2. For the lengths $a, b$ of the legs of a right triangle, it holds that

$$
\log \frac{a-b}{2}=\frac{1}{2}(\log a+\log b-\log 2)
$$

Determine the acute angles of this triangle.
Solution.

The equality

$$
2 \log \frac{a-b}{2}=\log a+\log b-\log 2
$$

can be written in the form

$$
\begin{gathered}
\log \left(\frac{a-b}{2}\right)^{2}=\log \frac{a b}{2}, \quad \text{i.e.} \\
\left(\frac{a-b}{2}\right)^{2}=\frac{a b}{2}
\end{gathered}
$$

from which, by multiplication, we get

$$
a^{2}+b^{2}=4 a b
$$

Now, recall that we are dealing with a right triangle, so $a^{2}+b^{2}=c^{2}$, from which we obtain $c^{2}=4 a b$. Furthermore, since $\sin \alpha=\frac{a}{c}, \cos \alpha=\frac{b}{c}$, we have

$$
4 a b=4 \cdot c \sin \alpha \cdot c \cos \alpha=4 c^{2} \sin \alpha \cdot \cos \alpha
$$

so we can write the last equality in the form

$$
\begin{aligned}
& 4 \sin \alpha \cos \alpha=1 \\
& \text { i.e. } \quad \sin 2 \alpha=\frac{1}{2}
\end{aligned}
$$

This implies $2 \alpha=30^{\circ}$ or $2 \alpha=150^{\circ}$, i.e., $\alpha=15^{\circ}$ or $\alpha=75^{\circ}$.

Since $\frac{a-b}{2}>0$ (so that the logarithm at the beginning of the problem is defined), we conclude $a>b$, i.e., $\alpha>\beta=90^{\circ}-\alpha$, so the only solution that fits is

$$
\alpha=75^{\circ}, \beta=15^{\circ}
$$

(5 points)