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An Archimedean solid is a convex uniform polyhedron, excluding the Platonic solids and the infinite families of uniform prisms and antiprisms. Here uniform means that every face is a regular polygon and that the symmetry group acts transitively on the vertices of a polyhedron: a symmetry of the entire solid carries any chosen vertex of a polyhedron to any other. Merely having equal numbers of faces of each type at every vertex of a polyhedron is a weaker local condition.^[ref]

There are thirteen types under this convention, counting a mirror-image pair as one type. The infinite prism and antiprism families are excluded by convention even though their uniform members satisfy the same regular-face and global vertex-equivalence conditions.^[ref]

For example, the truncated octahedron has six square faces and eight regular hexagon faces. Counting face-edge incidences gives $(6\cdot 4+8\cdot 6)/2=36$ edge of a polyhedron; Euler characteristic then gives $36-14+2=24$ vertices of a polyhedron. At each vertex of a polyhedron, one square and two regular hexagons meet. The icosidodecahedron, with twenty equilateral triangles and twelve regular pentagons, gives a different regular-face combination.

For a regular pentagon, take $R$ as rotation through $2\pi /5$ and $m$ as reflection in the horizontal axis through vertex zero. The distinct elements areEvery nonidentity rotation has order five. The reflections are the five order-two elements, since $(R^{j}m{)}^2=R^j{R}^{-j}=I$.

The reflection $R^{j}m$ has axis at angle $j\pi /5$ to the horizontal, understood modulo $\pi$. If vertices have angles $2k\pi /5$, it sends vertex $k$ to vertex $j-k(\mathrm{mod}5)$. Each axis passes through one vertex and the midpoint of the opposite edge; that vertex stays fixed and the other four vertices exchange in two pairs.

True. Label the pentagon polygon vertices clockwise from the upper left asUse $a$ for the upward single arrows and $b$ for the rightward double arrows in the original diagram. Their endpoint identifications areAll five polygon vertices are therefore identified. The remaining diagonal polygon edge, from $v_3$ to $v_4$, has its endpoints identified and is the only boundary component of a surface.

The paired polygon edges join half-disc neighbourhoods into disc neighbourhoods. At the common vertex, the five corner sectors join in one chain whose two free ends lie on the unpaired polygon edge: its vertex link in a polygonal schema is a closed interval. Hence that point has a half-disc neighbourhood. This verifies that the quotient space is a surface with boundary, rather than merely counting its CW cells. The paired polygon edges are oppositely traversed around the polygon, so this surface is orientable.

Its CW cells comprise one vertex, three edges (the $a$ pair, the $b$ pair, and the free polygon edge), and one face. ThereforeThe classification of compact surfaces identifies it as a torus with an open disc removed.

For five points, partition the unit square into four squares of side $1/2$. Assign boundary points to just one neighbouring piece. By the pigeonhole principle, two of the points lie in one piece, whose diameter is its diagonal, $\sqrt{(1/2{)}^2+(1/2{)}^2}=\sqrt{2}/2$.

For eight points, use seven smaller-diameter pieces rather than eight rectangles. Start with the convex pentagonTake its four reflected copies at the corners of the square. The remaining pieces are the left triangle with vertices $(0,1/4),(1/4,1/2),(0,3/4)$, its right-hand reflection, and the central diamond with vertices $(1/4,1/2),(1/2,1/4),(3/4,1/2),(1/2,3/4)$. These seven polygons cover the square with disjoint interiors. To verify coverage directly, for $0\le y\le 1/4$ the two bottom pentagons split the horizontal section at $x=1/2$. For $1/4\le y\le 1/2$, the five successive sections have endpoints $0,y-1/4,3/4-y,y+1/4,5/4-y,1$ and belong respectively to the left triangle, bottom-left pentagon, diamond, bottom-right pentagon and right triangle. Reflection gives the top half.

The diameter of a finite convex hull is the greatest distance between its vertices: if two points are convex combinations of vertices, their difference is a convex combination of vertex differences, and the triangle inequality bounds its norm by the largest such distance. Checking the five vertices of $P$ gives diameter $\sqrt{(1/2{)}^2+(1/4{)}^2}=\sqrt{5}/4$; reflections preserve that value. Each remaining triangle and the diamond has diameter $1/2$, which is smaller. After assigning shared boundaries to one piece, eight points in these seven pieces force a pair in a common piece. Hence some pair is at distance at most $\sqrt{5}/4$.

Both arguments are instances of a pigeonhole distance bound from a partition.

A point group is a group of spatial symmetry operations fixing at least one common point. Choosing that point as the origin represents its operations by orthogonal matrices, so the point group is a subgroup of the orthogonal group. A pure nonzero translation cannot belong to it.^[ref]

A crystallographic point group additionally preserves a full-rank Euclidean lattice and is finite. For a crystallographic space group, the group of orthogonal linear parts is its quotient by the translation subgroup; it preserves the translation Euclidean lattice. In two dimensions the crystallographic restriction theorem allows only rotation orders $1,2,3,4,6$.

The broader definition is needed for molecular point groups. A regular planar pentagon has a fivefold rotation, so its finite point group is not crystallographic. An ideal centrosymmetric linear molecule has the infinite D infinity h molecular point group, including rotations through arbitrary angles about its axis. Thus neither finiteness nor lattice preservation belongs to the unrestricted definition.