What are the requirements of the six edges $a,b,c,d,e,$ and $f$ of a tetrahedron?
I know for each face of the tetrahedron, the three edges must satisfy the triangle condition. I also know by just satisfying all four faces does not meet the requirement of forming a tetrahedron. The requirements looks complicated. Can somebody helps me out?
The answer is beautifully given in the paper "Edge lengths determining tetrahedrons" by Wirth and Dreiding, Elemente der Mathematik, 64, 160 (2009). Roughly speaking, the Cayley-Menger determinant must be positive.
The condition is indeed quite complicated, too much so for me to be tempted to write it down here, but if you have access to Mathematica or similar software, here is a simple method to deduce it in a few minutes. We suppose that the tetrahedron is ABCD with $a$, $b$, $c$, $d$, $e$ and $f$ the lengths of $AB$, $BC$, $CA$, $AD$, $BD$ and $CD$ in that order. By a simple argument, one can choose coordinates so that the vertices are $(0,0,0)$, $(a,0,0)$, $(p,q,0)$, and $(r,s,t)$. Then if you use the command
Solve[{(p-a)^2+q^2==b^2,p^2+q^2==c^2,r^2+s^2+t^2==d^2,(r-a)^2+s^2+t^2==e^2,(r-p)^2+(s-q)^2+t^2==f^2},{p,q,r,s,t}]
this will not only solve your problem but it will give you coordinates for the required tetrahedron. The condition for existence is simply the positivity of the terms which occur under a square root sign in the solution.
