The follow is the process I did to set up my bibliography.

1. Type in all the cited articles into notepad with @article{}..and then rename as .bib file ( i am not sure whether I can did this way. but did not find online how to make the database)

2. \bibliographystyle{plain}
\bibliography{my bib filename}

3. than cite use \cite{cite_key)

I did not get the output of bibliography. but where I have \cite{cite_key} show up with an entry having border line..

It pops several errors as :   You are missing an entry type--line 3 of file  ch1citation.bib

Could anyone tell me how to do with it?

[quote]You are missing an entry type--line 3 of file ch1citation.bib [/quote]

Check that line; do you have proper syntax?

Here's a sample entry if you're using plain BibTeX:

[code]@misc{oconnor1999hipp,
	author       = "J. J. O'Connor and E. F. Robertson",
	title        = "Hipparchus of Rhodes",
	year         = "April 1999",
	note         = "Accessed 22 Jan. 2009",
	url          = "http://www-history.mcs.st-andrews.ac.uk/Biographies/Hipparchus.html"
	}[/code]


inequalities
geometry
circumcircle
inequalities proposed

Given a triangle $ ABC$ with three sides: $ BC\equal{}a,CA\equal{}b,AB\equal{}c$.Let $ r_{a},r_{b},r_{c}$ denote the $ A$-exradius,$ B$-exradius and $ C$-exradius respectively,let $ R,r$ be respectively the circumradius and the inradius.Prove that:
$ \frac{r_{a}\plus{}r_{b}\plus{}r_{c}}{a\plus{}b\plus{}c} \ge \sqrt{1\minus{}\frac{r}{2R}}$

[quote="quykhtn-qa1"]Given a triangle $ ABC$ with three sides: $ BC = a,CA = b,AB = c$.Let $ r_{a},r_{b},r_{c}$ denote the $ A$-exradius,$ B$-exradius and $ C$-exradius respectively,let $ R,r$ be respectively the circumradius and the inradius.Prove that:
$ \frac {r_{a} + r_{b} + r_{c}}{a + b + c} \ge \sqrt {1 - \frac {r}{2R}}$[/quote]

$ r_{a} + r_{b} + r_{c} = 4R + r$  and   $ a + b + c = 2s$

So we must prove : ${ 1 - \frac {2s^2(2R - r)}{R(4R + r)^2})} \ge 0$

If $ H$ is $ orthocenter$ of $ \triangle ABC$ and $ G_e$ is $ Gergonne$ $ Point$ of  $ \triangle ABC$ then :
\[ \boxed{HG_e^2 = 4R^2(1 - \frac {2s^2(2R - r)}{R(4R + r)^2})}
\]
:)


function
calculus
algebra
domain
calculus computations

I was just wondering if you guys could give me some tips for turning equations into graphs.  Especially composite, multiplied, added, divided, etc. functions.  Sorry if I am being n00bish, but I really need some help.  I put this in the Calculus forum because I am working with these things in the review part at the beginning of the course, if a MOD chooses to move it, that's fine.

Thanks

Paul

Well I'm in AB and here's what you need (I hear)
Domain
Range
Asyptotes
Extrema
Points of Inflection
Interval a Function is Increasing / Decreasing
Interval a Function is Concave Up / Down


probability
function
combinatorics theorems
combinatorics

I was thinking about the following problem: what is the probability that two randomly chosen $ n$-element permutations commute?  The problem can be solved by considering the group $ S_n$ acting on itself by conjugation and applying Burnside's Lemma.  The odd thing is that Burnside's Lemma is being used "backwards": normally we use it to find the number of orbits, but in this case we know the number of orbits and we are trying to find the average size of the stabilizer.  So I'm wondering: are there other interesting problems that use Burnside's Lemma "backwards"?

Sure.  One application of Burnside's lemma is what I call [url=http://qchu.wordpress.com/2009/06/16/gila-iii-the-orbit-counting-lemma-and-baby-polya/]baby Polya[/url]; if a finite group $G$ acts on a set $ X$, then the number of orbits of functions $ X \to \{ 1, 2, ... n \}$ is $ \frac{1}{|G|} \sum_{g \in G} n^{c(g)}$, where $ c(g)$ is the number of cycles in the cycle decomposition of $ g$.  One corollary is that when $ n$ is large the number of orbits is approximately $ \frac{1}{|G|} n^{|G|}$; in other words, the average function is asymmetric, or the average orbit is full.

Isn't that a "forward" use of Burnside though?

Well, the point is that the dominant term gives you information about the sizes of the orbits rather than their number.


Is there something wrong with the solution of question 2?
[url=http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-06Spring-2005/270FAFD7-7CA5-49C2-909D-6B2E919D0879/0/quiz2_1806_s05.pdf]Problems[/url]
[url=http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-06Spring-2005/5258219C-25B0-428F-8F70-1B177B31D1EE/0/exam2sol.pdf]Solutions[/url]

this is probably some engineering exam, but there is nothing wrong with it otherwise  :lol:

In order to convince yourself look at the differentiable function $ f(x)\equal{}\parallel{}b\minus{}ax\parallel{}^2\equal{}\parallel{}b\parallel{}^2\minus{}2x \,a^T\cdot b\plus{}x^2\parallel{}a\parallel{}^2$ and determine the obvious minimum

But if $ \mathbf a\not\|\mathbf b$ then $ \mathbf b\minus{}x\mathbf a$ must not be zero... :maybe:

The problem asked for minimization and not zero search:

$ f(x)\equal{}\parallel{}b\minus{}ax\parallel{}^2\equal{}\parallel{}a\parallel{}^2\left(x\minus{}\frac{a^T\cdot b}{\parallel{}a\parallel{}^2}\right)^2\plus{}\parallel{}b\parallel{}^2\minus{}\left(\frac{a^T\cdot b}{\parallel{}a\parallel{}}\right)^2$ has a unique minimum at $ x\equal{}\frac{a^T\cdot b}{\parallel{}a\parallel{}^2}\equal{}\frac{a^T\cdot b}{a^T\cdot a}$

[quote="hjbrasch"]The problem asked for minimization and not zero search:

$ f(x) \equal{} \parallel{}b \minus{} ax\parallel{}^2 \equal{} \parallel{}a\parallel{}^2\left(x \minus{} \frac {a^T\cdot b}{\parallel{}a\parallel{}^2}\right)^2 \plus{} \parallel{}b\parallel{}^2 \minus{} \left(\frac {a^T\cdot b}{\parallel{}a\parallel{}}\right)^2$ has a unique minimum at $ x \equal{} \frac {a^T\cdot b}{\parallel{}a\parallel{}^2} \equal{} \frac {a^T\cdot b}{a^T\cdot a}$[/quote]
Yes, I do it with the same method as you. I just want to know why he wrote $ \mathbf a x\equal{}\mathbf b$.


Kan noen finne feilen her...
Oppgave:
 Gitt at $x+y=2$ og $x^5+y^5=992$, kva er då alle mogelege verdiar til$x$ og $y$ ($x$ og $y$ er heile tal)?

La oss definere $T_n = x^n+y^n$.  Vi vet at $T_0 = x^0+y^0 = 2$ og $T_1 = x+y = 2$
Så $T_0 = T_1 = 2$
$(x+y)(x^n + y^n) = x^{n+1}+y^{n+1} + xy (x^{n-1}+y^{n-1})$  Dette kan omskrives sånn :
$x^{n+1}+y^{n+1} = (x+y)(x^n + y^n) - xy (x^{n-1}+y^{n-1})$ Og det er samme som :
$T_{n+1} = 2T_n - xy T_{n-1}$  - dette er en rekursiv tallfølge med $T_0 = T_1 = 2$
Ved å bruke den kan vi finne ut $T_5$
$T_2 = 2 T_1 -xy T_0 = 4 - 2xy$
$T_3 = 8- 6xy$
$T_4 = 2(8-6xy) - xy (4-2xy) = 16 - 16xy - 2x^2y^2$
$T_5 = 2(16-16xy - 2x^2y^2) - xy(8-6xy) = 32 -40xy + 2 x^2y^2$
$T_5 = 992$  gir oss
$992 = 32 - 40 xy + 2 x^2 y^2$ og det er jo en andregradsligning
$(xy)^2-20xy-480= 0$
Men denne gir ikke noen heltallige løsninger så jeg vet ikke ...

Det er en feil på linja med $T_4$, det skal være $+2x^2y^2$ i stedet for $-2x^2y^2$ (for siste gang, minus og minus blir pluss! :P ). Hvis du retter opp det, får du
\[ (xy)^2-4xy-96 = 0 \Rightarrow xy = -8 \ \mathrm{ eller }\ xy =12. \]
Setter vi inn $y = 2-x$ får vi enten
\[ x(2-x) = -8 \Rightarrow\ x^2-2x-8 = 0 \Rightarrow x = -2 \ \mathrm{ eller }\ x = 4, \]
eller
\[ x(2-x) = 12 \Rightarrow x^2-2x+12 = (x-1)^2 + 11 = 0, \]
som ikke har noen løsning. Sjekker vi $x = -2$ og $x = 4$ ser vi at dette er løsninger, så $(-2, 4)$ og $(4, -2)$ er løsninger.

:blush:  takk for hjelpen
"The  olympic village" - den var god  :)
Har du prøvd å løse oppgave 2 fra IMO 2000 ? Det er en fin Schur øvelse...
Oppgave 1 (geometri) fra samme året er også "lett". Prøv deg!

Her er ei sjølvlaga oppgåve:

Vis at for alle positive reelle x, er $3+x^4\geq 4x$

Hint:
[hide]AMGM[/hide]
Løysing:
[hide]Skriv det som $\frac{1+1+1+x^4}{4}\geq\sqrt[4]{1\cdot1\cdot1\cdot x^4}$. Dette er AM-GM og held for alle positive reelle x.[/hide]

En annen løsning:
[hide] Se på funksjonen $f(x) = x^4 - 4x + 3$. Vi skal finne minimum av funksjonen $f(x)$
$f'(x) = 4x^3 - 4$. Sjekker at minimum oppnås når $x=1$
Nå kan vi finne minimum av funksjonen. Den oppnås når $x=1$
$f(1) = 1^4-4+3 = 0$ dvs. $f(x) = x^4 -4x +3 \geq 0$ og $x^4+3 \geq 4x$[/hide]

kan ikkje du jørgen legge ut oppgåva som du gav meg på fysikken, kanskje atle eller aslan greier den.(eg har gløymt den)  ellers er det kult med andre oppgåver. forresten, veit noken om framgangsmåten for å rekne ut kor mange sudokar du kan putte inn på et 9x9 brett?:P Trur den er ganske hard, noke for jørgen?

Ny oppgave:

I ein trekant med vinklar $\alpha,\beta,\gamma\leq90$, vis at:

$sin\alpha+sin\beta+sin\gamma \geq sin2\alpha+sin2\beta+sin2\gamma$

Den oppgaven var henta fra Mathlinks da, men uansett... :P
For $a,b,c\geq 0$ og $n \geq 1$, vis at
\[ \frac{a^n}{b+c} + \frac{b^n}{a+c} + \frac{c^n}{a+b} \geq \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}. \]

Og her er en anna ganske fin oppgave fra Problem Solving Strategies:
For reelle tall $x, y, z$, vis at
\[ \frac{|x-y|}{\sqrt{1+x^2}\sqrt{1+y^2}} + \frac{|y-z|}{\sqrt{1+y^2}\sqrt{1+z^2}} \geq \frac{|x-z|}{\sqrt{1+x^2}\sqrt{1+z^2}}. \]

De som fant antall mulige kombinasjoner i sudoku brukte et dataprogram og masse innvikla logikk, kan lese en del her: [url]http://www.afjarvis.staff.shef.ac.uk/sudoku/[/url] men det er ikke så fryktelig spennende...

eller ei anna frå PSS. vis for alle reelle tal a og b at følgende gjelder:

$\frac{|a+b|}{1+|a+b|} \leq \frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$

Greide den netopp;)

Hei! Jørgen sin oppgave er Nesbitt med n=1 og Nordisk 2005 med n=2  :D (jeg lurer på om det går an å løse den ved å bruke rearrangement direkte... $\frac{1}{b+c}$ mot $\frac{1}{2a}$ ??
 :idea:  Jeg foreslår også at vi åpner en ny topic hver gang vi har en ny oppgave (da er det lettere å diskutere )

Sjekk denne linken her:  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22364&postorder=asc&start=20]www.mathlinks.ro/Forum/viewtopic.php?t=22364&postorder=asc&start=20[/url]
Utrolig artig metode  :P

Alopex jeg løste oppgaven din med "brute force", jeg vil gjerne se på en finere løsning da...

kva kallte du det? "brute force", kva er det? Løysinga eg hadde var ikkje særlig fin, eg delte opp oppgåva og alt vart ganske langt.  Men eg kan legge ut løysing viss du trur du kan lære noke.

Når det gjeld ulikskapen til volder så fekk eg ut rearangement til slutt, men det var lettare å gjere det på en anna måte, den som volder viste.

Har dokke noken generelle tips på oppgåver som inneheld absoluttverdi? kor bør en starte på slike? Vanlige reknereglar gjeld ikkje alltid...

Hm, en måte, som er fryktelig tåpelig men som gjør at det i hvert fall ikke er med noe absoluttverditegn er å bytte ut alle $|x|$ med $\sqrt{x^2}$. Men jeg har ikke sett at noen har brukt det på noen oppgave, så det er sikkert ganske meningsløst :roll: ...

En anna måte er å dele opp ulikheten i forskjellige tilfeller ettersom om det som står inne i absoluttverditegnene er positivt eller negativt, og så fjerne absoluttverditegnene i hvert av de tilfellene. (Hvis du f. eks. sjekker tilfellet hvor $x < 0$, bytter du ut $|x|$ med $-x$ og setter $x > 0$.) Da får du flere ulikheter med bare positive variable, som sikkert er kjempetrivielle :P .

Ellers funker det ofte å bruke trekantulikheten, siden den er med absoluttverdier ($|x| + |y| \geq |x+y|$).

noken som greier ulikskapen til volder? ikkje eg i alle fall...


$ ABC$ is an isosceles triangle$ ($ $ AB$ $ \equal{}$ $ BC$ $ )$.$ AD$ is bisector.The area of $ ABD$ and $ ADC$ triangles are $ S _ 1$
and $ S _ 2$ respectively.Find $ AC$.

Let the base angle of the isosceles triangle = 2r.

let (line-AB) = c, (line-BC) = a, (line-AC) = b, Let (line-AD) = d
for isosceles triangle,  c = a.

We are asked to solve for b.

S1 = (1/2)cdsin(r)
S2 = (1/2)bdsin(r)

Therefore

(S1/S2) = c/b    (1)

or c*c = (b*b)(S1*S1)/(S2*s2)		(2)

The perimeter of the triangle

p = 2c + b

The semiperimeter s = (1/2)p = c + (1/2)b

s-a = (1/2)b
s-c = (1/2)b
s-b = c - (1/2)b

the total area of the triangle

S1 + S2 = SQRT[s(s-a)(s-b)(s-c)]
        = SQRT{[c + (1/2)b]*(1/2)b*(1/2)b*[c - (1/2)b]}
	= (1/2)b*SQRT[c*c - (1/4)b*b]				(3)

Square bith sides of (3), substitute (2) into (3) and rearrange

The answer:

(line-AC) = b = 2*SQRT[S2*(S1+S2)]/SQRT{SQRT[4*S1*S1 - S2*S2]}


Is the following statement equivalent to the Goldbach Conjecture? Every even positive integer can be expressed as the difference of two primes?

I'm pretty sure of it, but just need some confirmations :D.

Are you sure?  It seems to me a distinct result from the Goldbach Conjecture.  That is, I don't think either can be proven from the other.

They are may be equavalent, if both statments are true. But you must proof both for it.  :D
Let ${S=\{n^{2}+n+1}$, then for any even number 2m exist two members S, suth that $2m=(m^{2}+m+1)-[(m-1)^{2}+(m-1)+1]$,
but $12\not =s_{1}+s_{2}$.

So they aren't equivalent. How do you prove it?

Rust, I didn't understand what do you mean.

If both statments A and B are true, it mean, that $A\to B, B\to A$ or $A$ and $B$ equavalent.
But for proof it you must proof both: A - true and B- true.

i know one simple equivalente see here  http://www.mathlinks.ro/Forum/viewtopic.php?t=88854  does any one know an other proprety?

[quote="Rust"]Let ${S=\{n^{2}+n+1}$, then for any even number 2m exist two members S, suth that $2m=(m^{2}+m+1)-[(m-1)^{2}+(m-1)+1]$,
but $12\not =s_{1}+s_{2}$.[/quote]
I didn't understand what you mean with this part.

S satisfyed property: for any even m, exist $s_{1},s_{2}\in S$, suth that $s_{1}-s_{2}=m$,
but don't satisfyed property: for any even m, exist $s_{1},s_{2}\in S$, suth that $s_{1}+s_{2}=m$.


How does one go about using an ordinary calculator and logarithms to find the first several digits of a huge number?. For instance, the newest prime discovered is 2^25,964,951-1. How can the methods I mentioned above be used to find its first 3 digits?.

Thank you,
Cody

All I know how to do w/log and very large numbers is finding how many digits is in the number, but I don't know if it's possible to find the first digits with just an ordinary calculator.  Here is how to find the number of digits:[hide]Since $\log 2\approx.3010299957$, and we want to find $2^{25964951}-1$, we can write the number as $\left(10^{.3010299957}\right)^{25964951}=10^{7816229.087}$, so the number has 7816230 digits.[/hide] Okay, that didn't really answer your question but I just felt like posting that  :P

Don't post a topic in multiple forums.  I just answered this in the number theory wing.


[quote="mathswizard1992"]hi!!!!!!! i am a new kid on the block
i scored 76th rank this year...........[/quote]

In what?? IIT-JEE? or NSTSE? Or NCO? or NSO?

I would like to know that too.. And btw, you got the spelling of Hogwarts wrong.........

:lol: actually in nstse...my score was 84/100....
and about the spellings of hogwarts .. i don't think i have to copy anyone this is my location
what about you and something about your school or studies...
by the way ...
happy world maths day!!!!!!!

As you say, "Hogwards" may not be wrong. But "quiet observant" surely is! Must be quite.

forget about the vocabulary..
we are concerned about maths...... :roll:

Doesnt mean you shouldnt be concerned about your language :) But anyhow, what exactly do you want to know? We are mostly a bunch of high school seniors, who supposedly have an exam day after tomorrow, and are still online without any thoughts whatsoever regarding the afforementioned exam :D

[quote="shruthibaskaran"] We are mostly a bunch of high school seniors, who supposedly have an exam day after tomorrow, and are still online without any thoughts whatsoever regarding the afforementioned exam :D[/quote]

Thats because the above mentioned person has finished 5 revisions for school exam and 3 for JEE and right now does not know what to do.

And yes, which year in Hogwards are you in?? First year?

in my campus or imagination(whatever) , there are no years there are profs, i am in third . :D
don't mind my language mistakes but i will definitely try to correct it :roll:

Coming back to the main discussion, in which exam did I get 76th rank?

sorry, did YOU get 76th rank?
@shruthibaskaran
How did u manage 3times for JEE revision???

If you had known the guy who posts as Cyber Shreyas, you would have known that he has an uncanny ability to over exaggerate some stuff. And for the record, he was JOKING!!!!!!!!!!!!!!!!!!!

(He is the one who finished revising 5 times for JEE... )

Dont take us seriously man, we just like to engage in friendly threats like these often.....

[quote="Aravind Srinivas L"]@shruthibaskaran
How did u manage 3times for JEE revision???[/quote]

Well u see dear fella people here are nerds to the core and here are some statistics:
shre finished 5 times
shruthi finished 3 times
chappli finished 10 times
vihang finished 12 times n is seeing oscar movies  :rotfl:
hash_include finished 20 times
chemrock 50 times
anit sahu 10000000000000000000000000000000000000000 times
and ofc pardesi is doing his 100000000000000000000000000000000000001th rev so he isnt online  :D

And i was the only sane person here until u came  :D

[quote="madness"][quote="Aravind Srinivas L"]@shruthibaskaran
How did u manage 3times for JEE revision???[/quote]

Well u see dear fella people here are nerds to the core and here are some statistics:
shre finished 5 times
shruthi finished 3 times
chappli finished 10 times
vihang finished 12 times n is seeing :rotfl:
hash_include finished 20 times
chemrock 50 times
anit sahu 10000000000000000000000000000000000000000 times
and ofc pardesi is doing his 100000000000000000000000000000000000001th rev so he isnt online  :D

And i was the only sane person here until u came  :D[/quote]

Wow, what a perfect example of irony you are speaking like this.. :D

Ha!!!! And chappli has only finished 9 times, he would be visibly upset to see you lying like this.,

[quote="shruthibaskaran"]What madness meant was that chappli is presently doing his 10th revision :D[/quote]

Check your english dear topper. "Finished" is surely past tense.. Or am I getting my tenses wrong?

No dear fellow topper.. You didnt get your tenses wrong, I was more inclined to think that you got your intentions misplaced..... ;) May the Maldivian Gods guide you in your quest to find truth, whoever they are.....

[quote="madness"][quote="shruthibaskaran"]No dear fellow topper.[/quote]

This phrase quite suggests a talk between equals dont u think? And if one is called a topper, it applies to the other by the principal of mathematical induction  :P   :rotfl:

EDIT: [b]AND OFCOURSE VARUN ARASU FINISHED 10000000000000000000000000000 TIMES FOR IIT!!![/b]  :omighty:[/quote]
go to hell da madness, i havent even started my board prep, let alone iit. probably u r referring to iraqi institute of terrorism. in dat case ur quite rite.(hope no one from the iraq forum sees this.)

[quote="VARUNARASU"]i just realizedthat  there is no iraq forum at all. :P[/quote]

Good for you. Maybe you were referring to the

International Institute of Typewriting which if is affiliated to the Massachusetts Institute of Typewriting and you can do a B.Type and do your masters in MIT>

[quote="Cyber-Shreyas"][quote="VARUNARASU"]i just realizedthat  there is no iraq forum at all. :P[/quote]

Good for you. Maybe you were referring to the

International Institute of Typewriting which if is affiliated to the [b]Massachusetts Institute of Typewriting [/b]and you can do a B.Type and do your masters in MIT>[/quote]

Rest in peace shre u might die after exactly 15 mins  :D   :P  :rotfl:

Results are coming out on the Sat, so I am sure ppl would be more tensed and anxious about that.

[quote="shruthibaskaran"]Tensed is not the word Shre, more like freaking nervously and anxiously waiting for life to spiral out of control :( I am just going mad...

@madness: I wont kill him just yet da.... He needs to be tortured first...[/quote]

surely you'll get i swear by the maldivian gods  :D

I hope the Maldivian Gods are not playing a cruel joke on me...

God-O-God, it would be a brutal joke

Yeah, yeah whatever. Do not try to act smart please! :D

What baby meant was that, it was Shre's much used dialogue, and I had used it.. SO give it a rest :D


MIT
college
abstract algebra
inequalities
linear algebra
linear algebra unsolved

Let $A,B \in \mathcal{M}_{n}(\mathbb{F})$.
Suppose that $A$ is nilpotent and commutes with $AB-BA$.

Show that $AB$ is nilpotent.

I want to use a lemma to prove that if $A$ commutes with $AB-BA$ then the latter is nilpotent. Lemma. If $\forall k\ge 1\; tr(A^{k})=0$ then $A$ is nilpotent. Ok, now one can use it to show that $\forall k\ge 1\; tr(AB-BA)^{k}=0$. The result follows almost readily from the fact that $A$ is nilpotent.

That works in zero characteristic. However, there are matrices $A$ which are not nilpotent and such that $tr(A^{k})=0$ for all $k$ if the characteristic is not $0$;

Thank you harazi for mentioning that. I knew it and just supposed that we have such a restriction on $\mathbb{F}$. Can you prove it without that assumption or it may have a counterexample then?

I cannot prove it in non-zero characteristic, but knowing that JC Math posted it, I assume it is true.  :)

Here F is 2-characteristic.
A=[0,1,0,0] (row by row) is a (2,2) nilpotent matrix.
Let B=[0,0,1,0]; AB=[1,0,0,0]; BA=[0,0,0,1];
A commutes with AB-BA=I.
AB and AB-BA aren't nilpotent matrices!

[quote="treegoner"]Can you explain to me more specifically. Why is $ tr(AB-BA)^{k}= 0$ ?[/quote]
$ tr(AB-BA)^{k}=tr[(AB-BA)(AB-BA)^{k-1}]=tr[AB(AB-BA)^{k-1}-BA(AB-BA)^{k-1}]=tr[AB(AB-BA)^{k-1}-B(AB-BA)^{k-1}A]$
But
$ tr[B(AB-BA)^{k-1}A]=tr[A B(AB-BA)^{k-1}]$

$ AB$ is nilpotent is a result due to Kostant; the following is the proof that you can find p.40 in this book (in french):"réduction des endomorphismes" by  R. Mneimné; ed:  Calvage et Mounet. This strange book is very unmethodical but filled with interesting results.
 Here I give a sketch of the proof in which I corrected an error of detail.
We assume $ carac(F)=0$; then $ AB-BA$ is nilpotent.$ ker(A)\subset{ker(A^{2})}\subset\cdots\subset{ker(A^{m})}=E$ where $ F^{n}=E$.
1) $ A^{k}B-BA^{k}=k(ABA^{k-1}-BA^{k})$. Then $ ker(A^{k})$ is $ AB$-stable. Let $ G=ker(A)\oplus(ker(A^{2})/ker(A))\oplus\cdots\oplus(ker(A^{m})/ker(A^{m-1}))$, $ g: (u_{k})_{k\leq{m}}\in{G}\rightarrow(ABu_{k})_{k\leq{m}}\in{G}$, $ v: (x_{k})_{k\leq{m}}\in{E^{m}}\rightarrow(k(AB-BA)x_{k}))_{k\leq{m}}\in{E^{m}}$, $ h: (u_{k})_{k\leq{m}}\in{G}\rightarrow(A^{k-1}u_{k})_{k\leq{m}}\in{E^{m}}$. $ h$ is well defined and one to one, $ spectrum(v)=\{0\}$ and $ v\circ{h}=h\circ{g}$. Moreover $ charpoly(g)=charpoly(AB)$.
$ g(u)=\lambda{u}$ implies $ v(h(u))=\lambda{h(u)}$. If $ \lambda\not=0$ then $ h(u)=0$ and $ u=0$.  QED
2) Remark: $ AB$ is nilpotent implies that $ BA$ is nilpotent.

Hi Moobi.
Kostant (MIT) has proved many great theorems (Mathematical carreer: 1954-1993), for example:
1) The universal enveloping algebra of a semi simple complex Lie algebra is a free module over its center.
2) The generalization of the Schur-Horn convexity theorem to the case of semisimple Lie groups.
3) (With Kirilov and Souriau) a new structure in symplectic geometry.
4) A theorem about projection of orbits onto a maximal abelian subalgebra.
    Here we study a little theorem which is linked to the filtered vectorspaces; I don't see a direct link with the great theorems.


trigonometry
calculus
geometry
topology
probability
real analysis
complex analysis

have just finished teaching myself College Algebra using the "Blitzer Algebra & Trigonometry 3e" Textbook. After finishing it, I asked the professors at my school if I could be aloud to take the final exams for both Algebra and Trig...

...I passed with a perfect scores :) . Now my question is, what's the next step? Would it be Calculus? Is Calculus the basis for the other 'maths'?

My goal is to become a Mathematician or [theoretical] Physicist. Simply put, I want to work specifically with math and theory's.

Currently I'm in a Community College taking General Studies. I'm switching to Basic Engineering next spring, and then transferring to a University in around 1-2 years. I know its laughable, but its all I've got right now.

So what do you guys suggest?

Thanks

[quote="Bictor717"]For physics, you'll definitely need at least multivariable calculus and differential equations. Real analysis might be unnecessary, but complex analysis could be helpful for things having to do with electricity.
The short answer is take calculus as soon as you can. Also, linear algebra may turn out to be helpful for some of the topics I listed above.[/quote]

I disagree. If your intention is theoretical physics or mathematics, there's much overlapping between the math subjects you should take a look at. Particularly real analysis and complex variable are a must have in my opinion. I don't know what is the level of that book of algebra you say but begin studying linear algebra and proceed to abstract algebra. And analysis. Of the typical courses you take in a math career I would say that if you do th. ph. you probably won't need much Discrete Math, (although it is Ok to know some)  and almost none Number Theory(maybe congruences come up from time to time). Algebraic NT doesn't appear but is so beautiful it is worth studying it. Besides that I would recomend almost every other course, Functional Analysis, Diff Geometry, Measure Theory, Topology, Algebra, Probability theory and related are recurrent subjects in theoretical physics.

Ursula

If you are going to teach yourself calculus, I highly recommend Tom Apostol's [i]Calculus[/i]. It is much more rigorous than the standard calculus textbook, and the exercises are useful for an aspiring theorist. After that, you should be ready to go into some serious real analysis and abstract algebra, maybe even some [u]fun[/u] point-set topology!

Edit: Just say no to engineering!  :P

[quote="CarlosRamos"]Thanks for the input guys, I really love math so I wish to eventually study through as much as possible.  However, my next step [i]would[/i] be to take Calculus right?

Oh and Curl, a professor at school also suggested that book...despite the fact that she doesn't teach from it.  Thanks, if Calculus is the next step, I'm definitely getting that one.[/quote]

I think you should take Calculus. That's what I would do...

Apostols book is good for basic calculus, you might want to try Hardy's A course in Pure Mathematics , or for a more updated but classic text 'An introduction to Real Analysis' by A. Kolmogorov and S. Fomin for Real Analysis.

For complex analysis Ahlfors and Rudin's books might not be the most accesible. Try Needham's book or Conway's book, but I recommend some firm grounding in real analysis and geometry first.
Then you can progress onto differential equations, etc.


induction
modular arithmetic
number theory

Can you show this by induction?  :huh:

$ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equiv a_1^2 \plus{} a_2^2 \plus{} \cdots \plus{} a_n^2 \pmod{2}$

Presumably some set of integers.

[hide="Fortunately..."] Induction isn't necessary.  We have $ a_i^2 \equiv a_i \bmod 2$ for any integer $ a_i$.  Either $ a_i$ is even (in which case so is $ a_i^2$) or $ a_i$ is odd (in which case so is $ a_i^2$).  Alternately, this is [url=http://mathworld.wolfram.com/FermatsLittleTheorem.html]Fermat's little theorem[/url] for $ p \equal{} 2$. [/hide]

duh! how did i forget? the $ a_i$'s are integers ... :0) Thanks t0r  :P

Is it true that $ 4^n \equiv 1 \pmod{9}$? why?

Not really, no.  $ 4^1 \equiv 4 \bmod 9$ and $ 4^2 \equiv 7 \bmod 9$.  It is true that $ 4^{3n} \equiv 1 \bmod 9$, though.

and also that $ 6\cdot 4^n\equiv 6\pmod{9}$. I thought this could somehow follow from the untrue $ 4\equiv 1 \pmod{9}$.

What are you talking about?  Neither of those statements are true.

Note that it is true that $ 4 \equiv 1 \bmod 3$.  Moreover, $ 3 | 6$.  In other words, because $ 3 | 4^n \minus{} 1$ and $ 3 | 6$ we conclude that $ 9 | 6(4^n \minus{} 1) \Leftrightarrow 6 \cdot 4^n \equiv 6 \bmod 9$.

We cannot divide by $ 6$ to conclude that $ 4^n \equiv 1 \bmod 9$, however.  In the language of rings, this is because $ 6$ is a [url=http://en.wikipedia.org/wiki/Zero_divisor]zero divisor[/url] (in particular, $ 6 \cdot 3 \equiv 0 \bmod 9$).


I have received my copy of "Divine Proportions: Rational Trigonometry to Universal Geometry" by Dr. Norman Wildberger, a professor that gave a lecture at my city's university this past summer, where I first became intrigued by his theories. I thought his approach to trigonometry was elegant and decided to order the book to examine his views more thoroughly. I am currently only half way through chapter 3 (I am a slow and very analytical reader) but am really enjoying the book and believe that his ideas really have value in simplifying trigonometry. I hope that I'll be able to understand the material in further chapters... it looks like it gets pretty complex, but I'll try the best I can. Anyways, as always, I'm interested in new ideas and approaches to math and would be happy to discuss this book and its theories with anyone else that may have stumbled across it. Let me know what you think of it!
Thanks,
Krista

First of all:

Hello!!

I have read about the book in a magazine and got immediately interested in it;
i mean hey, the author said that trigonometry was basically a big missunderstanding,
and that there are easier approaches to problems with angles.
So I went to the webpage and read a little more about it, to figure out what exactly his new idea is,
and as far as i got it he says that

          sin²a = AB² / BC²

(i figured that out by drawing a little picture, and thinking about it *proud*)

that seems logical to me, but i dont quite see how this should make life easier...
I think his approach to look at these squares instead of angles is quite good though.
But since you would have the problem of converting all the problems with angles to his format,
i think it doesn't really pay off, does it?

Well, i look forward to reading from you! :)

Wiedersehen,
JAN


I just started a Big Ten boarding schools' forum, and I want to make it popular, what should I do?
By the way, here is the link: http://www.bigtenschool.com

you obviously have to find people that are interested in it, most likely kids who attend boarding schools, but that's easier said than done. Posting here like you did might help you.

Keeping it well-moderated (removing spam, etc., etc.) and generating a little initial content (so the first people to come don't just see a blank page) should also help.  Having patience is probably key, though.

1.) Post on other popular forums and have the link on your signature.

2.) Put it on Google!


Hello everyone, I am new here. I would like to know what you think about my attempt to prove Goldbach's conjecture that you can read in the following link (I'm French, and the link is in French. If you don't understand my native language, please tell me) :

[url]http://les-mathematiques.u-strasbg.fr/phorum5/read.php?5,463546,page=2[/url]

If you ever have any questions, don't hesitate to ask me.
Thank you in advance.

I think you should translate to english if you want people here to read it.  It seems like you're already got feedback on it.

Ok. So let's consider a composite natural number n greater or equal to 4. We aim at proving that there is at least one natural number r such as (n-r) and (n+r) are both primes. For obvious reasons r<n-2. Such a number r will be called a "primality radius" of n.

Now let's define the number ord[C](n), which depends on n, in the following way :
ord[C](n):=pi(sqrt(2n-3)), where pi(x) is the number of primes less or equal to x, and sqrt(x) is the square root of x.
(n+r) is a prime only if for all prime p less or equal to sqrt(2n-3), p doesn't divide (n+r). There are exactly ord[C](n) such primes. The number ord[C](n) will be call the "natural configuration order" of n.
Now let's define the "k-order configuration" of an integer m, denoted C[k](n), as the sequel (m mod 2, m mod 3,...,m mod p[k]).
For example C[4](10)=(10 mod 2, 10 mod 3, 10 mod 5, 10 mod 7)=(0,1,0,3).
We call C[ord[C](n)](n) the "natural configuration" of n.

A sufficient condition to make r a primality radius of n is that for all integer i such that 0<i<1+ord[C](n), (n-r) mod p[i] differs from 0 and (n+r) mod p[i] differs from 0. If this statement is true, r will be called a "potential typical primality radius" of n.
More over, if r<n-2, then r will be called a "typical primality radius" of n.

Let's define N[1](n) as the number of potential typical primality radii of n less than P[ord[C](n)], where P[ord[C](n)]=2*3*...*p[ord[C](n)]. It is easy to give the expression of N[1](n).

Now let's define N[2](n) as the number of typical primality radii of n.

One can show that N[2](n)=n.N[1](n)/P[ord[C](n)]+O(1) (Landau's notation).

The boy whose nickname is qg77 in the link I gave earlier (a brilliant highschool pupil) showed that :

n.N{1](n)/P[ord[C](n)]>(c.n/(log(n))²)(1+o(1)), where c is a positive constant.

So that N[2](n) tends to infinity as n grows, and the Goldbach's conjecture is true for sufficiently large numbers.

What is your opinion about this?


induction
algebra
partial fractions
Putnam
college contests

Use partial fraction decomposition and it will telescope.

I tried partial fractions but was not able to find numerators that worked.

side note: Kalva used induction which was somewhat depressing and didnt explain how they saw the patter for the partial sums.

thanks but could you explain how you got that? or was it just guess and check?

Oh srry ! I was on haste this morning ! I did partial fraction, everything seemed fine but double check got wrong answer. Still got stuck !!


1) Find the inverse of the relation {(-14, 18), (3, -18), (-2, -1)}

2)  Find the value of the expression log 9 81  (9 is the base)

3)  Express as a product ln y52

Try using LaTeX, your questions are hard to understand.

[hide="hints"]1) Inverse function is the original function with the inputs and outputs reversed.

2) Hint: $81=9^{2}$.

3) Hint: $\ln(a^{b}) = b\ln(a)$[/hide]

Thanks for the hints.  Does anyone know the answers?

He is implying that you should be able to solve the questions from the hints given...

Show us what you have, then we can help you more than the hints.

I don't want to post the answers unless you can prove these aren't homework questions.


$21$ distinct numbers are chosen from the set $\{1,2,3,\ldots,2046\}.$ Prove that we can choose three distinct numbers $a,b,c$ among those $21$ numbers such that
\[bc<2a^2<4bc\]

Let the elements of $S$ be $a_1, a_2, \ldots, a_{21}$ where $a_1 < a_2 < \ldots < a_{21}$. Assume there does not exist three elements $a, b, c \in S$ for which $bc < 2a^2 < 4bc$.

Now consider terms $a_i < a_{i+1} < a_{i+2} \in S$.

Edit: The way I used the cases was definitely unclear. Let me rewrite this.

Assume $a_i < a_{i+1} < a_{i+2} \le 2a_i$. Then, $a_i a_{i+2} \le 2a_i^2$ but $a_i < a_{i+1}$ implies $2a_i^2 < 2a_{i+1}^2$ so $a_i a_{i+2} < 2a_{i+1}^2$. Since we are assuming there do not exist three elements $a, b, c \in S$ for which $bc < 2a^2 < 4bc$, and since $a_i a_{i+2} < 2a_{i+1}^2,$ that implies $2a_{i+1}^2 \ge 4bc$ which implies $2a_{i+1}^2 \ge 4a_i a_{i+2}$ which implies $\displaystyle \frac{a_{i+1}}{a_i} > \frac{2a_{i+2}}{a_{i+1}}$. But since $a_{i+2} > a_{i+1}, \displaystyle \frac{2a_{i+2}}{a_{i+1}} > 2$ which implies $a_{i+1} > 2a_i$ but $a_{i+1} < a_{i+2} \le 2a_i$ contradicts this.

From here we see that for $bc < 2a^2 < 4bc$ to never be true, $2a_i < a_{i+2}$ is always true, and from here we consider the minimal element of $S$, $a_1$.

$2a_1 < a_3, 2a_3 < a_5, \ldots 2a_{19} < a_{21}$, so clearly

$1024a_1 < a_{21}$

Since $a_{21} \le 2046$, this implies $a_1 = 1$.

However, since $2a_i$ is STRICTLY less than $a_{i+2}$ for all $i$, we have
$a_1 = 1$ implies $a_3 \ge 3$ implies $a_5 \ge 7$ $\ldots$ implies $a_{21} \ge 2047$.

But the maximal possible element of $S$ is $2046$, so this is a contradiction and we are done.

Forgive me, but i have a little problem understanding your solution. Since you showed that a_21 >= 2047, and that contradicts the assumption that 2046 is the largest element. Then that implies a_(i+2) <= 2a_i for some i. How would we make use of that fact to show that there must exist bc < 2a^2 < 4bc?

Sorry to bother you,

You're right, the previous solution was incoherent, as I didn't bother to draft up an on paper solution before trying to write an online one, I managed to leave out a chunk where I explain how I applied the cases.

Since I had to edit the solution anyway, I figured I might as well write it out the more elegant way.

The cases themselves were incorrectly stated (the way I had it typed up, with the additional information I had on paper it made sense).

Partition $S$ into $10$ subsets as following:
$S_i =\{2^{i+1} -2 ,2^{i+1} -3,\ldots,2^{i+1} - (2^i +1) = 2^i -1\}$ for all $i = 1,2,\ldots,10$.

All $21$ elements are in the $10$ sets, so there exist $3$ elements $a <b<c$ within one set.
Now it's very easy to prove that  $ac <2  b^2 <4ac$.
 :)

[quote="namanhams"]Divide S into 10 subsets as following:
S_i ={2^(i+1) -2 ,.....,2^i -1 } for all i = 1,2,....10

All 21 elements are in 10 sets ,so there exist 3 elements a <b<c which are in one set .
Now it's very easy to prove that  ac <2  b^2 <4ac
 :)[/quote]
I'll be happy if you prove it...
Thanks

It's really great, namanhams. Congratulation.

Partition $S$ into $10$ subsets as following:
$S_1 = \{1, 2, 3\}$
$S_i = \{2^{i},\ldots,2^{i+1}-1\}, i=2,\ldots,9$
$S_{10} = \{2^{10},\ldots,2^{11}-2\}$

Lemma: if $a, b, c \in S_i$ and $b<a<c$ then $bc<2a^2<4bc$

Proof:
$a, b, c \in S_i$ and $b<a<c$ $\Rightarrow c<2a$, and $(b<a, c<2a) \Rightarrow bc<2a^2 $ (*)

Similarily, $a, b, c \in S_i$ and $b<a<c$ $\Rightarrow a{\le}2b$, and $(a{\le}2b, a<c) \Rightarrow a^2<2bc \Rightarrow 2a^2<4bc$ (**)

(*) and (**) $\Rightarrow bc<2a^2<4bc$

If we select 21 members of the set $\{1,2,3,\ldots,2046\}$ at least 3 of them will be from one of the $S_i$ sets. According to the lemma above the inequality holds for these three elements.
Q.E.D.

I agree!

Same solution, but worded differently:

Let $S$ be the chosen set, and let set $T$ ($|T|=11$)consist of the smallest number in $S$, the third smallest number in $S$, the fifth smallest, etc. We will prove that there must be a pair of numbers $(m,n)$ in $T$ such that $m<n\leq 2m$. Suppose for the sake of contradiction that every pair of numbers in $T$ with $m<n$ has $n>2m$. Then, let $t_i$ be the $i$th smallest element in $T$. We have that $t_1\geq 2^1-1$, and then by an easy induction we obtain that $t_i\geq 2^i-1$. But this implies that $t_{11}\geq 2047$, contradiction.

Thus, take one such pair $(m,n)$, and let $A$ be the number in $S$ greater than $m$ and less than $n$. Then, we have
\[\frac n2\leq m<A< n\leq 2m,\]
so $(a,b,c)=(A,m,n)$ is one such triple, as desired.


inequalities
calculus
rearrangement inequality
inequalities proposed

Suppose that n-distinct natural numbers $x_1,x_2,\cdots ,x_n$ satisfies $1\leqq x_k\leqq n\ (1\leqq k\leqq n)$.
Prove the following inequality.

\[\sum_{k=1}^n\frac{x_k^2}{k^2+1}>n-\frac{8}{5}\]

Clearly the numbers must be $1,...,n$ in some order. Let
\[E=\sum_{k=1}^n \frac{x_k^2}{k^2+1}\]
By the rearrangement inequality $E \geq \sum_{k=1}^n \frac{k^2}{k^2+1}=n-\sum_{k=1}^n \frac{1}{k^2+1}$. So we need to prove that $\sum_{k=1}^n \frac{1}{k^2+1} \leq \frac{8}{5}$. Applying the inequality $\frac{1}{k^2+1} \leq \frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}$ for $k \geq 2$, we obtain the slightly stronger bound $\frac{3}{2}$.

Thank you for your reply, Fiachra.

You are right. In using calculus,we will be able to obtain the more stronger bound $\frac{\pi}{2}$.

kunny

I must have made some mistake then, since $\frac{\pi}{2} > \frac{3}{2}$ !


algebra
polynomial
MIT
college
geometry
geometric transformation

Prove that for any odd prime number p, the polynomial f(x) = 2(1 + x^(1/2*(p^n+1)) + (1-x)^(1/2*(p^n+1))) is congruent mod p to the square of a polynomial with integer coefficients.

[i]Mikls Schweitzer competition 1955 problem 7[/i]

Let g(x) = 1 + x (pn + 1)/2  - (1-x) (pn + 1)/2 .
Clearly, g(x) is a polynomial with integer coefficients and g(0) = 0. Then h(x) = 2g(x)/x is a polynomial with integer coefficients.

Moreover f(x)h(x) = 4( (1 + x (pn + 1)/2 ) 2  - (1-x) (pn + 1)  )/x.

But, using that k*C(p n ,k) = p n *C(p n -1 ,k-1), it follows that C(p n , k) = 0 mod[p] when 0 < k < p n . Thus : (1-x) pn  = 1 - x pn  mod[p].
It follows that f(x)h(x) =  4( (1 + x (pn + 1)/2 ) 2  - (1-x)(1 - x pn  ) )/x = 4(1 + x (pn - 1)/2 ) ) 2.

Now, it suffices to prove that f(x) and h(x) are relatively prime in Q_p, that is there is no integer a such that f(a) = h(a) = 0 mod[p].
Thus : Suppose that such an integer a does exist.
Then 4( 1 - a)(pn + 1)/2 ) = f(a) - ag(a) = 0 mod[p], from which it follows that a = 1 mod[p]. Then f(a) = 4 =/= 0 mod[p]. A contradiction, and we are done.

Pierre.

This problem is related to a theorem of L.Rdei

L.Rdei, ber einige merkwrdige Polynome in endlichen Krpen mit
Zahlentheoretischen Beziehungen, Acta Sci.Math. 11 (1946-48), p.42

Could you be more explicit, please?

Pierre.

The reference  Acta.Sci.Math, I found it in the book
Contest in Higher Mathematics, Schweitzer 1949-1961
in math research library.But I did not look the article of Rdei.
If any members can get a copy of this article and explain what is it
about it will be nice, I think it written in german.

Hi Moubinool,

At first I wondered about your cited source as I knew it from somewhere but I could not remember where I saw it. But then I realized that it was the one which is given in the book "Mikls Schweitzer 1949-1961".

Zeitschrift:  Acta scientiarum mathematicarum / Institutum Bolyaianum
Universitatis Szegediensis
ISSN:  0001-6969

It is available in the state library in Berlin. In my local math library in Potsdam there are only records back until 1986.

Sorry Orl, my english is poor
I read your post I don't see the question.What is the question ?

Hi Moubinool,

I just wanted to indicate that I could get the material. But it would mean that I had to re-subscribe to the state library Berlin (my account has expired). Moreover I had to go to Eastern Berlin by "S-Bahn" which takes me one hour each. So I will prefer only to get there if you are really interested in the article. So my question was somehow hidden.

Well, I think the paper can probably be find in our usual library here in Paris. So, don't go before we check if I'm right Orl. Thanks for the proposition.

Pierre.

Hi orl,

Don't go to Berlin nice city only for the article to long your journey 1hour.
Math library is not far from my place in Paris maybe 20' or 30'.
I will try get a photocopy someday, but I 'll need your help for translation
german->english.

thx
Moubinool


This is Miu...who screwed up the USAMO poll twice in a row...The second poll was screwed by a mistake of the forum though><

Anyway, about the S-III form for USAMO, If I live in Canada, do I leave the second part BLANK? or I just say "no" to all of them?

Thanks~

Please direct all informational and procedural queries about AMC forms, awards, certificates, procedures etc directly to the AMC office at amcinfo@maa.org, or call our toll-free number at 800-527-3690. It makes much more sense to contact us directly with a question and have it answered by the responsible staff member than to post your question here for people who are not reponsible for executing the procedures to speculate about.

Unless you are an American Citizen or Legal Resident, leave it blank.

Steve Dunbar
AMC Director

Ifyou're a legal resident of canada, then just fill it out anyways.


function
modular arithmetic
algebra unsolved
algebra
Fixed point
set
functional equation

Let $ X\equal{}\{1,2,3,4,...,1996\}$ and $ f: X\rightarrow X$ be a function for which $ f(f(f(f(f(x)))))\equal{}x ,\forall x \in X$. Prove that there is $ x_0 \in X$ with $ f(x_0)\equal{}x_0.$

All points in $ X$ are fixed points for the $ 5^{\textrm{th}}$ iterate of $ f$, so they are partitioned in cycles for $ f$ of (orbit) lengths $ \lambda_i \mid 5$. But $ 1996 \equiv 1 \pmod{5}$, so there must exist at least one cycle of length $ 1$, i.e. a fixed point $ x_0$ for $ f$.
In fact, the number of fixed points for $ f$ is thus  forced to be $ \mathcal{M}5 \plus{} 1$.


RTP: $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geq\frac{3}{2}$ for positive real numbers $a, b, c$

this is the nesbitt inequality, it has been posted a billion times

A proof without AM-HM, Rearrangement:
\[ \sum \frac{a}{b+c} - \sum \frac{1}{2} = \sum \frac{(a-b)^2}{2(b+c)(a+c)} \geq 0 \]


A Mathlon is a competition in which there are $M$ athletic events. Such a competition was held in which only $A,$ $B,$ and $C$ participated. In each event $p_{1}$ points were awarded for first place, $p_{2}$ for second and $p_{3}$ for third, where $p_{1}>p_{2}>p_{3}>0$ and $p_{1},$ $p_{2},$ $p_{3}$ are integers. The final score for $A$ was $22,$ for $B$ was $9,$ and for $C$ was also $9.$ $B$ won the $100$ metres. What is the value of $M$ and who was second in the high jump?

I worked through this problem with intuition, and some trial and error. Is there a more systematic way to do these types of problems?

[hide="Result"]$p_{1}=5,$ $p_{2}=2,$ $p_{3}=1.$ $M=5,$ and $C$ was second in the high jump.

$A$ won 4 times, got second once
$B$ won once, got last 4 times
$C$ got second 4 times, last one time

$A$ must've won every game excluding the one that $B$ won, so he must've gotten second in the $100$ metres. That leaves $C$ as the only person to get second in the high jump.
[/hide]


I know any poll like this pigeonholes people unrealistically, but I'm just wondering.

At the Political Compass, my dot is right on top of the Dalai Lama's.

I do have to wonder if this is slanted at all.

Your political compass
Economic Left/Right: -4.63
Social Libertarian/Authoritarian: -2.56

i'm a bit to the right of ghandi

Your political compass
Economic Left/Right: 6.88
Social Libertarian/Authoritarian: -0.56

I'm not as libertarian as Badnarik according to that site, but I do believe that I share his economic views for the most part.

Economic Left/Right: 2.00
Social Libertarian/Authoritarian: 2.00

On Hitler and Thatcher's side

I think it's interesting that if we total in either category (according to the poll I mean), we get
-------------
Social
liberal 22
centrist 5
conservative 7
Fiscal
liberal 10
centrist 9
conservative 15
-------------
That is, the demographic of AoPS users seems to be a bit skewed in favor of social liberalism, but fairly even on fiscal matters. Is this because most AoPSers are too young to be well educated about economics and fiscal policy (so our demographic would be biased there too, if more older people were voting), or because we are actually pretty evenly distributed there?

[quote="JBL"]Can we please [i]not[/i] turn this into a thread in which everyone just lists their results from taking the political compass test?  That's the sort of topic I usually dispatch to the G&FF or the "polls" forum.  The Round Table is for [i]discussion[/i], so if you're going to discuss, discuss, and if you're just going to give a data point, do it somewhere else.[/quote]
Well you are right.
For me the test is pretty useless anyway (when those tests ask me "does the life of a USA citizen matter more" what am I  supposed to say?  :roll: )

I've said this before, but words like "liberal" mean completely different things in other places, here it's about the most right wing you can be (apart from far right).

I was about to say that in other parts of the world , there are different issues and therefore different things at stake when elections are held.
But wouldn't the same be true in a big country like the USA? I can imagine someone from Chicago being less worried about immigration than someone from San Diego?

hmmm.

Why is there no option for those who simply don't care about politics?

I was both socially and fiscally liberal, and I was pretty close to the "gandhi" range.  And in response to treething's post, I don't think this is a test to find out how much you care about politics, it is a test to find where your beliefs lie with respect to politics: a political blogger and an apathetic student can have the same basic values, but their interest is obviously completely different.  Thus, you don't have to care in order to have a position on the spectrum.

Or maybe I'm just blathering.


Numbers $1,2,3.....1994$ are written down on a black board.Two numbers are chosen,erased and their difference is written up instead.This continues till there is only one number left on the black board.Is the final number odd or even?
 :)

An invariant! Nice.

[hide="Hint"]Examine the parity (even or odd) of the difference. Will this parity always be the same as a different operation we can do?[/hide]

[quote="chesspro"]An invariant! Nice.

[/quote]

agreed!

[hide="solution"]let the sum be S, easy evaluation gives S is odd

consider all partity subtractions

even - even=even
even-odd=odd
odd-odd=even

in all of those the sum is the same partity mod 2 as the difference (i.e. odd or even)

therefore, the sum will always be odd, and we can eliminate all the terms, leaving the last term odd[/hide]


[quote="daermon"]is it [hide]34?[/hide][/quote]

proof or hint?

[quote="daermon"]is it [hide]34?[/hide][/quote]

yes it is. can you prove it?  :)

[quote="daermon"]actually, I kinda just plugged it into my calculator  :blush:[/quote]

haha. must be some calculator...  :lol:

actually, its a pretty average calculator.  not even scientific

oh so you looked at the scientific notation...  :)

heres a hint...

[hide]use scientific notation and inequalities[/hide]

also...

[hide]let one number be x and the other be y.[/hide]

[quote="fireemblem13"]or u could use Mathematica
but my Mathematica doesn't work  :huh:[/quote]

It's quite easy with the hints buzzer gave you?

Why doesn't your mathematica work? why are you using it in the first place?

A really expensive calculator for your computer.

idk. i was just suggesting a method of arriving at the answer. you wouldn't really use mathematica. ok. when i click the mathematica icon, nothing happens :huuh:

[quote="Totally Zealous"][quote="buzzer11"]How many digits are in the product of 3,659,893,456,789,325,678 and 342,973,489,379,256?

source: AHSME 1969

 :P[/quote]

[hide]$3.7\cdot10^{18}\times3.4\times10^{15}=12.58\cdot10^{33}\approx1.3\times10^{34}.$ So 34.[/hide]

What's mathematica?[/quote]

Actually the way that you did this shows that their are 35 digits in the product.  It is 13 followed by 33 zeroes.

$35$

Did I do anything wrong?

[quote="t0rajir0u"]Computations are almost entirely unnecessary.  You only need this fact:

$32^{2}= 1024 \implies \sqrt{10}< 3.2$[/quote]

Thats exactly what i did.

I counted digits after first digit and got 14 and 18.

14+18+1 (because first two digits were greater than 32) + 1 = 34.

okay, as all of you know the answer is:

[hide]34[/hide]

and here is the given solution:

[hide]Let the numbers be $x$ and $y$, respectively. Hence, $3.7(10)^{18}$ is greater than $x$ is greater than $3.6(10)^{18}$ and $3.5(10)^{14}$ is greater than $y$ is greater than $3.4(10)^{14}$. Thus $3.7(10)^{18}* 3.5(10)^{14}= 1.295(10)^{33}$, which is greater than $xy$ which is greater than $3.6(10)^{18}* 3.4(10)^{14}= 1.224(10)^{33}$. Both of these limits have 34 digits, so the product $xy$ has $\boxed{34}$ digits.[/hide]

:blush::blush:
How come moderators can make up to 8 simileys?

How much does mathematica cost?

http://www.wolfram.com/products/mathematica/index.html
You will be shocked.  :lol:

You get a discount if you're a student but you get an 'incomplete' version


Let $ x_1,x_2$ are solutions of $ x^2 \minus{} 6x \plus{} 1 \equal{} 0$. Prove $ \forall n \in \mathbb{Z}$: $ x_{1}^n \plus{} x_{2}^n \ne 5K, K\in \mathbb{Z}$

$ x_{1,2} \equal{} 3\pm 2\sqrt {2}$. since $ x_1, x_2$ are reciprocals of each other, we need only consider $ n > 0$. consider the Pell equation $ r^2 \minus{} 2s^2 \equal{} 1$. its smallest solution is $ (3, 2)$ and so all solutions are given by $ r_n \plus{} s_n\sqrt {2} \equal{} (3 \plus{} 2\sqrt {2})^n$. Then $ x_1^n \plus{} x_2^n \equal{} (3 \plus{} 2\sqrt {2})^n \plus{} (3 \minus{} 2\sqrt {2})^n \equal{} r_n \plus{} s_n\sqrt {2} \plus{} r_n \minus{} s_n\sqrt {2} \equal{} 2r_n$, so we need to show that $ 2r_n\not\equiv 0 \pmod{5}$, i.e. $ r_n\not\equiv 0\pmod{5}$. but if $ r_n\equiv 0\pmod{5}$, then we'd have $ 2s_n^2 \plus{} 1\equiv 0 \pmod{5}$, a contradiction because the latter has no solutions in integers.


Zdaj veste kako je imeti priprave na Bledu in v Ljubljani. Kje vam je bolj všeč in zakaj?

Tako Bled kot Ljubljana imata svoje prednosti (pa tukaj ne mislim Taroka, da se razumemo). Meni je dosti vseeno kje so sama predavanja, če so le ta zanimiva :) (kar MMO priprave nedvomno so), sicer pa se mi Bled nekako bolj dopade, ker je cel teden posvečen matematiki. Vendar ne moremo odmisliti prisotnost šole (bolje rečeno odsotnost) in šolskih obveznosti, ki jih moramo prav tako izpolnjevati in nadoknaditi zamujeno čez vikend, kar zna biti nekoliko moteče, še posebej, če se predela ogromno pomembne snovi. Sicer pa jaz zase lahko povem, da so priprave kar velik zalogaj. Mislim, da je količina fint, ki smo jih slišali na Bledu kar obilna. Moj način dela je pač tak, da vsako stvar potem doma še enkrat prežvečin v svojem lastnem tempu in zadeve še enkrat premeljem :huh: pa tudi koncentracija lahko popusti, če delaš matematiko cele dneve... moje skromno mnenje.

Se strinjam, da je teden priprav velik zalogaj, ki potrebuje nekaj časa. Lepo bi bilo, če bi kaj od stvari spoznavali že v šoli, ampak program je pač tak kot je. Je pa lepo slišat, da obstaja nekdo, ki te naloge še enkrat prebavlja doma. Sem pa opazila tudi, da se zelo pozna kako si stvari zapisujete v zvezke že med samimi predavanji, urejenost se potem prenaša na reševanje testov, to pa povzroča veliko zadovoljstvo popravljalca in hitro popravljanje. :D

Vikend je pa prost zato, da lahko poskušate nadoknaditi zamujeno v šoli. :) Iz lastnih izkušenj pa lahko povem, da ni nobena katastrofa, če kdaj človek manjka v šoli. Jaz sem v četrtem letniku zaradi priprav in dobršne mere bolezni manjkala skoraj tretjino leta! Zdaj se mi to nikjer ne pozna (vsaj meni se ne zdi). :) Upam, da se tudi tebi ne bo preveč.

Se samo meni zdi, al po tem forumu veliko manj ljudi piše kakor po prejšnjem počasnem?  :juggle:

Ja verjetno je bil bolj znan, pa se link nanj smo imeli na strani dmfa.


Fie a,b,c si d lungimile laturilor unui patrulater ABCD.
Daca $ \frac{2(a\plus{}c)(b\plus{}d)}{a\plus{}b\plus{}c\plus{}d}\equal{}\frac{ab}{a\plus{}b}\plus{}\frac{bc}{b\plus{}c}\plus{}\frac{cd}{c\plus{}d}\plus{}\frac{da}{d\plus{}a}$ atunci ABCD este paralelogram
G.M. 5/2007

Vom folosi urmatoarea:
[color=blue]LEMA[/color]
\[ \frac{x}{a\plus{}x}\plus{}\frac{y}{a\plus{}y}\leq\frac{2(x\plus{}y)}{2a\plus{}x\plus{}y},\forall a,x,y>0.\]

Demonstratia lemei este simpla. Prin reducere, ajungem la forma echivalenta $ a(x\minus{}y)^{2}\geq 0$, ceea ce spune totodata ca egalul se atinge daca si numai daca $ x\equal{}y$.

Atunci, in baza lemei, avem
\[ \frac{ab}{a\plus{}b}\plus{}\frac{bc}{b\plus{}c}\plus{}\frac{cd}{c\plus{}d}\plus{}\frac{da}{d\plus{}a}\equal{}a\Bigg(\frac{b}{b\plus{}a}\plus{}\frac{d}{d\plus{}a}\Bigg)\plus{}c\Bigg(\frac{b}{b\plus{}c}\plus{}\frac{d}{d\plus{}c}\Bigg)\leq\]
\[ \leq\frac{2a(b\plus{}d)}{2a\plus{}b\plus{}d}\plus{}\frac{2c(b\plus{}d)}{2c\plus{}b\plus{}d}\equal{}(b\plus{}d)\Bigg(\frac{2a}{2a\plus{}b\plus{}d}\plus{}\frac{2c}{2c\plus{}b\plus{}d}\Bigg)\leq\]
\[ \leq(b\plus{}d)\cdot\frac{2(2a\plus{}2c)}{2(a\plus{}b\plus{}c\plus{}d)}\equal{}\frac{2(a\plus{}c)(b\plus{}d)}{a\plus{}b\plus{}c\plus{}d}.\]
Egalul are loc daca si numai daca $ a\equal{}c$ si $ b\equal{}d$.


geometry
rectangle
combinatorics unsolved
combinatorics

Given a finite string $S$ of symbols $X$ and $O$, we write $@(S)$ for the number of $X$'s in $S$ minus the number of $O$'s. (For example, $@(XOOXOOX) =-1$.) We call a string $S$ [b]balanced[/b] if every substring $T$ of (consecutive symbols) $S$ has the property $-2 \leq @(T) \leq 2$. (Thus $XOOXOOX$ is not balanced since it contains the sub-string $OOXOO$ whose $@$-value is $-3$.) Find, with proof, the number of balanced strings of length $n$.

Define $facevalue$ of $X$ as $1$
 and that of $0$ as $-1$.

 Let $S_{k}$ be the sum of the facevalues of the first $k$ elements of the string.
 A string is balanced if and only if for any two naturals, [i]l [/i]and [i]m[/i],
  $S_{l}-S_{m}$ $\in$ $\{-2, 2 \}$.

  Clearly, $S_{k}$ and $S_{k+1}$ can differ only by $1$.
  So if we write a sequence for $S_{i}$,
  say $1,0,{-1},0,{-1},0,1$,
 we are actually referring to the sequence XOOXOXX.
 which satisfies the given conditions.

 So if we write such  a balanced sequence of $S_{i}$, we have effectively written a balanced
 string.

 Use the following rules to write a sequence of $S_{i}$, and then count the total number
 of strings that you can write:

 1. Assume that the first character of the balanced string is X. At the end double your answer.
     Each $S_{i}$ string thus starts with $1$

 2.  Note that in each string of $S_{i}$, there can only appear $3$ different characters.

     These may be $\{-1, 0, 1\}$ or $\{0, 1, 2\}$
      $-2$ can not appear as $1$
      exists in the string.

 3. Consider each of the above two cases and count the number of possible strings in each
     case. The easiest way is to find manually for trivial cases ,then prove by induction.

This was as I solved it in the TST,
Map each string to a diagonal lattice path starting at the origin.Say X corresponds to a step of length sqrt (2) at 45 degrees to the positive x axis and O to a -45.
Then every balanced string of length  is bounded by a 2*n rectangle and vice versa.Thus we have a bijection,count the number of such paths,
(easy recursive calculation).

Cheers
Bhargav

This problem was also on Croatian MEMO Selection Test 2007. It was one of the most difficult problems.


[b]Ziua 1[/b]

[b]Subiectul 1[/b]: Sa se determine toate numerele prime $ p$ pentru care numarul $ 1 \plus{} p2^{p}$ este patrat perfect. ([b]Romania [/b])

[b]Subiectul 2[/b]: In triunghiul $ ABC$, bisectoarea unghiului $ A$ intersecteaza latura $ [BC]$ in punctul $ A_{1}$,iar cercul circumscris triunghiului $ ABC$ in punctul $ A_{2}$.Analog se obtin punctele $ B_{1}$ si $ B_{2}$, respectiv $ C_{1}$ si $ C_{2}$.Sa se arate ca:
\[ \frac {A_{1}A_{2}}{BA_{2} \plus{} A_{2}C} \plus{} \frac {B_{1}B_{2}}{CB_{2} \plus{} B_{2}A} \plus{} \frac {C_{1}C_{2}}{AC_{2} \plus{} C_{2}B} \geq \frac {3}{4}
\]
([b]Republica Moldova[/b])

[b]Subiectul 3[/b]: Fie $ 0 \leq x \leq 2\pi$. Sa se demonstreze ca :
\[ \sqrt {\frac {\sin^{2}x}{1 \plus{} \cos^{2}x}} \plus{} \sqrt {\frac {\cos^{2}x}{1 \plus{} \sin^{2}x}}\geq 1
\]
([b]Spania[/b])
 [hide="Hint"]Inegalitatea este destul de slabuta , mi-a dat ceva de genul $ MS > \frac {2}{\sqrt {3}} > 1$  :D [/hide]

[b]Ziua 2[/b]

[b]Problema 4[/b]: Se considera tetraedrul arbitrar $ ABCD$.Fie $ E$ si $ F$ mijloacele muchiilor $ AB$ , respectiv $ CD$.Daca $ \alpha$ este unghiul format de muchiile $ AD$ si $ BC$ , sa se calculeze $ \cos \alpha$ in functie de lungimile segmentelor $ [EF],[AD]$ si $ [BC]$. ([b]Romania[/b])

[b]Problema 5[/b]: Diagonalele patrulaterului inscriptibil  $ ABCD$ se intersecteaza in punctul $ E$. Punctele $ K$ si $ M$ sunt mijloacele laturilor $ AB$ , respectiv $ CD$.Punctele $ L$ si $ M$ sunt proiectiile punctului $ E$ pe laturile $ BC$ , respectiv $ AD$. Demonstrati ca dreptele $ KM$ si $ LN$ sunt perpendiculare. ([b]Republica Moldova[/b])

[b]Problema 6[/b]: Se considera multimea de numere naturale $ U \equal{} \{ 1,2,3,...,6024\}$.Demonstrati ca pentru orice partitie a multimii $ U$ in trei submultimi cu cate $ 2008$ elemente fiecare, putem alege cate un numar din fiecare submultime astfel incat unul dintre numere sa fie suma celorlalte  doua numere . ([b]Spania[/b])

[hide="Ziua 1 ,Subiectul 3"]
Fie functia $ f: (0,\infty) \to \mathbb{R}$ , $ f(x) \equal{} \frac {1}{\sqrt {x}}$ , deoarece $ f '(x) > 0$ avem f convexa .
Inegalitatea de demonstrat este echivalenta cu $ \sin^2x \cdot f(\sin^2x \plus{} \sin^2x \cos^2x) \plus{} \cos^2x \cdot f(\cos^2x \plus{} \sin^2x \cos^2x) \geq 1$
Deoarece $ \sin^2x \plus{} \cos^2x \equal{} 1$ putem aplica inegalitatea lui Jensen si rezulta :

$ \sin^2x f(\sin^2x \plus{} \sin^2x \cos^2x) \plus{} \cos^2x f(\cos^2x \plus{} \sin^2x \cos^2x) \geq f \left( \sin^4x \plus{} \cos^4x \plus{} \sin^2x \cos^2x\right)$ .
Asadar mai ramane de demonstrat ca :

$ \frac {1}{\sqrt {\sin^4x \plus{} \cos^4x \plus{} \sin^2x \cos^2x}} \geq 1$ $ \Leftrightarrow$ $ \sin^2x(\sin^2x \minus{} 1) \leq 0$.[/hide]

Nu cred ca e bine ceea ce ai gasit tu ca egalitatea nu are loc .ia de exemplu $ x \equal{} \frac {\pi}{2}$ si vezi ca are loc egalitate.

[quote]Nu cred ca e bine ceea ce ai gasit tu ca egalitatea nu are loc .ia de exemplu $ x=\frac{\pi}{2}$  si vezi ca are loc egalitate.[/quote]
Nu-ti face griji, avem egalitate si pentru $ x=0$.:D
Sa luam functia $ f: (0,1)\to \mathbb{R}_{+}$ , $ f(x)=\frac{1-x}{1+x}$ si avem ca $ f'(x)<0$ si $ f''(x)>0$.
Obtinem ca si functia $ \sqrt{f(x)}$ este de asemenea [b]strict convexa[/b] [hide="deoarece:"] \[ (\sqrt{f(x)})''=\frac{1}{2}(\frac{f''(x)}{\sqrt{f(x)}}+\frac {-f'(x)}{2\sqrt{f(x)}f(x)}) >0\][/hide]
Aplicand inegalitate lui Jensen pentru $ x\in (0,\frac{\pi}{2})$ avem :
\[ f(\sin^{2}x)+f(\cos^{2}x) > 2 f(\frac{\sin^{2}x+\cos^{2}x}{2})=2f(\frac{1}{2})=\frac{2}{\sqrt{3}}>1\]
Deci am aratat ca pentru $ x\in (0,\frac{\pi}{2})$ inegalitatea este [b]stricta[/b], si cum avem egalitate pentru $ x\in\{0,\frac{\pi}{2}\}$ problema este incheiata
[hide="Srry"]Scuze ca nu am specificat la inceput ca era stricta pe $ x\in (0,\frac{\pi}{2})$ :D[/hide]

Pentru [b]problema 6 , ziua a doua [/b], asa cum presupun ca toti ati observat, am fost surprins de faptul ca Spaniolii au propus o problema care are o oarecare asemanare cu problema care a fost data la [url=http://www.mathlinks.ro/Forum/resources.php?c=114&cid=75&year=2008]Moldova TST,Problema 2,Ziua 1 (2008)[/url].

Poate cineva sa posteze problemele pentru Juniori...??
Multumesc..

I will try to post the junior problems, if you asked me that. I don't know romanian.
Btw, I saw the subjects for seniors and the first day is quite easy (maybe an exception is the second issue, but this is no hard also :wink: )
The second day is harder, but I have a question to Romanian organizators:
[b]What is the use of the problem nr.4 ?[/b]
This is an international IMO preparing contest, but not a romanian district round. :spam:

Problem J1
Let $ x_1,x_2,...,x_n\geq0$ s.t. $ \sum{x_i}\equal{}1$.
Prove that:
$ (a)$ $ \sum{\sqrt{\dfrac{x_i}{2\minus{}x_i}}}\geq1$ for $ n\equal{}2$.
$ (b)$ $ \sum{\sqrt{\dfrac{x_i}{2\minus{}x_i}}}\geq1$ for $ n>2$.

Very easy....

[quote][color=darkred][b][u]Problema 2[/u][/b]. In triunghiul $ ABC$, bisectoarea unghiului $ A$ intersecteaza latura $ [BC]$ in punctul $ A_{1}$ si cercul circumscris triunghiului in punctul $ A_{2}$ .

Analog se definesc $ B_{1}$ si $ B_{2}$ respectiv $ C_{1}$ si $ C_{2}$ . Sa se arate ca $ \frac {A_{1}A_{2}}{BA_{2} + A_{2}C} + \frac {B_{1}B_{2}}{CB_{2} + B_{2}A} + \frac {C_{1}C_{2}}{AC_{2} + C_{2}B} \geq \frac {3}{4}\ (*)$ ([b]Republica Moldova[/b]).[/color][/quote]

[color=darkblue][b][u]Demonstratie.[/u][/b] Se stie ca $ \left\|\begin{array}{c} A_2B = A_2C = A_2I \\
 \\
\frac {A_1B}{c} = \frac {A_1C}{b} = \frac {a}{b + c}\end{array}\right\|$ . Din asemanarea $ \triangle A_1BA_2\ \sim\ \triangle A_1AC$ retinem $ \frac {A_1A_2}{BA_2} = \frac {CA_1}{AC}$ , adica $ \frac {A_1A_2}{BA_2} = \frac {a}{b + c}$ .

In concluzie, inegalitatea $ (*)$ este echivalenta cu $ \sum\frac {a}{b + c}\ge\frac 32$ care este evidenta (culmea, poarta si un nume !) Cam "ratacit" acest subiect la [b]IMAC[/b].[/color]

=============================================================================================================

[quote][color=darkred][b][u]Problema 3[/u][/b]. Fie $ 0 \leq x \leq 2\pi$. Sa se arate ca $ \sqrt {\frac {\sin^{2}x}{1 + \cos^{2}x}} + \sqrt {\frac {\cos^{2}x}{1 + \sin^{2}x}}\geq 1$ ([b]Spania[/b])[/color][/quote]
[size=117][color=darkblue][b]Iata o usoara extindere a[/b][/color] [color=darkred][b]problemei 3[/b][/color] [color=darkblue][b]propusa de Spania :[/b] [/color][/size]

[quote="Virgil Nicula"]${ a\ge 1\ \wedge\ x\ge 0\ \implies\ \frac {1}{\sqrt a}\le \sqrt {\frac {\sin^{2}x}{a + \cos^{2}x}} + \sqrt {\frac {\cos^{2}x}{a + \sin^{2}x}}\leq \frac {2}{\sqrt {2a + 1}}\ }\ \ (*)$ . [/quote]

[color=darkblue][b][u]Demonstratie.[/u][/b] Se observa ca $ \sqrt {\frac {\sin^{2}x}{a + \cos^{2}x}} + \sqrt {\frac {\cos^{2}x}{a + \sin^{2}x}} =$ $ \sqrt {\frac {\sin^{2}x}{a\left(\sin^2x + \cos^2x\right) + \cos^{2}x}} + \sqrt {\frac {\cos^{2}x}{a\left(\sin^2x + \cos^2x\right) + \sin^{2}x}} =$

$ \sqrt {\frac {\sin^{2}x}{a\sin^2x + (a + 1)\cos^2x}} + \sqrt {\frac {\cos^{2}x}{(a + 1)\sin^2x + a\cos^2x}}$ . Notam $ \tan^2x = t\ge 0$ . Inegalitatea $ (*)$ devine :

$ \frac {1}{\sqrt a}\le\sqrt {\frac {t}{at + (a + 1)}} + \sqrt {\frac {1}{(a + 1)t + a}}\le\frac {2}{\sqrt {2a + 1}}$ pentru orice $ t\ge 0$ , adica $ f(0)\le f(t)\le f(1)$ pentru orice $ t\ge 0$ ,

unde $ \boxed {\ f(t) = \sqrt {\frac {t}{at + (a + 1)}} + \sqrt {\frac {1}{(a + 1)t + a}}\ }$ . Se observa ca $ f\left(\frac 1t\right) = f(t)$ pentru orice $ t > 0$ . Deci este suficient sa aratam ca

$ \boxed {f(0)\le f(t)\le f(1)\ ,\ (\forall)\ t\in[0,1]}$ . Se arata usor ca $ f'(t)\mathrm {\ .a.s.\ }[(a + 1)t + a]^3 - t[at + (a + 1)]^3$ , adica

$ f'(t)\mathrm {\ .a.s.\ }\left(1 - t^2\right)\left[a^3t^2 + (a + 1)(2a^2 - 2a - 1)t + a^3\right]$ . Insa cel de-al doilea factor de gradul doi are discriminantul $ \Delta \le 0$ deoarece

$ \Delta = - (3a + 1)(a - 1)(2a + 1)^2\le 0$ . Prin urmare, $ f'(t)\ge 0$ pentru orice $ t\in [0,1]$ , adica functia $ f\ : \ [0,1]\ \rightarrow\mathbb R$ este strict crescatoare.

In concluzie, $ \boxed {f(0)\le f(t)\le f(1)\ ,\ (\forall)\ t\in[0,1]}$ . Am folosit notatia $ X\mathrm {\ .a.s.\ }Y\ \Longleftrightarrow\ X = Y = 0\ \vee\ XY > 0$ ("[b]a[/b]celasi [b]s[/b]emn").[/color]

=============================================================================================================

[quote][color=darkred][b][u]Problema 5.[/u][/b] Fie patrulaterului inscriptibil  $ ABCD$ . Notam $ E\in AC\cap BD$ , mijloacele $ K$ si $ M$ ale laturilor $ AB$ , $ CD$ respectiv

si proiectiile $ L$ , $ M$ ale punctului $ E$ pe laturile $ BC$ , $ AD$ respectiv. Sa se arate ca $ KM\perp LN$ ([b]Republica Moldova[/b]).[/color][/quote]

[color=darkblue][b][u]Demonstratie.[/u][/b] Aplicam de doua ori cazul particular al lemei de mai jos astfel :

triunghiului $ ABE$ in afara caruia avem  triunghiurile dreptunghice asemenea $ BEL$ , $ AEN$ si obtinem $ KL = KN$ ;

triunghiului $ CDE$ in afara caruia avem triunghiurile dreptunghice asemenea $ CEL$ , $ DEN$ si obtinem $ ML = MN$ .

In concluzie, punctele $ K$ , $ M$ apartin mediatoarei segmentului $ [LN]$ , adica $ KM\perp LN$ .

[b]Vezi[/b][/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=213068][color=red][b]aici[/b][/color][/url] [color=darkblue]si alte demonstratii ale acestei probleme[/color].

[quote] [color=darkred][b][u]Lema.[/u][/b] Pe laturile $ \triangle ABC$ construim in afara acestuia $ \triangle ABF$ , $ \triangle ACE$ , $ \triangle BCD$ astfel incat  $ DB = DC$ , $ m(\widehat {BCD}) = m(\widehat {CBD}) = z$ ,

${ m(\widehat {BAF}) = m(\widehat CAE}) = x$ , $ m(\widehat {ABF}) = m(\widehat {ACE}) = y$ si $ x + y + z = 90^{\circ}$ . Sa se arate ca  $ DE = DF$ and $ m(\widehat {EDF}) = 2y$ .

[b]Caz particular.[/b] Pentru $ z = 0$ ( $ x + y = 90^{\circ}$ ). $ \left\|\begin{array}{c} D\in (BC)\ \ ,\ \ DB = DC \\
 \\
FA\perp FB\ \ ,\ \ EA\perp EC \\
 \\
\widehat {BAF}\equiv \widehat {CAE}\end{array}\right\|\ \implies\ \left\|\begin{array}{c} DE = DF \\
 \\
m(\widehat {EDF}) = 2y\end{array}\right\|$ .[/color]

 [color=darkblue][b]Vezi[/b][/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61873][color=red][b]aici[/b][/color][/url] [color=darkblue][b]demonstratia lemei.[/b][/color] [/quote]

Avem:
$ cos \alpha \equal{} \frac {\overrightarrow{AD}.\overrightarrow{BC}}{|AD|.|BC|} \equal{} \frac {1}{|AD|.|BC|}.\overrightarrow{AD}.(\overrightarrow{AC} \minus{} \overrightarrow{AB}) \equal{} \\
\frac {1}{|AD|.|BC|}.(\overrightarrow{AD}.\overrightarrow{AC} \minus{} \overrightarrow{AD}.\overrightarrow{AB}) \equal{} \frac {|AC|^2 \plus{} |BD|^2 \minus{} |AB|^2 \minus{} |CD|^2}{2.|AD|.|BC|}.$
Pe de alta parte,
$ |EF|^2 \equal{} \frac {|AC|^2 \plus{} |AD|^2 \plus{} |BC|^2 \plus{} |BD|^2 \minus{} |AB|^2 \minus{} |CD|^2}{4}\Rightarrow \\
\Rightarrow|AC|^2 \plus{} |BD|^2 \minus{} |AB|^2 \minus{} |CD|^2 \equal{} 4.|EF|^2 \minus{} (|AD|^2 \plus{} |BC|^2),\dots$

[b]Coincide cu Problema 2037 din Kvant nr.2/2007; vezi:[/b]
http://www.mathlinks.ro/viewtopic.php?t=184253

[color=darkred][b][u]Lema (identitatea lui Euler).[/u][/b] Fie mijloacele $ E$, $ F$ ale muchiilor $ [AB]$ , $ [CD]$ in tetraedrul $ ABCD$ .

Atunci exista relatia $ \boxed {\ AB^2 + CD^2 + 4\cdot EF^2 = (AC^2 + AD^2) + (BC^2 + BD^2)\ }$ .[/color]

[color=darkblue][u][b]Demonstratie.[/u][/b] Aplicam [b]teorema medianei[/b] in triunghiurile $ ACD$ , $ BCD$ si $ ABF$ :

$ \left\|\begin{array}{ccc} 4\cdot AF^2 + CD^2 = 2\cdot\left(AC^2 + AD^2\right) & \\
 \\
4\cdot BF^2 + CD^2 = 2\cdot\left(BC^2 + BD^2\right) & \\
 \\
4\cdot EF^2 + AB^2 = 2\cdot \left(AF^2 + BF^2\right) & \bigodot 2\end{array}\right\|\bigoplus$ $ \implies$ $ \boxed {\ AB^2 + CD^2 + 4\cdot EF^2 = (AC^2 + AD^2) + (BC^2 + BD^2)\ }$ .

[b][u]Cazuri  particulare in plan.[/u][/b]

$ \blacksquare$ In patrulaterul convex $ ABCD$ notam mijloacele $ M$ , $ N$ ale diagonalelor $ [AC]$ , $ [BD]$ respectiv

si mijloacele $ X$ , $ Y$ , $ Z$ , $ T$ ale laturilor $ [AB]$ , $ [BC]$ , $ [CD]$ , $ [DA]$ respectiv. Atunci exista [b]relatiile lui Euler[/b] :

$ \blacktriangleright\ AC^2 + BD^2 + 4\cdot MN^2 = AB^2 + BC^2 + CD^2 + DA^2$ .

$ \blacktriangleright\ \left\|\begin{array}{c} AB^2 + CD^2 + 4\cdot XZ^2 = (BC^2 + AD^2) + (AC^2 + BD^2) \\
 \\
AD^2 + BC^2 + 4\cdot YT^2 = (AB^2 + CD^2) + (AC^2 + BD^2)\end{array}\right\|$ .

$ \blacksquare$ In trapezul $ ABCD\ ,\ AB\parallel CD$ exista [b]relatia lui Euler[/b] $ AC^2 + BD^2 = AD^2 + BC^2 + 2\cdot AB\cdot CD$ .

$ \blacksquare$ In paralelogramul $ ABCD$ exista [b]relatia lui Euler[/b] $ AC^2 + BD^2 = 2\cdot\left(AB^2 + AD^2\right)$ .

$ \blacksquare$ Notam mijlocul $ D$ al laturii $ [AC]$ din triunghiul $ ABC$ . Daca aplicam [b]relatia lui Euler[/b]

patrulaterului degenerat $ ABCD$ obtinem teorema medianei : $ 4\cdot m_b^2 + b^2 = 2\cdot\left(a^2 + c^2\right)$ .[/color]

[color=darkred][b]Problema 4 .[/b] Se considera tetraedrul arbitrar $ ABCD$ . Fie $ E$ si $ F$ mijloacele muchiilor $ AB$ , $ CD$ respectiv. Daca $ \phi$ este masura

unghiului format de muchiile $ AD$ si $ BC$ , sa se calculeze $ \cos \phi$ in functie de lungimile segmentelor $ [EF],[AD]$ si $ [BC]$ ([b]Romania[/b]).[/color]

[color=darkblue][b][u]Demonstratie[/u] (pentru ... juniori !).[/b] Notam proiectiile $ X$ , $ Y$ pe dreapta $ BC$ ale varfurilor $ A$ , $ D$ .[/color]

$ \left\{\begin{array}{ccccc} \blacktriangleright\ \left\|\begin{array}{c} BX^2 - XC^2 = AB^2 - AC^2 \\
 \\
\overline {BX} + \overline {XC} = \overline {BC}\end{array}\right\| & \implies & \left\|\begin{array}{c} \overline {BX} + \overline {XC} = \overline {BC} \\
 \\
\overline {BX} - \overline {XC} = \frac {AB^2 - AC^2}{\overline {BC}}\end{array}\right\| & \implies & \overline {BX} = \frac {BC^2 + AB^2 - AC^2}{2\cdot\overline {BC}} \\
 \\
\blacktriangleright\ \left\|\begin{array}{c} BY^2 - YC^2 = DB^2 - DC^2 \\
 \\
\overline {BY} + \overline {YC} = \overline {BC}\end{array}\right\| & \implies & \left\|\begin{array}{c} \overline {BY} + \overline {YC} = \overline {BC} \\
 \\
\overline {BY} - \overline {YC} = \frac {DB^2 - DC^2}{\overline {BC}}\end{array}\right\| & \implies & \overline {BY} = \frac {BC^2 + DB^2 - DC^2}{2\cdot\overline {BC}}\end{array}\right\|$ $ \implies$

$ \left\{\begin{array}{c} \overline {XY} = \overline {BY} - \overline {BX} = \frac {DB^2 - DC^2 - AB^2 + AC^2}{2\cdot\overline {BC}} \\
 \\
AD^2 + BC^2 + BD^2 + AC^2 = AB^2 + CD^2 + 4\cdot EF^2 \\
 \\
\mathrm {( identitatea\ lui\ EULER )}\end{array}\right\|$ $ \implies$ ${ \left\|\begin{array}{c} XY = \frac {\left|AD^2 + BC^2 - 4\cdot EF^2\right|}{2\cdot BC} \\
 \\{
(\ XY = AD\cdot |\cos\phi |\ }\end{array}\right\|$ $ \implies$

$ \boxed {\ |\cos\phi | = \frac {\left|AD^2 + BC^2 - 4\cdot EF^2\right|}{2\cdot BC\cdot AD}\ }$ . [color=darkblue]Banuiesc ca se cunoaste la nivelul clasei a VIII - a [b]identitatea lui Euler[/b]

intr-un tetraedru (vezi lema de mai sus !) cu cele doua cazuri particulare intr-un patrulater convex ![/color]


pigeonhole principle
number theory
relatively prime

For which primes p ≠ 2 or 5 is there a multiple of p whose digits are all 9? For example, 999999 = 13·76923.

So Fermat trivializes a Spanish MO problem?

[quote="PenguinIntegral"]Spanish MO problem[/quote]

 :rotfl:

So it would appear!

(Of course, you can also do a similar kind of reasoning with Pigeonhole; in particular, showing that the values of $10^{n}\bmod p$ must repeat, so some $10^{n}-10^{m}\equiv 0 \bmod p$.)

It's a weaker claim; we don't make any statements about the length of the period, only that the sequence must repeat some values at some point.

Edit:  What I mean is, it is not actually necessary to prove that the sequence $r_{n}= 10^{n}\bmod p$ is periodic (although it is pretty obvious from the argument that [i]is[/i] necessary), but because it only has a finite range there must exist $i \neq j$ such that $r_{i}= r_{j}$. )

:rotfl:

This entire problem is just uncool.

And I don't mean uncool like getting your wallet stolen. I mean uncool like walking into your dorm room to find your roomate playing with hairspray and matches and "accidently" destroying your entire collection of mathbooks.

When i came across this problem first, i didnt know FLT. Ig to an interesting solution, which is a small bit generalised.

Say $a$ is relatively prime to $10$ then $\frac1a$ is a repeating decimal. Cut this decimal off after the first period, and mutiply by $a$ you get a string of $9$'s after a zero, and you can just multiply by the appropriate power of 10.

EG choose the mumber $21$ then $\frac{1}{21}=0.04761904761904761904761904761905.....$ so we get that

$(21)(47619)=999999$


When are the statistics for the AIME supposed come out?  I imagine it's sometime after the AIME II.

I think March 28th? Though it'll make more sense if they gave the stats to AIME I seperate.

It's too bothersome to find out who got what in my state by looking at state statistics on the website... but I am curious about this year's AIME... which I think will come soon. But I'm still waiting for OSU to put this year's results... with names on their website...
(I think this might have been a little off-topic)

Just a quick question: Does they post statistics for AIME by state, because I want to see how much I need to get on the AIME to qualify by having the highest USAMO index in my state.  I tried searching Ohio AIME statistics on google, but I couldn't find anything.

I don't think they list AIME stats by state, but they do list USAMO qualifiers by state - if there's typically exactly one from Ohio, at least you'd know that you could probably qualify by highest-index-in-state rather than top-160-indices.

AMC doesn't post state statistics for AIME, but sometimes, a major college will post the results (with names). In Oklahoma, OSU, starting from 2002, has posted AMC and AIME results (indiv. and team) for the state.

The problem is: I'm not certain most other states have that  :?
Well, good luck in your searching!

It's unlikely that Ohio will have only one qualifier, though.  Not only is it a big state, but a review of the past few years have shown that they usually have at least three qualifiers (last year, there were 5; the year before, 3, before that, 6).  Three of last year's five did not graduate yet, so they will be back.

This may have been asked before, but searching it's pretty tough. Do the AMC people count the state you live in or the state your school's in? It doesn't apply to me, but I'm curious.

[quote="Chinaboy"]AMC doesn't post state statistics for AIME, but sometimes, a major college will post the results (with names). In Oklahoma, OSU, starting from 2002, has posted AMC and AIME results (indiv. and team) for the state.

The problem is: I'm not certain most other states have that  :?
[/quote]

KS does a great job in posting the statistics.  You can find out school winners, AIME qualifiers, honor, merit and achievement roll status - the whole nine yards.

http://www.math.ksu.edu/main/events/hscomp/amc.htm

[quote="yif man12"][quote="filletwho"]Your school should be in the state you live in, shouln't it?[/quote]

Not all the time.[/quote]

I think it would be state you live in, because you can go to a weaker state.
I wish school... then I can go to a school in Wyoming to take the test.  :)

Wow. That site does have a lot of stuff. Our site just gives the scores. You have to figure the rest (which isn't much, and it is better than just looking at AMC's statistics).

[url]http://www.math.okstate.edu/~berthol/[/url]

Yep, there are just a few states with only a single USAMO qualifier each year, and Utah is probably the one with the greatest population. In all other states, you have to assume that you will only get in "the hard way" by actually earning a qualifying index or getting the floor value AIME score in tenth grade or below.


search
inequalities proposed
inequalities

[quote="BlnGcc"]APMO 2005 (Hojoo Lee)[/quote]

It is not mine. If I am not wrong, it was proposed by USA.
(I am not 100% sure.)

[quote="mathboy_cntHDVN"]It is APMO 2005[/quote]
See at http://www.mathlinks.ro/Forum/viewtopic.php?search_id=241572567&t=30952


algebra
polynomial
AMC
USA(J)MO
USAMO
Olimpiada de matematicas

1. Sea $ABC$ un triángulo isósceles con $BC =AC$. Sea $P$ un punto de su circunferencia
circunscrita situado en el arco $AB$ que no contiene a $C$. Sea $D$ el pie de la perpendicular trazada desde $C$ a la recta $PB$. Demostrar que $PA+PB = 2PD$

(Problema cambiado)

2. Sean $a,b$ enteros positivos tales que $a|b^{2}, b^{2}|a^{3},a^{3}|b^{4}...$ y asi sucesivamente. Pruebe que $a=b$.

(Problema Cambiado)

3. Halle todos los polinomios $P(n)$ tales que existen complejos $z_{1},z_{2}...,z_{k}$ tales que $\sum_{i=1}^{k}z_{i}^{m}=0$ si y sólo si $m\neq P(n)$ para todo $n$.

4. Se tienen un número finito de segmentos en una linea de modo que entre cualesquier $m+1$ segmentos hay dos que se intersectan. Pruebe que se pueden seleccionar $m$ puntos en la linea de modo que cualesquier segmento contiene al menos uno de estos puntos.

Ola Pascual, esta vez si voa entrar al reto, ya me salieron dos problemas!!!
Donde pongo las soluciones???

Grax

[color=blue]¿Existe algún material útil para iniciarse con polinomios? En mi país hemos dado una breve noción acerca de cómo trabajar con ellos, mas resulta insuficiente.
Por supuesto no quiero que la información necesaria para trabajar con ellos esté servida, pero me agradaría contar con un buen material de consulta.[/color]

Las soluciones deben enviarlas a mi correo: pascual2k1@hotmail.com

Ahora estoy corrigiendo las soluciones de la cuarta edición, entonces tenganme paciencia.

Respecto a lo de polinomios, puedes buscar en internet hay muchas elcturas, inclusive aqui en el foro colgue un pdf de polinomios. Sino un buen libro puede ser polynomials de Barbeau.

Suerte

[quote="Pascual2005"]1. Sea $ABC$ un triángulo isósceles con $BC =AC$. Sea $P$ un punto de su circunferencia
circunscrita situado en el arco $AB$ que no contiene a $C$. Sea $D$ el pie de la perpendicular trazada desde $C$ a la recta $PB$. Demostrar que $PA+PB = 2PD$

[/quote]

Por fin encontre donde postean el reto de la semana!  :D ... lastima que el problema de geo sea tan facil  :(

Por Ptolomeo $AC.PB+CB.AP=AB.CP$ con lo que $AC(AP+PB)=AB.CP$. Es obvio que $AB=2AC.cos\angle CAB$ y $PD=APcos\angle CPB$. Desde que $m\angle CAB=m\angle CPB$ la conclusion es inmediata.

Aki mi solucion:
Supongamos que $a>b$, entonces como son enteros positivos tenemos que $a^{n}>b^{n}$ de aqui que la division de$a^{n-1}/b^{n}$ es un entero menor que $a$. Pero ahora, dado que $a$ es un entero positivo, entonces entre $1$ y $a$ existen $a-1$ diferentes enteros, los cuales es claro que son finitos.
Ahora, debido a que es fnito, en algun momento, alguna de esas divisiones va a tener que ser igual a otra, supongamos que si $1\leq k<a$ y que $a^{n-1}/b^{n}= a^{m-1}/b^{m}= k$ de ai deducimos que como $a$ es entero positivo, entonces $a=b$ $CONTRADICCION$
Ahora, si empezamos suponiendo que $b>a$ usamos un argumento similar, solo que ahora usaremos las divisiones $b^{n-1}/a^{n}$ y de ahi llegamos nuevamente a que $a=b$ otra $contradiccion$.
De manera que la unica forma es que a=b.

NO CHICOS! no leyeron las reglas? los problemas no los deben discutir, en vista de que ya dieron soluciones a estos, los cambiaré! la idea es que me envien sus soluciones y dentro de una semana ya los pueden discutir,e sto es para ponerlo cierto toque competitivo al asunto.

1. Pruebe que la ecuación $2x^{2}-3x = 3y^{2}$ tiene infinitas soluciones enteras.

2. Sea $ABC$ un triángulo para el cual existe un punto $F$ tal que $\angle AFB=\angle BFC=\angle CFA$.  Las líneas $BF$ y $CF$ intersectan a los lados
$AC$ y $AB$ en los puntos $D$ y $E$, respectivamente.  Demostrar que \[AB+AC\geq 4DE.\]

BUENO YA QUE NO RESPETARON LAS REGLAS, Y PUBLICARON SUS SOLUCIONES, Y ESOS PROBLEMAS NO VALEN PUBLICARE MIS SOLUCIONES DEL 1 YA Q EL PROBLEMA 2 LO ISE = QUE CONEJITA  :D
PROBLEMA 1..
PROLONGAMOS $PB$ HASTA $E$ TALQUE $BD=DE$, ENTONCES TENEMOS $AC=CB=CE$, SOLO FALTARIA PROBAR QUE $<AEP=<ABP$.
  UNIMOS $AE \Rightarrow <CAE=<CEA$, SEA $<AEP=b$, POR EL ISOCELES $(BCE), <CBE=<CEA+b$, AHORA POR EL INSCRIPTIBLE $ABCD$, $<CAP=<CBE$, CON LO CUAL SE DEMUESTRA QUE $<EAP=<AEP$, POR LO TANTO ISOCELES,$AP+PB=2PD$

[color=blue]Pascual2005, ¿podrías facilitarme la dirección del archivo .pdf de poliinomios que colgaste?

Muchas gracias.[/color]

Pascual, ya vas a poner las soluciones de los problemas del reto??
Me interesa la del punto de Fermat....
GRAX :lol:

ya pueden comentar los problemas, como es usual si nadie resuelve alguno tratare de poner mi solución. Animo y espero 4 bellas soluciones!

Aun no cuelguen las soluciones, si me sale o no el 2, el domingo envio mis soluciones.

En el segundo la idea es que $P$ es un polinomio con coeficientes enteros y $m,n$ son enteros. Chen, puedes colgar tus soluciones por favor ya que segun el mail que recibi las tienes todas?

1.- Una solucion es $x_{1}=-3, y_{1}=3$ Construímos las sucesiones ${x_{n}}$ y ${y_{n}}$
    Así: $x_{k+1}=5x_{k}-6y_{k}-3$ $y_{k+1}=3-4x_{k}+5y_{k}$ por induccion x_n es siempre negativo y y_n es siempre positivo, entonces $x_{k+1}<x_{k}$ y $y_{k+1}>y_{k}$ Por induccion tbn podemos demostar que (x_n, y_n) es una solucion del sistema, luego tenemos infintas soluciones distintas.

2.-Hay que considerar que F esta dentro de ABC, sino se puede encontrar un contraejemplo.
Sean: s=FE, t=FD. del triangulo ABF el angulo BAF es menor que 60, entonces en el triangulo AFE AF>FE, es decir x>s, analogamente x>t.
$AB+AC\ge 2\sqrt{AB.AC}$. Luego basta probar que $AB.AC\ge 4{DE}^{2}$
Para el triangulo AFE, Fb es bisectriz exterior entonces AB es como x, y BE es como s, luego AB/AE=x/(x-s). AE del teorema de cosenos es la raiz cuadrada de x^2-xs+s^2={x-s}^2+(x-s)s+s^2\ge 3{\frac{(x-s)+s}{2}}^2. Luego $AE\ge \frac{x\sqrt{3}}{2}$, entonces $AB\ge \frac{x^{2}\sqrt{3}}{2(x-s)}$.
Analogamente $AC\ge \frac{x^{2}\sqrt{3}}{2(x-t)}$.
Entonces $AB.AC\ge \frac{3x^{4}}{4(x-s)(x-t)}$. Ademas del teorema de cosenos, $DE^{2}=s^{2}+st+t^{2}$
Sea g(x)=$\frac{x^{4}}{(x-s)(x-t)}$

Si s=t entonces $(x-2s)^{2}\ge 0$, luego $x^{2}\ge 4s(x-s)$
Entonces $x^{4}\ge 16s^{2}(x-s)^{2}$ ademas DE^2=3s^2
entonces $AB.AC\ge 3/4 .g(x)\ge 4DE^{2}$.
Si s y t son distintos:
Sean: a=$\sqrt{s^{2}+st+t^{2}}$, 2b=s+t.
Si $x\le a+b$:
x>s,x>t, sumando: 2x>2b.
Luego $0<x-b\lea$, entonces $(x-b)^{2}\le a^{2}$.
Pero la sgte desigualdad, despues de dividir todo entre a^2-3b^2( que es ((s-t)^2)/4>0) es equivalente a la anterior.
$(x-s)(x-t)a^{2}\le (x-b)^{2}3b^{2}$
Analogamente al caso s=t, podemos demostrar que
$x^{4}\ge 16b^{2}(x-b)^{2}$, el termino de la derecha es mayor o igual que 16/3. $(x-s)(x-t)a^{2}$. Luego $AB.AC\ge 3/4 g(x)\ge 4a^{2}=4DE^{2}$
Si $x\ge a+b$:
g'(x)=(x^3)/2(x-s)(x-t))^2  .  (4x^2-12xb+32b^2-8a^2).
El termino de la izquierda es positivo pues x>a. y a>s,t,0
El termino de la derecha es (2x-3b)^2+23b^2-8a^2
Como $x\ge a+b$: $2x-3b\ge 2a-b$ a>s,a>t, entonces 2a>2b, luego 2a-b=a+(a-b) es positivo. luego $(2x-3b)^{2}\ge (2a-b)^{2}$.
Luego el termino de la derecha mencionado, es mayor o igual que 24b^2-4ab-4a^2=4(3b+a)(2b-a). (s+t)^2>a^2, entonces 2b>a.
Luego g'(x)>0, mientras $x\ge a+b$
entonces es creciente en el intervalo de extremos a+b e infinito positivo.
Luego $g(x)\ge g(a+b); x\ge a+b$
Notar que $g(a+b)=\frac{{a+b}^{4}}{3b^{2}}$
pero $(a+b)^{2}\ge 4ab$ entonces $(a+b)^{4}\ge 16a^{2}b^{2}$.
Luego $g(a+b)\ge 16/3 . a^{2}$.
$AB.AC\ge 3/4 . g(x)\ge 3/4 . g(a+b)\ge 4a^{2}=4DE^{2}$.

3.- una cosa mas  :P, para que no esté 0^0, los complejos son no nulos, m es natural, o solo diferente de cero?

4.- Considerando intervalos cerrados a los segmentos
Por induccion:
   para m=1. cada 2 segmentos se intersecan, entonces tomemos el segmento con extremo izquierdo, lo mayor posible., los demas segmentos lo intersecan en al menos el extremo izquierdo, sino no lo intersecarian. Ese es el punto, tal que cualquier segmento lo contiene.
Supong. cierta la afirmacion para m=n:
Entonces Si tenemos que entre cualesquiera n+2 segmentos hay 2 que se intersecan. Sea s el segmento con mayor extremo izquierdo, si los demas segmentos lo intersecan entonces contienen al extremo izquierdo, luego basta que este punto este entre los n+1 puntos que escojamos, si hay por lo menos un segmento que no lo interseca, Sea C el conjunto formado por todos los segmentos de este tipo, si en C tomamos n+1 segmentos, con s, tendriamos n+2 segmentos, entonces  hay 2 de ellos que se intersecan(q y r) ni q ni r pueden ser s, entonces q y r son 2 segmentos de C que se intersecan, de la hipotesis inductiva concluimos que en C hay n puntos tales que cualquier segmento de C con tiene a uno de estos n puntos, para completar esta induccion basta tomar dichos n puntos y el extremo izquierdo de s( pues los segmentos que intersedcan a s, lo hacen en ese extremo izquierdo).

[quote="Pascual2005"]1. Sea $ ABC$ un triángulo isósceles con $ BC \equal{} AC.$ Sea $ P$ un punto de su circunferencia circunscrita situado en el arco $ AB$ que no contiene a $ C.$ Sea $ D$ el pie de la perpendicular trazada desde $ C$ a la recta $ PB.$ Demostrar que $ PA \plus{} PB \equal{} 2PD.$[/quote]
Sea $ CA \equal{} CB \equal{} L.$  Aplicando teorema de Ptolomeo al cuadrilátero cíclico $ CAPB$ resulta

$ PA \cdot L \plus{} PB \cdot L \equal{} PC \cdot AB  \Longrightarrow \ PA \plus{} PB \equal{} \frac {PC \cdot AB}{L} \ (*)$

Por otro lado si $ H$ es el pie de la perpendicular de $ C$ a $ AB$ (punto medio de AB) se tiene que

$ \angle CAH \equal{} \angle CPD \Longrightarrow \ \triangle CPD \sim \triangle CAH  \Longrightarrow \ \frac {2PD}{AB} \equal{} \frac {PC}{L}$

Combinando esta ultima con $ (*)$ se obtiene $ PA \plus{} PB \equal{} 2PD.$

[quote="Pascual2005"]2. Sean $a,b$ enteros positivos tales que $a|b^{2}, b^{2}|a^{3},a^{3}|b^{4}...$ y asi sucesivamente. Pruebe que $a=b$.
[/quote]

Sea $c(n)$ la cantidad de divisores primos de $n$. Primero, es sencillo ver que $c(n^k)=c(n)$ para todo $k\in \mathbb {N}$. Como cada divisor primo de $a$ divide a $b^2$, entonces $b^2$ contiene todos los divisores primos de $a$. Esto implica que $c(a)\leq c(b^2)=c(b)$. Con el mismo argumento, obtenemos que $c(a)=c(a^3)\ge c(b^2)=c(b)$ y por ende $c(a)=c(b)$. Es más, $a,b$ tienen los mismos divisores primos.

Escribamos $a=\prod_{k=1}^{r} p_k^{\alpha_k}$ y $b=\prod_{k=1}^r p_k^{\beta_k}$ donde $p_i<p_j$ siempre que $i<j$ y cada $\alpha_k, \beta_k$ es un natural. Para cada $n$ natural, tenemos que $a^{2n-1}|b^{2n}$, esto implica que $(1-\frac{1}{2n})\alpha_k\leq \beta_k$. Si $n\to \infty$, tenemos que $1-\frac{1}{2n}\to 1$ y por ende $\alpha_k\leq \beta_k$ para todo $1 \leq k\leq r$. Por otra parte, como $b^{2n}|a^{2n+1}$, tenemos que $(1+\frac{1}{2n})\alpha_k\ge \beta_k$. Esta vez, si $n\to \infty$, tenemos que $1+\frac{1}{2n}\to 1$ y esto significa que $a_k\ge \beta_k$. Por lo tanto $\alpha_k=\beta_k$ para cada $k\in \{1,2,...,r\}$, de donde se deduce que $a=b$.


The following links contain the math chats from previous 5 weeks.

http://dhost.info/jwec05/Silverfalcon/t1.PDF
http://dhost.info/jwec05/Silverfalcon/t2.PDF
http://dhost.info/jwec05/Silverfalcon/t3.PDF
http://dhost.info/jwec05/Silverfalcon/t4.PDF
http://dhost.info/jwec05/Silverfalcon/t5.PDF

thx for the collection.  Hoping to find them.  Now i can go through the problems.  God news i got the AoPS book series and its great.  Hope to learn a lot from it :D


hello, do you meant $ 2x\plus{}y\plus{}5\equal{}0$ and $ 6x\plus{}2y\equal{}\minus{}5$?
Sonnhard.

Oh im sorry, i meant 2x=y+5 and the other equation was correct.

[quote="uanaka"]At which point do the graph of 2x+y+5 and 6x+2y=-5 intersect?

I WILL STUMP YOU!!![/quote]
\[ (1): 2x \plus{} y \equal{} 5
\]

\[ (2): 6x \plus{} 2y \equal{} \minus{} 5
\]

\[ (1): \minus{} 3(2x \plus{} y) \equal{} \minus{} 3(5)
\]

\[ (1): \minus{} 6x \minus{} 3y \equal{} \minus{} 15
\]

\[ (1) \plus{} (2): \minus{} y \equal{} \minus{} 20
\]

\[ y \equal{} 20
\]

\[ (1): 2x \plus{} 20 \equal{} 5
\]

\[ (1): x \equal{} \minus{} 7.5
\]
Thus the point of intersection is $ ( \minus{} 7.5, 20)$.

EDIT: oops, used wrong equation for (1). I thought it was $ 2x \plus{} y \equal{} 5$.

The intersection must have the same $ x$ and $ y$ values.  So this is like a system of equations:

$ 2x \equal{} y \plus{} 5$

$ 6x \plus{} 2y \equal{} \minus{}5$

Firstly, check if there are infinite or no solutions.  There is only one solution.

Subtract $ 2y$ from the second equation.

$ 2x \equal{} y \plus{} 5$

$ 6x \equal{} \minus{}2y \minus{} 5$

Multiply the first equation by $ 3$.

$ 6x \equal{} 3y \plus{} 15$

$ 6x \equal{} \minus{}2y \minus{} 5$

By the transitive property:

$ 3y \plus{} 15 \equal{} \minus{}2y \minus{} 5$

$ 5y \plus{} 15 \equal{} \minus{}5$

$ 5y \equal{} \minus{}20$

$ y \equal{} \minus{}4$

Plugging back into the first equation, we get:

$ 2x \equal{} \minus{}4 \plus{} 5$

$ 2x \equal{} 1$

$ x \equal{} \frac12$

So our answer is $ \boxed{(\frac12,\minus{}4)}$.

hello,multiplying the first equation by $ 3$ we get $ 6x\equal{}3y\plus{}15$, substituting this in the second one we get $ 3y\plus{}15\plus{}2y\equal{}\minus{}5$, from here we get $ y\equal{}\minus{}4$,and with this we get $ 2x\equal{}\minus{}4\plus{}5 \Leftrightarrow x\equal{}\frac{1}{2}$, so our solution is
$ L\equal{}\{(x,y)|x\equal{}\frac{1}{2},y\equal{}\minus{}4\}$
Sonnhard.

This forum is not for stumping people.  If it's your homework, you can openly say so...


inequalities
linear algebra
linear algebra unsolved

Let $ A,B \in M_n(C)$ such that$ A\plus{}B\equal{}AB$ .Prove that $ rank(A)\equal{}rank(B)$

$ (A\minus{}I)(B\minus{}I) \equal{} I$ meaning that both of them are invertible, but then $ B \equal{} A(B\minus{}I)$, so they have the same rank.

[quote="Soarer"]$ (A \minus{} I)(B \minus{} I) \equal{} I$ meaning that both of them are invertible, but then $ B \equal{} A(B \minus{} I)$, so they have the same rank.[/quote]
sry , but i don't think you're wright.. why does $ B \equal{} A(B \minus{} I)$ involve that A,B are invertible?
chose A=0 and B=0..they're not invertible :wink:

I mean that A-I, B-I are both invertible.

i thought you said A,B invertible.my mistake :blush:
B-I invertible involves A,B have the same rank .here is the solution $ \Rightarrow rank(B\minus{}I)\equal{}n$
from sylvester inequality we have
 $ rank(A)\plus{}rank(B\minus{}I)\minus{}n\leq rank(A(B\minus{}I))\equal{}rank(B) \leq min \{ rank(A),rank(B\minus{}I) \}$
$ \Rightarrow  rank(A)\leq rank(B)\leq rank(A)$ $ \Rightarrow rank(A)\equal{}rank(B)$


Demonstrati ca in orice triunghi avem:
R/r>b/a + a/b, daca triunghiul nu este echilateral.

$a=2RsinA,b=2RsinB$ deci inecuatia devine: $RsinAsinB>r$.
Folosind formulele de calcul a ariei triunghiului gasim:
$sinA=\frac{2pr}{bc}$ si $sinB=\frac{2pr}{ac}$ deci inecuatia devine: $\frac{4p^{2}r^{2}R}{abc^{2}}>r$ echivalent cu:
$\frac{4p^{2}rR}{abc^{2}}>1$ dar $prR=\frac{abc}{4}$ deci gasim
$\frac{p}{c}>1$ ceea ce evident adevarat in triunghi neechilateral:)


geometry
MIT
college
HCSSiM
CMU
Carnegie Mellon

Hey, is anyone going to be going to their respective Governor's Schools this summer?

If you have attended in the past (especially in PA), what did you think?  How hard are the courses, especially if you've had exposure to the subject areas (Honors/AP classes) at school already?

I'm 99% sure that I will be accepting my invitation to PA Gov School for the Sciences at Carnegie Mellon since I didn't get into RSI.

I was just wondering if a) anyone else will be going to a Gov School and b) what other people's experiences have been like.

i will be going to mine (South Carolina) :)

i have a couple of friends (who have many friends at MIT throught PA's Gov School) who really liked the program (one actually liked it more than HCSSiM and felt he learned more there and is going as a counslor this summer)

enjoy!

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81784]www.artofproblemsolving.com/Forum/viewtopic.php?t=81784[/url]
Haven't been there before, but i've heard it's just oodles of fun.  None of the ones I spoke to went to the Sciences school though.  It can't be much different i suppose.

EDIT: I don't know about other states, but i don't think TN has a great sciences offering.  I changed my mind about enrolling, and plan instead on working on a research project. hopefully I can get one done for siemens or intel by the deadline.

Iversonfan, have fun at yours!  I know a couple people who went to PA Gov School for the Arts and loved it (and are all planning on majoring in music).

Mariag, I'm glad to hear that you know some people who loved it as well.  I'm looking forward to going, if nothing else, I'll get to meet lots of nerds  :D

I dont know if this is too late, but I've had some friends attend Gov Schools in PA (especially for sciences) and they seemed to enjoy the experience.  They are all attending top colleges now so I suppose it is pretty good for college applications as well (if that's a concern).


integration
trigonometry
calculus
calculus computations

I would be grateful for some help with this question.

Bank A offers to pay continuous interest on your savings with the time dependent interest rate $ r(t)\equal{} 0.001(1\plus{}sin{\pi t})$. Bank B with time dependent rate $ r(t)\equal{}0.07 \exp^{0.1t}$ and C with constant rate $ r(t)\equal{}0.68$ where t denotes time in units of years. you want to fix your saving for one year. which offer should you choose? for the best offer, what do you have in your account after 1 year if you intially deposited £1000?

In general, for \$1, the amount in the bank at the end of the year is $ e^{\int_0^1r(t)\,dt}$

So for
Bank A: $ 1000\text{exp}[\displaystyle\int_0^1 0.001(1 \plus{} \sin\pi t)\, dt]$
Bank B: $ 1000\text{exp}[\displaystyle\int_0^1 0.07e^{0.1t}\, dt]$
Bank C: $ 1000\text{exp}[\displaystyle\int_0^1 0.68\, dt]$

The maximum of these is your answer.


Evaluate the following sums:

$ S_1\equal{}\sum_{k\equal{}1}^n\left\lfloor{k\plus{}3\sqrt{k} \over k}\right\rfloor,$

$ S_2\equal{}\sum_{k\equal{}1}^{n^2}\left\lfloor\sqrt{k}\right\rfloor,$

$ S_3\equal{}\sum_{k\equal{}0}^n\left\lfloor{a\plus{}k \over n}\right\rfloor,$

in which $ n \in \mathbb N^*, a \in \mathbb R$.

For $ S_2$

[hide]
$ S_2 = \sum_{k = 1}^{n^2}\lfloor\sqrt {k}\rfloor,$

We have that for consecutive squares, $ n^2$ and $ (n-1)^2$ a difference of $ 2n-1$. This means the we can express this as

$ S_2 = \sum_{k = 1}^{n-1}(2n+1)n+n$
[/hide]