Difference between revisions of "2008 IMO Problems/Problem 2"
(→Solution) |
(→Solution) |
||
Line 29: | Line 29: | ||
with equality holding iff <math>q = 0</math>. That proves part '''(i)''' and points us in the direction of looking for rational <math>\alpha,\beta,\gamma</math> for which <math>q=0</math> and (hence) <math>p=-1</math>, that is: | with equality holding iff <math>q = 0</math>. That proves part '''(i)''' and points us in the direction of looking for rational <math>\alpha,\beta,\gamma</math> for which <math>q=0</math> and (hence) <math>p=-1</math>, that is: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \alpha+\beta+\gamma & = 1\\ | + | \alpha+\beta+\gamma & = -1\\ |
\alpha\beta+\beta\gamma+\gamma\alpha & = 0 | \alpha\beta+\beta\gamma+\gamma\alpha & = 0 | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | 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> | ||
+ | as the sole equation we need to satisfy in rational numbers. |
Revision as of 20:56, 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 .
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 equation we need to satisfy in rational numbers.