Track / Stage 5 / 185 of 400 #1265 of 2444

Problem 1265

AIME late

Algebra Difficulty 5.3 Prove it Mathematical Olympiad in Ukraine, Fourth Round (March 23, ) · Ukraine · 2010

Parabola $y = ax^2 + bx + c$ passes through the points $A(-2, 1)$ and $B(2, 9)$ , and does not intersect $x$ -axis. Find all possible values of $x$ coordinate of the vertex of the parabola.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

We first write analytically the conditions that our parabola passes through the given points:
$\begin{cases} 4a - 2b + c = 1, \\ 4a + 2b + c = 9, \end{cases}$
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 $x_v = -\frac{b}{2a} = -\frac{1}{a} \Rightarrow x_v \in (-4, -1)$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.