# Difference between revisions of "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

Solution by e_power_pi_times_i $4n = m$, as stated in the question. In the line $L_1$, draw a triangle with the coordinates $(0,0)$, $(1,0)$, and $(1,m)$. Then $m = \tan(\theta_1)$. Similarly, $n = \tan(\theta_2)$. Since $4n = m$ and $\theta_1 = 2\theta_2$, $\tan(2\theta_2) = 4\tan(\theta_2)$. Using the angle addition formula for tangents, $\dfrac{2\tan(\theta_2)}{1-\tan^2(\theta_2)} = 4\tan(\theta_2)$. Solving, we have $\tan(\theta_2) = 0, \dfrac{\sqrt{2}}{2}$. But line $L_1$ is not horizontal, so therefore $(m,n) = (2\sqrt{2},\dfrac{\sqrt{2}}{2})$. Looking at the answer choices, it seems the answer is $(2\sqrt{2})(\dfrac{\sqrt{2}}{2}) = \boxed{\text{(C)} \ 2}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 