TITLE: Partitions of convex planar regions into zonogons
QUESTION [6 upvotes]: A zonogon is a centrally symmetric convex polygon.

Are there convex non-zonogons that can be partitioned into a finite number of (convex) zonogons?

Same as 1 with the pieces allowed to be nonconvex but centrally symmetric polygons.

REPLY [2 votes]: I extend on the answer of Yaakov Baruch.
Choose a generic direction as being "down". We can then speak of bottom edges and top edges of the polygon and its tiles.
Suppose that the polygon $P$ is partitioned into convex tiles, so that there exists a perfect matching between the tiles that are not centrally symmetric and matched tiles are related by a central point reflection (see the figure below).
Assume further that the partition is edge-to-edge, that is, no vertex of one tile meets an edge of another tile (Yaakov explained why this is no restriction).
Let $e_0$ be a top edge.
Note that every edge is the top-edge of at most one tile.
We construct a sequence of edges $e_1,...,e_m$ of edges as follows:

if $e_i$ is the top edge of a centrally symmetric tile $T$, then let $e_{i+1}$ be the "opposite" bottom edge of $T$.
if $e_i$ is the top edge of not-centrally symmetric tile $T$, then let $e_{i+1}$ be the corresponding edge in the matched tile to $T$. $e_{i+1}$ is then a bottom edge of the matched tile.
if $e_i$ is not the top edge of any tile, then $e_i$ is a bottom edge of $P$.

Note that all edges in this sequence are of the same length and have the same orientation, and so each top edge of $P$ corresponds to an equally long and equally oriented bottom edge. Since $P$ is convex, the order of these edges at the bottom must be inverse to the order at the top. This shows that $P$ must be centrally symmetric.

Note that this formulation includes the case of non-convex tiles.
Every centrally symmetric tile can be decomposed into a set of convex centrally symmetric tiles and a set of pairs of matched convex tiles that are not necessarily centrally symmetric: