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

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__TOC__
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== Problem ==
 
== Problem ==
 
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.
 
The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.
  
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__TOC__
+
== Solution 1 ==
−
== Solution ==
 
−
=== Solution 1 ===
 
 
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to
 
First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x</math>, which reduces to
 
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath>
 
<cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath>
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Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.
 
Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.
  
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=== Solution 2 ===
+
== Solution 2 ==
 
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>.   
 
First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>.   
  
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Clearly we can discard the positive root, so <math>e = 58</math>.
 
Clearly we can discard the positive root, so <math>e = 58</math>.
  
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=== Solution 3 ===
+
== Solution 3 ==
 
<!-- some linear algebra -->
 
<!-- some linear algebra -->
 
We first note (as before) that the number not in the range of
 
We first note (as before) that the number not in the range of
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We may represent the real number <math>x/y</math> as
 
We may represent the real number <math>x/y</math> as
−
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]
+
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[vector|column vectors]]
 
considered equivalent if they are scalar multiples of each other.  Similarly,
 
considered equivalent if they are scalar multiples of each other.  Similarly,
 
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix
 
we can represent a function <math>F(x) = \frac{Ax + B}{Cx + D}</math> as a matrix
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our answer.
 
our answer.
  
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=== Solution 4 ===
+
== Solution 4 ==
 
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.
 
Any number that is not in the domain of the inverse of <math>f(x)</math> cannot be in the range of <math>f(x)</math>. Starting with <math>f(x) = \frac{ax+b}{cx+d}</math>, we rearrange some things to get <math>x = \frac{b-f(x)d}{f(x)c-a}</math>. Clearly, <math>\frac{a}{c}</math> is the number that is outside the range of <math>f(x)</math>.
  
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This solution follows in the same manner as the last paragraph of the first solution.
 
This solution follows in the same manner as the last paragraph of the first solution.
  
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=== Solution 5 ===
+
== Solution 5 ==
 
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.
 
Since <math>f(f(x))</math> is <math>x</math>, it must be symmetric across the line <math>y=x</math>. Also, since <math>f(19)=19</math>, it must touch the line <math>y=x</math> at <math>(19,19)</math> and <math>(97,97)</math>. <math>f</math> a hyperbola that is a scaled and transformed version of <math>y=\frac{1}{x}</math>. Write <math>f(x)= \frac{ax+b}{cx+d}</math> as <math>\frac{y}{cx+d}+z</math>, and z is our desired answer <math>\frac{a}{c}</math>. Take the basic hyperbola, <math>y=\frac{1}{x}</math>. The distance between points <math>(1,1)</math> and <math>(-1,-1)</math> is <math>2\sqrt{2}</math>, while the distance between <math>(19,19)</math> and <math>(97,97)</math> is <math>78\sqrt{2}</math>, so it is <math>y=\frac{1}{x}</math> scaled by a factor of <math>39</math>. Then, we will need to shift it from <math>(-39,-39)</math> to <math>(19,19)</math>, shifting up by <math>58</math>, or <math>z</math>, so our answer is <math>\boxed{58}</math>. Note that shifting the <math>x</math> does not require any change from <math>z</math>; it changes the denominator of the part <math>\frac{1}{x-k}</math>.
  
−
=== Solution 6===
+
== Solution 6 (Short) ==
−
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+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 have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. 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})=0</math> which means <math>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>.
 +
~solution by mathleticguyyy
 +
 
 +
== Solution 7 ==
 +
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\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}</math>, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get <math>116c=a-d</math>. Coincidentally, then <math>116c+d=a</math>, which is familiar because <math>f(116)=\frac{116a+b}{116c+d}</math>, and since <math>116c+d=a</math>, <math>f(116)=\frac{116a+b}{a}</math>. Also, <math>f(f(116))=\frac{a(\frac{116a+b}{a})+b}{c(\frac{116a+b}{a})+d}=116</math>, due to <math>f(f(x))=x</math>. This simplifies to <math>\frac{116a+2b}{c(\frac{116a+b}{a})+d}=116</math>, <math>116a+2b=116(c(\frac{116a+b}{a})+d)</math>, <math>116a+2b=116(c(116+\frac{b}{a})+d)</math>, <math>116a+2b=116c(116+\frac{b}{a})+116d</math>, and substituting <math>116c+d=a</math> and simplifying, you get <math>2b=116c(\frac{b}{a})</math>, then <math>\frac{a}{c}=58</math>. Looking at <math>116c=a-d</math> one more time, we get <math>116=\frac{a}{c}+\frac{-d}{c}</math>, and substituting, we get <math>\frac{-d}{c}=\boxed{58}</math>, and we are done.
 +
 
 +
== Solution 8 (shorter than solution 6) ==
 +
Because there are no other special numbers other than <math>19</math> and <math>97</math>, take the average to get <math>\boxed{58}</math>. (Note I solved this problem the solution one way but noticed this and this probably generalizes to all <math>f(x)=x, f(y)=y</math> questions like these)
 +
 
 +
== Solution 9 (Simple)==
 +
By the function definition, <math>f(f(x))</math>, <math>f</math> is its own inverse, so the only value not in the range of <math>f</math> is the value not in the domain of <math>f</math> (which is <math>-d/c</math>).
 +
 
 +
Since <math>f(f(x))</math>, <math>f(f(0)=0</math> (0 is a convenient value to use).  
 +
<math>f(f(0))=f(f(\tfrac{b}{d})=\dfrac{a\cdot\tfrac{b}{d}+b}{c\cdot\tfrac{b}{d}+d}=\dfrac{ab+bd}{bc+d^2}=0 \Rightarrow ab+bd=0</math>.
 +
 
 +
Then
 +
<math>ab+bd=b(a+d)=0</math> and since <math>b</math> is nonzero, <math>a=-d</math>.
 +
 
 +
The answer we are searching for, <math>\dfrac{-d}{c}</math> (the only value not in the range of <math>f</math>), can now be expressed as <math>\dfrac{a}{c}</math>.
 +
 
 +
We are given <math>f(19)=19</math> and <math>f(97)</math>, and they satisfy the equation <math>f(x)=x</math>, which simplifies to <math>\dfrac{ax+b}{cx+d}=x\Rightarrow x(cx+d)=ax+b\Rightarrow cx^2+(d-a)x+b=0</math>. We have written this quadratic with roots <math>19</math> and <math>97</math>.
 +
 
 +
By Vieta, <math>\dfrac{-(d-a)}{c}=\dfrac{-(-a-a)}{c}=\dfrac{2a}{c}=19+97</math>.
 +
 
 +
So our answer is <math>\dfrac{116}{2}=\boxed{058}</math>.
 +
 
 +
~BakedPotato66
 +
 
 +
==Solution 9 (30-sec solve)==
 +
Notice that the function is just an [[involution]] on the real number line. Since the involution has two fixed points, namely <math>19</math> and <math>97</math>, we know that the involution is an [[circular inversion|inversion]] with respect to a circle with a diameter from <math>19</math> to <math>97</math>. The only point that is undefined under an inversion is the center of the circle, which we know is <math>\frac{19+97}{2}=\boxed{58}</math> in both <math>x</math> and <math>y</math> dimensions.
 +
 
 +
~kn07
 +
 
 +
 
 +
Or if you don't think about inversion: A linear rational function like this is <math>f= a/c + (b-d)/(cx+d)</math>, and so has asymptotes at <math>x=-d/c</math> and <math>y=a/c</math>, and these values must be equal because <math>f</math> is an "involution", its own inverse. (Reflecting over <math>x=y</math> does not change <math>f</math>).
  
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=== Solution 7===
+
By self-inverse symmetry, both asymptotes are equidistant to the graph points <math>(19,19)</math> and <math>(97,97)</math>, so they must intersect at the mean of <math>19</math> and <math>97</math>, which is <math>58</math>.
−
Begin by finding the inverse function of <math>f(x)</math>, which turns out to be <math>f^{-1}(x)=\frac{19d-b}{a-19c}</math>. Since <math>f(f(x))=x</math>, <math>f(x)=f^{-1}(x)</math>, so substituting 19 and 97 yields the system, <math>\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}</math>, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get <math>116c=a-d</math>. Coincidentally, then <math>116c+d=a</math>, which is familiar because <math>f(116)=\frac{116a+b}{116c+d}</math>, and since <math>116c+d=a</math>, <math>f(116)=\frac{116a+b}{a}</math>. Also, <math>f(f(116))=\frac{116(\frac{116a+b}{a})+b)}{116(\frac{116a+b}{a})+b)}</math>
 
  
 
== See also ==
 
== See also ==

Latest revision as of 16:53, 8 September 2024

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 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 (Short)

From $f(f(x))=x$, it is obvious that $\frac{-d}{c}$ is the value not in the range. First notice that since $f(0)=\frac{b}{d}$, $f(\frac{b}{d})=0$ which means $a(\frac{b}{d})+b=0$ 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 $\boxed{058}$. 

~solution by 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.

Solution 8 (shorter than solution 6)

Because there are no other special numbers other than $19$ and $97$, take the average to get $\boxed{58}$. (Note I solved this problem the solution one way but noticed this and this probably generalizes to all $f(x)=x, f(y)=y$ questions like these)

Solution 9 (Simple)

By the function definition, $f(f(x))$, $f$ is its own inverse, so the only value not in the range of $f$ is the value not in the domain of $f$ (which is $-d/c$).

Since $f(f(x))$, $f(f(0)=0$ (0 is a convenient value to use). $f(f(0))=f(f(\tfrac{b}{d})=\dfrac{a\cdot\tfrac{b}{d}+b}{c\cdot\tfrac{b}{d}+d}=\dfrac{ab+bd}{bc+d^2}=0 \Rightarrow ab+bd=0$.

Then $ab+bd=b(a+d)=0$ and since $b$ is nonzero, $a=-d$.

The answer we are searching for, $\dfrac{-d}{c}$ (the only value not in the range of $f$), can now be expressed as $\dfrac{a}{c}$.

We are given $f(19)=19$ and $f(97)$, and they satisfy the equation $f(x)=x$, which simplifies to $\dfrac{ax+b}{cx+d}=x\Rightarrow x(cx+d)=ax+b\Rightarrow cx^2+(d-a)x+b=0$. We have written this quadratic with roots $19$ and $97$.

By Vieta, $\dfrac{-(d-a)}{c}=\dfrac{-(-a-a)}{c}=\dfrac{2a}{c}=19+97$.

So our answer is $\dfrac{116}{2}=\boxed{058}$.

~BakedPotato66

Solution 9 (30-sec solve)

Notice that the function is just an involution on the real number line. Since the involution has two fixed points, namely $19$ and $97$, we know that the involution is an inversion with respect to a circle with a diameter from $19$ to $97$. The only point that is undefined under an inversion is the center of the circle, which we know is $\frac{19+97}{2}=\boxed{58}$ in both $x$ and $y$ dimensions.

~kn07


Or if you don't think about inversion: A linear rational function like this is $f= a/c + (b-d)/(cx+d)$, and so has asymptotes at $x=-d/c$ and $y=a/c$, and these values must be equal because $f$ is an "involution", its own inverse. (Reflecting over $x=y$ does not change $f$).

By self-inverse symmetry, both asymptotes are equidistant to the graph points $(19,19)$ and $(97,97)$, so they must intersect at the mean of $19$ and $97$, which is $58$.

See also

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

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