## Subject II. (20 points)

Let $\mathrm{x}, \mathrm{y} \in \mathbb{R}$ that satisfy the relation: $x^{2}-2 \sqrt{3} x+y^{2}+4 y+3=0$. Determine $\mathrm{x}, \mathrm{y} \in \mathbb{R}$ for which $[\mathrm{x}]=[\mathrm{y}]$, where $[\mathrm{a}]$ represents the integer part of the real number a.

Prof. Violin Gorcea, Avram Iancu Theoretical High School, Cluj-Napoca
## Subject II.

The relationship from the statement becomes: $x^{2}-2 \sqrt{3} x+3+y^{2}+4 y+4=4 \Leftrightarrow(x-\sqrt{3})^{2}+(y+2)^{2}=4$.

Then $(x-\sqrt{3})^{2} \leq 4$ and $(y+2)^{2} \leq 4$.

$(x-\sqrt{3})^{2} \leq 4 \Leftrightarrow|x-\sqrt{3}| \leq 2 \Leftrightarrow x \in[-2+\sqrt{3}, 2+\sqrt{3}]=S_{1}$.

If $(y+2)^{2} \leq 4$ we get $-2 \leq y+2 \leq 2 \Leftrightarrow y \in[-4,0]=S_{2}$.

From $S_{1} \cap S_{2}=[-2+\sqrt{3}, 0]$ we can have $[\mathrm{x}]=[\mathrm{y}]=-1$ or $[\mathrm{x}]=[\mathrm{y}]=0$.

If $[\mathrm{x}]=[\mathrm{y}]=-1$ we get $x \in[-2+\sqrt{3}, 0)$ and $y \in[-1,0)$.

If $[\mathrm{x}]=[\mathrm{y}]=0$ then $x \in[0,1)$ and $y=0$.