In general, a tetrahedron is a
three-dimensional object with four faces, six edges, and four vertices. It can be considered as
pyramid whenever one of its faces can be considered as the
base. Its
1-skeleton can be generally seen as a
graph by
Steinitz's theorem, known as
tetrahedral graph, one of the
Platonic graphs. It is
complete graph K_4 because every pair of its vertices has a unique edge. In a plane, this graph can be regarded as a triangle in which three vertices connect to its fourth vertex in the center, known as the
universal vertex; hence, the tetrahedral graph is a
wheel graph. The tetrahedron is one of the polyhedra that does not have
space diagonal; the other polyhedra with such a property are
Császár polyhedron and
Schonhardt polyhedron. It is also known as
3-simplex, the generalization of a triangle in multi-dimension. It is
self-dual, meaning its
dual polyhedron is a tetrahedron itself. Many other properties of tetrahedra are explicitly described in the following sections.
Volume The volume of a tetrahedron is given by the formula: V = \frac{1}{3}Ah. where A is the
base's area and h is the height from the base to the apex. This applies to 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.
Algebraic approach A
linear algebra approach is an alternative way by the given vertices in terms of
vectors as: \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} In terms of a
determinant, the volume of a tetrahedron is \frac{1}{6} \det(\mathbf{a} - \mathbf{d}, \mathbf{b} - \mathbf{d}, \mathbf{c} - \mathbf{d}) , one-sixth of any
parallelepiped's volume sharing three converging edges with it. Similarly by the given vertices, another approach is by the
absolute value of the scalar triple product, representing the absolute values of determinants 6 \cdot V = \begin{vmatrix} \mathbf{a} & \mathbf{b} & \mathbf{c} \end{vmatrix}. 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}. Here \mathbf{a} \cdot \mathbf{b} = ab\cos{\gamma} , \mathbf{b} \cdot \mathbf{c} = bc\cos{\alpha} , and \mathbf{a} \cdot \mathbf{c} = ac\cos{\beta}. The variables a , b , and c denotes each
norm of a vector \mathbf{a} , \mathbf{b} , and \mathbf{c} respectively. This gives V = \frac {abc} {6} \sqrt{1 + 2\cos{\alpha}\cos{\beta}\cos{\gamma}-\cos^2{\alpha}-\cos^2{\beta}-\cos^2{\gamma}}, \, where the
Greek lowercase letters denotes the plane angles occurring in vertex \mathbf{d} : the angle \alpha is an angle between the two edges connecting the vertex \mathbf{d} to the vertices \mathbf{b} and \mathbf{c} ; the angle \beta does so for the vertices \mathbf{a} and \mathbf{c} ; while the angle \gamma is defined by the position of the vertices \mathbf{a} and \mathbf{b} . Considering that \mathbf{d} = 0 , then 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|\,. 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 \{\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}\} , and d_{ij} is the pairwise distance between them, 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.
Other approaches Let a , b , and c be the lengths of three edges that meet at a point, and x , y , and z be those of the opposite edges. The volume of the tetrahedron V is: 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} where \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} 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}} The volume of a tetrahedron can be ascertained by using the Heron formula. Suppose U , V , W , u , v , and w are the lengths of the tetrahedron's edges as in the following image. Here, the first three form a triangle, with u opposite U , v opposite V , and w opposite W . Then, V = \frac{\sqrt {\,( - p + q + r + s)\,(p - q + r + s)\,(p + q - r + s)\,(p + q + r - s)}}{192\,u\,v\,w} where \begin{align} p = \sqrt{xYZ}, &\quad q = \sqrt{yZX}, \\ r = \sqrt{zXY}, &\quad s = \sqrt {xyz}, \end{align} and \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} Any plane containing a bimedian (connector of opposite edges' midpoints) of a tetrahedron
bisects the volume of the tetrahedron. For tetrahedra in
hyperbolic space or in three-dimensional
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, after Jun Murakami and Masakazu Yano. 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 d be the distance between the skew lines formed by opposite edges a and \mathbf{b} - \mathbf{c} as calculated
here. Then another formula for the volume of a tetrahedron V is given by V = \frac {d |(\mathbf{a} \times \mathbf{(b-c)})|}{6}.
Analogous to 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.
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. 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. If the tetrahedron's altitudes do intersect, then the Monge point and the orthocenter coincide to give the class of
orthocentric tetrahedron. 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, a tetrahedron has four medians and three bimedians. These seven line segments are all
concurrent at a point called the
centroid of the tetrahedron. In addition,
the four medians are divided in a 3:1 ratio by the centroid. 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. 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. Additionally, 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 :\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 . 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 \pi
sr, and at which the angles subtended by opposite edges are equal. A solid angle of \pi 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 \pi sr, O lies inside the tetrahedron. Because the sum of distances from O to the vertices is a minimum, O coincides with the
geometric median M of the vertices. If the solid angle at one of the vertices v measures exactly \pi sr, then O and M coincide with v . However, if a tetrahedron has a vertex v with solid angle greater than \pi sr, M still corresponds to v , but O lies outside the tetrahedron.
Trigonometry and the space of all shapes A corollary of the usual
law of sines is that in a tetrahedron with vertices O , A , B , C , is given by \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. Hence, common factors are cancelled from both sides, and 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? 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 5-dimensional. Let P_1 , P_2 , P_3 , P_4 be the points of a tetrahedron. Let \Delta_i be the area of the face opposite vertex P_i and let \theta_{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 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: \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, 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: 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_1 for i=1,2,3,4, then: \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, then: V = \frac13A_1r+\frac13A_2r+\frac13A_3r+\frac13A_4r .
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 :R=\frac{\sqrt{(aA+bB+cC)(aA+bB-cC)(aA-bB+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 of a tetrahedron with vertices ,,, can be formulated as matrix-vector product: :\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} Unlike obtuse triangle, the circumcenter may not always lie outside of an obtuse tetrahedron (i.e. when one or more
dihedral angles are greater than \pi/2), nor does it necessarily lie inside for an acute tetrahedron (i.e. when all dihedral angles are less than \pi/2) ==Integer tetrahedra==