Let us define a solid angle at the a vertex of a polyhedron (a 3-D solid) to be "acute" if no two sides meeting at that vertex from an angle of at least $90^\circ$. For example, by this definition a regular tetrahedron has four acute vertices.
I suspect that no convex polyhedron can have more than four acute vertices. Can you prove this conjecture?
(In 2-D, no convex polygon can have more than three acute vertices, and an arbitrary $n$-gon must have less than $\frac{2n+4}{3}$ acute angles.)
Now if we loosen the definition of "acute" to say that the solid angle formed by the sides meeting at the vertex subtends no more than $\pi$ steradians, does that increase the largest possible number of acute vertex angles in a convex polyhedron?