How many parallelograms can be placed at most in a regular hexagon with a side length of 3?
Let's call \( s \) the area of an equilateral triangle with side 1. The regular hexagon can be divided into 6 equilateral triangles with sides 3, thus each having an area of \( 9s \): its total area is \( 54s \). Now, each parallelogram has an area of \( 4s \): since \( \frac{54s}{4s} > 13 \), we should be able to place 13 of them. 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, we can place a maximum of 12 parallelograms, but we still need to prove that it is indeed possible to place 12, which is relatively easy (by not covering the central white hexagon at all, for example).

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