Paradise Trinity Day

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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Tripartite Ratios design
“Once in a blue moon” :farao:

:idea: Given: Squared circles with
Diameters = 2, sqrt(Pi), Pi/2

Means/Extremes Property of
Pi/2 : sqrt(Pi) ~ sqrt(Pi) : 2
2(Pi/2) = sqrt(Pi)^2
Therefore, Pi = Pi :shock:

;) About 2(sqrt(1/Pi))

2 / 1.5707963267948966192313216916398..
= 1.2732395447351626861510701069801..

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

:!: Ratios (calculate backward/forward from <>)

= 2.0
x 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
<> 1.7724538509055160272981674833411.. sqrt(Pi)
/ 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
= 1.5707963267948966192313216916398.. Pi/2

Rod :D
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Golden Rays 'n Pi (new design concept)

Intriguing correlation of Pi/2, sqrt(Pi), and 2
in a composition of 7 circles, all squared.

Rod :D
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Golden Rays 'n Pi design
"Ingredients: Pi/2, sqrt(Pi), 2"

Featuring unique symmetry of squared circles. 8)

Rod ... :bike: ...
(shopping "à la mode" for supercool Rum Raisin)
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Golden Rays 'n Pi design
"Ingredients: Pi/2, sqrt(Pi), 2"

Why highlight the trident? Because it's there! 8)
The highest authority of Heaven? Perhaps ...
but in this Cartesian neighborhood?

Rod :D
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Golden Rays 'n Pi design (updated*)
"Ingredients: Pi/2, sqrt(Pi), 2"

Re: https://www.tillamook.com/recipes/apple ... crust.html :roll
:idea: Soul mates; serve on Leftover Pi Day, Nov. 10 (314th day of 2015).

* http://aitnaru.org/images/Tripartite_Soul.pdf

Rod :D
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Golden Rays 'n Pi design
"Ingredients: Pi/2, sqrt(Pi), 2"

:idea: Apparently, this geometry is meant to highlight the
influence of the Pythagorean Theorem (a^2 + b^2 = c^2)
in these 7 integrated circles* all squared.

* two displayed as arcs; complete squares not shown.

:geek: Supporting correspondence:

1.5707963267948966192313216916398.. Pi/2
/2 = 0.78539816339744830961566084581988.. Pi/4

0.88622692545275801364908374167057.. sqrt(Pi)/2
^2 = 0.78539816339744830961566084581988.. Pi/4

Rod ... :bike: ...
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Golden Rays 'n Pi design
"Ingredients: Pi/2, sqrt(Pi), 2"

:idea: Hmmm ... now a five-pronged spear (instead of trident)!
Allusion to "fire fishing" (late shift) on the Sea of Galilee? ;)

:geek: The geometry proves equilibrium between the two sides of the circle's diameter
(center "ray", D = 2), with sqrt(Pi) representing the left side and Pi/2 the right side.

Rod
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Golden Rays 'n Pi design
"Ingredients: Pi/2, sqrt(Pi), 2"

:geek: More about geometric correspondence
between the two sides of this circle's diameter
(one representing sqrt(Pi), the other Pi/2):

Pi = 3.1415926535897932384626433832795..
Pi/2 = 1.5707963267948966192313216916398..
sqrt(Pi) = 1.7724538509055160272981674833411..

1.7724538509055160272981674833411..
/ 2 = 0.56418958354775628694807945156077..
x 2 = 1.1283791670955125738961589031221..

2 / 1.5707963267948966192313216916398..
= 1.2732395447351626861510701069801..
sqrt(1.2732395447351626861510701069801..)
= 1.1283791670955125738961589031221..

Thus, 1.128.. identifies direct relationship: :roll
sqrt(Pi) is part of the circle-squaring right triangle;
Pi/2 is directly related to this circle-squraing magic.

Rod :D
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Golden Rays 'n Pi design (all the numbers*)
"Ingredients: Pi/2, sqrt(Pi), 2"

* relationship of each side (sqrt(Pi), Pi/2) to diameter
and relationship of the two sides to each other.

:geek: More about geometric correspondence
between the two sides of this circle's diameter
(one representing sqrt(Pi), the other Pi/2):

Pi = 3.1415926535897932384626433832795..
Pi/2 = 1.5707963267948966192313216916398..
sqrt(Pi) = 1.7724538509055160272981674833411..

1.7724538509055160272981674833411..
/ 2 = 0.56418958354775628694807945156077..
x 2 = 1.1283791670955125738961589031221..

2 / 1.5707963267948966192313216916398..
= 1.2732395447351626861510701069801..
sqrt(1.2732395447351626861510701069801..)
= 1.1283791670955125738961589031221..

1.7724538509055160272981674833411..
/ 1.5707963267948966192313216916398..
= 1.1283791670955125738961589031215..

Thus, 1.128.. confirms the direct relationship: :roll
sqrt(Pi) is part of the circle-squaring right triangle;
Pi/2 is directly related to this circle-squaring magic.

And, whatever the number of decimal digits of Pi,
squared-circle geometry must exist! (IMO)

But, how to get there from here (the Greek challenge)
is a different matter. :roll:

Rod ... :bike: ...
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: 3D of GRoP design

"Correlation of diameters Pi/2, 2.0, sqrt(Pi)
displayed as the Golden Rays of Pi."

An addendum to the Golden Rays of Pi geometry. :roll

Rod :D
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: 3D of GR'nP design (geometry of squared circles)
in http://www.aitnaru.org/images/Tripartite_Soul.pdf

"Correlation of diameters Pi/2, 2.0, sqrt(Pi),
in the geometry of Golden Rays 'n Pi."

:cheers: After several weeks of almost daily changes,
the most important geometric lines are now identified!

Note: Magenta circle on the right (D = Pi/2) is a sibling
of magenta circle on the left with the same diameter.
D = sqrt(Pi) of next largest magenta circle.
D = 2 of largest blue circle.

Rod ... :bike: ...
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: 3D of GR'nP design
This design may reveal the prompt-inspired "Texas 'T'": 8)

Two inscribed right triangles, adjoined on their hypotenuses,
with length of hypotenuses = sqrt(Pi) (= diameter of circle)
and length of each of the long sides = Pi/2.

A straight line between the 90-degree angles completes the 'T'.
:idea: Hmmm ... a^2 + b^2 = c^2 comes to mind (twice)
... and the 'T' ('t') has symbolic familiarity. :hithere

Rod :D
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: 3D of GR'nP design (revised)
A challenge for serious geometers ...

Unlock the secret of squared circles by evaluating the scalene triangle
in the geometry of the small blue circle (also squared) at the bottom
of this design ... assuming that such an "impossible" secret exists. ;)

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Re: Paradise Trinity Day

Post by Amigoo »

:geek: About the 3D of GR'nP "challenge for serious geometers" ...
[regarding scalene triangle inscribed in circle of D = 2]

1.7724538509055160272981674833411.. sqrt(Pi) = side of small blue circle's square (area)
/ 1.2533141373155002512078826424061.. sqrt(Pi/2) = side of right triangle inscribed in circle
= 1.4142135623730950488016887242097.. sqrt(2) = side of circle's inscribed square (one side displayed)

Rod :D
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: 3D of GR'nP design (one more revision:
highlights features of squared small blue circle; D = 2)

:geek: About the 3D of GR'nP "challenge for serious geometers" ...
[regarding scalene triangle inscribed in circle of D = 2]

1.7724538509055160272981674833411.. sqrt(Pi) = side of small blue circle's square (area)
/ 1.2533141373155002512078826424061.. sqrt(Pi/2) = side of right triangle inscribed in circle
= 1.4142135623730950488016887242097.. sqrt(2) = side of circle's inscribed square (one side displayed)

;) Hint: SoIS = D(sqrt(2))
If SoIS = 2(sqrt(2)), D = 4

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Re: Paradise Trinity Day

Post by Amigoo »

Hint: SoIS = D(sqrt(2)), If SoIS = 2(sqrt(2)), D = 4

:stars: What happened to the slanted line (/) ?

Better hint: SoIS = D/sqrt(2)
If SoIS = 2(sqrt(2)), D = 4

SoIS = Side of Inscribed Square ;)

Rod :roll:
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Re: Paradise Trinity Day

Post by Amigoo »

:cheers: To complete the story ...

1.7724538509055160272981674833411.. = sqrt(Pi) = side of small blue circle's square (area)
/ 1.2533141373155002512078826424061.. = sqrt(Pi/2) = side of right triangle inscribed in circle
= 1.4142135623730950488016887242097.. = sqrt(2) = side of circle's inscribed square (one side displayed)

:geek: Therefore ...

1.4142135623730950488016887242097.. ratio of sqrt(Pi) to sqrt(Pi/2)
x 1.4142135623730950488016887242097.. sqrt(2)
= 2.0 = diameter of circle squared by sqrt(Pi)

Rod :D
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Re: Paradise Trinity Day

Post by Amigoo »

:geek: Re: http://aitnaru.org/images/Tripartite_Soul.pdf

Untitled design (internal file name is "Untitled") provides another perspective
of lines in a squared circle. Longest, horizontal red line has length = sqrt(Pi).
The two longest, diagonal golden lines have length = sqrt(2).
Diameter of largest circle has length = 2. ;)

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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Untitled design (improved)
Both red "crosses" (one is horizontal) have the same dimensions. ;)

:idea: Hmmm ...
Same 2 strokes as Chinese character for 10 ("ten; complete; perfect").
A "1010" midwayer prompt may suggest time/space synchronization.

Rod :D
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Untitled2 design
(exploration of Untitled geometry)

:geek: For a circle having a diameter of length = 2, hypotenuse of right triangle
defining circle's square exists in continuum of hypotenuses from sqrt(2) to 2.0
and is mathematically positioned between these two values (incr = increment):

hyp = 2.0
incr = 1.1283791670955125738961589031216.. 2(sqrt(1/Pi))
hyp = 1.7724538509055160272981674833411.. sqrt(Pi)
incr = 1.2533141373155002512078826424054.. sqrt(Pi/2)
hyp = 1.4142135623730950488016887242097.. sqrt(2)

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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Untitled2 design (improved again)

Now, a design worthy of "synchronization of time and space" ...
but drawn to highlight the red circle-squaring right triangles
at the top: ratio of hypotenuse and long side = sqrt(Pi). 8)

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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Untitled2 design (an algebraic slice of Pi)
Given: 2(sqrt(1/Pi)) = sqrt(Pi)/(Pi/2) 8)

:geek: (sqrt(Pi)/(Pi/2)) x sqrt(Pi) = Pi/(Pi/2) = 2

1.7724538509055160272981674833404.. sqrt(Pi)
/ 1.5707963267948966192313216916398.. Pi/2
= 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
x 1.7724538509055160272981674833404.. sqrt(Pi)
= 2.0

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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: Untitled2 design

Untitled2 found a name (an acronym): VoSoTaS
... but the meaning of the letters is a secret
... since "A secret shared is an oxymoron." ;)

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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: VoSoTaS design (simplified math)
Calculate backward/forward from > sqrt(Pi).

[ = 1.4142135623730950488016887242097.. ] sqrt(2)
[ / 1.2533141373155002512078826424055.. ] sqrt(Pi/2)
[ > 1.7724538509055160272981674833404.. ] sqrt(Pi)
[ x 1.1283791670955125738961589031215.. ] 2(sqrt(Pi))/Pi
[ = 2.0 ]

"Pi are square" comes to mind. ;)

Rod
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Re: Paradise Trinity Day

Post by Amigoo »

:flower: Re: VoSoTaS design

(2(sqrt(Pi))/Pi)(sqrt(2)) equals length of diameter of circle
that is squared by inscribed square in circle of D = 2 8)
(length of side of inscribed square = sqrt(2).

[ > 1.4142135623730950488016887242097.. ] sqrt(2)
[ x 1.1283791670955125738961589031215.. ] 2(sqrt(Pi))/Pi
[ = 1.595769121605730711759784239736... ] (2(sqrt(Pi))/Pi)(sqrt(2))
[ / 1.2533141373155002512078826424055.. ] sqrt(Pi/2)
[ > 2.0 ]

Rod :D
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