Page **142** of **163**

### Re: Paradise Trinity Day

Posted: **Thu Mar 07, 2019 5:33 am**

by **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

### Re: Paradise Trinity Day

Posted: **Thu Mar 07, 2019 7:11 pm**

by **Amigoo**

Re: Quadrature 360 design

(

http://aitnaru.org/images/Khristos_Voskrese.pdf )

Speaking of the House of Pi where Pi are Square ...

Pi are square

d-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)

### Re: Paradise Trinity Day

Posted: **Fri Mar 08, 2019 10:33 am**

by **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

### Re: Paradise Trinity Day

Posted: **Fri Mar 08, 2019 6:28 pm**

by **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

### Re: Paradise Trinity Day

Posted: **Fri Mar 08, 2019 8:38 pm**

by **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 ...")

### Re: Paradise Trinity Day

Posted: **Mon Mar 11, 2019 6:33 pm**

by **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 ...

...

### Re: Paradise Trinity Day

Posted: **Tue Mar 12, 2019 12:48 pm**

by **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

### Re: Paradise Trinity Day

Posted: **Tue Mar 12, 2019 8:55 pm**

by **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 ...

...

### Re: Paradise Trinity Day

Posted: **Wed Mar 13, 2019 5:22 am**

by **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

### Re: Paradise Trinity Day

Posted: **Wed Mar 13, 2019 3:14 pm**

by **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

### Re: Paradise Trinity Day

Posted: **Wed Mar 13, 2019 6:33 pm**

by **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 ...

...

### Re: Paradise Trinity Day

Posted: **Thu Mar 14, 2019 11:33 am**

by **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

### Re: Paradise Trinity Day

Posted: **Thu Mar 14, 2019 6:45 pm**

by **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

### Re: Paradise Trinity Day

Posted: **Fri Mar 15, 2019 1:59 pm**

by **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

### Re: Paradise Trinity Day

Posted: **Fri Mar 15, 2019 2:53 pm**

by **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

### Re: Paradise Trinity Day

Posted: **Fri Mar 15, 2019 3:55 pm**

by **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)

### Re: Paradise Trinity Day

Posted: **Fri Mar 15, 2019 8:01 pm**

by **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

### Re: Paradise Trinity Day

Posted: **Fri Mar 15, 2019 10:44 pm**

by **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

### Re: Paradise Trinity Day

Posted: **Sat Mar 16, 2019 3:45 pm**

by **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

### Re: Paradise Trinity Day

Posted: **Sat Mar 16, 2019 7:27 pm**

by **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

### Re: Paradise Trinity Day

Posted: **Sat Mar 16, 2019 10:55 pm**

by **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 ...

...

### Re: Paradise Trinity Day

Posted: **Sun Mar 17, 2019 5:11 pm**

by **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)

### Re: Paradise Trinity Day

Posted: **Mon Mar 18, 2019 3:22 am**

by **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)

### Re: Paradise Trinity Day

Posted: **Mon Mar 18, 2019 3:39 pm**

by **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

### Re: Paradise Trinity Day

Posted: **Mon Mar 18, 2019 4:33 pm**

by **Amigoo**

Re: Spiraling Squares design*

"Inward pursuit of the Big Bang of Quadrature"

* updated in:

http://aitnaru.org/images/Khristos_Voskrese.pdf
Rod ...

...