11:11 Angels Message Board

Re: MC Thrice Circular design
"(sqrt(Pi) + sqrt(4-Pi)) / d = sqrt(2)"

Combined lengths of short side and long side
of the inscribed circle-squaring right triangle,
when divided by length of diagonal (red line),
= sqrt(2).

"Say what?!"
Sqrt(2) rules the Cartesian Neighborhood

Rod ... ...

Re: MC Thrice Circular design*
Another red diagonal with sqrt(2) ambience!

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

Rod

Re: MC POPT design
(Proof Of Pythagorean Theorem)

This Cartesian compostion features three similar right triangles,
all circle-squaring! Large one contains the same area as the other two
(visual confirmation of the Pythagorean Theorem since the other two
relate by one of these sqrt factors: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 ).

aka "Morbus Cyclometricus POPT"

Rod ... ...

Re: MC POPT design
(Proof Of Pythagorean Theorem)

Geometer's secret about proving the Theorem ...

1. All triangles are enclosed in the evenly divided square.

2. The three similar right triangles prove 1:2 area relationship
via a sqrt factor (2, 3, 4, 5, 6, 7, 8, 9, 10, or 11 ).

3. The three similar isosceles right triangles prove
a^2 + b^2 = c^2 via 1/4 of those squares and via 1:2 separation
of the triangles into opposite sides of the divided square.

4. This makes a geometric proof super easy to write ...
according to a defunded "math professor" at the bar.

Rod

Re: MC POPT design
(Proof Of Pythagorean Theorem)

MC POPT 2X Who knew

"Morbus Cyclometricus" is all about drawing lines
'til you're too pooped to POPT (twice)

Rod

Re: MC POPT design
(Proof Of Pythagorean Theorem)

MC POPT 2X~Q3X
(POPT twice, Quadrature thrice)

"Say what?!"
Ask a geometer of Quadraturial persusasion.

Rod ... ...

Re: MC POPT design
(Proof Of Pythagorean Theorem)

Better color balance!

'69" means "upside down and backward"
... like a parallel universe?

Rod

Re: MC POPT design*

Alas, the "Smile of Pythagoras" returned.

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

Rod

Re: MC POPT design

the "Smile of Pythagoras" returned

Geometer's secret ...
Given large circle's Diameter = 2.0,
areas of related isosceles right triangles:

1.0 (sqrt(2)^2) / 2
= 0.78539816339744830961566084581988.. (sqrt(Pi)^2) / 4
+ 0.21460183660255169038433915418012.. (sqrt(4-Pi)^2) / 4

Rod ... ...

Re: MC POPT design

the "Smile of Pythagoras" returned

Given large circle's Diameter = 2.0,
areas of related circle-squaring right triangles =

sqrt(Pi)(sqrt(4-Pi))/2
= 2((sqrt(Pi)/2)(sqrt(4-Pi))/2))

0.82109168386816193825721376980014..
= 2(0.41054584193408096912860688490007..)

Rod

Re: MC POPT design
(better identification of sqrt(2))

Given large circle's Diameter = 2.0,
Area of large circle-squaring right triangle
= sqrt(Pi)(sqrt(4-Pi)) / 2

1.6421833677363238765144275396003..
/ 2 = 0.82109168386816193825721376980014..

Area of one of two smaller circle-squaring right triangles
= (sqrt(Pi)/sqrt(2) (sqrt(4-Pi)/sqrt(2))) / 2

1.2533141373155002512078826424055.. sqrt(Pi) / sqrt(2)
x 0.65513637756203355309393588562466.. sqrt(4-Pi) / sqrt(2)
= 0.41054584193408096912860688490006..

x 2 = 0.82109168386816193825721376980014..

Rod

Re: MC POPT design
(better identification of sqrt(2))

Circle-squaring right triangle, side c = 2.0, hypotenuse
1.7724538509055160272981674833411.. sqrt(Pi), side a
+ 0.92650275035220848584275966758914.. sqrt(4-Pi), side b
/ sqrt(2) = 1.9084505148775338043018185280302..

2.0 / 1.9084505148775338043018185280302..
= 1.3494783006288622565704635754652..

1.9084505148775338043018185280302..
/ 1.4142135623730950488016887242097.. sqrt(2)
= 1.0479705836796826616103553929819..

1.3494783006288622565704635754652..
x 1.0479705836796826616103553929819..
= 1.4142135623730950488016887242097.. sqrt(2)

Proof that Pi is evenly divisible by sqrt(2)?

Rod

Re: MC POPT design
(better identification of sqrt(2))

Geometer's secret (where D = 2.0) ...

Lengths of sides of circle-squaring right triangle:
2.0, sqrt(Pi), sqrt(4-Pi)

Lengths of sides of circle-squaring scalene triangle ...
sqrt(2), sqrt(Pi), (sqrt(Pi)+sqrt(4-Pi))/sqrt(2)

Rod

Re: MC POPT design*

More thinkin' outside the box.

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

Rod

Re: MC POPT design

better identification of sqrt(2)

Geometer's secret ...

Only astute geometers (of Quadraturial persuasion)
can identify the sqrt(2)-associated quadrilaterals
that show replication of circle-squaring objects.

"What are those quadrilaterals and objects?"
Ask a real geometer ... of Quadraturial persuasion.

Rod ... ...

Re: MC POPT design

More tweaking of the colors to better reveal the objects of Quadrature
... and gives greater meaning to the whimsical "Morbus Cyclometricus".

Rod

Re: https://temporarytemples.co.uk/project/ ... ndley-2020

Very interesting discussion: The Flower Piano, by Karen Alexander

.... and the Pythagorean Comma now suggests:
"I think, therefore I am ... thinking ... I think."

Rod

Re: https://temporarytemples.co.uk/project/ ... ndley-2020

I think, therefore I am ... thinking ... I think.

The Pythagorean Comma (music frequencies not evenly tempered)
hints that the limitation is how sound travels through the air ...
and this hints that the circle can be squared only in theory ...
since lines in Cartesian space have similar temperament.

"Say what?!" Dunno! Might have been a daydream.

Rod ... ...

Re: MC POPT design
aka "Pi in the LOO" (sqrt(2) Linkage Of Objects)

Geometer's secret ...

Lines of similar, inscribed, circle-squaring quadrilaterals maintain sqrt(2) association
... including the respective lines representing a Side of Circle's Square (SoCS).

Rod

Re: MC POPT design*
aka "Pi in the LOO" (sqrt(2) Linkage Of Objects)

Geometer's secret about the spirally levered LOO ...
is too esoteric to discuss in social circles not squared.

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

Rod ... ...

Re: MC Spirally design, developed from MC POPT
aka "Pi in the LOO" (sqrt(2) Linkage Of Objects)

Geometer's secret about the spirally-levered LOO,
too esoteric to explain in social circles not squared:

For the large circle where D =2, SoCS = sqrt(Pi),
for the circle starting next sqrt(2) spiral
D = 0.125, SoCS = sqrt(Pi) / 16

"So, Pi is constrained by sqrt(2)?"
There's always a higher authority.

Rod

Re: MC Spirally design, developed from MC POPT
aka "Pi in the LOO" (sqrt(2) Linkage Of Objects)

Geometry nerds (of Quadraturial persuasion) were quick to note
that sqrt(Pi) is line length and sqrt(Pi)^2 equals area of circle's square.

What's difference in area between largest and smallest square?
3.1415926535897932384626433832795.. Pi
/ 0.01227184630308512983774470071594... (sqrt(Pi)/16)^2
= 256 = (16)^2

Rod

Re: MC Spirally design, developed from MC POPT
aka "Pi in the LOO" (sqrt(2) Linkage Of Objects)

A major clue about Quadrature may be that
relationship of circle's circumference to its diameter
remains constant in the sqrt(2) continuum ...
and Pi/2 ,4 ,16, 256 ... highlight this!

Rod ... ...

Re: Quadraturial Concentricity design*
Our universe is best perceived from "outside the box"

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

Geometer's secret ...
This composition highlights another ratio of Quadrature:
sqrt(Pi) / sqrt(4-Pi) = 1.9130583802711007947403078280205..

sqrt(Pi) = long side of circle-squaring right triangle.
sqrt(4-Pi) = short side; circle's diameter = 2.
D = 2 for each of the four golden circles.

From center of design, area of squares
increases by sqrt(Pi)/sqrt(4-Pi).

Rod ... ...

Re: Quadraturial Concentricity (PI) design

Geometer's secret ...

Who knew The Portal to Infinity is new PI,
perceived from "outside the box", but entered
by Quadraturial concentricity (dimensions*).
Go explore! ... the Cartesian boundaries.

* https://www.youtube.com/watch?v=If3SXJeZzMQ

Rod ... ...