Mathematical Olympiad in Ukraine

Since $0 \leq (1 - x^2) \leq 1$, $0 \leq (1 - y^2) \leq 1$, we get that
$$$$
$\sqrt{1-x^2} + \sqrt{1-y^2} \geq$ (1-x^2) + (1-y^2), $\text{}$ and so the equality holds if }
$$\begin{cases} \sqrt{1-x^2} = 1-x^2 \\ \sqrt{1-y^2} = 1-y^2 \end{cases}$$ $\Rightarrow$
$$\begin{cases} x^2 \in \{0;1\} \\ y^2 \in \{0;1\} \end{cases}$$.
$\text{}$ Thus we get the points from the answer.}