Does the Delaunay graph (dual of Voronoi) always contain a hamiltonian path (traceable)? I know the answer is negative for hamiltonian cycles since there are several examples such as the one gave by Dillencourt.
Reference: Michael B. Dillencourtt. Toughness and Delaunay Triangulations. Discrete Comput Geom 5:575-601 (1990)
You probably already solved this by now, but the answer is no. A counter-example exists with only 4 triangles. Consider an equilateral triangle split up into 4 smaller triangles of equal size (like the Legend of Zelda triforce). It's easy to see this is a valid Delaunay triangulation. The spanning tree has one triangle with a degree of 3 (the center vertex/triangle).