, 2010

Let the L-shape be placed so that the two longer sides meet at the top left corner. Starting from the top left we place on it $k$ unit L-shapes diagonally with the same orientation as the large L-shape (Fig. 12). The rest consists of two equal staircase-like parts; it is enough to show that one of them, e.g. the lower part, can be covered. The staircase has $k$ stairs, the lowest one at the height $k-1$ and the highest at the height $2k-2$.

In case $k=1$ the staircase is empty, in case $k=2$ it can be covered with one unit L-shape. Assume that the claim holds for the staircase with $k$ stairs and consider the staircase with $k+2$ stairs. Separate a strip of width 2 from the left and bottom. The rest can be covered by the induction assumption. The topmost part of the strip is covered with one unit L-shape. Now we have

Fig. 12

Fig. 13

Fig. 14

Fig. 15

to cover the rest of the strip whose lower and left sides have correspondingly the lengths $k+2$ and $2k$.

* If $k$ is divisible by 3, then cut the figure into two strips of sizes $2 \times 2k$ and $k \times 2$ and cover both of them with $2 \times 3$ rectangles consisting of two unit L-shapes (Fig. 13) and we are done.
* If $k \equiv 1 \pmod{3}$, then cut the figure into two strips of sizes $2 \times (2k-2)$ and $(k+2) \times 2$ and cover both of them with $2 \times 3$ rectangles (Fig. 14). This is possible because $2k-2$ and $k+2$ are divisible by 3.
* If $k \equiv 2 \pmod{3}$, then cut the figure into two strips of sizes $2 \times (2k-4)$ and $(k-2) \times 2$, and a corner part, which is a L-shape with $k=2$. Both strips can be covered by $2 \times 3$ rectangles since $2k-4$ and $k-2$ are divisible by 3; the corner part can be covered by induction basis (Fig. 15).