2014 AMC 12B Problem 17

Attempt Problem 17 of the 2014 AMC 12B below, then check your answer against the professionally curated solution 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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17.

Let PP be the parabola with equation y=x2y = x^2 and let Q=(20,14).Q = (20, 14). There are real numbers rr and ss such that the line through QQ with slope mm does not intersect PP if and only if r<m<s.r \lt m \lt s. What is r+s?r + s?

11

2626

4040

5252

8080

Answer: E
Concepts:parabolaquadraticVieta’s Formulas
Difficulty rating: 2010
Solution:

The line through QQ is y=m(x20)+14.y = m(x-20) + 14. Substituting into y=x2y = x^2 gives x2mx+(20m14)=0. x^2 - mx + (20m - 14) = 0.

There is no intersection exactly when this has no real root, i.e. when the discriminant m24(20m14)m^2 - 4(20m-14) =m280m+56= m^2 - 80m + 56 is negative. That happens between the two roots rr and ss of m280m+56=0.m^2 - 80m + 56 = 0.

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

Thus, the correct answer is E.

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