Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 12A Problems/Problem 11"

m (→‎Video Solution by Math4All999)
m (→‎Solution 5 (Complex Numbers))
Line 44: Line 44:
 
Z_2 &= Z_1 \cdot re^{i\theta} \\
 
Z_2 &= Z_1 \cdot re^{i\theta} \\
 
1+2i&=(3+i) \cdot re^{i\theta} \\
 
1+2i&=(3+i) \cdot re^{i\theta} \\
−
1+2i&=(3 + i) \cdot r(cos\theta + isin\theta) \\
+
1+2i&=(3 + i) \cdot r(\cos\theta + i\sin\theta) \\
−
1+2i&=3rcos\theta - rsin\theta  + 3risin\theta + ricos\theta \\
+
1+2i&=3r\cos\theta - r\sin\theta  + 3ri\sin\theta + ri\cos\theta \\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
 
From this we have:
 
From this we have:
 
<cmath>\begin{align}
 
<cmath>\begin{align}
−
1 &= 3rcos\theta - rsin\theta \\
+
1 &= 3r\cos\theta - r\sin\theta \\
−
2 &= rcos\theta + 3rsin\theta  
+
2 &= r\cos\theta + 3r\sin\theta  
 
\end{align}</cmath>
 
\end{align}</cmath>
  
Line 64: Line 64:
 
<math>3\cdot(2) - (1)</math>
 
<math>3\cdot(2) - (1)</math>
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
5 &= 10rsin\theta \\
+
5 &= 10r\sin\theta \\
−
\frac{1}{2r} &= sin\theta \\
+
\frac{1}{2r} &= \sin\theta \\
−
\frac{1}{2\frac{\sqrt{2}}{2}} &= sin\theta \\
+
\frac{1}{2\frac{\sqrt{2}}{2}} &= \sin\theta \\
−
\frac{\sqrt{2}}{2} &= sin\theta \\
+
\frac{\sqrt{2}}{2} &= \sin\theta \\
 
\theta &= \boxed{\textbf{(C)}  45^\circ} \\
 
\theta &= \boxed{\textbf{(C)}  45^\circ} \\
 
\end{align*}</cmath>
 
\end{align*}</cmath>

Revision as of 00:51, 25 December 2023

Problem

What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$?

$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$

Solution 1

Remind that $\text{slope}=\dfrac{\Delta y}{\Delta x}=\tan \theta$ where $\theta$ is the angle between the slope and $x$-axis. $k_1=2=\tan \alpha$, $k_2=\dfrac{1}{3}=\tan \beta$. The angle formed by the two lines is $\alpha-\beta$. $\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\dfrac{2-1/3}{1+2\cdot 1/3}=1$. Therefore, $\alpha-\beta=\boxed{\textbf{(C)} 45^\circ}$.

~plasta

Solution 2

We can take any two lines of this form, since the angle between them will always be the same. Let's take $y=2x$ for the line with slope of 2 and $y=\frac{1}{3}x$ for the line with slope of 1/3. Let's take 3 lattice points and create a triangle. Let's use $(0,0)$, $(1,2)$, and $(3,1)$. The distance between the origin and $(1,2)$ is $\sqrt{5}$. The distance between the origin and $(3,1)$ is $\sqrt{10}$. The distance between $(1,2)$ and $(3,1)$ is $\sqrt{5}$. We notice that we have a triangle with 3 side lengths: $\sqrt{5}$, $\sqrt{5}$, and $\sqrt{10}$. This forms a 45-45-90 triangle, meaning that the angle is $\boxed{45^\circ}$.

~lprado

Solution 3 (Law of Cosines)

Follow Solution 2 up until the lattice points section. Let's use $(0,0)$, $(2,4)$, and $(9,3)$. The distance between the origin and $(2,4)$ is $\sqrt{20}$. The distance between the origin and $(9,3)$ is $\sqrt{90}$. The distance between $(2,4)$ and $(9,3)$ is $\sqrt{50}$. Using the Law of Cosines, we see the $50 = 90 + 20 - 2\times\sqrt{20}$ $\times\sqrt{90}$ $\times\cos(\theta)$, where $\theta$ is the angle we are looking for.

Simplifying, we get $-60 = -2\times(\sqrt{20}) \times(\sqrt{90}) \times\cos(\theta)$.

$30 = \sqrt{1800} \times\cos(\theta)$.

$30 = 30\sqrt{2} \times\cos(\theta)$.

$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}= \cos(\theta)$.

Thus, $\theta = \boxed{\textbf{(C)} 45^\circ}$

~Failure.net

Solution 4 (Vector Bash)

We can set up vectors $\vec{a} = <1,2>$ and $\vec{b} = <3,1>$ to represent the two lines. We know that $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \cos \theta$. Plugging the vectors in gives us $\cos \theta = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$. From this we get that $\theta = \boxed{\textbf{(C)} 45^\circ}$.

~middletonkids

Solution 5 (Complex Numbers)

Let $Z_1 = 3 + i$ and $Z_2 = 1 + 2i$ \begin{align*} Z_2 &= Z_1 \cdot re^{i\theta} \\ 1+2i&=(3+i) \cdot re^{i\theta} \\ 1+2i&=(3 + i) \cdot r(\cos\theta + i\sin\theta) \\ 1+2i&=3r\cos\theta - r\sin\theta + 3ri\sin\theta + ri\cos\theta \\ \end{align*}

From this we have: \begin{align} 1 &= 3r\cos\theta - r\sin\theta \\ 2 &= r\cos\theta + 3r\sin\theta \end{align}

To solve this we must compute $r$ \begin{align*} r &= \frac{|Z_2|}{|Z_1|} \\ &= \frac{\sqrt{5}}{\sqrt{10}} \\ &= \frac{\sqrt{2}}{2} \end{align*}

Using elimination we have: $3\cdot(2) - (1)$ \begin{align*} 5 &= 10r\sin\theta \\ \frac{1}{2r} &= \sin\theta \\ \frac{1}{2\frac{\sqrt{2}}{2}} &= \sin\theta \\ \frac{\sqrt{2}}{2} &= \sin\theta \\ \theta &= \boxed{\textbf{(C)} 45^\circ} \\ \end{align*}

~luckuso

Solution 6

The lines $y = 2x, y = \frac {1}{3}x$, and $x = 3$ form a large right triangle and a small right triangle. Call the angle that is formed by the x-axis and the line $y = 2x$ $\alpha$, and call the angle that is formed by the x-axis and the line $y = \frac {1}{3}x$ $\beta$. We try to find $\sin (\alpha - \beta)$ first, and then try to see if any of the answer choices match up.

$\sin (\alpha - \beta)$ = $\sin \alpha$ $\cos \beta$ - $\sin \beta$ $\cos \alpha$.

Using soh-cah-toa, we find that $\sin \alpha = \frac {2}{\sqrt 5}, \sin \beta = \frac {1}{\sqrt 10}, \cos \alpha = \frac {1}{\sqrt 5},$ and $\cos \beta = \frac {3}{\sqrt 10}$.

Plugging it all in, we find that $\sin (\alpha - \beta) = \frac {5}{\sqrt {50}}$, which is equivalent to $\frac {\sqrt 2}{2}$. Since $\sin (\alpha - \beta) = \frac {\sqrt 2}{2}$, we get that $\alpha - \beta = 45^{\circ}$. Therefore, the answer is $\boxed {\textbf {(C)} 45^{\circ}}$.

~Arcticturn

Solution 7 (Cheese)

Using graph paper, as well as the protractor you bought beforehand, draw an accurate to-scale diagram. You can do this by simply drawing the two lines such that they intersect at the origin. Then, measure the angle with a protractor to conclude that they form a 45 degree angle. The answer is $\boxed{45^\circ}$.

~InstallHelp_Hex

Video Solution (Under 4 minutes)

https://youtu.be/PWBEUXTtUxI

~Education, the Study of Everything


Video Solution

https://youtu.be/kPcsTZpFzTY

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Math4All999 (pretty easy)

https://youtu.be/sa2HHgMZjSg?feature=shared

See also

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_11&oldid=208345"