Polyhedron with 4 faces
Regular tetrahedron

(Click here for rotating model)

Type 
Platonic solid

shortcode 
3> 2z

Elements 
F = 4, E = 6 V = 4 (χ = 2)

Faces by sides 
4{3}

Conway notation 
T

Schläfli symbols 
{3,3}

h{4,3}, s{2,4}, sr{2,2}

Face configuration 
V3.3.3

Wythoff symbol 
3  2 3  2 2 2

Coxeter diagram 
=

Symmetry 
T_{d}, A_{3}, [3,3], (*332)

Rotation group 
T, [3,3]^{+}, (332)

References 
U_{01}, C_{15}, W_{1}

Properties 
regular, convexdeltahedron

Dihedral angle 
70.528779° = arccos(1⁄3)

3.3.3 (Vertex figure)

Selfdual (dual polyhedron)

Net

3D model of regular tetrahedron.
In geometry, a tetrahedron (plural: tetrahedra or tetrahedrons), also known as a triangular pyramid, is a polyhedron composed of four triangular faces, six straight edges, and four vertex corners. The tetrahedron is the simplest of all the ordinary convex polyhedra and the only one that has fewer than 5 faces.^{[1]}
The tetrahedron is the threedimensional case of the more general concept of a Euclidean simplex, and may thus also be called a 3simplex.
The tetrahedron is one kind of pyramid, which is a polyhedron with a flat polygon base and triangular faces connecting the base to a common point. In the case of a tetrahedron the base is a triangle (any of the four faces can be considered the base), so a tetrahedron is also known as a "triangular pyramid".
Like all convex polyhedra, a tetrahedron can be folded from a single sheet of paper. It has two such nets.^{[1]}
For any tetrahedron there exists a sphere (called the circumsphere) on which all four vertices lie, and another sphere (the insphere) tangent to the tetrahedron's faces.^{[2]}
Regular tetrahedron
A regular tetrahedron is a tetrahedron in which all four faces are equilateral triangles. It is one of the five regular Platonic solids, which have been known since antiquity.
In a regular tetrahedron, all faces are the same size and shape (congruent) and all edges are the same length.
Five tetrahedra are laid flat on a plane, with the highest 3dimensional points marked as 1, 2, 3, 4, and 5. These points are then attached to each other and
a thin volume of empty space is left, where the five edge angles do not quite meet.
Regular tetrahedra alone do not tessellate (fill space), but if alternated with regular octahedra in the ratio of two tetrahedra to one octahedron, they form the alternated cubic honeycomb, which is a tessellation. Some tetrahedra that are not regular, including the Schläfli orthoscheme and the Hill tetrahedron, can tessellate.
The regular tetrahedron is selfdual, which means that its dual is another regular tetrahedron. The compound figure comprising two such dual tetrahedra form a stellated octahedron or stella octangula.
Coordinates for a regular tetrahedron
The following Cartesian coordinates define the four vertices of a tetrahedron with edge length 2, centered at the origin, and two level edges:
 $\left(\pm 1,0,{\frac {1}{\sqrt {2}}}\right)\quad {\mbox{and}}\quad \left(0,\pm 1,{\frac {1}{\sqrt {2}}}\right)$
Expressed symmetrically as 4 points on the unit sphere, centroid at the origin, with lower face level, the vertices are:
$v_{1}=\left({\sqrt {\frac {8}{9}}},0,{\frac {1}{3}}\right)$
$v_{2}=\left({\sqrt {\frac {2}{9}}},{\sqrt {\frac {2}{3}}},{\frac {1}{3}}\right)$
$v_{3}=\left({\sqrt {\frac {2}{9}}},{\sqrt {\frac {2}{3}}},{\frac {1}{3}}\right)$
$v_{4}=(0,0,1)$
with the edge length of ${\sqrt {\frac {8}{3}}}$.
Still another set of coordinates are based on an alternated cube or demicube with edge length 2. This form has Coxeter diagram and Schläfli symbol h{4,3}. The tetrahedron in this case has edge length 2√2. Inverting these coordinates generates the dual tetrahedron, and the pair together form the stellated octahedron, whose vertices are those of the original cube.
 Tetrahedron: (1,1,1), (1,−1,−1), (−1,1,−1), (−1,−1,1)
 Dual tetrahedron: (−1,−1,−1), (−1,1,1), (1,−1,1), (1,1,−1)
Regular tetrahedron ABCD and its circumscribed sphere
Angles and distances
For a regular tetrahedron of edge length a:
Face area

$A_{0}={\frac {\sqrt {3}}{4}}a^{2}\,$

Surface area^{[3]}

$A=4\,A_{0}={\sqrt {3}}a^{2}\,$

Height of pyramid^{[4]}

$h={\frac {\sqrt {6}}{3}}a={\sqrt {\frac {2}{3}}}\,a\,$

Centroid to vertex distance

${\frac {3}{4}}\,h={\frac {\sqrt {6}}{4}}\,a={\sqrt {\frac {3}{8}}}\,a\,$

Edge to opposite edge distance

$l={\frac {1}{\sqrt {2}}}\,a\,$

Volume^{[3]}

$V={\frac {1}{3}}A_{0}h={\frac {\sqrt {2}}{12}}a^{3}={\frac {a^{3}}{6{\sqrt {2}}}}\,$

Facevertexedge angle

$\arccos \left({\frac {1}{\sqrt {3}}}\right)=\arctan \left({\sqrt {2}}\right)\,$ (approx. 54.7356°)

Faceedgeface angle, i.e., "dihedral angle"^{[3]}

$\arccos \left({\frac {1}{3}}\right)=\arctan \left(2{\sqrt {2}}\right)\,$ (approx. 70.5288°)

VertexCenterVertex angle,^{[5]} the angle between lines from the tetrahedron center to any two vertices. It is also the angle between Plateau borders at a vertex. In chemistry it is called the tetrahedral bond angle. This angle (in radians) is also the arclength of the geodesic segment on the unit sphere resulting from centrally projecting one edge of the tetrahedron to the sphere.

$\arccos \left({\frac {1}{3}}\right)=2\arctan \left({\sqrt {2}}\right)\,$ (approx. 109.4712°)

Solid angle at a vertex subtended by a face

$\arccos \left({\frac {23}{27}}\right)$ (approx. 0.55129 steradians) (approx. 1809.8 square degrees)

Radius of circumsphere^{[3]}

$R={\frac {\sqrt {6}}{4}}a={\sqrt {\frac {3}{8}}}\,a\,$

Radius of insphere that is tangent to faces^{[3]}

$r={\frac {1}{3}}R={\frac {a}{\sqrt {24}}}\,$

Radius of midsphere that is tangent to edges^{[3]}

$r_{\mathrm {M} }={\sqrt {rR}}={\frac {a}{\sqrt {8}}}\,$

Radius of exspheres

$r_{\mathrm {E} }={\frac {a}{\sqrt {6}}}\,$

Distance to exsphere center from the opposite vertex

$d_{\mathrm {VE} }={\frac {\sqrt {6}}{2}}a={\sqrt {\frac {3}{2}}}a\,$

With respect to the base plane the slope of a face (2√2) is twice that of an edge (√2), corresponding to the fact that the horizontal distance covered from the base to the apex along an edge is twice that along the median of a face. In other words, if C is the centroid of the base, the distance from C to a vertex of the base is twice that from C to the midpoint of an edge of the base. This follows from the fact that the medians of a triangle intersect at its centroid, and this point divides each of them in two segments, one of which is twice as long as the other (see proof).
For a regular tetrahedron with side length a, radius R of its circumscribing sphere, and distances d_{i} from an arbitrary point in 3space to its four vertices, we have^{[6]}
 ${\begin{aligned}{\frac {d_{1}^{4}+d_{2}^{4}+d_{3}^{4}+d_{4}^{4}}{4}}+{\frac {16R^{4}}{9}}&=\left({\frac {d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}}{4}}+{\frac {2R^{2}}{3}}\right)^{2};\\4\left(a^{4}+d_{1}^{4}+d_{2}^{4}+d_{3}^{4}+d_{4}^{4}\right)&=\left(a^{2}+d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}\right)^{2}.\end{aligned}}$
Isometries of the regular tetrahedron
The proper rotations, (order3 rotation on a vertex and face, and order2 on two edges) and reflection plane (through two faces and one edge) in the symmetry group of the regular tetrahedron
The vertices of a cube can be grouped into two groups of four, each forming a regular tetrahedron (see above, and also animation, showing one of the two tetrahedra in the cube). The symmetries of a regular tetrahedron correspond to half of those of a cube: those that map the tetrahedra to themselves, and not to each other.
The tetrahedron is the only Platonic solid that is not mapped to itself by point inversion.
The regular tetrahedron has 24 isometries, forming the symmetry group T_{d}, [3,3], (*332), isomorphic to the symmetric group, S_{4}. They can be categorized as follows:
 T, [3,3]^{+}, (332) is isomorphic to alternating group, A_{4} (the identity and 11 proper rotations) with the following conjugacy classes (in parentheses are given the permutations of the vertices, or correspondingly, the faces, and the unit quaternion representation):
 identity (identity; 1)
 rotation about an axis through a vertex, perpendicular to the opposite plane, by an angle of ±120°: 4 axes, 2 per axis, together 8 ((1 2 3), etc.; 1 ± i ± j ± k/2)
 rotation by an angle of 180° such that an edge maps to the opposite edge: 3 ((1 2)(3 4), etc.; i, j, k)
 reflections in a plane perpendicular to an edge: 6
 reflections in a plane combined with 90° rotation about an axis perpendicular to the plane: 3 axes, 2 per axis, together 6; equivalently, they are 90° rotations combined with inversion (x is mapped to −x): the rotations correspond to those of the cube about facetoface axes
Orthogonal projections of the regular tetrahedron
The regular tetrahedron has two special orthogonal projections, one centered on a vertex or equivalently on a face, and one centered on an edge. The first corresponds to the A_{2} Coxeter plane.
Cross section of regular tetrahedron
A central cross section of a
regular tetrahedron is a
square.
The two skew perpendicular opposite edges of a regular tetrahedron define a set of parallel planes. When one of these planes intersects the tetrahedron the resulting cross section is a rectangle.^{[7]} When the intersecting plane is near one of the edges the rectangle is long and skinny. When halfway between the two edges the intersection is a square. The aspect ratio of the rectangle reverses as you pass this halfway point. For the midpoint square intersection the resulting boundary line traverses every face of the tetrahedron similarly. If the tetrahedron is bisected on this plane, both halves become wedges.
A tetragonal disphenoid viewed orthogonally to the two green edges.
This property also applies for tetragonal disphenoids when applied to the two special edge pairs.
Spherical tiling
The tetrahedron can also be represented as a spherical tiling, and projected onto the plane via a stereographic projection. This projection is conformal, preserving angles but not areas or lengths. Straight lines on the sphere are projected as circular arcs on the plane.
Helical stacking
Regular tetrahedra can be stacked facetoface in a chiral aperiodic chain called the Boerdijk–Coxeter helix. In four dimensions, all the convex regular 4polytopes with tetrahedral cells (the 5cell, 16cell and 600cell) can be constructed as tilings of the 3sphere by these chains, which become periodic in the threedimensional space of the 4polytope's boundary surface.
Other special cases
Tetrahedral symmetry subgroup relations

Tetrahedral symmetries shown in tetrahedral diagrams

An isosceles tetrahedron, also called a disphenoid, is a tetrahedron where all four faces are congruent triangles. A spacefilling tetrahedron packs with congruent copies of itself to tile space, like the disphenoid tetrahedral honeycomb.
In a trirectangular tetrahedron the three face angles at one vertex are right angles. If all three pairs of opposite edges of a tetrahedron are perpendicular, then it is called an orthocentric tetrahedron. When only one pair of opposite edges are perpendicular, it is called a semiorthocentric tetrahedron. An isodynamic tetrahedron is one in which the cevians that join the vertices to the incenters of the opposite faces are concurrent, and an isogonic tetrahedron has concurrent cevians that join the vertices to the points of contact of the opposite faces with the inscribed sphere of the tetrahedron.
Isometries of irregular tetrahedra
The isometries of an irregular (unmarked) tetrahedron depend on the geometry of the tetrahedron, with 7 cases possible. In each case a 3dimensional point group is formed. Two other isometries (C_{3}, [3]^{+}), and (S_{4}, [2^{+},4^{+}]) can exist if the face or edge marking are included. Tetrahedral diagrams are included for each type below, with edges colored by isometric equivalence, and are gray colored for unique edges.
Tetrahedron name

Edge equivalence diagram

Description

Symmetry

Schön.

Cox.

Orb.

Ord.

Regular tetrahedron


Four equilateral triangles It forms the symmetry group T_{d}, isomorphic to the symmetric group, S_{4}. A regular tetrahedron has Coxeter diagram and Schläfli symbol {3,3}.

T_{d} T 
[3,3] [3,3]^{+} 
*332 332 
24 12

Triangular pyramid


An equilateral triangle base and three equal isosceles triangle sides It gives 6 isometries, corresponding to the 6 isometries of the base. As permutations of the vertices, these 6 isometries are the identity 1, (123), (132), (12), (13) and (23), forming the symmetry group C_{3v}, isomorphic to the symmetric group, S_{3}. A triangular pyramid has Schläfli symbol {3}∨( ).

C_{3v} C_{3} 
[3] [3]^{+} 
*33 33 
6 3

Mirrored sphenoid


Two equal scalene triangles with a common base edge This has two pairs of equal edges (1,3), (1,4) and (2,3), (2,4) and otherwise no edges equal. The only two isometries are 1 and the reflection (34), giving the group C_{s}, also isomorphic to the cyclic group, Z_{2}.

C_{s} =C_{1h} =C_{1v} 
[ ] 
* 
2

Irregular tetrahedron (No symmetry)


Four unequal triangles
Its only isometry is the identity, and the symmetry group is the trivial group. An irregular tetrahedron has Schläfli symbol ( )∨( )∨( )∨( ).

C_{1} 
[ ]^{+} 
1 
1

Disphenoids (Four equal triangles)

Tetragonal disphenoid


Four equal isosceles triangles
It has 8 isometries. If edges (1,2) and (3,4) are of different length to the other 4 then the 8 isometries are the identity 1, reflections (12) and (34), and 180° rotations (12)(34), (13)(24), (14)(23) and improper 90° rotations (1234) and (1432) forming the symmetry group D_{2d}. A tetragonal disphenoid has Coxeter diagram and Schläfli symbol s{2,4}.

D_{2d} S_{4} 
[2^{+},4] [2^{+},4^{+}] 
2*2 2× 
8 4

Rhombic disphenoid


Four equal scalene triangles
It has 4 isometries. The isometries are 1 and the 180° rotations (12)(34), (13)(24), (14)(23). This is the Klein fourgroup V_{4} or Z_{2}^{2}, present as the point group D_{2}. A rhombic disphenoid has Coxeter diagram and Schläfli symbol sr{2,2}.

D_{2} 
[2,2]^{+} 
222 
4

Generalized disphenoids (2 pairs of equal triangles)

Digonal disphenoid


Two pairs of equal isosceles triangles This gives two opposite edges (1,2) and (3,4) that are perpendicular but different lengths, and then the 4 isometries are 1, reflections (12) and (34) and the 180° rotation (12)(34). The symmetry group is C_{2v}, isomorphic to the Klein fourgroup V_{4}. A digonal disphenoid has Schläfli symbol { }∨{ }.

C_{2v} C_{2} 
[2] [2]^{+} 
*22 22 
4 2

Phyllic disphenoid


Two pairs of equal scalene or isosceles triangles
This has two pairs of equal edges (1,3), (2,4) and (1,4), (2,3) but otherwise no edges equal. The only two isometries are 1 and the rotation (12)(34), giving the group C_{2} isomorphic to the cyclic group, Z_{2}.

C_{2} 
[2]^{+} 
22 
2

General properties
Volume
The volume of a tetrahedron is given by the pyramid volume formula:
 $V={\frac {1}{3}}A_{0}\,h\,$
where A_{0} is the area of the base and h 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 apexes to the opposite faces are inversely proportional to the areas of these faces.
For a tetrahedron with vertices
a = (a_{1}, a_{2}, a_{3}),
b = (b_{1}, b_{2}, b_{3}),
c = (c_{1}, c_{2}, c_{3}), and
d = (d_{1}, d_{2}, d_{3}), the volume is 1/6det(a − d, b − d, c − d), or any other combination of pairs of vertices that form a simply connected graph. This can be rewritten using a dot product and a cross product, yielding
 $V={\frac {(\mathbf {a} \mathbf {d} )\cdot ((\mathbf {b} \mathbf {d} )\times (\mathbf {c} \mathbf {d} ))}{6}}.$
If the origin of the coordinate system is chosen to coincide with vertex d, then d = 0, so
 $V={\frac {\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} )}{6}},$
where a, b, and c represent three edges that meet at one vertex, and a · (b × c) is a scalar triple product. Comparing this formula with that used to compute the volume of a parallelepiped, we conclude that the volume of a tetrahedron is equal to 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:
 $6\cdot V={\begin{Vmatrix}\mathbf {a} &\mathbf {b} &\mathbf {c} \end{Vmatrix}}$ or $6\cdot V={\begin{Vmatrix}\mathbf {a} \\\mathbf {b} \\\mathbf {c} \end{Vmatrix}}$ where ${\begin{cases}\mathbf {a} =(a_{1},a_{2},a_{3}),\\\mathbf {b} =(b_{1},b_{2},b_{3}),\\\mathbf {c} =(c_{1},c_{2},c_{3}),\end{cases}}$ are expressed as row or column vectors.
Hence
 $36\cdot V^{2}={\begin{vmatrix}\mathbf {a^{2}} &\mathbf {a} \cdot \mathbf {b} &\mathbf {a} \cdot \mathbf {c} \\\mathbf {a} \cdot \mathbf {b} &\mathbf {b^{2}} &\mathbf {b} \cdot \mathbf {c} \\\mathbf {a} \cdot \mathbf {c} &\mathbf {b} \cdot \mathbf {c} &\mathbf {c^{2}} \end{vmatrix}}$ where ${\begin{cases}\mathbf {a} \cdot \mathbf {b} =ab\cos {\gamma },\\\mathbf {b} \cdot \mathbf {c} =bc\cos {\alpha },\\\mathbf {a} \cdot \mathbf {c} =ac\cos {\beta }.\end{cases}}$
which gives
 $V={\frac {abc}{6}}{\sqrt {1+2\cos {\alpha }\cos {\beta }\cos {\gamma }\cos ^{2}{\alpha }\cos ^{2}{\beta }\cos ^{2}{\gamma }}},\,$
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.
Given the distances between the vertices of a tetrahedron the volume can be computed using the Cayley–Menger determinant:
 $288\cdot V^{2}={\begin{vmatrix}0&1&1&1&1\\1&0&d_{12}^{2}&d_{13}^{2}&d_{14}^{2}\\1&d_{12}^{2}&0&d_{23}^{2}&d_{24}^{2}\\1&d_{13}^{2}&d_{23}^{2}&0&d_{34}^{2}\\1&d_{14}^{2}&d_{24}^{2}&d_{34}^{2}&0\end{vmatrix}}$
where the subscripts i, j ∈ {1, 2, 3, 4} represent the vertices {a, b, c, d} and d_{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.^{[8]}
Denote a, b, c be three edges that meet at a point, and x, y, z the opposite edges. Let V be the volume of the tetrahedron; then^{[9]}
 $V={\frac {\sqrt {4a^{2}b^{2}c^{2}a^{2}X^{2}b^{2}Y^{2}c^{2}Z^{2}+XYZ}}{12}}$
where
 ${\begin{aligned}X&=b^{2}+c^{2}x^{2},\\Y&=a^{2}+c^{2}y^{2},\\Z&=a^{2}+b^{2}z^{2}.\end{aligned}}$
The above formula uses six lengths of edges, and the following formula uses three lengths of edges and three angles.
 $V={\frac {abc}{6}}{\sqrt {1+2\cos {\alpha }\cos {\beta }\cos {\gamma }\cos ^{2}{\alpha }\cos ^{2}{\beta }\cos ^{2}{\gamma }}}$
Herontype formula for the volume of a tetrahedron
Six edgelengths of Tetrahedron
If U, V, W, u, v, w are lengths of edges of the tetrahedron (first three form a triangle; u opposite to U and so on), then^{[10]}
 $V={\frac {\sqrt {\,(p+q+r+s)\,(pq+r+s)\,(p+qr+s)\,(p+q+rs)}}{192\,u\,v\,w}}$
where
 ${\begin{aligned}p&={\sqrt {xYZ}},&q&={\sqrt {yZX}},&r&={\sqrt {zXY}},&s&={\sqrt {xyz}},\end{aligned}}$
 ${\begin{aligned}X&=(wU+v)\,(U+v+w),&x&=(Uv+w)\,(vw+U),\\Y&=(uV+w)\,(V+w+u),&y&=(Vw+u)\,(wu+V),\\Z&=(vW+u)\,(W+u+v),&z&=(Wu+v)\,(uv+W).\end{aligned}}$
Volume divider
A plane that divides two opposite edges of a tetrahedron in a given ratio also divides the volume of the tetrahedron in the same ratio. Thus any plane containing a bimedian (connector of opposite edges' midpoints) of a tetrahedron bisects the volume of the tetrahedron.^{[11]}^{[12]}^{:pp.89–90}
NonEuclidean volume
For tetrahedra in hyperbolic space or in threedimensional elliptic geometry, the dihedral angles of the tetrahedron determine its shape and hence its volume. In these cases, the volume is given by the Murakami–Yano formula.^{[13]} However, in Euclidean space, scaling a tetrahedron changes its volume but not its dihedral angles, so no such formula can exist.
Distance between the edges
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 d be the distance between the skew lines formed by opposite edges a and b − c as calculated here. Then another volume formula is given by
 $V={\frac {d(\mathbf {a} \times \mathbf {(bc)} )}{6}}.$
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 center and points such as a centroid. However, there is generally no orthocenter in the sense of intersecting altitudes.^{[14]}
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 at a point called the centroid of the tetrahedron.^{[15]} 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 ninepoint circle of the general triangle has an analogue in the circumsphere of a tetrahedron's medial tetrahedron. It is the twelvepoint 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.^{[16]}
The center T of the twelvepoint 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 twelvepoint center lies midway between the Euler point orthogonal line and the centroidal orthogonal line. Furthermore, for any face, the twelvepoint center lies at the midpoint of the corresponding Euler point and the orthocenter for that face.
The radius of the twelvepoint 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^{[17]}
 ${\begin{vmatrix}1&\cos {(\alpha _{12})}&\cos {(\alpha _{13})}&\cos {(\alpha _{14})}\\\cos {(\alpha _{12})}&1&\cos {(\alpha _{23})}&\cos {(\alpha _{24})}\\\cos {(\alpha _{13})}&\cos {(\alpha _{23})}&1&\cos {(\alpha _{34})}\\\cos {(\alpha _{14})}&\cos {(\alpha _{24})}&\cos {(\alpha _{34})}&1\\\end{vmatrix}}=0\,$
where α_{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 sr, and at which the angles subtended by opposite edges are equal.^{[18]} 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 3simplex. 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 3dimensional space).
A tetrahedron is a triangular pyramid, and the regular tetrahedron is selfdual.
A regular tetrahedron can be embedded inside a cube in two ways such that each vertex is a vertex of the cube, and each edge is a diagonal of one of the cube's faces. For one such embedding, the Cartesian coordinates of the vertices are
 (+1, +1, +1);
 (−1, −1, +1);
 (−1, +1, −1);
 (+1, −1, −1).
This yields a tetrahedron with edgelength 2√2, centered at the origin. For the other tetrahedron (which is dual to the first), reverse all the signs. These two tetrahedra's vertices combined are the vertices of a cube, demonstrating that the regular tetrahedron is the 3demicube.
The volume of this tetrahedron is onethird the volume of the cube. Combining both tetrahedra gives a regular polyhedral compound called the compound of two tetrahedra or stella octangula.
The interior of the stella octangula is an octahedron, and correspondingly, a regular octahedron is the result of cutting off, from a regular tetrahedron, four regular tetrahedra of half the linear size (i.e., rectifying the tetrahedron).
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 compound of five cubes gives two more regular compounds, containing five and ten tetrahedra.
Regular tetrahedra cannot 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.
However, several irregular tetrahedra are known, of which copies can tile space, for instance the disphenoid tetrahedral honeycomb. The complete list remains an open problem.^{[19]}
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 sidenote: these two kinds of tetrahedron have the same volume.)
The tetrahedron is unique among the uniform polyhedra in possessing no parallel faces.
A law of sines for tetrahedra and the space of all shapes of tetrahedra
A corollary of the usual law of sines is that in a tetrahedron with vertices O, A, B, C, we have
 $\sin \angle OAB\cdot \sin \angle OBC\cdot \sin \angle OCA=\sin \angle OAC\cdot \sin \angle OCB\cdot \sin \angle OBA.\,$
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 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 5dimensional.^{[20]}
Law of cosines for tetrahedra
Let {P_{1} ,P_{2}, P_{3}, P_{4}} be the points of a tetrahedron. Let Δ_{i} be the area of the face opposite vertex P_{i} and let θ_{ij} be the dihedral angle between the two faces of the tetrahedron adjacent to the edge P_{i}P_{j}.
The law of cosines for this tetrahedron,^{[21]} which relates the areas of the faces of the tetrahedron to the dihedral angles about a vertex, is given by the following relation:
 $\Delta _{i}^{2}=\Delta _{j}^{2}+\Delta _{k}^{2}+\Delta _{l}^{2}2(\Delta _{j}\Delta _{k}\cos \theta _{il}+\Delta _{j}\Delta _{l}\cos \theta _{ik}+\Delta _{k}\Delta _{l}\cos \theta _{ij})$
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_{a}, F_{b}, F_{c}, and F_{d}. Then^{[22]}^{:p.62,#1609}
 $PA\cdot F_{\mathrm {a} }+PB\cdot F_{\mathrm {b} }+PC\cdot F_{\mathrm {c} }+PD\cdot F_{\mathrm {d} }\geq 9V.$
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^{[22]}^{:p.226,#215}
 $PA+PB+PC+PD\geq 3(PJ+PK+PL+PM).$
Inradius
Denoting the inradius of a tetrahedron as r and the inradii of its triangular faces as r_{i} for i = 1, 2, 3, 4, we have^{[22]}^{:p.81,#1990}
 ${\frac {1}{r_{1}^{2}}}+{\frac {1}{r_{2}^{2}}}+{\frac {1}{r_{3}^{2}}}+{\frac {1}{r_{4}^{2}}}\leq {\frac {2}{r^{2}}},$
with equality if and only if the tetrahedron is regular.
If A_{1}, A_{2}, A_{3} and A_{4} denote the area of each faces, the value of r is given by
 $r={\frac {3V}{A_{1}+A_{2}+A_{3}+A_{4}}}$.
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 $V={\frac {1}{3}}A_{1}r+{\frac {1}{3}}A_{2}r+{\frac {1}{3}}A_{3}r+{\frac {1}{3}}A_{4}r$.
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^{[23]}^{[24]}
 $R={\frac {\sqrt {(aA+bB+cC)(aA+bBcC)(aAbB+cC)(aA+bB+cC)}}{24V}}.$
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 C of a tetrahedron with vertices x_{0},x_{1},x_{2},x_{3} can be formulated as matrixvector product:^{[25]}
 ${\begin{aligned}C&=A^{1}B&{\text{where}}&\ &A=\left({\begin{matrix}\left[x_{1}x_{0}\right]^{T}\\\left[x_{2}x_{0}\right]^{T}\\\left[x_{3}x_{0}\right]^{T}\end{matrix}}\right)&\ &{\text{and}}&\ &B={\frac {1}{2}}\left({\begin{matrix}\x_{1}x_{0}\^{2}\\\x_{2}x_{0}\^{2}\\\x_{3}x_{0}\^{2}\end{matrix}}\right)\\\end{aligned}}$
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 computes as the arithmetic mean of its four vertices, see Centroid.
Faces
The sum of the areas of any three faces is greater than the area of the fourth face.^{[22]}^{:p.225,#159}
Integer tetrahedra
There exist tetrahedra having integervalued edge lengths, face areas and volume. These are called Heronian tetrahedra. One example has one edge of 896, the opposite edge of 990 and the other four edges of 1073; two faces are isosceles triangles with areas of 436800 and the other two are isosceles with areas of 47120, while the volume is 124185600.^{[26]}
A tetrahedron can have integer volume and consecutive integers as edges, an example being the one with edges 6, 7, 8, 9, 10, and 11 and volume 48.^{[27]}
Related polyhedra and compounds
A regular tetrahedron can be seen as a triangular pyramid.
A regular tetrahedron can be seen as a degenerate polyhedron, a uniform digonal antiprism, where base polygons are reduced digons.
A regular tetrahedron can be seen as a degenerate polyhedron, a uniform dual digonal trapezohedron, containing 6 vertices, in two sets of colinear edges.
A truncation process applied to the tetrahedron produces a series of uniform polyhedra. Truncating edges down to points produces the octahedron as a rectified tetrahedron. The process completes as a birectification, reducing the original faces down to points, and producing the selfdual tetrahedron once again.
Family of uniform tetrahedral polyhedra

Symmetry: [3,3], (*332)

[3,3]^{+}, (332)

















{3,3}

t{3,3}

r{3,3}

t{3,3}

{3,3}

rr{3,3}

tr{3,3}

sr{3,3}

Duals to uniform polyhedra









V3.3.3

V3.6.6

V3.3.3.3

V3.6.6

V3.3.3

V3.4.3.4

V4.6.6

V3.3.3.3.3

This polyhedron is topologically related as a part of sequence of regular polyhedra with Schläfli symbols {3,n}, continuing into the hyperbolic plane.
The tetrahedron is topologically related to a series of regular polyhedra and tilings with order3 vertex figures.
*n32 symmetry mutation of regular tilings: {n,3}

Spherical

Euclidean

Compact hyperb.

Paraco.

Noncompact hyperbolic













{2,3}

{3,3}

{4,3}

{5,3}

{6,3}

{7,3}

{8,3}

{∞,3}

{12i,3}

{9i,3}

{6i,3}

{3i,3}

An interesting polyhedron can be constructed from five intersecting tetrahedra. This compound of five tetrahedra has been known for hundreds of years. It comes up regularly in the world of origami. Joining the twenty vertices would form a regular dodecahedron. There are both lefthanded and righthanded forms, which are mirror images of each other. Superimposing both forms gives a compound of ten tetrahedra, in which the ten tetrahedra are arranged as five pairs of stellae octangulae. A stella octangula is a compound of two tetrahedra in dual position and its eight vertices define a cube as their convex hull.
The square hosohedron is another polyhedron with four faces, but it does not have triangular faces.
Applications
Numerical analysis
An irregular volume in space can be approximated by an irregular triangulated surface, and irregular tetrahedral volume elements.
In numerical analysis, complicated threedimensional shapes are commonly broken down into, or approximated by, a polygonal mesh of irregular tetrahedra in the process of setting up the equations for finite element analysis especially in the numerical solution of partial differential equations. These methods have wide applications in practical applications in computational fluid dynamics, aerodynamics, electromagnetic fields, civil engineering, chemical engineering, naval architecture and engineering, and related fields.
Structural engineering
A tetrahedron having stiff edges is inherently rigid. For this reason it is often used to stiffen frame structures such as spaceframes.
Aviation
At some airfields, a large frame in the shape of a tetrahedron with two sides covered with a thin material is mounted on a rotating pivot and always points into the wind. It is built big enough to be seen from the air and is sometimes illuminated. Its purpose is to serve as a reference to pilots indicating wind direction.^{[28]}
Chemistry
The tetrahedron shape is seen in nature in covalently bonded molecules. All sp^{3}hybridized atoms are surrounded by atoms (or lone electron pairs) at the four corners of a tetrahedron. For instance in a methane molecule (CH^{}
_{4}) or an ammonium ion (NH^{+}
_{4}), four hydrogen atoms surround a central carbon or nitrogen atom with tetrahedral symmetry. For this reason, one of the leading journals in organic chemistry is called Tetrahedron. The central angle between any two vertices of a perfect tetrahedron is arccos(−1/3), or approximately 109.47°.^{[5]}
Water, H^{}
_{2}O, also has a tetrahedral structure, with two hydrogen atoms and two lone pairs of electrons around the central oxygen atoms. Its tetrahedral symmetry is not perfect, however, because the lone pairs repel more than the single O–H bonds.
Quaternary phase diagrams of mixtures of chemical substances are represented graphically as tetrahedra.
However, quaternary phase diagrams in communication engineering are represented graphically on a twodimensional plane.
Electricity and electronics
If six equal resistors are soldered together to form a tetrahedron, then the resistance measured between any two vertices is half that of one resistor.^{[29]}^{[30]}
Since silicon is the most common semiconductor used in solidstate electronics, and silicon has a valence of four, the tetrahedral shape of the four chemical bonds in silicon is a strong influence on how crystals of silicon form and what shapes they assume.
Color space
Tetrahedra are used in color space conversion algorithms specifically for cases in which the luminance axis diagonally segments the color space (e.g. RGB, CMY).^{[31]}
Games
The Royal Game of Ur, dating from 2600 BC, was played with a set of tetrahedral dice.
Especially in roleplaying, this solid is known as a 4sided die, one of the more common polyhedral dice, with the number rolled appearing around the bottom or on the top vertex. Some Rubik's Cubelike puzzles are tetrahedral, such as the Pyraminx and Pyramorphix.
Geology
The tetrahedral hypothesis, originally published by William Lowthian Green to explain the formation of the Earth,^{[32]} was popular through the early 20th century.^{[33]}^{[34]}
Weaponry
Modern caltrop with hollow spikes to puncture selfsealing rubber tyres
Some caltrops are based on tetrahedra as one spike points upwards regardless of how they land and can be easily made by welding two bent nails together.
Contemporary art
The Austrian artist Martina Schettina created a tetrahedron using fluorescent lamps. It was shown at the light art biennale Austria 2010.^{[35]}
It is used as album artwork, surrounded by black flames on The End of All Things to Come by Mudvayne.
Popular culture
Stanley Kubrick originally intended the monolith in 2001: A Space Odyssey to be a tetrahedron, according to Marvin Minsky, a cognitive scientist and expert on artificial intelligence who advised Kubrick on the HAL 9000 computer and other aspects of the movie. Kubrick scrapped the idea of using the tetrahedron as a visitor who saw footage of it did not recognize what it was and he did not want anything in the movie regular people did not understand.^{[36]}
In Season 6, Episode 15 of Futurama, named "Möbius Dick", the Planet Express crew pass through an area in space known as the Bermuda Tetrahedron. Many other ships passing through the area have mysteriously disappeared, including that of the first Planet Express crew.
In the 2013 film Oblivion the large structure in orbit above the Earth is of a tetrahedron design and referred to as the Tet.
Tetrahedral graph
The skeleton of the tetrahedron (comprising the vertices and edges) forms a graph, with 4 vertices, and 6 edges. It is a special case of the complete graph, K_{4}, and wheel graph, W_{4}.^{[37]} It is one of 5 Platonic graphs, each a skeleton of its Platonic solid.
3fold symmetry

See also
References
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