2008 AMC 12A Problems/Problem 9

Problem

Older television screens have an aspect ratio of $4: 3$ . That is, the ratio of the width to the height is $4: 3$ . The aspect ratio of many movies is not $4: 3$ , so they are sometimes shown on a television screen by "letterboxing" - darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of $2: 1$ and is shown on an older television screen with a $27$ -inch diagonal. What is the height, in inches, of each darkened strip?

$[asy] unitsize(1mm); filldraw((0,0)--(21.6,0)--(21.6,2.7)--(0,2.7)--cycle,grey,black); filldraw((0,13.5)--(21.6,13.5)--(21.6,16.2)--(0,16.2)--cycle,grey,black); draw((0,0)--(21.6,0)--(21.6,16.2)--(0,16.2)--cycle); [/asy]$

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 2.25 \qquad \textbf{(C)}\ 2.5 \qquad \textbf{(D)}\ 2.7 \qquad \textbf{(E)}\ 3$

Solution

Let the width and height of the screen be $4x$ and $3x$ respectively, and let the width and height of the movie be $2y$ and $y$ respectively.

By the Pythagorean Theorem, the diagonal is $\sqrt{(3x)^2+(4x)^2}=5x = 27$ . So $x=\frac{27}{5}$ .

Since the movie and the screen have the same width, $2y = 4x \Rightarrow y=2x$ .

Thus, the height of each strip is $\frac{3x-y}{2} = \frac{3x-2x}{2} = \frac{x}{2} = \frac{27}{10} = 2.7 \Rightarrow D$ .

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2008 AMC 12A Problems/Problem 9

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