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Difference between revisions of "2023 AMC 12A Problems/Problem 11"
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~plasta
 
~plasta
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==Solution 2==
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We can take any two lines of this form, since the angle between them will always be the same. Let's take <math>y=2x</math> for the line with slope of 2 and <math>y=\frac{1}{3}x</math> for the line with slope of 1/3. Let's take 3 lattice points and create a triangle. Let's use <math>(0,0)</math>, <math>(1,2)</math>, and <math>(3,1)</math>. The distance between the origin and <math>(1,2)</math> is <math>\sqrt{5}</math>. The distance between the origin and <math>(3,1)</math> is <math>\sqrt{10}</math>. The distance between <math>(1,2)</math> and <math>(3,1)</math> is <math>\sqrt{5}</math>. We notice that we have a triangle with 3 side lengths: <math>\sqrt{5}</math>, <math>\sqrt{5}</math>, and <math>\sqrt{10}</math>. This forms a 45-45-90 triangle, meaning that the angle is <math>\boxed{45^\circ}</math>.
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==See also==
 
==See also==
 
{{AMC12 box|year=2023|ab=A|num-b=10|num-a=12}}
 
{{AMC12 box|year=2023|ab=A|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:41, 9 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

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

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