Ingeometry, themidsphere orintersphere of aconvex polyhedron is asphere which istangent to everyedge of the polyhedron. Not every polyhedron has a midsphere, but theuniform polyhedra, including theregular,quasiregular andsemiregular polyhedra and theirduals (Catalan solids) all have midspheres. The radius of the midsphere is called themidradius. A polyhedron that has a midsphere is said to bemidscribed about this sphere.^[1]

When a polyhedron has a midsphere, one can form two perpendicularcircle packings on the midsphere, one corresponding to the adjacencies between vertices of the polyhedron, and the other corresponding in the same way to itspolar polyhedron, which has the same midsphere. The length of each polyhedron edge is the sum of the distances from its two endpoints to their corresponding circles in this circle packing.

Every convex polyhedron has a combinatorially equivalent polyhedron, thecanonical polyhedron, that does have a midsphere, centered at thecentroid of the points of tangency of its edges. Numericalapproximation algorithms can construct the canonical polyhedron, but its coordinates cannot be represented exactly as aclosed-form expression. Any canonical polyhedron and its polar dual can be used to form two opposite faces of a four-dimensionalantiprism.

A midsphere of a three-dimensionalconvex 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 theinscribed 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 atangential polygon), and all of these inscribed circles must belong to a single sphere. For example, arectangular 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 aunit cube centered at theorigin of theCartesian coordinate system, with vertices at the eight points$(\pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2})$, the midpoints of the edges are at distance$\frac{\sqrt{2}}{2}$ from the origin. Therefore, for this cube, the midsphere is centered at the origin, with radius$\frac{\sqrt{2}}{2}$. This is larger than the radius of theinscribed sphere,$\frac{1}{2}$, and smaller than the radius of thecircumscribed sphere,$\frac{\sqrt{3}}{2}$. More generally, for anyPlatonic solid of edge length${\displaystyle \ell }$, the midradius is^[4]

• ${\displaystyle \frac{\sqrt{2}}{4}\ell }$ for a regulartetrahedron,
• ${\displaystyle \frac{1}{2}\ell }$ for a regularoctahedron,
• ${\displaystyle \frac{\sqrt{2}}{2}\ell }$ for a regular cube,
• ${\displaystyle \frac{\phi }{2}\ell }$ for a regularicosahedron, where${\displaystyle \phi }$ denotes thegolden ratio, and
• ${\displaystyle \frac{{\phi }^2}{2}\ell }$ for a regulardodecahedron.

Theuniform polyhedra, including theregular,quasiregular andsemiregular polyhedra and theirduals all have midspheres. In the regular polyhedra, the inscribed sphere, midsphere, and circumscribed sphere all exist and areconcentric,^[5] and the midsphere touches each edge at its midpoint.^[6]

Not every irregulartetrahedron 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]

IfO is the midsphere of a convex polyhedronP, then the intersection ofO with any face ofP is a circle that lies within the face, and is tangent to its edges at the same points where the midsphere is tangent. Thus, each face ofP has an inscribed circle, and these circles are tangent to each other exactly when the faces they lie in share an edge. (Not all systems of circles with these properties come from midspheres, however.)^[1]

Dually, ifv is a vertex ofP, then there is acone that has its apex atv and that is tangent toO in a circle; this circle forms the boundary of aspherical cap within which the sphere's surface isvisible from the vertex. That is, the circle is thehorizon of the midsphere, as viewed from the vertex. The circles formed in this way are tangent to each other exactly when the vertices they correspond to are connected by an edge.^[9]

If a polyhedronP has a midsphereO, then thepolar polyhedron with respect toO also hasO as its midsphere. The face planes of the polar polyhedron pass through the circles onO that are tangent to cones having the vertices ofP as their apexes.^[2] The edges of the polar polyhedron have the same points of tangency with the midsphere, at which they are perpendicular to the edges ofP.^[10]

For a polyhedron with a midsphere, it is possible to assign areal number to each vertex (thepower of the vertex with respect to the midsphere) that equals the distance from that vertex to the point of tangency of each edge that touches it. For each edge, the sum of the two numbers assigned to its endpoints is just the edge's length. For instance, Crelle's tetrahedra can be parameterized by the four numbers assigned in this way to their four vertices, showing that they form a four-dimensional family.^[11]

As an example, the four points (0,0,0), (1,0,0), (0,1,0), and (0,0,1) form one of Crelle's tetrahedra, with three isosceles right triangles and one equilateral triangle for a face. These four points are the centers of four pairwise tangent spheres, with radii${\displaystyle \frac{1}{2}\sqrt{2}\approx 0.707}$ for the three nonzero points on the equilateral triangle and${\displaystyle 1-\frac{1}{2}\sqrt{2}\approx 0.293}$ for the origin. These four numbers (three equal and one smaller) are the four numbers that parameterize this tetrahedron. Three of the tetrahedron edges connect two points that both have the larger radius; the length of these edges is the sum of these equal radii,${\displaystyle \sqrt{2}}$. The other three edges connect two points with different radii summing to one.

When a polyhedron with a midsphere has aHamiltonian cycle, the sum of the lengths of the edges in the cycle can be subdivided in the same way into twice the sum of the powers of the vertices. Because this sum of powers of vertices does not depend on the choice of edges in the cycle, all Hamiltonian cycles have equal lengths.^[12]

One stronger form of thecircle packing theorem, on representing planar graphs by systems of tangent circles, states that everypolyhedral 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, bystereographic projection, into a circle packing in theEuclidean plane whoseintersection 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 sameface lattice and the same midsphere can be transformed into each other by aprojective 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 aMöbius 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 thecentroid 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 thecanonical polyhedron, with the property that all combinatorially equivalent polyhedra will produce the same canonical polyhedra as each other, up tocongruence.^[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 inlinear time, and the canonical polyhedron defined in this alternative way hasmaximalsymmetry 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 acuboid, 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]

A numerical approximation to the canonical polyhedron for a givenpolyhedral graph can be constructed by representing the graph and itsdual graph as perpendicularcircle packings in the Euclidean plane,^[20] applying a stereographic projection to transform it into a pair of circle packings on a sphere, searching numerically for a Möbius transformation that brings the centroid of the crossing points to the center of the sphere, and placing the vertices of the polyhedron at points in space having the dual circles of the transformed packing as their horizons. However, the coordinates and radii of the circles in the circle packing step can benon-constructible numbers that have no exactclosed-form expression using arithmetic andnth-root operations.^[21]

Alternatively, a simpler numerical method for constructing the canonical polyhedron proposed byGeorge W. Hart works directly with the coordinates of the polyhedron vertices, adjusting their positions in an attempt to make the edges have equal distance from the origin, to make the points of minimum distance from the origin have the origin as their centroid, and to make the faces of the polyhedron remain planar. Unlike the circle packing method, this has not been proven to converge to the canonical polyhedron, and it is not even guaranteed to produce a polyhedron combinatorially equivalent to the given one, but it appears to work well on small examples.^[19]

The canonical polyhedron and its polar dual can be used to construct a four-dimensional analogue of anantiprism, one of whose two opposite faces is combinatorially equivalent to any given three-dimensional polyhedron. It is unknown whether every three-dimensional polyhedron can be used directly as a face of a four-dimensional antiprism, without replacing it by its canonical polyhedron, but it is not always possible to do so using both an arbitrary three-dimensional polyhedron and its polar dual.^[1]

The midsphere in the construction of the canonical polyhedron can be replaced by any smoothconvex body. Given such a body, every polyhedron has a combinatorially equivalent realization whose edges are tangent to this body. This has been described as "caging an egg": the smooth body is the egg and the polyhedral realization is its cage.^[22] Moreover, fixing three edges of the cage to have three specified points of tangency on the egg causes this realization to become unique.^[23]

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

• Polyhedra
• Spheres
• Circle packing

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