Paradise Trinity Day

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Re: Paradise Trinity Day
Re: Two Similar design
Who knew that the isosceles right triangle precisely "calculates"
sqrt(2) when math requires millions of decimal digits Then isosceles
proves its calculation by defining a line having precisely half the length
of the first triangle's hypotenuse.
What this suggests is that math is not yet accurate enough
to define the magic of geometry, mystical magic that permits
the circle to be squared when math says it's impossible.
Rod
Who knew that the isosceles right triangle precisely "calculates"
sqrt(2) when math requires millions of decimal digits Then isosceles
proves its calculation by defining a line having precisely half the length
of the first triangle's hypotenuse.
What this suggests is that math is not yet accurate enough
to define the magic of geometry, mystical magic that permits
the circle to be squared when math says it's impossible.
Rod

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Re: Paradise Trinity Day
Re: Two Similar design*
More busyness (lower left quadrant of design),
showing sqrt(2) propagation for circlesquaring right
and isosceles right triangles. Patterns prevail
* updated in: http://aitnaru.org/images/Khristos_Voskrese.pdf
Rod ... ...
More busyness (lower left quadrant of design),
showing sqrt(2) propagation for circlesquaring right
and isosceles right triangles. Patterns prevail
* updated in: http://aitnaru.org/images/Khristos_Voskrese.pdf
Rod ... ...

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Re: Paradise Trinity Day
Re: Two Similar design
Regarding Quadrature ...
Geometers in higher society say "loquacious" ...
but geometers of the people say "chatty" ...
however, "verbose" is also notable
Rod
Regarding Quadrature ...
Geometers in higher society say "loquacious" ...
but geometers of the people say "chatty" ...
however, "verbose" is also notable
Rod

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Re: Paradise Trinity Day
Re: Two Similar design
Regarding sqrt(2) ...
The isosceles right triangle presents intrigue about math
and sqrt(2)'s millions of decimal digits: Since srt(2)^2 = 2
(re: adjoined isosceles rights), this suggests that, while
the interim value cannot be known, beginning and ending
values are known ... precisely
This hints that Pi suffers the same unknown
... but only "interim" values of Pi.
Rod
Regarding sqrt(2) ...
The isosceles right triangle presents intrigue about math
and sqrt(2)'s millions of decimal digits: Since srt(2)^2 = 2
(re: adjoined isosceles rights), this suggests that, while
the interim value cannot be known, beginning and ending
values are known ... precisely
This hints that Pi suffers the same unknown
... but only "interim" values of Pi.
Rod

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Re: Paradise Trinity Day
Re: Two Similar design
Regarding sqrt(2) ...
On second thought, to say that Pi must end in an even digit
is like saying that a diameter can only have an even length.
But the "magic" of sqrt(2) and isosceles right triangle still remains;
geometry can show what math proves "impossible" ... apparently.
Maybe geometry is the great rounder upper or downer.
Who knew The lesson is that infinity is a reality!
Welcome to the real inhabited universe(s).
Rod
Regarding sqrt(2) ...
On second thought, to say that Pi must end in an even digit
is like saying that a diameter can only have an even length.
But the "magic" of sqrt(2) and isosceles right triangle still remains;
geometry can show what math proves "impossible" ... apparently.
Maybe geometry is the great rounder upper or downer.
Who knew The lesson is that infinity is a reality!
Welcome to the real inhabited universe(s).
Rod

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Re: Paradise Trinity Day
To summarize ...
 Math says that the circle cannot be squared.
 Geometry shows that squared circles exist.
 Sqrt(2) (math) and isosceles triangle (geometry)
prove that Pi can be divided by 2 ... precisely.
"Say what?!"
There is no "remainder" in infinity.
Who knew
Rod
 Math says that the circle cannot be squared.
 Geometry shows that squared circles exist.
 Sqrt(2) (math) and isosceles triangle (geometry)
prove that Pi can be divided by 2 ... precisely.
"Say what?!"
There is no "remainder" in infinity.
Who knew
Rod

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Re: Paradise Trinity Day
Re: iQ by Two design
Another Cartesian doodle.
Rod ... ...
Another Cartesian doodle.
Rod ... ...

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Re: Paradise Trinity Day
Re: iQ by Two design ("Cartesian doodle")
Three concentric 2circle pairs, separated by sqrt(2)
with the 2circle pair separated by 2/sqrt(Pi).
Obviously, the outer pair of circles have twice
the dimensions of the inner pair (re: sqrt(2)^2).
Rod
About the iQ geometry of circles squared ...The circle cannot be squared, but squared circles exist
Three concentric 2circle pairs, separated by sqrt(2)
with the 2circle pair separated by 2/sqrt(Pi).
Obviously, the outer pair of circles have twice
the dimensions of the inner pair (re: sqrt(2)^2).
Rod

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Re: Paradise Trinity Day
Re: iQ by Two design ("Cartesian doodle")
In this concentric CSC continuum (CCSCC),
line length ratio of circumference to diameter remains Pi,
but length of SoCS (Side of Circle's Square) changes by sqrt(2)
... when the circles exist in CSC relationship.
If a diameter equals 2 times an increment of 10,
what's a Pi ending in zero? "Pizero" comes to mind.
And what's this? Infinity minus 1
However, a Pizero occurs only when a sufficient
number of zeros are used as the diameter's increment.
CSC geometry seems to represent this Pizero moment,
giving infinity minus 1, minus 2, minus 3 ...
Rod
About the geometric magic of iQ by Two ...The circle cannot be squared, but squared circles exist
In this concentric CSC continuum (CCSCC),
line length ratio of circumference to diameter remains Pi,
but length of SoCS (Side of Circle's Square) changes by sqrt(2)
... when the circles exist in CSC relationship.
If a diameter equals 2 times an increment of 10,
what's a Pi ending in zero? "Pizero" comes to mind.
And what's this? Infinity minus 1
However, a Pizero occurs only when a sufficient
number of zeros are used as the diameter's increment.
CSC geometry seems to represent this Pizero moment,
giving infinity minus 1, minus 2, minus 3 ...
Rod

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Re: Paradise Trinity Day
Re: iSosceles Rights design
Featuring two Siamese Pythagorean Right Triangles
and a divisional 2.0 within their sqrt(2) continuum.
"Say what?!" Quadrature by the numbers (there's more!)
"What other numbers?!" Go figure! (using these:)
= 2.506628274631000502415765284811.. 2(sqrt(Pi)(sqrt(2))
x 1.2533141373155002512078826424055.. sqrt(Pi)(sqrt(2))
2.0
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.7724538509055160272981674833411.. sqrt(Pi)
2.506628274631000502415765284811.. 2(sqrt(Pi)(sqrt(2))
/ 1.7724538509055160272981674833411.. sqrt(Pi)
= 1.4142135623730950488016887242097.. sqrt(2)
Rod ... ... (cruisin' by the numbers)
Featuring two Siamese Pythagorean Right Triangles
and a divisional 2.0 within their sqrt(2) continuum.
"Say what?!" Quadrature by the numbers (there's more!)
"What other numbers?!" Go figure! (using these:)
= 2.506628274631000502415765284811.. 2(sqrt(Pi)(sqrt(2))
x 1.2533141373155002512078826424055.. sqrt(Pi)(sqrt(2))
2.0
/ 1.1283791670955125738961589031215.. 2/sqrt(Pi)
= 1.7724538509055160272981674833411.. sqrt(Pi)
2.506628274631000502415765284811.. 2(sqrt(Pi)(sqrt(2))
/ 1.7724538509055160272981674833411.. sqrt(Pi)
= 1.4142135623730950488016887242097.. sqrt(2)
Rod ... ... (cruisin' by the numbers)

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Re: Paradise Trinity Day
Re: iSosceles Rights design
when two adjoined circlesquaring scalene triangles announce their presence!
Who knew A Squared Circles Soirée!
Rod
So, I attempt to simplify the geometry and coloridentify the "divisional 2.0"and a divisional 2.0 within their sqrt(2) continuum
when two adjoined circlesquaring scalene triangles announce their presence!
Who knew A Squared Circles Soirée!
Rod

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Re: Paradise Trinity Day
Re: iSoscalene Rights design
(morphed from "iSosceles Rights")
... because of the presence of circlesquaring scalene triangleS.
Rod ... ... (cruisin' the Rights ... and Lefts)
(morphed from "iSosceles Rights")
... because of the presence of circlesquaring scalene triangleS.
Rod ... ... (cruisin' the Rights ... and Lefts)

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Re: Paradise Trinity Day
Re: iSqrt(2)^2 design
More "geometry that speaks for itself" ...
in the esoteric language of Quadrature.
Rod
More "geometry that speaks for itself" ...
in the esoteric language of Quadrature.
Rod

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Re: Paradise Trinity Day
Re: iSqrt(2)^2 design*
Geometers' secret:
Largest squared circle has precisely 4 times the area
of the smallest squared circle ... gracias a sqrt(2).
* updated in: http://aitnaru.org/images/Khristos_Voskrese.pdf
Rod ... ...
Geometers' secret:
Largest squared circle has precisely 4 times the area
of the smallest squared circle ... gracias a sqrt(2).
* updated in: http://aitnaru.org/images/Khristos_Voskrese.pdf
Rod ... ...

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Re: Paradise Trinity Day
Re: iSqrt(2)^2 design
Now, a design that complements the PDF name.
Rod
Now, a design that complements the PDF name.
Rod

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Re: Paradise Trinity Day
Re: Center of Infinity design
Who knew that such a center exists ...
identified by 3 overlapping, circlesquaring scalene triangles
in a sqrt(2)dominated Cartesian Neighborhood (of course).
Rod
Who knew that such a center exists ...
identified by 3 overlapping, circlesquaring scalene triangles
in a sqrt(2)dominated Cartesian Neighborhood (of course).
Rod

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Re: Paradise Trinity Day
Re: Center of Infinity design
area that is the eternal residence of a First Source ...
according to the revelation of iQuadrature.
Rod ... ... (cruisin' toward the Center)
The Center of Infinity is an area  not a point;Who knew that such a center exists.
area that is the eternal residence of a First Source ...
according to the revelation of iQuadrature.
Rod ... ... (cruisin' toward the Center)

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Re: Paradise Trinity Day
Re: Center of Infinity design
but the reality: at the start of infinity there is nothing (no point).
Note: "Ceneter of Infinity" (in PDF) sounds like "senator"
and is whimsical wordplay about nothing.
Rod
One more line tries to insist on a single starting point of infinity,The Center of Infinity is an area  not a point
but the reality: at the start of infinity there is nothing (no point).
Note: "Ceneter of Infinity" (in PDF) sounds like "senator"
and is whimsical wordplay about nothing.
Rod

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Re: Paradise Trinity Day
Re: Center of Infinity design
Red crossed lines of equal length, connecting similar angles of two overlapping scalenes,
do not define the geometric center  that is identified by the small blue cross.
Rod
Geometers' secret:Before infinity, there is no point.
Red crossed lines of equal length, connecting similar angles of two overlapping scalenes,
do not define the geometric center  that is identified by the small blue cross.
Rod

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Re: Paradise Trinity Day
Re: Center of Infinity design
in a convincing Cartesian Neighborhood, nigh "impossible" ...
A new arc completes the overlapping, circlesquaring scalene triangles;
each triangle inscribed in an adjoined pair of similar circles, adjoined
on a shared side of their inscribed squares.
Obviously (to astute geometers), these two complex pairs
of geometric objects are defined by their sqrt(2) association.
Of course, such complex geometry is best comprehended
by a stroll along the ER Bridge on a moonless night.
"Say What Where (and when) is such a moon?"
Theoretical ... since it's "impossible" to observe.
Rod
Geometers' secret of Quadraturial representationBefore infinity, there is no point.
in a convincing Cartesian Neighborhood, nigh "impossible" ...
A new arc completes the overlapping, circlesquaring scalene triangles;
each triangle inscribed in an adjoined pair of similar circles, adjoined
on a shared side of their inscribed squares.
Obviously (to astute geometers), these two complex pairs
of geometric objects are defined by their sqrt(2) association.
Of course, such complex geometry is best comprehended
by a stroll along the ER Bridge on a moonless night.
"Say What Where (and when) is such a moon?"
Theoretical ... since it's "impossible" to observe.
Rod

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Re: Paradise Trinity Day
Re: Divisible Pi design
Focus on the divisibility of Pi by 2 and sqrt(2).
In this geometric sqrt(2) spiral, long side of circlesquaring Pythagorean right triangles
are contrasted as they progress from sqrt(Pi) for D = 2.0 (the hypotenuse) to sqrt(Pi)/2.
While each circle's ratio of circumference to diameter remains Pi, the beginning long side
has length equal to a precise sqrt(Pi) and downsizes by sqrt(2) to sqrt(Pi)/2 ...
according to the simple geometry in this Cartesian Neighborhood.
Rod
Focus on the divisibility of Pi by 2 and sqrt(2).
In this geometric sqrt(2) spiral, long side of circlesquaring Pythagorean right triangles
are contrasted as they progress from sqrt(Pi) for D = 2.0 (the hypotenuse) to sqrt(Pi)/2.
While each circle's ratio of circumference to diameter remains Pi, the beginning long side
has length equal to a precise sqrt(Pi) and downsizes by sqrt(2) to sqrt(Pi)/2 ...
according to the simple geometry in this Cartesian Neighborhood.
Rod

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Re: Paradise Trinity Day
Re: Divisible Pi design
Focus on the divisibility of Pi by 2 and sqrt(2).
Given a diameter of 2.0, a diameter divisible by 2 and sqrt(2)
and directly associated with Pi (therefore divisible by 2 and sqrt(2)),
the significance of Pi's decimal digits decreases significantly
as the diameter decreases by the duo of 2 (IMO).
Thrice divided Pi is not as nice as twice.
Rod
Focus on the divisibility of Pi by 2 and sqrt(2).
Given a diameter of 2.0, a diameter divisible by 2 and sqrt(2)
and directly associated with Pi (therefore divisible by 2 and sqrt(2)),
the significance of Pi's decimal digits decreases significantly
as the diameter decreases by the duo of 2 (IMO).
Thrice divided Pi is not as nice as twice.
Rod

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Re: Paradise Trinity Day
Re: Divisible Pi design
Focus on the divisibility of Pi by 2 and sqrt(2).
"Similar scalenity in a sqrt(2) spiral
with quadraturial progression."
Rod ... ...
Focus on the divisibility of Pi by 2 and sqrt(2).
"Similar scalenity in a sqrt(2) spiral
with quadraturial progression."
Rod ... ...

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Re: Paradise Trinity Day
Re: Divisible Pi design
Focus on the divisibility of Pi by 2 and sqrt(2).
Geometers' secret:
Dark blue arc (1/4 of a circle) connects two SoCS
sides (long side of circlesquaring right triangles),
one having twice the length of the other.
Rod
Focus on the divisibility of Pi by 2 and sqrt(2).
Geometers' secret:
Dark blue arc (1/4 of a circle) connects two SoCS
sides (long side of circlesquaring right triangles),
one having twice the length of the other.
Rod