Difference between revisions of "2024 AMC 12A Problems/Problem 8"
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| − | ==Solution== | + | ==Solution 1== |
Note that this is equivalent to <math>\sin(3\theta)\cos(2\theta)=1</math>, which is clearly only possible when <math>\sin(3\theta)=\cos(2\theta)=\pm1</math>. (If either one is between <math>1</math> and <math>-1</math>, the other one must be greater than <math>1</math> or less than <math>-1</math> to offset the product, which is impossible for sine and cosine.) They cannot be both <math>-1</math> since we cannot take logarithms of negative numbers, so they are both <math>+1</math>. Then <math>3\theta</math> is <math>\dfrac\pi2</math> more than a multiple of <math>2\pi</math> and <math>2\theta</math> is a multiple of <math>2\pi</math>, so <math>\theta</math> is <math>\dfrac\pi6</math> more than a multiple of <math>\dfrac23\pi</math> and also a multiple of <math>\pi</math>. However, a multiple of <math>\dfrac23\pi</math> will always have a denominator of <math>1</math> or <math>3</math>, and never <math>6</math>; it can thus never add with <math>\dfrac\pi6</math> to form an integral multiple of <math>\pi</math>. Thus, there are <math>\boxed{\textbf{(A) }0}</math> solutions. | Note that this is equivalent to <math>\sin(3\theta)\cos(2\theta)=1</math>, which is clearly only possible when <math>\sin(3\theta)=\cos(2\theta)=\pm1</math>. (If either one is between <math>1</math> and <math>-1</math>, the other one must be greater than <math>1</math> or less than <math>-1</math> to offset the product, which is impossible for sine and cosine.) They cannot be both <math>-1</math> since we cannot take logarithms of negative numbers, so they are both <math>+1</math>. Then <math>3\theta</math> is <math>\dfrac\pi2</math> more than a multiple of <math>2\pi</math> and <math>2\theta</math> is a multiple of <math>2\pi</math>, so <math>\theta</math> is <math>\dfrac\pi6</math> more than a multiple of <math>\dfrac23\pi</math> and also a multiple of <math>\pi</math>. However, a multiple of <math>\dfrac23\pi</math> will always have a denominator of <math>1</math> or <math>3</math>, and never <math>6</math>; it can thus never add with <math>\dfrac\pi6</math> to form an integral multiple of <math>\pi</math>. Thus, there are <math>\boxed{\textbf{(A) }0}</math> solutions. | ||
~Technodoggo | ~Technodoggo | ||
| + | ==Solution 1.1 (less words)== | ||
| + | <cmath>log(sin(3\theta))+log(cos(2\theta))=0</cmath> | ||
| + | <cmath>log(sin(3\theta)cos(2\theta))=0</cmath> | ||
| + | <cmath>sin(3\theta)cos(2\theta)=1</cmath> | ||
| + | <cmath>\text{Since } -1\le sin(x),cos(x)\le 1 \Rightarrow sin(3\theta)=cos(2\theta)= \pm1</cmath> | ||
| + | BUT note that <math>log(-1)</math> is not real | ||
| + | <cmath>\Rightarrow sin(3\theta)=cos(2\theta)= 1</cmath> | ||
| + | <cmath>3\theta=\frac{\pi}{2}+2\pi n; \space 2\theta=2\pi m \space (m,n \in \mathbb{Z})</cmath> | ||
| + | <cmath>\theta=\frac{\pi}{6}+\frac{2\pi n}{3}; \space \theta=\pi m</cmath> | ||
| + | <cmath>\Rightarrow \theta\text { has no solution}</cmath> | ||
| + | Giving us <math> \fbox{(A) 0}</math> | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=A|num-b=7|num-a=9}} | {{AMC12 box|year=2024|ab=A|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:37, 8 November 2024
Problem
How many angles 
with 
satisfy 
?
Solution 1
Note that this is equivalent to 
, which is clearly only possible when 
. (If either one is between 
and 
, the other one must be greater than or less than to offset the product, which is impossible for sine and cosine.) They cannot be both since we cannot take logarithms of negative numbers, so they are both 
. Then 
is 
more than a multiple of 
and 
is a multiple of , so is 
more than a multiple of 
and also a multiple of 
. However, a multiple of will always have a denominator of or 
, and never 
; it can thus never add with to form an integral multiple of . Thus, there are 
solutions.
~Technodoggo
Solution 1.1 (less words)
![\[log(sin(3\theta))+log(cos(2\theta))=0\]](http://latex.artofproblemsolving.com/5/5/0/550f13e643a071224c23be32a2b791182e9f34c4.png)
![\[log(sin(3\theta)cos(2\theta))=0\]](http://latex.artofproblemsolving.com/9/3/f/93ffc1c2266d71b89d19bd8fc2086f24df57714d.png)
![\[sin(3\theta)cos(2\theta)=1\]](http://latex.artofproblemsolving.com/9/5/7/9575d1d8a7310c570b50d3426df489309384baf7.png)
![\[\text{Since } -1\le sin(x),cos(x)\le 1 \Rightarrow sin(3\theta)=cos(2\theta)= \pm1\]](http://latex.artofproblemsolving.com/7/4/c/74c25aceeefdde10f94c1ef5d99827f3e8dc1d0e.png)
BUT note that 
is not real
![\[\Rightarrow sin(3\theta)=cos(2\theta)= 1\]](http://latex.artofproblemsolving.com/d/9/c/d9cf6aa421dd7792622b4ee732acdcf8f4226ba9.png)
![\[3\theta=\frac{\pi}{2}+2\pi n; \space 2\theta=2\pi m \space (m,n \in \mathbb{Z})\]](http://latex.artofproblemsolving.com/3/b/a/3ba11370c8c85671049db5c7ec6aa915d5736e47.png)
![\[\theta=\frac{\pi}{6}+\frac{2\pi n}{3}; \space \theta=\pi m\]](http://latex.artofproblemsolving.com/b/8/4/b84f1ec0578bc52c1d81e6ebc1e56b8274779a2c.png)
![\[\Rightarrow \theta\text { has no solution}\]](http://latex.artofproblemsolving.com/4/2/e/42ebae291ec5bd000b0092d3f6d02aea38fa4861.png)
Giving us 
See also
| 2024 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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. 
