
Re: "Just Puzzling" design,
updated in:
http://aitnaru.org/images/Yin_Yang_Pi.pdf

Quadrature's infinite nibbles of Pi by geometric Pi-ranhas.

Essential formulary of the Pi Corral:
2 / 1.7724538509055160272981674833411.. ( 2/sqrt(Pi) )
= 1.1283791670955125738961589031215.. (defines unique right triangle)
^2 = 1.2732395447351626861510701069801..

Lotsa numbas for circle-squaring right triangle
inscribed in a circle whose Circumference = 4.0
a = 1.1283791670955125738961589031215.. 2/sqrt(Pi)
^2 = 1.27323954473516268615107010698..
b = 0.63661977236758134307553505349006.. ((2/sqrt(Pi))^2)/2
x 0.92650275035220848584275966758914.. (sqrt(4-Pi))^2
= 0.58982997002716101129132484048001..
^2 = 0.34789939354224165695100130437553..
c = 1.27323954473516268615107010698.. (2/sqrt(Pi))^2
^2 = 1.6211389382774043431020714113555..
1.27323954473516268615107010698.. a^2
+ 0.34789939354224165695100130437553.. b^2
= 1.6211389382774043431020714113555.. c^2

Pop Quiz (from the School of Pi-ranhas)
Q: When is area of a circle equal to its diameter?
A: When circle's circumference = 4 = Pi(D)
When Circumference = 4.0
Diameter = (2/sqrt(Pi))^2
SoCS = 2/sqrt(Pi)
Area = (2/sqrt(Pi))^2
= Diameter
Rod ...

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