Paradise Trinity Day

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Re: Paradise Trinity Day
Re: MC Thrice Circular design
"(sqrt(Pi) + sqrt(4Pi)) / d = sqrt(2)"
Combined lengths of short side and long side
of the inscribed circlesquaring right triangle,
when divided by length of diagonal (red line),
= sqrt(2).
"Say what?!"
Sqrt(2) rules the Cartesian Neighborhood
Rod ... ...
"(sqrt(Pi) + sqrt(4Pi)) / d = sqrt(2)"
Combined lengths of short side and long side
of the inscribed circlesquaring right triangle,
when divided by length of diagonal (red line),
= sqrt(2).
"Say what?!"
Sqrt(2) rules the Cartesian Neighborhood
Rod ... ...

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Re: Paradise Trinity Day
Re: MC Thrice Circular design*
Another red diagonal with sqrt(2) ambience!
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod
Another red diagonal with sqrt(2) ambience!
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod

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Re: Paradise Trinity Day
Re: MC POPT design
(Proof Of Pythagorean Theorem)
This Cartesian compostion features three similar right triangles,
all circlesquaring! 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 ... ...
(Proof Of Pythagorean Theorem)
This Cartesian compostion features three similar right triangles,
all circlesquaring! 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 ... ...

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Re: Paradise Trinity Day
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
(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

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Re: Paradise Trinity Day
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
(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

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Re: Paradise Trinity Day
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 ... ...
(Proof Of Pythagorean Theorem)
MC POPT 2X~Q3X
(POPT twice, Quadrature thrice)
"Say what?!"
Ask a geometer of Quadraturial persusasion.
Rod ... ...

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Re: Paradise Trinity Day
Re: MC POPT design
(Proof Of Pythagorean Theorem)
Better color balance!
'69" means "upside down and backward"
... like a parallel universe?
Rod
(Proof Of Pythagorean Theorem)
Better color balance!
'69" means "upside down and backward"
... like a parallel universe?
Rod

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Re: Paradise Trinity Day
Re: MC POPT design*
Alas, the "Smile of Pythagoras" returned.
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod
Alas, the "Smile of Pythagoras" returned.
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod

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Re: Paradise Trinity Day
Re: MC POPT design
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(4Pi)^2) / 4
Rod ... ...
Geometer's secret ...the "Smile of Pythagoras" returned
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(4Pi)^2) / 4
Rod ... ...

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Re: Paradise Trinity Day
Re: MC POPT design
areas of related circlesquaring right triangles =
sqrt(Pi)(sqrt(4Pi))/2
= 2((sqrt(Pi)/2)(sqrt(4Pi))/2))
0.82109168386816193825721376980014..
= 2(0.41054584193408096912860688490007..)
Rod
Given large circle's Diameter = 2.0,the "Smile of Pythagoras" returned
areas of related circlesquaring right triangles =
sqrt(Pi)(sqrt(4Pi))/2
= 2((sqrt(Pi)/2)(sqrt(4Pi))/2))
0.82109168386816193825721376980014..
= 2(0.41054584193408096912860688490007..)
Rod

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Re: Paradise Trinity Day
Re: MC POPT design
(better identification of sqrt(2))
Given large circle's Diameter = 2.0,
Area of large circlesquaring right triangle
= sqrt(Pi)(sqrt(4Pi)) / 2
1.6421833677363238765144275396003..
/ 2 = 0.82109168386816193825721376980014..
Area of one of two smaller circlesquaring right triangles
= (sqrt(Pi)/sqrt(2) (sqrt(4Pi)/sqrt(2))) / 2
1.2533141373155002512078826424055.. sqrt(Pi) / sqrt(2)
x 0.65513637756203355309393588562466.. sqrt(4Pi) / sqrt(2)
= 0.41054584193408096912860688490006..
x 2 = 0.82109168386816193825721376980014..
Rod
(better identification of sqrt(2))
Given large circle's Diameter = 2.0,
Area of large circlesquaring right triangle
= sqrt(Pi)(sqrt(4Pi)) / 2
1.6421833677363238765144275396003..
/ 2 = 0.82109168386816193825721376980014..
Area of one of two smaller circlesquaring right triangles
= (sqrt(Pi)/sqrt(2) (sqrt(4Pi)/sqrt(2))) / 2
1.2533141373155002512078826424055.. sqrt(Pi) / sqrt(2)
x 0.65513637756203355309393588562466.. sqrt(4Pi) / sqrt(2)
= 0.41054584193408096912860688490006..
x 2 = 0.82109168386816193825721376980014..
Rod

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Re: Paradise Trinity Day
Re: MC POPT design
(better identification of sqrt(2))
Circlesquaring right triangle, side c = 2.0, hypotenuse
1.7724538509055160272981674833411.. sqrt(Pi), side a
+ 0.92650275035220848584275966758914.. sqrt(4Pi), 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
(better identification of sqrt(2))
Circlesquaring right triangle, side c = 2.0, hypotenuse
1.7724538509055160272981674833411.. sqrt(Pi), side a
+ 0.92650275035220848584275966758914.. sqrt(4Pi), 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

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Re: Paradise Trinity Day
Re: MC POPT design
(better identification of sqrt(2))
Geometer's secret (where D = 2.0) ...
Lengths of sides of circlesquaring right triangle:
2.0, sqrt(Pi), sqrt(4Pi)
Lengths of sides of circlesquaring scalene triangle ...
sqrt(2), sqrt(Pi), (sqrt(Pi)+sqrt(4Pi))/sqrt(2)
Rod
(better identification of sqrt(2))
Geometer's secret (where D = 2.0) ...
Lengths of sides of circlesquaring right triangle:
2.0, sqrt(Pi), sqrt(4Pi)
Lengths of sides of circlesquaring scalene triangle ...
sqrt(2), sqrt(Pi), (sqrt(Pi)+sqrt(4Pi))/sqrt(2)
Rod

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Re: Paradise Trinity Day
Re: MC POPT design*
More thinkin' outside the box.
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod
More thinkin' outside the box.
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod

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Re: Paradise Trinity Day
Re: MC POPT design
Only astute geometers (of Quadraturial persuasion)
can identify the sqrt(2)associated quadrilaterals
that show replication of circlesquaring objects.
"What are those quadrilaterals and objects?"
Ask a real geometer ... of Quadraturial persuasion.
Rod ... ...
Geometer's secret ...better identification of sqrt(2)
Only astute geometers (of Quadraturial persuasion)
can identify the sqrt(2)associated quadrilaterals
that show replication of circlesquaring objects.
"What are those quadrilaterals and objects?"
Ask a real geometer ... of Quadraturial persuasion.
Rod ... ...

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Re: Paradise Trinity Day
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
More tweaking of the colors to better reveal the objects of Quadrature
... and gives greater meaning to the whimsical "Morbus Cyclometricus".
Rod

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Re: Paradise Trinity Day
Re: https://temporarytemples.co.uk/project/ ... ndley2020
Very interesting discussion: The Flower Piano, by Karen Alexander
.... and the Pythagorean Comma now suggests:
"I think, therefore I am ... thinking ... I think."
Rod
Very interesting discussion: The Flower Piano, by Karen Alexander
.... and the Pythagorean Comma now suggests:
"I think, therefore I am ... thinking ... I think."
Rod

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Re: Paradise Trinity Day
Re: https://temporarytemples.co.uk/project/ ... ndley2020
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 ... ...
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 ... ...

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Re: Paradise Trinity Day
Re: MC POPT design
aka "Pi in the LOO" (sqrt(2) Linkage Of Objects)
Geometer's secret ...
Lines of similar, inscribed, circlesquaring quadrilaterals maintain sqrt(2) association
... including the respective lines representing a Side of Circle's Square (SoCS).
Rod
aka "Pi in the LOO" (sqrt(2) Linkage Of Objects)
Geometer's secret ...
Lines of similar, inscribed, circlesquaring quadrilaterals maintain sqrt(2) association
... including the respective lines representing a Side of Circle's Square (SoCS).
Rod

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Re: Paradise Trinity Day
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 ... ...
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 ... ...

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Re: Paradise Trinity Day
Re: MC Spirally design, developed from MC POPT
aka "Pi in the LOO" (sqrt(2) Linkage Of Objects)
Geometer's secret about the spirallylevered 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
aka "Pi in the LOO" (sqrt(2) Linkage Of Objects)
Geometer's secret about the spirallylevered 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

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Re: Paradise Trinity Day
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
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

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Re: Paradise Trinity Day
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 ... ...
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 ... ...

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Re: Paradise Trinity Day
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(4Pi) = 1.9130583802711007947403078280205..
sqrt(Pi) = long side of circlesquaring right triangle.
sqrt(4Pi) = 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(4Pi).
Rod ... ...
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(4Pi) = 1.9130583802711007947403078280205..
sqrt(Pi) = long side of circlesquaring right triangle.
sqrt(4Pi) = 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(4Pi).
Rod ... ...

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Re: Paradise Trinity Day
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 ... ...
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 ... ...