2014 AMC 12B Problems/Problem 11

Problem

Let $\mathcal P$ 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 $\mathcal P$ if and only if $r < m < s$ . What is $r + s$ ?

\[\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}}\ 52\qquad\textbf{(E)}\ 80\] (Error compiling LaTeX. Unknown error_msg)

Solution

The line through $Q$ only meets the parabola if the system of equations $y = x^2$ and $y = m(x - 20) + 14$ has one or more solution(s) (the second equation comes from point-slope form). We can equate $y$ and get the equation $x^2 = mx - 20m + 14$ . Now if we move all the terms to the left, we get $x^2 - mx - (20m - 14) = 0$ . We want this equation to have no solutions, and are trying to find the bounds of $r$ and $s$ for $m$ such that this is unsolvable. A quadratic is unsolvable across the reals if its discriminant is less than 0. So, we set the discriminant to be less than 0: $\sqrt{m^2 - 80m - 56} \le 0$ We can square each side, keeping track of extraneous solutions: the resulting equation is not just $m^2 - 80m - 14 = 0$ , but $\pm(m^2 - 80m - 66) = 0$ . The solutions to this equation are $r$ and $s$ : to get their sum we use Vieta's Formulas and get $r + s = \pm80$ . Because $-80$ is not a valid answer choice, we can know for certain that the correct answer is $\boxed{(\textbf{E})\:80}$ .

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2014 AMC 12B Problems/Problem 11

Problem

Solution