Inscribed parallelotope in a $d$-simplex The problem setup is simple: Given a $d$-simplex $\Delta_d:=\{(x_1,\cdots,x_d):x_i\geq 0,\sum_i x_i\leq 1\}$, can we construct a finite sequence of parallelotope $A_i$ so that $\Delta_d=\cup_{i=1}^N A_i$, where $N$ may depend optimally on $d$ in the sense any sequence of such covers must have cardinality lower bounded by $N$ up to a universal constant.
The problem in dimension $d=1,2$ is straightforward. However even in $d=3$, it is not clear to me if there is such a construction.
My main motivation for such a result is trying to reduce certain problem on simplex to parallelotope, which is easier to handle.
Any comment shall be greatly appreciated.
 A: Edit 1: Indeed, my previous suggestion was not good. As it has been pointed out to me by  Dongryul Kim, the parallelotopes should be finitely many. I have overlooked that. Maybe this could work? 
Construction 1: 
In the case of a triangle $ABC$ we define a parallelogram at vertex $A$ to be a parallelogram constructed as follows: pick a point $Q$ on $BC$ and draw the two lines parallel to $AB$ and $AC$.
Induction: Take a $d-1$ face $f$ of the $d$-simplex and the opposite vertex $v$ (the one vertex which is not on $f$). Take a vertex $w$ of $f$. Look at the edge $vw$. Take a point $p$ on $vw$ and draw the hyper-plane $L$ through $p$ parallel to $f$. $L$ intersects the simplex in a $d-1$ simplex $f'$ similar to $f$ (actually homothetic from vertex $v$). Draw the parallelotope on $f'$ with vertex $p$ (see first step above or inductive step). Translate it to $f$ by the vector $\overrightarrow{pw}$ to obtain the desired $d$-dimensional parallelotope at vertex $w$.
Construction 2:
A parallelogram sitting on the edge $AB$ of triangle $ABC$ is a parallelogram defined as follows: Take midpoint $M$ on $AB$ and pick a point $Q$ on $CM$. Draw line parallel to $AB$ to form a segment parallel to $AB$ and then translate it down to $AB$ with vector $\overrightarrow{QM}$. 
Induction: Pick a face $d-1$ dimensional face $f$ of the $d$-simplex and let $v$ be the vertex opposite to it (like before). Let $M$ be the barycenter of $f$ and choose a point $p$ on $vM$. Draw a $d-1$ hyperplane $L$ through $p$ parallel to $f$. $L$ intersects the $d$-simplex in a $d-1$ simplex $f'$ homothetic to $f$ (and parallel). choose a $d-2$ face $f''$ of $f'$ and draw the parallelotope sitting on $f''$ by induction. Now, translate it via the vector $\overrightarrow{pM}$ to $f$. Thus, one obtains a $d$-parallelotope sitting on $f$.     
Seem like using these two procedures one can cover the simplex with finitely many parallelotopes for large enough $N$, by various combinations of face $f$, vertex $w \in f$ and point $p \in vw$ choices. Like various midpoints / barycenters for instance. Maybe only the first procedure is enough, but the second may lower the number of parallelotopes. I guess it depends on the problem. Do we need an estimate on the number $N$ in terms of $d$? 
A: EDIT: This is an answer to the previous version of the question, asking for an existence of a finite number of required polytopes.
The answer is obviously yes by a compactness argument:
Consider $\Delta^d$ as a (compact) topological space. For any $x \in \Delta^d$, there is a parallelotope $P_x$ containing $x$ in $P_x'$ where $P_x'$ is the interior of $P_x$ in $\Delta^d$. (Note that the interior is indeed taken in $\Delta^d$ and not in $\mathbb{R}^d$.)
Thus $P'_x$ form an open cover of $\Delta^d$ and there is a finite subcover $\{P'_{x_i}\}$. The required set of parallelotopes is then $\{P_{x_i}\}$. 
This reasoining does not say what are the required parallelotopes. But I believe that it is actually not very hard to find the required finite set (not too big) of them by choosing $x_i$ and $P_{x_i}$ suitably. 
