Inspired by this recent MSE problem I am working on the problem of solving the tetrahedron, given outward normal vectors to its faces, and the area of one face.
That is, I want to find all the edges of the tetrahedron given the above-mentioned information.
A detailed solution on how to do this is highly appreciated.
On
Let's say the vertices are $\vec{v}_1 = (x_1, y_1, z_1)$, $\vec{v}_2 = (x_2, y_2, z_2)$, $\vec{v}_3 = (x_3, y_3, z_3)$, and $\vec{v}_4 = (x_4, y_4, z_4)$. Then, the four outwards-facingnd normal vectors are $$\begin{aligned} \vec{n}_{123} &= (\vec{v}_2 - \vec{v}_1) \times (\vec{v}_3 - \vec{v}_1) \\ \vec{n}_{124} &= -(\vec{v}_2 - \vec{v}_1) \times (\vec{v}_4 - \vec{v}_1) \\ \vec{n}_{134} &= (\vec{v}_3 - \vec{v}_1) \times (\vec{v}_4 - \vec{v}_1) \\ \vec{n}_{234} &= -(\vec{v}_3 - \vec{v}_2) \times (\vec{v}_4 - \vec{v}_2) \\ \end{aligned} \tag{1}\label{1}$$ where $\times$ denotes vector cross product, and only the direction of the normal vector matters. Their magnitudes are twice the area of the corresponding face, i.e. $$A_{123} = \frac{1}{2}\left\lVert \vec{n}_{123} \right\rVert, ~ ~ ~ A_{124} = \frac{1}{2}\left\lVert \vec{n}_{124} \right\rVert, ~ ~ ~ A_{134} = \frac{1}{2}\left\lVert \vec{n}_{134} \right\rVert, ~ ~ ~ A_{234} = \frac{1}{2}\left\lVert \vec{n}_{234} \right\rVert$$ because the magnitude of a cross product of two vectors is the area of the corresponding parallelogram, which in turn has exactly twice the area of a triangle with two of the edges matching the two vectors.
I would start by scaling the given normal vectors so that their magnitude matches twice the area of that face; if $\vec{\eta}$ is one of the given normal vectors (with unspecified magnitude), and $\lambda$ the area of that face, then use $$\vec{n} = 2 \lambda \frac{\vec{\eta}}{\left\lVert\vec{\eta}\right\rVert} = 2 \lambda \frac{\vec{\eta}}{\sqrt{\vec{\eta}\cdot\vec{\eta}}}$$ for the corresponding normal vector; it embodies all the data needed for that particular face.
Then, because of how we chose the magnitude, $\eqref{1}$ is exactly true.
Next, I would arbitrarily choose the first vertex to be at origin, $\vec{v}_1 = \vec{0} = (0, 0, 0)$, as the normals or the areas do not define its location; it can be translated if need be later on.
At this point, we have a system of nine unknowns, and $3 \times 4 = 12$ linear equations. Write them out component-wise, i.e. $$\left\lbrace \begin{aligned} x_{123} &= (y_2 - y_1)(z_3 - z_1) - (z_2 - z_1)(y_3 - y_1) \\ y_{123} &= (z_2 - z_1)(x_3 - x_1) - (x_2 - x_1)(z_3 - z_1) \\ z_{123} &= (x_2 - x_1)(y_3 - y_1) - (y_2 - y_1)(x_3 - x_1) \\ ~ & ~ ~ \vdots \\ \end{aligned} \right.$$ where $\vec{n}_{123} = (x_{123}, y_{123}, z_{123})$, and start solving. Good luck!
On
Number the faces $1, 2, 3, 4$. The corresponding unit normal vectors are $ n_1, n_2, n_3, n_4 $
Assume the area of the first face $F_1$ is known, then from Minkowski's theorem we know that
$ F_1 n_1 + F_2 n_2 + F_3 n_3 + F_4 n_4 = 0 $
Which is a linear system of $3 $ equations in the $3$ unknowns $F_2, F_3, F_4$ and can be solved readily.
Now we have $F_1, F_2, F_3, F_4 $ all known.
Labelling the edges from the faces it lies on, we have
$F_1$ contains edges $e_{12}, e_{13}, e_{14} $
etc.
Define the unit vectors $w_{ij} = \dfrac{n_i \times n_j} {\| n_i \times n_j \| } $
Then $ e_{ij} $ is along the unit vector $w_{ij}$
Now
$F_1 = \dfrac{1}{2} \ e_{12} e_{13} \ \| w_{12} \times w_{13} \| = \dfrac{1}{2} \ e_{12} \ e_{14} \ \| w_{12} \times w_{14} \| = \dfrac{1}{2} \ e_{13} \ e_{14} \ \| w_{13} \times w_{14} \| $
It follows that
$\dfrac{1}{e_{14}} \| w_{12} \times w_{13} \| = \dfrac{1}{e_{13}} \| w_{12} \times w_{14} \| = \dfrac{1}{e_{12}} \| w_{13} \times w_{14} \| = \dfrac{1}{t_1} $
Which gives
$(e_{12}, e_{13}, e_{14}) = t_1 ( \| w_{13} \times w_{14} \|, \| w_{12} \times w_{14} \| , | w_{12} \times w_{13} \| ) $
now we can solve for $t_1$ because
$F_1 = \dfrac{1}{2} \ t_1^2 \ \| w_{12} \times w_{13} \| \ \| w_{13} \times w_{14} \| \ \| w_{12} \times w_{14} \| $
so $t_1$ is now known, and so are $e_{12}, e_{13}, e_{14}$.
Similarly,
$F_2 = \dfrac{1}{2} \ t_2^2 \ \| w_{21} \times w_{23} \| \ \| w_{21} \times w_{24} \| \ \| w_{23} \times w_{24} \| $
hence , $e_{23}, e_{24} $ can be found as follows:
$(e_{23}, e_{24}) = t_2 ( \| w_{21} \times w_{24} \| , \| w_{21} \times w_{23} \| )$
and finally,
$F_3 = \dfrac{1}{2} \ t_3^2 \ \| w_{31} \times w_{32} \| \ \| w_{31} \times w_{34} \| \ \| w_{32} \times w_{34} \| $
From which,
$e_{34} = t_3 \ \| w_{31} \times w_{32} \|$
Thus, all the edge lengths have been found.
Consider a tetrahedron with concurrent edges $a$, $b$, $c$ opposite respective edges $d$, $e$, $f$. Define faces $W:=\triangle def$ (whose area we'll consider known), $X:=\triangle dbc$, $Y:=\triangle aec$, $Z:=\triangle abf$, and let $A$, $B$, $C$, $D$, $E$, $F$ be the dihedral angles along respective edges.
Let $w$, $x$, $y$, $z$ be the known unit outward normals to faces $W$, $X$, $Y$, $Z$. Then we have $$\begin{align} \cos A = -y\cdot z \qquad \cos B &= -z\cdot x \qquad \cos C = -x\cdot y \\ \cos D = -w\cdot x \qquad \cos E &= -w\cdot y \qquad \cos F = -w\cdot z \end{align}$$ So, all dihedral angles are known.
Let $O$ be the vertex opposite $W$. Project $O$ to $O'$ in that face, and project $O'$ to $O''$ on (possibly-extended) edge $d$. Then $\angle OO''O'$ matches dihedral angle $D$. Defining $h:=|OO''|$, we can write $|OO'|=h\sin D$ and $|O'O''|=h\cos D$. The latter is the altitude to base $d$ of a subtriangle of face $W$; this subtriangle has area $\frac12dh\cos D$. But $h$ itself is the altitude to base $d$ of face $X$, so that $X=\frac12dh$; thus, the subtriangle has area $X\cos D$. Likewise $O'$ determines subtriangles along edges $e$ and $f$ with areas $Y\cos E$ and $Z \cos F$; thus, we have a decomposition of face $W$: $$W = X \cos D + Y \cos E + Z \cos F$$ (The reader can verify that this makes sense even with obtuse dihedral angles.) In a similar fashion, abbreviating $\cos\theta$ as $\ddot{\theta}$, we have $$\begin{align} X &= W\ddot{D} + Y \ddot{C} + Z \ddot{B} \\[4pt] Y &= X\ddot{C} + W \ddot{E} + Z \ddot{A} \\[4pt] Z &= X\ddot{B} + Y \ddot{A} + W \ddot{F} \end{align}$$ Solving the linear system for $X$, $Y$, $Z$ gives $$\begin{align} X &= W\;\frac{\ddot{A}\;(-\ddot{A}\ddot{D}+\ddot{B}\ddot{E}+\ddot{C}\ddot{F}) + \ddot{D}+\ddot{B}\ddot{F}+\ddot{C}\ddot{E}}{1-2\ddot{A} \ddot{B} \ddot{C} - \ddot{A}^2-\ddot{B}^2-\ddot{C}^2}\\[8pt] Y &= W\;\frac{\ddot{B}\;(\phantom{-}\ddot{A}\ddot{D}-\ddot{B}\ddot{E}+\ddot{C}\ddot{F}) + \ddot{E}+\ddot{C}\ddot{D}+\ddot{A}\ddot{F}}{1-2\ddot{A} \ddot{B} \ddot{C} - \ddot{A}^2-\ddot{B}^2-\ddot{C}^2}\\[8pt] Z &= W\;\frac{\ddot{C}\;(\phantom{-}\ddot{A}\ddot{D}+\ddot{B}\ddot{E}-\ddot{C}\ddot{F}) + \ddot{F}+\ddot{A}\ddot{E}+\ddot{B}\ddot{D}}{1-2\ddot{A} \ddot{B} \ddot{C} - \ddot{A}^2-\ddot{B}^2-\ddot{C}^2} \end{align}$$ So, all face-areas are known. Then also volume, via this formula: $$V^4 = \frac{4}{81} X^2 Y^2 Z^2 \left(\; 1-2\ddot{A} \ddot{B} \ddot{C} - \ddot{A}^2-\ddot{B}^2-\ddot{C}^2\;\right)$$
We can also calculate volume from face $W$ and altitude $OO'$: $$V = \frac13 W h\sin D = \frac13 W\frac{2X}{d}\sin D \qquad\to\qquad d = \frac23 \frac{WX}{V}\sin D$$
So, edge-length $d$ is known. Likewise, $$ a = \frac23\frac{YZ}{V}\sin A \qquad b = \frac23\frac{ZX}{V}\sin B \qquad c = \frac23\frac{XY}{V}\sin C$$ $$e = \frac23\frac{WY}{V}\sin E \qquad f = \frac23\frac{WZ}{V}\sin F$$ The tetrahedron is solved. $\square$