2023 AMC 12A Problems/Problem 11

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 $Z_2 = 1 + 2i$ $\begin{align*} Z_2 &= Z_1 \cdot re^{i\theta} \\ 1+2i&=(3+i) \cdot re^{i\theta} \\ 1+2i&=(3 + i) \cdot r(cos\theta + isin\theta) \\ 1+2i&=3rcos\theta - rsin\theta + 3risin\theta + ricos\theta \\ \end{align*}$

From this we have: $\begin{align} 1 &= 3rcos\theta - rsin\theta \\ 2 &= rcos\theta + 3rsin\theta \end{align}$

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

~luckuso

Solution 6

Video Solution

https://youtu.be/kPcsTZpFzTY

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

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

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