Is There a Self-Dual Hendecahedron?

Change history

7 Mar 2019: Added subheadings.

30 Jan 2017: Added canonical form, minor cleanup.


The principle of duality states that for every polygon there is a dual, or reciprocal, polygon, whose edges correspond to the vertices (corner points) of the original and whose vertices likewise correspond to its edges. Similarly in three dimensions, for every polyhedron there is a dual, or reciprocal, polyhedron, whose faces correspond to the vertices (corner points) of the original and whose vertices correspond to its faces. The principle is well known - for duals of many common polyhedra, see [1].

Figure 1   Reciprocation

To obtain the dual of a polygon, the original is reciprocated with respect to a circle. If the dual is then reciprocated, the original polygon is obtained. If any regular polygon is reciprocated with respect to the appropriate circle, a congruent polygon is obtained and the polygon may be said to be self-dual.

To obtain the dual of a polyhedron, the original is reciprocated with respect to a sphere. If the dual is then reciprocated, the original polyhedron is obtained.

In Figure 1, A and B' are vertices. For polygons, a' and b are edges. For polyhedra, a' and b are sections through faces, and the circle is a section through the reciprocating sphere.

A and a' are reciprocal. For a polygon or polyhedron with vertex A, its dual has an edge or face a'. Likewise b and B' are reciprocal.

Note that OA . OB' = r2.
So, for r = 1, OA = 1 / OB'
ie the lengths OA and OB' are reciprocal.


A few polyhedra are self-dual, that is to say the dual is congruent to the original. Examples are to be found in the family of pyramids, including the regular tetrahedron. In a self-dual polyhedron the vertices and faces have a one-to-one correspondence, so it must have the same number of each (eg four in the case of the regular tetrahedron). Another characteristic involves the vertex figures: for our purposes, a vertex figure is the polygon formed by the cut surface when a given vertex is sliced off. For self-duality, the faces and vertex figures have a particular correspondence, in that each vertex figure is the dual polygon of the corresponding face.

To ensure that the dual is the same size as the original, the centre and radius of the reciprocating sphere must be carefully chosen. For the regular tetrahedron the intersphere is used, that is the sphere which has its centre at the same point as the tetrahedron, and touches the edges at their mid-points. The compound of two regular tetrahedra in such a dual relationship is known as the stella octangula (which is Latin for an eight-pointed star).

The hendecahedron

Figure 2   The bisymmetric

A few years ago [2] I came across a hendecahedron (an eleven-sided polyhedron) with the same number of vertices as faces. By distorting it I obtained different versions with different characteristics. The most symmetrical was the bisymmetric hendecahedron shown in Figure 2 and in more detail here, so called because it has two planes of symmetry.

I recently noticed that the vertex figures and faces appear to have the basic correspondence required for self-duality. The two planes of symmetry would be interchanged in the dual, so that it stands on edge, and the dual polyhedron is oriented back to front. For example the face CGJF reciprocates to a vertex congruent to E and located a little outwards from CGJF, and likewise the vertex E reciprocates to a face congruent to CGJF and located a little inwards from E. But in the hendecahedron shown, the vertex figure of E has the wrong proportions to be the planar dual of CGJF – it is rectangular, whereas the dual of CGJF is square. Is it possible to distort the polyhedron into a version with the correct proportions to be self-dual? And what are the location and radius of the reciprocating sphere? Or are there a whole range of self-dual arrangements, or none?

Approaches to the problem

There are two possible approaches to the problem, analytical or numeric. Either way, the fact that the polyhedron has two planes of symmetry allows the problem to be simplified a little – the solution to part of the polyhedron may be found and then by symmetry applied to the remainder of it.

Analytically, where each of the planes of symmetry bisects the polyhedron, the cut face is a pentagon. Because the two planes are interchanged by the reciprocation, for self-duality these pentagons must be planar duals of eachother. However, I do not think they will be regular. This is as far as I have been able to analyse.

Numerically, the best approach seems to be successive approximations. I would suggest the following algorithm:
(1) Guess the answer,
(2) Find its dual, by reciprocating with respect to a sphere of radius 1 and centre 0,0 as in steps (a) to (e) below,
(3) Take a polyhedron intermediate between the two (I am not sure of the best way to do this. The important thing is to ensure that the faces are flat),
(4) Use this as the next guess,
(5) And so on, until the difference between successive guesses is suitably small.

The first guess must be reasonably close (eg vertices and faces in a correct relationship to the reciprocating sphere). Different guesses could be tried, to see whether they all find the same solution or a range of solutions.

To find the dual of a polyhedron, I would suggest:
(a) For each vertex, find the equation of the radius joining the centre to the vertex,
(b) If the distance of the vertex from the centre is v and r = 1, then the distance of the reciprocal plane from the centre = 1/v ,
(c) The plane is orthogonal to the radius, so its equation can now be found,
(d) The line of intersection between two planes is the line in which their common edge lies,
(e) The point of intersection between three or more edges is a vertex of the dual polyhedron.

This is as far as I have got with a numerical approach. I have been unable to solve the problem either way, though obviously an analytical solution is preferable.

The canonical form

There is at least one solution, called the canonical form of the polyhedron. It is a theorem that every convex polyhedron can be morphed into a canonical form. This form has an intersphere such that every edge is tangent to it, the centre of gravity of the tangent points is the sphere centre, and every edge of the dual polyhedron shares the same tangent points. Is dual is called its canonical dual.

In the case of the canonical hendecahedron, the dual pair are congruent and therefore the canonical hendecahedron is self-dual.

But is the canonical figure the only solution? Might there be other arrangements where the dual figures are congruent but not canonical?


[1] Cundy, H.M. and Rollett, A.P.; Mathematical Models, OUP, 1961

[2] G. Inchbald, G.; "Five Space-Filling Polyhedra", The Mathematical Gazette, 80 (1996), 466-475.

[3] Hart, G. W.; "Calculating Canonical Polyhedra." Mathematica Educ. Res.[/I] 6, 5-10, Summer 1997.