Measurements Every ideal polyhedron with n vertices has a surface that can be subdivided into 2n-4
ideal triangles, each with area \pi. Therefore, the surface area is exactly (2n-4)\pi. In an ideal polyhedron, all face angles and all solid angles at vertices are zero. However, the
dihedral angles on the edges of an ideal polyhedron are nonzero. At each vertex, the
supplementary angles of the dihedral angles incident to that vertex sum to exactly 2\pi. This fact can be used to calculate the dihedral angles themselves for a regular or
edge-symmetric ideal polyhedron (in which all these angles are equal), by counting how many edges meet at each vertex: an ideal regular tetrahedron, cube or dodecahedron, with three edges per vertex, has dihedral angles 60^\circ=\pi/3=\pi(1-\tfrac{2}{3}), an ideal regular octahedron or
cuboctahedron, with four edges per vertex, has dihedral angles 90^\circ=\pi/2=\pi(1-\tfrac{2}{4}), and an ideal regular icosahedron, with five edges per vertex, has dihedral angles 108^\circ=3\pi/5=\pi(1-\tfrac{2}{5}). The volume of an ideal
tetrahedron can be expressed in terms of the
Clausen function or
Lobachevsky function of its dihedral angles, and the volume of an arbitrary ideal polyhedron can then be found by partitioning it into tetrahedra and summing the volumes of the tetrahedra. The
Dehn invariant of a polyhedron is normally found by combining the edge lengths and dihedral angles of the polyhedron, but in the case of an ideal polyhedron the edge lengths are infinite. This difficulty can be avoided by using a
horosphere to
truncate each vertex, leaving a finite length along each edge. The resulting shape is not itself a polyhedron because the truncated faces are not flat, but it has finite edge lengths, and its Dehn invariant can be calculated in the normal way, ignoring the new edges where the truncated faces meet the original faces of the polyhedron. Because of the way the Dehn invariant is defined, and the constraints on the dihedral angles meeting at a single vertex of an ideal polyhedron, the result of this calculation does not depend on the choice of horospheres used to truncate the vertices.
Combinatorial structure As proved, the
maximum independent set of any ideal polyhedron (the largest possible subset of non-adjacent vertices) must have at most half of the vertices of the polyhedron. It can have exactly half only when the vertices can be partitioned into two equal-size independent sets, so that the graph of the polyhedron is a balanced
bipartite graph, as it is for an ideal cube. More strongly, the graph of any ideal polyhedron is
1-tough, meaning that, for any k, removing k vertices from the graph leaves at most k connected components. For example, the
rhombic dodecahedron is bipartite, but has an independent set with more than half of its vertices, and the
triakis tetrahedron has an independent set of exactly half the vertices but is not bipartite, so neither can be realized as an ideal polyhedron. ==Characterization and recognition==