Kvant Math Problem 119
The vectors described in the statement depend only on the face areas and outward unit normals.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m24s
Source on kvant.digital
Problem
Prove that if, on each face of a convex polyhedron, one chooses a point and draws from that point a vector perpendicular to the corresponding face, directed outward, whose length is equal to the area of that face, then the sum of all such vectors is equal to zero.
N. B. Vasiliev
Exploration
The vectors described in the statement depend only on the face areas and outward unit normals. The chosen points on the faces are irrelevant to vector addition, since vectors are free vectors. Thus the problem asks to prove that
$$\sum_i S_i\mathbf n_i=\mathbf 0,$$
where $S_i$ is the area of the $i$-th face and $\mathbf n_i$ is its outward unit normal.
A tetrahedron provides a useful test case. For each face, the vector $S_i\mathbf n_i$ is the area vector of that face. In elementary vector geometry, the oriented area vectors of the four faces of a tetrahedron sum to zero. The statement therefore holds for tetrahedra.
A natural attempt is to project all these vectors onto an arbitrary direction $\mathbf e$. If the projection of the sum onto every direction is zero, then the sum itself is zero. The projection of $S_i\mathbf n_i$ onto $\mathbf e$ equals
$$S_i(\mathbf n_i\cdot\mathbf e).$$
This quantity has a geometric meaning: it is the signed area of the orthogonal projection of the face onto a plane perpendicular to $\mathbf e$.
The crucial point is to prove that the total signed projected area of all faces is zero. Looking at the polyhedron from the direction $\mathbf e$, every line parallel to $\mathbf e$ that meets the interior enters the polyhedron through one face and leaves through another. The projected region is covered once with positive sign and once with negative sign. Hence the positive and negative contributions cancel.
The step most likely to conceal an error is the passage from face projections to the projected image of the whole polyhedron. One must justify carefully that the signed projected areas cancel pointwise over the projection plane.
Problem Understanding
We are given a convex polyhedron. For each face, a vector is drawn perpendicular to that face, directed outward, with magnitude equal to the area of the face. The task is to prove that the vector sum of all these face vectors is zero.
This is a Type B problem, a pure proof.
The core difficulty is to connect the geometric data of all faces simultaneously. The natural quantity attached to a face is its area vector $S_i\mathbf n_i$, and the statement becomes the assertion that the sum of all area vectors of a closed convex polyhedral surface vanishes.
Proof Architecture
Consider the vectors $S_i\mathbf n_i$, where $S_i$ is the area of a face and $\mathbf n_i$ its outward unit normal; proving their sum is zero is equivalent to the statement of the problem.
For an arbitrary unit vector $\mathbf e$, the scalar product $S_i\mathbf n_i\cdot\mathbf e$ equals the signed area of the orthogonal projection of the face onto a plane perpendicular to $\mathbf e$, because orthogonal projection multiplies area by the cosine of the angle between the normals.
The sum of the signed projected areas of all faces onto a plane perpendicular to $\mathbf e$ is zero, because every point of the projection of the polyhedron corresponds to exactly one visible face and one opposite face, contributing equal areas with opposite signs.
If the scalar product of a vector with every unit vector $\mathbf e$ is zero, then the vector itself is zero.
The most delicate lemma is the cancellation of the signed projected areas. Any imprecision about multiplicities in the projection could invalidate the argument.
Solution
Let the faces of the convex polyhedron be $F_1,\dots,F_m$. Denote by $S_i$ the area of $F_i$ and by $\mathbf n_i$ the outward unit normal to that face. The vector attached to $F_i$ is
$$\mathbf v_i=S_i\mathbf n_i.$$
We must prove that
$$\sum_{i=1}^{m}\mathbf v_i=\mathbf 0.$$
Let
$$\mathbf V=\sum_{i=1}^{m}S_i\mathbf n_i.$$
It suffices to show that $\mathbf V=\mathbf 0$.
Choose an arbitrary unit vector $\mathbf e$ and let $\Pi$ be the plane perpendicular to $\mathbf e$. For each face $F_i$, consider its orthogonal projection onto $\Pi$.
If $\theta_i$ denotes the angle between $\mathbf n_i$ and $\mathbf e$, then the area of the projection of $F_i$ onto $\Pi$ equals
$$S_i|\cos\theta_i|.$$
Assign to this projected area the sign of $\cos\theta_i$. The resulting signed area is
$$S_i\cos\theta_i = S_i(\mathbf n_i\cdot\mathbf e).$$
Hence
$$\mathbf V\cdot\mathbf e = \sum_{i=1}^{m}S_i(\mathbf n_i\cdot\mathbf e),$$
which is exactly the sum of the signed projected areas of all faces onto $\Pi$.
We now compute this sum geometrically.
Project the entire polyhedron orthogonally onto $\Pi$. Because the polyhedron is convex, every line parallel to $\mathbf e$ intersects it either in a segment or not at all. Consider a point of the projection lying in the interior of the projected image and not on the projection of an edge. The line through that point parallel to $\mathbf e$ meets the boundary of the polyhedron at exactly two points: one on the side facing the direction $\mathbf e$, and one on the opposite side.
These two boundary points belong to two faces. The face on the side facing $\mathbf e$ has outward normal with positive scalar product with $\mathbf e$ and contributes positively to the signed projected area. The opposite face has outward normal with negative scalar product with $\mathbf e$ and contributes negatively.
For the chosen point of the projection plane, these two contributions have equal magnitude, since both project onto the same area element of $\Pi$, and opposite signs. Thus they cancel.
This cancellation occurs for every interior point of the projected image except a set of measure zero coming from projected edges and vertices. Integrating over the projection plane, the total signed projected area of all faces is therefore zero. Consequently,
$$\sum_{i=1}^{m}S_i(\mathbf n_i\cdot\mathbf e)=0.$$
Hence
$$\mathbf V\cdot\mathbf e=0$$
for every unit vector $\mathbf e$.
A vector whose scalar product with every unit vector is zero must itself be the zero vector. Therefore
$$\mathbf V=\mathbf 0.$$
Since $\mathbf V=\sum_{i=1}^{m}\mathbf v_i$, we obtain
$$\sum_{i=1}^{m}\mathbf v_i=\mathbf 0.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the identification of
$$S_i(\mathbf n_i\cdot\mathbf e)$$
with a signed projected area. Let $\alpha$ be the angle between the planes of $F_i$ and $\Pi$. Then the angle between their normals is also $\alpha$, so orthogonal projection multiplies area by $|\cos\alpha|$. Since $\alpha$ is the angle between $\mathbf n_i$ and $\mathbf e$, the projected area is $S_i|\mathbf n_i\cdot\mathbf e|$. Introducing the sign of $\mathbf n_i\cdot\mathbf e$ yields the signed area $S_i(\mathbf n_i\cdot\mathbf e)$.
The second delicate step is the cancellation over the projection plane. A careless argument might count projected regions multiple times. For a convex polyhedron, every line parallel to $\mathbf e$ intersects the boundary in exactly two points whenever it passes through the interior. Thus each generic projected point corresponds to exactly one front face and one back face. Their signed contributions are $+dA$ and $-dA$, where $dA$ is the corresponding area element in the projection plane. No additional intersections occur because convexity excludes them.
The final step is the deduction that $\mathbf V=\mathbf0$ from $\mathbf V\cdot\mathbf e=0$ for every unit vector $\mathbf e$. Taking $\mathbf e$ parallel to $\mathbf V$ would give
$$\mathbf V\cdot\mathbf e=|\mathbf V|.$$
Since every such scalar product is zero, $|\mathbf V|=0$, hence $\mathbf V=\mathbf0$.
Alternative Approaches
A different proof uses the divergence theorem. Let $\mathbf c$ be any constant vector field. Its divergence is zero. Applying the divergence theorem to the polyhedron gives
$$0=\iiint_{\Omega}\operatorname{div}\mathbf c,dV =\iint_{\partial\Omega}\mathbf c\cdot\mathbf n,dS.$$
Since $\mathbf c$ is constant,
$$0 = \mathbf c\cdot \left( \iint_{\partial\Omega}\mathbf n,dS \right).$$
Because this holds for every constant vector $\mathbf c$,
$$\iint_{\partial\Omega}\mathbf n,dS=\mathbf0.$$
For a polyhedron the normal is constant on each face, so
$$\iint_{\partial\Omega}\mathbf n,dS = \sum_i S_i\mathbf n_i.$$
Hence the desired sum is zero.
The main proof is preferable in the context of a school mathematics problem because it uses only elementary geometry of projections and convexity, avoiding vector calculus and the divergence theorem.