AMC12 2008 A

AMC12 2008 A · Q9

AMC12 2008 A · Q9. 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?

老式电视屏幕的长宽比为 $4: 3$，即宽与高之比为 $4: 3$。许多电影的长宽比不是 $4: 3$，因此有时会用“信箱式画面”(letterboxing) 在电视屏幕上播放——在屏幕顶部和底部各加一条等高的黑边，如图所示。若某电影的长宽比为 $2: 1$，并在一台对角线为 $27$ 英寸的老式电视上播放，则每条黑边的高度（英寸）是多少？

(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

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

设屏幕的宽和高分别为 $4x$ 和 $3x$，电影画面的宽和高分别为 $2y$ 和 $y$。由勾股定理，对角线为 $\sqrt{(3x)^2+(4x)^2}=5x = 27$，所以 $x=\frac{27}{5}$。由于电影与屏幕宽度相同，$2y=4x\Rightarrow y=2x$。因此每条黑边的高度为 $\frac{3x-y}{2}=\frac{3x-2x}{2}=\frac{x}{2}=\frac{27}{10}=2.7\Longrightarrow\mathrm{(D)}$。

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