Difference between revisions of "2023 AMC 12A Problems/Problem 11"
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+ | ==Solution 4 (Vector Bash)== | ||
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+ | 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 this in, we get that <math>\cos \theta = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}} </math> From this we get that Thus, <math>\theta = \boxed{\textbf{(C)} 45^\circ}</math> | ||
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+ | ~middletonkids | ||
==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 00:22, 10 November 2023
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 .
.
.
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Thus,
~Failure.net
Solution 4 (Vector Bash)
We can set up vectors and to represent the two lines. We know that . Plugging this in, we get that From this we get that Thus,
~middletonkids
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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