2007 AMC 12B Problem 19

Question

Rhombus

ABCD,

with side length

6,

is rolled to form a cylinder of volume

6

by taping

\overline{AB}

to

\overline{DC}.

What is

\sin(\angle ABC)?

\dfrac{\pi}{9}

\dfrac{1}{2}

\dfrac{\pi}{6}

\dfrac{\pi}{4}

\dfrac{\sqrt{3}}{2}

Taping

\overline{AB}

to

\overline{DC}

makes a side of length

6

the base circumference, so the radius is

\frac{3}{\pi}

The cylinder’s height is the rhombus altitude

6\sin\theta;

set the volume equal to

6

Let

\theta=\angle ABC.

The base circle has circumference

6,

so its radius is

\dfrac{6}{2\pi}=\dfrac3\pi.

The height of the cylinder is the rhombus altitude

6\sin\theta.

The volume is

\pi\left(\dfrac3\pi\right)^2(6\sin\theta)=\dfrac{54}{\pi}\sin\theta=6,

so

\sin\theta=\dfrac{\pi}{9}.

Thus, the correct answer is A.

Problem 19 in other years