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Difference between revisions of "1997 AIME Problems/Problem 12"

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=== Solution 6 (Shortest) ===
 
=== Solution 6 (Shortest) ===
  
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From <math>f(f(x))=x</math>, it is obvious that <math>\frac{-d}{c}</math> is the value not in the range. First notice that since <math>f(0)=\frac{b}{d}</math>, <math>f(\frac{b}{d})=a(\frac{b}{d})+b=0</math> so <math>a=-d</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we obtain <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>, so the answer is <math>\boxed{058}</math>. - mathleticguyyy
+
From <math>f(f(x))=x</math>, it is obvious that <math>\frac{-d}{c}</math> is the value not in the range. First notice that since <math>f(0)=\frac{b}{d}</math>, <math>f(\frac{b}{d})=a(\frac{b}{d})+b=0</math> so <math>a=-d</math>. Using <math>f(19)=19</math>, we have that <math>b=361c+38d</math>; on <math>f(97)=97</math> we obtain <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>, so the answer is <math>\boxed{058}</math>. - mathleticguyyy
  
 
=== Solution 7===
 
=== Solution 7===

Revision as of 12:47, 19 October 2019

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 $(97,97)$. $f$ 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 $\frac{1}{x-k}$.

Solution 6 (Shortest)

From $f(f(x))=x$, it is obvious that $\frac{-d}{c}$ is the value not in the range. First notice that since $f(0)=\frac{b}{d}$, $f(\frac{b}{d})=a(\frac{b}{d})+b=0$ so $a=-d$. Using $f(19)=19$, we have that $b=361c+38d$; on $f(97)=97$ we obtain $b=9409c+194d$. Solving for $d$ in terms of $c$ leads us to $d=-58c$, so the answer is $\boxed{058}$. - mathleticguyyy

Solution 7

Begin by finding the inverse function of $f(x)$, which turns out to be $f^{-1}(x)=\frac{19d-b}{a-19c}$. Since $f(f(x))=x$, $f(x)=f^{-1}(x)$, so substituting 19 and 97 yields the system, $\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}$, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get $116c=a-d$. Coincidentally, then $116c+d=a$, which is familiar because $f(116)=\frac{116a+b}{116c+d}$, and since $116c+d=a$, $f(116)=\frac{116a+b}{a}$. Also, $f(f(116))=\frac{a(\frac{116a+b}{a})+b}{c(\frac{116a+b}{a})+d}=116$, due to $f(f(x))=x$. This simplifies to $\frac{116a+2b}{c(\frac{116a+b}{a})+d}=116$, $116a+2b=116(c(\frac{116a+b}{a})+d)$, $116a+2b=116(c(116+\frac{b}{a})+d)$, $116a+2b=116c(116+\frac{b}{a})+116d$, and substituting $116c+d=a$ and simplifying, you get $2b=116c(\frac{b}{a})$, then $\frac{a}{c}=58$. Looking at $116c=a-d$ one more time, we get $116=\frac{a}{c}+\frac{-d}{c}$, and substituting, we get $\frac{-d}{c}=\boxed{58}$, and we are done.

See also

1997 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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