In the convex quadrilateral $A B C D$, we divided the sides into $n \geq 2$ equal parts. On the sides $A B, B C, C D, D A$, let the $k$-th division points from the vertices $A, B, C, D$ be $A_{k}, B_{k}, C_{k}, D_{k}$. For which pairs $(n, k)$ does it hold that the quadrilateral $A B C D$ is a parallelogram if and only if the quadrilateral $A_{k} B_{k} C_{k} D_{k}$ is also a parallelogram?
Solution. Parallelograms can be easily characterized using vectors. Let's start vectors from a fixed point $O$, and denote these vectors with the corresponding lowercase letters, for example, let $\overrightarrow{O P}=\mathbf{p}$. First, we provide a vector description of the parallelograms. We will prove the following statement.

A convex quadrilateral EFGH is a parallelogram if and only if

$$
\mathbf{e}+\mathbf{g}=\mathbf{f}+\mathbf{h}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_00c40da4f1579bc05d0dg-1.jpg?height=265&width=545&top_left_y=434&top_left_x=779)

Figure 1

Indeed, a quadrilateral is a parallelogram if and only if one of its pairs of opposite sides is parallel and equal in length, i.e., if $\overrightarrow{E F}=\overrightarrow{H G}$. This, however, holds if and only if $\overrightarrow{O F}-\overrightarrow{O E}=\overrightarrow{O G}-\overrightarrow{O H}$, i.e., if $\mathbf{f}-\mathbf{e}=\mathbf{g}-\mathbf{h}$, from which we obtain our statement by rearrangement. (It also follows from the proof that the validity of the vector equality is independent of the choice of $O$ - although the individual vectors depend on $O$.)

![](https://cdn.mathpix.com/cropped/2024_05_02_00c40da4f1579bc05d0dg-1.jpg?height=364&width=545&top_left_y=978&top_left_x=774)

Figure 2

After this, the solution to the original problem is a simple calculation. Since the vertices of the new quadrilateral divide the sides of the original quadrilateral in the ratio $k:(n-k)$, we have

$$
\begin{aligned}
& \mathbf{a}_{\mathbf{k}}=\frac{k \mathbf{b}+(n-k) \mathbf{a}}{n}, \quad \mathbf{b}_{\mathbf{k}}=\frac{k \mathbf{c}+(n-k) \mathbf{b}}{n} \\
& \mathbf{c}_{\mathbf{k}}=\frac{k \mathbf{d}+(n-k) \mathbf{c}}{n} \quad \text { and } \quad \mathbf{d}_{\mathbf{k}}=\frac{k \mathbf{a}+(n-k) \mathbf{d}}{n}
\end{aligned}
$$

Based on the above, the quadrilateral $A_{k} B_{k} C_{k} D_{k}$ is a parallelogram if and only if $\mathbf{a}_{\mathbf{k}}+\mathbf{c}_{\mathbf{k}}=\mathbf{b}_{\mathbf{k}}+\mathbf{d}_{\mathbf{k}}$, i.e., if

$$
\frac{n-2 k}{n}(\mathbf{a}+\mathbf{c}-\mathbf{b}-\mathbf{d})=\mathbf{0}
$$

holds.

If $n \neq 2 k$, then this implies $\mathbf{a}+\mathbf{c}=\mathbf{b}+\mathbf{d}$, i.e., the convex quadrilateral $A B C D$ is also a parallelogram (and vice versa). If, however, $n=2 k$, then this condition always holds.

The latter expresses the well-known fact that the midpoints of the sides of any convex quadrilateral form a parallelogram. (Using the notation of Figure 3: $A_{k} B_{k}$ is the midline of triangle $A B C$, and $C_{k} D_{k}$ is the midline of triangle $A D C$, both of which are parallel to $A C$ and half as long as $A C$. Therefore, the quadrilateral $A_{k} B_{k} C_{k} D_{k}$ is a parallelogram.)

![](https://cdn.mathpix.com/cropped/2024_05_02_00c40da4f1579bc05d0dg-1.jpg?height=420&width=588&top_left_y=2235&top_left_x=755)

Figure 3

Thus, the statement of the problem holds for all pairs $(n, k)$ such that $n \neq 2 k$.