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Difference between revisions of "1980 AHSME Problems/Problem 12"

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== Solution ==
 
== Solution ==
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<math>4n = m</math>, as stated in the question. In the line <math>L_1</math>, draw a triangle with the coordinates <math>(0,0)</math>, <math>(1,0)</math>
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Solution by e_power_pi_times_i
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<math>4n = m</math>, as stated in the question. In the line <math>L_1</math>, draw a triangle with the coordinates <math>(0,0)</math>, <math>(1,0)</math>, and <math>(1,m)</math>. Then <math>m = \tan(\theta_1)</math>. Similarly, <math>n = \tan(\theta_2)</math>. Since <math>4n = m</math> and <math>\theta_1 = 2\theta_2</math>, <math>\tan(2\theta_2) = 4\tan(\theta_2)</math>. Using the angle addition formula for tangents, <math>\dfrac{2\tan(\theta_2)}{1-\tan^2(\theta_2)} = 4\tan(\theta_2)</math>. Solving, we have <math>\tan(\theta_2) = 0, \dfrac{\sqrt{2}}{2}</math>. But line <math>L_1</math> is not horizontal, so therefore <math>(m,n) = (2\sqrt{2},\dfrac{\sqrt{2}}{2})</math>. Looking at the answer choices, it seems the answer is <math>(2\sqrt{2})(\dfrac{\sqrt{2}}{2}) = \boxed{\text{(C)} \ 2}</math>.
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1980|num-b=11|num-a=12}}   
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{{AHSME box|year=1980|num-b=11|num-a=13}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:02, 20 March 2018

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 $(2\sqrt{2})(\dfrac{\sqrt{2}}{2}) = \boxed{\text{(C)} \ 2}$.

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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