2020 AMC 12B Problem 7

Question

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

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|

With slopes

m

and

6m,

set

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

and maximize

6m^2

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.

Problem 7 in other years