Selection Examinations for the IMO

We will show that an $n$-gon with this property exists for $n$ even, but not for $n$ odd.
Assume that a convex $n$-gon can be divided into finitely many parallelograms. Denote one of the sides by $a$. There exists a parallelogram with one side lying on $a$. Denote the opposite side of this parallelogram by $b_1$. If $b_1$ does not lie on some side of the $n$-gon, then there exists a parallelogram that shares a segment with $b_1$. Denote the opposite side of this parallelogram by $b_2$. Repeat.

The parallelograms obtained in this way are all distinct. Since there are only finitely many parallelograms altogether, we eventually (after a finite number of steps) get to a side $b_k$, which lies on some side $c$ of the $n$-gon.
Since the segments $b_i$ and $b_{i+1}$ are parallel for all $i$, $a$ is parallel to $b_1$ and $c$ is parallel to $b_k$, we conclude that $a$ and $c$ are also parallel. Obviously, $c \neq a$. We have shown that for each side of our $n$-gon we can find another side parallel to the first.
Since the $n$-gon is convex no three of its sides are parallel. Indeed, denote the vertices of the $n$-gon by $A_1, A_2, \dots, A_n$. Because of the convexity we have
$$\begin{aligned} 0 < & \angle(\overrightarrow{A_1A_2}, \overrightarrow{A_2A_3}) < \angle(\overrightarrow{A_1A_2}, \overrightarrow{A_3A_4}) < \\ & \vdots \\ < & \angle(\overrightarrow{A_1A_2}, \overrightarrow{A_{n-1}A_n}) < \angle(\overrightarrow{A_1A_2}, \overrightarrow{A_nA_1}) < 2\pi \end{aligned}$$
(the angles are measured in the positive direction from the first vector to the second). Hence, there is only one number $i$ such that $\angle(\overrightarrow{A_1A_2}, \overrightarrow{A_iA_{i+1}}) = \pi$. We conclude that the segment $A_iA_{i+1}$ is parallel to $A_1A_2$. This $n$-gon has pairs of parallel sides. In particular, the number of the sides is even, so $n$ is even. We have hereby shown that an $n$-gon with the required property does not exist for $n$ odd.

Now, let us prove by induction that for all even $n \ge 3$ every convex $n$-gon consisting of pairs of parallel segments of equal length can be divided into finitely many parallelograms. If $n = 4$, then this 4-gon is a parallelogram and
such a splitting exists. Now, assume that we already have the division of an $n$-gon into finitely many parallelograms for some even $n$.
Consider a $(n+2)$-gon consisting of pairs of parallel segments of equal length. Denote its vertices by $A_1, A_2, \dots, A_{n+2}$. Assume that $A_1A_2$ and $A_k, A_{k+1}$ are parallel and of equal length. Let $\tau$ be the translation by $\overrightarrow{A_2A_1}$. Denote $A'_i = \tau(A_i)$ for all $2 \le i \le k$. We have $A'_2 = A_1$ and $A'_k = A_{k+1}$.
Since $\tau$ is a translation, the quadrilateral $A'_iA_iA_{i+1}A'_{i+1}$ is a parallelogram for all $i = 2, 3, \dots, k-1$. At the same

time, $A'_2A'_3\dots A'_kA_{k+2}A_{k+3}\dots A_{n+2}$ is a convex $n$-gon consisting only of pairs of parallel sides of equal length. By the induction hypothesis it can be divided into finitely many parallelograms. Hence, the $(n+2)$-gon can also be divided into finitely many parallelograms.