Paradise Trinity Day

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Re: Paradise Trinity Day
Re: Tri Phi Pi design
Behold the geometric lock
Small, golden plus sign (+) identifies center of circle where D = 2.0;
D = 4(sqrt(1/Pi)) identifies hypotenuse of large right triangle.
Rod ... ...
Behold the geometric lock
Small, golden plus sign (+) identifies center of circle where D = 2.0;
D = 4(sqrt(1/Pi)) identifies hypotenuse of large right triangle.
Rod ... ...

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Re: Paradise Trinity Day
Re: Tri Phi Pi design (for the record)
Wiggly Numbers of The Right Triangle(s)
Given: SoCS = Side of Circle's Square,
3.1415926535897932384626433832795.. Pi
1.7724538509055160272981674833411.. sqrt(Pi)
1.9130583802711007947403078280203.. iPhi, longsidetoshortside
1.1283791670955125738961589031215.. 2(sqrt(1/Pi)), hypotenusetolongside
Calculate short side a1 of largest right triangle:
Diameter = 2.256758334191025147792317806243.. 4(sqrt(1/Pi)),
side c1, hypotenuse of right triangle
SoCS = 2.0, side b1, long side of right triangle
From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a1 = 1.0454464017541266302735942239052..
iPhi = b1/a1 = 1.9130583802711007947403078280203..
Calculate short side a2 of next largest right triangle:
Diameter = 1.27323954473516268615107010698.. 2(sqrt(1/Pi))^2,
side c2, hypotenuse of right triangle
SoCS = 1.1283791670955125738961589031215.. 2(sqrt(1/Pi)),
side b2, long side of right triangle
From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a2 = 0.58982997002716101129132484047986..
iPhi = b2/a2 = 1.9130583802711007947403078280203..
Calculate line length ratio of short side a1 to a2:
short side a1 / short side a2
= 1.0454464017541266302735942239052..
/ 0.58982997002716101129132484047986..
= 1.7724538509055160272981674833411.. sqrt(Pi)
Rod (Tri Phi Pi overload)
Wiggly Numbers of The Right Triangle(s)
Given: SoCS = Side of Circle's Square,
3.1415926535897932384626433832795.. Pi
1.7724538509055160272981674833411.. sqrt(Pi)
1.9130583802711007947403078280203.. iPhi, longsidetoshortside
1.1283791670955125738961589031215.. 2(sqrt(1/Pi)), hypotenusetolongside
Calculate short side a1 of largest right triangle:
Diameter = 2.256758334191025147792317806243.. 4(sqrt(1/Pi)),
side c1, hypotenuse of right triangle
SoCS = 2.0, side b1, long side of right triangle
From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a1 = 1.0454464017541266302735942239052..
iPhi = b1/a1 = 1.9130583802711007947403078280203..
Calculate short side a2 of next largest right triangle:
Diameter = 1.27323954473516268615107010698.. 2(sqrt(1/Pi))^2,
side c2, hypotenuse of right triangle
SoCS = 1.1283791670955125738961589031215.. 2(sqrt(1/Pi)),
side b2, long side of right triangle
From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a2 = 0.58982997002716101129132484047986..
iPhi = b2/a2 = 1.9130583802711007947403078280203..
Calculate line length ratio of short side a1 to a2:
short side a1 / short side a2
= 1.0454464017541266302735942239052..
/ 0.58982997002716101129132484047986..
= 1.7724538509055160272981674833411.. sqrt(Pi)
Rod (Tri Phi Pi overload)

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Re: Paradise Trinity Day
Re: Tri Phi Pi design (for the record, expanded)
Wiggly Numbers of The Right Triangle(s)
Given: SoCS = Side of Circle's Square,
3.1415926535897932384626433832795.. Pi
1.7724538509055160272981674833411.. sqrt(Pi)
1.9130583802711007947403078280203.. iPhi, longsidetoshortside
1.1283791670955125738961589031215.. 2(sqrt(1/Pi)), hypotenusetolongside
Calculate short side a1 of largest right triangle (1 of 3):
Diameter = 2.256758334191025147792317806243.. 4(sqrt(1/Pi)),
side c1, hypotenuse of right triangle
SoCS = 2.0, side b1, long side of right triangle
From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a1 = 1.0454464017541266302735942239052..
iPhi1 = b1/a1 = 1.9130583802711007947403078280203..
Calculate short side a2 of next largest right triangle (2 of 3):
Diameter = 1.27323954473516268615107010698.. 2(sqrt(1/Pi))^2,
side c2, hypotenuse of right triangle
SoCS = 1.1283791670955125738961589031215.. 2(sqrt(1/Pi)),
side b2, long side of right triangle
From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a2 = 0.58982997002716101129132484047986..
iPhi2 = b2/a2 = 1.9130583802711007947403078280203..
Calculate short side a3 of smallest right triangle (3 of 3):
Diameter = 0.71834848850066624675632793451077.. 4(sqrt(1/Pi)) / Pi
side c3, hypotenuse of right triangle
SoCS = 0.63661977236758134307553505349006.. 2/Pi
side b3, long side of right triangle
From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a3 = 0.33277592515360954394708556243436..
iPhi3 = b3/a3 = 1.9130583802711007947403078280203..
Calculate line length ratio of short side a1 to a2:
short side a1 / short side a2
= 1.0454464017541266302735942239052..
/ 0.58982997002716101129132484047986..
= 1.7724538509055160272981674833411.. sqrt(Pi)
Calculate line length ratio of short side a2 to a3:
short side a2 / short side a3
= 0.58982997002716101129132484047986..
/ 0.33277592515360954394708556243436..
= 1.7724538509055160272981674833411.. sqrt(Pi)
About the resonance of the two portals (re: SoCS) ...
2.0 / 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
= 1.7724538509055160272981674833411.. sqrt(Pi)
Rod (Tri Phi Pi overload ^2)
Wiggly Numbers of The Right Triangle(s)
Given: SoCS = Side of Circle's Square,
3.1415926535897932384626433832795.. Pi
1.7724538509055160272981674833411.. sqrt(Pi)
1.9130583802711007947403078280203.. iPhi, longsidetoshortside
1.1283791670955125738961589031215.. 2(sqrt(1/Pi)), hypotenusetolongside
Calculate short side a1 of largest right triangle (1 of 3):
Diameter = 2.256758334191025147792317806243.. 4(sqrt(1/Pi)),
side c1, hypotenuse of right triangle
SoCS = 2.0, side b1, long side of right triangle
From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a1 = 1.0454464017541266302735942239052..
iPhi1 = b1/a1 = 1.9130583802711007947403078280203..
Calculate short side a2 of next largest right triangle (2 of 3):
Diameter = 1.27323954473516268615107010698.. 2(sqrt(1/Pi))^2,
side c2, hypotenuse of right triangle
SoCS = 1.1283791670955125738961589031215.. 2(sqrt(1/Pi)),
side b2, long side of right triangle
From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a2 = 0.58982997002716101129132484047986..
iPhi2 = b2/a2 = 1.9130583802711007947403078280203..
Calculate short side a3 of smallest right triangle (3 of 3):
Diameter = 0.71834848850066624675632793451077.. 4(sqrt(1/Pi)) / Pi
side c3, hypotenuse of right triangle
SoCS = 0.63661977236758134307553505349006.. 2/Pi
side b3, long side of right triangle
From the Pythagorean Theorem (a^2 + b^2 = c^2):
short side a3 = 0.33277592515360954394708556243436..
iPhi3 = b3/a3 = 1.9130583802711007947403078280203..
Calculate line length ratio of short side a1 to a2:
short side a1 / short side a2
= 1.0454464017541266302735942239052..
/ 0.58982997002716101129132484047986..
= 1.7724538509055160272981674833411.. sqrt(Pi)
Calculate line length ratio of short side a2 to a3:
short side a2 / short side a3
= 0.58982997002716101129132484047986..
/ 0.33277592515360954394708556243436..
= 1.7724538509055160272981674833411.. sqrt(Pi)
About the resonance of the two portals (re: SoCS) ...
2.0 / 1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
= 1.7724538509055160272981674833411.. sqrt(Pi)
Rod (Tri Phi Pi overload ^2)

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Re: Paradise Trinity Day
Re: Power of 2 design
Visual antidote to "Tri Phi Pi overload ^2"
with impressive display of similar isoceles trapezoids.
Two pairs of SoCS red lines have 2.0 or 1.0 length;
big D = 4(sqrt(Pi)), little D = 2(sqrt(Pi))
Rod (with little 'd')
Visual antidote to "Tri Phi Pi overload ^2"
with impressive display of similar isoceles trapezoids.
Two pairs of SoCS red lines have 2.0 or 1.0 length;
big D = 4(sqrt(Pi)), little D = 2(sqrt(Pi))
Rod (with little 'd')

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Re: Paradise Trinity Day
Re: Power of 2 II design
(further exploration of the Po2 geometry)
Better integration of circlesquaring scalene and right triangles
... and brings the newly revealed iPhi back into focus
A very interesting pattern:
These are the angles of the two geometric objects:
62.403.., 27.597.., 90.0 right triangle
62.403.., 72.597.., 45.0 scalene triangle
The only difference in these sets of angles is the 45 degree
"borrowing" by 72.597.. from 90. Such numeric assocation
may be a clue to "squaring the circle"
Rod
(further exploration of the Po2 geometry)
Better integration of circlesquaring scalene and right triangles
... and brings the newly revealed iPhi back into focus
A very interesting pattern:
These are the angles of the two geometric objects:
62.403.., 27.597.., 90.0 right triangle
62.403.., 72.597.., 45.0 scalene triangle
The only difference in these sets of angles is the 45 degree
"borrowing" by 72.597.. from 90. Such numeric assocation
may be a clue to "squaring the circle"
Rod

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Re: Paradise Trinity Day
Re: Power of 2 II design
( http://aitnaru.org/images/The_Right_Triangle.pdf )
Now more esoteric to show that the circlesquaring scalene and right triangles
in the large circle are sqrt(2) larger than these triangles in the small circle
(where its diameter has sqrt(2) relationship to the large circle).
Rod ... ...
( http://aitnaru.org/images/The_Right_Triangle.pdf )
Now more esoteric to show that the circlesquaring scalene and right triangles
in the large circle are sqrt(2) larger than these triangles in the small circle
(where its diameter has sqrt(2) relationship to the large circle).
Rod ... ...

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Re: Paradise Trinity Day
Re: iCorrelation design
(impossible Correlation)
This sqrt(2)dominated Cartesian neighborhood, a convincing "Pi Corral",
highlights influence of two conjoined trapezoids, one set per CSC circle.
Note tight integration within the trapezoidal sets of the overlapping,
circlesquaring right and scalene triangles (in each trapezoidal set).
Cartesian quadrature extraordinaire
Rod
(impossible Correlation)
This sqrt(2)dominated Cartesian neighborhood, a convincing "Pi Corral",
highlights influence of two conjoined trapezoids, one set per CSC circle.
Note tight integration within the trapezoidal sets of the overlapping,
circlesquaring right and scalene triangles (in each trapezoidal set).
Cartesian quadrature extraordinaire
Rod

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Re: Paradise Trinity Day
Re: Concentric Too design
(aka "Concentric 2")
With the inclusion of one more circle (creates CSCSC), this "Pi Corral"
shouts that the dimensions of the largest circle's objects are double
the dimensions of the smallest circle's (CSCSC Geometry 101).
example: (8/sqrt(2)) / sqrt(2) = 4.
From this perspective, transcendental Pi is twice as nice
... and evenly divisible by two.
Rod ... ...
(aka "Concentric 2")
With the inclusion of one more circle (creates CSCSC), this "Pi Corral"
shouts that the dimensions of the largest circle's objects are double
the dimensions of the smallest circle's (CSCSC Geometry 101).
example: (8/sqrt(2)) / sqrt(2) = 4.
From this perspective, transcendental Pi is twice as nice
... and evenly divisible by two.
Rod ... ...

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Re: Paradise Trinity Day
Re: CSCSCalenity design
(pronounced "CSC Scalenity")
Sublime geometric balance in the Pi Corral.
"Lines and triangles and squares! Oh, Pi !"
Rod ... ...
(pronounced "CSC Scalenity")
Sublime geometric balance in the Pi Corral.
"Lines and triangles and squares! Oh, Pi !"
Rod ... ...

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Re: Paradise Trinity Day
Re: CSCSCalenity design (new insight)
"Geometric quiescence in the Pi Corral (the circle is squared)"
Rod
"Geometric quiescence in the Pi Corral (the circle is squared)"
Rod

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Re: Paradise Trinity Day
Re: Pi Double Quadrature design
(aka "PDQ Yes!")
Consistent line length ratios in two concentric circles, both squared
... and with convincing presence of their geometric progeny!
Rod ... ... (cruisin' Cartesian corridors PDQ)
(aka "PDQ Yes!")
Consistent line length ratios in two concentric circles, both squared
... and with convincing presence of their geometric progeny!
Rod ... ... (cruisin' Cartesian corridors PDQ)

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Re: Paradise Trinity Day
Re: Pi Double Quadrature design
(aka "PDQ, Yes!") ... and created PDQ
This geometry is all about ratio 2(sqrt(1/Pi))
( = 2/sqrt(Pi) = sqrt(Pi)/(Pi/2) )
Rod
(aka "PDQ, Yes!") ... and created PDQ
This geometry is all about ratio 2(sqrt(1/Pi))
( = 2/sqrt(Pi) = sqrt(Pi)/(Pi/2) )
Rod

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Re: Paradise Trinity Day
Re: Foursquare Plus Four design
Four concentric* circles all squared, each with a sibling (circled radius)
... and tight integration of this entire Cartesian family!
* from the center, diameters increase by 2(sqrt(1/Pi))
= 2/sqrt(Pi) = 1.1283791670955125738961589031215..
"foursquare  marked by boldness and conviction"
Rod
Four concentric* circles all squared, each with a sibling (circled radius)
... and tight integration of this entire Cartesian family!
* from the center, diameters increase by 2(sqrt(1/Pi))
= 2/sqrt(Pi) = 1.1283791670955125738961589031215..
"foursquare  marked by boldness and conviction"
Rod

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Re: Paradise Trinity Day
Re: Foursquare Plus Four design
"Final, final, final" ... esoterically mesmerizing ...
as would any geometry proven impossible.
Being esoterically mesmerized, now elevate your mesmery ...
and consider ratio of diameters of the two golden circles.
Rod
"Final, final, final" ... esoterically mesmerizing ...
as would any geometry proven impossible.
Being esoterically mesmerized, now elevate your mesmery ...
and consider ratio of diameters of the two golden circles.
Rod

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Re: Paradise Trinity Day
Re: Foursquare Plus Four design
Blah! blah! blah! This geometry only chatters about 2(sqrt(1/Pi))
( = 2/sqrt(Pi) = sqrt(Pi)/(Pi/2) = 1.1283791670955125738961589031215.. )
IMO, if you've seen one squared circle, you've seen foursquare!
Rod ... ... (runnin' in foursquare circles)
Blah! blah! blah! This geometry only chatters about 2(sqrt(1/Pi))
( = 2/sqrt(Pi) = sqrt(Pi)/(Pi/2) = 1.1283791670955125738961589031215.. )
IMO, if you've seen one squared circle, you've seen foursquare!
Rod ... ... (runnin' in foursquare circles)

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Re: Paradise Trinity Day
Re: Foursquare Plus Four design
"Four concentric circles all squared,
each with a sibling (circled radius);
quadrarture proven by 2(sqrt(1/Pi))"
Rod
"Four concentric circles all squared,
each with a sibling (circled radius);
quadrarture proven by 2(sqrt(1/Pi))"
Proving quadrature geometrically is noteworthy chatter!This geometry only chatters about 2(sqrt(1/Pi))
Rod

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Re: Paradise Trinity Day
Re: Foursquare Plus Four design
Geometry now so esoteric, that a disclaimer is required:
This was not found on an alien spacecraft!
Rod ... ... (once cruisin' in foursquare circles
... at warp speed )
Geometry now so esoteric, that a disclaimer is required:
This was not found on an alien spacecraft!
Rod ... ... (once cruisin' in foursquare circles
... at warp speed )

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Re: Paradise Trinity Day
Re: Foursquare Plus Four design
"quadrarture proven by 2(sqrt(1/Pi))
( = 2/sqrt(Pi) = sqrt(Pi)/(Pi/2) )"
Recolored to emphasize simplicity of a circle squared:
Geometers need only know diameters of two golden circles:
small golden circle  2(sqrt(1/Pi)); large circle  2.0
The rest of the composition is Geometry 101
Rod ... ...
"quadrarture proven by 2(sqrt(1/Pi))
( = 2/sqrt(Pi) = sqrt(Pi)/(Pi/2) )"
Recolored to emphasize simplicity of a circle squared:
Geometers need only know diameters of two golden circles:
small golden circle  2(sqrt(1/Pi)); large circle  2.0
The rest of the composition is Geometry 101
Rod ... ...

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Re: Paradise Trinity Day
A quiet sevenyear anniversary in the local Cartesian neighborhoodIf not already declared On High, 101010 presents favorable symbolism for a Paradise Trinity Day.
(which explains the 3:33 AM wakeup call to review the past)
Rod

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Re: Paradise Trinity Day
Re: QT1010 design (Quadrature Trinity)
Simplification of Foursquare Plus Four design, highlighting
squaredcircles trinity of 2(sqrt(1/Pi)), 2.0, and sqrt(Pi)
(1.1283791670955125738961589031215.., 2.0,
and 1.7724538509055160272981674833411..).
... on the PTD 10/10/17 sevenyear anniversary.
Note: 2(sqrt(1/Pi)) = 2/sqrt(Pi) = sqrt(Pi)/(Pi/2)
Rod
Simplification of Foursquare Plus Four design, highlighting
squaredcircles trinity of 2(sqrt(1/Pi)), 2.0, and sqrt(Pi)
(1.1283791670955125738961589031215.., 2.0,
and 1.7724538509055160272981674833411..).
... on the PTD 10/10/17 sevenyear anniversary.
Note: 2(sqrt(1/Pi)) = 2/sqrt(Pi) = sqrt(Pi)/(Pi/2)
Rod

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Re: Paradise Trinity Day
Re: QT1010 design (Quadrature Trinity)
Long story short: 2(sqrt(1/Pi)) is the increment/decrement
from the center of the squared circles' concentricity (2.0):
x 2(sqrt(1/Pi)) =
x 2(sqrt(1/Pi)) = 2(2(sqrt(1/Pi))^2)
x 2(sqrt(1/Pi)) = 4(sqrt(1/Pi))
2.0
/ 2(sqrt(1/Pi)) = sqrt(Pi)
/ 2(sqrt(1/Pi)) = Pi/2
/ 2(sqrt(1/Pi)) =
Rod
Long story short: 2(sqrt(1/Pi)) is the increment/decrement
from the center of the squared circles' concentricity (2.0):
x 2(sqrt(1/Pi)) =
x 2(sqrt(1/Pi)) = 2(2(sqrt(1/Pi))^2)
x 2(sqrt(1/Pi)) = 4(sqrt(1/Pi))
2.0
/ 2(sqrt(1/Pi)) = sqrt(Pi)
/ 2(sqrt(1/Pi)) = Pi/2
/ 2(sqrt(1/Pi)) =
Rod

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Re: Paradise Trinity Day
Re: QT1010 design (Quadrature Trinity)
(numeric continuum of both sets here below):
Set 1
2.8733939540026649870253117380431..
                        
2.5464790894703253723021402139602.. 2(2(sqrt(1/Pi)))^2)
2.2567583341910251477923178062431.. 4(sqrt(1/Pi))
2.0
1.7724538509055160272981674833411.. sqrt(Pi)
1.5707963267948966192313216916398.. Pi/2
                        
1.3920819992079269613212044955297..
Set 2
1.4366969770013324935126558690215..
                        
1.2732395447351626861510701069801.. (2(sqrt(1/Pi)))^2
1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
1.0
0.88622692545275801364908374167057.. sqrt(Pi)/2
0.78539816339744830961566084581988.. Pi/4
                        
0.69604099960396348066060224776486..
Rod ... ... (all set for more casual cruisin')
Added another set of 3 concentric circles,Long story short: 2(sqrt(1/Pi)) is the increment/decrement
from the center of the squared circles' concentricity
(numeric continuum of both sets here below):
Set 1
2.8733939540026649870253117380431..
                        
2.5464790894703253723021402139602.. 2(2(sqrt(1/Pi)))^2)
2.2567583341910251477923178062431.. 4(sqrt(1/Pi))
2.0
1.7724538509055160272981674833411.. sqrt(Pi)
1.5707963267948966192313216916398.. Pi/2
                        
1.3920819992079269613212044955297..
Set 2
1.4366969770013324935126558690215..
                        
1.2732395447351626861510701069801.. (2(sqrt(1/Pi)))^2
1.1283791670955125738961589031215.. 2(sqrt(1/Pi))
1.0
0.88622692545275801364908374167057.. sqrt(Pi)/2
0.78539816339744830961566084581988.. Pi/4
                        
0.69604099960396348066060224776486..
Rod ... ... (all set for more casual cruisin')

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Re: Paradise Trinity Day
Re: QT1010 design (Quadrature Trinity)
In the local universe ...
squared circles do not exist without pattern
How to know the circle is squared?
Look for the Smile of Pythagoras!
(turn design clockwise 90 degrees)
Rod
In the local universe ...
squared circles do not exist without pattern
How to know the circle is squared?
Look for the Smile of Pythagoras!
(turn design clockwise 90 degrees)
Rod

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Re: Paradise Trinity Day
Re: Tenetcy of Quadrature design
... highlights the other squares (inscribed),
objects in the defining set of principles.
"Lines and triangles and squares! Oh, Pi!"
Rod
... highlights the other squares (inscribed),
objects in the defining set of principles.
"Lines and triangles and squares! Oh, Pi!"
Rod

 Family
 Posts: 3809
 Joined: Fri Mar 17, 2006 8:32 pm
 Please type in these numbers: 46373: 0
 Please type in these numbers:91294: 0
 Location: Dallas, TX
Re: Paradise Trinity Day
Re: Tenetcy of Quadrature design (aka "TQ Waldo?")
The circlesquaring scalene triangle (red) can be found in every circle,
albeit geometry expertise is required for the second largest circle.
So, where's TQ Waldo?
Speaking of "TQ Waldo?" ... Who knew
There are 8 similar scalene triangles in this composition!
Note: "Tenetcy" is a real word
... in socially squared circles.
Rod ... ...
Lines and triangles and squares! Oh, Pi!
The circlesquaring scalene triangle (red) can be found in every circle,
albeit geometry expertise is required for the second largest circle.
So, where's TQ Waldo?
Speaking of "TQ Waldo?" ... Who knew
There are 8 similar scalene triangles in this composition!
Note: "Tenetcy" is a real word
... in socially squared circles.
Rod ... ...