Decomposition of hyper-rectangles into congruent simplices Let $(a_1, \ldots, a_d) \in \mathbb{N}_+^d$ be positive integers and the semi-axes of the $d$-dimensional $\ell_1$-ellipse 
$$ E_{\bf a} := \{{\bf x} \in \mathbb{R}_{\geq 0}^d: \sum_{j=1}^d \frac{x_j}{a_j} \leq 1 \}, $$
which also is a simlex with the $(d+1)$ integer vertices $(0,\ldots,0), (a_1,0,\ldots,0)$, $(0,a_2,0,\ldots, 0), \ldots,  (0,\ldots,0,a_d)$ .
Now, $E_{\bf a}$ is contained in $R_{\bf a} := \{{\bf x} \in \mathbb{R}_{\geq 0}^d: \frac{x_j}{a_j} \leq 1 \}$, i.e. the multi-dimensional rectangle anchored at ${\bf 0}$ that has side-lengths $a_j$. We can assume that the a_j are ordered, i.e. $a_1 \geq a_2 \geq \ldots \geq a_d \geq 1$.
In the two-dimensional setting $d=2$, we can decompose the rectangle $R_{\bf a}$ into
$$ R_{\bf a} =  E_{\bf a} \cup \hat{E}_{\bf a},$$
where 
$$\hat{E}_{\bf a} := \{ (a_1 - x_1, a_2 - x_2) | (x_1,x_2) \in E_{\bf a}  \}$$
also has integer vertices. Moreover, $E_{\bf a} \cap \hat{E}_{\bf a}$ is a null set and $\hat{E}_{\bf a}$ is congruent to $E_{\bf a}$.
Now comes my question: Is such a construction also possible in the higher dimensional setting, i.e. $d\geq 3$? I suppose (by comparing volumes) one needs at least $d!$ (congruent) copies of $E_{\bf a}$? Unfortunately, I have severe difficulties with spatial visualization and already for $d=3$, I was not able to figure out a valid decomposition of $R_{\bf a}$. 
I hope someone can guide me to a solution or tell me the vertices of the $6$? tetrahedrons. Thanks!
 A: For $d = 3$, you can choose the vertices $(0,0,0)$ and $(a_1,a_2,a_3)$ to be common vertices of each tetrahedron, and then choose one vertex that has one non-zero entry, say in the $i$th position ($i=1,2,3$) and then choose one vertex that has one additional non-zero entry, say in the $j$th position where $j \neq i$. 
This gives $6 = 3!$ simplexes, just like you suspected, and clearly they are all of the same volume, namely $a_1a_2a_3/3!$ because each is obtained by a permutation matrix multiplied by a shear matrix that has all $1$ on the diagonal and all $1$ above the diagonal, so the transformation matrix has determinant absolute value 1 and hence is volume preserving. Also the 6 simples have disjoint interiors, i.e. they form a valid triangulation. 
However I don't believe these simplices are congruent unless you have a cube, i.e. all $a_i$ equal. This is because the 6 side lengths in $L_2$ norm are $|a_i|$,$\sqrt{a_i^2+a_j^2}$,$\sqrt{a_i^2+a_j^2+a_k^2}$,$|a_j|$,$\sqrt{a_j^2+a_k^2}$,$|a_k|$. This is not invariant under permuting the $i,j,k$ assuming the side lengths $a_i,a_j,a_k$ are distinct. I think it may be impossible to insist that the simplices are all congruent and also all have vertices that are vertices of the original box, which is traditionally how a triangulation works (i.e. all simplexes have vertices that are vertices of the original box).
In any event, clearly you can extend this decomposition idea to get equal volume simplexes that all have vertices that are vertices of the box, in any dimension, simply by taking each ordering of axes and then adding the sides in order starting at $0$ and ending at the opposite corner, for each possible permutation of sides. There are $d!$ distinct ways to do this, giving $d!$ equal volume simplices whose interiors are disjoint.
