User:Tet-Math/sandbox

Calculating a,b,c from x,y,z is easy. Deriving x,y,z from a,b,c is more complex. These 5 internal relationships (K) equal each other BUT only in Tri-Rectangular Tetrahedrons. Where K is simply a declared equation constant.

$K\ =\ {\Biggl [}{\frac {a^{2}+b^{2}+c^{2}}{2}}{\Biggr ]}\ =\ {\Bigl [}x^{2}+y^{2}+z^{2}{\Bigr ]}\ =\ {\Bigl [}a^{2}+z^{2}{\Bigr ]}={\Bigl [}b^{2}+y^{2}{\Bigr ]}\ =\ {\Bigl [}c^{2}+x^{2}{\Bigr ]}$

$\ \ \ \ Given\ a,b,c\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Given\ x,y,z$

$K={\Biggl [}{\frac {a^{2}+b^{2}+c^{2}}{2}}{\Biggr ]}\ \ \ \ \ \ \ \ \ \ \ K={\Bigl [}x^{2}+y^{2}+z^{2}{\Bigr ]}$

$\ x={\sqrt {K-c^{2}\ }}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c={\sqrt {K-x^{2}\ }}$

$\ y={\sqrt {K-b^{2}\ }}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b={\sqrt {K-y^{2}\ }}$

$\ z={\sqrt {K-a^{2}\ }}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a={\sqrt {K-z^{2}\ }}$

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$a={\sqrt {x^{2}+y^{2}\ }}\ \ \ \ \ \ \ x={\sqrt {\frac {\ a^{2}+b^{2}-c^{2}\ }{2}}}\ \ \ \ \ \ \ y={\sqrt {\frac {\ a^{2}+c^{2}-b^{2}\ }{2}}}$

$b={\sqrt {x^{2}+z^{2}\ }}\ \ \ \ \ \ \ x={\sqrt {\frac {\ a^{2}+b^{2}-c^{2}\ }{2}}}\ \ \ \ \ \ \ z={\sqrt {\frac {\ b^{2}+c^{2}-a^{2}\ }{2}}}$

$c={\sqrt {y^{2}+z^{2}\ }}\ \ \ \ \ \ \ y={\sqrt {\frac {\ a^{2}+c^{2}-b^{2}\ }{2}}}\ \ \ \ \ \ \ \ z={\sqrt {\frac {\ b^{2}+c^{2}-a^{2}\ }{2}}}$