Problem 1285

We start with the observation that if in space a convex planar polygon $S$ and a plane $\pi$ are given, and if $\overrightarrow{n}$ is a unit vector perpendicular to the plane of the polygon $S$, and $\overrightarrow{w}$ is a unit vector perpendicular to the plane $\pi$, then denoting by $W$ the orthogonal projection of the polygon $S$ onto the plane $\pi$, we have the following relationship between the areas of the polygons $S$ and $W$:

$$\text{Area}(W) = |\overrightarrow{n} \cdot \overrightarrow{w}| \cdot \text{Area}(S)$$

(The dot between the vectors denotes the dot product.)
Formula (I) is obvious in the case where the considered planes are parallel or perpendicular (because in these cases, respectively, $|\overrightarrow{n} \cdot \overrightarrow{w}|$ equals $1$ or $0$). In the remaining case, when the planes intersect along a line $l$ forming an angle $\varphi (0 < \varphi < \pi/2)$, formula (1) follows from the following observations:
If $S$ is a rectangle with one side parallel and the other perpendicular to the line $l$, its projection $W$ onto the plane $\pi$ is also such a rectangle. The length of the side parallel to $l$ does not change during projection, while the length of the side perpendicular to $l$ is shortened by a factor of $|\cos \varphi|$. The area of the rectangle changes by the same factor; thus, for a rectangle $S$, we obtain formula (1), because $|\overrightarrow{n} \cdot \overrightarrow{w}| = |\cos \measuredangle (\overrightarrow{n}, \overrightarrow{w})| = |\cos \varphi|$.
From this, it immediately follows that (1) is true for any right triangle with one leg parallel and the other perpendicular to $l$, because such a triangle can be completed to a rectangle (whose two sides are the legs of the given triangle). The area of the rectangle changes during projection by a factor of $|\cos \varphi| = |\overrightarrow{n} \cdot \overrightarrow{w}|$, so the same happens to the area of the triangle, which is half the area of the rectangle (Figure 8).
It is enough to notice that any convex polygon $S$ is the sum of a finite number of right triangles with legs parallel and perpendicular to the line $l$, with disjoint interiors. The area of $S$ is the sum of the areas of these triangles, and the area of $W$ (the projection of $S$ onto $\pi$) is the sum of the areas of the projections of the individual triangles (Figure 8). The area of the projection of each of these triangles equals the area of the triangle multiplied by the same factor equal to $|\overrightarrow{n} \cdot \overrightarrow{w}|$, and by summing up, we obtain the formula (1) to be proved.
We proceed to the proof of the theorem. Let

$$\overrightarrow{v} = \sum_{i=1}^{k} \text{Area}(S_i) \overrightarrow{n_i}$$

We need to prove that $\overrightarrow{v}$ is the zero vector.
Choose any unit vector $\overrightarrow{w}$ in space. Let $\pi$ be any plane perpendicular to the vector $\overrightarrow{w}$. The orthogonal projection of the considered polyhedron onto the plane $\pi$ is a convex polygon $W$. The projection of any face $S_i$ is a convex polygon $W_i$ (degenerate to a segment if $\overrightarrow{n_i} \bot \overrightarrow{w}$).
We partition the set of indices $\{1, \ldots, k\}$ into three subsets:

$$I_+ = \{i : \overrightarrow{n_i} \cdot \overrightarrow{w} > 0\}$$
$$I_- = \{i : \overrightarrow{n_i} \cdot \overrightarrow{w} < 0\}$$
$$I_0 = \{i : \overrightarrow{n_i} \cdot \overrightarrow{w} = 0\}$$

(Looking at the polyhedron from the outside, in the direction of the vector $\overrightarrow{w}$, we see the faces $S_i$ with indices $i \in I_-$, and we do not see the faces with indices $i \in I_+$) (Figure 9). The projections of the faces with indices $i \in I_+ \cup I_0$ have no common interior points and fill the entire polygon $W$. The same can be said about the projections of the faces with indices $i \in I_- \cup I_0$. Therefore,

$$\sum_{i \in I_+ \cup I_0} \text{Area}(W_i) = \text{Area}(W)$$

and simultaneously,

$$\sum_{i \in I_- \cup I_0} \text{Area}(W_i) = \text{Area}(W)$$

Notice now that according to formula (1),

$$\text{Area}(W_i) = |\overrightarrow{n_i} \cdot \overrightarrow{w}| \cdot \text{Area}(S_i)$$

Therefore,

$$\sum_{i \in I_+ \cup I_0} |\overrightarrow{n_i} \cdot \overrightarrow{w}| \cdot \text{Area}(S_i) = \text{Area}(W)$$

and

$$\sum_{i \in I_- \cup I_0} |\overrightarrow{n_i} \cdot \overrightarrow{w}| \cdot \text{Area}(S_i) = \text{Area}(W)$$

Since the vector $\overrightarrow{w}$ was chosen arbitrarily, we have thus shown that the vector $\overrightarrow{v}$ is perpendicular to any unit vector. It is therefore the zero vector.
(Figure 9)