Difference between revisions of "1997 AIME Problems/Problem 12"
(→Solution 1) |
(a slightly more sophisticated way of abbreviating algebraic manipulations) |
||
Line 5: | Line 5: | ||
== 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 < | + | 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 < | + | 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>. | ||
Line 16: | Line 20: | ||
(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{ | + | <cmath>\begin{align*} |
− | 17 &= | + | 17 &= \frac{b - \frac da}{17 - e} + e\\ |
− | 97 &= | + | 97 &= \frac{b - \frac da}{97 - e} + e\\ |
− | b - \frac da &= | + | b - \frac da &= (17 - e)^2 = (97 - e)^2\\ |
− | 17 - e &= | + | 17 - e &= \pm (97 - e) |
− | \end{ | + | \end{align*}</cmath> |
− | Clearly we can discard the positive root, so <math>e = \ | + | 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 defined by , where ,, and are nonzero real numbers, has the properties , and for all values except . Find the unique number that is not in the range of .
Solution
Solution 1
First, we use the fact that for all in the domain. Substituting the function definition, we have , which reduces to In order for this fraction to reduce to , we must have and . From , we get or . The second cannot be true, since we are given that are nonzero. This means , so .
The only value that is not in the range of this function is . To find , we use the two values of the function given to us. We get and . Subtracting the second equation from the first will eliminate , and this results in , 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 by using the fact that .
Solution 2
First, we note that 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 will be . . Without loss of generality, let , so the function becomes .
(Considering as a limit) By the given, . , so . as reaches the vertical asymptote, which is at . Hence . Substituting the givens, we get
Clearly we can discard the positive root, so .
Solution 3
We first note (as before) that the number not in the range of is , as is evidently never 0 (otherwise, would be a constant function, violating the condition ).
We may represent the real number as , with two such column vectors considered equivalent if they are scalar multiples of each other. Similarly, we can represent a function as a matrix . Function composition and evaluation then become matrix multiplication.
Now in general, In our problem . It follows that for some nonzero real . Since it follows that . (In fact, this condition condition is equivalent to the condition that for all in the domain of .)
We next note that the function evaluates to 0 when equals 19 and 97. Therefore Thus , so , our answer.
See also
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 | ||
All AIME Problems and Solutions |