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> Interno Takmicenje, Rezultati

m_r
post Mar 24 2008, 06:34 PM
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Evo rezultata internog takmicenja za 1. razred.

Da podsetim da je interno takmicenje za 3-4. razred u utorak od 10.

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Hannibal Lecter
post Mar 24 2008, 07:48 PM
Post #2





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e je l' ima neko za drugi biggrin.gif?

cestitam svima, vidim da svi imaju 50+


--------------------
And as we wind on down the road
Our shadows taller than our soul
There walks a lady we all know
Who shines white light and wants to show
How everything still turns to gold
And if you listen very hard
The truth will come to you at last
When all are one and one is all
To be a rock and not to roll


Svi me žele, a ja sam nedodirljiv!
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Hannibal Lecter
post Mar 24 2008, 08:16 PM
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Zadaci za II godinu:

1.Izracunati sumu

 \sum_{i=0}^{2007}  \frac{Xi^3}{1-3Xi+3Xi^2}

gde je  Xi=i/2007 za i=0...2007

2.Neka su a,b,c pozitivni realni brojevi tako da  a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}
Dokazati nejednakost  a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}

3.Unutar trougla ABC nalazi se tacka M takva da je
 \angle MAB = \angle MBC+\angle MCA = \phi
Dokazati da je
 \frac{1}{sin^2\phi}=\frac{1}{sin^2\alpha}+\frac{1}{sin^2\beta}+\frac{1}{sin^2\gamma}

4.Krugovi C1 i C2 seku se u tackama P i Q. Zajednica tanegenta ovih krugova bliza tacki P dodiruje krugove u tackama A i B, redom. Tangenta kruga C1 u tacki P sece krug C2 u tacki E, razlicitoj od , dok tangenta kruga C2 u P sece C1 u tacki F razlicitoj od P. Neka su H i N tacke na duzima AF i BE, redom, takve da ja AH=AP i BK=BP. Dokazati da su tacke A, H, Q, K, B konciklicne.

5.Dato je 8 jedinicnih kocki, pri cemu su 24 strane obojene u plavo i 24 u crveno. Dokazati da od njih mozemo sastaviti kocku dimenzija 2*2*2 koja ima jednak broj plavih i crvenih jedinicnih kvadrata na povrsini

smile.gif

Uradio sam prva dva i treci nesto muljo

I da tako sam pro za Mimetex soproud.gif

This post has been edited by Noddy: Mar 24 2008, 08:17 PM


--------------------
And as we wind on down the road
Our shadows taller than our soul
There walks a lady we all know
Who shines white light and wants to show
How everything still turns to gold
And if you listen very hard
The truth will come to you at last
When all are one and one is all
To be a rock and not to roll


Svi me žele, a ja sam nedodirljiv!
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Wolfbrother
post Mar 24 2008, 08:18 PM
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Zahvaljujem. Sad cu da probam ovu nejednakost :S


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"It's a known fact... math is the spawn of satan. Algebra? SATANIC! Pre-cal? SATANIC! Calculus? SATANIC! it is all horrible!"
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Hannibal Lecter
post Mar 24 2008, 08:18 PM
Post #5





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Nejednakost se radi preko najseljackije fore XD.gif


--------------------
And as we wind on down the road
Our shadows taller than our soul
There walks a lady we all know
Who shines white light and wants to show
How everything still turns to gold
And if you listen very hard
The truth will come to you at last
When all are one and one is all
To be a rock and not to roll


Svi me žele, a ja sam nedodirljiv!
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Wolfbrother
post Mar 24 2008, 08:34 PM
Post #6





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Ja sam je isterao preko celog lista mada moram da proverim biggrin.gif

btw pisem bas krupno XD.gif

Edit: Sad cu postovati, trebalo bi dobro, ali valja proveriti.

This post has been edited by Wolfbrother: Mar 24 2008, 08:36 PM


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Wolfbrother
post Mar 24 2008, 08:49 PM
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Iz
 a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}
Sledi  \frac{a+b+c}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\geq 1

E sad homogenizujemo nejednakost koju dokazujemo pa treba da dokazemo:

 \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{3}{a+b+c} + 2\frac{a+b+c}{abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}

Cauchy-jem  \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\geq \frac{3}{a+b+c}

Sada jos treba dokazati:  \frac{2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}{3}\geq \frac{2(a+b+c)}{ab+bc+ac}
Sto je ekvivalentno:
 \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}\geq a+b+c
To se izmnozi i trivijalno Mjurhedom.

Valja li? Koji je taj seljacki nacin?

This post has been edited by Wolfbrother: Mar 24 2008, 08:50 PM


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Hannibal Lecter
post Mar 24 2008, 09:00 PM
Post #8





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a+b+c \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}  /  +a+b+c
<=> 2*(a+b+c) \geq (a+\frac{1}{a})+(b+\frac{1}{b})+(c+\frac{1}{c})
x+\frac{1}{x}) \geq 2 => a+b+c \geq 3

/me ovde gubi zivce za ovo sranje dry.gif

pa onda imamo AG za (a+b+c)/3>=treci sqrt(abc)
=> 3/(a+b+c) <= 1/treci sqrt(abc) pa posto a+b+c>=3 onda je 3/(a+b+c)<=1 pa je i 1<=1/treci sqrt(abc) pa odatle dobijes biggrin.gif

This post has been edited by pyost: Mar 25 2008, 12:29 PM


--------------------
And as we wind on down the road
Our shadows taller than our soul
There walks a lady we all know
Who shines white light and wants to show
How everything still turns to gold
And if you listen very hard
The truth will come to you at last
When all are one and one is all
To be a rock and not to roll


Svi me žele, a ja sam nedodirljiv!
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Wolfbrother
post Mar 24 2008, 09:04 PM
Post #9





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Vidis nisam znao ovu foru da kad imas zbir >= od zbira reciprocnih mozes da odredis uslov za a+b+c. ql


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Hannibal Lecter
post Mar 24 2008, 09:11 PM
Post #10





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ma ja moram da se slusim tim jer znam jedino H<G<A<Q nejednakosti biggrin.gif


--------------------
And as we wind on down the road
Our shadows taller than our soul
There walks a lady we all know
Who shines white light and wants to show
How everything still turns to gold
And if you listen very hard
The truth will come to you at last
When all are one and one is all
To be a rock and not to roll


Svi me žele, a ja sam nedodirljiv!
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SerGrimReaper
post Mar 25 2008, 12:21 PM
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QUOTE(Noddy @ Mar 24 2008, 09:00 PM)
a+b+c \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}  /  +a+b+c
<=> 2*(a+b+c) \geq (a+\frac{1}{a})+(b+\frac{1}{b})+(c+\frac{1}{c})
x+\frac{1}{x}) \geq 2 => a+b+c \geq 3

/me ovde gubi zivce za ovo sranje dry.gif

pa onda imamo AG za (a+b+c)/3>=treci sqrt(abc)
=> 3/(a+b+c) <= 1/treci sqrt(abc) pa posto a+b+c>=3 onda je 3/(a+b+c)<=1 pa je i 1<=1/treci sqrt(abc) pa odatle dobijes biggrin.gif
*


Ovo boldirano ti nije tacno :/
Radio sam slicno kao wulf... Ustvari sam razlagao na dva slucaja... abc>=1 i abc<=1. Za abc>=1 je trivijalno kao sto je Noddy pokazao... Za abc<1:
imas da je trazena nejednakost (u obliku 1/a+1/b+1/c >= 3/(a+b+c) +2/abc) kad se izmnozi ekvivalentna sa (ab+bc+ac)(a+b+c)>=2a+2b+2c+3abc. Iz pocetnog uslova imas:
(a+b+c)ab>=ab/c+b+a
(a+b+c)ac>=a+c+ac/b
(a+b+c)bc>=b+c+bc/a
kad se saberu imas:
(a+b+c)(ab+bc+ac)>=2a+2b+2c+ab/c+ac/b+bc/a>=2a+2b+2c+3treci sqrt(abc)>=(zbog abc<1) 2a+2b+2c+3abc sto je trebalo i dokazati.

This post has been edited by SerGrimReaper: Mar 25 2008, 12:36 PM
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Hannibal Lecter
post Mar 25 2008, 01:01 PM
Post #12





Group: Članovi
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e jebajiga XD.gif


--------------------
And as we wind on down the road
Our shadows taller than our soul
There walks a lady we all know
Who shines white light and wants to show
How everything still turns to gold
And if you listen very hard
The truth will come to you at last
When all are one and one is all
To be a rock and not to roll


Svi me žele, a ja sam nedodirljiv!
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Hannibal Lecter
post Mar 25 2008, 02:13 PM
Post #13





Group: Članovi
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49 dance13fm0uvnh9.gif

Ocekivao sam manje, ali se ispostavilo da mi je treci dosta dobro uradjen soproud.gif

Rezultati za drugu godinu u attachmentu.


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Attached File  Rezultati_1_.xls ( 13.5k ) Number of downloads: 155


--------------------
And as we wind on down the road
Our shadows taller than our soul
There walks a lady we all know
Who shines white light and wants to show
How everything still turns to gold
And if you listen very hard
The truth will come to you at last
When all are one and one is all
To be a rock and not to roll


Svi me žele, a ja sam nedodirljiv!
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^_NiN0_^
post Mar 25 2008, 02:40 PM
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Sta bi za 3. i 4. godinu?


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Yo!hambina, Yo!hambin
Yo-yo, yo-yo, yo
Yo!hambin,
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Cekaaa
post Mar 25 2008, 03:03 PM
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tek ce da izadje u toku dana... mada se niko nije bas proslavio.. teodor/luka bi trebalo da budu najbolji
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^_NiN0_^
post Mar 25 2008, 03:20 PM
Post #16


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a tu bilo vise drugaka, prvaka i osmaka nego trecaka i cetvrtaka ?


--------------------
Yo!hambin,
Yo!hambina, Yo!hambin
Yo-yo, yo-yo, yo
Yo!hambin,
Yo!hambina, Yo!hambiiina
(Mala matura deca kokaina®
Velika matura deca Yo!hambina)
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Cekaaa
post Mar 25 2008, 05:05 PM
Post #17





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dap... mislim da je bio jedan cetvrtak, onaj neki plavi decko... i bila je jos jedna trecakinja/cetvrtakinja koja je otisla mnooogo ranije. Poered njih bilo je 4 drugaka i dva prvaka
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^_NiN0_^
post Mar 25 2008, 05:21 PM
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pwned ... mislim da nasa generacija ne zasluzuje da se takmici ... nije to za nas tongue.gif


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Yo!hambin,
Yo!hambina, Yo!hambin
Yo-yo, yo-yo, yo
Yo!hambin,
Yo!hambina, Yo!hambiiina
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Velika matura deca Yo!hambina)
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