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

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by the Pythagorean identity. Since we can pair up <math>1</math> with <math>89</math> and keep going until <math>44</math> with <math>46</math>, we get
 
by the Pythagorean identity. Since we can pair up <math>1</math> with <math>89</math> and keep going until <math>44</math> with <math>46</math>, we get
 
<cmath>x_1+x_2+\dots+x_{90}=44+x_{45}+x_{90}=44+(\frac{\sqrt{2}}{2})^2+1^2=\frac{91}{2} </cmath>
 
<cmath>x_1+x_2+\dots+x_{90}=44+x_{45}+x_{90}=44+(\frac{\sqrt{2}}{2})^2+1^2=\frac{91}{2} </cmath>
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Hence the mean is
+
Hence the mean is <math>\boxed{\textbf{(E) }\frac{91}{180}}</math>
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<cmath>\frac{91}{180} \text{ or } \boxed{\textbf{(E) }\frac{91}{180}}</cmath>
 

Revision as of 00:32, 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 $\boxed{\textbf{(E) }\frac{91}{180}}$

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