2008 AMC 12A Problem 9

Attempt Problem 9 of the 2008 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

9.

Older television screens have an aspect ratio of 4:3.4:3. That is, the ratio of the width to the height is 4:3.4:3. The aspect ratio of many movies is not 4:3,4:3, 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 2:12:1 and is shown on an older television screen with a 2727-inch diagonal. What is the height, in inches, of each darkened strip?

22

2.252.25

2.52.5

2.72.7

33

Answer: D
Concepts:Pythagorean Tripleratio and proportion
Difficulty rating: 1410
Solution:

Since the sides and diagonal are in ratio 3:4:5,3:4:5, the height is 3527=16.2\tfrac{3}{5} \cdot 27 = 16.2 inches and the width is 4527=21.6\tfrac{4}{5} \cdot 27 = 21.6 inches.

The movie has aspect ratio 2:1,2:1, so its height is 21.62=10.8\tfrac{21.6}{2} = 10.8 inches.

Each darkened strip therefore has height 16.210.82=2.7 \dfrac{16.2 - 10.8}{2} = 2.7 inches.

Thus, D is the correct answer.

← Problem 8#8
Full Exam

Problem 9 in Other Years