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, $c\ne a$. 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 $$\begin{array}{rl}0<& \angle (\overrightarrow{A_1{A}_2},\overrightarrow{A_2{A}_3})<\angle (\overrightarrow{A_1{A}_2},\overrightarrow{A_3{A}_4})<\\ & ⋮\\ <& \angle (\overrightarrow{A_1{A}_2},\overrightarrow{A_{n-1}{A}_n})<\angle (\overrightarrow{A_1{A}_2},\overrightarrow{A_n{A}_1})<2\pi \end{array}$$ (the angles are measured in the positive direction from the first vector to the second). Hence, there is only one number $i$ such that $\angle (\overrightarrow{A_1{A}_2},\overrightarrow{A_i{A}_{i+1}})=\pi$. We conclude that the segment $A_i{A}_{i+1}$ is parallel to $A_1{A}_2$. 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 $A_1{A}_2$ and $A_k,A_{k+1}$ are parallel and of equal length. Let $\tau$ be the translation by $\overrightarrow{A_2{A}_1}$. Denote $A_i^{\prime }=\tau (A_i)$ for all $2\le i\le k$. We have $A_2^{\prime }=A_1$ and $A_k^{\prime }=A_{k+1}$. Since $\tau$ is a translation, the quadrilateral $A_i^{\prime }{A}_i{A}_{i+1}{A}_{i+1}^{\prime }$ is a parallelogram for all $i=2,3,\dots ,k-1$. At the same time, $A_2^{\prime }{A}_3^{\prime }\dots A_k^{\prime }{A}_{k+2}{A}_{k+3}\dots A_{n+2}$ 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.