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

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== Problem ==
 
== Problem ==
  
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The equations of <math>L_1</math> and <math>L_2</math> are <math>y=mx</math> and <math>y=nx</math>, respectively. Suppose <math>L_1</math> makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis ) as does <math>L_2</math>, and that <math>L_1</math> has 4 times the slope of <math>L_2</math>. If <math>L_1</math> is not horizontal, then <math>mn</math> is
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The equations of <math>L_1</math> and <math>L_2</math> are <math>y=mx</math> and <math>y=nx</math>, respectively. Suppose <math>L_1</math> makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis) as does <math>L_2</math>, and that <math>L_1</math> has <math>4</math> times the slope of <math>L_2</math>. If <math>L_1</math> is not horizontal, then <math>mn</math> is
  
 
<math>\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}</math>
 
<math>\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}</math>
  
  
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== Solution ==
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== Solution 1 (Trigonometry) ==
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Solution by e_power_pi_times_i
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Since <math>L_1</math> has <math>4</math> times the slope of <math>L_2</math>, <math>m = 4n</math>.  Let <math>O = (0,0)</math>, <math>A = (1,0)</math>, <math>B = (1,n)</math>, <math>C = (1,m)</math>, and <math>\theta = \angle AOB = \angle BOC</math>.  Then <math>m = \tan 2\theta</math> and <math>n = \tan \theta</math>. 
  
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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>.
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Since <math>m = 4n</math>, <math>\tan 2\theta = 4\tan \theta</math>. By the tangent double-angle formula, <math>\dfrac{2\tan \theta}{1-\tan^2 \theta} = 4\tan \theta</math>. Cross-multiplying and collecting terms on one side yields <math>4\tan^3 \theta - 2\tan\theta = 0</math>, which factors as <math>2\tan \theta(2\tan^2\theta - 1) = 0</math>. Substituting <math>\tan\theta = n</math> yields <math>2n(2n^2-1)=0</math>.
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Since line <math>L_1</math> is not horizontal, <math>n \neq 0</math>.  So <math>2n^2 - 1 = 0</math>, and thus <math>n^2 = \frac{1}{2}</math>. Therefore, <math>mn = 4n^2 = 4 \cdot \frac{1}{2} = \boxed{(\textbf{C}) \ 2}</math>.
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-e_power_pi_times_i, edited by j314andrews
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== Solution 2 (Angle Bisector Theorem) ==
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Since <math>L_1</math> has <math>4</math> times the slope of <math>L_2</math>, <math>m = 4n</math>.  Let <math>O = (0,0)</math>, <math>A = (1,0)</math>, <math>B = (1,n)</math>, <math>C = (1,m) = (1, 4n)</math>. By the Angle Bisector Theorem, <math>\frac{OC}{OA} = \frac{BC}{AB}</math>. That is, <math>OC = \frac{3n}{n} = 3</math>.
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By the Pythagorean Theorem, <math>AC = \sqrt{3^2-1^2} = 2\sqrt{2}</math>. So <math>m = 2\sqrt{2}</math> and <math>n = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}</math>. Therefore, <math>mn = \left(2\sqrt{2}\right)\left(\frac{\sqrt{2}}{2}\right) = \boxed{(\textbf{C}) \ 2}</math>.
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-j314andrews
  
 
== See also ==
 
== See also ==

Latest revision as of 15:05, 14 August 2025

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 1 (Trigonometry)

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 $m = 4n$, $\tan 2\theta = 4\tan \theta$. By 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

Solution 2 (Angle Bisector Theorem)

Since $L_1$ has $4$ times the slope of $L_2$, $m = 4n$. Let $O = (0,0)$, $A = (1,0)$, $B = (1,n)$, $C = (1,m) = (1, 4n)$. By the Angle Bisector Theorem, $\frac{OC}{OA} = \frac{BC}{AB}$. That is, $OC = \frac{3n}{n} = 3$.

By the Pythagorean Theorem, $AC = \sqrt{3^2-1^2} = 2\sqrt{2}$. So $m = 2\sqrt{2}$ and $n = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}$. Therefore, $mn = \left(2\sqrt{2}\right)\left(\frac{\sqrt{2}}{2}\right) = \boxed{(\textbf{C}) \ 2}$.

-j314andrews

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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