Re: "PT 'n Q" design,
in:
http://aitnaru.org/images/Yin_Yang_Pi.pdf
Color coded to complement proof that Pythagorean Theorem
is true even in "impossible" Quadrature ... gracias a sqrt(2).
Also, two circle-squaring trapezoids where large one
has dimensions twice the size of the small one.
Long story short (re: 1/4 of PT's 3 squares,
where a^2 + b^2 = c^2 )
PT = Pythagorean Theorem
C-SRT = Circle-Squaring Right Triangle
IRT = Isosceles Right Triangle
2.0 = hypotenuse of C-SRT
1.7724538509055160272981674833411.. sqrt(Pi), long side of C-SRT
x 0.92650275035220848584275966758914.. sqrt(4-Pi), short side of C-SRT
= 1.6421833677363238765144275396003..
/ 2 = 0.82109168386816193825721376980014.. area of C-SRT
0.92650275035220848584275966758914.. sqrt(4-Pi), short side of C-SRT
/ 1.4142135623730950488016887242097.. sqrt(2)
= 0.65513637756203355309393588562466..
1.7724538509055160272981674833411.. sqrt(Pi), long side of C-SRT
/ 1.4142135623730950488016887242097.. sqrt(2)
= 1.2533141373155002512078826424055..
x 0.65513637756203355309393588562466..
= 0.82109168386816193825721376980014.. area of 2 similar RTs
Therefore, area of large IRT on right
equals area of 2 smaller IRTs on left,
proving a^2 + b^2 = c^2
0.78539816339744830961566084581988.. (sqrt(Pi)^2)/4, area of large IRT on left
+ 0.21460183660255169038433915418012.. (sqrt(4-Pi)^2)/4, area of small IRT on left
= 1.0 = (2^2)/4, area of IRT on right
Rod