2024 AMC 12B Problem 11

Attempt Problem 11 of the 2024 AMC 12B below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

11.

Let

x_n = \sin^2(n^\circ).

What is the mean of

x_1,

x_2,

x_3,

\ldots,

x_{90}?

\dfrac{11}{45}

\dfrac{22}{45}

\dfrac{89}{180}

\dfrac{1}{2}

\dfrac{91}{180}

Answer: E

Concepts:trigonometric identity mean symmetry

Difficulty rating: 1610

Small hint:

Use

\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}

to rewrite the sum

Big hint:

The cosine terms

\cos 2^\circ, \cos 4^\circ, \ldots, \cos 180^\circ

cancel in pairs via

\cos(180^\circ - \theta) = -\cos\theta,

leaving only

\cos 180^\circ = -1

Solution:

Using

\sin^2\theta = \dfrac{1 - \cos 2\theta}{2},

\begin{aligned} \sum_{n=1}^{90} \sin^2(n^\circ) &= \frac{90}{2} \\ &\quad {}- \frac12 \sum_{n=1}^{90}\cos(2n^\circ). \end{aligned}

In the cosine sum, the terms for

n

and

90 - n

satisfy

\cos(2n^\circ) + \cos(180^\circ - 2n^\circ) = 0,

and

\cos 90^\circ = 0,

so everything cancels except

\cos 180^\circ = -1.

Hence the sum is

45 - \tfrac12(-1) = 45.5,

and the mean is

\dfrac{45.5}{90} = \dfrac{91}{180}.

Thus, the correct answer is E.

Problem 11 in other years