Show that the area vectors for a general tetrahedron sum to zero Using vector addition and multiplication, it is possible to show that the sum of the area vectors for a general closed tetrahedron in $\mathbb{R}^3$ (3-space) is zero.
Hint: start by writing down three vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$ and derive relationships for the other sides in terms of $\vec{a}$, $\vec{b}$, and $\vec{c}$.
 A: So for a general tetrahedron, we can derive that three of the sides are described by the vectors: $\vec{a}$, $\vec{b}$ and $\vec{c}$.
We can then write the area vector – using the outward pointing convention – for the first side as: $$\vec{A}_{ab} = \frac{1}{2}\vec{a}\times\vec{b}$$
Similarly, two of the remaining three area vectors can be written: $$\vec{A}_{bc} = \frac{1}{2}\vec{b}\times\vec{c}$$
$$\vec{A}_{ca} = \frac{1}{2}\vec{c} \times \vec{a}$$
The final area can be found using the fact that the vectors that define the sides define a closed figure and thus must sum to zero as suggested in the first hint by Blue:  $$\vec{A}_{(c-a)(b-a)}=\frac{1}{2}(\vec{c}-\vec{a})\times(\vec{b}-\vec{a})$$
Summing all of these we find that: $$ \frac{1}{2}\vec{a}\times\vec{b} + \frac{1}{2}\vec{b}\times\vec{c}+\frac{1}{2}\vec{c} \times \vec{a}+\frac{1}{2}(\vec{c}-\vec{a})\times(\vec{b}-\vec{a}) = \frac{1}{2}\left(\vec{a}\times\vec{b} + \vec{b}\times\vec{c}+\vec{c} \times \vec{a}+(\vec{c}-\vec{a})\times(\vec{b}-\vec{a})\right) = \vec{0}$$
by applying the properties of distribution, simplifying the cross products and canceling.
A: If $\vec{n}$ is outward unit normal to a face $F$ in then the (signed) area of the projection of $F$ onto a plane with unit normal $\vec{p}$ is $\vec{n}\cdot\vec{p}$ times the area of $F$. For a closed polyhedron, the projection of the faces pointing in one direction exactly match the projection of the faces pointing in the opposite direction.
Consider any polygonal prism with axis parallel to $\vec{p}$ whose intersection with a polyhedron are the red and green faces.
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The area of the green face times $\vec{p}\cdot\vec{n}_g$ is the cross-sectional area of the prism. The area of the red face times $\vec{p}\cdot\vec{n}_r$ is the negative of the cross-sectional area of the prism. Their sum is $0$ and so is the sum over all prisms parallel to $\vec{p}$.
Since this is true for any $\vec{p}$, the sum of the area vectors for a general closed polyhedron is $0$.
A: Idea based on Stokes' Theorem
Sum of area vectors vanish for any solid (generalisable to solids bounded by non-planar surfaces):
Imagine a constant flux field. Then the surface integral of the normal vector dotted into the flux direction equates to the null accumulation inside the solid. 
We can choose any orientation for the flux field, so the "average normal vector" is identically zero.
This is handwavy, but you can't beat it for succinctness.
A: Suppose you start with a triangulated surface forming a closed and orientable surface of any genus. Taking a typical triangle, as the surface is orientable, we can choose the coordinate vectors of the vertices a,b,c in a 'clockwise' direction. The area of triangle is then given by -(a^b+b^c+c^a)/6. Now take (without loss of generality) the adjacent triangle BAD also described in a clockwise direction. There will always be an adjacent triangle because the surfce is closed. The area is then -(b^a+a^d+d^b)/6. If these are summed the contribution from the adge AB cancels out as a^b+b^a=0. Likewise all the contributions from all the edges cancel down to zero. In conclusion the vector area of any triangulated closed and orientable surface is zero. Note, a Klein Bottle is not orientable.
