Additive combinatorics
combinatorics
Additive Number Theory
Extremal combinatorics
Subsets
IMO Shortlist

Let $ k,m,n$ be integers such that $ 1 < n \leq m \minus{} 1 \leq k.$ Determine the maximum size of a subset $ S$ of the set $ \{1,2,3, \ldots, k\minus{}1,k\}$ such that no $ n$ distinct elements of $ S$ add up to $ m.$

First, if $m<n(n+1)/2$ then the answer is clearly $k$.

Now suppose that $m\ge n(n+1)/2$, and note that $f(k,m,n)=k-(m-1)+f(m-1,m,n)$, so we just need to find $f(m-1,m,n)$.

If some $n$ elements in $S$ do sum up to $m$, then
\[n\min{S}+\frac{n(n-1)}{2}\le m\implies \min{S}\le\left\lfloor\frac{m}{n}-\frac{n-1}{2}\right\rfloor,\]so we have
\[f(m-1,m,n)\ge m-1-\left\lfloor\frac{m}{n}-\frac{n-1}{2}\right\rfloor=X.\]Now assume for the sake of contradiction that we have a set with $|S|\ge X+1$ and no $n$ distinct elements adding up to $m$ containing smallest elements
\[1\le a_1<a_2<\cdots<a_{n-2}\le\frac{m}{n}+\frac{n-7}{2}\](otherwise we clearly cannot fit $X+1-(n-2)$ numbers greater than $a_{n-2}$). Let
\[s=a_1+\cdots+a_{n-2}\le (n-2)a_{n-2}-\frac{(n-2)(n-3)}{2}\le\frac{(n-2)m}{n}-2(n-2).\]If $m-s=u+v$ for some $a_{n-2}<u<v<m-s-a_{n-2}$, then at most one of $u,v$ can be in $S$. We can check that $m-s-2a_{n-2}\ge n+3\ge0$, so
\begin{align*}
m-\left\lfloor\frac{m}{n}-\frac{n-1}{2}\right\rfloor = X+1 \le |S|
&\le n-2+\left\lfloor\frac{m-s-2a_{n-2}}{2}\right\rfloor+m-1-(m-s-a_{n-2}-1) \\
&= n-2+\left\lfloor\frac{m+s}{2}\right\rfloor \\
&\le \left\lfloor\frac{(n-1)m}{n}\right\rfloor,
\end{align*}a clear contradiction.


abstract algebra
number theory unsolved
number theory

How to prove that multiplicative group modulo p is isomorphical to the additive group modulo (p-1)? I would really appreciate elementary solution because i'm a bit stupid and i don't understand some operations with symbols like $\forall$ and $\exists$.

This is the simplest solution that I know:

First begin with the Fundamental Theorem for Finitely Generated Abelian Groups and notice that

$\left (\mathbb{Z}/p\mathbb{Z}\right )^\times \cong \mathbb{Z}/n_{1}\mathbb{Z}\times\cdots\times\mathbb{Z}/n_{t}\mathbb{Z}$

for some integer $t$.  We also know that $n_{1}|p-1$.  Now consider any $x$ that is in the set $\left (\mathbb{Z}/p\mathbb{Z}\right )^\times$.  Then from above, we also have that $x^{n_{1}}=1$, and so by the Fundamental Theorem of Algebra we know that $x^{n_{1}}-1=0$ has $p-1$ roots, and so $n_{1}\ge p-1$.  But this implies that $n_{1}=p-1$, and so we have the isomorphism as desired.


Why is $ G_m (g,P) \equal{} G_m^\circ(g) \plus{} RT\ln P$?

I'm reading Atkins, where they use this relation to show $ G_{vap} \equal{} \minus{}RTlnP$ and then Clausius-Clapeyron, but I don't remember reading about it.

lol im pretty sure atkins just pulled off a "it can be thermodynamically proven that..."

http://en.wikipedia.org/wiki/Gibbs-Helmholtz_equation
Wikipedia has a calculus-based derivation.


superior algebra
superior algebra solved

let a,b be elements of group G show order (ab)=order(ba)


calculus
integration
real analysis
real analysis theorems

Hi Everybody .I  need some infomation about Abel and Dirichclet standard .Can you give me   proof of two standard?

for Abel use this link [url=http://mathworld.wolfram.com/search/index.cgi?q=Abel+]Abel[/url]

Do you mean by Abel's Test and Dirichlet's Test for series and improper integrals?


I live in Peru, so I probably won't be competing against any of you. But I'll have to compete in my third language, Spanish, so it'll be tough on me none the less.

Anyhow, what kind of preparations should I do for the math, chem, and physics olympiads? I'm most worried about math because I've seen some of the sample problems, and they seemed very hard. I've not had the chance to do many proof problems, so they seem very difficult for me. However, I have about five months to prepare for the next national competition and I'm willing to work hard everyday so I can make the international teams.

So for a new guy like me, does anyone have any tips? What books should I get? What material should I have covered in school? Is completion or enrollment in calculus absolutely necessary?

What about for Chemistry? I'm less worried about this because I figure that they will mostly be lab and essay type questions right?

What about for Physics?

Keep in mind that I want to make it to the international olympiads in these areas. I probably won't make it to the international chemistry or physics olympiads for 2004, but hopefully I'll make it for 2005.

The Math competition is on June, and I'm pretty sure that the Chemistry and Physics competition is around August or September.

I can buy some books, but I think most are too expensive for me. The ones I've seen on Amazon are around 50-60$, and I can't really afford anything above 15-20$ =(.

[quote="Binary Star"]I live in Peru. . . . I can buy some books, but I think most are too expensive for me. The ones I've seen on Amazon are around 50-60$.[/quote]

Here's a tip I learned when living overseas: try looking for bookstores near the best university you can reach by convenient transportation, and ask for titles you've heard about through online searches. You may find out that they are available in Peruvian-market-only editions that are much less expensive than the editions in the original country of publication. In some poor countries, international publishers sell famous textbooks at a small percentage of the usual retail cost in the United States, because cost of printing and distribution is low in such countries, authors' royalties (by contract) often are payable only for copies of the book sold in the original country market, and the publisher figures it is better to sell some books at an affordable price than no books at all at an expensive price. Students in universities in MANY countries around the world study the same required and optional books about mathematics as the students in the developed English-speaking world, but pay less for the books. It's worth a look--I did Amazon searches when I last lived abroad, and discovered very shortly before I left to return to the States that many of the books I could only dream of buying at United States prices were affordable on an overseas income at the prices on offer in that country. Try it; you might save some money, and get the books after having paged through them to see which you like best.

Do you have to have Peruvian permanent residency status to be able to join those teams?

Good luck!

Weird, I'm living overseas also, but I can take the American competitions.  Do you go to a Peruvian school or is it an American or international school?

I go to an American school, but I'm not an American citizen, so I can't participate in American competitions.

Now I understand.  I think there may be people at my school who aren't American citizens also (I'm American though).

What American competitions can you participate in? There are American competitions overseas?

Yep.  I'm taking the AMC 10 and AMC 12 this year.

Nani?  Hai boku wa wakarimasu nihongo desu.  Hajimemashite de nihongo desuka?

I can understand Japanese (you asked if I can understand the language in your previous post), but I'm horrible at speaking and writing Japanese, I can understand and speak it relatively good for an American citizen I would say.  I am by no means fluent though :-(.  Hmm...by any chance is Japanese another language of yours?  Because if so, that's awesome.
And well that's all for now.  Sayonara.  Oyasuminasai!

You don't need to know calculus to do well on the math olympiads.  Sometimes it can be valuable but there are a lot more topics as well that you will need to look over, number theory's one example.

[quote="Binary Star"]Oh lucky you. I wish I had a green card or something, heh.

One of the things I don't like being an international is because I'm not eligible for RSI and things like that. Plus the olympiads I will participate in will be way hard for me since I have to take it in Spanish <_<.[/quote]
May I ask why you have to take it in spanish? Maybe you're not talking about the math olympiad, I'm not sure, because in that one you are allowed to get copies of the paper in like almost as many languages as you want (ok theres probably a limit of 3 or something)..

I don't speak Japanese, but I'm in the process of learning it right now ^_^. It's coming along pretty easily because I speak Korean.

I asked the guy in charge of the math olympiad if I could take it in Spanish. I think the reason I have to take it in Spanish is everyone else does, and I doubt they will get a translator and stuff just to accomodate me.

Usually they print them in more than one language.


I guess it's easier to see if you note that $ 123\equiv\minus{}6\pmod{129}$.

Then it's just multiplying over and over again...

[quote="cdymdcool"]what's 123^131 mod 129[/quote]
[hide]
We use Carmichael Function $ \lambda(129)\equal{}42$

Then $ 123^{42}\equiv 1 \pmod {129}$

$ 123^{131}\equal{}123^{42\times 3}\cdot 123^{5}\equiv (\minus{}6)^5 \pmod {129}$

And $ (\minus{}6)^4\equal{}1296 \equiv 6 \pmod {129}$

$ \therefore (\minus{}6)^5 \equiv \minus{}36 \equiv \boxed {93} \pmod {129}$

Reference of  Carmichael Function :

[url]http://en.wikipedia.org/wiki/Carmichael_function[/url]  [/hide]

[quote="math154"]The problem is that $ \gcd(123,129)\neq1$, so the Carmichael Function is invalid.[/quote]

Sorry ,it is not correct ! :lol:

Do it again,

$ 123 \equiv \minus{}6 \pmod  {129}$

$ (\minus{}6 )^4\equal{}1296\equiv 6 \pmod {129}$

$ \therefore 123^4 \equiv  6 \pmod {129}$

$ 123^{131}\equal{}123^{128}\cdot 123^3 \equal{}(123^4)^{32}\cdot 123^3$

$ (123^4)^{32}\equiv  6^{32} \equiv ((6^4)^4)^2\equiv 6^2\pmod {129}$

Thus $ 123^{131}\equiv 6^2\cdot (\minus{}6)^3\equiv \minus{}36 \equiv \boxed{93}\pmod {129}$

Right? :P   :D

[quote="vinskman"][quote="math154"]The problem is that $ \gcd(123,129)\neq1$, so the Carmichael Function is invalid.[/quote]

Sorry ,it is not correct ! :lol:

Do it again,

$ 123 \equiv - 6 \pmod {129}$

$ ( - 6 )^4 = 1296\equiv 6 \pmod {129}$

$ \therefore 123^4 \equiv 6 \pmod {129}$

$ 123^{131} = 123^{128}\cdot 123^3 = (123^4)^{32}\cdot 123^3$

$ (123^4)^{32}\equiv 6^{32} \equiv ((6^4)^4)^2\equiv 6^2\pmod {129}$

Thus $ 123^{131}\equiv 6^2\cdot ( - 6)^3\equiv - 36 \equiv \boxed{93}\pmod {129}$

Right? :P   :D[/quote]

This is correct. :)

[quote="vinskman"][quote="cdymdcool"][quote="math154"][hide="Bashed Solution"]
Notice that $ 123^7\equiv123\pmod{129}$, so

$ 123^{131}\equiv(123^7)^{18}\cdot123^5\equiv123^{23}\equiv(123^7)^3\cdot123^2\equiv123^5\equiv\boxed{93}\pmod{129}$.
[/hide]
[hide="CRT Solution"]
$ 129 = 3\cdot43$, so we can just find $ 123^{131}\pmod3$ and $ 123^{131}\pmod{43}$ to find $ 123^{131}\pmod{129}$.

$ 123^{131}\equiv3^{131}41^{131}\equiv0\pmod3$, and since $ \phi(43) = 42$,

$ 123^{131}\equiv37^{131}\equiv37^5\equiv( - 6)^5\equiv - 7776\equiv7\pmod{43}$.

Solving the system $ x = 3a = 43b + 7$ by taking both sides $ \bmod3$, we get

$ 3a\equiv0\equiv43b + 7\equiv b + 1\pmod3\implies b\equiv2\pmod3\implies b = 3c + 2$

$ \implies x = 129c + 93\implies x\equiv\boxed{93}\pmod{129}$.
[/hide][/quote]

how did you get the first step in your bashed solution?[/quote]

I wonder what means  "CRT"  solution.[/quote]

CRT is the [url=http://www.artofproblemsolving.com/Wiki/index.php/Chinese_Remainder_Theorem]Chinese Remainder Theorem[/url].

[quote="math154"][/quote][quote="vinskman"][quote="math154"]The problem is that $ \gcd(123,129)\neq1$, so the Carmichael Function is invalid.[/quote]

Sorry ,it is not correct ! :lol:

Do it again,

$ 123 \equiv - 6 \pmod {129}$

$ ( - 6 )^4 = 1296\equiv 6 \pmod {129}$

$ \therefore 123^4 \equiv 6 \pmod {129}$

$ 123^{131} = 123^{128}\cdot 123^3 = (123^4)^{32}\cdot 123^3$

$ (123^4)^{32}\equiv 6^{32} \equiv ((6^4)^4)^2\equiv 6^2\pmod {129}$

Thus $ 123^{131}\equiv 6^2\cdot ( - 6)^3\equiv - 36 \equiv \boxed{93}\pmod {129}$

Right? :P   :D[/quote]

This is correct. :)

[quote="vinskman"][quote="cdymdcool"][quote="math154"][hide="Bashed Solution"]
Notice that $ 123^7\equiv123\pmod{129}$, so

$ 123^{131}\equiv(123^7)^{18}\cdot123^5\equiv123^{23}\equiv(123^7)^3\cdot123^2\equiv123^5\equiv\boxed{93}\pmod{129}$.
[/hide]
[hide="CRT Solution"]
$ 129 = 3\cdot43$, so we can just find $ 123^{131}\pmod3$ and $ 123^{131}\pmod{43}$ to find $ 123^{131}\pmod{129}$.

$ 123^{131}\equiv3^{131}41^{131}\equiv0\pmod3$, and since $ \phi(43) = 42$,

$ 123^{131}\equiv37^{131}\equiv37^5\equiv( - 6)^5\equiv - 7776\equiv7\pmod{43}$.

Solving the system $ x = 3a = 43b + 7$ by taking both sides $ \bmod3$, we get

$ 3a\equiv0\equiv43b + 7\equiv b + 1\pmod3\implies b\equiv2\pmod3\implies b = 3c + 2$

$ \implies x = 129c + 93\implies x\equiv\boxed{93}\pmod{129}$.
[/hide][/quote]

how did you get the first step in your bashed solution?[/quote]

I wonder what means  "CRT"  solution.[/quote][quote="math154"]

CRT is the [url=http://www.artofproblemsolving.com/Wiki/index.php/Chinese_Remainder_Theorem]Chinese Remainder Theorem[/url].[/quote]

Thanks for your instructions.


Hi guys, this is the first time I've EVER been in a forum, so I don't know how this works, but I just post a problem, right? How hard can it be??
This was a problem from the 2007 AMC 12 B. For some reason, I cannot find any links between the information given.

14. Point P is inside equilateral triangle ABC. Points Q, R, and S are the feet of the perpendiculars from P to AB, BC, and CA, respectively. Given that PQ = 1, PR = 2, PS = 3, what is AB?
(A) 4
(B) [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/84507545f131b7108866546c72caaa20.gif[/img]
(C) 6
(D) [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/9f525fcad8e41db98394ae9281e4da10.gif[/img]
(E) 9

Thanks a lot if you can help!
darkPrince

I already made this thread  :|

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135691

Oh, I'm sorry! I didn't know!
Thanks for telling me about this t0rajir0u.
darkPrince


Why are aqua complexes generally pink in colour? (I dunno the answer)  :D

It is because the net dipole effect of the H2O group and the complex tend to balcance so make it look pink(Only a guess). :maybe:

what does dipole effect have to do with colour?  :o   :maybe:

copper aqua complexes are blue. can you tell me what aqua complexes other than that of manganese are pink

Ti complex and also something else i dont remember  :P

that doesnt mean that aqua complexes are generally pink. its just three cases against so many :roll:

I think only aqua complexes of Manganese are pink.........

[hide]i think this is because Mn having 5 unpaired electrons absorbs almost all of the radiation in the visible spectrum and thus emits colour complementary to it i.e white/pale pink...[/hide]

But im sure its true for Ti also  :D
And many of such compounds are also pink. That's why i asked this  :D

Can you tell me what comp. of Ti has pink colour??

obviously since we have been talking abt aqua complexes $ \ce{Ti(H2O)6}$

Aqua complexes are not in general pink. In fact, they couldn't have all the same colour.

So is their colour the genl colour of the cation?  :)

I think so because the cause of the colour is the no of unpaired electrons.

truly speaking i hate such questions to the core...
why are transition metals generally paramagnetic
why are they generally coloured
why are they....
and CBSE Answers
1)because they have unpaired electrons :o  so why have transition metals
2)because of d-d transition...all colours are due to transitions

the best answer i would suggest is experimentally observed if the question is true :)

I have 2 more examples for u:-

1.[Co(H2O)6]2+ - Pink
2.[Fe(H2O)6]2+ - Yellow

So ur question is not very well framed.


Bonne année à cette partie française du forum ...

Visiblement, il n'y a pas foule !

Salut Alain,

Une bonne année à toi, aux autres membres de ce forum et des autres  :)

t�T�

Une bonne année à tous aussi :)

Je parle un peu de français :P mais je fais beaucoup d'erreurs ...

Bonne annee tout le monde! :) (faut bien peupler ce forum!)

Bonne année à tous.
Il semble qu'il y ait finalement quelques français alors ici...

Pierre.

Aller, on se lance !!

Calculer
$\int_0^{+\infty} \frac{\cos(t)}{1+t^2}dt$

CA marche !!

Bon, heu,...je ne voudrais pas tre pnible tout de suite, mais je pense que si l'on veut que cela soit facile  suivre, il soit prfrable de poster des exos dans des fils de discussion spars, genre nouvel exo = nouveau fil, sauf s'il s'agit d'une continuit vidente du fil...Cela vitera les fils super longs ou plus personne ne sait qui rpond  quoi ou  qui et d'ailleurs ne lit plus, et facilitera aussi la recherche des rponses.

D'autre part, je rappelle que les forums 'principaux', genre advanced section, College playground...ne sont pas destins  tre remplacs, sans quoi chacun va se retrouver dans sa propre communaut, ce qui ne me semble pas trop l'objectif  ;)

Pierre, dsol d'tre super ch... ds le dpart.

Il ne faut pas tre dsol pour si peu !

Mon premier "message" tait l pour voir si tout fonctionnait normalement.

A la limite, en le dplaant aussi vite, tu as montr tout ton savoir-faire et ta capacit de raction  :)

[b]Bonne Anne 2005 !!!!!!![/b]

Salut,

je suis nouveau sur ce forum.
Je vous adresse  tous mes meilleurs voeux pour la nouvelle anne.

A bientt.

Dcidement, c'est plus une auberge espagnole qu'un forum franais ce truc  :D

Pierre.

[quote="harazi"]Alors j'imagine que je ne peux pas intervenir, n'est pas?  :P .[/quote]Mon dieu, beacoup de personne conne francais ici :P

'connaissent' Valentin  :D
'conne' a une autre signification en français  :D  :D  :D  :D

Wow! Harazi!!!!

Pierre.

Malhereusement, je connais la signification du mot utilise par Valentin!  ;) . Tout ce que j'espere c'est qu'il connait pas le bon mot. N'est pas, Valentin?  :P

[quote="harazi"]N'est pas, Valentin?  :P[/quote]Bien sur :P

[quote="Rodolphe2005"]Merci pour l'insulte valentin !!  :lol:[/quote]Avec plaisir :P

J'ai changé les fonts dans unicode ainsi certains des messages ne peuvent être lus plus  ;) Si vous voulez vous pouvez éditer les messages ...

Je vois ce texte sans tous artefacts

Translated with Promt 7 SE :)

bonjour à tous  :D  bonne année  2005 !!! meme si on est déjà à la sainte geneviève (khrââââsssss)  :lol:

Chrostiphe, que content de son premier message sur mathlinks  :)

[quote="Valentin Vornicu"]Mon dieu, beacoup de personne conne francais ici :P[/quote]

Sache Valentin qu'en français on accorde en général les choses (comme en roumain d'ailleurs) . Après correction ça donnerait :
"Mon dieu, beaucoup de personnes connes françaises ici"  :P  Ca sonnerait mieux si l'on invertissait  "connes" et "françaises", mais pour ce forum ça ira comme ça :P

 :lol:  Une excellente année 2005 quand même !! :lol:


At Tomi's middle school, 40\% of the students are boys.  Fifty
percent of the boys chose blue as their favorite color, while
20\% of the girl students chose blue as their favorite color.
Of the students who chose blue as their favorite color, what
fraction are boys?  Express your answer as a common fraction.

Let's say there are 100 students. That means that 40 are boys and 60 are girls. Half the boys like blue, so 20 boys like blue. 20% of the girls like blue, so 12 like blue. That means 20 + 12 = 32 kids like blue, and 20 are boys. Thus, $ \boxed{\frac{5}{8}}$ is our answer.


trigonometry
quadratics
modular arithmetic

Find all solutions to $ (1\minus{}\tan x)(1\plus{}\sin 2x)\equal{}1\plus{}\tan x$

[quote="cancer"]Find all solutions to $ (1\minus{}\tan x)(1\plus{}\sin 2x) \equal{} 1\plus{}\tan x$[/quote]

[hide="solution"]$ (1\minus{}\tan x)(1\plus{}2\sin x\cos x)$

$ \equal{}1\minus{}\tan x\plus{}2\sin x\cos x\minus{}2\sin^{2}x\equal{}1\plus{}\tan x$

$ \Rightarrow\sin x\cos x\minus{}\sin^{2}x\equal{}\frac{\sin x}{\cos x}$

$ \Rightarrow\cos x\minus{}\sin x\equal{}\frac{1}{\cos x}$

$ \cos^{2}x\minus{}1\equal{}\sin x\cos x$

$ \minus{}\sin^{2}x\equal{}\sin x\cos x$

$ \tan x\equal{}\minus{}1$

$ x\equal{}\frac{3}{4}\pi,\frac{7}{8}\pi...$[/hide]

$ \implies\sin{\pi/2}\plus{}\sin{2x}\equal{}\frac{\tan{\pi/4}\plus{}\tan{x}}{1\minus{}\tan{\pi/4}\tan{x}}$.
$ \implies 2\sin(\pi/4\plus{}x)\cos(\pi/4\minus{}x) \equal{}\tan(\pi/4\plus{}x)$.
$ \implies 2\cos(\pi/4\minus{}x) \equal{} 1/\cos(\pi/4\plus{}x)$. :maybe:

EDIT:
$ \implies 2\cos(\pi/4\minus{}x)\cos(\pi/4\plus{}x) \equal{} 1$
$ \implies\cos(2x)\plus{}\cos(\pi/2) \equal{} 1$
$ \implies\cos(2x) \equal{} 1$. $ x \equal{} k\pi$?

EDIT2:
Division by 0 in going from step 2 to 3, if $ \sin(\pi/4\plus{}x)\equal{}0\implies x\equal{}3\pi/4\plus{}k\pi$.

$ \boxed{x\equal{}k\pi, 3\pi/4\plus{}k\pi}$, for integer k :) .

[quote="cancer"][quote="jli"]

$ \Rightarrow\sin x\cos x\minus{}\sin^{2}x \equal{}\frac{\sin x}{\cos x}$

$ \Rightarrow\cos x\minus{}\sin x \equal{}\frac{1}{\cos x}$

[/quote]

Can't $ \sin x$ be $ 0$ here?[/quote]

Indeed, $ 0$ is also a solution. Nice catch  :D

Here's how I factored it:
[hide]$ \left(1\minus{}\frac{\sin x}{\cos x}\right)(1\plus{}2\sin x\cos x) \equal{} 1\plus{}\frac{\sin x}{\cos x}$
$ 1\minus{}\frac{\sin x}{\cos x}\plus{}2\sin x\cos x\minus{}2\sin^{2}x \equal{} 1\plus{}\frac{\sin x}{\cos x}$
$ 0 \equal{}\frac{\sin x}{\cos x}\plus{}\sin^{2}x\minus{}\sin x\cos x$
$ 0 \equal{}\sin x\plus{}\sin^{2}x\cos x\minus{}\sin x\cos^{2}x$
$ 0 \equal{}\sin x(1\plus{}\sin x\cos x\minus{}\cos^{2}x)$
$ 0 \equal{}\sin x(\sin^{2}x\plus{}\sin x\cos x)$
$ \boxed{0 \equal{}\sin^{2}x (\sin x\plus{}\cos x)}$
[/hide]
Now I think you can actually solve it by making a quadratic out of the second factor, and doing whatever for the first one.

[quote="archimedes1"]$ \boxed{0 \equal{}\sin^{2}x (\sin x\plus{}\cos x)}$[/quote]
From this we have two equations which are valid. $ \sin^{2}x\equal{}0$ and $ \sin x\plus{}\cos x\equal{}0$

The first is easy to work with:
$ \sin^{2}x\equal{}0\rightarrow\sin x\equal{}0$
That's it, just find where $ \sin x\equal{}0$ and you're done.
General Solution: $ x\equal{}n\cdot\pi,\text{ }n\text{ is an integer.}$
Within the Unit Circle: $ x\equal{}0,\pi$

Slightly more complicated:
$ \sin x\plus{}\cos x\equal{}0\rightarrow\sin x\equal{}\minus{}\cos x$
$ \sin x$ only equals $ \minus{}\cos x$ at the axises and in the $ II$ and $ IV$ quadrants at their own $ \frac{\pi}{4}$.
General Solutions: $ \boxed{x\equal{}\frac{n\cdot\pi}{2}\text{ and }x\equal{}\frac{3\pi}{4}\plus{}n\cdot\pi,\text{ }n\text{ is an integer.}}$
Within the Unit Circle: $ \boxed{x\equal{}0,\frac{\pi}{2},\frac{3\pi}{4},\pi,\frac{3\pi}{2},\frac{7\pi}{4}}$

All of the solutions given in case 1 are included in case 2, so there is no reason to go any farther.

[hide]
Use $ \sin 2x\equal{}\frac{2\tan x}{1\plus{}\tan^{2}x}$ to get
\[ 1\plus{}\tan x\equal{}(1\minus{}\tan x)\cdot\frac{(1\plus{}\tan x)^{2}}{1\plus{}\tan^{2}x}\]
\[ \Rightarrow\tan^{2}x(1\plus{}\tan x)\equal{}0\]
\[ \Rightarrow x\equiv 0,\frac{3\pi}{4}\pmod{\pi}\]
[/hide]

[quote]Here's how I factored it:
$ \left(1\minus{}\frac{\sin x}{\cos x}\right)(1\plus{}2\sin x\cos x) \equal{} 1\plus{}\frac{\sin x}{\cos x}1\minus{}\frac{\sin x}{\cos x}\plus{}2\sin x\cos x\minus{}2\sin^{2}x \equal{} 1\plus{}\frac{\sin x}{\cos x}0 \equal{}\frac{\sin x}{\cos x}\plus{}\sin^{2}x\minus{}\sin x\cos x 0 \equal{}\sin x\plus{}\sin^{2}x\cos x\minus{}\sin x\cos^{2}x 0 \equal{}\sin x(1\plus{}\sin x\cos x\minus{}\cos^{2}x) 0 \equal{}\sin x(\sin^{2}x\plus{}\sin x\cos x)\boxed{0 \equal{}\sin^{2}x (\sin x\plus{}\cos x)}$

Now I think you can actually solve it by making a quadratic out of the second factor, and doing whatever for the first one.[/quote]

You can use the sum identity:
$ \sin{a}\plus{}\sin{b}\equal{} 2\sin{\frac{a\plus{}b}{2}}\cos{\frac{a\minus{}b}{2}}$:

convert $ \cos{x}$ into $ \sin{\frac{\pi}{2}\minus{}x}$
then use the identity and you have all solutions :D


superior algebra
superior algebra theorems

let G is a group  that order (G )= p.q  , p and q are prime numbers  .prove that G is  a group solution

Let $ p<q$. Use Sylows theorems for the prime $ q$.


Let f(x,y) be a function defined as

$ f(x,y) \equal{}\frac{x^{p}y^{q}}{x^{2}\plus{}xy\plus{}y^{2}}$ if $ (x,y)\neq (0,0)$; $ f(x,y) \equal{} 0$ if $ (x,y) \equal{} (0,0)$.

Find the relation between the real numbers p and q such that f(x,y) is continuous at (0,0).


probability
combinatorics unsolved
combinatorics

[quote="malcolm"]If I randomly choose $ n$ points on a stick, and break the stick at those points, what is the probability the pieces can form an $ n \plus{} 1$-gon?[/quote]

The pieces cannot form an $ n \plus{} 1$-gon
$ \Leftrightarrow$ the longest piece is longer than sum of all other sticks
$ \Leftrightarrow$ the $ n$ points are all on one side of the stick.

The probability for which $ \equal{}2 \times \frac{1}{2^n}\equal{}\frac{1}{2^{n\minus{}1}}$

So the required probability $ \equal{}\frac{2^{n\minus{}1}\minus{}1}{2^{n\minus{}1}}$

[quote="stephencheng"]the longest piece is longer than sum of all other sticks
$ \Leftrightarrow$ the $ n$ points are all on one side of the stick.[/quote]  This is obviously false.  See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5715 for a solution to a much more general problem (note that we go from circle to stick just by treating the first cut as the endpoints of the stick).

Interestingly, while I can find dozens of copies of the solution with "polygon" replaced by "triangle," and several threads in which this problem is mentioned, I wasn't able to find a thread with the solution to this problem.


calculus
integration
trigonometry
calculus computations

Find all real numbers $ x$ such that $ \int_0^x t^2\sin (x\minus{}t)\ dt\equal{}x^2$.

Chemistry?!

I don't see any clever way to do it, though some integrating by parts causes the left hand side to evaluate rather nicely to $ x^2 \plus{} 2\cos x \minus{}2$. (no Mathematica, I swear! :lol: )

That's so nice, though, it makes me feel like there is some really elegant way to do the problem...

[quote="kunny"]Find all real numbers $ x$ such that $ \int_0^x t^2\sin (x \minus{} t)\ dt \equal{} x^2$.[/quote]
hmmm.... how about if i take $ \frac {d}{dx}$ of both sides

$ \Rightarrow x^2 \sin \,(x \minus{} x)\; 1 \; \equal{} \; 2x$

but this only gives $ x \equal{} 0$  :(

maybe this is not the right way to do it...

Hey, let's solve $ x\equal{}1$ the same way. :)

$ 1\equal{}0$ ?!?

Good one kunny. I enjoyed it.  :D
[hide="solution"]\begin{eqnarray*}I
&=&\int t^2\sin(x-t)dt\\
&=&t^2\cos{(x-t)}-\int 2t\cos{(x-t)}dt\\
&=&t^2\cos{(x-t)}+2t\sin{(x-t)}-\int 2\sin{(x-t)}dt\\
&=&\boxed{t^2\cos{(x-t)}+2t\sin{(x-t)}-2\cos{(x-t)}+C}\\
\end{eqnarray*}
After applying the limits of integration we obtain $ x^2=x^2+2\cos{x}-2\Leftrightarrow \cos{x}=1$. Thus all desired values of $ x$ are of the form $ x=2\pi n: n\in\mathbb{Z}$.[/hide]


probability
combinatorics proposed
combinatorics

Let there are $ n$ kind of  candies with prices 1,2,3,...,n in the shop. Bob bought $ k$ candies without see the price. ( It is possible too buy more than one candies from one group). Also Alice bought $ k$ candies without see the prices. Find the probability that Alice and Bob end up with the same total!

The question as written is not exactly coherent -- do you mean, "Find the probability that Alice and Bob end up with the same total?"  (I suspect that you do, because you've also posted this problem (in generatingfunctional form) in another part of the forum: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=271330[/url])

Also, in the future, please give your posts more informative names.  For example, this one could have been titled "Choosing candy of same price."


Let$ a,b,c,d,x,y,z,t$ be positive numbers such that:$ ax \plus{} by \plus{} cz \plus{} dt \equal{} xyzt$.Prove that:
$ x \plus{} y \plus{} z \plus{} t \plus{} 1 > \sqrt {1 \plus{} \frac {8}{\sqrt {3}} (\sqrt {a \plus{} b} \plus{} \sqrt {b \plus{} c} \plus{} \sqrt {c \plus{} d} \plus{} \sqrt {d \plus{} a} \plus{} \sqrt {a \plus{} c} \plus{} \sqrt {b \plus{} d})}$

No one interested in this inequality? :(


I need help with the following question:

Find three pairs of integers x and y such that xy – yx  = -1  (the first y and second x are raised powers)

Thanks for any help

[quote="EricLeafs"]I need help with the following question:

Find three pairs of integers x and y such that $ x^y \minus{} y^x\equal{} \minus{} 1$  (the first y and second x are raised powers)

Thanks for any help[/quote]

$ (1,2)$
$ (2,3)$
$ (0,1)$
$ (0,2)$
$ (0,3)$
$ (0,4)$
...


Euler
inequalities
geometry
circumcircle
geometric transformation
homothety
cyclic quadrilateral

The points $M, N, P$ are chosen on the sides $BC, CA, AB$ of a triangle $\Delta ABC$, such that the triangle $\Delta MNP$ is acute-angled. We denote with  $x$ the length of the shortest altitude of the triangle $\Delta ABC$, and with $X$ the length of the longest altitudes of the triangle $\Delta MNP$. Prove that $x \leq 2X$.

For simplifying the notations, denote $y$ the longest altitude of $\triangle{MNP}$.

Also denote $H$ the orthocenter of $\triangle{MNP}$ and $\triangle{A'B'C'}$ the pedal triangle of $H$.

Thus $\frac{\sum{HA'}}{x}\geq \sum{\frac{HA'}{h_{a}}}= \sum{\frac{[HBC]}{[ABC]}=1}$

Therefore $x\leq \sum{HA'}$ $(1)$.

Now let's see that $\sum{HA'}\leq \sum{HM}$, and try to prove that $\sum{HM}\leq 2y$ $(2)$.

Suppose that $MM_{1}$ is the longest altitude of $\triangle{MNP}$, therefore $(NP)$ is the shortest side.

Now call $D$ the symmetrical point of $H$, w.r.t to $M_{1}$. Thus $D$ is on the circumcircle of $\triangle{MNP}$.

Observe that $\sum{HM}= HM+DN+DP$, and rewrite $(2)$ as:

$DN+DP \leq MD$, which is obvious by the Ptolemy theorem in the cyclic quadrilateral $MNDP$
($MD \cdot NP = DN \cdot MP+DP \cdot MN$, but $NP = \min\left\{NP,PM,MN\right\}$), so $(2)$ is proved.

Thus by $(1)$ and $(2)$ we have that $x\leq 2y$.

Remark: Another nice solution involves an homothety and a translation, I will post it after a month.

An approach without using orthocenters...
[hide]Consider XYZ the midpoint triangle of ABC, that is, where X is the midpoint of BC, Y the midpoint of AC, and Z the midpoint of AB. Obviously, XYZ is 1/4 of the area of ABC.
Lemma 1: twice any height of XYZ >= the shortest height of ABC
Proof:[hide]pretty straightforward: XYZ's longest height is half of ABC's longest height, so twice its longest height must be longer than ABC's shortest height. In fact, the twice of any height of XYZ is greater than or equal to the height of ABC.[/hide]

Proceed by contradiction. suppose 2X < x
We know that M and N cannot both be on CY and CX respectively, or else the height from P to MN is longer than the height from Z to YX, which would make our supposition false, by Lemma 1. Likewise, both M and P cannot both be on BX and BZ, and both N and P cannot both be on CX and CY respectively. Thus, M N and P are either on CX, AY, and BZ respectively, or they are on CY AZ and BX respectively. Let's call triangles happy if they meet this criterion.

Lemma 2: happy triangles have a larger area than XYZ.
Proof: [hide]Let MNP be happy. Then [MNP]>=[XNP]>=[XYP]=[XYZ], because it is clear that the height from M to NP is less than the height from X to NP, and so forth. Thus [MNP] >  = [XYZ].[/hide]

Lemma 3: The longest height of any triangle of area [MNP] is greater than sqrt(3*[MNP])
Proof: [hide]Straightforward using areas.[/hide]

Lemma 4: The shortest height of any triangle of area [ABC] is less than sqrt(3*[ABC]).
Proof: [hide]straightforward with areas.[/hide]

MNP is happy, so 2X > = 2 sqrt(3*[MNP])>=2sqrt(3*[XYZ]) =  sqrt(3*[ABC]) >= x --> 2X >=x, and so we have contradiction with our initial assumption. Thus we are done.

P.S: pohoatza, it's been way over a month :P[/hide]

[quote="Scrambled"]
We know that M and N cannot both be on CY and CX respectively, or else the height from P to MN is longer than the height from Z to YX, which would make our supposition false, by Lemma 1.
[/quote]
But I think the counterexample is easy to come up with :( Am I right?
Dear pohoata,
Could you post the solution involves an homothety and a translation?
I think this problem is interesting and hard :) Thank you!

Ok, I will post it as soon as I will remember it   :D (it has been a while).

Here is a different approach...

Let $D,E,F$ be the midpoints of $BC,CA,AB$. Assume that $P,N$ lie on segments $AF,AE$, respectively. If $MH$ is the altitude from $M$ to $NP$ then $H$ lies on $NP$ and therefore $H$ lies inside triangle $AFE$, Therefore, if $MH \cap FE = H'$, then $MH > MH' $. Then the distance from $M$ to $NP$ is greater than the distance from $M$ to $FE$, which equals half of the distance from $A$ to $BC$ and we're done. Therefore we can assume
$P \in AF, N \in EC, M \in BD$ (or $P \in BF, N \in EA, M \in DC$, but this is analogous).

Fix triangle $ABC$, and consider the triangle right-angled or acute $MNP$ (satisfying the above condition) which minimizes $X$, and assume $X < x/2$. This is possible since we are taking $M,N,P$ in closed intervals. If, say, $M=D$, then notice the altitude from $P$ to $MN$ is greater than half of the altitude from $C$ to $AB$, so we're done. If, say, $M=B$, then the altitude from $N$ to $MP$ is greater than or equal to that of $E$ to $BA$ which is half of the altitude from $C$ to $AB$, so we're done. Otherwise, we can assume that $M,N,P$ are different from $A,B,C,D,E,F$.

If we start to move $N$ towards $E$ (unless $ \angle N = 90$), then $X$ will increase, by definition, but the altitude from $P$ to $MN$ decreases and so does the altitude from $N$ to $MP$. Therefore, the altitude from $M$ to $PN$ must equal $X$ (otherwise, the resulting triangle $MN'P$ will have a smaller $X$, since the altitude from $M$ to $PN'$ increased but is still smaller than the original $X$). Similarly, the altitude from $N$ to $PM$ and the altitude from $P$ to $MN$ equal $X$. Therefore, $MNP$ is equilateral, or right-angled and isosceles.

If it is right-angled and isosceles (with side $h$), say at $M$, notice that the distance from $P$ to $BC$ is at least half of that from $A$ to $BC$, and so it is $<h$. But it is also at least $PM = h$, contradiction!

So proving it for $MNP$ equilateral is enough. Let $h$ be its altitude. Notice that, WLOG, $\angle A \ge 60$, and thus $A$ lies inside the reflection of the circumcircle of $MNP$ across $NP$. This implies that $A$ lies inside the circle centered at $M$ with radius $2h$ (since this circle engulfs the previous one). Thus, the altitude from $A$ to $BC$ is less than or equal to $AM$, which is less than or equal to $2h$.

Thus, we're done.


integration logarithms calculus trigonometry function algebra domain

Why the hell do olympiad kattans like chappli n soumya all get set 3??? :mad: :mad:

God-O-God this is a brutal joke :mad:

Plz be free to give in your comments here and no filters here for a while that's the min i can say as a consolation

[color=blue]FORTUNE FAVOURS THE BRAVE[/color]

Dei i might wanna kill you now but i cant but atleast pardesi will :mad: :mad: :mad:

no i am happy with my paper i think i got everything else right and will get agrace in a 6 mark question...and that $ 1$ has to be changed to $ \infty$ and the answer is $ \frac{\pi^{2}}{8}$..we all have written the answer to the corrected question :)

WHO TOLD YA THAT THAT WAS THE CORRECTED QUESTION???

no actually that's the only way something like this can be integrated :P  :rotfl:  :rotfl:

with the actual question, one would have to compute the value of $ \int_0^{ \frac {\pi}{4}} \log(\sec x) dx$

BUT HOW CAN YOU ASSUME THAT THAT WAS THE ACTUAL QUESTION??? WHO TOLD YOU THAT U CAN DO STH LIKE THAT?? WILL THEY GIVE MARKS EVEN FOR TRYING TO DO THIS SUM TILL THIS INTEGRAL???

yes that's why i said $ \infty$ would giev the upper limit as $ \frac {\pi}{2}$ so that we would have to get $ \int_{0}^{\frac {\pi}{2}} \ln \sec x dx = \int_{0}^{\frac {\pi}{2}} - \ln \cos x dx$

yes and that would be another question in the paper :D :D that of logsinx

but answer my other part to the best of ur knowledge???

that was actually.....[hide="......."][size=200][color=blue]CHUMMA JUJUBI [/color][/size]  :rotfl:  :rotfl:  :rotfl:  :D [/hide]

YES I AGREE :) BUT CAN YOU TELL ME THE ANSWER TO MY QUESTION THAT DO YOU KNOW ANYTHING ABT THE MARKING PATTERN FOR THAT??? ARE U SURE???

WOLFRAM HAS CONFIRMED THAT THAT CANNOT BE DOME INDEFINITELY PHEW :D

Vijeyandran says he has solved
 :rotfl:  :rotfl:  :rotfl:
(That boy is a boy in our school)

no we found out his mistake in the bus itself :D :D
n srinath can brag that he got it howmuchever he wants but the truth is that if u do parts 4 times like he said, u will get 0 = 0:rotfl:

[quote="Cyber-Shreyas"]*Shre Swtiches to Srinism/Anitism

Are you guys referring to me? i am total waste in Computer Science and a big zero in C++. I got through to the Info TC on pure fluke and also by looking at the monitor of the person near me. I am praying that I will atleast pass in C++ and am planning to put a rs.500 note inside my paper.

I hope I will pass. I have already started my preparation. I am now only learning C++ from the beginning cout and cin... etc.

*Back to Shre Mode

Good question
1) What was the initial name of C++?
2) [u][b]What are the 2 languages on which C is based on? [/b][/u]
3) Why do you think OOP is a total waste?

My Ans for 3rd: Simply because it gives more importance to data rather than logic making it useless in olympiads. And after all, if your logic is good and you have a little tech-sense, you can master any language.[/quote]

U asked for C da :rotfl:

[quote="Cyber-Shreyas"]

B is named after some idiots wife's pet name. I forget who. It was a quiz question.
[/quote]
i thought that was quincy  :rotfl:

[quote="apeksharma"]i think b can take values of R+ as the expression (x+bx^2) should be +ve.
and so answer should be like
If b=0 :a=-2 ,c=0
if b>0:  a=-1.5,c=.5[/quote]
how is the funcion defined if $ b\equal{}0$??
also as i explained before it doesn't matter whether $ b>0$ or $ <0$ the function is always defined n an $ \epsilon^{\plus{}}$ neighbourhood of 0

but the fact is you arent a damn bit bothered for other parts of the fn are you? :)

EDIT why does d a m n come out to be a damn?  :huh:

now hy shud i be...
see if i choose $ x < \frac{1}{b'}$ where $ b\equal{}\minus{}b'$ then we have the function defined and for a limit to exist at a point the function ahs only to be defined in some $ \epsilon$ neghbourhood not necessarily everywhere

[quote="SOUMYASHANT NAYAK"]We were not given domain and codomain of function....

So for a truce, we assume it to be from C-> C...
Then by definition of continuity for complex functions,
we find the necessary (although may not be sufficient) condition for continuity...
(as we would have to come to zero from all directions in the plane....but by slight hard work I think such a function to suppport our ans can be constructed....)

But b=0 is definitely and positively [b]wrong[/b]...[/quote]
why ??? when you substitute B=0 you get 0/0 form whcih is resolved by multiplying and dividing by the requiired conjugate and removing b whats wrong or you say when b=0 the value is not a definite one and thus cannot be evaluated thats what you say no??/ but i think CBSE will award marks for all i think as chappli is awaiting a centum this time and he will even approach ashok ganguly to provbe this  to get his hundred!!! :D

[quote="Albert Einstien"]
why ??? when you substitute B=0 you get 0/0 form whcih is resolved by multiplying and dividing by the requiired conjugate and removing b whats wrong or you say when b=0 the value is not a definite one and thus cannot be evaluated thats what you say no??/ but i think CBSE will award marks for all i think as chappli is awaiting a centum this time and he will even approach ashok ganguly to provbe this  to get his hundred!!! :D[/quote]

OMG, has it come to this????

Division by zero is [b]not defined[/b]
(not yet defined with a productive logical consequence by any mathematician)

[b]0/0 is not indeterminate[/b]
[b]It is not defined[/b] (as plain and straight as that...)
You cannot resolve a 0/0 or anything like that by whatever you wish to do....

You are confusing '0/0 form' with '0/0'...
In the former limits are involved in which case your normal algebraic operations can be performed....
Please throw your NCERT book out of the window and study a good book on calculus or better real analysis ... :rotfl:  :rotfl:

@Srini: Look da. you just cannot substitute b=0. If you substitute b=0, then the function's definition for certain intervals become problematic.

Note this:

f(x) = $ \frac{x^2 \minus{} 3x \plus{} 2}{x\minus{}1}$
is not the same as
$ f(x) \equal{} x\minus{}2$

oh thanks for clearing my misconception guys!! yeah you all are rite too much of nving the ncert book the last two days has changed my functioning of my brain!!!! got to recharge!!! :D

[quote="Cyber-Shreyas"]
f(x) = $ \frac {x^2 \minus{} 3x \plus{} 2}{x \minus{} 1}$
is not the same as
$ f(x) \equal{} x \minus{} 2$[/quote]
y it is not the same??????

Because the domain of definition of the 2 functions are not the same. You cannot cancel x-1 whenever you want. Its only when $ x\neq1$. They behave the same when $ x\neq1$.

[quote="Cyber-Shreyas"]Because the domain of definition of the 2 functions are not the same. You cannot cancel x-2 whenever you want. Its only when $ x\neq2$. They behave the same when $ x\neq2$.[/quote]

Dont you mean $ x\neq1$???  :)

We arent going to get 6 marks for that question Cha Why did i miss such an easy subsn!!! Wth cha! nooooooooo i learnt that soooooo long ago in 1st std :( Why didnt i think of that??

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1072333#1072333]Sappa Answer[/url]

[hide="Why the hell did i miss that"] Cha all my teachers of the primary would have been sooo disappointed that i didnt get that :rotfl: [/hide]

Oh my god the answer is so simple ah???? cha dont worry rohit but for the steps you have written i think u deserve atleast 7??? :rotfl:


Let $ x_1,x_2,...,x_n$ be positive real numbers. Prove that
$ \left.\frac {x^5_1}{x^4_1 \plus{} x^2_1x^2_2 \plus{} 4x^4_2} \plus{} \frac {x^5_2}{x^4_2 \plus{} x^2_2x^2_3 \plus{} 4x^4_3} \plus{} \cdots \plus{} \frac {x^5_n}{x^4_n \plus{} x^2_nx^2_1 \plus{} 4x^4_1}\geq\frac {x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n}{6}\right.$

(Thanks, [b]Abdek[/b]! I've made the correction, and now it's OK. It was a typo, copy&paste, and finally... the mistake. Sorry!)

[quote="marin.bancos"]Let $ x_1,x_2,...,x_n$ be positive real numbers. Prove that
$ \left.\frac {x^5_1}{x^4_1 \plus{} x^2_1x^2_2 \plus{} 8x^4_2} \plus{} \frac {x^5_2}{x^4_2 \plus{} x^2_2x^2_3 \plus{} 8x^4_3} \plus{} \cdots \plus{} \frac {x^5_n}{x^4_n \plus{} x^2_nx^2_1 \plus{} 8x^4_1}\geq\frac {x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n}{6}\right.$[/quote]

Maybe  you mean  $ \frac {1}{10}$ in the $ RHS$ because when we take $ x_1 \equal{} x_2 \equal{} x_3 \equal{} ... \equal{} x_n \equal{} 1$ we get $ \frac {n}{10} \ge \frac {n}{6}$ which is not true  :)

Dear [b]Abdek[/b],
Thanks for your solution. You  said that your solution is not very nice. But you know that, all the time: [b]"My solution to the problem is the best"[/b] .
In this respect, I can assure you that my solution (for the general case) is shorter and simpler. I'm still waiting for new ideas.

Abdek's idea is just so right and he was nearly close to the solution of general problem.

[i][u][b]Solution[/b][/u][/i]

We have

\[ x_1 \minus{} \frac{x_1^5}{x_1^4\plus{}x_1^2x_2^2\plus{}4x_2^4} \le \frac{x_1x_2^2(x_1^2\plus{}4x_2^2)}{3(x_2^4\plus{}3x_1^2x_2^2)} \equal{} \frac{x_1}{3}\plus{}\frac{x_1x_2^2}{x_1^2\plus{}x_2^2} \le \frac{x_1}{3}\plus{}\frac{x_2}{2}\]

We are done!

This solution is dedicated to Abdek.

[quote="great math"]Abdek's idea is just so right and he was nearly close to the solution of general problem.

[i][u][b]Solution[/b][/u][/i]

We have
\[ x_1 \minus{} \frac {x_1^5}{x_1^4 \plus{} x_1^2x_2^2 \plus{} 4x_2^4} \le \frac {x_1x_2^2(x_1^2 \plus{} 4x_2^2)}{3(x_2^4 \plus{} x_1^2x_2^2)} \equal{} \frac {x_1}{3} \plus{} \frac {x_1x_2^2}{x_1^2 \plus{} x_2^2} \le \frac {x_1}{3} \plus{} \frac {x_2}{2}\]
We are done!

This solution is dedicated to Abdek.[/quote]
Dear [b]Great math[/b] Thank you Very much for your nice solution Which is  more nicer and simpler than mine  :lol:

Dear [b]great math[/b] and [b]Abdek[/b],
Thank you for your solutions! [b]Great math[/b]'s solution is also my solution to the problem. Congratulations my friends!


geometry
trapezoid
perimeter
ratio
MATHCOUNTS
Pythagorean Theorem

Let $a$ be the side of the equilateral triangle on top (the small one).  Let $b$ be the left and right sides of the trapezoid.  By equating the perimeters you have:

$a+a+a=a+\left(a+b\right)+b+b$

$a=3b\implies \frac{a}{b}=\boxed{3}$

Finishing what was done above:

[hide]The area of the smaller triangle is $\frac{a^2\sqrt{3}}{4}=\frac{9b^2\sqrt{3}}{4}$. The area of the trapezoid is $\frac{1}{2}h(a+(a+b))$ where $h$ is the height.

To find $h$, drop altitudes from the trapezoid, splitting the bottom into one piece with length $a$ and two with length $b/2$. It follows from Pythagorean Theorem that $h=\frac{b\sqrt{3}}{2}$.

The area of the trapezoid is then $\frac{1}{2}*\frac{b\sqrt{3}}{2}(2a+b)=\frac{7b^2\sqrt{3}}{4}$ and the ratio is $\boxed{\frac{9}{7}}$.[/hide]

[quote="gmslinger"]A line intersects two sides of an
equilateral triangle and is parallel to
the 3rd side. If the smaller triangle
and the trapezoid formed have equal
perimeters, determine the ratio of
the area of the smaller triangle to that
of the trapezoid. Express your answer
as a fraction a b.
 :huh:
**I have spent many hours on this problem, and I finally think I found the answer, but i want to see what you guys get.**[/quote]

Draw an equilateral triangle. Label the sides a. Draw the line parallel to one of the sides of the triangle. Label this b. You now have an equilateral triangle with sides b, and a trapezoid with bases, b and a, and sides, a-b. Equating the perimeters of these two:

$3b=b+a+2a-2b$
$4b=3a$
$b=\frac{3a}{4}$

The area of the smaller equilateral triangle is $(\frac3{4})^2=\frac9{16}$ the area of the equilateral triangle with side length a.
The area of the trapezoid is $1-(\frac3{4})^2=\frac7{16}$ the area of the equilateral triangle with side length a.
This makes the ratio $\frac{\frac9{16}}{\frac7{16}}=\frac9{7}$

Edit: Thinking about it again, this kind of resembles a MATHCOUNTS problem.

[hide]I said, let a side of the smaller triangle be $a$, and let the original triangle have side length $1$.  Then, the perimeter of the smaller triangle is $3a$, while that of the trapezoid is $1+a+1-a+1-a=3-a$.  Hence, $3a=3-a \implies a=\frac{3}{4}$.  So then, the smaller triangle has area $\frac{9}{16}$ of the original triangle, and thus, the ratio of the areas of the small triangle and the trapezoid is $\frac{9}{7}$.[/hide]

Oh, right.  For some reason I thought you wanted the ratio of the sides, my bad.

[quote="gmslinger"]How on earth did you get that so fast!!!  :o It took me forever to come up with the correct way to do it. Anyway, good job the answer is 9/7.[/quote]

Did you not use algebra?  ;)  Just let a side of the original triangle be 1, and a side of the small triangle be some variable.  Then express each perimeter in terms of the variable, that is, $3a$ and $3-a$.  Solving for $a$ gives you $\frac{3}{4}$, in which case, you just plug in and get the answer.  It's a pretty basic geometry question...One unknown, one equation.
Which part took you the longest?  You have to use what you know and apply it to what you don't know to figure out those things you don't know.

Well first of all I was using 1/2 base times height, and instead of solving for the area of the larger triangle and subtracting the area of the smaller triangle, I was attempting to find the area of the trapezoid itself. However I did finally think to do it the way you guys did, however it took a lot of trial and error. Thanks for the help though!!!! :coolspeak:

You don't even need to find the actual area.  Just use ratios between area: that's all one cares about in this problem.
Why use trial and error?  Just use algebra...


[quote="jclarkemathL314159"]$2v+w+x+y+z=4$
$v+2w+x+y+z=5$
$v+w+2x+y+z=6$
$v+w+x+2y+z=7$
$v+w+x+y+2z=8$

Find $(v,w,x,y,z)$.

Hint[hide]notice the symmetry[/hide]
Huge hint[hide]add everything[/hide]
Hugest hint[hide]divide by 6[/hide][/hide][/quote]

haha.
[hide]$v=-1$
$w=0$
$x=1$
$y=2$
$z=3$[/hide]

I say... your hints are more like directions.  :lol:

[hide="Solution"]Add up all the equations to get $6(v+w+x+y+z)=30$.  So $v+w+x+y+z=5$.  Now just substitute in with the original five equations to get $\boxed{(-1,0,1,2,3)}$.[/hide]

oh I forgot. I had to find $(v,w,x,y,z)$. Okay, so:

[hide]$(-1,0,1,2,3)$[/hide]

 :)

[hide]Adding the equations and dividing by 6 gives v+w+x+y+z=5. Subtacting from each of the original equations gives (v,w,x,y,z)=(-1,0,1,2,3).[/hide]

[hide]

first add everything up to get 6(v+w+x+y+z)=30

Then divide by six to find that v+w+x+y+z=5

5+v=4
5+w=5
5+x=6
5+y=7
5+z=8

v=-1, w=0, x=1, y=2, z=3[/hide]

using your hints  :rotfl: [hide](-1,0,1,2,3)[/hide]


[quote="happyme"]A ladder is leaning against a house with its bottom 15 feet from the house. When the bottom of the ladder is pulled 9 feet further away from the house, the upper end slides 13 feet down. How many feet long is the ladder?[/quote]

[hide]Let the length of the ladder $x$. Let the original height of the ladder (from the ground) be $y$. From Pythagorean Theorem,

$x^{2}=15^{2}+y^{2}$
$x^{2}=(15+9)^{2}+(y-13)^{2}=24^{2}+y^{2}-26y+169$

$225+y^{2}=576+y^{2}-26y+169$
$y=\frac{576+169-225}{26}=20$

Therefore $x=\sqrt{225+400}=\boxed{25}$.[/hide]

k, here is how I do it.

[hide="clicky clicky"]
x = length of ladder
a = distance from top of the ladder to the ground

first of all, the ladder is 15 feet from the wall, and the distance from the top is unknown so

$a^{2}+15^{2}= x^{2}$

then the ladder is moved 9 more feet away from the wall, and it slides down 13 feet closed to the ground so
$(a-13)^{2}+24^{2}= x^{2}$

therefore

$a^{2}+15^{2}= (a-13)^{2}+24^{2}$
$26a = (-13)^{2}+24^{2}-15^{2}$
$a = 20$

sub our now known value of a into our first equation

$20^{2}+15^{2}= x^{2}$
$x = \sqrt{625}$
$x = 25$[/hide]

hehe sorry.. mistake on my part  :oops:

EDIT: yeah, i didn't sub my $x$ back in..

[hide]We have 2 right triangles of equal hypotenuses, so if we call y the height of the right triangle in the first triangle and x the hypotenuse of the triangles, we have the second triangle with leg lengths 24 and y-13, so thus:

$24^{2}+(y-13)^{2}=15^{2}+y^{2}$

$576+y^{2}-26y+169=225+y^{2}$

$576-26y+169=225$

$520=26y$

$y=20$

Now we plug in y in any of the right triangles and find [u][b]x=25[/b][/u].[/hide]

[quote="happyme"]A ladder is leaning against a house with its bottom 15 feet from the house. When the bottom of the ladder is pulled 9 feet further away from the house, the upper end slides 13 feet down. How many feet long is the ladder?[/quote]
[hide="answer"]
set up 2 quadratic equations, simplify, then substitute.
25 feet[/hide]


Find all quadruples $(a,b,m,n)$ of nonnegative integers that satisfy $a^{n}b^{m}=a^{m}b^{n}+a^{m+n}+b^{n}$.

See here http://www.mathlinks.ro/Forum/viewtopic.php?t=143486

Please don't post repeat!  :wink: Topic locked.


Let the sequence $ B_n$ have the property that $ B_0\equal{}B_1\equal{}1$, and $ B_{x\plus{}1}^{x\plus{}1}\equal{}B_x^x\plus{}B_{x\minus{}1}^{x\minus{}1}$ for all positive integers x. Find an explicit formula for $ B_n$.


Find all polynomial $P(x)$ with real coefficients such that :

$p(x^2)=p(x)^2=p(p(x))$

And, of course, when you do that you should also see what happens if we only have two of those three expressions equal.


Prove that ([n!*e]-1)/n is an integer for all n in N*.([x] is the integer part of x). :D

The result is trivial for n = 1. So let's assume that n > 1.

Using that e =  \sum 1/k!   for k=0 to +oo, we have :
n!*e - 1 = n*( \sum (n-1)!/k!) +  \sum n!/k!
where the first sum, say S, is over k = 0,...,n-1 and therefore is an integer,
and the second sum, say S', is over k >= n+1.
But 0 < S' < 1/(n+1) + 1/(n+1) 2 +.... = 1/n < 1.

Then, for n > 1, we have ([n!*e]-1)/n = S.

Pierre.


I made an image and I saved it in a PNG extension, then I converted it into a EPS one.

When compiling the document the image does not equal to the original one, how can I fix this?

Why not leave the image as a PNG? You can compile a document with this image format using pdflatex so the quality should remain the same. Don't forget that you will get a worse result if you try to scale the image, so ensure that it stays the same size.

stevem, can you please compile the following image, attach the PDF and put the preamble you used?

[img]http://img140.imageshack.us/img140/2152/fsdfq.png[/img]

Also, I'm using LaTeX => PS => PDF.

[quote="Null"]I'm using LaTeX => PS => PDF.[/quote]Ah, you must use LaTeX => PDF if you want to use PNG. Is there a reason that you need PS? You only need that if there are \specials or you are using PSTricks for example, otherwise there's no need for all the extra processing.

If you must have PS then it's easy to copy the PNG into OpenOffice Draw and save as EPS. OpenOffice gives a better quality output as attached PDF shows. The code is simple:
[code]\documentclass{article}
\usepackage{graphicx}
\begin{document}
\includegraphics{triangle.eps}
\end{document}[/code]

Okay, I've tried the methods you suggested but I'm not getting what I want.

Let's try with a different image:

[img]http://img182.imageshack.us/img182/2282/p100archimedeancirclear.png[/img]

I used OpenOffice Draw, copied the image and then I saved it as a EPS, but, when compiling on my latex document, the imagen did look horrible, which is not what I wanted.

I've tried so many times to get full quality but I didn't succeed... How, how can you put exactly that image into a PDF with the same quality?? :(

Like this. LaTeX1.pdf uses LaTeX=>PDF and image2.png whereas LaTeX2.pdf uses LaTeX=>PS=>PDF and image2.eps, converted by OpenOffice.

Ohh, thanks for all your help!

Well, definitely looks better latex1 than latex latex2, besides I modified the first one at 125% and that showed the quality that I was looking for.

I tried to do the same by myself, considering the image in post #5, I did

[code]\documentclass{article}
\usepackage{graphicx}
\begin{document}
\includegraphics{image.png}
\end{document}[/code]

but that produces a bigger image, a bigger one than what you did as in LaTeX1.pdf which uses LaTeX=>PDF and image2.png.

Why am I producing different results than you?

My result is attached below.

[quote="Null"]but that produces a bigger image, a bigger one than what you did as in LaTeX1.pdf which uses LaTeX=>PDF and image2.png.

Why am I producing different results than you?[/quote]I didn't have the image so I copied it from your post, pasted into OpenOfficeDraw and then saved as png with Compression 0 and got image2.png below. If however I just saved your image from imageshack and used that I get the larger image3.png below. Somewhere in the first process the png was resized.

Attached is the pdf showing the various effects.

For some reason, I try to do the same PDF as yours but I can't get something like that. I'm assuming I'm messing it somewhere, probably with the Open Office... Alright, I'll describe my process, from here we'll see if my procedure agrees with yours:

1) I have an image, consider the same located on post #5.
2) I copy the image and then I paste it into the Open Office, this is what I get:

[img]http://h.imagehost.org/0779/procedure.png[/img]

3) Go to File -> Export, then I choose the extension PNG, then a box appears and I put Compression in 0 and there's another option which says "entwine," (I think so, since I'm using the software in other language) dunno if I leave it activated or not, then I proceed and I get the image. (Saved it where is located my TEX file.)

4) The image exported looks like this:

[img]http://img521.imageshack.us/img521/2812/sdfg.png[/img]

5) Finally, I put on my latex file
[code]\documentclass{article}
\usepackage{graphicx}
\begin{document}
\includegraphics{image.png}
\end{document}[/code]

and the attached file is what I get.

-------------

I should get the same as you did in your last file, I should get the first image, but I'm not getting it.

Thanks for answering.

That has happened because you have exported the whole page as the image so your page image is on the second page of the pdf. Click on the diagram in OpeOffice so there are some green squares surrounding it showing that you have selected it (see below). Then export and you will get just the image itself.

Apart from this, all your steps are correct; I left "interlaced" ticked but I don't know whether that makes any difference at all.

Yes, I did that, I clicked on the image and those green squares appeared, then I exported but I still keep the same big image.

Don't know really why this is not working. :(

Edit: I realized when exporting, below says "selection," ahaha, I finally made it, but I'm not getting anyway the same image as you did on your last PDF.

Are you sure that [code]\includegraphics{image.png}[/code] is including the correct png?
View image.png in a program such as Paint. Is it the correct size? What size does Windows Explorer say it is? The images I saved are 442 x 267.

Well...

I picked the image from post #5, pasted on Open Office, exported the selection as PNG. Size of the original image is 15 KB, however, size of the exported image is 462 KB.

I compiled with the image exported, but that doesn't produce the same result as yours, but now, I scaled the image with 0.8 and set the percentage in 125% and that shows the quality I was looking for.

The thing that annoys me is that I definitely couldn't get the same as you, for example, when you took the image in post #5, and exported as PNG and finally compiled in the TEX, I only had to set the percentage as 125% and that showed the quality immediately.

[quote="Null"]I picked the image from post #5, pasted on Open Office, exported the selection as PNG. Size of the original image is 15 KB, however, size of the exported image is 462 KB.[/quote]That's because compression is turned off
[quote="Null"]and set the percentage in 125%[/quote]What do you mean? Are you talking about the percentage in the PDF viewer? If so, that is irrelevant because you will get different quality viewing in different
 viewers, different screen resolutions, different computers, different types of monitors etc. You may be able to influence the scaling in the viewer preferences but it could well look different on other setups.

[quote="Null"]The thing that annoys me is that I definitely couldn't get the same as you, for example, when you took the image in post #5 and exported as PNG and finally compiled[/quote]Attached is the result of doing this but exporting with different compressions and interlacing. On my viewer, the second image looks poor until I zoom in and then it looks fine, but it could well look different on your setup.


geometry parallelogram probability rotation

The square shown is constructed with the seven \textit{tangram} pieces. Two of the seven tangram pieces are small isosceles right triangles. Three other tangram pieces---a square, a  parallelogram, and another isosceles right triangle---are
equal in area to two of the smaller triangles. The final two tangram pieces---the two large isosceles right triangles---are equal in area to four of the small triangles. What is the
probability that a randomly chosen point inside the square will lie in the shaded region? Express your answer as a common fraction.

[asy]size(100); import patterns; currentpen = linewidth(1);add("hatch",hatch(1mm,E,linewidth(.5)));
path largeisoceles = (0,0)--(sqrt(2),0)--(0,sqrt(2))--cycle;
path mediumisoceles = (0,0)--(1,0)--(0,1)--cycle;
path smallisoceles = (0,0)--(1/sqrt(2),0)--(0,1/sqrt(2))--cycle;
path square = (0,0)--(1/sqrt(2),0)--(1/sqrt(2),1/sqrt(2))--(0,1/sqrt(2))--cycle;
path parallelogram = (0,0)--1/sqrt(2)*expi(asin(1/sqrt(2)))--(1,0)+1/sqrt(2)*expi(asin(1/sqrt(2)))--(1,0)--cycle;

draw(mediumisoceles); draw(shift(1*right)*rotate(45)*square); draw(shift((1-1/sqrt(2))*right)*rotate(-135,(1/sqrt(2),0))*smallisoceles);
filldraw(shift(2*up)*rotate(-90)*parallelogram,pattern("hatch"));
draw(shift(1*up)*shift(1*right)*rotate(135)*smallisoceles);
filldraw(shift(1*up)*shift(1*right)*rotate(45)*largeisoceles,pattern("hatch"));
draw(shift(1*up)*shift(1*right)*rotate(-45)*largeisoceles);[/asy]

let total area of square equal 1.  then big triangle shaded=1/4, and parallelogram==1/8.  so 3/8 is the probability


For positive numbers $ x$ and $ y$, $ x^2\plus{}2xy\plus{}y^2 \equal{} 36$. What is the mean of $ x$ and $ y$?

$ x^2\plus{}2xy\plus{}y^2 \implies (x\plus{}y)^2\equal{}36 \implies x\plus{}y\equal{}6$

x+y is not -6 since they are both positive integers.

so the mean of x and y are 6/2=3.

answer : 3


geometry
3D geometry
symmetry
calculus
calculus computations

Hey,

[b]1. The problem statement, all variables and given/known data[/b]
Find the surface area of a (right) cone, with base radius $ {r}$ and altitude $ {h}$ by applying Pappus's Centroid Theorem.

[b]2. Relevant equations[/b]
Pappus's Centroid Theorem
[hide="Theorem 1"]
Theorem 1 - The surface area ($ {S}$) of a - surface of revolution ($ {u}$) - generated by rotating a plane curve ($ {C}$) about an axis external to $ {C}$ and on the same plane is,
$ {S_{u}} = {{s_{{}_{C}}}{{d}_{{\overline{c}}_{C}}}}$

Where, $ {{d}_{{\overline{c}}_{C}}}$ is the distance traveled by the: centroid of [u]only[/u] the curve $ {C}$ ($ {{\overline{c}}_{C}}$).
$ {{d}_{{\overline{c}}_{C}}} = {{2}{\pi}{{\overline{c}}_{C}}}$
[/hide]
[b]3. The attempt at a solution[/b]
[hide="My attempt at a solution,"]
Let,
$ {{S}_{{u}_{cone}}} = {{S}_{u}}$

Let the surface area of the cone be,
$ {S_{cone}} = {{{S}_{u}}{ + }{{A}_{c}}}$

Where,
$ {{A}_{c}} = {{\pi}{{r}^{2}}}$

Also,
$ {{S}_{u}} = {{{(}{\sqrt {{{r}^{2}} + {{h}^{2}}}}{)}}{{{{d}_{\overline{x}}}}}}$

Where,
$ {{d}_{\overline{x}}} = {{2}{\pi}{\overline{x}}}$

However, I am not able to figure out how to find $ {\overline{x}}$, such that I arrive at the answer,
$ {S_{cone}} = {{{\pi}{r}{\sqrt {{{r}^{2}} + {{h}^{2}}}}}{ + }{{{\pi}{{r}^{2}}}}}$

As the surface area of the cone. How do I find, $ {\overline{x}}$?
[/hide]
Thanks,

-APSStudent

$ \overline x$ is simply equal to $ \frac{1}{2}r$ by symmetry :|

see http://mathworld.wolfram.com/PappussCentroidTheorem.html

Hey,
[quote="misan"]
$ \overline x$ is simply equal to $ \frac {1}{2}r$ by symmetry :|

see http://mathworld.wolfram.com/PappussCentroidTheorem.html
[/quote]
Yea, I just realized I was misinterpreting the centroid $ {\overline{x}}$: to refer to the centroid of the area of the triangle as opposed to the correct interpretation as the centroid $ {\overline{x}}$ of (only) the curve $ {C}$ (in other words the centroid referred to in Pappus's Theorem 1 is the centroid of [u]only[/u] the curve $ {C}$). Took me a while to realize that.

Thanks,

-APSStudent


Ross Mathematics Program
summer program
PROMYS
HCSSiM

I would like to know if a F1 visa is required to attend programs such as Ross, Promys or HCSSIM? Some friends of mine say that a tourist visa is fine.

Are there any international student from outside US who went and attended these programs to help me?

Best is to ask the particular program. The Website generally has good information about foreign students. Or ask them.

General answer: Depends on which country you are applying. For some, (say Canada) you may not need any kind of visa, for others (say Cuba) it may be all but impossible.

Even if you have F1 visa (valid for say  regular semester in a university), you may have to do some paper work, (eg register etc) but it would be valid to  go to program like PROYMS.

For tourist visa, as long as  your primary purpose was  technically " a tourist"  you would be okay.. (Eg you may still like get out of your dorm room  to check out  the City of Bostonn  <smile> )  (you may still have to register or inform that you would be attending Proyms etc..) but you can attend a short program like this without any problem.  - In short the  practical answer is  yes.

As I said, consult with the program  (they know all the practical rules etc) or US consulate to find all the requirement. For most of the countries  (eg most Europian ones, India, or any "friendly" country ) the process at present is pretty easy, and other than, may be some paperwork,  any visa would be okay. -- BUT don't take my word for this just ask the consulate. (Rules depends on many things)

thanks a lot for your help! Once I get a positive decision I will ask the program coordinator what I need for a visa.

Let me chime in here---don't wait until you've been invited to a summer program to find out the visa requirements!  It takes many weeks to get a non-tourist visa, even in the simplest of cases.  Also, while I can't speak for other summer program coordinators, I don't know what the visa requirements are for foreign students attending summer programs in the US.


Salam Be hame Kasi Hast komak kone ?

1) dar mosalase ABC nokhteye D roye emtedade zel'e AB , az tarafe nokhteye B , tori ekhtiyar mishavad ke |BD|=|CB| . be hamin nahve , bar emtedade zel'e CB az tarafe B , nokhteye F tori ekhtiyar mishavad ke |BF|=|AB| , sabet konid nokhtehaye A , C , D , F bar yek dayere ke markazash roye dayereye mohiti mosalase ABC ghara darad , Vagheand .

2) Farza konid yek mosalase hadeolzaviye bashad ke BC > CA va O ra markaze dayereye mohiti an begirid . farze konid H markaze ertfa'eye An va F paye Ertefa'e Ch bashad .amod bat Of dar F ,
zele'e AC radar P ghate mikonad . sabet konid . \[ \widehat{FHP} \equal{} \widehat{BAC}
\]

3) moslase hadeolzaviye ABC Dadeh shodeh Ast . noghate A1 va A2 roye zel'e BC ( A2 beyne A1 va C) va B1 va B2 roye AC (B2 beyne B1 va A) Va C1 va C2 roye AB ( C2 beyne C1 va B) tori entekhab shodehand Ke : \[ \widehat{AA1A2} \equal{} \widehat{AA2A1} \equal{} \widehat{BB1B2} \equal{} \widehat{BB2B1} \equal{} \widehat{CC1C2} \equal{} \widehat{CC2C1}
\]

sebt konid mahale talaghi khotote AA1 , BB1 va CC1 be hamrahe 3 noghteye khotote AA2 , BB2 va CC2 roye yek dayerehand .

[quote="SoNYDvDPro"]1) dar mosalase ABC nokhteye D roye emtedade zel'e AB , az tarafe nokhteye B , tori ekhtiyar mishavad ke |BD|=|CB| . be hamin nahve , bar emtedade zel'e CB az tarafe B , nokhteye F tori ekhtiyar mishavad ke |BF|=|AB| , sabet konid nokhtehaye A , C , D , F bar yek dayere ke markazash roye dayereye mohiti mosalase ABC ghara darad , Vagheand . [/quote]

khob baraye inke sabet konim 4 noghte $ A,C,D,F$ rooye ye dayeran,kafie sabet konim $ ACDF$ mohatie ya be ebarati bayad sabet konim $ BC.BF \equal{} BA.BD$ ke inam vazehe cherake $ BF \equal{} BA,BD \equal{} BC$ pas 4 noghte $ A,C,D,F$ rooye ye dayere gharar daran,esme in dayeraro bezarid $ w$ va esme markazesho bezarid $ O$...
khob hala mosalase $ \triangle BCD$ motasaviosagheine va zavie $ \angle ABC$ zavie khareji in mosalase pas $ \angle BDC \equal{} \frac {\angle B}2$ ya be ebarati $ \angle ADC \equal{} \frac {\angle B}2$...

amma zavie $ \angle ADC$ ye zavie mohatie tooye $ w$ va barabare ba nesfe kamane $ AC$ az in dayeras,pas zavie markazie rooberoo be in kaman dobarabare in zavias yeni $ \angle AOC \equal{} 2\angle ADC \equal{} \angle B$
pas darim:

$ \angle ABC \equal{} \angle AOC$ yeni 4zelie $ ACOB$ mohatie yeni $ O$ rooye dayere mohitie mosalase $ \triangle ABC$ vaghe ast...

[quote="SoNYDvDPro"]Salam Be hame Kasi Hast komak kone ?

2) Farza konid yek mosalase hadeolzaviye bashad ke BC > CA va O ra markaze dayereye mohiti an begirid . farze konid H markaze ertfa'eye An va F paye Ertefa'e Ch bashad .amod bat Of dar F ,
zele'e AC radar P ghate mikonad . sabet konid .
\[ \widehat{FHP} \equal{} \widehat{BAC}
\]
.[/quote]
vasate zele $ AC$ ra $ M$ benamid,pas $ OM\perp AC$ khob pas too 4zelie $ PFOM$ darim:

$ \angle PFO\equal{}\angle OMP\equal{}90$

pas in 4zeli mohatieo noghate $ P,F,O,M$ roo ye dayere gharar daran,esme in dayeraro bezarid $ w$,khob pas $ w$ ye dayeras be ghotre $ OP$ dge,pas markaze in dayere noghte vasate $ OP$ hastesh,esme in noghtaro bezarid $ O'$...

khob hala age markaze dayere 9noghtaro ba $ N$ neshoon bedim oon moghe,in dayere az $ M,F$ migzare dge,hamchenin hamoontori ke didim $ w$ ham az $ M,F$ migzasht,pas dar vaghe $ MF$ vatar moshtarake in do dayeras,pas be khattol markazeine shoon amoode yeni:

$ MF\perp O'N$ (*)

amma az tarafi chon $ N$ vasate $ OH$ va $ O'$ vasate $ OP$e pas tebghe tales khahim dasht $ O'N\parallel PH$ pas tebghe (*) natije mishe $ PH\perp FM$...

khob pas chon $ \angle PFO\equal{}90$ ba yekhorde deghat natije mishe $ \angle FHP\equal{}\angle PFM\plus{}\angle HFO$ amma az tarafi azlae zavie $ \angle HFO$ be azlae zavie $ \angle PFA$ amoodan,pas $ \angle HFO\equal{}\angle PFA$ pas:

$ \angle FHP\equal{}\angle PFM\plus{}\angle HFO\equal{}\angle PFM\plus{}\angle PFA\equal{}\angle MFA$

pas bayad sabet konim $ \angle MFA\equal{}\angle BAC$ ke inam be sadegi natije mishe cherake $ F$ rooye dayere noh noghtas pas age vasate do zele digaro ba $ M',M''$ neshoon bedim $ MFM'M''$ mohatie yeni $ \angle MFA\equal{}\angle MM'M''$ ke inam tebghe tales mosavie zavie $ \angle A$ hastesh...


[quote="sylow_theory"]You have 10 colored stones, all distinct colors. How man differenent ways are there to place the stones (in a line) so that the black and the white are together and the the red and blue are not?[/quote]

[hide="Maybe?"]So first take the white and black stone, and stick them together so as they move on the line they will move together. Then ignore the red and blue stones so that there are 7 'stones' left, and there are $ 7!$ ways to order those 'stones'. Then take either the blue or red stone, there are eight spaces you can put that stone in, and seven spaces you can put the other stone in because two spaces are added when the first stone goes in but those two cannot be occupied and the original space the stone was in was occupied, so the first stone takes away three spaces, and in total takes away one. So then it (possibly) is $ 7!(8\cdot 7)$ but then you have to multiply by 2 for the black and white stones and you get $ 2\cdot 7! (8\cdot 7)$? [/hide]

See this is the problem with counting problems --  I do them two different ways and get two different answers but I can't figure out why.

[hide="Blech"]As before, $ 2!$ for the black and white stones, then fix the red and blue, and multiply by $ 2$ later. Then you take the partitions of 7 that have three or less elements which are $ 7$, $ 6\plus{}1$, $ 5\plus{}2$, $ 5\plus{}1\plus{}1$, $ 4\plus{}3$, $ 4\plus{}2\plus{}1$, $ 3\plus{}3\plus{}1$, and $ 3\plus{}2\plus{}2$. Then one of those numbers in each partition has to be the number of stones between the red and blue stones, so for the partitions with 3 different numbers, there are six ways to put those before red, between red and blue, and after blue. Then three for the partitions with either two numbers, or three numbers of which two are the same...and the partition with only 7 is just one...so then the total is $ (2!)(2)(7!)(1\plus{}2\plus{}2\plus{}3\plus{}2\plus{}6\plus{}3\plus{}3)\equal{}(2!)(7!)(44)$...which is different? [/hide]

[hide="Hint - solution farther inside"]Consider this similar problem:

We have 6 gray stones, 2 white stones, and 2 black stones. In how many ways can they be arranged in a line so the white stones are adjacent but the black ones are not?

[hide="Solution to that problem and thus the original one"]
Ignoring the white stones, we have 6 grays and 2 blacks. There are $ {8\choose 2}\equal{} 28$ ways to arrange them. Of those arrangements, $ 7$ contain adjacent black stones. In these $ 7$ arrangements we must place the white stones in between the black stones. In the other $ 21$ the two white stones can be in any of 9 positions. So we have $ 7\cdot 1\plus{}21\cdot 9 \equal{} 196$ arrangements.

Now convert back to the original problem. Let the white stones be the white-black pair, the black stones be the red-blue pair, and the gray stones be other 6 colors. Any of the cases in the equivalent problem can have white-black or red-blue interchanged, so we multiply by 4. Additionally, the 6 gray stones when colored can be interchanged in any way, so we also multiply by $ 6!$. Thus, the answer is $ 196\cdot 4\cdot 6!$, whatever that is.
[/hide][/hide]

[hide="For the previous post"]
The answer I got is the one you got in your first answer. I think the second solution's mistake is missing the case that only the white-black stones are in between the red-blue pair, but I didn't understand what you did perfectly.
[/hide]

[quote="ch1n353ch3s54a1l"]...so then the total is $ (2!)(2)(7!)(1\plus{}2\plus{}2\plus{}3\plus{}2\plus{}6\plus{}3\plus{}3) \equal{} (2!)(7!)(44)$...which is different? [/quote]
That's how I did it! I get as my final answer $ 2\cdot 9!\minus{}4\cdot 8!$.

[hide="Step 1"]First find the # black stones together with white stones. There are $ 1\plus{}2\plus{}...\plus{}2\plus{}1 \equal{} 2\cdot 9$ ways to places these stones together. And for each one arrangement there are $ 8!$ other premutations of the remaining stones so in total $ 2\cdot 9\cdot 8! \equal{} 2\cdot 9!$.
[/hide]

[hide="Step 2"]Now find # of blue stones together with red stones. This is pure case work from here. But the pattern emerges as ch1n353ch3s54a1l did. Kinda boring. But here I get $ 2\cdot 8!$ going half way into this problem, because the other half is symettric and in total $ 4\cdot 8!$.[/hide]

[hide="Step 3"]Now remove #2 from #1 because its unnecesarry so there the answer.
[/hide]

Hmm ... I guess this generalizes. :) Let there be $ n$ stones (where $ n$ is big number) of different colors find the number of arrangments given the same conditions.


Do I need to prove that
\[ 2.2< ln(10)\]

Or can I just say
\[ 2.2< 2.30259 \approx ln(10) \Rightarrow 2.2< ln(10)\]

You know how to approximate natural logarithms off the top of your head?

I'm not offering a proof because if this is for a contest, I don't want to provide help.

Well actually, yes I can... but the graders most likely can't. :wink:

I'm wondering if I need to prove that my calculator does logs right.

Rule #9:
[quote]Except where otherwise noted in problems, participants may use calculators and computers to solve problems. In this case, the participant must explain the program that was used to solve the problem, and must prove that the method used in the program does solve the problem. Symbolic manipulation calculators and mathematical programs such as Mathematica, Maple, and Matlab are considered programs for this purpose. Therefore, if a key step of a participant's solution uses these tools, the participant must demonstrate that the tool completed the step correctly.[/quote]

It says "if a key step".... So I guess I'm good.  :)

Theres no way you need to prove that 2<ln(10).

even if that was a key step, I doubt you would need to prove it. Its not a symbolic manipulation of a calculator or anything.


The combustion of titanium with oxygen produces titanium dioxide. When 0.721 g of titanium is combusted in a bomb calorimeter, the temperature of the calorimter increases from 25.00 to 53.80 degrees Celcius. In a seperate experiment, theheat capacity of the calorimeter is measured to be 9.84 kJ/K. What is the heat of reaction for the combustion of one mole of Ti in this calorimeter?

Instead of simply copying the problems from your worksheet (or textbook or whatever) why don't you provide a solution by yourself? Then it would be much more easy to us to see where are your doubts and guide you in the right direction.


What do think if Nicolas Sarkozy is elected as polician.Will it change the european Nationalism ?
I mean it has been for centuries that no Forgien Laguange speaker is not elected as President.
Do think that this would bring changes to Europeans ?

[quote="roh3x2n"]What do think if Nicolas Sarkozy is elected as polician.Will it change the european Nationalism ?
[b]I mean it has been for centuries that no Forgien Laguange speaker is not elected as President.[/b]
Do think that this would bring changes to Europeans ?[/quote]
I don't understand that sentence?

I think it is very likely Sarkozy will enter a second round with Royal and then become president.
I don't know if you meant that he spoke another language, since I don't think he does.  His father was Hungarian and his mother French, I think he was raised in French (I have never seen him speaking Hungarian)

I don't see what you mean by European nationalism?

It's sort of a shame, this topic could be an interesting one (and we need those) but the original poster hasn't even clarified his questions.

Oh well, the first round was this Sunday, and as expected, no candidate got more than 50% of the votes.  Royal and Sarkozy will face each other within two weeks in a second round.

The results can be seen here :
[url]http://en.wikipedia.org/wiki/French_presidential_election,_2007#Results[/url]

Top candidates:
Sarkozy : 31.18% (right wing)
Royal : 25.87% (socialist party)
Bayrou: 18.57%(centrum party)
Le Pen : 10.44% (far right)

Le Pen lost a lot of votes, compared to his stunt in 2002 when it was him who faced Chirac in a second round.
I think Sarkozy managed to get the far-right wing voters back to his camp by creating a strong image of a man who wants to do something about the criminality.

Now those right-wing will almost certainly support Sarkozy in the second round, so that gives Sarkozy at least about 41.62%
Apart from Bayrou, all other candidates are leftist just like Royal, so we can estimate she will get at least 39.81%.

So this is exciting! It can be go both ways, the Bayrou voters will decide.  Bayrou himself could speak up and say which candidate he supports now, but he won't (probably because he himself wants to go at it again in 2012)

I think it will be Sarkozy :wink: , but who knows (we must keep in mind that it's likely that a lot more people will only vote in the second round)

Interesting Topic.

Here is his speech in English.

I think Roh3x2n wanted to know Sarkozy's his Political Views on Europe.
Here what Sarkozy said about his European policy
[quote]
I want to be the president of a new Europe: a Europe with an identity and borders, a Europe that acts in a concrete way, a Europe that protects its interests and its peoples in a globalized world. With 27 member-countries, Europe cannot go on functioning as it did when it had just 6 member-countries. I will suggest to our partners a treaty limited to institutional questions which will enable Europe to operate. I will oppose Turkey’s integration. (...)[/quote]

http://www.support-sarkozy-france.com/nicolas_sarkozy/speech.php

[quote="most-wanted"]
I think Roh3x2n wanted to know Sarkozy's his Political Views on Europe.
Here what Sarkozy said about his European policy
[quote]
I want to be the president of a new Europe: a Europe with an identity and borders, a Europe that acts in a concrete way, a Europe that protects its interests and its peoples in a globalized world. With 27 member-countries, Europe cannot go on functioning as it did when it had just 6 member-countries. I will suggest to our partners a treaty limited to institutional questions which will enable Europe to operate. I will oppose Turkey’s integration. (...)[/quote]
[/quote]

It might be interesting to mention that Sarkozy, just like president Chirac, advised people to vote yes in 2005 when a referendum about accepting a Constitution for the European Union was held.  Just like in the Netherlands, the result was "no".  Only Spain and Luxemburgh held a succesful referendum, other countries just accepted it without holding a referendum, or simply haven't accepted it (yet).

I am a bit wary when Sarkozy says "cannot go on functioning as it did when it had just 6 members".... That seems to hint at more power for the bigger members like France, the UK and Germany. :|

[quote="Xevarion"]My roommate and I attempted to poll the people living around us in our dorm to see what their thoughts about Sarkozy were. It turned out that my roommate and I were the only ones who knew who Sarkozy was at all.  :o That's the USA for you...[/quote]
Well that's a shame, it's always good to understand the politics of other countries.  For instance, there was some fuss about Turkey invading the Kurdish part of Iraq in the near future and I bought it...until some Turks, living in Dutchspeaking areas, pointed out that this is in fact essentially part of an internal power struggle. I mean : people shouldn't blame every inhabitant of a country for one or two things one of their leaders has done.

It's only normal though, it just depends in what sphere of influence you live.  In the case of my country, this is particularly interesting. Francophone media treats the French elections as if it's a domestic affair. The Flemish media(that I watch) tend to focus on Dutch politics a lot more.
Presidential and parliamentary elections are followed closely, but for instance, I can only name two US governors (Schwarzenegger, I also knew Jeb Bush but he quit)

Within two days, on Wednesday, there will be an important debate between him and Royal.  We'll see. :huh:

[quote="fredbel6"]It's only normal though, it just depends in what sphere of influence you live.  In the case of my country, this is particularly interesting. Francophone media treats the French elections as if it's a domestic affair. The Flemish media(that I watch) tend to focus on Dutch politics a lot more.
Presidential and parliamentary elections are followed closely, but for instance, I can only name two US governors (Schwarzenegger, I also knew Jeb Bush but he quit)[/quote]
But governors don't have much power. In most states the governor is not the most powerful official in the state. For example, in Texas (where I live), the governor is actually the sixth most powerful person in the state government (in terms of authority), and he's actually behind the lieutenant governor!
On that theme, when I made fun of my other roommate Alex (who is studying finance) about not knowing who Sarkozy was, he asked me if Sarkozy was someone important. I answered that he might be the next president of France, and Alex laughed at me and walked away. :P

[url]http://news.bbc.co.uk/2/hi/europe/6611539.stm[/url]
Le Pen has urged his voters....not to vote in the second round.  Of course, we can't predict how many of them will actually do this, but this isn't good news for Sarkozy. :maybe:

It might be important to note that there will be parliamentary elections in France as well, on the 10th of June, with a second round on the 17th of June

[quote="Kent Merryfield"]One oddity: I have a brother-in-law named Sarkozi. His family has been in the U.S. for several generations, but that must be the same family name. (He - the brother-in-law - has done some acting. In [url=http://www.imdb.com/title/tt0293346/][i]Legend of the Chupacabra[/i][/url] he's the first to die.)[/quote]
That's rather odd indeed! :o
[quote]
But governors don't have much power. In most states the governor is not the most powerful official in the state. For example, in Texas (where I live), the governor is actually the sixth most powerful person in the state government (in terms of authority), and he's actually behind the lieutenant governor! [/quote]
Well yeah, but I just tested myself, and it turns out that the only people I know in the Bush
 administration are Bush, Cheney, Rice and Gates.

[quote]On that theme, when I made fun of my other roommate Alex (who is studying finance) about not knowing who Sarkozy was, he asked me if Sarkozy was someone important. I answered that he might be the next president of France, and Alex laughed at me and walked away. [/quote]
Why did he laugh at you? :|

[quote="Xevarion"]Alex meant "how could you possibly even think I would consider the president of France to be important?"[/quote]
Well that's a shame, I think the "France never wins" campaign from 2003 managed to portray France as if it's like Moldova :roll:  or something.  Yes they're not among the richest in Europe, yes they're not a world power anymore like they used to be, but...they are still one of the five nuclear powers, they still have one of the most reliable commercial launching systems,..

I just read that they asked the 29 parliamentary members from the centrist party UDF whom they support. 15 said they support Sarkozy, 10 (including Bayrou himself) didn't want to answer.  Now that Bayrou's votes can flow to both candidates,and that Le Pen asked not to vote, this is highly unpredictable.

But as I said, there will be parliamentary elections, so those parliamentary members have good reasons not to say whom they support, since those elections use a slightly different two-round system : in each region, if no candidate got more than 50% of the votes, all candidates who get more than 12.5% meet each other in a second round, and they often negotiate between those two rounds to make alliances....... :ninja:

Well Wednesday there was the big debate between Sarkozy and Royal on French television.
The last time the French had a debate between two presidential candidates was in 1995 (Chirac versus Jospin), since Chirac refused to do this with Le Pen in 2002.

Bayrou has given some sort of "weasel advice" by saying he wouldn't vote for Sarkozy.
[url]http://www.youtube.com/watch?v=llBdfrgNHjw&mode=related&search=[/url] shows the first ten minutes.

Sarkozy proposed limiting the number of presidential mandates to two, but as far as I know, nobody has been the president of France three times anyway. :|

[quote="Xevarion"]Hm Sarkozy said that he didn't think the result of that debate would really have much impact on the choices of French voters. But d'Estaing said he thought his presidential debate was an important contribution to his victory back ages ago whenever he was president... Does Sarkozy dislike how the debate went and he hopes to control damage? Or is that really his belief and he was just commenting?

(Personally I also believe in the power of televised debates. For example consider Nixon vs Kennedy in 1960..)[/quote]
To be honest with you, I haven't heard anything about his own reaction to the debate. However, most polls show that he has about 53/54% of the votes... :huh:  Royal is said to have made quite a desperate move, since she "attacked" by saying that voting for Sarkozy could be dangerous and lead to insecurity.  She is referring to the notorious French riots in 2005 (which even spread-in much lighter form- to my believed city of Ghent  :ninja: ). It's the age old debate : can you stop crime by taking a firm stance or by expressing tolerance.  Sarkozy definitely stands for the first and I think that's why many people will choose him.

d'Estaing was elected in 1974.
[url]http://en.wikipedia.org/wiki/French_presidential_election,_1974[/url]

This second round will be extremely exciting. Last time, in 2002, it was just "pro forma" (Chirac got more than 80%). Everyone who wasn't far right voted Chirac.  I remember images of socialist voters who went to vote for Chirac and then "cleaned" themselves symbolically afterwards. Tomorrow, the overseas regions like French Polynesia, French Guyana, ... will already vote.... :ninja:

There are no official results yet, but reliable exit polls show that Sarkozy has gained 53% of the votes.  It is thus almost certain he is the next president of France.

Well this wasn't really an elections with lots of surprises , many people expected Sarkozy and Royal to go to a second round months ago, and many expected Sarkozy to win. Even the number 53% was predicted. It's a sharp contrast with the last one in 2002, when the second round wasn't exciting at all.

Thousands of policemen are on alert right now, because it could become a stormy night.  Many youth gangs have already declared that they cannot accept Sarkozy as a president. An event like in October 2005 could happen again. :maybe:

[quote="spix"]:clap: Word on the news is that Sarkozy won. :jump:  :clap:[/quote]
You support him?

I actually find it interesting to see how many prominent politicians were not born in France or have foreign roots...

Giscard d'Estaing was born in Koblenz, Germany...
de Villepin : born in Rabat, Morocco
Sarkozy: Hungarian father
Royal : born in Dakar, Senegal

Sincerely, before the debate I liked Royal more, but her non-diplomatic atittude made me change my mind totally. As for Sarkozy's roots, I noticed that there are not so many pure french people in France.
For example the football team :)(90% of the players have foreign roots)

[quote="Xevarion"]France has amazing voter turnout. 85% in this election!!  :o  :o

I wish USA could accomplish such amazing feats. I think it would have a significant impact on how politicians here behave...[/quote]
[url]http://en.wikipedia.org/wiki/Voter_turnout[/url] covers that topic somewhat.
I have to vote. However, voter turnout is about 91% , in some provinces even 88%.

[quote="fredbel6"]Well this wasn't really an elections with lots of surprises , many people expected Sarkozy and Royal to go to a second round months ago, and many expected Sarkozy to win. Even the number 53% was predicted. It's a sharp contrast with the last one in 2002, when the second round wasn't exciting at all.  [/quote]
What do you mean here? Hopefully I'm not coming across as nitpicky, but I'm not sure what you intended to say here. It seems like [i]this[/i] year's second round wasn't exciting, so did you mean the 2002 election's second round wasn't... expected?

[i]Bad BBCode fixed by moderator.[/i]

[quote="Klebian"]
What do you mean here? Hopefully I'm not coming across as nitpicky, but I'm not sure what you intended to say here. It seems like [i]this[/i] year's second round wasn't exciting, so did you mean the 2002 election's second round wasn't... expected?[/quote]
Forgive if I was vague.
I meant that people were really expecting to see Chirac and Jospin face each other in a second round in 2002, but it was Le Pen who faced Chirac. And then of course the second round in that election was not exciting, since about everyone who didn't vote Le Pen went to vote for Chirac to stop a far right victory.

This second round was exciting, it still could have gone both ways.  But like half a year ago, I'd have predicted all of this.

[quote="roh3x2n"]He won ! :D[/quote]
Not yet!  :wink:

As I said, legislative elections are ahead.  Should that mean that the prime minister is not from the same party, then the president does lose a considerable amount of power, as there will have to be "une cohabitation".


No, no hay latinos en AoPS. Pero, yo hablo espanol! :lol:

yo también hablo español. pero no soy latino tampoco =/

No soy español ni latino, pero hablo español !

Ning'un español sobre un forum español !

Com'o se hacen los acentos y los ? ?

Si trabajas con el Windows, están en el programa "C:\WINDOWS\system32\charmap.exe". De ahí puedes copiarles en tu texto.

y ¡perdonad lo mío mediocrísimo español! , por favor.

Yo recomiendo que use éstos formularios para formar los acentos y la puntuación invertida:

á = Alt + 0225
é = Alt + 0233
í = Alt + 0237
ó = Alt + 0243
ú = Alt + 0250
ñ = Alt + 0241
Ñ = Alt + 0209
ö = Alt + 0246
ü = Alt + 0252
? = Alt + 0193
É = Alt + 0201
? = Alt + 0205
Ó = Alt + 0211
Ú = Alt + 0218
Ö = Alt + 0214
Ü = Alt + 0220
¡ = Alt + 0161
¿ = Alt + 0191

Donde aprendieron español la mayoría de ustedes?? solos?

Yo lo aprendí por estudiar (cuatro años) en escuela.

Tambien hablo español. Lo estudio en mi escuela.

[quote="manuel"]No hay latinos aki???[/quote]

Correct me I'm wrong but isn't it:

No hay latinos AQUI, not ak???

My #1st language isn't spanish so I'm not expert but that's what I learned.. Maybe it differs from country to country.

And...

MUCHO GUSTO! :lol:

Para responder a Mildorf estudio Español en la escuela...

akí no existe ; es : aquí
Pero ambas las palabras se pronuncian de la misma manera...

Um..

I'm not expert on spanish so..

aki = aqui?

Is that what you said??? :?

Lo que pasa es que los latinos cuando chateamos usamos -aki- como un "shortcut" para escribir -aqui-.
Y muchas veces en ves de "Que" escrbimos "ke" ...pero perdonen, no esta bien escrito..

Silverfalcon: en ingles, se puede usar "u" en placer de "you". En español, se puede usar "aki" en placer de "aqui" tambien.

Ok. ;)

I think you must meant something like aki can replace aqui in spanish compared to english or something.

(sorry for using ingles in here but I'm not expert on spanish :( )

hehe, spanish in the age of instant messaging. I just found that pretty amusing.

I find it slightly unusual as well - I think its because you are so used to doing it in your own language, but when you learn a foreign language you always learn everything strictly to the rules, and don't often realise that of course you can do it in that language as well.. thats when having a forum like this where you can see someone else fluent at the language is useful.

Yo puedo hablar un poco espanol tambien porque yo lo estudio en mi escuela.

Buenas, soy nuevo en el forum. Estoy estudiando matematicas en la universidad de la florida. Me parece bien interesante este lugar. Espero poder contribuir a los forums.
Con mucho gusto,
Demis

DemisM que area de las matematicas te interesa?

Hola manuel yo soy latino :D  :D  mentiras obviamente suted ya sabe, el caso es que ahi van todos bien con su español, y pues si quieren practicar aqui nos tienen, suerte a todos!

Casi soy Latina... pq he vivido mas de la mitad de mi vida en Latinoamerica
(3 anos en Argentina, 3 en la Republica Dominicana, y 3 en Mexico)...jaja

mis amigos que pasan? pues, soy un poco latino si chino es un tipo de latino.  sin embargo, me gusta hablar en espanol mucho y me encanta conocer mas amigos como ustedes.

Espero que además de deciros HOLA os propongais problemas y demas, yo soy Roberto de Sevilla y me gustaría participar en el foro.

Un saludo cordial  a todos y todas.

Roberto

Hey, que mas, yo si soy latino! :). Bueno, usualmente no me meto mucho en los foros porque pierdo mucho tiempo pero me gusta mucho la matematica y la musica. Se tocar guitarra y violin (aunque el hace tiempo que no toco violin). Hablamos por ahi. Saludos a todos

Bueno,

Parece que mucho menos de la mitad de los que ponemos algo aqui somos originalmente hispanos, latinos, o como nos quieran llamar. Yo por mi lado soy oriundo de Costa Rica, aunque acutalmente vivo en USA, donde estoy haciendo mi doctorado.

Alguien pidio un problema, que les parece este (bastante facil)

Tenemos S un conjunto finito no vacio, y sea f: P(S) -> N una funcion definida en los subconjuntos de S, tal que para cualesquiera dos subconjuntos A y B de S se cumple que

1. f(A) = f(S\A)
2. max{f(A), f(B)}  >= f(A :cup: B)

Pruebe que |Rango(f)| <= |S|.

Me parece que el problema es originalmente hungaro, al menos fue un amigo hungaro quien me lo dio.

Nos vemos,

David

Bueno, justo vi este thread, asi que aprovecho para presentarme: soy de Buenos Aires, Argentina; y me gusta sobre todo la rama de matematica sobre teoría de números, y jugar al go (aunque recien empiezo :( ). Actualmente estudio cs. de la computación, y me interesa mucho la aplicación de la matemática en la creación de nuevos algoritmos.

salu2, Laucha :D

[quote="TripleM"]Silverfalcon: en ingles, se puede usar "u" en [b]placer[/b] de "you". En español, se puede usar "aki" en placer de "aqui" tambien.[/quote]

Por cierto TripleM, creo que queres decir [b]lugar[/b] en vez de [b]placer[/b].

[b]Lugar:[/b] [i]Place[/i]
[b]Placer:[/b] [i]Pleasure[/i]

Jeje, solo me parecio entretenido.

Solo por casualidad, ¿Hay algún otro mexicano por aquí?

[quote="manuel"]Dave, tu eres un profesor o estudiante?
Hay otro mexicano por ahi .  marko avila.[/quote]

Depende. Durante un par de anios trabaje como profesor de secundaria. Desde hace unos cuatro anios, hay semestres en los que imparto algun curso en la universidad en la que este (primero en la Universidad de Costa Rica y en el Instituto Tecnologico de Costa Rica, y ahora en el Georgia Institute of Technology), frecuentemente algo entre precalculo a ecuaciones diferenciales.

Tambien soy un estudiante de doctorado en el Georgia Institute of Technology. Por que la pregunta?

[quote="manuel"]David no era a ti era a Dave el de mexico a ver quien es ![/quote] :blush: ups!

Perdon, pero desde que estoy en gringolandia, me he acostumbrado a que me digan Dave!!

[quote="Dave Vather"]Estoy concursando todavía para representar a México en la próxima IMO.
Por cierto, tengo curiosidad: ¿Cómo son los procesos para seleccionar a los que representan a sus países en las olimpiadas internacionales?[/quote]

En el caso de Costa Rica, las olimpiadas Nacionales tienen tres niveles, A para 7mo y 8vo, B para 9no y 10mo, y C para 11mo y 12mo. Ahora, en niveles A y B, no podes participar un anio si has ganado medalla en ese nivel en algun anio anterior (por eso hay gente que esta en nivel C desde 9no!). Ahora, todo el que tenga medalla en A y B (si tiene la edad) o estubo muy cerca del corte, puede participar en la seleccion para la Centro. Para la Ibero, bueno, necesitas haber ganado medalla en nivel B, o haber estado muy cerca de ganar medalla en nivel C. Ahora, cada uno de estos niveles tiene algunas clases durante varios meses (cuatro tres veces por semana durante enero, dos de febrero a Junio), y cada mes se hace un examen de clasificacion. El promedio de los examenes decide!

YO  si soi latino de mexiko soi. y ustedes?? debe aber mas latinos. a yalguien se prepara para la imo?' va ser en donde yo vivo en merida. jejeje  y alguien q estudia pa su olimpiada nacional kon esperas de irse a eslovenia o vietnam komo yo???

marko avila..
yo pienso ir a eslovenia tambien!

En teoría yo voy a esta IMO... (Hasta donde me doy cuenta, aquí se han hecho los exámenes con mucha anticipación) Y, por cierto, si las cosas siguen bien económicamente hablando, espero ir a Eslovenia y Vietnam... Denlo por seguro!!!! jajajajaja (la ventaja de vivir en un país pequeño :D, aquí los "pre-olímpicos" no son más de 20, contando a los que "ocupan espacio", osea los que no son para nada buenos)


I was wondering whether anyone knew of a good Pre-Calculus textbook that could be purchased at bookstores and taught the curriculum well. (It would help if it's not too pricey :D )

Thanks

Why don't you take a look at AoPS 2 ? I also suggest you read the calculus trap article if you have not already done so. Your postings in HSB make me think you should concentrate on the depth of your studies, not on the breadth of your upper level knowledge.

If you study AoPS and similar books deeply enough you should be able to derive nearly all of the pre-Calculus curriculum :)  (Besides the parts that introduce calculus.  Those are a little tough to come up with your own without the proper motivation.)