, 2008

Solution:
We will prove that desired covering is impossible.
Let us assume the opposite, i.e., a square with side length $a$ can be tiled with $k$ congruent right-angled triangles, whose sides are of lengths $b$, $b \sqrt{3}$, and $2b$.

Then the area of such a triangle is $\frac{b^{2} \sqrt{3}}{2}$.
And the area of the square is
$$S_{sq} = k b^{2} \frac{\sqrt{3}}{2}$$
Furthermore, the length of the side of the square, $a$, is obtained by the contribution of an integer number of lengths $b$, $2b$, and $b \sqrt{3}$, hence
$$a = m b \sqrt{3} + n b$$
where $m, n \in \mathbb{N} \cup \{0\}$, and at least one of the numbers $m$ and $n$ is different from zero. So the area of the square is
$$S_{sq} = a^{2} = (m b \sqrt{3} + n b)^{2} = b^{2} \left(3 m^{2} + n^{2} + 2 \sqrt{3} m n\right)$$
Now because of (1) and (2) it follows $3 m^{2} + n^{2} + 2 \sqrt{3} m n = k \frac{\sqrt{3}}{2}$, i.e.
$$6 m^{2} + 2 n^{2} = (k - 4 m n) \sqrt{3}$$
Because $3 m^{2} + n^{2} \neq 0$ and from the equality (3) it follows $4 m n \neq k$. Using once more (3), we get
$$\sqrt{3} = \frac{6 m^{2} + 2 n^{2}}{k - 4 m n}$$
which contradicts the fact that $\sqrt{3}$ is irrational, because $\frac{6 m^{2} + 2 n^{2}}{k - 4 m n}$ is a rational number.

Finally, we have obtained a contradiction, which proves that the desired covering is impossible.