2021 AMC 12A Spring Problem 20

Below is the professionally curated solution for Problem 20 of the 2021 AMC 12A Spring, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Spring solutions, or check the answer key.

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Concepts:parabola distance formula Vieta’s Formulas

Difficulty rating: 2300

20.

Suppose that on a parabola with vertex $V$ and a focus $F$ there exists a point $A$ such that $AF = 20$ and $AV = 21.$ What is the sum of all possible values of the length $FV?$

$13$

$\dfrac{40}{3}$

$\dfrac{41}{3}$

$14$

$\dfrac{43}{3}$

Solution:

Let $V = (0, 0),$ focus $F = (0, f),$ and directrix $y = -f,$ where $f = FV.$ A point $A = (x, y)$ on the parabola satisfies $x^2 = 4fy$ and $AF = y + f = 20,$ so $y = 20 - f.$ Also $AV^2 = x^2 + y^2 = 4fy + y^2 = 441.$

Substituting $y = 20 - f:$ $4f(20 - f) + (20 - f)^2 = 441 \;\Longrightarrow\; 3f^2 - 40f + 41 = 0.$ By Vieta's formulas, the sum of the two possible values of $f$ is $\dfrac{40}{3}.$

Thus, the correct answer is B.

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