Stellation of Polyhedra
This worksheet describes the stellating process to create new polyhedra from a given polyhedron. The stellate function is part of the geom3d package.

The Process of Stellation


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$\mathrm{restart}$$\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{with}\left(\mathrm{geometry}\right)\:$

Let S be a rotation through angle $\frac{2\mathrm{\pi}}{p}$, and let ${A}_{0}$ be any point not on the axis of S. Then the points
${A}_{i}\={A}_{0}^{\left({S}^{i}\right)}$, i = ...,2,1,0,1,2,...
are the vertices of a regular polygon {p}, whose sides are the segments ${A}_{0}{A}_{1}\,{A}_{1}{A}_{2}\,{A}_{2}{A}_{3}$,...
But the polygon can be closed without p being integral: All that is required is that the period of S be finite; in other words, that p be rational.
When the rational number p is expressed as a fraction in its lowest terms, we denote its numerator and denominator by ${n}_{p}$ and ${d}_{p}$. Thus $p\=\frac{{n}_{p}}{{d}_{p}}$, where ${n}_{p}$ and ${d}_{p}$ are coprime. The regular polygon {p} is traced out by a moving point that continuously describes equal chords of a fixed circle and that returns to its original position after describing ${n}_{p}$ chords and making ${d}_{p}$ revolutions around the center. The following exhibits some instances of {p}'s:
The general regular polygon {p} can be derived from the convex polygon $\left\{{n}_{p}\right\}$ by either one of two reciprocal processes: stellating and faceting.
To stellate a polyhedron, we have to extend its faces symmetrically until they again form a polyhedron. To investigate all possibilities, we consider the set of lines in which the plane of a particular face would be cut by all the other faces (sufficiently extended), and try to select regular polygons bounded by a set of these lines.
For a given starpolyhedron or compound, the core is the largest convex solid that can be drawn inside of it. And the compound or starpolyhedron may be constructed by stellating its core (which has the same face planes).

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$\mathrm{restart}$$\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{with}\left(\mathrm{geom3d}\right)\:$


The stellate Command in geom3d


Maple currently supports stellation of the five Platonic solids and of the two quasiregular solidsthe cuboctahedron and the icosidodecahedron. To stellate a given polyhedron, use the Maple command stellate(gon,core,n) where gon is the name of the stellated polyhedron to be created, core the core polyhedron, and n a nonnegative integer.
For the tetrahedron {3,3} and the cube {4,3}, the only lines are the sides of the face itself. Therefore, the only possible value of n is 0.
For the octahedron {3,4}, the eight facial planes enclose not only the original octahedron, but also other portions of space exterior to this octahedron. For a particular face ABC, the faces opposite to those that immediately surround ABC meet the plane in a larger triangle MNP. The eight large triangles so derived from all the faces form the stella octangula. Therefore, possible values of n are 0, 1.
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$\mathrm{octahedron}\left(\mathrm{t3}\,\mathrm{point}\left(o\,0\,0\,0\right)\,1.\right)\:$$\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{stellate}\left(\mathrm{st3\_1}\,\mathrm{t3}\,1\right)\:\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}$$\mathrm{draw}\left(\mathrm{st3\_1}\,\mathrm{cutout}=\frac{7}{8}\,\mathrm{lightmodel}=\mathrm{light4}\,\mathrm{title}=''first\; stellation:\; stella\; octangula''\right)$

 
Let us now stellate the dodecahedron {5,3}, and denote one face as 11111. By stellating this pentagon, we obtain the pentagram 22222, which is a face of {5/2,5}. The large pentagon that has the same vertices 22222 is a face of {5,5/2}. By stellating this pentagon, we obtain the large pentagram 33333, which is a face of {5/2,3}. The process now terminates, because the 10 lines account for all the other faces of {5,3}, the 12th face being parallel to 11111. Therefore, possible values of n are 0,1,2,3.
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$\mathrm{dodecahedron}\left(\mathrm{t4}\,o\,1.\right)\:$$\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{stellate}\left(\mathrm{st4\_2}\,\mathrm{t4}\,2\right)\:\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}$$\mathrm{draw}\left(\mathrm{st4\_2}\,\mathrm{cutout}=\frac{7}{8}\,\mathrm{lightmodel}=\mathrm{light4}\,\mathrm{title}=''second\; stellation:\; \{5,5/2\}''\right)$



Stellations of the Icosahedron


For the case of the icosahedron {3,5}, its stellations are so numerous and complicated that some care is needed to state precisely what varieties shall be considered properly significant and distinct.
In The FiftyNine Icosahedra, Coxeter, Du Val, Flather, and Petrie developed a complete enumeration of stellated icosahedra by considering the possible faces (first method), and by considering solid cells (second method), and were able to prove that 59 icosahedra exist, based on the following set of five restricted rules suggested by J.C.P. Miller:
1. The faces must lie in twenty planes, namely, the bounding planes of the regular icosahedron.
2. All parts composing the faces must be the same in each plane, although they might be quite disconnected.
3. The parts included in any one plane must have trigonal symmetry, with or without reflection. This secures icosahedral symmetry for the whole solid.
4. The parts included in any plane must all be "accessible" in the complete solid (that is, they must be on the "outside").
5. We exclude from consideration cases where the parts can be divided into two sets, each giving a solid with as much symmetry as the whole figure; but we allow the combination of an enantiomorphous pair having no common part.
Of the 59 icosahedra, 32 have full icosahedral symmetry, and 27 are enantiomorphous forms with twisted appearance. Below, we demonstrate one stellation of each type.
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$\mathrm{icosahedron}\left(\mathrm{ic}\,\mathrm{point}\left(o\,0\,0\,0\right)\,1.\right)\:$$\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathbf{for}i\mathbf{in}\left[26comma;34\right]\mathbf{do}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{2.0em}{0.0ex}}\mathrm{stellate}\left(\mathrm{cat}\left(\mathrm{ic}comma;i\right)comma;\mathrm{ic}comma;i\right);\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{pic}\left[i\right]\u2254\mathrm{draw}lpar;\mathrm{ic}\Vert icomma;\mathrm{style}=\mathrm{patch}comma;\mathrm{lightmodel}=\mathrm{light4}comma;\mathrm{title}=\mathrm{sprintf}\left(''stellation\; \%d\; of\; icosahedron''comma;i\right)rpar;\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathbf{end}\mathbf{do}colon;$

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$\mathrm{pic}\left[26\right]$

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$\mathrm{pic}\left[34\right]$



The KeplerPoinsot Polyhedra


It can be proved that only three of the stellated dodecahedra {5/2,5}, {5/5/2}, {5/2,3} and one stellated icosahedron {3,5/2} are regular polyhedra. They are called KeplerPoinsot polyhedra. The two with star facesthe two stellated dodecahedrawere found by Kepler (15711630); the others with regular faces and star verticesthe great icosahedron and the great dodecahedronby Poinsot (1777  1859). These four polyhedra, together with the five Platonic solids known to the ancient world, form the set of nine regular polyhedra. The other stellated polyhedra are compound polyhedra.


Stellations of the Archimedean Solids


In order to stellate the Archimedean solids, each facial plane must be extended indefinitely to generate cells exterior to the original solid. Using these cells as building blocks, one can form many new solids. Complete enumerations of all possible stellations (even with uniqueness restrictions) is still a topic of investigation.
The following shows the stellation process being applied to the two Archimedean solids: the cuboctahedron and the icosidodecahedron. The set of polyhedra shown
Of the Archimedean solids, geom3d can stellate the cuboctahedron and the icosidodecahedron (the two quasiregular polyhedra). The system of enumeration is that described in Polyhedron Models by Magnus J. Wenninger, which gives four stellated cuboctahedra and nineteen stellated icosidodecahedra. Below, we demonstrate one stellation of each polyhedra.
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$\mathrm{cuboctahedron}\left(\mathrm{cu}\,\mathrm{point}\left(o\,0\,0\,0\right)\,1.\right)\:\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}$$\mathrm{stellate}\left(\mathrm{cu4}\,\mathrm{cu}\,4\right)\:$$\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{draw}\left(\mathrm{cu4}\,\mathrm{cutout}=\frac{7}{8}\,\mathrm{lightmodel}=\mathrm{light4}\,\mathrm{title}=''stellation\; 4\; of\; cuboctahedron''\right)$

 
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$\mathrm{icosidodecahedron}\left(\mathrm{ic}\,\mathrm{point}\left(o\,0\,0\,0\right)\,1.\right)\:\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}$$\mathrm{stellate}\left(\mathrm{ic9}\,\mathrm{ic}\,9\right)\:$$\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\phantom{\rule[0.0ex]{0.0em}{0.0ex}}\mathrm{draw}\left(\mathrm{ic9}\,\mathrm{style}=\mathrm{patch}\,\mathrm{lightmodel}=\mathrm{light4}\,\mathrm{title}=''stellation\; 9\; of\; icosidodecahedron''\right)$


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