Difference between revisions of "2024 AMC 12B Problems/Problem 11"
Kafuu chino (talk | contribs) (Created page with "==Problem== Let <math>x_n = \sin^2(n^{\circ})</math>. What is the mean of <math>x_1,x_2,x_3,\dots,x_{90}</math>? <math>\textbf{(A) } \frac{11}{45} \qquad\textbf{(B) } \frac{2...") |
Kafuu chino (talk | contribs) (→Solution 1) |
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<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> | ||
Hence the mean is | Hence the mean is | ||
| − | <cmath>\frac{91}{180} \text{ or } \textbf{(E)}</cmath> | + | <cmath>\frac{91}{180} \text{ or } \boxed{\textbf{(E) }0}</cmath> |
Revision as of 00:30, 14 November 2024
Problem
Let 
. What is the mean of 
?
Solution 1
Add up 
with 
, 
with 
, and 
with 
. Notice
![\[x_i+x_{90-i}=\sin^2(i^{\circ})+\sin^2((90-i)^{\circ})=\sin^2(i^{\circ})+\cos^2(i^{\circ})=1\]](http://latex.artofproblemsolving.com/9/3/b/93b7240a2833c84ab1628a99965d3794bbfbd6e6.png)
by the Pythagorean identity. Since we can pair up 
with 
and keep going until 
with 
, we get
![\[x_1+x_2+\dots+x_{90}=44+x_{45}+x_{90}=44+(\frac{\sqrt{2}}{2})^2+1^2=\frac{91}{2}\]](http://latex.artofproblemsolving.com/1/8/c/18c524b9d37a09a5a5801e960739ae7ea9b3d14f.png)
Hence the mean is
![\[\frac{91}{180} \text{ or } \boxed{\textbf{(E) }0}\]](http://latex.artofproblemsolving.com/6/1/e/61e0e486a5b7f9520726feb2fc98a88250c89311.png)
