Coloring triangles of triangulations in $\mathbb{R}^2$ with permutations, s. t. opposing vertices get the same number I wonder if it is always possible and if there is a known algorithm for the following problem:
Given:
A triangulation $\mathcal{K}$ of a domain $\Omega \subseteq \mathbb{R}^2$ with Lipschitz boundary
consisting of triangles $T_k$ with $\{\bar{T}_k\}_{k=0}^{k=N} = \Omega$.
We define the set of vertices $\mathcal{V}$ of the triangulation as the points $v_{k,l}$ in
$\mathbb{R}^2$ being vertices of the triangles $T_k$.
Because my motivation comes from a numerical analysis setting, we ignore all weird corner cases.
Every edge is straight and not curved. 
Every vertex belongs to at least one triangle.
Every intersection of the interiors of two triangles is always empty.
Every intersection of the closures of two triangles is either empty, consists of a vertex or the common edge of the two triangles.
There are no hanging nodes (numerical analysis speak).
Problem:
Is it always possible to color with the numbers $0, 1, 2$
the vertices of the triangles in such a way, that opposing vertices of two triangles get the same number?
(Here opposing vertices of two neighbour triangles mean the two vertices not belonging to the common edge.)
By coloring the vertices of the triangles, its meant to put the three different numbers $0, 1, 2$
into the three different corners of each of the triangles.
The following picture shows the situation for an example triangulation, where pairs of opposing vertices get one of the three numbers $0, 1, 2$. I have marked pairs of opposing vertices with different "RGB colors" here red, dark red, blue and dark blue.
These "RGB colors" are NOT part of the problem.

On the left it can be seen that one can't simultaneously
color the vertices such that all triangles are equally oriented (pluses and minuses).
The bottom triangle would get two times the same number to satisfy
the opposing vertex criterion,
but three different numbers per triangle are required.
On the right the two criteria are satisfied.
Questions:

*

*Is this a well known problem?

*What is the situation in dimension 3 or higher dimensions where one uses d-simplices?

*Are there known algorithms for this problem?

What I've done:
I haven't found a counter example yet.
I haven't tried some random triangulation generator yet.
I have written a small C++ program for finding possible orientation patterns for a triangulated $n$-sided convex polygon.
Looks like this is related to https://oeis.org/A001045
 A: Not a full solution, but perhaps a useful transformation?
Since you require opposing vertices $\angle BAC, \angle BDC$ of two edge-sharing triangles $ABC, DBC$ to be same-colored, we might as well use that color to color the common edge $BC$.  Then your problem simply becomes:
Is there a way to color all edges of the triangulation s.t. every triangle has one edge of each color?
Sadly, I don't know the answer to the transformed problem.

UPDATE 3/30 to address some comments:
@andre:
If I understand the OP problem correctly, my above transformation is bijective: Given a (vertex or more properly angle) coloring required by OP, it is trivial to transform it into an edge coloring meeting my criterion.  Conversely, given an edge coloring meeting my criterion, it is trivial to transform it into a coloring required by OP.
Specific to the issue of "unshared" edges: A particular angle (what the OP calls a vertex) e.g. $\angle BAC$ only has one constraint: it must be same-colored with the other opposing angle $\angle BDC$ in the other triangle that shares edge $BC$, if such a triangle $\triangle BDC$ exists.  If there is no such opposing angle, then the color of $\angle BAC$ can be whatever color unused among $\{\angle ABC, \angle ACB\}$ -- just as the edge $BC$ can be whatever color unused among edges $\{AB, AC\}$.  In other words, "unopposed" angles and "unshared" edges both can be colored trivially.
@Max:
If I understand you correctly, by "dualizing" you mean this:

*

*Create a new graph where each original triangle maps to a new node, and two new nodes have an edge (graph-theoretic sense, not geometric sense) between them iff the two original triangles share an edge. (Sharing a vertex does not count -- @domotorp)  The new graph is planar since the old triangulation is planar.


*Then 3-color the new graph, and if a new edge gets color $X$ the original shared edge gets the same color $X$.


*Thus, any 3-coloring of the new graph (where no two edges of same color meet at a node) becomes a 3-coloring of all the SHARED edges of the old triangulation (where no two edges of same color belong to the same triangle).  The unshared edges can now be colored trivially.


*FCT implies all tri-valent planar graphs are 3-edge-colorable.  The new graph is not trivalent, because some triangles can have < 3 shared edges.  UPDATE: This actually means the FCT result doesn't apply (directly) here, and we cannot conclude a 3-coloring is always possible.

UPDATE 3/31
My 3/30 update was wrong, and I have since partially deleted it.  Here's the current status AFAICT:

*

*My original transformation is valid.  I.e. the OP problem is 3-colorable iff the original triangulation can be 3-edge-colored s.t. every triangle has 3 different colored edges.


*Max's suggested dualization is also valid.  I.e. the OP problem is 3-colorable iff the new graph is 3-edge-colorable.  Note: Every triangulation can be dualized into a graph, but not every (planar) graph (with node degree $\le 3$) can be transformed back into a triangulation, see below.


*If we do not represent the outer region as a node, then the new graph is NOT tri-valent (each outermost triangle has degree $< 3$), so FCT does not apply.  So we cannot (directly) conclude the OP problem is always 3-colorable.

*

*Yesterday I mistakenly thought since every node has degree $\le 3$, then it is a subgraph of a tri-valent graph, and the FCT result applies, but that was a mistake.  In the footnote below, I show a simple graph where all node degrees $\le 3$ but is not 3-colorable.


*However, AFAICT this example CANNOT be transformed back into a triangulation.  So this does not provide a counterexample to the OP problem.




*Alternatively, we can have a new node represent the outer region. So, if the outer region (technically its complement) is a triangle, then the new graph is trivalent and FCT does apply and the new graph (hence OP triangulation) is 3-colorable.  But of course, this is a small subcase of the OP question.

*

*Here's what I am unsure about: Suppose the original triangulation does not have a triangular outer region.  Can we enclose it in a bigger triangle, then add arbitrary edges (or even vertices) between the bigger triangle and the original outer vertices, to fully triangulate the picture?  If so, then dualizing the enlarged triangulation will result in a trivalent graph and FCT will apply, right?  This will mean a 3-coloring exists on the enlarged graph, and therefore a 3-coloring exists on the enlarged triangulation (and therefore also the original triangulation).  Is this right?



Footnote (not relevant to the main discusion): Example showing why tri-valency is required for FCT result to apply:
P-------Q
|\      |
| 1     |
|  \    |
2   R   |
|  / \  |
| 3   2 |
|/     \|
S---1---T

This graph has $5$ nodes $PQRST$, and the colors around $R$ are forced.  This in turn forces colors for $PS$ and $ST$.  These in turn force $PQ$ and $QT$ to both be colored $3$, which is infeasible.  So this small graph cannot be 3-colored.  However, I cannot find any way to transform this back into a triangulation.
A: This answers the original formulation of the question, in which a vertex of a triangle was allowed to fall onto the interior of an edge of a neighboring triangle.
A simple case analysis shows that this is a counterexample (after adding a bounding square to it):
https://photos.app.goo.gl/5CX4fQcdGMibGgdP8
