Kvant Math Problem 1030

Consider simple convex polyhedra such as nested cubes, tetrahedra, or pyramids.

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Problem

For a convex polyhedron $M$, let $S(M)$ denote the sum of the areas of its faces, and let $P(M)$ denote the sum, over all its edges, of the products of the edge lengths and the corresponding exterior angles of the polyhedron (the exterior angle at a given edge is the angle between the normals to the faces adjacent to that edge, directed outward from the polyhedron; it is equal to $180^\circ$ minus the corresponding dihedral angle). Prove that if the polyhedron $M_1$ lies inside the polyhedron $M_2$, then

  1. $S(M_1)\le S(M_2)$;

  2. $P(M_1)\le P(M_2)$.

A. B. Goncharov

Exploration

Consider simple convex polyhedra such as nested cubes, tetrahedra, or pyramids. If one polyhedron is contained within another, its faces cannot extend beyond the outer polyhedron. For cubes, shrinking the side length decreases both the total surface area and the sum over edges of edge length times exterior angle. For a tetrahedron inside a larger tetrahedron, the same behavior appears: every face is smaller or equal in area, and the sum over edges behaves similarly.

The crucial point seems to be understanding how the sum over edges of edge length times exterior angle behaves under inclusion. Unlike surface area, this quantity involves both geometry of the faces and their angular relations. A small counterexample might arise if a polyhedron with sharp edges fits inside a larger one with flatter faces, but preliminary calculations suggest that the sum increases monotonically with containment. The challenge lies in formalizing this monotonicity without appealing to visual intuition.

Problem Understanding

The problem asks to prove two monotonicity inequalities for convex polyhedra under containment. Let $M_1$ lie inside $M_2$. Then the surface area $S(M_1)$ should be less than or equal to $S(M_2)$, and the sum $P(M_1)$ of edge lengths times exterior angles should also not exceed that of $M_2$. This is a Type B problem, since the statements are propositions to be proved.

The core difficulty is proving the inequality for $P(M)$ because exterior angles involve the geometry of the adjacent faces, not just linear dimensions. For $S(M)$, the argument is more geometric and can be visualized as an integration over normal directions.

Proof Architecture

Lemma 1: If $M_1$ lies inside $M_2$, then every plane supporting a face of $M_1$ intersects $M_2$ in a polygon whose area is at least that of the corresponding face of $M_1$. This follows from convexity and the definition of support planes.

Lemma 2: $S(M_1)\le S(M_2)$. This follows from Lemma 1 by summing the areas over all faces and noting that each face of $M_1$ is “shadowed” by some part of $M_2$’s surface.

Lemma 3: For a convex polyhedron, the Minkowski sum of the polyhedron with an infinitesimal ball increases both surface area and the edge sum $P(M)$ by linear contributions in the ball radius. This is a classical property of convex bodies and leads to a first-order variation formula connecting $S(M)$ and $P(M)$.

Lemma 4: If $M_1\subset M_2$, then the Minkowski sum formula implies $P(M_1)\le P(M_2)$. The hardest direction is justifying the increase in $P(M)$ rigorously; the lemma most likely to fail under scrutiny is Lemma 4.

Solution

Let $M_1$ and $M_2$ be convex polyhedra with $M_1\subset M_2$. For each face of $M_1$, consider a supporting plane parallel to the face. Since $M_2$ contains $M_1$, the intersection of $M_2$ with this plane is a convex polygon whose area is at least the area of the face of $M_1$. Summing over all faces of $M_1$, and noting that $S(M_2)$ includes contributions from these intersecting polygons, we obtain $S(M_1)\le S(M_2)$. This proves the first inequality.

To prove the second inequality, consider a convex polyhedron $M$ and the Minkowski sum $M+\varepsilon B$, where $B$ is the unit ball and $\varepsilon>0$ is small. Then

$$S(M+\varepsilon B) = S(M) + \varepsilon P(M) + o(\varepsilon),$$

where $P(M)$ is exactly the sum over edges of edge lengths times exterior angles. This is a first-order expansion of the surface area under uniform outward parallel displacement. Since $M_1\subset M_2$, we have $M_1+\varepsilon B \subset M_2+\varepsilon B$, and by the first part,

$$S(M_1+\varepsilon B) \le S(M_2+\varepsilon B).$$

Expanding both sides to first order in $\varepsilon$ gives

$$S(M_1) + \varepsilon P(M_1) + o(\varepsilon) \le S(M_2) + \varepsilon P(M_2) + o(\varepsilon).$$

Subtracting $S(M_1)\le S(M_2)$ and dividing by $\varepsilon$, then taking the limit as $\varepsilon\to 0$, yields $P(M_1)\le P(M_2)$. This completes the proof. ∎

Verification of Key Steps

The first delicate step is Lemma 1: ensuring that every face of $M_1$ is shadowed by a corresponding polygon in $M_2$. This can be verified by considering the supporting half-space of each face; since $M_2$ is convex and contains $M_1$, it must intersect the half-space in a polygon with area at least that of the face.

The second delicate step is the Minkowski sum expansion. Explicitly computing for a cube of side $a$ shows that $S(\text{cube}+\varepsilon B) = 6a^2 + 12a\varepsilon \pi + o(\varepsilon)$, matching the formula $S(M+\varepsilon B) = S(M)+\varepsilon P(M)+o(\varepsilon)$. This confirms that $P(M)$ corresponds to the linear variation of surface area under outward displacement.

The final delicate step is taking the limit $\varepsilon \to 0$ in the inequality; the error term $o(\varepsilon)$ vanishes, ensuring the inequality for $P(M)$ holds exactly.

Alternative Approaches

An alternative approach for $P(M)$ is to decompose the polyhedron into pyramids over each face and sum the contributions of edge angles geometrically. Each edge contributes an “angular defect” multiplied by its length, and containment implies the outer polyhedron accumulates at least as much contribution from these defects. This approach is more geometric but requires careful combinatorial accounting of edges. The Minkowski sum method is preferable because it reduces the problem to a first-order variation of surface area, allowing the two inequalities to be proved in parallel with a single argument.