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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==
+
==Solution 1 (complex)==
−
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>.
+
Let the intersection point be the origin. Let <math>(a,b)</math> be a point on the line of lesser slope. The mutliplication of <math>a+bi</math> by cis 45.
 +
<math>(a+bi)(\frac{1}{\sqrt 2 }+i*\frac{1}{\sqrt 2 })=\frac{1}{\sqrt 2 }((a-b)+(a+b)*i)</math> and therefore <math>(a-b, a+b)</math> lies on the line of greater slope.
 +
 
 +
Then, the rotation of <math>(a,b)</math> by 45 degrees gives a line of slope <math>\frac{a+b}{a-b}</math>.
 +
 
 +
We get the equation <math>\frac{6b}{a}=\frac{a+b}{a-b}\implies a^2-5ab+6b^2=(a-3b)(a-2b)=0</math> and this gives our answer to be <math>\mathbf{(C)} \frac{3}{2}</math>.
 +
 
 +
~jeffisepic
 +
 
 +
==Solution 2 (vector products)==
 +
Intersect at the origin and select a point on each line to define vectors <math>\mathbf{v}_{i}=(x_{i},y_{i})</math>.
 +
Note that <math>\theta=45^{\circ}</math> gives equal magnitudes of the vector products
 +
<cmath>\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 .</cmath>
 +
 
 +
Substituting coordinate expressions for vector products, we find
 +
<cmath>\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} .</cmath>
 +
Divide this equation by <math>x_{1}x_{2}</math> to obtain
 +
<cmath>1+m_{1}m_{2} = m_{2}-m_{1} ,</cmath>
 +
where <math>m_{i}=y_{i}/x_{i}</math> is the slope of line <math>i</math>.
 +
Taking <math>m_{2}=6m_{1}</math> , we obtain
 +
<cmath>6m_{1}^{2}-5m_{1}+1 = 0
 +
\ \implies\ m_{1} \in \{\frac{1}{3},\frac{1}{2}\} .</cmath>
 +
The latter solution gives the largest product of slopes <math>m_{1}m_{2} = 6m_{1}^2 = \frac{3}{2} . \quad \boxed{\textbf{(C)}}</math>
  
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~JHawk0224
+
~fairpark
  
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==Solution 2 (bash)==
+
==Solution 3 (bash)==
 
Place on coordinate plane.
 
Place on coordinate plane.
 
Lines are <math>y=mx, y=6mx.</math> The intersection point at the origin.
 
Lines are <math>y=mx, y=6mx.</math> The intersection point at the origin.
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~ccx09
 
~ccx09
  
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==Solution 3 (complex)==
+
==Solution 4 (tan)==
−
Let the intersection point is the origin. Let <math>(a,b)</math> be a point on the line of lesser slope. The mutliplication of <math>a+bi</math> by cis 45.
+
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>.
−
<math>(a+bi)(\frac{1}{\sqrt 2 }+i*\frac{1}{\sqrt 2 })=\frac{1}{\sqrt 2 }((a-b)+(a+b)*i)</math> and therefore <math>(a-b, a+b)</math> lies on the line of greater slope.
 
  
−
Then, the rotation of <math>(a,b)</math> by 45 degrees gives a line of slope <math>\frac{a+b}{a-b}</math>.
+
~JHawk0224
  
−
We get the equation <math>\frac{6b}{a}=\frac{a+b}{a-b}\implies a^2-5ab+6b^2=(a-3b)(a-2b)=0</math> and this gives our answer.
+
==Solution 5 (matrix transformation)==
−
 
 
−
~jeffisepic
 
−
 
 
−
==Solution 4 (matrix transformation)==
 
 
Multiply by the rotation transformation matrix  
 
Multiply by the rotation transformation matrix  
 
<math>
 
<math>
Line 39: Line 58:
 
where <math>\theta = 45^{\circ}.</math>
 
where <math>\theta = 45^{\circ}.</math>
  
−
==Solution 5 (Cheating)==
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==Solution 6 (Cheating)==
 
Let the smaller slope be <math>m</math>, then the larger slope is <math>6m</math>. Since we want the greatest product we begin checking each answer choice, starting with (E).  
 
Let the smaller slope be <math>m</math>, then the larger slope is <math>6m</math>. Since we want the greatest product we begin checking each answer choice, starting with (E).  
  
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Using this same method for the other answer choices, we eventually find that the answer is <math>\boxed{\textbf{(C)}\  \frac32}</math> since our slopes are <math>\frac12</math> and <math>3</math> which forms a perfect 45 degree angle.
 
Using this same method for the other answer choices, we eventually find that the answer is <math>\boxed{\textbf{(C)}\  \frac32}</math> since our slopes are <math>\frac12</math> and <math>3</math> which forms a perfect 45 degree angle.
 +
 +
==Solution 7==
 +
If you have this formula memorized then it will be very easy to do: If <math>\theta</math> is the angle between two lines with slopes <math>m_1</math> and <math>m_2</math>, then <math>\tan(\theta)=\frac{m_1-m_2}{1+m_1m_2}</math>.
 +
 +
Now let the smaller slope be <math>m</math>, thus the other slope is <math>6m</math>. Using our formula above: <cmath>\tan(45^\circ)=\frac{6m-m}{1+6m^2} \implies 6m^2-5m+1=0 \implies (2m-1)(3m-1)=0.</cmath> Therefore the two possible values for <math>m</math> are <math>\tfrac{1}{2}</math> and <math>\tfrac{1}{3}</math>. We choose the larger one and thus our answer is <cmath>6m^2=6 \cdot \frac{1}{4} = \frac{3}{2} \implies \boxed{\mathbf{(C)}}.</cmath>
 +
 +
~AngelaLZ
 +
 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/MLyMs8w1xEg
 +
 +
~Education, the Study of Everything
 +
 +
 +
 +
 +
 +
  
 
==Video Solution==
 
==Video Solution==
−
Two solutions
 
 
https://youtu.be/6ujfjGLzVoE
 
https://youtu.be/6ujfjGLzVoE
−
 
−
~IceMatrix
 
  
 
==See Also==
 
==See Also==

Latest revision as of 22:54, 1 October 2024

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 be 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

See Also

2020 AMC 12B (Problems • Answer Key • Resources)
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Problem 8
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