Regular tetrahedron

 ( Deltahedra ) 

A regular tetrahedron is a polyhedron with four equilateral triangular faces.

---

Description

---

Classifications A regular tetrahedron is a tetrahedron (that is, a four-sided polyhedron) in which all four faces are equilateral triangles. In other words, all of its faces are the same size and shape (congruent), and all edges are the same length. As a convex polyhedron, the regular tetrahedron is the deltahedron with the smallest number of vertices and faces out of eight convex deltahedra. Being a deltahedron means that all faces of a polyhedron are equilateral triangles. Like the other pyramids and all tetrahedra in general, the regular tetrahedron is self-dual, meaning its Dual polyhedron is a regular tetrahedron itself.

The regular tetrahedron is also one of the five regular —polyhedra in which all of their faces are regular polygons. Known since antiquity, Platonic solids are named after the Greek philosopher Plato, who associated four of those solids with fundamental natural elements. He assigned the regular tetrahedron to the classical element of fire, because its corner is sharpest and most penetrating. Following its attribution with nature by Plato, Johannes Kepler in his Harmonices Mundi sketched each of the Platonic solids. In his Mysterium Cosmographicum, Kepler also proposed the Solar System by using the Platonic solids, setting one inside another and separating them with six spheres resembling the six planets. The ordered solids started from the innermost to the outermost: regular octahedron, regular icosahedron, regular dodecahedron, regular tetrahedron, and cube.

Regular tetrahedron is the degenerated antiprism.

---

Cartesian coordinates One way to construct a regular tetrahedron is by using the following Cartesian coordinates, defining the four vertices of a tetrahedron with edge length 2, centered at the origin, and two-level edges:$$\backslash left(\backslash pm\; 1,\; 0,\; -\backslash frac\{1\}\{\backslash sqrt\{2\}\}\backslash right)\; \backslash qquad\; \backslash left(0,\; \backslash pm\; 1,\; \backslash frac\{1\}\{\backslash sqrt\{2\}\}\backslash right)$$

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$$\backslash begin\{align\}$$

(1,1,1), &\quad (1,-1,-1), \\
(-1,1,-1), &\quad (-1,-1,1).
     

\end{align} This yields a tetrahedron with edge-length $2\; \backslash sqrt\{2\}$, centered at the origin. For the other tetrahedron (which is dual polyhedron 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-Demihypercube, a polyhedron that is by alternating a cube. This form has Coxeter diagram and Schläfli symbol $\backslash mathrm\{h\}\backslash \{4,3\backslash \}$.

---

Properties

---

Measurement Let $a$ be the edge length of a regular tetrahedron. The height of an equilateral triangle, the segment line is drawn from any vertex to the midpoint of an edge, and by the calculation of Pythagorean theorem, the height of any equilateral triangle is $\backslash frac\{1\}\{2\}a\backslash sqrt\{3\}$. The distance between the centroid and a vertex has two different lengths; the longest is two-thirds of the height of an equilateral triangle. By the Pythagorean theorem again, one can find the height of a regular tetrahedron: $$h\; =\; \backslash frac\{\backslash sqrt\{6\}\}\{3\}a\; \backslash approx\; 0.816a.$$ Since the shortest distance between the centroid and a vertex is one-third of the height of an equilateral triangle, the dihedral angle (i.e., an angle between two triangular faces) of a regular tetrahedron is $\backslash arccos\; \backslash left(1/3\; \backslash right)\; =\; \backslash arctan\backslash left(2\backslash sqrt\{2\}\backslash right)\; \backslash approx\; 70.529^\backslash circ$.

The surface area of a regular tetrahedron is four times area of an equilateral triangle:$$A\; =\; 4\; \backslash cdot\; \backslash left(\backslash frac\{\backslash sqrt\{3\}\}\{4\}a^2\backslash right)\; =\; a^2\; \backslash sqrt\{3\}\; \backslash approx\; 1.732a^2.$$

The volume is one-third of the base times the height, the general formula for a pyramid. This can also be found by dissecting a cube into a tetrahedron and four triangular pyramids.$$V\; =\; \backslash frac\{1\}\{3\}\; \backslash cdot\; \backslash left(\backslash frac\{\backslash sqrt\{3\}\}\{4\}a^2\backslash right)\; \backslash cdot\; \backslash frac\{\backslash sqrt\{6\}\}\{3\}a\; =\; \backslash frac\{a^3\}\{6\backslash sqrt\{2\}\}\; \backslash approx\; 0.118a^3.$$

The radii of its circumsphere $r\_\backslash mathrm\{ci\}$ (a sphere that contains the regular tetrahedron and touches its vertices.), insphere $r\_\backslash text\{in\}$ (a sphere within a regular tetrahedron and touches to its faces, midsphere $r\_\backslash mathrm\{mi\}$ (a sphere that touches its edges), and exsphere (a sphere outside, touches the face of a regular tetrahedron and the planes defined by extending the adjacent faces outwards) $r\_\backslash mathrm\{ex\}$ are: $$\backslash begin\{align\}$$

r_\mathrm{ci} = \frac{\sqrt{3}}{2\sqrt{2}} \approx 0.612a, &\qquad r_\mathrm{in} = \frac{1}{3}r_\mathrm{ci} \approx 0.204a, \\
r_\mathrm{mi} = \sqrt{r_\mathrm{in} r_\mathrm{ci}} \approx 0.354a, &\qquad r_\mathrm{ex} = \frac{a}{\sqrt{6}} \approx 0.408a.
     

\end{align} For a regular tetrahedron with side length $a$ and circumsphere radius $R$, the distances $d\_i$ from an arbitrary point in 3-space to its four vertices satisfy the equations:

With respect to the base plane the slope of a face (2) is twice that of an edge (), 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 ).

Its solid angle at a vertex subtended by a face is $\backslash arccos\backslash left(\backslash frac\{23\}\{27\}\backslash right)\; =\; \backslash frac\{\backslash pi\}\{2\}\; -\; 3\backslash arcsin\backslash left(\backslash frac\{1\}\{3\}\backslash right)\; =\; 3\backslash arccos\; \backslash left(\backslash frac\{1\}\{3\}\backslash right)-\backslash pi,$ or approximately 0.55129 or 1809.8 .

---

Symmetry The regular tetrahedron has a three-dimensional symmetry group known as full tetrahedral symmetry $\backslash mathrm\{T\}\_\backslash mathrm\{d\}$. This symmetry group has 24 isometries, containing seven rotational axes and six reflectional planes. The seven rotational axes are the four axes of three-fold rotational symmetry (0°, 120°, and 240°) passing through a vertex to the centroid of an equilateral triangular face, and the three axes of two-fold rotational symmetry (0° and 180°) passing through the midpoint of two edges. This point group has rotational tetrahedral symmetry $\backslash mathrm\{T\}$. The six reflections in a plane perpendicular to an edge, six reflections in a plane combined with 90° rotation about an axis perpendicular to the plane, consisted of three axes, two per axis, together six (equivalently, they are 90° rotations combined with inversion).

The vertices of a cube can be grouped into two groups of four, each forming a regular tetrahedron, 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 not mapped to itself by point inversion.

---

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 A₂ Coxeter plane.

+ Orthographic projection !Centered by !Face/vertex !Edge

---

Related figures

---

Compounds The stellated octahedron is constructed by Stellation to form equilateral triangles on each regular octahedron's face. In the perspective of polyhedral compounds, this figure comprises two such dual tetrahedra, a self-dual because of sharing a common intersphere in the center. Another interesting polyhedral compound involves five intersecting tetrahedra, which 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 of each other. Superimposing both forms gives a compound of ten tetrahedra, in which the ten tetrahedra are arranged as five pairs of stella octangula.

---

Construction of polyhedra Many polyhedra are constructed involving regular tetrahedra. By truncating the vertices of a regular tetrahedron, it becomes a truncated tetrahedron. The dual of this solid is the triakis tetrahedron, a regular tetrahedron with four triangular pyramids attached to each of its faces. i.e., its Kleetope. Some Johnson solid such as elongated triangular pyramid and elongated triangular bipyramid are constructed by attaching one and two regular tetrahedra onto the bases of a triangular prism; the triangular bipyramid is constructed by attaching two regular tetrahedra face-to-face.

Regular tetrahedra can be stacked face-to-face in a chiral aperiodic chain called the Boerdijk–Coxeter helix.

---

Four-dimensional figures The pentachoron is a Polytope, a generalization of a tetrahedron in four-dimensional space. It is bounded by five regular tetrahedra, known as cells.

In four dimensions, all the convex regular 4-polytopes with tetrahedral cells (the 5-cell, 16-cell and 600-cell) can be constructed as tilings of the 3-sphere by these chains, which become periodic in the three-dimensional space of the 4-polytope's boundary surface.

---

Dehn invariant and honeycombs The Dehn invariant of a regular tetrahedron can be defined as a Tensor product of the edge length and the dihedral angle of a regular tetrahedron $6a\; \backslash otimes\; \backslash arccos\; \backslash frac\{1\}\{3\}$, which is non-zero, where $a$ is the edge length of a regular tetrahedron. The Dehn invariant was originally dated from Hilbert's third problem, Hilbert problems by David Hilbert, asking whether, given any two polyhedra with the same volume, the first polyhedron may be dissected into pieces, which then reassembled into the second polyhedron. His student, Max Dehn, provided the negative answer by showing his invariant.

Every polyhedron with a Dehn invariant of zero can tile a space with its copy by attaching its faces to another, forming a honeycomb. Since the regular tetrahedron has a non-zero Dehn invariant, it cannot alone do such. A related result that two different polyhedra that join to tile the space can have the Dehn invariant of zero. In the case of a regular tetrahedron, it can be alternated with regular octahedra in the ratio of two tetrahedra to one octahedron, forming the alternated cubic honeycomb.

---

Spherical tetrahedron The tetrahedron can also be represented as a spherical tiling (of spherical triangles), and projected onto the plane via a stereographic projection. This projection is Conformal map, preserving angles but not areas or lengths. Straight lines on the sphere are projected as circular arcs on the plane.

---

Cross-section of a regular tetrahedron 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. 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.

---

Footnotes

---

Bibliographies

• (2026). 9781614442165, Mathematical Association of America. ISBN 9781614442165
• Mark A. Armstrong (1988). 9781475740349, Springer. . ISBN 9781475740349
• (2026). 9783319939483, Springer International Publishing. ISBN 9783319939483
• Peter R. Cromwell (1997). 9780521554329, Cambridge University Press. . ISBN 9780521554329
• Robin Hartshorne (2026). 9780387986500, Springer-Verlag, New York. ISBN 9780387986500
• Mario Livio (2026). 9780767908160, Random House. . ISBN 9780767908160
• (2026). 9783030880590 . ISBN 9783030880590
• (2026). 9789811237126, World Scientific. . ISBN 9789811237126
• Anthony Pugh (1976). 9780520030565, University of California Press. . ISBN 9780520030565
• Larisa V. Shavinina (2026). 9780203387146, Routledge. ISBN 9780203387146
• Robert Williams (1979). 9780486237299, Dover Publications, Inc.. . ISBN 9780486237299

Post Comment