Difference between revisions of "2008 IMO Problems/Problem 2"
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Expressing <math>\alpha</math> from the first equation and substituting into the second, we get | Expressing <math>\alpha</math> from the first equation and substituting into the second, we get | ||
<cmath>\beta\gamma + ( \beta+\gamma ) ( -1 - \beta -\gamma ) = 0</cmath> | <cmath>\beta\gamma + ( \beta+\gamma ) ( -1 - \beta -\gamma ) = 0</cmath> | ||
− | as the sole | + | as the sole condition we need to satisfy in rational numbers. |
If <math>\beta = \frac{b}{m}</math> and <math>\gamma = \frac{c}{m}</math> for some integers <math>b</math>,<math>c</math>,and <math>m</math>, they would need to satisfy | If <math>\beta = \frac{b}{m}</math> and <math>\gamma = \frac{c}{m}</math> for some integers <math>b</math>,<math>c</math>,and <math>m</math>, they would need to satisfy | ||
<cmath>bc = m(b+c)+(b+c)^2 \Leftrightarrow m = \frac{bc}{b+c} - (b+c).</cmath> | <cmath>bc = m(b+c)+(b+c)^2 \Leftrightarrow m = \frac{bc}{b+c} - (b+c).</cmath> | ||
− | + | For <math>m</math> to be integer, we would like <math>b+c</math> to divide <math>bc</math>. | |
Consider the example | Consider the example | ||
<cmath>b=t, c=t^2-t, m = t-1-t^2,</cmath> | <cmath>b=t, c=t^2-t, m = t-1-t^2,</cmath> | ||
where <math>b+c = t^2</math> divides <math>bc = t(t^2-t)</math> for any integer <math>t \ne 0</math>. Substituting back, that gives us | where <math>b+c = t^2</math> divides <math>bc = t(t^2-t)</math> for any integer <math>t \ne 0</math>. Substituting back, that gives us | ||
− | <cmath>\beta = \frac{t}{t-1-t^2}, \gamma = \frac{t^2- | + | <cmath>\beta = \frac{t}{t-1-t^2},\quad |
+ | \gamma = \frac{t^2-t}{t-1-t^2},\quad | ||
+ | \alpha = \frac{1-t}{t-1-t^2}.</cmath> | ||
+ | A simple check shows that <math>\alpha,\beta,\gamma</math> are rational and well defined and that <math>p=-1</math> and <math>q=0</math> for ''any'' integer <math>t</math> (even for <math>t=0</math>). | ||
+ | |||
+ | Moreover, from <math>\lim_{t\rightarrow +\infty} \beta = 0</math> and <math>\beta < 0</math> for large <math>t</math>, we see that infinitely many <math>t</math> generate infinitely many ''different'' triplets of <math>\alpha</math>, <math>\beta</math>, and <math>\gamma</math>. That completes the proof of part '''(ii)'''. |
Revision as of 22:35, 4 September 2008
Problem 2
(i) If ,
and
are three real numbers, all different from
, such that
, then prove that
.
(With the
sign for cyclic summation, this inequality could be rewritten as
.)
(ii) Prove that equality is achieved for infinitely many triples of rational numbers ,
and
.
Solution
Consider the transormation defined by
and put
. Since
is also one-to one from
to
, the problem is equivalent to showing that
subject to
and that equallity holds for infinitely many triplets of rational
.
Now, rewrite (2) as and express it as
where
and
. Notice that (1) can be written as
But from
, we get
with equality holding iff
. That proves part (i) and points us in the direction of looking for rational
for which
and (hence)
, that is:
Expressing
from the first equation and substituting into the second, we get
as the sole condition we need to satisfy in rational numbers.
If and
for some integers
,
,and
, they would need to satisfy
For
to be integer, we would like
to divide
.
Consider the example
where
divides
for any integer
. Substituting back, that gives us
A simple check shows that
are rational and well defined and that
and
for any integer
(even for
).
Moreover, from and
for large
, we see that infinitely many
generate infinitely many different triplets of
,
, and
. That completes the proof of part (ii).