1997 AIME Problems/Problem 12

Problem

The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$ , where $a$ , $b$ , $c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$ , $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$ . Find the unique number that is not in the range of $f$ .

The only value that is not in the range of this function is $\lim_{x\to \infty}f(x) = \frac {a}{c}$ . To find $\frac {a}{c}$ , we use the two values of the function given to us. We get $2(97)a + b = 97^2c$ and $2(19)a + b = 19^2c$ . Subtracting the second equation from the first will eliminate $b$ , and this results in $2(97 - 19)a = (97^2 - 19^2)c$ , so ${\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 .$

Alternatively, we could have found out that $a = -d$ by using the fact that $f(f(-b/a))=-b/a$ .

First, we note that $e = \frac ac$ is the horizontal asymptote of the function, and since this is a linear function over a linear function, the unique number not in the range of $f$ will be $e$ . $\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}$ . Without loss of generality, let $c=1$ , so the function becomes $\frac{b- \frac{d}{a}}{x+d} + e$ .

Clearly we can discard the positive root, so $e = 58$ .

We first note (as before) that the number not in the range of $f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d}$ is $a/c$ , as $\frac{b-ad/c}{cx+d}$ is evidently never 0 (otherwise, $f$ would be a constant function, violating the condition $f(19) \neq f(97)$ ).

We may represent the real number $x/y$ as $\begin{pmatrix}x \\ y\end{pmatrix}$ , with two such column vectors considered equivalent if they are scalar multiples of each other. Similarly, we can represent a function $F(x) = \frac{Ax + B}{Cx + D}$ as a matrix $\begin{pmatrix} A & B\\ C& D \end{pmatrix}$ . Function composition and evaluation then become matrix multiplication.