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The Icosian game made the vertex-visit puzzle tangible. The report of its 1857 presentation describes a plane projection of a regular dodecahedron, with holes at the vertices for ivory pins. The Trinity College Dublin archive connects this game with the search for closed routes visiting each vertex exactly once. This puzzle supplies a historical illustration of the Hamilton cycle requirement.^[ref]

The icosidodecahedron has 20 triangular and 12 pentagonal faces, with the faces alternating around each of its 30 vertices. It can be obtained by cutting a dodecahedron through the midpoints of its edges.

For a spherical triangle on a sphere of radius $R$, bounded by shorter great circle arcs and with interior angles $\alpha ,\beta ,\gamma$, the spherical excess formula, the triangular form of the Gauss-Bonnet theorem, isHere is an area proof. On the unit sphere, a spherical lune of angle $\alpha$ has area $2\alpha$: rotating around its vertices shows it occupies a fraction $\alpha /(2\pi )$ of the sphere when one uses the full angular width, or directly integrates ${\int }_0^{\alpha }{\int }_0^{\pi }\sin \theta d\theta d\phi$. The three side great circles of the triangle divide the sphere into eight regions, including the triangle $T$ and its antipodal triangle $-T$. Choose, at each vertex, the lune containing $T$, and also its antipodal lune. Each of $T$ and $-T$ is covered three times by these six lunes; each other region is covered once. Thus, with $A$ the area of $T$,This proves $A=\alpha +\beta +\gamma -\pi$; scaling lengths by $R$ scales area by $R^2$.

Radially project each face of the regular dodecahedron from its centre onto the unit sphere. Each edge and the origin lie in a plane, so its image is a shorter great circle arc. The twelve faces give congruent convex regular spherical polygons partitioning the sphere. At each projected vertex three congruent face angles fill the tangent plane. Hence each spherical pentagon has angle $2\pi /3$ at each of its five vertices. Splitting a pentagon into three spherical triangles and summing the spherical excess formula givesThis also agrees with $4\pi /12$, as required by the spherical partition.

Work first on the unit sphere. A spherical lune of angle $\alpha$ has area $2\alpha$: its fraction of sphere area is $\alpha /(2\pi )$. The three great circles extending the sides of a nondegenerate spherical triangle divide the sphere into four antipodal pairs of triangular regions. Call the desired region $T$, of area $A$, and its three adjacent regions $T_1,T_2,T_3$, of areas $A_1,A_2,A_3$. These four regions contain one region from each antipodal pair, so $A+A_1+A_2+A_3=2\pi$. The lune at each vertex consists of $T$ and one of these adjacent regions. Therefore, if the three interior angles are $\alpha ,\beta ,\gamma$,which proves the spherical excess formula, the spherical-triangle form of the Gauss-Bonnet theorem:For a sphere of radius $s$, multiply this area by $s^2$.

Triangulate a convex spherical $n$-gon by diagonals from a vertex. There are $n-2$ spherical triangles, and their angles add to the polygon's interior angle sum. Thus its unit-sphere area isFor a nondegenerate convex regular polygon in an open hemisphere, area is positive and each angle is less than $\pi$, giving $(n-2)\pi /n<\alpha <\pi$. Every angle in this interval occurs. Indeed place $n$ vertices at equal longitude spacings and colatitude $t\in (0,\pi /2)$ about a pole, and join consecutive vertices by shorter great circle arcs. The polygon is convex and regular. Split a central isosceles spherical triangle along the perpendicular to its edge. The resulting right spherical triangle has angles $\pi /n$, $\alpha /2$, $\pi /2$, and hypotenuse $t$. The angular spherical law of cosines yieldsso $\cot (\alpha /2)=\cos t\tan (\pi /n)$. As $t$ increases from $0$ to $\pi /2$, $\alpha$ increases continuously through exactlyThe endpoints are degenerate limits, not strict convex polygons.

For the spherical triangle reflection group, positive area requires $1/p+1/q+1/r>1$. Each angle is less than $\pi$, so the integer denominators are at least two. Arrange $p\le q\le r$. If $p\ge 3$ the sum is at most one, hence $p=2$. If $q\ge 4$ the sum is again at most one; thus either $q=2$ and any $r\ge 2$ works, or $q=3$ and $r=3,4,5$. These triangles exist, for example as the reflection chambers of a regular dihedral configuration, tetrahedron, cube, and dodecahedron, respectively.

To count group elements using the assumed tessellation, first label the original triangle's three vertices by their types and propagate these labels through side reflections. Around a type-$p$ vertex there are $2p$ sectors; the associated alternating reflections make a full turn and return the labels unchanged. The same holds at the other types. Every closed path in the dual tessellation is generated by these circuits around vertices, since the sphere is simply connected. Thus the labels are globally consistent, including when two angle denominators coincide. Every generated isometry preserves these labels. An element stabilizing a triangle therefore fixes its three vertices and is the identity. Consequently group elements correspond bijectively to triangles, and area givesThe possibilities, including permutations of the entries, are

Here is an explicit dodecahedron construction for the last case, avoiding an assumption about which vertices of the tessellation to join. Put $\phi =(1+\sqrt{5})/2$, and take the convex hull of the twenty pointswith independent signs. They all have squared radius three. The supporting plane $\phi y+z={\phi }^2$ contains the five verticesin cyclic order. Consecutive distances are $2/\phi$, and nonconsecutive distances are $2$, so this is a regular pentagon. Changing the two signs in its normal $(0,\phi ,1)$ and cyclically permuting coordinates gives twelve supporting planes, all with a congruent pentagon. Each listed vertex lies on three of these faces; the sixty face-edge incidences pair into thirty edges. The faces form a closed convex polyhedral surface with $20-30+12=2$, hence exhaust the boundary of the convex hull. This constructs a regular dodecahedron.

Choose the spherical directions of a face center $F=(0,\phi ,1)$, the midpoint $M=(0,\phi ,0)$ of its edge between $(\pm {\phi }^{-1},\phi ,0)$, and that edge's endpoint $V=({\phi }^{-1},\phi ,0)$. Normalize these to the unit sphere. The three side planes of triangle $FMV$ are $x=0$, $z=0$, and $\mathbf{n}\cdot \mathbf{x}=0$ with $\mathbf{n}=(-\phi ,{\phi }^{-1},-1)$. Their reflections are coordinate sign changes in the first two cases andin the third, since $\Vert \mathbf{n}{\Vert }^2=4$. Substitution in the twenty listed vertices, using ${\phi }^2=\phi +1$, shows that all three reflections permute the vertex set. The side-plane angles give interior angles $\pi /2$ at $M$, $\pi /3$ at $V$, and $\pi /5$ at $F$ because $|n_z|/\Vert \mathbf{n}\Vert =1/2$ and $|n_x|/\Vert \mathbf{n}\Vert =\phi /2=\cos (\pi /5)$. Thus this triangle is congruent to $\Delta$, and its side-generated group is $G$, conjugated by the aligning isometry.

It already has $120$ elements. Conversely, a symmetry of the dodecahedron is determined by the image of one face, one vertex on that face, and the orientation along its boundary: there are at most $12\cdot 5\cdot 2=120$ choices. An isometry fixing these choices fixes that face pointwise and must preserve the polyhedron's interior side, hence is the identity. Therefore the complete symmetry group of the constructed dodecahedron is $G$, including orientation-reversing symmetries.

The Euler formula for a sphere, applied to the surface of a convex polyhedron, is $V-E+F=2$. In a regular polyhedron let $q\ge 3$ be the common number of edges incident at a vertex. Counting vertex-edge and face-edge incidences, pentagonal faces give $qV=2E$ and $5F=2E$. Hencewhose right-hand side can be positive only if $q<10/3$. Thus $q=3$. Substitution then yields $2=E/15$, provingThese are the counts of the dodecahedron. Here regular polyhedron is understood in the convex, ordinary-face sense used by the stated Euler formula; self-intersecting star constructions do not share these hypotheses.

Work on the unit sphere with geodesic edges. A spherical lune of angle $\alpha$ has area ${\int }_0^{\alpha }{\int }_0^{\pi }\sin \theta d\theta d\phi =2\alpha$. The three great circles through the sides of a spherical triangle partition the sphere into four antipodal pairs of triangles. If the original triangle has area $A$ and the other pair representatives have areas $B,C,D$, then $A+B+C+D=2\pi$. The three lunes at its vertices have areas $A+B,A+C,A+D$, soThis proves the spherical excess formula $\boxed{A=\alpha +\beta +\gamma -\pi }$. Triangulate a convex spherical $k$-gon from one vertex into $k-2$ triangles; summing areas and vertex angles givesFor a decomposition of the whole sphere, angles around every vertex sum to $2\pi$ and each edge belongs to two faces. Summing the polygon area formula over all faces therefore givesNow $F=m+n+p$, $2E=4m+5n+6p$, and $3V=2E$, since three edges meet at each vertex. Substituting into Euler's formula yields $2m+n=12$. Nonnegative integer solutions therefore give at most seven pairs:At least three distinct pairs are realized: a dodecahedron gives $(0,12)$, a pentagonal prism gives $(5,2)$, and a cube gives $(6,0)$. Center each convex polyhedron around an interior origin and radially project its faces onto the sphere; edges become great-circle arcs and the resulting spherical faces are convex, preserving the incidence counts. This supplies the required actual decompositions.

Each nonidentity element of the special orthogonal group $\mathrm{SO}(3)$ is a rotation fixing exactly the two unit vectors along its axis. Since $G$ is a finite group, $\Omega$ is finite. If $v\in \Omega$ is fixed by $g\ne 1$ and $h\in G$, then $hv$ is fixed by $hg{h}^{-1}\ne 1$. Hence $G$ acts on $\Omega$ by a group action.

Let $N=|G|>1$, let $r$ be the number of orbits of a group action, and choose representatives with stabilizer subgroups of orders $n_1,\dots ,n_r$, each at least two. Count pairs $(g,v)$ with $g\ne 1$ and $gv=v$. Counting by $g$ gives $2(N-1)$. Counting by group orbits and using the orbit-stabilizer theorem givesEvery summand is at least $1/2$, so $r<4$: there are at most three orbits. For the trivial group, $\Omega$ is empty and the assertion also holds.

For the rotation group of a regular dodecahedron, $N=60$. The three group orbits are the unit vectors towards the 12 face centres, 20 vertices and 30 edge midpoints, whose stabilizer subgroup orders are respectively $5,3,2$. Rotations about these axes exhaust the nonidentity rotations. Consequently the orbit sizes are

The sum of the degrees of all faces is $2e$. Each of the $t$ triangular faces contributes three. Because the graph is bridgeless and has no four-cycle, every other face has degree at least five, and therefore

No edge can border two triangular faces: two distinct triangles sharing that edge would have their other two edges form a four-cycle, while the same triangle on both sides would force the connected bridgeless graph to be that triangle, contrary to $n\ge 4$. Thus the $3t$ edge incidences belonging to triangular faces use distinct edges, soCombining the inequalities givesand hence

The Euler formula for a connected planar graph now yieldsso the triangle-pentagon planar edge bound is

Equality is possible. Cut each corner of a dodecahedron through the midpoints of its incident edges. The resulting icosidodecahedral graph haswith every edge incident to one triangular and one pentagonal face. It has no four-cycle, and

The rotations in three dimensions preserving a dodecahedron form a group of order $60$. There are $24$ nonidentity rotations about the six axes through opposite face centres, $20$ about the ten axes through opposite vertices, and $15$ half-turns about the fifteen axes through opposite edge midpoints. Its nonidentity conjugacy classes have sizes $12,12,20,15$: the centralizer of a rotation of order five or three consists of rotations about its oriented axis and has order five or three, respectively, while all edge half-turns are conjugate. A normal subgroup is a union of whole conjugacy classes containing the identity. No proper nontrivial sum of these class sizes plus one divides $60$, proving that it is a simple group. The centralizers of the fifteen involutions are five distinct Klein four-groups, each containing three involutions. Conjugation permutes these five subgroups nontrivially. The kernel is a normal subgroup, so simplicity makes this nontrivial action injective. Its sign map to a group of order two must be trivial, since simplicity would otherwise force an impossible injection of sixty elements into two. Thus its image lies in the alternating group $A_5$, and equality of orders makes the image all of $A_5$.