Packing of parallelograms with sides $1$ and $\sqrt{2}$ and angle $45^{\circ}$ in a rectangular container If $ABCD$ is a parallelogram such that $\angle BAD=\frac{\pi}{4}$, $AB=\sqrt{2}$, $AD=1$ then we shall say that $ABCD$ is a good parallelogram. Parallelograms can be rotated by any angle.
a) Prove that in a rectangular container $2 \times n$ ($n \ge 2$) cannot be packed more than $2n-2$ good parallelograms (packing must be without overlaps between parallelograms).
b) Prove that in a rectangular container $4 \times 4$ cannot be packed more than $12$ good parallelograms (packing must be without overlaps between parallelograms).
My work. It is easy to pack $2n-2$ good parallelograms into rectangular container $2 \times n$. The area of the good parallelogram is $1$. The area of the rectangular container $2 \times n$ is $2n$. Obviously, parallelograms will not be able to cover every region of the rectangular container. Therefore, in a rectangular container $2 \times n$ ($n \ge 2$) cannot be packed more than $2n-1$ good parallelograms. But I have no idea how to improve the evaluation for one parallelogram.
 A: My proof for $2\times n$ by induction: 
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Let us show the statement is true for $2\times 2$: (The proof includes geometric figures can be seen at the link: https://www.geogebra.org/classic/khfaznnu .) Showing $3$ parallelograms cannot fit into $2\times 2$ rectangle will be enough. Lets consider barycentres of these parallelograms. In order to fit a parallelogram, the distance between its barycentre and vertices of rectangle must be at least $\dfrac{\sqrt5}{2}$. Also consider that the distance between barycentres of $2$ parallelogram is at least $\dfrac{\sqrt2}{2}$. So, if we can fit $3$ parallelograms, we can fit their barycentres inside $EHFG$ (I mean it is the figure outside of $4$ circles. I don't know if the figure has a name. If has, a comment  will be appreciated.). However, as can be seen at the link, we cannot fit an triangle with sides are length at least $\dfrac{\sqrt2}{2}$ because drawing a circle with center $G$ and radius $\dfrac{\sqrt2}{2}$ shows us the best proper triangle is $GPQ$ but $|PQ|\le\dfrac{\sqrt2}{2}$  (note that the most distant point of a point inside $EHFG$ is a vertex of $EHFG$ so drawing such circle is logical), thus we cannot fit $3$ parallelograms into $2\times 2$ rectangle (all sides of the triangle has some region outside of $EHFG$ thus any rotating or movement cannot fit it.).
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Let it is also true for $2\times k$ , $k>1$. In order to prove it for $2\times (k+1)$ , it is enough to show that we cannot fit $3$ new "good parallelograms" that has some region in added new $2\times 1$ rectangle. Suppose we can fit, then at least $2$ of "good parallelograms" must also have some region in rectangle $2\times k$ but in that case we can not fit the third "good parallelogram" into new $2\times 1$ rectangle because if we can, then the last parallelogram must cover half of the longer sides of the $2\times 1$ rectangle, so it intersects one of the $2$ other parallelograms. (Note that these $2$ parallelograms cannot both in upper half or lower half of the $2\times 1$ rectangle) Also we can not fit the last parallelogram so that it has some region in rectangle $2\times k$ because in that case we would have $3$ "good parallelograms" which covers at least $1$ unit height inside $2$ unit height (here I suppose in a rectangle $2\times n$, $2$ is height and $n$ as width). 
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Thus, we can add "at most" $2$ "good parallelograms" which has some region in $2\times 1$ rectangle and it shows that the statement is also true for $2\times (k+1)$
A: In general the packing problem is a very difficult problem. If you want to give an inductive proof, you have to eliminate the following possibility, in which 3 parallelograms have some region in the new added 2×1 rectangle.

