Maths Olympiad Prep

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Geometry Difficulty 6.2 National olympiad Prove it

4.52. Prove that the difference between the sum of the solid angles of the dihedral angles at the edges of a polyhedron and the sum of the solid angles of the polyhedral angles at its vertices is 2π(Γ2)2 \pi(\Gamma-2), where Γ\Gamma- is the number of faces of the polyhedron.

## Problems for Independent Solving

Solution

4.52. The solid angle at the ii-th vertex of a polyhedron is σi(ni2)π\sigma_{i}-\left(n_{i}-2\right) \pi, where σi\sigma_{i} is the sum of the dihedral angles along the edges emanating from it, and nin_{i} is the number of these edges (see problem 4.44). Since each edge emanates from exactly two vertices, ni=2P\sum n_{i}=2 \mathrm{P}, where P\mathrm{P} is the number of edges. Therefore, the sum of the solid angles of the polyhedral angles is 2σ2(PB)π2 \sigma-2(\mathrm{P}-\mathrm{B}) \pi, where σ\sigma is the sum of the dihedral angles, and B\mathrm{B} is the number of vertices. It remains to note that PB=Γ2\mathrm{P}-\mathrm{B}=\Gamma-2 (problem 8.14).

V. V. Prasolov, I. F. Sharygin

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