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

Revision as of 04:16, 21 July 2022

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

@@ Line 1: / Line 1: @@
+__TOC__
 == 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>.
-__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
 <cmath>\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\frac {px + q}{rx + s} = x. </cmath>
@@ Line 14: / Line 14: @@
 Alternatively, we could have found out that <math>a = -d</math> by using the fact that <math>f(f(-b/a))=-b/a</math>.
-=== 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>.
@@ Line 28: / Line 28: @@
 Clearly we can discard the positive root, so <math>e = 58</math>.
-=== Solution 3 ===
+== Solution 3 ==
 <!-- some linear algebra -->
 We first note (as before) that the number not in the range of
@@ Line 60: / Line 60: @@
 our answer.
-=== 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>.
@@ Line 70: / Line 70: @@
 This solution follows in the same manner as the last paragraph of the first solution.
-=== Solution 3 ===
+== 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>.
-=== Solution 6 (Short) ===
+== Solution 6 (Short) ==
 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===
+== 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) ===
+== 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)===
+== 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>).

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