## Task A-3.2.

Let $a, b$, and $c$ be positive real numbers such that

$$
a^{2}+b^{2}-c^{2}=\sqrt{3} a b, \quad a^{2}-b^{2}+c^{2}=\sqrt{2} a c
$$

Determine the ratio $b: c$.
## First solution.

By adding the equations, we get

$$
2 a^{2}=\sqrt{3} a b+\sqrt{2} a c
$$

and since $a \neq 0$, dividing by $2 a$ gives us

$$
a=\frac{\sqrt{3} b+\sqrt{2} c}{2}
$$

Now substitute the obtained expression into the first equation. We have

$$
\frac{3 b^{2}+2 \sqrt{6} b c+2 c^{2}}{4}+b^{2}-c^{2}=\frac{3 b^{2}+\sqrt{6} b c}{2}
$$

from which, after multiplying by 4, we get

$$
7 b^{2}+2 \sqrt{6} b c-2 c^{2}=6 b^{2}+2 \sqrt{6} b c
$$

which implies $b^{2}=2 c^{2}$.

Since $b$ and $c$ are positive, taking the square root gives $b=\sqrt{2} c$, so $b: c=\sqrt{2}: 1$.