Olympiad Algebra Tournament

Yongyi781

It appears as though the maximum occurs when $x_1=x_2=\dots=x_{2010}$ , which will make the resulting value look slightly ugly.

**Posted:** Sun Jun 28, 2009 5:46 pm

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"quoted and followed by itself is a quine" quoted and followed by itself is a quine.
"yields falsehood when preceded by its quotation" yields falsehood when preceded by its quotation.
(Almost) Red MOP 2009

CatalystOfNostalgia

Guys guys see USAMO 1998 problem 3 for that part of the problem.

also yongyi for that functional equation dude you probably can't just randomly assume something is continuous? (edit: ok i just realized you saw this too) often this trivializes stuff (for example, if you get to something like f(x+y)=f(x)+f(y), from which you can immediately conclude f(x)=kx if f is continuous), and it doesn't give any insight as to how the proof would work if you don't make such assumptions. in general, partial credit is only given to work that is, no matter how nontrivial or interesting, actually useful for completing a solution (i may be quoting zhai but that is probably ok). that is, if you're going to get five points, you should be able to just extend the same ideas a little further to finish the solution.

just my two cents on that...but i'm not in charge.

**Posted:** Sun Jun 28, 2009 5:53 pm

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~Carl 白人看不懂

MELODIOUS

ocha · Yang-Mills Theory Offline Joined: 23 Sep 2008 Posts: 556

**Posted:** Sun Jun 28, 2009 6:13 pm

Altheman

wow I was stupid, I thought that 4 was hard because you couldn't maximize the terms individually. So I just sorta gave up. I got the two upper bounds that ocha got but I thought they were useless because that wouldn't contribute to the upper bound on the product.

Then the problem is pretty straightforward.

Let $y_i=\frac{x_i^2}{x_i^2+1}$ for $i=1,2,...,n=2010$ Then $y_1+...+y_n=1$ ; $y_i\ge 0$ . WLOG $x_i\ge 0$ . Then $x_i=\sqrt{\frac{y_i}{1-y_i}}$ .

Then $\prod x_i^2=\prod\frac{y_i}{1-y_i}=\prod_i\frac{y_i}{\sum_{j\ne i}y_j}\le \prod_i \frac{y_i}{(n-1)\sqrt[n-1]{\prod_{j\ne i}y_...$

durr for assuming the problem was harder than it was...

**Posted:** Sun Jun 28, 2009 7:12 pm

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-Alex
Altheman's Problem Column

CatalystOfNostalgia

Oh hmm I seem to have misread it...never mind.

**Posted:** Sun Jun 28, 2009 7:28 pm

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~Carl 白人看不懂

MELODIOUS

Differ

**Posted:** Sun Jun 28, 2009 9:12 pm

jevgeniy

ghjk, did you receive my second PM about Problem 3? maybe

**Posted:** Sun Jun 28, 2009 9:53 pm

ghjk

Problems for Round 2:

Problem 1:
Given $1 > {a_1} > {a_2} > {a_3} > ... > {a_n} > 0$ .
Prove that: $\frac {a^2_1}{1 - a_1} + \frac {a^2_2}{a_1 - a_2} + \frac {a^2_3}{a_2 - a_3} + ... + \frac {a^2_n}{a_{n - 1} - a_n} > \fra...$

Problen 2:
Solve the following system of equations in real numbers:
$\left\{\begin{aligned}x = \frac {2y}{1 + y^{2}} \\ y = \frac {2z}{1 + z^{2}} \\ z = \frac {2x}{1 + x^{2}}\end{aligned}\right$

Problem 3:
Given the quadratic equation $f(x) = x^2 + ax + b$ . We know that with every real numbers $x$ , we always can find a real number y such that $f(y) = f(x) + y$ . Find maximum value of $a$

Problem 4:
Given x,y,z are positive real numbers satisfying: $xyz = x + 2y + 2z$
Find the minimum value of: $A = x + y + z$

The deadline of submission is on Friday next week! Remember that if you can generalize the problems and prove them successfully, you get 2 bonus points for each one.Good luck and enjoy the problems!I probably will make another thread and have moderator poll it for me as a place to discuss solutions and comments/suggestions. Anyway, hopefully more than 44 people will submit solutions this time. Smile

@jevgeniy: I'm sorry for miscalculating your points on Round 1. You should get 16 points instead of 14. But I couldn't edit my post anymore apparently. So pleased wait for moderators to edit it(I'll PM to let them know)

**Posted:** Sun Jun 28, 2009 11:27 pm

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Try your best....do over your best!

enndb0x

ghjk ,you did not post the first round solutions yet.

**Posted:** Mon Jun 29, 2009 3:59 am

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Inequalities Marathon

CatalystOfNostalgia

**Posted:** Mon Jun 29, 2009 8:08 am

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~Carl 白人看不懂

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Differ

Yongyi781

Blah I was being stupid.... Neutral

*smacks self into wall*

**Posted:** Mon Jun 29, 2009 11:53 am

_________________
"quoted and followed by itself is a quine" quoted and followed by itself is a quine.
"yields falsehood when preceded by its quotation" yields falsehood when preceded by its quotation.
(Almost) Red MOP 2009

1=2

It's not healthy to smack yourself into walls...

Also problem 2 inspired me to make a similar problem.

**Posted:** Mon Jun 29, 2009 12:17 pm

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Hmm... apparently quoting Disney movies gets me muted on FTW. Razz

zapi2007

sorry ghjk, i'll have to pull out
i'm gone way too much (as in i missed last week (no internet, out of town) will miss this week, and will miss the next 1 and 1/2 weeks (out of town again)) Sad

**Posted:** Tue Jun 30, 2009 3:17 pm

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"There is only one difference between a madman and me. The madman thinks he is sane. I know I am mad." -Salvador Dali

anonyme93

hello!!
i'm a new member....i'd like to participate in this competition ...so can i be with you???

**Posted:** Wed Jul 01, 2009 7:06 am

CatalystOfNostalgia

**Posted:** Thu Jul 02, 2009 7:59 pm

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~Carl 白人看不懂

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Differ

**Posted:** Thu Jul 02, 2009 11:10 pm

hurdler

Just to clarify, is each week's solutions due at midnight of Friday (which would be in 3 1/2 hours right now)?

**Posted:** Fri Jul 03, 2009 5:35 pm

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goals: usamo (if made) - have fun

echinodermata · Poincare Conjecture Offline Joined: 27 Feb 2009 Posts: 183

That would still depend on time zone.

**Posted:** Fri Jul 03, 2009 6:19 pm

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nihil maiestati simplicitate detrahitur
Nothing is detracted from majesty by simplicity.

ghjk

The deadline of submission is at midnight today(you just need to follow your local time). And I need much much more submissions than the number I have now. Common people! Signed up but not join is failure! Sad

**Posted:** Fri Jul 03, 2009 7:26 pm

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Try your best....do over your best!