# Tetrahedron

Not to be confused with tetrahedroid.
For the academic journal, see Tetrahedron (journal).
Regular Tetrahedron

Type Platonic solid
Elements F = 4, E = 6
V = 4 (χ = 2)
Faces by sides 4{3}
Schläfli symbol {3,3} and s{2,2}
Wythoff symbol 3 | 2 3
| 2 2 2
Coxeter diagram

Symmetry Td, A3, [3,3], (*332)
Rotation group T, [3,3]+, (332)
References U01, C15, W1
Properties Regular convex deltahedron
Dihedral angle 70.528779° = arccos(1/3)

3.3.3
(Vertex figure)

Self-dual
(dual polyhedron)

Net

In geometry, a tetrahedron (plural: tetrahedra) is a polyhedron composed of four triangular faces, three of which meet at each vertex. It has six edges and four vertices. The tetrahedron is the only convex polyhedron that has four faces.[1]

The tetrahedron is the three-dimensional case of the more general concept of a Euclidean simplex.

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 nets.[1]

For any tetrahedron there exists a sphere (the circumsphere) such that the tetrahedron's vertices lie on the sphere's surface.

## Regular tetrahedron

Main article: Regular tetrahedron

A regular tetrahedron is one in which all four faces are equilateral triangles, and is one of the Platonic solids.

### Formulas for a regular tetrahedron

The following Cartesian coordinates define the four vertices of a tetrahedron with edge-length 2, centered at the origin:

(±1, 0, -1/√2)
(0, ±1, 1/√2)

Another set of coordinate are based on an alternated cube. Inverting these coordinates generates another dual tetrahedron, with edge length $\sqrt{2}$;

(1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1)

For a regular tetrahedron of edge length a:

Base plane area                                                                                           $A_0={\sqrt{3}\over4}a^2\,$
Surface area[2]                                                                                           $A=4\,A_0={\sqrt{3}}a^2\,$
Height[3]                                                                                                 $H={\sqrt{6}\over3}a=\sqrt{2\over3}\,a\,$
Volume[2]                                                                                                 $V={1\over3} A_0h ={\sqrt{2}\over12}a^3={a^3\over{6\sqrt{2}}}\,$
Angle between an edge and a face                                                                          $\arccos\left({1 \over \sqrt{3}}\right) = \arctan(\sqrt{2})\,$ (approx. 54.7356°)
Angle between two faces[2]                                                                                $\arccos\left({1 \over 3}\right) = \arctan(2\sqrt{2})\,$ (approx. 70.5288°)
Angle between the segments joining the center and the vertices,[4] also known as the "tetrahedral angle"  $\arccos\left({-1\over3}\right ) = 2\arctan(\sqrt{2})\,$ (approx. 109.4712°)
Solid angle at a vertex subtended by a face                                                               $\arccos\left({23\over27}\right)$ (approx. 0.55129 steradians)
Radius of circumsphere[2]                                                                                 $R={\sqrt{6}\over4}a=\sqrt{3\over8}\,a\,$
Radius of insphere that is tangent to faces[2]                                                            $r={1\over3}R={a\over\sqrt{24}}\,$
Radius of midsphere that is tangent to edges[2]                                                           $r_M=\sqrt{rR}={a\over\sqrt{8}}\,$
Radius of exspheres                                                                                       $r_E={a\over\sqrt{6}}\,$
Distance to exsphere center from a vertex                                                                 $d_{VE}={\sqrt{6}\over2}a=\sqrt{3\over2}\,a\,$

Note that with respect to the base plane the slope of a face ($\scriptstyle 2 \sqrt{2}$) is twice that of an edge ($\scriptstyle \sqrt{2}$), 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).

### Orthogonal projections

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 A2 Coxeter plane.

Orthogonal projection
Centered by Edge Face/vertex
Image
Projective
symmetry
[4] [3]

## Other special cases

An isosceles tetrahedron, also called a disphenoid, is a tetrahedron where all four faces are congruent isosceles triangles. A space-filling tetrahedron packs with congruent copies of itself to tile space, like the disphenoid tetrahedral honeycomb.

Name Coxeter
Diagram
Faces Symmetry
Schönflies Coxeter Orbifold Order
Digonal disphenoid Two types of
isosceles triangles
D1h [2] (*22) 4
Rhombic disphenoid
(scalene tetrahedron)
Identical
scalene triangles
D2 [2,2]+ (222) 4
Tetragonal disphenoid
(isosceles tetrahedron)
Identical
isosceles triangles
D2d [2+,4] (2*2) 8
Regular tetrahedron Equilateral triangles Td [3,3] (*332) 24

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 semi-orthocentric 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.

## Volume

The volume of a tetrahedron is given by the pyramid volume formula:

$V = \frac{1}{3} A_0\,h \,$

where A0 is the area of the base and h 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 = (a1, a2, a3), b = (b1, b2, b3), c = (c1, c2, c3), and d = (d1, d2, d3), the volume is (1/6)·|det(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 triple scalar can be represented by the following 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 $\mathbf{a} = (a_1,a_2,a_3) \,$ is expressed as a row or column vector etc.

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 $\mathbf{a} \cdot \mathbf{b} = ab\cos{\gamma}$ etc.

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 $i,\,j\in\{1,\,2,\,3,\,4\}$ represent the vertices {a, b, c, d} and $\scriptstyle d_{ij}$ 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.[5]

### Heron-type formula for the volume of a 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[6]

$\text{volume} = \frac{\sqrt {\,( - a + b + c + d)\,(a - b + c + d)\,(a + b - c + d)\,(a + b + c - d)}}{192\,u\,v\,w}$

where

$\begin{align} a & = \sqrt {xYZ} \\ b & = \sqrt {yZX} \\ c & = \sqrt {zXY} \\ d & = \sqrt {xyz} \\ X & = (w - U + v)\,(U + v + w) \\ x & = (U - v + w)\,(v - w + U) \\ Y & = (u - V + w)\,(V + w + u) \\ y & = (V - w + u)\,(w - u + V) \\ Z & = (v - W + u)\,(W + u + v) \\ z & = (W - u + v)\,(u - v + W). \end{align}$

## 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

$V = \frac {d |(\mathbf{a} \times \mathbf{(b-c)})| } {6}.$

## Properties of a general tetrahedron

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. The circumsphere of the medial tetrahedron is analogous to the triangle's nine-point circle, but does not generally pass through the base points of the altitudes of the reference tetrahedron.[7]

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.[8] 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, 1/3 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.[9]

The center T of the twelve-point sphere also lies on the Euler line. Unlike its triangular counterpart, this center lies 1/3 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 1/3 of the circumradius of the reference tetrahedron.

There is a relation among the angles made by the faces of a general tetrahedron given by [10]

$\begin{vmatrix} -1 & \cos{(\alpha_{12})} & \cos{(\alpha_{13})} & \cos{(\alpha_{14})}\\ \cos{(\alpha_{12})} & -1 & \cos{(\alpha_{23})} & \cos{(\alpha_{24})} \\ \cos{(\alpha_{13})} & \cos{(\alpha_{23})} & -1 & \cos{(\alpha_{34})} \\ \cos{(\alpha_{14})} & \cos{(\alpha_{24})} & \cos{(\alpha_{34})} & -1 \\ \end{vmatrix} = 0\,$

where $\alpha_{ij}$ is the angle between the faces i and j.

## More vector formulas in a general tetrahedron

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If OABC forms a general tetrahedron with a vertex O as the origin and vectors a, b and c represent the positions of the vertices A, B, and C with respect to O, then the radius of the insphere is given by[citation needed]:

$r= \frac {6V} {|\mathbf{b} \times \mathbf{c}| + |\mathbf{c} \times \mathbf{a}| + |\mathbf{a} \times \mathbf{b}| + |(\mathbf{b} \times \mathbf{c}) + (\mathbf{c} \times \mathbf{a}) + (\mathbf{a} \times \mathbf{b})|} \,$

and the radius of the circumsphere is given by:

$R= \frac {|\mathbf{a^2}(\mathbf{b} \times \mathbf{c}) + \mathbf{b^2}(\mathbf{c} \times \mathbf{a}) + \mathbf{c^2}(\mathbf{a} \times \mathbf{b})|} {12V} \,$

which gives the radius of the twelve-point sphere:

$r_T= \frac {|\mathbf{a^2}(\mathbf{b} \times \mathbf{c}) + \mathbf{b^2}(\mathbf{c} \times \mathbf{a}) + \mathbf{c^2}(\mathbf{a} \times \mathbf{b})|} {36V} \,$

where:

$6V= |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|. \,$

In the formulas throughout this section, the scalar a2 represents the inner vector product a·a; similarly b2 and c2.

The vector positions of various centers are as follows:

The centroid

$\mathbf{G} = \frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{4}. \,$

The incenter

$\mathbf{I}= \frac{ |\mathbf{b}\times \mathbf{c}| \, \mathbf{a} + |\mathbf{c}\times \mathbf{a}| \, \mathbf{b} + |\mathbf{a}\times \mathbf{b}| \, \mathbf{c} }{ |\mathbf{b}\times \mathbf{c}| + |\mathbf{c}\times \mathbf{a}| + |\mathbf{a}\times \mathbf{b}| + |\mathbf{b}\times \mathbf{c} + \mathbf{c}\times \mathbf{a} + \mathbf{a}\times \mathbf{b}| }. \,$

The circumcenter

$\mathbf{O}= \frac {\mathbf{a^2}(\mathbf{b} \times \mathbf{c}) + \mathbf{b^2}(\mathbf{c} \times \mathbf{a}) + \mathbf{c^2}(\mathbf{a} \times \mathbf{b})} {2\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})}. \,$

The Monge point

$\mathbf{M} = \frac {\mathbf{a} \cdot (\mathbf{b} + \mathbf{c})(\mathbf{b} \times \mathbf{c}) + \mathbf{b}\cdot (\mathbf{c} + \mathbf{a})(\mathbf{c} \times \mathbf{a}) + \mathbf{c} \cdot (\mathbf{a} + \mathbf{b})(\mathbf{a} \times \mathbf{b})} {2\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})}. \,$

The Euler line relationships are:

$\mathbf{G} = \mathbf{M} + \frac{1}{2} (\mathbf{O}-\mathbf{M})\,$
$\mathbf{T} = \mathbf{M} + \frac{1}{3} (\mathbf{O}-\mathbf{M})\,$

where T is twelve-point center.

Also:

$\mathbf{a} \cdot \mathbf{O} = \frac {\mathbf{a^2}}{2} \quad\quad \mathbf{b} \cdot \mathbf{O} = \frac {\mathbf{b^2}}{2} \quad\quad \mathbf{c} \cdot \mathbf{O} = \frac {\mathbf{c^2}}{2}\,$

and:

$\mathbf{a} \cdot \mathbf{M} = \frac {\mathbf{a} \cdot (\mathbf{b} + \mathbf{c})}{2} \quad\quad \mathbf{b} \cdot \mathbf{M} = \frac {\mathbf{b} \cdot (\mathbf{c} + \mathbf{a})}{2} \quad\quad \mathbf{c} \cdot \mathbf{M} = \frac {\mathbf{c} \cdot (\mathbf{a} + \mathbf{b})}{2}.\,$

## 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).

A tetrahedron is a triangular pyramid, and the regular tetrahedron is self-dual.

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 edge-length $\scriptstyle 2 \sqrt{2}$, 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 3-demicube.

The stella octangula.

The volume of this tetrahedron is 1/3 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, 5 is the minimum number of tetrahedra required to compose a cube.

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.[11]

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 polyhedra in possessing no parallel faces.

### Related polyhedra

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 self-dual tetrahedron once again.

Family of uniform tetrahedral polyhedra
Symmetry: [3,3], (*332) [3,3]+, (332)
{3,3} t0,1{3,3} t1{3,3} t1,2{3,3} t2{3,3} t0,2{3,3} t0,1,2{3,3} s{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.

{3,3}  {3,4}  {3,5}  {3,6}  {3,7}  {3,8}  {3,9}  ...  (3,∞}

The tetrahedron is topologically related to a series of regular polyhedra and tilings with order-3 vertex figures.

Polyhedra Euclidean Hyperbolic tilings

{2,3}

{3,3}

{4,3}

{5,3}

{6,3}

{7,3}

{8,3}
...
(∞,3}

Compounds:

### Intersecting tetrahedra

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 left-handed and right-handed forms, which are mirror images of each other.

## Isometries

### Isometries of regular tetrahedra

The proper rotations, (order-3 rotation on a vertex and face, and order-2 on 2 edges) and reflection plane (through 2 faces and one edge) in the symmetry group of the regular tetrahedron
Tetrahedral symmetry subgroup relations

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 Td, [3,3], (*332), isomorphic to the symmetric group, S4. They can be categorized as follows:

• T, [3,3]+, (332) is isomorphic to alternating group, A4 (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 face-to-face axes

### 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 3-dimensional point group is formed. Two other isometries (C3, [3]+), and (S4, [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, forming the symmetry group Td, isomorphic to the symmetric group, S4.
Td
T
[3,3]
[3,3]+
*332
332
24
12
Trigonal pyramid An equilateral triangle base and three isosceles triangle sides 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 C3v, isomorphic to the symmetric group, S3.
C3v
C3
[3]
[3]+
*33
33
6
3
Tetragonal disphenoid
Isosceles tetrahedron
Four congruent isosceles triangles gives 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 D2d.
D2d
S4
[2+,4]
[2+,4+]
2*2
2×
8
4
Rhombic disphenoid
Scalene tetrahedron
Four congruent scalene triangles gives 4 isometries. The isometries are 1 and the 180° rotations (12)(34), (13)(24), (14)(23). This is the Klein four-group V4 or Z22, present as the point group D2.
D2 [2,2]+ 222 4
Digonal disphenoid Two pairs of congruent 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 C2v, isomorphic to the Klein four-group V4.
C2v
=D1h
[2] *22 4
Sphenoid Two unequal isosceles 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 Cs, also isomorphic to the cyclic group, Z2.
Cs
=C1h
=C1v
[ ] * 2
Half turn tetrahedron Two pairs of congruent scalene 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 C2 isomorphic to the cyclic group, Z2.
C2
=D1
[2]+ 22 2
Irregular tetrahedron No edges equal, four unequal scalene triangles, so that the only isometry is the identity, and the symmetry group is the trivial group.
C1 [ ]+ 1 1

## 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 in a sense 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. One reason to be interested in this "independence" relation is this: It is widely known that 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, not from 8 down to 4, but only from 8 down to 5, since the fourth constraint is not independent of the first three. Thus the space of all shapes of tetrahedra is 5-dimensional.[12]

## Applications

The ammonium+ ion is tetrahedral

### Numerical analysis

In numerical analysis, complicated three-dimensional 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.

### Chemistry

Main article: Tetrahedral molecular geometry

The tetrahedron shape is seen in nature in covalent bonds of molecules. All sp3-hybridized atoms are surrounded by atoms lying in each corner of a tetrahedron. For instance in a methane molecule (CH4) or an ammonium ion (NH4+), 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. See also tetrahedral molecular geometry. The central angle between any two vertices of a perfect tetrahedron is $\arccos{\left(-\tfrac{1}{3}\right)}$, or approximately 109.47°.

Water, H2O, 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 in chemistry are represented graphically as tetrahedra.

However, quaternary phase diagrams in communication engineering are represented graphically on a two-dimensional plane.

### Electricity and electronics

Main articles: 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.[13][14]

Since silicon is the most common semiconductor used in solid-state 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.

### Games

Main article: Game

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 4-sided dice, one of the more common polyhedral dice, with the number rolled appearing around the bottom or on the top vertex. Some Rubik's Cube-like puzzles are tetrahedral, such as the Pyraminx and Pyramorphix.

The net of a tetrahedron also makes the famous Triforce from Nintendo's The Legend of Zelda franchise.

### Color space

Main article: 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).[15]

### Contemporary art

Main article: Contemporary art

The Austrian artist Martina Schettina created a tetrahedron using fluorescent lamps. It was shown at the light art biennale Austria 2010.[16]

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.[17]

In Season 6, Episode 15 of Futurama, aptly named Möbius Dick, the Planet Express crew pass through an area in space known as the Bermuda Tetrahedron. Where 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.

### Geology

Main article: Geology

The tetrahedral hypothesis, originally published by William Lowthian Green to explain the formation of the Earth,[18] was popular through the early 20th century.[19][20]

### 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. It is mounted on a rotating pivot and always points into the wind. It's construction to be 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.[21][22]

## See also

• Boerdijk–Coxeter helix
• Caltrop
• Demihypercube and simplex - n-dimensional analogues
• Hill tetrahedron
• Schläfli orthoscheme
• Tetra Pak
• Tetrahedral kite
• Tetrahedral number
• Tetrahedron packing
• Triangular dipyramid – constructed by joining two tetrahedra along one face
• Trirectangular tetrahedron

## References

1. ^ a b
2. Coxeter, Harold Scott MacDonald; , Methuen and Co., 1948, Table I(i)
3. Havlicek, Hans; Weiß, Gunter (2003). "Altitudes of a tetrahedron and traceless quadratic forms". 110 (8): 679–693. doi:10.2307/3647851. JSTOR 3647851.
4. Leung, Kam-tim; and Suen, Suk-nam; "Vectors, matrices and geometry", Hong Kong University Press, 1994, pp. 53–54
5.
6.
7. Senechal, Marjorie (1981). "Which tetrahedra fill space?". (Mathematical Association of America) 54 (5): 227–243. doi:10.2307/2689983. JSTOR 2689983
8. Rassat, André; Fowler, Patrick W. (2004). "Is There a "Most Chiral Tetrahedron"?". Chemistry: A European Journal 10 (24): 6575–6580. doi:10.1002/chem.200400869
9. Klein, Douglas J. (2002). "Resistance-Distance Sum Rules" (PDF). Croatica Chemica Acta 75 (2): 633–649. Retrieved 2006-09-15.
10. Vondran, Gary L. (April 1998). "Radial and Pruned Tetrahedral Interpolation Techniques" (PDF). HP Technical Report. HPL-98-95: 1–32.
11. "Marvin Minsky: Stanley Kubrick Scraps the Tetrahedron". Web of Stories. Retrieved 20 February 2012.
12. Green, William Lowthian (1875). Vestiges of the Molten Globe, as exhibited in the figure of the earth, volcanic action and physiography. Part I. London: E. Stanford. OCLC 3571917.
13. Holmes, Arthur (1965). Principles of physical geology. Nelson. p. 32.
14. Hitchcock, Charles Henry (January 1900). "William Lowthian Green and his Theory of the Evolution of the Earth's Features". In Winchell, Newton Horace. The American Geologist XXV (Geological Publishing Company). pp. 1–10.
15. "Tetrahedron (wind Direction Indicator)". datwiki.net. Retrieved 2013-03-19. []
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