Do we have an analogue to triangle inequality in 3-D say tetrahedron inequality(or any other relation), which once satisfied by any six real numbers; implies an existence of a tetrahedron with those side lengths?

  • $\begingroup$ added geometry tag. I don't think algebraic geometers studied this sort of thing. $\endgroup$ – achille hui Oct 3 '13 at 5:46

On any non-degenerate tetrahedron, pick any vertex and let $a, b, c$ be the length of edges connected to it. Let $\tilde{a},\tilde{b},\tilde{c}$ be the length of corresponding opposite edges. The edges on all six faces will satisfy triangular inequalities. These inequalities can be summarized in a single statement: $$\min(a+b+c,a+\tilde{b}+\tilde{c},\tilde{a}+b+\tilde{c},\tilde{a}+\tilde{b}+c) > \max(a+\tilde{a},b+\tilde{b},c+\tilde{c})\tag{*1}$$ Furthermore, the corresponding Calyer-Menger determinant is positive. Up to a scaling factor, it means: $$\left|\begin{matrix}0 & 1 & 1 & 1 & 1\cr 1 & 0 & {a}^{2} & {b}^{2} & {c}^{2}\cr 1 & {a}^{2} & 0 & \tilde{c}^{2} & \tilde{b}^{2}\cr 1 & {b}^{2} & \tilde{c}^{2} & 0 & \tilde{a}^{2}\cr 1 & {c}^{2} & \tilde{b}^{2} & \tilde{a}^{2} & 0\end{matrix}\right| > 0\tag{*2}$$

Given any 6 positive numbers, if $(*1)$ and $(*2)$ are satisfied, then it can be realized as the edge lengths of a non-degenerate tetrahedron.

For a proof of this, please see the paper Edge lengths determining tetrahedrons by Karl Wirth and Andre S. Dreiding.

Side Notes

About the inequalities of face areas mentioned in Glen O's answer. It is also a sufficient condition. More precisely, give any 4 positive numbers $A_1, A_2, A_3, A_4$. If can be realized as the face areas of a non-degenerate tetrahedron if and only if the following 4 inequalities are satisfied:

$$\begin{cases} A_2 + A_3 + A_4 > A_1\\A_3 + A_4 + A_1 > A_2\\ A_4 + A_1 + A_2 > A_3\\ A_1 + A_2 + A_3 > A_4\end{cases}$$

For a proof of this, please see answers in a related question (Inequality for each $a, b, c, d$ being each area of four faces of a tetrahedron).

  • $\begingroup$ The paper refers to address the issue comprehensively $\endgroup$ – ARi Oct 3 '13 at 8:12

There won't be a single inequality between the six edges (there'll be a set of four inequalities, one for each side), but there should be an inequality between the four faces.

That is,

$$ A_1 + A_2 + A_3 \geq A_4 $$ Where 1, 2, 3, and 4 are the four faces.

This would resolve as a set of cross products - if you have the origin as one point, and the other three points as $A$, $B$, and $C$, then you have

$$ |A\times B| + |A\times C| + |B\times C| \geq |(A-C)\times(B-C)| = \left|(A\times B) - (A\times C) - (B\times C)\right|$$


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