Kvant Math Problem 14

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Source on kvant.digital

Problem

Fig. 2

A convex white polyhedron has some of its faces 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:

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

then it is impossible to inscribe a sphere in this polyhedron.

Exploration

The condition that no two black faces share an edge means that every edge of the polyhedron is incident with at most one black face. If a sphere is inscribed, every face is tangent to the sphere. For a tangential polyhedron there is a standard relation: if a face is tangent to the sphere at a point $T$, then the area of each small triangle formed by $T$ and an edge of the face equals $\frac r2$ times the length of that edge, where $r$ is the inradius. Summing around the face gives

$$\operatorname{area}(F)=\frac r2, p(F),$$

where $p(F)$ is the perimeter of the face.

Thus areas are proportional to perimeters. The problem should reduce to a statement about perimeters.

Let $P_B$ be the sum of perimeters of black faces and $P$ the sum of perimeters of all faces. Since no edge belongs to two black faces, every edge contributes at most once to $P_B$. On the other hand, every edge contributes exactly twice to $P$, because it belongs to exactly two faces. Hence

$$P_B\le E,\qquad P=2E,$$

where $E$ is the total length of all edges. Therefore

$$P_B\le \frac P2.$$

If a sphere is inscribed, areas satisfy the same inequality:

$$S_B=\frac r2 P_B\le \frac r2\cdot \frac P2=\frac S2.$$

This immediately contradicts condition 2.

Condition 1 concerns the number of faces rather than area. Since each face area equals $\frac r2$ times its perimeter, every face has positive perimeter. Let $W$ be the set of white faces. Because every edge adjacent to a black face is adjacent on its other side to a white face, assigning each edge of a black face to the white face sharing that edge gives

$$P_B\le P_W.$$

Since each face area is proportional to perimeter, we also get

$$S_B\le S_W.$$

If the number of black faces exceeded the number of white faces, then among all faces there would be more black than white. Yet each black face must be paired through its boundary with white faces carrying at least as much total perimeter. The counting version should come from the same perimeter argument applied to face perimeters. More directly, let $b$ and $w$ be the numbers of black and white faces. The sum of perimeters of black faces does not exceed the sum of perimeters of white faces. Hence the average perimeter of a black face is at most $\frac{P_W}{b}$. If $b>w$, then

$$\frac{P_W}{b}<\frac{P_W}{w},$$

so some white face has perimeter larger than the average black perimeter. This alone does not yield a contradiction.

A different idea is needed. Since each black face can share edges only with white faces, the dual graph of the polyhedron has black vertices forming an independent set. The dual graph of a convex polyhedron is planar. A planar graph is properly $4$-colorable, but that does not imply an independent set has size at most half the vertices. For example, a star graph has a larger independent set. So planarity alone is insufficient.

The tangential condition must be used. For a tangential polyhedron, to each face corresponds a polygon circumscribed about the contact circle on the sphere. Let $t_e$ be the tangent length associated with an edge. The perimeter of a face equals twice the sum of the tangent lengths at its sides. Summing over black faces counts each edge at most once; summing over white faces counts every edge not counted by black, and possibly more. Hence $P_B\le P_W$.

Now suppose $b>w$. Since $P_B\le P_W$, the average perimeter of a black face is strictly smaller than the average perimeter of a white face. This is perfectly possible, so still no contradiction.

The intended route is probably contraposition: if a sphere is inscribed, then neither condition 1 nor condition 2 can hold. Condition 2 is already proved. For condition 1, observe that every face area is positive and $S_B\le S_W$. If $b>w$, then at least one black face has area not exceeding the average white area. That still does not force anything.

A stronger fact is available. For a tangential polyhedron, each face corresponds on the sphere to a spherical polygon. The area formula above shows that the measure associated to perimeter behaves additively. The black faces form an independent set in the dual graph. In a tangential polyhedron, assigning to each edge its tangent segment length $t_e$, define the weight of a face as the sum of its incident $t_e$'s. Then every weight is positive, black-face total weight does not exceed white-face total weight. If there were more black faces than white faces, some black face would have weight below the white average, but that is not impossible.

The statement only requires proving impossibility if either condition holds. Since condition 2