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 $\frac{91}{180} \text{ or } \boxed{\textbf{(E) }0}$

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2024 AMC 12B Problems/Problem 11

Problem

Solution 1