AMC12 2020 B

AMC12 2020 B · Q7

AMC12 2020 B · Q7. It mainly tests Quadratic equations, Trigonometry (basic).

Two nonhorizontal, nonvertical lines in the xy-coordinate plane intersect to form a 45° 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?

在 xy 坐标平面上有两条非水平、非垂直的直线相交形成 45° 角。其中一条直线的斜率是另一条直线的斜率的 6 倍。两条直线斜率的乘积的最大可能值为多少？

(A) \frac{1}{6} \frac{1}{6}

(B) \frac{2}{3} \frac{2}{3}

(D) 3 3

(E) 6 6

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let $\theta_1$ and $\theta_2$ be the angles that the two lines make with the rightward-pointing $x$-axis, and let $m$ and $6m$ be their slopes. Without loss of generality, $\tan \theta_1 = m$ and $\tan \theta_2 = 6m$. It follows from the identity for the tangent of the difference of two angles that $$1 = \tan 45^\circ = \tan(\theta_1 - \theta_2) = \frac{m - 6m}{1 + 6m^2}$$ or $$1 = \tan 45^\circ = \tan(\theta_2 - \theta_1) = \frac{6m - m}{1 + 6m^2}.$$ Therefore $6m^2 \pm 5m + 1 = 0$, and $m = \pm \frac{1}{2}$ or $\pm \frac{1}{3}$. Then $6m = \pm 3$ or $\pm 2$, respectively. The product of the slopes is either $(\pm \frac{1}{2})( \pm 3) = \frac{3}{2}$ or $(\pm \frac{1}{3})( \pm 2) = \frac{2}{3}$, so the greatest possible value of the product of the slopes is $\frac{3}{2}$. For example, the lines could have equations $y = \frac{1}{2}x$ and $y = 3x$.

设 $\theta_1$ 和 $\theta_2$ 是两条直线与向右的 x 轴的夹角，斜分别为 $m$ 和 $6m$。不失一般性，$\tan \theta_1 = m$，$\tan \theta_2 = 6m$。由两角差的正切公式得 $$1 = \tan 45^\circ = \tan(\theta_1 - \theta_2) = \frac{m - 6m}{1 + 6m^2}$$ 或 $$1 = \tan 45^\circ = \tan(\theta_2 - \theta_1) = \frac{6m - m}{1 + 6m^2}$$ 因此 $6m^2 \pm 5m + 1 = 0$，$m = \pm \frac{1}{2}$ 或 $\pm \frac{1}{3}$。则 $6m = \pm 3$ 或 $\pm 2$。斜率乘积为 $(\pm \frac{1}{2})( \pm 3) = \frac{3}{2}$ 或 $(\pm \frac{1}{3})( \pm 2) = \frac{2}{3}$，故最大值为 $\frac{3}{2}$。例如，直线方程可为 $y = \frac{1}{2}x$ 和 $y = 3x$。

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