Difference between revisions of "1997 AIME Problems/Problem 12"
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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>. | ||
− | + | == 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>. | ||
− | + | == 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>. | ||
− | + | == 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. | ||
− | + | == 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. | ||
− | + | == 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 (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})=a(\frac{b}{d})+b=0</math> so <math>a=-d</math>. Using <math>f(19)=19</math>, we have that | + | 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. | 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 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>). | ||
+ | |||
+ | 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>. | ||
== See also == | == See also == |
Latest revision as of 22:59, 21 October 2023
Contents
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 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
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.
Solution 4
Any number that is not in the domain of the inverse of cannot be in the range of
. Starting with
, we rearrange some things to get
. Clearly,
is the number that is outside the range of
.
Since we are given , we have that
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
.
This solution follows in the same manner as the last paragraph of the first solution.
Solution 5
Since is
, it must be symmetric across the line
. Also, since
, it must touch the line
at
and
.
a hyperbola that is a scaled and transformed version of
. Write
as
, and z is our desired answer
. Take the basic hyperbola,
. The distance between points
and
is
, while the distance between
and
is
, so it is
scaled by a factor of
. Then, we will need to shift it from
to
, shifting up by
, or
, so our answer is
. Note that shifting the
does not require any change from
; it changes the denominator of the part
.
Solution 6 (Short)
From , it is obvious that
is the value not in the range. First notice that since
,
which means
so
. Using
, we have that
; on
we obtain
. Solving for
in terms of
leads us to
, so the answer is
.
~solution by mathleticguyyy
Solution 7
Begin by finding the inverse function of , which turns out to be
. Since
,
, so substituting 19 and 97 yields the system,
, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get
. Coincidentally, then
, which is familiar because
, and since
,
. Also,
, due to
. This simplifies to
,
,
,
, and substituting
and simplifying, you get
, then
. Looking at
one more time, we get
, and substituting, we get
, and we are done.
Solution 8 (shorter than solution 6)
Because there are no other special numbers other than and
, take the average to get
. (Note I solved this problem the solution one way but noticed this and this probably generalizes to all
questions like these)
Solution 9 (Simple)
By the function definition, ,
is its own inverse, so the only value not in the range of
is the value not in the domain of
(which is
).
Since ,
(0 is a convenient value to use).
.
Then
and since
is nonzero,
.
The answer we are searching for, (the only value not in the range of
), can now be expressed as
.
We are given and
, and they satisfy the equation
, which simplifies to
. We have written this quadratic with roots
and
.
By Vieta, .
So our answer is .
~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 and
, we know that the involution is an inversion with respect to a circle with a diameter from
to
. The only point that is undefined under an inversion is the center of the circle, which we know is
in both
and
dimensions.
~kn07
Or if you don't think about inversion: A linear rational function like this is , and so has asymptotes at
and
, and these values must be equal because
is an "involution", its own inverse. (Reflecting over
does not change
).
By self-inverse symmetry, both asymptotes are equidistant to the graph points and
, so they must intersect at the mean of
and
, which is
.
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 |
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