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Given a tetrahedral $\rho$ with faces $\Xi, \Pi, \Gamma, \Delta$ with areas $\xi , \pi, \gamma , \delta$, respectively, assign a normal vector to each face such that $\mid \mid \hat{\xi} \mid \mid = \xi, \mid \mid \hat{\pi} \mid \mid = \pi, \mid \mid \hat{\gamma} \mid \mid = \gamma, \mid \mid \hat{\delta} \mid \mid = \delta$
Prove that the sum of the vectors is the zero using vector cross product.

Anthony Peter
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    I don't think you can assume the angles between $\hat{\xi}, \hat{\pi}, \hat{\gamma}$ are $\frac{\pi}{3}$. In a regular tetrahedron, the angles would be $\pi - \arccos \frac 13 \approx 109.5^\circ$. The problem statement seems to consider general tetrahedra. – Ross Millikan Jul 24 '13 at 20:45
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    Put one vertex at the origin, and let the edges determine vectors $\hat{a}$, $\hat{b}$, $\hat{c}$. The vector $\hat{a}\times\hat{b}$ is perpendicular to the face bounded by $\hat{a}$ and $\hat{b}$ ... and ... it magically(!) has length equal to the twice the area of that face. So, three of the normal vectors are easy to determine; the last one (for the face opposite the vertex at the origin) is the (only-slightly) tricky one, but you'll find that the formula for it immediately proves that the sum of all four vectors is zero. – Blue Jul 24 '13 at 21:22
  • I salute you for removing the "what have you tried" material that clutters up too many postings here. It is easier to read the question without it. And leaving a copy of that extra content in the Revisions list for the posting is a nice way to get comments on that work (if one wants them), by indicating that the proof attempts can be found in the edit history, while keeping it separated from the question. I hope other users follow your lead. – zyx Jul 25 '13 at 00:26

1 Answers1

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I happened to draw some TikZ illustration for the some computational geometry slides, which may be handy in explaining. Just to continue Blue's argument in the comments.

tetra

Say $\newcommand{\b}{\boldsymbol}\b{a} = \vec{V_1 V_2}$, $\b{b} = \vec{V_1 V_4}$, $\b{c} = \vec{V_1 V_3}$. Then $$ \b{a}\times \b{b}, \; \b{b}\times \b{c}, \; \b{c}\times\b{a}\tag{1} $$ will produce the vectors normal to the triangular faces $F_3$, $F_2$, $F_4$ opposite to the vertex $V_3$, $V_2$, $V_4$ respectively, with length $2|F_3|$, $2|F_2|$, and $2|F_4|$ (absolute value just denotes the area). Also notice that all of them point outward with respect to this tetrahedron.

For the last vector normal to the face $F_1$ opposite to $V_1$, notice that

$$ \vec{V_3 V_4} = \vec{V_1 V_4} - \vec{V_1 V_3} = \b{b}-\b{c}, \\ \vec{V_4 V_2} = \vec{V_1 V_2} - \vec{V_1 V_4} = \b{a}-\b{b}. $$ Then the last vector $\b{n}_{F_1}$ is produced by $$ (\b{b}-\b{c}) \times (\b{a}-\b{b}).\tag{2} $$ Adding everything factored by $1/2$ in (1) and (2), using the anti-commutativity of the cross product, see what happens.

Another nice proof I like is to use the Barycentric coordinate function, the vector normal to a triangular face opposite to the vertex $V$, having the scale of the area of that face, happens to be a constant about the volume times the gradient of barycentric coordinate $\lambda_V$ of this very vertex $V$. $\lambda_V$ is a linear function valued $1$ at $V$, and linearly decreases to $0$ to other three vertices. The sum of the normal being zero is just the the gradient of the sum of all four barycentric coordinates being zero (barycentric coordinates sum to $1$).

Shuhao Cao
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