2024 AMC 12B Problems/Problem 11: Difference between revisions

Revision as of 13:49, 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 $\boxed{\textbf{(E) }\frac{91}{180}}$

~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

~Kathan

@@ Line 19: / Line 19: @@
 ~tsun26
-==See also==
+==Solution 3==
-{{AMC12 box|year=2024|ab=B|num-b=10|num-a=12}}
+[[Image: 2024_AMC_12B_P11.jpeg|thumb|center|600px|]]
+~Kathan
-{{MAA Notice}}

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