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

(Solution 1)
(a slightly more sophisticated way of abbreviating algebraic manipulations)
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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== 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 <math>{\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} = \frac {ex + f}{gx + h} = x}</math>. In order for this fraction to reduce to <math>x</math>, we must have <math>f = g = 0</math> and <math>e = h\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.
+
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 {ex + f}{gx + h} = x}. </cmath>
 +
In order for this fraction to reduce to <math>x</math>, we must have <math>f = g = 0</math> and <math>e = h\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>.
  
The only value that is not in the range of this function is <math>\lim_{x\to \infty}f(x) = \frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so <math>{\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = \boxed{058}</math>.
+
The only value that is not in the range of this function is <math>\lim_{x\to \infty}f(x) = \frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so
 +
<cmath>{\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 . </cmath>
  
 
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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(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get  
 
(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[Asymptote (Geometry)|asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get  
  
<cmath>\begin{eqnarray*}
+
<cmath>\begin{align*}
17 &=& \frac{b - \frac da}{17 - e} + e\\
+
17 &= \frac{b - \frac da}{17 - e} + e\\
97 &=& \frac{b - \frac da}{97 - e} + e\\
+
97 &= \frac{b - \frac da}{97 - e} + e\\
b - \frac da &=& (17 - e)^2 = (97 - e)^2\\
+
b - \frac da &= (17 - e)^2 = (97 - e)^2\\
17 - e &=& \pm (97 - e)
+
17 - e &= \pm (97 - e)
\end{eqnarray*}</cmath>
+
\end{align*}</cmath>
  
Clearly we can discard the positive root, so <math>e = \boxed{58}</math>.
+
Clearly we can discard the positive root, so <math>e = 58</math>.
 +
 
 +
=== Solution 3 ===
 +
<!-- some linear algebra -->
 +
We first note (as before) that the number not in the range of
 +
<cmath> f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d} </cmath>
 +
is <math>a/c</math>, as <math>\frac{b-ad/c}{cx+d}</math> is evidently never 0 (otherwise, <math>f</math>
 +
would be a constant function, violating the condition <math>f(19) \neq f(97)</math>).
 +
 
 +
We may represent the real number <math>x/y</math> as
 +
<math>\begin{pmatrix}x \\ y\end{pmatrix}</math>, with two such [[column vectors]]
 +
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
 +
<math>\begin{pmatrix} A & B\\ C& D \end{pmatrix}</math>.  Function composition and
 +
evaluation then become matrix multiplication.
 +
 
 +
Now in general,
 +
<cmath>  f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} =
 +
\frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .</cmath>
 +
In our problem <math>f^2(x) = x</math>.  It follows that
 +
<cmath> \begin{pmatrix} a & b \\ c& d \end{pmatrix} = K
 +
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} , </cmath>
 +
for some nonzero real <math>K</math>.  Since
 +
<cmath> \frac{a}{d} = \frac{b}{-b} = K, </cmath>
 +
it follows that <math>a = -d</math>.  (In fact, this condition condition is equivalent
 +
to the condition that <math>f(f(x)) = x</math> for all <math>x</math> in the domain of <math>f</math>.)
 +
 
 +
We next note that the function
 +
<cmath> g(x) =  x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d} </cmath>
 +
evaluates to 0 when <math>x</math> equals 19 and 97.  Therefore
 +
<cmath> \frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}. </cmath>
 +
Thus <math>-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}</math>, so <math>a/c = (19+97)/2 = 58</math>,
 +
our answer. <math>\blacksquare</math>
  
 
== See also ==
 
== See also ==

Revision as of 01:12, 17 May 2009

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 $\lim_{x\to \infty}f(x) = \frac {a}{c}$. 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

\[{\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 .\] (Error compiling LaTeX. ! Missing } inserted.)

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*} 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{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. $\blacksquare$

See also

1997 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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