Editing Tetrahedron (section)

== General properties ==
===Volume===
The volume of a tetrahedron can be obtained in many ways. It can be given by using the formula of the pyramid's volume: 
<math display="block"> V = \frac{1}{3}Ah. </math>
where <math> A </math> is the [[base (geometry)|base]]' area and <math> h </math> is the height from the base to the apex. This applies for each of the four choices of the base, so the distances from the apices to the opposite faces are inversely proportional to the areas of these faces. Another way is by dissecting a triangular prism into three pieces.{{sfn|Alsina|Nelsen|2015|p=[https://books.google.com/books?id=FEl2CgAAQBAJ&pg=PA67 67]}}

Given the vertices of a tetrahedron in the following:
<math display="block"> \begin{align}
 \mathbf{a} &= (a_1, a_2, a_3), \\
 \mathbf{b} &= (b_1, b_2, b_3), \\
 \mathbf{c} &= (c_1, c_2, c_3), \\
 \mathbf{d} &= (d_1, d_2, d_3).
\end{align} </math>
The volume of a tetrahedron can be ascertained in terms of a [[determinant]] <math display="inline"> \frac{1}{6} \det(\mathbf{a} - \mathbf{d}, \mathbf{b} - \mathbf{d}, \mathbf{c} - \mathbf{d}) </math>,{{sfn|Fekete|1985|p=[https://books.google.com/books?id=3_AIXBO11bEC&pg=PA68 68]}} or any other combination of pairs of vertices that form a simply connected [[Graph (discrete mathematics)|graph]]. Comparing this formula with that used to compute the volume of a [[parallelepiped]], we conclude that the volume of a tetrahedron is equal to {{sfrac|1|6}} of the volume of any parallelepiped that shares three converging edges with it.

The absolute value of the scalar triple product can be represented as the following absolute values of determinants:
:<math>6 \cdot V =\begin{Vmatrix}
\mathbf{a} & \mathbf{b} & \mathbf{c}
\end{Vmatrix}</math>{{pad|2em}}or{{pad|1em}}<math>6 \cdot V =\begin{Vmatrix}
\mathbf{a} \\ \mathbf{b} \\ \mathbf{c}
\end{Vmatrix}</math>{{pad|2em}}where{{pad|1em}}<math>\begin{cases}\mathbf{a} = (a_1,a_2,a_3), \\
 \mathbf{b} = (b_1,b_2,b_3), \\
 \mathbf{c} = (c_1,c_2,c_3), \end{cases}</math>{{pad|1em}}are expressed as row or column vectors.

Hence
:<math>36 \cdot V^2 =\begin{vmatrix}
\mathbf{a^2} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\
\mathbf{a} \cdot \mathbf{b} & \mathbf{b^2} & \mathbf{b} \cdot \mathbf{c} \\
\mathbf{a} \cdot \mathbf{c} & \mathbf{b} \cdot \mathbf{c} & \mathbf{c^2}
\end{vmatrix}</math>{{pad|1em}}where{{pad|1em}}<math>\begin{cases}\mathbf{a} \cdot \mathbf{b} = ab\cos{\gamma}, \\ \mathbf{b} \cdot \mathbf{c} = bc\cos{\alpha}, \\ \mathbf{a} \cdot \mathbf{c} = ac\cos{\beta}. \end{cases}</math>

where <math>a = \begin{Vmatrix} \mathbf{a} \end{Vmatrix} </math>, <math>b = \begin{Vmatrix} \mathbf{b} \end{Vmatrix} </math>, and <math>c = \begin{Vmatrix} \mathbf{c} \end{Vmatrix} </math>, which gives
:<math>V = \frac {abc} {6} \sqrt{1 + 2\cos{\alpha}\cos{\beta}\cos{\gamma}-\cos^2{\alpha}-\cos^2{\beta}-\cos^2{\gamma}}, \,</math>
where ''α'', ''β'', ''γ'' are the plane angles occurring in vertex '''d'''. The angle ''α'', is the angle between the two edges connecting the vertex '''d''' to the vertices '''b''' and '''c'''. The angle ''β'', does so for the vertices '''a''' and '''c''', while ''γ'', is defined by the position of the vertices '''a''' and '''b'''.

If we do not require that '''d''' = 0 then
:<math>6 \cdot V = \left| \det \left( \begin{matrix}
a_1 & b_1 & c_1 & d_1 \\
a_2 & b_2 & c_2 & d_2 \\
a_3 & b_3 & c_3 & d_3 \\
 1  &  1  &  1  &  1
\end{matrix} \right) \right|\,.</math>

Given the distances between the vertices of a tetrahedron the volume can be computed using the [[Cayley–Menger determinant]]:
:<math>288 \cdot V^2 =
\begin{vmatrix}
  0 & 1        & 1        & 1        & 1        \\
  1 & 0        & d_{12}^2 & d_{13}^2 & d_{14}^2 \\
  1 & d_{12}^2 & 0        & d_{23}^2 & d_{24}^2 \\
  1 & d_{13}^2 & d_{23}^2 & 0        & d_{34}^2 \\
  1 & d_{14}^2 & d_{24}^2 & d_{34}^2 & 0
\end{vmatrix}</math>
where the subscripts {{nowrap|''i'', ''j'' ∈ {1, 2, 3, 4}|}} represent the vertices {{nowrap|{'''a''', '''b''', '''c''', '''d'''}|}} and ''d{{sub|ij}}'' is the pairwise distance between them – i.e., the length of the edge connecting the two vertices. A negative value of the determinant means that a tetrahedron cannot be constructed with the given distances. This formula, sometimes called [[Tartaglia's formula]], is essentially due to the painter [[Piero della Francesca]] in the 15th century, as a three-dimensional analogue of the 1st century [[Heron's formula]] for the area of a triangle.<ref>[http://www.mathpages.com/home/kmath664/kmath664.htm "Simplex Volumes and the Cayley-Menger Determinant"], MathPages.com</ref>

Let <math> a </math>, <math> b </math>, and <math> c </math> be the lengths of three edges that meet at a point, and <math> x </math>, <math> y </math>, and <math> z </math> be those of the opposite edges. The volume of the tetrahedron <math> V </math> is:{{sfn|Kahan|2012|p=11}}
<math display="block"> V = \frac{\sqrt{4 a^2 b^2 c^2-a^2 X^2-b^2 Y^2-c^2 Z^2+X Y Z}}{12}</math>
where
<math display="block"> \begin{align}
 X &= b^2+c^2-x^2, \\
 Y &= a^2+c^2-y^2, \\
 Z &= a^2+b^2-z^2.
\end{align}</math>
The above formula uses six lengths of edges, and the following formula uses three lengths of edges and three angles.{{sfn|Kahan|2012|p=11}}
<math display="block"> V = \frac{abc}{6} \sqrt{1 + 2\cos{\alpha}\cos{\beta}\cos{\gamma}-\cos^2{\alpha}-\cos^2{\beta}-\cos^2{\gamma}}</math>

[[Image:Six edge-lengths of Tetrahedron.png|class=skin-invert-image|right|thumb|240px|Six edge-lengths of Tetrahedron]]
The volume of a tetrahedron can be ascertained by using the Heron formula. Suppose <math> U </math>, <math> V </math>, <math> W </math>, <math> u </math>. <math> v </math>, and <math> w </math> are the lengths of the tetrahedron's edges as in the following image. Here, the first three form a triangle, with <math> u </math> opposite <math> U </math>, <math> v </math> opposite <math> V </math>, and <math> w </math> opposite <math> W </math>. Then,
<math display="block"> V = \frac{\sqrt {\,( - p + q + r + s)\,(p - q + r + s)\,(p + q - r + s)\,(p + q + r - s)}}{192\,u\,v\,w}</math>
where
<math display="block"> \begin{align}
 p = \sqrt{xYZ}, &\quad q = \sqrt{yZX}, \\
 r = \sqrt{zXY}, &\quad s = \sqrt {xyz},
\end{align} </math>
and
<math display="block"> \begin{align}
 X = (w - U + v)(U + v + w), &\quad x = (U - v + w)(v - w + U), \\
 Y = (u - V + w)(V + w + u), &\quad y = (V - w + u)(w - u + V), \\
 Z = (v - W + u)(W + u + v), &\quad z = (W - u + v)\,(u - v + W).
\end{align}
</math>

Any plane containing a bimedian (connector of opposite edges' midpoints) of a tetrahedron [[bisection|bisects]] the volume of the tetrahedron.{{sfn|Bottema|1969}}

For tetrahedra in [[hyperbolic space]] or in three-dimensional [[elliptic geometry]], the [[dihedral angle]]s of the tetrahedron determine its shape and hence its volume. In these cases, the volume is given by the [[Murakami–Yano formula]], after Jun Murakami and Masakazu Yano.{{sfn|Murakami|Yano|2005}} However, in Euclidean space, scaling a tetrahedron changes its volume but not its dihedral angles, so no such formula can exist.

Any two opposite edges of a tetrahedron lie on two [[skew lines]], and the distance between the edges is defined as the distance between the two skew lines. Let <math> d </math> be the distance between the skew lines formed by opposite edges <math> a </math> and <math> \mathbf{b} - \mathbf{c} </math> as calculated [[Skew lines#Distance between two skew lines|here]]. Then another formula for the volume of a tetrahedron <math> V </math> is given by
<math display="block"> V = \frac {d |(\mathbf{a} \times \mathbf{(b-c)})|}{6}. </math>

===Properties analogous to those of a triangle===
The tetrahedron has many properties analogous to those of a triangle, including an insphere, circumsphere, medial tetrahedron, and exspheres. It has respective centers such as incenter, circumcenter, excenters, [[Spieker circle|Spieker center]] and points such as a centroid. However, there is generally no orthocenter in the sense of intersecting altitudes.<ref>{{Cite journal |last1=Havlicek |first1=Hans |last2=Weiß |first2=Gunter |title=Altitudes of a tetrahedron and traceless quadratic forms |journal=[[American Mathematical Monthly]] |volume=110 |issue=8 |pages=679–693 |year=2003 |url=http://www.geometrie.tuwien.ac.at/havlicek/pub/hoehen.pdf |doi=10.2307/3647851 |jstor=3647851 |arxiv=1304.0179 }}</ref>

[[Gaspard Monge]] found a center that exists in every tetrahedron, now known as the '''Monge point''': the point where the six midplanes of a tetrahedron intersect. A midplane is defined as a plane that is orthogonal to an edge joining any two vertices that also contains the centroid of an opposite edge formed by joining the other two vertices. If the tetrahedron's altitudes do intersect, then the Monge point and the orthocenter coincide to give the class of [[orthocentric tetrahedron]].

An orthogonal line dropped from the Monge point to any face meets that face at the midpoint of the line segment between that face's orthocenter and the foot of the altitude dropped from the opposite vertex.

A line segment joining a vertex of a tetrahedron with the [[centroid]] of the opposite face is called a ''median'' and a line segment joining the midpoints of two opposite edges is called a ''bimedian'' of the tetrahedron. Hence there are four medians and three bimedians in a tetrahedron. These seven line segments are all [[Concurrent lines|concurrent]] at a point called the ''centroid'' of the tetrahedron.<ref>Leung, Kam-tim; and Suen, Suk-nam; "Vectors, matrices and geometry", Hong Kong University Press, 1994, pp. 53–54</ref> In addition the four medians are divided in a 3:1 ratio by the centroid (see [[Commandino's theorem]]). The centroid of a tetrahedron is the midpoint between its Monge point and circumcenter. These points define the ''Euler line'' of the tetrahedron that is analogous to the [[Euler line]] of a triangle.

The [[nine-point circle]] of the general triangle has an analogue in the circumsphere of a tetrahedron's medial tetrahedron. It is the '''twelve-point sphere''' and besides the centroids of the four faces of the reference tetrahedron, it passes through four substitute ''Euler points'', one third of the way from the Monge point toward each of the four vertices. Finally it passes through the four base points of orthogonal lines dropped from each Euler point to the face not containing the vertex that generated the Euler point.<ref>{{Cite book |last1=Outudee |first1=Somluck |last2=New |first2=Stephen |title=The Various Kinds of Centres of Simplices |publisher=Dept of Mathematics, Chulalongkorn University, Bangkok |url=http://www.math.sc.chula.ac.th/ICAA2002/pages/Somluck_Outudee.pdf |url-status=bot: unknown |archive-url=https://web.archive.org/web/20090227143222/http://www.math.sc.chula.ac.th/ICAA2002/pages/Somluck_Outudee.pdf |archive-date=27 February 2009}}</ref>

The center ''T'' of the twelve-point sphere also lies on the Euler line. Unlike its triangular counterpart, this center lies one third of the way from the Monge point ''M'' towards the circumcenter. Also, an orthogonal line through ''T'' to a chosen face is coplanar with two other orthogonal lines to the same face. The first is an orthogonal line passing through the corresponding Euler point to the chosen face. The second is an orthogonal line passing through the centroid of the chosen face. This orthogonal line through the twelve-point center lies midway between the Euler point orthogonal line and the centroidal orthogonal line. Furthermore, for any face, the twelve-point center lies at the midpoint of the corresponding Euler point and the orthocenter for that face.

The radius of the twelve-point sphere is one third of the circumradius of the reference tetrahedron.

There is a relation among the angles made by the faces of a general tetrahedron given by<ref>{{cite web |first=Daniel |last=Audet |title=Déterminants sphérique et hyperbolique de Cayley-Menger |url=http://archimede.mat.ulaval.ca/amq/bulletins/mai11/Chronique_note_math.mai11.pdf |publisher=Bulletin AMQ |date=May 2011 }}</ref>
:<math>\begin{vmatrix}  -1 & \cos{(\alpha_{12})} & \cos{(\alpha_{13})} & \cos{(\alpha_{14})}\\
\cos{(\alpha_{12})} & -1 & \cos{(\alpha_{23})} & \cos{(\alpha_{24})} \\
\cos{(\alpha_{13})} & \cos{(\alpha_{23})} & -1 & \cos{(\alpha_{34})} \\
\cos{(\alpha_{14})} & \cos{(\alpha_{24})} & \cos{(\alpha_{34})} & -1 \\ \end{vmatrix} = 0\,</math>

where ''α{{sub|ij}}'' is the angle between the faces ''i'' and ''j''.

The [[geometric median]] of the vertex position coordinates of a tetrahedron and its isogonic center are associated, under circumstances analogous to those observed for a triangle.  [[Lorenz Lindelöf]] found that, corresponding to any given tetrahedron is a point now known as an isogonic center, ''O'', at which the solid angles subtended by the faces are equal, having a common value of π [[Steradian|sr]], and at which the angles subtended by opposite edges are equal.<ref>{{cite journal |last1=Lindelof |first1=L. |title=Sur les maxima et minima d'une fonction des rayons vecteurs menés d'un point mobile à plusieurs centres fixes |journal=Acta Societatis Scientiarum Fennicae |date=1867 |volume=8 |issue=Part 1 |pages=189–203}}</ref>  A solid angle of π sr is one quarter of that subtended by all of space.  When all the solid angles at the vertices of a tetrahedron are smaller than π sr, ''O'' lies inside the tetrahedron, and because the sum of distances from ''O'' to the vertices is a minimum, ''O'' coincides with the [[geometric median]], ''M'', of the vertices.  In the event that the solid angle at one of the vertices, ''v'', measures exactly π sr, then ''O'' and ''M'' coincide with ''v''.  If however, a tetrahedron has a vertex, ''v'', with solid angle greater than π sr, ''M'' still corresponds to ''v'', but ''O'' lies outside the tetrahedron.

===Geometric relations===
A tetrahedron is a 3-[[simplex]]. Unlike the case of the other Platonic solids, all the vertices of a regular tetrahedron are equidistant from each other (they are the only possible arrangement of four equidistant points in 3-dimensional space, for an example in [[electromagnetism]] cf. [[Thomson problem]]).

The above embedding divides the cube into five tetrahedra, one of which is regular. In fact, five is the minimum number of tetrahedra required to compose a cube. To see this, starting from a base tetrahedron with 4 vertices, each added tetrahedra adds at most 1 new vertex, so at least 4 more must be added to make a cube, which has 8 vertices.

Inscribing tetrahedra inside the regular [[Polyhedral compound|compound of five cubes]] gives two more regular compounds, containing five and ten tetrahedra.

Regular tetrahedra cannot [[Honeycomb (geometry)|tessellate space]] by themselves, although this result seems likely enough that [[Aristotle]] claimed it was possible. However, two regular tetrahedra can be combined with an octahedron, giving a [[rhombohedron]] that can tile space as the [[tetrahedral-octahedral honeycomb]].

On otherhand, several irregular tetrahedra are known, of which copies can tile space, for instance the [[#Orthoschemes|characteristic orthoscheme of the cube]] and the [[#Disphenoid|disphenoid]] of the [[disphenoid tetrahedral honeycomb]]. The complete list remains an open problem.<ref>{{Cite journal
 | doi = 10.2307/2689983
 | last = Senechal | first = Marjorie | author-link = Marjorie Senechal
 | title = Which tetrahedra fill space?
 | year = 1981
 | journal = [[Mathematics Magazine]]
 | volume = 54
 | issue = 5
 | pages = 227–243
 | publisher = Mathematical Association of America
 | jstor = 2689983
 }}</ref>

If one relaxes the requirement that the tetrahedra be all the same shape, one can tile space using only tetrahedra in many different ways. For example, one can divide an octahedron into four identical tetrahedra and combine them again with two regular ones. (As a side-note: these two kinds of tetrahedron have the same volume.)

The tetrahedron is unique among the [[uniform polyhedron|uniform polyhedra]] in possessing no parallel faces.

===A law of sines for tetrahedra and the space of all shapes of tetrahedra===
[[Image:tetra.png|class=skin-invert-image|248px]]

{{main|Trigonometry of a tetrahedron}}

A corollary of the usual [[law of sines]] is that in a tetrahedron with vertices ''O'', ''A'', ''B'', ''C'', we have
:<math>\sin\angle OAB\cdot\sin\angle OBC\cdot\sin\angle OCA = \sin\angle OAC\cdot\sin\angle OCB\cdot\sin\angle OBA.\,</math>

One may view the two sides of this identity as corresponding to clockwise and counterclockwise orientations of the surface.

Putting any of the four vertices in the role of ''O'' yields four such identities, but at most three of them are independent: If the "clockwise" sides of three of them are multiplied and the product is inferred to be equal to the product of the "counterclockwise" sides of the same three identities, and then common factors are cancelled from both sides, the result is the fourth identity.

Three angles are the angles of some triangle if and only if their sum is 180° (π radians). What condition on 12 angles is necessary and sufficient for them to be the 12 angles of some tetrahedron? Clearly the sum of the angles of any side of the tetrahedron must be 180°. Since there are four such triangles, there are four such constraints on sums of angles, and the number of [[Degrees of freedom (statistics)|degrees of freedom]] is thereby reduced from 12 to 8. The four relations given by this sine law further reduce the number of degrees of freedom, from 8 down to not 4 but 5, since the fourth constraint is not independent of the first three. Thus the space of all shapes of tetrahedra is 5-dimensional.<ref>{{Cite journal |title=Is There a "Most Chiral Tetrahedron"? |first1=André |last1=Rassat |first2=Patrick W. |last2=Fowler |journal=Chemistry: A European Journal |volume=10 |issue=24 |pages=6575–6580 |year=2004 |doi=10.1002/chem.200400869 |pmid=15558830 }}</ref>

=== Law of cosines for tetrahedra ===
{{main|Trigonometry of a tetrahedron}}

Let <math> P_1 </math>, <math> P_2 </math>, <math> P_3 </math>, <math> P_4 </math> be the points of a tetrahedron. Let <math> \Delta_i </math> be the area of the face opposite vertex <math> P_i </math> and let <math> \theta_{ij} </math> be the dihedral angle between the two faces of the tetrahedron adjacent to the edge <math> P_i P_j </math>. The [[law of cosines]] for a tetrahedron, which relates the areas of the faces of the tetrahedron to the dihedral angles about a vertex, is given by the following relation:{{sfn|Lee|1997}}
<math display="block"> \Delta_i^2 = \Delta_j^2 + \Delta_k^2 + \Delta_l^2 - 2(\Delta_j\Delta_k\cos\theta_{il} + \Delta_j\Delta_l \cos\theta_{ik} + \Delta_k\Delta_l \cos\theta_{ij})</math>

=== Interior point ===
Let ''P'' be any interior point of a tetrahedron of volume ''V'' for which the vertices are ''A'', ''B'', ''C'', and ''D'', and for which the areas of the opposite faces are ''F''<sub>a</sub>, ''F''<sub>b</sub>, ''F''<sub>c</sub>, and ''F''<sub>d</sub>. Then<ref name="Crux">''Inequalities proposed in “[[Crux Mathematicorum]]”'', [http://www.imomath.com/othercomp/Journ/ineq.pdf].</ref>{{rp|p.62,#1609}}

:<math>PA \cdot F_\mathrm{a} + PB \cdot F_\mathrm{b} + PC \cdot F_\mathrm{c} + PD \cdot F_\mathrm{d} \geq 9V.</math>

For vertices ''A'', ''B'', ''C'', and ''D'', interior point ''P'', and feet ''J'', ''K'', ''L'', and ''M'' of the perpendiculars from ''P'' to the faces, and suppose the faces have equal areas, then<ref name=Crux/>{{rp|p.226,#215}}

:<math>PA+PB+PC+PD \geq 3(PJ+PK+PL+PM).</math>

===Inradius===

Denoting the inradius of a tetrahedron as ''r'' and the [[inradius|inradii]] of its triangular faces as ''r''<sub>''i''</sub> for ''i'' = 1, 2, 3, 4, we have<ref name=Crux/>{{rp|p.81,#1990}}

:<math>\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r_4^2} \leq \frac{2}{r^2},</math>

with equality if and only if the tetrahedron is regular.

If ''A''<sub>''1''</sub>, ''A''<sub>''2''</sub>, ''A''<sub>''3''</sub> and ''A''<sub>''4''</sub> denote the area of each faces, the value of ''r'' is given by

:<math>r=\frac{3V}{A_1+A_2+A_3+A_4}</math>.

This formula is obtained from dividing the tetrahedron into four tetrahedra whose points are the three points of one of the original faces and the incenter. Since the four subtetrahedra fill the volume, we have <math>V = \frac13A_1r+\frac13A_2r+\frac13A_3r+\frac13A_4r</math>.

===Circumradius===

Denote the circumradius of a tetrahedron as ''R''.  Let ''a'', ''b'', ''c'' be the lengths of the three edges that meet at a vertex, and ''A'', ''B'', ''C'' the length of the opposite edges.  Let ''V'' be the volume of the tetrahedron.  Then<ref>{{cite book |chapter-url=https://archive.org/stream/sammlungmathemat01crel#page/104 |chapter=Einige Bemerkungen über die dreiseitige Pyramide |title=Sammlung mathematischer Aufsätze u. Bemerkungen 1 |publisher=Maurer |location=Berlin |last=Crelle |first=A. L. |year=1821 |pages=105–132 |language=de |access-date=7 August 2018}}</ref><ref>{{citation|first=I.|last= Todhunter|title=Spherical Trigonometry: For the Use of Colleges and Schools|year=1886|page= 129 |url=//www.gutenberg.org/ebooks/19770}} ( Art. 163 )</ref>

:<math>R=\frac{\sqrt{(aA+bB+cC)(aA+bB-cC)(aA-bB+cC)(-aA+bB+cC)}}{24V}.</math>

===Circumcenter===
The circumcenter of a tetrahedron can be found as intersection of three bisector planes. A bisector plane is defined as the plane centered on, and orthogonal to an edge of the tetrahedron.
With this definition, the circumcenter {{math|''C''}} of a tetrahedron with vertices {{math|''x''<sub>0</sub>}},{{math|''x''<sub>1</sub>}},{{math|''x''<sub>2</sub>}},{{math|''x''<sub>3</sub>}} can be formulated as matrix-vector product:<ref>{{citation
 | last1 = Lévy | first1 = Bruno
 | last2 = Liu | first2 = Yang
 | doi = 10.1145/1778765.1778856
 | issue = 4
 | journal = ACM Transactions on Graphics
 | pages = 119:1–119:11
 | title = {{math|''L''<sub>p</sub>}} centroidal Voronoi tessellation and its applications
 | volume = 29
 | year = 2010}}</ref>
:<math>\begin{align}
C &= A^{-1}B &
\text{where} & \ &
A = \left(\begin{matrix}\left[x_1 - x_0\right]^T \\ \left[x_2 - x_0\right]^T \\ \left[x_3 - x_0\right]^T \end{matrix}\right) & \ &
\text{and} & \ &
B = \frac{1}{2}\left(\begin{matrix} \|x_1\|^2 - \|x_0\|^2 \\ \|x_2\|^2 - \|x_0\|^2 \\ \|x_3\|^2 - \|x_0\|^2 \end{matrix}\right) \\
\end{align}
</math>
In contrast to the centroid, the circumcenter may not always lay on the inside of a tetrahedron.
Analogously to an obtuse triangle, the circumcenter is outside of the object for an obtuse tetrahedron.

===Centroid===
The tetrahedron's center of mass can be computed as the [[arithmetic mean]] of its four vertices, see [[Centroid#Of a tetrahedron and n-dimensional simplex|Centroid]].

===Faces===

The sum of the areas of any three faces is greater than the area of the fourth face.<ref name=Crux/>{{rp|p.225,#159}}