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Difference between revisions of "2008 AMC 12A Problems/Problem 9"

(New page: ==Problem== Older television screens have an aspect ratio of <math>4: 3</math>. That is, the ratio of the width to the height is <math>4: 3</math>. The aspect ratio of many movies is not ...)
 
(Duplicate, AMC 10 box, style, answer choices)
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{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #9]] and [[2008 AMC 10A Problems/Problem 14|2008 AMC 10A #14]]}}
 
==Problem==  
 
==Problem==  
 
Older television screens have an aspect ratio of <math>4: 3</math>. That is, the ratio of the width to the height is <math>4: 3</math>. The aspect ratio of many movies is not <math>4: 3</math>, 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 <math>2: 1</math> and is shown on an older television screen with a <math>27</math>-inch diagonal. What is the height, in inches, of each darkened strip?
 
Older television screens have an aspect ratio of <math>4: 3</math>. That is, the ratio of the width to the height is <math>4: 3</math>. The aspect ratio of many movies is not <math>4: 3</math>, 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 <math>2: 1</math> and is shown on an older television screen with a <math>27</math>-inch diagonal. What is the height, in inches, of each darkened strip?
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<center>
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<asy>unitsize(1mm);
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<asy>
 
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unitsize(1mm);
 
 
filldraw((0,0)--(21.6,0)--(21.6,2.7)--(0,2.7)--cycle,grey,black);
 
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);
 
filldraw((0,13.5)--(21.6,13.5)--(21.6,16.2)--(0,16.2)--cycle,grey,black);
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draw((0,0)--(21.6,0)--(21.6,16.2)--(0,16.2)--cycle);
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draw((0,0)--(21.6,0)--(21.6,16.2)--(0,16.2)--cycle);</asy>
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</asy>
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<math>\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 2.25\qquad\mathrm{(C)}\ 2.5\qquad\mathrm{(D)}\ 2.7\qquad\mathrm{(E)}\ 3</math>
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</center>
 
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<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 2.25 \qquad \textbf{(C)}\ 2.5 \qquad \textbf{(D)}\ 2.7 \qquad \textbf{(E)}\ 3</math>
 
  
 
==Solution==  
 
==Solution==  
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By the [[Pythagorean Theorem]], the diagonal is <math>\sqrt{(3x)^2+(4x)^2}=5x = 27</math>. So <math>x=\frac{27}{5}</math>.  
 
By the [[Pythagorean Theorem]], the diagonal is <math>\sqrt{(3x)^2+(4x)^2}=5x = 27</math>. So <math>x=\frac{27}{5}</math>.  
  
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Since the movie and the screen have the same width, <math>2y = 4x \Rightarrow y=2x</math>.  
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Since the movie and the screen have the same width, <math>2y=4x\Rightarrow y=2x</math>.  
  
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Thus, the height of each strip is <math>\frac{3x-y}{2} = \frac{3x-2x}{2} = \frac{x}{2} = \frac{27}{10} = 2.7 \Rightarrow D</math>.  
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Thus, the height of each strip is <math>\frac{3x-y}{2}=\frac{3x-2x}{2}=\frac{x}{2}=\frac{27}{10}=2.7\Longrightarrow\mathrm{(D)}</math>.  
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=A|num-b=8|num-a=10}}
 
{{AMC12 box|year=2008|ab=A|num-b=8|num-a=10}}
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{{AMC10 box|year=2008|ab=A|num-b=13|num-a=15}}

Revision as of 23:57, 25 April 2008

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 (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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