china will win IMO for next 5 years at least. they r committed to excellence and i am sure that she is not gonna give its place to anyone.
   and well iran might get 2nd this year.

Iran will be first iff there isn't China .Like 1999

Well, what i know is that, there'll be no chance for any country to beat China in 2007
a simple reason, a guy named Zhen Xu from Shanghai will sure be there. He's so brilliant. plus there's at least three other guys from ShenZhen city. They will all get a mark around 40. and i'm pretty sure that if you let Xu Zhen do IMO problems, chance that he don't get them all correct is less than 1%

[quote="Kalimdor"]Well, what i know is that, there'll be no chance for any country to beat China in 2007
a simple reason, a guy named Zhen Xu from Shanghai will ne there. [/quote]
Why are you so sure with that? The TST hasn't been held yet. And as we all know, everything can in the team selecting test.

[quote="Mamat"]Does anyone know how many students participate in mathematical olympiads in China  :?: ?[/quote]
Uncoutable... ...

u mean infinity??? or limit approaches to inifinity......hehe.....is india have the same situation????

[quote="Loser"]Hmmm, I wonder who's this guy  Zhen Xu and how can you be so sure that he will take 42.
I mean, probably he's very very good but IMO problems often are very tough and it's quite hard to obtain 42 (I myself will be glad if I score points enough for a medal so... maybe I ain't the right person to  comment it :) )[/quote]

When students can get a perfect score at CMO, you know IMO means less to them. I believe many guys in China has the ability to get a very high score or even a perfect score, but I don't really think you can guarantee that a guy can enter the China team... I think, although it maybe exaggerated, that entering China team is harder than getting a perfect score at IMO.

The situation in India is totally different. We're not encouraged to put Olympiads first on our list of priorities. In addition, the standard as well as the number of people who compete  in India is far less than that of China. I sure hope that changes though...

Well, Soarer, I agree with you because I've seen problems from CMO and they're really tough.
But, on the other hand, some of the members of the Chinese team at the IMO can't get full scores and in 2003 our small country was even in front of Cnina :)

[quote="pavel kozlov"]Where I can find problem from 2006 China Olympiad?[/quote]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=69697]here[/url] :)

[quote="Arne"][quote="fredbel6"]There were also allegations, never proven, about Cyprus IMO2002.
[/quote]

Wasn't that in 2001, when they all solved problem 2 in the same way?[/quote]

Well, that would be equal to cheating (because I never saw a problem solved in an exactly way here in our olympiad, and we're a lot of participants) but, how could they possible do it?? I mean, they did not sit together, or maybe they knew the problem before the IMO started??

Coincidence happens.

The accusation is of course that their team leader, who is not with the contestants before the exams, and who has helped choosing questions, phoned them the question along with the solution that came along with it.

Another quite likely scenario is that they have seen a similar problem, and all went straight ahead with the same technique.

[quote="darktreb"]Don't the coaches get the IMO long list ahead of time or something? It seems to me you could train specifically from that.... Or do the coaches have contact with their students after they see the problems? After all we are in the age of instant communication....[/quote]

Offer: please read the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]rough version of my IMO article.[/url] No, the team leaders do not see the ILL at all. And the team leaders do not see the ISL prior to arriving at the IMO location when it is handed out in the IMO jury sessions. Leaders and participants are kept separated until after the second day of the contest but as you mentioned communications devices are smart today so we must rely on trust and honour.  :)

Well, I would like to add something.

At the opening ceremony in 2002, IMO leaders were present, but they sat high up in a balcony, and we were strictly separated from them all times.  Weird : our best guy was the son of our team leader (and his mother was with the team leaders as well!).  I think it were the Australians who joked they could deduce the type of questions from the waving (waving was allowed).

And we were not allowed to know the location of the team leaders before the exam.  And we were asked to hand over cell phones (that could have been an after cyprus 2001 regulation).  But still : IMO still depends on trust :)

Bulgarians are strong, and problem 3 is not that difficult. If problem 3 is really difficult, why is it solved by 2 contestants (honourable mention) who scored lower than the cutoff score for bronze medal?

bronze medal = 13
ARG3 Deymonnaz, Alejandro 017310  12
GER5 Schreiber, Alex 227010  12

Bulgarians win the first place through hard work, and they win it fair and square. I think Bulgarians and the two honourable mention have the same style of thinking, that's why they can solve problem 3 easily.

[quote="darktreb"]That was a super strong team. On a related note, don't count out the USA's squad in 2007 which looks to be our strongest one since 04 :)[/quote]

w0rd

[quote="Arne"][quote="mozilla"]Bulgarians are strong, and problem 3 is not that difficult. If problem 3 is really difficult, why is it solved by 2 contestants (honourable mention) who scored lower than the cutoff score for bronze medal?

bronze medal = 13
ARG3 Deymonnaz, Alejandro 017310  12
GER5 Schreiber, Alex 227010  12
[/quote]

In my opinion, this problem is in the top three of hardest IMO problems ever - together with 1996/5 and 2006/6.

On this question, only 33 contestants managed to get any marks at all, and 24 contestants solved it. The average score for this question is 0.405. The average score for 1996/5 is 0.491, and the average score for 2006/6 is 0.19...[/quote]

That statistic is pretty interesting. I find #3 to be a sort of conundrum. To be honest, I didn't notice the factorization of the LHS when I tried the problem (silly..). However, once someone told me about that, I found the problem quite straightforward. Ironically, once I saw the factorization of the LHS I immediately recognized a way to "factor" the RHS because I had seen it factored in that way before. From there problem is almost completely standard (quick smooth, then AM-GM). It's funny how a problem that pretty much relied on basic factorization and rearrangement of terms turned out to be so hard.

[quote="Arne"]In fact the problem 3 we were discussing is problem 3 from IMO [b]2003[/b].[/quote]

Ohh my mistake for not reading more carefully. Yes, that was quite a hard problem.

[quote="Arne"] I mean, they are good, but many people consider it as very unlikely for them to become THAT good, all of a sudden...[/quote]
Why all of a sudden. Rosen Kralev was the only participant in bulgariant team for who this was the first IMO and the other boys had all gold medals. The team was very strong and thats why they were the best

As far as I know, Zhiyu Liu (perf. scorer) has participated in RusMO 2005 (N.Novgorod). He has got 53/56 and 1st diploma, but not for 11th grade... I may be wrong, but, as I remember, the only Chinese 1st diploma was even for 9th grade. So Chinese may lose points on easy problems.

----------
IMO2006 -- Inequality Math Olympiad (V.Voronovich, BLR Leader). Don't you think it's a bonus for China?

According to one of the books published in China: There were 6 students from HuBei province of China who participated in the RMO 2005. 2 9th grade and 4 10th grade. Zhiyu Liu ranked No.1 in the competition of the 10th grade.


Prove that $ n^7$+7 is never a perfect square for n integer


Evaluate:

$\sum^{2006}_{x=1}\frac{1}{x}$

That probably will be in some competitions...

My mom came up with that, and I don't have a clue of what to do..

Help!  :(

I believe it can't be calculated in simple terms for there is no explicit formula for sum of terms of an HArmonic progression.... although I don't know a proof of this fact!!!

Sorry but this cannot be in competitions. There is no short formula for sum of reciprocals

[quote="chess64"]:huuh:
...
why is your signature a quote? :P
...

[hide="the answer is"]
TI-83+ Program:
[code]
:ClrHome:0$\rightarrow \theta$
:For(X,1,2006)
:1/X+$\theta \rightarrow \theta$
:End:Disp $\theta$
[/code]
...after about a minute or so...
$\theta = \boxed {8.181362865...}$
[/hide][/quote]

Wow....

The quote has to do with my screen name. Mathnerd314 told me it.

Can you use the frac button on the TI 83+ to change that into a fraction?

Actually, I was referring to 10000th user's sig.

The denominator is so huge that the >Frac button doesn't work. :D

[quote="Drunken_Math"]

$\sum^{2006}_{x=1}\frac{1}{x}$

[/quote]
can somebody explain to me what  that means? :huh:

oh thanks Rushil
my way of doing this would be to find a common denominator and add everything up but that would take a year to do.

omg :o
how did you do that
what did you use!!!!!????!!?!?!?!!? :o

No, mine is right. anirudh is WRONG. :D The calculator can never be wrong. It looks like he just made that up. :P

i've found a problem that's sort of similar to this one from 1989 USAMO. problem 1 at this site :
http://www.unl.edu/amc/a-activities/a7-problems/questions/q-usamo/-pdf/usamo1989.pdf

same idea i guess.

Anirudh has a problem: (1+2+3...2006)/2006! = (2006*2007/2)/2006! = 2007/2005!2. 2005! has 5753 digits. ...

Anirudh also has a problem: The denominator should indeed be 2006!. However, when each of the fractions is expressed with a denominator of 2006!, each of the numerators becomes 2006!/(original numerator). So it should not be 1+2+3...2006, it should be 2006!+2006!/2+2006!/3...2005!

Oh, and IT is equal to 8.1813628649577925. I found that by direct addition(by computer program ;) , not by hand!)

So chess64's answer was correct  :D
But this sum doesn't approach a certain number?

cuz he used a comp programming, which i didn't. i just used a calc.

[quote="Hamster1800"]The sum, if continued, will approach $\infty$. I came up with a way to get the common fraction form, although for a number as big as 2006 it would be impractical. Here it is:

$\frac{\sum_{x=1}^{k} \frac{k!}{x}}{k!}$ where k is the number of terms. You could, with quite a bit of computer time, get the whole fraction form, but that would be extremely long. Let's stick with the decimal.

EDIT: while this form is only about 2 steps away from the original, all of the numbers in the numerator will be whole numbers and there is a nice common denominator.[/quote]
no ,it doesn't approach $\infty$. It definitely limits at an irrational and probably transadental number that is probably pretty low, not $\infty$

[quote="nat mc"]no ,it doesn't approach $\infty$. It definitely limits at an irrational and probably transadental number that is probably pretty low, not $\infty$[/quote]

There was a contest problem involving this, and it does go to infinity. I don't see how though...

Take the first term. $\frac 11$ is 1. If you take the next three terms, $\frac 12 + \frac 13 + \frac 14$, the sum is greater than 1. The sum of the next 8 terms, $\frac 15 + \frac 16 + \frac 17 + \frac 18 + \frac 19 + \frac 1{10} + \frac 1{11} + \frac 1{12}$ is also greater than 1. You can continue finding sections that are greater than 1 in this way, and since you have an infinity of terms, you have an infinity of 1s, which means that the sum $\sum_{x=1}^{\infty} \frac 1x$ approaches infinity.

If the top bound is infinity, yes, it will go to infinity, although it will result in a smaller infinity than the infinity that you're using as the upper bound. However, the problem we are given has the top bound as $2006$, not $\infty$.


What are the products of the pyrolysis of $ (NH_{4})_{2}SO_{4}$?

Pyrolysis and combustion are two different things.

yes pyrolysis means simple heating so what are  the products

I have no idea how to choose between these (hope that at least one of them is right)
$ 3(NH_{4})_{2}SO_{4}\longrightarrow 4NH_{3} \plus{} N_{2}\plus{}3SO_{2}\plus{}6H_{2}O$
or
$ (NH_{4})_{2}SO_{4}\longrightarrow 2NH_{3}\plus{}SO_{2}\plus{}H_{2}O\plus{}\frac{1}{2}O_{2}$

I don't think oxygen to be a product of a pyrolysis.

So ...?
What to choose or both of them are wrong?  :D

@Carcul: Can you explain a little bit about your post?  :roll:

Well, I said above that oxygen is not very likely a possible product of a pyrolysis simply because I have never heard of a heating reaction that liberates oxygen; otherwise, it would be a combustion.
Regarding the heating of ammonium sulphate, here's a hint: what products do we obtain when ammonium chloride is heated?

For $ NH_{4}Cl$, we will get $ NH_{3}$ and $ HCl$.
However, I think our problem is different since we have $ S^{\plus{}6}$ can be deoxidized downto $ \plus{}4$

There is the possible non-redox answer of ammonia, water, and $ \text{SO}_3,$ all of them gases.

[quote="Bibobeo"]can be deoxidized downto[/quote]

Or reduced. However, for that to occur we would need a reducing agent.
The answer of Professor Merryfield is essentially correct, since the actual products are ammonia gas and sulphuric acid vapour. However, at temperatures above ~350ºC an equilibrium is established between sulphuric acid, sulphur trioxide, and water (this because sulphur trioxide hydration is highly exothermic - as a side note, how is then sulphuric acid produced in industry?).

Now a tougher problem: what products are obtained when ammonium nitrate is heated?

How much blast protection do we want before we attempt that?

"We" are not going to attempt nothing (at least I will not). However, whether there is a blast or not, we have a reaction occurring, and I asked what the products of that reaction are.

I was taught that it would give $ N_{2}O$ and $ H_{2}O$ with $ t^{o}\equal{}250^{0}C$, or $ N_{2}$, $ O_{2}$ and $ H_{2}O$ with $ t^{o}\equal{}300^{0}C$.

And one more thing is how I get my first $ (NH_{4})_{2}SO_{4}$ pyrolysis.
$ (NH_{4})_{2}SO_{4}\longrightarrow 2NH_{3} \plus{} H_{2}SO_{4}$
$ H_{2}SO_{4}\longrightarrow H_{2}O \plus{} SO_{3}$
$ 2NH_{3} \plus{} 3SO_{3}\longrightarrow N_{2} \plus{} 3SO_{2}\plus{}3H_{2}O$
(All of them are heated)
Then sum up, and I got mine.
Mr.Kent Merryfield's answer is just my first 2 reactions.

That's right, in the case of ammonium nitrate there is first a redox reaction. Who is the oxidizing agent?

Oxidizing agent is $ N^{\plus{}5}$ and Deoxidizing agent is $ N^{\minus{}3}$.
Is that right?

Those are not the oxidizing/reducing agents, at most they are the oxidation numbers of nitrogen on those agents. But what are the species that act as oxidant and reductor?

I hope I understand what you mean
Both oxidant and reductor is $ NH_{4}NO{3}$

[quote="Carcul"]No. With heating the NH4NO3 decomposes first to two compounds, one of them will then act as an oxidizing agent and the other as the reducing agent. Who are they?[/quote]
Well, I guess $ N_{2}O$ is both oxidizing and reducing agent
$ N^{\plus{}1}$ downto $ N^{0}$, and $ O^{\minus{}2}$ to $ O^{0}$

No, N2O is one of the final products. Let me try to explain in a different way:

$ NH_4NO_3 \longrightarrow A \plus{} B \longrightarrow N_2O \plus{} H_2O$

where A and B are two intermediate species. You are to find A and B.

Oh  :blush: I'm sorry for not understanding your question
After reading you reply, I immediately realize that I have no idea about the answer :oops:  Can you post it?

[hide="Click here"]A and B are ammonia and nitric acid. Now it would be a good exercise to write the half-reactions.[/hide]


Fie sirul $ (a_n)_{n\ge1}$, unde $ a_1>0$ si $ a_{n\plus{}1}\equal{}a_n\plus{}\frac2{a_n}$, pentru orice $ x$ in $ N^*$. Calculati $ \displaystyle\lim_{n\to\infty}\frac{a_n}{\sqrt{n}}$.
RMT 3/2007.

Tot din RMT, cls a 11a

Se considera sirul $ (x_n)_n$ definit prin: $ x_0\equal{}a\le1, x_1\equal{}b\ge1$ si $ x_{n\plus{}2}\equal{}\frac23x_{n\plus{}1}\plus{}\frac13min\{1,x_n\}$ pentru orice $ n$ natural. Sa se arate ca sirul $ (x_n)_n$ este convergent si determinati $ \displaystyle\lim_{n\to\infty}x_n$.

De fapt aceasta este problema la care as vrea sa stiu rezultatul, cand am scris problema anterioara uitasem ca am rezolvat-o acum cateva zile :p. dar merita si ea incercata ;)

Pentru prima problema: exista o metoda prin care putem gasi care e comportarea asimptotica a sirului.
Astfel, sa consideram o functie $ f: R\rightarrow R$ astfel ca $ a_{n}\equal{}f\left(
n\right)  .$ Atunci $ a_{n\plus{}1}\minus{}a_{n}\equal{}f\left(  n\plus{}1\right)  \minus{}f\left(  n\right)
\equal{}\frac{f\left(  n\plus{}1\right)  \minus{}f\left(  n\right)  }{n\plus{}1\minus{}n}.$ Aceasta fractie ne
sugereaza teorema lui Lagrange si poate fi aproximata cu $ f^{\prime}\left(
n\right)$. De aceea, putem gandi ca $ f$ verifica o egalitate de tipul
$ f^{\prime}\left(  x\right)  \equal{}\frac{2}{f\left(  x\right)  }$ sau $ f\left(
x\right)  f^{\prime}\left(  x\right)  \equal{}2,$ de unde $ \left(  \frac{1}{2}
f^{2}\left(  x\right)  \right)  ^{\prime}\equal{}2.$ Deducem ca $ f\left(  x\right)
\equal{}\sqrt{4x\plus{}c},$ deci $ a_{n}\simeq\sqrt{4n\plus{}c},$ iar limita ceruta este egala cu 2.

Riguros, acum: observam mai intai ca sirul e strict crescator si, deci, exista
$ \lim a_{n}\equal{}L\in\overline{R}.$ Daca $ L\neq\plus{}\infty,$ trecand la limita in
relatia de recurenta, obtinem $ L\equal{}L\plus{}\frac{2}{L},$ absurd. Deci $ a_{n}
\rightarrow\plus{}\infty.$

Acum calculam cu Stolz-Cesaro $ \lim\frac{a_{n}^{2}}{n}\equal{}\lim\frac{a_{n\plus{}1}
^{2}\minus{}a_{n}^{2}}{n\plus{}1\minus{}n}\equal{}\lim\left(  \left(  a_{n}\plus{}\frac{2}{a_{n}}\right)
^{2}\minus{}a_{n}^{2}\right)  \equal{}\lim\left(  4\plus{}\frac{4}{a_{n}^{2}}\right)  \equal{}4,$ de unde concluzia.


whoa thats cool
I read it like really easily.
except this word rdgnieg. typo i presume?

[quote="mustafa"]The phaonmneel pweor of the hmuan mnid: I cdnuolt blveiee taht I cluod aulaclty uesdnatnrd waht I was radneig. Aoccdrnig to a rscheearcr at Cmagbride Uinervtisy, it deosn't mttaer in waht oredr the ltteers of a wrod are, the olny iprmoatnt tihng is taht the frist and lsat ltteer be in the rghit pclae.

The rset can be a taotl mses and you can sitll raed it wouthit a porbelm. Tihs is bcuseae the huamn mnid deos not raed ervey lteter by istlef, but the wrod as a wlohe. Amzanig huh?[/quote]
My 8th grade English teacher showed us this exact same thing :D

pisoinokonaclovociliscipocsorcimartluonomuens :D

somewhere i saw this...but you know what?! i also see this in hogwarts when people eat acid pops but it is way harder to decipher...

[quote="lotrgreengrapes7926"]pisoinokonaclovociliscipocsorcimartluonomuens :D[/quote]

LOL... and pneumonoultramicroscopicsilicovolcanokoniosis isn't a real word; it was actually made up by a prankster and somehow made its way into the Oxford. The actual name of the disease is pneumonokoniosis.

My mom e-malied taht to me a few mhtnos ago. I unstooderd all of it.

[quote="mustafa"]The phaonmneel pweor of the hmuan mnid: I cdnuolt blveiee taht I cluod aulaclty uesdnatnrd waht I was radneig. Aoccdrnig to a rscheearcr at Cmagbride Uinervtisy, it deosn't mttaer in waht oredr the ltteers of a wrod are, the olny iprmoatnt tihng is taht the frist and lsat ltteer be in the rghit pclae.

The rset can be a taotl mses and you can sitll raed it wouthit a porbelm. Tihs is bcuseae the huamn mnid deos not raed ervey lteter by istlef, but the wrod as a wlohe. Amzanig huh?[/quote]

thats awesome. even though i read another version of this... alsofrom canbridge university

wow this is very intersting
it was very easy to read
why does the brain work that way?

Wow, that is cool!  :D
has that been proven, or is it just some whacko that doesnt have anything better to do?  :rotfl:
At first i thought it would be hard to read, but it wasnt, very cool!

it might be the wacko idealism. I could read it, but some words took a little time to find out?

i read it easily...i guess the words quickly form in my mind.

the words could be more scrambled, then I wouldn't be able to read it  :P


the perimeter of an equilateral triangle is 12 times the square root of 3. The perimeter of a square is 16. What is the ratio of the area of the square to the area of the triangle?

This is a very interesting question

by the way, this problem has to go to other sections because it's from a math tournament.

and also, it shoud have the star sign in order to question people

[hide]perimeter of the triangle = 12 * square root of 3= 2^2 * 3 *square root of 3 = 2^2

so the perimeter is 4

so the base of this triangle should be 4/3.......

to get the area, you have to find the heigh of it...
by using the pythagoris theorim
the height is 12/9

the area of this triangle is, 12/9 * 4/3 / 2 =8/9 square units

the area of the square is 16

the ratio would be, 8/9 : 16

I probably got it wrong..Because I doubt my answer...[/hide]

[quote]

perimeter of the triangle = 12 * square root of 3= 2^2 * 3 *square root of 3 = 2^2

[/quote]

that's not exactually right because when the 2^2*3*the square root of 3 is all under a square root sign, it would be the square root of 144*3 which is not 6

when I said one of the math tournaments, I meant like not a national tournament or anything.... just a problem that we use in math tournaments for MIDDLE SCHOOL people. so I thought that was okay, and it's only for seventh grade anyways. and besides, one of KBabe's problem was from the same sheet of paper, the one about 25, 26, 29...

[quote]that's not exactually right because when the 2^2*3*the square root of 3 is all under a square root sign, it would be the square root of 144*3 which is not 6[/quote]

but you said, it was[quote]equilateral triangle is 12 times the square root of 3[/quote] wich means square root of 3 and multiplying 12 are the separate stuff.

which means it does not go under the square root

well, I'm not sure whether or not it's because I can't put a square root sign or that you are not getting the point of the problem but the perimeter is 12*the square root of 3, and you were right, the 12 isn't under the square root sign right now. But what you did still can't be done as far as I know because 2^2*3*the square root of 3 is NOT equal to 4, which is what you are saying.  i'm not sure how to explain this so can somebody explain it?

[quote="rrusczyk"]I've moved this into The Basics; please only use Classroom Math for math from classes, not 'contest math'.[/quote]

I'm not sure whether or not you know what math team is, but is math team considered "math from class"

[quote="shinwoo"]Wait, your right

it can't be done

I was supposed to say, 2 * 2 * 3 * square root of 3

=

2 * 2 * 3 * 3^1/2

=

6

I'm sorry
LuCy4EvA

so, my final answer would be...

1.5 : 16[/quote]
 how did u get that?

Just to let you guys know, $\sqrt{3}\cdot 3 \neq 1$, $3\cdot \frac{1}{3} = 1$.

And shinwoo, I don't think there's any way to simplyfy $12\sqrt{3}$ further.

And here's a hint if you need it
[hide=" HINT"]
$\frac{x\cdot \sqrt{3}}{3} = \frac{x}{3} \cdot \sqrt{3}$[/hide]

[quote]And shinwoo, I don't think there's any way to simplyfy  and further.[/quote]

I think she meant like 12 * square root of 3
I don't think she meant 12throot of 3

but when we prime factorize(?) it it turns to be like 2*2*3*squareroot of 3 and it would turn out like 2 * 3 right?

am I wrong?

ahh!!!!

Try putting that into a calculator, and see if it works.  And look at my hint if you need to.

Shinwoo, why do you think the $\sqrt{3}$ would cancel out one of the other factors?

I think Shinwoo is not getting the problem because he's misunderstanding the problem, can someone with Latex retype the problem for me?

Shinwoo, maybe you should ask your teacher to show you how 12 times the square root of 3 is not 6, it's kinda hard to explain it on the internet.

but if you square 12 times the square root of 3, you would get 144*3 which is 432, and the square root of 432 is NOT 6. I don't know if you now how to work with square roots, but say you change the 3 to a 4, it would be 12*2 which is 24, that's not the same as the square root of 48.

....

I better go look up on AoPS textbook again..

ahhhhhhhhhhhhhhhh!!

Shinwoo -- can you explain, in detail, what you were trying to do?  You said, [quote="shinwoo"]but when we prime factorize(?) it it turns to be like 2*2*3*squareroot of 3 and it would turn out like 2 * 3 right?[/quote]

Why did you think that was true?

[quote]Shinwoo, maybe you should ask your teacher to show you how 12 times the square root of 3 is not 6, it's kinda hard to explain it on the internet.
[/quote]

I self taught math,( had coaches few times, but the all stink )

and the school math teacher is really dumb

Hint:

[hide]The area of an equilateral triangle is always $\frac{s^2\sqrt{3}}{4}$. Try deriving this by dropping an altitude in an equilateral triangle.[/hide]


For any point $ P$ on a segment $ AB$, isosceles and right-angled triangles $ AQP$ and $ PRB$ are constructed on the same side of $ AB$, with $ AP$ and $ PB$ as the bases. Determine the locus of the midpoint $ M$ of $ QR$ when $ P$ describes the segment $ AB$.

Let X,Y be midponts of AP and BP. M' the orthogonal projection of M on the line AB.  QRYX is a trapezoid and MM' is the medium parallel to its bases ---> MM'=1/2(QX+RY)=1\2(AX+BY)=1\4AB=constant. The locus of M is the parallel in the same side of Q,R  whose distance to AB is 1\4 AB.

It is also interesting that the envelope of the line QR is a parabola.


pigeonhole principle
combinatorics unsolved
combinatorics

Given 5 straight line segments such that triangles can be constructed from any three of them .
schow that among them there is at least one acute-angled triangle.

pigeonhol principal and we are done thanks i've find the unswer.


algebra
polynomial
Vieta
number theory proposed
number theory

Let $ a$, $ b$, $ c$, $ m$ be four positive integers satisfying $ a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} 1 \equal{} abcm$. Prove that $ m \equal{} 4$.

[quote="pohoatza"]Let $ a$, $ b$, $ c$, $ m$ be four positive integers satisfying $ a^{2} + b^{2} + c^{2} + 1 = abcm$. Prove that $ m = 4$.[/quote]
The first ,we will prove that $ m\geq 4$ ,it means that the equation has no solution in cases $ m=2,3$
If $ m=2$ then we have the equation
\[ a^2+b^2+c^2+1=2abc
\]
Therefore in the set ${ \{a,b,c}$ there is an odd number which is odd.
If have only one number is odd then we have $ a^2+b^2+c^2+1\equiv 2 (\mod 4)$ but $ 4|2abc$ ,it gives contradiction. If all a,b,c is odd then the left divide by 4 ,but the right is not . Therefore the equation has no solution when n=2 .
It is similar to prove $ m=3$ doesn't satisfy by considering mod 4  .
Now suppose that exist $ m>4$ such that the equation has natural solution . Call $ (a_0,b_0,c_0)$ is root such that $ a_0+b_0+c_0$ is minimum . We can assume that $ a=\max\{a,b,c\}$
If $ a_0=b_0=c_0$ then we have $ 3a_{0}^2+1=ma_{0}^3$ ,it gives $ a_0=1,m=4$ (contradiction ) .Therefore at least one of two number $ b_$ or $ c_0$ is smaller than $ a_0$ . Assume this number is $ b$ .
The equation also has an other solution $ a_1$ such that
\[ a_1=\frac{b_0^2+c_0^2+1}{a_0}
\]
We have $ a_0^2=ma_{0}.b_{0}.c_{0}-b_{0}^2-c_{0}^2-1$
Because $ m>4$ and $ a_0>b_0$ so
$ ma_0b_0c_0=\frac{m-1}{2}a_{0}b_{0}c_{0}+\frac{m-1}{2}a_{0}b_{0}c_{0}+a_{0}b_{0}c_{0}>b_{0}^2+c_{0}^2+2$
Therefore $ 0<a_1<a_0$ ,we have root $ (a_1,b_0,c_0)$ ,it gives contradiction .
Therefore m must be equal 4 .

[quote="TTsphn"][quote="pohoatza"]
The equation also has an other solution $ a_1$ such that
\[ a_1 \equal{} \frac {b_0^2 \plus{} c_0^2 \plus{} 1}{a_0}
\]
[/quote][/quote]

How do you know that $ a_1$ is an integer

[quote="ricardo4"][quote="TTsphn"][quote="pohoatza"]
The equation also has an other solution $ a_1$ such that
\[ a_1 \equal{} \frac {b_0^2 \plus{} c_0^2 \plus{} 1}{a_0}
\]
[/quote]

How do you know that $ a_1$ is an integer[/quote][/quote]
Because $ a_1\plus{}a_0\equal{}mb_0c_0$ and it is clear that a is a positive number .

Very nice ,TTsphn,but my proof is not same to you(although i also use Vietta law)
And the same if $ a,b,c\in N^*$ $ 0\le A\equal{}a^2\plus{}b^2\minus{}abc\le c$ Prove that $ A$ is a perfect square
The same feature that I mention is the way to kill it


Yes, it's true. Posted: http://www.mathlinks.ro/Forum/viewtopic.php?t=56071

[quote="shuchang"]Let $ a,b,c > 0,$ prove that

$ \frac {a}{\sqrt {a \plus{} b}} \plus{} \frac {b}{\sqrt {b \plus{} c}} \plus{} \frac {c}{\sqrt {c \plus{} a}}\geq\frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}}{\sqrt2}.$[/quote]Substitute $ a \equal{} x^2,b \equal{} y^2,c \equal{} z^2.$ Using the Hölder inequality,

$ \left(\sum_{cyc}\frac {x^2}{\sqrt {x^2 \plus{} y^2}}\right)^2\sum_{cyc}{x^2(x^2 \plus{} y^2)(x \plus{} y \plus{} 2z)^3}\geq\left[\sum_{cyc}x^2(x \plus{} y \plus{} 2z)\right]^3,$

we need to prove

$ \left[\sum_{cyc}x^2(x \plus{} y \plus{} 2z)\right]^3\geq\frac {(x \plus{} y \plus{} z)^2}{2}\sum_{cyc}{x^2(x^2 \plus{} y^2)(x \plus{} y \plus{} 2z)^3}.$

But $ 2\left[\sum_{cyc}x^2(x \plus{} y \plus{} 2z)\right]^3 \minus{} (x \plus{} y \plus{} z)^2\sum_{cyc}{x^2(x^2 \plus{} y^2)(x \plus{} y \plus{} 2z)^3}$

$ \equal{} \sum_{cyc}(y \minus{} z)^2\Big[2xz^2(46x^3y \plus{} 27x^3z \plus{} 2x^2y^2 \plus{} 24x^2yz \plus{} 3xz^3 \plus{} 5yz^3 \plus{} z^4)$

$ \plus{} (x \minus{} z)^2(8x^4z \plus{} 12x^3z^2 \plus{} 14xy^4 \plus{} 26xy^2z^2 \plus{} xz^4 \plus{} 2y^5 \plus{} 8y^4z \plus{} y^3z^2 \plus{} 13y^2z^3$

$ \plus{} 6yz^4 \plus{} z^5)\Big]\geq0.$


p,q,r>0 , p+q+r=1 Prove that

7(pq+qr+pr) \leq 2+9prq

This follows easily from adding the inequalities: 4(pq+qr+pr) \leq 1+9pqr and 3(pq+qr+pr) \leq (p+q+r)^2=1.

this inequality is one of the typical inequalities where one can normalize (homogenize ) and use muirhead and stuff
cheers

I have three different solutions for this inequality.

The first one is the "brute force" solution (homogenizing) ;

The second one is due to John Scholes (http://www.kalva.demon.co.uk/bmo/bsoln/bsol993.html) ;

The third one is as follows.
The inequality is obviously true if one of p, q, r exceeds 7/9.
So suppose p, q, r < 7/9.
Put 7(pq + qr + rp) - 9pqr = A. We want to prove that A <= 2.
Then (7 - 9p)(7 - 9q)(7 - 9r) = 7 - 9.7 + 7.9(pq + pr + qr) - 9pqr = 81A - 98.
Now if p, q, r < 7/9 then (7 - 9p)(7 - 9q)(7 - 9r)  <=  4 = 64 by AM-GM.
So 81A - 98 <= 64 gives A <= 2. QED

PS. pcalin, why is 4(pq + qr + rp) <= 1 + 9pqr so obvious ?

We have that (1-2p)(1-2q)(1-2r)<=pqr. If one of the numbers p,q,r is grater than 1/2 then the inequality is obvious. If p,q,r<=1/2 then we can apply the AM-GM inequality: (1-2p)(1-2q)<=r^2 and so on and by multiplying these inequalities we obtain the desired inequality. But now our inequality is equivalent with 4(pq+qr+rp)<=1+9pqr.


prove that one can divide any equilateral triangle into nsquare congruent triangles.
                                                    amir siddique

isn't it just equally dividing each side into n parts?

Divide each side into n equal segments and draw lines parallel to the triangle sides through these dividing points. Assume the side $BC$ is horizontal and the vertex $A$ above it. The 1st row of triangles from the top contains 1 equilateral triangle, the 2nd row 3 equilateral triangles, ..., and the last bottom row 2n - 1 equilateral triangles.

$1 + 3 + ... + 2n - 1 = \sum_{k=1}^n (2k - 1) = n^2$

The proof is by induction:

[color=blue][b](1)[/b][/color] [color=white].[/color] $1 = 1^2$

[color=blue][b](2)[/b][/color] [color=white].[/color] $\sum_{k=1}^{n+1} (2k - 1) = \sum_{k=1}^n (2k - 1) + 2(n + 1) - 1 = n^2 + 2n + 1 = (n + 1)^2$

yetti u r exactly right.but u have started k from zero,but it shud start from 1.
     anyhow there is no need to prove that the sum of first n odd integers is equal to n-square.anyhow if it is compulsury we can use sum of arithematic series.
      there is also very interesting prove by using equilateral triange arguements.let an equilateral triangle with side equal to n(n is an integer >1).divide all its sides into n equal parts and by drawing paralle lines to its sides passing through these points we will get small equilateral triangles in the big equilateral.the total triangles in the big triangles r =1+3+5+..........+(2n-1)
  the area of big triangle=area of one small triangle *total number of small trianghles.
   √3/4 *n-square= √3/4*total no. of triangles
   total no. of triangles =n-square
  but total number of triangles =1+3+5+............+2n-1=n-square
 hence proved


Something happened, i don't know what... I was on an onlilne class and suddenly I just got quit out... I didn't exit the site... and I couldn't get back on because it said somebody was already logged on as my username...

someone already logged in as your username? That can't be...unless u've given your pw out...and even so, I've been able to log on as myself on two different windows...although if you were at a different IP and such...im not sure...

Very weird.  I strongly suspect what happened there was your internet connection dropped briefly.  When you came back in, your 'ghost' was still in the room.  After a couple minutes, the room kicked the ghost out (realizing you weren't actually connected), then you were able to get back in.  You didn't have any trouble once you got back in, right?


calculus
derivative
inequalities
inequalities proposed

[u][b]The author of this posting is : Megus[/b][/u]
____________________________________________________________________

Using sum of derivatives - we assume:

$f\left(a,b,c\right) = \frac{1}{4} - \left(\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ca}{b+1}\right) \geq 0$

I. Suppose a=0

$f\left(0,b,c\right) = \frac{1}{4} - bc \geq 0$
$b+c \geq 4bc$

what seems obvious but maybe not...  :?  [this is no 100% solution]

II. now we have to prove $f^{\prime}\left(a,b,c\right) \geq 0$
$\sum_{sym}\frac{b}{c+1}+\frac{c}{b+1}-\frac{bc}{\left(a+1\right)^2} \geq 0$
now we can divide into 3 symmetrical groups:
$\sum_{sym}\frac{\left(a+b\right)\left(c+1\right)-ab}{\left(c+1\right)^2} \geq 0$

what is true due to goven conditions.

Sorry for this proof being so imperfect, but I'm still learning to use sum of derivatives properly

[u][b]The author of this posting is : Lagrangia (edited by Darij Grinberg)[/b][/u]
____________________________________________________________________

In "Old and New Inequalities" by T. Andreescu, G. Dospinescu, V. Cirtoaje and M. Lascu you will find the following problem:

35. Let $a$, $b$, $c$ be positive real numbers. Prove that:

$\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b} \leq \frac{1}{4}\left(a+b+c\right)$.

This one is, of course, equivalent to the inequality posted by ThAzN1 (it is actually the dehomogenized version of that inequality).

[u][b]The author of this posting is : Lagrangia (edited by Darij Grinberg)[/b][/u]
____________________________________________________________________

[quote="Lagrangia"]35. Let $a$, $b$, $c$ be positive real numbers. Prove that:

$\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b} \leq \frac{1}{4}\left(a+b+c\right)$.[/quote]

I'm posting two solutions I had for this problem. If you are bored try reading them! :D

[b]$1^{st}$ solution:[/b] Because the inequality is homogenous then we can consider without loss of generality that: $a+b+c=1$ and so the inequality is equivalent to: $\sum \frac{1}{a\left(a+1\right)} \leq \frac{1}{4abc}$. We have that $\frac{1}{t\left(t+1\right)}=\frac{1}{t}-\frac{1}{t+1}$ so the inequality is equivalent to: $\sum \frac{1}{a} \leq \sum \frac{1}{a+1} + \frac{1}{4abc}$. We will now prove the following intercalation: $\sum \frac{1}{a} \leq \frac{9}{4} + \frac{1}{4abc} \leq \sum \frac{1}{a+1} + \frac{1}{4abc}$. The last inequality is obvious from Cauchy Buniakowsky Schwartz $\left(\sum \frac{1}{a+1}\right)\left(\sum \left(a+1\right)\right) \geq 9$, but because $\sum \left(a+1\right) = 4$ we get that: $\sum \frac{1}{a+1} \geq \frac{9}{4}$. For the first inequality we denote $a=\frac{x}{x+y+z}$, $b=\frac{y}{x+y+z}$ and $c=\frac{z}{x+y+z}$ so the inequality is equivalent to: $\left(\sum x\right)^{3} + 9xyz \geq 4\left(\sum x\right)\left(\sum xy\right)$ but this inequality reduces itself to the well known inequality: $\prod \left(-x+y+z\right) \leq \prod x$. So the inequality $\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b} \leq \frac{1}{4}\left(a+b+c\right)$ is true for all positive reals a,b and c.

[b]$2^{nd}$ solution:[/b] Because the inequality is homogenous then we can consider without loss of generality that: $a+b+c=1$ and so the inequality is equivalent to: $\sum ab - abc \left(\sum \frac{1}{a+1}\right) \leq \frac{1}{4}$. Because the harmonic mean is less or equal with the arithmetic mean we have that: $\frac{1}{\sum \frac{1}{a+1}} \leq \frac{\sum \left(a+1\right)}{3} = \frac{4}{3}$ or $\sum \frac{1}{a+1} \geq \frac{9}{4}$. So it is sufficient to prove that $E\left(a,b,c\right) = \sum ab - \frac{9abc}{4} \leq \frac{1}{4}$. We will prove this inequality using Sturm's method. We consider that c is fixed and we vary a and b without changing their sum. We have that: $E\left(a+x,b-x,c\right)-E\left(a,b,c\right) = x\left(b-a-x\right)\left(1-\frac{9}{4}c\right)$. If $a<b$ and $0<x<b-a$ then by making a really close to b we get that $E\left(a,b,c\right)$ increases its value if $c < \frac{4}{9}$ and decreases if $c > \frac{4}{9}$. We consider that $a \leq b \leq c$. If $c > \frac{4}{9}$ then by making a really far from b we get that: $E\left(a,b,c\right) \leq E\left(0,1-c,c\right)$. Now by making $1-c$ really close to c we find that: $E\left(0,1-c,c\right) \leq E\left(0,\frac{1}{2},\frac{1}{2}\right) = \frac{1}{4}$. If $c < \frac{4}{9}$ then by making a really close to b we get that the value of $E\left(a,b,c\right)$ increases and because $a \leq \frac{1}{3}$ and $b \geq \frac{1}{3}$, and now we get that: $E\left(a,b,c\right) \leq E\left(\frac{1}{3},b+a-\frac{1}{3},c\right)$. Now by fixing $c=\frac{1}{3}<\frac{4}{9}$ and by making $b+c-\frac{1}{3}$ really close to c, we get that $E\left(a,b,c\right) \leq E\left(\frac{1}{3},b+a-\frac{1}{3},c\right) \leq E\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right) = \frac{1}{4}$. Following the above judgement we can see that the maximum is obtained if $\left(a,b,c\right) = \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$ or $\left(a,b,c\right) = \left(0,\frac{1}{2},\frac{1}{2}\right)$.

cheers! :D  :D

[quote="darij grinberg"]In "Old and New Inequalities" by T. Andreescu, G. Dospinescu, V. Cirtoaje and M. Lascu you will find the following problem:

35. Let $a$, $b$, $c$ be positive real numbers. Prove that:

$\frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b} \leq \frac{1}{4}\left(a+b+c\right)$.[/quote]

This one has also been solved in the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=14289 . Harazi's wonderful solution in this thread is the same as the solution in the book ("Old and New Inequalities").

  Darij

A much more simple solution:

$ \frac{1}{4} \left( \frac{ab}{1-b}+\frac{ab}{1-a} \right) \geq \frac{ab}{c+1}\\$
Adding the ineq we get:
$\sum{\frac{1}{4}(\frac{ab}{a+c}+\frac{bc}{c+a})}=\frac{1}{4} \sum a$

Sorry Darij I forgot about that topic and I posted the solution just after your last post so I didn't see it. It's exactly what Harazi wrote.


Let p be a prime and p=12k+7.Prove that:p=$ u^{2}\plus{}uv\plus{}v^{2}$ (u,v are natural number)

Let $ p \equiv 1 \mod 3$. See http://www.mathlinks.ro/viewtopic.php?t=55714 (for Thue, see http://www.mathlinks.ro/Forum/viewtopic.php?t=107844 ) to get that $ p\equal{}a^2\plus{}3b^2$ for some $ a,b$.
Set $ x\equal{}a\minus{}b,y\equal{}2b$, then $ x^2\plus{}xy\plus{}y^2 \equal{} a^2\minus{}2ab\plus{}b^2\plus{}2ab\minus{}2b^2\plus{}4b^2 \equal{} a^2\plus{}3b^2\equal{}p$.
If $ x,y$ have different signs, w.l.o.g. assume $ |x| \leq |y|$ and replace $ x \to \minus{}x, y \to x\plus{}y$, as then still $ (\minus{}x)^2\plus{}(\minus{}x)(x\plus{}y)\plus{}(x\plus{}y)^2 \equal{} x^2\plus{}xy\plus{}y^2 \equal{} p$. But if the signs where different, they now are equal (we consider $ 0$ to have both signs).
If now both $ x,y<0$, replace them by $ \minus{}x,\minus{}y$, so we reached $ x,y\geq 0$.
Surely $ x,y$ can't be $ 0$ (as then $ p$ is a perfect square), thus we got positive $ x,y$.

The less elementary way: let $ \zeta_3$ be a root of $ x^2\plus{}x\plus{}1\equal{}0$. Then consider $ \mathbb Z[\zeta_3]$, the ring of integers of $ \mathbb Q[\zeta_3]$; it is factorial. By standard algebraic number theory, the prime $ p$ splits iff $ x^2\plus{}x\plus{}1 \equiv 0$ has distinct solutions $ \mod p$. As this happens for $ p \equiv 1 \mod 3$, all such primes split, thus there are prime elements $ q_1,q_2 \in \mathbb Z[\zeta_3]$ such that $ q_1 q_2 \equal{} p$. Taking norms, $ N(q_1)N(q_2) \equal{} N(p) \equal{} p^2$. As the norms are nonnegative and not $ \equal{}1$ (as then the correspinding element is a unit), we get that $ N(q_1)\equal{}p$. Now $ q_1 \equal{} x\minus{}y\zeta_3$ for some integers $ x,y$, thus $ p\equal{}N(q_1) \equal{} (x\minus{}y\zeta_3)(x\minus{}y\zeta_3^{\minus{}1}) \equal{} x^2\plus{}xy\plus{}y^2$. For positivity, either do as before, or multiply $ q_1$ with some power of $ \zeta_3$.


quadratics
calculus
integration
algebra
quadratic formula

Let a,b two nonzero integers then prove the following statement:

$ x^2 \plus{} ax \plus{} b \equal{} 0$ have an integer root $ \Leftrightarrow$ The discriminent is a perfect square.

why not use quadratic formula?

the roots are
$ x \equal{} \frac { \minus{} a\pm\sqrt {a^2 \minus{} 4b}}{2}$

now for integer roots discriminant must be a perfect square
otherwise the equation will not have rational roots(for roots to be integral,they ought to be rational)

[quote]now for integer roots discriminant must be a perfect square
otherwise the equation will not have rational roots[/quote]

this is true, but you need to prove that a perfect square discriminant will lead to integer $ x$ too, right?

[quote="Daan"]
this is true, but you need to prove that a perfect square discriminant will lead to integer $ x$ too, right?[/quote]

Right. $ a$ has the same parity as $ \minus{}a$ and $ a^2\minus{}4b$ which then has the same parity as its square root, if that root is an integer. So the sum/diff will be divisible by 2.

[quote]Right.  has the same parity as  and  which then has the same parity as its square root, if that root is an integer. So the sum/diff will be divisible by 2.[/quote]

Indeed. aviateurpilot's solution was ok too.

sorry
i missed to correctly interprate the question


function
search
algebra proposed
algebra

For which integers $k$ does there exist a function $f : N \longrightarrow Z$ such that
$(a)\ \ f(1995) = 1996$, and
$(b)\ \ f(xy) = f(x) + f(y) + kf(gcd(x,y))$ for all $x, y \in N$?

hello mathx

would you tell me what is $(gcd(x,y))$

thanks... :huh:  :huh:

The greatest common divisor of $x$ and $y$.

I'm pretty sure this has been posted before, by the way.

This problem is in Czech-Slovak Contest 96-97 . I have a solution for this problem but I think you can search for it  ;)


Prove that
$ \frac{1}{2\plus{}a}\plus{}\frac{1}{2\plus{}b}\plus{}\frac{1}{2\plus{}c}\leq1$  if  $ a,b,c>0$ and $ abc\equal{}1$.

[quote="Armenia"]Prove that
$ \frac {1}{2 \plus{} a} \plus{} \frac {1}{2 \plus{} b} \plus{} \frac {1}{2 \plus{} c}\leq1$  if  $ a,b,c > 0$ and $ abc \equal{} 1$.[/quote]

multiplying by $ (2\plus{}a)(2\plus{}b)(2\plus{}c)$ we get:

$ 12\plus{}4a\plus{}4b\plus{}4c\plus{}ab\plus{}bc\plus{}ca \le 8 \plus{} 4a \plus{} 4b \plus{} 4c \plus{} 2ab \plus{} 2bc \plus{} 2ca \plus{} abc$

$ 4 \le ab\plus{}bc\plus{}ca\plus{}abc$

but $ abc\equal{}1$

so we have to prove: $ 3 \le ab\plus{}bc\plus{}ca$

which can be done using AM-GM inequality:

$ ab\plus{}bc\plus{}ca \ge 3\sqrt[3]{a^2b^2c^2} \equal{} 3$

Also we can see that inequality 2. combined with already proved inequality proposed by $ Armenia$ gives us inequality 1.

Inequality 2. is well known problem. It is Bulgaria 1997. and it was discussed in this forum.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=243[/url]

Inequality 1 is easier than 2. It may be proved without using 2. I saw proof in Math Excalibur.

1.
let $ \sum_{cyc} \equiv \sum$
let $ a \equal{} \frac {x}{y}$
$ b \equal{} \frac {y}{z}$
$ c \equal{} \frac {z}{x}$
the inequality becomes
$ \sum \frac {xy}{x^{2} \plus{} y(x \plus{} z)} \leq 1$
normalize to $ \sum x \equal{} 1$
thsu we need to prove
$ \sum xf( \minus{} \frac {x^{2} \plus{} y(1 \minus{} y)}{y}) \leq 1$
where $ f(t) \equal{} \frac { \minus{} 1}{t}$
observe that $ f$ is concave and increasing so we need to prove

$ \sum xf( \minus{} \frac {x^{2} \plus{} y(1 \minus{} y)}{y}) \leq 1$
$ \Rightarrow f(\sum x ( \minus{} \frac {x^{2} \plus{} y(1 \minus{} y)}{y})) \leq f( \minus{} 1)$
$ \Rightarrow \sum \frac {x(x^{2} \plus{} y(1 \minus{} y))}{y} \geq 1$
$ \Rightarrow \sum \frac {x^{3}}{y} \plus{} \sum x \minus{} \sum xy \geq 1$
which is true since by cauchy schwartz
$ \sum \frac {x^{3}}{y} \geq \frac {(\sum x^{2})^{2}}{\sum xy} \geq \sum xy$

1.
Let $ a\equal{}x^{3},b\equal{}y^{3},c\equal{}z^{3},xyz\equal{}1$.We have
$ \sum \frac{1}{1\plus{}a\plus{}b}\equal{}\sum \frac{1}{xyz\plus{}x^{3}\plus{}y^{3}}
\leq \sum \frac{1}{xyz\plus{}x^{2}y\plus{}xy^{2}}\equal{}\sum \frac{1}{xy(x\plus{}y\plus{}z)}\equal{}\frac{1}{x\plus{}y\plus{}z}(\frac{1}{xy}\plus{}\frac{1}{xz}\plus{}\frac{1}{yz})\equal{}1$

[quote="borislav_mirchev"]If: $ a, b, c > 0; abc \equal{} 1$ Prove:
1. $ \frac {1}{1 \plus{} a \plus{} b} \plus{} \frac {1}{1 \plus{} b \plus{} c} \plus{} \frac {1}{1 \plus{} c \plus{} a} \leq 1$
2. $ \frac {1}{1 \plus{} a \plus{} b} \plus{} \frac {1}{1 \plus{} b \plus{} c} \plus{} \frac {1}{1 \plus{} c \plus{} a} \leq \frac {1}{2 \plus{} a} \plus{} \frac {1}{2 \plus{} b} \plus{} \frac {1}{2 \plus{} c}$[/quote]
2/Let $ x \equal{} a \plus{} b \plus{} c;y \equal{} ab \plus{} bc \plus{} ca$
Inequality becomes:
$ \frac {x^2 \plus{} 4x \plus{} y \plus{} 3}{x^2 \plus{} 2x \plus{} y \plus{} xy} \leq \frac {12 \plus{} 4x \plus{} y}{9 \plus{} 4x \plus{} 2y}$
<=>$ \frac {2x \plus{} 3 \minus{} xy}{x^2 \plus{} 2x \plus{} y \plus{} xy} \leq \frac {3 \minus{} y}{9 \plus{} 4x \plus{} 2y}$
Note that $ x\geq3;y \geq 3;x^2 \geq 3y$
so we have $ \frac {5}{3}x^2y\geq 5x^2;\frac {x^2y}{3}\geq y^2;xy^2 \geq 9x;5xy \geq 15x;xy \geq 3y;x^2y \geq 27$
Done!


function
real analysis
real analysis unsolved

this problem has two parts

i) A function $ f$ from from $ S \subset \mathbb{R^{n}}$ into $ \mathbb{R^{m}}$ is called a Lipschitz function if there is a constant $ C$ such that

$ \parallel{}f(x)\minus{}f(y)\parallel{} \le C\parallel{}x\minus{}y\parallel{}$ for all $ x,y \in S$.

The Lipschitz constant of $ f$ is the smallest choice of $ C$ for which the previous condition holds. Show that every Lipschitz function is continuous.

ii) Let $ \mathcal{F}$ be the set of all Lipschitz functions $ [0,1]\rightarrow \mathbb{R}$ whose Lipschitz constant is at most $ 47$. Show that $ \mathcal{F}$ is uniformly equicontinuous on $ [0,1]$.

for first part, I just took $ \parallel{}x\minus{}y\parallel{} \le \frac{\epsilon}{C}$.

my question is on the second part, where the answer is just to take $ C \equal{}47$ from part i), BUT why does that show "uniformly equicontinuous" and not simply just "continuous"? Does the "equi-" part come from the fact that $ \mathcal{F}$ is a family of Lipschitz functions? and if so, then why is it also "uniform" and not just equi-cont.?

A set of functions $ \mathcal{F} \equal{} \{f\colon \ldots\}$ is uniformly equicontinuous if $ \forall \epsilon > 0 \, \exists \delta > 0 \, \forall f\, \forall x \cdots$. In words, $ \delta$ is chosen independently of both $ f$ and $ x$. Independence of $ f$ is the "equi-" part, independence of $ x$ is the "uniformly" part.

A lot of times people will say "equicontinuous" and mean "uniformly equicontinuous". It's just because you don't often see a set of functions that is equicontinuous at one point only. ($ \{x\sin(nx)\colon n\ge 1\}$ would be an example of this kind)


I'm in 8th grade now, and I'm gonna be taking the AMC 10/12 next year. I took the AMC 8 and 10 this year, I got a perfect score on 8, but I only got 117 on AMC 10.

What are some good courses and texts that are useful for the AMC tests?

Thanks in advance.

use the search function... aops 1 for 10 and 2 for 12 and aime and usamo basics...


logarithms
number theory proposed
number theory

Let \[ f(n)=  \sum_{p|n , \;p^{\alpha} \leq n < p^{\alpha+1} \ } p^{\alpha} .\] Prove that \[ \limsup_{n \rightarrow \infty}f(n) \frac{ \log \log n}{n \log n}=1 .\]

[i]P. Erdos[/i]

I am able to show that $$\limsup_{n\to\infty}f(n)\dfrac{\log\log n}{n\log n}\le 1.$$ I will continue to work on finding the lower bound.

[hide="Upper Bound"]
We note that the condition $p^{\alpha}\le n\le p^{\alpha+1}\Rightarrow \alpha=\lfloor \log_p n\rfloor.$ Hence, $$f(n)=\sum_{p|n} p^{\lfloor \log_p n\rfloor}\le n\omega(n),$$ where $\omega(n)$ denotes the number of distinct prime factors of $n$. Hence, $$\limsup_{n\to\infty}f(n)\dfrac{\log\log n}{n\log n}\le \limsup_{n\to\infty}\dfrac{\omega(n)\log\log n}{\log n}.$$ Hence, we need to investigate this last term. Let $\text{rad}(n)$ denote the product of all the distinct prime factors of $n$ so that $\omega(\text{rad} n)=\omega(n).$ Hence, $$\dfrac{\omega(n)\log\log n}{\log n}\le \dfrac{\omega(\text{rad} n)\log\log \text{rad} n}{\log \text{rad} n}.$$ It follows that in our investigation of $$\limsup_{n\to\infty}\dfrac{\omega(n)\log\log n}{\log n}$$ we should only consider where $n$ is square-free. Now, define the function $\theta(n)=\sum_{p\le n}\log p.$ It is well known that the Prime Number Theorem is equivalent to $$\lim_{n\to\infty}\frac{\theta(n)}{n}=1.$$ Letting $n=p_k$ where $p_k$ denotes the $k$'th prime, we get $$p_1p_2...p_k\sim e^{p_k}.$$ Hence, If $n=p_1p_2...p_k$ (i.e. we look at the particular sequence when $n$ is primorial) then $$\dfrac{\omega(n)\log\log n}{\log n}\sim \dfrac{k\log p_k}{p_k}\to 1$$ by Prime Number Theorem. It follows that $$\limsup_{n\to\infty} f(n)\dfrac{\log\log n}{n\log n}\le 1.$$ [/hide]


quadratics
LaTeX
algebra
quadratic formula

[quote="Singular"]I think this is intermediate?  Pretty heavy computationally.

I propose simplified problem:

$\sqrt{x-5} + \sqrt{x-8} - \sqrt{x} = 0$[/quote]

Yes, Eminem's problem was way too confusing and would take a long time, so I'll just use your simplified version singular :lol:

$(sqrt(x-5)+sqrt(x-8))^2=sqrt(x)^2$
$x-5+2sqrt(x-5)(x-8)+x-8=x$
$2x-13+2sqrt(x-5)(x-8)=x$
$2sqrt(x-5)(x-8)=-x+13$
square both sides, and you get:
$4(x^2-13x+40)=x^2-26x+169$
bring all the like terms together and you get:
$3x^2-26x-9=0$
Use the quadratic formula to get:

$x=-1/3,9$

[quote="Bob"]In eminem's problem, substitution will greatly simplify the work. Let $y=2x^2-3x$. Then the equation becomes $\sqrt{4y-39}+\sqrt{y-15}-\sqrt{y+20}=0$ which is not that bad[/quote]

Yes, you are right, wow, I shoulda thought of that :blush: And then when you get what y equals, then you can find x.

[quote="236factorial"][quote="chess64"]$\frac{-1}{3}$ is extraneous, so $\fbox {9}.$
[/quote]

How is -1/3 extraneous??[/quote]

Oh, I just realized that...because it would give you a negative in the square root which makes imaginary numbers.

Original problem has no solutions for x?

[quote="Stan17"]original problem has 2 solutions for x...try letting a=2x^2 - 3x -15...then you have sqrt(4a+21) + sqrt(a) = sqrt(a+35)[hide]solving for a...you get a=1...then for x...we get x= (3,plus-minus, sqrt(137))/2...[/hide][/quote]
let me $\LaTeX$ his post
solve $\sqrt{8x^2-12x-39}+\sqrt{2x^2-3x-15}-\sqrt{2x^2-3x+20}=0$
[hide="solution"]like Stan17 suggested, let $y=2x^2-3x-15$
lets cut it short and save our breath... from Stan17 we have
\[\sqrt{4y+21}+\sqrt{y}=\sqrt{y+35}\]
if solved in the same way as from previous posts, this time is nice and we only get $y=1$
now we have $y=2x^2-3x-15=1$ and solving we find that $x=\frac{3\pm\sqrt{137}}{4}$
[/hide]
its so much easier to read when i $\LaTeX$ his post ;)

thanx  :D  guys for your cool solutions!


probability
ratio
geometry
combinatorial geometry
probability and stats

Four points are randomly placed in a triangle, find out the probability that the convex hull of these four points is not a triangle. (answer: 2/3)

You are right. Though it is easy to convert the original probability problem to this expectation problem, it is still difficult to calculate the expectation of the small triangle's area, right?

[quote="jmerry"]Don't expect too much elegance. Variations of this can be [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122451]absurdly difficult[/url].

The simplest approach is to ask this question: after placing three points to form a triangle, what is the probability that the fourth point is inside that triangle? It's the ratio of that triangle's area to the big triangle's area; we just have to find the average of this.[/quote]

Thanks a lot. :)

[quote="roah"]Ibm ponder this challenge:
[url]http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/Challenges/October2006.html[/url][/quote]


In triangle ABC, let the A-excircle touch BC at D. Let the B-excircle of triangle ABD touch AD at P and let the C-excircle of triangle ACD touch AD at Q. Is P=Q true for all triangles ABC?


geometry
inradius
ratio
trigonometry
incenter
angle bisector

For the orthocenter $ H$ of $ \triangle ABC$ denote $ \|\begin{array}{c}
D\in AH\cap BC\\\\
E\in BH\cap CA\\\\
F\in CH\cap AB\end{array}$ .

Define the orthocenter $ H_1$ of $ \triangle AEF$ and $ D_1\in AH_1\cap EF$ .

Prove that $ H_1D_1$ is equally with the inradius of $ \triangle DEF$ .

[hide="solution"]$ \triangle EFH$ and $ \triangle EFH_1$ are equal cause $ HEFH_1$ is a parallelogram. Then the thesis is equivalent to show that the altitude from $ H_1$ to EF i sequal to the altitude from H to EF that is true.[/hide]

[hide="alternate approach"]
$ \triangle AEF\sim\triangle ABC$ by ratio $ \cos\angle A$, so $ H_1D_1=HD\cos\angle A$. But since $ \angle HDF=\angle HBF=90-\angle A=\angle HCE=\angle HDE$, we have that $ DH$ is an angle bisector of $ \angle FDE$. Similarily, $ FH$ and $ EH$ are angle bisectors, and $ H$ is the incenter of $ \triangle DEF$. Draw the perpendicular from $ H$ to $ DF$; this altitude is the radius $ r$ of the incenter of $ \triangle DEF$. Then $ r=HD\sin(90-\angle A)=HD\cos\angle A$, and hence, $ H_1D_1=r$.
[/hide]


trigonometry
ratio
geometry
geometry proposed

Let $ ABC$ be a triangle. Consider three points $ D$ , $ E$ , $ F$ which belong to the [u]sidelines[/u] $ BC$ , $ CA$ , $ AB$

respectively. Denote $ P\in AD\cap EF$ . Prove that there is the relation $ \boxed {\frac {PF}{PE}\equal{}\frac {DB}{DC}\cdot\frac {AF}{AE}\cdot \frac {AC}{AB}}$ .

Draw parallels to $ BC$ through $ E$ and $ F$, intersecting $ AD$ at  $ E'$ and $ F'$ respectively.
Then $ \frac{EP}{PF} \equal{} \frac {EE'}{FF'}$ ( 1 ), but $ \frac {FF'}{BD} \equal{} \frac {AF}{AB}$ ( 2 ) and $ \frac {CD}{EE'} \equal{} \frac {AC}{AE}$ ( 3 ).
Substituting in (1) the values of $ EE'$ and $ FF'$ from (2) & (3), we get the required relation.

Best regards,
sunken rock

[hide]Let $ \angle DAB\equal{}\alpha_1,\angle DAC\equal{}\alpha_2,\angle APF\equal{}\theta,\angle ADB\equal{}\gamma$. We have $ \frac{PF}{AF}\cdot\frac{AE}{PE}\equal{}\frac{\sin\alpha_1}{\sin\theta}\cdot\frac{\sin\theta}{\sin\alpha_2}\equal{}\frac{\sin\alpha_1}{\sin\alpha_2}$ and $ \frac{DB}{AB}\cdot\frac{AC}{DC}\equal{}\frac{\sin\alpha_1}{\sin\gamma}\cdot\frac{\sin\gamma}{\sin\alpha_2}\equal{}\frac{\sin\alpha_1}{\sin\alpha_2}$. So $ \frac{PF}{AF}\cdot\frac{AE}{PE}\equal{}\frac{DB}{AB}\cdot\frac{AC}{DC}$ and hence $ \frac{PF}{PE}\equal{}\frac{DB}{DC}\cdot\frac{AF}{AE}\cdot\frac{AC}{AB}$. Q.E.D.[/hide]

[quote="Virgil Nicula"] [color=darkred]Let $ ABC$ be a triangle. Consider three points $ D$ , $ E$ , $ F$ which belong to the [u]sidelines[/u] $ BC$ , $ CA$ , $ AB$

respectively. Denote $ P\in AD\cap EF$ . Prove that there is the relation $ \boxed {\frac {PF}{PE} \equal{} \frac {DB}{DC}\cdot\frac {AF}{AE}\cdot \frac {AC}{AB}}$ .[/color] [/quote]
[color=darkblue][b][u]Proof[/u].[/b] Denote $ \left\|\begin{array}{c}
m(\angle DAB)\equal{}m(\angle PAF)\equal{}x\\\\
m(\angle DAC)\equal{}m(\angle PAE)\equal{}y\end{array}\right\|$ . Using well-known relations

$ \left\|\begin{array}{c}
\frac {DB}{DC}\equal{}\frac {AB}{AC}\cdot\frac {\sin\widehat {DAB}}{\sin\widehat {DAC}}\equal{}\frac {AB}{AC}\cdot\frac {\sin x}{\sin y}\\\\
\frac {PE}{PF}\equal{}\frac {AE}{AF}\cdot\frac {\sin\widehat {PAE}}{\sin\widehat {PAF}}\equal{}\frac {AE}{AF}\cdot\frac {\sin y}{\sin x}\end{array}\right\|\ \bigodot\ \implies\ \frac {PF}{PE} \equal{} \frac {DB}{DC}\cdot\frac {AF}{AE}\cdot \frac {AC}{AB}$ .[/color]

This is a nice and useful relation mr Virgil , May I ask? I'm wondering what application do you have in mind?. This relation seems like having a conection with Cristea's Theorem.

$ \vec{AD} \equal{} \frac {BD}{BC}.\vec{AC} \plus{} \frac {CD}{BC}.\vec{AB}\Leftrightarrow \frac {AP}{AD}.\vec{AP} \equal{} \frac {BD}{BC}.\frac {AC}{AE}.\vec{AE} \plus{} \frac {CD}{BC}.\frac {AB}{AF}.\vec{AF}$
On the other side, $ \vec{AP} \equal{} \frac {PF}{EF}.\vec{AE} \plus{} \frac {EP}{EF}.\vec{AF}$
Therefore $ \dfrac{\frac {PF}{EF}}{\frac {BD}{BC}.\frac {AC}{AE}} \equal{} \dfrac{\frac {EP}{EF}}{\frac {CD}{BC}.\frac {AB}{AF}} \Rightarrow QED$

The more powerful result with directed ratio is also right.

Let $ \overline{S_{XYZ}}$ denotes the directed area of $ \triangle XYZ$. We have:

$ \frac{\overline{PF}}{\overline{PE}} \equal{} \frac{\overline{S_{APF}}}{\overline{S_{APE}}} \equal{} \frac{AF}{AE} \cdot \frac{\sin{(AP,AF)}}{\sin{(AP,AE)}} \equal{}$
$ \equal{} \frac{AF}{AE} \cdot \frac{AC}{AB} \cdot \frac{AB \cdot \sin{(AP,AB)}}{AC \cdot \sin{(AP,AC)}} \equal{} \frac{AF}{AE} \cdot \frac{AC}{AB} \cdot \frac{\overline{S_{ADB}}}{\overline{S_{ADC}}} \equal{} \frac{AF}{AE} \cdot \frac{AC}{AB} \cdot \frac{\overline{DB}}{\overline{DC}}$

Mr.[b]Tiger100[/b], could you stated the Cristea's theorem?


Two particles, each of mass $ m$, are connected by a light string of length $ 2L$.  A steady force $ \textbf{F}$ is applied at the midpoint of the string ($ x\equal{}0$) at a right angle to the inital position of the string.  Show that the acceleration of each mass in the direction $ 90^{\circ}$ to $ \textbf{F}$ is given by
$ a_{x}\equal{}\frac{F}{2m}\frac{x}{\sqrt{L^{2}\minus{}x^{2}}}$
in which $ x$ is the perpendicular distance of one of the particles from the line of action of $ \textbf{F}$.

Consider $ x>0$
Let $ T$ be tension in string, $ \theta$ the angle between string and line of force. We get two equations,
$ 2T\cos\theta\equal{}F$
$ T\sin\theta\equal{}ma_x$
Using $ \tan\theta\equal{}\frac{x}{\sqrt{L^2\minus{}x^2}}$ we get the answer.

I really do not know how to deal with $ x\equal{}0$  :(


From the [url=http://www.unl.edu/amc/e-exams/e9-imo/e9-1-imoarchive/2005-ia/2005imoteam.html]AMC site[/url]:

Robert Cordwell
Grade 12
Manzano High School
Albuquerque, NM

Sherry Gong
Grade 10
Philips Exeter Academy
Exeter, NH

Hyun Soo Kim
Grade 12
Academy for the Advancement of Science and Technology
Hackensack, NJ

Brian Lawrence
Grade 10
Montogomery Blair High School
Silver Spring, MD

Thomas Mildorf
Grade 12
Thomas Jefferson High School of Science and Technology
Alexandria, VA

Eric Price
Grade 12
Thomas Jefferson High School of Science and Technology
Alexandria, VA

I see two who took a lot of AoPS classes and another very esteemed community member  :coolspeak:

Congratulations everyone! :)

Good luck at the IMO, make America proud! lol.

Congrats to all, especially HSK and Robert from the Clay Academy  :lol:

[quote="rrusczyk"]I'll note that we had four Clay Academy scholars in the Olympiad Problem Solving class in the spring (at least).[/quote]

Who is it?  :)

[quote="orl"][quote="rrusczyk"]I see two who took a lot of AoPS classes and another very esteemed community member  :coolspeak:[/quote]

The community guy is Mildorf it seems. And who are the others?  :)[/quote]

Hyun Soo Kim (Stryker) posted fairly regularly in the olympiad number theory and olympiad inequalities sections, or at least he did in the past.

[quote="chess64"]Go Brian!!! (MD pride, lol  :P ) Wasn't Zach Abel expected to be on the team?  :?[/quote]

Yay! MD pride!

Does anyone know what Zach scored on the TST? (psstt Zach, what'd you score?) :P

Didn't he decide to go to RSI instead? :? (though I don't actually know if he took the TST or not)

The TSTs are never returned, so I don't know my score. Any number I might give would be only speculation. And yes, I decided to go to RSI. Sorry to dissapoint you, chess64 ;) Maybe next year....

Sorry to give such an un informative answer :(

[quote="mna851"]Zack, do you think anyone will ever tell you if you would have made IMO had you decided not to go to RSI?  If you would have, would you regret going to RSI, instead of IMO?  Sorry to make this seem like an interview.[/quote]

No, they won't tell me, and I wouldn't want them to. Of course I would be sad about missing such a wonderful opportunity (I think anyone would), but I have no regrets about my choice.

Geez so many specific questions about the test and everything else. Relax everyone. And of course best of luck to the team.

Good luck at the IMO team USA !!! :P  :lol:

Just a brief aside to prevent further name ambiguity - I do prefer being called Thomas over being Tom, and they have since adjusted the web site.


geometry
geometric transformation
function
vector
linear algebra
linear algebra unsolved

Show that the curve which has the property that all the normals passes through a fix point is a circle.

Assume to start with that we are working in $ \mathbb{R}^{2}.$

Relabeling by translation if necessary, assume that the point through which all of the normals pass is the origin. Assume we can parameterize the curve by a differentiable function $ \mathbf{r}(t).$ At any given point, the vector $ \mathbf{r}'(t)$ lies in the direction of the tangent to the curve. Any normal $ \mathbf{n}$ must be perpendicular to the tangent, hence $ \mathbf{n}\cdot\mathbf{r}'(t)\equal{}0.$ But if the normal line passes through the origin, we must have that the normal direction $ \mathbf{n}$ lies in the same (or opposite) direction as $ \mathbf{r}(t).$ Hence, $ \mathbf{r}(t)\cdot\mathbf{r}'(t)\equal{}0$ for all $ t.$

Multiply by $ 2$ to get

$ 2\mathbf{r}(t)\cdot\mathbf{r}'(t)\equal{}\frac{d}{dt}\,\mathbf{r}(t)\cdot\mathbf{r}(t)\equal{}\frac{d}{dt}\,\|\mathbf{r}(t)\|^{2}\equal{}0.$

Integrate that to find that $ \|\mathbf{r}(t)\|$ is constant. So the curve lies at a constant distance from the origin and hence is contained in some circle. (We can't say that it is all of that circle, of course.)

We can find meaning in some analogous statements in $ \mathbb{R}^{n}$ for $ n\ge3,$ but we have to choose our descriptions with more care than the statement here.

I didn't understand why you said that n lies in the same direction as r(t). $ r(t)*r'(t)\equal{}0$ ?? They are not perpendicular.

The normal to the curve passes through the origin, so the normal to the curve is parallel to the position vector of the curve.  The normal and tangent of any curve are perpendicular, by definition.  Thus the position vector and tangent vector are perpendicular.


function
calculus
derivative
real analysis
real analysis unsolved

Find all the functions f:[0, \pi ]->R ,indefinite derivable and with the properties:
[(-1)^n][f^(2n)(x)]>=0 for every x
f^(2n)(0)=f^(2n)( \pi )=0 for every natural n.

(f^(2n) means the 2n derivative of f )


Let $ ABC$ be an equilateral triangle and $ \Gamma$ the semicircle drawn exteriorly to the triangle, having $ BC$ as diameter. Show that if a line passing through $ A$ trisects $ BC,$ it also trisects the arc $ \Gamma.$

Let's note D the point on (BC) trisecting it, i.e. BC = 3 BD, F the middle of AB and O the middle of BC. Obviously, the triangles ADC and FDB are similar, as AC/BF=CD/BD=2 and, consequently, <ADC=<FDB, therefore, AD is the symmetrical of DF w.r.t. BC. As F lies on the circle (O,OB), it's symmetrical w.r.t. BC will be E, the intersection of AD with the semicircle; further, as <BOF = 60 deg, we get <BOE=60 deg too, hence AD trisects the semicircle.

Best regards,
sunken rock

Complete the circle $\omega$ with diameter $BC$.
Let, $AC \cap \omega=F$, $D$ be the point on $\omega$ diametrically opposite to $F$, and, $BT=\frac{1}{3}BC$ where, $T\in BC$. Since, $BC$ is a diameter of $\omega$ & $\angle{BCF}=60^\circ$, $F$ is the vertex of a regular hexagon, and, as a result, so is $D$.
$\therefore$ $D$ trisects the semi-circular arc $\Gamma$. So, it suffices to show that, $D$ lies on $AT$.
Now,
($\because BF \perp AC$), $BD=CF=\frac{1}{2}AC$ $\Rightarrow \frac{BD}{AC}=\frac{1}{2}=\frac{BT}{TC}$ & $\angle{DBT}=\angle {TCA}=60^\circ$ (Regular-Hexagon Property). $\therefore \bigtriangleup{TBD}\sim \bigtriangleup{TBC} \Rightarrow \angle{BTD}=\angle{CTA} \Rightarrow$ $A$, $T$, $D$ are co-linear.


algebra
polynomial
advanced fields
advanced fields unsolved

I just read the following in AMM, and I found it beautiful.
Show that every integer appears as the coefficient of a cyclotomic polynomial.
(I don't know how difficult it is to prove, I just couldn't find a proof :( )


Okay, I don't know how to solve theses can somone post a detailed/explained answer for these? Thanks any help is good.

Suppose that function S is the inverse of function R and that R(5)=11.
a)R o S(5)                b)Find S o R(5)

Note:I captilized the r and s to make the o look smaller which its suppose to be the r and the s are lower case even though I don't think it makes a difference I put this note here

(o means you switch the things or whatever like F o G and you got function f(x)=x^2+1 and g(x)=Sqroot(x-1) then you can do this like f(g(x)=x because you get sqroot(x-1) in the function f(x) instead of a single number you get (sqrt(x-1))^2 you then get x-1+1=x)

Next Problem

Suppose that f(n)=4+f(n-1) and that f(1)=12

a)Find f(2)      B)Find f(3).     c) f(4)

hello, to the seconde problem
we have
$ f(2)\equal{}4\plus{}f(1)\equal{}4\plus{}12\equal{}16$
$ f(3)\equal{}4\plus{}f(2)\equal{}4\plus{}16\equal{}20$
$ f(4)\equal{}4\plus{}f(3)\equal{}4\plus{}20\equal{}24$
Sonnhard.

[quote="pcvrx540"]
Suppose that function S is the inverse of function R and that R(5)=11.
a)R o S(5)                b)Find S o R(5)

[/quote]

Since $ S(x)$ is the inverse of $ R(x)$, we have that $ R(S(x)) \equal{} R o S(x) \equal{} x$.

a) $ R o S(5) \equal{} 5$

b) $ S o R(5) \equal{} 5$

[quote="pcvrx540"]
Next Problem

Suppose that f(n)=4+f(n-1) and that f(1)=12

a)Find f(2)      B)Find f(3).     c) f(4)[/quote]

now for  such problems  u need to keep on plugging  the  values   and  find  any relation  or  anything  $ interesting$  if u  can.

there is  no general way to these questions.  well this  was  only  a  $ suggestion$

[quote]a)R o S(5) b)Find S o R(5)

Note:I captilized the r and s to make the o look smaller which its suppose to be the r and the s are lower case even though I don't think it makes a difference I put this note here[/quote]

To do this, you can use: $ R\circ S$

[code]$R\circ S$[/code]

Alternatively, it would be easier to just say R(S(5)).

The second problem is just a recursion, so you just add 4 each time you increment x by 1.

And btw, why does the title say "trig"? There does not appear to be any trig.


Nice inequality, Potla. :)

$ (b \plus{} c)^2\geq 4bc\Longleftrightarrow (b \minus{} c)^2\geq 0$, since $ a > 0,\ b > 0,\ c > 0$, we have $ \frac {a(1 \minus{} a)^2}{bc}\equal{}\frac {a(b \plus{} c)^2}{bc}\geq 4a$.

Similarly we have $ \frac {b(c \plus{} a)^2}{ca}\geq 4b,\ \frac {c(a \plus{} b)^2}{ab}\geq 4c$. Adding these inequality of both sides gives

$ \frac {a(1 \minus{} a)^2}{bc} \plus{} \frac {b(1 \minus{} b)^2}{ca} \plus{} \frac {c(1 \minus{} c)^2}{ab}$

$ \equal{} \frac {a(b \plus{} c)^2}{bc} \plus{} \frac {b(c \plus{} a)^2}{ca} \plus{} \frac {c(a \plus{} b)^2}{ab}\geq 4(a \plus{} b \plus{} c)\geq 4\ \because a \plus{} b \plus{} c \equal{} 1$.

Did   not have to toil much.However carry on. Hope ur 2nd one wil be a bit tougher :first:

Actually I had arrived here from:
$ (\frac {1}{a} \plus{} \frac {1}{b})^2\ge \frac {4}{ab} \equal{} (\frac {1}{a} \plus{} \frac {1}{b})^2 \plus{} (\frac {1}{a} \minus{} \frac {1}{b})^2$.
Summing up yields :
$ \sum_{cyc}(\frac {1}{a} \plus{} \frac {1}{b})^2\ge \frac {4}{abc}(a \plus{} b \plus{} c)$
$ \Longrightarrow \sum_{cyc}(\frac {a \plus{} b}{ab})^2\ge \frac {4}{abc}$
$ \Longrightarrow \sum_{cyc} \frac {c(1 \minus{} c)^2}{ab} \ge 4$
Hence the result. :)
I had inserted $ a \plus{} b \plus{} c \equal{} 1$  as to get a bit further.... :lol:

[quote="kunny"]
$ \equal{} \frac {a(b \plus{} c)^2}{bc} \plus{} \frac {b(c \plus{} a)^2}{ca} \plus{} \frac {c(a \plus{} b)^2}{ab}\geq 4(a \plus{} b \plus{} c)\geq 4 \because a \plus{} b \plus{} c \equal{} 1$[/quote]

The last part is not quite true, is it?
$ 4(a \plus{} b \plus{} c) \equal{} 4$, not $ \ge$. :lol:

However, my inequality was too childish for words! :|
Still, I felt adventurous while discovering/inventing this inequality........ :)

[quote="Potla"]Prove that for all $ a, b, c\in\mathbb{R}^{ + }$, such that $ a + b + c = 1$, we can have:
$ \frac {a(1 - a)^2}{bc} + \frac {b(1 - b)^2}{ca} + \frac {c(1 - c)^2}{ab}\ge 4$[/quote]
Multiplying both side by $ abc$ we get,
\begin{align*}&\sum_{cyc}a^2(1-a)^2\geq 4abc\\ \iff &\sum_{cyc} a^2-2\sum_{cyc}a^3+3\geq 4abc\end{align*}
Using AM-GM we get, $ \sum_{cyc}a^4+1\geq 4abc$
So we need to prove that, $ 2\geq 2\sum_{cyc}a^3 \iff 1\geq \sum_{cyc}a^3$
And $ 1=(a+b+c)^3\geq \sum_{cyc}a^3$
So, the proof is complete. Nice problem. :)

[quote="Potla"]Hello, this is my own first inequality which, of course is too loose for words, and also I do not think it worthwhile presenting here! :(
But still, here goes:
Prove that for all $ a, b, c\in\mathbb{R}^{ \plus{} }$, such that $ a \plus{} b \plus{} c \equal{} 1$, we can have:
$ \frac {a(1 \minus{} a)^2}{bc} \plus{} \frac {b(1 \minus{} b)^2}{ca} \plus{} \frac {c(1 \minus{} c)^2}{ab}\ge 4$

Please do not laugh at me for such a childish inequality.  :noo:
But, after all, it is my first one! :lol:[/quote]

[hide]
$ \frac {a(1 \minus{} a)^2}{bc} \plus{} \frac {b(1 \minus{} b)^2}{ca} \plus{} \frac {c(1 \minus{} c)^2}{ab}\iff \frac {a(b\plus{}c)^2}{bc} \plus{} \frac {b(c\plus{}a)^2}{ca} \plus{} \frac {c(a\plus{}b)^2}{ab}$

Apply AM._GM. for

$ \frac {a(b\plus{}c)^2}{bc} \plus{} \frac {b(c\plus{}a)^2}{ca} \plus{} \frac {c(a\plus{}b)^2}{ab}\ge \frac {4abc}{bc} \plus{} \frac {4abc}{ca} \plus{} \frac {4abc}{ab}$

$ \equal{}4a\plus{}4b\plus{}4c\equal{}4(a\plus{}b\plus{}c)\equal{}4$

$ \therefore \frac {a(1 \minus{} a)^2}{bc} \plus{} \frac {b(1 \minus{} b)^2}{ca} \plus{} \frac {c(1 \minus{} c)^2}{ab}\ge 4$
[/hide]

Actually Cauchy works here very well.

The inequality is equivalent to
$ \sum \frac {a(b \plus{} c)^2}{bc} \ge 4 < \equal{} \equal{} \equal{} > {4(ab \plus{} bc \plus{} ca)^2} \ge  12 abc \equal{} 12abc(a \plus{} b \plus{} c) < \equal{} \equal{} \equal{} > \sum (ab \minus{} bc)^2 \ge 0$ which is truly true. :wink:
Not a bad one Potla!!!

Thank you to you all for your interest and your kind coordination :blush:
Potla.

[quote="kunny"][quote="Potla"][quote="kunny"]
$ \equal{} \frac {a(b \plus{} c)^2}{bc} \plus{} \frac {b(c \plus{} a)^2}{ca} \plus{} \frac {c(a \plus{} b)^2}{ab}\geq 4(a \plus{} b \plus{} c)\geq 4 \because a \plus{} b \plus{} c \equal{} 1$[/quote]

The last part is not quite true, is it?
$ 4(a \plus{} b \plus{} c) \equal{} 4$, not $ \ge$. :lol: [/quote]

You are right.

Ptola, your solution is nice, but you seemed to make a typo in the first line.[/quote]
Yeah, I meant my first line to be:
$ (\frac{1}{a}\plus{}\frac{1}{b})^2\ge\frac{4}{ab}\equal{}(\frac{1}{a}\plus{}\frac{1}{b})^2\minus{}(\frac{1}{a}\minus{}\frac{1}{b})^2$ :blush:

[quote="vinskman"][quote="Potla"]Hello, this is my own first inequality which, of course is too loose for words, and also I do not think it worthwhile presenting here! :(
But still, here goes:
Prove that for all $ a, b, c\in\mathbb{R}^{ \plus{} }$, such that $ a \plus{} b \plus{} c \equal{} 1$, we can have:
$ \frac {a(1 \minus{} a)^2}{bc} \plus{} \frac {b(1 \minus{} b)^2}{ca} \plus{} \frac {c(1 \minus{} c)^2}{ab}\ge 4$

Please do not laugh at me for such a childish inequality.  :noo:
But, after all, it is my first one! :lol:[/quote]

[hide]
$ \frac {a(1 \minus{} a)^2}{bc} \plus{} \frac {b(1 \minus{} b)^2}{ca} \plus{} \frac {c(1 \minus{} c)^2}{ab}\iff \frac {a(b \plus{} c)^2}{bc} \plus{} \frac {b(c \plus{} a)^2}{ca} \plus{} \frac {c(a \plus{} b)^2}{ab}$

Apply AM._GM. for

$ \frac {a(b \plus{} c)^2}{bc} \plus{} \frac {b(c \plus{} a)^2}{ca} \plus{} \frac {c(a \plus{} b)^2}{ab}\ge \frac {4abc}{bc} \plus{} \frac {4abc}{ca} \plus{} \frac {4abc}{ab}$

$ \equal{} 4a \plus{} 4b \plus{} 4c \equal{} 4(a \plus{} b \plus{} c) \equal{} 4$

$ \therefore \frac {a(1 \minus{} a)^2}{bc} \plus{} \frac {b(1 \minus{} b)^2}{ca} \plus{} \frac {c(1 \minus{} c)^2}{ab}\ge 4$
[/hide][/quote]

Isn't that my solution? :wink:


prove that exist a graph with  order 4k that has this property:
the complement of graph is homomorphism with graph.

can we say this for a graph with order 4k+1?

Here's a silly construction:  the unique graph with zero vertices and the unique graph with one vertex both are isomorphic to their complements.  Suppose that $ G$ is isomorphic to the complement of $ G$ with isomorphism $ \alpha$.  Then define a new graph $ G'$ such that $ V(G') \equal{} V(G) \cup \{t, u, v, w\}$ and $ E(G') \equal{} E(G) \cup (V(G) \times \{t, w\}) \cup \{tu, uv, vw\}$, i.e., we add four new vertices forming a three-path such that the endpoints of the path are connected to every preexisting vertex of $ G$ and the middle vertices of the path are not connected to any vertex of $ G$.  It's easy to check that this graph is isomorphic to its complement: the map $ \alpha'$ defined by $ \alpha'(x) \equal{} \alpha(x)$ if $ x \in V(G)$, $ \alpha'(t) \equal{} u, \alpha'(u) \equal{} t, \alpha'(v) \equal{} w, \alpha'(w) \equal{} v$ is an isomorphism bewteen $ G'$ and the complement of $ G'$.  This gives us a graph isomorphic to its complement of order $ 4k$ and $ 4k \plus{} 1$ for every $ k$.

Of course, no such graph can exist of order $ 4k \plus{} 2$ or $ 4k \plus{} 3$ because $ \binom{4k \plus{} 2}{2}$ and $ \binom{4k \plus{} 3}{2}$ are both odd.


hey. stop!

we have a paradox!!!

let's call it "Turtle paradox" because i love turtles :)

Provided that you aren't blind nor got something really annoying tapping your eyes, it's easy to see that:

$8+3 = 11$
$1+1 = 83$

Hence $11+83 = 8131$

That's why we can assume that $\infty+m = minfinity = n$, with $n$ tending to a Random Number.

Ok, $\infty$ is $8$ rotated 90º, roughly, and $m$ is like $3$ rotated 90º, so we must rotate the $11$ 90º to get $=$.

If that was what the intended solution was, it might be better to use $\omega$ than $m$

[quote="Fabiomr1"]That's why we can assume that $\infty+m = minfinity = n$,[/quote]

Therefore, it's the minimum finity, so that's the smallest known negative. :wink:

[quote="yagaron"]It's well-know that $8+3 =11$

but, how can i calculate $\infty+m = ?$[/quote]

you don't need to dude
prove compactness of proposition spaces, or get yourself a pizza

dear The Zuton Force

the equation it's a little bit non-sense like you said

$8+m = =$ double equal?

maybe we can try put this equation in latin, i hope if it is a good idea


analytic geometry
geometry
3D geometry
function
combinatorics unsolved
combinatorics

Suppose that $A$ and $B$ are two distinct lattice points in $\mathbb{R}^n$ with non-negative integer coordinates. We say that $A$ dominates $B$ (denote by $A > B$) if all the components of $A - B$ are nonnegative. For example $(3, 5, 4) > (2,5,1)$, but $(3,5,4) \ngtr (1,6,3)$.

(1) Suppose that $S$ is a infinite set of lattice points in $\mathbb{R}^n$ with non-negative coordinates. Must there exist an infinte sequence of points, $A_1, A_2, A_3, \ldots \in S$, such that $A_1 < A_2 < A_3 < \cdots$?

(2) Suppose that $T$ is a set of lattice points in the box $[0, x_1] \times [0, x_2] \times \cdots [0, x_n]$ (where $x_1, x_2, \ldots, x_n$ are fixed nonnegative integers). It is known that no element in $T$ dominates another element. What is the maximum value of $|T|$?

The answer to the first question is yes (what can I do? I'm solving the easy one :)).

There exists a subsequence $A_{k_1},A_{k_2},\ldots$ s.t. the sequence formed by the first coordinates of $A_{k_i}$ is non-decreasing. We repeat the procedure on this sequence, looking at the second coordinate, then at the third, and so on until we reach the $n$'th.

What is the answer for the cube $[0,n]^3$ for the (2)-nd question? :?

I think something like it has been posted before, right? :? I seem to remember a proof that for any antichain with $k$ elements we can find an antichain with $k$ elements s.t. the sum of the coordinates is the same for all of the $k$ points.

I think it was problem about scores: some students were solving 3 problems and each problems is scored from 0 to 7. it is known that thre is no dominating student. prove that the number of students is at most 49.

i think that the following solves the problem(in principle):(hopefully there aren't (m)any mistakes!).
if we have ranked posets $\mathcal{P},\mathcal{Q}$ then we can form the ranked poset $\mathcal{P}\times\mathcal{Q}:=\{(p,q)|p \in \mathcal{P},q \in \mathcal{Q}\}$ and the order induced in the natural way:$(p_1,q_1) < (p_2,q_2)$ iff $p_1<p_2$ and $q_1<q_2$ and further,for the ranks we have $R=r_{\mathcal{P}}+r_{\mathcal{Q}}$.Here in the present problem,the poset is $\mathcal{P}=\mathcal{P}_1\times \mathcal{P}_2\times\cdots \mathcal{P}_n$ where $\mathcal{P}_i=\{0,1,\ldots,n_i\}$,thus each of these is simply a chain.the rank function for $\mathcal{P}$ is $R(x_1,x_2,\ldots,x_n)=x_1+x_2+\cdots x_n$. We wish to find the size of a maximal antichain of $\mathcal{P}$.
firstly as grobber observed,elements of the same rank obviously form an antichain.thus if $n_r=|\{(p,q)|R(p,q)=r\}|$ then $m=\max\{n_r,r\geq 0\}$ is a lower bound for the size of a maximal antichain.thus if we invoke Dilworth's theorem,it suffices to get a decomposition of $\mathcal{P}$ into $m$ chains.
we do the following:suppose we have a decomposition of a ranked poset $\mathcal{P}$ into symmetric chains,i.e. chains $C_i=c_0<c_1<\cdots <c_k$ which further satisfy $R(c_i)+R(c_{k-i})=R(\mathcal{P}),R(c_{i+1})=R(c_i)+1\ \forall\ i$,and likewise for the poset $\mathcal{Q}$, then we can obtain such a symmetric chain decomposition for $\mathcal{P}\times \mathcal{Q}$ as well.Suppose $\mathcal{P}=C_0\cup C_1\cup \cdots \cup C_k$ and $\mathcal{Q}=D_0\cup D_1\cup \cdots \cup D_l$ where $C_i,D_j$ are all symmetric chains, then we obtain such a chain decomposition for $\mathcal{P}\times \mathcal{Q}$ as follows:
   take $C_i,D_j$.write the set $C_i\times D_j$ as the array
$\begin{array}{cccc}(c_0,d_0)&(c_1,d_0)&\cdots&(c_k,d_0)\\(c_0,d_1)&(c_1,d_1)&\cdots&(c_k,d_1)\\ \\(c_0,d_l)&(c_1,d_l)&\cdots&(c_k,d_l) \end{array}$

Now read off the chains as $(c_0,d_0)<(c_1,d_0)<\cdots (c_k,d_0)<(c_k,d_1)<\cdots (c_k,d_l)$ and so on(basically read starting with the ith row and read horizontally till you reach an already listed element and then read vertically after that).it is easy to check that this is again a symmetric chain.(there is a corresponding result of de Bruijn for the poset of all subsets of a set;i am just using the same proof here as well.possibly even this result is well known).

Thus since each $\mathcal{P}_i$ is a symmetric chain to start with,it follows inductively that $\mathcal{P}$ as defined above also admits a symmetric chain decomposition.
now finally, since in each symmetric chain there is exactly one element of rank=$[R(\mathcal{P})/2]$,it follows that the number of chains = number of elements of rank= $[R(\mathcal{P})/2]$.
finally it is easy(i guess so!) to check that $m=[R(\mathcal{P})/2]$.


search
combinatorics unsolved
combinatorics

Define $ f(n,m)$ (with $ n$ and $ m$ being positive integers) to be the length of the longest possible sequence of real numbers so that sum of any consecutive $ n$ terms is positive and sum of any consecutive $ m$ terms is negative. Find a closed form for $ f(n,m)$.

Imo 1977 prob 2: [url]http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1977&p=367411[/url]

and your generalization at  [url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1728742661&t=61033[/url]


geometry
calculus
AMC
AIME
MATHCOUNTS
number theory
puzzles

Random, probably easy, and pointless.
942, 12, 109, 294, 630...
What are the next several terms?

On-Line Encyclopedia of Integer Sequences doesn't have this :( .

[quote="LynnelleYe"]Random[/quote]
I agree. :)

 [quote="LynnelleYe"]probably easy[/quote]
I strongly disagree. :(

[quote="LynnelleYe"]and pointless. [/quote]
:?:

[hide="Hint"]This may be easier for those who fit into the following categories:
1. You live in the U.S. (Mainly. Canada and a few other countries might work too, depending on circumstances)
2. You are in middle school or high school.
3. The problems in the Getting Started forum are generally too simple for you.[/hide]

1. I lived most of my life in korea.
2. I am 2 weeks away from graduating high school.

[quote="LynnelleYe"]
1. You live in the U.S. (Mainly. Canada and a few other countries might work too, depending on circumstances).[/quote]
I lived in Canada for 3 years.

[quote="LynnelleYe"]2. You are in middle school or high school.[/quote]
I'm finishing grade 11.

[quote="LynnelleYe"]3. The problems in the Getting Started forum are generally too simple for you.[/quote]
I haven't been at the forum.

I fit perfectly! How come I cannot solve this? :(

In math, you can't figure out with definite certainty what the next term is. All you can ask for is the most logical next term. I once read of a professor, who was given the first five terms of a certain sequence, was asked for the next, and said something like "ten trillion and four". When asked the sequence he had in mind, he simply said "the one with the first five terms equal to those you mentioned and all the rest equal to ten trillion and four".  :D  :D  :D

But for the most logical term...

942, 12, 109, 294, 630... :huuh:

Please, please, please, as a hint: What fields of math does it apply? Geometry, algebra, number theory, calculus (eek!!!)...

Help!!!

[quote="LynnelleYe"]It isn't really a math question. I mean it's related, but you don't need to do any math to figure out the answer. You could say it's a math [i]contest[/i] question, by which I mean a question about math contests. By which I mean one math contest in particular. Note that said contest must have some relevance to me, a former 7th grader in the U.S. I think I've said enough.[/quote]

Haha. I thought that 942 looked familiar...

[hide]45[/hide]

I fit perfectly too!!! But I still don't understand it. :huh:  :huh:  :huh:  :huh:

[quote="Geehoon"]No wonder! I got #3 wrong![/quote]

So did I. Haha, but when LynnelleYe mentioned a certain math contest in the US I immediately thought AMC and 942 rang a bell.

[quote="LynnelleYe"]$a_n$ is the correct answer to question #n on this year's AIME I. :lol:[/quote]
I never looked at AIME. No wonder I couldn't solve this. :D

Me neither. I was thinking of something subtle like Mathcounts ("Middle School") and wondered what "high school" was doing there.

I never looked at AIME.


[quote="MithsApprentice"]Lol. In the spirit of this topic, I propose 100x100 pixels on avatars. :D[/quote]

What does "lol" mean???

Hmm....yes, it's laugh out loud --- And that's the reason you get this smiley --  :lol: when you give lol within :'s

i actually said 'lol' when i saw the topic. lol.

lol is, like someone said above, laugh out loud. you said it when something is funny.

[quote="filletwho"]I've heard people say "lol" in real life. It's sorta scary.[/quote]

Those ppl are eager chatters  :roll:

[quote="puuhikki"]lol is the absolute value of o.[/quote]

u`re very funny  :rotfl:  good joke  :clap:

You can say it either way. I've heard looole before.
My brother still thinks it means loser.

[quote="mysmartmouth"]When you say it out loud, do you say "low-l" or "ell-ow-ell"   :?:[/quote]

Iff I say it, which is unusual(I can't remember when I did), I pronounce it lahl.

i don't say it in conversation... i think it when i see something silly. say law-l.

Yeah i "think" is ell-oh-ell.  I never actually say it.  LOL.

el-oh-el...but it seems weird in conversation

:rotfl:

My principal (first year at the school, poor guy) was trying to be...you know, cool or something...the other day, during an assembly. He was talking about acronyms, as an introduction to some inspirational word/acronym he'd come up with*, and he said some approximately like the following:

"And for those of you who are savvy with computers and the Internet, LOL means 'Lots Of Love.'"

Unsurprisingly, everyone burst out laughing. I think he thought we were laughing with him, as opposed to at him.

*Our esteemed principal is notoriously bad at coming up with inspirational acronyms. At the beginning of the year, he tried to introduce "S.T.R.E.A.K." as the school word.

I never do use lol.If I'm laughing,I let my emotions flow ..Aaaaaaahahaha :D

lol is dutch for fun. :) so we actually say it in real life.

[quote="kyyuanmathcount"]Lol is the most overused word.

Please restrain from using it, except when the joke is ACTUALLY funny.[/quote]

LOL!

No seriously, people say it after jokes now a days? I've pretty much gotten accustomed just to someone randomly out of the blue saying lol after something that wasn't even a joke.

I use "Heh" for something that's not really funny but..meh, can't explain it

[quote="colts18"][quote="t0rajir0u"][quote="mysmartmouth"]When you say it out loud, do you say "low-l" or "ell-ow-ell"   :?:[/quote]

Normally:  "Loll."  (Rhymes with "Ball.")

Sarcastically:  "El-oh-el."

The way I say it when I don't feel like saying it the other two ways:  "Lole!"  (Rhymes with "Bowl.")[/quote]

wow, some ppl r just sad. how do we stoop down to these levels saying lol? its just saddening... y r we even discussing this?[/quote]
y mus ppl uze chatspeeeek? it makez it soz hard to undastand!

[quote="colts18"]wow, some ppl r just sad. how do we stoop down to these levels saying lol? its just saddening... y r we even discussing this?[/quote]
Of course, using ppl for people, r for are, and y for why is completely normal. This isn't a telegram, you don't have to pay per letter.

It means " I am stupid and should not be given respect".

how does one say lol...
like lawl
or spell it out like l-o-l

you spell it out like L-O-L.

Lol also stands for Lots of Lunatics

Lebecca of LunnybookFarm

some of you might get that

would "Loads of Lollipops" also work as lol?

lasy old Lancelot

the prince liked to sing and dance a lot........

A poem of LOL!

Last One Living

very scary......

Leprechauns On Lakes

Lovable old Labrador (that site can't have everything!)

i've been using "lol" so much now that when my friend told a funny story i said "lol".  I'm weird.

Just a sign how man's natural laughter is dying away in this age of IT!

Not me though! I can still laugh my lungs out at something that strikes the right chord! And I also have a circle of friends with the same natural flair. (Boy, were we a nuisance in the class!!) :rotfl:


How good are HMMT problems as practice for the AIME? Does it make sense for me to practice HMMT problems now for the AIME?

I personally used some of them to prepare for the AIME.
They are among the harder problems on the AIME usually, like equal to numbers 6-15.
They are all very high quality problems as well.

[quote="solafidefarms"]From what I've seen of HMMT problems, they're hard. So if you're going for the last 5 or so AIME problems, which I think you are, it makes sense I think. However, don't forget that Altheman was 3rd in Alg. at HMMT this year yet didn't do well on the AIME...[/quote]

That's true, but Alex got pwned because of the Combinatorics problems (correct me if I'm wrong, Alex) and made careless errors... which really isn't Algebra. Besides, this AIME was "quite a turkey," as Arnav would say.

I'd do the problems...they're good practice.

Good thing I made most of my careless mistakes at HMMT instead of AIME though...

Just keep in mind that HMMT is very focused on specific topics, so it's hard to tell how someone will do on the AIME based on an HMMT test. However, I suppose the General test might give some indication.

I also feel that the HMMT requires a little more intuition, i.e. guessing factorizations and stuff. They're good practice if you have time, but I doubt you have time to do all the practice AIME problems already available, so you should probably start there.

I'd say HMMT is reasonably good practice for AIME.  I used to use it when training for ARML, and I'd say the difficulties of the two are reasonably comparable.  Do keep in mind, though, that HMMT (like ARML) gives you 5 minutes per problem on average, whereas AIME gives you 12.

[quote="Pakman2012"]While I admit alot of the problems on HMMT are pretty good, there are so many questions where the real question is just figuring out what you need to do. The problems are often too difficult to be in the first 10 on the AIME. If you're running out of problems, and you're trying to get those last 5 on the aime.. i guess its good practice. I personally wouldn't try it.[/quote]

Actually, I'd say that 11-15 on the AIME (with the exception of these past few years) give me way more trouble than most HMMT problems.

[quote="Altheman"]hmmt is harder to approach, like there was a simpson line problem, either you know it or you don't...aime is more approachable in my opinion[/quote]

That's probably true. Something I like about the AIME is that it emphasizes insight more than straight-out knowledge of theorems. HMMT reminds me of a harder, free-response version of Mu Alpha Theta.

[quote="Pakman2012"]HMMT to me is more intimidating as a short contest. Its alot more about whether you can sift through the important information in a quick amount of time and see the clever solution. The AIME is definitely alot more clear in what they want you to do, and at that point, it becomes about being able to do that. HMMT is pretty vague. The problems have super slick solutions, that are really hard to see during the contest. I'm just a little bitter, I suppose. They're really not testing the same skills though.[/quote]

So then the HMMT problems are ~10-15 on the AIME? Or at least around that level?

[quote="Phelpedo"]Well, the first few HMMT problems are at or below AIME level.[/quote]

That's very true... how about questions 5 - 10?

Those would probably be the later AIME problems. I guess it all depends on your definition of difficulty. As has been said, some HMMT problems require the use of an obscure theorem, which could make them quite difficult in terms of the number of people who can come up with a solution.

The first few problems in HMMT are usually easy enough to be lower-medium AMC problems. Then it gets harder, but I'd say that #14-15 on the AIME is about the same difficulty as #10 at HMMT. AIME certainly gives you more time, but when I do HMMT problems I don't really find the solutions particularly "slick," but often obscure. They both have their pros and cons, but I've found that early AIME questions are more easily brute-forced and later HMMT problems are more easily solved by making general assumptions (that you may not necessarily be able to prove but still have the hunch that it's right - especially true for geometry).

I don't know that much about the HMMT Geo section, but for Comb and esp Alg, I think the solutions are not very computationally intensive and instead focus on problem solving strategy.

There are some HMMT problems that end up being obscure (a foreseeable problem) as a direct result of this (such as #10 on calc), but for the most part, I find HMMT problems more interesting.

Doing HMMT problems will help you with your strategy, but ultimately it will be practice AIME's that will allow you to develop your application, since in the end, strategic approach is only one of the many skills required for the AIME.


let a,b,c be 3 sides of ABC triangle. prove that
$ \frac{a.A\plus{}b.B\plus{}c.C}{a\plus{}b\plus{}c}\geq \frac{\pi }{2}$

$ \frac {\pi}{3}\le \frac {aA \plus{} bB \plus{} cC}{a \plus{} b \plus{} c} < \frac {\pi}{2}.$
[b]Proof:[/b]
[b]1). [/b]
$ (a\minus{}b)(A\minus{}B)\plus{}(b\minus{}c)(B\minus{}C)\plus{}(c\minus{}a)(C\minus{}A)>\ge 0\Rightarrow \\
\Rightarrow a(2A\minus{}B\minus{}C)\plus{}b(2B\minus{}A\minus{}C)\plus{}c(2C\minus{}A\minus{}B)\ge 0\Rightarrow \\
\Rightarrow a(3A\minus{}\pi)\plus{}b(3B\minus{}\pi)\plus{}c(3C\minus{}\pi)\ge 0\Rightarrow\\
\Rightarrow 3(aA\plus{}bB\plus{}cC)\ge(a\plus{}b\plus{}c).\pi\Rightarrow \frac{\pi}{3}\le \frac {aA \plus{} bB \plus{} cC}{a \plus{} b \plus{} c}.$
[b]2). [/b]
$ (b\plus{}c\minus{}a)A\plus{}(a\plus{}c\minus{}b)B\plus{}(a\plus{}b\minus{}c).C>0\Rightarrow \\
\Rightarrow a(B\plus{}C\minus{}A)\plus{}b(A\plus{}C\minus{}B)\plus{}c(A\plus{}B\minus{}C)>0\Rightarrow\\
\Rightarrow a(\pi\minus{}2A)\plus{}b(\pi\minus{}2B)\plus{}c(\pi\minus{}2C)>0\Rightarrow \\
\Rightarrow \pi.(a\plus{}b\plus{}c)>2(aA\plus{}bB\plus{}cC)\Rightarrow \frac{aA \plus{} bB \plus{} cC}{a \plus{} b \plus{} c} < \frac {\pi}{2}.$

[quote="viebboy"]let a,b,c be 3 sides of ABC triangle. prove that
$ \frac {a.A \plus{} b.B \plus{} c.C}{a \plus{} b \plus{} c}\geq \frac {\pi }{2}$[/quote]

A>B>C => a>b>c
Then, we USE  Chebyshev  ....

See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=306151]here[/url] , Applications #1 :oops:

thanks for all your help:D . everyday i have a lot of excercises and i have found that this forum is really intersting. here i can found many solutions for my excercises:)
1) a,b,c>0. prove that
$ \sum \frac {1}{a^{3} \plus{} b^{3} \plus{} abc}\leq \frac {1}{abc}$
2) let a,b,c be the 3 sides of a triangle. prove that
$ \sum a^{3}(b^{2} \minus{} c^{2}) < 0$
3) given $ 0\leq a,b,c\leq 2$
prove: $ 2(a \plus{} b \plus{} c) \minus{} (ab \plus{} bc \plus{} ac)\leq 4$
4) $ \sum_{i \equal{} 2}^{n}\frac {n \minus{} 1}{n!} < 1$
5) let a,b>0 prove that
$ \left ( 1 \plus{} \frac {a}{b} \right )^{n} \plus{} \left ( 1 \plus{} \frac {b}{a} \right )^{n}\geq 2^{n \plus{} 1}$
6) $ \frac {\left | a \minus{} b \right |}{1 \plus{} \left | a \minus{} b \right |}\leq \frac {\left | a \right |}{1 \plus{} \left | a \right |} \plus{} \frac {\left | b \right |}{1 \plus{} \left | b \right |}$
7) given x,y satisfy x+y>0
prove $ \frac {1}{1 \plus{} 4^{x}} \plus{} \frac {1}{1 \plus{} 4^{y}}\geq \frac {2}{1 \plus{} 2^{x \plus{} y}}$
8) $ \sum_{i \equal{} 1}^{n}\frac {1}{1 \plus{} a}\geq \frac {n}{1 \plus{} \sqrt [n]{a_{1}.a_{2}...a_{n}}}$
9) let k,m be integer and 0<k<m. prove
$ \sqrt [k]{\frac {1}{n}\sum_{i \equal{} 1}^{n}a_{i}^{k}}\leq \sqrt [m]{\frac {1}{n}\sum_{i \equal{} 1}^{n}a_{i}^{m}}$
with $ a_{i}\geq 0$
10) given  m\leq a,b,c,d\leq M and M>m>0
prove that $ \sum a\sum \frac {1}{a}\leq 16 \plus{} 4(\sqrt {\frac {m}{M}} \minus{} \sqrt {\frac {M}{m}})^{2}$