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

Revision as of 00:12, 10 November 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, $\theta = \boxed{\textbf{C} 45}$ ~Failure.net

@@ Line 16: / Line 16: @@
 ~lprado
+==Solution 3 (Law of Cosines)==
+Follow Solution 2 up until the lattice points section. Let's use <math>(0,0)</math>, <math>(2,4)</math>, and <math>(9,3)</math>. The distance between the origin and <math>(2,4)</math> is <math>\sqrt{20}</math>. The distance between the origin and <math>(9,3)</math> is <math>\sqrt{90}</math>. The distance between <math>(2,4)</math> and <math>(9,3)</math> is <math>\sqrt{50}</math>. Using the Law of Cosines, we see the <math>50 = 90 + 20 - 2\times\sqrt{20}</math> <math>\times\sqrt{90}</math> <math>\times\cos(\theta)</math>, where <math>\theta</math> is the angle we are looking for.
+Simplifying, we get <math>-60 = -2\times(\sqrt{20}) \times(\sqrt{90}) \times\cos(\theta)</math>.
+<math>30 =  sqrt{1800} \times\cos(\theta)</math>.
+<math>30 =  30\sqrt{2} \times\cos(\theta)</math>.
+<math>\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}= \cos(\theta)</math>.
+Thus, <math>\theta = \boxed{\textbf{C} 45}</math>
+~Failure.net
 ==See also==
 {{AMC12 box|year=2023|ab=A|num-b=10|num-a=12}}
 {{MAA Notice}}

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