How to prove a tiling of a hexagon must form a 3D cubic stack? How to prove that a tiling of a big hexagon consisting of triangles, using only $2$-triangle tiles (three possible orientations), must resemble a continuous, convex (for each small cube), manifold, $3$D cubic stack (i.e. a 3D staircase) if we color each orientation differently?
The original question comes from Mathologer's video from Youtube, in which he asks the question why there must be an equal number of tiles for each orientation. The answer is simple by simply projecting each faces to the "wall", if we assume all tilings can resemble a convex continuous manifold $3$D cubic stack. However this assumption does not seem obvious to me. There might be some tiling that leads to strange shapes which we cannot do the $3$D projection and we need to prove that's never happening.


 A: This discrete surface "contained" into an hexagonal box with sidelength $N$ (see an example on Fig. 1) is perfectly known in terms of a set of regularly spaced level lines at heights $k+1/2, k=0,1,... N-1$ (see dotted lines on figure 2, in connection with figure 1).

Fig. 1: A realization.

Fig. 2: The associated paths.
In fact, we can think in terms of an algorithm able to build any such set of regularly spaced level lines, and associate to this set the adequate lozenges (or rhombi), providing the final 3D visual interpretation.
Here is the principle of this algorithm for a "side-$N$" hexagon (here $N=8$):

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*Initialize the hexagon with a "all blue color".


*Build $N$ paths connecting the LHS to the RHS ; a path being a broken line with slopes $\pm 1/2$ with two constraints a) they must be non-intersecting b) for each path, the $k$-th line segment on the LHS is connected to the $k$-th line segment on the right (otherwise said, each broken line has the sum of its $\pm 1/2$ slopes equal to $0$).


*Now the graphical part: for each path, in the case of a $+1/2$ slope (resp. $-1/2$ slope) associate to it an ascending yellow rhombus (resp. a descending red rhombus).
Almost magically, in fact very naturally, the blue sectors appear as flat "plateaus" or "mesas".
Remarks:

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*This issue is connected to important combinatorial problems, for example the so-called "Aztech diamond" configuration; see the very interesting slideshow here where our algorithm is described as well.


*I am indebted to one of my students who has written on my demand the program that has given these figures... funnily just some days ago!


*The site "Art Of Problem Solving" uses this kind of "cube piles" as a (default) visual code for each of its participants. See for example [here] (https://artofproblemsolving.com/community/c3h1170419p5617947).


*Working on hexagonal grids necessitates sometimes specifing coding means: have a look at this site.
