= Solution
Work on the unit sphere with geodesic edges. A <spherical lune> of angle $\alpha$ has area $\int_0^\alpha\int_0^\pi\sin\theta\,d\theta\,d\varphi=2\alpha$. 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 $A+B,A+C,A+D$, so
$$
2(\alpha+\beta+\gamma)=3A+B+C+D=2A+2\pi.
$$
This 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 gives
$$
\boxed{A_k=\sum_{j=1}^k\alpha_j-(k-2)\pi.}
$$
For 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 gives
$$
4\pi=2\pi V-\pi(2E-2F),\qquad
\boxed{F-E+V=2}.
$$
Now $F=m+n+p$, $2E=4m+5n+6p$, 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:
$$
\boxed{(m,n)=(0,12),(1,10),(2,8),(3,6),(4,4),(5,2),(6,0).}
$$
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.
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