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Difference between revisions of "2024 AMC 12B Problems/Problem 11"

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==Solution 5 (Calculus)==
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==Solution 5 (Integral Approximation)==
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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>\left[\frac{\pi}{180}, \frac{\pi}{2}\right]</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>  
  
−
While there are 90 discrete terms meaning \( f(n)\) isn't actually continuous, the interval isn't very large and 90 is a lot of values so calculating the average over EVERY term in the interval will be very close to calculating the average over 90 terms, so we can instead calculate the average value from <math>\frac{\pi}{180}</math> to <math>\frac{\pi}{2}</math>
+
While there are 90 discrete terms meaning \( f(n)\) isn't actually continuous, the interval is very small compared to 90, so calculating the average over EVERY term in the interval will be very close to calculating the average over 90 terms, so we can instead calculate the average value from <math>\frac{\pi}{180}</math> to <math>\frac{\pi}{2}</math>
  
 
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>
 
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>

Latest revision as of 23:47, 2 August 2025

Problem

Let $x_n = \sin^2(n^{\circ})$. What is the mean of $x_1,x_2,x_3,\dots,x_{90}$?

$\textbf{(A) } \frac{11}{45} \qquad\textbf{(B) } \frac{22}{45} \qquad\textbf{(C) } \frac{89}{180} \qquad\textbf{(D) } \frac{1}{2} \qquad\textbf{(E) } \frac{91}{180}$

Solution 1

Add up $x_1$ with $x_{89}$, $x_2$ with $x_{88}$, and $x_i$ with $x_{90-i}$. Notice \[x_i+x_{90-i}=\sin^2(i^{\circ})+\sin^2((90-i)^{\circ})=\sin^2(i^{\circ})+\cos^2(i^{\circ})=1\] by the Pythagorean identity. Since we can pair up $1$ with $89$ and keep going until $44$ with $46$, we get \[x_1+x_2+\dots+x_{90}=44+x_{45}+x_{90}=44+\left(\frac{\sqrt{2}}{2}\right)^2+1^2=\frac{91}{2}\] Hence the mean is $\boxed{\textbf{(E) }\frac{91}{180}}$

~kafuu_chino

Solution 2

We can add a term $x_0$ into the list, and the total sum of the terms won't be affected since $x_0=0$. Once $x_0$ is added into the list, the average of the $91$ terms is clearly $\frac{1}{2}$. Hence the total sum of the terms is $\frac{91}{2}$. To get the average of the original $90$, we merely divide by $90$ to get $\frac{91}{180}$. Hence the mean is $\boxed{\textbf{(E) }\frac{91}{180}}$

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: $x_n=\sin^2\left(\frac{n\pi}{2\times90}\right)$. Notice that $90$ have no specialty beyond any other integers, so we can use some inductive processes.

If we change $90$ to $2$: \[\frac{\sin^2\left(\frac{\pi}{4}\right)+\sin^2\left(\frac{2\pi}{4}\right)}{2}=\frac{\left(\frac{1}{\sqrt{2}}\right)^2+\left(1\right)^2}{2}=\frac{\frac{1}{2}+1}{2}=\frac{3}{4}\,.\]

If we change $90$ to $3$: \[\frac{\sin^2\left(\frac{\pi}{6}\right)+\sin^2\left(\frac{2\pi}{6}\right)+\sin^2\left(\frac{3\pi}{6}\right)}{3}=\frac{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2+\left(1\right)^2}{3}=\frac{\frac{1}{4}+\frac{3}{4}+1}{3}=\frac{2}{3}=\frac{4}{6}\,.\]

By intuition, although not rigorous at all, we can guess out the solution if we change $90$ into $k$, we get $\frac{k+1}{2k}$. Thus, if we plug in $k=90$, we get $\frac{90+1}{2\times90}=\boxed{\mathbf{(E)}\,\frac{91}{180}}$

~Prof. Joker

Solution 4

2024 AMC 12B P11.jpeg

~Kathan

Solution 4

Note that $\sin^2(x) = \frac{1 - \cos(2x)}{2}$. We want to determine $\frac{1}{90}\sum_{n = 1}^{90} \sin^2(n^{\circ})$.

\begin{align*} &= \frac{1}{90} \sum_{n = 1}^{90} \frac{1 - \cos(2n)}{2} \\ &= \frac{1}{2} -\frac{1}{180}\sum_{n = 1}^{90} \cos(2n) \\ \end{align*}

Graphing $\cos(x)$, we can pair $\cos(2^{\circ}) + \cos(178^{\circ}) = 0$ and so on. We are left with $\cos(90^{\circ}) + \cos(180^{\circ}) = -1$.

Our answer is $\frac{1}{2} + \frac{1}{180} = \boxed{\textbf{(E) }\frac{91}{180}}$

~vinyx


Solution 5 (Integral Approximation)

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: $\left[\frac{\pi}{180}, \frac{\pi}{2}\right]$

While there are 90 discrete terms meaning \( f(n)\) isn't actually continuous, the interval is very small compared to 90, so calculating the average over EVERY term in the interval will be very close to calculating the average over 90 terms, so we can instead calculate the average value from $\frac{\pi}{180}$ to $\frac{\pi}{2}$

The average value of \( f(n) \) is $\frac{1}{\frac{\pi}{2} - \frac{\pi}{180}} \int_{1\cdot\frac{\pi}{180}}^{90\cdot\frac{\pi}{180}} \sin^{2}(x) \, dx = \frac{180}{89\pi} \int_{\pi/180}^{\pi/2} \sin^2(x) \, dx$


We can solve this integral using the power-reduction identity: $\sin^2(x) = \frac{1 - \cos(2x)}{2}$

$\frac{180}{89\pi} \int_{\pi/180}^{\pi/2} \frac{1 - \cos(2x)}{2} \, dx$ $= \frac{180}{89\pi} \left.\left(\frac{x}{2} - \frac{\sin(2x)}{4}\right)\right|_{\pi/180}^{\pi/2}$ $= \frac{180}{89\pi} \left[ \left(\frac{\pi}{4} - 0\right) - \left(\frac{\pi}{360} - \frac{\sin\left(\frac{\pi}{90}\right)}{4}\right) \right]$ $= \frac{180}{89\pi} \left( \frac{89\pi}{360} + \frac{\sin\left(\frac{\pi}{90}\right)}{4} \right)$


The taylor series of $\sin(\theta) = \theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \cdots$

Because $\sin\left(\frac{\pi}{90}\right)$ is a small angle, we can use a first-order taylor approximation and approximate $\sin\left(\frac{\pi}{90}\right) \approx \frac{\pi}{90}$

Hence, our result is $\frac{180}{89\pi}\left(\frac{89\pi}{360} + \frac{\pi}{360}\right) = \frac{180}{89\pi}\cdot\frac{90\pi}{360} = \frac{180}{89\pi}\cdot\frac{\pi}{4} = \frac{45}{89}$


While $\frac{45}{89}$ isn't an option, it is greater than $\frac{1}{2}$, and out of the given options, only $\frac{91}{180}$ is greater than $\frac{1}{2}$ (We approximated $\sin\left(\frac{\pi}{90}\right)$ with a first-order taylor approximation and calculated the average value over the interval instead of over 90 discrete values, but it's close enough to $\frac{91}{180}$).

So the answer is $\boxed{\textbf{(E) }\frac{91}{180}}$

-sourodeepdeb

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

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