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Olympiad Algebra Tournament
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198 Posts • Page 8 of 10 • Previous 1, 2, 3, ..., 6, 7, 8, 9, 10 Next
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1=2
Birch & Swinnerton Dyer
Birch & Swinnerton Dyer


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#141
I didn't think you were being unpolite, I was just vehemently disagreeing with your statement that they were very very easy. Smile
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PostPosted: Sat Jun 27, 2009 10:42 am  Back to top 
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enndb0x
Yang-Mills Theory
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#142
Why ghjk is not posting the results Shocked
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PostPosted: Sat Jun 27, 2009 11:02 am  Back to top 
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hasan4444
Riemann Hypothesis
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#143
I think he would take some time to post it

PostPosted: Sat Jun 27, 2009 11:11 am  Back to top 
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Xantos C. Guin
Navier-Stokes Equations
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#144
enndb0x wrote:
Why ghjk is not posting the results Shocked


It takes a lot of time to grade all the submissions and compile the results, so be patient. However, I think he should have at least posted something about the deadline expiring, who submitted stuff, the next problem set, and/or something else indicating that this contest is still going on.
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PostPosted: Sat Jun 27, 2009 11:11 am  Back to top 
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ghjk
Yang-Mills Theory
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#145
Here's the list of people who submitted and their scores:

\begin{tabular}{|c|c|c|c|c|c|} \hline {\bf Round 1} & {\bf Username} & {\bf Total Score} \\ \hline 1 & nayel &...

A sad fact: Nobody is able to solve completely #4, which is pretty unexpected since that problem is not that hard. I'll post solutions to all of the problem tomorrow.
The total people who submit is only 23/66, which is extremely disappointed Sad . I think people who signed up must be responsible for submitting their solutions. Remember one thing: Your effort+My effort=this tournament. I hope the 2nd round will have at least 44/66 submissions. Round 2 will be posted tomorrow, after the solutions being posted.
The new helper of this tournament is: Vworldv. since calc cruz is way too busy with his business. Vworldv will propose problems and I'll check to see if they suit the difficulty of this tournament. He will also help me grade the solutions too!
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Last edited by ghjk on Sun Jun 28, 2009 9:32 am; edited 1 time in total 
PostPosted: Sat Jun 27, 2009 10:14 pm  Back to top 
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addikaye03
P versus NP
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#146
i deserve 15, OP is an idiot. You know why OP

PostPosted: Sat Jun 27, 2009 10:20 pm  Back to top 
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1=2
Birch & Swinnerton Dyer
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#147
I sent in only one problem, I get 7/7 points on it, and I end up with double that?
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PostPosted: Sun Jun 28, 2009 2:14 am  Back to top 
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Ahwingsecretagent
Hodge Conjecture
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#148
That's because ghjk mentioned that everyone gets 7 free points for the original no.2.

PostPosted: Sun Jun 28, 2009 3:31 am  Back to top 
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enndb0x
Yang-Mills Theory
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#149
I am a bit surprised by Altheman , I thought we will take 30/28 Mr. Green
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PostPosted: Sun Jun 28, 2009 4:03 am  Back to top 
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Ahwingsecretagent
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#150
Can someone outline their solution to Problem 2 please? I was not able to make much progress at all.

PostPosted: Sun Jun 28, 2009 4:24 am  Back to top 
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1=2
Birch & Swinnerton Dyer
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#151
I was wondering how people did number 3...
1
Note the trivial solution (1,1,1). Then prove if one value is less than 1 the universe breaks down (one equation is not satisfied). Then prove if two values are less than 1 the universe breaks down (one equation is not satisfied). If all values are less than 1 then the universe breaks down (all equations are not satisfied). Therefore all numbers are at least 1, and if any number is greater than 1 the universe breaks down again (all equations are not satisfied), so they all must be 1, and (1,1,1) is the only solution, and the universe stays intact.
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Hmm... apparently quoting Disney movies gets me muted on FTW. Razz

PostPosted: Sun Jun 28, 2009 6:31 am  Back to top 
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Yongyi781
Navier-Stokes Equations
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#152
The way I proved number 3 was to prove that \max(f(x)/x)=\min(f(x)/x), thus f(x)=kx, and it's pretty easy to solve for k from there. However, I only got 5/7 because, unfortunately, that relied on the fact that f(x) was continuous.
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PostPosted: Sun Jun 28, 2009 7:23 am  Back to top 
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justinahmann
Poincare Conjecture
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#153
It seems a lot of people are not submitting, so can I join in one of their places?

PostPosted: Sun Jun 28, 2009 9:48 am  Back to top 
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ffao
Poincare Conjecture
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#154
I did submit, even though I submitted a blank PM Rotfl

PostPosted: Sun Jun 28, 2009 10:34 am  Back to top 
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ghjk
Yang-Mills Theory
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#155
justinahmann wrote:
It seems a lot of people are not submitting, so can I join in one of their places?

Yes you can, but you have to show me your participation first on the 2nd Round, which is probably posted tonight. Smile
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PostPosted: Sun Jun 28, 2009 11:12 am  Back to top 
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justinahmann
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#156
Thanks! Smile Can I have 7 points from the first round due to the unsolvable sine problem?

PostPosted: Sun Jun 28, 2009 11:30 am  Back to top 
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Xantos C. Guin
Navier-Stokes Equations
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#157
Yongyi781 wrote:
The way I proved number 3 was to prove that \max(f(x)/x) = \min(f(x)/x), thus f(x) = kx, and it's pretty easy to solve for k from there. However, I only got 5/7 because, unfortunately, that relied on the fact that f(x) was continuous.


Could you elaborate on how you did this?
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PostPosted: Sun Jun 28, 2009 12:02 pm  Back to top 
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Yongyi781
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#158
Easy. Since f is continuous (by assumption, since I couldn't do the discontinuous case) and monotonic it is bijective. This means that \max(f(x)/x)=\max(f(f(x))/f(x))=\max(1+2x/f(x)) and then we have M=1+2/m and m=1+2/M therefore M=m and the result follows.

The way ghjk did it was to solve the recurrence relation in f^n(x) using characteristic polynomials to get f^n(x)=a(x)2^n+b(x)(-1)^n and stuff, and prove that one of a(x) and b(x) must be 0.
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PostPosted: Sun Jun 28, 2009 1:23 pm  Back to top 
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BOGTRO
Birch & Swinnerton Dyer
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#159
Sorry, forgot about this. Will do rd 2.
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PostPosted: Sun Jun 28, 2009 5:30 pm  Back to top 
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ocha
Yang-Mills Theory
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#160
Any Ideas for number 4? All I had was;


After factoring we are trying to maximise,
\frac{1}{2}(x_1+x_2+...+x_{2010})^2 - \frac{3}{2}(x_1^2+...+x_{2010}^2) + x_1x_2\cdot\cdot\cdot x_{2010}

From the condition
\sum_{k = 1}^{2010}\frac{1}{x_{k}^{2}+1}= 2009 \Leftrightarrow \sum_{k = 1}^{2010}\frac{x_i^2}{x_{k}^{2}+1}= 1

So by Cauchy
(\sum x_1^2 + 1)\left( \sum_{k = 1}^{2010}\frac{x_i^2}{x_{k}^{2}+1}\right)\ge (x_1+...+x_{2010})^2

\therefore (x_1+...+x_{2010})^2 - (x_1^2 +...+x_{2010}^2) \le 2010

And also by cauchy
(\sum x_1^2 + 1)\left( \sum_{k = 1}^{2010}\frac{1}{x_{k}^{2}+1}\right) \ge 2010^2

\therefore \sum x_i^2 \ge \frac{2010}{2009}

Then I was stuck on how to maximise the product x_1x_2...x_{2010}

PostPosted: Sun Jun 28, 2009 5:36 pm  Back to top 
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