9.2. On a circle, 2012 points are marked, dividing it into equal arcs. From these, $k$ points are chosen and a convex $k$-gon is constructed with vertices at the chosen points. What is the largest $k$ for which it could turn out that this polygon has no parallel sides?
Answer. For $k=1509$.

Solution. Let $A_{1}, A_{2}, \ldots, A_{2012}$ be the marked points in the order of traversal (we will assume that $A_{2013}=A_{1}, A_{2014}=$ $A_{2}$). We will divide them into quadruples of points $\left(A_{1}, A_{2}, A_{1007}, A_{1008}\right)$, $\left(A_{3}, A_{4}, A_{1009}, A_{1010}\right), \ldots, \left(A_{1005}, A_{1006}, A_{2011}, A_{2012}\right)$. If among the selected $k$ points, all points of some quadruple $\left(A_{2 i-1}, A_{2 i}, A_{2 i+1005}, A_{2 i+1006}\right)$ are found, then in the resulting polygon, there will be two sides $A_{2 i-1} A_{2 i}$ and $A_{2 i+1005} A_{2 i+1006}$, which are symmetric relative to the center of the circle and therefore parallel. This is impossible; hence, in each of the 503 quadruples, no more than three vertices will be marked, that is, $k \leqslant 503 \cdot 3=1509$.

It remains to provide an example of a 1509-gon without parallel sides with vertices at the marked points. For example, the polygon $A_{1} A_{2} \ldots A_{1006} A_{1008} A_{1010} \ldots A_{2012}$ (its vertices are all points with numbers from 1 to 1006 and all points with even numbers from 2008 to 2012) fits. Indeed, the sides $A_{2012} A_{1}, A_{1} A_{2}, \ldots, A_{1005} A_{1006}$ lie on one side of the diameter $A_{2012} A_{1006}$ and therefore cannot be parallel; similarly, the sides $A_{1006} A_{1008}, \ldots, A_{2010} A_{2012}$ are pairwise non-parallel. Finally, the small diagonal $A_{j} A_{j+2}$ of a regular 2012-gon is not parallel to its sides; hence, no two sides of the form $A_{i} A_{i+1}$ and $A_{j} A_{j+2}$ can be parallel.