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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 $P$.

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 $P$.

The answer to the first question is yes. Actually it follows pretty quickly from Minkowski's existence theorem for convex polyhedra that convex polyhedra with rational face areas are dense in the space of 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 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 $P'$ is isomorphic to $P$.

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 $P$.

The answer to the first question is yes. Actually it follows pretty quickly from Minkowski's existence theorem for convex polyhedra that convex polyhedra with rational face areas are dense in the space of 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 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 $P'$ is isomorphic to $P$.

The answer to the first question is yes. Actually it follows pretty quickly from Minkowski's existence theorem for convex polyhedra that convex polyhedra with rational face areas are dense in the space of 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 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 $P'$ is isomorphic to $P$.