XLII OM - I - Problem 8

Determine the largest natural number $ n $ for which there exist in space $ n+1 $ polyhedra $ W_0, W_1, \ldots, W_n $ with the following properties:
(1) $ W_0 $ is a convex polyhedron with a center of symmetry,
(2) each of the polyhedra $ W_i $ ($ i = 1,\ldots, n $) is obtained from $ W_0 $ by a translation,
(3) each of the polyhedra $ W_i $ ($ i = 1,\ldots, n $) has a point in common with $ W_0 $,
(4) the polyhedra $ W_0, W_1, \ldots, W_n $ have pairwise disjoint interiors.
Suppose that polyhedra $W_0, W_1, \ldots, W_n$ satisfy the given conditions. Polyhedron $W_1$ is the image of $W_0$ under a translation by a certain vector $\overrightarrow{\mathbf{v}}$ (condition (2)). Let $O_0$ be the center of symmetry of polyhedron $W_0$ (condition (1)); the point $O_1$, which is the image of $O_0$ under the translation by $\overrightarrow{\mathbf{v}}$, is the center of symmetry of $W_1$. Figure 3 illustrates a planar variant of the considered problem (a representation of the spatial configuration would obscure this illustration); polyhedra $W_0$ and $W_1$ are depicted as centrally symmetric polygons.
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By condition (3), polyhedra $W_0$ and $W_1$ have common points (possibly many). Let $K$ be a common point of $W_0$ and $W_1$ (arbitrarily chosen). Denote by $L$ the image of point $K$ under the central symmetry with respect to $O_1$; thus, $L \in W_1$. Let $N$ be a point such that $\overrightarrow{NL} = \overrightarrow{\mathbf{v}}$ and let $M$ be the midpoint of segment $NK$ (Figure 4). Therefore, $N \in W_0$. According to condition (1), the set $W_0$ is convex; this means that with any two points belonging to $W_0$, the entire segment connecting these points is contained in $W_0$. Since $K \in W_0$ and $N \in W_0$, it follows that $M \in W_0$. Segment $MO_1$ connects the midpoints of segments $KN$ and $KL$, and thus $\overrightarrow{MO_1} = \frac{1}{2} \overrightarrow{NL} = \frac{1}{2} \overrightarrow{\mathbf{v}}$, which means $M$ is the midpoint of segment $O_0O_1$.
Let $U$ be the image of polyhedron $W_0$ under a homothety with center $O_0$ and scale factor 3. We will show that $W_1 \subset U$. Take any point $P \in W_1$: let $Q \in W_0$ be a point such that $\overrightarrow{QP} = \overrightarrow{\mathbf{v}}$ and let $S$ be the center of symmetry of parallelogram $O_0O_1PQ$. The medians $O_0S$ and $QM$ of triangle $O_0O_1Q$ intersect at a point $G$ such that $\overrightarrow{O_0G} = \frac{2}{3}\overrightarrow{O_0S} = \frac{1}{3}\overrightarrow{O_0P}$ (Figure 5). This means that $P$ is the image of point $G$ under the considered homothety. Since $G$ is a point on segment $QM$ with endpoints in the set (convex) $W_0$, it follows that $G \in W_0$. Therefore, $P \in U$ and from the arbitrariness of the choice of point $P \in W_1$ we conclude that $W_1 \subset U$.
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In the same way, we prove that each of the sets $W_i (i=1, \ldots, n)$ is contained in $U$. Of course, also $W_0 \subset U$. Thus, the set $W_0 \cup W_1 \cup \ldots \cup W_n$ is a polyhedron contained in $U$. By conditions (2) and (4), its volume equals the volume of $W_0$ multiplied by $n+1$. On the other hand, the volume of $U$ equals the volume of $W_0$ multiplied by 27. Therefore, $n \leq 26$.
It remains to note that the value $n = 26$ can be achieved (example realization: 27 cubes $W_0, \ldots, W_{26}$ arranged like a Rubik's cube). Thus, the sought number is $26$.
Note. We obtain the same result assuming that $W_0, \ldots, W_n$ are any bounded, closed convex bodies (not necessarily polyhedra), with non-empty interiors, satisfying conditions (1)-(4); the reasoning carries over without any changes. Moreover, condition (4) turns out to be unnecessary. This was proven by Marcin Kasperski in the work 27 convex sets without a center of symmetry, awarded a gold medal at the Student Mathematical Paper Competition in 1991; a summary of the work is presented in Delta, issue 3 (1992).