### If you want to kill your productivity, move to a new house!

I know things have been extremely slow lately, but I was moving last week (and well, that means packing everything, moving things to the new house, cleaning the old house, unpacking things at the new one… well, you get the picture).

My goal for this week is presenting Aleksandrov’s solution to  the Minkowski problem (see an earlier post I did introducing this problem). So I am going to leave you a problem as a preview, it is a sort of discrete version of the Minkowski problem:

Let $n_1,...,n_k$  be a family of non-coplanar unit vectors in $\mathbb{R}^3$ and let $\alpha_1,...,\alpha_k$ be positive numbers such that

$\sum \limits_{i=1}^k \alpha_in_i=0$

Then, show that there exists a convex closed polyhedron with exactly $k$ faces with normal vectors given by $n_1,...,n_k$ and corresponding areas $\alpha_1,...,\alpha_k$. Plus, this polyhedron is unique up to translation.

This is fact is not surprising (since it is not hard to check that any polyhedron has this property), but the proof is far from trivial. As you may guess, the proof cannot be constructive, it will use a continuity argument to show that there must be at least one such polyhedron. I will present this in my next post.