A 12-volt lead storage battery with a 20 ampere-hour capacity rating has a mass of 10 kg.
   (a) How many kilograms of lead sulfate is formed when this battery is discharged (molecular weight of $ \mbox{PbSO}_{4}$)?
   (b) How many kilograms of batteries of this type would be required to store the energy derived from 1 kg of gasoline by an engine of 20 percent efficiency (Heat of combustion of gasoline: $ 4.5\times 10^{4}$ joules/gm)?


Prove that if $ n$ is odd, then the odd prime divisors of $ F_n$ are $ \equiv 1 \mod 4$


geometry
geometric transformation
dilation

I just had a little trouble with this problem:

Write the net ionic equation for the reaction between aqueous solutions of HF and Barium hydroxide.

$ 2\,\,HF(aq) \plus{} Ba(HO)_2(aq) \longrightarrow BaF_2(aq) \plus{} 2\,\,H_2O(l)$

Is that a net ionic equation? I think Barium would be spectator ion.
HF   +     OH-     =   F-   + HOH
this would be an ionic equation, right?

rawr

2HF + OH- = F- + HF + HOH = HF2- + HOH

-__-

I think $ Ba(OH)_2$ is considered insoluble so barium would not be a spectator ion.

hydroxide salts are insoluble in water unless paired up with an alkali metal or barium.

HSAB (kinda) strikes again!

I am also going to assume it's not a trick question.
Then the net ionic reaction is H+ + OH-  = H2O.
Once you start with aqueous solutions of the base and acid, we assume they are dissolved. The barium hydroxide must be dilute enough to be in solution. The base and acid are going to be at least partly ionized. Barium fluoride is even a bit less soluble, but "soluble" and "insoluble" are relative terms. Their meaning depends on what you are trying to do. If the solutions were concentrated enough, some barium fluoride could precipitate, but I don't think that would be the intention of such a problem.

i don't think you can separte HF to H+ and F- because HF is not a strong acid. it's a weak acid.

actually you can. It is an equilibrium system, and only a portion of HF will dissociate into H+ and F-. (well, also HF2-...but...)

Explanation:
Assume a static system. The reason why a "weak" (Bronsted...easier to explain) acid is "weak" is because the dissociation of its conjugate base and proton is not complete. In other words, the original number of molecules of weak acid is not equal to the present number of molecules of conjugate base. That is what the acid dissociation constant, pKa, gives.

$ K_\mathrm{a}\equal{}\frac{[\mathrm{H}^\plus{}][\mathrm{base}^\minus{}]}{[\mathrm{acid}]}$

So given a concentration of acid and a pH, there is bound to be a certain concentration of conjugate base.

[quote="Jdlkem"]The reason why a "weak" (Bronsted...easier to explain) acid is "weak" is because the dissociation of its conjugate base and proton is not complete.[/quote]

Now, that's a five star explanation. And why the dissociation is not complete? That's the question you should answer.

Ah, before I forget, I'd like to add that a strong (Bronsted) acid is strong is because the resulting conjugate base is "strong" enough to remain in solution without being directly attached to its proton.

@ Carcul:
The gen chem answer: because there is an equilibrium constant small enough that there is a notable concentration of both acid and conjugate base.
The inorganic chem answer: HSAB, yo.  :lol:
The Hard and Soft Acids and Bases theory places cations/anions into different categories: hard, soft, and in between. The best explanation I've heard so far is that "soft" acids/bases are "squishy" in a way. The opposite can be said about hard acids/bases. "Soft" cations react with "Soft" anions to make a strongly bonded molecule, and "Hard" cations react with "hard" anions to also make a strongly bonded molecule. If the cation/anion is of a different type, then bonding is not strong. Examples include LiF ("hard" and "hard";not very soluble), HI ("hard" and "soft; very soluble), and BaSO4 ("soft" and "soft"; not very soluble). There is a "hardness" quantity, but I'm not too knowledgeable in the details, and it seemed to me kind of .... I'd have to look in my inorganic book to check it out.

So in this example with HF, H and F are both "hard" ions, and thus the bonding is strong.  :P

[quote="Jdlkem"]a strong (Bronsted) acid is strong is because the resulting conjugate base is "strong" enough to remain in solution without being directly attached to its proton.[/quote]

This could be shortened to "a strong (Bronsted) acid is strong because the resulting conjugate base is weak", and this happens when some special factor (like ressonance for example) "anormally" stabilizes the conjugate base.


Forgive me it was my book's mistake  :maybe:

Maybe this is your problem
http://www.mathlinks.ro/Forum/viewtopic.php?p=124421#p124421 ($x,y,z>0$).


Since i'm new in this forum, and my English is not so good, i'll try to explain my prob.

Here it is:
[img]http://img204.imageshack.us/img204/584/hinhveaq2.png[/img]

AB, AC are tangents
AMN, APQ are secants
A is chosen randomly (outside of the circle)

Prove that BC, MQ, NP go together

Please help. Thanks in advance ^^

What do you mean by "go together"?  Intersect at the same point?


Let $ a, b$ be integers greater than $ 2$. Prove that there exists a positive integer $ k$ and a finite sequence $ n_1, n_2, ..., n_k$ of positive integers such that:
$ n_1 \equal{} a, n_k \equal{} b$;  and $ n_i.n_{i\plus{}1}$is divisible by $ n_i \plus{} n_{i\plus{}1}$for each $ i (1 \geq i < k).$

We define an equivalence relation on $ \mathbb{N}\setminus\{1,2\}$ via $ a\sim b$, iff $ \exists n_1,\ldots,n_k\in\mathbb{N}\setminus\{1,2\}$ with $ k>1$ such that $ n_1\equal{}a,n_k\equal{}b$ and for all $ i\equal{}1,\ldots,k\minus{}1$ we have $ \frac{n_in_{i\plus{}1}}{n_i\plus{}n_{i\plus{}1}}\in\mathbb{N}$. Symmetry and transitivity are obvious and $ n_1\equal{}a,n_2\equal{}a(a\minus{}1),n_3\equal{}a$ verifies reflexivity. Moreover, note that with $ x\sim y$ we get $ xt\sim yt$ by multiplying the corresponding sequence $ \{n_i\}_{i\equal{}1,\ldots,k}$ of $ (x,y)$ with $ t$. Denote equivalence classes by $ [x]\equal{}\{y\in\mathbb{N}\setminus\{1,2\}|x\sim y\}$. For example the sequence $ n_1\equal{}3,n_2\equal{}6,n_3\equal{}12,n_4\equal{}4$ shows that $ 4,6,12\in [3]$.

[b]Claim: $ [3]\equal{}\mathbb{N}\setminus\{1,2\}$[/b] (this is equivalent to the assertion of the original post)

[i]Proof:[/i] We assume the existence of a minimal counterexample $ m>4$ and aim to derive a contradiction.

[b]Case 1[/b]: For $ m$ composite, let $ d>1$ be its smallest divisor. Since $ \frac{m}{d}\equal{}2$ would imply $ d\equal{}2,m\equal{}4$, we must have $ \frac{m}{d}\geq 3$. Therefore, by minimality of $ m$ we have $ 3\sim\frac{m}{d}$ and $ 3d\sim m$. If $ 3d<m$, we will get the contradiction $ 3\sim 3d\sim m$ and $ m\in [3]$. Hence, we have $ m\equal{}3d$, i.e. $ m\equal{}6,9$ which implies $ m\equal{}9$. Due to $ n_1\equal{}9,n_2\equal{}72,n_3\equal{}24,n_4\equal{}8$ we conclude $ 9\in [8]$ and by minimality of $ m$ we have $ [8]\equal{}[3]$, a contradiction.

[b]Case 2[/b]: For $ m$ prime observe that $ m\sim m\plus{}1$ holds due to $ n_1\equal{}m, n_2\equal{}m(m\minus{}1), n_3\equal{}m(m\plus{}1),n_4\equal{}m\plus{}1$. Since $ 3\leq \frac{m\plus{}1}{2}<m$ holds, we get by minimality of $ m$ the relation $ 3\sim\frac{m\plus{}1}{2}$ and hence, $ 6\sim m\plus{}1$, i.e. $ 3\sim 6\sim m\plus{}1\sim m$, a contradiction.


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This is the first of two polls. Please vote in both.

What do you consider "better"?  I honestly voted "Possibly", because it depends what aspect you look at.

I would guess that some females would say that males and females are equal.

It really makes no difference if you are a girl or boy. The fact the boys are better is mainly assumed because it's mainly boys in the competitions.
Yes, this is sexist.

i'm a girl, and i feel it has a lot to do with where u live and how u were brought up.

in majority of places, boys are given much more encouragement in maths and sciences.   girls are enouraged in arts and music and such.  i know im being sterotypical, but its true

where i live, i would say that it is equal, because both are encouraged equally, and there is no discrimination.  but i realize that it is not the same in many other areas.

[quote]Does anybody else feel that this is pretty sexist (or at least, is anybody else wondering if this is pretty sexist), or is everybody else looking at this topic without so much as batting an eye?[/quote]

this is sexist. the whole topic/s is/are.

[quote="neelnal"]Notice something, 5 females in the other poll. 3 votes for females are better at math =P.

Unless a male voted for that.

Probably TreeThing did that.[/quote]

How sexist to assume that females naturally think their gender is better and imply that they are wrong but that the males are and also how Treethingist to assume Treething would vote opposite.
Also, this poll is gender biased to start with with males being better as the "Yes" reply and only ONE females are better response while two options for males are better.

[quote="neelnal"][quote="K81o7"]And, as of yet, no one has defined "better" for me.[/quote]

Intuitively.

Go to dictionary.com  :rotfl:[/quote]

I think the point I made which you did not understand is that we have not defined better in the context of mathematics.  Does it mean more creative?  More efectively able to formulate and prove ideas?  More competition-geared?  More enthusiastic?  More knowledgeable?

I would say that you can't really say who is "better". It varies person to person.

[quote="star99"]it is not sexist.  it is a valid poll, exploring opinions, and this topic is for civil arguments.  support your opinion, dont just say guys are better than math if you are a guy.  think about that a little more.[/quote]
agree

Well, there are more male mathemathicians, than female mathemathicians, I think that that is somehow, a prove that [b]in general[/b] guys are better at maths than girls, as for example, I think that women are better at doing several things at the same time. This is, of course, only in general

(also: I think that this is only true in math in general. I have heard that in geometry, girls defeathed guys at IMO in Mexico. Not sure about in Slovenia)

[quote="bpms"]Isn't this poll slanted as it has 2 (3 if you count possibly) that favor guys, and 1 for girls?[/quote]

I clicked enter topic before clicking the "add option" button for the last option which was "females are def. better" by mistake.

Please stop lambasting the poll. People were asking me to make it in the first place.

[quote="Ignite168"]The question already slants the poll as it asks if the male gender is better than the female gender which already implies that the female gender is in someway inferior or equal at best.[/quote]

Well, actually, phrasing it as a question, implies that the opposite statement is more widely accepted...so I might argue it's slanted the other way (if at all)...

there were only like 50 girls at nats last year, none in the CD round

I put that they are equal. If you look at the whole picture, you see that the number of males in each mathematical standard and the number of females in each mathematical standard are equal. Besides, not all people at IMO are females. :wink:  (I'm a male)

[quote="kstan013"]females are more consistent, males are the best and the worst at math (if you know what I mean)[/quote]

To be honest, that seems to be true in many practices...
Males are always the crazy weird ones too.

i agree.  females are pretty consistent, and get better consistently.  generally males are kinda jumpy

Well,for the general case, yes, definately.I don't remember very well where i have read this, but  males tend to develor the brain part connected with exact sciences (Maths,Physics,Computer Science,etc) and girls tend to develop the creativity part (Litterature, Arts, etc).

However, it depends a lot from case to case.There are a lot of girls who know math better than me  :blush:.

this is my point of view. a girl [size=200]CAN[/size] achieve anything a boy can. The only example in which this is harder is for certain [size=200]physical[/size] areas in life.
-jorian

And btw, in the high school at my town's graduation last year...the top 5 grades were totally dominated by girls

[quote="jhredsox"]
And btw, in the high school at my town's graduation last year...the top 5 grades were totally dominated by girls[/quote]

My response is more in reference to the first part of this sentence.

Grades are a very limited measure of intelligence. You need a certain amount of good smartness to be the top in your school grade-wise, but I think in the end, it comes down to being willing to work hard to the bone and many times, give up social life and/or health.

For instance, at my school, this one girl that has the highest Grade Point Average took the AIME's with us. Did that make her better at math or smarter because she had a much,much higher GPA and much,much better grades? Apparently not. A majority of the 16 people taking the AIMEs from my school beat her.

GPA =/= intelligence only

However, sometimes, GPA can show intelligence. It is a vague area.

I also believe in the fact that girls are pretty consitent in their GPA range and SAT scores while boys tend to be the two ends of the spectrum.

Referring to another post, yes, we may have to wait for a very,very long time before we can be sure of the intuitive characteristics of each gender.

ive met some pretty smart feminine math geeks so (and also not to be sexist) i think it was a possibility

http://www.swedish.org/111149.cfm

That article describes some of the differences in brain structure and development and IQ/math test scores between males and females. If you're interested, read it.

We say sexist like it's a bad thing, but nature is sexist. Females are different from males. We're all human, all contributing and valuable members of society, but we can't keep pretending that we're the same. It's no different than other genetic and developmental distinctions either--some people are naturally faster, stronger, more intelligent, or more focused, for example. I doubt you would deny this, but many would try to deny that gender is a factor in various abilities. True, culture--"nurture"--is part of the experimental variance between men and women, but it's a stetch to say there's a 100% correlation, with nothing owing to the genetic differences.

As some people mentioned, yes, men have a greater standard deviation of intelligence than women, with women more likely to be closer to average, and men more likely to have low intelligence or high intelligence rather than average.

I don't think these facts should have any impact on an individual's life, though. I'm female and, stereotypically, strongly slanted to verbal over mathematical ability. Still stereotypically, I have fairly high intelligence, but not as much as many males (or females, for that matter, but I'll bet that the majority of people smarter than me are males). However, I still have greater mathematical ability than my brother, and people recognize that.

I know that I'm guilty of seeing the world through gender-biased eyes, and sometimes it suprises me when someone's abilities and inclinations don't match the stereotypes, but I'm working on seeing beyond that, and not just for gender stereotypes: I'm trying to project myself into other people's minds and see themselves the way they see themselves--what their words, gestures, and actions appear from their side, what they look like and the difference between that and how they wish they looked, etc.

Please forgive this post--I kept shifting my topic about half a degree every clause. Read the article, at least, and perhaps there are some valuable ideas in my rant. My brain's all tied up around various ideas and incapable of sorting out what I said right now--we all think differently--but I don't want to delete everything and end up saying nothing. Criticisms and corrections welcome; if you hate my guts for posting dumb, off-topic nonsense now, give me a chance to change.

[quote="akkaren"]I'm female and, stereotypically, strongly slanted to verbal over mathematical ability.[/quote]

...How did you end up here?  :huh:

I totally agree, though; I never see any reason to worry about these differences.  If I want to know how good someone is at math, I actually talk to the person: I don't make judgements beforehand based on gender.

[quote="K81o7"][quote="akkaren"]I'm female and, stereotypically, strongly slanted to verbal over mathematical ability.[/quote]

...How did you end up here?  :huh:
[/quote]
I'm an angsty writer who ends up lots of places in search of some sort of certainty.

Once, math was the only thing I had to hold onto... And now I'm caught in a sort of desire to finish things, to yield to the pressure of old friends and the world of competitive math I briefly lived in... But I'm unsure.

I think any answer other than "possibly" is a bad answer, if you add the word "intrinsically" (i.e. is one gender intrinsically better at math?)

This is because we are totally unable to answer this sort of question given our current societal conditions.  There are many basic cultural biases which clearly affect everyone's lives and their relation to, involvment with, and ability in mathematics.

HOWEVER....

if the question is simply are males better at math (which would I guesse imply "currently") then the answer is a definitive yes.  Similarly, The Asian-American population os currently better at math than the Caucasian-American population, which is better than the African-American population.  This is based on SAT results (I believe they have several other racial categories which they define on teh test to get this info, but my point is made)

However, I'm guessing that the heart of the question was the first one I posed and the answer to that question is an equally definitive "it is not currently knoweable" for all questions about the inate differences in the mathematical abilities of people of different genders or races.

Guys are certainly more represented at math competitions; however, girls can do equally well if they are determined to do so.

The question should be: Are [b]most[/b] guys better than [b]most[/b] girls in math? (ignoring the ambiguity of the word 'better') In that case the answer is yes, in my experience, that's probably true. does this mean that given any girl and any guy, it would be fair to assume that the guy is 'better' at math than the girl? No.
There is a spectrum of mathematical ability, both boys and girls have the full range of that spectrum, and I don't think this question is trying to ask whether a male or a female holds the highest position on that spectrum, nor do I think information like that would even be useful.

I's also like to point out that most of the examples cited here adress the population of people who are good enough that they can [b]get[/b] to AIME or IMO, while below that threshold girls may be equal to or better than boys in mathematical ability.

I think that yes, but only slightly. There is only one girl in our 1 grade up math class, and 8 boys. But, me and that girl do about as well as each other, same with math team, and we both do better then all of the other kids in math so its about even but as some of u guys said, men are better at math generally.

[quote="lyra"]I feel that it is impossible to determine the answer to this question at this point.  For generations, females have been excluded and discouraged for pursuing the fields of mathematics and science.  Although this problem is beginning to be fixed, the majority of students who participate in math competitions and delve into higher mathmatics are male.  Whether males are intuitivley better than females is impossible to determine at this point, and would only be determinable after long and gruleling research.

~lyra
(I am female)[/quote]

I agree pretty strongly with this.  Also, I think bringing up the issue over and over again, saying "maybe girls just aren't as good" puts females off.
(They start to wonder if they are any good, and lack of confidence can undermine their success in a field where you really need to trust your own logic).

I'm a female mathematician (well, almost - I've nearly finished my PhD).  I never did math competitions because I absolutely hate [b]competitive[/b] activities.  All the talk about who is better than who just seemed to distract from the beauty of the subject. So I think maybe (and I'm stereotyping wildly) girls don't like competitions, but this doesn't mean they don't like, or can't do math.

And there's also the whole question of nurture versus nature (with both competitive activities and math).  My parents certainly encouraged my brothers to do things like make machines and write computer programs, and expected them to do well at maths and science; whereas on the other hand, my sister and I got dolls and play-dough and paint-sets and were encouraged in English and Art, etc.  When I read about math, my Dad would make comments about how I was just trying to be like my brothers, but I just wasn't as smart at those things, so should just give up, etc., etc.
(This wasn't in the far distant past, but in the eighties and nineties).  I think part of the reason I ended up doing a PhD in math is because I have a rebellious streak and don't like being told what I can't do.

(Also, I should add - my Dad made those comments despite the fact that I got better grades in math than all my brothers).

(1) This is a flawed argument:
[quote="capybaralet"]
if the question is simply are males better at math (which would I guesse imply "currently") then the answer is a definitive yes.  Similarly, The Asian-American population os currently better at math than the Caucasian-American population, which is better than the African-American population.  This is based on SAT results [/quote]
Despite its name, the SAT doesn't actually test aptitude, but basic knowledge.  It doesn't indicate how good someone is at mathematics, because a multiple-choice exam can't measure creativity, originality, depth of thought process, etc.

(2) Two days ago, one of my students (female, senior math major) told me that over Thanksgiving break one of her relatives said, "Math?  Isn't that a guy major?"

So it's not just a 1980's or 1990's thing...

Nothing bad 2 to the girls but male minds r more adapted (on average) to things like math and science where girls r better at english or arts. Its been proven.

I just can say that there are more males than females like math.

[quote="now a ranger"]Some females are better than males at math. But a lot more males are better than females.

But it depends.[/quote]
that's only since there are more males in math right? :lol:
[quote="Lucy"]
I'm a female mathematician (well, almost - I've nearly finished my PhD). I never did math competitions because I absolutely hate competitive activities. All the talk about who is better than who just seemed to distract from the beauty of the subject. So I think maybe (and I'm stereotyping wildly) girls don't like competitions, but this doesn't mean they don't like, or can't do math. [/quote]
Well you should see all the girls in sports. Girls are so competitive it kills me. :lol:


Hi Everyone,

In another post I made recently, the topic of mathematics related distributed computing projects came up.   Many of you have probably heard of the Great Internet Mersenne Prime Search < [url]http://www.mersenne.org/prime.htm[/url] >.  There are other distributed computing projects that are currently active that one can participate in.  For example, I participate in the ZetaGrid project < [url]http://www.zetagrid.net/[/url] >.  The project's aim is to either find statistics about the zeroes of the Reimann zeta-function that may contribute ideas on how to prove the Reimann hypothesis is correct or to actaully find a zero that does not lie on the line [tex]1/2+bi [/tex] in the complex plane.   My computer has computed over 170 million zeroes of the Reimann zeta function to date, running the ZetaGrid program as a screensaver.   For a good list of other mathematics related distributed computing project, check out < [url]http://www.aspenleaf.com/distributed/ap-math.html[/url] >.  Most of these programs run using "wasted" CPU time and hence do not effect the performance of one's computer too much.

Yours truly,

mathfreak

I've analyzed 1137 data units for SETI@Home, at a cost of over 6478 hours, and I don't even have a flying saucer to show for it.

[quote="mathfreak"]Most of these programs run using "wasted" CPU time and hence do not effect the performance of one's computer too much.[/quote]

I dunno... when I let my pc alone for a while I want the CPU to have cooled a bit... if it's on and working all day, it'd get way too hot! :?


I need to use the curly lower case l (the uppercase is "L"), the one written in script (I think that's the correct term) with the loop in it. If someone knows the code off the top of his/her head, please post it here. Thanks.

This is the only place on the internet that I found a quick and simple answer to how to place a "curly l" in a Latex document. Thank you.


AMC
AIME
trigonometry
percent
USA(J)MO
USAMO
number theory
\/closed

Is it too hard to take WOOT while in high school?  I heard there's some work, and in high school there's a lot more homework...

I was thinking about taking WOOT next year, as I can consistently score over 5 on practice AIME's...because it's probably a bit too hard for now, but I can always review transcripts, etc. instead of taking it as a freshman, with more homework at school.

However, my first priority with classes next year is definitely Intermediate Trig/Complex Numbers, and the AIME Problem Series A class...so is it too much of a workload?

Suggestions/comments please!!!!

A lot of this depends on how you budget your time, and if you are careful to avoid overloading yourself with extracurricular activities you don't really care about.  (That is, don't just join a bunch of activities for college applications -- focus on the ones you enjoy.  You'll be happier, and you'll learn more.)

As for whether or not you're ready, a lot of that will depend on how you progress in the next 6 months.  If you keep working the way you have, you'll be surprised at the leaps you'll make over that time.

WOOT + Trig + AIME might be overdoing a bit, but we haven't set the fall schedule yet.  It's possible there won't me too great an overlap between the Trig and AIME class in terms of time.  Also, WOOT is more spread out than our other courses.

Thanks a bunch!

Another question:  When will the Trig book be out?  I've heard discussion about it, that it would be out "soon", but is there an estimated date?  In time for the fall class?

We won't hold the trig class again without the book, so, yes, the book will be out in time for the fall class.  (We'll probably change the name of the class to Precalculus, but that's not set in stone.  Also, it will likely be an 18-week class.)

I hope to have the book finished by the end of the summer.

perhaps you can manage your time better than I can, but:

I am taking woot and pretty much no extracurriculars except an hour of piano everyday. (well also some boy scout stuff olympiad geo). I had to quit running this year to make some time for woot. I would go ahead and take woot (but this is just me...)and try to learn most of the stuff by reviewing if you plan on taking woot at all. Even though I quit running, I still have trouble finding time to even half the message board problems. I plan on doing a ton reviewing of woot stuff in the summer, when I will have much more time for math. It is definitely harder (definitely more homework in high school compared to middle school, even if you try to do it all during school), although not impossible by any means, to do woot in high school in addition to other EC stuff you may do, than if you do it in middle school.

really it boils down to how you want to budget your time. I think it would be a lot of work doing all three classes, but I suppose you can drop woot after a few weeks into it or something if you don't think you have time.

edit: just to confirm: we can access the forums from woot after we take for forever right? (i know we cant post anything, but we can look at what others have posted during the year, right?) and we can access the handouts whenver, right? Or will we need to download some stuff before woot ends?

i really wish i could take woot this year, which i can't really do because of a few monetary issues. but, this may turn out to be a good thing; waiting for next year, when i'm more prepared. as of now, i fail at math.

but i'm wondering what is the average grade level of the participants? specifically, what percentage of them are freshmen and what percentage is sophomores?

[quote="hurdler"]

edit: just to confirm: we can access the forums from woot after we take for forever right? (i know we cant post anything, but we can look at what others have posted during the year, right?) and we can access the handouts whenver, right? Or will we need to download some stuff before woot ends?[/quote]

You will have indefinite access to the forums, but not to the handouts and transcripts -- these you should save yourself.  The course homepage will be taken down about a month after WOOT ends.

[quote="hurdler"]my guess/estimate for this year: a good chunk are freshman/sophomore, maybe about 25-40 percent.[/quote]
Wait, so does that mean the majority of those left over are also in high school?

[quote="hurdler"]ok thanks Mr. Rusczyk

qwerty: yes, I know of 3 7th graders and maybe few more than 3 eighth graders (I think). My guess is that  around 90 percent of WOOTers are high schoolers[/quote]

Also, last year there were 2 6th graders who took WOOT.

Would WOOT cover a lot of the same material as is in some of the other courses such as Intermediate Algebra and Intermediate Number Theory?
Thanks!

It has small amounts of overlap with the more advanced parts of those courses, but not a whole lot of overlap.

For me I found the classes to be really hard to attend, since it just eats a huge chunk out of your day...For me WOOT has boiled down to mass reading the handouts once in a while, doing the Olympiads and POTDs.

Hey richard, do you have any WOOT stats?

Like how many people that take WOOT qualify for the USAMO? or get one full question right? or something like that.

Well, the stats are bound to be off, because they have nearly all the MOP qualifiers from the previous year :)

There's heavy selection bias in the WOOT stats, both because there's a big bias in who enrolls, and there's a bias in who reports their results.  I guesstimate that at least 3/4 of the WOOTers qualify for the USAMO, and that many of them get at least one problem right.    But again, these are hardly reliable stats considering the reporting and bias issues.  Typically, many of the winners are WOOTers, but that's to be expected considering the fact that we bring in the MOPpers.


1) A positive integer n, when written in base b, has the form 211. When n is written in base b+2, it has the form 110. Determine n and b (in base 10).

2) n how many ways can the numbers 1, 2, 3, 4, 5, 6 be arranged as a sequence u, v, w, x, y, z such that u + x = v + y = w + z?

3) How many distinct (that is, noncongruent) trangles are there with all sides of integer length, with the longest side (or sides) having length 2005?

Thanks a lot.

[hide="1"]We have $ 2b^2\plus{}b\plus{}1\equal{}(b\plus{}2)^2\plus{}b\plus{}2$. Expanding and simplyfying, we have
$ (b\minus{}5)(b\plus{}1)\equal{}0$, and since b>0, b is 5. N is then 56.[/hide]

[hide="Hint for 2"]What is $ u\plus{}x$?[/hide]

[hide="Solution for 2"]Since $ u\plus{}v\plus{}w\plus{}x\plus{}y\plus{}z\equal{}21$, $ u\plus{}x\equal{}v\plus{}y\equal{}w\plus{}z\equal{}7$.  There are 6 possible values for $ u$, which determines $ x$.  That leaves 4 values for $ v$, which determines $ y$.  Finally, there are 2 values left for $ w$, which determines $ z$.  Therefore there are $ 6\cdot4\cdot2$ or $ \boxed{48}$ ways.[/hide]

[hide="1"]$ 2b^2\plus{}b\plus{}1\equal{}b^2\plus{}5b\plus{}6$

$ b^2\equal{}4b\plus{}5$

$ b\equal{}\frac{4\pm 6}{2}\equal{}\minus{}1,5$

Since b cannot be -1, b must be 5. And n must be 56.[/hide]

[hide="2"]$ u\plus{}x\equal{}v\plus{}y\equal{}w\plus{}z\equal{}\dfrac{1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6}{3}\equal{}7$

For every choice u, there is only one choice for x. Then for every choice v, there is only one choice for y. And for every choice w, there is only one choice for z. There are 6 choices for u, and then 4 choices for v, and then 2 choices for w. $ 6*4*2\equal{}48$[/hide]

[hide="3"]Let the side lengths be (a,b,c), with a≤b≤c. c=2005. Now we find the number of triangles for some a:

a=1: From the [[triangle inequality]], the only possible value of b is 2005.

a=2: The only values of b there are are 2004 and 2005.(2006 isn't possible, since b≤c)

.
.
.
.
.
.

a=1002: There are 1002 values: 1004 through 2005.

a=1003: There are 1003 values: 1003 through 2005.

a=1004: Instead of 1004 values, there are only 1002 values: 1004 through 2005, since a≤b.

And so on.

The sum is $ \dfrac{1003*1004}{2}\plus{}\dfrac{1002*1003}{2}\equal{}\dfrac{1003*2006}{2}\equal{}1003^2\equal{}1006009$[/hide]


Please pm me if you are interested.

Sorry if this isn't supposed to be here, but I thought it would get more attention as its own thread.

I am interested in being a replacement in the Mole game.

[quote="perfect628"]Please pm me if you are interested.[/quote]

Hmm...

I WILL BE A REPLACEMENT

BUT THAT WOULD BE KINDA UNFAIR AS I KNOW WHO THE MOLE IS


[quote="boxedexe"]By the binomial theorem, your sum is equal to $(1-1)^{n}=0$.[/quote]

Yes of course, thanks  :oops: .

[quote="Rzeszut"][quote="color"]Sorry if this isn't the right place to post this.

Show that for $n>0$ one has
\[\sum_{j=0}^{n}(-1)^{n}\left( \begin{array}{c}n \\ j \end{array}\right)=0 \]
[/quote]Of course it should be $\sum_{j=0}^{n}(-1)^{j}\binom{n}{j}$.[/quote]

Yes youre right, my mistake.


integration
function
calculus
calculus computations

$w(x)=\frac{1}{2 \pi} \sqrt{4-x^2} 1_{[-2;+2]}$ the distribution of the semi-circle law. Show that $M_{2n}(f)=\int_{\mathbb{R}} x^{2n} f(x) dx = \frac{C_{2n}^n}{n+1}$ for $n \geq 1$.

Catalan numbers!  (no, I don't know an answer, just wondering if that is relevant at all)

if you know the generating function of the Catalan numbers, maybe you will see some similarities  :)


calculus
limit
Inequality
integration
area
IMO
IMO 1970

The real numbers $a_0,a_1,a_2,\ldots$ satisfy $1=a_0\le a_1\le a_2\le\ldots. b_1,b_2,b_3,\ldots$ are defined by $b_n=\sum_{k=1}^n{1-{a_{k-1}\over a_k}\over\sqrt a_k}$.

[b]a.)[/b] Prove that $0\le b_n<2$.

[b]b.)[/b] Given $c$ satisfying $0\le c<2$, prove that we can find $a_n$ so that $b_n>c$ for all sufficiently large $n$.

[quote="Yimin Ge"]Hi,
I think I've got a solution for part a). The interesting is that I actually did not need $a_0 \leq a_1 \leq ...$, only that $a_0=1$ and $a_k > 0$

I proved that for all $n \geq 1:$   $b_n \leq 2-2\cdot\frac{1}{\sqrt{a_n}}$ holds, which is a very simple induction proof.....

As to part b)...I'm still working on it....[/quote]
sorry,I’m not very sure about your solution for part a)...I think the  $a_0 \leq a_1 \leq ...$ do need.
For part b):juat take $a_n=x^n$ let $x \to 1^{+}$ :)

Well, according to my gut feeling, it is possible because if you take a number smaller than the previous one, you have a negative summand and therfore you somehow "ease" the inequality (though the following summand will be greater but this might cancel) but well, I'm talking to mathematicians and not to physicians so I'll have to prove my claim :lol:

I'm going to prove that $b_n\leq 2-2\cdot\frac{1}{\sqrt{a_n}}$ by induction.
Let $u_n = \sqrt{a_n}$ for all $n\geq 0$ (just because I dont want to write so many $\sqrt{\ }$. :P
$\Rightarrow b_n = \sum_{k=1}^n(1-\frac{u_{k-1}^2}{u_k^2})\frac{1}{u_k}$
i.b.: For $n=1$ we have $b_1=(1-\frac{1}{u_1^2})\frac{1}{u_1} \leq 2-2\frac{1}{u_1}$ which is after some standard transforming equivalent to $0\leq 2u_1^3-3u_1^2+1=(u_1-1)^2(2u_1+1)$
Now we deduce from n to n+1
i.p.: $b_n=\sum_{k=1}^n(1-\frac{u_{k-1}^2}{u_k^2})\frac{1}{u_k} \leq 2-2\frac{1}{u_n}$
i.c.: $b_{n+1}=\sum_{k=1}^{n+1}(1-\frac{u_{k-1}^2}{u_k^2})\frac{1}{u_k} \leq 2-2\frac{1}{u_{n+1}}$
Using the i.p. we obtain $\sum_{k=1}^{n+1}(1-\frac{u_{k-1}^2}{u_k^2})\frac{1}{u_k} \leq 2-2\frac{1}{u_n} + (1-\frac{u_n^2}{u_{n+1}^2})\frac{1}{u_{n+1}}$
so it remeans to be shown that $2-2\frac{1}{u_n} + (1-\frac{u_n^2}{u_{n+1}^2})\frac{1}{u_{n+1}} \leq 2-2\frac{1}{u_{n+1}}$. After some simplifying and some standard transforming we obtain
$0\leq 2u_{n+1}^3-3u_{n+1}^2u_n + u_n^3 = (u_{n+1}-u_n)^2(2u_{n+1}+u_n)$.

Assuming that I did not miscalculate (which would be very embarrassing though) $u_{n+1} \geq u_n$ is not necessary as we have $(u_{n+1}-u_n)^2$.....

yes,you are right.$1=a_0\le a_1\le a_2\le\ldots$ for part a) and  b) :)
just use $a_i \ge 0 (i=1,2,\ldots)$
and this problem is very strange,for part b) we must fix $x$ first ,then fix $n$.
you will easily see $\lim_{n \to \infty}b_n=1+ \sqrt{x}$ :)

frill, you could've simplified it:

$\begin{aligned}\left(1-\dfrac{a_{k-1}}{a_k}\right)\cdot\frac1{\sqrt{a_k}}&=\dfrac{\sqrt{a_k}-\sqrt{a_{k-1}}}{\sqrt{a_k}}\cdot\dfrac{\sqrt{a_k}+\sqrt{a_{k-1}}}{\sqrt{a_k}}\cdot\dfrac1{\sqrt{a_k}}\\&\le\dfrac{\sqrt{a_k}-\sqrt{a_{k-1}}}{\sqrt{a_k}}\cdot\dfrac{2\sqrt{a_k}}{\sqrt{a_k}}\cdot\dfrac1{\sqrt{a_{k-1}}}\\&=2\dfrac{\sqrt{a_k}-\sqrt{a_{k-1}}}{\sqrt{a_k}\cdot\sqrt{a_{k-1}}}\\&=2\left(\frac1{\sqrt{a_{k-1}}}-\frac1{\sqrt{a_k}}\right)\end{aligned}$

However, this only proves $2$ is an upper bound, not $\sup\{x\in\Bbb R^+\mid x=b_n,n\in\Bbb N_0\}$. Now, we should find the optimal sequence so that $c=2-\varepsilon$ for some arbitrarily small $\varepsilon>0$ and $b_n\in(c,2)$ for sufficiently large $n\in\Bbb N$.


Points $A$ and $B$ are 5 units apart. How many lines in a given plane containing $A$ and $B$ are 2 units from $A$ and 3 units from $B$?

$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{more than 3}$

How can a line always be $2$ units from $A$ and $3$ from $B$? :huh:

[quote="Altheman"][hide="solution"]the locus of points a distance from a point is a circle, looking at the distances, the circles are tangent, 1 pt: B[/hide][/quote]

There are three common tangents - two outer and one inner - hence the answer is D.

wow, i did not read the problem..."[b]lines[/b]"


geometry
conics
hyperbola
geometry proposed

I am sure this problem has been posted before, but i just cant find it, thats why a link to it would be nice. Here it goes:
Let ABC be a triangle and X, Y, Z points on its bisectors IA, IB, IC, such that IA/IX=IB/IY=IC/IZ. Prove, that  XD, YE, ZF concur, where D E and F are the touching point of the incircle of ABC with BC, AC, AB.

XYZ and ABC are homothetic with center I and so we can easily see that XYZ and DEF are orthologic triangles with the 2 orthologic center coincident so the are also perspective. It could also be prove that the locus of the perspector center is the Jerabek Hyperbola of DEF.


The two diameters shown are each 10 centimeters long, and they intersect at a right angle. What is the number of square centimeters in the area of the unshaded region? Use 3.14 as an approximation for $ \pi$, and express your answer as a decimal to the nearest tenth.

[asy]path A = (0,0)--dir(45)--dir(135)--cycle;
path B = (0,0)--dir(-45)--dir(-135)--cycle;
draw(A);draw(B);draw(Circle((0,0),1));
fill(A,grey);fill(B,grey);[/asy]

Hello
Area of shaded region =$ 2*\frac{5*5}{2}\equal{}25$sq. cm
Area of unshaded region =$ 25(\pi\minus{}1)$=$ 53.5$sq.cm
Thank u


logarithms
number theory proposed
number theory

given a natural number $N$.prove that there exist a prime which begins with$N$.(for example $17$begins with $1$.)

N is any number, not just ranging from 1-9. This seems tricky....

as seamusoboyle mentioned $N$ is any natural number:
sketch:
the sequense {$logp$} is densite.({} is the fraqtional part.)

Proving its desne is another matter. log is certainly dense for integers, but for primes only?

seamusoboyle, i know it is densite ,but why? i do not know.

Dense means you can get as close as you want to any number.

This would follow from any theorem of the kind "there is always a prime between $n^k$ and $(n+C)^k$". I know that for $k=2, C=1$ it is open, but aren't there proven weaker versions?

EDIT: I had said $k=1, C=2$; of course I meant it the other way around. I've edited it.

[quote="Severius"]This would follow from any theorem of the kind "there is always a prime between $n^k$ and $(n+C)^k$". I know that for $k=1, C=2$ it is open, but aren't there proven weaker versions?[/quote]

Probably it is open when $k=2, C=1$.

$p$ starts with $a$ iff there exists an $n$ such that:

$a.10^n<p<(a+1).10^n$

iff

$n + \log_{10}{a}<\log_{10}{p}<n + \log_{10}{(a+1)}$

iff

$\log_{10}{a}<\log_{10}{p}-n < \log_{10}{(a+1)}$

So it suffices to show that $\{\log_{10}{p} \}$ is dense in [0,1].

I have a shaky solution to this, using derichlets theorem, but im in a rush for a lecture so someone else can sweat it out!

Hoping i put the problem on more stable ground, Shea.


My coach did not give us a copy of the stats sheets at Nationals (meaning the ones that show question difficulty, how many people got certain scores, and how many people got certain problems right). If so, can anyone here please post them here (if possible) or PM/email them to me? Thanks. :)

Yeah, but our coach promised us that he'll email each of us a copy, which he didn't. :( That's why I'm requesting it here, if anyone has it.


Prove or disprove

$36^{36}+41^{41}$ is divisible by $77$


inequalities
LaTeX
logarithms
algebra
polynomial
inequalities proposed

Problem1:  [tex] a,b,c [tex]>0 such that abc=1. Find the maximum value of the following inequality:

      P=[tex] 1/(a+2bc)+1/(b+2ca)+1/(c+2ab) [tex]

  Problem2: Let [tex]a,b,c[tex]>0 such that abc=1. Prove that:
          [tex] a/(a+2bc)+b/(b+2ca)+c/(c+2ab) <=(a+b+c)/3 [tex]

[b]dduclam[/b] wrote:
Problem1: $ a,b,c >0$ such that $ abc\equal{}1$. Find the maximum value of the following inequality:

$ P\equal{} \frac{1}{a\plus{}2bc}\plus{}\frac{1}{b\plus{}2ca}\plus{}\frac{1}{c\plus{}2ab}$

Problem2: Let $ a,b,c>0$ such that  $ abc\equal{}1$ . Prove that:
$ \frac{a}{a\plus{}2bc}\plus{}\frac{b}{b\plus{}2ca}\plus{}\frac{c}{c\plus{}2ab}\leq\frac{a\plus{}b\plus{}c}{3}$

problem1 : $ P\equal{} \frac{1}{a\plus{}2bc}\plus{}\frac{1}{b\plus{}2ca}\plus{}\frac{1}{c\plus{}2ab}\leq\frac{1}{2\sqrt{2abc}}\plus{}\frac{1}{2\sqrt{2abc}}\plus{}\frac{1}{2\sqrt{2abc}}\equal{}\frac{3}{2\sqrt{2}}$----->
$ a\equal{}2bc,b\equal{}2ac,c\equal{}2ab$------> $ a\equal{}b\equal{}c\equal{}\sqrt{2}$------> $ max(P)<\frac{3}{2\sqrt{2}}$

[quote="zaya_yc"]
Problem2: Let $ a,b,c > 0$ such that  $ abc \equal{} 1$ . Prove that:
$ \frac {a}{a \plus{} 2bc} \plus{} \frac {b}{b \plus{} 2ca} \plus{} \frac {c}{c \plus{} 2ab}\leq\frac {a \plus{} b \plus{} c}{3}$
[/quote]

Wrong $ (\frac{1}{2},\frac{1}{2},4)$  :wink:

For these problems, I just let $ a\equal{}e^x ; b\equal{}e^y; c\equal{}e^z$ then $ x\plus{}y\plus{}z\equal{}1$ and then analyse the analogous functions.

[quote="Vasc"]I conjecture that the following statement holds:

If $ a,b,c$ are positive numbers such that $ abc \equal{} 1$, then
$ \frac 1{a \plus{} 5bc} \plus{} \frac 1{b \plus{} 5ca} \plus{} \frac 1{c \plus{} 5ab} \le \frac 1{2}$.[/quote]

For $ \lambda > 0$, we can conjecture that $ \frac 1{a \plus{} \lambda bc} \plus{} \frac 1{b \plus{} \lambda ca} \plus{} \frac 1{c \plus{} \lambda ab} \le \frac {3}{1 \plus{} \lambda}$.

[quote="mathangel"] I just let $ a \equal{} e^x ; b \equal{} e^y; c \equal{} e^z$ then $ x \plus{} y \plus{} z \equal{} 1$ and then analyse the analogous functions.[/quote]
I don't think this method solves the first inequality, because $ f$ is not left concave function.

[quote="kunny"]For $ \lambda > 0$, we can conjecture that $ \frac 1{a \plus{} \lambda bc} \plus{} \frac 1{b \plus{} \lambda ca} \plus{} \frac 1{c \plus{} \lambda ab} \le \frac {3}{1 \plus{} \lambda}$.[/quote]
The inequality is not true for $ \lambda > 5.1$ and for $ \lambda < \frac 1{5.1}$.

[quote="FOURRIER"][quote="zaya_yc"]
Problem2: Let $a,b,c > 0$ such that  $abc = 1$ . Prove that:
$\frac {a}{a + 2bc} + \frac {b}{b + 2ca} + \frac {c}{c + 2ab}\leq\frac {a + b + c}{3}$
[/quote]

Wrong $(\frac {1}{2},\frac {1}{2},4)$  :wink:[/quote]
  No! It's right  :lol:
$(\frac {1}{2},\frac {1}{2},4)$, LHS=$\frac{10}{9}<$RHS=$\frac{15}{9}$  :lol:

Sorry! I cann't write the math formulas...
 Can you helf me?

The verb is to help not to helf  :lol:

If you want to write math formulas in latex you insert a dollar then the lawex code then another dollar

Please Look here for more information
http://www.mathlinks.ro/index.php?f=123
 :wink:

Ok,thanks a lot!
[quote="dduclam"][quote="FOURRIER"][quote="zaya_yc"]
Problem2: Let $ a,b,c > 0$ such that  $ abc \equal{} 1$ . Prove that:
$ \frac {a}{a \plus{} 2bc} \plus{} \frac {b}{b \plus{} 2ca} \plus{} \frac {c}{c \plus{} 2ab}\leq\frac {a \plus{} b \plus{} c}{3}$
[/quote]

Wrong $ (\frac {1}{2}, \frac {1}{2},4)$  :wink:[/quote]
  No! It's right  :lol:
$ (\frac {1}{2},\frac {1}{2},4)$, LHS=$ \frac{10}{9}<$ RHS= $ \frac{15}{9}$  :lol:[/quote]

[quote="dduclam"]
  No! It's right  :lol:
$ (\frac {1}{2},\frac {1}{2},4)$, LHS=$ \frac {10}{9} <$ RHS= $ \frac {15}{9}$  :lol:[/quote]

Sorry I made a mistake

[quote="Vasc"]I conjecture that the following statement holds:

If $ a,b,c$ are positive numbers such that $ abc \equal{} 1$, then
$ \frac 1{a \plus{} 5bc} \plus{} \frac 1{b \plus{} 5ca} \plus{} \frac 1{c \plus{} 5ab} \le \frac 1{2}$.[/quote]
Can you post your solusion,VASC? :?:

[quote="dduclam"]Problem1:  $ a,b,c > 0$ such that $ abc \equal{} 1.$ Find the maximum value of the following expression:
\[ P \equal{} \frac {1}{a \plus{} 2bc} \plus{} \frac {1}{b \plus{} 2ca} \plus{} \frac {1}{c \plus{} 2ab}
\]
[/quote]
$ P\leq1\Leftrightarrow\sum_{cyc}f(a)\leq1,$ where $ f(a) \equal{} \frac {a}{a^2 \plus{} 2}.$
We obtain: $ f'(a) \equal{} \frac {2 \minus{} a^2}{(a^2 \plus{} 2)^2}.$ Hence, $ \max_{a > 0}f \equal{} f(\sqrt2) \equal{} \frac {\sqrt2}{4}.$
$ P\leq1\Leftrightarrow\sum_{cyc}\left(\frac {a}{a^2 \plus{} 2} \minus{} \frac {1}{3} \minus{} \frac {\ln a}{9}\right)\leq0.$
Let $ g(a) \equal{} \frac {a}{a^2 \plus{} 2} \minus{} \frac {1}{3} \minus{} \frac {\ln a}{9}.$ Then $ g'(a) \equal{} \frac {(1 \minus{} a)(a^3 \plus{} 10a^2 \plus{} 14a \minus{} 4)}{9a(a^2 \plus{} 2)^2}.$
$ g(0.06738) < 0$ and $ g(1)\equal{}0.$ Hence, $ \sum_{cyc}g(a)\leq0$ if $ \min\{a,b,c\}\geq0.06738.$
Let $ \min\{a,b,c\} < 0.06738.$ Then $ \sum_{cyc}f(a) < f(0.06738) \plus{} 2f(\sqrt2) < 1.$
Thus, for positive $ a,$ $ b$ and $ c$ we obtain: $ \max_{abc \equal{} 1}P \equal{} 1.$

[quote="arqady"][quote="Zhas"]:?: Help me! I can't solve Inequality from Costa-Rica 2006:
a,b,c-are the sides of a triangle and prove that:[/quote]
Maybe it needs to prove that $ \frac {3(a^4 \plus{} b^4 \plus{} c^4)}{(a^2 \plus{} b^2 \plus{} c^2)^2} \plus{} \frac {ab \plus{} ac \plus{} bc}{a^2 \plus{} b^2 \plus{} c^2}\geq2$  $ ?$[/quote]
 Standardize: $ a^2\plus{}b^2\plus{}c^2\equal{}3$. We have
 $ a^4\plus{}b^4\plus{}c^4\plus{}ab\plus{}bc\plus{}ca \geq 6 \equal{}\frac2{3} (a^2\plus{}b^2\plus{}c^2)^2$
$ \Leftrightarrow a^4\plus{}b^4\plus{}c^4 \geq a^2b^2\plus{}b^2c^2\plus{}c^2a^2$
 done :D

So,this problem? Can you help me?
[quote="dduclam"]

  Problem2: Let $ a,b,c>0$ such that $ abc\equal{}1$. Prove that:
          $ \frac a{a\plus{}2bc}\plus{} \frac b{b\plus{}2ca}\plus{} \frac c{c\plus{}2ab} \leq \frac {a\plus{}b\plus{}c}3$[/quote]

[quote="Vasc"]I conjecture that the following statement holds:

If $ a,b,c$ are positive numbers such that $ abc \equal{} 1$, then
$ \frac 1{a \plus{} 5bc} \plus{} \frac 1{b \plus{} 5ca} \plus{} \frac 1{c \plus{} 5ab} \le \frac 1{2}$.[/quote]

It clearly holds for $ y\equal{}z$ using polynomial factoring and such.

If we do $ \sum \frac{1}{a\plus{}\frac{5}{a}}$ and apply the lagrange multiplier method, we get that the extrema must occur when

$ < \frac{a^3\minus{}5a}{(a^2\plus{}5)^2},...> \equal{}k*<1,1,1>$

I think that this can be reduced by subtracting, factoring, and repeating, I will look at it more later.

[quote="arqady"][quote="dduclam"]Problem1:  $ a,b,c > 0$ such that $ abc \equal{} 1.$ Find the maximum value of the following expression:
\[ P \equal{} \frac {1}{a \plus{} 2bc} \plus{} \frac {1}{b \plus{} 2ca} \plus{} \frac {1}{c \plus{} 2ab}
\]
[/quote]
$ P\leq1\Leftrightarrow\sum_{cyc}f(a)\leq1,$ where $ f(a) \equal{} \frac {a}{a^2 \plus{} 2}.$
We obtain: $ f'(a) \equal{} \frac {2 \minus{} a^2}{(a^2 \plus{} 2)^2}.$ Hence, $ \max_{a > 0}f \equal{} f(\sqrt2) \equal{} \frac {\sqrt2}{4}.$
$ P\leq1\Leftrightarrow\sum_{cyc}\left(\frac {a}{a^2 \plus{} 2} \minus{} \frac {1}{3} \minus{} \frac {\ln a}{9}\right)\leq0.$
Let $ g(a) \equal{} \frac {a}{a^2 \plus{} 2} \minus{} \frac {1}{3} \minus{} \frac {\ln a}{9}.$ Then $ g'(a) \equal{} \frac {(1 \minus{} a)(a^3 \plus{} 10a^2 \plus{} 14a \minus{} 4)}{9a(a^2 \plus{} 2)^2}.$
$ g(0.06738) < 0$ and $ g(1) \equal{} 0.$ Hence, $ \sum_{cyc}g(a)\leq0$ if $ \min\{a,b,c\}\geq0.06738.$
Let $ \min\{a,b,c\} < 0.06738.$ Then $ \sum_{cyc}f(a) < f(0.06738) \plus{} 2f(\sqrt2) < 1.$
Thus, for positive $ a,$ $ b$ and $ c$ we obtain: $ \max_{abc \equal{} 1}P \equal{} 1.$[/quote]

The following unexpected is a bit simpler; $ \sum_{cyc}f(a)\leq\sum_{cyc} \frac {a}{2a \plus{} 1}\leq1$ but this does not work for any $ k < 1/2$ or $ k > 2$ when considering sum of $ a/(a^2 \plus{} k)$

Thank you ywl and all! Finally,this problem
[quote="dduclam"]Problem1: $ a,b,c > 0$ such that $ abc \equal{} 1$. Find the maximum value of the following inequality:

$ P \equal{} \frac {1}{a \plus{} 2bc} \plus{} \frac {1}{b \plus{} 2ca} \plus{} \frac {1}{c \plus{} 2ab}$  [/quote]
solved -a very nice solution :D
$ P \equal{} \sum\frac a{a^2 \plus{} 2}\leq\sum\frac a{2a \plus{} 1} \leq1$
$ \Leftrightarrow \sum_{cyc}a(2b \plus{} 1)(2c \plus{} 1)\leq(2a \plus{} 1)(2b \plus{} 1)(2c \plus{} 1)$
$ \Leftrightarrow a \plus{} b \plus{} c\geq3$ true by AM-GM :D
Thus Max $ P \equal{} 1$ !

And [b]Problem 2[/b]???

[quote="arqady"][quote="dduclam"]

And [b]Problem 2[/b]???[/quote]
[quote="dduclam"]
 Problem2: Let $ a,b,c > 0$ such that $ abc \equal{} 1.$ Prove that:
$ a/(a \plus{} 2bc) \plus{} b/(b \plus{} 2ca) \plus{} c/(c \plus{} 2ab) < \equal{} (a \plus{} b \plus{} c)/3$[/quote]
Try $ a \equal{} \frac {16}{25}$ and $ b \equal{} c \equal{} \frac {5}{4}.$ :wink:[/quote]
Thanks,arqady!This problem is false :D But we have a very easy problem(true)
[b]Problem 3[/b]:Let $ a,b,c > 0$ such that  $ abc \equal{} 1$ . Prove that:
$ \frac {a}{a \plus{} bc} \plus{} \frac {b}{b \plus{} ca} \plus{} \frac {c}{c \plus{} ab}\leq\frac {a \plus{} b \plus{} c}{2}$

and a very hard problem
[b]Problem 4[/b]:Let $ a,b,c > 0$ such that  $ abc \equal{} 1$  Find the best constant for following inequality:
$ \frac {a}{a \plus{} kbc} \plus{} \frac {b}{b \plus{} kca} \plus{} \frac {c}{c \plus{} kab}\leq\frac {a \plus{} b \plus{} c}{1\plus{}k}$

[quote="dduclam"]
[b]Problem 3[/b]:Let $ a,b,c > 0$ such that  $ abc \equal{} 1$ . Prove that:
$ \frac {a}{a \plus{} bc} \plus{} \frac {b}{b \plus{} ca} \plus{} \frac {c}{c \plus{} ab}\leq\frac {a \plus{} b \plus{} c}{2}$
[/quote]
We have:
$ \sum_{cyc} \frac{a}{a\plus{}bc} \leq \sum_{cyc} \frac{a}{2\sqrt{abc}} \equal{}\sum_{cyc} \frac{a}{2}\equal{}\frac{a\plus{}b\plus{}c}{2}$

[quote="Vasc"]I conjecture that the following statement holds:

If $ a,b,c$ are positive numbers such that $ abc \equal{} 1$, then
$ \frac 1{a \plus{} 5bc} \plus{} \frac 1{b \plus{} 5ca} \plus{} \frac 1{c \plus{} 5ab} \le \frac 1{2}$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=333667


Prove that in any triangle  $ ABC$ exists the relation  $ \boxed {\ \max\ \left\{\ \frac {a}{p\minus{}a}\ ,\ \frac {b}{p\minus{}b}\ ,\ \frac {c}{p\minus{}c}\ \right\}\ \ge \ \frac Rr\ }$ .

let's denote
$ p\minus{}a\equal{}x,p\minus{}b\equal{}y, p\minus{}c\equal{}z$ and assume an ordering $ x\geq y\geq z$
then we are trying to prove
$ \frac{x\plus{}y}{z}\geq \frac{(x\plus{}y)(y\plus{}z)(x\plus{}z)}{4xyz}$
$ \iff 4xy\geq (x\plus{}z)(y\plus{}z)$
$ \iff x(y\minus{}z)\plus{}y(x\minus{}z)\plus{}(xy\minus{}z^2)\geq 0$ which is true...

[quote="Virgil Nicula"]Prove that in any triangle  $ ABC$ exists the relation  $ \boxed {\ \max\ \left\{\ \frac {a}{p \minus{} a}\ ,\ \frac {b}{p \minus{} b}\ ,\ \frac {c}{p \minus{} c}\ \right\}\ \ge \ \frac Rr\ }$ .[/quote]

Let : $ \alpha \ge \beta \ge \gamma \equal{}> sin(\frac{\alpha} {2}) \ge \frac{1}{2}$

$ \max({\frac {a}{p \minus{} a}\ ,\ \frac {b}{p \minus{} b}\ ,\ \frac {c}{p \minus{} c}}) \equal{} \frac{2Rsin(\alpha) }{r ctg(\frac{\alpha}{2})}\equal{}4\frac{R}{r}sin^2(\frac{\alpha}{2}) \ge \frac{R}{r}$


Show that the number $ 3^{{4}^{5}} \plus{} 4^{{5}^{6}}$ can be expresed as the product of two integers greater than $ 10^{2009}$

Write it as $ 3^{4^5} \plus{} 4\cdot4^{5^6 \minus{} 1}$ and use Sophie Germain.

Precising that the two terms should be integers because of the parity of $5^6+1$, which makes $2^{5^6+1}$ to be an integer, and therefore, the hole expressions.

We can write $3^{4^5} + 4 \cdot 4^{5^6-1}$ as $m^4 + 4n^4$ where $m = 3^{4^4}$  and $n = 2^{\frac{5^6-1}{2}}$. Clearly, by the Sophie Germain identity, we have that:
$$m^4+4n^4 = (m^2-2mn+2n^2)(m^2+2mn+2n^2)$$
All that is left to show is that $(m^2-2mn+2n^2) =  (m-n)^2+n^2 \ge 10^{2009}$.
Clearly, $n^2 = 2^{5^6-1}$. Recalling that $2^{10} = 1024$, we have that $\log_{10}{n^2} \ge \frac{3}{10} \cdot \left(5^6-1 \right) \ge 2009$.

As a sidenote, I’m curious why they made the bounds (10^2009) so loose, when they could’ve made it tighter!

The first part is, yes.
[url]https://artofproblemsolving.com/wiki/index.php/Sophie_Germain_Identity[/url]


Can the following equation be solved as a Pell equation? If it can,how?

$ 12y^2\minus{}x^2\equal{}1$

Note: For $ x$ and $ y$ positive integers

[quote="Chalao"]Can the following equation be solved as a Pell equation? If it can,how?

$ 12y^2 \minus{} x^2 \equal{} 1$

Note: For $ x$ and $ y$ positive integers[/quote]
[hide]Taking it in $ \bmod{12}$ we get that $ x^{2}\equiv 11\bmod{12}$, but there are no square residues with $ 11$ in $ \bmod{12}$, so there are no solutions.[/hide]
I think this is right...unless my reasoning is flawed...

ok...but consider the case of $ dx^2 \minus{}y^2\equal{}a$ could it be solved as a Pell equation?

I'm sorry I can't contribute much to this, but here's a link with lots of info.

http://mathworld.wolfram.com/PellEquation.html

[quote="Chalao"]ok...but consider the case of $ dx^2 \minus{} y^2 \equal{} a$ could it be solved as a Pell equation?[/quote]
I'm pretty sure there are cases where you can solve it. If you take that equation in $ \bmod{d}$, then find the quadratic residues of $ d$ and hope that $ a$ is one of them, then it could work out.

How can i solve Pell equations which minimum solution isn't easy to find? For example:
$ x^2\minus{}12y^2\equal{}4$ I assume (4,1) is a trivial solution

I got the solutions (x,y)= (7,4) and you can prove that these are the only solutions according to the proof that  ckck gave you. Hope this will help you
the solutions i got this way:
x^2 -12 y^2 =1  according to the pell equation , since this equation is from this form u^2-dv^2=1 , the minimum solution is 7,4


let a,b two positive integers such that a b no=1
find all integer values of
$\frac{a^{2}+a b+b^{2}}{a b-1}$

[hide]
Let $n=\frac{a^{2}+ab+b^{2}}{ab-1}$. When $a=b=2$ and $a=11,\ b=2$, we find $n=\boxed{4,7}$. We now show that no other $n$ are possible. Suppose some $a,b$ produced an integer $n\neq 4,7$. Then consider such a solution with least value of $\max \{ a,b \}$.

If $a=b$, then $n=\frac{3a^{2}}{a^{2}-1}=3+\frac{3}{a^{2}-1}$. Then $a^{2}-1$ divides 3, so $a^{2}-1\leq 3\Rightarrow a\leq 2$. But $a$ is positive and not 1, so $a$ must equal 2 and $n=4$, a contradiction.

Otherwise, $a\neq b$. WLOG let $a>b$. Our expression for $n$ can be written as
\[a^{2}-b(n-1)a+(b^{2}+n)=0 \]
a quadratic in $a$. Let $c$ be the other root of the quadratic. Then from Vieta's, $c=b(n-1)-a$, an integer, and $c=\frac{b^{2}+n}{a}$, which is positive. We now show $c<a$ which is equivalent to $b^{2}+n<a^{2}$. After writing $n$ in terms of $a$ and $b$ and multiplying by $\frac{ab-1}{a}$, it's equivalent to
\[b(a^{2}-b^{2})>2a+b \]
Now $a>b$, so $b\leq a-1$ and $b(a^{2}-b^{2})\geq b(a^{2}-(a-1)^{2})=b(2a-1)$. Finally, $b(2a-1)>2a+b$ is equivalent to $(a-1)(b-1)>1$. But as $b\geq 2$ and $a\geq 3$, this is clearly true.

Now that $c<a$, consider the ordered pair $(b,c)$. Since $\max \{ b,c\}<a=\max \{ a,b\}$, this ordered pair cannot be one that satisfies $\frac{b^{2}+bc+c^{2}}{bc-1}\neq 4,7$ if $c\neq 1$. In this case $\frac{b^{2}+bc+c^{2}}{bc-1}$ either equals 4 or 7. But using the quadratic equation with $a$ and $c$ as roots, we find $\frac{b^{2}+bc+c^{2}}{bc-1}=n$, so $n=4$ or $n=7$, a contradiction.

Otherwise, $c=1$ so that
\[n=\frac{b^{2}+b+1}{b-1}=b+2+\frac{3}{b-1}\]
Then $b-1$ divides 3, so $b=2$ or $b=4$. Either way, $n=7$, a contradiction.

So in any case, no $n$ other than 4 or 7 are possible, as desired.
[/hide]


November 1-8 of this year in Russia held the seventh Kolmogorov's Cup. Here is one nice graph problem taken from this quite prestige tournament

In a country there are 100 towns, some of them are connected by roads with one way traffic. It is known that from every town we can
reach every other town. But if we close any road this property will not be valid anymore. What is the maximum number of roads in this
country ? (Any two towns are connected by at most one road).

Namdung
----------
For russianspoken

(В стране 100 городов, некоторые пары городов соединены дорогой с односторонним движением. Известно, что из любого города можно доехать до любого другого, но при закрытии любой дороги это условие нарушается. Какое наибольшее число дорог может быть в этой стране? (Любые два города соединены не более чем одной дорогой.)

I will reformulate the problem as follows:
Consider oriented graph $ G$ with $ 100$ vertexes,that satisfies to the following properties:
a)$ G$ is connected,i.e for every two vertexes $ a,b\in G$,there exist a path between them.
b)If we delete any edge,then $ G$ will be not connected anymore.
c)If $ a\rightarrow b$ then $ b\not\rightarrow a$.
Find the maximum number of edges in $ G$.
Answer:$ 197$.
Solution:
Since our graph is connected and satisfies to condition $ (c)$, we conclude that $ G$ contains a cycle of a length $ n\geq 3$:
$ A \equal{} A_1\rightarrow A_2\rightarrow\dots\rightarrow A_n$.
If there exist an edge $ A_i\rightarrow A_j$,then we can delete it,but our graph is still connected,contradiction,hence there is no edges between vertexes from our cycle.Also if there is a vertex $ B\in G$ and $ B\not\in A$,then $ B$ isn't connected with more than  one vertex from $ A$,i.e if $ B\rightarrow A_i$ then $ B\not\rightarrow A_j$,where $ i\neq j$,because if $ B\rightarrow A_i$ and $ B\rightarrow A_j$,then we can delete one of them and $ G$ is still connected(just replace a path $ BA_i$ by $ BA_jA_{j \plus{} 1}\dots A_i$).The same reason gives us that if $ A_i$ is connected with some vertex $ C\not\in A$,i.e $ A_i\rightarrow C$,then $ A_j\not\rightarrow C$.
So we could replace our cycle $ A$ by vertex $ A'$ and if $ B\not\in A$ and $ B\rightarrow A_i$,then $ B\rightarrow A'$,also if $ A_i\rightarrow C$,where $ C\not\in A$,then $ A'\rightarrow C$.Name our new graph $ G'$.It contains $ 100 \minus{} m \plus{} 1$ vertexes and the diffrence between $ G$ and $ G'$ is that $ G'$ possibly doesn't satisfies to the condition $ (c)$,i.e there may not exist a cycle in $ G'$.
Now we will prove that if some oriented graph $ G$ with $ k$ vertexes satisfies to conditions $ (a)$ and $ (b)$( and not necessery to $ (c)$), then $ G$ contains at most $ 2k \minus{} 2$ edges.
Suppose for the contrary, that $ G$ contains at least $ 2k \minus{} 1$ edges.Number our vertexes arbitrarly:$ X_1,X_2,\dots,X_k$ and if there exist and edge $ X_i\rightarrow X_j$,where $ i > j$,then we delete it.If now $ G$ contains less than $ k$ edges,then we delete other edges,i.e that goes from vertex with less index number to greater index number.
In any case one of the new graphes,contains at least $ k$ edges and $ k$ vertexes,also it has no cycles,but then it has at most $ k \minus{} 1$ edges,contradiction.
So using the obtained result to graph $ G'$ we obtain that $ G'$ contains at most $ 200 \minus{} 2m$ edges,so $ G$ contains at most
$ 200 \minus{} m\leq 200 \minus{} 3$ edges.
Now it is enough to construct an example,when the needed result achieved:
Let $ X$ be connected with $ Y$ and $ Y$ is connected with $ Z_1,Z_2,\dots Z_{98}$ also each of the vertexes $ Z_1,Z_2,\dots,Z_{98}$ is connected with $ X$.It is easy to check that the example is correct.


symmetry function calculus derivative algebra polynomial trigonometry

Find the maximum value of the expression $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz$ where $ x^2\plus{}y^2\plus{}z^2\equal{}1$ and $ x,y,z$ are real numbers

[hide="Failed Algebraic Manipulation"]
$ (x \plus{} y \plus{} z)^{3} \equal{} x^{3} \plus{} y^{3} \plus{} z^{3} \plus{} 3 x^{2} y \plus{} 3 x^{2} z \plus{} 3 x y^{2} \plus{} 3 x z^{2} \plus{} 3 y^{2} z \plus{} 3 y z^{2} \plus{} 6 x y z$
$ (x \plus{} y \plus{} z)(x^{2} \plus{} y^{2} \plus{} z^{2}) \equal{} x^{3} \plus{} y^{3} \plus{} z^{3} \plus{} x^{2}y \plus{} x^{2}z \plus{} xy^{2} \plus{} xz^{2} \plus{} yz^{2} \plus{} y^{2}z$

$ 3(x \plus{} y \plus{} z)(x^{2} \plus{} y^{2} \plus{} z^{2}) \minus{} (x \plus{} y \plus{} z)^{3} \equal{} 2x^{3} \plus{} 2y^{3} \plus{} 2z^{3} \minus{} 6xyz$
$ \frac {3}{2}(x \plus{} y \plus{} z)(x^{2} \plus{} y^{2} \plus{} z^{2}) \minus{} \frac {1}{2}(x \plus{} y \plus{} z)^{3} \equal{} x^{3} \plus{} y^{3} \plus{} z^{3} \minus{} 3xyz$

Since $ x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} 1$, we have
$ \frac {3}{2}(x \plus{} y \plus{} z) \minus{} \frac {1}{2}(x \plus{} y \plus{} z)^{3} \equal{} x^{3} \plus{} y^{3} \plus{} z^{3} \minus{} 3xyz$
$ \frac {1}{2}(x \plus{} y \plus{} z)(3 \minus{} (x \plus{} y \plus{} z)^{2}) \equal{} x^{3} \plus{} y^{3} \plus{} z^{3} \minus{} 3xyz$
$ \frac {1}{2}(x \plus{} y \plus{} z)(2 \minus{} 2xy \minus{} 2xz \minus{} 2yz) \equal{} x^{3} \plus{} y^{3} \plus{} z^{3} \minus{} 3xyz$
$ (x \plus{} y \plus{} z)(1 \minus{} xy \minus{} xz \minus{} yz) \equal{} x^{3} \plus{} y^{3} \plus{} z^{3} \minus{} 3xyz$
Not sure what to do from here...
[/hide]
[hide="Lagrange Multipliers"]
We want to maximize $ f(x,y,z) \equal{} x^{3} \plus{} y^{3} \plus{} z^{3} \minus{} 3xyz$ with constraint $ g(x,y,z) \equal{} x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} 1 \equal{} 0$.
$ 3x^{2} \minus{} 3yz \equal{} \lambda (2x)$
$ 3y^{2} \minus{} 3xz \equal{} \lambda (2y)$
$ 3z^{2} \minus{} 3xy \equal{} \lambda (2z)$
This will be really messy to solve, and may not even give us an answer...
[/hide]

Edit: slight error in the the algebraic manipulation part...

[quote="SarahAhmed"]What grade are you supposed to learn this in?!?!?!?

 :|[/quote]

Well, the algebra is just Algebra 2/Precalc stuff, while the Lagrange Multipliers is multivariable calculus.  Of course, no one will teach you this kind of factoring in school, so you have to learn it yourself...

Continuing in the Lagrange Multiplier vein:

Isolating the $ \lambda$ and getting rid of the fractional stuff:

$ \frac{x^2\minus{}yz}{x}\equal{}\frac{y^2\minus{}xz}{y}\equal{}\frac{z^2\minus{}xy}{z}$

Looking at the first two:

$ x^2y\minus{}y^2z\equal{}xy^2\minus{}x^2z \implies (x^2\minus{}y^2)z\equal{}xy(y\minus{}x) \implies z\equal{}\frac{\minus{}xy}{x\plus{}y}$

Now looking at our 2nd and 3rd thingies:

$ \frac{y^2\plus{}\frac{x^2y}{x\plus{}y}}{y}\equal{}\frac{\frac{x^2y^2}{(x\plus{}y)^2}\minus{}xy}{\frac{\minus{}xy}{x\plus{}y}}$

Getting rid of stuff:

$ y(x\plus{}y)\plus{}x^2\equal{}\frac{x^2y^2\minus{}xy(x\plus{}y)^2}{\minus{}xy}$

$ xy(x^2\plus{}xy\plus{}y^2)\equal{}x^3y\minus{}2x^2y^2\plus{}xy^3\minus{}x^2y^2$

$ x^3y\plus{}x^2y^2\plus{}xy^3\equal{}x^3y\minus{}3x^2y^2\plus{}xy^3$

$ 4x^2y^2\equal{}0$

Which means either x or y =0. However if that is the case, we get that x=y=z=0 which clearly doesn't work. Thus, we made a mistake somewhere and divided by zero somewhere. The only place that could be is when we were solving for z, and divided by x-y. Thus, the only thingy that works is when x=y (=z by symmetry)=rt(1/3).

That's not rigorous but it gets the job done.

I do not understand the Lagrange Multiplier thing, but the results do not seem accurate
First, if $ x\equal{}y\equal{}z\equal{}\frac 13$, then the sum of the variables squared does not equal 1.
It would make more sense though if it was $ \frac {\sqrt 3}3$, however that yields an answer of 0.
A random trial of $ (1, 0, 0)$ produces a greater value...

[quote="RobRoobiks"]I do not understand the Lagrange Multiplier thing, but the results do not seem accurate
First, if $ x \equal{} y \equal{} z \equal{} \frac 13$, then the sum of the variables squared does not equal 1.
It would make more sense though if it was $ \frac {\sqrt 3}3$, however that yields an answer of 0.
A random trial of $ (1, 0, 0)$ produces a greater value...[/quote]

The method of Lagrange multipliers merely finds the critical points of this function, not necessarily the maxima/minima. So we need to check the boundary points as well. Oops.

@Rob: Lagrange multipliers method - you setup a function which is the sum of yours and $ \lambda$ times the constraint (if the condition is $ a\plus{}b\equal{}1$, then you get $ \lambda(a\plus{}b\minus{}1)$), and look for partial derivatives of that.

Maxima/minima should be on the boundary points or in points where those partial derivatives are 0.

I'll rewrite the problem using a more effective set of variables -- the [url=http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial]elementary symmetric polynomials[/url]. [hide="Click to reveal brief background"]The elementary symmetric polynomials in three variables are:
$ \sigma_0 \equal{} 1$
$ \sigma_1 \equal{} x \plus{} y \plus{} z$
$ \sigma_2 \equal{} xy \plus{} yz \plus{} zx$
$ \sigma_3 \equal{} xyz$

You might recognize these from Viete's sums. Basically, a set of three roots defines a unique set of coefficients of a monic cubic, and vice versa. Note that, as a result, the symmetric polynomials are independent variables (why?). This makes these polynomials an effective change of variables.[/hide]

Here are the two relevant polynomials in terms of elementary symmetric polynomials.
$ F \equal{} (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx) \equal{} (x \plus{} y \plus{} z)(1 \minus{} xy \minus{} yz \minus{} zx) \equal{} \sigma_1 (1 \minus{} \sigma_2)$
$ G \equal{} x^2 \plus{} y^2 \plus{} z^2 \equal{} \sigma_1^2 \minus{} 2 \sigma_2$

We are given $ G \equal{} 1 \implies \sigma_2 \equal{} \frac {\sigma_1 ^2 \minus{} 1}{2}$

Substituting, $ F \equal{} \sigma_1 (1 \minus{} \frac {\sigma_1 ^2 \minus{} 1}{2}) \equal{} \frac {3\sigma_1 \minus{} \sigma_1 ^3}{2}$

For $ G \equal{} 1$, $ \sigma_1$ ranges from $ \minus{} \sqrt {3}$ to $ \sqrt {3}$, where the extrema occur when $ x \equal{} y \equal{} z$.

Over this interval, $ F$ takes maximum value at $ \sigma_1 \equal{} 1$, where $ F \equal{} 1$.

Let $ u = x + y + z$ and $ v = xy + yz + zx$ so $ u^2 - 2v = x^2 + y^2 + z^2 = 1$, then by AM-GM,
\begin{align*} & (x^3 + y^3 + z^3 - 3xyz)^2 \\
= & (x + y + z)^2(x^2 + y^2 + z^2 - xy - yz - zx)^2 \\
= & (1 + 2v)(1 - v)(1 - v) \\
\le & \left(\frac {(1 + 2v) + (1 - v) + (1 - v)}{3}\right)^3 = 1, \end{align*}
which is valid since $ 1 - v\ge0\Leftrightarrow(x - y)^2 + (y - z)^2 + (z - x)^2\ge xy + yz + zx$ and $ 1 + 2v\ge0\Longleftrightarrow(x + y + z)^2\ge0$.

Of course from [b]TZF[/b]'s $ F = (3u - u^3)/2 = u(3 - u^2)/2$ we can first note that $ 3 - u^2\ge0\Longleftrightarrow3(x^2 + y^2 + z^2)\ge(x + y + z)^2$ and then do the substitution $ u = \sqrt {3}\sin\theta$ from which we get $ F^2 = 27\sin^2\theta\cos^4\theta/4 = (27/8)(2\sin^2\theta)(1 - \sin^2\theta)(1 - \sin^2\theta)$
$ \le(27/8)(((2\sin^2\theta) + (1 - \sin^2\theta) + (1 - \sin^2\theta))/3)^3 = 1$.

Or realize that $ 1-F=(2-3u+u^3)/2=(2+u)(1-u)^2/2\ge0$.

TZF, can you explain the last two lines?

The second to last line says that if you're on the unit sphere, then $ \sigma_1$ is at least $ \minus{}\sqrt{3}$ and at most $ \sqrt{3}$. Note that the set of points satisfying $ \sigma_1 \equal{} r$ for some real number $ r$ is a plane parallel to the plane $ x\plus{}y\plus{}z\equal{}0$. The plane shifts as you increase $ r$. What we want to do is to find the greatest $ r$ such that there is a point on the unit sphere with $ \sigma_1 \equal{} r$. If you visualize the plane and the sphere, you will realize that this happens at the point $ \left(\frac{1}{\sqrt{3}} , \; \frac{1}{\sqrt{3}} , \; \frac{1}{\sqrt{3}} \right)$. Alternately, you can find this by solving for when the system $ x^2\plus{}y^2\plus{}z^2 \equal{} 1$, $ x\plus{}y\plus{}z\equal{}r$ has only one solution, and these will be the extreme values of $ \sigma_1$.

The last line says that on the interval $ [\minus{}\sqrt{3},\sqrt{3}]$ function $ F(\sigma_1) \equal{} \frac{3 \sigma_1 \minus{} \sigma_1 ^3}{2}$ is maximized at $ \sigma_1 \equal{} 1$, and if you plug that in you get $ F \equal{} 1$.


function
integration
advanced fields
advanced fields unsolved

$f: \mathbb{R}\to \mathbb{R}$ a $C^{\infty}$ function such that $f(x)=0$ if $x \not \in [0;1]$, and $\Omega$ a smooth connected bounded open subset of $\mathbb{R}^{n}$. Suppose that $u: \Omega \to \mathbb{R}$ is a $C^{2}$ function, continuous on $\overline{\Omega}$, that satisfies $-\Delta(u) = f(u)$ on $\Omega$ and $u(x)=0$ if $x$ is on the boundary of $\Omega$. Prove that if $u$ is not non-identically null on $\Omega$, then it is strictly positive on $\Omega$

Let me slightly generalize the problem:

Let $\Omega$ be a smooth, open, bounded subset of $\mathbf R^{n}$, $f \in C^{1}(\mathbf R)$ be such that $Df \in L^\infty(\mathbf R)$ and $\mathrm{supp}\, f \subset [0,\infty)$. Show that if $u \in C^{2}(\Omega) \cap C^{0}(\overline\Omega)$ is a solution to the problem
	\[\begin{cases}-\Delta u = f(u) & \text{ in }\Omega \\ u = 0 & \text{ on }\partial\Omega, \end{cases}\]
	then either $u=0$, or $u>0$ in $\Omega$.

[hide]

Multiply both sides of the equation by $u^{-}$, then it follows from the inclusion $\mathrm{supp}\, f \subset [0,\infty)$ that
	\[\int_{\Omega}|\nabla u^{-}|^{2}= \int_{\Omega}f(u)u^{-}= 0,\]
	yielding that $u^{-}=0$. Thus all solutions are $\ge0$.

	Take arbitrary $a \ge \|Df\|_\infty$, then	the function $s \mapsto f(s)+as$ is increasing and is $\ge0$ for all $s\ge0$ (because $f(0)=0$).

	Now if $u$ is a solution then since $u\ge0$, it satisfies
	\[\begin{cases}-\Delta u+au = f(u)+au \ge 0& \text{ in }\Omega \\ u = 0 & \text{ on }\partial\Omega, \end{cases}\]
	hence by Strong Maximum Principle, either $u=0$, or $u>0$ in $\Omega$ (and $\frac{\partial u}{\partial\nu}< 0$ on $\partial\Omega$).

[/hide]


please make your titles more specific ;)

[title changed from "uh math" to current]


Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number.  This number is exactly nine times the product Sarah should have obtained.  What is the sum of the two-digit number and the three-digit number?

[hide="Solution"]Let $N$ be the two digit number and $M$ be the three digit number. From the requirements of the problem, we have \[ 9\cdot N\cdot M=1000\cdot N+M.  \] Thus, \[ 81\cdot N\cdot M-9000\cdot N-9\cdot M=0,  \] \[ 81\cdot N\cdot M-9000\cdot N-9\cdot M+1000=1000,  \] \[ (9\cdot N-1)(9\cdot M-1000)=1000,  \] $N$ and $M$ must both be integers and $1\equiv1000\equiv1\pmod{9}$. Hence, if $(9\cdot N-1)=x$ and $(9\cdot M-1000)=y$, then $x\equiv y\equiv8\pmod{9}$ for $N$ and $M$ to both be integers. Over the factors of $1000$, only $8$ and $125$ are congruent to $8\pmod{9}$. Since $N$ must have two digits and $M$ must have three, we conclude \[ 9\cdot N-1=125,\qquad 9\cdot M-1000=8,  \] \[ N=14,\qquad M=112.  \] Indeed, $14\cdot 112\cdot 9=14112$. Therefore, the desired answer is $14+112=\boxed{126}$.[/hide]


So, Ravi B. and I just noticed that I can post things in Intermediate using the "Source" field, but that he cannot.  (Apparently, it doesn't exist for him when he posts.)  I haven't yet found out whether this is a skin difference since I don't know what skin he's using (I'm using AoPS), but, is this supposed to happen?  Am I special or something? ;)

By the way, I had posted a related question earlier at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23778]source field?[/url].  That's where Valentin had said that the source field does not appear in the lower-level forums, which I assumed included Intermediate.

(Joel, I switched between AoPS and MathLinks skins, and it didn't seem to make a difference.)

I have explained this one time in the topic Ravi B posted a link for. Please continue there the discussion (if there is anything that left is to be discussed further).


[quote="236factorial"]Convert $46365_8$ into base 10.[/quote]

I'm not too sure about my answer... I think it's correct. I made a topic on this and learned how to convert into base 10 real easily:

[hide]$8^0*5=5$
$8^1*6=48$
$8^2*3=192$
$8^3*6=3072$
$8^4*4=16384$

$16384+3072+192+48+5=\boxed{19701}$[/hide]

Right?

[quote="236factorial"]You made a tiny multiplication mistake. See if you can catch it.[/quote]

Alright! Thanks! I think I fixed it!  :)

[quote="anirudh"][quote="236factorial"]Convert $46365_8$ into base 10.[/quote]

$8^0=1$
$8^1=8$
$8^2=64$
$8^3=512$
$8^4=4096$
$8^5=32768$

And now i don't know what to ddo.[/quote]

Multiply the one's digit by 8^0, ten's digit by 8^1, etc, and then add them together.

Check your digits please. You copied them wrong.


compute $ f(x)$ when $ x \to 0$ ,with  $ f(x) \equal{} \sqrt {x}ln(sh(\frac {\sqrt {x \plus{} 1} \plus{} 1}{2\sqrt {x}}))$.


function
algebra
domain
algebra unsolved

f is real function and is not constant and:

$ f(x\plus{}y) \equal{} f(x) \plus{}f(y) \minus{} 2f(xy)$

find $ f(1383)$ and $ f(2004)$

[quote="behdad.math.math"]f is real function and is not constant and:

$ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \minus{} 2f(xy)$

find $ f(1383)$ and $ f(2004)$[/quote]

Pleaaaase, once again, try to give a full exact problem statement (including the domain where equation is exact) !

1) If the equation is true for any $ x,y\in\mathbb{R}$ there are no non constant solution :
Let $ P(x,y)$ the property $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\minus{}2f(xy)$ $ \forall x,y\in\mathbb{R}$
$ P(0,0)$ $ \implies$ $ f(0)\equal{}0$
$ P(x,1)$ $ \implies$ $ f(x\plus{}1)\equal{}f(1)\minus{}f(x)$ and so $ f(x\plus{}2)\equal{}f(x)$ and so $ f(2)\equal{}0$
$ P(x,2)$ $ \implies$ $ f(x\plus{}2)\equal{}f(x)\plus{}f(2)\minus{}2f(2x)\equal{}f(x)\minus{}2f(2x)$ and so $ f(2x)\equal{}0$ $ \forall x\in\mathbb{R}$
And so the only solution is $ f(x)\equal{}0$

2) If the equation is true for any $ x,y\in\mathbb{Z}$  :
We again have :
$ f(x\plus{}2)\equal{}f(x)$, $ f(0)\equal{}0$ and $ f(1)\equal{}a$
And it's easy to verify that these mandatory conditions are enough.

So $ f(2004)\equal{}0$ and $ f(1383)$ may be any value.


MIT
college
Harvard
Princeton
Putnam
algorithm
college contests

I analyzed Putnam results to draw conclusions about mathematics at MIT, Harvard and Princeton in my essay:
http://blog.tanyakhovanova.com/?p=139

As JBL already noted, your conclusion doesn't necessarily follow as the algorithm is flawed anyway.  Also, I don't see how you can use Putnam results as indication of how seriously a school treats its math program.  If a student at Harvard is in Math 55 but chooses not to compete in Putnam, is he any less serious about math than his counterpart at MIT who chooses to participate in Putnam?  Many people don't like competition math and therefore, it's impossible to completely gauge a student body.

[quote="Tanya Khovanova"]MIT shouldn’t assume that those three continue to be their strongest competitors. Instead they should probably test their students right before the Putnam competition, because if you look at MIT’s top individual performers, had they been on a team together, they would have won.[/quote]

this seems pretty silly to me.  one of the biggest problems with using the previous year's putnam as the basis for choosing the next year's team is the signal-to-noise ratio.  giving them another test does not do anything to mitigate this concern, it just replaces one flawed data set with another data set, although you seem to implicitly assume that testing the students right before the putnam (when they have more pressing concerns like final exams) will magically determine the 'best' contestants.

moreover, it takes a lot of time and effort to produce a test which is capable of distinguishing between putnam fellow-caliber students, and it's not at all clear from what you wrote that this is worth it.

in my opinion, the most likely explanations for MITs failure to 'win' the putnam is as follows:

1) there's a significant amount of randomness involved in putnam performance, especially since a single bad score can effectively kill your team
2) although MIT dominates the HM or better count, it's not at all clear that this is particularly relevant to 'winning' the putnam.  What really matters is the quality of the top 5 or so undergraduate mathematics majors at the university, and it's not clear that [i]for the purposes of winning the Putnam[/i] that MIT has the best team.  as long as harvard and princeton replenish their pool of real superstars (say, people who would be in the top 20 Putnam competitors consistently), it doesn't really matter how many people they have outside this range.  of course if you change the putnam team competition to the top 20 students, MIT will win every year since the effects of randomness are less and since the 11th-20th best students at MIT are probably significantly better than the 11th-20th best students at Harvard/Princeton.

note an interesting ramification of 1) + 2) -- due to randomness and the depth of MITs teams, it is likely that in at least one of the last few years, a student who probably was the 4th-10th best putnam competitor at MIT outperformed expectations and was a Putnam fellow.  since the putnam team competition is only about the top 3 scores, this doesn't matter; it's just 'noise'.  but it does two things; it creates the impression that MIT is unlucky when in fact it's more likely to have outliers doing unexpectedly well due to the depth of their talent pool, and it also weakens their team in future years (assuming that relative ability levels stay constant) because really, you can't not put a putnam fellow on your team the following year if it is possible.

[quote="JRav"] If a student at Harvard is in Math 55 but chooses not to compete in Putnam,[/quote]

haha okay this really doesn't happen much so it's basically a straw man.

e: for an annoying typo

But using a combination of several exams should increase the signal-noise ratio.  I suspect a bigger obstacle is that a lot of potential team members wouldn't be willing to sit through an extra 6-hour exam just to secure a seat on the team (especially near the end of the semester, etc.).

[quote="blahblahblah"]a single bad score can effectively kill your team[/quote]  Actually, even a single very good-but-not-quite-amazing score can kill your team if you're talking about winning a top-three spot.

[quote="blahblahblah"]It does two things; it creates the impression that MIT is unlucky when in fact it's more likely to have outliers doing unexpectedly well due to the depth of their talent pool, and it also weakens their team in future years (assuming that relative ability levels stay constant) because really, you can't not put a putnam fellow on your team the following year if it is possible.[/quote]  That's a really good insight.


algebra
polynomial
number theory unsolved
number theory

For all positive integers $ n$, let $ s_n$ be the smallest integer such that there exists a polynomial $ P$ with integer coefficients and degree $ s_n$ and $ P(1), P(2), \cdots , P(n)$ are all prime. Is the sequence $ \{s_i\}_{i\equal{}1}^{\infty}$ bounded?

Yes; $ s_n \equal{} 1$ for all $ n$ by the [url=http://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem]Green-Tao theorem[/url].


[hide="hint maybe"]In 1992 Frank Hollad [1] stated the following inequality:

$ \sqrt[n]{a_1 \frac{a_1\plus{}a_2}{2} ... \frac{a_1\plus{}a_2\plus{} ... \plus{}a_n}{n}}  \ge  \frac{a_1 \sqrt{a_1a_2} \sqrt[3]{a_1a_2a_3} ...  \sqrt[n]{a_1a_2 ... a_n}}{n}$

Maybe the solution of that problem is related to the above inequality.[/hide]

[quote="hasan4444"]$ \frac {a_{1} \plus{} \sqrt {a_{1}a_{2}} \plus{} \sqrt [3]{a_{1}a_{2}a_{3}} \plus{} \cdots \plus{} \sqrt [n]{a_{1}a_{2}\cdots a_{n}}}{n}\leq\sqrt [n]{a_{1}\cdot\frac {a_{1} \plus{} a_{2}}{2}\cdots\frac {a_{1} \plus{} \cdots \plus{} a_{n}}{n}}$ with the $ a_i\geq 0$[/quote]
This is a very nice and very hard inequality. See the attachment below for the proof of Prof. Kiran Kedlaya:


Let a word have an "alphabet" number explaining how many letters from the beggining of the alphabet it has.  For example, [b]cab[/b] and [b]back[/b] have alphabet numbers of 2 be cause they both have an [b]a, b, [/b]and [b]c[/b].  [b]Backed[/b] has an alphabet number of 5.  Can you find a word with an alphabet number of 6?

[hide="Not sure if this is a word"]
Fabricade or Fabricized or some other "fabric" word with an ED ending.
[/hide]

By your own definition, the words "cab" and "back" have
alphabet numbers of 3, not 2 as you gave it.  Then I agree with you that the word "backed" has an alphabet number of 5.

A word with an alphabet number of 6?  I didn't know it, but I looked at your hidden choices, but they came up as non-words
in a brief computer search.

Along your lines of suggestion, here is one, but there may be at
least another correct one:
[hide]fabricated[/hide]


Let $n$ be a positive integer such that the sum of all positive divisors of $n$ (except $n$) plus
the number of these divisors is equal to$n$. Prove that $n = 2m^2$ for some integer$m$.

You want $(\sigma(n) - n) + (d(n) - 1) = n$, i.e. $\sigma(n) + d(n) = 2n + 1$. So let $n = 2^t \cdot q$, where $t \in \mathbb{N}$ and $q \in 2\mathbb{N}+1$. If $q = 1$, this leads to $(2^{t+1} - 1) + (t + 1) = 2^{t+1} + 1$, which is true iff $t = 1$, making $n = 2 \cdot 1^2$. Then suppose $q > 1$ and put $q = \prod_{i=1}^r p_i^{a_i}$, where $r$ is a positive integer, $p_1, p_2, \ldots, p_r$ are pairwise distinct natural odd primes and $a_i \in \mathbb{Z}^+$, for any $i = 1, 2, \ldots, r$. You get $\sigma(n) + d(n) = 2n + 1$ iff $(2^t-1) \cdot \prod_{i=1}^r (1 + p_i + \ldots + p_i^{a_i}) + (1+t)\cdot \prod_{i=1}^r (1 + a_i) = 2n + 1$, arguing you need $t \cdot \prod_{i=1}^r (1+a_i) \equiv 1 \bmod 2$. Therefore you have $t \equiv 1 \bmod 2$ and $a_i \equiv 0 \bmod 2$, for any $i = 1, 2, \ldots, r$. This means $q =b^2$ and $t = 2s + 1$, where $s, b\in\mathbb{N}$, that's to say $n = 2m^2$, with $m = 2^s \cdot b$.


Let $n>1$ be integer. Show that every continous $f: \mathbb{S}^{n}\to\mathbb{S}^{1}$ is a nullhomotopy.

$f$ induces a homomorphism $f_{*}$ of fundamental groups. Now use the following property: if $p: \widetilde X\to X$ is a covering map and $f: Y\to X$ is a map such that $f_{*}(\pi_{1}(Y))\subseteq p_{*}(\pi_{1}(\widetilde X))\ (*)$, then $f$ lifts to $\widetilde X$, i.e. there is a map $\tilde f: Y\to\widetilde X$ such that $f=p\circ\tilde f$ (all of this if the spaces $X,\widetilde X,Y$ are "nice enough", i.e. have all the connectedness properties you could wish for).

We apply the above to $X=\mathbb S^{1},\widetilde X=\mathbb R,Y=\mathbb S^{n}$. Because $\mathbb S^{n},\ n>1$ is simply connected, $(*)$ holds, so $f$ factors through $\mathbb R$. Since it's clear that every map from $\mathbb S^{n}$ to $\mathbb R$ is nullhomotopic, the conclusion follows.

because $\mathbb{S}^{n}$ is simply connected for $n \geq 1$, you can find a lifting $\hat f : \mathbb{S}^{n}\to \mathbb{R}$ such that $f=p \circ \hat f$. And because $\mathbb{R}^{n}$ is convex (hence contractile), $\hat f$ is null homotopic. Hence $f$ is null homotopic.


Let P be a regular 20-gon which is inscribed in a circle of radius 1. Find the sum of the squares of the lengths of all of the sides and diagonals of P.

Sorry. That's incorrect. Check your work, and remember you're looking for the sum of the [i]squares[/i] of the sides and diagonals.

EEK! AND diagonals... i totally missed that part of the question. I gave you only the sides :)

LoL.

By the way ComplexZeta, I'm sure there is a better way to do it, but would getting the diagonals require Ptolemy's, or just taking small n-gons and looking for a pattern?

I wouldn't suggest finding the lengths of diagonals that you don't just know instantly (such as diameters). Don't go finding cos(18) or anything like that. You don't need Ptolemy, and it would be better for you not to look at small cases. You get a better insight into the nature of the problem if you can do it without relying on small cases. Having said that, when you are actually taking a competition, solve it in any way you can. But right now it's a challenge problem on the forum, so try not to look for patterns.

hmm this looked interesting....

[hide]

we can pair off every diagonal/side except for the main ones going through the center of the circle. pythagoras' theorem yields the sum of squares for each pair is equal to the main diagonal squared = 4.

taking the main diagonals (all 10 of them) separate:

mains: 10 * 4 = 40

others: (20 C 2) - 10 sides/diags = 180 => 90 pairs => 360

40 + 360 = 400.

[/hide]

ooooo
niceness
i shulda thought of that one

Sorry for being slow to pick up on this new concept... Can you specify how you are pairing these diagonals and how you get the inequalities?

Those aren't inequalities. => means implies. The thing you use here is that if an angle is inscribed in a semicircle, then it is a right angle. That's how you pair them.

And he's not actually pairing diagonals. He's pairing each diagonal with the side adjaccent to it such that they form a right triangle. with the hypotenuse being the diameter of the circle.

two diagonals can be paired together...

but i do emphasize can. its not always the case; however it is irrelevant to the problem as long as the pairings of line segments do exist, which is fairly clear as any angle with vertex on a circle subtending half the circle is 90 degrees in measure.


[url=http://www.isid.ac.in/~rbb/olympiads.html]www.isid.ac.in/~rbb/olympiads.html[/url]

This site is emerging as a great site for Indian Olympiads--- webpage of NBHM head!

Its got all latest results, solutions to latest CRMOs (the national ones! -- some regions do this while others preapre own papers) and the INMOs -- even those for 2006!!
Check it out!!!

Thanks a lot....that site has plenty of rather interesting links......And just by the way, do you know any place where I can find question papers for the previous indian physics olympiads and the previous IAO's {Indian Astronomy Olympiads} ???


function
limit
logarithms
quadratics
number theory
number theory solved

Could somebody suggest me a number theory function which:
1) has not too complicated definition;
2) does not have a useful closed formula;
3) $\lim_{n\to\infty}\dfrac{f(n)}n=0$ what is neither too trivial nor too hard to prove?

To better clarify my question, consider the function $\pi(n)$ (the number of primes up to $n$).
1) It really does have a very simple definition;
2) It does not have a useful closed formula;
3) It indeed holds $\pi(n)=o(n)$ which is neither too obvious nor too hard to prove.

So, $f(n)=\pi(n)$ would be a perfect choice, but unfortunately I cannot use this function because I already used it at an other point (I can't explain better, just assume that $\pi(n)$ and its variants are not permited). Any other suggestions are welcomed. Thank you.

Replace "primes" with "sums of two squares" and you get another such function.

Is $\mathrm{A}_2 (n) : = 4 \cdot \sum\limits_{\substack{ d \in \mathbb{N} \\ d|n \\ d \equiv 1 \mod 2}} (-1)^{\dfrac{d-1}{2}} = 4 \cdot \prod_{\substack{p \in \mathbb{P} \\ p\equiv 1 \mod 4}} (v_p(n)+1)$ as the number of representations of $n$ as sum of two squares really not useful¿

But other functions are very 'complicated':
- Van mangold function (but the limit is obvious)
- representatations as sum of three squares (not totally sure about the limit)
- the class number of $\mathfrak{o} ( \mathbb{Z} [\sqrt n])$ should also do it

If you let Von Mangoldt lambda in the set of suitable functions, then what about Moebius $\mu$? It checks trivially all the properties Bojan Basic is asking for. Sure! But... Mindflyer, my dear friend, why don't you tell us that ooold story about closed formulae? I think it could help... Don't you?

[quote="ZetaX"]Is $\mathrm{A}_2 (n) : = 4 \cdot \sum\limits_{\substack{ d \in \mathbb{N} \\ d|n \\ d \equiv 1 \mod 2}} (-1)^{\frac{d-1}{2}} = 4 \cdot \prod_{\substack{p \in \mathbb{P} \\ p\equiv 1 \mod 4}} (v_p(n)+1)$ as the number of representations of $n$ as sum of two squares really not useful¿[/quote]

It is too much and too little, if one wants the asymptotic number of integers between $1$ and $n$ which are sums of two squares.   The proof that this is proportional to $n / \sqrt{\log n}$ does not use $A_2(n)$, as far as I can remember, but rather an analysis of $\sqrt{\zeta (s)}$ near $s=1$.

[quote]- representatations as sum of three squares (not totally sure about the limit) [/quote]

The limit of what function?

[quote="fleeting_guest"][quote="ZetaX"]Is $\mathrm{A}_2 (n) : = 4 \cdot \sum\limits_{\substack{ d \in \mathbb{N} \\ d|n \\ d \equiv 1 \mod 2}} (-1)^{\frac{d-1}{2}} = 4 \cdot \prod_{\substack{p \in \mathbb{P} \\ p\equiv 1 \mod 4}} (v_p(n)+1)$ as the number of representations of $n$ as sum of two squares really not useful¿[/quote]

It is too much and too little, if one wants the asymptotic number of integers between $1$ and $n$ which are sums of two squares.   The proof that this is proportional to $n / \sqrt{\log n}$ does not use $A_2(n)$, as far as I can remember, but rather an analysis of $\sqrt{\zeta (s)}$ near $s=1$.[/quote]
Sorry, I thought you meant $f=A_2$. Yes, your function fulfills the condition.

[quote="fleeting_guest"]
[quote]- representatations as sum of three squares (not totally sure about the limit) [/quote]

The limit of what function?[/quote]
The one given in 3) ;)

If you mean that $f(n)$ = number of representations of $n$ as sum of 3 squares, there are formulas for this coming from the theory of quadratic forms, so it does not satisfy the conditions of the problem.  I'm pretty sure that $f(n)/n \to 0$ and that this is not difficult to prove.


find out all the function for: f(f(x))=x+2005

and similar one:
f(f(x))=2005x

What is the range of the function? $\mathbb {N}, \mathbb {R}, \cdots$?

Pierre.

Well it was not stated where i got it. So can if u can then plz give me solution for both cases.

(I solved for N :D )

Well, if the function is from $\mathbb {N}$ into itself, it is the famous problem of IMO 1987-4 (with $2005$ instead of $1987$ but the only relevant hyopthesis is to use an odd integer).
If the function is from $\mathbb {R}$ into itself then there are an infinite number of solutions. Let $a,b$ be two real numbers. Choose that $f(a)=b$ and then $f^{2n}(a) = a+2005n$ and $f^{2n+1}(a) = b+ 2005n.$
This define $f$ on the set of the numbers which belongs to at least one of the infinite (both directions) arithmetical progressions with ratio $2005$ and 'first' term $a$ or $b$. Then, choose another pair with no member already in the above progressions, and so on...

Pierre.


MATHCOUNTS
geometry
calculus
quadratics
number theory
prime factorization
AMC 10

I desire calculator programs for Mathcounts.  Unfortunately I do not have any access to a connectivity cable to a computer, so i would need the souce code. Can anyone help me?

if you are saying that you need to be able to write your own programs then there are afew good sites for this, the one listed below is how i learned and even though it says that its for the TI-83 the programming code is basically the same for all TI calculators except the TI-86.  It is really good for the basics.  I use it and create great programs for MATHCOUNTS.  Hope its helpfull

[url]http://www.ticalc.org/programming/columns/83plus-bas/cherny/[/url]

A couple good programs that i made are Herons formula and Shoestring.

Time spent looking up/writing programs or learning how to is time wasted solving problems.

I have no idea how to write a program (well, I can sort of use Mathematica) on a calculator, and I've done just fine in MathCounts.

All that I can program well are games, and I've learned that this year. The only one program I might suggest getting is a prime factorization one-it's the only one I ever found remotely useful.

[quote="unimpossible"]Well i would suggest you to look it up on the web or something.

but i DO NOT encourage using programs in mathcounts.  It will just increase your dependency on unnecesarry calculators, and you should note that higher contests are all banning calculators now.[/quote]

aka amc 10

WHat's harder, AMC 10 or MC?
I am a 5th grader and i did OK on MC and made it to state so i am thinking of taking the AMC 10 next year

You're in 5th grade. You can't go to MC officially lol.  And to take AMC 10 you have to pass AMC 8. Ask your teachers. Unless they're totally incompetent, they'll tell you the exact same things. AMC 10 is harder though.

mmmm..... No way. AMC 10 requires knowledge of some high school basics. You need to know at most geometry for mathcounts (anything).

the time limit in mathcounts evens it out

at least for normal people...(not you =P)

[quote="mathcrazed"]What's that supposed to mean?  :o Are you calling me stupid!! I get it now!! Mark Sellke is god. Almost. Time limit... hmmm... Never noticed that...[/quote]

amc should be everything up to geometry (pre_algebra, algebra I, and geometry) because a normal 10th grader (not someone whos taking calculus as a 9th grader...) is expected to have learned everything up to geometry.

How do u program the shoestring formula?

i program distance formula, quadratic (although I barely use it), and the midpoint formula.

they saved a lot of time for me.


Solve in $ M_2(\mathbb{Z})$ the equation: $ X^{n\plus{}2}\minus{}X^{n\plus{}1}\plus{}X^n\equal{}\left(\begin{array}{cc}1 & 1\\ \ 0 & 1\end{array}\right), n \in \mathbb{N}^*.$

$ X$ cannot be diagonalizable, since if it was, the LHS would be diagonalizable, but the RHS is not. So $ X$ has an eigen value $ \lambda\in \mathbb C$ of algebraic multiplicity 2. Thus the characteristic polynomial of $ X$ has the form $ T^2\minus{}2\lambda T\plus{}\lambda^2\in\mathbb Z[T]$ which implies $ \lambda\in\mathbb Z$. The given equation implies $ \lambda^n(\lambda^2\minus{}\lambda\plus{}1)\equal{}1$ which yields $ \lambda\equal{}1$, because each of the factors must be either $ \plus{}1$ or $ \minus{}1$. Hence $ X\equal{}E\plus{}N$ where $ N\in M_2(\mathbb Z)$ with $ N^2\equal{}0$. By Cayley-Hamilton, it holds $ X^2\minus{}X\plus{}E\equal{}X$, thus $ X^n(X^2\minus{}X\plus{}E)\equal{}X^{n\plus{}1}$ is equal to the RHS of the given equation. Expanding $ X^{n\plus{}1}\equal{}(E\plus{}N)^{n\plus{}1}\equal{}E\plus{}(n\plus{}1)N\equal{}RHS$ yields that $ n\equal{}0$ and $ X\equal{}RHS$. In this case it is easy to check that $ X\equal{}RHS$ fulfils the equation.

Result: $ n\equal{}0$ and $ X\equal{}RHS$.

Den von dir gegenenen Lösungsweg find ich genial!


Positive real numbers a, b, c satisfy 21ab + 2bc + 8ca \leq 12. Find the least possible value of 1/a + 2/b + 3/c.

[i]Source: MOSP Homework 2003[/i]

This problem is similar to one posted by Namdung, which was given in a team selection test in Vietnam. I strongly believe that the solution posted by me can be adapted to solve this problem ,too.

Yes, it is exactly the Vietnamese 2001 IMO Team Selection. I changed the condition and posted in forum.

Namdung

Hello. I'm sorry, but could you direct me to the similar problem please? :D

It is on the Proposed Problems-Discussed Topics and it is called Minima without calculus.

[quote="StRyKeR"]Positive real numbers a, b, c satisfy 21ab + 2bc + 8ca \leq 12. Find the least possible value of 1/a + 2/b + 3/c.[i]Source: MOSP Homework 2003[/i][/quote]
Let $ a, b, c $ be ositive real numbers satisfy $ 21ab + 2bc + 8ca \leq 12$. Find the least possible value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$


Let $x=2^{2976221}-1$ a prime number. Find the number of digits of $x$.

If necessary, consider $\log_{10}2 \approx 0.3$.

That approximation isn't good enough, since $N=1+[\log_{10}(2^{2,976,221}-1)]=1+[2,976,221\log_{10} 2]$. The exact value for $N$ is $895,932$, whereas the given approximation yields $N^{*}=892,867$ - more than $3,000$ digits less.

I agree with you [b]Farenhajt[/b]. But ...

[hide="Answer"]$N=892,866$[/hide]