dihedral angles – Elite Math Program

A tetrahedron is the simplest three-dimensional polyhedron: four vertices, six edges, and four triangular faces. Yet it hides some remarkable relationships. We will discover why its four face-area vectors add to zero, derive its volume from coordinates, prove a three-dimensional Pythagorean theorem, and uncover identities connecting face areas and dihedral angles. Four discovery problems, with solutions hidden at the end, invite you to go further.

Figure 1. A tetrahedron with vertices P, Q, R, and S. Dashed edges indicate edges that may be hidden in the three-dimensional view.

1. The four faces and their area vectors

For each face, choose the unit normal pointing outward from the solid. Multiply that normal by the area of the face. The resulting vector is called its outward area vector. Denote the four vectors by N₁, …, N₄, and their lengths (the face areas) by A₁, …, A₄.

N_{1} + N_{2} + N_{3} + N_{4} = 0

Geometric proof by projection

Choose any direction, represented by a unit vector u. The signed area of a face projected onto a plane perpendicular to u is the dot product of its outward area vector with u. Looking through the tetrahedron along u, each projected interior point is entered through a face and exited through another. The positive and negative projected contributions cancel. Thus the sum of the four dot products with u is zero. Since this is true for every direction u, the sum of the area vectors itself must be zero.

Algebraic proof by cross products

Write a = Q − P, b = R − P, and c = S − P. Assume the orientation is chosen so that det(a, b, c) is positive. The outward area vectors for the faces opposite S, R, Q, and P, respectively, are

\begin{array}{l} N_{S} = - \frac{1}{2} (a \times b), & N_{R} = \frac{1}{2} (a \times c), \\ N_{Q} = - \frac{1}{2} (b \times c), & N_{P} = \frac{1}{2} (b \times c - a \times c + a \times b) . \end{array}

Adding these four expressions gives the zero vector. If the determinant is negative, all four displayed normals reverse direction; the sum remains zero.

2. The volume of a tetrahedron

Let the vertices have position vectors P, Q, R, S. Set a = Q − P, b = R − P, c = S − P. The parallelepiped spanned by these vectors has volume equal to the absolute value of their scalar triple product. A tetrahedron occupies one-sixth of that parallelepiped:

V = \frac{1}{6} | det (a, b, c) | = \frac{1}{6} | (a \times b) \cdot c | .

Why one-sixth? The base triangle has half the area of the parallelogram spanned by a and b. A pyramid has one-third the volume of a prism with the same base and height. Multiplying these factors gives one-sixth.

Figure 2. Base PQR and perpendicular altitude SH. Rear edges PR, RQ, and RS are dashed; the internal altitude is dashed as well.

A new numerical example

Consider the tetrahedron with vertices P = (0, 0, 0), Q = (4, 0, 0), R = (0, 3, 0), and S = (1, 1, 6). This is different from the example in the textbook. The three edge vectors based at P are (4, 0, 0), (0, 3, 0), and (1, 1, 6). Therefore

V = \frac{1}{6} | det (\begin{matrix} 4 & 0 & 1 \\ 0 & 3 & 1 \\ 0 & 0 & 6 \end{matrix}) | = \frac{72}{6} = 12 .

We can verify this without determinants: triangle PQR is right-angled, with area (4 × 3)/2 = 6. Since S has third coordinate 6, its perpendicular distance from the plane z = 0 is 6. Thus V = (1/3)(6)(6) = 12 cubic units.

3. The three-dimensional Pythagorean theorem

Suppose the three edges meeting at S are mutually perpendicular. Such a vertex is called tri-rectangular. Let A, B, C denote the areas of the three faces meeting at S, and D the area of the opposite face PQR. Then

D^{2} = A^{2} + B^{2} + C^{2} .

This is de Gua’s theorem, a three-dimensional counterpart of the Pythagorean theorem.

Figure 3. Three mutually perpendicular edges meet at S. The rear edge QR is dashed where it passes behind the front faces. Projected angles need not appear to be 90°.

Proof using area vectors

The three faces meeting at S lie in mutually perpendicular planes. Their outward area vectors are therefore pairwise perpendicular. By Section 1, the outward area vector of PQR is the negative of their sum. Squaring its length and using the Pythagorean theorem for orthogonal vectors gives D² = A² + B² + C².

Independent coordinate proof

Place S at the origin and the other vertices at P = (a, 0, 0), Q = (0, b, 0), R = (0, 0, c), with positive a, b, c. The three faces at S have areas ab/2, ac/2, and bc/2. The area of PQR is half the length of (Q − P) × (R − P), which equals (bc, ac, ab). Consequently

D^{2} = \frac{a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2}}{4} = A^{2} + B^{2} + C^{2} .

4. A surprising inequality for the four face areas

Let A₁, A₂, A₃, A₄ be the areas of the faces of any nondegenerate tetrahedron. Since N₁ = −(N₂ + N₃ + N₄), the triangle inequality gives

A_{1} = ‖ N_{2} + N_{3} + N_{4} ‖ \leq A_{2} + A_{3} + A_{4} .

In a nondegenerate tetrahedron the inequality is strict: equality in the vector triangle inequality would require all three outward normals on the right to point in the same direction, impossible for three distinct faces of a genuine tetrahedron. The same argument applies to every face.

Immediate test: no tetrahedron can have face areas 2, 3, 4, and 10, because 10 is greater than 2 + 3 + 4.

5. The angle between two faces

Two faces sharing an edge meet at an interior dihedral angle θ. The angle between their outward normals is π − θ, not θ. Thus, if their outward area vectors have lengths Aᵢ and Aⱼ, the dot product formula gives

N_{i} \cdot N_{j} = A_{i} A_{j} cos (π - θ_{i j}) = - A_{i} A_{j} cos θ_{i j} .

Figure 4. Cross-section perpendicular to the common edge. The blue rays enclose the interior dihedral angle θ. The orange outward normals are perpendicular to their respective rays; their angle is π − θ.

For a regular tetrahedron, the interior dihedral angle satisfies cos θ = 1/3. The outward normals therefore have dot product −A²/3, where A is their common face area.

6. An identity involving all six dihedral angles

Start with the zero-sum identity N₁ + N₂ + N₃ + N₄ = 0. Take the squared length of both sides:

0 = \sum_{i = 1}^{4} A_{i}^{2} + 2 \sum_{i < j} N_{i} \cdot N_{j} .

Substitute the dot-product formula from Section 5 and rearrange:

\sum_{i = 1}^{4} A_{i}^{2} = 2 \sum_{i < j} A_{i} A_{j} cos θ_{i j} .

The second sum runs over the six unordered pairs of faces. This identity holds for every nondegenerate tetrahedron, whether or not any of its angles are right angles. De Gua’s theorem is a special orthogonal configuration of the same area-vector principle.

Discovery problems

Try these before opening the solutions. The problems become progressively more demanding.

Problem 1. The regular tetrahedron

A regular tetrahedron has six edges of length a. Find (i) the area of one face, (ii) its altitude, (iii) its volume, and (iv) its interior dihedral angle.

Problem 2. Which face areas are possible?

Can a nondegenerate tetrahedron have face areas 2, 3, 4, and 10? Prove your answer. Then decide whether the four numbers 2, 3, 4, and 8 are ruled out by the same argument.

Problem 3. An inverse problem

If the areas of all four faces are known, must the tetrahedron’s volume be uniquely determined? Prove your answer by constructing two tetrahedra with the same four face areas but different volumes.

Problem 4. Higher-dimensional Pythagoras

In n-dimensional Euclidean space, let an n-simplex have n mutually perpendicular edges meeting at one vertex. Prove that the square of the (n − 1)-dimensional volume of its opposite facet equals the sum of the squares of the (n − 1)-dimensional volumes of its other n facets.

Solutions to the discovery problems

Each solution is hidden until you choose to expand it.

Solution 1 — Regular tetrahedron

Each face is equilateral, so its area is √3 a²/4. The foot of an altitude is the centroid of the opposite equilateral triangle. The distance from that centroid to a vertex is a/√3, so the altitude is √(a² − a²/3) = a√(2/3). Hence V = (1/3)(√3 a²/4)(a√(2/3)) = √2 a³/12.

All four outward area vectors have the same length A and sum to zero. By symmetry, the dot product between any two distinct outward area vectors is a common value c. Squaring their sum gives 0 = 4A² + 12c, so c = −A²/3. Therefore −A² cos θ = −A²/3 and θ = arccos(1/3) (approximately 70.53°).

Solution 2 — Which face areas are possible?

The largest face cannot exceed the sum of the other three, and in a genuine tetrahedron the inequality must be strict. Since 10 > 2 + 3 + 4 = 9, the first collection is impossible. For 2, 3, 4, 8, the largest area is 8 and the sum of the other three is 9, so the same necessary inequality does not rule it out. Passing this test alone does not construct a tetrahedron.

Solution 3 — Same face areas, different volumes

Consider the four vertices (x,y,z), (x,−y,−z), (−x,y,−z), (−x,−y,z), with x,y,z > 0. Each face has the same area

A = 2 \sqrt{x^{2} y^{2} + x^{2} z^{2} + y^{2} z^{2}},

while its volume is 8xyz/3. First choose x = y = z = 1. Every face has area 2√3 and the volume is 8/3. Next choose x = y = 1/√2 and z = √11/2. Then x²y² + x²z² + y²z² = 1/4 + 11/8 + 11/8 = 3, so again every face has area 2√3. But the volume is (8/3)(1/2)(√11/2) = 2√11/3, different from 8/3. Thus the four face areas do not determine the volume.

Solution 4 — The n-dimensional Pythagorean theorem

Put the right-angle vertex at the origin and the other vertices at a₁e₁, …, aₙeₙ, where the eᵢ form an orthonormal basis and each aᵢ > 0. The (n − 1)-dimensional volume Fᵢ of the facet opposite aᵢeᵢ (which contains the origin) is

F_{i} = \frac{\prod_{j \neq i} a_{j}}{(n - 1)!} .

Let F₀ denote the volume of the opposite facet. The full simplex has n-volume V = (a₁⋯aₙ)/n!. Its opposite facet lies in the hyperplane x₁/a₁ + ⋯ + xₙ/aₙ = 1, whose distance from the origin is h = 1/√(Σᵢ 1/aᵢ²). The pyramid formula V = F₀h/n gives

F_{0} = \frac{a_{1} \dots a_{n}}{(n - 1)!} \sqrt{\sum_{i = 1}^{n} \frac{1}{a_{i}^{2}}} .

Squaring and distributing the product shows F₀² = Σᵢ Fᵢ², as required. For n = 3 this is de Gua’s theorem; for n = 2 it is the ordinary Pythagorean theorem.

Final perspective

The central insight is that face areas of a tetrahedron are not merely four unrelated numbers: they are the lengths of four outward vectors whose sum is zero. That simple fact produces an area inequality, de Gua’s theorem, and a six-angle identity. The same ideas extend to simplices in higher dimensions, where the geometry of facet normals continues to organize the geometry of the whole solid.

The tetrahedron is the simplest three-dimensional simplex. For a related exploration of simplex geometry, see The Steiner Inellipse and Its Area Characterization .

Area vectors and surface normals also play an important role in integration over spheres. For another application, see our article on integrals over the sphere .

Tag: dihedral angles

The Geometry of a Tetrahedron: From Pythagoras to Vector Identities