Does every convex polyhedron have a combinatorially isomorphic counterpart whose faces all have rational areas? 
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*Does every convex polyhedron have a combinatorially isomorphic counterpart whose faces all have rational areas?

*Does every convex polyhedron have a combinatorially isomorphic counterpart whose edges all have rational lengths?

*Does every convex polyhedron have a combinatorially isomorphic counterpart whose vertices all have rational $x,y,z$ coordinates?


Can multiple conditions above be combined?
Update: all polyhedra in question are in $\mathbb{R}^3$.
 A: The answer to the first question is yes for simple polyhedra (those with degree $3$ at each vertex). Actually it follows pretty quickly from Minkowski's existence theorem for convex polyhedra that simple convex polyhedra with rational face areas are dense in the space of simple convex polyhedra.
To start, let us recall Minkowski's theorem: there exists a convex polyhedron $P$ with face areas $a_i$ and outward unit normals $n_i$ if and only if 
$$
\sum_{i=1}^k a_i n_i=0,
$$ where we assume that $n_i$ are distinct and span $R^3$. Further $P$ is unique, up to a rigid motion. The proof of Minkowski's theorem also makes it clear that $P$ depends continuously on $a_i$ and $n_i$.
Now suppose we are given a simple convex polyhedron $P$ with face areas $a_i$ and outward unit normals $n_i$, $i=1,\dots,k$. Let $a_i'$ be rational numbers with $|a_i-a_i'|\leq\epsilon$, and set
$$
n_i':=n_i \;\text{for}\; i<k,\quad\quad\text{and}\quad\quad n'_k:=\frac{-1}{a_k'}\sum_{i=1}^{k-1}a_i'n_i'.
$$
Then we have
$$
\sum_{i=1}^k a_i' n_i'=0.
$$
So there exists a convex polyhedron $P'$ with (rational) face areas $a_i'$ and outward unit normals $n_i'$. Since $a_i'$, $n_i'$ are close to $a_i$, $n_i$, it follows (from the uniqueness part of Minkowski's theorem) that the planes of the faces of $P'$ are close to those of $P$. Hence, since $P$ is simple, it follows that $P'$ is isomorphic to $P$.
A: The answer to the second question is yes for simplicial polyhedra, i.e., those with triangular faces. Indeed one may use Alexandrov's isometric embedding theorem to show that simplicial polyhedra with rational edge lengths are dense in the space of simplicial polyhedra, as discussed below.
Let $\ell_i$ be the edge lengths of a given simplicial convex polyhedron $P$ and $\ell_i'$ be rational numbers with $|\ell_i'-\ell_i|\leq\epsilon$. For each face $F_i$ of $P$ let $F_i'$ be the triangle with edge lengths $\ell_{i1}',\ell_{i2}',\ell_{i3}'$, where $\ell_{i1},\ell_{i2},\ell_{i3}$ are edge lengths of $F_i$. Assuming $\epsilon$ is small, the sum of the angles of $F_i'$ around each vertex will be less than $2\pi$ if $F_i'$ are glued together the same way that $F_i$ are. Thus, by Alexandrov's theorem, gluing $F_i'$ yields a convex polyhedron $P'$. By the uniqueness part of Alexandrov's theorem, $P'$ is close to $P$, after a rigid motion, and is therefore isometric to it since $P$ is simplicial. This in turn yields that the edges of $P'$ coincide with those of $F_i'$, because edges of $F_i'$ are the unique geodesics in $P'$ connecting their end points. So $P'$ has rational edge lengths as desired.
A: The answer to the third question is no.  This is a rather counter-intuitive discovery of Micha Perles from the sixties.  See this paper of Ziegler, for a simpler construction and other pertinent information.  However, for polytopes in dimension $3$, the answer is yes, as mentioned in the same paper of Ziegler. 
