Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Do you demand that the polyhedron be convex? Otherwise, there's a really boring $k = 1$ solution (for the weaker notion of convergence): rasterize the sphere by approximating it as the union of many tiny cubes, and then erect a shallow square-based pyramid on every square face that results.
I'm afraid I can't answer the comment [about the 46-variable quadratic form]: the runtime will depend on both the quadratic form itself (not just the number of variables) and the solver you're using, so the best way to answer that question is to experiment yourself.
@mathworker21 Equation (5) contains a nonlinear constraint $x^T \lambda = 0$ (because neither $x$ nor $\lambda$ is a constant), so it's not a MILP. Equation (6) has purely linear constraints and objective, so it's in the correct form for you to input it into a MILP solver.
Thanks! Yes, it looks like the space is called DIII in Cartan's classification of compact Riemannian symmetric spaces. (I'll accept this if you post it as an answer.)