Problem 1246

Let $G$ be a convex $n$-gon that can be divided [into parallelograms]. For any edge $a$ on $G$, first rotate it to a horizontal position, then starting from $a$ we can find, going upward step by step, a sequence of parallelograms, each having two sides parallel to $a$ and connected in sequence.

(P): There exists a unique edge on $G$ such that the top of the last parallelogram in the above sequence is attached to this edge.

The reason is: if this sequence does not touch the boundary of $G$, or touches some edge of $G$ but not in the manner of the above-mentioned parallel attachment, then we can continue upward to find the next parallelogram; moreover, since the division uses infinitely many parallelograms, property (P) holds. If it were not unique (there could be other different sequences), then it could not be a convex polygon. We call the edge described by property (P) the parallel opposite edge of $a$, and denote it by $O(a)$. Clearly $O(O(a)) = a$, that is, $O(\cdot)$ is a bijection from edges to edges, and is also an involution. Since $O(a) = a$ is impossible, this involution forms a complete pairing from edges to edges, so $n$ is even; in the following assume $n = 2m$.

In counterclockwise order, $a_0, a_1, \dots, a_{2m-1}$ are the edges of the convex $2m$-gon, and $\vec{a}_i$ is exactly the counterclockwise vector formed by edge $a_i$. Rotate $\vec{a}_0$ to horizontal, then consider the angle $\text{( a i)}$ between each vector and the $x$-axis. Suppose $O(a_0) = a_k$, then
$$\text{0 = ( a 0) < ( a 1) < < ( a k) = 180 < ( a k+1 ) < < ( a 2n-1 ) < 360 .}$$
From this we can see: the parallel pairing is between $a_1, \dots, a_{k-1}$ and $a_{k+1}, \dots, a_{2m-1}$, so the two sets have the same number of edges, hence $O(a_0) = a_m$. Similarly $O(a_j) = a_{j+m}$ (indices taken $\mod 2m$ as appropriate). (PS. This paragraph can be skipped without proof, because for a convex $2m$-gon formed by $m$ pairs of parallel lines, this is naturally the case.)

(1) Looking from the left boundary, none will protrude to the left;
(2) The left boundary goes all the way from the tail of $\vec{a}$ to the head of $O(\vec{a})$.

If (1) or (2) is not true, then conversely we can find, from top to bottom, a sequence of parallelograms (which of course has no intersection with the previous sequence), and the last one should attach to $a$; however, $A$ is the leftmost parallelogram attached to $a$, and the new sequence is to the left of the previous sequence, so the last one of the new sequence cannot attach to $a$. We call this left broken-line boundary $\ell_1$. Similarly we can obtain the right broken-line boundary $\ell_2$, connecting from the head of $\vec{a}$ to the tail of $O(\vec{a})$. Inside the region surrounded by $a, \ell_1, O(a), \ell_2$, there exist finitely many degree-3 vertices of the types $\sqcup$ and $\perp$; for a $\sqcup$ we can draw a parallel line all the way to $a$, and for a $\perp$ we can also draw a parallel line all the way to $O(a)$. After drawing these, it is still a parallelogram division. At this point we can see: edge $a$ is divided into finitely many segments, each segment having a stacked sequence of parallelograms; and for each sequence, its parallelograms all have equal width; there may be other spaces interspersed between adjacent sequences, but that is fine. In this way we go all the way from $a$ to $O(a)$. Correspondingly, $O(a)$ also has this property. The conclusion is that the finitely many segments into which $a$ and $O(a)$ are divided form a bijection, and each corresponding segment has the same length, so naturally $a$ and $O(a)$ have the same edge length.

The above conclusion is: if a convex $n$-gon can be completely divided into finitely many parallelograms, then $n$ is even and every edge has an equal-length parallel opposite edge.

Consider an edge $a$ and its equal-length opposite edge $O(a)$. Going counterclockwise from $a$ along the edges of the convex $n$-gon to reach $O(a)$, draw $(n-2)/2$ parallelograms along the way, letting the widths of these parallelograms all be equal in length to $a$. After removing these parallelograms, what remains is a convex $(n-2)$-gon, and it still has the property that every edge has an equal-length parallel opposite edge; by induction, this convex $(n-2)$-gon can be completely divided into finitely many parallelograms; hence the given convex $n$-gon can also be completely divided into finitely many parallelograms. QED!