Thickness of complete graph on 12 vertex minus Hamiltonian cycle The thickness of a graph $G$, denoted by $\Theta(G)$, is the minimum number of planar spanning subgraphs into which the graph can be decomposed. Let $K_{12}$ is a complete graph on $12$ vertex and $C$ is a Hamiltonian cycle of $K_{12}$. Put $G=K_{12}\setminus C$. Is it true $\Theta(G)>2$?
 A: No.  In fact we can add six edges to $G$ to get the complement of a 
"$1$-factor" (degree-$1$ subgraph), which still has thickness $2$:
it can be decomposed into two icosahedra, with the one-factor
joining each vertex to its antipode.
Here's one nice way to see that the complement of the icosahedron
is the union of the six diameters and another icosahedron.
Recall that we can decompose the twelve vertices into
three pairwise orthogonal golden rectangles $R$, with all edge lengths
equal to the shorter side of $R$:

We can choose coordinates so that the vertices are cyclic permuttions of
$(0,\pm1,\pm\varphi)$ where $\varphi=(1+\sqrt{5})/2$ is the golden ratio;
then edges have length $2$, while non-edges are
diameters and pairs of vertices at distance $2\varphi$.
Changing $\varphi$ to its algebraic conjugate $1-\varphi$
we again obtain three pairwise orthogonal golden rectangles,
but with shorter and longer edges switched, and thus
edges of the icosahedron switched with the non-edges other than diameters.
