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2024 AMC 12B Problem 11

Below is the professionally curated solution for Problem 11 of the 2024 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12B solutions, or check the answer key.

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Concepts:trigonometric identitymeansymmetry

Difficulty rating: 1610

11.

Let xn=sin⁡2(n∘).x_n = \sin^2(n^\circ). What is the mean of x1,x2,x3,…,x90?x_1, x_2, x_3, \ldots, x_{90}?

1145\dfrac{11}{45}

2245\dfrac{22}{45}

89180\dfrac{89}{180}

12\dfrac{1}{2}

91180\dfrac{91}{180}

Solution:

Using sin⁡2θ=1−cos⁡2θ2,\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}, ∑n=190sin⁡2(n∘)=902−12∑n=190cos⁡(2n∘).\sum_{n=1}^{90} \sin^2(n^\circ) = \frac{90}{2} - \frac12 \sum_{n=1}^{90}\cos(2n^\circ). In the cosine sum, the terms for nn and 90−n90 - n satisfy cos⁡(2n∘)+cos⁡(180∘−2n∘)=0,\cos(2n^\circ) + \cos(180^\circ - 2n^\circ) = 0, and cos⁡90∘=0,\cos 90^\circ = 0, so everything cancels except cos⁡180∘=−1.\cos 180^\circ = -1.

Hence the sum is 45−12(−1)=45.5,45 - \tfrac12(-1) = 45.5, and the mean is 45.590=91180.\dfrac{45.5}{90} = \dfrac{91}{180}.

Thus, the correct answer is E.

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