Regular Tetrahedron — padho.wiki

In short

A regular tetrahedron is a solid with four equilateral triangular faces, four vertices, and six edges, all of equal length. The angle between any edge and the opposite face is \arctan(\sqrt{2}) \approx 54.74°. The dihedral angle between any two faces is \arccos(1/3) \approx 70.53°. For edge length a, its height is a\sqrt{2/3} and its volume is \frac{a^3}{6\sqrt{2}}.

Pick up a samosa — the triangular, deep-fried kind, not the cone-shaped one. If the samosa were perfect — every edge exactly the same length, every face a perfect equilateral triangle — it would be a regular tetrahedron: the simplest possible solid that encloses space with flat faces.

A triangle is the simplest polygon — three vertices, three edges. A regular tetrahedron is the 3D analogue: the simplest polyhedron. It has four vertices, six edges, and four faces, and every single edge has the same length. Every face is an equilateral triangle, and every vertex looks exactly the same as every other.

These symmetries make it a beautiful object to study. Every measurement — angle, height, volume — is determined entirely by one number: the edge length a. The rest of this article derives all of them from first principles, using coordinates and vectors.

Setting up coordinates

The cleanest way to work with a regular tetrahedron is to place it inside a coordinate system. There is a particularly elegant placement that exploits the connection between a regular tetrahedron and a cube.

A cube has eight vertices. Pick any vertex and take the three vertices that are diagonally opposite on the three faces meeting at that vertex. Those four vertices form a regular tetrahedron. The tetrahedron sits inside the cube, with its edges running along the face diagonals.

A regular tetrahedron inscribed inside a cube of side 1. The four red vertices $A(0,0,0)$, $B(1,1,0)$, $C(1,0,1)$, $D(0,1,1)$ are alternating vertices of the cube. Each edge of the tetrahedron is a face diagonal of the cube, so every edge has length $\sqrt{2}$.

Place the cube with one corner at the origin and side length 1. The eight vertices of the cube are all points (x, y, z) where each coordinate is 0 or 1. The four vertices that form a regular tetrahedron are:

A = (0, 0, 0), \quad B = (1, 1, 0), \quad C = (1, 0, 1), \quad D = (0, 1, 1)

Check the edge lengths. The distance from A to B is \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}. The distance from A to C is \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}. The distance from B to C is \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}. Every edge has length \sqrt{2} — so the edge length is a = \sqrt{2}.

This coordinate system makes every calculation clean. Whenever we need a general formula in terms of edge length a, we substitute a = \sqrt{2} at the end and simplify.

Properties of a regular tetrahedron

Before computing angles and volumes, collect the basic properties.

Faces: 4, each an equilateral triangle.
Edges: 6, all equal.
Vertices: 4, each meeting 3 edges.
Symmetry: 12 rotational symmetries (the rotation group of the tetrahedron). Every vertex can be mapped to every other vertex by a symmetry, which is why all angles and measurements computed at one vertex hold at every vertex.
Euler's formula check: V - E + F = 4 - 6 + 4 = 2. Correct for any convex polyhedron.

The centroid (centre of mass) of the regular tetrahedron ABCD is the average of all four vertices:

G = \frac{A + B + C + D}{4} = \frac{(0,0,0) + (1,1,0) + (1,0,1) + (0,1,1)}{4} = \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)

The centroid sits at the exact centre of the cube — another consequence of the symmetry.

Angle between an edge and a face

Take edge AD and the face BCD (the face opposite to vertex A). The angle between an edge and a face is the angle between the edge and its projection onto the plane of the face.

Step 1: Direction of edge AD.

\vec{AD} = D - A = (0, 1, 1) - (0, 0, 0) = (0, 1, 1)

Step 2: Find the equation of the plane through B, C, D.

Two vectors in the plane are:

\vec{BC} = C - B = (0, -1, 1), \quad \vec{BD} = D - B = (-1, 0, 1)

The normal to the plane is:

\mathbf{n} = \vec{BC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{vmatrix}

= \hat{i}((-1)(1) - (1)(0)) - \hat{j}((0)(1) - (1)(-1)) + \hat{k}((0)(0) - (-1)(-1))

= \hat{i}(-1) - \hat{j}(1) + \hat{k}(-1) = (-1, -1, -1)

So the normal to face BCD is \mathbf{n} = (-1, -1, -1), or equivalently (1, 1, 1).

Step 3: Compute the angle between edge \vec{AD} and the plane.

The angle \alpha between a line with direction \mathbf{d} and a plane with normal \mathbf{n} satisfies:

\sin\alpha = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}

Here \mathbf{d} = (0, 1, 1) and \mathbf{n} = (1, 1, 1):

\mathbf{d} \cdot \mathbf{n} = 0 + 1 + 1 = 2

|\mathbf{d}| = \sqrt{0 + 1 + 1} = \sqrt{2}, \quad |\mathbf{n}| = \sqrt{1 + 1 + 1} = \sqrt{3}

\sin\alpha = \frac{2}{\sqrt{2} \cdot \sqrt{3}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}

Therefore:

\alpha = \arcsin\sqrt{\frac{2}{3}} = \arctan\sqrt{2} \approx 54.74°

The last equality follows because if \sin\alpha = \sqrt{2/3}, then \cos\alpha = \sqrt{1/3}, so \tan\alpha = \sqrt{2/3}/\sqrt{1/3} = \sqrt{2}.

Angle between edge and face

In a regular tetrahedron, the angle between any edge and the face it does not belong to is

\alpha = \arctan(\sqrt{2}) \approx 54.74°

This angle is the same for every edge-face pair, by symmetry.

This angle is not 60° — a common misconception. Even though every face is equilateral (all angles 60°), the angle between an edge and a non-adjacent face is a different measurement and turns out to be \arctan\sqrt{2}.

The angle between edge $AD$ and the base face $BCD$. The edge makes an angle $\alpha = \arctan\sqrt{2} \approx 54.74°$ with its projection on the base. The right angle at the foot of the perpendicular confirms this is the angle between the edge and the plane.

Angle between two faces (dihedral angle)

The dihedral angle is the angle between two adjacent faces, measured along their common edge. Take faces ABC and BCD, which share edge BC.

Step 1: Find the normals to both faces.

For face ABC, use vectors in the plane:

\vec{AB} = B - A = (1, 1, 0), \quad \vec{AC} = C - A = (1, 0, 1)

\mathbf{n}_1 = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{vmatrix} = \hat{i}(1-0) - \hat{j}(1-0) + \hat{k}(0-1) = (1, -1, -1)

For face BCD, we already computed the normal: \mathbf{n}_2 = (-1, -1, -1), or equivalently (1, 1, 1).

Step 2: Compute the angle between the normals.

\cos\phi = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}

With \mathbf{n}_1 = (1, -1, -1) and \mathbf{n}_2 = (1, 1, 1):

\mathbf{n}_1 \cdot \mathbf{n}_2 = 1 - 1 - 1 = -1

|\mathbf{n}_1| = \sqrt{1 + 1 + 1} = \sqrt{3}, \quad |\mathbf{n}_2| = \sqrt{3}

\cos\phi = \frac{|-1|}{\sqrt{3}\cdot\sqrt{3}} = \frac{1}{3}

The dihedral angle is either \arccos(1/3) or \pi - \arccos(1/3). To determine which, note that the normals point outward from their respective faces — and the two outward normals make an obtuse angle (their dot product is negative). The dihedral angle is the supplement of this obtuse angle, which is \arccos(1/3).

Dihedral angle of a regular tetrahedron

The angle between any two adjacent faces of a regular tetrahedron is

\phi = \arccos\left(\frac{1}{3}\right) \approx 70.53°

This is the same for all six edges, by symmetry.

Notice that this is significantly less than 90° — the faces of a regular tetrahedron lean inward quite sharply. Two faces of a cube meet at 90°; two faces of a regular tetrahedron meet at about 70.5°.

Cross-section perpendicular to a shared edge. The two adjacent faces meet at the dihedral angle $\arccos(1/3) \approx 70.53°$. The dashed line is the altitude from the apex to the base — it sits exactly in the bisector of this dihedral angle.

Height of a regular tetrahedron

The height is the perpendicular distance from one vertex to the opposite face. Take vertex A = (0,0,0) and the face BCD.

Step 1: Find the equation of the plane through B, C, D.

The normal to face BCD is (1, 1, 1). The plane passes through B = (1, 1, 0), so its equation is:

1(x - 1) + 1(y - 1) + 1(z - 0) = 0 \quad \Rightarrow \quad x + y + z = 2

Step 2: Compute the distance from A = (0,0,0) to this plane.

h = \frac{|0 + 0 + 0 - 2|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{2}{\sqrt{3}}

Step 3: Express in terms of the edge length a.

In our coordinates, a = \sqrt{2}. So:

h = \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{1}{1} = \frac{2}{\sqrt{3}}

To express this in terms of a: since a = \sqrt{2}, we have a^2 = 2, so 2 = a^2 and \sqrt{3} = \sqrt{3}. Then:

h = \frac{a^2}{\sqrt{3}} \cdot \frac{1}{a/\sqrt{2}} \;?

A cleaner approach: compute h/a directly.

\frac{h}{a} = \frac{2/\sqrt{3}}{\sqrt{2}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{4}{6}} = \sqrt{\frac{2}{3}}

So h = a\sqrt{2/3}.

Height of a regular tetrahedron

For a regular tetrahedron with edge length a:

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

Both forms are equivalent. The second follows from rationalising: \sqrt{2/3} = \sqrt{6}/3.

For a = 1: h = \sqrt{6}/3 \approx 0.816. The tetrahedron is not as tall as it is wide — it is slightly squat.

Volume of a regular tetrahedron

The volume of any tetrahedron (not just a regular one) is \frac{1}{6}|\vec{AB} \cdot (\vec{AC} \times \vec{AD})| — one-sixth of the absolute value of the scalar triple product.

Step 1: Compute the vectors from A.

\vec{AB} = (1, 1, 0), \quad \vec{AC} = (1, 0, 1), \quad \vec{AD} = (0, 1, 1)

Step 2: Compute \vec{AC} \times \vec{AD}.

\vec{AC} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(0-1) - \hat{j}(1-0) + \hat{k}(1-0) = (-1, -1, 1)

Step 3: Compute the scalar triple product.

\vec{AB} \cdot (\vec{AC} \times \vec{AD}) = (1)(-1) + (1)(-1) + (0)(1) = -2

Step 4: Compute the volume.

V = \frac{1}{6}|{-2}| = \frac{1}{3}

Step 5: Express in terms of edge length a.

With a = \sqrt{2}, we have a^3 = 2\sqrt{2}. So:

V = \frac{1}{3} = \frac{a^3}{3 \cdot 2\sqrt{2}} \cdot 2\sqrt{2} \cdot \frac{1}{2\sqrt{2}} \;?

More directly: V/a^3 = (1/3)/(2\sqrt{2}) = 1/(6\sqrt{2}). So:

V = \frac{a^3}{6\sqrt{2}}

Volume of a regular tetrahedron

For a regular tetrahedron with edge length a:

V = \frac{a^3}{6\sqrt{2}} = \frac{a^3\sqrt{2}}{12}

Both forms are equivalent. The second follows from rationalising the denominator.

Verification by base-area method. The volume of a tetrahedron is also \frac{1}{3} \times \text{base area} \times \text{height}. The base is an equilateral triangle with side a = \sqrt{2}, so its area is \frac{\sqrt{3}}{4}(\sqrt{2})^2 = \frac{\sqrt{3}}{2}. The height is \frac{2}{\sqrt{3}}. Then:

V = \frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot \frac{2}{\sqrt{3}} = \frac{1}{3} \quad \checkmark

Both methods agree. The scalar triple product method is more direct; the base-times-height method is a good cross-check.

Example 1: Find the height and volume of a regular tetrahedron with edge length 6

Step 1. Record the edge length: a = 6.

Why: all measurements of a regular tetrahedron are determined by a single number — the edge length.

Step 2. Compute the height using h = \frac{a\sqrt{6}}{3}.

h = \frac{6\sqrt{6}}{3} = 2\sqrt{6} \approx 4.899

Why: the formula h = a\sqrt{2/3} = a\sqrt{6}/3 was derived from the perpendicular distance of a vertex to the opposite face.

Step 3. Compute the volume using V = \frac{a^3\sqrt{2}}{12}.

V = \frac{6^3 \cdot \sqrt{2}}{12} = \frac{216\sqrt{2}}{12} = 18\sqrt{2} \approx 25.456

Why: the volume formula gives cubic units, consistent with the cube of a length.

Step 4. Cross-check with base-area method. The base is an equilateral triangle with side 6, so its area is \frac{\sqrt{3}}{4}(36) = 9\sqrt{3}.

V = \frac{1}{3} \times 9\sqrt{3} \times 2\sqrt{6} = \frac{18\sqrt{18}}{3} = \frac{18 \cdot 3\sqrt{2}}{3} = 18\sqrt{2} \quad \checkmark

Why: the base-area method confirms the scalar triple product result. Two independent methods giving the same answer is strong evidence of correctness.

Result: Height = 2\sqrt{6} units, Volume = 18\sqrt{2} cubic units.

The regular tetrahedron with edge 6. The dashed line from the apex to the centroid of the base has length $2\sqrt{6} \approx 4.9$ — about 82% of the edge length. The volume is $18\sqrt{2} \approx 25.5$ cubic units.

For comparison, a cube with edge 6 has volume 216. The tetrahedron inscribed in it has volume 18\sqrt{2} \approx 25.5, which is about 11.8\% of the cube's volume. A regular tetrahedron uses its bounding box very inefficiently — most of the space inside the cube is outside the tetrahedron.

Example 2: Find the centroid and circumradius of a regular tetrahedron with vertices at (1,1,1), (1,−1,−1), (−1,1,−1), (−1,−1,1)

Step 1. Verify the tetrahedron is regular. Compute one edge length:

|AB| = |(1 - 1, -1 - 1, -1 - 1)| = |(0, -2, -2)| = \sqrt{0 + 4 + 4} = 2\sqrt{2}

Check another: |AC| = |(-2, 0, -2)| = \sqrt{4 + 0 + 4} = 2\sqrt{2}. And |BC| = |(-2, 2, 0)| = 2\sqrt{2}.

All edges are 2\sqrt{2}, so the tetrahedron is regular with a = 2\sqrt{2}.

Why: always verify regularity before using the special formulas. A tetrahedron with four vertices need not be regular.

Step 2. Find the centroid.

G = \frac{(1,1,1) + (1,-1,-1) + (-1,1,-1) + (-1,-1,1)}{4} = \frac{(0, 0, 0)}{4} = (0, 0, 0)

Why: for a regular tetrahedron, the centroid is equidistant from all four vertices and lies at the average of their coordinates.

Step 3. Compute the circumradius — the distance from the centroid to any vertex.

R = |GA| = |(1, 1, 1)| = \sqrt{1 + 1 + 1} = \sqrt{3}

Why: by symmetry, the circumcentre coincides with the centroid for a regular tetrahedron, so the circumradius is just the distance from the centroid to any vertex.

Step 4. Express R in terms of edge length a. With a = 2\sqrt{2}:

\frac{R}{a} = \frac{\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{3}}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}}{4}

So R = \frac{a\sqrt{6}}{4} — this is the circumradius formula for a regular tetrahedron.

Result: Centroid = (0, 0, 0), Circumradius = \sqrt{3} = \frac{a\sqrt{6}}{4} where a = 2\sqrt{2}.

Projection of the regular tetrahedron onto the $xy$-plane. All four vertices project to the corners of a square. The dashed circle has radius $\sqrt{3}$ — the circumradius — and passes through all four projected points. The centroid $G$ sits at the origin.

The xy-projection shows the symmetry clearly: the four vertices form a square in projection, and the centroid lands right at the centre. The circumradius \sqrt{3} is the distance from the centre to each vertex.

Common confusions

"All four faces are equilateral, so the dihedral angle must be 60°." The dihedral angle and the face angle are different measurements. The faces have angles of 60°, but the angle between two faces (the dihedral angle) is \arccos(1/3) \approx 70.53°.
"The height of a regular tetrahedron is a\sqrt{3}/2." That is the height of an equilateral triangle with side a, not the height of the tetrahedron. The tetrahedron's height is a\sqrt{6}/3, which is less than a\sqrt{3}/2.
"The centroid, circumcentre, incentre, and orthocentre of a regular tetrahedron are all different points." They are all the same point. By symmetry, the unique centre of a regular tetrahedron serves all four roles simultaneously. This is analogous to an equilateral triangle in 2D, where all four centres coincide.
"I can compute the volume as \frac{1}{3} \times base area \times edge length." The factor of \frac{1}{3} goes with the height, not the edge length. The height of a regular tetrahedron is shorter than the edge, so using the edge instead gives a volume that is too large.

Going deeper

If you came here for the formulas and the worked examples, you have them. The rest is for readers who want to understand why the cube-inscription trick works and what the regular tetrahedron's place is among the Platonic solids.

Why the cube-inscription works

A cube has eight vertices. Colour them black and white like a 3D chessboard: two vertices sharing an edge get different colours. Each vertex is adjacent to three others, and those three are all the opposite colour. There are exactly 4 black vertices and 4 white vertices.

The 4 black vertices form a regular tetrahedron. The 4 white vertices form another regular tetrahedron. The two tetrahedra together (called the stella octangula) have the cube as their convex hull and a regular octahedron as their intersection.

The reason the black vertices form a regular tetrahedron is that every pair of black vertices is connected by a face diagonal of the cube (since they are never on the same edge). Every face diagonal of a cube with side s has length s\sqrt{2}, and there are exactly \binom{4}{2} = 6 such pairs — the same as the number of edges in a tetrahedron.

The inradius

The centroid $G$ divides the altitude from vertex to base in the ratio $3:1$ from the vertex. The circumradius $R$ (dashed, vertex to $G$) is exactly three times the inradius $r$ (solid, $G$ to base).

The inradius r of a regular tetrahedron is the radius of the largest sphere that fits inside it, tangent to all four faces. For a regular tetrahedron with edge length a:

r = \frac{a}{2\sqrt{6}} = \frac{a\sqrt{6}}{12}

Notice that the circumradius R = \frac{a\sqrt{6}}{4} is exactly 3r. This R = 3r relation is specific to the regular tetrahedron. For comparison, in an equilateral triangle R = 2r. The ratio R/r grows as you go from 2D to 3D regular simplices.

Platonic solids

The regular tetrahedron is one of exactly five Platonic solids — convex polyhedra with all faces congruent regular polygons and the same number of faces meeting at each vertex. The five are: tetrahedron (4 triangular faces), cube (6 square faces), octahedron (8 triangular faces), dodecahedron (12 pentagonal faces), and icosahedron (20 triangular faces).

The fact that there are exactly five is a theorem — proved by Euclid, and it is one of the oldest impossibility results in mathematics. The proof uses Euler's formula and the constraint that the angles meeting at a vertex must sum to less than 360°. With equilateral triangles, you can have 3, 4, or 5 at a vertex (giving tetrahedron, octahedron, icosahedron). With squares, only 3 (giving the cube). With regular pentagons, only 3 (giving the dodecahedron). With regular hexagons, 3 meet at 360° exactly — that gives a flat tiling, not a solid. So the list stops.

Indian mathematical and architectural traditions have long known these shapes. The Shulba Sutras — ancient Sanskrit texts on altar construction — describe precise geometric constructions for 3D shapes, and the symmetries of regular polyhedra appear in temple architecture from Hoysala to Chola periods.

Where this leads next

3D Coordinates — the coordinate system that made all the calculations in this article possible.
Dot Product — the tool used to compute angles between edges, faces, and normals.
Cross Product — how to find normal vectors to faces and areas of triangular faces.
Scalar Triple Product — the single formula that gives the volume of any tetrahedron, regular or not.
Plane: Basic Equations — the equation of the face of a tetrahedron as a plane, which feeds into distance and angle calculations.