1987 AMC 12 第 18 题

Question

某书架恰好可由

A

本代数书（厚度均相同）与

H

本几何书（厚度均相同，且比代数书厚）共同放满。另有

S

本代数书与

M

本几何书也能放满同一书架。最后，单独用

E

本代数书也能放满这个书架。已知

A

、

H

、

S

、

M

、

E

是互不相同的正整数，则

E

为

It takes

A

algebra books (all the same thickness) and

H

geometry books (all the same thickness, which is greater than that of an algebra book) to completely fill a certain shelf. Also,

S

of the algebra books and

M

of the geometry books would fill the same shelf. Finally,

E

of the algebra books alone would fill this shelf. Given that

A,

H,

S,

M,

E

are distinct positive integers, it follows that

E

is

\frac{AM+SH}{M+H}

\frac{AM^2+SH^2}{M^2+H^2}

\frac{AH-SM}{M-H}

\frac{AM-SH}{M-H}

\frac{AM^2-SH^2}{M^2-H^2}

答案：D

知识点：linear equations rates and mixtures algebraic elimination

难度评级：1980

小提示：

设两种书的厚度分别为

a

和

g

，并将书架长度定为

1

Let

a

and

g

be the two book thicknesses and normalize the shelf length to

1

大提示：

从

Aa+Hg=1

与

Sa+Mg=1

中消去几何书的厚度

Eliminate the geometry-book thickness from

Aa+Hg=1

and

Sa+Mg=1

解答：

令书架长度为

1

，代数书和几何书的厚度分别为

a,g

。则

\begin{aligned} Aa+Hg&=1,\\ Sa+Mg&=1,\qquad Ea=1 \end{aligned}\text{。}

将第一个方程乘以

M

，第二个方程乘以

H

，再相减，得到

(AM-SH)a=M-H\text{。}

因此

E=\frac1a=\frac{AM-SH}{M-H}\text{。}

所以正确答案是 D。

Let the shelf length be

1,

and let

a,g

be the algebra- and geometry-book thicknesses. Then

\begin{aligned} Aa+Hg&=1,\\ Sa+Mg&=1,\qquad Ea=1. \end{aligned}

Multiplying the first equation by

M,

the second by

H,

and subtracting gives

(AM-SH)a=M-H.

Hence

E=\frac1a=\frac{AM-SH}{M-H}.

Thus the correct answer is D.

1987 AMC 12 第 18 题

答案：D

难度评级：1980

小提示：

大提示：

解答：

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