Difference between revisions of "1997 AIME Problems/Problem 12"
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=== Solution 5 === | === 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 | + | 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>. |
== See also == | == See also == |
Revision as of 06:46, 15 August 2018
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
.
Contents
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
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
.
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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