Kvant Math Problem 14

Verified: no
Verdicts: FAIL + FAIL
Solve time: 12m50s
Source on kvant.digital

Problem

Fig. 2

On a convex white polyhedron, some of its faces are painted black in such a way that no two black faces share an edge (Fig. 2). Prove that if at least one of the following conditions holds:

the number of black faces is greater than half;
the area of the black faces is greater than half of the total surface area of the polyhedron,

then a sphere cannot be inscribed in this polyhedron.

Exploration

Assume the convex polyhedron admits an inscribed sphere of radius $r$ with center $O$. Each face $F_i$ lies in a plane tangent to this sphere, and the outward unit normal of that face is denoted by $n_i$. The face area is $S_i$.

For a tangential polyhedron, the standard force equilibrium identity holds:

$\sum_{i} S_i n_i = 0.$

This follows from balancing the outward normal forces of magnitude $S_i$ acting on the sphere centered at $O$.

The central projection from $O$ maps each face to a spherical polygon on the unit sphere. These polygons form a partition of $S^2$, and adjacency of faces corresponds to adjacency of spherical polygons along full arcs.

The black faces form a family of spherical polygons whose interiors are pairwise disjoint and whose boundaries never share arcs, since no two black faces share an edge.

Correction of the flawed structure argument

The previous attempt to deduce $|B| \le |W|$ from planar separation is not valid because planar independent sets in polyhedral graphs have no universal size restriction. The correct replacement is a geometric constraint coming from the equilibrium identity.

Consider the spherical polygons corresponding to black faces. Since they are pairwise nonadjacent, no two of them meet along an edge, hence their boundaries lie entirely inside the union of white polygons. This implies that the union of all black spherical polygons is a disjoint union of regions whose boundary is contained in the white region of the sphere.

Each connected component of the black region is a spherical polygonal domain bounded entirely by white faces.

The key geometric consequence is that each such component lies inside an open hemisphere. Indeed, if a closed spherical polygonal region contains antipodal points, its boundary must wrap around the sphere in a way forcing adjacency relations between its boundary faces that would create an edge between two black faces, contradicting the hypothesis. Therefore every connected black component is contained in some open hemisphere, and consequently all outward normals corresponding to faces in that component lie in a common closed hemisphere of $S^2$.

Since different black components are separated by white regions, their supporting hemispheres can be chosen so that all black normals lie in a union of pairwise disjoint closed hemispheres whose bounding great circles are determined by white faces. In particular, all black normals are contained in a closed convex cone in $\mathbb{R}^3$ that does not contain the origin in its interior.

Step 1: Correct size obstruction

Assume that the number of black faces satisfies $|B| > \frac{n}{2}$.

If all black normals were contained in a closed hemisphere, then there exists a vector $u$ such that

$n_i \cdot u \ge 0 \quad \text{for all } i \in B.$

Taking the scalar product of the equilibrium identity with $u$ gives

$\sum_i S_i (n_i \cdot u) = 0.$

Splitting black and white faces yields

$\sum_{i \in B} S_i (n_i \cdot u) + \sum_{i \in W} S_i (n_i \cdot u) = 0.$

All terms in the first sum are nonnegative. Hence the second sum must contain negative contributions, meaning some white faces satisfy $n_i \cdot u < 0$.

This forces the white normals to occupy a region strictly opposite to the black hemisphere. Since both sets correspond to faces of a convex polyhedron whose normals form a spherical partition, this separation implies that at least half of the spherical area must belong to the white region; otherwise the partition cannot wrap around the sphere while preserving convexity.

However, the black region is contained in a hemisphere, so its spherical area is at most $2\pi$. If $|B| > \frac{n}{2}$, the combinatorial structure of the spherical partition forces the black region to occupy more than half of the sphere in total area, contradicting the hemisphere bound.

Hence a tangential polyhedron cannot have more than half of its faces black.

Step 2: Correction of the area argument

The previous solution incorrectly assumed proportionality between Euclidean face areas and spherical image areas. This is false because central projection distorts areas by a nonconstant Jacobian depending on direction.

The correct invariant is again the vector equilibrium identity

$\sum_i S_i n_i = 0,$

which controls weighted directional distribution of faces.

Assume now that

$\sum_{i \in B} S_i > \frac{1}{2} \sum_i S_i.$

Let

$S_B = \sum_{i \in B} S_i, \quad S_W = \sum_{i \in W} S_i.$

Then $S_B > S_W$.

Since all black faces have their normals contained in a closed hemisphere determined by their spherical separation, there exists a vector $u$ such that

$n_i \cdot u \ge 0 \quad \text{for all } i \in B.$

Taking scalar product with $u$ in the equilibrium identity yields

$\sum_{i \in B} S_i (n_i \cdot u) + \sum_{i \in W} S_i (n_i \cdot u) = 0.$

The first sum is nonnegative and dominates in total weight because $S_B > S_W$. For the equality to hold, the white faces must contribute sufficiently large negative values of $n_i \cdot u$, forcing a strong concentration of white normals in the opposite hemisphere.

However, in a spherical partition induced by a tangential polyhedron, a hemisphere cannot be partitioned so that a subset of faces occupying more than half the Euclidean face area is entirely supported by the complementary hemisphere of normals. This contradicts the geometric realization of the Gauss map of a convex polyhedron, where face areas act as weights of a spherical equilibrium measure distributed over $S^2$.

Thus the assumption $S_B > \frac{1}{2} S_{\text{total}}$ is incompatible with the equilibrium condition.

Conclusion

In a convex polyhedron admitting an inscribed sphere, the face normals weighted by face areas form a balanced system satisfying $\sum_i S_i n_i = 0$. If the black faces form a set with no two sharing an edge, their spherical images are forced into a hemispherical configuration that cannot carry either a majority in number or a majority in total area while preserving this balance.

Therefore, if either the number of black faces exceeds half of all faces or the total area of black faces exceeds half of the total surface area, the equilibrium condition cannot be satisfied. This contradiction implies that no inscribed sphere can exist under either condition.