2020 AMC 12B Problem 7

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?

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Solution

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,

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

giving

6m^2 - 5m + 1 = 0

6m^2 + 5m + 1 = 0.

The first yields

m = \tfrac12

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.