Difference between revisions of "2023 AMC 12A Problems/Problem 11"
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~middletonkids | ~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 + isin\theta) \\ | ||
+ | 1+2i&=3rcos\theta - rsin\theta + 3risin\theta + ricos\theta \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | From this we have: | ||
+ | <cmath>\begin{align} | ||
+ | 1 &= 3rcos\theta - rsin\theta \\ | ||
+ | 2 &= rcos\theta + 3rsin\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 &= 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*}</cmath> | ||
+ | |||
+ | ~luckuso | ||
==Video Solution== | ==Video Solution== |
Revision as of 13:53, 11 November 2023
Contents
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:
~luckuso
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.