How many different ways can octahedron be inscribed in icosahedron so that all vertices of octahedron are selected vertices of icosahedron?
Can it even be done? There are 4 edges in the middle of octahedron. If I put one of them on an edge in icosahedron, it doesn't seem to me like 2 vertices (of the "middle square") would be touching anything. Is my imagination wrong? Why?
On
Another geometrical relation does allow you to inscribe an octahedron inside an icosahedron. The vertices of the octahedron lie on edge centers, not on vertices, of the icosahedron.
Select any vertex $A$ of the icosahedron. The five nearest vertices to $A$ form a regular pentagonal ring; render these vertices $B,C,D,E,F$ in rotational order around the ring.
Now connect, let us say, the midpoint of the "spoke" $\overline{AB}$ with the midpoint of the opposite ring edge $\overline{DE}$. This connection spans exactly $90°$ of arc on the inscribed or circumscribed sphere and thus qualifies as an octahedral edge.
Since there are five ways to orient this connection around each of twelve vertices of the icosahedron and the octahedron has twelve edges, there must be five inscribed octahedra defined in this way. In the above construction each of the possible choices for $B$ corresponds to a different octahedron. This set of five octahedra is also inscribed in a second regular icosahedron which is obtained by interchanging spokes with ring edges at the endpoints of each octahedral-edge connection.
According to this page the answer is no, you can't inscribe an octahedron in an icosahedron so the vertices coincide. You can do it, however, to make the faces coincide.
If group theory doesn't worry you, the wikipedia page on icosahedral symmetry has information about the symmetry subgroups of the icosahedron, and confirms that you can inscribe tetrahedra and cubes in icosahedra so the vertices coincide.