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Let $A$ be a rectangular parallelepiped with edges of lengths $15, 20, 30$. Let $B$ be a tetrahedron on four non-adjacent vertices of $A$ (i.e no two vertices of $B$ share a common edge of $A$). Compute the volume of $B$.

This site gives a way to calculate but there's gotta be a closed form elegant formula for this.

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Let the rectangular parallelepiped measurements be $a,\,b,\,c$ then the mixed product of the three vectors is $6$ times the volume of the desired tetrahedron $$V=\frac16\operatorname{abs} \left| \begin{array}{ccc} a&b&0\\ a&0&c\\ 0&b&c \end{array} \right|=\frac13 abc$$ Thus the desired volume is $\frac13\cdot 15\cdot 20\cdot 30=3000$.

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    $\begingroup$ More simply (to my mind, anyway), each of the tetrahedra at the vertices $A', B, C', D$ has volume $\tfrac16abc,$ so the tetrahedron $AB'CD'$ has volume $abc - \frac23abc = \frac13abc.$ The question Volume of a tetrahedron whose 4 faces are congruent is relevant, but has more detail than is needed here. $\endgroup$ Aug 11 '20 at 21:00

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