# Difference between revisions of "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 $f$.

## Solution

### Solution 1

First, we use the fact that $f(f(x)) = x$ for all $x$ in the domain. Substituting the function definition, we have ${\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x}$, which reduces to ${\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} = \frac {ex + f}{gx + h} = x}$. In order for this fraction to reduce to $x$, we must have $f = g = 0$ and $e = h\not = 0$. From $c(a + d) = b(a + d) = 0$, we get $a = - d$ or $b = c = 0$. The second cannot be true, since we are given that $a,b,c,d$ are nonzero. This means $a = - d$, so $f(x) = \frac {ax + b}{cx - a}$.

The only value that is not in the range of this function is $\lim_{x\to - d/c}f(x) = \frac {a}{c}$. 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 ${\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = \boxed{058}$ (Error compiling LaTeX. ! Missing } inserted.).

### Solution 2

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 $\frac{b- \frac{d}{a}}{x+d} + e$.

(Considering $\infty$ as a limit) By the given, $f(f(\infty)) = \infty$. $\lim_{x \rightarrow \infty} f(x) = e$, so $f(e) = \infty$. $f(x) \rightarrow \infty$ as $x$ reaches the vertical asymptote, which is at $-\frac{d}{c} = -d$. Hence $e = -d$. Substituting the givens, we get $\begin{eqnarray*} 17 &=& \frac{b - \frac da}{17 - e} + e\\ 97 &=& \frac{b - \frac da}{97 - e} + e\\ b - \frac da &=& (17 - e)^2 = (97 - e)^2\\ 17 - e &=& \pm (97 - e) \end{eqnarray*}$

Clearly we can discard the positive root, so $e = \boxed{58}$.