I have recently noticed that Bourbaki's proof (Chapter 6, section 4, no.1, Lemma 10, on p205) of the classification of finite Coxeter groups uses the inequality
$$ \frac{1}{1+p} + \frac{1}{1+q} + \frac{1}{1+r} > 1,\qquad p \geq q \geq r \geq 1$$
to classify the finite Coxeter graphs who have branch points in their dynkin diagrams. Namely, the solutions are
$$ (p,q,r) \in \bigcup_{i=1}^\infty \{(1,1,i)\} \cup \{(1,2,2), (1,2,3), (1,2,4)\}$$
which correspond to $D_n, E_6, E_7, E_8$ respectively, where the tuple describes the length of the chains attached to the branch point. This comes about from an orthogonal projection involving the distance of some vector to the space spanned by three orthonormal basis vectors.
Also, the solutions to this equation give the triangles on a 2-sphere with interior angles submultiples of $\pi$, as follows. Davis shows that every regular polytope corresponds to a regular tiling of the n-sphere. Such a tiling is constructed out of simplices with dihedral angles of m(i,j), so that these triangles would have angles $$ \frac{\pi}{m(1,2)} + \frac{\pi}{m(1,3)} + \frac{\pi}{m(2,3)} > \pi.$$
Again, the solutions are given by
$$ (m(1,2), m(2,3), m(1,3)) \in \bigcup_{i=2}^\infty \{(2,2,i)\} \cup \{(2,3,3), (2,3,4), (2,3,5)\}$$
corresponding to the regular $n$-gon with symmetry group $I_2(n)$ (cross $\mathbb{Z}_2$), the tetrahedron with symmetry group $A_n$, the cube/octahedron with symmetry group $B_n$, and the dodecahedron/icosahedron with symmetry group $H_3$.
This is, to me, quite surprising. Is there any deeper reason behind this?