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Difference between revisions of "2014 AMC 12B Problems/Problem 11"

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

Revision as of 19:18, 20 February 2014

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 $\boxed{(\text{E}) 80}$.

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