Proving the "centroid" property and the existence of corresponding convex polyhedron in Minkowski Problem Assume $P$ is a convex polyhedron embedded in $\mathbb{R}^{3}$, the faces are $\left\{F_{1}, F_{2}, \cdots, F_{k}\right\}$, the unit normal vector to the face $F_{i}$ is $\mathbf{n}_{i}$, the area of $F_{i}$ is $A_{i}, 1 \leq i \leq k$.
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*

*Show that
$$
A_{1} \mathbf{n}_{1}+A_{2} \mathbf{n}_{2}+\cdots A_{k} \mathbf{n}_{k}=\mathbf{0}
$$

*Given $k$ unit vectors $\left\{\mathbf{n}_{1}, \mathbf{n}_{2}, \cdots, \mathbf{n}_{k}\right\}$ which can not be contained in any half space, and $k$ real positive numbers $\left\{A_{1}, A_{2}, \cdots, A_{k}\right\}, A_{i}>0$, and satisfying the above condition, show that there exists a convex polyhedron $P$, whose face normals are $\mathbf{n}_{i}$ 's, face areas are $A_{i}$ 's.$$$$
In my opinion the easy case of this problem is when it's the $\mathbb{R}^{2}$ case, the sum of area of the triangle times the corresponding unit vector is $0$ by properties of the centroid, but I have no idea of how to extend this property in to 3-dimension. Is there any hint for continuing here?


The identity
$ \displaystyle \sum_{i=1}^k A_i \mathbf{n}_i = 0 $
can be proven using the divergence theorem which states that the surface integral of a vector field $\mathbf{F}(x,y,z)$ over a closed surface, which is called the flux through the surface, is equal to the volume integral of the divergence of $\mathbf{F}$ over the region inside the surface.
This is expressed in mathematical terms as follows
$ \displaystyle \int_D \nabla \cdot \mathbf{F} dV = \int_S \mathbf{F} \cdot \mathbf{n} dS $
Now if we select $\mathbf{F} = \mathbf{c}$ , a constant vector, then the divergence of this constant vector field is $0$, hence
$ 0 = \displaystyle \int_S \mathbf{c} \cdot \mathbf{n} \ dS = \mathbf{c} \cdot \int_S \mathbf{n} \ dS $
Since we can choose $\mathbf{c}$ to point in any direction, it follows that it must be the case that
$ \displaystyle \int_S \mathbf{n} \ d S = \mathbf{0} $
Finally, in our case this last surface integral is just the summation
$\displaystyle \sum_{i=1}^k A_i \mathbf{n}_i  $
Thus we have shown that it is the zero vector for any polyhedron.
This answers the first part of your question.  I wasn't able to come up with anything concerning the second part.
