Suppose we have a region $\Omega \subset \mathbb{R}^n$ that is a polytope, i.e. a non-empty intersection of half-spaces. Visually this would be a polygon in 2D or a polyhedron in 3D – “polytope” is a single term to describe such objects regardless of the number of dimensions. The important feature is that the boundary of the region can be divided into flat pieces – the edges of a 2D polygon, or the facets of a 3D polyhedron. Consider the sum:
$$\displaystyle S = \sum_f A_f \hat{n}_f $$
where $A_f$ is the area of face $f$ and $\hat{n}_f$ is the outward-pointing unit normal vector to $f$. We can rewrite the sum as a surface integral and then use the gradient theorem to convert it to a volume integral:
$$\displaystyle \begin{aligned} \sum_f A_f \hat{n}_f &= \int_{\partial\Omega} \hat{n}\,dS \\ &= \int_\Omega \nabla 1 \,dx \\ &= 0 \end{aligned} $$
since the gradient of a constant is zero. We have arrived at a non-obvious identity valid for all polytopes, which I’ll write again for emphasis:
$$\displaystyle \sum_f A_f \hat{n}_f = 0 $$
As a side note, Minkowski’s theorem establishes the converse: any positive numbers $A_f$ and vectors $\hat{n}_f$ that satisfy the above relation also define a polytope.
Now let’s attempt to generalize this identity. If we add some coefficients $b_f$:
$$\displaystyle \begin{aligned} \sum_f b_f A_f\hat{n}_f &= \int_{\partial\Omega} b(x)\hat{n}\,dS \\ &= \int_\Omega \nabla b(x)\,dx\end{aligned} $$
where we have defined $b_f A_f := \int_{\partial\Omega_f} b(x)\,dS $ where $\partial\Omega_f$ is the face $f$ of the polytope $\Omega$. We see that if we can concoct a function $b(x)$ with a gradient whose integral over the whole domain $\Omega$ can be found, we will obtain some identity relating the face areas, face normals, and possibly other quantities. The identity we have already found corresponds to $b(x) = c$ for any constant $c$.
Generalization 1: Dot Product with a Constant Vector
A natural choice is to choose $b(x) = u^T x$ for some fixed vector $u$, since in this case the gradient $\nabla b = u$ is constant. From this choice we get:
$$\displaystyle \begin{aligned} Vu &= \int_\Omega u\,dx \\ &= \int_\Omega \nabla (u^T x)\,dx \\ &= \int_{\partial\Omega} \hat{n}_f u^Tx\,dS \\ &= \sum_f \hat{n}_f u^T \int_{\partial\Omega_f} x\,dS \\ &= \sum_f \hat{n}_f A_f u^Tx_f \end{aligned} $$
where the centroid $x_f$ of face $f$ satisfies $A_f x_f = \int_{\partial\Omega_f}$ and $V$ is the volume of $\Omega$. Since $u^Tx_f$ is a scalar, we may equally well write:
$$\displaystyle \begin{aligned} Vu &= \sum_f A_f \hat{n} u^Tx_f \\ &= \sum_f A_f \hat{n}_fx_f^T u \\ &= \left( \sum_f A_f \hat{n}_f x_f^T \right)u \end{aligned} $$
and since this is true for any vector $u$, it must be that:
$$\displaystyle \sum_f A_f \hat{n}_f x_f^T = V\mathcal{I}$$
where $\mathcal{I}$ is the identity matrix. This is another non-obvious polytope identity.
Generalization 2: Distance from Centroid
Another choice is to set
$$\displaystyle b(x) = |x – x_c|^2 $$
where $x_c$ is the centroid of $\Omega$ meaning that
$$\displaystyle Vx_c = \int_\Omega x\,dx $$
Note that the gradient of $b(x)$ is:
$$\displaystyle \nabla b(x) = 2(x – x_c)$$
So we get:
$$\displaystyle \begin{aligned} \int_\Omega \nabla b(x)\,dx &= \int_{\partial\Omega} b(x)\hat{n}\,dS \\ 2\int_\Omega x – x_c \,dx &= \int_{\partial\Omega} |x – x_c|^2 \hat{n} \,dS \\ 0 &= \sum_f \int_{\partial\Omega_f} |x – x_c|^2\hat{n}_f\,dS \end{aligned} $$
This is already one identity, but we can take it further by rewriting the integrands in terms of the face centroids $x_f$:
$$\displaystyle \begin{aligned} |x – x_c|^2 &= (x – x_f + x_f – x_c)\cdot(x – x_f + x_f – x_c) \\ &= |x-x_f|^2 + 2(x-x_f)\cdot(x_f – x_c) + |x_f – x_c|^2 \end{aligned} $$
Plugging this into the sum:
$$\displaystyle 0 = \sum_f \int_{\partial\Omega_f} \left( |x-x_f|^2 + 2(x-x_f)\cdot(x_f – x_c) + |x_f – x_c|^2 \right) \hat{n}_f\,dS $$
Since $x_f$ is the face centroid and $x_f – x_c$ is a constant on each face, the middle term integrates to zero leaving us with:
$$\displaystyle \begin{aligned} 0 &= \sum_f \int_{\partial\Omega_f} \left( |x-x_f|^2 + |x_f – x_c|^2 \right) \hat{n}_f\,dS \\ &= \sum_f \left[ \int_{\partial\Omega_f} |x-x_f|^2\,dS + A_f |x_f – x_c|^2 \right]\hat{n}_f \end{aligned}$$
This identity is less convenient than our previous two since it involves an integral which is not always easily evaluated. But we shall soon see that in a particular case, it can be.
The Simpler Case of the Simplex
A simplex in $D$ dimensions is a polytope with $D+1$ faces: in 2D, a triangle, in 3D, a tetrahedron.
In fact, each of the $D+1$ faces is itself a $(D-1)$-dimensional simplex. The faces of a tetrahedron, for example, are all triangles. Recall our first identity:
$$\displaystyle \sum_f A_f\hat{n}_f = 0 $$
If the polytope in question is a simplex, each $\hat{n}_f$ is a vector with $D$ components, there are $D+1$ of them, and they are linearly independent if the simplex has nonzero volume. By the rank-nullity theorem it follows that there is exactly one set of coefficients $c_f$ such that $\sum_f A_f \hat{n}_f = 0$. In fact, we’ve already found them – they are just the face areas $A_f$. But since there is exactly one such set, the third identity
$$\displaystyle \sum_f \left[ \int_{\partial\Omega_f} |x-x_f|^2\,dS + A_f |x_f – x_c|^2 \right]\hat{n}_f = 0 $$
since it is also just a linear combination of the face normals $\hat{n}_f$, must be just a restatement of the first. Meaning that for a simplex:
$$\displaystyle \int_{\partial\Omega_f} |x – x_c|^2\,dS = \int_{\partial\Omega_f} |x-x_f|^2\,dS + A_f |x_f – x_c|^2 = KA_f $$
for each face $f$ and some constant $K$ that is the same for all faces. So that we can evaluate the integrals in at least one example, suppose that the simplex is in fact a triangle. Then, parametrizing the edge $f$ by $u$:
$$\displaystyle \begin{aligned} \int_{\partial\Omega_f} |x – x_f|^2\,dS &= A_f\int_{-\frac12}^\frac12 A_f^2 u^2\,du \\ &= \frac{1}{12}A_f^3 \end{aligned} $$
So for any triangle:
$$\displaystyle \begin{aligned} \frac{1}{12}A_f^3 + A_f |x_f – x_c|^2 &= KA_f \\ \frac{1}{12}A_f^2 + |x_f – x_c|^2 &= K \end{aligned} $$
Summary
Putting together everything in one place, for any polytope we have three identities:
$$\displaystyle \sum_f A_f \hat{n}_f = 0 $$
$$\displaystyle \sum_f A_f \hat{n}_f x_f^T = V\mathcal{I}$$
$$\displaystyle \sum_f \int_{\partial\Omega_f} |x – x_c|^2\hat{n}_f\,dS = \sum_f \left[ \int_{\partial\Omega_f} |x-x_f|^2\,dS + A_f |x_f – x_c|^2 \right]\hat{n}_f = 0 $$
For any simplex:
$$\displaystyle \frac{1}{A_f}\int_{\partial\Omega_f} |x – x_c|^2\,dS = \frac{1}{A_f}\int_{\partial\Omega_f} |x-x_f|^2\,dS + |x_f – x_c|^2 = K $$
For any triangle:
$$\displaystyle \begin{aligned} \frac{1}{12}A_f^3 + A_f |x_f – x_c|^2 &= KA_f \\ \frac{1}{12}A_f^2 + |x_f – x_c|^2 &= K \end{aligned} $$
None of these are particularly obvious at first glance, but all of them are easily derived (well, I thought it was easy) from basic vector calculus and linear algebra facts. If you find another choice of $b(x)$ that leads to a polytope identity, sound off in the comments below.