For Triangle Centers, as seen at the Encyclopedia of Triangle Centers, the various centers each have a triangle center function $f(a,b,c)$ that is homogeneous, bisymmetric, and cyclic within barycentric or trilinear coordinates. Here are functions for centers known to Euclid.
name trilinear barycentric
X_1 incenter I 1 a angle bisectors
X_2 centroid G 1/a 1 medians
X_3 circumcenter O cos(A) a^2(b^2+c^2-a) perpendicular bisectors
X_4 orthocenter H sec(A) tan(A) altitudes
Are there tetrahedron center functions similar to the triangle center functions in barycentric or trilinear coordinates? The barycentric centroid and trilinear incenter are known.
I've made an elaborate Tetrahedron Centers demonstration, and posted code showing $4 \pi =2 \sum dihedral – \sum solid $ . At fermat point I give exact coordinates of centers for a specific tetrahedron. At Dihedral Constant Center some exact coordinates are calculated for a new tetrahedron center.
Some available items for tetrahedron $ABCD$ are :
volume, total surface area, total perimeter, dihedral constant.
solid angle $A$, face area $A$ ($\triangle BCD$), perimeter $A$ … $B$ … $C$ … $D$
edge length $ab$, dihedral angle $ab$ … $ac$ … $ad$ … $bc$ … $bd$ … $cd$
I'd like to get a list of tetrahedron center functions started.
Best Answer
For the comparatively-easy ones ...
As in my other answer, given tetrahedron $PABC$, I use barycentric coordinates $\rho$, $\alpha$, $\beta$, $\gamma$ to parameterize the point of interest as $$\frac{\rho\,P + \alpha\,A + \beta\, B + \gamma\,C}{\rho + \alpha + \beta + \gamma}$$
We'll take $V$ to be the volume of the tetrahedron. Other elements (side-lengths, face-areas, dihedral angles) are as labeled in the figure:
Of note:
See the addendum to this recent answer about how to convert edge-length expressions into hedronometric (face-area-based) forms, which may be more appropriate in this context. (Finding the "best" hedronometric form is something of a challenge.)
Centroid ($G$)
$$\rho : \alpha : \beta : \gamma \;=\; 1 : 1 : 1 : 1$$
Incenter ($I$)
$$\rho : \alpha : \beta : \gamma \;=\; W : X : Y : Z$$
Circumcenter ($O$)
$$\begin{align}\rho &=\phantom{+} a^2 d^2 \left(-d^2 + e^2 + f^2 \right) \\ &\phantom{=}+ b^2 e^2 \left(\phantom{-}d^2 - e^2 + f^2 \right) \\[4pt] &\phantom{=}+ c^2 f^2 \left(\phantom{-}d^2 + e^2 - f^2 \right) \\[4pt] &\phantom{=}- 2 d^2 e^2 f^2 \\[8pt] &= 18 V^2 - a^2\,W X \cos D - b^2\,W Y \cos E - c^2\,W Z \cos F \end{align}$$
Nine/Twelve-Point Center ($T$)
(I'm taking this to mean the center of sphere through the centroids of the faces, which also contains other points of interest, as described in the Wikipedia "Tetrahedron" entry.)
$$\begin{align}\rho \;=\; &\phantom{-\;} 2 a^2 d^2 \left(-a^2 + b^2 + c^2 \right) + a^2 d^2 \left(-d^2 + e^2 + f^2 \right) \\[4pt] &+ 2 b^2 e^2 \left(\phantom{-}a^2 - b^2 + c^2 \right) + b^2 e^2 \left(\phantom{-}d^2 - e^2 + f^2 \right) \\[4pt] &+ 2 c^2 f^2 \left(\phantom{-}a^2 + b^2 - c^2 \right) + c^2 f^2 \left(\phantom{-}d^2 + e^2 - f^2 \right) \\[4pt] &- 2 b^2 c^2 d^2 - 2 a^2 c^2 e^2 - 2 a^2 b^2 f^2 \end{align}$$
Isogonal Conjugates, and the Symmedian Point ($K$)
According to the Forum Geometricorum article "Isogonal Conjugates in a Tetrahedron" (PDF) by Sadek, et al., if isogonal conjugates have barycentric coordinates $(\rho,\alpha,\beta,\gamma)$ and $(\rho^\prime, \alpha^\prime, \beta^\prime, \gamma^\prime)$, then
$$\frac{\rho\rho^\prime}{W^2} = \frac{\alpha\alpha^\prime}{X^2} = \frac{\beta\beta^\prime}{Y^2} = \frac{\gamma\gamma^\prime}{Z^2}$$
(This result reconfirms that the Incenter ($I$) is its own isogonal conjugate. But I digress ...) Thus, $$\rho^\prime : \alpha^\prime : \beta^\prime : \gamma^\prime \;=\; \frac{W^2}{\rho} : \frac{X^2}{\alpha} = \frac{Y^2}{\beta} = \frac{Z^2}{\gamma}$$
Since the Symmedian Point ($K$) is the isogonal conjugate of the centroid ($G$), its barycentric coordinates satisfy
$$\rho : \alpha : \beta : \gamma \;=\; W^2 : X^2 : Y^2 : Z^2$$
Monge Point ($M$)
$$\begin{align} \rho &=\phantom{+}a^2d^2 \left(-a^2 + b^2 + c^2\right) \\[4pt] &\phantom{=\,}+ b^2 e^2 \left(\phantom{-}a^2 - b^2 + c^2\right) \\[4pt] &\phantom{=\,}+ c^2 f^2 \left(\phantom{-}a^2 + b^2 - c^2\right) \\[4pt] &\phantom{=\,}+ d^2 e^2 f^2 - b^2 c^2 d^2 - a^2 c^2 e^2 - a^2 b^2 f^2 \end{align}$$
Spieker Point ($S$)
The Spieker Point is the incenter of the medial tetrahedron (ie, the tetrahedron whose vertices are the centroids of the faces) of $PABC$.
By the Incenter formula above, $S$ is given by
$$\frac{W^\prime\,P^\prime + X^\prime\,A^\prime + Y^\prime\,B^\prime + Z^\prime\,C^\prime}{W^\prime + X^\prime + Y^\prime + Z^\prime}$$
for the medial tetrahedron with vertices $P^\prime = \frac13(A+B+C)$, etc, and face-areas $W^\prime = \frac19 W$, etc. This can be re-written as
$$\frac{\frac13(X+Y+Z)\,P + \frac13(W+Y+Z)\,A + \frac13(W+X+Z)\,B + \frac13(W+X+Y)\,C}{W + X + Y + Z}$$ from which we observe $$\rho : \alpha : \beta : \gamma \;=\; X + Y + Z : W + Y + Z : W + X + Z : W + Z + Y $$
Orthocenter ($H$)
In general, a tetrahedron does not have an orthocenter. Equivalent conditions that force a tetrahedron to be orthocentric are $$a^2 + d^2 = b^2 + e^2 = c^2 + f^2 $$ $$\cos A \cos D = \cos B \cos E = \cos C \cos F$$ $$\overrightarrow{PA}\perp\overrightarrow{BC} \qquad \overrightarrow{PB}\perp\overrightarrow{CA} \qquad \overrightarrow{PC}\perp\overrightarrow{AB}$$
Note: If any two orthogonality conditions hold, then the third one rides for free.
Symmetric-looking barycentric coordinates for the orthocenter of such a tetrahedron are a bit elusive. I'll have to come back to this.