2021 AMC 12B Fall Problem 13

Question

Let

c = \dfrac{2\pi}{11}.

What is the value of

\small \dfrac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?

-1

-\dfrac{\sqrt{11}}{5}

\dfrac{\sqrt{11}}{5}

\dfrac{10}{11}

1

Each angle is a multiple of

\dfrac{2\pi}{11}

; reduce the numerator angles modulo

2\pi

Use

\sin(2\pi - x) = -\sin x

to relate leftover numerator factors to denominator factors

Write each angle as

kc = \dfrac{2\pi k}{11}.

Reducing modulo

2\pi,

\sin 12c = \sin c

and

\sin 15c = \sin 4c.

So the numerator is

\sin 3c \cdot \sin 6c \cdot \sin 9c

\cdot \sin c \cdot \sin 4c.

Cancelling the common factors

\sin c,

\sin 3c,

\sin 4c

leaves

\dfrac{\sin 6c \cdot \sin 9c}{\sin 2c \cdot \sin 5c}.

Now

\sin 9c = \sin\left(2\pi - 2c\right) = -\sin 2c

and

\sin 6c = \sin\left(2\pi - 5c\right) = -\sin 5c,

so the ratio equals

\dfrac{(-\sin 5c)(-\sin 2c)}{\sin 2c \cdot \sin 5c} = 1.

Thus, the correct answer is E.

Problem 13 in other years