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

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Hence the mean is <math>\boxed{\textbf{(E) }\frac{91}{180}}</math>
 
Hence the mean is <math>\boxed{\textbf{(E) }\frac{91}{180}}</math>
  
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==Solution 2==
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We can add a term <math>x_0</math> into the list, and the total sum of the terms won't be affected since <math>x_0=0</math>. Once <math>x_0</math> is added into the list, the average of the <math>91</math> terms is clearly <math>\frac{1}{2}</math>. Hence the total sum of the terms is <math>\frac{91}{2}</math>. To get the average of the original <math>90</math>, we merely divide by <math>90</math> to get <math>\frac{91}{180}</math>. Hence the mean is <math>\boxed{\textbf{(E) }\frac{91}{180}}</math>
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~tsun26
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Notice that $x_0+x_1+\cdots+x_{90}
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=B|num-b=10|num-a=12}}
 
{{AMC12 box|year=2024|ab=B|num-b=10|num-a=12}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:59, 14 November 2024

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 $\boxed{\textbf{(E) }\frac{91}{180}}$

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 


Notice that $x_0+x_1+\cdots+x_{90}

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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