juxtaposition design

Obviously (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.

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

Rod

[Amigoo]

Re: Quadraturial Juxtaposition design

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

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

[Amigoo]

Re: Juxta14159 design

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

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.

Rod

[Amigoo]

Re: Juxta14159 design

"My point ... and I do have one" ...
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

[Amigoo]

Re: Juxta14159 design

"My point ... and I do have one" ...

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 ...

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 ... ...

[Amigoo]

Re: Juxta14159 design

"My point ... and I do have one" ...

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 ...

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 ...

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

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

Rod ... ... (need mental gas to go further)

[Amigoo]

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

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 ...

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 ...

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

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

Even further ...

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 ... ... (obviously, enough gas for one day)

[Amigoo]

Re: Spiraling Squares design

"My point ... and I do have one" ...

A quick assembly of Pi leftovers, geometrically speaking,
and suggested for Leftover Pi Day (Julian 314, Nov. 10).

Rod

[Amigoo]

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

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

Rod ... ...