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

Revision as of 14:19, 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 22: / Line 22: @@
 [[Image: 2024_AMC_12B_P11.jpeg|thumb|center|600px|]]
 ~Kathan
+==See also==
+{{AMC12 box|year=2024|ab=B|num-b=10|num-a=12}}
+{{MAA Notice}}

Page

Toolbox

Search