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I'm looking for a formula to find the volume of a tetrahedron from four "heights"—not the edges or vertices, but from the distance of the vertices from a common origin. (I know this is not an actual height, would like to know the right word…)

For instance, if I have the input "heights" of 3, 1, 4, and 4 (see image) can use these to find the volume of the tetrahedron?

A drawing. I have the lengths of the dotted lines but not of the edges.

Ultimately, I hope to be able to use a formula in Excel to be able to input 4 "heights" and receive an area easily…

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  • $\begingroup$ There is no such formula, as different tetrahedrons with different volumes can have the same lengths. $\endgroup$ – Ivan Neretin Jun 4 '16 at 21:40
  • $\begingroup$ What are the restrictions on your "common origin"? Clearly distances of the four vertices to an arbitrary point do not suffice to find a tetrahedron volume. $\endgroup$ – hardmath Jun 4 '16 at 21:42
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Imagine your reference point, and 4 nested spheres of the specified lengths. Each vertex of the tetrahedron is an unconstrained choice on the surface of the relevant sphere. Even if all your "heights" are the same, so there is only one determining sphere, it's clear by sliding the choice of the vertices around the sphere that the volume is still highly variable. Even if there is a requirement to keep the reference origin inside the tetrahedron, for example, the points can still be chosen to make the tetrahedron arbitrarily thin and the volume small.

So, no: there is no such formula.

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