2020 AMC 12B 第 7 题

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

小提示：

斜率为

m_1, m_2

的两直线夹角的正切为

\left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|

。

The tangent of the angle between lines of slopes

m_1, m_2

is

\left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|

大提示：

设斜率为

m

和

6m

，令

\left|\dfrac{5m}{1 + 6m^2}\right| = 1

，再求

6m^2

的最大值。

With slopes

m

and

6m,

set

\left|\dfrac{5m}{1 + 6m^2}\right| = 1

and maximize

6m^2

解答：

设两条直线的斜率为

m

和

6m

。它们的夹角满足

\left|\frac{6m - m}{1 + 6m^2}\right| = \tan 45^\circ = 1\text{，}

所以

5m = \pm(1 + 6m^2)

，从而

6m^2 - 5m + 1 = 0

或

6m^2 + 5m + 1 = 0

。

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,