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

Revision as of 14:03, 8 June 2023

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 1 (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 to be $\mathbf{(C)} \frac{3}{2}$ .

~jeffisepic

Solution 2 (vector products)

Intersect at the origin and select a point on each line to define vectors $\mathbf{v}_{i}=(x_{i},y_{i})$ . Note that $\theta=45^{\circ}$ gives equal magnitudes of the vector products $\mathbf{v}_1\cdot\mathbf{v}_2 = v_{1}v_{2}\cos\theta \quad\mathrm{and}\quad |\mathbf{v}_1\times\mathbf{v}_2| = v_{1}v_{2}\sin\theta .$

Substituting coordinate expressions for vector products, we find $\mathbf{v}_1\cdot\mathbf{v}_2 = |\mathbf{v}_1\times\mathbf{v}_2| \ \implies\ x_{1}x_{2}+y_{1}y_{2} = x_{1}y_{2}-x_{2}y_{1} .$ Divide this equation by $x_{1}x_{2}$ to obtain $1+m_{1}m_{2} = m_{2}-m_{1} ,$ where $m_{i}=y_{i}/x_{i}$ is the slope of line $i$ . Taking $m_{2}=6m_{1}$ , we obtain $6m_{1}^{2}-5m_{1}+1 = 0 \ \implies\ m_{1} \in \{\frac{1}{3},\frac{1}{2}\} .$ The latter solution gives the largest product of slopes $m_{1}m_{2} = 6m_{1}^2 = \frac{3}{2} . \quad \boxed{\textbf{(C)}}$

~fairpark

Solution 3 (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 4 (tan)

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 5 (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 6 (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.

Solution 7

If you have this formula memorized then it will be very easy to do: If $\theta$ is the angle between two lines with slopes $m_1$ and $m_2$ , then $\tan(\theta)=\frac{m_1-m_2}{1+m_1m_2}$ .

Now let the smaller slope be $m$ , thus the other slope is $6m$ . Using our formula above: $\tan(45^\circ)=\frac{6m-m}{1+6m^2} \implies 6m^2-5m+1=0 \implies (2m-1)(3m-1)=0.$ Therefore the two possible values for $m$ are $\tfrac{1}{2}$ and $\tfrac{1}{3}$ . We choose the larger one and thus our answer is $6m^2=6 \cdot \frac{1}{4} = \frac{3}{2} \implies \boxed{\mathbf{(C)}}.$

~AngelaLZ

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/MLyMs8w1xEg

~Education, the Study of Everything

Video Solution

https://youtu.be/6ujfjGLzVoE

@@ Line 75: / Line 75: @@
 ~AngelaLZ
+==Video Solution (HOW TO THINK CREATIVELY!!!)==
+https://youtu.be/MLyMs8w1xEg
+~Education, the Study of Everything
 ==Video Solution==
-https://youtu.be/6ujfjGLzVoE (two solutions)
+https://youtu.be/6ujfjGLzVoE
-~IceMatrix
 ==See Also==

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