Reduce 4 barycentric coordinates to 3, outside and inside of tetrahedron For any point inside of a tetrahedron, it is possible to describe it in 3 barycentric parameters (the 4th can be deduced from the other 3). Is this reduction from 4 to 3 coordinates also possible for any other point outside the tetrahedron? If so, how do I find which 3 barycentric coordinates describe this point and how I can deduce the 4th?
 A: The three-dimensional barycentric coordinates of a point $P$, with respect of tetrahedron $ABCD$, are either:

*

*the four volumes $[ABCP], [ABPD], [APCD], [PBCD]$, which add to $[ABCD]$ for any point $P$ inside the tetrahedron, or

*the normalized versions, obtained by dividing by $[ABCD]$, which add up to $1$ for any point $P$ inside the tetrahedron.

We can generalize either of these definitions to points outside the tetrahedron if we use a signed volume. There are many equivalent definitions. For example, assuming that the signed volume of $ABCD$ is positive, we can define the signed volume of $ABCP$ to be positive if $P$ is on the same side of plane $ABC$ as $D$, and negative if it is on the other side. Do the same for the other volumes.
Then, the un-normalized signed volumes will still sum to $[ABCD]$, and the normalized signed volumes will still sum to $1$, and so any three of the coordinates tell you the fourth under either convention.
More algebraically, the volume of tetrahedron $ABCD$ is given by the formula
$$
   \frac16 \det\begin{bmatrix}1 & x_a & y_a & z_a \\ 1 & x_b & y_b & z_b \\ 1 & x_c & y_c & z_c \\ 1 & x_d & y_d & z_d\end{bmatrix}.
$$
Usually, we take the absolute value, but if we don't, then we get one notion of signed volume. Then the coordinates for point $P$ are determined by the four volumes given by the four determinants
$$
\scriptsize
   \left(\frac16 \det\begin{bmatrix}1 & x_a & y_a & z_a \\ 1 & x_b & y_b & z_b \\ 1 & x_c & y_c & z_c \\ 1 & x_p & y_p & z_p\end{bmatrix}, \frac16 \det\begin{bmatrix}1 & x_a & y_a & z_a \\ 1 & x_b & y_b & z_b \\ 1 & x_p & y_p & z_p \\ 1 & x_d & y_d & z_d\end{bmatrix}, \frac16 \det\begin{bmatrix}1 & x_a & y_a & z_a \\ 1 & x_p & y_p & z_p \\ 1 & x_c & y_c & z_c \\ 1 & x_d & y_d & z_d\end{bmatrix}, \frac16 \det\begin{bmatrix}1 & x_p & y_p & z_p \\ 1 & x_b & y_b & z_b \\ 1 & x_c & y_c & z_c \\ 1 & x_d & y_d & z_d\end{bmatrix}\right).
$$
We know that the four volumes add up to the volume of $ABCD$ inside the tetrahedron, because the four tetrahedra $ABCP, ABPD, APCD, PBCD$ partition $ABCD$. But the sum of these four determinants is a polynomial in $x_p, y_p, z_p$ (taking the points $A,B,C,D$ as constant). If this polynomial has a constant value for all $(x_p, y_p, z_p)$ inside the tetrahedron, then it must be constant on all of $\mathbb R^3$, which is why the signed volumes continue to behave outside the tetrahedron.
