2024 AMC 12B Problems/Problem 11

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 $x_1+x_2+\dots+x_{90}=44+x_{45}+x_{90}=44+(\frac{\sqrt{2}}{2})^2+1^2=\frac{91}{2}$ 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 $\boxed{\textbf{(E) }\frac{91}{180}}$

~tsun26

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

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~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

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=gJq7DhLNnZ4&t=0s

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2024 AMC 12B Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

Solution 3 (Inductive Reasoning)

Solution 4

Solution 4

Video Solution 1 by SpreadTheMathLove

See also