Paradise Trinity Day

[Amigoo]

Re: "PB & J" design

Oops! Even more lines ...
but highlighting Quadraturial patterns
in this sqrt(2)-hosted "Pi Corral".

Ro ... ...

[Amigoo]

Re: "Lock 'n Mach" design

An "impossible" Quadraturial riddle,
sans material notes but neuronal prelude
not touted in Lock 'n Mach, Texas.

Question clue about the Q ...
"What must we do to change the '2',
diameter untouchable in this stew
(of geometric juxtaposition).

Rod

[Amigoo]

Re: "Lock 'n Mach" design,
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf

(renamed from "Mock" to retain similar sound
but also allude to the metaphor about speed:
faster re-evaluation of "transcendental" Pi)

Better highlighting of the geometry-controlling,
Pythagorean a^2 + b^2 = c^2 in this Cartesian courtyard.

Pi / (4-Pi) = 3.6597923663254876944787072692571..
= (1.9130583802711007947403078280205..)^2
(area ratio that must exist between a^2 and b^2
when a Pythagorean triangle squares the circle).

"Say what " Dunno! It's all Greek to me!
And who knows why? on some Cartesian courtyards
of Quadraturial persuasion, '2' is a "divinity" ratio.

Ultimately, the Pythagorean Theorem forecasts that
there is indeed finity to Pi's transcendental decimal digits!
Who knew

Ro ... ...

[Amigoo]

Re: "L 'n M Simplified" design,
added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf

Allusion to no guessing required for "What's where?!"
in this Quadraturial composition with Pythagorean perspective.

Rod

[Amigoo]

Re: "impossibly Pythagorean" design

Generous simplification of a^2 + b^2 = c^2,
precisely split across sqrt(2)-hosted composition
of Quadraturial ambience, featuring 2/sqrt(Pi)
= 1.1283791670955125738961589031215..

Rod

[Amigoo]

Re: "impossibly Pythagorean" design*
OMG! Pi are square! Just ask sqrt(2).

Geometer's secret about the Pythagorean Wave:

This Cartesian composition clearly shows via sqrt(2) that the areas
of the Pythagorean right and isosceles right triangles are perfectly balanced
on both sides of this geometric equation. And geometry nerds easily notice
that the Pythagorean squares are divided by 4 for the central square
surrounding the enclosed circle also squared.

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

Ro ... ...

[Amigoo]

Re: "impossibly Pythagorean" design
OMG! Pi are square! Just ask sqrt(2).

Long whispered in private conversations of geometry nerds of Quadraturial persuasion,
Cartesian PMS (Pythagorean Morbus Symmetry) apparently does indeed exist ... albeit
this was long ago proven impossible to construct "inside the box".

("morbus" from "morbus cyclometricus")

Rod

[Amigoo]

Re: https://www.theguardian.com/science/202 ... ulating-pi
"Swiss researchers have spent 108 days calculating Pi to a new record accuracy of 62.8tn digits."

"Mathematicians have estimated that an approximation of Pi to 39 digits is sufficient for most cosmological calculations –
accurate enough to calculate the circumference of the observable universe to within the diameter of a single hydrogen atom."

This seems a clue that Pi is a "twilight zone" constant,
since by design, apparently, can never be a final calculation!

A circle having diameter = 2.0 has area square = Pi.

Square root of Pi must be between sqrt(2) and 2.0,
creating the dividing angles 17.403.. and 27.597..

Therefore, these angles can also be represented
as having 62.8tz digits! Who knew

Rod

[Amigoo]

Re: "Equilibrium Squares iQ" design*
(iQ = impossibly Quadraturial)

Not how to get there from here, but what there looks like.
Pack well! You will not return here once you get there (they say).

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

Rod

[Amigoo]

Re: "Equilibrium Squares iQ" design
(iQ = impossibly Quadraturial)

OMG! Now with sqrt(2) "baby bump"
... geometrically speaking.

However, a sonogram did not identify a "he" or "she",
but "they" seems accurate for these sqrt(2) "rabbits".

Rod ... ...

[Amigoo]

Re: "Equilibrium Squares iQ" design*
"Lines and triangles and squares, oh my!"

Geometer's secrets, rarely revealed ...
Geometry nerds of Quadraturial persuasion know that
tandem sqrt(Pi) + sqrt(2) spirals are the foundation of iQ.

Every two line segments, parallel and 180 degrees apart on both spirals,
maintain 4:1 (or 1:4) length relationship regardless of Pi's decimal digits.
Apparently, either sqrt(2) is transcendental or Pi is not.

Small circle-squaring right triangle, 180 degrees opposite
large circle-squaring right triangle, has precisely 1/16
the area of the large triangle.

√2 controls the spectrum of Diameter:Circumference/2
(obvious to geometry nerds of Quadraturial persuasion):
4:2π, 2√2:√2(π), 2:π, √2:π/√2, 1:π/2, √2/2:π/(2√2), .5:π/4
That 4/.5 = 8 and 2π/(π/4) = 8 is a Big Point

A related natural sequence: 4,5,3,7,-1,15, ?

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

Rod ... ...

[Amigoo]

Sweet 16 of Quadrature where E = S,iQ
(Equilibrium = Squares, impossibly Quadraturial)

This particular right triangle incorporates sqrt(2) via triangle's hypotenuse (circle's diameter)
and sqrt(Pi) via triangle's long side (Side of Circle's Square (SoCS)). As similar right triangles,
with area of small triangle precisely 1/16 the area of the large triangle, sqrt(2) and sqrt(Pi)
integrate to prove geometrically that ... "Either sqrt(2) is transcendental or Pi is not."

Now with "Pi Fork" where paired lines in right triangle
have either sqrt(Pi) or 2/sqrt(Pi) length relationship.

Bon Appétit! (Quadraturially speaking)

Rod

[Amigoo]

Re: "Equilibrium Squares iQ" design*
"Lines and triangles and squares, oh my!"

Geometer's secret ...

"Pi Fork" contains 2 circle-squaring right triangles
where sqrt(2) is the hypotenuse*1 in one triangle
and Side of Circle's Square (SoCS)*2 in the other,
giving area squares of Pi/2 and 2.

*1 circle's diameter, *2 triangle's long side
where diameter/long side = 2/sqrt(Pi)
= 1.1283791670955125738961589..

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

Rod

[Amigoo]

Re: "7 to 7 to 7" design

A riddle ...

The House of Pi is open
weekdays from 7 to 7 to 7,
weekends from 11 to 11 to 11,
holidays from 1 to 1 to 1,
5 to 5 to 5 when calculated
the last decimal digit of Pi.

Pop quiz ...

Inner circles have area squares equal to Pi or 4.
Which square associates with which time?
(answer will be available 5 to 5 to 5)

Rod ... ...

[Amigoo]

Re: "7 to 7 to 7" design*
“Two Pi in a Pod”

Most people comprehend that the House of Pi is open all hours,
but geometers of Quadraturial persuasion know to finesse the hours
for a transcendental shopping experience.

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

Suddenly, there's another Pi in the Cartesian sandbox!
I don't know when this happened ... '7 to 7' or '7 to 7'?
Might be expected of two Pi of the opposite sex.

Then sqrt(2) does rope tricks in this Pi Corral.

Rod

[Amigoo]

Re: "Obviously Pythagorean!" design

Who knew
In the days when a^2 + b^2 = c^2
Quadrature is Obviously Pythagorean!

Geometers of Quadraturial persuasion insist that the Pythagogon
is required in a geometric proof (8 equal sides, four 55.194.. angles,
four 214.806.. angles, 1080 total degrees).

Who knows?!
It's all Greek to me!

Rod ... ...

[Amigoo]

Re: "Obviously Pythagorean!" design

Simplified with Pi Fork to authenticate Pythagorean Pi,
where line length ratios = 2, 2/sqrt(Pi), and sqrt(Pi)
where the areas of squares = 4, 4-Pi, and Pi

Before dawn, Pythagoras projected a "Bat-Signal" in the sky over North Texas,
then texted "Can you hear me now?" (This B-S still looks Greek to me!)

Something about "Lines and triangles and squares, oh my!"

Rod

[Amigoo]

Re: "Obviously Pythagogon!" design,
added to: http://aitnaru.org/images/Yin_Yang_Pi.pdf

Quadraturial simplicity with Pythagorean Pi.

Rod ... ...

[Amigoo]

Re: "Obviously Pythagogon!" design

Geometers of Quadraturial persuasion clamor to know
the five line lengths of the pentagons in the Pythagogon:

Four lines have length = 1.0, the fifth has length
= 1.3102727551240671061878717712493..

1.2533141373155002512078826424055.. (sqrt(Pi)(sqrt(2))/2)
/ 1.9130583802711007947403078280205.. (sqrt(Pi)/sqrt(4-Pi))
= 0.65513637756203355309393588562465..
x 2 = 1.3102727551240671061878717712493..

The five angles (clockwise from "top")
= 145.194.., 90, 107.403.., 107.403.., 90
= 540 degrees.


Geometers of Quadraturial persuasion clamor to know
the two angle<>angle line lengths of the small parallelograms
(with four sides having similar lengths = 1.0):

short angle<>angle length =
1.2533141373155002512078826424055..
- 0.65513637756203355309393588562465..
= 0.59817775975346669811394675678086..

long angle<>angle length =
1.2533141373155002512078826424055..
+ 0.65513637756203355309393588562465..
= 1.9084505148775338043018185280301..

The four angles (clockwise from "top")
= 34.806.., 145.194.., 34.806.., 145.194..
= 360 degrees.

Of course, measurements are rounded.


Rod

[Amigoo]

Re: "Obviously Pythagogon!" design,
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf

Simplification of the "pro-teen spyke" ...
that causes young minds of Quadraturial persuasion
to be more vulnerable to Morbus Cyclometricus!

Rod

[Amigoo]

Re: "MyOPIC" design, "goal to go",
updated in: http://aitnaru.org/images/Yin_Yang_Pi.pdf

... with visual proof that Pi is evenly divisible by sqrt(2)
(note similar quadrilaterals with two 90 degree angles).

"Twas sticks 'n stones thence Quadrature tome
midst proof transcendental, math pheromone."

Rod ... ...

[Amigoo]

Re: "MyOPIC" design, "goal to go"
Even better sqrt(2) color identification!

Now a familiar tune comes to mind:
"If Pi were a carpenter ..."

Rod

[Amigoo]

Re: "MyOPIC" design*
A Cartesian cacophony, coalesced!

Who knew?! Nine out of ten geometers
of Quadraturial persuasion prefer a rounded bottom!
(but might require a standup desk)

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

Rod ... ...

[Amigoo]

Re: "PT 'n Q" design
(a bit esoteric even for geometers of Quadraturial persuasion)

Proof that the Pythagorean Theorem is true even in Quadrature!
(note sqrt(2) effect on the circle-squaring right triangle)

Clue: The Pythagorean Theorem of this geometry
is focused on 1/4 of the triangle's three squares.

Rod

[Amigoo]

Re: "PT 'n Q" design,
in: http://aitnaru.org/images/Yin_Yang_Pi.pdf

Color coded to complement proof that Pythagorean Theorem
is true even in "impossible" Quadrature ... gracias a sqrt(2).
Also, two circle-squaring trapezoids where large one
has dimensions twice the size of the small one.

Long story short (re: 1/4 of PT's 3 squares,
where a^2 + b^2 = c^2 )

PT = Pythagorean Theorem
C-SRT = Circle-Squaring Right Triangle
IRT = Isosceles Right Triangle
2.0 = hypotenuse of C-SRT

1.7724538509055160272981674833411.. sqrt(Pi), long side of C-SRT
x 0.92650275035220848584275966758914.. sqrt(4-Pi), short side of C-SRT
= 1.6421833677363238765144275396003..
/ 2 = 0.82109168386816193825721376980014.. area of C-SRT

0.92650275035220848584275966758914.. sqrt(4-Pi), short side of C-SRT
/ 1.4142135623730950488016887242097.. sqrt(2)
= 0.65513637756203355309393588562466..

1.7724538509055160272981674833411.. sqrt(Pi), long side of C-SRT
/ 1.4142135623730950488016887242097.. sqrt(2)
= 1.2533141373155002512078826424055..
x 0.65513637756203355309393588562466..
= 0.82109168386816193825721376980014.. area of 2 similar RTs

Therefore, area of large IRT on right
equals area of 2 smaller IRTs on left,
proving a^2 + b^2 = c^2

0.78539816339744830961566084581988.. (sqrt(Pi)^2)/4, area of large IRT on left
+ 0.21460183660255169038433915418012.. (sqrt(4-Pi)^2)/4, area of small IRT on left
= 1.0 = (2^2)/4, area of IRT on right

Rod