50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010)

We first write analytically the conditions that our parabola passes through the given points: $$\{\begin{array}{l}4a-2b+c=1,\\ 4a+2b+c=9,\end{array}$$ We can now find $b=2$ and $4a+c=5$. From the condition, that our parabola does not have real zeros we get $D=b^2-4ac=4-4a(5-4a)<0$, thus, the following inequalities hold: $\frac{1}{4}<a<1$. The only thing that is left now is to solve the inequality for $x$ coordinate of the vertex of the parabola. Due to the fact that $x_v=-\frac{b}{2a}=-\frac{1}{a}\Rightarrow x_v\in (-4,-1)$.