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Difference between revisions of "2020 AMC 12B Problems/Problem 7"

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<math>\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\  \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6</math>
 
<math>\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\  \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6</math>
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==Solution==
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Let one of the lines have equation <math>y=ax</math>. Let <math>\theta</math> be the angle that line makes with the x-axis, so <math>\tan(\theta)=a</math>. The other line will have a slope of <math>\tan(45^{\circ}+\theta)=\frac{\tan(45^{\circ})+\tan(\theta)}{1-\tan(45^{\circ})\tan(\theta)} = \frac{1+a}{1-a}</math>. Since the slope of one line is <math>6</math> times the other, and <math>a</math> is the smaller slope, <math>6a = \frac{1+a}{1-a} \implies 6a-6a^2=1+a \implies 6a^2-5a+1=0 \implies a=\frac{1}{2},\frac{1}{3}</math>. If <math>a = \frac{1}{2}</math>, the other line will have slope <math>\frac{1+\frac{1}{2}}{1-\frac{1}{2}} = 3</math>. If <math>a = \frac{1}{3}</math>, the other line will have slope <math>\frac{1+\frac{1}{3}}{1-\frac{1}{3}} = 2</math>. The first case gives the bigger product of <math>\frac{3}{2}</math>, so our answer is <math>\boxed{\textbf{(C)}\  \frac32}</math>.
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~JHawk0224
  
 
==See Also==
 
==See Also==

Revision as of 22:20, 7 February 2020

Problem

Two nonhorizontal, non vertical lines in the $xy$-coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?

$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\ \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$

Solution

Let one of the lines have equation $y=ax$. Let $\theta$ be the angle that line makes with the x-axis, so $\tan(\theta)=a$. The other line will have a slope of $\tan(45^{\circ}+\theta)=\frac{\tan(45^{\circ})+\tan(\theta)}{1-\tan(45^{\circ})\tan(\theta)} = \frac{1+a}{1-a}$. Since the slope of one line is $6$ times the other, and $a$ is the smaller slope, $6a = \frac{1+a}{1-a} \implies 6a-6a^2=1+a \implies 6a^2-5a+1=0 \implies a=\frac{1}{2},\frac{1}{3}$. If $a = \frac{1}{2}$, the other line will have slope $\frac{1+\frac{1}{2}}{1-\frac{1}{2}} = 3$. If $a = \frac{1}{3}$, the other line will have slope $\frac{1+\frac{1}{3}}{1-\frac{1}{3}} = 2$. The first case gives the bigger product of $\frac{3}{2}$, so our answer is $\boxed{\textbf{(C)}\ \frac32}$.

~JHawk0224

See Also

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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
All AMC 12 Problems and Solutions

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