Question

直线 $y = x$ 与 $y = 3x$ 的图像在原点形成的锐角的角平分线方程为 $y = kx$ 。 $k$ 是多少？

The angle bisector of the acute angle formed at the origin by the graphs of the lines $y = x$ and $y = 3x$ has equation $y = kx.$ What is $k?$

$\dfrac{1 + \sqrt{5}}{2}$

$\dfrac{1 + \sqrt{7}}{2}$

$\dfrac{2 + \sqrt{3}}{2}$

$2$

$\dfrac{2 + \sqrt{5}}{2}$

答案：A

知识点：角平分线向量分母有理化

难度评级：1660

解答：

角平分线沿两条直线的单位方向向量之和： $\dfrac{(1,1)}{\sqrt2} + \dfrac{(1,3)}{\sqrt{10}}$ 。它的斜率为 $k = \frac{\tfrac{1}{\sqrt2} + \tfrac{3}{\sqrt{10}}}{\tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt{10}}} = \frac{\sqrt5 + 3}{\sqrt5 + 1}.$

分子分母同乘 $\sqrt5 - 1$ ，得 $\dfrac{(\sqrt5 + 3)(\sqrt5 - 1)}{4}$ $= \dfrac{2 + 2\sqrt5}{4}$ $= \dfrac{1 + \sqrt5}{2}$ 。

所以正确答案是 A。

The bisector points along the sum of the unit vectors of the two lines: $\dfrac{(1,1)}{\sqrt2} + \dfrac{(1,3)}{\sqrt{10}}.$ Its slope is $k = \frac{\tfrac{1}{\sqrt2} + \tfrac{3}{\sqrt{10}}}{\tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt{10}}} = \frac{\sqrt5 + 3}{\sqrt5 + 1}.$

Multiplying numerator and denominator by $\sqrt5 - 1$ gives $\dfrac{(\sqrt5 + 3)(\sqrt5 - 1)}{4}$ $= \dfrac{2 + 2\sqrt5}{4}$ $= \dfrac{1 + \sqrt5}{2}.$

Thus, the correct answer is A.

2021 AMC 12A Fall 第 13 题

答案：A

难度评级：1660

解答：

其他年份的第 13 题