Let $\mathcal{O}$ be a regular octahedron. How many lines are there such that a rotation of at most $180^{\circ}$ about them can map $\mathcal{O}$ onto itself?
Solution. Since in a rotation about an axis, only the points on the axis are fixed, and the center $K$ of the octahedron is also fixed, the possible axes must pass through the center of the octahedron. Thus, every such axis intersects the surface of the octahedron at exactly two points ($P$ and $P'$) symmetrically placed with respect to $K$. We distinguish three cases based on whether these two points are vertices, interior points of edges, or interior points of faces.

1. Case, if $P$ is a vertex. There are three such lines, which connect pairs of opposite vertices. The $90^\circ$ rotations about these lines are suitable. It is sufficient to verify this on the vertices: $P$ and $P'$ remain fixed, while the remaining four vertices form the four vertices of a square, and the $90^\circ$ rotation about the $P P'$ axis acts as a $90^\circ$ rotation about the center of the square, thus the vertices are indeed cyclically mapped to each other.
2. Case, if $P$ is an interior point of an edge $AB$. The image of the edge $AB$ under rotation is also an edge. Furthermore, since $P$ is fixed during the rotation, the image of the edge $AB$ also passes through $P$. This can only be true if the image of $AB$ is itself, so the image of $A$ is $B$ (since $A$ cannot be fixed). This implies that $P$ must necessarily be the midpoint of the edge $AB$.

There are 6 lines connecting the midpoints of opposite edges, and every $180^\circ$ rotation about these lines is suitable. Indeed, the endpoints of the edge $AB$ containing $P$, and the edge $A'B'$ containing $P'$, are swapped by the rotation, while the segment connecting the remaining two vertices intersects $P P'$ perpendicularly at $K$, so the rotation also maps these two vertices to each other.

3. Case, if $P$ is an interior point of the equilateral triangle face $ABC$. Similarly to the second case, it can be shown that the image of the triangle $ABC$ must necessarily be itself. Since the vertices of $ABC$ are not fixed, there are two possibilities: $A' = B, B' = C$, and $C' = A$; or $A' = C, C' = B$, and $B' = A$. In both cases, the fixed point $P$ must be equidistant from all three vertices, so $P$ is the center of the face $ABC$.

There are 4 lines connecting the centers of opposite faces, and the $120^\circ$ rotations about these lines are suitable. Indeed, the axis $P P'$ intersects the face $ABC$ perpendicularly, so the $120^\circ$ rotation about $P P'$ acts as a $120^\circ$ rotation about $P$ in the plane of $ABC$, thus mapping the vertices $A, B$, and $C$ to each other. Similar reasoning applies to the opposite face.

In total, there are 13 suitable axes.