Final Round

Answer: $2\sqrt{2}$.

Let $A(a;a^2)$, $B(b;b^2)$, $C(c;c^2)$. Without loss of generality we assume that $a<c$. Since $M$ is the midpoint of $AC$, we have $M\left(\frac{a+c}{2};\frac{a^2+c^2}{2}\right)$. Since $BM\parallel Oy$, we see that the abscissae of $M$ and $B$ are equal, i.e. $\frac{a+c}{2}=b$, so $BM=\frac{a^2+c^2}{2}-b^2$. Therefore $$\begin{array}{rl}2& =\frac{a^2+c^2}{2}-b^2=\frac{a^2+c^2}{2}-\frac{(a+c{)}^2}{4}\\ & =\frac{(a-c{)}^2}{4}.\end{array}$$ Hence, $(a-c{)}^2=8$, i.e. $c-a=2\sqrt{2}$.

$$S(ABC)=2S(ABM)=2\cdot 0.5AK\cdot BM,$$



where $AK$ is the altitude of the triangle $ABM$. Since $BM\parallel Oy$, we have $AK\parallel Ox$, so the length of the segment $AK$ is equal to the difference of the abscissae of $A$ and $C$, i.e. $c-a=2\sqrt{2}$.

Therefore, $$S(ABC)=2\cdot 0.5\cdot 2\sqrt{2}\cdot 2=2\sqrt{2}.$$