1980 AHSME Problems/Problem 12

Problem

The equations of $L_1$ and $L_2$ are $y=mx$ and $y=nx$ , respectively. Suppose $L_1$ makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis) as does $L_2$ , and that $L_1$ has $4$ times the slope of $L_2$ . If $L_1$ is not horizontal, then $mn$ is

$\text{(A)} \ \frac{\sqrt{2}}{2} \qquad \text{(B)} \ -\frac{\sqrt{2}}{2} \qquad \text{(C)} \ 2 \qquad \text{(D)} \ -2 \qquad \text{(E)} \ \text{not uniquely determined}$

Solution

Since $L_1$ has $4$ times the slope of $L_2$ , $m = 4n$ . Let $O = (0,0)$ , $A = (1,0)$ , $B = (1,n)$ , $C = (1,m)$ , and $\theta = \angle AOB = \angle BOC$ . Then $m = \tan(2\theta)$ and $n = \tan(\theta)$ .

Since $m = 4n$ , $\tan(2\theta) = 4\tan(\theta)$ . Using the tangent double-angle formula, $\dfrac{2\tan(\theta)}{1-\tan^2(\theta)} = 4\tan(\theta)$ . Cross-multiplying and collecting terms on one side yields $4\tan^3 \theta - 2\tan\theta = 0$ , which factors as $2\tan \theta(2\tan^2\theta - 1) = 0$ . Substituting $\tan\theta = n$ yields $2n(2n^2-1)=0$ .

Since line $L_1$ is not horizontal, $n \neq 0$ . So $2n^2 - 1 = 0$ , and thus $n^2 = \frac{1}{2}$ . Therefore, $mn = 4n^2 = 4 \cdot \frac{1}{2} = \boxed{(\textbf{C}) \ 2}$ .

-e_power_pi_times_i, edited by j314andrews

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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1980 AHSME Problems/Problem 12

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