Paradise Trinity Day

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Re: Paradise Trinity Day
Re: "2(A=2)" design,
added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
If the area of each of the two largest circles = 2,
what is length of diagonal blue line in the center?
"Which blue line?"
Yes (since "center" is qualifier)
Tip: Have Pi while you consider sqrt(2).
Rod ... ...
added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
If the area of each of the two largest circles = 2,
what is length of diagonal blue line in the center?
"Which blue line?"
Yes (since "center" is qualifier)
Tip: Have Pi while you consider sqrt(2).
Rod ... ...

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Re: Paradise Trinity Day
Re: "QQuadrangles of 2" design*
(Quadrangles of Quadrature with sqrt(2) nesting)
Michelangelo was hired to paint the ceiling of the QQuadrangles of 2,
but was counseled to wait until the project was proven impossible.
He is still waiting ... apparently.
Geometer's secret ...
The two sets of golden lines have length relationship.
Typically, only geometry nerds of Quadraturial persuasion
discover the relationship, permitting portal passage
and proving that they "think outside the box".
So, you're thinkin' outside the box?!
Where's sqrt(1) in this geometry
A Quiddle of Impossible
When sticks not stones evoke leg bones,
afore some sticks align on plate,
twas sqrt(1), thence sqrt(8).
"Why did the chicken cross the road?"
To take the Sqrt(2) Eggspress through the twilight
lest oneway transit require another dispensation.
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
with adjoining: http://aitnaru.org/images/A_Quiddle.pdf
Rod
(Quadrangles of Quadrature with sqrt(2) nesting)
Michelangelo was hired to paint the ceiling of the QQuadrangles of 2,
but was counseled to wait until the project was proven impossible.
He is still waiting ... apparently.
Geometer's secret ...
The two sets of golden lines have length relationship.
Typically, only geometry nerds of Quadraturial persuasion
discover the relationship, permitting portal passage
and proving that they "think outside the box".
So, you're thinkin' outside the box?!
Where's sqrt(1) in this geometry
A Quiddle of Impossible
When sticks not stones evoke leg bones,
afore some sticks align on plate,
twas sqrt(1), thence sqrt(8).
"Why did the chicken cross the road?"
To take the Sqrt(2) Eggspress through the twilight
lest oneway transit require another dispensation.
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
with adjoining: http://aitnaru.org/images/A_Quiddle.pdf
Rod

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Re: Paradise Trinity Day
Re: "cLife Squared" design
Comprehending Quadrature is like opening your own Book of Life!
The final chapter is amazing! (when you write it)
and it's always being written (if that be your style
... and eternity preference).
Rod ... ...
Comprehending Quadrature is like opening your own Book of Life!
The final chapter is amazing! (when you write it)
and it's always being written (if that be your style
... and eternity preference).
Rod ... ...

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Re: Paradise Trinity Day
Re: "cLife Squared" design,
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Geometry nerds of Quadraturial persuasion can be so exasperating!
You would think that Quadrature was all about geometric precision.
(two small lines were needed for sqrt(2) balance).
Geometer's secret ...
Large scalene triangle is easily divided
into four similar scalene triangles.
Rod
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Geometry nerds of Quadraturial persuasion can be so exasperating!
You would think that Quadrature was all about geometric precision.
(two small lines were needed for sqrt(2) balance).
Geometer's secret ...
Large scalene triangle is easily divided
into four similar scalene triangles.
Rod

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Re: Paradise Trinity Day
Re: "cLife Entangled" design,
updated in: updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Before the Big Bang, prematter was already entangled!
... according to the Book of Eternal Life.
Only in Quadrature exists a dualsided key ...
that unlocks the portal when the time is right.
Quadrature tells the story of eternal partnership
... as symbolized by its dualsided ringed key.
In the Book of Eternal Life,
there are many authors ... apparently.
Why not contribute a chapter?!
Rod ... ...
updated in: updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Before the Big Bang, prematter was already entangled!
... according to the Book of Eternal Life.
Only in Quadrature exists a dualsided key ...
that unlocks the portal when the time is right.
Quadrature tells the story of eternal partnership
... as symbolized by its dualsided ringed key.
In the Book of Eternal Life,
there are many authors ... apparently.
Why not contribute a chapter?!
Rod ... ...

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Re: Paradise Trinity Day
Re: "Fermat's Last Theorem", by Simon Singh, 2011, p.51
"Pi = 4(1/11/3+1/51/7+1/91/11+1/131/15..)"
Apparently,
Pi = (1/11/3+1/51/7+1/91/11+1/131/15..)^2
Since 4/Pi
= 4/3.1415926535897932384626433832795..
= 1.2732395447351626861510701069801..
= (1.1283791670955125738961589031215..)^2
= (2/sqrt(Pi))^2
Restated,
4 = Pi x (1/11/3+1/51/7+1/91/11+1/131/15..)^2
4 = Pi x (2/sqrt(Pi))^2
Who knew
Pi finally resolves with finite decimal digits.
Rod
"Pi = 4(1/11/3+1/51/7+1/91/11+1/131/15..)"
Apparently,
Pi = (1/11/3+1/51/7+1/91/11+1/131/15..)^2
Since 4/Pi
= 4/3.1415926535897932384626433832795..
= 1.2732395447351626861510701069801..
= (1.1283791670955125738961589031215..)^2
= (2/sqrt(Pi))^2
Restated,
4 = Pi x (1/11/3+1/51/7+1/91/11+1/131/15..)^2
4 = Pi x (2/sqrt(Pi))^2
Who knew
Pi finally resolves with finite decimal digits.
Rod

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Re: Paradise Trinity Day
Relationship of Circumference/2 to Diameter (C/2 : D)
3.1415926535897932384626433832795.. Pi
/ 2.0 = Pi/2
2.2214414690791831235079404950303.. (Pi(sqrt(2))/2
/ 1.4142135623730950488016887242097.. = Pi/2
1.5707963267948966192313216916398.. Pi/2
/ 1.0 = Pi/2
D = 1 x sqrt(2) x sqrt(2) = 2
C/2 = Pi/2 x sqrt(2) x sqrt(2) = Pi
Therefore, in all circles ...
Pi and sqrt(2) maintain fixed relationship
Rod
3.1415926535897932384626433832795.. Pi
/ 2.0 = Pi/2
2.2214414690791831235079404950303.. (Pi(sqrt(2))/2
/ 1.4142135623730950488016887242097.. = Pi/2
1.5707963267948966192313216916398.. Pi/2
/ 1.0 = Pi/2
D = 1 x sqrt(2) x sqrt(2) = 2
C/2 = Pi/2 x sqrt(2) x sqrt(2) = Pi
Therefore, in all circles ...
Pi and sqrt(2) maintain fixed relationship
Rod

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Re: Paradise Trinity Day
Re: "Go Figure!" design*
A study of Pi as circumference of circle.
Pop Quiz:
If Pi is the length of every red line,
what are the diameters of the 4 circles
(geometry nerds of Quadraturial persuasion
usually ask next "What's the test ?!")
* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod ... ...
A study of Pi as circumference of circle.
Pop Quiz:
If Pi is the length of every red line,
what are the diameters of the 4 circles
(geometry nerds of Quadraturial persuasion
usually ask next "What's the test ?!")
* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod ... ...

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Re: Paradise Trinity Day
Re: "Go Figure!" design
A study of Pi as circumference of circle.
Esoteric geometry that proves only that Pi remains
directly related to sqrt(2) regardless of circle's diameter.
Of what value is this?! Go figure!
Perhaps, sqrt(2) is transcendental or Pi is not.
Rod
A study of Pi as circumference of circle.
Esoteric geometry that proves only that Pi remains
directly related to sqrt(2) regardless of circle's diameter.
Of what value is this?! Go figure!
Perhaps, sqrt(2) is transcendental or Pi is not.
Rod

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Re: Paradise Trinity Day
Re: "Go Figure!" design
A study of Pi as circumference of circle.
Secret about "Three Moonsa P'eye":
Astrological phenomenon; occurs only when
sqrt(Pi) and sqrt(2) are in conjunction.
Regarding circlesquaring triangle:
2/sqrt(Pi) defines a circle (diameter)
and its Side of Circle's Square (SoCS).
And 2/sqrt(Pi) = sqrt(Pi)/(Pi/2).
"What about the short side?"
sqrt(4Pi) re: Pythagorean Theorem
Some believe Quadrature's "Three Moonsa P'eye"
heralds several nearterm "Disclosure" events,
but others believe they're just relatives.
Go Figure!
Rod ... ...
A study of Pi as circumference of circle.
Secret about "Three Moonsa P'eye":
Astrological phenomenon; occurs only when
sqrt(Pi) and sqrt(2) are in conjunction.
Regarding circlesquaring triangle:
2/sqrt(Pi) defines a circle (diameter)
and its Side of Circle's Square (SoCS).
And 2/sqrt(Pi) = sqrt(Pi)/(Pi/2).
"What about the short side?"
sqrt(4Pi) re: Pythagorean Theorem
Some believe Quadrature's "Three Moonsa P'eye"
heralds several nearterm "Disclosure" events,
but others believe they're just relatives.
Go Figure!
Rod ... ...

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Re: Paradise Trinity Day
Re: "Go Figure!" design
Apparently, "Pi are square!"
Geometer's secrets ...
(Morbus Cyclometricus in deed!)
Geometry nerds of Quadraturial persuasion know that
"Either sqrt(2) is transcendental or Pi is not!" refers to
the tandem geometric spirals of sqrt(Pi) and 2.
Regarding long side and hypotenuse of right triangle
in foundational pairing of sqrt(Pi) and 2, these lengths
increase by a precise factor of 8 in one 360 degree
extension of the two tandem spirals.
2 / sqrt(Pi) = 8(2) / 8(sqrt(Pi))
= 1.1283791670955125738961589031215..
Re: "iiSpiral" design
(further exploration of "Go Figure!")
Transcendence of Pi renews every 360 degrees
Don't believe it? Go Figure!
Rod
Apparently, "Pi are square!"
Geometer's secrets ...
(Morbus Cyclometricus in deed!)
Geometry nerds of Quadraturial persuasion know that
"Either sqrt(2) is transcendental or Pi is not!" refers to
the tandem geometric spirals of sqrt(Pi) and 2.
Regarding long side and hypotenuse of right triangle
in foundational pairing of sqrt(Pi) and 2, these lengths
increase by a precise factor of 8 in one 360 degree
extension of the two tandem spirals.
2 / sqrt(Pi) = 8(2) / 8(sqrt(Pi))
= 1.1283791670955125738961589031215..
Re: "iiSpiral" design
(further exploration of "Go Figure!")
Transcendence of Pi renews every 360 degrees
Don't believe it? Go Figure!
Rod

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Re: Paradise Trinity Day
Re: "iQspiral" design
As a tee, no one needs to know that 'iQ' refers to Quadrature  not intelligence,
but it takes geometric IQ to comprehend how '8' relates to Quadrature
(serve square Pi to introduce the concept).
Geometer's secret iQ ...
A circle's square rests upon only eight points of the circle
And it takes super geometric IQ to comprehend how Pi relates to "Sweet 16"
(smallest circlesquaring right triangle that begins the inner 360 degree spiral
has precisely 1/256 the area of the triangle beginning the outer spiral).
Tasty Sqrt(2) Pi Sandwich (aka "Pi Corral")
for circle where diameter = 2
Perimeters of squares divided ...
8.0 / 5.6568542494923801952067548968388.. enclosing/inscribed
= 1.4142135623730950488016887242097.. sqrt(2)
Circumference vs perimeters (calc up & down) ...
= 8.0 enclosing perimeter
x 1.2732395447351626861510701069801.. (2/sqrt(Pi))^2
6.283185307179586476925286766559.. circumference = Pi(D)
/ 1.1107207345395915617539702475152.. (Pi/2)/sqrt(2)
= 5.6568542494923801952067548968388.. inscribed perimeter
Multiplier x divisor of calculated perimeters ...
1.2732395447351626861510701069801.. (2/sqrt(Pi))^2
x 1.1107207345395915617539702475152.. (Pi/2)/sqrt(2)
= 1.4142135623730950488016887242097.. sqrt(2)
Tasty Sqrt(2) Pi Sandwich, (aka "short stack")
for circle where diameter = 2 (calc up & down):
= 2.0 circle's diameter
x 1.2732395447351626861510701069801.. (2/sqrt(Pi))^2
1.5707963267948966192313216916398.. circumference/4
/ 1.1107207345395915617539702475152.. (Pi/2)/sqrt(2)
= 1.4142135623730950488016887242097.. side of inscribed square
1.2732395447351626861510701069801.. (2/sqrt(Pi))^2
x 1.1107207345395915617539702475152.. (Pi/2)/sqrt(2)
= 1.4142135623730950488016887242097.. sqrt(2)
Thus, Pi is directly associated with sqrt(2).
Go Figure!
Rod
As a tee, no one needs to know that 'iQ' refers to Quadrature  not intelligence,
but it takes geometric IQ to comprehend how '8' relates to Quadrature
(serve square Pi to introduce the concept).
Geometer's secret iQ ...
A circle's square rests upon only eight points of the circle
And it takes super geometric IQ to comprehend how Pi relates to "Sweet 16"
(smallest circlesquaring right triangle that begins the inner 360 degree spiral
has precisely 1/256 the area of the triangle beginning the outer spiral).
Tasty Sqrt(2) Pi Sandwich (aka "Pi Corral")
for circle where diameter = 2
Perimeters of squares divided ...
8.0 / 5.6568542494923801952067548968388.. enclosing/inscribed
= 1.4142135623730950488016887242097.. sqrt(2)
Circumference vs perimeters (calc up & down) ...
= 8.0 enclosing perimeter
x 1.2732395447351626861510701069801.. (2/sqrt(Pi))^2
6.283185307179586476925286766559.. circumference = Pi(D)
/ 1.1107207345395915617539702475152.. (Pi/2)/sqrt(2)
= 5.6568542494923801952067548968388.. inscribed perimeter
Multiplier x divisor of calculated perimeters ...
1.2732395447351626861510701069801.. (2/sqrt(Pi))^2
x 1.1107207345395915617539702475152.. (Pi/2)/sqrt(2)
= 1.4142135623730950488016887242097.. sqrt(2)
Tasty Sqrt(2) Pi Sandwich, (aka "short stack")
for circle where diameter = 2 (calc up & down):
= 2.0 circle's diameter
x 1.2732395447351626861510701069801.. (2/sqrt(Pi))^2
1.5707963267948966192313216916398.. circumference/4
/ 1.1107207345395915617539702475152.. (Pi/2)/sqrt(2)
= 1.4142135623730950488016887242097.. side of inscribed square
1.2732395447351626861510701069801.. (2/sqrt(Pi))^2
x 1.1107207345395915617539702475152.. (Pi/2)/sqrt(2)
= 1.4142135623730950488016887242097.. sqrt(2)
Thus, Pi is directly associated with sqrt(2).
Go Figure!
Rod

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Re: Paradise Trinity Day
Re: "de Triomphe" design*
(that's what they said)
Go Figure!
A bit of enlightenment ...
The mystery dissolves a bit (for geometers of Quadraturial persuasion)
with esoteric presence of the right triangle's "Pi Fork" paired lines
(paired lines have either sqrt(Pi) or 2/sqrt(Pi) length relationship).
Today's Menu (a riddle)
Big Bang, Pi Fork, Oyster, Pearl
Bon Appétit
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod ... ...
(that's what they said)
Go Figure!
A bit of enlightenment ...
The mystery dissolves a bit (for geometers of Quadraturial persuasion)
with esoteric presence of the right triangle's "Pi Fork" paired lines
(paired lines have either sqrt(Pi) or 2/sqrt(Pi) length relationship).
Today's Menu (a riddle)
Big Bang, Pi Fork, Oyster, Pearl
Bon Appétit
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Rod ... ...

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Re: Paradise Trinity Day
Re: "de Triomphe" design
(that's what they said)
With area of smallest circle's square equal to 1/16 area
of largest circle's square (D = 2, A = Pi), there is no doubt ...
that Pi is evenly divisible by a sweet 16!
... and that Quadrature is impregnated with sqrt(2).
Rod
(that's what they said)
With area of smallest circle's square equal to 1/16 area
of largest circle's square (D = 2, A = Pi), there is no doubt ...
that Pi is evenly divisible by a sweet 16!
... and that Quadrature is impregnated with sqrt(2).
Rod

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Re: Paradise Trinity Day
Re: "TOI Box" design*
Within Quadrature's Triangles Of Interest box
one can assemble a collection of triangles to prove
"Once there was A point to Morbus Cyclometricus."
Re: https://notes.misentropy.com/post/18903 ... ngdisease
"The ancient Greeks used the word τετραγωνιζειν (tetragonidzein),
which translates 'to occupy oneself with the quadrature'.
* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Ro ... ...
Within Quadrature's Triangles Of Interest box
one can assemble a collection of triangles to prove
"Once there was A point to Morbus Cyclometricus."
Re: https://notes.misentropy.com/post/18903 ... ngdisease
"The ancient Greeks used the word τετραγωνιζειν (tetragonidzein),
which translates 'to occupy oneself with the quadrature'.
* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Ro ... ...

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Re: Paradise Trinity Day
Re: "TOI Box" design
Geometer's secret ...
The union of sqrt(Pi) and sqrt(2) is symbolized by a certain quadrilateral,
easily visible in the upper circle, with Pi and 2 being infinite eternals.
The quadrilateral? Go Figure!
Rod
Geometer's secret ...
The union of sqrt(Pi) and sqrt(2) is symbolized by a certain quadrilateral,
easily visible in the upper circle, with Pi and 2 being infinite eternals.
The quadrilateral? Go Figure!
Rod

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Re: Paradise Trinity Day
Re: "TOI Box" design*
Say what
There's a similar quadrilateral in the larger circle?
And its dimensions associate by sqrt(2) ?
Who told you ?! That cheeky Pi ?
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Geometer's secret ...
In each quadrilateral, sqrt(2) length relationship exists
between the lines connecting their 90degree angles.
Ro ... ...
Say what
There's a similar quadrilateral in the larger circle?
And its dimensions associate by sqrt(2) ?
Who told you ?! That cheeky Pi ?
* updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Geometer's secret ...
In each quadrilateral, sqrt(2) length relationship exists
between the lines connecting their 90degree angles.
Ro ... ...

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Re: Paradise Trinity Day
Re: "El Soluna" design
(from "el sol" + "la luna")
"Impossible" confluence of the circlesquaring
right and scalene triangles having sqrt(2) association
from their respective sqrt(2)related circles.
Rod
(from "el sol" + "la luna")
"Impossible" confluence of the circlesquaring
right and scalene triangles having sqrt(2) association
from their respective sqrt(2)related circles.
Rod

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Re: Paradise Trinity Day
Re: "GR Squared" design*
(Quadrature's Golden Ratio, squared)
* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Featuring the identifying ratios ...
1.9130583802711007947403078280204.. sqrt(Pi) / sqrt(4Pi)
1.1283791670955125738961589031215.. 2 / sqrt(Pi)
1.2732395447351626861510701069801.. (2 / sqrt(Pi)) ^2
Dimensions of largest inscribed right triangle ...
a = 1.7724538509055160272981674833411.. sqrt(Pi)
b = 0.92650275035220848584275966758914.. sqrt(4Pi)
c = 2.0
Pop Quiz ...
1. What are the line length ratios of the red diagonals?
2. When the Geiger counter triggers the switch that causes the flask to break,
is the testy cat of Quadraturial persuasion real or transcendental? or both?
"What's the bib for ?!"
The next Pi eating contest! (of mathematical persuasion)
A Cartesian cat that has Morbus Cyclometricus
... might need a flee collar!
Ro ... ...
(Quadrature's Golden Ratio, squared)
* added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Featuring the identifying ratios ...
1.9130583802711007947403078280204.. sqrt(Pi) / sqrt(4Pi)
1.1283791670955125738961589031215.. 2 / sqrt(Pi)
1.2732395447351626861510701069801.. (2 / sqrt(Pi)) ^2
Dimensions of largest inscribed right triangle ...
a = 1.7724538509055160272981674833411.. sqrt(Pi)
b = 0.92650275035220848584275966758914.. sqrt(4Pi)
c = 2.0
Pop Quiz ...
1. What are the line length ratios of the red diagonals?
2. When the Geiger counter triggers the switch that causes the flask to break,
is the testy cat of Quadraturial persuasion real or transcendental? or both?
"What's the bib for ?!"
The next Pi eating contest! (of mathematical persuasion)
A Cartesian cat that has Morbus Cyclometricus
... might need a flee collar!
Ro ... ...

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Re: Paradise Trinity Day
Re: "PB & J" design,
added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Long story short about "Pi + (4Pi) = 4" ...
Locals knew that a nearby carpenter gifted baby Pythagoras
with random Blocks of wood for his playroom. Only later
did historians comprehend why Pythagoras favored three,
especially when they had precise Juxtaposition. However,
"PB&J" long remained a red herring since peanut butter
was not invented until 1895 by Dr. Kellogg (after 1882
when Ferdinand savored transcendental Pi).
Geometer's secrets ...
The signature sandwich (hor d'oeuvre) at the annual Squared Circles Soirée
has one slice a right triangle and the other an isosceles right triangle
(similar to the red triangles in the center of this composition).
Know one knows that baby Pythagoras observed from Quadrature
(know to him only as geometric shapes with unique juxtaposition)
that parallel universes would be the same but different.
Rod
added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf
Long story short about "Pi + (4Pi) = 4" ...
Locals knew that a nearby carpenter gifted baby Pythagoras
with random Blocks of wood for his playroom. Only later
did historians comprehend why Pythagoras favored three,
especially when they had precise Juxtaposition. However,
"PB&J" long remained a red herring since peanut butter
was not invented until 1895 by Dr. Kellogg (after 1882
when Ferdinand savored transcendental Pi).
Geometer's secrets ...
The signature sandwich (hor d'oeuvre) at the annual Squared Circles Soirée
has one slice a right triangle and the other an isosceles right triangle
(similar to the red triangles in the center of this composition).
Know one knows that baby Pythagoras observed from Quadrature
(know to him only as geometric shapes with unique juxtaposition)
that parallel universes would be the same but different.
Rod

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Re: Paradise Trinity Day
Re: "PB & J" design,
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
In both universes, parallel, ...
Most people see a box and ask "whY"
but geometry nerds of Quadraturial persuasion
(in both universes, parallel) see a box
and ask "X, Y, and Z"
Ro ... ...
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf
In both universes, parallel, ...
Most people see a box and ask "whY"
but geometry nerds of Quadraturial persuasion
(in both universes, parallel) see a box
and ask "X, Y, and Z"
Ro ... ...