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A line segment along which two faces of a polyhedron meet. A rectangular parallelepiped has twelve edges, four parallel to each of its three side directions.

At an edge of a convex polyhedron, its interior dihedral angle is the angle of the wedge inside the solid, measured in a plane perpendicular to the edge. If the two incident faces have outward unit normal vectors ${\hat{\mathbf{n}}}_1,{\hat{\mathbf{n}}}_2$, then their normal angle is supplementary to the interior angle. Thus $\cos \theta =-{\hat{\mathbf{n}}}_1\cdot {\hat{\mathbf{n}}}_2$. Equivalently, subtract from each vector pointing into one face its orthogonal projection onto the common edge; the angle between the resulting inward face directions is the interior angle. This distinguishes an interior angle from the acute angle convention for two unoriented planes.

A linear polyhedron is an intersection of finitely many closed affine half-spaces in finite-dimensional real space. It may be empty, unbounded, or lower-dimensional. This usage differs from a three-dimensional geometric polyhedron. The strict separation of disjoint linear polyhedra follows from linear programming duality.

An octahedron is a polyhedron with eight faces. The regular octahedron has six vertices, twelve equal edges and eight equilateral triangular faces. One realization has vertices $\pm {\mathbf{e}}_1,\pm {\mathbf{e}}_2,\pm {\mathbf{e}}_3$; two vertices are adjacent exactly when they are not an opposite pair. Its edge graph is the octahedral graph.

A parallelepiped is a polyhedron with six parallelogram faces in three opposite pairs. Three generating vectors determine its edges; the absolute value of their determinant gives its volume in orthonormal Cartesian coordinates.

For an open cover $X=A\cup B$, the Mayer–Vietoris theorem gives the long exact sequenceThe same sequence holds for a union of simplicial subcomplexes. All homology groups here have integer coefficients.

An explicit triangulation of $X_q$ is as follows. Take vertices $b_0,b_1,b_2$ forming a target circle, a circle of $3q$ vertices $a_0,\dots ,a_{3q-1}$, and a cone vertex $c$. With subscripts on $a$ interpreted modulo $3q$ and on $b$ modulo three, include trianglesfor every $j$, together with all their faces. The first two types triangulate the mapping cylinder of the degree-$q$ covering $a_j\mapsto b_j$; the last type fills the inner circle by a disc. A filled disc with an attached annular collar is still a disc, and identifying the outer collar boundary by this covering gives exactly the source quotient. Distinct inner vertices ensure this is a genuine simplicial complex, not just a one-cell presentation.

Let $A$ be the mapping cylinder and $B$ the coned disc. Then $A$ retracts to the target circle, $B$ is contractible and $A\cap B$ is the inner circle. Its inclusion into $A$ induces multiplication by $q$ on $H_1$. The exact sequence therefore reduces toConnectedness and the absence of higher-dimensional simplices giveThis is a two-dimensional Moore space with torsion in degree one.

The suspension of a connected complex is the union of two cones whose intersection retracts to the original complex. Reduced Mayer-Vietoris givesIn particular its $H_1$ vanishes. For the wedge of two connected polyhedra, choose neighborhoods retracting onto the two summands with contractible intersection at the wedge point. ThenDecompose each prescribed $G_i$ into cyclic summands. Realize each infinite cyclic summand in degree $i$ by $S^i$, and each finite nontrivial summand $\mathbb{Z}/q$ by $S^{i-1}{X}_q$, meaning $i-1$ successive suspensions. Take their finite wedge, or a point if there are no summands. The suspension formula shifts the torsion to degree $i$ and creates no other positive-degree homology, while the wedge formula adds summands. Thus the resulting connected polyhedron satisfies $\boxed{H_0(X)=\mathbb{Z},H_i(X)=G_i(1\le i\le n),H_i(X)=0(i>n)}$. The finite triangulations above, their simplicial suspensions and wedges provide an actual polyhedron.

For every nonempty simplex $\sigma$ of $K$, let $b_{\sigma }$ be its barycentre. The first barycentric subdivision $K^{\prime }$ has the $b_{\sigma }$ as vertices, andis a simplex exactly when, after reordering, ${\sigma }_0⊊\cdots ⊊{\sigma }_q$. Its realization is the same polyhedron as that of $K$. The iterated barycentric subdivision is defined by

A seven-vertex polyhedron consisting of a five-membered equatorial ring and two axial vertices, each joined to every ring vertex. Removing an equatorial vertex leaves the six-vertex pentagonal-pyramidal skeleton used for some nido clusters.

A polyhedron with two congruent parallel polygonal bases whose corresponding vertices are joined by parallel edges. A right prism has edges between the bases perpendicular to their planes. The optical use of a transparent prism is a separate application of this geometric shape.

A pyramid is a polyhedron with a polygon as its base and triangular faces joining the base edges to one apex outside the base plane. If its base area is $A$ and perpendicular height is $h$, its volume is $Ah/3$. At height $z$ above the base, similarity gives cross-sectional area $A(1-z/h{)}^2$. Integrating this area from $0$ to $h$ proves the formula.

For the faces of a closed polyhedron, take outward unit normal vectors. Applying the divergence theorem to an arbitrary constant vector field $c$ givesSince $c$ is arbitrary, the vector areas sum to zero. This is an oriented statement; the positive scalar surface areas do not cancel. The same proof works for every piecewise smooth closed boundary to which the divergence theorem applies.