Difference between revisions of "1997 AIME Problems/Problem 12"
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− | 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>. | + | 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 | ~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 its own inverse ( | + | 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>. | 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>. |
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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