Editing Star polygon (section)

===Construction via vertex connection===
Regular star polygons can be created by connecting one [[Vertex (geometry)|vertex]] of a regular ''p''-sided simple polygon to another vertex, non-adjacent to the first one, and continuing the process until the original vertex is reached again.<ref>{{cite book |last=Coxeter |first=Harold Scott Macdonald |title=[[Regular polytopes (book)|Regular polytopes]] |publisher=Courier Dover Publications |page=[https://archive.org/details/regularpolytopes0000coxe/page/93 93] |date=1973 |isbn=978-0-486-61480-9 }}</ref> Alternatively, for integers ''p'' and ''q'', it can be considered as being constructed by connecting every ''q''th point out of ''p'' points regularly spaced in a circular placement.<ref>{{MathWorld |urlname=StarPolygon |title=Star Polygon}}</ref> For instance, in a regular pentagon, a five-pointed star can be obtained by drawing a line from the 1st to the 3rd vertex, from the 3rd to the 5th vertex, from the 5th to the 2nd vertex, from the 2nd to the 4th vertex, and from the 4th to the 1st vertex.

If ''q'' ≥ ''p''/2, then the construction of {''p''/''q''} will result in the same polygon as {''p''/(''p'' − ''q'')}; connecting every third vertex of the pentagon will yield an identical result to that of connecting every second vertex. However, the vertices will be reached in the opposite direction, which makes a difference when retrograde polygons are incorporated in higher-dimensional polytopes. For example, an [[antiprism]] formed from a prograde pentagram {5/2} results in a [[pentagrammic antiprism]]; the analogous construction from a retrograde "crossed pentagram" {5/3} results in a [[pentagrammic crossed-antiprism]]. Another example is the [[tetrahemihexahedron]], which can be seen as a "crossed triangle" {3/2} [[cuploid]].

====Degenerate regular star polygons====
If ''p'' and ''q'' are not coprime, a degenerate polygon will result with coinciding vertices and edges. For example, {6/2} will appear as a triangle, but can be labeled with two sets of vertices: 1–3 and 4–6. This should be seen not as two overlapping triangles, but as a double-winding single unicursal hexagon.<ref>[http://www.math.washington.edu/~grunbaum/Your%20polyhedra-my%20polyhedra.pdf Are Your Polyhedra the Same as My Polyhedra?] {{Webarchive|url=https://web.archive.org/web/20160803160413/http://www.math.washington.edu/~grunbaum/Your%20polyhedra-my%20polyhedra.pdf |date=2016-08-03 }}, Branko Grünbaum</ref><ref>Coxeter, The Densities of the Regular Polytopes I, p. 43:<BR>If ''q'' is odd, the truncation of {''p''/''q''} is naturally {2''p''/''q''}. But if ''q'' is even, the truncation of {''p''/''q''} consists of two coincident {{nowrap|{''p''/(''q''/2)}'s;}} two, because each side arises once from an original side and once from an original vertex. Since {{nowrap|2(''q''/2) {{=}} ''q'',}} the density of a polygon is never altered by truncation.</ref>
:[[File:Doubly wound hexagon.svg|160px]]