Volume of a tetrahedron, given areas of 6 triangles Given an arbitrary tetrahedron, I always have an inscribed sphere inside, which contacts the tetrahedron at four new points (one on each of the four faces). Each face, which is itself a triangle, is now divided into $3$ small triangles, so I have $12$ such small triangles. It turns out that each pair of triangles with a common edge also has a common area; so I have $6$ different areas for these small triangles (one area corresponding to each edge).
The question is: If I start with the areas of those small triangles, is my tetrahedron fixed? If so, is there a simple expression for the volume of the tetrahedron in terms of those $6$ small areas?
Thank you so much for any attention to this problem.

 A: Introducing conventions familiar to me, I'll consider tetrahedron $OABC$ to have edges and face-angles about $O$, and face-areas
$$a := |OA| \qquad b := |OB| \qquad c := |OC| \\[6pt]
\alpha := \angle BOC \qquad \beta := \angle COA \qquad \gamma := \angle AOB \\[6pt]
W := |\triangle ABC| \qquad X := |\triangle OBC| \qquad Y := |\triangle OCA| \qquad Z:=|\triangle OAB|$$
Without fear of confusion, I'll also use $A$, $B$, $C$ to refer to the dihedral angles along edges $a$, $b$, $c$.
As OP notes, insphere-determined sub-triangles that share an edge have the same area. Let $\sigma_a$, $\sigma_b$, $\sigma_c$ be the areas of the triangles sharing respective edges $a$, $b$, $c$; further, let $\sigma_d$, $\sigma_e$, $\sigma_f$ be the areas of the triangles sharing edges $|BC|$, $|CA|$, $|AB|$. (I typically denote those edges $d$, $e$, $f$, but that's not important here.) Of course, the full face-areas are simple sums of these sub-areas:
$$W = \sigma_d + \sigma_e + \sigma_f \qquad X = \sigma_d+\sigma_b+\sigma_c \qquad \text{etc} \tag1$$
Below I show that $\cos A$, $\cos B$, $\cos C$ are also expressible in terms of the $\sigma$s, which in turn guarantees a unique volume, via
$$81V^4 = 4X^2Y^2Z^2(1-2\cos A\cos B\cos C-\cos^2 A-\cos^2 B-\cos^2 C) \tag2$$

To get at the cosine relations, let's coordinatize:
$$O=(0,0,0) \qquad A = (a,0,0) \qquad B = (b \cos\alpha,b\sin\alpha, 0) \\[6pt]
C = (c\cos\beta,c\sin\beta\cos A,c\sin\beta\sin A)$$
It is "known" that the incenter of $OABC$ is given by
$$I = \frac{W O + X A + Y B + Z C}{W+X+Y+Z} \tag3$$
If $P$ is the point where the insphere touches face $Z$ (the $xy$-plane), then $$\begin{align}
\sigma_a := |\triangle OAP| &= \frac12|OA|P_y = \frac12 a \frac{Yb\sin\gamma+ Z c \sin\beta \cos A}{W+X+Y+Z} \\[8pt]
&= \frac{YZ(1+\cos A)}{W+X+Y+Z} \tag4
\end{align}$$
Likewise,
$$\sigma_b = \frac{ZX(1+\cos B)}{W+X+Y+Z} \qquad \sigma_c = \frac{X Y(1+\cos C)}{W+X+Y+Z} \tag5$$

Substituting $X$, $Y$, $Z$, $\cos A$, $\cos B$, $\cos C$ in terms of the $\sigma$s into $(2)$, we find

$$\begin{align}
81 V^4 =\, &16\,(\sigma_a + \sigma_b + \sigma_c + \sigma_d + \sigma_e + \sigma_f)^2 \\ &\cdot(p+q+r) (-p+q+r) (p-q+r) (p+q-r)\end{align} \tag{$\star$}$$
where
$$p := \sqrt{\sigma_a\sigma_d} \qquad q := \sqrt{\sigma_b\sigma_e} \qquad r := \sqrt{\sigma_c\sigma_f}$$

As a sanity check, a regular tetrahedron with side-length $s$ has equilateral faces of area $W=X=Y=Z=\frac14s^2\sqrt{3}$ and sub-faces of area $\sigma_a=\cdots=\sigma_f = \frac1{12}s^2\sqrt{3}=p=q=r$. As a result, $(\star)$ yields
$$V^4 = \frac{16}{81} \left(6\cdot \frac1{12}s^2\sqrt{3}\right)^2\left(3\cdot \frac1{12}s^2\sqrt{3}\right)\left(\frac1{12}s^2\sqrt{3}\right)^3 = \left(\frac1{12} \sqrt{2} s^3\right)^4$$
as expected.
