AMC12 2014 B

AMC12 2014 B · Q17

AMC12 2014 B · Q17. It mainly tests Quadratic equations, Coordinate geometry.

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 $r < m < s$. What is $r + s$?

设 $P$ 是方程 $y = x^2$ 的抛物线，$Q = (20, 14)$。存在实数 $r$ 和 $s$，使得通过 $Q$ 且斜率为 $m$ 的直线不与 $P$ 相交当且仅当 $r < m < s$。求 $r + s$。

(A) 1 1

(B) 26 26

(D) 52 52

(E) 80 80

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): The line passing through point $Q=(20,14)$ with slope $m$ has equation $y-14=m(x-20)$. The requested values for $m$ are those for which the system \[ \left\{ \begin{array}{l} y-14=m(x-20)\\ y=x^2 \end{array} \right. \] has no solutions. Solving for $y$ in the first equation and substituting into the second yields $m(x-20)+14=x^2$, which reduces to $x^2-mx+(20m-14)=0$. This equation has no solution for $x$ when the discriminant is negative, that is, when $m^2-4\cdot(20m-14)=m^2-80m+56<0$. This quadratic in $m$ is negative between its two roots $40\pm\sqrt{40^2-56}$, which are the required values of $r$ and $s$. The requested sum is $r+s=2\cdot40=80$.

答案（E）：过点 $Q=(20,14)$ 且斜率为 $m$ 的直线方程为 $y-14=m(x-20)$。所求的 $m$ 的取值是使得方程组 \[ \left\{ \begin{array}{l} y-14=m(x-20)\\ y=x^2 \end{array} \right. \] 无解的那些 $m$。由第一式解出 $y$ 并代入第二式，得 $m(x-20)+14=x^2$，化简为 $x^2-mx+(20m-14)=0$。当判别式为负时，该方程对 $x$ 无解，即当 $m^2-4\cdot(20m-14)=m^2-80m+56<0$。这个关于 $m$ 的二次式在其两根 $40\pm\sqrt{40^2-56}$ 之间为负，这两根就是所求的 $r$ 和 $s$。因此所求和为 $r+s=2\cdot40=80$。

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