Paradise Trinity Day

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Re: Paradise Trinity Day

Post by Amigoo » Thu Aug 03, 2017 12:48 am

:flower: Re: Golden rPi design

Re: The Golden Ratio by Mario Livio, 2002, p. 85
"The Golden Rectangle is the only rectangle with the property
that cutting a square from it produces a similar rectangle."

:bana: Impressive, however ...
The rPi Golden Rectangle is the only rectangle with the property
that its diagonal has length equal to the diameter of a circle and
sides a+b have length equal to a side of that circle's square.

Given that the diagonal is side d, d / (a + b) = 2(sqrt(1/Pi)),
consistent ratio of these sides in all related right triangles. 8)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Thu Aug 03, 2017 1:23 pm

:flower: Re: Golden rPi design

I finally comprehend :duh that this geometry is revealing a new "golden ratio(s)"
with another Cartesian dimension beyond the line-segmented Golden Ratio: :shock:
relationships of the lines* in a circle-squaring right triangle ... lines that
define a new spiral for consumption in believing math circles. ;)

* one line is segmented (to allude to the old Golden Ratio?) 8)

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sat Aug 05, 2017 1:55 am

:flower: Re: Golden rPi design

Who knew :?: :!: That a new Phi ratio would exist for the rPi rectangle!
(repeating ratio observable in similar objects/lines of Golden rPi)

Current Phi = 1.618033988749894848204586834 ..

:hithere New iPhi = 1.913067540823067208163119533.. (approx)
'i' alludes to "impossible" squaring of the circle. :roll:

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Sat Aug 05, 2017 3:55 am

:flower: Re: Golden rPi design

:hithere New iPhi = 1.913058380271100794740307828.. (approx)

Revised calculation starting with hypotenuse of right triangle
= 2,000,000,000 units and downsizing via 2(sqrt(1/Pi))
and the Pythagorean Theorem to obtain line lengths
of the golden cross in the largest rectangle.

Say what :?: :!:
Just ask a squared circles geometer. ;)

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Sat Aug 05, 2017 4:44 am

:flower: Re: Golden rPi design (the wiggly numbers) :roll:
New iPhi = 1.913058380271100794740307828.. (approx)

:geek: downsize right triangles via 2(sqrt(1/Pi)):

:arrow: Find length of shortest side of largest right triangle
2,000,000,000.0 = largest hypotenuse (length of longer golden line)
/ 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
= 1772453850.9055160272981674833411.. sqrt(Pi) x 1,000,000,000

Apply Pythagorean Theorem
2,000,000,000.0^2 - 1772453850.9055160272981674833411..^2
= 858407346410206761.53735661672066..
sqrt( ) = 926502750.35220848584275966758923.. = shortest side of large right triangle

:arrow: Find length of shorter golden line (another hypotenuse)
926502750.35220848584275966758923.. = shortest side of large right triangle
x 2(sqrt(1/Pi)) = 1045446401.7541266302735942239055.. length of smaller golden line

:arrow: Calculate new iPhi ratio (approx)
2,000,000,000.0 / 1045446401.7541266302735942239055..
= 1.9130583802711007947403078280203.. iPhi (approx) :hithere

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Sat Aug 05, 2017 9:11 pm

:flower: Re: Golden rPi design

HCIT! After calculating the iPhi value, use it to confirm
length of shortest side from the Pythagorean Theorem:

Apply Pythagorean Theorem :roll
2,000,000,000.0^2 - 1772453850.9055160272981674833411..^2
= 858407346410206761.53735661672066..
sqrt( ) = 926502750.35220848584275966758923.. = shortest side of large right triangle

Confirm PT calculation using iPhi 8)
1772453850.9055160272981674833411.. length of long side
/ 1.9130583802711007947403078280203.. iPhi constant
= 926502750.35220848584275966758919.. = shortest side of large right triangle

The difference between ..8923.. and ..8919.. ?
4 (early rounding of squared circle corners) :roll:

Rod ... :bike: ... (bumping curb on quick right turns)

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Re: Paradise Trinity Day

Post by Amigoo » Tue Aug 08, 2017 8:37 am

:flower: Re: Golden rPi design

Signature iPhi added to design (lower left, green lines) showing
two circle-squaring right triangles adjoined at their 90 degree angle
with all line pair lengths having iPhi ratio, including hypotenuse
(blue lines) of the two triangles. 8)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Tue Aug 08, 2017 5:55 pm

:flower: Re: Golden rPi design ("final, final")

Golden rPi “Alpha to Omega” (Aom)
"Get a clue and devour the concept" ;)

:geek: New millennium “golden rectangle” featuring
the circle-squaring ratio 2(sqrt(1/Pi)) = d/(a+b)
and supporting ratio iPhi = c/a = (a+b)/c
= 1.91305838027110079474030782802..

About the "critter" in lower left of design ...
"The bulldog who ate Pi" (almost daily for years) :roll:

:cheers: Current collection:
http://aitnaru.org/images/Squarely_Entwined.pdf

Rod ... :bike: ... (off to get a looking glass)

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Re: Paradise Trinity Day

Post by Amigoo » Tue Aug 08, 2017 10:37 pm

:flower: Re: Golden rPi design ("final, final")
"The bulldog who ate Pi"
Trimmed its tail a bit to not look like a cat's. :roll:

About the blue trapezoid in the lower right of design ...
Probably, a doghouse from which to escape occasional
brimstone attacks on the concept of squared circles.

Rod :rambo:

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Re: Paradise Trinity Day

Post by Amigoo » Wed Aug 09, 2017 3:22 pm

:flower: Re: Golden rPi design

A quick study of the iPhi concept:
http://aitnaru.org/images/Golden_rPi.pdf

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Thu Aug 10, 2017 3:53 am

:flower: Re: Golden rPi design
(quick study of the iPhi concept)

Another page added to the PDF showing the mathematical correlation
of the circle-squaring triangle's line lengths (hypotenuse-long_side vs
hypotenuse-short_side). Who knew :?: :!: ... that 2(sqrt(1/Pi))
would give the most convincing evidence of this correlation!

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Thu Aug 10, 2017 1:33 pm

:flower: Re: Golden rPi design
(quick study of the iPhi concept)

Golden line 'e' is now identified. Despite all this Cartesian busyness,
the circle is not squared until a geometric proof is created.

The chess pieces are positioned - let the game begin! :roll

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Fri Aug 11, 2017 10:22 am

:flower: Re: Golden rPi design
(quick study of the iPhi concept)

Line 'f' is now identified in Golden rPi Simplified design
(to illuminate the golden diagonals object d/e):
a = f = (a+b)/(iPhi^2). Say what :?: :!: iPhi^2 :!:

:geek: iPhi = c/a = (a+b)/c = d/e
= 1.9130583802711007947403078280203..
^2 = 3.6597923663254876944787072692565..

Rod :stars:

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Re: Paradise Trinity Day

Post by Amigoo » Fri Aug 11, 2017 9:27 pm

:flower: Re: GPSR Calculator design
(Golden Pi Square Root Calculator from Golden Pi rectangle geometry)
"Get a clue and devour the concept"

The new clue? The d/e object giving (a+b)/(iPhi^2) 8)
:idea: iPhi supports a geometric square root calculator!
(that "calculates" both square and square root) :shock:

:scratch: While GPSR geometry shows that such a calculator could exist,
its operation is a mystery (easier than creating a geometric proof
of a squared circle?) Let this next game begin! :roll

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sat Aug 12, 2017 1:17 am

:flower: Re: GPSR Calculator design
its operation is a mystery

Research indicates that its existence is also questionable;
the pattern locks on iPhi^2 and not just any square/root.
"Say what :?: :!: " A flash in the pan. ;)

Rod :duh

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Re: Paradise Trinity Day

Post by Amigoo » Sat Aug 12, 2017 6:46 am

:flower: Re: GPSR Calculator design
"Purpose and operation unknown!
"Begs the question: 'What's the point?'"

Indeed :!:

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sat Aug 12, 2017 2:46 pm

:flower: Re: GPSR Calculator design II
Purpose and operation unknown!

Begs the question: "What's the point?"
... but answers: "What's the point?"

Rod :D OIC

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Re: Paradise Trinity Day

Post by Amigoo » Sun Aug 13, 2017 6:06 am

:flower: Re: Golden rPi design
the circle is not squared until a geometric proof is created
:bana: Actually, this is a two-part challenge:
1. Prove that squared circles exist (geometrically)*
2. Square the circle by the Greek rules.

Methinks that Part 1 is easier than Part 2 ...
and should have the greater priority. ;)

* Not to worry that, mathematically, this is "impossible" :roll:
(making Part 2 a futile journey ... also).

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Sun Aug 13, 2017 6:44 am

:flower: Re: Golden rPi design

Actually, this is a two-part challenge:
1. Prove that squared circles exist (geometrically)
2. Square the circle by the Greek rules.

Constant 2(sqrt(1/Pi))* suggests that sqrt(Pi) can be
represented as the long side of a circle-squaring
right triangle whose hypotenuse = 2.0 (diameter).

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

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Sun Aug 13, 2017 3:37 pm

:flower: Re: The Right Triangle design

Long Story Short :!: (aka, "Period. End of Story"):

If this right triangle does not exist in a squared circle,
the circle is not squared :shock: according to this geometry.

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Mon Aug 14, 2017 2:22 pm

:flower: Re: The Right Triangle design

Embellished design with geometric LOIN (Lines Of Interest Noteworthy) :roll:
since the integrated scalene triangle also squares the circle.

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Tue Aug 15, 2017 5:43 am

:flower: Re: Three Pi Vise design

"See how they run! ... in the 'impossible' House of Pi.
Diameters = Pi/2, sqrt(Pi), 2.0; Paramount ratio: 2(sqrt(1/Pi))"

:roll Also online: http://aitnaru.org/threepoints.html

"Answers 'What’s the point?' (about squaring circles). Right triangle (3 points)
contains defining hypotenuse-to-long-side ratio 1.128379167095512573896..
= 2(sqrt(1/Pi)) = sqrt(Pi)/(Pi/2)"

Golden diagonals of flutterby also have 2(sqrt(1/Pi)) relationship. 8)

Rod ... :bike: ...

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Re: Paradise Trinity Day

Post by Amigoo » Tue Aug 15, 2017 2:17 pm

:flower: Re: Three Pi Vise design
"The crown awaits the king (of a geometric proof?)"

:geek: Geometers' tip:
These line length ratios exist in the "crown":
1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
^2 = 1.2732395447351626861510701069801..
^3 = 1.4366969769991930158060821722332..

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Tue Aug 15, 2017 5:44 pm

:flower: Re: Three Pi Vise design
"The crown awaits the king (of a geometric proof?)"

:geek: Geometers' tip #2:

With Diameters = Pi/2, sqrt(Pi), 2.0
and the recently revealed constants:
1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
1.9130583802711007947403078280203.. "iPhi"
and the Pythagorean Theorem: a^2 + b^2 = c^2

Length of flutterby diagonals can be calculated
with their ratio then proven mathematically :roll
= 1.1283791670955125738961589031215..

Easy enough once the right triangles are selected
for this short mathematical journey! ;)

Rod :D

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Re: Paradise Trinity Day

Post by Amigoo » Wed Aug 16, 2017 12:55 am

:flower: Re: Three Pi Vise design
"The crown awaits the king (of a geometric proof?)"

:scratch: Who knew :?: :!:
The longer diagonal is the diameter of a circle
and the shorter diagonal has length equal
to a side of that circle's square!

:idea: That's why their length ratio is 2(sqrt(1/Pi)) 8)

"Three Pi Vise ... See how they run!" :bana:
Did you ever see such a sight in your life!

Rod :D

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