# Tetrahedron Centers

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.

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:

• Opposite edge-pairs are $$(a,d)$$, $$(b,e)$$, $$(c,f)$$.
• Edges surrounding vertices $$P$$, $$A$$, $$B$$, $$C$$ are $$(a,b,c)$$, $$(a,e,f)$$, $$(d,b,f)$$, $$(d,e,c)$$.
• Edges surrounding faces $$W$$, $$X$$, $$Y$$, $$Z$$ are $$(d,e,f)$$, $$(d,b,c)$$, $$(a,e,c)$$, $$(a,b,f)$$.

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.

Edit. (17 October, 2023. Five years later!) I'm expanding this compilation with a couple of items from OP's recent questions.

Tangential Tetrahedron (question)

Let $$P'A'B'C'$$ be a tetrahedron tangent to the circumsphere of $$PABC$$ at its vertices, with $$P'$$ opposite the face tangent to $$P$$, etc. Barycentric coordinates of $$P'$$ are ...

\begin{align} -2d^2e^2f^2 &:\,d^2(-a^2d^2+b^2e^2+c^2f^2) \\ &:\,e^2(\phantom{-}a^2d^2-b^2e^2+c^2f^2) \\ &:f^2(\phantom{-}a^2d^2+b^2e^2-c^2f^2) \end{align}

Extangents Tetrahedron (mentioned in comments to this question)

Let $$P'A'B'C'$$ be the "outer" tetrahedron such that plane $$A'B'C'$$ is tangent to the exspheres of $$PABC$$ opposite vertices $$A$$, $$B$$, $$C$$; etc. Barycentric coordinates of $$P'$$ are ...

\begin{align} - W(3 + \cos A + \cos B + \cos C ) &: X(1 + \cos D + \cos B + \cos C ) \\ &: Y(1 + \cos A + \cos E + \cos C ) \\ &: Z(1 + \cos A + \cos B + \cos F ) \end{align}

The shaded tetrahedron is the extangents tetrahedron:

• These are brilliant. I usually substitute Monge for the Orthocenter. I believe when H=M when H exists. The Fermat point is on my want list. Interesting that K is the 2nd power point. We can calculate the Euler line. In a triangle, the incenter can be added and the Euler line and 3 induced Euler lines concur. Are there points like that in a Tetrahedron? Aug 15, 2018 at 19:26
• Also need to look at anticomplementary, contact, excentral, extangents, extouch, incentral, medial, orthic, and tangential tetrahedra. Aug 15, 2018 at 19:31
• So there exists a tetrahedral analogue to the “isogonal conjugate”, but what about the “isotomic conjugate” (as well as the “cyclocevian conjugate”)? Feb 15 at 4:53
• @user688486: The Wolfram Function Repo defines IsotomicConjugate. In bary-coords, the conj of $U=(r:s:t:u)$ is $U'=(1/r:1/s:1/t:1/u)$. Geometrically, cevians through $U$ and $U'$ from a tet vertex meet the opposite face in pts that are in turn isotomic conjugates with respect to that face. (This is analogous to the triangle case, where the cevians from a vertex meet the opposite edge in "isotomic conjugates" (having reciprocal bary-coords) with respect to that edge.) ... I haven't checked about cyclocevians.
– Blue
Feb 15 at 9:23
• Thanks very much! Given that the orthocentre (point of concurrence of three lines) has to be generalised to the Monge point (intersection of six or four planes), I reckon that the concept of planes effectively makes more sense than that of lines in modern tetrahedron geometry. Feb 16 at 20:54

Inspired by this previous question by OP, I investigated "hedronometric" (face-area-based) counterparts of a triangle's Congruent Isoscelizers Point and Congruent Parallelians Point. I wrote a note about them.

In the following, barycentric coordinates $\rho$, $\alpha$, $\beta$, $\gamma$ describe the point $$\frac{\rho\,P + \alpha\,A + \beta\,B + \gamma\,C}{\rho + \alpha + \beta + \gamma} \tag{\star}$$ The corresponding trilinear (er, um, tetraplanar) coordinates of the point derive by dividing each barycentric by the area of the appropriate opposite face (here, $W$, $X$, $Y$, $Z$):

$$\frac{\rho}{W} : \frac{\alpha}{X} : \frac{\beta}{Y} : \frac{\gamma}{Z} \tag{\star\star}$$

Equal-Area Parallelians Point

Definition and Theorem. A parallelian of tetrahedron $PABC$ with respect to vertex $P$ is a triangle, with vertices on the edge-lines through $P$, in a plane parallel to that of $\triangle ABC$. Any non-degenerate tetrahedron admits a unique point common to the planes of four equal-area parallelians (one for each vertex); this is the Equal-Area Parallelians Point (EAPP).

The barycentric coordinates of the EAPP are $$\rho =-\frac{2}{\sqrt{W}}+\frac{1}{\sqrt{X}}+\frac{1}{\sqrt{Y}}+\frac{1}{\sqrt{Z}}\qquad \alpha =\phantom{-}\frac{1}{\sqrt{W}}-\frac{2}{\sqrt{X}}+\frac{1}{\sqrt{Y}}+\frac{1}{\sqrt{Z}}\qquad\text{etc} \tag{1}$$

Equal-Area Trisohedralizers Point

Definition and Theorem. A trisohedralizer of a tetrahedron $PABC$ with respect to vertex $P$ is a triangle $\triangle A^\prime B^\prime C^\prime$, with vertices on the edge-lines through $P$, that serves as a base for a "trisohedral" tetrahedron. (That is, $|\triangle PB^\prime C^\prime| = |\triangle P C^\prime A^\prime| = |\triangle P A^\prime B^\prime |$.) Any non-degenerate triangle admits a unique point common to the planes of four equal-area trisohedralizers (one for each vertex); this is the Equal-Area Trisohedralizers Point (EATP).

The barycentric coordinates of the EATP are $$\rho = W \left(\;2\sqrt{W/\lambda_P} - \sqrt{X/\lambda_A} - \sqrt{Y/\lambda_B} - \sqrt{Z/\lambda_C}\;\right) \qquad \text{etc} \tag{2}$$ where $$\lambda_V^2 \;=\; 3 - 2\;\sum_{\theta} \cos\theta \tag{3}$$ with the sum taken over the three dihedral angles along edges emanating from vertex $V$.

• Neat! How about I O H T K S X M? Jul 26, 2018 at 21:52