Difference between revisions of "2024 AMC 12B Problems/Problem 11"
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Let \( f(n) = \sin^2(n^\circ) \) | Let \( f(n) = \sin^2(n^\circ) \) | ||
− | The function \( f(n) = \sin^2(n) \) is continuous for all real values of \( n \). We want to find the average value of \( f(n) \) over the interval: [<math>\frac{\pi}{180}</math>, <math>\frac{\pi}{2}</math>] | + | The function \( f(n) = \sin^2(n) \) is continuous for all real values of \( n \). We want to find the average value of \( f(n) \) over the interval: <math>\left[\frac{\pi}{180}, \frac{\pi}{2}\right]</math> |
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+ | |||
+ | The average value of \( f(n) \) is <math>\frac{1}{\frac{\pi}{2} - \frac{\pi}{180}} \int_{1\cdot\frac{\pi}{180}}^{90\cdot\frac{\pi}{180}} \sin^{2}(x) \, dx = \frac{180}{89\pi} \int_{\pi/180}^{\pi/2} \sin^2(x) \, dx</math> | ||
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+ | We can solve this integral using the power-reduction identity: <math>\sin^2(x) = \frac{1 - \cos(2x)}{2}</math> | ||
+ | |||
+ | <math>\frac{180}{89\pi} \int_{\pi/180}^{\pi/2} \frac{1 - \cos(2x)}{2} \, dx</math> | ||
+ | <math>= \frac{180}{89\pi} \left.\left(\frac{x}{2} - \frac{\sin(2x)}{4}\right)\right|_{\pi/180}^{\pi/2}</math> | ||
+ | <math>= \frac{180}{89\pi} \left[ \left(\frac{\pi}{4} - 0\right) - \left(\frac{\pi}{360} - \frac{\sin\left(\frac{\pi}{90}\right)}{4}\right) \right]</math> | ||
+ | <math>= \frac{180}{89\pi} \left( \frac{89\pi}{360} + \frac{\sin\left(\frac{\pi}{90}\right)}{4} \right)</math> | ||
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+ | The taylor series of <math>\sin(\theta) = \theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \cdots</math> | ||
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+ | Because <math>\sin\left(\frac{\pi}{90}\right)</math> is a small angle, we can use a first-order taylor approximation and approximate <math>\sin\left(\frac{\pi}{90}\right) \approx \frac{\pi}{90}</math> | ||
+ | |||
+ | Hence, our result is <math>\frac{180}{89\pi}\left(\frac{89\pi}{360} + \frac{\pi}{360}\right) = \frac{180}{89\pi}\cdot\frac{90\pi}{360} = \frac{180}{89\pi}\cdot\frac{\pi}{4} = \frac{45}{89}</math> | ||
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+ | While <math>\frac{45}{89}</math> isn't an option, it is greater than <math>\frac{1}{2}</math>, and out of the given options, only <math>\frac{91}{180}</math> is greater than <math>\frac{1}{2}</math> (we don't get <math>\frac{91}{180}</math> exactly because we approximated <math>\sin\left(\frac{\pi}{90}\right)</math> via first-order Taylor approximation, but it's close enough to <math>\frac{91}{180}</math>). | ||
+ | |||
+ | So the answer is | ||
==Video Solution 1 by SpreadTheMathLove== | ==Video Solution 1 by SpreadTheMathLove== |
Revision as of 19:52, 2 August 2025
Contents
Problem
Let . What is the mean of
?
Solution 1
Add up with
,
with
, and
with
. Notice
by the Pythagorean identity. Since we can pair up
with
and keep going until
with
, we get
Hence the mean is
~kafuu_chino
Solution 2
We can add a term into the list, and the total sum of the terms won't be affected since
. Once
is added into the list, the average of the
terms is clearly
. Hence the total sum of the terms is
. To get the average of the original
, we merely divide by
to get
. Hence the mean is
This method is called constructing a variable (although most of you already know).
~tsun26, ShortPeopleFartalot
Solution 3 (Inductive Reasoning)
If we use radians to rewrite the question, we have: . Notice that
have no specialty beyond any other integers, so we can use some inductive processes.
If we change to
:
If we change to
:
By intuition, although not rigorous at all, we can guess out the solution if we change into
, we get
. Thus, if we plug in
, we get
~Prof. Joker
Solution 4
~Kathan
Solution 4
Note that . We want to determine
.
Graphing , we can pair
and so on. We are left with
.
Our answer is
~vinyx
Solution 5 (Calculus)
Let \( f(n) = \sin^2(n^\circ) \)
The function \( f(n) = \sin^2(n) \) is continuous for all real values of \( n \). We want to find the average value of \( f(n) \) over the interval:
The average value of \( f(n) \) is
We can solve this integral using the power-reduction identity:
The taylor series of
Because is a small angle, we can use a first-order taylor approximation and approximate
Hence, our result is
While isn't an option, it is greater than
, and out of the given options, only
is greater than
(we don't get
exactly because we approximated
via first-order Taylor approximation, but it's close enough to
).
So the answer is
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=gJq7DhLNnZ4&t=0s
See also
2024 AMC 12B (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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.