Does a planar triangulation always contain a Hamiltonian path? What about a Hamiltonian path in a triangulation of an n-gon? If not, how long is the longest path?
 A: There are two planar triangulations on 14 vertices without hamiltonian paths. This is the smallest size. For sure these are well-known.

Those answer the question for $n=3$. For $n=4$ the first examples appear at 12 vertices. Same for $n=5$. For $n=6$ one with 10 vertices. For $n=7$ one with 11 vertices.  And so forth.
A: It is a famous theorem of
Whitney (1931*) that
a $4$-connected planar triangulation has a Hamiltonian cycle.

          


          

Example of non-Hamiltonian triangulation from Joseph Malkevitch, 

              
obviously not $4$-connected: removing $3$ vertices (surrounding one) disconnects.

             
(A graph is $4$-connected if it requires removal of $4$ vertices to disconnect it.) 



*
  H Whitney.
  A theorem on graphs.
  Ann. of Math., 32 (1931), pp. 378–390.

In response to the OP's query:

          


          

A connected triangulated graph with no Hamiltonian path.


Perhaps the OP may be interested in this paper:

Arkin, Esther M., Martin Held, Joseph SB Mitchell, and Steven S. Skiena. "Hamiltonian triangulations for fast rendering." The Visual Computer. 12, No. 9 (1996): 429-444.
  
  (Springer link.)

A: See the answer to this question. The magic word is "Kleetope". For more information than you imagined possible, see Guido Helden's Thesis.
