Balkan Olympiads 2000

How many $1 \times 10 \sqrt{2}$ rectangles can be placed in a $50 \times 90$ rectangle without overlapping, with the sides of the smaller rectangles parallel to those of the larger one?
The area of the large rectangle is between 318 and 319 times that of a small one. On the other hand, if we place 50 packages of 6 elongated small rectangles in the large one, a rectangle of size $50 \times 90-60 \sqrt{2}$ remains. This latter can accommodate $3 \times 5$ small rectangles, which totals 315, but it is difficult to see how to do better.

Let's rework the upper bound: in a standard orthonormal coordinate system, we place the large rectangle so that its vertices are at $(0,0),(90,0),(50,90)$, and $(0,50)$. For $k \in \mathbb{N}^{*}$, we consider the lines within the large rectangle with the equation $x+y=10 k \sqrt{2}$. Each small rectangle whose sides are parallel to the large one intersects some (one or two) of these lines over a total length of $\sqrt{2}$. The sum of the lengths of these lines is $570 \sqrt{2}-360$, which is strictly less than $316 \sqrt{2}$, allowing us to conclude.