AMC12 2024 B

AMC12 2024 B · Q11

AMC12 2024 B · Q11. It mainly tests Averages (mean), Trigonometry (basic).

Let $x_n = \sin^2(n^{\circ})$. What is the mean of $x_1,x_2,x_3,\dots,x_{90}$?

设 $x_n = \sin^2(n^{\circ})$。$x_1,x_2,x_3,\dots,x_{90}$ 的平均值是多少？

(A) \frac{11}{45} \frac{11}{45}

(B) \frac{22}{45} \frac{22}{45}

(D) \frac{1}{2} \frac{1}{2}

(E) \frac{91}{180} \frac{91}{180}

Answer

Correct choice: (E)

正确答案：（E）

Solution

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 \[x_1+x_2+\dots+x_{90}=44+x_{45}+x_{90}=44+\left(\frac{\sqrt{2}}{2}\right)^2+1^2=\frac{91}{2}\] Hence the mean is $\boxed{\textbf{(E) }\frac{91}{180}}$

将 $x_1$ 与 $x_{89}$、$x_2$ 与 $x_{88}$、$x_i$ 与 $x_{90-i}$ 两两相加。注意到 \[x_i+x_{90-i}=\sin^2(i^{\circ})+\sin^2((90-i)^{\circ})=\sin^2(i^{\circ})+\cos^2(i^{\circ})=1\] 由勾股恒等式可得。由于可以将 $1$ 与 $89$ 配对，一直到 $44$ 与 $46$，我们得到 \[x_1+x_2+\dots+x_{90}=44+x_{45}+x_{90}=44+\left(\frac{\sqrt{2}}{2}\right)^2+1^2=\frac{91}{2}\] 因此平均值为 $\boxed{\textbf{(E) }\frac{91}{180}}$

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