AMC10 2008 A

AMC10 2008 A · Q14

AMC10 2008 A · Q14. It mainly tests Pythagorean theorem, Coordinate geometry.

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?

老式电视屏幕的宽高比是4:3，即宽度与高度之比为4:3。许多电影的宽高比不是4:3，因此有时通过“画中画”方式在电视屏幕上显示——在屏幕顶部和底部加等高的黑色条带，如图所示。假设一部电影宽高比为2:1，在对角线为27英寸的老式电视屏幕上播放。每条黑色条带的高度是多少英寸？

(A) 2 2

(B) 2.25 2.25

(D) 2.7 2.7

(E) 3 3

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $h$ and $w$ be the height and width of the screen, respectively, in inches. By the Pythagorean Theorem, $h:w:27=3:4:5$, so $$ h=\frac{3}{5}\cdot 27=16.2 \quad \text{and} \quad w=\frac{4}{5}\cdot 27=21.6. $$ The height of the non-darkened portion of the screen is half the width, or $10.8$ inches. Therefore the height of each darkened strip is $$ \frac{1}{2}(16.2-10.8)=2.7 \text{ inches.} $$

答案（D）：设屏幕的高和宽分别为 $h$ 和 $w$（单位：英寸）。由勾股定理，$h:w:27=3:4:5$，所以 $$ h=\frac{3}{5}\cdot 27=16.2 \quad \text{且} \quad w=\frac{4}{5}\cdot 27=21.6. $$ 屏幕未变暗部分的高度是宽度的一半，即 $10.8$ 英寸。因此，每条变暗带的高度为 $$ \frac{1}{2}(16.2-10.8)=2.7 \text{ 英寸。} $$

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