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2020 AMC 12B Problem 7

Problem 7 of 25EasierAlgebraProblem-Solving Techniques

Two nonhorizontal, non-vertical lines in the xyxy-coordinate plane intersect to form a 45∘45^\circ angle. One line has slope equal to 66 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 mm and 6m.6m. The angle between the lines satisfies ∣6m−m1+6m2∣=tan⁡45∘=1,\left|\frac{6m - m}{1 + 6m^2}\right| = \tan 45^\circ = 1, so 5m=±(1+6m2),5m = \pm(1 + 6m^2), giving 6m2−5m+1=06m^2 - 5m + 1 = 0 or 6m2+5m+1=0.6m^2 + 5m + 1 = 0. The first yields m=12m = \tfrac12 or m=13;m = \tfrac13; the second yields the negatives of these. The product of the slopes is 6m2,6m^2, which is largest when m=12,m = \tfrac12, giving 6⋅14=32.6 \cdot \tfrac14 = \tfrac32. Thus, the correct answer is C.
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