Building a polyhedron from areas of its faces Is there a known algorithm which, given a finite multiset (unordered list) of integers $A$, returns a yes/no answer  for "Is there a polyhedron such that the multiset of areas of all its faces is exactly $A$?"? 
Is there a known general algorithm for $n$-dimensional polytopes?
 A: I can answer your question with the specialization to convex polyhedra and polytopes.
Specializing further to $\mathbb{R}^3$, the result is that

$n \ge 4$ positive real numbers are the face areas of a convex polyhedron 
  if and only if the largest number is not more than the sum of the others.

I wrote up a short note
establishing this: "Convex Polyhedra Realizing Given Face Areas," arXiv:1101.0823.
The result relies on Minkowski's 1911 theorem, which perhaps you know:

Theorem (Minkowski). Let $A_i$ be positive faces areas and $n_i$ distinct, 
  noncoplanar unit face normals,
  $i=1,\ldots,n$.
  Then if $\sum_i A_i  n_i = 0$, there is a closed convex polyhedron
  whose faces areas uniquely realize those areas and normals.

This theorem reduces the problem to finding orientations $n_i$ so that vectors of
length $A_i$ at those orientations sum to zero.  And this is not difficult.
Here is Figure 3 from my note from which you can almost infer the construction:
           

Minkowski's theorem generalizes to $\mathbb{R}^d$ and so does an analog of the above claim
(but I did not work that out in detail in the arXiv note).
In terms of an algorithm, the decision question is linear in the number $n$ of facet areas,
and even constructing the polyhedron is linear in $\mathbb{R}^3$,
and likely $O(dn)$ in $\mathbb{R}^d$ (but again, I didn't work that out).
But you don't mention the word "convex" in your post, so perhaps you are interested
in nonconvex polyhedra and polytopal complexes?  
