Midsphere

A midsphere of a three-dimensional convex polyhedron is defined to be a sphere that is tangent to every edge of the polyhedron. That is to say, each edge must touch it, at an interior point of the edge, without crossing it. Equivalently, it is a sphere that contains the inscribed circle of every face of the polyhedron.[2] When a midsphere exists, it is unique. Not every convex polyhedron has a midsphere; to have a midsphere, every face must have an inscribed circle (that is, it must be a tangential polygon), and all of these inscribed circles must belong to a single sphere. For example, a rectangular cuboid has a midsphere only when it is a cube, because otherwise it has non-square rectangles as faces, and these do not have inscribed circles.[3]

For a unit cube centered at the origin of the Cartesian coordinate system, with vertices at the eight points ${\textstyle (\pm {\tfrac {1}{2}},\pm {\tfrac {1}{2}},\pm {\tfrac {1}{2}})}$, the midpoints of the edges are at distance ${\displaystyle 1/{\sqrt {2}}}$ from the origin. Therefore, for this cube, the midsphere is centered at the origin, with radius ${\textstyle 1/{\sqrt {2}}}$. This is larger than the radius of the inscribed sphere, ${\textstyle 1/2}$, and smaller than the radius of the circumscribed sphere, ${\textstyle {\sqrt {3/4}}}$. More generally, for any Platonic solid of edge length ${\displaystyle \ell }$, the midradius is[4]

• ${\displaystyle {\tfrac {1}{2{\sqrt {2}}}}\ell }$ for a regular tetrahedron,
• ${\displaystyle {\tfrac {1}{2}}\ell }$ for a regular octahedron,
• ${\displaystyle {\tfrac {1}{\sqrt {2}}}\ell }$ for a regular cube,
• ${\displaystyle {\tfrac {\varphi }{2}}\ell }$ for a regular icosahedron, where ${\displaystyle \varphi }$ denotes the golden ratio, and
• ${\displaystyle {\tfrac {\varphi ^{2}}{2}}\ell }$ for a regular dodecahedron.

The uniform polyhedra, including the regular, quasiregular and semiregular polyhedra and their duals all have midspheres. In the regular polyhedra, the inscribed sphere, midsphere, and circumscribed sphere all exist and are concentric,[5] and the midsphere touches each edge at its midpoint.[6]

The centers of four pairwise tangent spheres form the vertices of a Crelle's tetrahedron. Here, four equal spheres form a regular tetrahedron. The midsphere passes through the six points of tangency of these spheres, which in this case form a regular octahedron.

Not every irregular tetrahedron has a midsphere. The tetrahedra that have a midsphere have been called "Crelle's tetrahedra"; they form a four-dimensional subfamily of the six-dimensional space of all tetrahedra (as parameterized by their six edge lengths). More precisely, Crelle's tetrahedra are exactly the tetrahedra formed by the centers of four spheres that are all externally tangent to each other. In this case, the six edge lengths of the tetrahedron are the pairwise sums of the four radii of these spheres.[7] The midsphere of such a tetrahedron touches its edges at the points where two of the four generating spheres are tangent to each other, and is perpendicular to all four generating spheres.[8]

A circle packing in the plane (blue) obtained by stereographically projecting the horizon circles on the midsphere of an octahedron. The yellow vertices and red edges represent the octahedron itself, centrally projected onto the midsphere and then stereographically projected onto the plane.

One stronger form of the circle packing theorem, on representing planar graphs by systems of tangent circles, states that every polyhedral graph can be represented by the vertices and edges of a polyhedron with a midsphere. Equivalently, any convex polyhedron can be transformed into a combinatorially equivalent form, with corresponding vertices, edges, and faces, that has a midsphere. The horizon circles of the resulting polyhedron can be transformed, by stereographic projection, into a circle packing in the Euclidean plane whose intersection graph is the given graph: its circles do not cross each other and are tangent to each other exactly when the vertices they correspond to are adjacent.[13] Although every polyhedron has a combinatorially equivalent form with a midsphere, some polyhedra do not have any equivalent form with an inscribed sphere, or with a circumscribed sphere.[14]

Any two convex polyhedra with the same face lattice and the same midsphere can be transformed into each other by a projective transformation of three-dimensional space that leaves the midsphere in the same position. This transformation leaves the sphere in place, but moves points within the sphere according to a Möbius transformation.[15] Any polyhedron with a midsphere, scaled so that the midsphere is the unit sphere, can be transformed in this way into a polyhedron for which the centroid of the points of tangency is at the center of the sphere. The result of this transformation is an equivalent form of the given polyhedron, called the canonical polyhedron, with the property that all combinatorially equivalent polyhedra will produce the same canonical polyhedra as each other, up to congruence.[16] A different choice of transformation takes any polyhedron with a midsphere into one that maximizes the minimum distance of a vertex from the midsphere. It can be found in linear time, and the canonical polyhedron defined in this alternative way has maximal symmetry among all combinatorially equivalent forms of the same polyhedron.[17] For polyhedra with a non-cyclic group of orientation-preserving symmetries, the two choices of transformation coincide.[18] For example, the canonical polyhedron of a cuboid, defined in either of these two ways, is a cube, with the distance from its centroid to its edge midpoints equal to one and its edge length equal to ${\displaystyle {\sqrt {2}}}$.[19]

• Ideal polyhedron, a hyperbolic polyhedron in which each vertex lies on the sphere at infinity