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

Revision as of 17:24, 9 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

Solution 2 (bash)

Place on coordinate plane. Lines are $y=mx, y=6mx.$ The intersection point at the origin. Goes through $(0,0),(1,m),(1,6m),(1,0).$ So by law of sines, $\frac{5m}{\sin{45^{\circ}}} = \frac{\sqrt{1+m^2}}{1/(\sqrt{1+36m^2})},$ lettin $a=m^2$ we want $6a.$ Simplifying gives $50a = (1+a)(1+36a),$ so $36a^2-13a+1=0 \implies 36(a-1/4)(a-1/9)=0,$ so max $a=1/4,$ and $6a=3/2 \quad \boxed{(C)}.$

Law of sines on the green triangle with the red angle (45 deg) and blue angle, where sine blue angle is $1/(\sqrt{1+36m^2})$ from right triangle w vertices $(0,0),(1,0),(1,6m).$

~ccx09

Solution 3 (complex)

Let the intersection point is the origin. Let $(a,b)$ be a point on the line of lesser slope. The mutliplication of $a+bi$ by cis 45. $(a+bi)(\frac{1}{\sqrt 2 }+i*\frac{1}{\sqrt 2 })=\frac{1}{\sqrt 2 }((a-b)+(a+b)*i)$ and therefore $(a-b, a+b)$ lies on the line of greater slope.

Then, the rotation of $(a,b)$ by 45 degrees gives a line of slope $\frac{a+b}{a-b}$ .

We get the equation $\frac{6b}{a}=\frac{a+b}{a-b}\implies a^2-5ab+6b^2=(a-3b)(a-2b)=0$ and this gives our answer.

~jeffisepic

Solution 4 (matrix transformation)

Multiply by the rotation transformation matrix $\begin{bmatrix} \cos \theta & - \sin \theta\\ \sin \theta & \cos \theta \end{bmatrix}$ where $\theta = 45^{\circ}.$

Solution 5 (Cheating)

Let the smaller slope be $m$ , then the larger slope is $6m$ . Since we want the greatest product we begin checking each answer choice, starting with (E).

$6m^2=6$ .

$m^2=1$ .

This gives $m=1$ and $6m=6$ . Checking with a protractor we see that this does not form a 45 degree angle.

Using this same method for the other answer choices, we eventually find that the answer is $\boxed{\textbf{(C)}\ \frac32}$ since our slopes are $\frac12$ and $3$ which forms a perfect 45 degree angle.

Video Solution

Two solutions https://youtu.be/6ujfjGLzVoE

~IceMatrix

@@ Line 30: / Line 30: @@
 ~jeffisepic
-==Solution 4==
+==Solution 4 (matrix transformation)==
 Multiply by the rotation transformation matrix
 <math>

2020 AMC 12B (Problems • Answer Key • Resources)
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