1987 AMC 12 Problem 18

Question

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}

Let

a

and

g

be the two book thicknesses and normalize the shelf length to

1

Eliminate the geometry-book thickness from

Aa+Hg=1

and

Sa+Mg=1

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.

Problem 18 in other years