Difference between revisions of "2008 IMO Problems/Problem 2"
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<cmath>\beta\gamma + ( \beta+\gamma ) ( -1 - \beta -\gamma ) = 0</cmath> | <cmath>\beta\gamma + ( \beta+\gamma ) ( -1 - \beta -\gamma ) = 0</cmath> | ||
as the sole equation we need to satisfy in rational numbers. | as the sole equation 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 | ||
+ | <cmath>bc = m(b+c)+(b+c)^2 \Leftrightarrow m = \frac{bc}{b+c} - (b+c).</cmath> | ||
+ | That means we would like <math>b+c</math> to divide <math>bc</math>. | ||
+ | Consider the example | ||
+ | <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 | ||
+ | <cmath>\beta = \frac{t}{t-1-t^2}, \gamma = \frac{t^2-1}{t-1-t^2}, \alpha = \frac{1-t}{t-1-t^2},</cmath> |
Revision as of 21:14, 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 equation we need to satisfy in rational numbers.
If and for some integers ,,and , they would need to satisfy That means we would like to divide . Consider the example where divides for any integer . Substituting back, that gives us