2008 AMC 12A Problems/Problem 9

The following problem is from both the 2008 AMC 12A #9 and 2008 AMC 10A #14, so both problems redirect to this page.

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

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 $\frac{3x-y}{2}=\frac{3x-2x}{2}=\frac{x}{2}=\frac{27}{10}=2.7\Longrightarrow\mathrm{(D)}$.

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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