Given any convex uniform polyhedron, denote all angles formed at a vertex $\alpha_1, \alpha_2, \dots$, I find that, \begin{equation*} \text{the number of vertices in the polyhedron}=\frac{4\pi}{2\pi-\sum_i \alpha_i}. \end{equation*} But I'm not sure how to prove it or what the reason behind this is.
For example, any vertex of a Tetrahedron forms 3 angles: $\alpha_1=\alpha_2=\alpha_3=\pi/3$ and $4\pi/(2\pi-3(\pi/3))=4$,
any vertex of a Cube forms 3 angles: $\alpha_1=\alpha_2=\alpha_3=\pi/2$ and $4\pi/(2\pi-3(\pi/2))=8$,
any vertex of a Octahedron forms 4 angles: $\alpha_1=\alpha_2=\alpha_3=\alpha_4=\pi/3$ and $4\pi/(2\pi-4(\pi/3))=6$,
any vertex of a Truncated icosahedron (soccer ball pattern) forms 3 angles: $\alpha_1=\alpha_2=120\text{ deg}=2\pi/3$, $\alpha_3=108\text{ deg}=3\pi/5$ and $4\pi/(2\pi-(2(2\pi/3)+3\pi/5))=4\pi/(\pi/15)=60$.
I tried several other examples in the link below, they all seem to obey the equation. https://en.wikipedia.org/wiki/List_of_uniform_polyhedra
The expression $2\pi - \sum \alpha_i$ is called the "angular defect" of the vertex. There are several nice theorems about the total defects of shapes, and they imply that the total defect of a convex polyhedron is $4\pi$. Look up Descartes' Angular Defect theorem and/or the Gauss-Bonnet theorem to see more about it.
So your equation is saying that the total defect is equal to the number of vertices times the defect for each vertex (because the vertices are equivalent).