I've recently come across some mid-sized 3-manifolds that I think are likely hyperbolic, but SnapPea has some trouble with them. This is related to my previous question

can you fool SnapPea?

but in this case I'm dealing with closed, orientable 3-manifolds instead of knot and link complements.

The 3-manifolds I've come across have 11, 12 and 13 tetrahedra in their triangulations. One of them SnapPea finds a solution to the gluing equations but it has "negatively oriented tetrahedra". Does this mean what I think it means -- that once you've put the geometric structure on the tetrahedra, you have a tetrahedron folded-over? If you view the gluing equations from the upper half-space model, they say that the sum of a bunch of angles should be $2\pi$. Is this the case where one of those angles is negative?

The other two triangulations SnapPea finds geometric structures with degenerate tetrahedra. In the upper half-space model this is where one of the angles is zero, I believe. Is there a way to fix this, so that I could get a Dirichlet domain, drill and fill, etc? This is my primary question.

Here are the triangulations, both in SnapPea format and Regina format.

Regina file, all triangulations

11-tet, SnapPea

12-tet, SnapPea

13-tet, SnapPea

With the 12-tetrahedron example, SnapPea can generate a Dirichlet Domain. This one has what looks almost like bevelled-edges. Is there a way to find a better basepoint to grow the Dirichlet domain from?

12-tet Dirichlet domain http://dl.dropbox.com/u/46424505/triangulations/pic.png

edit: After working a bit with Nathan's answer, I've identified these three manifolds and got a little closer to understanding how to work SnapPea to maximal advantage.

tri11, as Nathan mentioned, is hyperbolic. It has a fairly pretty Dirichlet domain.

Tri11 Dirichlet Domain http://dl.dropbox.com/u/46424505/triangulations/tri11.jpg

Another common name for this manifold would be the 0-surgery on the 2-component link $7a_6$ (in the Thistlethwaite table). Similarly tri12 can be identified as Nathan says.

After playing around with Regina a bit I found an incompressible torus in tri13 that splits tri13 into the union of an orientable $I$-bundle over the Klein bottle and a figure-8 complement. So this answers the core of my question.