TITLE: Does knowing the surface area of all faces uniquely determine a tetrahedron?
QUESTION [24 upvotes]: I was wondering if the four areas of a tetrahedron faces were sufficient information to uniquely determine its shape. For example, is it true to say that if the surface areas are equal then the solid must be a regular tetrahedron?
If the answer is negative, then what else we need to fully determine the shape of the tetrahedron in space?

REPLY [8 votes]: The answer is "no," which can be demonstrated by counting degrees of freedom.
A tetrahedron has 4 vertices in 3-dimensional space, and is therefore defined by 12 independent parameters, i.e. the $x$, $y$, $z$ coordinates of each vertex.
Now consider what we mean by two tetrahedra having "the same shape". The orientation in 3-D space doesn't matter, and that eliminates 6 of the 12 parameters (there are 3 rigid body translations, and 3 rotations).
But the tetrahedron only has four faces, so fixing the area of each face still leaves two independent parameters to vary its shape in an arbitrary way.
If you want to pick two more quantities to fix the shape of the tetrahedron, that could be done in many different ways, but it would be nice to do it in a way that is independent of any coordinate system used to describe it.
One natural parameter could be the volume of the tetrahedron, but finding a second one is not so "obvious". 
The radius of the inscribed and circumscribed spheres might be interesting choices. Proving that they do uniquely define the tetrahedron in combination with the area of the faces is left as an exercise for the reader :)
Another option might be to fix the lengths of two edges. Choosing a pair of edges which do not share a vertex has a nice symmetry about it.