# Is there a convex polyhedron all of whose 2D polygonal cross sections are asymmetrical?

Define a nondegenerate convex n-polytope as the convex hull of a finite number of points in Euclidean n-space, such that its interior is nonempty.

It seems true that every nondegenerate convex polyhedron has a 2D cross section that is asymmetrical -- that is, a cross section whose only symmetry in its affine 2D subspace is the identity transformation. For example, the regular tetrahedron is quite symmetrical, but it's easy to find a cross section that produces a scalene triangle.

Here is a much harder problem: is there a nondegenerate convex polyhedron where all its 2D cross sections that are nondegenerate polygons are asymmetrical? The nondegeneracy of the cross-sectional polygon excludes cross sections that are empty, or just contain a single point or line segment.

• By "asymmetrical" you mean devoid of any symmetries? I guess (though I don't know how to make this rigorous right away) that a random polytope (convex hull of a random sample of points in, say, the sphere) has this property almost surely. May 29, 2023 at 12:38
• Actually, it seems you need that all vertices have degree $\ge 5$. Because if there is a vertex of degree 3 or 4, then its vertex figure is a triangle or quadrangle, and for those it is know that any two are projectively equivalent. Note that you can get any projective transformation of a vertex figure by choosing the cutting plane appropriately. So you could try a deformed icosahedron. May 29, 2023 at 13:10