The reciprocation (or "polar reciprocation") with respect to a sphere S changes a point M in its polar plane (locus of the points N such as segment [MN] is diameter of a sphere orthogonal to S). Conversely, a plane is changed in a point, its pole. If M belongs to S, then its polar plane is the tangent plane at M. Therefore the conjugation changes a polyhedron into an other polyhedron, its dual.
To be more precise if S is the sphere with radius r and center the origin O and D a line going through M and cutting S in A and B then the locus of N, harmonic conjugate of M with respect to A and B (i.e. AM/AN=BM/BN) is a plane P (polar plane of M).
To a point M(a,b,c) corresponds the plane with equation ax+by+cz=r² and to the plan with equation ux+vy+wz=t corresponds the point N(ur²/t,vr²/t,wr²/t). The line OM is orthogonal to the polar plane P of M, and the distances from O to the point M and to the plane P are "inverse" (relatively to r): d(O,M)/r=r/d(O,P).
In a plane Π going through O and M the conjugation is reduced to "the polar inversion" with center O whose inversion circle is the intersection of the sphere S with Π; the polar of M is the intersection line of its polar plane with Π.
If we don't have better we may use the sphere with center the isobarycentre of the vertices and radius the average of their distances to the center.
Duality does not only concern convex polyhedra, but be careful! the pole of a plane going through the center of the sphere is at infinity.
This figure can be modified dynamically (see LiveGraphics3D help): the two big points (blue and green) may be moved with the mouse pointer; it is interesting to move them through the sphere to observe the behavior of the polar plane.
A polar plane and its pole are represented with the same color (blue and green). The intersection of the two polar planes is a line (in red) orthogonal to the segment (in red) joining the two points. And if the segment is tangent to the sphere then it is perpendicular to the intersection of the planes at the contact point. If these two points are two vertices of a polyhedron, linked by an edge, then the corresponding faces of the dual are in the polar planes, and their common edge belongs to the intersection line of the two planes. 

Duality exchanges the number of faces and of vertices and maintains the number of edges; to a face corresponds a vertex of same order and vice versa. To a regular face (equal sides and equal angles) corresponds a regular vertex (equal dihedral angles and equal edges angles) of same order. To two adjacent faces correspond two vertices linked by an edge, and two corresponding edges are orthogonal.
Duality preserves convexity and symmetry (axes, planes and a possible center).
Here are four examples of pairs of dual polyhedra: two hexahedra (7 and 6 vertices) and two pyramids (quadrangular and pentagonal).
One of the polyhedra is shown with its faces (light blue) and the other only with its edges (dark blue). The two orange edges correspond each other; they are orthogonal. The center of the conjugation's sphere is the isobarycentre of the vertices; it's the orange point which may be visualized by erasing the faces with "f".




To be self dual a polyhedron must have the same number of faces and vertices of each order.
A typical example is the npyramid: the base is a ngon and the n other faces are triangular, its main vertex is of order n and the n base vertices are of order three. This necessary condition is not sufficient; to be geometrically self dual a pyramid must be regular and its main vertex must lay at a suitable distance from the base (in fact it must be canonical).
And here are the self dual canonical forms of the three last examples above (each one has a symmetry axis, respectively of order 2, 4 and 5).



To conjugate a regular or semiregular polyhedron, the most interesting is to use its circumscribed sphere; its dual has then an inscribed sphere. Note that if we can use the sphere tangent to the edges of the initial polyhedron, then the edges of its dual polyhedron are perpendicular to the edges of the initial polyhedron (intersection points on the sphere).
The Archimedes' polyhedra, the prisms and antiprisms have regular faces (of two or three orders) and superimposable (but not regular) vertices; their duals, the Catalan's polyhedra, the diamonds and antidiamonds have superimposable (but not regular) faces and regular vertices (of two or three orders).
The duals of the regular polyhedra are regular: cube and octahedron, dodecahedron and icosahedron are dual, the tetrahedron is self dual.
Remark: the cube is a 4prism and an 3antidiamond; its dual, the octahedron, is a 4diamond and an 3antiprism.



By adjusting the sizes (or using the sphere tangent to the edges) we can check that the edges of two duals are orthogonal; so we get very attractive compound polyhedra.



By removing the little pyramids (or by cutting the "peaks") of this three polyhedra we get the intersections of the two regular polyhedra which compose them: respectively a cuboctahedron, a regular octahedron and an icosidodecahedron. Therefore the three pictures show stellated polyhedra. Their convex envelops (polyhedra with vertices the tops of the pyramids) are respectively a rhombic dodecahedron, a cube and a rhombic triacontahedron, duals of the intersections.
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