1967 AMC 12 第 19 题

Question

一个矩形加长

2\dfrac12

英寸并变窄

\dfrac23

英寸，或缩短

2\dfrac12

英寸并变宽

\dfrac43

英寸时，面积都保持不变。它的面积（单位：平方英寸）为：

The area of a rectangle remains unchanged when it is made

2\dfrac12

inches longer and

\dfrac23

inch narrower, or when it is made

2\dfrac12

inches shorter and

\dfrac43

inch wider. Its area, in square inches, is:

30

\dfrac{80}{3}

24

\dfrac{45}{2}

20

设原来的长和宽为

l

和

w

Let the original length and width be

l

and

w

展开两个面积相等的方程；

lw

项会相消

Expand both equal-area equations; the

lw

terms cancel

两个条件为

\begin{aligned} lw&=\left(l+\frac52\right)\left(w-\frac23\right),\\ lw&=\left(l-\frac52\right)\left(w+\frac43\right) \end{aligned}\text{。}

展开后，得到

-4l+15w=10

与

8l-15w=20

。因此

l=\frac{15}{2}

、

w=\frac{8}{3}

，且

lw=20

。

因此，正确答案是 E。

The two conditions are

\begin{aligned} lw&=\left(l+\frac52\right)\left(w-\frac23\right),\\ lw&=\left(l-\frac52\right)\left(w+\frac43\right). \end{aligned}

After expanding, these give

-4l+15w=10

and

8l-15w=20.

Hence

l=\frac{15}{2},

w=\frac{8}{3},

and

lw=20.

Therefore, the correct answer is E.

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