Tessellating the sphere It is a famous result that the plane can be tessellated by regular triangles, squares, and hexagons.
Which regular polygons can tessellate the sphere?
 A: Suppose we tesselate the sphere using $F$ polygons, all congruent, each with $p$ vertices and $p$ edges, and that at each vertex $q$ polygons meet Since every edge of the final picture belongs to two polygons, there are in all $E=pF/2$ edges. Similarly, each vertex in the final picture belongs to $q$ polygons, so there are $V=pF/q$ vertices in all.
In view of Euler's formula, we must have $$2=V-E+F=pF/q-pF/2+F=(p/q-p/2+1)F.$$ Dividing by $pF$ we see that $$\frac{2}{pF}+\frac{1}{2}=\frac{1}{q}+\frac{1}{p}$$ so that $$\frac{1}{q}+\frac{1}{p}>\frac{1}{2}.$$
Let's solve this inequality, forgetting for now that $p$ and $q$ are what they are. Suppose that $p\geq q$, as we may. If $q\geq 4$, then $1/p+1/q\leq1/4+1/4=1/2$, contradicting our inequality. It follows that $q\leq 3$. If $q=3$, then $1/p>1/2-1/q=1/6$, and $p\leq 5$. In this way we find as candidates for $(p,q)$ the pairs $(3,3)$, $(4,3)$ and $(5,3)$, and the three are actually solutions. Suppose now that $q=2$: then $1/p>1/2-1/2=0$, and there is no constrain on $p$ except being at least $2$, so we get the solutions $(p,2)$ with $p\geq2$.
We see that we have the solutions listed in the first two columns of the following table, and they actually occur, as we have the tesselations described in the third column.
$$
\begin{array}{ccc}
p & q \\ \hline
3 & 3 & \text{tetrahedron}\\
4 & 3 & \text{cube} \\
3 & 4 & \text{octahedron} \\
3 & 5 & \text{icosahedron} \\
5 & 3 & \text{dodecahedron} \\
n & 2 & \text{two $n$-gons} \\
2 & n & \text{$n$ digons} 
\end{array}
$$
Here $n$ is an integer greater or equal to $2$. Notice that the last two rows are somewhat degenerate cases.
One can also prove that the tesselations are in fact unique in each case. Indeed, from $p$ and $q$ one can compute $F$, the number of faces, and then the area of each of the polygons. On a unit sphere, the Gauss-Bonnet theorem tells us that the angles of a regular geodesic polygon are determined by the area, and then so are the length of the sides. It follows that in each case the actual polygon is completely determined by the pair $(p,q)$. As they only fit in one way, the tesselation is also determined.

In the case of the plane, the computation is similar, by the way. Suppose we tesselate the plane by regular polygons, each with $p$ vertices and edges and that at each vertex $q$ of them meet. The internal angle at each vertex in any one of the polygons is $\pi(1-2/p)$. Since at each vertex we have $q$ poygons meeting and their internal angles there have to sum up to $2\pi$, we have $$\pi(1-2/p)q=2\pi$$, which can be rewritten  as $$\frac1p+\frac1q=\frac12.$$ Since this inequality is again symmetric in $p$ and $q$, we may assume for the moment that $p\geq q$. If $q>4$, then $1/p+1/q<1/4+1/4=1/2$, and this is impossible. It follows that $q\leq 3$. If $q=4$, then $1/p=1/2-1/q=1/4$ and $p=4$; if $q=3$, then $1/p=1/2-1/q=1/6$ and $p=6$; if $q=2$, then $1/p=1/2-1/2=0$, and this is absurd. We see that we have the solutions are therefore $(4,4)$, $(3,6)$ and $(6,3)$, which correspond to using squares, triangles and hexagons. If we insist, we can interpret the degenerate case as a $(\infty,2)$, giving the tesselation of the plane by two zilchgons or $(2,\infty)$, giving the tesselation by infinitely many digons with vertices at infinity.
A: If the following restrictions are satisfied:


*

*All polygons are regular

*Only one type of polygon is permitted and all instances are congruent

*The entire surface of the sphere is tessellated exactly once


then I believe the only tessellations that are admitted are precisely the Platonic solids (projected onto their circumscribing sphere) and no others.
If only condition (2) is relaxed to allow distinct but finitely many polygons, then this allows Archimedean solids and the regular prisms and antiprisms, but it may also allow other non-polyhedral solutions--I'm not sure.
If only condition (1) is relaxed, then Archimedean duals will work as well as the prism and antiprism duals.  Again, there may be additional solutions.
If conditions (2) and (3) are relaxed, then this can admit the projections of non-convex uniform polyhedra.

Indeed, there is a case that I had not previously considered, that satisfies all three conditions stated above.  Consider the projection of a regular octahedron onto its circumscribing sphere.  This projection divides the sphere along three mutually intersecting great circles.  For any such great circle, rotate the hemispheres on either side by $\pi/4$ relative to each other.  The result is a (distinct) tessellation of the sphere by equilateral triangles, but it does not correspond to the projection of a polyhedron.
A: Tessellations by Platonic solids into (4,6,8,20,12) number of regular triangles,squares and pentagons. The icosi-dodecahedron I like in particular because regular pentagons and equilateral triangles are enclosed between five great circles.
