Difference between revisions of "1973 IMO Problems/Problem 5"
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Using our first observation, <math>\frac{a_2 b_1 + b_2 - a_1 b_2 - b_1}{a_1 a_2}=0</math>. Rearranging, we get <math>\frac{b_1}{a_1-1}=\frac{b_2}{a_2-1}</math>. Therefore, the fixed point of <math>f_1</math> equals the fixed point of <math>f_2</math>. Since we made no assumptions about <math>f_1</math> and <math>f_2</math>, this is true for all <math>f</math> in <math>G</math>. | Using our first observation, <math>\frac{a_2 b_1 + b_2 - a_1 b_2 - b_1}{a_1 a_2}=0</math>. Rearranging, we get <math>\frac{b_1}{a_1-1}=\frac{b_2}{a_2-1}</math>. Therefore, the fixed point of <math>f_1</math> equals the fixed point of <math>f_2</math>. Since we made no assumptions about <math>f_1</math> and <math>f_2</math>, this is true for all <math>f</math> in <math>G</math>. | ||
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+ | Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln735.html |
Revision as of 19:17, 25 October 2014
is a set of non-constant functions of the real variable
of the form
and
has the following properties:
(a) If and
are in
, then
is in
; here
.
(b) If is in
, then its inverse
is in
; here the inverse of
is
.
(c) For every in
, there exists a real number
such that
.
Prove that there exists a real number such that
for all
in
.
Solution
First, observe that for each function in
, if
then
. This is a result of (c); for example,
could not be in
because it does not have a fixed point. Or if
, then every point is a fixed point.
Also, for each function in
, if
then the fixed point of
is where
intersects
, namely where
.
Now, take and
, both in
. By (a),
and
must also both be in
. By (b),
must also be in
. Finally, by (a),
must also be in .
Using our first observation, . Rearranging, we get
. Therefore, the fixed point of
equals the fixed point of
. Since we made no assumptions about
and
, this is true for all
in
.
Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln735.html