Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 4.8 AIME Prove it Ukraine

A line $l_1$ intersects the parabola $y = ax^2 + bx + c$ ( $a \neq 0$ ) at the two points $A$ and $B$ . A line $l_2$ is parallel to the line $l_1$ and tangent to this parabola at the point $C$ . Prove that the arithmetic mean of abscissas of points $A$ and $B$ equals the abscissa of point $C$ .

Solution

Let the equation of the line $l_1$ be $y = kx + d$ . Then abscissas of points $A$ and $B$ are defined using the equality $ax^2 + bx + c = kx + d$ . Therefore these abscissas $x_1, x_2$ satisfy that quadratic equation, then according to Vieta's formula we have equality $x_1 + x_2 = \frac{-b + k}{a}$ .

The abscissa of point $C$ is defined using the same equality

$ax^2 + bx + c = kx + d$

on condition that the line $l_2$ is tangent to this parabola. Then the abscissa of point $C$ is $\frac{k-b}{2a} = \frac{x_1 + x_2}{2}$ , which completes the proof.

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