Difference between revisions of "2008 IMO Problems/Problem 2"
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'''(ii)''' Prove that equality is achieved for infinitely many triples of rational numbers <math>x</math>, <math>y</math> and <math>z</math>. | '''(ii)''' Prove that equality is achieved for infinitely many triples of rational numbers <math>x</math>, <math>y</math> and <math>z</math>. | ||
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== Solution == | == Solution == | ||
Consider the transormation <math>f:\mathbb{R}/\{1\} \rightarrow \mathbb{R}/\{-1\}</math> defined by <math>f(u) = \frac{u}{1-u}</math> and put <math>\alpha = f(x), \beta = f(y), \gamma = f(z)</math>. Since <math>f</math> is also one-to one from <math>\mathbb{Q}/\{1\}</math> to <math>\mathbb{Q}/\{-1\}</math>, the problem is equivalent to showing that | Consider the transormation <math>f:\mathbb{R}/\{1\} \rightarrow \mathbb{R}/\{-1\}</math> defined by <math>f(u) = \frac{u}{1-u}</math> and put <math>\alpha = f(x), \beta = f(y), \gamma = f(z)</math>. Since <math>f</math> is also one-to one from <math>\mathbb{Q}/\{1\}</math> to <math>\mathbb{Q}/\{-1\}</math>, the problem is equivalent to showing that | ||
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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. |
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+ | 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> | ||
+ | For <math>m</math> to be integer, 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},\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>). | ||
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+ | 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)'''. | ||
+ | --[[User:Vbarzov|Vbarzov]] 03:03, 5 September 2008 (UTC) | ||
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+ | ==See Also== | ||
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+ | {{IMO box|year=2008|num-b=1|num-a=3}} |
Latest revision as of 00:09, 19 November 2023
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). --Vbarzov 03:03, 5 September 2008 (UTC)
See Also
2008 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |