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 $\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x.$ In order for this fraction to reduce to $x$ , we must have $q = r = 0$ and $p = s\not = 0$ . 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 $\frac {a}{c}$ . 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 $f(f(-b/a))=-b/a$ .

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

$\begin{align*} 19 &= \frac{b - \frac da}{19 - e} + e\\ 97 &= \frac{b - \frac da}{97 - e} + e\\ b - \frac da &= (19 - e)^2 = (97 - e)^2\\ 19 - e &= \pm (97 - e) \end{align*}$

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

Solution 3

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.

Now in general, $f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} = \frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .$ In our problem $f^2(x) = x$ . It follows that $\begin{pmatrix} a & b \\ c& d \end{pmatrix} = K \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} ,$ for some nonzero real $K$ . Since $\frac{a}{d} = \frac{b}{-b} = K,$ it follows that $a = -d$ . (In fact, this condition condition is equivalent to the condition that $f(f(x)) = x$ for all $x$ in the domain of $f$ .)

We next note that the function $g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d}$ evaluates to 0 when $x$ equals 19 and 97. Therefore $\frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}.$ Thus $-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}$ , so $a/c = (19+97)/2 = 58$ , our answer.

Solution 4

Any number that is not in the domain of the inverse of $f(x)$ cannot be in the range of $f(x)$ . Starting with $f(x) = \frac{ax+b}{cx+d}$ , we rearrange some things to get $x = \frac{b-f(x)d}{f(x)c-a}$ . Clearly, $\frac{a}{c}$ is the number that is outside the range of $f(x)$ .

Since we are given $f(f(x))=x$ , we have that $x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}$ $cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)$ All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that $a = -d$ .

This solution follows in the same manner as the last paragraph of the first solution.

Solution 5

Since $f(f(x))$ is $x$ , it must be symmetric across the line $y=x$ . Also, since $f(19)=19$ , it must touch the line $y=x$ at $(19,19)$ and $(97,97)$ . $f{x}$ a hyperbola that is a scaled and transformed version of $y=\frac{1}{x}$ . Write $f(x)= \frac{ax+b}{cx+d}$ as $\frac{y}{cx+d}+z$ , and z is our desired answer $\frac{a}{c}$ . Take the basic hyperbola, $y=\frac{1}{x}$ . The distance between points $(1,1)$ and $(-1,-1)$ is $2\sqrt{2}$ , while the distance between $(19,19)$ and $(97,97)$ is $78\sqrt{2}$ , so it is $y=\frac{1}{x}$ scaled by a factor of $39$ . Then, we will need to shift it from $(-39,-39)$ to $(19,19)$ , shifting up by $58$ , or $z$ , so our answer is $\boxed{58}$ . Note that shifting the $x$ does not require any change from $z$ ; it changes the denominator of the part $\frac{1}{x-k}$ .

1997 AIME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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