I want to hold some ARML Practices through AIM. Basically, it's just going over some ARML problems in an hour but since I'm not that strong enough to solve all ARML problems that I have now, I figure that solving with other people will help me.

The practice will be held via AIM in every Saturday 9:00 to 10:00 PM EST. If you're interested, please post here with your sn or contact me through PM or jwec05 on IM.

If you're way experienced, I discourage this. I'd like to have people around my level (who can do "okay" on ARML and like to have score high on ARML). Most problems will be Individual and Team but there may be couple relay problems.

Thanks!

Sounds cool.  If I'm around on AIM at that time, send me an invite.  I'll leave the easier problems for less experienced people to try instead of spoiling them.  Also, I'm sure some of my fellow SC ARML teammates would love to practice as well :D

i would like to do that as well...
although it may sound wierd but i don't have AIM, hopefullyi can download it somehow

also i agree with Repmax, sunday sounds better than Saturday!

Well, Sunday evening is the practice for Lehigh Valley (4-6). So, I don't really want to do "too" much ARML stuff on Sunday since I have to go to school next day.

BTW, Rep, I don't remember your sn though. You need to tell me by IMing or something because I don't think I have it anymore.

Well, I've already practiced with Silverfalcon a few times, so I'm in. :)

I'm in, even though I'm not going to ARML. Practice can never hurt.

If I'm on invite me: filletwho

I think Saturday is a good time. Sunday is when I do hw.

Like Saturday WAY BETTER than Sunday

sn: pkerichang

I'll be pretty busy these upcoming weekends, but this sounds really fun. Please invite me if I'm available. Either day works for me.

i would like to participate also, it sounds like fun!
aim:
altheman1

[quote="Elemennop"]Oh yes, I'm in.

I agree with Joml that Sunday'll be better.[/quote]

Max recommended Sunday :D  Personally, it doesn't really matter too much to me.

Amir, just go to aim.com and you should be able to easily download AIM.  Another option is to use AIM Express which doesn't require you do download anything.

Actually, you know what.  I'll be in charge of Sunday practices and silverfalcon will be in charge of Saturday practices.

Oh, my bad, I don't really look at the names that often so I guessed who said it

and my AIM's elemmnop

I'd like to do this!  If I'm on AIM at the time, please invite me.  My sn is [b]cheesybread92[/b].

s/n = xstatik08

Invite if I'm online Saturday.

[quote="joml88"]
Actually, you know what.  I'll be in charge of Sunday practices and silverfalcon will be in charge of Saturday practices.[/quote]

Great! I don't think I can participate one in Sunday so can you send me the transcript through IM?

g89unit@gmail.com

Even you can't participate in the ARML discussion, I'll post the transcript on my blog. It's in my signature.

Yeah, I'll probably also post the transcript somewhere else where everyone on the forums can view it.

If you post the transcript Joe, can you Latex it up nicely?  AIM is bad for math because of the lack of latex capabilities.

StellarSupremacy

IM me, y'know, whenever ... I'll probably be on and I may or may not answer depending on whether I'm actually AT the computer

I'll participate as well.

screenname: jen7math

Here is a list of people who posted that they wanted to participate, if you want to join, please kindly add your name to the list in the same format.

[i]Aops name[/i] - aim name

[i]Silverfalcon[/i] - jwec05
[i]joml88[/i] - BaseballJoe2002
[i]Rep123max[/i] - MaxHarris08
[i]amirhtlusa[/i] -
[i]4everwise[/i] - Myths Apprentice
[i]Pakman2012[/i] - Pakman2012
[i]Elemennop[/i] - elemmnop
[i]filletwho[/i] - filletwho
[i]pkerichang[/i] - pkerichang
[i]towersfreak2006[/i] - towersfreak2006
[i]Altheman[/i] - altheman1
[i]mysmartmouth[/i] - cheesybread92
[i]dark_sunburst[/i] - xstatik08
[i]MysticTerminator[/i] - StellarSupremacy
[i]Adunakhor[/i] - Adunakhor I
[i]jen7[/i] - jen7math

Sometimes I'm busy on the weekends, but I'd like to participate as well; invite me if I'm online?

[i]mathwhiz2004[/i] - tocolorthesky

Thanks!

[i]minsoens[/i] - aimslkg
thanks to all who sponsor these practices! :)

I'll join to see what it's like.

[i]ccy[/i] - theccy

Thanks!

hey silverfalcon, i wanna practice for ARML too! aim=raininyang

erm...i was there yesterday for a short period of time

SN: Shobhit2005

[quote="nat mc"]I'll do it.

AIM: natmcAoPS

So where are we doing it, what chatroom, or will you invite us all?[/quote]

It's on AIM usually at 9 PM ET on Saturdays (silverfalcon) and Sundays (me).  We'll invite you...but IM us if you're on at that time and we haven't invited you yet.

i need some practice, it's my first time with arml

sn: the3nabil

AIM: speedykrab

might not be on, on saturday

so it is tomorrow at 9:00 eastern? So 5:00 AoPS time?

Due to personal reason, I won't be able to host one tomorrow. Thanks.

Tonight?  Um, yeah...sure, I'll fill in for Silverfalcon tonight.  Hopefully my power/cable don't get knocked out (bad weather around here). So if I'm not on, that's probably the reason :P

So tonight at 9:00 PM EST I will host an ARML practice session over AIM.  I'll probably start inviting people a little bit before 9 into the room.  So if it's getting close to 9 and you haven't gotten a chat invite from me, send me a message.

I'll still host one tomorrow as well.

Guess I gotta go prepare some material now :D

Sorry...but I don't think I'll be able to do it tonight :(

Sorry, there wasn't any today. I was out of the town until now (it's nearly 1:00) and I didn't get to use computer early this morning (and I didn't know I was going to be this late).

I'm not sure how many I can do for sure. My grade isn't that great now  :blush: so I need to work on school more so let see.. I'll hopefully let you know.

i will do it if it is sunday.

SN bigspikeeee

[quote="Silverfalcon"]After thinking this for long time, I came to conclusion: I can't do this math meet through AIM every Saturday. First of all, time is a big issue. There are times that I'm not home and obviously, doing only 5-6 problems in an hour is not very effective. So, I won't be doing this anymore.

I don't know if Joe will do it. I know I'm not.

But don't be sad. I will start another activity for ARML. I'll make a new thread about this.[/quote]

I just randomly started one yesterday night....

Yeah, I think we'll still have practices on Sundays.  We'll just do where someone posts a problem.  Everyone works on that one.  Once we get an answer someone else posts a problem.  I like it this way because it doesn't put all the work of preparing problems on one person and you get a better diversity of problems usually.  So try and have a problem or two prepared for Sunday if you wish to participate (this is not mandatory though).


algorithm
induction
number theory
Diophantine equation
number theory theorems

If you want to show that some diophantine equation has no solutions, you might be able to show that [u]if[/u] there exists a solution, then a "smaller" solution must exist too. This way you can construct a sequence of "decreasing" solutions, and such a sequence cannot exist in the set of positive integers.

Do you have an article about infinite descent? or do this problem by use infinite descent, $a^2-abq+b^2+q=0$ then $q=5$. Please explain clearly. I don't understand Myth's solution to this problem at http://www.mathlinks.ro/Forum/viewtopic.php?p=146009#p146009 .

That's not really the traditional way of doing descent. In the descent method I was talking about, you keep "decreasing" solutions. Myth actually did some descent with nonincreasing solutions: the process does not necessarily terminate, but that only happens in from very specific cases (in which case you can find all solutions by performing the descent backwards).

Let me give a very simple example of traditional descent:

Solve in positive integers: $x^3 + 2y^3 + 4z^3 = 4xyz$.

Let $(x',y',z')$ be the smallest solution (to make this rigorous: let this be the solution for which $x' + y' + z'$ is minimal, for example). From this equation it is obvious that $x'$ is even. Write $x' = 2x$, then we get $y'^3 + 2z'^3 + 4x^3 = 4xy'z'$ and hence $(y',z',x)$ solves the equation too! Since $x + y' + z' < x' + y' + z'$, there are no solutions, by infinite descent.

Infinite descent is reverse form of induction, pioneered by Fermat who is consider father of moder number thoery. It is very useful technique in number theory as Myth and Arne have shown. Also, can also be used to prove Fermat's Last Theorem and results on the sum of squares representation for primes.
It is versatile as well, many theoremns out there use infinite descent.
To mathpk: as an exercise, try proving x^3+y^3=z^3 has no solutions in integers over positive domain.

Bomb


function
calculus
derivative
trigonometry
logarithms
limit
calculus computations

Help Please

Help with any of these problems would be greatly appreciated, i have a test tomarrow and these are the study problems

Suppose that f(x)=ln(2+cosx) on the interval (0, 2?)
calculate f'(x) and f''(x)
Find the intervals on which the function f is concave up

Use chain rule to verify that every function of the form y=asin(5t)+bcos(5t) is a solution to the differential equation y''=-25y. Then use this fact to find the solution which also satisfies the initial conditions: y(0)= and y'(0)=0

The differential equation y''= -ky-cy' is used to model a motion of a mass on a spring with damping, where k is the spring constant and c is the damping coefficient.
a) show that a function y= (e^-t)cos3t satisfies the differential equation y''= -10y-2y
b) sketch the graph of y(t)

find the limit lim x-->0 ((3e^2x)-3)/x by recognizing it as a derivative f'(a) of the appropriate function at a suitable value of a. Please specify the function f in your answer, numerical computation of the limit is not sufficient.

find the limit lim x--->infinity cos(x)/x can l'Hopital's rule be used in this problem? explain why or why not

Help with any of these problems would be greatly appreciated, i have a test tomarrow and these are the study problems

HELP PLEASE!!

[hide="Maybe its good to work on it first before you ask it"]
$f(x)= \ln(2+ \cos x)$
Use chain-rule, that is $u'(v(x)) = u'(v(x)) v'(x)$
Here $u(x)= \ln (x)$ and $v(x)=2 + \cos x$ so $f(x)=u(v(x))$.
$u'(x)=\frac{1}{x}$ and $v'(x)= - \sin x$.
So $u'(v(x))= \frac{1}{v(x)} = \frac{1}{2+ \cos x}$.
And $f'(x)=\frac{- \sin x}{2+ \cos x}$
For $f''(x) = \frac{-2 \cos x - \cos^2 x - \sin^2 x}{(2 + \cos x)^2} = \frac{-2 \cos x - 1}{(2 + \cos x)^2}$.
To determine where the function is concave, find the interval for which $f''(x)<0$ (striclty concave) or $f'(x) \leq 0$ (concave). that is when $\cos x \leq \frac{1}{2}$
$\frac{\pi}{3} \leq x < 2$.
----
$y= a \sin 5t + b \cos 5t$.   $y' = 5a \cos 5t - 5b \sin 5t$.
\begin{eqnarray*}y'' &=& -25 a \sin 5t - 25b \cos 5t \\ &=& -25 (a \sin 5t + b \cos 5t) \\ &=& -25 y' \end{eqnarray*}

$y(0)=0 \Longrightarrow a \sin 0 + b \cos 0 = 0 \Longrightarrow b=0$
$y'(0)=0 \Longrightarrow 5a \cos 0 - 5b \sin 0 = 0 \Longrightarrow a=0$

---
$y=e^{-t} \cos 3t$.
$y'=-e^{-t} \cos 3t -3 e^{-t} \sin 3t = -e^{-t}(\cos 3t + 3 \sin 3t)$
So
\begin{eqnarray*}y''&=&e^{-t}(\cos 3t + 3 \sin 3t)-e^{-t}(-3 \sin 3t + 9 \cos 3t)\\ &=& -8 e^{-t} \cos 3t + 6 e^{-t} \sin 3t \\ &=& -10e^{-t} \cos 3t + 2 e^{-t} \cos 3t + 6 e^{-t} \sin 3t \\ &=& -10e^{-t} \cos 3t - 2(- e^{-t} \cos 3t - 3 e^{-t} \sin 3t) \\ &=& -10y -2y' \end{eqnarray*}

---
$\lim_{x \rightarrow 0} \frac{3e^{2x} - 3}{x}$
Let $f(x)=3e^{2x}$ we have $f(0)=0$.And
$f'(0)= \lim_{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} = \lim_{x \rightarrow 0} \frac{e^{2x}-3}{x}$.
we have $f'(x) = 6 e^{2x}$ so that $f'(0)=\lim_{x \rightarrow 0} \frac{e^{2x}-3}{x}=6$.

---
No, because it is not in the form $\frac{0}{0}$ , $\frac{\infty}{\infty}$. The function $\cos x$ is at most $1$. it can reach the value $\infty$, even though $x \rightarrow \infty$.
[/hide]  :)


Let a_1, a_2, a_3,... be a sequence of positive numbers. If there exists a positive number M such that for n = 1,2,3,...,

$a^{2}_{1}+a^{2}_{2}+...+a^{2}_{n}< Ma^{2}_{n+1}$

then prove that there exist a positive number M' such that for every n = 1,2,3,...,

$a_{1}+a_{2}+...+a_{n}< M'a_{n+1}$

By Cauchy we have that
$Ma_{n+1}^{2}>a_{1}^{2}+...+a_{n}^{2}\geq \frac{(a_{1}+...+a_{n})^{2}}{n}$

If we take square roots we have $a_{1}+...+a_{n}<\sqrt{nM}a_{n+1}$

So choose $M'=\sqrt{nM}$ and we are done

Sorry but the problem is to find a constant $M'$, independent of $n$. :(
For this problem, no one in Hong Kong but davidlam was able to come up with a solution, so David please post it here. :)
If I remember correctly, his answer is $M'=\sqrt{M(M+1)}(\sqrt{M}+\sqrt{M+1})$


Sezona je drzavnih takmicenja. Tako da bi bilo zanimljivo da ovdje napisete sastave ekipa za IMO... Da vidimo ko ove godine jurisa na zlato u Spaniji  :wink:

Hrvatska ekipa (abecednim redom):

[b]Nina Kamčev[/b], 3.r., XV. gimnazija, Zagreb (prošle godine bronca pojedinačno na MEMO-u)
[b]Adrian Satja Kurdija[/b], 2.r., V. gimnazija, Zagreb
[b]Ines Marušić[/b], 4.r., V. gimnazija, Zagreb (prošle godine bronca pojedinačno na MEMO-u)
[b]Sara Muhvić[/b], 4.r. III. gimnazija, Osijek
[b]Melkior Ornik[/b], 4.r. XV. gimnazija, Zagreb (prošle godine srebro pojedinačno na MEMO-u) (WEEE!)
[b]Goran Žužić[/b], 3.r., V. gimnazija, Zagreb (prošle godine bronca pojedinačno na MEMO-u)

Kao što vidite, nema nikoga tko je prije bio na IMO-u, a i poprilično smo mlada ekipa. Uglavnom, kako bi Ines (a vidim i Adrian na jednom topicu) rekli, sigurno smo najljepša ekipa - 3 pripadnice ljepšeg spola (navodno ljepšeg- nikada to neću službeno priznati  :D ) .
U vezi rezultata, ima jedna lijepa poslovica iz doba stoika: "Ničemu se ne nadam, ničega se ne bojim - slobodan sam." BTW, na veliko veselje cijele hrvatske ekipe (a, nadam se, i ostalih) nosim udžbenik iz filozofije u Madrid (da vas razveselim s par izabranih citata Heraklita, Heideggera i drugih).  :D

Ostali?

[size=59]P.S. Još uvijek ne mogu vjerovati kako sam razvalio malu olimpijadu![/size]

Cestitke morniku i djuri  :lol:

Za informacije o nasoj ekipi cete se nacekati, ako cemo prema tradiciji... A Srbija bira ekipu ovaj vikend, ako se ne varam. Za ostale ne znam...

[size=59]P. S. Gdje mogu pogledati zadatke sa Male olimpijade?[/size]

Hvala na čestitkama. Pogledaj PM.

Čim netko sazna nešto iz ex-Yu zemalja, nek javi, makar znam da je ovo dosta rano. Čudim se što Srbija ima državno tako rano... naše je tradicionalno u 5. mjesecu, ali je ove godine (koliko znam) ministarstvo isforsiralo travanj.

Uvijek su hrvatska i bh Olimpijada bile posljednje. Jer ne saljemo ekipe na Balkanijadu, pa se ne zurimo.

P. S. (pogledao, hvala jos jednom)

A jel mozemo i mi da dobijemo te zadatke sa Male olimpijade i Drzavnog takmicenja?

Srpska olimpijada odrzava se danas i sutra i sve informacije moci cete da nadjete na http://srb.imomath.com ili http://www.dms.org.yu

Može s Male. S državnog (bar za 4.r.) ih se ne mogu svih sjetiti, ali probat ću bar neke.

Dakle, [b]Mala olimpijada[/b]:

1. Odredi minimalnu vrijednost od:
a) $ \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz}$
b) $ \frac {x^2 \plus{} y^2 \plus{} 2z^2}{xy \plus{} yz}$
x, y, z su pozitivni realni brojevi.

2. Za koje prirodne brojeve n postoje racionalni (koji nisu prirodni) a i b takvi da je i $ a \plus{} b$ i $ a^n \plus{} b^n$ prirodan broj?

3. Na stranici BC trokuta ABC odabrana je točka M tako da težište $ T_c$ trokuta ABM leži na kružnici opisanoj trokutu ACM i analogno težište $ T_b$ od ACM na kružnici opisanoj trokutu ABM. Dokaži da su duljine težišnica od ABM i ACM povučenih iz vrha M jednake.

4. Neka je S skup svih neparnih prirodnih brojeva koji nisu višekratnici od 5 i koji su manji od 30m (m je neki zadani prirodni broj). Koji je najmanji broj k (prirodan broj, jasno, ovisan o m) takav da se u svakom k-članom podskupu elemenata iz S nalaze dva različita broja od kojih jedan dijeli drugi?

[b]Državno za 4.r.[/b]:

1. Dokaži da za sve prirodne m i n vrijedi $ (\frac {1}{m})^\frac {1}{n} \plus{} (\frac {1}{n})^\frac {1}{m} > 1.$

2. Odredi zbroj $ [\sqrt{1}]\plus{}[\sqrt{2}]\plus{}...\plus{}[\sqrt{n^2\minus{}1}]$, pri čemu je [r] najveće cijelo broja r ([3.2]=3, npr.). n je prirodan broj.

3. Imamo trokut ABC. Nad stranicama AB i BC napravljeni su kvadrati ABKL i BCMN (s vanjske strane trokuta).
a) Ako je točka D takva da je ABCD paralelogram, dokaži da je trokut DAB sukladan trokutu NBK.
b) Dokaži da je četverokut s vrhovima u: središtu kvadrata ABKL, polovištu AC, središtu kvadrata BCMN i polovištu KN kvadrat.

4. U prostoru postoji 6 točaka: O, $ T_1, T_2, ..., T_5$. Dokaži da moraju postojati neke dvije različite točke $ T_i$ i $ T_j$ takve da je kut $ T_i O T_j$ manji ili jednak 90.

5. Imamo pravokutnik podijeljen na n*p manjih jediničnih kvadrata. Svaki od tih kvadratića je ili crn ili bijel. Ako je bijel, on može postati crn tako da ima dva susjedna (s zajedničkim bridom) crna kvadratića. Odredi najmanji k takav da postoji neko bojanje s k crnih kvadratića i (n*p-k) bijelih kvadratića u kojem je moguće da nakon nekog konačnog broja poteza svi kvadratići budu crni.

Evo, ipak sam ih se svih sjetio (mislim, nemojte me zatuć ako sam pogriješio negdje). Dobru zabavu i puno sreće na natjecanju!

EDIT: Hahahahaaaa... glupi LaTeX.
EDIT2: Glup sam. 2. zadatak s državnog ipak nije bio toliko trivijalan.

Hvala lepo!

Evo i zadataka sa prvog dana SMO-a:

1. U skupu celih brojeva resiti jednacinu \[ 12^x\plus{}y^4\equal{}2008^z.\]

2. Dat je trougao $ ABC$. Neka su tacke $ D$ i $ E$ na pravoj $ AB$ takve da je $ D\minus{}A\minus{}B\minus{}E$, $ AD\equal{}AC$ i $ BE\equal{}BC$. Simetrale unutrasnjih uglova kod temena $ A$ i $ B$ seku naspramne stranice u tackama $ P$ i $ Q$, redom, a opisanu kruznicu oko trougla $ ABC$ u tackama $ M$ i $ N$, redom. Prave koje spajaju tacke $ A$ i $ B$ sa centrima kruznica opisanih oko trouglova $ BME$ i $ AND$ seku se u tacki $ X$. Dokazati da je $ CX\perp PQ$.

3. Dokazati da za pozitivne realne brojeve $ a$, $ b$ i $ c$, takve da je $ a\plus{}b\plus{}c\equal{}1$, vazi nejednakost \[ \frac{1}{bc\plus{}a\plus{}\frac1a}\plus{}\frac{1}{ca\plus{}b\plus{}\frac1b}\plus{}\frac{1}{ab\plus{}c\plus{}\frac1c}\leqslant \frac{27}{31}.\]

Cini mi se da su zadaci dosta teski...

Ovak, na prvi pogled, čine mi se bolesni. Puno puno puno (jesam li rekao puno?) teži od naše Male... Probat ću riješiti prvi, ostala dva mi se čine podosta beznadnima. Pogledat ću, doduše.

Prvi sa SMO(r)a sam rijesio... dalje necu :D

(Nije da nisam pokusao rijesiti nejednakost, a geometriju nisam ni dotakao...)

Jesam ga! Prvi, to jest. Čak nije ni toliko težak koliko sam mislio - nadam se samo da nisam napravio neku grešku.
Drugi i treći nisam nešto pokušavao (imam lošu naviku (na sreću, ne na natjecanjima) odustajanja od geometrije kad se nemam osjećaj da se može riješiti analitički ili trigonometrijski). Možda probam u nekoj doglednoj budućnosti.

Koliko su inače ljudi riješili?

za sve postoji dokaz. ovdje je dokaz onoga sto ines kaze...
[img]http://photos-e.ak.facebook.com/photos-ak-sf2p/v236/82/14/1067888778/n1067888778_50324_6500.jpg[/img]
slijeva nadesno: prof. dr. sc. Neven Elezovic, Sara Muhvic, Ines Marusic, Goran Zuzic, Nina Kamcev, Adrian Satja Kurdija, Melkior Ornik

Hm, hm. Imam ja bolju.

[img]http://img145.imageshack.us/img145/975/dsc00056co2.jpg[/img]

No, da ne zaspamamo topic (samo) svojim slikama, također očekujem slike drugih ekipa kad budu izabrane.  :)

Zgodne majice (MIOC Tim... vidim da inspiraciju crpite sa Iblerovog trga :) )

Je, je.

Mogu ti reći da imamo i dobar natpis s druge strane:

"i) Idem na državno iz matematike.
 ii) I moj brat također.

OSTVARENO!
[b]
[i]idemo dalje[/i][/b]"

Pa, ako si upoznat s HDZ-ovim sloganima za izbore... a ako ne: http://www.youtube.com/watch?v=U1PmDvHQBMg , pa se upoznaj.  :D

Ja sam uradio zadatak.
Trebao sam dobiti ili 6 ili 7 bodova na njega.
Ja sam ispitao extremni slucaj kad je najbolji ucenik uradio 5 zadataka (tj svi su po 5 zadataka uradili)
Krivo mi je jer mi nisu [b]nijedan[/b] bod priznali na zalbi, samo onaj jedan od ranije.

Ako ti se rješenje svodi na 14 < 15 kao meni, onda ti je dobro. :)

Možda je problem to što si možda samo napisao da je to najnepovoljniji slučaj, a nisi dokazao da je to doista tako (iako se po rješenju vidi da doista jest).

Javite kako je prošlo razigravanje i zadatke.

I što se tiče tog 4. zadatka, sigurno bi dobio sve bodove da si razmatrao i (lagane) slučajeve kad je najbolji uradio 6, 7 ili 8 zadataka. Mislim da si zakinut.

[quote="djuro"] Pa kad se već zadaci
prepisuju sa stranih natjecanja neka bar prepišu koje valjaju. Zanima me da li BiH i Srbija sami sastavljaju zadatke za svoja takmičenja?[/quote]

Dok cekam rezultate razigravanj, da ti usput odgovorim i na ovo pitanje (barem sto se BiH tice):
Zadaci uglavnom nisu originalni, ali zna se desiti da se pojavi nesto novo, originalno. Tako je npr, ako se ne varam, 1/2006 bio originalan (M. Treml, cini mi se), 1/2007 je bio nenamjerno originalan  :lol: ... Za ovu godinu ne znam (ono sto sa sigurnoscu mogu tvrditi da je 2008/3 iz P. Mladenovic: Kombinatorika (no tamo je naveden samo rezultat), te da su 2008/2 i 2008/5 poznate ideje, tako da ne vjerujem da su originalni. Ah da, 2008/6 je iranski).

(edit: kad sam govorio o poznatim zadacima, umjesto 2008/4 sam napisao 2008/5. Nije velika greska, jer ni 2008/5 ne daje neku novu ideju...)

[quote="hsiljak"][quote="djuro"]te da su 2008/2 i 2008/5 poznate ideje, tako da ne vjerujem da su originalni. Ah da, 2008/6 je iranski).[/quote][/quote]

Da, zadaci 2008/2 i 2008/5 su bas poznati.

Sto se tice Srpske olimpijade, komisija se trudi da zadaci budu originalni ili barem vrlo nepoznati, ali i tu dodje do propusta (2007/2 i 2007/5 su bas poznati). U svakom slucaju svake godine predlozi se zavidan broj solidnih originalnih zadataka, tako da je situacija poslednjih godina mnogo bolja nego pre nekih 5-6 (to verovatno ima veze sa podmladjivanjem komisije).

Ekskluzivno: sesti clan ekipe BiH je Jelena Radovic. Cestitke!!!

Pa to je sjajno  :lol:  !  A zar nisi rekao da je razigravanje izmedju trojice ? Cini mi se...

Ja nisam nista rekao... to je Vedran sirio informacije. Ali na BHMO jeste receno, na ceremoniji zatvaranja, da je razigravanje izmedju Uljarevica, Causevica i Besirevica. Vise cemo valjda saznati od Vedrana kad se vrati iz Lukavice :)

Vratio se Vedran kao samoprozvani observer današnjeg razigravanja. Naime, u vrijeme dok smo mi diskutovali o ovoj trojici kao konkurentima za 6.mjesto izvršen je neki tip revizije takmičenja kojime je odlučeno da je Jelena trebala imati 13 a ne 10 bodova i pozvana je na razigravanje...Malo čudno jer dosad su revizije poslije takmičenja donosile samo izvinjenja a ne i promjene u poretku...No, kako god rezultati današnjeg rada
Jelena Radović 7 7 4 0 18
Admir Beširević 0(zbog ovog ću ga prebiti) 7 0 7 14
Vlado Uljarević 7 3 0 0 10
Haris Čaušević 0 0 0 0 0

 Zadaci fini, pravi takmičarski....uskoro će biti na odgovarajućim linkovima  :P

[quote]Vratio se Vedran kao samoprozvani observer današnjeg razigravanja.[/quote]

Eto sta bude sa nasim momcima kad je Vedran observer.  :P

Da li sam ja bio uradio bi moj Sura, pored ova dva, jos i geometriju.

A ova revizija, ma to je nama svima jasno o cemu se radi.

Usput, prema mojim inf. , zadaci 1. i 3. su iz RSa (pogledajte bodovanje :wink: )

A Tucakovicu, hajd molim te konkretno nam napisi tamo u forumu (stavio sam ja zadatak pod combinatorics /proposed and own problems), ili ovdje, to tvoje rjesenje. Meni Elmir rece da je kod vas extreman slucaj bio kad su svi po 6 zad. uradili. :?:

Ma Elmir je nesto zeznuo
Meni je extremni slucaj kad su svi po 5 uradili.
Sad mi se ne da pisati rjesenje, ne sjecam se bas najbolje.
Spremam se da bih mogao ici na IMO2009 - Bremen :D

[quote="delegat"][quote]
A ova revizija, ma to je nama svima jasno o cemu se radi.
[/quote][/quote]

Zasto uvijek sve moramo gledati na taj nacin? Zar ne mozemo pretpostaviti da je konacno bilo malo pravde na Drzavnom? Da je ovo neki pomak naprijed...

A sto se toga ko je predlozio koji zadatak tice... zadaci su, objektivno receno, dosta lagani. Tako da mi se ne cini to toliko bitnim... Naravno, ostavljam ti pravo da imas svoje misljenje o ovim desavanjima, ali mislim da nije ni vrijeme ni mjesto za ovu vrstu rasprave.  :wink:

Ma sam nacin izbora ekipe nije ni blizu da se moze nazvati "dobrim" Interesantne su ove revizije :wink:
mada nemam nista protiv. Uostalom, Damir je i sam prihvatio ovu zadnju.

Pa i pricam o revizijama. Izbor ekipe nije bio na nivou prethodnih izbora, jer prosle dvije godine (za ta dva Drzavna mogu govoriti, jer sam ucestvovao) nije bilo ovoliko zalbi (skoro da uopce nije bilo zalbi-okace se rezultati i hajdemo odmah na ceremoniju), rezultata pod siframa... Samim tim, ne iznenadjuju ovakve revizije, kad je vec toliko narusen integritet komisije koja je pregledala (narusen je i cinjenicom da se otkrilo da su jednog takmicara ( ;) ) zakinuli za najmanje 7 bodova). Helem nejse... Da se ostavimo toga i pocnemo pripreme za Spaniju...


I have lost the source. Probably it is from an Indian olympiad

Let $ DE\cap BC\equal{}M,\ BE\cap AD\equal{}N$
We have $ DE\perp BC$.
Triangles $ AOD$ and $ BEC$ are isosceles and quadrilater $ BNMD$ is cyclic.
Then $ \widehat{DAO}\equal{}\widehat{ADO}\equal{}\widehat{EBC}\equal{}\widehat{ECB}\equal{}\widehat{ADB}\Rightarrow AO\parallel BD$

From the isosceles triangles $ \bigtriangleup OAD,$ $ \bigtriangleup DBE,$ we have that

$ \angle BDA \equal{}\angle ADO \equal{}\angle ODA$ $ \Longrightarrow$ $ BD\parallel AO$ and the proof is completed.

Kostas vittas.

[color=darkblue][b]Proof.[/b] I used only the well-known relation $ m(\widehat{ODA}) \equal{} m(\widehat{OAD}) \equal{}\frac{B\minus{}C}{2}$.
$ \boxed{\ BE\perp AD\ }\Longleftrightarrow\widehat{BDA}\equiv\widehat{EDA}\equiv\widehat{ODA}$ $ \Longleftrightarrow$ $ \widehat{BDA}\equiv\widehat{OAD}$ $ \Longleftrightarrow$ $ \boxed{\ BD\parallel AO\ }$ $ \Longleftrightarrow$ $ C \equal{}\frac{B\minus{}C}{2}$ $ \Longleftrightarrow$ $ \boxed{\ B \equal{} 3C\ }$.

[b]Remark.[/b] In this case $ C<45^{\circ}$ and if denote the second intersection $ F$ of the circumcircle with the line $ BE$,
then  $ AF\parallel BC$, $ AF\equal{}DB\equal{}DC$, $ AB\parallel DF$, $ AD\parallel CF$ and $ AC\perp FD$.[/color]


logarithms
calculus
calculus computations

Suupose $ a_n>\minus{}1$ for all $ n$.  By using a suitable application of taylor's thm to $ \log{1\plus{}x}$ or $ e^x$, show that

a)  $ \sum a_n$ is absolutely convergent iff $ \sum \log{(1\plus{}a_n)}$ is absolutely convergent

b)  Let $ a_n\equal{}\frac{(\minus{}1)^{n\plus{}1}}{\sqrt{n}}$. then $ \sum_1^{\infty}a_n$ is conditionally convergent but $ \sum_1^{\infty}\log{(1\plus{}a_n)}$ diverges.

Da(n) <= log (1 + a(n)) <= Ca(n) for some scalars D and C (assume C > 0)

Sum and realize that AC means partial sums are bounded.

The idea is really the same as that for the proof of Cauchy's condensation test, just to kind of sandwich things so that the boundedness of one is equivalent to the boundedness of the other.

Of course, the application is to convergence of infinite products; $ \prod (1\plus{}a_n)$ converges absolutely if and only if $ \sum a_n$ converges absolutely.

i dont understand §outh§tar's proof... :maybe:


The diagonal $AC$, $BD$ of convex quadrilateral $ABCD$ are perpendicular. Through the midpoints of $AC$, $BD$ drop perpendiculars to $CD$, $BC$ are drawn. Prove that these lines intersect at a point on $AC$

Prove that for any positive integer n, there exists an integer k which can be expressed as a sum of two
squares and such that $n \leq k \leq n+3(n)^{\frac{1}{4}}$.

Let $x_{1},...,x_{n}$ be a sequence of decreasing positive numbers. Prove:

$( \sum^{n}_{k=1}x^{2}_{k})^{\frac{1}{2}}\leq \sum^{n}_{k=1}\frac{x_{k}}{k^{\frac{1}{2}}}$

[quote="LPaladin9"]Let $x_{1},...,x_{n}$ be a sequence of decreasing positive numbers. Prove:

$( \sum^{n}_{k=1}x^{2}_{k})^{\frac{1}{2}}\leq \sum^{n}_{k=1}\frac{x_{k}}{k^{\frac{1}{2}}}$[/quote]
[hide]The sign should be reversed. To prove the corrected version, divide both sides by $k^{\frac12}$ and you will get just $\mathrm{QM-AM}$. Of course the assumption that the sequence is decreasing is unnecessary.  :D [/hide]


function
quadratics
algebra
quadratic formula

Solve in the set of the real numbers this equation :

$ 3\plus{}\sqrt{3\plus{}x\sqrt{x}}\equal{}x^3$

[hide="Solution"]
The LHS of the original equation is positive, so $ x^3 > 0$, implying $ x > 0$.
We can let $ u \equal{} x\sqrt {x}$, $ u > 0$. The original equation becomes
$ 3 \plus{} \sqrt {3 \plus{} u} \equal{} u^2$ or
$ \sqrt {3 \plus{} u} \equal{} u^2 \minus{} 3$.
Let $ f(u) \equal{} \sqrt {3 \plus{} u}$ and $ g(u) \equal{} u^2 \minus{} 3$, so $ f(u) \equal{} g(u)$.
We will show that $ g(u) \equal{} f^{ \minus{} 1}(u)$.
$ u \equal{} \sqrt {3 \plus{} [f^{ \minus{} 1}(u)]}$
$ u^2 \equal{} 3 \plus{} f^{ \minus{} 1}(u)$
$ u^2 \minus{} 3 \equal{} f^{ \minus{} 1}(u) \equal{} g(u)$.
Since $ f$ and $ g$ are inverse functions $ f(u) \equal{} g(u)$ holds if and only if $ g(u) \equal{} u$, so
$ u^2 \minus{} 3 \equal{} u$
$ u^2 \minus{} u \minus{} 3 \equal{} 0$
and by the quadratic formula, $ u \equal{} \dfrac{1\pm \sqrt {13}}{2}$.
The root $ u \equal{} \dfrac{1 \minus{} \sqrt {13}}{2} < 0$ is extraneous so
$ u \equal{} \dfrac{1 \plus{} \sqrt {13}}{2}$
$ x^\frac {3}{2} \equal{} \dfrac{1 \plus{} \sqrt {13}}{2}$
$ x \equal{} \left(\dfrac{1 \plus{} \sqrt {13}}{2}\right)^{\frac {2}{3}}$[/hide]

hello, another approach, squaring, simplifying and factoring we get $ (x\minus{}1)(x^2\plus{}x\plus{}1)(x^6\minus{}7x^3\plus{}9)(x^3\minus{}4)\equal{}0$. From here we have the only real solution $ x\equal{}\frac{1}{4}(4\plus{}4\sqrt{13})^{\frac{2}{3}}$
Sonnhard.


blogs Gauss search AMC AIME USAMTS geometry

Here's how the game works:

Each person is to go through the smilies in order where they are listed and tell a story with them. The words you put in the story must relate to the way the smilies look. Also, the story must be one central idea and not something as random as the Missouri forum.

I'll start:

 :D One day, gauss1181 was so happy that she got obssessed with aleph0's avatar.

[quote="evilhamster"]:D One day, gauss1181 was so happy that she got obssessed with aleph0's avatar.
 :lol: aleph0 gave a very big grin, because he liked it when people got obsessed with his avatars. He wished that people would get obsessed with Wicked's though.
 :bomb: When nobody did, aleph0 became very mad. Very, very mad.[/quote]
Hey, wrong order. Go through the smilies in which you click them on the list.

And you guys don't need to copy paste the previous poster's parts.

 :maybe: But he sighed, looking doubtful. His emoness made him feel that way.

gosh, you copied this from the Missouri Forum... :P

Of course he did! Didn't you read the first sentence?

:D One day, gauss1181 was so happy that she got obssessed with aleph0's avatar.
 :lol: aleph0 gave a very big grin, because he liked it when people got obsessed with his avatars. He wished that people would get obsessed with Wicked's though.
 :bomb: When nobody did, aleph0 became very mad. Very, very mad.
 :blush:  aleph0 blushed. He was amazed at how his emotions changed so quickly.
 :maybe: Dojo looked at aleph0 in a curious way.

Wait, isn't the smiley limit 3 per post?

:D One day, gauss1181 was so happy that she got obssessed with aleph0's avatar.
 :lol: aleph0 gave a very big grin, because he liked it when people got obsessed with his avatars. He wished that  people would get obsessed with Wicked's though.
  :bomb: When nobody did, aleph0 became very mad. Very, very mad.
  :blush: aleph0 blushed. He was amazed at how his emotions changed so quickly.
  :maybe: Dojo looked at aleph0 in a curious way.
 :) aleph0 smiled innocently. however, something was behind that innocent smile...
 :( Dojo realized aleph0 was hiding something, and became sad.
 :| then pacman tried to eat aleph0, but he couldn't open his mouth at all.

:D One day, gauss1181 was so happy that she got obssessed with aleph0's avatar.
 :lol: Very Happy aleph0 gave a very big grin, because he liked it when people got obsessed with his avatars. He wished that people would get obsessed with Wicked's though.
 :bomb: When nobody did, aleph0 became very mad. Very, very mad.
 :blush: aleph0 blushed. He was amazed at how his emotions changed so quickly.
 :maybe: Dojo looked at aleph0 in a curious way.
 :) aleph0 smiled innocently. however, something was behind that innocent smile...
 :( Dojo realized aleph0 was hiding something, and became sad.
 :| then pacman tried to eat aleph0, but he couldn't open his mouth at all.
 :rotfl: ak909090 started rolling around laughing like a maniac (rallam) at pacman's predicament
*error- this post has reached the max amount of smilies*

:ninja:  was next. But I have no idea what is going on.

so  tcjv did this to gaussy :omighty:  and gaussy like [b]THAT[/b] So, gaussy let tcjv play nhl

:yoda: And everything got too confusing, and so, Yoda comes and slices everyone up with his lightsaber.

:D Everyone turned green because of Yoda.

:blush: The other people blushed because they could never be as  :D as Yoda.

And they blushed  :blush: in shame-they could never be as  :diablo: as GaussWannabe.

well,  :maybe: they could, said a random person from the crowd.

:) And Arvind, who was, indeed, as  :diablo: as GaussWannabe, smiled knowingly.

everybody in the crowd then :( , for no apparent reason (for now!)

:o They started screaming at Gaussy :o because of her blog :o

the were  :mad: because her background hurt their eyes

THE STORY IS GETTING REVAMPED!!! ARGH!
I drowned in :D's
Guassy smiled because of the U KNOW WHATS. :D

WHAT?! TOO MANY SMILEYS?!  :mad:


AMC USA(J)MO USAMO AIME ARML geometry college

I know someone out there (Dr. Merryfield, maybe?  I don't remember) compiles a cumulative list on a regular basis.  But in any case, here's a list of all schools with 3 or more qualifiers this year:

Schools with 3 or more USAMO qualifiers, 2007

20:
5% Club, Los Angeles

16:
Phillips Exeter

12:
TJHSST

9:
Montgomery Blair
AAST

8:
Phillips Academy (Andover, MA)

6:
Clements HS (Sugar Land, TX)

5:
Vestavia Hills
Lynbrook (San Jose)
Iowa City West HS
Stuyvesant

4:
Gunn HS (Palo Alto, CA)
Saratoga HS (CA)
Naperville North (IL)
Lexington HS (MA)
ICAE (Troy, MI)
Klein HS (TX)
Don Mills CI (Don Mills, ON)
Olympiads School (North York, ON)
University of Toronto Schools

3:
Carlmont HS (Belmont, CA)
Choate Rosemary Hall (Wallingford, CT)
Mission San Jose HS
Palo Alto HS (CA)
Redwood MS (Saratoga, CA)
F.W. Buchholz HS (Gainesville, FL)
West Lafayette HS (IN)
William Henry Harrison HS (West Lafayette, IN)
Boston Latin School
Detroit County Day School (Beverly Hills, MI)
Wylie E. Groves HS (Beverly Hills, MI)
William G. Enloe HS (Raleigh, NC)
A&M Consolidated HS (College Station, TX)
Longfellow MS (Falls Church, VA)
Lakeside School (Seattle)
Stephen Leacock CI (Agincourt, ON)
Vincent Massey SS (Windsor, ON)

Notably absent: Illinois Math and Science Academy (only 2 this year, compared to 5 last year, 5 in 2005, 6 in 2004, 6 in 2003, 9 in 2002, and 8 in 2001).

If I had qualified my school would be on the list. Oh well.

>_<
look for a resurgence from gunn next year :D

3 - Longfellow MS (Falls Church, VA)

I have no real comments at this point, though I'm interested to see the cumulative list to see how expansion has effected everything.

Not me. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31725]Ravi B.[/url] keeps the list you're talking about.

Thanks for the numbers, Sly Si.  I will try to post the cumulative list soon.

What is this 5% Club.  The names all look somewhat Korean.  DON'T MESS WITH THOSE ASIANS MAN.

What, a 6th grader?  I must find out what these people eat.

The "5% Club" is not a school. On another thread, kohjhsd posted the information that it is a mechanism for Korean students to take the AIME; it is unclear why it is listed as being in Los Angeles, as I doubt that any of those students reside in L.A.

If you go to the AMC website, the USAMO pages, and click on "Archive" you get a rather complete year-by-year listings of individuals who have taken the USAMO for as far back as the AMC has records. Within each year, you get each individual's complete record, so there's a great deal of overlap between adjacent years. In a smaller project, I took all of these charts, dumped them into Excel spreadheets, did a lot of sorting, and fished out just those listings for Southern California. (The school championship there for the full history belongs to Palos Verdes Peninsula HS, with or without including Palos Verdes HS and Rolling Hills HS which merged to become Peninsula HS - and have since split again.)

There's one flaw in that archive in that both the 1993 and 1994 links point to the 1994 data - but with the overlaps from adjacent years, there may not be all that much missing.

To some extent, I think it is the fact that many talented students are recruited/attracted to these schools.  Also, being around a community of other talented mathematicians helps (these are both reasons why AoPS accounts for so many USAMO qualifiers  :D )

[quote="rnwang2"]I would really like to know how Phillips Exeter and TJHSST train their students so well.  :o[/quote]

I think I'm going to steal JSteinhardt's role this time and post this link:

[url]http://activities.tjhsst.edu/vmt/pages/tjusamo/tjusamo2k7/index.phtml[/url]

ICAE really isn't a school too because people go there on Saturdays to do more.

Really there are 10 ICAE people for we all went to ICAE so time ago...but the 4 listed went there to take the test.

[quote="rnwang2"]I would really like to know how Phillips Exeter and TJHSST train their students so well.  :o[/quote]
Phillips Exeter Academy (PEA) has some of the well known math teachers in the nation. One for example is Dr. Zuming Feng, who is a regular USAMO coach. PEA is also one of the richest schools in the nation, having more than 1 billion dollars in endowments last year. They recruit the best faculty, usually with graduate degrees from ivy colleges. I recently visited PEA and the campus is so huge and impressive, it is like a big university. I was told that their library is the largest for a high school in the world.

It is also in my backyard (not really... 50 miles away).  :)  I am planning to go there next year. My current school produced only 1 USAMO qualifier this year...that is me.

[quote="Hamster1800"][url]http://activities.tjhsst.edu/vmt/pages/tjusamo/tjusamo2k7/index.phtml[/url][/quote]

Except that is only this year's -- go up a few directories to [url]http://activities.tjhsst.edu/vmt/pages/tjusamo/[/url] in order to see the lectures from multiples years (though that does not take into account ARML lectures and such).

EDIT: And having Longfellow MS in our district doesn't hurt either.

Slight correction: West Lafayette technically only has 2 qualifiers.  One was from West Lafayette Junior High.

please edit the list on the top of this page - look at AMCDirector's new post and you will see why  [hide=":lol:"]the 5% Club USAMO qualifiers have been disqualified![/hide]

[quote="JSteinhardt"]Except that is only this year's -- go up a few directories to [url]http://activities.tjhsst.edu/vmt/pages/tjusamo/[/url] in order to see the lectures from multiples years (though that does not take into account ARML lectures and such).

EDIT: And having Longfellow MS in our district doesn't hurt either.[/quote]
Whoa! Who grades all those USAMOs??

I wish my school was as serious about math competitions. We don't even have a real math club or anything.  :(

Haitao and I grade them =). Though not all of them are contests, a lot are just problem sessions. We've had 3 large practice USAMOs so far this year (i.e. many people), and then 1 small one (just a few people, but 4.5 hours long), and will probably have at most one more by the end of the year. We probably don't grade as rigorously as the actual USAMO people though (i.e. we probably won't spend 15 minutes reading a bad solution trying to give it a point or two of partial credit, and we probably won't 5 extra minutes reading a good solution seeing if there is any place to take off points...unless it is really ugly). But it certainly gives some perspective when writing up a proof having had to read some yourself.

Corrections to the original list (I can't edit the post anymore):

Remove 5% club.

Fairview HS (Boulder, CO) now has 3 qualifiers.

Oak Park & River Forest HS (IL) now has 3 qualifiers.

Detroit County Day School now has 4 qualifiers.

ICAE now has 5 qualifiers.

AAST now has 10 qualifiers.

yay clements...how many of these schools are like normal public schools that arent magnet/boarding/private

lol
most of the good schools are all private
but i dont know, i guess at least some of the better ones probably are public
100% of people from my school made it!!! (i.e. 1 out of 1 XD)

[quote]most of the good schools are all private [/quote]

Dealing only with the part of the list with which I have the most familiarity:

Of the 14 qualifiers in the greater Los Angeles metropolitan area (that is, the territory of the Southern California ARML team), one is homeschooled, one attends a private school, and the other 12 attend public schools. (One of these is, I think, a charter school operating within a public school district.)

Of the 47 qualifiers from the SFBA area:

3 have non-school addresses on the AMC website (EPGY, Berkeley Math Circle); on another topic, one of those was identified as a homeschooler.

9 come from schools that I suspect from the name of being private schools (Harker, College Prep. School, Crystal Springs Uplands, Marin Academy, Athenian School, Menlo School).

That leaves 35 from what I think are public schools. Note in particular the cluster that wraps around the the southern end of the bay of public high schools (and their feeder middle schools) with multiple qualifiers each. Zenith Xan commented on a middle school in Virginia (a feeder to TJ?) that had 3 - but notice also Redwood MS in Saratoga (a feeder to Saratoga HS?) with 3

So was the 5% club some sort of Korean intermediary for the USAMO?  I'm confused.

[quote="Kent Merryfield"]
9 come from schools that I suspect from the name of being private schools (Harker, College Prep. School, Crystal Springs Uplands, Marin Academy, Athenian School, Menlo School).
[/quote]

Yes, all of the schools above are private. I happen to know the ones who qualified at Crystal and Menlo.


function
real analysis
real analysis unsolved

let $f: S^{1}\rightarrow \mathbb{R}$ be any continuous map. $S^{1}$ is the unit circle in the plane. Let $A=\{(x,y)\in S^{1}\times S^{1}: x\neq y, f(x)=f(y)\}$ .Is A non- empty? If yes, is the cardinality countable or uncountable?

It is uncountable.  $S^{1}$ is compact, so any continuous function has a maximum and a minimum.  If they are the same, the function is constant and so the cardinality is clearly uncountable.  If they are different, note that the two open arcs between the point at which the minimum and maximum are achieved (if there are many such points, select a pair) include preimages of all the values between the minimum and maximum value.  These two arcs are disjoint, so for each value in this open interval (uncountably many) we get at least one pair.  Thus, again, uncountable.


[b]Slightly Different:[/b]

We first make the substitution:
\[ \{\begin{array} {ccc}a=x+y\\\\b=y+z\\\\c=z+x\end{array}\] transforming the inequality to:

\[\sum_{cyc}\frac{x+y}{z}\geq6\] noting that $\[\sum_{cyc}\frac{x+y}{z}=\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}$ the inequality is true by AM-GM.

in a triangle,
2s = a + b + c
2s - 2b = a + c - b
similarly , 2s - 2a = b + c - a
and 2s - 2c =  a + b - c
so the given inequality becomes,
1/2[a/(s-b)] + 1/2[b/(s-c)] + 1/2[c/(s-a)]
taking 1/2 as common,
1/2[a/(s-b) + b/(s-c) + c/(s-a)]
the value in the bracket is always greater than or equal to 6.
so the whole equation is greater than or equal to 3.

[quote=Fermat -Euler]Let $x=c+a-b$
$y=a+b-c$
$z=b+c-a$
then $a=\frac{x+y}{2} ;b=\frac{y+z}{2} ;c=\frac{x+z}{2}$
 $\frac{x+y}{2x} + \frac{y+z}{2y} + \frac{x+z}{2z}$

$=\frac{3}{2} +\frac{1}{2}(\frac{y}{x}+ \frac{z}{y} +\frac{x}{z})$
$\geq \frac{3}{2} +\frac{3}{2} = 3$[/quote]

Wonderful solution.. Thanks

[quote=Rushil]If $a,b,c$ are sides of a triangle, prove that \[ \frac{a}{c+a-b} + \frac{b}{a+b-c} + \frac{c}{b+c-a} \geq 3. \][/quote]
Given $a,b,c>0,b+c-\lambda a>0,c+a-\lambda b>0,a+b-\lambda c>0$ $(-2<\lambda\leq 0).$Prove that$$\frac{a}{c+a+\lambda b} + \frac{b}{a+b+\lambda  c} + \frac{c}{b+c+\lambda a} \geq \frac{3}{\lambda+2},$$$$\frac{a}{b+c+\lambda a} + \frac{b}{c+a+\lambda  b} + \frac{c}{a+b+\lambda c} \geq \frac{3}{\lambda+2}.$$
[url=https://artofproblemsolving.com/community/c6h501238p2816500]here[/url]

[quote=ZombieNine][s]I-I'm sure the person from 2005 still needs help. =p[/quote]

It is not just about helping.

[quote=Combigeontal231]Just multiply numerator and denominator  of each term by numerator  itself and use titus lemma.
Then apply $a^2+b^2+c^2\geq $ ab+bc+ca.[/quote]

Could you provide a solution? Ji tried it but could not get anything

BTW, I have a proof different than the ones provided, something involving AM-GM and Geometric Inequalities. Give a look through it, and plz upvote if you like it [s]and even if you don't like it[/s] :P

Let $\Delta$ denote area of triangle
By some simplification in Heron's formula,
$\frac{16\Delta^2}{a+b+c}=(a+b-c)(a-b+c)(-a+b+c)$
By AM-GM, $LHS\geq 3\sqrt[3]{\frac{abc}{(a+b-c)(a-b+c)(-a+b+c)}}=3\sqrt[3]{\frac{abc(a+b+c)}{16\Delta^2}}\geq 3\sqrt[3]{\frac{R}{2r}}\geq 3$
As $R=\frac{abc}{4\Delta}$, $r=\frac{2\Delta}{a+b+c}$ and $R\geq 2r$
[Hide=motivation] The expression in the denominators reminded me of that simplification of Heron's formula which I usually used to derive as a time pass.
As I needed a multiplication of those, I applied AM-GM, and then things turned out to be smooth and simpler than expected, one I noticed the formulae of R and r in that cube root[/hide]

[quote=xzlbq][quote=Rushil]If $a,b,c$ are sides of a triangle, prove that \[ \frac{a}{c+a-b} + \frac{b}{a+b-c} + \frac{c}{b+c-a} \geq 3. \][/quote]

stronger is:

\[ \sum(\frac{a}{c+a-b} \cos^4{\frac{B-C}{2}}) \geq 3. \][/quote]

This should be it.

\[ \sum(\frac{a}{b+c-a} \cos^4{\frac{B-C}{2}}) \geq 3. \]

\[ \sum(\frac{a}{c+a-b} \cos^2{\frac{B-C}{2}}) \geq 3. \]

[quote=Math-wiz][quote=xzlbq][quote=Rushil]If $a,b,c$ are sides of a triangle, prove that \[ \frac{a}{c+a-b} + \frac{b}{a+b-c} + \frac{c}{b+c-a} \geq 3. \][/quote]

stronger is:

\[ \sum(\frac{a}{c+a-b} \cos^4{\frac{B-C}{2}}) \geq 3. \][/quote]
I am not much sure, but
Won't that be the same whole proving?
As I think there is a conditional identity for $A+B+C=\pi$ that $\Pi cos^4\frac{B-C}{2}=1$
I am not that sure[/quote]
Thank you!
 I was wrong.

[quote=Combigeontal231]Just multiply numerator and denominator  of each term by numerator  itself and use titus lemma.
Then apply $a^2+b^2+c^2\geq $ ab+bc+ca.[/quote]

Could you show your proof plz?
@above wowwww. Great solution :)

[quote=WolfusA]$\sum \frac{a^2}{ab+ac-a^2}\ge \frac{(a+b+c)^2}{2(ab+ac+bc)-a^2-b^2-c^2}\ge3$[/quote]

But how is $\frac{a^2}{ab+bc-a^2}=\frac{a}{c+a-b}$

My mistake. It's impossible to solve it in a way I tried.

[quote=fjm30]Use Ravi's substitution, then simplify the expression, then the question is trivial by AM-GM.[/quote]

Woww. Great solution. Ravi Substitution helps in many ways :)


Let ABC be an acute triangle.
Prove that:

sinA/cosA+ sinB/cosB+sinC/cosC \geq 27S/ \sum ma 2

This is quite trivial. Use use that  \sum a 2  \geq 4 \sqrt 3*S   and  \sum tgA \geq 3 \sqrt 3 and that  \sum (m_a) 2 =k* \sum a 2 , where k is a constant that I'm too bored to compute.

[quote]...that I'm too bored to compute.[/quote]
 :D  :D  :D Yeah I like that idea harazi  :P


function
ratio
calculus
derivative
quadratics
analytic geometry
calculus computations

A quartic function with a "W" shape has two points of inflection, Q and R.  Then we draw the line QR and it will pass through the quartic graph twice more, calling those points P and S.

Now, this was an old IB portfolio question.

It involves 6steps. The first four steps were easy, I found that
the ratio $ PQ: QR: RS \equal{} 1: \phi: 1$ through examination of two different W shaped quartic functions.

However, for number 5, I have to prove this analytically not just with examples.

Im kind of lost here.

I set the function as $ ax^{4} \plus{} bx^{3} \plus{} cx^{2} \plus{} dx \plus{} e$, then for the second derivative tofind the points of inflection I get $ 12ax^{2} \plus{} 6bx \plus{} 2c$.  Now, the two points of inflection are easy tofind through the quadratic formula.  However, inorder to prove the ratio, Iwill also need they-coordinates of the points and I have no idea how to do that clealy.  Iwill also need to find P and S and I need to find the equation of the line going through Q and R but I dontknow how to do that.  To top it off, I have to use the distanceformula tocompute the ratios of the lenghs.

I have a feeling this is not the way to go so can someone help me a bit.  Thanks.

[i]Im trying to prove $ PQ: QR: RS \equal{} 1: \phi: 1$where Q and R are inflection points of any quartic function with a "W" shape and P and S are points where the line QRintersects the graph again.[/i] and when I say PQ i mean the length of that line segment and so forth.

My thoughts:
1) You can assume that $ a \equal{} 1$, because rescaling does not change the ratios in question.
2) You don't need the distance formula: the distance between two points on a line is equal to |slope| times the difference of their x-coordinates. So it's just the x-coordinates of P and S that you need.
3) The equation of line QR is $ y \equal{} f(x_Q) \plus{} \frac {y_R \minus{} y_Q}{x_R \minus{} x_Q}(x \minus{} x_Q)$.
4) Equating the line equation to $ f$, you end up with a quartic equation for them, which is scary. But you already know two roots of that equation: $ x_Q$ and $ x_R$. That means that the equation can be divided by $ 12x^2 \plus{} 6bx \plus{} 2c$ without a remainder, leaving you with a quadratic.
5) There ought to be a better way to do this.  :maybe:

In my experience, whenever you have a line intersecting a quartic, you should subtract the line.  (In other words, note that the values of $ d$ and $ e$ are also irrelevant to the outcome -- adding lines to each other doesn't change ratios of distances between points at fixed $ x$-coordinates.  Thank goodness for Euclidean geometry.)

[hide="Solution"]Subtracting, shifting, and rescaling, we get the points of inflection to be $ (\pm 1,0)$. With the $ a \equal{} 1$ normalization we have $ f''(x) \equal{} 12x^2 \minus{} 12$, which means $ f(x) \equal{} x^4 \minus{} 6x^2 \plus{} dx \plus{} e$. On the other hand, $ f \equal{} (x^2 \minus{} 1)q(x)$, where $ q$ is a quadratic polynomial. The only possibility is $ q(x) \equal{} x^2 \minus{} 5$, hence the other points of intersection are $ \pm\sqrt {5}$.
Neat.  :) [/hide]


Pentru cei interesati, va postez problemele care s-au dat la barajul de la Arhi .
[b]Subiectul 1[/b] Sa se determine toate multimile de $8$ numere intregi consecutive cu proprietatea ca suma cuburilor lor este cub perfect.

[b]Subiectul 2[/b] Sa se demonstreze ca pentru orice $a,b,c \in R$ avem inegalitatea
\[a^{3}+b^{3}+c^{3}-3abc\leq (a^{2}+b^{2}+c^{2}) \sqrt{ a^{2}+b^{2}+c^{2}}\].

[b]Subiectul 3[/b] Fie $ABCD$ un patrat cu lungimea laturilor egala cu $1$.Fie $E$ pe $[BA]$ si $F$ pe $[BC]$ cu proprietatea ca $BE=BF$.Fie $N$ proiectia punctului $B$ pe $CE$.Sa se calculeze minimul si maximul expresiei $NF^{2}+ND^{2}+NC^{2}$, cand punctul $E$ se plimba pe latura $[AB]$.

[b]Subiectul 4[/b] Fie $n\geq 2$ un numar natural si $\lambda\in (0,\infty)$.Sa se demonstreze ca urmatoarele doua afirmatii sunt echivalente :
$1)$  $\left | \sum_{j=1}^{n}a_{j}b_{j}\right|^{\lambda}\leq (\sum_{j=1}^{n}\left | a_{j}\right |^{\lambda})(\sum_{j=1}^{n}\left | b_{j}\right |^{\lambda})$ , pentru orice $a_{j},b_{j}\in C , j\in \left\{ 1,2,...n \right\}$.
$2)$ $\lambda \in (0,2]$

[b]Subiectul 5[/b] Fie $p$ un numar prim , $k,n\in N^{*}$ si $z_{1},z_{2},..,z_{n}\in C$ cu urmatoarele proprietati :
$1)$ $\sum_{k=1}^{n}z_{k}= 0$
$2)$ $z_{j}^{p^{k}}=1$ pentru orice $j\in \left \{ 1,2,..n \right \}$.
Sa se demonstreze ca $p | n$.

[hide="S2"]
Inegalitatea este echivalenta cu:
\[\left(\sum a^{3}-3abc\right)^{2}\leq (\sum a^{2})^{3}\]
\[\Leftrightarrow\]
\[\sum a^{6}+2\sum a^{3}b^{3}+9a^{2}b^{2}c^{2}-6abc\sum a^{3}\leq \sum a^{6}+3\sum_{sim}a^{4}b^{2}+6a^{2}b^{2}c^{2}\]
\[\Leftrightarrow\]
\[2\sum a^{3}b^{3}+3a^{2}b^{2}c^{2}\leq 6\sum a^{4}bc+3\sum_{sim}a^{4}b^{2}\]
\[\Leftrightarrow\]
$\sum_{sim}a^{3}b^{3}+\frac{1}{2}\sum_{sim}a^{2}b^{2}c^{2}\leq 3\sum_{sim}a^{4}bc+3\sum_{sim}a^{4}b^{2}$ ,evident conform Muirhead.(sau AM-GM).Egalitatea are loc $\Leftrightarrow$ $a=b=c=0$
O sa caut si o solutie "frumoasa", cu siguranta exista pentru ca e destul de slaba inegalitatea daca se merge "straightforward"
[/hide]

[hide="tot la 2"]folosim formula $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac)$ si ridicam la patrat. notam $m=a^{2}+b^{2}+c^{2}$ si $n=ab+bc+ac$. inegalitatea se reduce chiar la $m \geq n$, evident adevarata.[/hide]

[hide="S3"]
Consideram sistemul de axe cu originea in $B$ si $BA=Ox$ si $BC=Oy$ astfel $B(0,0)$,$C(0,1)$ si $D(1,1)$
De asemenea fie $\frac{BE}{BA}=t$ $\Leftrightarrow$ $BE=t$ $t\in [0,1]$ $\rightarrow$ $E(t,0)$ si $F(0,t)$
Cu punctele astfel definite avem ecuatiile dreptelor
$(CE): y=-\frac{1}{t}x+1$
$(BN): y=tx$
Astfel avem $BN\cap CE=N\left( \frac{t}{1+t^{2}},\frac{t^{2}}{1+t^{2}}\right)$
Cunoscand si coordonatele punctului $N$ avem:
$A=NF^{2}+ND^{2}+NC^{2}=2\left(\frac{t}{1+t^{2}}\right)^{2}+\left( \frac{t^{2}}{1+t^{2}}\right)^{2}+\left(\frac{t}{1+t^{2}}-1\right)^{2}+2\left( \frac{1}{1+t^{2}}\right)^{2}$
Fie functia $f: [0,1]\to \mathbb R$ ,$f(t)=A$ , $f$ strict descrescatoare (suma de functii descrescatoare) ,deci
$A\in [f(1),f(0)]$ $\Leftrightarrow$ $A\in \left[\frac{3}{2},3\right]$

[/hide]

[hide="S1"]
Fie $a\in \mathbb Z$ cel mai mic dintre cele 8 numere consecutive.
atunci avem ecuatia:
$a^{3}+\sum_{k=1}^{7}(a+k)^{3}=n^{3}$ ,$n \in \mathbb Z$ $\Leftrightarrow$
$(2a+7)^{3}+126a+441=n^{3}$
Dar $2a+4\leq n\leq 2a+10$ altfel obtinem contradictie.
Ne ramane de analizat cele 7 cazuri de unde obtinem $a\in \{-5,-4,-3,-2\}$
[/hide]

La $S_{2}$ eu cred ca avem egalitate pentru $\sum ab =0$
Dupa ce discutam semnul lui $\sum a$ apoi ridicam la patrat inegalitatea.
Dupa notatiile lui $Svejk$ avem ca inegalitatea este echivalenta cu $n^{2}[2(m-n)+m]\geq 0$ care este evident adevarat deoarece $2(m-n)+m=\sum (a-b)^{2}+\sum a^{2}\geq 0$
Egalitate avem daca $n=0$ adica $\sum ab=0$ sau $\sum (a-b)^{2}+\sum a^{2}=0$ adica $a=b=c=0$.Deci este suficient ca $\sum ab =0$ pentru a avea egalitate.
Corectati-ma daca gresesc :D .

[i]Subiectul 5 - Soluție[/i]

[hide="Capitolul 1."]
Fie $f(X) = X^{p-1}+X^{p-2}+\ldots+X+1$ și $f_{k}(X) = f \left( X^{p^{k-1}}\right), \, \forall k \geq 1$.
Voi demonstra că $g_{k}(X) = f_{k}(X+1)$ are toți coeficienții, mai puțin cel dominant, divizibili cu $p$.
Mai întâi observăm că $f_{k}(X) = \frac{X^{p^{k}}-1}{X^{p^{k-1}}-1}$, deci
\[g_{k}(X) = \frac{(X+1)^{p^{k}}-1}{(X+1)^{p^{k-1}}-1}. \ \ \ \ \ \left( \star \right) \]
O proprietate destul de cunoscută spune că $(X+1)^{p^{j}}-1 \equiv X^{p^{j}}\left( \bmod \, p \right)$.
Reducând toți coeficienții $\mod p$ în relația $\left( \star \right)$, obținem $X^{p^{k-1}}\cdot g_{k}(X) = X^{p^{k}}$, în $\mathbb Z \slash p \mathbb Z [X]$. Cum factorizările sunt unice în inelul de polinoame modulo $p$, rezultă că $g_{k}(X) \equiv X^{p^{k}-p^{k-1}}$. Întorcându-ne peste $\mathbb Z$, deducem afirmația făcută la început.
Dar termenul liber al lui $g_{k}$ este $p$, deci, conform criteriului lui Eisenstein, $g_{k}$ este ireductibil.
Așadar, $f_{k}$ este ireductibil, deci este polinomul minimal al tuturor rădăcinilor sale.
[/hide]
[hide="Capitolul 2."]
Acuma să presupunem că avem $z_{1}, \ldots, z_{n}$ ca-n problemă.
Fie $\omega$ o rădăcină primitivă de ordinul $p^{k}$ al unității. Observăm că fiecare $z_{i}$ poate fi scris sub forma $\omega^{k_{i}}$. Condiția $\sum_{j=1}^{n}z_{j}= 0$ devine $\sum_{j=0}^{p^{k}-1}a_{j}\omega^{j}= 0$, unde $\sum_{j=0}^{p^{k}-1}a_{j}= n$. Dacă notăm $g(X) = \sum_{j=0}^{p^{k}-1}a_{j}X^{j}\in \mathbb Z[X]$, atunci știm că $g \left( \omega \right) = 0$ și vrem să demonstrăm că $n = g(1)$ e divizibil cu $p$.
Dar, cum am zis mai înainte, $f_{k}$ e polinomul minimal al lui $\omega$, deci $g = f_{k}h$.
Ultimul pas: $g(1) = f_{k}(1) \cdot h(1) = p \cdot h(1)$, care-i divizibil cu $p$.
[/hide]

@[b]Vrajitoarea Andrei[/b]: Cred că ai dreptate... În orice caz, soluția lui [i]Svejk[/i] nu-i tocmai bună în forma în care e acuma. Problema e că Muirhead "merge" doar pentru $a,b,c > 0$, și aici avem $a,b,c \in \mathbb R$.

[quote="perfect_radio"]
@[b]Vrajitoarea Andrei[/b]: Cred că ai dreptate... În orice caz, soluția lui [i]Svejk[/i] nu-i tocmai bună în forma în care e acuma. Problema e că Muirhead "merge" doar pentru $a,b,c > 0$, și aici avem $a,b,c \in \mathbb R$.[/quote]

Da solutia mea e gresita.Am rezolvat inegalitatea cu impresia ca $a,b,c\geq 0$.Altfel nici nu puteam ridica la patrat ,dar sa mai aplic Muirhead. :blush:

Am  o solutie pentru problema a patra , dar nu stiu daca este corect.Corectati-ma daca gresesc :D
[hide="S4"] Din inegalitatea triunghiului :
\[\left | \sum_{j=1}^{n}a_{j}b_{j}\right |^{\lambda}\leq \left (\sum_{j=1}^{n}\left | a_{j}b_{j}\right | \right )^{\lambda}\]
Din Cauchy:
\[\left (\sum_{j=1}^{n}\left | a_{j}\right |^{\lambda}\right )\left (\sum_{j=1}^{n}\left | b_{j}\right |^{\lambda}\right )=\left (\sum_{j=1}^{n}(\left | a_{j}\right |^{\frac{\lambda}{2}})^{2}\right )\left (\sum_{j=1}^{n}(\left | b_{j}\right |^{\frac{\lambda}{2}})^{2}\right )\geq \left (\sum_{j=1}^{n}\left | a_{j}b_{j}\right |^{\frac{\lambda}{2}}\right )^{2}\]
Este suficient sa aratam ca
\[\left (\sum_{j=1}^{n}\left | a_{j}b_{j}\right |^{\frac{\lambda}{2}}\right )^{2}\geq \left (\sum_{j=1}^{n}\left | a_{j}b_{j}\right | \right )^{\lambda}\]

\[\Leftrightarrow \]

\[\sum_{j=1}^{n}\left | a_{j}b_{j}\right |^{\frac{\lambda}{2}}\geq \left (\sum_{j=1}^{n}\left | a_{j}b_{j}\right | \right )^{\frac{\lambda}{2}}\]
Daca notam $\frac{\lambda}{2}=\alpha$ ,unde $\alpha>0$, si $\left | a_{j}b_{j}\right |=x_{j}$ cu $x_{j}> 0$ atunci problema se rezuma la a arata ca :
\[\sum_{j=1}^{n}x_{j}^{\alpha}\geq\left( \sum_{j=1}^{n}x_{j}\right )^{\alpha}\]

\[\Leftrightarrow \]

\[\alpha \in (0,1) \]
Inegalitatea este echivalenta cu :
\[\sum_{j=1}^{n}\left ( \frac{x_{j}}{\sum_{j=1}^{n}x_{j}}\right )^{\alpha}\geq 1 \]
Observam ca pentru $\alpha=1$ avem conditia de egalitate ( adica $\lambda=2$).
Daca $\alpha > 1$ atunci
\[\sum_{j=1}^{n}\left ( \frac{x_{j}}{\sum_{j=1}^{n}x_{j}}\right )^{\alpha}< \sum_{j=1}^{n}\frac{x_{j}}{\sum_{j=1}^{n}x_{j}}=1 \]
Daca $\alpha < 1$ atunci
\[\sum_{j=1}^{n}\left ( \frac{x_{j}}{\sum_{j=1}^{n}x_{j}}\right )^{\alpha}> \sum_{j=1}^{n}\frac{x_{j}}{\sum_{j=1}^{n}x_{j}}=1 \]
Din cele demonstrate rezulta concluzia :D
[/hide]

Stie cineva ce s-a intamplat la concursul de saptamana trecuta (miercuri si joi)? Eram curios sa vad ce probleme s-au dat.

Gasesti subiectele si rezultatele aici :) http://www.arhimede.ro/first.html


number theory
greatest common divisor
number theory proposed

Prove that for each positive integer $ n > 2$ and $ 0<i<j<n$ we have $ \gcd(\binom{n}{i},\binom{n}{j}) > 1$

one of my friend gave me a solution based on this idea: for each $ i$ and $ j$, $ \binom{n}{i}$ and $ \binom{n}{j}$ are divisable by  greater than half of distinct prime divisors of $ n$. so we can directly deduce that their gcd is greater than 1. but now i see that's not true :o  so plz help me...(i've only proved the problem for these cases: $ \gcd(i,n)\equal{}1,\gcd(j,n)\equal{}d$ and if
$ \gcd(\binom{n}{i},\binom{n}{j},n)\equal{}1$ then for any prime $ p$ with $ p^{a}\parallel{}n$ then we should have at least one of this divisibility: $ p^{a}|i$ or $ p^{a}|j$)

A wonderful solution to this problem is to consider the following identity: (the solution is not mine)

$ \dbinom{n}{i} \dbinom{n\minus{}i}{j}\equal{} \dbinom{n}{j} \dbinom{n\minus{}j}{i}$

The rest is easy, try finishing the proof by yourself, if you want I can write the other part

[quote="lambruscokid"]A wonderful solution to this problem is to consider the following identity: (the solution is not mine)

$ \dbinom{n}{i} \dbinom{n \minus{} i}{j} \equal{} \dbinom{n}{j} \dbinom{n \minus{} j}{i}$

The rest is easy, try finishing the proof by yourself, if you want I can write the other part[/quote]
thx lambruscokid your solution was NICE :first:


inequalities
calculus
ratio
triangle inequality
inequalities unsolved

where a,b,c are sides of triangle
1) prove [tex]\frac {a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>1.4545[/tex]
2) find a tight minimum and maximum for the above expression

In general, if S=a_1+...+a_n for a_i in R, sum(a_i/(S-a_i))<=n/(n-1)

Proof: Let b_i = S - a_i implying that a_i = S - b_i and that sum(b_i)=(n-1)S.

Then sum(a_i/(S - a_i))=sum((S - b_i)/b_i)=Ssum(1/b_i) - n = 1/(n-1) sum(b_i) sum(1/b_i) - n >= n^2 / (n-1) - n = n/(n-1).  Equality holds iff a_1=...=a_n.  n=3 gives us the bound of interest: a/(b+c) + b/(c+a) + c/(a+b) >= 3/2 (interestingly, this does not even require a, b, and c to be sides of a triangle).

For an upper bound, note that the triangle inequality dictates that the semiperimeter of the a-b-c triangle exceeds each edge.  Consequently, a/(b+c) + b/(c+a) + c/(a+b) < (a+b+c)/s = 2.  We easily see that this upper bound is tight.  Setting a = b = x and c = y gives a/(b+c) + b/(c+a) + c/(a+b) = x/(x+y) + x/(x+y) + y/2x = 2x/(x+y) + y/2x.  Letting y approach 0, the above expression can approach arbitrarily close to 2.

We conclude that 3/2 <= a/(b+c) + b/(c+a) + c/(a+b) < 2 are the tightest bounds possible.

Please look again at the formula in the statement. It's another expression!

This problem is discussed [url=http://65.36.154.75/Forum/viewtopic.php?t=5506]here[/url].
The answer I found (using calculus, somewhat straightforward, but tedious) is

[tex]\displaystyle{\frac{{5\sqrt 5  - 8\sqrt 2  + 9}}{6}<\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}<\frac{{ - 5\sqrt 5  + 8\sqrt 2  + 9}}{6}}[/tex]

With equality when the following ratio is approached
[tex]\displaystyle{a:b:c = \frac{{1 + \sqrt 5  - 2\sqrt 2 }}{2}:\frac{{1 - \sqrt 5  + 2\sqrt 2 }}{2}:1}[/tex]
or
[tex]\displaystyle{a:b:c = \frac{{1 - \sqrt 5  + 2\sqrt 2 }}{2}:\frac{{1 + \sqrt 5  - 2\sqrt 2 }}{2}:1}[/tex]

I would be very impressed if someone could find a nice solution to this problem, considering how messy the answer is.


trigonometry AMC USA(J)MO USAMO FTW Mafia AIME

Okay, I am starting a new game, and in it, each user that joins starts off with 10,000, and they start off by taking half of my money away from me. Then the next user that posts gets to take away half of the money away from somebody that has already posted and then it just continues. You can post more than once, and in order to win as much money as possible, you can post as many times as you want, but not twice in a row. There has to be a user that submits inbetween all your posts. So this is an example, A, B, C, D, and E are random user names that play the game;

Wickedestjr: 10,000

A takes half of the money away from Wickedestjr;

A: 15,000
Wickedestjr: 5,000

B takes half of the money away from A;

B: 17,500
A: 7,500
Wickedestjr: 5,000

C takes half of the money away from B ;

C: 18,750
B: 8,750
A: 7,500
Wickedestjr: 5,000

D takes away half of the money away from B because he doesn't want to have the highest amount of money, because then he would be a bigger target;

and so on;

I start off with 10,000 of course;

Wickedestjr: 10,000

What did you do? You are only supposed to take half away from one person. We're going back to where we were before. Please say who you are taking half away from.

Wickedestjr: 12,500
splatyango: 7,500

Crazychinesemaniac: 16250
Splatyango: 7500
Wickedjstr: 6250

*robs CCM*

PI-Dimension: 18125
CCM:8125
Splatyango:7500
Wickedestjr:6250

CCM!!!

I'll be nice and take the small half

PI-Dimension: 18125
Yes_trig: 14062
Splatyango:7500
Wickedestjr:6250
CCM:4063

/me is angry at $ \pi$ for no partivular reason and robs him

PI-Dimension: 962.5
Yes_trig: 14062
Splatyango:9462.5
Wickedestjr:6250
CCM:4063

(was my math right?)

Yes_trig for completing the double rob.

PI-Dimension: 962.5
Yes_trig: 7031
Splatyango:9462.5
Wickedestjr:6250
CCM:11091

[quote="Wickedestjr"]What did you do? You are only supposed to take half away from one person. We're going back to where we were before. Please say who you are taking half away from.[/quote]

The posts must have been seconds apart since I received no warning that a new reply was posted, and the times say that they are one minute apart.

Oh well.

PI-Dimension: 481.25
Yes_trig: 7031
Splatyango:9462.5
Wickedestjr:6250
CCM:11091
1=2: 10481.25

I think that I am going to take half away from 1=2 to get;

Wickedestjr: 11,490.625
CCM: 11,091
Splatyango: 9462.5
Yes_trig: 7031
1=2: 5240.625
PI-Dimension: 481.25

Wickedestjr: 5,745.3125
CCM: 11,091
Splatyango: 9462.5
Yes_trig: 7031
1=2: 5240.625
PI-Dimension: 481.25
ernie: 15,745.3125

Hmmm, this has turned to nasty division xD.

Shoulda started at some power of 2.

Wickedestjr: 5,745
CCM: 18964
Splatyango: 9463
Yes_trig: 7031
1=2: 5241
PI-Dimension: 481
ernie: 7873

Robs Ernie for creating a largely unneccary decimal. And b/c he has more pts than wickedestjr. How about we round from now on...here I'll round everyones score

Wickedestjr: 5,745
CCM: 9482
Splatyango: 9463
Yes_trig: 7031
1=2: 5241
PI-Dimension: 481
ernie: 7873
Lotsofmath: 19482

chenhsi: 11509
1=2: 3477
BoesFX: 17884
cf249: 26210
CA Math: 3345
dragon96: 27880
ernie: 16320
james4l: 9993
Wickedestjr: 21429
Xantos C. Guin: 7388
AIME15: 14049
ak909090: 15917
mathelete: 16686
SuperMathBoy: 16012
mathking: 13006
jjx1: 32035

Hey jjx1, do you want to form an alliance with me and chenhsi?

Somebody, SOMEBODY, forgot me. Well ill rob Wickedestjr for revenge.

chenhsi: 11501
1=2: 3677
BoesFX: 17884
cf249: 37994
CA Math: 3695
dragon96: 13040
ernie: 16330
jhredsox93: 37392
james4l: 9993
Wickedestjr: 16372
Xantos C. Guin: 3722
AIME15: 14049
ak909090: 15487
mathelete: 13046
SuperMathBoy: 16532
mathking: 16006
jjx1: 24889
qntty: 19553
MathTwo: 16372


1.Let T be a linear operator on a finite-dimensional inner product space V
If T is an orthogonal projection, prove that ||T(x)||<= x
and IF T is a projection such that ||T(x)||<=x for all x,
prove that T is an orthonomal projection

2.Let T be a normal operator on a finite dimensional complex inner product space V.
a. for all x belong to V, if T^n(x)=0 for some n, then T is a zero transformation.
b.T is invertible if and only if all eigenvalue of T is non zero
c there exits a normal operator U on V such that U^2=T

[quote="TeeNKiT"]1.Let T be a linear operator on a finite-dimensional inner product space V
If T is an orthogonal projection, prove that ||T(x)||<= x
and IF T is a projection such that ||T(x)||<=x for all x,
prove that T is an orthonomal projection[/quote]

first should follow from definition
the second : let U and W be complementary spaces, we project onto W

suppose u and w are not perpendicular

then u+w has as projection w

now your condition implies

$0\leq ||u||^2+2 \Re (<u,w>)$

you can manipulate w now, by taking scalar (real) multiples, such that is contradicted

[quote] 2.Let T be a normal operator on a finite dimensional complex inner product space V.
a. for all x belong to V, if T^n(x)=0 for some n, then T is a zero transformation.
b.T is invertible if and only if all eigenvalue of T is non zero
c there exits a normal operator U on V such that U^2=T[/quote]

a) follows immediately from diagonalisation of normal operators
b) is true for all matrices over any field, just use definition!
3) follows again from diagonalisation


The length of the side of the smallest square which will enclose three non-overlapping discs, each of radius 1 unit is?

Look at the picture [url=http://hydra.nat.uni-magdeburg.de/packing/csq/csq3.html]here[/url].
- The triangle formed by the three centers of the circles is equilateral
(with three equal sides of length $2r$,
if $r$ is the radius of the triangle).
- The side of the square must be
$s=r+p+r$,
where $p$ is the horizontal projection of the side of the triangle,
$p=2r\cos\alpha$.
- Since the triangle is equilateral
(three equal angles of $60^\circ$),
and the diagonal of the square is a symmetry axis,
$\alpha+60^\circ+\alpha=90^\circ$,
$\alpha=15^\circ$.
- Finally, the side of the square is
$s=r(2+2\cos 15^\circ)$.

You asked for $s/r=2+2\cos 15^\circ$.
The web [url=http://www.packomania.com/]Packomania[/url] gives $r/s$ for $N$ circles. In this case $N=3$, $r/s=0.25433309\dots$.

I used to have a teacher that wanted us to make informatic programs to improve the results that appear at that web.

[I edited this post to make it clearer.]


inequalities
function
geometry
3D geometry
sphere
geometry solved

In the plane there is a worm of length 1. Prove that it can be always covered by means of half of a circular disk of diameter 1.

[i]Note.[/i] Under a "worm", we understand a continuous curve. The "half of a circular disk" is a semicircle including its boundary.

Lemma :
Let $P,Q,X$ be three points of the plane, and $M$ be the midpoint of the segment $[PQ]$.
Then $MX \leq \frac{1}{2}(PX+QX)$.

Proof;
Let $X'$ be the symmetric of $X$ with respect to $M$. Then, using the triangular inequality, we have :
$2MX = XX' \leq PX+PX' = PX+QX$ hence result.

Now, let's give a continuous curve $\gamma$ with length 1.
Adding at most two points, we may assume that $\gamma$ is closed, in a topological sense (that is it contains its endpoints : we may consider $\gamma$ as the image of $[0;1]$ in the plane, by a continuous function $f$. It is not assumed that $\gamma$ is closed in the geometrical sense, that is $f(0)=f(1)$) which clearly does not change the length of $\gamma$.

- If $\gamma$ is closed (in the geometrical sense, see above) :
Let $d$ be any line, which does not meet $\gamma$ (which is possible, otherwise $\gamma$ would not have a finite length). Since the distance to $d$ is a continuous function, there exists a point $R \in \gamma$ for which the distance to $d$ is minimal (among the distance between a point of $\gamma$ and $d$). Let $d_{R}$ be the line parallel to $d$ through $R$. Then $\gamma$ lies entirely in a closed half-plane with boundary $d_{R}$.
Let $X$ be any point on $\gamma$. We then have $XR \leq \frac{1}{2}$ because the lengths of the two arcs of $\gamma$ connecting $X$ to $R$ cannot be both with length greater than $\frac{1}{2}$.
Thus, an half-disk with center $R$ and diameter 1 covers $\gamma$.

- If $\gamma$ is an open curve (in the geometrical sense) with endpoints $A,B$ :
We procced as above but with an intial line $d$ which is parallel to the line $AB$. The point $R$ is chosen as above.
Let $A',B'$ be the symmetrics of $A,B$ with respect to $d_{R}$ respectiveley.
Then $AA'B'B$ is a rectangle. Let $O$ be its center. Clearly $O \in d_{R}$.
Let $X \in \gamma$.
If $X$ belongs to the arc $AR$ of $\gamma$ : then using the lemma, we have
$OX \leq \frac{1}{2}(AX+B'X) \leq \frac{1}{2}(AX+B'R+RX) = \frac{1}{2}(AX+BR+RX) \leq \frac{1}{2}$ since $AX+XR+RB$ is not greater than the length of $\gamma$ which connects $A,X,R,B$ in that order.
We have a similar inequality if $X$ belongs to the arc $BR$.
Then, $\gamma$ is covered by an half-disk with center $O$ and diameter 1.

Pierre.

Two remarks :

1) The result is optimal by considering a curve $\gamma$ which is a line segment with length 1.

2) The proof above can be used words for words in higher dimensions, just replacing 'line' by 'hyperplane' to get the following result :
In $R^n$ any continuous curve with length $L$ can b covered by an $n$-dimensional half-sphere with diameter $L$.

Pierre.


function
modular arithmetic
algebra unsolved
algebra

Find all functions $f : \mathbb Z \to \mathbb Z$ such that $f(f(n))+f(n+1) = 2n+3$ and $f(n) \geq n-1$, $ \forall n \in \mathbb Z$.

[quote="Svejk"]If $f(n)\geq n+2$ then we have $f(f(n))\geq f(n)+2\geq n+4$ and $f(n+1)\geq n+3 \Rightarrow 2n+3 \geq 2n+7$ , contradiction $\Rightarrow f(n)\leq n+1$ So it's obvious that $f(n)=n+1$[/quote]

read more carefully : $f(n)=n+1 \Rightarrow f(f(n))=f(n+1)=n+2$ and $f(n+1)=n+2$

$\Rightarrow f(f(n))+f(n+1)=2n+4$ not $2n+3$ :!:

[quote="Svejk"]If $f(n)\geq n+2$ then we have $f(f(n))\geq f(n)+2\geq n+4$ and $f(n+1)\geq n+3 \Rightarrow 2n+3 \geq 2n+7$ , contradiction $\Rightarrow f(n)\leq n+1$ So it's obvious that $f(n)=n+1$[/quote]
[quote="Svejk"]:oops: It means that it doesn't exist such function ? because if $f(n)\geq n-1$ and $f(n)\leq n+1$ $\Rightarrow$ f(n) $\in$ {n-1,n,n+1} ,and neither of this solutions are good ??  :?:[/quote]

if you mean: there exist $n$ such that $f(n)\geq n+2$ we can have $fof(n)\leq f(n)+2$

if you mean: $\forall n$ $f(n)\geq n+2$ $\Rightarrow$ $f(n+1)+fof(n)=2n+7\ned 2n+3$ contradiction
then you can't say that $\forall n;f(n)\in \{n-1,n,n+1\}$
but you can say that there existe $n$ such that $f(n)\in \{n-1,n,n+1\}$

From $f(n)\ge n-1 \ \forall n$ we get $f(n)\le n+4 \ \forall n$.
It had many solutions. For example periodic for $g(n)=f(n)-n$ solutions $f(n)=n+4$ if $n=k(mod 5)$, else $f(n)=n-1$.

Long and tedious case work give exactly 3 solutions $f_{1}$,$f_{2}$,$f_{3}$ with $f_{k}(n)=\{\begin{array}{cc}n+4 & \text{ for }n\equiv k \pmod{3}\\ n-1 & \text{otherwise}\end{array}$.

Say $f(n)\leq n$ for all $n\in \mathbb{Z}$.
Then $2n+3=f(f(n))+f(n+1)\leq f(n)+n+1\leq 2n+1$. Contradiction!
So $f(n)\geq n+1$ for some $n\in \mathbb{Z}$.

Case 1: $f(n)=n+1$ for some $n\in \mathbb{Z}$
Then $2n+3=f(f(n))+f(n+1)=2f(n+1)$ is even. Contradiction!

Case 2: $f(n)=n$ and $f(n+1)=n+1$ for some $n\in \mathbb{Z}$
Then $2n+3=f(f(n))+f(n+1)=n+(n+1)$. Contradiction!

Case 3: $f(n)=n+2$ for some $n\in \mathbb{Z}$
Then $2n+3=f(n+2)+f(n+1)$ with $f(n+1)\geq n$ and $f(n+2)\geq n+1$.
So $(f(n+1),f(n+2))=(n,n+3)$ or $(n+1,n+2)$ or $(n+2,n+1)$.
But then we are back to Case 1 or 2. Contradiction!

Case 4: $f(n)=n+3$ for some $n\in \mathbb{Z}$
Then $2n+3=f(n+3)+f(n+1)$ with $f(n+1)\geq n$ and $f(n+3)\geq n+2$.
So $(f(n+1),f(n+3))=(n,n+3)$ or $(n+1,n+2)$.
But then we have $f(n+2)=2(n+1)+3-f(f(n+1))=n+2$ or $n+4$
and we are back in Case 2 or 3. Contradiction!

So we have $f(n)\geq n+4$ for some $n\in \mathbb{Z}$.
But then $2n+3=f(f(n))+f(n+1)\geq (f(n)-1)+n\geq n+4-1+n=2n+3$ and the equality implies $(f(n),f(n+1))=(n+4,n)$.
And then $f(n+2)=2(n+1)+3-f(f(n+1))=n+1$ and $f(n+3)=2(n+2)+3-f(f(n+2))=n+7$
And now repeating this argument we get for all $k\in \mathbb{Z},k\geq 0$:
$f(n+3k)=(n+3k)+4$
$f(n+3k+1)=(n+3k+1)-1$
$f(n+3k+2)=(n+3k+2)-1$
Similarly we have
$2(n-1)+3=f(f(n-1))+f(n)\geq f(n-1)-1+f(n)\geq (n-1)-1-1+f(n)=2(n-1)+3$
and this equality implies $f(n-1)=n-2$ and $f(n-2)=f(f(n-1))=f(n-1)-1=n-3$.
And then $f(n-3)=f(f(n-2))=2(n-2)+3-f(n-1)=n+1$.
Again repeating this argument we get for all $k\in \mathbb{Z},k\geq 0$:
$f(n-3k)=(n-3k)+4$
$f(n-3k-1)=(n-3k-1)-1$
$f(n-3k-2)=(n-3k-2)-1$
So $f=f_{1},f_{2}$ or $f_{3}$, depending on the residue of $n \pmod{3}$.


function
parameterization
topology
geometry
3D geometry
sphere
geometric transformation

Suppose $f$ is a smooth function $\mathbb{R}^{2}\to \mathbb{R}$ and consider the level set $f(x_{1},x_{2})=0$. Often it will be apparent from drawing an accurate picture (with computer) that the level set defines one or several smooth curves. We know by the implicit function theorem that if $\nabla f \neq 0$ this is true locally on the curve. But can one somehow prove that all of the level set consists of some curves with smooth parametrization?

As an example , let $f(x,y) =ye^{-y}-x^{-1}e^{x}$ for $x,y > 0$.  I know from something else that it defines a single closed smooth curve in the first quadrant, but I wouldn't have any clue as how to directly prove it.

The implicit function theorem tells you that once all critical points of $f$ are removed, each level set is a union of one-dimensional differentiable manifolds. You want to prove that each such manifold is diffeomorphic to $\mathbb R$. This belongs more to differential topology than to calculus.

A vague idea of a proof (which is probably tedious to write down): in one dimension, charts with nonempty overlap can be glued together.

To show that the level set $\{f=0\}$ is a smooth curve or a union of several smooth curves, it suffices to show that the equations $f=0$ and $\nabla f=0$ cannot hold simultaneously. To determine the number of components is a trickier business. For smooth functions on the sphere you can get some information from investigation of critical points of $f$. The idea is that if you look at the level sets $f=c$ and start to change $c$ from $-\infty$ to $+\infty$ in a continuous manner, each time when you pass through a non-degenerated local minimum, one component is born and each time you pass through a non-degenerated local maximum, one component dies. The problem is that when you pass through a non-degenerate saddle point, you may have either one component splitting into two or two components merging with each other and there is no easy way to tell which one is the case. If you have degenerated critical points, the picture gets even more complicated. On the plane this idea works too if the function has some reasonable behavior near infinity (say, $f(x)\to+\infty$ as $x\to\infty$)

In your particular example, $f(x,y)=0$ defines the empty set because $ye^{-y}<1<x^{-1}e^{x}$, so, probably, you meant something else :?.

[quote="fedja"]To show that the level set $\{f=0\}$ is a smooth curve or a union of several smooth curves, it suffices to show that the equations $f=0$ and $\nabla f=0$ cannot hold simultaneously[/quote]
Why is this? I only know the implicit function theorem, and I don't see why this is true.

Being a smooth curve on the plane or on the sphere is a local property: if $M\subset \mathbb R^{2}$ and for every point $p$ on the plane there is a neighborhood $U\ni p$ such that $M\cap U$ is a single smooth curve (or the empty set), then $M$ is a union of smooth curves. I admit that it is not obvious but it is true. If you want a proof, we should first decide exactly what definition of a smooth curve you would like to use.

It seems to me that Kalle wanted a smooth parametrization of a level set, i.e. a diffeomorphism to $\mathbb R^{1}$ or $S^{1}$. I never saw a proof that these are the only $1$-dimensional manifolds (w/o boundary). Tried to look it up in one book and got "it can be shown..."

OK, let's follow mlok's idea and define a union of smooth curves on the plane as a collection of (images of) smooth mappings $f_{i}: \mathbb R\to \mathbb R^{2}$ such that each mapping is either periodic and 1 to 1 on the period, or is 1 to 1 on the entire line and tends to infinity as $t\to\infty$, and such that for every $R>0$ there exists $\delta>0$ such that if $x\in f_{i}(\mathbb R)$, $y\in f_{j}(\mathbb R)$, $|x|,|y|\le R$ and $i\ne j$, then $|x-y|\ge \delta$.

One possible idea that allows to represent the level set of $f$ as a union of smooth curves in the above sense is to consider the differential equation $\frac d{dt}X=U\nabla f(X)$ where $U$ is the counterclockwise rotation by $90$ degrees. If you start on the set $\{f=0\}$, you'll stay on that set forever. Each two trajectories are either disjoint, or coinside. The standard theorems from differential equations allow to conclude that each trajectory is a smooth curve in the above sense. The implicit function theorem (with estimates) allows to conclude that if two points $x,y\in \{f=0\}$ are not too far from the origin and close enough to each other, then they are on the same trajectory. That's besically all. Try to see if you can follow this scheme.


[code]a b c d
e f g h
i j k l
m n o p[/code]

Is the determinant of this matrix:

$ a (fkp \minus{} flo \minus{} gjp \plus{} gln \plus{} hjo \minus{} hkn) \minus{} b (ekp \plus{} glm \plus{} hio \minus{} gip \minus{} hkm \minus{} elo) \plus{} c (ejp \plus{} flm \plus{} hip \minus{} eln \minus{} fip \minus{} hjm) \minus{} d (ejo \plus{} fkm \plus{} gin \minus{} gjm \minus{} fio \minus{} enk)$?

No it is not. The first $ hjp$ should be $ hjo$.

That was a typo.  But is the rest of it right???

What's the point of this? Just apply [url=http://en.wikipedia.org/wiki/Laplace_expansion]Laplace's expansion technique[/url].

Considering the fact that I didn't know that, there is no point anymore.

In my math class, the teacher simply taught the mnemonic way--"diagonals". He said the Laplace expansion (called expansion by minors in my textbook) was too difficult to remember.  :|

On a related note, the aforementioned technique can be improved by using this powerful tool called [url=http://en.wikipedia.org/wiki/Gaussian_elimination]Gaussian elimination[/url].

BTW, my Algebra 2 class never teach us anything about calculating determinants other than the 2x2...

Well, in my class, we started by learning the determinant of a 2 X 2 (easy enough). Then you can find the determinant of a 3 X 3 in terms of 2 X 2s. Then you can repeat to get 4 X 4s, etc.

[quote="Yongyi781"]The technique described is in fact very common - it is taught in almost all Algebra II classes as the method of obtaining determinants of 3x3 and greater matrices. They were just not called Laplace's expansion technique.[/quote]

I'm learning this by myself because I'm gonna miss a week of Alg II.  :P


geometry
cyclic quadrilateral
geometry proposed

Let $ ABC $ be a triangle , $ \angle B $ is not 90 degree . $ M \in AB ; N \in BC $ . $ AN $ intersects $ CM $ at $ I $ . $ H $ is the orthocenter of $ \triangle ABC $ . $ H' $ is the orthocenter of $ \triangle BMN $  Prove that : $ AMNC $ is a cyclic quadrilateral  $ \iff $ $ H , I , H' $ are collinear .  ;)

Well, at least one direction (quadrilateral AMNC is cyclic $\Longrightarrow$ H, I, H' are collinear) is equivalent to the problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=29893 . The other direction can be proven by just reversing the arguments.

  darij


inequalities
function
integration
Functional Analysis
real analysis
real analysis unsolved

Show that for real-valued $ f\notin L^{p}(E)$, there exists a function $ g\in L^{q}(E)$ ($ p$ and $ q$ are Holder conjugates so their reciprocals sum to 1) such that $ fg\notin L^{1}(E)$.

The suggestion is to construct $ g$ of the form $ \sum a_{k}g_{k}$ for appropriate $ a_{k}$ and $ g_{k}$, with $ g_{k}$ satisfying $ \int_{E}fg_{k}\rightarrow\infty$.

take $ f$ positive to make the things a little bit easier, but tha's the same for general $ f$. Since $ f$ is not in $ L^p$, you can find a sequence of $ g_k \in L^q$ such that $ \|g_k\|_{L^q}\equal{}1$, positive, and $ \int f \cdot g_k \geq k$ (Hahn-Banach). Now take your favourite positive sequence $ a_k$ that satisfies $ \sum a_k < \infty$ but $ \sum k a_k \equal{} \plus{} \infty$ and put $ g\equal{}\sum a_k g_k$.

My favorite sequence that does that?  I can't really think of too many off the top of my head other than $ a_{k}\equal{}\frac{1}{k}$ if $ k$ is a power of 2 and 0 otherwise  :P .

We are not making any specifications on $ f$ and $ g_{k}$, right?  Other than the fact that $ \left\|g_{k}\right\|_{q}\equal{}1$ and $ g_{k}\in L^{q}$ and $ f\notin L^{p}$?

[quote="JRav"]My favorite sequence that does that?  I can't really think of too many off the top of my head other than $ a_{k} \equal{} \frac {1}{k}$ if $ k$ is a power of 2 and 0 otherwise  :P .
[/quote]
what about $ a_k \equal{} \frac{1}{k^{\alpha}}$ for $ < \alpha \leq 2$, which is a little bit more easier isn't it ?

sorry to bump this up, but I would like to know how we can get the existence of sequence of $g_k\in L^q(E)$ such that

$ \int f\cdot g_{k}\geq k $

(How is Hahn-Banach playing a role here?)

[quote="shyong"] how can we get the existence of a sequence $g_k\in L^q(E)$ such that $\int f\cdot g_{k}\geq k$ and $\|g_k\|_{L^q}=1$ ? [/quote]

You might know that $\|f\|_{L^p} = \sup \; \int f(x) \, g(x) \, dx$ where the supremum is over all the functions satisfying $\|g\|_{L^q}=1$ ? Does it help ?


calculus
integration
function
trigonometry
real analysis
real analysis theorems

This problem - on Integration By Parts – in the attached figure, may appear trivial ; however, the two final results - by taking alternatively, one of the fns as the "first fn", and the other as the "second fn" - do not tally !
Would appreciate if someone could point out [color=red]where my mistake is.[/color]

Thanks
...

[quote="t0rajir0u"]$ \text{CosIntegral}$ is an even function.  The difference of the two terms is zero.[/quote]
The thought that something like this should first be verified, did strike me too.
So I had tested out in Mca - but the results are NOT encouraging !, [color=blue]else Mca itself would have probably given the diff as 0 ![/color]

Pl consider these Mca results :

[b]Let z = 0.25 + j 0.75[/b] [You could use "i" or "j" as we Electrical Engrs like to.]

CosIntegral [+z] = 0.468333 [color=red][b]+[/b][/color] j 1.15134
CosIntegral [-z]  = 0.468333 [color=red][b]-[/b][/color] j 1.99026

// [b]So, there is a [color=red]diff of j Pi[/color][/b]

**********

However, SinIntegral [-z]  =  -SinIntegral [+z]  =  -(0.273088 + j 0.765602)

So, once again, therefore :

[b]Would appreciate if someone could point out [color=red]where my mistake is[/color][/b] - in my calculations Fig IBP-Ver2_AOPS.JPG ?

Thanks
...

Hmm.  I based my previous assertion on the [url=http://mathworld.wolfram.com/CosineIntegral.html]identity[/url]

$ \text{Ci}(x) \equal{} \frac {\text{Ei}(ix) \plus{} \text{Ei}( \minus{} ix)}{2}$

where $ \text{Ei}(z)$ is the [url=http://mathworld.wolfram.com/ExponentialIntegral.html]exponential integral[/url].  It would appear that these functions are multivalued, which isn't entirely surprising; the integral definitions they're based off of have singularities at the origin.  I'm reasonably certain this is just a complication in the definition of the functions alone, so you shouldn't worry about it.

[quote="t0rajir0u"]... [b]so you shouldn't worry about it....[/b][/quote]
I beg to differ ; in Maths, we can't leave things like this ! - when things don't add up (as in this case, when the two IBP methods yield diff results), we definitely need to "worry", and get to the bottom of it. That is why, to help all the readers, I have sent the file showing all the steps that I followed.

[quote="t0rajir0u"]... I'm reasonably certain this is [b]just a complication in the definition[/b] of the functions ...[/quote]
So, what is that "complication in the definition", here ?

***********

Finally, there ought to be a reason - either a mistake in my computation steps, or, as you say, "a complication in the definition of the functions".
In Maths, we must know whatever that is.

Can other readers too pl look into this ?

Thanks.
...

Upon closer examination, it's much more likely that the problem with this computation is that you didn't specify constants correctly.  When you computed the integration by parts, you plugged in indefinite integrals without checking if everything satisfied the same initial conditions.

[quote="t0rajir0u"]Upon closer examination, it's much more likely ... When you computed the integration by parts, you plugged in indefinite integrals [b]without checking if everything satisfied the [u]same initial conditions ??[/u][/b].[/quote]
Could you pl be more specific, and let me know the correct solution.
[b]Exactly, what initial conditions do I need to apply, at what step ?[/b]

It may be noted that for [b]ALL the 3 integrals : IDirect[/b], as well as [b]BOTH IBPA and IBPB[/b] (just a matter of juxtaposing Fn 1 and Fn 2), I have only compared the results of the [b]Indefinite Integral[/b]. [b]So, should all the 3 results ought not to match ?[/b]

To "t0rajir0u" and all readers,
This correction will not only enable me to understand the nuances, but also prevent me from making such errors in future.

Can we have more inputs on this from other readers too ? And expert moderators too pl ?

Thanks.
...

[quote="Sundar_AOPS"][b]Exactly, what initial conditions do I need to apply, at what step ?[/b][/quote]
This is your responsibility, not mine.  I've told you the step that's wrong (the definitions of IBPA, IBPB in terms of indefinite integrals alone); inspect it!

It might help if you see how this discrepancy is possible for a simpler case.  Suppose I wanted to compute the indefinite integral of $ \sin x \cos x$.  I could do so in two ways:

$ \int \sin x \cos x \, dx \equal{} \sin^2 x \minus{} \int \sin x \cos x \, dx \implies$
$ \int \sin x \cos x \, dx \equal{} \frac{1}{2} \sin^2 x$

or I could compute

$ \int \cos x \sin x \, dx \equal{} \minus{}\cos^2 x \minus{} \int \cos x \sin x \, dx \implies$
$ \int \cos x \sin x \, dx \equal{} \minus{} \frac{1}{2} \cos^2 x$.

The two answers differ by a constant $ 1$ here; why?  I have been sloppy with integration by parts!  Properly stated, it only applies to definite integrals; when applied to indefinite integrals, you need to add constants appropriately.  And since you also need to add constants when evaluating $ v \equal{} \int dv$, instead of being off by a constant you've ended up being off by an entire function.


geometry
geometric transformation
algebra solved
algebra

Kolga-Jaani Mathematical Olympiad 2003, problem nr. 1:
There is a set of integers S = {2;3;4...;1110;1111}. Find the greatest natural number n, for which we can choose n elements from the set S so, that their product is a^n for some integer a, which isn't a power of any other integer and belongs to the set S.

It's a nice problem. To "guest", just to be sure I get it (I should probably work my english instead of maths :( )   :

2^2 * (3*2^4) * (3^3*2^8) * (2^10) = (3*2^6)^4 works for n=4 (there are 4 numbers, 3*2^6 <= 1111,  3*2^6 is different from 2^2, (3*2^4), (3^3*2^8), (2^10), and 3*2^6 is not a power). Ok ? that's all the conditions ?

yes, you got it right.
Maybe that wasn't clear from my post, but all these n numbers should be different and also "a" should differ from these n numbers. I apologize for my translation to English as well.

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I have four pairs of socks to be hung on a straight clothesline. The socks themselves are identical but each pair of socks come in different colors. How many different arrangements are possible if no sock can be next to its mate?

This is good casework because but I think the corner socks are the key...

[hide="Solution and Generalization"]Proceed by recursion.

Let $a_{n}=$the number of arrangements for $n$ pairs of socks.  We will assume for now that a pair of socks consists of two distinguishable socks.

Letting X and x denote a pair of socks, all possible arrangements when $n=2$ are:
ABab
AbaB
aBAb
abAB
BAba
BabA
bABa
baBA
So $a_{2}=8$.

If another pair of socks is added to form $n$ pairs of socks, there are $2n-1$ ways to place the first sock among the others and $2n-2$ ways to place the second one.  So $a_{n+1}=a_{n}(2n-1)(2n-2)$.

So $a_{n}=8((4)(5)(6)(7)\ldots(2n-1))$
$a_{n}=8\cdot\frac{(2n-1)!}{3!}$
$a_{n}=\frac{4(2n-1)!}{3}$

However, each pair of socks consists of two nondistinguishable socks.  So we must divide by $2^{n}$, yielding:
$a_{n}=\frac{(2n-1)!}{3\cdot 2^{n-2}}$

The problem asks for $a_{4}$.  So the answer is $\frac{7!}{3\cdot 2^{2}}$ or $\boxed{420}$.[/hide]

[quote="lingomaniac88"]

If another pair of socks is added to $n$ pairs of socks, there are $2n-1$ ways to place the first sock among the others and $2n-2$ ways to place the second one. [/quote]

why 2n-1 ways?

lingomaniac's answer is based on the assumption that in each pair of socks, the two socks are distinguishable, and it doesn't say anywhere in the problem that theu are
EDIT: nevermind but i get 180
Edit:nevermind again, arithmetic error

[quote="lingomaniac88"]Sorry.  I meant "to form $n$ pairs of socks."

And here's why:

We have $2n-2$ socks.  We can place another sock between two socks or at the edges.  That leaves $2n-1$ places.  And then the second sock can go into any of the other $2n-2$ spots.  Try it.[/quote]
[quote="lingomaniac88"]

If another pair of socks is added to form $n$ pairs of socks, there are $2n-1$ ways to place the first sock among the others and $2n-2$ ways to place the second one.  So $a_{n+1}=a_{n}(2n-1)(2n-2)$.[/quote]

I still dont get it.  If its to form n pairs of socks shouldnt the formula be
$a_{n}=a_{n-1}(2n-1)(2n-2)$ not $a_{n+1}=a_{n}(2n-1)(2n-2)$
Furthermore I think theres a flaw in this way

Take in case:

BaAb alright we put C at the end BaAbC
so your telling me we only have 4 spots:
_BaABC
B_aABC
Ba_ABC
BaA_BC to place the little c.

Thats not correct

we can actually place it like this
BaABcC, although we have the two Cs next to eachother right now, once we add the D we can add a D inbetween the c and C satisfying the condition.

[quote="pda"]The socks themselves are [size=200]identical[/size]...[/quote]
:)

Then it's:
\[\frac{8!}{(2!)^{4}}-\dbinom{4}{1}\left( \frac{7!}{(2!)^{3}}\right)+\dbinom{4}{2}\left( \frac{6!}{(2!)^{2}}\right)-\dbinom{4}{3}\left( \frac{ 5!}{2!}\right)+4!=\boxed{864}\](edited: corrected method.)

[quote="da ban man"][quote="cincodemayo5590"][quote="pda"]The socks themselves are [size=200]identical[/size]...[/quote]
:)

Then it's just $\frac{8!}{(2!)^{4}}-4!=\boxed{2496}$.[/quote]

why minus $4!$?  you dont count cases like RRBWGBWG where only the reds are touching eachother..[/quote]
Oops, right. Ha. OK, I'll go back to it. I think we have to PIE it.  :)

The sequence in the OEIS (which doesn't have anything smarter than the PIE argument cincodemayo used) is http://www.research.att.com/~njas/sequences/A114938

[quote="cincodemayo5590"][quote="pda"]The socks themselves are [size=200]identical[/size]...[/quote]
:)

Then it's:
\[ \frac{8!}{(2!)^{4}}-\dbinom{4}{1}\left( \frac{7!}{(2!)^{3}}\right)+\dbinom{4}{2}\left( \frac{6!}{(2!)^{2}}\right)-\dbinom{4}{3}\left( \frac{ 5!}{2!}\right)+4!=\boxed{864}\]
(edited: corrected method.)[/quote]

I"m confused. How did you get that answer? I don't get what or how you are using PIE on.

Hes finding how many have pairs together (adding for 1, subtracting for 2, adding for 3, subtracting for 4).  Then he takes that amount and subtracts it from the total possiblities.


Euler
geometry
circumcircle
geometric transformation
reflection
power of a point
radical axis

[color=darkblue][b]Let $ ABC$ be an acute triangle with $ (O)$ its circumcircle and $ M_b,M_c$ respectively be the midpoints of $ AC,AB$. Let $ H$ be the orthocenter of $ \triangle AM_bM_c$. Denote $ A_1$ be an arbitrary point on the $ 9 \minus{}$ point circle wrt $ \triangle ABC$. Through $ A_1$, let $ d$ be the line which is perpendicular to $ OA_1$, $ d$ intersects $ (O)$ at $ B_1,C_1$. Prove that $ H$ also lies on the $ 9 \minus{}$ point circle of $ \triangle AB_1C_1$.[/b][/color]

Best regard,
[b][i]mathVNpro[/i][/b]

Nice result dear friend   :)
[b]My solution:[/b]
Denote $ K,H'$ the orthocenters of triangle $ ABC, AB_1C_1$, respectively, $ E,E'$ the centers of Nine-point circles of triangles $ ABC, AB_1C_1$, respectively, $ M_a$ the midpoint of $ BC$.
$ H\in (E') \Leftrightarrow EE'\perp HA_1$
Since $ EE'$ is the midline of triangle $ KOH'$ then $ EE'//H'K$. But $ HM_a$ is a diameter of $ (E)$ thus $ HA_1\perp M_aA_1$.
So we need to show that $ M_aA_1//KH'$.
Let $ S$ be the reflection of $ A$ wrt $ O$ then $ A_1, M_a$ is the midpoints of $ H'S, KS$ therefore $ M_aA_1//KH'$.
We are done.

Very nice solution, dear [b]Linh[/b]. I will present mine.

[i]Proof[/i]

Let $ O_1,O_2$ repsectively be the $ 9 \minus{}$ point circle center of $ \triangle ABC$ and $ \triangle AB_1C_1$ and $ N_c,N_b,M_a$ be the midpoint of $ AB_1,AC_1,BC$. Denote $ H_c,H_b,P_c,P_b$ respectively by the projections of $ C,B,C_1,B_1$ onto $ AB,AC,$ $ AB_1,AC_1$. Then $ H_bH_c\|P_cP_b$ (since both of them $ \perp AO$). But since $ H,O_1,M_a$ are collinear and $ HM_a\perp H_bH_c$ then $ O_1M_a\equiv O_1H\perp H_bH_c$. Similarly, $ O_2A_1\perp P_bP_c$. Therefore, $ O_1H\| O_2A_1$ $ (1)$. In the other hand, it is easy to prove that $ N_c,N_b\in ([AO])$ and $ R_{([AO])} \equal{} R_1$, where $ R_i$ is the radii of $ (O_i)$ $ \Longrightarrow$ $ R_1 \equal{} R_2$, which implies $ O_1H \equal{} O_2A_1$. Combine with $ (1)$, we conclude $ O_1HO_2A_1$ are parallelogram. Therefore, $ O_2H \equal{} O_1A_1 \equal{} R_1 \equal{} R_2$ $ \Longrightarrow$ $ H\in (N_cN_bA_1)$, which leads to the result of the problem. Our proof is completed then. $ \square$

Best regard
[b][i]mathVNpro[/i][/b]

Let $ H_1,H'_1$ be the orthocenter of $ \triangle ABC$ and $ \triangle AB_1C_1$, $ E,E'$ be the centre of the Euler's nine-point cirlce of $ \triangle ABC$ and $ \triangle AB_1C_1$. Denote $ H'\equiv AH_1 \cap (E')$. We have $ H,H'$ are the midpoint of $ AH_1,AH'_1$, respectively and hence $ \overrightarrow{HH'} \equal{} \overrightarrow{EE'} \equal{} \frac {1}{2}\overrightarrow{H_1H'_1} \equal{} \overrightarrow{M_aA_1}$, where $ M_a$ is the midpoint of $ BC$. On the other hand, we have a well-known fact that $ M_aH$ is the diameter of $ (E)$, so $ A_1H \perp A_1M_a$. This yields $ A_1H \perp EE'$. But $ A_1 \in (E) \cap (E')$ (because $ A_1$ is the midpoint of $ B_1C_1$) so $ A_1H$ is the radical axis of $ (E)$ and $ (E')$. However, $ H\in (E)$ so $ H\in (E')$, QED.

My solution:

Let $ R $ be the radius of $ \odot (O) $ .
Let $ H' $ be the orthocenter of $ \triangle ABC $ .
Let $ N $ be the nine point center of $ \triangle AB_1C_1 $ .
Let $ O' $ be the reflection of $ O $ with respect to $ d $ .

Since the nine point circle is the pedal circle of Macbeath inellipse,
so $ d $ is tangent to Macbeath inellipse. i.e. $ O' $ lies on $ \odot (H',R) $

Since $ N $ is the midpoint of $ AO' $ ,
so $ N $ lies on $ \odot (H, R/2) $ (i.e. $ NH $ = $ R/2 $) ,
hence we get $ H $ lie on the nine point circle of $ \triangle AB_1C_1 $ .

Q.E.D


outside the triangle ABC construct rectangles ABLK,BCNM,ACQP.Prove that the perpendiculars dropped from A to KQ amd the similar lines intersect

http://www.mathlinks.ro/Forum/viewtopic.php?t=22321 post #4 Lemma 6


It took Jim 1$ \frac{1}{3}$ hours to walk $ \frac{4}{9}$ of the way home. At the same rate, how many hours would it take Jim to walk all the way home?

Jim has to walk 9/4 of the distance that he walked in 4/3 hours. Therefore it will take him (9*4)/(4*3)=3 hours to get all of the way home.