Given a $n$-faced polyhedron, associate to each face an outward-pointing normal vector with length equal to the area of that face. Show that the sum of these $n$ vectors is zero.
I've already proved the cases $n=4, n=6$, that is when the polyhedron is respectively a tetrahedron and a six-faced polyhedron (the last one is a combination of two tetrahedra with two congruent faces). Is there a way to show this using combinations of tetrahedra?
I'm a bit buzzed after an evening of jazz, so I swat this with a cannonball.
This follows from the divergence theorem. Let $\vec{u}$ be any constant vector. Let $V$ be that polyhedron, $\partial V$ its boundary, and $\vec{S}_i, i=1,\ldots, N,$ the vectors that you listed.
By the divergence theorem the sum $$ \vec{u}\cdot\sum_i\vec{S}_i=\oint_{\partial V}\vec{u}\cdot d\vec{S}=\int_V\nabla\cdot\vec{u}\,dV. $$ But because $\vec{u}$ is constant its divergence vanishes. Therefore so does this integral. Consequently $$ \vec{u}\cdot\sum_i\vec{S}_i=0. $$ But here $\vec{u}$ was arbitrary, so this can happen only if $$\sum_i\vec{S}_i=\vec{0}.$$