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> Mgf Mathletics, NERDZ only! XD

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Wolfbrother
post Feb 29 2008, 08:17 AM
Post #81





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Bravo! Ti si sledeci.


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Nostradamus
post Feb 29 2008, 03:32 PM
Post #82





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Dat je krug i dva disjunktna kruga upravna na njega. Koliko ima krugova upravnih na ta tri kruga?

This post has been edited by Nostradamus: Feb 29 2008, 03:34 PM


--------------------
 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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Nostradamus
post Mar 2 2008, 04:23 PM
Post #83





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Pa dobro, ima li neko neku ideju ili da stavljam novi zadatak?


--------------------
 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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Wolfbrother
post Mar 2 2008, 04:23 PM
Post #84





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Pa ja ga bas ne kuzim. Nisam pre video nista slicno.


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Nostradamus
post Mar 2 2008, 04:25 PM
Post #85





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Jel znas sta je inverzija?


--------------------
 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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Wolfbrother
post Mar 2 2008, 04:27 PM
Post #86





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Izometrijska transformacija? Nisam je jos radio


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Nostradamus
post Mar 2 2008, 04:29 PM
Post #87





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QUOTE(Wolfbrother @ Mar 2 2008, 04:27 PM)
Izometrijska transformacija? Nisam je jos radio
*



Jeste transformacija, ali nije izometrijska. Mislim da preko toga moze da se uradi. Da budem iskren ja ne znam kako se ovaj zadatak radi u okvirima Euklidske geometrije...


--------------------
 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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username
post Mar 2 2008, 04:32 PM
Post #88





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odgovor je broj od 0-4 a mozda i beskonacno.. jesam li pogodio smile.gif
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Nostradamus
post Mar 2 2008, 04:33 PM
Post #89





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QUOTE(username @ Mar 2 2008, 04:32 PM)
odgovor je broj od 0-4 a mozda i beskonacno.. jesam li pogodio smile.gif
*


Jesi smile.gif . Mislim mogu i da kazem resenje, to nece mnogo pomoci. Bitan je postupak sleep.gif


--------------------
 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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Wolfbrother
post Mar 2 2008, 04:40 PM
Post #90





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Bah ne kuzim ja ovo.


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Nostradamus
post Mar 2 2008, 04:48 PM
Post #91





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Evo, ako neko zna Hiperbolicku geometriju, zadatak je jedan i po red.

Ako ne zna, trebalo bi da se radi preko pramenova krugova i inverzije


--------------------
 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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username
post Mar 2 2008, 04:55 PM
Post #92





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koliko sam shvatio ugao izmedju 2 kruga je onaj koji zakalapaju njihove tangente u jednoj tacki njihovog preseka, a ako je taj ugao prav onda su krugovi normalni
neka su 2 kruga k1 i k2, sa centrima O1 i O2, a seku se u tacki A (i mozda jos nekoj tacki)
ako su normalni onda tangenta kruga k1 u tacki A sadrzi O2, a tangenta kruga k2 u tacki A sadrzi O1
slicno uvedemo novi krug k3 sa centrom O3 koji se sece sa k2 u tacki B a nema zajednickih tacaka sa k1 (da bi bili disjunktni)
dakle za svaki trazeni krug (neka je to krug k) vazi: sece k1, k2, k3 u nekim tackama tako da tangente krugova k1,k2,k3 iz tih tacaka sadrze centar k, i tangente kruga k sadrze centre prvobitnih krugova
znaci pitanje je koliko ima tacaka P takvih da: kad se iz njih povuku tangente na k1, k2 i k3 i normale na te tangente (iz tacke preseka) one(normale) sadrze centre prvih krugova(tacke O1, O2 i O3) (sto je uvek tacno), a tacke preseka (tangenta povucenih iz tacke P) moraju biti isto udaljene od te tacke P

razumecu ako niko ne shvata sta sam hteo da kazem smile.gif

This post has been edited by username: Mar 2 2008, 09:49 PM
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Nostradamus
post Mar 3 2008, 02:42 PM
Post #93





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Mislim da tako neces doci do resenja


--------------------
 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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username
post Mar 3 2008, 05:20 PM
Post #94





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verovatno...
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Nostradamus
post Mar 5 2008, 06:04 PM
Post #95





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Posto mi se cini da ovo niko nece resiti, da postavljam novi zadatak?


--------------------
 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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Wolfbrother
post Mar 5 2008, 07:32 PM
Post #96





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Ja nemam dovoljno teorije za ovaj XD.gif. A ionako ne razumem zadatak u potpunosti tako da sam za novi


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Nostradamus
post Mar 7 2008, 02:02 AM
Post #97





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Pa dobro, postoji jedan krug, zato sto svaka 2 kruga pripadaju 1 pramenu, samim tim ona 2 disjunktna pripadaju hiperbolickom pramenu. Kako na svaki pramen postoji jedinstven upravan pramen, a za hiperbolicki to je elipticki, i prvi krug ce pripadati tom pramenu, zadatak se svodi na to koliko ima krugova upravnih na krug u date 2 tacke. Ako te tacke nisu dijametralno suprotne 1, ako jesu nijedan (u sustini tada ce prava biti upravna na taj krug)

Ovo sad bi trebalo da bude jednostavno

Neka su a, b, c \in \mathbb{R}+. Pokazati da vazi

\frac{1}{a+b} + \frac{1}{b+c} +\frac{1}{c+a} \leq \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c}

Edit: Ovde se nesto gadno izbagovao

This post has been edited by Nostradamus: Mar 7 2008, 02:03 AM


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 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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Wolfbrother
post Mar 7 2008, 01:04 PM
Post #98





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Poznat mi ovaj zadatak. Sad cu da pokusam.


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Wolfbrother
post Mar 7 2008, 01:10 PM
Post #99





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\frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c}= \frac{1}{4a}+ \frac{1}{4b} +\frac{1}{4b}+ \frac{1}{4c}+ \frac{1}{4c}+ \frac{1}{4a}
Cauchyjem:
\frac{1}{4a}+ \frac{1}{4b} \geq \frac{1}{a+b}
\frac{1}{4b}+ \frac{1}{4c} \geq \frac{1}{b+c}
\frac{1}{4c}+ \frac{1}{4a} \geq \frac{1}{c+a}
To spojimo i dobijemo trazenu nejednakost.
\frac{1}{a+b} + \frac{1}{b+c} +\frac{1}{c+a} \leq \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c}

This post has been edited by Wolfbrother: Mar 7 2008, 07:41 PM


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Nostradamus
post Mar 7 2008, 08:51 PM
Post #100





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Dobro je, naravno


--------------------
 (a,b) \sim (c,d) \Leftrightarrow (a - c) \in \mathbb{Z}  \wedge  (b - d) \in \mathbb{Z}

\mathbb{T}^2\equiv\mathbb{R}^2/\sim
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