Paradise Trinity Day

[Amigoo]

Re: The PoiT design (pronounced "The Poet")
"The Pi of impossible Triangulation"

Juxtaposition of triangles with sqrt(2) ambience.

Rod ... ...

[Amigoo]

Re: The PoiT design* (pronounced "The Poet")
"The Pi of 'impossible' Triangulation"

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

Rod

[Amigoo]

Re: The PoiT design (pronounced "The Poet")
"The Pi of 'impossible' Triangulation"

The Poit forecasts that Quadrature reveals a chrysallis
... but what of universe significance will soon emerge

Rod ... ...

[Amigoo]

Re: The PoiT design (pronounced "The Poet")
"The Pi of 'impossible' Triangulation"

Geometer's secret ...

Small irregular pentagon (golden) has precisely 1/4 area of large irregular pentagon
and each similar side of large pentagon has twice the length of the small's side.

Also, each pentagon has a similar collection of circle-squaring objects:
scalene triangles, right triangles, isosceles trapezoids.

E = MC^2 (Extreme Morbus Cyclometricus Squared)

Rod

[Amigoo]

Re: The PoiT design (pronounced "The Poet")
E = MC^2 (Extreme Morbus Cyclometricus Squared)

"Moons of Pi" embellishment and a familiar tune:
(myth records that these were Martin's original lyrics,
revealing a geometer's first love of Quadrature)

Re: https://www.youtube.com/watch?v=5Bt_uJvlKK8
"When the moon hits your eye like a big piece of Pi ..."

Rod ... ...

[Amigoo]

Re: The PoiTT design (pronounced "The Poet Tee")
"The Pi of 'impossible' Triangulation Trapezoidal"

Focus on enclosing trapezoid and its 4 similar trapezoids inscribed.

Rod

[Amigoo]

Re: The PoiTT design* (pronounced "The Poet Tee")
"The Pi of 'impossible' Triangulation Trapezoidal"

Focus on enclosing trapezoid and its 4 similar trapezoids inscribed.

"Where's the fourth trapezoid?" Go figure!

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

Rod ... ...

[Amigoo]

Re: The PoiT design (pronounced "The Poet")
E = MC^2 (Extreme Morbus Cyclometricus Squared)

Salient numbers ...
(re: circumference, diameter/sqrt(2) progression)

2/sqrt(Pi) = sqrt(Pi)/(Pi/2) = 2(1/sqrt(Pi))
= 1.1283791670955125738961589031215..
^2 = 1.2732395447351626861510701069801..
= (2/sqrt(Pi))^2

2 = 2.0
sqrt(2) = 1.4142135623730950488016887242097..
1 = 1.0
sqrt(2)/2 = 0.70710678118654752440084436210485..

Pi/1 = 3.1415926535897932384626433832795..
Pi/2 = 1.5707963267948966192313216916398..
Pi/4 = 0.78539816339744830961566084581988..
Pi/8 = 0.39269908169872415480783042290994..

Pi/1 / 2 = 1.5707963267948966192313216916398..
Pi/2 / sqrt(2) = 1.1107207345395915617539702475152..
Pi/4 / 1 = 0.78539816339744830961566084581988..
Pi/8 / sqrt(2)/2 = 0.55536036726979578087698512375759..

2 / 1.5707963267948966192313216916398.. = (2/sqrt(Pi))^2
sqrt(2) / 1.1107207345395915617539702475152.. = (2/sqrt(Pi))^2)
1 / 0.78539816339744830961566084581988.. = (2/sqrt(Pi))^2
sqrt(2)/2 / 0.55536036726979578087698512375759.. = (2/sqrt(Pi))^2

Rod

[Amigoo]

Re: The PoiT Ratios design
E = MC^2 (Extreme Morbus Cyclometricus Squared)

Ratios of sides of overlapping right triangles
(circle-squaring right triangle and isosceles rt)

Effective calculation of Ratio 1
2 / 1.7724538509055160272981674833411.. sqrt(Pi)
= 1.1283791670955125738961589031215.. 2/sqrt(Pi)

Effective calculation of Ratio 2
1.4142135623730950488016887242097.. sqrt(2)
x 1.7724538509055160272981674833411.. sqrt(Pi)
= 2.506628274631000502415765284811.. sqrt(2)sqrt(Pi)
/2 = 1.2533141373155002512078826424055.. (sqrt(2)sqrt(Pi))/2

Both ratios multiplied (= hyp/side of isosceles right triangle)

1.1283791670955125738961589031215.. 2/sqrt(Pi)
x 1.2533141373155002512078826424055.. (sqrt(2)sqrt(Pi))/2
= 1.4142135623730950488016887242096.. sqrt(2)

Rod

[Amigoo]

Re: The PoiT Ratios design,
added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf

One more Poet of "impossible" Quadrature.

Rod ... ...

[Amigoo]

Re: The PoiT Star design
(for the collection known as "Quadrature of PoiT")

Line length ratio of long to short golden lines in center of design
= line length ratio of long to short lines in circle-squaring right triangle
= sqrt(Pi) / sqrt(4-Pi) = 1.9130583802711007947403078280205..

Rod

[Amigoo]

Re: AESOP design (new concept)
"All Eyes Syndicate On Pi"

Cartesian Fable or Pythagorean Reality?

Rod

[Amigoo]

Re: AESOP design
"All Eyes Syndicate On Pi"

When Quadrature depicts the geometric scream:
"Move over Vesica! - there's new Pi in the Neighborhood."

See also: https://en.wikipedia.org/wiki/The_Scream
"I sensed an infinite scream passing through nature"

Rod ... ... (screechin' in square circles)

[Amigoo]

Re: AESOP design*
"All Eyes Syndicate On Pi"

Pythagorean Quadrature for D = 2,
where Pi + (4-Pi) = 4 since a^2 + b^2 = c^2

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

Rod

[Amigoo]

Re: Peace of Pi design

Ultimately, in Pythagorean Quadrature,
the Fork is the memorable Peace of Pi.

Rod ... ...

[Amigoo]

Re: Peace of Pi design

Ultimately, in Pythagorean Quadrature,
the Fork is the most memorable Peace of Pi.

Non-geometers believe smaller utensil in this Cartesian Place Setting is salad fork,
but it's usually provided for petite bites of the square Pi or for sharing a taste.

Note: Napkins (isosceles right triangles) adjoin perfectly aligned Forks.

Rod

[Amigoo]

Re: Squared Circles design

Just circles squared ... by a unique scalene triangle.

Rod ... ...

[Amigoo]

Re: Squared Circles design

Is Pi/2 ... precisely ... "transcendental"?
Is sqrt(2) in Quadrature "transcendental"

Rod

[Amigoo]

Re: Squared Circles design*

Who knew
Pi is directly related to sqrt(2):
For a circle with Diameter = 2, Circumference = Pi(D)
therefore C/2 = Pi and Pi/2 relates to D/sqrt(2).

"Say what?"
Both Diameter and Pi divide by 2, precisely ...
and remain mathematically tandem.

"Say what?"
Take a look from outside the box.

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

Rod

[Amigoo]

Re: Squared Circles design

Pi is directly related to sqrt(2

Obviously (within the domain of Cartesian geometry),
sqrt(Pi) and sqrt(2) have similar mathematical properties.
That one should be transcendental and the other not
gives evidence of conceptual ledgerdemain.

Rod

[Amigoo]

Re: Behind the 8 design

Behind the 8 ... may linger a 7

"Go for it!" (they say)

Rod ... ...

[Amigoo]

Re: Behind the 8 design

Behind the 8 ... may linger a 7

Persistently Proportional Pi
[columns: / sqrt(2), / 2, * sqrt(2) ]

2 / Pi = 0.63661977236758134307553505349006..
sqrt(2) / (Pi/2) = 0.90031631615710606955519919100674..
1 / (Pi/4) = 1.2732395447351626861510701069801..
(sqrt(2)/2) / (Pi/8) = 1.8006326323142121391103983820135..
.5 / (Pi/16) = 2.5464790894703253723021402139602..

Rod ^2

[Amigoo]

Re: Behind the 8 design

Behind the 8 ... may linger a 7

Persistently Proportional Pi
[columns: / sqrt(2), / 2, * sqrt(2), * sqrt(2) ]

2 / Pi = 0.63661977236758134307553505349006..
sqrt(2) / (Pi/2) = 0.90031631615710606955519919100674.. sqrt(2)
1 / (Pi/4) = 1.2732395447351626861510701069801.. 2
(sqrt(2)/2) / (Pi/8) = 1.8006326323142121391103983820135.. 2(sqrt(2))
.5 / (Pi/16) = 2.5464790894703253723021402139602.. 4
(sqrt(2)/4) / (Pi/32) = 3.6012652646284242782207967640269.. 4(sqrt(2))
.25 / (Pi/64) = 5.0929581789406507446042804279207.. 8

Geometers' secret about PPP Table
(Diameter = 2, half of Circumference = Pi)

Table gives relationship of D to C/2
as D decreases by sqrt(2), C2 by 2

Rod

[Amigoo]

Re: Behind the 8 design

Behind the 8 ... may linger a 7

... or three.

Rod ... ...

[Amigoo]

Re: Behind the 8 design*
"Behind the 8 ... may linger a 7"

Geometer's secret ...
Only sqrt(2) knows the family secret of the diagonal dark blue line
in the lower right quadrant of this composition ... since all new lines
associated with this line were later removed for design simplification.

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

Rod