
  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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