2021 AMC 12A Fall 第 13 题

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}

角平分线方向是两条直线方向单位向量之和：

\dfrac{(1,1)}{\sqrt2} + \dfrac{(1,3)}{\sqrt{10}}

A bisector direction is the sum of unit vectors along the two lines:

\dfrac{(1,1)}{\sqrt2} + \dfrac{(1,3)}{\sqrt{10}}

k

是（第二个分量）/（第一个分量）；再有理化结果

k

is the ratio (second component)/(first component); rationalize the result

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}\text{。}

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.

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