Editing Tait's conjecture (section)

===The counterexample===
[[Image:PlanarNonHamil.png|right]]

The fragment can then be used to construct the non-Hamiltonian '''[[Tutte graph]]''', by putting
together three such fragments as shown on the picture. The "compulsory" edges of the fragments, that must be part of any Hamiltonian path through the fragment, are connected at the central vertex; because any cycle can use only two of these three edges, there can be no Hamiltonian cycle.

The resulting [[Tutte graph]] is [[graph connectivity|3-connected]] and [[planar graph|planar]], so by [[Steinitz' theorem]] it is the graph of a polyhedron. In total it has 25 faces, 69 edges and 46 vertices.
It can be realized geometrically from a tetrahedron (the faces of which correspond to the four large faces in the drawing, three of which are between pairs of fragments and the fourth of which forms the exterior) by multiply truncating three of its vertices.