How many parallelograms with sides 1 and 2, and angles $60^{\circ}$ and $120^{\circ}$, can be placed at most in a regular hexagon with side 3?
Let $s$ be the area of an equilateral triangle with side 1. A regular hexagon can be divided into 6 equilateral triangles with sides of 3, thus each having an area of $9s$: its total area is $54s$. Now, each parallelogram has an area of $4s$: since $\frac{54s}{4s} > 13$, one might think it is possible to place 13 parallelograms. However, this is not possible: it suffices to color the hexagon with 12 black triangles and 42 white triangles such that any parallelogram, regardless of the four cells it occupies, will necessarily cover one and only one black triangle. Therefore, the maximum number of parallelograms that can be placed is 12, but it still needs to be proven that 12 can indeed be placed (for example, by not covering the white hexagon at the center at all).

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