Difference between revisions of "2023 AMC 12A Problems/Problem 11"
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− | + | ==Problem== | |
+ | What is the degree measure of the acute angle formed by lines with slopes <math>2</math> and <math>\frac{1}{3}</math>? | ||
+ | |||
+ | <math>\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | Remind that <math>\text{slope}=\dfrac{\Delta y}{\Delta x}=\tan \theta</math> where <math>\theta</math> is the angle between the slope and <math>x</math>-axis. <math>k_1=2=\tan \alpha</math>, <math>k_2=\dfrac{1}{3}=\tan \beta</math>. The angle formed by the two lines is <math>\alpha-\beta</math>. <math>\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\dfrac{2-1/3}{1+2\cdot 1/3}=1</math>. Therefore, <math>\alpha-\beta=\boxed{\textbf{(C)} 45^\circ}</math>. | ||
+ | |||
+ | ~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 <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>. | ||
+ | |||
+ | ~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^\circ}</math> | ||
+ | |||
+ | ~Failure.net | ||
+ | |||
+ | ==Solution 4 (Vector Bash)== | ||
+ | |||
+ | We can set up vectors <math>\vec{a} = <1,2></math> and <math>\vec{b} = <3,1></math> to represent the two lines. We know that <math>\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \cos \theta</math>. Plugging the vectors in gives us <math>\cos \theta = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}</math>. From this we get that <math>\theta = \boxed{\textbf{(C)} 45^\circ}</math>. | ||
+ | |||
+ | ~middletonkids | ||
+ | |||
+ | ==Solution 5 (Complex Numbers)== | ||
+ | |||
+ | Let <math>Z_1 = 3 + i</math> and <math>Z_2 = 1 + 2i</math> | ||
+ | <cmath>\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*}</cmath> | ||
+ | |||
+ | From this we have: | ||
+ | <cmath>\begin{align} | ||
+ | 1 &= 3r\cos\theta - r\sin\theta \ | ||
+ | 2 &= r\cos\theta + 3r\sin\theta | ||
+ | \end{align}</cmath> | ||
+ | |||
+ | To solve this we must compute <math>r</math> | ||
+ | <cmath>\begin{align*} | ||
+ | r &= \frac{|Z_2|}{|Z_1|} \ | ||
+ | &= \frac{\sqrt{5}}{\sqrt{10}} \ | ||
+ | &= \frac{\sqrt{2}}{2} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Using elimination we have: | ||
+ | <math>3\cdot(2) - (1)</math> | ||
+ | <cmath>\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*}</cmath> | ||
+ | |||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
+ | |||
+ | ==Solution 5.b (Complex Numbers - Simpler)== | ||
+ | |||
+ | Let <math>Z_1 = 3 + i</math> and <math>Z_2 = 1 + 2i</math> | ||
+ | <cmath>\begin{align*} | ||
+ | Z_2 &= Z_1 \cdot re^{i\theta} \ | ||
+ | 1+2i &=(3+i) \cdot re^{i\theta} \ | ||
+ | \frac{1+2i}{3+i} &= re^{i\theta} \ | ||
+ | \frac{(1+2i)(3-i)}{(3+i)(3-i)} &= re^{i\theta} \ | ||
+ | \frac{3+2+6i-i}{(3+i)(3-i)} &= re^{i\theta} \ | ||
+ | tan(\theta) &= 5/5 \ | ||
+ | \theta &= \boxed{\textbf{(C)} 45^\circ} \ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
+ | |||
+ | ==Solution 6 == | ||
+ | The lines <math>y = 2x, y = \frac {1}{3}x</math>, and <math>x = 3</math> form a large right triangle and a small right triangle. Call the angle that is formed by the x-axis and the line <math>y = 2x </math> <math> \alpha</math>, and call the angle that is formed by the x-axis and the line <math>y = \frac {1}{3}x</math> <math>\beta</math>. We try to find <math>\sin (\alpha - \beta)</math> first, and then try to see if any of the answer choices match up. | ||
+ | |||
+ | <math>\sin (\alpha - \beta)</math> = <math>\sin \alpha</math> <math>\cos \beta</math> - <math>\sin \beta</math> <math>\cos \alpha</math>. | ||
+ | |||
+ | Using soh-cah-toa, we find that <math>\sin \alpha = \frac {2}{\sqrt 5}, \sin \beta = \frac {1}{\sqrt 10}, \cos \alpha = \frac {1}{\sqrt 5}, </math> and <math>\cos \beta = \frac {3}{\sqrt 10}</math>. | ||
+ | |||
+ | Plugging it all in, we find that <math>\sin (\alpha - \beta) = \frac {5}{\sqrt {50}}</math>, which is equivalent to <math>\frac {\sqrt 2}{2}</math>. Since <math>\sin (\alpha - \beta) = \frac {\sqrt 2}{2}</math>, we get that <math>\alpha - \beta = 45^{\circ}</math>. Therefore, the answer is <math>\boxed {\textbf {(C)} 45^{\circ}}</math>. | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | ==Solution 7 (Cheese)= | ||
+ | AMC 12 does not allow graph paper or protractor. https://amc-reg.maa.org/manual/AMC1012B.pdf | ||
+ | |||
+ | Using a makeshift ruler, 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 by eye or by folding paper to observe that they form a 45 degree angle. | ||
+ | The answer is <math>\boxed{45^\circ}</math>. | ||
+ | |||
+ | ~InstallHelp_Hex | ||
+ | |||
+ | ==Video Solution (easy to digest) by Power Solve== | ||
+ | https://youtu.be/YXIH3UbLqK8?si=Uh4K33tNEzBOXc_h&t=1406 | ||
+ | |||
+ | ==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== | ||
+ | {{AMC12 box|year=2023|ab=A|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Latest revision as of 22:33, 5 August 2024
Contents
[hide]Problem
What is the degree measure of the acute angle formed by lines with slopes and ?
Solution 1
Remind that where is the angle between the slope and -axis. , . The angle formed by the two lines is . . Therefore, .
~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 for the line with slope of 2 and for the line with slope of 1/3. Let's take 3 lattice points and create a triangle. Let's use , , and . The distance between the origin and is . The distance between the origin and is . The distance between and is . We notice that we have a triangle with 3 side lengths: , , and . This forms a 45-45-90 triangle, meaning that the angle is .
~lprado
Solution 3 (Law of Cosines)
Follow Solution 2 up until the lattice points section. Let's use , , and . The distance between the origin and is . The distance between the origin and is . The distance between and is . Using the Law of Cosines, we see the , where is the angle we are looking for.
Simplifying, we get .
.
.
.
Thus,
~Failure.net
Solution 4 (Vector Bash)
We can set up vectors and to represent the two lines. We know that . Plugging the vectors in gives us . From this we get that .
~middletonkids
Solution 5 (Complex Numbers)
Let and
From this we have:
To solve this we must compute
Using elimination we have:
Solution 5.b (Complex Numbers - Simpler)
Let and
Solution 6
The lines , and form a large right triangle and a small right triangle. Call the angle that is formed by the x-axis and the line , and call the angle that is formed by the x-axis and the line . We try to find first, and then try to see if any of the answer choices match up.
= - .
Using soh-cah-toa, we find that and .
Plugging it all in, we find that , which is equivalent to . Since , we get that . Therefore, the answer is .
~Arcticturn
=Solution 7 (Cheese)
AMC 12 does not allow graph paper or protractor. https://amc-reg.maa.org/manual/AMC1012B.pdf
Using a makeshift ruler, 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 by eye or by folding paper to observe that they form a 45 degree angle. The answer is .
~InstallHelp_Hex
Video Solution (easy to digest) by Power Solve
https://youtu.be/YXIH3UbLqK8?si=Uh4K33tNEzBOXc_h&t=1406
Video Solution (Under 4 minutes)
~Education, the Study of Everything
Video Solution
~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.