2008 AMC 10A Problem 14

1. First find the size of the screen. A width-to-height ratio of $4:3$ means the width, height and diagonal form a right triangle shaped like the $3$-$4$-$5$ triangle.
2. So the width, height and diagonal are $4$, $3$ and $5$ times some number $k$. The diagonal $5k$ is $27$, so $k = 27 \div 5 = 5.4$.
3. The screen is $4 \times 5.4 = 21.6$ inches wide and $3 \times 5.4 = 16.2$ inches tall.
4. The movie uses the full width of the screen, $21.6$ inches. Its width is twice its height, so its height is
$$21.6 \div 2 = 10.8 \text{ inches}.$$
5. The rest of the screen's height is dark: $16.2 - 10.8 = 5.4$ inches.
6. That dark height is split into two equal strips, one on top and one on the bottom:
$$5.4 \div 2 = 2.7 \text{ inches}.$$

The answer is $\boxed{\textbf{(D)}\ 2.7}$.

A ratio of sides plus a diagonal is a Pythagorean-theorem problem, and $4:3$ is the familiar $3$-$4$-$5$ triangle, so the diagonal gives the real sizes quickly. After that, notice which length the movie and the screen share (the width) and split the leftover height in two.

Stay with $k$ as long as possible. The screen is $4k$ wide and $3k$ tall. The movie is $4k$ wide, so it is $2k$ tall. The two strips together are $3k - 2k = k$ tall, so each strip is $\frac{k}{2}$. Since $5k = 27$, $k = 5.4$ and each strip is $2.7$.

A screen with width-to-height ratio $4:3$ has width $4k$, height $3k$ and diagonal $5k$ (a $3$-$4$-$5$ triangle). The diagonal is $27$, so $k = 5.4$: the screen is $21.6$ inches wide and $16.2$ inches tall.

The movie fills the full width $21.6$ and has aspect ratio $2:1$, so its height is $\frac{21.6}{2} = 10.8$ inches.

The leftover height is $16.2 - 10.8 = 5.4$ inches, shared equally by the top and bottom strips:
$$\frac{5.4}{2} = 2.7.$$

The answer is $\boxed{\textbf{(D)}\ 2.7}$.