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?
Consider the vertex set in $\mathbb R^3$ $$\{(a, b, 0), (a, -b, 0), (-a, 0, b), (-a, 0, -b)\}$$ for $a, b > 0$. Then it is clear that the faces determined by these vertices are all congruent isosceles triangles with side lengths $2b$, $\sqrt{4a^2+2b^2}$, $\sqrt{4a^2+2b^2}$ and common area $a \sqrt{4a^2+b^2}$. So by a suitable choice of $a$ and $b$ we can make the tetrahedron irregular but have faces of equal area; moreover, for a given area of a face, there is in general more than one choice of $(a,b)$, hence even for this very restricted type of tetrahedron, knowing the area of each face does not uniquely determine the tetrahedron.
A tetrahedron with vertices at $$ (0,0,0)\\ (1/2, 0,0)\\ (0,2,0)\\ (-0.0924127, 1.9387, 0.4913857) $$ will have area $1$ for all faces.
The coordinates of that last point are approximate (and of course you can freely swap the sign of the $z$-coordinate). The actual point is given as the solution to the three equations $$ \cases{x^2 + z^2 = 1/4\\ y^2 + z^2 = 4\\ \displaystyle\left(\frac4{\sqrt{17}}(x-1/2) -\frac1{\sqrt{17}}y\right)^2 + z^2 =\frac43} $$ I placed the three first vertices first, to make sure one face was non-equilateral and had area $1$, then these three equations are exactly the equations that ensure that the other three faces have area $1$. The left-hand sides are (the squares of) the altitudes of the three remaining faces if you put the fourth vertex at $(x, y, z)$, and the right-hand sides are what those altitudes must be to ensure that the area of the corresponding face is $1$.
As you can see, I am somewhat free to make the first face whatever shape I want, as long as its area is $1$, and then I just set up the three equations to find the fourth vertex. It's possible that some extreme versions of the first face causes the resulting three equations to not have any real solution, but I have demonstrated here that at least one non-regular tetrahedron may be generated this way.
A tetrahedron can be determined by the lengths of three edges emanating from a vertex and the three angles formed by pairs of those edges. Thus, a tetrahedron admits six degrees of freedom. Four face areas are not enough to determine the shape (or even the volume) of such a figure.
As an extreme example, you can consider any rectangle a "degenerate" tetrahedron with four congruent right-triangle faces – the volume is zero, which is decidedly different than that of a regular tetrahedron. In fact, one can construct "equihedral" (equal-face-area) tetrahedra with volumes anywhere from the minimum of zero to the regular tetrahedron's maximum.
As for what other information you can use, my favorite additional parameters are the areas of the tetrahedron's three (what I call) "pseudo-faces". You can think of these geometrically as the quadrilateral projections of the tetrahedron into planes parallel to a pair of opposite sides. Each pseudo-face area is calculated by $$\text{area} = \frac12 \text{side}\cdot\text{side} \cdot \sin \text{(angle)}$$ just like any other face, except here, the "$\text{side}$"s are opposite each other, and the "$\text{angle}$" is considered the angle between the corresponding direction vectors.
Four standard faces (say, $W$, $X$, $Y$, $Z$) and three pseudo-faces ($H$, $J$, $K$) make seven area parameters, which would seem to over-determine the figure. However, the sum-of-squares identity $$W^2+X^2+Y^2+Z^2=H^2+J^2+K^2 \tag{1}$$ introduces a dependency that reduces the degrees of freedom to the expected six.
Other pseudo-face relations include a strangely familiar-looking Law of Cosines. $$\begin{align} Y^2 + Z^2 - 2 Y Z \cos A &= H^2 = W^2 + X^2 - 2 W X \cos D \\ Z^2 + X^2 - 2 Z X \cos B &= \,J^2 = W^2 + Y^2 - 2 W Y \cos E \\ X^2 + Y^2 - 2 X Y \cos C &= K^2 = W^2 + Z^2 - 2 W Z \cos F \end{align} \tag{2}$$ where each of $A$, $B$, $C$, $D$, $E$, $F$ is the dihedral angle between appropriate pairs of faces ($A$ between $Y$ and $Z$, etc).
There's also this volume formula:
$$\begin{align} 81V^4 &= H^2 J^2 K^2 - 2 (W X-Y Z)(W Y-Z X)(W Z-X Y) \\ &-H^2(W X-YZ )^2-J^2(WY-ZX)^2-K^2(WZ-XY)^2 \end{align} \tag{3}$$
If the four face areas are equal, the formula reduces to $9V^2 = HJK$, which shows that the volume of an "equihedral" tetrahedron depends upon more than those face areas.
Anyway, you can read more about these kinds of relations in my Hedronometry notes. In particular, "Heron-Like Results for Tetrahedral Volume" (PDF) includes the stuff I've described above and a bit more.
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.
It's easy to construct a tetrahedron with sides congruent to any acute isosceles triangle.
Start by taking two copies of the triangle, glue them together at the base, and then pull the apices apart until the distance between them equals the length of the base. (You can always do this if the triangles are acute, since the base length of an acute isosceles triangle is less than twice the height.) Now the sides of the triangles, together with the line between their apices, form two additional isosceles triangles that are congruent with the original one, and you thus have a tetrahedron with four congruent faces. Cut some triangles out of paper and try it if you don't believe it!
In fact, essentially the same method actually works for any acute triangle, even if it's not isosceles! The only tricky part is that now there are two distinct possible ways to glue the first two copies of the triangle together at the base, depending on whether or not you flip one of them over first, and you need to choose the way that leaves the non-equal sides of the triangles adjacent. Then proceed just like above.
(See also the answers to this related question.)
Thus, in particular, for any acute triangle with area $A$ there exists a tetrahedron with all faces congruent to that triangle, and thus also with area $A$. Unless the triangle happens to be equilateral, this tetrahedron will not be regular.
You may be interested in my paper with Petr Lisonek, Metric Invariants of Tetrahedra via Polynomial Elimination, ISSAC 2000 conference, Aberdeen (Scotland), July 2000, 217-219. We show that in general the four areas, circumradius and volume do not determine a tetrahedron, but there exist non-regular tetrahedra that are determined by the four areas and the circumradius. For example, this is the case if one face is an equilateral triangle iscribed in a great circle of the circumscribed sphere.
