69th Belarusian Mathematical Olympiad

Answer: minimal possible $C$ equals $5$.

First we prove two lemmas.

Lemma 1: the given 14-gon has $7$ pairs of opposite parallel sides. Since the 14-gon is convex, it suffices to prove that each side has a side parallel to it. Consider an arbitrary side $AB$ of the given 14-gon. Draw the line perpendicular to $AB$ and define on this line the direction to the right in such way that the whole 14-gon lies to the right of $AB$ (it is possible since the 14-gon is convex). Consider a parallelogram $P_1$ with one (left) side lying on $AB$ and the opposite (right) side lies on another side or inside the 14-gon. If it lies inside the 14-gon, there is another parallelogram $P_2$, the left side of which intersect the right side of $P_1$, while the right side of $P_2$ lies on the side or inside the 14-gon. Continue choosing such parallelograms $P_1,P_2,\dots ,P_n$ until finally we find the parallelogram $P_n$, the right side of which lies on some side of the given 14-gon. This side is the required side, parallel to $AB$. Lemma 1 is proved.

Call the sequence $P_1,P_2,\dots ,P_n$ of the parallelograms, defined in the proof of the lemma 1 for an arbitrary side $AB$, the chain of the side $AB$.

Lemma 2: the chains of any two distinct non-opposite sides intersect (i.e. have a common parallelogram). Consider any two pairs $AB\parallel A^{\prime }{B}^{\prime }$ and $CD\parallel C^{\prime }{D}^{\prime }$ of opposite sides. Since the 14-gon is convex, if we bypass its perimeter (in some direction), the sides $AB$, $CD$, $A^{\prime }{B}^{\prime }$, $C^{\prime }{D}^{\prime }$ go in that order. Connect the midpoints of the parallel to $AB$ sides of the parallelograms from the chain of $AB$ to obtain a polyline (the ends of the polyline are the midpoints of $AB$ and $A^{\prime }{B}^{\prime }$). This polyline divides the 14-gon into two parts, with the sides $CD$ and $C^{\prime }{D}^{\prime }$ lying in distinct parts. Therefore, if we consider a similar polyline, corresponding to the chain of $CD$, these two polylines will intersect. This means that the chains of the sides $AB$ and $CD$ have a common parallelogram. Lemma 2 is proved.

Take any $7$ pairwise nonparallel sides of the 14-gon and to each of them (say, for $EF$) put in the correspondence the shortest possible integer vector $\vec{v}(EF)$. Call this vector to be directive for $EF$ and denote its coordinates by $x(EF)$ and $y(EF)$. The minimality of the length of $\vec{v}(EF)$ implies $\gcd (x(EF),y(EF))=1$. From Lemma 2 it follows that for any two non-opposite sides $AB$ and $CD$ there exists a parallelogram $P$ with the sides parallel to $AB$ and $CD$. The area of $P$ is divisible by the area of the parallelogram with the sides $v(AB)$ and $v(CD)$, since $P$ can be divided into such parallelograms.

Finally, we will prove that there exists a parallelogram with area divisible by $5$. The area $S(P)$ of the parallelogram $P$ equals $|x(AB)y(CD)-x(CD)y(AB)|$. If both $x(AB)$ and $x(CD)$ are divisible by $5$, then $S(P)$ is divisible by $5$. Suppose that not more than one directive vector has an $x$-coordinate which is divisible by $5$. Among at least $6$ directive vectors with nonzero $x$-coordinates there exist vectors with $\frac{y(AB)}{x(AB)}\equiv \frac{y(CD)}{x(CD)}(\mathrm{mod}5)$. Therefore, the area $S(P)$ of the corresponding parallelogram equals $$|x(AB)y(CD)-x(CD)y(AB)|=\left|\left(\frac{y(AB)}{x(AB)}-\frac{y(CD)}{x(CD)}\right)\cdot x(AB)x(CD)\right|\equiv 0(\mathrm{mod}5).$$