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I am looking for a formula: $V=f(S_1,S_2,S_3,S_4)$, where $S_1$, $S_2$, $S_3$, and $S_4$ are the areas of the four faces.

We know $V=\dfrac{S_1.h_1}{3}=\dfrac{S_2.h_2}{3}=\dfrac{S_3.h_3}{3}=\dfrac{S_4.h_4}{3}$, where $h_1$, $h_2$, $h_3$, and $h_4$ are the corresponding altitudes.

So we need to find

$h_1=g(S_2,S_3,S_4)$

$h_2=g(S_1,S_3,S_4)$

$h_3=g(S_1,S_2,S_4)$

$h_4=g(S_1,S_2,S_3)$

Also, if all points of the tetrahedron are on a plane, the volume should be zero. Thus

  • Firstly, if the projection of point is out of $S1$ Area, $S_1+S_2=S_3+S_4$ then $V=0$

  • If the projection of point is in $S1$ Area, $S_1=S_2+S_3+S_4$ then $V=0$

So can we determine from the areas if volume is zero or not ?

Is it possible to find $V=f(S_1,S_2,S_3,S_4)$? Is the surface information enough to create a unique closed volume?

Thanks a lot for advice and answers.

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This doesn't answer your question, but thinking about an analogy with Hereon's formula for the area of a triangle, I came across this on Wikipedia: en.wikipedia.org/wiki/… –  Old John Jun 19 '12 at 13:34
    
@OldJohn Indeed, it is possible to find the volume of an $n$-dimensional simplex from the edge lengths using the Cayley-Menger determinant. –  Jim Belk Jun 21 '12 at 16:56

1 Answer 1

up vote 15 down vote accepted

The volume of a tetrahedron cannot be determined from the surface areas of the faces. I shall provide a family of counterexamples.

A tetrahedron is called equifacial if its four faces are all congruent triangles. Equivalently, a tetrahedron is equifacial if its opposite edges have the same length.

Starting with any triangle in the plane, you can try to "glue" together four copies of the triangle in an obvious way to get a tetrahedron. This process has the following properties:

  • If you start with an equilateral triangle, the result is a regular tetrahedron.
  • If you start with an acute triangle, the result is an equifacial tetrahedron.
  • If you start with an right triangle, the result is a "degenerate" tetrahedron with zero volume.
  • If you start with an obtuse triangle, you can't make a tetrahedron.

Now, it is possible to continuously deform an equilateral triangle into a right triangle without changing the area. Therefore, it is possible to continuously deform a regular terahedron into a degenerate tetrahedron without changing the areas of the faces!

The following animation shows this process:

Folding tetrahedron animation

All of the tetrahedra shown in this animation have faces with area 1, but the volume decreases continuously from about $0.41$ to $0$.

Of course, equifacial tetrahedra aren't the only possible counterexample. Indeed, for any allowed quadruple $(S_1,S_2,S_3,S_4)$ of areas, there ought to be a whole interval of possible values for the volume.

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