Prove that a triangulated hexagon will have an equal number of double-triangle tiles in each direction. Good afternoon!
I got the problem below from my tutor, which should help with thinking about problems.

A regular hexagon is drawn on isometric paper, with each triangle having side length 1. Two of these triangles make up a tile, with a total of $3n^2$ tiles in any given hexagon. Using the attached images; show that for every possible tiling, of any sized hexagon, will result in an equal number of tiles in each direction.
The grid:

One possible tiling:


I have been given the following hints/advise:

*

*The images show an example for when $n=4$, the proof should work for any possible value of $n$.


*You may wish to think of it as 3-dimensional.


*You could answer with a series of structured sentences, which develop into a convincing proof.
I am finding it quite difficult, I can visualize it in 3D but am having trouble turning it into any kind of proof, I would appreciate any will be online as much as possible to reply and work with you!
 A: I'll give a few more hints and let you fill in the gaps, as I'm sure your tutor would not appreciate me spoiling the whole thing for you.
The first thing one should consider is that there will always exist vertices at which precisely one tile of every kind meets. You should prove that this is always the case. Notice though that there are two ways this can happen: either you have a little cubical 'indent' which goes away from your perspective, or one has a little cubical bump which comes towards you.
There are are two layouts of tiles in which you don't get both of these kinds of tile configurations at the same time, but instead only one of these tile configurations. What are they?
The next thing to notice is that, when one has such a vertex, one can perform a 'flip' of the tiles at this vertex. That is, make the little cube pop 'out' instead of in, or vice versa. Crucially, notice that such a flip does not change the number of tiles of each kind in our tiling.
Suppose that we decide to only do the flip 'in' $\rightarrow$ 'out'. Why can we always do such a flip? If we keep doing these flips, will we eventually run out of possible flips (hint: a flip will never give us a tiling that we have seen before. Why is that?)? When? If we run out of flips, what must the tiling look like? How many tiles of each type do we have in this final tiling? What does this say about the tiling we started with?
It's a bit like blowing up a cubical balloon bit by bit.
