2014 AMC 12B Problem 17

Below is the professionally curated solution for Problem 17 of the 2014 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 12B solutions, or check the answer key.

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Concepts:parabola quadratic Vieta’s Formulas

Difficulty rating: 2010

17.

Let $P$ be the parabola with equation $y = x^2$ and let $Q = (20, 14).$ There are real numbers $r$ and $s$ such that the line through $Q$ with slope $m$ does not intersect $P$ if and only if $r \lt m \lt s.$ What is $r + s?$

$1$

$26$

$40$

$52$

$80$

Solution:

The line through $Q$ is $y = m(x-20) + 14.$ Substituting into $y = x^2$ gives $x^2 - mx + (20m - 14) = 0.$

There is no intersection exactly when this has no real root, i.e. when the discriminant $m^2 - 4(20m-14) = m^2 - 80m + 56$ is negative. That happens between the two roots $r$ and $s$ of $m^2 - 80m + 56 = 0.$

By Vieta's formulas, $r + s = 80.$

Thus, the correct answer is E.

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