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Problem 580

AMC 10/12, early questions

Algebra Difficulty 3.7 Find the answer CEMC Fermat · Canada · 2026

Points $P$ and $Q$ are on the parabola with equation $y = -3x^2 + 4x + 27$ . The midpoint of
$PQ$ is $(0,0)$ . If $P$ lies above the $x$ -axis, what is the $y$ -coordinate of $P$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Suppose the midpoint of $P(a,b)$ and $Q(c,d)$ is $(0,0)$ . Then $\dfrac{a+c}{2}=0$ and $\dfrac{b+d}{2}=0$ , which implies that
$c=-a$ and $d=-b$ .

Thus, the points $P(a,b)$ and $Q(-a,-b)$ are both on the given parabola.
Substituting $(a,b)$ and $(-a,-b)$ into the equation for the
parabola gives the following two equations $\begin{align*} b &= -3a^2+4a+27 \\ -b &= -3a^2-4a+27\end{align*}$ Subtracting these two
equations gives $2b = 8a$ or $b=4a$ .

Substituting $b=4a$ into the first
equation gives $4a = -3a^2+4a+27$ ,
which can be simplified to $3a^2=27$
or $a^2=9$ . Therefore, $a=\pm 3$ , so $b=\pm 12$ . Since $P(a,b)$ is above the $x$ -axis, its $y$ -coordinate, $b$ , must be positive, so $b=12$ .

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.