Reply to this topicStart new topicStart Poll

Outline · [ Standard ] · Linear+

> Interno Takmicenje, Rezultati

m_r
post Mar 24 2008, 06:34 PM
Post #1





Group: Članovi
Joined: 23-April 06
Member No.: 56
Ime i prezime: Marko Radovanovic



Evo rezultata internog takmicenja za 1. razred.

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

Attached File(s)
Attached File  interno2007_08_2_rez.xls ( 15k ) Number of downloads: 228
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
Hannibal Lecter
post Mar 24 2008, 07:48 PM
Post #2





Group: Članovi
Joined: 15-October 06
From: People's Democratic Republic of Konjarnik
Member No.: 154
Status: Bivši učenik MGa
Ime i prezime: Ilija Ivanišević
Škola/Razred: Fizički fakultet, B smer, I godina,



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!
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
Hannibal Lecter
post Mar 24 2008, 08:16 PM
Post #3





Group: Članovi
Joined: 15-October 06
From: People's Democratic Republic of Konjarnik
Member No.: 154
Status: Bivši učenik MGa
Ime i prezime: Ilija Ivanišević
Škola/Razred: Fizički fakultet, B smer, I godina,



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!
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
Wolfbrother
post Mar 24 2008, 08:18 PM
Post #4





Group: Članovi
Joined: 26-December 07
Member No.: 825
Status: Bivši učenik MGa



Zahvaljujem. Sad cu da probam ovu nejednakost :S


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





Group: Članovi
Joined: 15-October 06
From: People's Democratic Republic of Konjarnik
Member No.: 154
Status: Bivši učenik MGa
Ime i prezime: Ilija Ivanišević
Škola/Razred: Fizički fakultet, B smer, I godina,



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!
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
Wolfbrother
post Mar 24 2008, 08:34 PM
Post #6





Group: Članovi
Joined: 26-December 07
Member No.: 825
Status: Bivši učenik MGa



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


--------------------
"It's a known fact... math is the spawn of satan. Algebra? SATANIC! Pre-cal? SATANIC! Calculus? SATANIC! it is all horrible!"
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
Wolfbrother
post Mar 24 2008, 08:49 PM
Post #7





Group: Članovi
Joined: 26-December 07
Member No.: 825
Status: Bivši učenik MGa



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


--------------------
"It's a known fact... math is the spawn of satan. Algebra? SATANIC! Pre-cal? SATANIC! Calculus? SATANIC! it is all horrible!"
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
Hannibal Lecter
post Mar 24 2008, 09:00 PM
Post #8





Group: Članovi
Joined: 15-October 06
From: People's Democratic Republic of Konjarnik
Member No.: 154
Status: Bivši učenik MGa
Ime i prezime: Ilija Ivanišević
Škola/Razred: Fizički fakultet, B smer, I godina,



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!
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
Wolfbrother
post Mar 24 2008, 09:04 PM
Post #9





Group: Članovi
Joined: 26-December 07
Member No.: 825
Status: Bivši učenik MGa



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


--------------------
"It's a known fact... math is the spawn of satan. Algebra? SATANIC! Pre-cal? SATANIC! Calculus? SATANIC! it is all horrible!"
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
Hannibal Lecter
post Mar 24 2008, 09:11 PM
Post #10





Group: Članovi
Joined: 15-October 06
From: People's Democratic Republic of Konjarnik
Member No.: 154
Status: Bivši učenik MGa
Ime i prezime: Ilija Ivanišević
Škola/Razred: Fizički fakultet, B smer, I godina,



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!
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
SerGrimReaper
post Mar 25 2008, 12:21 PM
Post #11





Group: Članovi
Joined: 31-January 06
Member No.: 9
Ime i prezime: Uglješa Stojanović



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
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
Hannibal Lecter
post Mar 25 2008, 01:01 PM
Post #12





Group: Članovi
Joined: 15-October 06
From: People's Democratic Republic of Konjarnik
Member No.: 154
Status: Bivši učenik MGa
Ime i prezime: Ilija Ivanišević
Škola/Razred: Fizički fakultet, B smer, I godina,



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!
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
Hannibal Lecter
post Mar 25 2008, 02:13 PM
Post #13





Group: Članovi
Joined: 15-October 06
From: People's Democratic Republic of Konjarnik
Member No.: 154
Status: Bivši učenik MGa
Ime i prezime: Ilija Ivanišević
Škola/Razred: Fizički fakultet, B smer, I godina,



49 dance13fm0uvnh9.gif

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

Rezultati za drugu godinu u attachmentu.


Attached File(s)
Attached File  Rezultati_1_.xls ( 13.5k ) Number of downloads: 153


--------------------
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!
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
^_NiN0_^
post Mar 25 2008, 02:40 PM
Post #14


Moderator
Group Icon

Group: Moderatori
Joined: 29-January 06
Member No.: 4
Status: Učenik MGa



Sta bi za 3. i 4. godinu?


--------------------
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)
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
Cekaaa
post Mar 25 2008, 03:03 PM
Post #15





Group: Članovi
Joined: 1-July 06
Member No.: 73
Status: Učenik MGa
Ime i prezime: Mihajlo Cekic
Škola/Razred: MG - IId



tek ce da izadje u toku dana... mada se niko nije bas proslavio.. teodor/luka bi trebalo da budu najbolji
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
^_NiN0_^
post Mar 25 2008, 03:20 PM
Post #16


Moderator
Group Icon

Group: Moderatori
Joined: 29-January 06
Member No.: 4
Status: Učenik MGa



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)
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
Cekaaa
post Mar 25 2008, 05:05 PM
Post #17





Group: Članovi
Joined: 1-July 06
Member No.: 73
Status: Učenik MGa
Ime i prezime: Mihajlo Cekic
Škola/Razred: MG - IId



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
User is offlineProfile CardPM
Go to the top of the page
+Quote Post
^_NiN0_^
post Mar 25 2008, 05:21 PM
Post #18


Moderator
Group Icon

Group: Moderatori
Joined: 29-January 06
Member No.: 4
Status: Učenik MGa



pwned ... mislim da nasa generacija ne zasluzuje da se takmici ... nije to za nas tongue.gif


--------------------
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)
User is offlineProfile CardPM
Go to the top of the page
+Quote Post

Reply to this topicTopic OptionsStart new topic
1 User(s) are reading this topic (1 Guests and 0 Anonymous Users)
0 Members: