Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 1 / 3G (Geometry)

For a geodesic triangle on the unit sphere with angles $\alpha ,\beta ,\gamma$, the Gauss-Bonnet theorem gives

For a geodesic triangulation with $V,E,F$ vertices, edges, and faces, summing the spherical excess gives $4\pi =2\pi V-\pi F$, because all angles around each vertex sum to $2\pi$. Also $3F=2E$. ThereforeThus every geodesic triangulation of the sphere has Euler number $2$.

If six triangles met at every vertex, counting vertex-face incidences would give $3F=6V$, hence $F=2V$. But part b gives $F=2V-4$, a contradiction. Therefore no such triangulation exists.