Shortest Hamilton Path Planar Problem I think that the problem to obtain the shortest path visiting once time each point (it is not needed to come back to start point so it is a Hamilton path), in its planar euclidean and symmetric version is an NP-complete problem. 
Wikipedia says:
"If the distance measure is a metric and symmetric, the problem becomes APX-complete"
I´m not sure if wikipedia is correct, please can anyone clarify if this problem is NP-complete, APX-complete or both? Thanks in advance.

This problem is polynomially equivalent to the (minimum) Traveling Salesman Problem (TSP). It cannot be NP-complete since it is not a decision problem, but it is surely NP-hard, because the Hamiltonian Cycle Problem is a particular case of TSP and is NP-complete.
$\Delta$TSP problem is TSP with metrics (triangle inequality). It is known to be APX-hard (the referred article is here, however authors don't operate terms of APX, since this class was introduced later). More precisely it is polynomially $\frac32$-approximable, but is NP-hard to be $(1+\varepsilon)$-approximated for small enough positive $\varepsilon$. Also $\Delta$TSP belongs to APX, so it is APX-complete.
So your problem is NP-hard and APX-complete, too. (By the way the first statement follows from the last one.)
