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 
. That is, the ratio of the width to the height is . The aspect ratio of many movies is not , 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 
and is shown on an older television screen with a 
-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]](http://latex.artofproblemsolving.com/0/e/0/0e0eb295ca042287f3d625b9ac6eed49532ef1ab.png)
Solution
Let the width and height of the screen be 
and 
respectively, and let the width and height of the movie be 
and 
respectively.
By the Pythagorean Theorem, the diagonal is 
. So 
.
Since the movie and the screen have the same width, 
.
Thus, the height of each strip is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
