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Home » Olympiad Section » High-School Contests » International Mathematical Olympiad » Previous IMOs » IMO 2005 Mexico » Results of the 46th IMO 2005 Mexico
 
Medal cutoffs
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Yossarian
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#1
Medal cutoffs
G35
S23
B12

Oh, and a Moldova contestant got a special prize for a really nice solution for Q3.
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Fine. Thanks. Overworked as usual. You?

PostPosted: Sat Jul 16, 2005 9:53 pm  Back to top 
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shobber
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#2
what is his solution? Confused

PostPosted: Sat Jul 16, 2005 10:24 pm  Back to top 
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LevonNurbekian
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#3
Dear Yossarian are these numbers official?
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Levon

PostPosted: Sat Jul 16, 2005 11:14 pm  Back to top 
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dblues
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#4
As far as i know, yes, these cut-offs are official.

PostPosted: Sun Jul 17, 2005 1:50 am  Back to top 
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oldman
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#5
The cut-off mentioned above reflects how the medal distributed among the contestants. See here:

http://erdos.fciencias.unam.mx/results.htm

It follows that the cut-off above is official Mr. Green Mr. Green Mr. Green
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PostPosted: Sun Jul 17, 2005 2:06 am  Back to top 
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Quique
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#6
It is oficial.
I was in the jury meeting when they decided the distribution.
There will also be a special solution prize.
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"It is impossible to be a mathematician without being a poet in soul." -Sofia Kovalevskaya

PostPosted: Sun Jul 17, 2005 12:06 pm  Back to top 
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Megus
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#7
Quique wrote:
It is oficial.
I was in the jury meeting when they decided the distribution.
There will also be a special solution prize.


Can anybody show this solution ? Confused
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Przemyslaw Chojecki

PostPosted: Sun Jul 17, 2005 10:28 pm  Back to top 
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Quique
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#8
Solution
Me
Megus wrote:
Quique wrote:
It is oficial.
I was in the jury meeting when they decided the distribution.
There will also be a special solution prize.


Can anybody show this solution ? Confused


\frac{x^5 - x^2}{x^5 + y^2 + z^2} - \frac{x^5 - x^2}{ x^3(x^2+y^2+z^2} = \frac{(x^3-1)^2 x^2(y^2+z^2)}{x^3(x^2+y^2+z^2)(x^5+y...
Therefore
\sum \frac{x^5-x^2}{x^5+y^2+z^2} \ge \sum\frac{x^5-x^2}{x^3(x^2+y^2+z^2)} = \frac{1}{x^2+y^2+z^2}\sum{(x^2 - \frac{1}{x})} \g...

The substitution (x^2 - \frac{1}{x} for (x^2 - yz) it's because of xyz \ge 1 and \sum{(x^2 - yz)} \ge 0 it's because \sum{(x^2 - yz)} = x^2 + y^2 + z^2 - xy - yz - xz result easy to show by factorizing that it is \ge 0
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PostPosted: Mon Jul 18, 2005 11:40 pm  Back to top 
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swelily
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#9
Why not putting the awarded solution to its home topic?

PostPosted: Wed Jul 20, 2005 2:11 am  Back to top 
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Yossarian
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#10
Ha ha! I rushed from the jury meeting to post the cutoffs before conking out in my hotel room (it was rather late!)

Wow. That was fun. I wanna do it again! But I doubt they'd let me. There're already 3+++? people who want to go next year as observer Bs.
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Fine. Thanks. Overworked as usual. You?

PostPosted: Fri Jul 22, 2005 2:14 am  Back to top 
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