Question

$xy$ -坐标平面中两条既不水平也不竖直的直线相交，形成一个 $45^\circ$ 角。其中一条直线的斜率等于另一条直线斜率的 $6$ 倍。这两条直线斜率乘积的最大可能值是多少？

Two nonhorizontal, non-vertical lines in the $xy$ -coordinate plane intersect to form a $45^\circ$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?

$\dfrac16$

$\dfrac23$

$\dfrac32$

$3$

$6$

答案：C

知识点：斜率二次方程最优化

难度评级：1410

解答：

设两条直线的斜率为 $m$ 和 $6m$ 它们的夹角满足所以 $5m = \pm(1 + 6m^2)$ 得 $6m^2 - 5m + 1 = 0$ 或 $6m^2 + 5m + 1 = 0$ 。 $\left|\frac{6m - m}{1 + 6m^2}\right| = \tan 45^\circ = 1,$

第一个方程给出 $m = \tfrac12$ 或 $m = \tfrac13$ 第二个方程给出它们的相反数。斜率乘积为 $6m^2$ 并在 $m = \tfrac12$ 时达到最大值 $6 \cdot \tfrac14 = \tfrac32$ 。

所以正确答案是 C。

Let the slopes be $m$ and $6m.$ The angle between the lines satisfies $\left|\frac{6m - m}{1 + 6m^2}\right| = \tan 45^\circ = 1,$ so $5m = \pm(1 + 6m^2),$ giving $6m^2 - 5m + 1 = 0$ or $6m^2 + 5m + 1 = 0.$

The first yields $m = \tfrac12$ or $m = \tfrac13;$ the second yields the negatives of these. The product of the slopes is $6m^2,$ which is largest when $m = \tfrac12,$ giving $6 \cdot \tfrac14 = \tfrac32.$

Thus, the correct answer is C.

2020 AMC 12B 第 7 题

答案：C

难度评级：1410

解答：

其他年份的第 7 题