Difference between revisions of "2024 AMC 12A Problems/Problem 8"
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+ | == Solution 2 == | ||
+ | Let <math>f(\theta)=\log(\sin(3\theta))</math> and let <math>g(\theta)=\log(\cos(2\theta))</math>. | ||
+ | |||
+ | Note that <math>-1\leq\sin(3\theta),\cos(2\theta)\leq1</math>. Because the logarithm of a nonpositive number is not real, the functions <math>f</math> and <math>g</math> only exist when <math>\sin(3\theta)</math> and <math>\cos(2\theta)</math> are positive, respectively. Furthermore, because the logarithm of any positive real number less than <math>1</math> is negative, the only case where the function <math>f+g</math> could equal <math>0</math> is if <math>f(\theta)=g(\theta)=0</math>, which only occurs when their respective sine and cosine expressions are both equal to <math>1</math>. | ||
+ | |||
+ | Thus, we have these two equations, where <math>m</math> and <math>n</math> are any integers: | ||
+ | \begin{align*} | ||
+ | \sin(3\theta)=1 \Rightarrow 3\theta = \frac\pi2+2\pi m &\Rightarrow \theta = \frac\pi6+\frac{2\pi}3 m = \frac{4m+1}6\pi\ | ||
+ | \cos(2\theta)=1 \Rightarrow 2\theta = 2\pi n &\Rightarrow \theta = \pi n | ||
+ | \end{align*} | ||
+ | |||
+ | Because <math>m</math> is an integer, <math>4m+1</math> is odd, and so <math>\frac{4m+1}6</math> is never an integer. Therefore, by the first equation, <math>\theta</math> can never be an integer multiple of pi. Thus, because the second equation requires that <math>\theta</math> be an integer multiple of pi, these two equations cannot both be satisfied, and so there are no solutions to the given equation. | ||
+ | |||
+ | Thus, we choose answer choice <math>\boxed{\textbf{(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 09:17, 9 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)
BUT note that is not real Giving us .
~Minor edits by evanhliu2009
Solution 2
Let and let .
Note that . Because the logarithm of a nonpositive number is not real, the functions and only exist when and are positive, respectively. Furthermore, because the logarithm of any positive real number less than is negative, the only case where the function could equal is if , which only occurs when their respective sine and cosine expressions are both equal to .
Thus, we have these two equations, where and are any integers:
Because is an integer, is odd, and so is never an integer. Therefore, by the first equation, can never be an integer multiple of pi. Thus, because the second equation requires that be an integer multiple of pi, these two equations cannot both be satisfied, and so there are no solutions to the given equation.
Thus, we choose answer choice .
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 |
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