the results for the romanian team are

1      2     3     4     5     6
------------------------------
7      7     3     7     7     7    38p ROM1 Caraiani Ana
5      1     0     7     7     1    21p ROM2 Kreindler Gabriel
7      7     0     0     1     0    15p ROM3 Manolescu Ioan
7      0     0     7     7     1    22p ROM4 Negut Andrei
7      7     0     7     1     0    22p ROM5 Raicu Claudiu
7      5     0     7     5     1    25p ROM6 Stefanescu Andrei

Bulgaria is the first Team

far behind comes China ( with one full score ) and USA ( no more than 36 p each ), followed by Vietnam ( 2 full scores !!! ).

Romania is looking at the 7th place.

I'll be back with more details (i.e. medal cut-offs, etc.) when they are available.

Valentin, thank you for an information. Does anybody know more? I guess, it is 11 PM in Japan now, so they should already sleep.

Cuts:

13 bronze
19 silver
29 gold

Pablo Dal Lago (an Argentinean excited as we`re getting our second gold medal in history. not me certainly :) )

1 gold(29), 2 silver(25,22),2 bronze(15,16) and 1HM(12)
so, we are 11th this year

Is Bulgaria the first place team? Does anyone know the Bulgarian score?

Congratulations to Bulgaria, Kazakhstan and who want them else. I kindly ask, just if it is possible to post Russian (shameful, I guess) results here. Thank you very much, Fedor.

Hi. I am from India . Can anybody post results of Indian Team and Names of perfect scorers. THAN S

Can anybody post results of Macedonian Team
Tank's.

Pretty pathetic...
vipul-23, shubhangi-22, swarnendu-21, nikhil-20, anand-17, bhaskar-12.
Rank is probably 16th.

Abhishek.

Hello. Can any body post names of perfect scorers at IMO 2003 and nationality.

What about Yugoslav, Serbomontenegrian, or how ever you call them, results? Members of the team apparently feign they 've never heard about such a thing like e-mail or internet - nobody here knows.
Congratulations to our neighbors Bulgarians (though 41+40+... is not 213), and to all 42's.

Sorry for the triple mail... I was reported an error and resended the message.

Are you sure about the medal cut-offs (bronze:13, silver:19, gold:29)??

PS: Congratulations tu Carlos Di Fiore (29pts), and to all the argentinean imo team!!

   RaMlaF.-

Felicitaciones Carlos!!!

Do you know any other results from Latin America?

Thanks, Javier Cppola (Uruguay)

What're about Vietnamese participants' score? Does anyone know the details?

Hi everybody!!

Why there isn't an official web page with the complete ranking and scores???
Past years there was! Or not?

However, could anyone post the scores of each member of Italian team?

Lot of thanks in advance.
Bye bye.

The first two results:

1. Bulgaria   227
2. China      217

The next teams are:
3. USA  (about 40 pts less than the winner)
4. Vietnam
5. Russia

The official IMO sites for 1995, 1997-2000, 2002 had full scores (not
always with names).  I don't know about 1996 (I could never access the
IMO96 site at the time).  2001 had the scores, but on one page per
contestant, not all linked to from anywhere.  When I get the scores I'll
put them up at [url]http://www.srcf.ucam.org/~jsm28/imo-scores/[/url].  I
expect them also to appear at the Estonian site
[url]http://www.math.ut.ee/olympiaadid/imo/[/url].  In the mean time, here
are details of the top 29 countries from Geoff Smith (UK Leader) in Japan:

[code]
1  Bulgaria   227
2  China      211
3  USA        188
4  Vietnam    172
5  Russia     167
6  Korea      157
7  Romania    143
8  Turkey     133
9  Japan      131
10 Hungary    128
   UK
12 Canada     119
   Kazakstan
14 Ukraine    118
15 India      115
16 Taiwan     114
17 Germany    112
   Iran
19 Belarus    111
   Thailand
21 Israel     103 (only 5 students)
22 Poland     102
23 Serbia & M 101
24 France      95
25 Mongolia    93
26 Australia   92
   Brazil
28 Argentina   91
   Hong Kong
[/code]

thanks guys for all the info, come and check out our french maths forum, we've got an imo topic but no one had anu results, come and say hello :)

http://mathsforum.membres.jexiste.org/forum.php?f=5&view=110

Hi. I would be obliged if any one post name of contestant who got special prize for most elegant solution and names of perfect scorers at IMO 2003. THAN S

I wonder why the japanese made this site: http://www.imo2003.com if they hadn't post the subjects, results and other information on time!?  :?

hope it will be done soon!

here is a nice german site: http://www.mathematik-olympiaden.de/

okay so at the special request (i have the list with all the student's ranks ) I can offer some details about that guys' score

in another order of ideas, I belive that it is good to mention here the 3 guys who scored 42 points:

1. Fu Yunhao - China
2. Hung Viet Bao Le - Vietnam
3. Trong Canh Nguyen - Vietnam

Following are

4. Kaloyan Slavov (41), Rumen Zarev (40) and Iliya Tsekov (38) from Bulgaria.

Next to them are Ana Caraiani (Romania) and Oleg Gol'berg (Russia), both of them having 38 points.

Valentin Vornicu:

Can you post the full score list? If you do not have time, please send it to me so that I can create the full score list of all students and their results and post them here.

Thanks.

I just checked the official website of the imo 2003 and noticed that they have posted the official resutls!

...but with country codes and no contestant names! Official solutions are yet to come, I guess. (check the NEWS NEWS link)

right - no names ! i have the actual paper list - it would take a lot of time to put it on the net, and I really don't have that time now

Here is the statistic of problem number and the number of contestant that solve the problem. (from http://village.infoweb.ne.jp/~fvgm9250/web/imo2003/contents/results_all.xls)

          -7-  -6-  -5-  -4-  -3-  -2-  -1-  -0-
problem 1 180    1   24    0    0  111   18  123
problem 2  67    1   20    0   93   40  119  117
problem 3  23    1    0    1    3    0    5  424
problem 4 272    2    3    1   35   23   31   90
problem 5  67    4    1    5    6   20  161  193
problem 6  24    1    1    0    1    9   64  357

I have now put the full results of IMO 2003, including contestant names, online at my site [url]http://www.srcf.ucam.org/~jsm28/imo-scores/[/url].

Additional older (pre-1995) results lists are always welcome to extend the site further back.

Does anyone know who have proposed (which countries) this years problems in Japan?

This year's problems were proposed by Brasil, Bulgaria, Poland, Finland, Ireland and France (in that order).

sal
I want to see the subjects from OIM 2003. Where can I find them??

here you can find them:

 http://www.kalva.demon.co.uk/imo/imo03.html

cheers!!!!!! :D

yeah well they are posted on this website also - just check the thread ahead :)

I would apreciate, as being the first guy in the world to put up the IMO 2003 questions to get some credit for that :D ...


function
real analysis
real analysis unsolved

If a real valued function on R real number , is uniform continuous , then there exists two positive real numbers a and b such that
Abs(f(x)) Less or equal than aAbs(x)+b


The real factors of $ x^2\plus{}4$ are:

$\textbf{(A)}\ (x^2+2)(x^2+2) \qquad
\textbf{(B)}\ (x^2+2)(x^2-2) \qquad
\textbf{(C)}\ x^2(x^2+4) \qquad\\
\textbf{(D)}\ (x^2-2x+2)(x^2+2x+2) \qquad
\textbf{(E)}\ \text{Non-existent}$

I think you mean $ x^4\plus{}4$ which has a famous factorization for D

Sorry to revive, but the original problem does say $ x^2 \plus{} 4$, and the answer is $ \boxed{\textbf{(E)}}$...

If what the guy with the monkey avatar is saying is correct then use Sophie Germain...

EDIT: Didn't see your post there math154...

i got e because the real way to factor is (x+2i)(x-2i)


MIT college Gauss Princeton Yale AMC USA(J)MO

did anyone from this site manage to get a copy of the acceptance letter to MIT? It seems that I did not.

MIT has a notoriously difficult to decipher admission process. I know quite a few (in my opinion) extremely talented individuals who did not get accepted to MIT. My theory is that they get so many talented math and science applicants that diverse or particularly unique talents that make you stand out become very important. Still, I think I've been more surprised by some of the MIT admission decisions than I have of any other school that I know of.

Regardless of where you get in though, the most important thing is to be happy at the school that you choose, and to do well there. The best way to show them they made a bad decision is to go somewhere else and be successful - and then donate a gabillion dollars to that school. :)

Listen bro, I totally know how you feel. Those fruitcakes rejected me too, and I was damn qualified. I think those tools needed to get their asses in gear. . . but yeah, do what that gauss man said, you'll be aight. I'm still pretty pissed though, but it'll be aight. As to the acceptance, I dunno, 6 really bright kids from our school applied, none got in.

MIT rejected me.  A few people from my school did get in, but they don't strike me as being particularly passionate about math or science.

[quote="blahblahblah"]I didn't apply to MIT.

But on the way home from the Princeton April Hosting weekend, I was talking with a girl, who told me that a guy from her school (in Florida), got into MIT with a 1260 SAT1.

So go figure.[/quote]

At the Yale information session they said their lowest SAT I score this year was an 1100.

[quote="probability1.01"]Yale, maybe, but MIT really shouldn't accept < 700 SAT math scores unless the guy made USAMO or something (not that they'd do this or consider the USAMO strongly).[/quote]

I disagree with this. First of all, while it's quite possible that the person in question did make a 700 on the SAT math; a sub-700 SAT math score alone shouldn't (and I'm sure doesn't) exclude someone from admission to MIT. The person's application could have other outstanding features that show that they could benefit from the school, be a benefit to the school, and be successful there.

Confuted could you give me some idea of your resume? You know like all your accomplishments.  I know that you took like 10 or more AP tests from another thread( I think that was you).
I was wondering because I am thinking about applying to MIT my senior year and I want to see what gets in and what doesn't. I know I can't just go on a schedule that will automatically let me make but I just wanted to get a feel for things.

MIT is actually more than just a math/engineering school.  They actually do have a great history department, and a great philosophy department.  Not everyone needs a 700 to get in, because there are going to be people who go there and hate math.

[quote="probability1.01"]Yeah I guess it shouldn't be a strict requirement, but there should be some justification for acceptance.  I wonder what that 1260 person had that distinguished her.  It's also possible that she scored high on math and low on verbal.[/quote]

*cough* under-represented minority? *cough*

At least that's the feeling I got from talking to her.

Frankly, I don't see how any student who wanted to go to an elite college with the academic reputation of MIT could score 1260, unless it was like some brilliant Bulgarian physics student who scored 800 math and 460 verbal.

Now, I hesitate to say that the SAT1s really mean anything, but really, 1260?  That's just too poor.

[quote="rlee"]It seems like MIT accepts more women than men. I think MIT has accepted four or five applicants from my high school in the past two years, ALL of them girls. Also, I don't like to say this, but some of the men were CLEARLY more qualified for MIT than the female applicants. Gender plays a huge role.[/quote]

I agree.  I was honestly shocked when I saw that they had a 60/40 ratio (I was expecting much further apart).  They clearly must go out of their way to recruit girls.  Though I suppose that that does make the entire experience much more enjoyable for the guys who are there.

I would be interested, as I'm sure most readers of this forum would be interested, in hearing what characteristics applicants can have that gain them admission despite low SAT I scores. The more detail, the better, for those of us who are planning ahead. Inasmuch as all the top schools every year reject some applicants with 1600s on the SAT I and top grade averages from their respective schools, we might as well learn as much as we can, as soon as we can, about what else makes an applicant an attractive candidate for admission.

[quote="blahblahblah"]Frankly, I don't see how any student who wanted to go to an elite college with the academic reputation of MIT could score 1260, unless it was like some brilliant Bulgarian physics student who scored 800 math and 460 verbal.[/quote]

How about all of the artists, musicians, writers, researchers, and brilliant students who are bad at testing?  A 1260 is in fact nearly 90th percentile on the SAT -- probably better than 5/6ths or more of all people who take the exam.  You are suffering from a very closed-minded vision of what gets people into colleges.

Furthermore, it's been shown that there's generally very little correlation between SAT scores and "success" in college. The Math SAT is really more like a placement exam. At most schools it's a good predictor of what math class you should be placed in as a freshman. But the best predictors of success throughout your college career are high school GPA, taking rigorous high school classes, and having good study skills. Those are the things that most schools look at first when they review your application - Grades, Transcript, Teacher Recommendations, and Personal Statement. SAT Scores, extracurricular activities, special awards, and other factors can get you noticed, or move you up or down in the pile from there.

[quote="JBL"]...You are suffering from a very closed-minded vision of what gets people into colleges.[/quote]

Actually, I don't really think that I am, seeing that I have gone through the whole admissions process myself.  Look, there obviously can be extenuating circumstances, and a 1260 SAT1 can just be an anomaly compared to the rest of the individual's academic record, extracurricular record, personal qualities, etc ...

However, from talking to the girl, she didn't mention anything else about the person - she just tossed out 'some guy from my school got into MIT with a 1260 SAT1'.  That, compounded with the very high underrepresented minority population of her state, makes me believe that this student is probably an underrepresented minority who got in largely due to that factor.

Let's put it this way - I would be very surprised if there was a single Asian-american male who got into MIT with a 1260 SAT1.  They just have too many qualified applicants to accept many 1260s.  And I could be wrong about this person, since I only know of them in the most indirect of manners, but I don't think that I am.

[quote="blahblahblah"]...Look, there obviously can be extenuating circumstances, and a 1260 SAT1 can just be an anomaly compared to the rest of the individual's academic record, extracurricular record, personal qualities, etc ...

However, from talking to the girl, she didn't mention anything else about the person - she just tossed out 'some guy from my school got into MIT with a 1260 SAT1'.  That, compounded with the very high underrepresented minority population of her state, makes me believe that this student is probably an underrepresented minority who got in largely due to that factor.[/quote]

So you don't know anything about the person other than their (alleged) SAT score, and from that you deduced that they must be an under-represented minority? That's a very very poor assumption.

I overspoke.

Look, it's hard to communicate the whole nuance of an actual, in-person conversation over an internet forum.  I could, for example, state that the alleged person goes to a high school with an over 80% Hispanic population, which was said to me.

But really, that's neither here nor there.  All I was saying, initially, is that getting into MIT has been a crapshoot.  We have one guy on this forum who made USAMO and USABO, and I personally know another guy who has excellent extracurricular activities, and 1600/800/800/800/800 SAT1/2s.

All it really says is that there are lots of qualified applicants who don't get in, and some applicants who are statistically unqualified that do get in.  I could also say that, for example, the admit rate is 28% for females to MIT and 12% for males.  So obviously, it is a crapshoot at best.

<---... I got a 1410  math, verbal (800/610) math iic, chem, writing (780,730,640)

Let me first note that I think the entire admissions process and criteria is wrong and corrupt in every college in America, and I think that the piece of paper you submit to a college tells them nothing about who you really are. Unfortunately, I have not come up with a better way myself, so I really have no basis for my argument, I just think somebody needs to come up with a better way to do it.

I think you picked the wrong school to discuss in this matter... I believe MIT has a much different way of admitting people than most colleges, and most people don't realize this. I'm speaking strictly regular admission here, early action is a different story. The first thing MIT does is goes through all of the applications quickly, looking at grades, test scores, work ethic etc... basically, everything that distinguishes the applicant's mental abilities and they separate them into two piles: those who could handle the workload and stress of MIT, and those could not. The ones that could not are rejected (after they review them all one final time to make sure no mistakes were made)... Then, they take the applications of applicants who could handle the work load and choose the admitted students from this pile, except all the judging is now based on "do we like this person? is this person the 'MIT Type'? Is this a person we would want to represent MIT?" Grades, scores, classes, all of that is moot at this point, the highest scores don't necessarily get you in, it's all based on how much they like you, and how inventive you are with your application. They like unique people... if you sent in the bare minimum app, I assure you, you won't get in.

I am not sure if many of you that applied did any research into MIT admission process, but if you did, you would know that the admission staff is a fairly young bunch, Amrys Williams, Matt Mcgann, and Juan Acosta are only a few years out of college, Armys in just her second year. They have been to MIT and know what they want people to be like. They try to choose students based on what they want the college to be like, it's very subjective. While some see this as unfair perhaps, because it is not based on "qualification," nevertheless, it's the way MIT does things, and it's been the way MIT has done it for 143 years, and will be the way MIT does it at least as long as Marilee Jones is the dean of admissions.

I have heard of some low SAT scores being admitted, but 1260 is awfully low, there must be some special reason why they would admit such a person, I'm sure there was a good one.

Look at my SAT1/2 scores... they are nothing special, matter of fact, they are pretty low by MIT standards, but does that not qualify me? Every person brings something unique to MIT, and that's what makes it such a special place, they don't admit based on money, you can't buy your way in, they admit exactly who they want to admit, and they are experts at it.

If anyone has any other questions, I might be able to provide some insight...

You seem to know a lot about MIT (especially the admissions).  I have found your comments helpful.  How do you know so much about it?

Finally, my question is were you accepted to MIT?

Haha.  How stupid do I feel.  Well it is in a weird font so it never really caught my eye.

Two summers ago, working as high school summer intern at Motorola, Inc, a few of my fellow high school interns were getting ready to enter their freshman year at MIT... we became great friends, some of the best friends I have ever had. They were some of the coolest people I had ever met, and I decided that MIT was the place I wanted to go. They have taught me a lot, specifically one named Ian, who makes it a point to meet all kinds of people. He is good friends with admissions, and knows exactly how it all works and functions. He told me a lot about what to do and what not to do.

Also, I learned a lot by actually calling them up and talking to them, it's really important to do. I called MIT admissions 12 times since January. They write comments about you and mark that you called, it shows them you have interest in their institution. Ask them questions and they will be more than happy to answer them. One can learn a lot this way. That is how I know many of them, we've spoken before. Talking with them also shows them that you are a normal competent human being, you'd be surprised at how many MIT students can't even hold a conversation with someone. They are delighted to know all about you, so call them up and tell them!

As I said in my first post, the application really tells them nothing about what type of person you are, talking to them does. Also, have a good interview (mine was fair, I think I bored the guy), it won't make or break your admission but it could inch you over someone else.

and btw, in case none of you missed my adding mistake, 800 + 610 = 1410, not 1490.... I'm not sure why I put 1490, perhaps it's because that is my girlfriends SAT score and I was thinking about her, who knows :-p.

Your addition mistake did not go unnoticed, but I wasn't going to say anything.

lol, sadly my SAT score was not very nice... my gf kicked my ass and my ex gf kicked my ass even more with a 1540... both of them are going to MIT, it should be really interesting!

Oh, seems like your ex-gf kicked my --- too.

My Score: [hide]1530[/hide]

she's a [tex]CENSORED[/tex] so you don't have to worry.

/\
|
|  Thousandth reply on the college board
|
|

[quote="Spoon"]she's a [censored] so you don't have to worry.[/quote]

As moderator, one of my jobs here is to help people stay out of embarrassing situations, so I'm going to ask you to edit your post to modify that statement, inasmuch as your identity (and thus hers) is ascertainable to a lot of people who know people in M.I.T.'s entering class.

Breaking up is hard to do . . .

sure thing... I was more curious whether people could get around the language control (which is too censored imo) by posting in [tex]\LaTeX[/tex]... which you can, interesting.

[quote="JBL"][quote="Sagar_Indurkhya"]I don't know about the actual process, but I have been asking around a lot, and from what I understand, the only thing that will put you into MIT is pure Passion. You should really check into RSI, ISEF, ITS, Siemmons-Westinghouse, etc. Even gaining regional recognition in these competitions is basically a guarantee admission to MIT, provided you have a decent app.
If you go through hs trying to get into MIT, you wont, but if you just persue your passion whatever it is, and have real motivation, you will. Scores don't matter. Anyone can plop down for 6 months, with Princeton Review, and score 1400+.[/quote]

Well, I have found a lot to disagree with in your comments.  Up front, you gave a list of things people should try to do to get into MIT and then said that if you try to get into MIT, you won't.  That is a logically difficult comment.  I think your suggestion that people focus on a small list of competitions completely misses the point -- for RSI and ITS particularly, you can't do well unless you've done well in other things.  A quick look down the list of Intel Finalists shows, in addition to researchers and perfect SAT scores and valedictorians, musicians, dancers, athletes, club presidents, etc., etc.  In other words, they pick people who already have acquired the traits that result in admissions at competitive universities.  So the fact that lots of them get into good schools has very little to do with the contests themselves and much more to do with similarities between the application processes.  So that piece of advice, to focus on those particular things, I think is particularly unhelpful.  I also think your assertation that they can guarantee you admission is flat-out false.

My final criticism is of a comment that seems, in various forms, to get tossed around a lot here: "Anyone can plop down for 6 months, with Princeton Review, and score 1400+."  That's just not true.  In fact, it's not even close to true.  There are millions of graduating high school seniors in this country who wouldn't, for a variety of reasons, be able to do that.

Okay, now I'm going to stop attacking you and actually agree with something you said, in regards to passion.  I think you are correct that probably the best way to increase your likelyhood of being admitted to a competitive college is to pursue your passions.  The benefits of working in areas that you enjoy are numerous -- you are more likely to spend your time for fun on such activities, you are more likely to enjoy the time you spend, you are more likely to do better than you would at other activities, and you will be more able to express to others (interviewers, essay readers, and people in your real life) how interesting you find what you're doing.  I don't think it can guarantee you admission to MIT (or any other school), but I think the potential benefits to you are very high.

Hmm, I just used "you" to refer to both the reader and the writer of the previous comment ... I hope everyone can figure it out.[/quote]

I didn't mean that those are the only competitions to concentrate on. Getting into these programs and doing well naturally involves doing well in everything else. If you get into RSI, then MIT knows you have what they are looking for. Check out CollegeConfidential.com forums. They have good stuff.

About the 'pursue your passion'...  I love physics but i got a late start, and sometimes even the pursuit of passion doesn't seem to be enough.  The valedictorian for the grade above me was in my calculus class and he has all these chem awards but he was waitlisted and ended up at harvard instead.

[quote="kentliu"]this is a very nice thread..

spoon.. so, wut was the main thing that got u into MIT? and for the phone calls, wut did u talk to them about.. also.. can u list a few other significant achievements in ur app. for example ur gpa..

edited: well.. i read anther thread and it seems like u studied physics c on ur own and probably got a 5.. thats good stuff.. might have something to do with getting you into MIT..
i took physics C in my school but they only teach mechanics there, so i learned E/M myself... pretty sure i'll be seeing some 5s..[/quote]

Umm... it's a hard question answer... as for my GPA, it was just over 5.0... I got 14 B/B+ total in high school, all of which came in 9th/10th grade. That was before I cared, tried or anything about school. That was the biggest mistake of my life... ANYWAY...

As far as talking to them goes... ask them first about the process, as as many questions as you can think of about the whole admission process, and ask them if calling helps. They should explain the green pen theory to you lol. After that, tell them about something interesting going on in your life at the moment. Always first ask if they have a moment to talk with them. After that, whatever comes to your mind. I have found that random things pop into my mind and I blurt them out as I went.

As for what probably got me in... well, I don't usually tell people this on a mass level because I don't want to cause a scene. But I am the co-owner and CEO of an investment company called To B Insight. I started it with a partner in 2003 after a lot of studying and planning. We have done fairly well to date. I'm hoping it stays that way. If it wasn't for that, I likely wouldn't have.

oh wow.. i wanted to start a biz too but i got no idea what to do?
what does ur company do and how much money have you invested into it?

Basically our business operations are simple.

We (my business partner and I) convince people that we know what we're talking about, and we convince them to give us some money. We try to deal with people who have a lot and if we lost all of their money, it wouldn't affect their lives in any way. We take their money and invest it for them. They sign an agreement saying we have full control. We (hopefully) return them a profit and take a percentage of the profit.

We are relatively small as of now, with 7 clients (a few of them part of extended family). We're more in it for the experience and "track record" so that in 5-6 years, when we get out of college, we will have an impressive resume.

This year the market has been a bit shaky, with the oil crisis and war going on, but things look to be doing ok. The most important part is that I have learned SO MUCH in the last 6 months, it's insane. HOpefully everything continues going well!

If you have any more specific questions, feel free to ask. You can also email me at don@tobinsight.com. I check it pretty much every minute.

[quote]http://www.prstats.com/2008/display.php?user=stefan1987

there u go (don't forget intl. kids!!)[/quote]

Damn, I wanted to post that.

http://www.pclaunch.com/~kayton/MIT/InsideAdmissions.htm

That site (along with the links it leads to) are very insightful about the MIT admissions process.

MIT class of 2010...

According to

http://www.techreview.com/articles/atwood1103.asp

[quote]According to Jones, theres been little change over the last 50 years in the qualities MIT seeks in its students. The kind of students who flourish here are people with very high levels of self-initiative, who are intensely curious by nature, and who have that additional follow-through, she says. They almost teach themselves because of their insatiable curiosity.[/quote]

Younger son going to MIT this year. (Older one,  chose not to  go there )


search geometry MATHCOUNTS USAMTS AMC AIME ARML

right now only competition i know is AMC.

is there any other mathematic competition i can apply for and not yet passed the due date??

by the way i'm in the united states

can u list the competitions and its due date if possible?

What grade are you in? google search and you'll get more than you want (university of south carolina, national olympiads, bay area math meet, mathcounts, etc.).

Here's a [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php]great list[/url] of competitions (with links!)

As you can see, there are tons of competitions ;)

[quote="yarakvar"]thx but that's too confusing

is there any competitions like amc 10/12??
or USAMTS?

where your school don't need to be  enrolled, don't require groups, or not too confusing to sign up.[/quote]

For US, I think there are AMC10/12, and if you are qualifed, you will get to write AIME then USAMO.

For other contests, you can check where you live, which state?

Go down the list, you will see all kinds of contests in each state. From that, click on the links and check the date for contests.

It is not really confusing though.

Good luck.

The Math League Competitions are alright. They aren't hard either.  :)

Also, you can do ARML and Mandelbrot, although I don't know if the deadline is passed.

USA Math Talent Search is ALWAYS open.  :)


Do you lack of something? If not, it will have infinite results. :wink:

Infinite results? No, no, only 1 of the 5 options  :D

Not infinite (but more than would be given)

[hide]$ 2^{8022} \plus{} 1 \equal{} (2^{2674}\plus{}1)(2^{5348}\plus{}2^{2674}\plus{}1)$[/hide]

[quote="SimonM"]Not infinite (but more than would be given)

[hide]$ 2^{8022} \plus{} 1 \equal{} (2^{2674} \plus{} 1)(2^{5348} \plus{} 2^{2674} \plus{} 1)$[/hide][/quote]

It is equal to (a),(b),(c),(d) or (e)?

In fact, it must have infinite result. If I assume that a,b is in C set, then we will get many solutions for this problem.
For example: when I give a=1, then b=2^(8022)+1 and a+b=[b]2^(8022)+2, which is not one of 5 answers above. Obviously, you lack of the restriction that a,b are integers(I just guess this).[/b]

Ok, sorry. This is a fault of the statement. It was posted in another forum and i think the author forgot the restriction. Consider $ a$ and $ b$ integers.

Now, you just do like Simon did above! This kind of problem is so popular :)

Either a restriction is missing, or this problem is no good. Simon's factorization was wrong, but going along those lines,
[hide]
$ 2^{8022} \plus{} 1 \equal{} (2^{2674} \plus{} 1)(2^{5348} \minus{} 2^{2674} \plus{} 1)$
$ 2^{8022}\plus{}1\equal{}(1)(2^{8022}\plus{}1)$

If we take sums of these pairs of factors, we get different answers.
[/hide]

[hide="Answer"]

$ \fbox{C}$

$ a = 2^{4011} + 2^{2006} + 1$

$ b = 2^{4011}^ - 2^{2006} + 1$[/hide]

I'm working on how to get that factorization.

Edit: Did a mistake in second exponent.

[hide="Ah!"] Recall the Sophie Germaine identity $ a^4 \plus{} 4b^4 \equal{} (a^2 \plus{} 2ab \plus{} 2b^2)(a^2 \minus{} 2ab \plus{} 2b^2)$.  Letting $ a \equal{} 1, b \equal{} 2^{2005}$ we have our desired factorization. [/hide]

I would take sometime to find that, thanks t0rajir0u! you got  :first:

You don't need a whole lot here since you are given the choices...

If $ a\plus{}b\equal{}s$ and $ ab\equal{}t$, then $ a,b\equal{}\frac{s\pm\sqrt{s^2\minus{}4t}}{2}$

So we just want $ s^2\minus{}4t$ to be an integer.

$ (2^n\plus{}2)^2\minus{}4(2^{8022}\plus{}1)\equal{}2^{2n}\plus{}4*2^{n}\minus{}4*2^{8022}$

If $ n\equal{}4012$, everything works out.

(obviously this factorization is not unique in any respects, so I don't think that there is much point to belabor this problem. If you match the 4*2^n factor instead, you get the factorization if you did 1 times whatever)


algebra
polynomial
function
real analysis
real analysis theorems

Mean value theorem for a one variable polynomial is as below,
$ f'(c) \equal{} \frac {f(b) \minus{} f(a)}{b \minus{} a}$ where $ a < b$ .

I see a document about Mean value theorem for multivariate polynomial
that has the following expression. Is this correct ?
If it is, could some professionals help me to point out why ?
$ f(x \plus{} \triangle x, y \plus{} \triangle y) \minus{} f(x,y \plus{} \triangle y) \equal{} \triangle x \frac {\partial f}{\partial x}\left(x \plus{} \theta \cdot \triangle x   , y\plus{}\triangle y \right)$
where
$ 0 < \theta < 1 \ and \ \frac {\partial f}{\partial x}$ is continuous .

Multivariate functions don't satisfy any obvious analogue of the mean value theorem.  The expression you've written seems like it wants to be $ \Delta x \frac{\partial f}{\partial x} (y \plus{} \theta \cdot \Delta y)$, although I am not sure if it's correct.

What you've written is just the regular old mean value theorem for the differentiable function $ g(x) \equal{} f(x,y \plus{} \Delta y)$, for fixed $ y\plus{}\Delta y$, on the interval $ \left[x,x \plus{} \Delta x\right]$.

$ \frac {g(x \plus{} \Delta x) \minus{} g(x)}{\Delta x} \equal{} \frac {dg}{dx}(x \plus{} \theta \cdot \Delta x) \equal{} \frac {\partial f}{\partial x}\left(x \plus{} \theta \cdot \triangle x , y \plus{} \triangle y \right)$, for $ 0 < \theta < 1$.

t0rajir0u, what you've written is not correct because $ \frac {\partial f}{\partial x}$ is still a function of two variables. wisheskernel's expression is correct as written.

It was written as a function of one variable at the time that I read the topic (in $ x$).  But I did indeed forget that it should be a function of two variables.

Dear gauss202,

I don't understand a portion of the experssion . Why is $ (x \plus{} \theta \cdot \triangle x)$ ?
Would you give me some suggestion ? Many thanks for your great help.

Perhaps I know where is my problem. The $ \theta$ doesn't mean a angle and it has other meaning.

Right. $ \theta$, with $ 0 < \theta < 1$, is just used to produce a number, ($ x \plus{} \theta \cdot \Delta x$), between $ x$ and $ x \plus{} \Delta x$.


$ x,y,z\in R^\plus{}$,$ x^2\plus{}y^2\plus{}z^2\equal{}1$

Try to prove that $ \frac{x^2}{1\minus{}x\plus{}y}\plus{}\frac{y^2}{1\minus{}y\plus{}x}\plus{}\frac{z^2}{1\minus{}y\plus{}z}\ge 1$

Follows CBS
$ \sum {\frac{x^2}{1\minus{}x\plus{}y}}\equal{}\sum {\frac{x^4}{x^2\minus{}x^3\plus{}x^2y}}\geq\frac{(\sum {x^2})^2}{\sum {(x^2\minus{}x^3\plus{}x^2y)}}\equal{}\frac{1}{1\minus{}\sum {x^3}\plus{}\sum {x^2y}}\geq1$

because $ \sum {x^3}\geq\sum {x^2y}$


[quote="ifai"]Let $n\in\mathbb{N}$ and $\forall 0\leq r\leq n$, $C_n^r=$the coefficient of $x^r$ in the expansion of $(1+x)^n$.

(1) Prove that $\forall 0\leq k\leq n$, the coefficient of $x^k$ in the expansion of $[1+(n+1)x](1+x)^{n-1}$ is $(k+1)C_n^k$.

(2) Prove that $\sum_{k=0}^n{(k+1)(C_n^k)^2}=(n+2)C_{2n-1}^n$.[/quote]

[hide="1"] ...the coefficient of $x^k$ in the expansion of $[1+(n+1)x](1+x)^{n-1}$ is $(k+1)C_n^k$ is the coefficient of $x^k$ in $(1+x)^{n-1}$ plus the coefficient of $x^{k-1}$ in $(1+x)^{n-1}$ times $(n+1)$.  This is written algebraicly as, $C_{n-1}^{k}+(n+1)C_{n-1}^{k-1} = C_n^k+nC_{n-1}^{k-1} = C_n^k+k \cdot (\frac{n}{k})C_{n-1}^{k-1} = C_n^k+kC_n^k = (1+k)C_n^k$[/hide]

Use $C_{n-k}^k = C_n^k$ and first result.


geometry
circumcircle
incenter
greatest common divisor
national olympiad

1 Given complex number $a,b,c$. Let $|a+b|=m,|a-b|=n$. Suppose $mn \neq0$. Prove that
\[\max\{|ac+b|,|a+bc|\}\geq\frac{mn}{\sqrt{m^{2}+n^{2}}}. \]
2.Prove:
(1) If $2n-1$ is a prime number, then for any n distinct positive integers $a_{1},a_{2},...,a_{n}$, there exist $i,j\in\{1,2,...,n\}$ such that
\[\frac{a_{i}+a_{j}}{(a_{i},a_{j})}\geq2n-1; \]
(2) If $2n-1$ is a composite number, then there exist n distinct positive integer $a_{1},a_{2},...,a_{n}$, such that for any $i,j\in\{1,2,...,n\}$,
\[\frac{a_{i}+a_{j}}{(a_{i},a_{j})}<2n-1, \]
where $(x,y)$ denotes the greatest common divisor of positive integer $x,y$.

3.Given distinct positive integers $a_{1},a_{2},...,a_{1}1$ with sum less than 2007. Write $1,2,...,2007$ on a blackboard in ascending order. 22 consecutive [i]operations[/i] are defined as an [i]operation group[/i]. The $i$th [i]operation[/i] in a [i]operation group[/i] is defined as follow: Choose any number on the blackboard, if $1\leq i \leq11$ then add $a_{1}$ to it, otherwise subtract $a_{i-11}$ from it.If the final result of an [i]operation group[/i] is an even permutation of $1,2,...,2007$, then we call this [i]operation group[/i] as even.If the final result of an [i]operation group[/i] is an odd permutation of $1,2,...,2007$, then we call this [i]operation group[/i] as odd. What is the difference between the number of even [i]operation groups[/i] and the number of odd [i]operation groups[/i]?
  [b]Remark.[/b] A permutation $x_{1},x_{2},...,x_{n}$ of $1,2,...,n$ is called an even permutation if $\prod_{i>j}(x_{1}-x_{j})$ is a positive number, otherwise, it is called a odd permutation.[/b]

[quote="jackhui"]
3.Given distinct positive integers $a_{1},a_{2},...,a_{1}1$ with sum less than 2007. Write $1,2,...,2007$ on a blackboard in ascending order. 22 consecutive [i]operations[/i] are defined as an [i]operation group[/i]. The $i$th [i]operation[/i] in a group is defined as follow: Choose any number on the blackboard, if $1\leq i \leq11$ then add $a_{1}$ to it, otherwise subtract $a_{i-11}$ from it.If the final result of an [i]operation group[/i] is an even permutation of $1,2,...,2007$, then we call this [i]operation group[/i] as even.If the final result of an [i]operation group[/i] is an odd permutation of $1,2,...,2007$, then we call this [i]operation group[/i] as odd. What is the difference between the number of even [i]operation groups[/i] and the number of odd [i]operation groups[/i]?
  [b]Remark.[/b] A permutation $x_{1},x_{2},...,x_{n}$ of $1,2,...,n$ is called an even permutation if $\prod_{i>j}(x_{1}-x_{j})$ is a positive number, otherwise, it is called a odd permutation.[/b][/quote]

I can't understand. Can you give an example?

[quote="jackhui"]In the 1st operation, we choose a number on the blackboard, for example, 10, then we add $a_{1}$ on it. After this operation, the numbers on the board become $1,2,3,4,5,6,7,8,9,10+a_{1},11,...,2007$.
The 1st to 11th operation is $+a_{i}$ the 12th to 22th operation is $-a_{11-i}$[/quote]

But so if, you can has not a permutation at END. Example $10+a_{1}=15$  :maybe:

[quote="N.T.TUAN"]
But so if, you can has not a permutation at END. Example $10+a_{1}=15$  :maybe:[/quote]
Then it won't be even nor odd.

[quote="jackhui"]
Then it won't be even nor odd.[/quote]

Now, I understand  :) But you not post these problems in subbox of Olympiad Box and create links here. Why?  :maybe:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=131230]Problem 1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=131233]Problem 2[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=131235]Problem 3[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=131237]Problem 4[/url]
Let $O$ and $I$ be the circumcenter and incenter of triangle $%Error. "bigtriangleuABC" is a bad command.
$ repectively. The incircle of $\bigtriangleup ABC$ touches sides $BC,CA,AB$ at $D,E,F$ respectively. Lines $FD$ and $CA$ intersect at $P$, $DE$ and $AB$ intersect at $Q$. $M,N$ are midpoints of line segments $PE$, and $QF$ respectively. Prove $OI\perp MN$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=131238]Problem 5[/url]
Suppose a bounded sequence $\{a_{n}\}_{n\geq1}$ satisfies that
\[a_{n}<\sum^{2n+2006}_{k=n}{\frac{a_{k}}{k+1}}+\frac{1}{2n+2007},n=1,2,3,....\]
Prove that $a_{n}<\frac{1}{n},\forall n \in N$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=131239]Problem 6[/url]
Find the least integer n satisfys the following condition;
For any n integer(not necessarily distinct) $a_{1},a_{2},...,a_{n}$, there exist 9 integer $1\leqq_{1}<i_{2}<...<i_{9}\leqq$ and $b_{i}\in\{4,7\}(i=1,2,...,9)$ such that $\sum_{j=1}^{9}{b_{j}a{}_{i}{}_{j}}$ is a multiple of 9

Thanks!

to Mods: Edit name this topic is Chinese Mathematical Olympiad 2007 , i think  :wink:


Consider the sequence $(-1)^{n}+1/n$. The first few terms are $0, 3/2,-2/3, 5/4,-4/5$.  I know that the infimum is the greatest lower bound and the supremum is the least upper bound. But in this question, I am confused.

It is evident that the negative numbers approach $-1$ and the positive numbers approach 1. Now, which is the infimum, supremum, limit of the supremum and limit of the infimum? Can you give another example which differentiates [b]inf [/b]from [b]lim inf[/b] and also [b]sup [/b]and [b]lim sup[/b]?

Thanks a lot!

For any sequence $(a_{n})$ one has $\inf a_{n}\le \liminf a_{n}\le \limsup a_{n}\le \sup a_{n}$.

Since the sequence $1/n$ converges to $0$, adding it does not change either $\limsup$ or $\liminf$. Thus $\limsup((-1)^{n}+1/n)=1$ and $\liminf((-1)^{n}+1/n)=-1$. In contrast with limit-like concepts, $\sup$ and $\inf$ can be affected by even a single term of a sequence. For example, $\sup$ is strictly greater than $\limsup$ if and only if there is a term of the sequence that is greater than $\limsup$ (and in this case $\sup$ is just the maximum of all such terms).


vector
algebra
polynomial
superior algebra
superior algebra unsolved

Suppose an extension $ E/K$ of degree $ 2^n$ is of the form $ E \equal{} K(\sqrt {a_1},\ldots,\sqrt {a_n})$ with $ a_i \in K$.
Prove that, if $ char K\neq 2$ then $ \sqrt {a_1} \plus{} \cdots \plus{} \sqrt {a_n}$ is a primitive element of $ E/K$.

Let $ \xi \equal{} \sqrt {a_{1}} \plus{} \ldots \plus{} \sqrt {a_{n}}$.
Obviously $ K(\xi) < E$
Let $ K_{j} \equal{} K(\sqrt {a_{1}}, \ldots,\sqrt {a_{j \minus{} 1}}, \sqrt {a_{j \plus{} 1}},\ldots, \sqrt {a_{n}})$.
We have $ [E : K_{j}] \equal{} 2$ (because $ [E\, : \, K] \equal{} 2^{n}$) and $ \{1, \sqrt {a_{j}}\}$ is an algebraic basis of $ E \equal{} K_{j}(\sqrt {a_{j}})$ as a vector space over $ K_{j}$. It is easy to conclude that $ \alpha \plus{} \beta \sqrt {a_{j}}\stackrel{\varphi_{j}}{\mapsto} \alpha \minus{} \beta \sqrt {a_{j}}$ is an $ K \minus{}$automorphism of $ E$.
Define $ \kappa: \{0,1\}^{n}\ni\varepsilon \equal{} (\varepsilon_{k})_{k \equal{} 1}^{n}\mapsto \varphi_{1}^{\varepsilon_{1}}\circ \ldots\circ\varphi_{n}^{\varepsilon_{n}}\in \operatorname{Aut} (E/K)$.
Observe that $ \operatorname{char} K \neq 2$ implies $ \varepsilon \neq \mu \implies (\kappa(\varepsilon))(\xi)\neq (\kappa(\mu))(\xi)$.
Now let $ p\in K[X]$ be the minimal polynomial of $ \xi$. Of course $ p((\kappa(\varepsilon))(\xi)) \equal{} 0$ for each $ \varepsilon \equal{} (\varepsilon_{k})_{k \equal{} 1}^{n}\in \{0,1\}^{n}$, so $ p$ has at least $ 2^{n}$ pairwise distinct roots in $ E$ and $ [K(\xi)\, : \, K] \equal{} \deg p \geqslant 2^{n}$, but $ K(\xi) < E$ and $ [E\, : \, K] \equal{} 2^{n}$, hence $ E \equal{} K(\xi)$.


Let 6 possitive real numbers such that $ a_{1}\plus{}a_{2}\plus{}a_{3}\equal{}b_{1}\plus{}b_{2}\plus{}b_{3}$
Prove that : $ \left(a_{1}\plus{}b_{1}\right) \cdot \left(a_{2}\plus{}b_{2}\right) \cdot \left(a_{3}\plus{}b_{3}\right) \ge 4(a_{1} \cdot a_{2} \cdot a_{3}\plus{}b_{1} \cdot b_{2} \cdot b_{3})$

[quote="tin_124"]Let 6 possitive real numbers such that $ a_{1} \plus{} a_{2} \plus{} a_{3} \equal{} b_{1} \plus{} b_{2} \plus{} b_{3}$
Prove that : $ \left(a_{1} \plus{} b_{1}\right) \cdot \left(a_{2} \plus{} b_{2}\right) \cdot \left(a_{3} \plus{} b_{3}\right) \ge 4(a_{1} \cdot a_{2} \cdot a_{3} \plus{} b_{1} \cdot b_{2} \cdot b_{3})$[/quote]
Try $ a_1\equal{}a_2\rightarrow0$ and $ b_1\plus{}b_2<b_3.$ :wink:


calculus
integration
analytic geometry
geometry unsolved
geometry

Does there exist an equilateral triangle
(a) on a plane; (b) in a 3-dimensional space;
such that all its three vertices have integral coordinates?

(a) Not exist.
Suppose the contrary, then we can translate the triangle till one of the vertices coincide with the origin while the other two still have integral coordinates.
Consider it on the Argand plane, let the two non-origins points as $ A\equal{}a\plus{}bi$ and $ B\equal{}c\plus{}di$, $ a,b,c,d$ are integers.
Then $ B \equal{} Ae^{\frac{i\pi}{3}}$, (or vice versa)
So $ (a\plus{}bi)(\frac{1}{2} \plus{}\frac{\sqrt{3}i}{2}) \equal{} (\frac{a}{2}\minus{}\frac{\sqrt{3}b}{2})\plus{}(\frac{b}{2}\plus{}\frac{\sqrt{3}a}{2})i$
So $ c\equal{}\frac{a}{2}\minus{}\frac{\sqrt{3}b}{2}$, $ d \equal{} \frac{b}{2}\plus{}\frac{\sqrt{3}a}{2}$,
i.e. $ 2c\minus{}a \equal{} \sqrt{3}b$ is integer, i.e. $ b \equal{} 0$ (coz $ c,a,b$ are integers), similarly $ a \equal{} 0$.
Thus the triangle becomes degenerated, which is not desirable.
So not exist.

(b) Consider $ (0,0,0), (7,1,0), (3,4,5)$, which forms an equilateral triangle of side length $ 5\sqrt{2}$


A set of n numbers are to be added together, and before they are added, each number is rounded off to the nearest integer, thus introducing an error whose distribution is uniform. Find the mean and the variance of the round-off error of the sum.
For the case of large n, find the least value of n for which there is a 50% chance that the round-off error of the sum exceeds 5. State any assumptions used.

We are to assume that fractional parts of the numbers to be added are independent random variables with uniform distribution in $ [0,1]$. Hence the round-off errors are independent RVs that are uniformly distributed in $ [\minus{}1/2,1/2]$. Use the CLT.


[quote="4865550150"]Let $ u,v \in R^ \plus{}$, prove or disprove:

$ 1 \plus{} 2 \frac {\sqrt {u} \plus{} \sqrt {v}}{u \plus{} v} \leq \frac {1 \plus{} u \plus{} v}{\sqrt {uv}}.$[/quote]
we have to prove that:
$ \sqrt{uv}\plus{}\frac{2\sqrt{u}v\plus{}2\sqrt{v}u}{u\plus{}v}\leq 1\plus{}u\plus{}v$
but $ \sqrt{uv}\leq\frac{u\plus{}v}{2}$

so we will prove the stronger:
$ u^2 \plus{} v^2 \plus{} 2uv \plus{} 2(u \plus{} v)\geq 4u\sqrt {v} \plus{} 4v\sqrt {u}$
but:
$ u^2 \plus{} v\plus{}  uv \plus{} u\geq 4u\sqrt {v}     ;    v^2 \plus{} u\plus{}uv \plus{} v\geq 4v\sqrt {u}$
adding these ineqs
Q>E>D

Your proof is very simple and nice. :D

My approach:

First, we need to prove: $ u^2 \plus{} v^2 \geq \sqrt {uv}(u \plus{} v)$ .

 [list] [u][b]Proving above inequality: [/b][/u]

By c-s inequality,

$ 2(u^2 \plus{} v^2) \geq (u \plus{} v)^2 \newline \Rightarrow u^2 \plus{} v^2 \geq \frac {u \plus{} v}{2} (u \plus{} v) \newline \Rightarrow u^2 \plus{} v^2 \geq \sqrt {uv}(u \plus{} v)$
[/list]
Hence, adding up $ u \plus{} v$, we have

$ u^2 \plus{} v^2 \plus{} u \plus{} v \geq (\sqrt {uv} \plus{} 1 )(u \plus{} v) \newline \Leftrightarrow \frac {u}{u \plus{} v} (1 \plus{} u) \plus{} \frac {v}{u \plus{} v}(1 \plus{} v) \geq \sqrt {uv} \plus{} 1 \newline \Leftrightarrow 2 \plus{} u \plus{} v \geq \sqrt {uv} \plus{} 1 \plus{} \frac {v}{u \plus{} v}(1 \plus{} u) \plus{} \frac {u}{u \plus{} v}(1 \plus{} v) \newline \left[ \textrm{Add both sides} \; \frac {v}{u \plus{} v}(1 \plus{} u) \plus{} \frac {u}{u \plus{} v}(1 \plus{} v) \right] \newline \Leftrightarrow 1 \plus{} u \plus{} v \geq \sqrt {uv} \plus{} \frac {v}{u \plus{} v} 2 \sqrt {u} \plus{} \frac {u}{u \plus{} v} 2 \sqrt {v} \; \; \; \; \; \; \; \; \; \; \; \; \; \;[\textrm{By am \minus{} gm}] \newline \Leftrightarrow \frac {1 \plus{} u \plus{} v}{\sqrt {uv}} \geq 1 \plus{} 2 \frac {\sqrt {u} \plus{} \sqrt {v}}{u \plus{} v}$
which is equivalent to  $ 1 \plus{} 2 \frac {\sqrt {u} \plus{} \sqrt {v}}{u \plus{} v} \leq \frac {1 \plus{} u \plus{} v}{\sqrt {uv}}$ .

The equality is $ u \equal{} v \equal{} 1$.


How do you quote someone and when its posted it says someone (their screename) said: quote?????? :|

That's not LaTeX.  Anyway, just enclose whatever you want quoted in [ quote] {put quote here [/ quote] (with no spaces).  If you click the quote button under a post, it will automatically put all of that post inside of quote tags.  There's also a quote button above where you type your message in.  If you highlight some text and click that it will put quote tags around what was highlighted.


The numbers from 1 to n > 1 are written down without a break. Can the resulting number be a palindrome (the same read left to right and right to left)? For example, if n was 4, the result would be 1234, which is not a palindrome.

[hide="Not sure if this works"]
Say that the longest string of consecutive zeroes in the number is $k$ zeroes long.  Since zero can never begin a number, such strings can only occur in $a*10^{k}$, for $a$ a digit.

Consider the first appearance of a string of $k$ zeroes.  This is in $10^{k}$, so it is preceded by a 1.  If the number is a palindrome, the last appearance of such a string must be succeeded by a 1.  Since this is the first digit of $a*10^{k}+1$, $a=1$.  So the only string of $k$ zeroes is in $10^{k}$.

But the number before $10^{k}$ consists of $k-1$ 9's, so the second digit before the string of $k$ zeroes is 9.  But the second digit of $10^{k}+1$ is 0.  So this is a contradiction and the number can never be a palindrome.

[/hide]

This has appeared on the forum before, but I'm having trouble finding it.

Edit: Thanks, Ravi.  I think that means it's appeared at least 2 or 3 times.


geometry
geometric transformation
rotation
combinatorics proposed
combinatorics

Two of the squares of a $ 7.7$ checkerboard are painted yellow and the rest are painted given. Two color schemes are equivalent if are can be obtained from the other by applying a rotation in the plane of the board. How many inequivalent color schemes are possible?

[quote="HUY VAN"]Two of the squares of a $ 7.7$ checkerboard are painted yellow and the rest are painted given. [/quote]

Is the rest of the board painted as a checkerboard white and black or all of the same color?

[quote="skuge"][quote="HUY VAN"]Two of the squares of a $ 7\times7$ checkerboard are painted yellow and the rest are painted given. [/quote]

Is the rest of the board painted as a checkerboard white and black or all of the same color?[/quote]
I think the rest of the board painted with same color.
[u][i]Solution:[/i][/u]
The number of way to select two yellow squares are $ \binom {49}{2}\equal{}1176$ and we can rotate the board $ 90$ degree ,(suppose clockwise), then the number of inequivalent colors are less than $ 1176$. We have two case :

1) [i]If two yellow squares are in same diagonal [/i]: It create two equivalent colors and there exist $ \frac{49\minus{}1}{2}\equal{}24$ yellow couples .

2) [i]If two yellow squares aren't in same diagonal [/i] : It create four equivalent colors.

Then the number of inequivalent colors $ \equal{} \frac {1176\minus{}24}{4}\plus{} \frac {24}{2}\equal{}300$


Solve the following system of equations:
\begin{align*}
3x-5y&=-1.5\\
7x+2y&=4.7
\end{align*}

Express your answer as an ordered pair $ (x,y)$ of decimals.

Multiply through and cancel terms.

$ -21x + 35y = 10.5$

$ 21x + 6y = 14.1$

$ 41y = 24.6$

$ y= 0.6$

$ 3x - 5(0.6) = -1.5$

$ 3x = 1.5$

$ x = 0.5$

$ (0.5, 0.6)$


Does anybody know the date of this year's ARML competition?  I'm hoping that I will not have to choose between ARML and graduating from high school, although it's likely that I will.

This years ARML competition will be on the weekend of June 5th. It's always on the first weekend after Memorial Day (which is always the last Monday in May).

Wow, you graduate early.  It also always overlaps with the SAT, but at my school at least they worked something out so we could take the SAT in a special session on the following Tuesday.  I think it was just like the religious exemption, except it was the math exemption.

well i know quite a few seniors from here have problems with going because of graduation too...

Blah, SAT date? Bummer, now I dunno whether I can go.

[quote="gauss202"]No, you guys graduate late actually. NY and Northest schools typically graduate later than the rest of the country. So ARML has always conflicted with the graduation dates of a lot of southern schools. There's been debate about moving it for years. But since the organizers are mostly from the northern states, no progress has been made.[/quote]

Where did you get this information Gauss?  I have never heard of any talks to move it, nor has any been suggested.  The ARML committees are very accomodating, but as you might all assume, it's very hard organize such a large event as to not inconvenient someone.

For those worried about SATs, there is a make-up date for those that needs to take it that month, but could not due to ARML.  I am currently organizing such a thing for my kids @ the Baltimore County area.  If anyone has a strong case for moving it due to graduation (my team lost a few kids last yr due to graduation, but keep in mind most kids would like to go on vacation once they graduate), I will bring it up with the organizers or during the ARML coaches meeting.

This's been an issue for my state for a long time, even going back to the early 90s when I was on the team. We traditionally lose about half of our seniors due to graduation conflicts each year; as do some other southern states, who's school years traditionally end before Memorial day (I think because of the heat).

I know it's been brought up in coaches meetings before, but I can understand the natural difficulty in moving it. You can't move it earlier into May, because the colleges that host it will still be in session, and that's AP season. Plus most of our students have their final exams in late May. Then moving it later into the Summer would conflict with the Northern States exams and graduations. Or, if you move it too late, students will start to dispurse for vacations and summer camps.

So changing it to suit everybody may not even be logistically possible. But the dates that are chosen now are mostly for the convenience of the northern states who were the first organizers of the contest. I think the issue should be revisited though, because possibly keeping it as either the first or second weekend in June (rather than the first weekend after Memorial Day) might be better. This year for instance, with it being in June, we have no conflicts with graduation.

That does, in fact, conflict with my graduation.  I'll probably end up missing ARML, then.  I'm from Massachusetts, which is generally not considered a Southern state.

I think NCSSM may have moved its graduation based on previous conflicts with ARML dates. I'm not sure though. But I do know the dates have conflicted in previous years.

Regarding the alternate SAT date, how does one go about organizing this?  Who do I talk to?  I've asked my guidance counselor in the past, and I was told that they never heard of such a thing.

There is a Sunday date for taking the SAT (generally the day after the Saturday test date), but I believe it is only for religious reasons.  See [url]http://www.collegeboard.com/student/testing/sat/reg/circum/sunday.html[/url].

Having said that, I know that one of my students arranged to take the SAT on a Sunday at his high school, even though he had no religious reason.  Don't know how he managed that.

Thanks for link.

Hm, is the only possible alternate date the sunday following the saturday administration?  I don't think that that Sunday would work for me, because I would get home from ARML at 4 a.m on Sunday, and I doubt i'd be in good enough condition to take the SAT's 4 hours later.

I would say call up the College Board, or, alternatively, try e-mailing one of the people in charge of the New York City Math Team or Stuyvesant math department.  But NYC does it every Tuesday after ARML.

I found out that the Eastern MA kids can take SAT I/II's the Wednesday (I think) after ARML weekend.

Why not change ARML from Saturday to Sunday?

Most participants will just be taking Monday instead of Friday off from school. SAT/graduation conflict may then become non issue.

It's not even too late to change for this year, is it?

Moving it to Sunday won't help much, because most teams have to start traveling the day before the competition, and then take a day to get back home. And the West Coast site begins the competition the night before the other two.

The ARML dates have always been a topic of discussion, for as long as I can remember. I don't think there is any date that can really satisfy everyone. It's way too late to change anything for this year though. That competition is such a huge logistical  feat. It's amazing that they run it so smoothly each year.

At the west coast location, we have to sleep over on Friday night. I just don't get that. Why can't we take all of it on saturday?

We'd have to do the power and team rounds at 6AM or something if we did. Also, the overnight thing is what makes ARML so much fun.

Everyone (at all three locations) stays over Friday night, with the possible exception of people who live very near either Penn State or Iowa.  (It is at Iowa, right?  One of those states out in that general direction ;))  We just have to start in the morning and go straight through, that's all.  You guys get to leave much earlier on Saturday than we do.

yes, its Iowa. Thats the one I go to... Who else does?

On Friday night, I always tell everyone that I don't want to discuss team and power round at all (to compare answers or solutions or anything). No one ever listens though.

Regarding the June SAT:

I have emailed collegeboard (no response).

I have asked my coach, who told me to talk to my principal.  I talked to my principal, who referred me to my guidance counselors, who in turn have never heard of such a thing.

I will email the NYC coach and the eastern MA coach.

Do not email the NYC coach....  he doesn't handle this, one of the other coaches does (Mr. Geller).

What happens is that a coach must talk to the SAT coordinator of the school taht they want to take the make up on, and most likely they'll say yes (because they know about this event).

Cheers,
Jack

The SAT coordinator is my guidance counselor, who has said that he's never heard of taking the SAT on an alternate date.  And I can't find any evidence from online or from books to show that it is possible...

Try this: [url=http://www-epgy.stanford.edu/arml/math-competitions/1999/990430.announcement.txt]Click here for evidence that SFBA ARML can get the SAT rescheduled[/url].

Thanks simon.

Looking at the counselors section instead of student, I found this:

http://tinyurl.com/yuott

"In the event that a school-sponsored activity unexpectedly prevents pre-registered students from attending a scheduled test administration, the school may request alternate testing arrangements. The principal, school counselor, or test center supervisor should contact us by email or telephone at 1-888-SAT-HELP (1-888-728-4357) no later than 10 working days before the test. Students will be charged the test center change fee of $17 plus $5 for switching to an alternate test date.
"

Now my question is, if I ask to get the date changed, does that mean all the other students(not going to arml) who were going to take it at my school can not take it on the original date?

[quote="cats..."]Now my question is, if I ask to get the date changed, does that mean all the other students(not going to arml) who were going to take it at my school can not take it on the original date?[/quote]

No.  They all get to take it Saturday as planned.

you don't have to choose between going to ARML and graduating, just going to graduation. I'll probably be skipping graduation to go to ARML.


group theory
abstract algebra
superior algebra
superior algebra unsolved

let $ G$ be a finite group and $ p_1,p_2....,p_n$ be distinct prime divisors of $ card(G)$,such that for each $ p_i$ there exist $ x_i,y_i$ in $ G$ with $ ord(x_i)\equal{}ord(y_i)\equal{}p_i$ and $ y_i$ is not a power of $ x_i$
prove that $ card(G)>\equal{}1\plus{}(p_1^2\minus{}1)....(p_n^2\minus{}1)$

Sylows theorem (in the right formulation) tells you, that the number of subgroups with order $ p_i$ is congruent to 1 mod $ p_i$. The condition that such x and y exist tells further, that this number is at least two, hence at least $ p_i\plus{}1$ for each $ i$.

If you cut off the identity, the subgroups of order $ p_i$ are disjoint, so there are at least $ (p_i\minus{}1)(p_i\plus{}1)\equal{}(p_i^2\minus{}1)$ elements of order $ p_i$. Sum this an you get $ |G|\geq 1\plus{}(p_1^2\minus{}1)\plus{}(p_2^2\minus{}1)\plus{}...\plus{}(p_n^2\minus{}1)$.

This also appeared [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=145110]here[/url], but it has been established there that the given solutions don't work.

With the sum, it's trivial. With the product, it's false.

A counterexample for the product version: $ A_4$. There are three elements of order 2, and 8 of order 3. $ 12<1\plus{}3\cdot 8$.

For the sum version, the argument I like is based of the fact that if $ m$ is squarefree and divides $ |G|$, the number of elements with order dividing $ m$ is divisible by $ m$. Applied to primes $ p_i$, the number of elements of order $ p_i$ is divisible by $ p_i\minus{}1$ and congruent to $ \minus{}1$ mod $ p_i$, so it is $ p_i\minus{}1\plus{}a_i(p_i^2\minus{}p_i)$ for some $ a_i$. By the hypotheses of the problem, $ a_i$ is nonzero.

[quote="Gockel"]Sylows theorem (in the right formulation) tells you, that the number of subgroups with order $ p_i$ is congruent to 1 mod $ p_i$. [/quote]

Sorry if it is a stupid question but whenever I found Sylow's Th it said that the number of p-Sylow subgroups is congruent to 1 mod p and maybe $ p_{i}$ divides $ n$ with a greater power than $ 1$, so does it remain true that the number of subgroups of order $ p_{i}$ (or maybe $ p_{i}^{k}$ for any possible $ k$) is congruent to $ 1$ mod $ p_{i}$? :maybe:
 Is this one the [i][b]right  formulation[/b][/i] for Sylow Theorem?

@mon: This is somewhat a generalisation, and yes it is true.

[quote="jmerry"]if $ m$ is squarefree and divides $ |G|$, the number of elements with order dividing $ m$ is divisible by $ m$. .[/quote]

Can you please give a proof?
Thanks.

:maybe: can it be proved without using actions and orbits?(i don't know this things... :( )


modular arithmetic
number theory
relatively prime

What is the remainder when $ 7^{251}\plus{}17^{122}$ is divided by 100?

(full solution plz)

how did u figure out all of those powers, and multiples??

There are several identities corresponding to modular arithmetic that can often save you from going through the rigorous process of actually working out all the powers/finding a big enough calculator.

lol, I know, I'm very new to modular arithmetic.
Anyone have a "prettier" solution?

Oh, @DaReaper:
The idea behind modular arithmetic is to work in terms of remainders instead of working things out.
I used the basic idea that $ n^k\equal{}n^{2*\frac{k}{2}}$ to whittle down the large number into a smaller one. One property of division by 100 is that the remainder is always going to be the last two digits of the dividend. In fact, $ n\mod{10^k}$ is always equal to the last $ k$ digits of $ n$ for $ k>0$.

Here's a very basic example. Let's find the last digit of $ 5^{10}$. Notice that the question is essentially asking "What is the remainder when $ 5^{10}$ is divided by $ 10$?"
We write this as $ 5^{10}\equiv{x}\pmod{10}$, and we are solving for $ x$. The strange equation form used here (read: five to the tenth is congruent to $ x$ modulo ten) simply implies that $ \frac{5^{10}\minus{}x}{10}$ is an integer.

To solve this problem, we can write $ 5^{10}$ as $ 25^5$. We also know that the remainder when 25 is divided by 10 is 5 (because 2*10=20). So we can write that $ 25\equiv{5}\pmod{10}$. Furthermore, this lets us do a cool little substitution: $ 5^{10}\equiv{25^5}\equiv{5^5}\pmod{10}$.
In similar fashion, we can break the number down.
$ 5^5\equiv{5*25^2}\equiv{5*5^2}\equiv{5*25}\equiv{5*5}\equiv{25}\equiv{5}\pmod{10}$.
Our answer is 5.

There also exist theorems and formulas related to this which can make your life easier (look up Fermat's Little Theorem, Chinese Remainder Theorem, etc, etc).

When you're working with remainders, it helps a lot to be comfortable with [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's Little Theorem[/url], which is generalized by [url=http://en.wikipedia.org/wiki/Euler%27s_totient_theorem]Euler's Totient Theorem[/url], and also the [url=http://en.wikipedia.org/wiki/Chinese_remainder_theorem]Chinese Remainder Theorem[/url].

The Chinese Remainder Theorem tells us that since $ 100 \equal{} 4 \cdot 25$, and $ 4$ and $ 25$ are relatively prime, the remainder when divided by $ 4$ and the remainder when divided by $ 25$ is enough information to find the remainder when divided by $ 100$. Let's check out the solution, which will be presented in 3 parts:

[hide="Remainder upon division by 4"]$ 7^{251} \plus{} 17^{122} \equiv ( \minus{} 1)^{251} \plus{} 1^{122} \equiv 0 \pmod{4}$

So, our remainder will be divisible by 4.[/hide]
[hide="Remainder upon division by 25"]For the first term: $ 7^{251} \equal{} 49^{125} \cdot 7 \equiv ( \minus{} 1)^{125} \cdot 7 \equiv \minus{} 7 \pmod{25}$

For the second term, we can invoke Euler's Totient Theorem:

$ \varphi (25) \equal{} 20$, so $ 17^{20} \equiv 1 \pmod{25}$.

$ \therefore 17^{122} \equiv (17^6)^{20} \cdot 17^2 \equiv 17^2 \equal{} 289 \equiv 14 \pmod{25}$

Adding the two terms, we get $ 7^{251} \plus{} 17^{122} \equiv \minus{} 7 \plus{} 14 \equiv 7 \pmod{25}$[/hide]
[hide="Putting the two together"]The remainder is the only number between $ 0$ and $ 99$ which is:
1. Divisible by 4
2. Has a remainder of 7 when divided by 25

Applying condition 2, the possibilities are 7, 32, 57, and 82. Applying condition 1, the remainder must be 32.
[/hide]

There were a number of small steps in the above. But note that our most intensive calculation was just $ 17^2 \equal{} 289$! And we didn't have to keep blindly multiplying numbers in hopes of finding one with a favorable remainder -- at every point, it was very clear what to do next.

Wow, that is very interesting.
I'm familiar with Chinese Remainder Theorem and Fermat's Little Theorem, but I didn't know about Euler's Totient Theorem.
And it is interesting to see the Chinese Remainder Theorem used like that...

I'll look up some problems and see if I can solve them similarly.


induction
inequalities unsolved
inequalities


calculus derivative quadratics algebra AMC

Find all pairs of real numbers (x,y) satisfying the following system of equations:

2 - x:^3: = y
2 - y:^3: = x

[hide]

Subtracting the 2 equations, we get

x-y=x^3-y^3.

Factoring the right side to be (x-y)(x^2+xy-y^2), we move the (x-y) to the left, so that we have the product of 2 expressions equalling 0. Set them each equal to zero, and then you are on your way to greatness!!

[/hide]

Uhh, maybe not greatness

It's not that easy

But wat you can do to solve the one with 2 variables, is to plot the graphs and look for the intersections...

[hide]Yes, I used a graphing approach. The first graph will be the normal cubic graph (whereby normal, I mean y=x^3) flipped upside down and shifted up 2. The second will be the normal cubic turned counterclockwise a quarter turn shifted right 2. You can see pretty easily that they only intersect at one place in the first quadrant. You get this solution by using darkquantum's factorization, setting x-y=0 in his factorization, and using the original equation to find x^3+x=2, one equation of which is obviously x=1, and since x=y, the ordered pair would be (1,1). I don't know how to show (other than just graphing it and seeing so) that there are not any other intersections. Perhaps by calculus? (this derivative is < that derivative, etc.)[/hide]

[quote="darkquantum"]Well, there are no solutions to the other factorization, x^2+y^2=xy, so that means there's only one solution.[/quote]

Not that easy. The actual  case turns out to be: x^2-y^2+xy=1. Well, let's use quadratic equation whereby we solve for x in terms of y. Regrouping into, x^2+xy+(1-y^2), the quadratic equation gives us: (-y :pm:  :sqrt: (5y^2-4))/2. Hmm....where to go from here?

Hmm. ya that would work. Bit messy though, no?

yes, is there a more elegant way though? for example, would my derivative idea work? anyone? (complexzeta?)


If x,a,y,b,z,c are in proportion. prove that: (x/a)^3 >= (x^3+y^3+z^3)/(3abc)
<{how to write this in latex?}>

$ \text{\LaTeX}$'d:

$ \left(\frac{x}{a}\right)^3\geq\frac{x^3+y^3+z^3}{3abc}$.

What do you mean by "$ x,a,y,b,z,c$ are in proportion"?

I'm guessing $ \frac {x}{a} \equal{} \frac {y}{b} \equal{} \frac {z}{c}$. But then the inequality is the other way around, since $ (\frac{x}{a})^3\equal{}\frac{xyz}{abc}$ and the result follows from AM-GM.

I'm not sure but maybe this works:

If in proportion is $ \dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k$ Then:

If $ x,a,y,b,z,c > 0$ then by AM-GM we have:

\begin{eqnarray*}
\dfrac{x^3+y^3+z^3}{3}&\ge& xyz\\
\dfrac{x^3+y^3+z^3}{3} \cdot \dfrac{1}{abc}&\ge& \dfrac{xyz}{abc}\\
\dfrac{x^3+y^3+z^3}{3} \cdot \dfrac{1}{abc}&\ge& k^3\\
\dfrac{x^3+y^3+z^3}{3abc}&\ge& \left(\dfrac{x}{a}\right)^3
\end{eqnarray*}

also given is.x>a. now it is trival i guess. ...


There is a house with $ n$ floors. We paint every floor red, blue or green. There may not be two red level beside each other, but the top floor has to be red.
What is the number of possible paintings?

[hide="possible answers"]Judging from my evaluation of $ n\equal{}1,2,\text{ or }3$ we have that the number of ways is $ \boxed{n!}$.

For $ n\equal{}1$, there is one way (red).

For $ n\equal{}2$, there are two ways (red over blue and red over green).

For $ n\equal{}3$, there are six ways (red/blue/red, red/green/red, red/blue/blue, red/blue/green, red/green/blue, and red/green/green).

Not sure if this continues to hold true however.[/hide]

Kyron, without any restrictions on red floors the answer is $ 3^{n}$, which means that your answer needs to be smaller than that for all $ n$; but $ n!$ clearly outgrows $ 3^{n}$.  In fact, since adding a new floor only gives at most $ 3$ new possibilities, there is no way that there are $ 4$ times more possibilities for $ n \equal{} 4$ than there are for $ n \equal{} 3$.

[hide="Solution"] Let $ r_{n}$ denote the number of paintings of $ n$ floors where the bottom floor is red, and let $ s_{n}$ denote the number of paintings of $ n$ floors where the bottom floor is not red.  The total number of paintings is $ p_{n}\equal{} r_{n}\plus{}s_{n}$.  Now, we have the initial conditions

$ r_{1}\equal{} 1, s_{1}\equal{} 0$

and the recursions

$ r_{n\plus{}1}\equal{} s_{n}$
$ s_{n\plus{}1}\equal{} 2r_{n}\plus{}2s_{n}$

(A painting of $ n\plus{}1$ floors with a red bottom is a painting of $ n$ floors, but with a red floor added, and a painting of $ n\plus{}1$ floors with a non-red bottom is a painting of $ n$ floors with either a blue or a green bottom added.)  Since $ r_{n\plus{}1}\equal{} s_{n}$ we have

$ s_{n\plus{}1}\equal{} 2s_{n}\plus{}2s_{n\minus{}1}$

which is a second-order linear recursion with characteristic equation

$ x^{2}\equal{} 2x\plus{}2\Leftrightarrow x \equal{} 1\pm\sqrt{3}$

Since $ r_{n\plus{}1}\equal{} s_{n}$, both have the same recursion and the same equation, and so does $ p_{n}$.  We therefore know that $ p_{n}$ is given by an equation of the form

$ p_{n}\equal{} A(1\plus{}\sqrt{3})^{n}\plus{}B(1\minus{}\sqrt{3})^{n}$

where $ A, B$ are easily calculated constants from the initial conditions.  After the calculation, we find that $ A \equal{}\frac{\sqrt{3}}{6}, B \equal{}\minus{}\frac{\sqrt{3}}{6}$, hence

$ p_{n}\equal{}\boxed{\frac{\sqrt{3}(1\plus{}\sqrt{3})^{n}\minus{}\sqrt{3}(1\minus{}\sqrt{3})^{n}}{6}}$ [/hide]
In fact, there are an infinitude of sequences of this form alone that begin with $ 1, 2, 6$.  Don't let the actual numbers guide your decision about the form of a sequence ;)

There is a house with $ n$ floors. We paint every floor red, blue or green. There may not be neither two red level nor two blue level beside each other.
What is the number of possible paintings?

The solution is in a similar vein.

[hide="Details"] The fundamental detail that makes the problem essentially the same as the other is to note that swapping the red and blue levels in a given painting produces a valid painting.

In other words, let $ r_{n}$ denote the number of paintings with a red bottom, $ b_{n}$ the number with a blue bottom, and $ g_{n}$ with a green bottom.  Then $ r_{n}\equal{} b_{n}$.  Moreover,

$ r_{1}\equal{} b_{1}\equal{} g_{1}\equal{} 1$
$ r_{2}\equal{} 2, b_{2}\equal{} 2, g_{2}\equal{} 3$

and

$ r_{n\plus{}1}\equal{} b_{n}\plus{}g_{n}$
$ b_{n\plus{}1}\equal{} r_{n}\plus{}g_{n}$
$ g_{n\plus{}1}\equal{} b_{n}\plus{}r_{n}\plus{}g_{n}$

Knowing that $ b_{n}\equal{} r_{n}$, we can rewrite as

$ r_{n\plus{}1}\equal{} r_{n}\plus{}g_{n}$
$ g_{n\plus{}1}\equal{} 2r_{n}\plus{}g_{n}$

The total number of paintings is $ r_{n}\plus{}b_{n}\plus{}g_{n}\equal{} 2r_{n}\plus{}g_{n}\equal{} p_{n}\equal{} g_{n\plus{}1}$.  Hence

$ p_{n\plus{}1}\equal{} 2r_{n}\plus{}2g_{n}\plus{}2r_{n}\plus{}g_{n}\equal{} 4r_{n}\plus{}3g_{n}\equal{} 2p_{n}\plus{}g_{n}\equal{} 2p_{n}\plus{}p_{n\minus{}1}$

which has characteristic equation

$ x^{2}\equal{} 2x\plus{}1\Leftrightarrow x \equal{} 1\pm\sqrt{2}$

We then have an explicit formula

$ p_{n}\equal{} A (1\plus{}\sqrt{2})^{n}\plus{}B (1\minus{}\sqrt{2})^{n}$

and initial conditions $ p_{1}\equal{} 3, p_{2}\equal{} 7$.  We calculate that $ A \equal{}\frac{1\plus{}\sqrt{2}}{2}, B \equal{}\frac{1\minus{}\sqrt{2}}{2}$, hence

$ p_{n}\equal{}\boxed{\frac{(1\plus{}\sqrt{2})^{n\plus{}1}\plus{}(1\minus{}\sqrt{2})^{n\plus{}1}}{2}}$ [/hide]


For all positive $a_i,i=1,2,\ldots,n$ whose sum is 1, prove that:
\[ \frac{a_1}{1+a_2+\cdots+a_n}+\cdots+\frac{a_n}{1+a_1+\cdots+a_{n-1}} \ge \frac{n}{2n-1}.  \]

\[ LHS \ge \frac{a_1+...+a_n}{n}\cdot \left( \frac{1}{1+a_2+\cdots+a_n}+\cdots+\frac{1}{1+a_1+\cdots+a_{n-1}}\right) \geq \frac{n}{2n-1} \]
By Chebyshev and AM-HM  :D

This problem was composed by Dana Piciu from Romania for a Romanian contest.I solved almost whith this idea.
Anywhay, I took the third place at this contest. :D

hi,

here is my solution :

$S\times(2n-1)\geq (\sqrt{a_1}+\dots+\sqrt{a_n})^2$

$S\times(2n-1)\geq n$

$\dots$ :D .

toetoe, please full solutions, not only some hints.I saw many post of you, but at all of them you didn't explain very well, only some dots...

ok,i am explaining :

$2n-1 = (2-a_1)+(2-a_2)+\dots+(2-a_n)$

by cauchy-Schwarz ,we have  :

$S\times(2n-1)\geq(\sqrt{a_1}+\dots+\sqrt{a_n})^2\geq n$

$S\geq \frac{n}{2n-1}$ :D .

:thumbup: OK!But I hope you will post full solutions not only at this problem.Have a nice day!


Let $ E$ be a compact nonempty subset of $ \mathbb{R}^k$, and let $ \delta \equal{}\sup\{d(x,y): x,y\in E\}$. Show that $ E$ contains points $ x_0,y_0$ such that $ d(x_0,y_0)\equal{}\delta$.

let $ \delta_n\equal{}\delta\minus{}\frac{1}{n}$

then we know since $ \sup\{d(x,y): x,y\in E\}\equal{}\delta$ there must be

$ (x_n,y_n)\in E^2$ such that $ d(x_n,y_n)>\delta_n$ for each n

then there must be a convergent subsequence $ \{x_{n_j}\}\subset \{x_n\}$

such that $ x_{n_j} \to x_0$ for some $ x_0\in E$ by compactness

then again there must be a convergent subsequence $ \{y_{n_{j_k}}\}\subset \{y_{n_j}\}$

such that $ y_{n_{j_k}}\to y_0\in E$ by compactness again

then we see that $ d(x_0,y_0)\equal{}\lim_{k\to \infty} d(x_{n_{j_k}},y_{n_{j_k}})$ by continuity of d, so were done


trigonometry
geometry
incenter
inradius
Euler
geometry unsolved

Show that for any triangle ABC, we have
2Rr= R^2 - (OI)^2   and   R^2 - (OG)^2= (a^2+b^2+c^2)/9

where O, G, I, R, r are the orthocentre, centroid, incentre, circumcentre and inradius respectively.

Pls explain to me with steps.. THZ A LOT!!

[quote="wailam010"]2Rr= R^2 - (OI)^2[/quote]

This is the famous Euler identity: http://www.mathlinks.ro/Forum/viewtopic.php?t=45138

[quote="wailam010"]R^2 - (OG)^2= (a^2+b^2+c^2)/9[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=17684 post #2
http://www.mathlinks.ro/Forum/viewtopic.php?t=64291

  Darij


geometry
geometric transformation
circumcircle
homothety
geometry unsolved

Let $H$ be the orthocenter of $\triangle ABC$ and $P$ - point inside the triangle. Denote $M$ midpoint of $PH$. Prove that Feuerbach circles of $\triangle AHP,\triangle BHP,\triangle CHP$ intersect at one point apart from $M$ and this point lies on the Feuerbach circle of $\triangle ABC$.

Does anyone have any solution?
This is an equivalent version of the problem:

Let $\triangle A'B'C'$ be the result of translation of $\triangle ABC$ and $H$ - the orthocenter of $\triangle ABC$. Prove that circumcircles of $\triangle AA'H,\triangle BB'H,\triangle CC'H$ intersect at one point apart from $H$ and this point lies on the circumcircle of $\triangle A'B'C'$.

First, I want to make sure that the Feurbach circle is the nine-point circle. If that is the case, for a two-line solution, angle-chase !!!! :P

[hide="Hint"] Call the second intersection of the nine-point circles of $AHP$ and $BHP$. Try to prove that this point lie in the nine-point circle of $ABC$ by noticing the midpoints of $AH$ and $BH$.

[/hide]

Yes indeed, Feuerbach circle is nine-point circle. Thanks for hint! The solution is very simple.

derezins, could you prove the equivalence of your first problem (with nine-point circles) and the second one (with $\triangle A'B'C'$)?

I did my best, but couldn't see it :blush:

Dear nxxx,
(using the notation from the first problem)
on the mentioned nine-point circles perform a homothety with center $H$ and scale $k=2$. $\triangle ABC$ becomes $\triangle A'B'C'$ in the second version and point $P$ becomes the orthocenter (point $H$ from the first problem disappeares). I hope it will be clear know. :yup:


Let $a,b,c \in Q $ such that $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$

Show that $A=\sqrt{a^2+b^2+c^2} \in Q$   !!!

 :)

excuse me, what do you mean of $Q$ ?, what kind of set it is?

We have:

$a^2 + b^2 + (\frac{ab}{a+b})^2$

where it remains to prove $\sqrt{(a^2 + b^2)(a+b)^2 + (ab)^2} \in Q$

Put s = a+b, p = ab.  It becomes $\sqrt{(s^2 - p)^2}$.


LaTeX
probability
AoPS Books
complex numbers

I was wondering what "generalizations" (or clever correspondances) you've encountered when factoring?

I'll start with the basics:

$a^2-b^2=(a+b)(a-b)$
$a^3+b^3=(a+b)(a^2-ab+b^2)$
$a^3-b^3=(a-b)(a^2+ab+b^2)$

$x^n-x^z=x^z(x-1)(x^{n-z-1}+x^{n-z-2}...+1)$
For $a^n+b^z$, it is $(x+1)$ and the signs between $x^{n-z-1}$... alternate. (I guess that's kind of an extension.)
Where $n>z$ and $n$ is an odd integer and $z$ is an even integer...

Well,

$a^2+2ab+b^2=(a+b)^2$

$a^2-2ab+b^2=(a-b)^2$

What's with the Latex these days?

[quote="LordoftheMorons"]$ac + ad + bc + bd = (a + b)(c + d)$[/quote]
good luck recognizing that when it's all bogged down with coefficients and exponents and stuff. I've done it before but it's hard...especially when it's not in that nice little order...

chapter 7 might help.

If you have book 2, chapter 13 has several factorizations.

Thanks. I really should go through these books (I forgot I had them ). :rotfl:

A. I am homeschooled.
B. I'm ONLY working on mathematics. :rotfl:

Nonono, subjets like language, geography, chemistry, etc..........

does someone have more complicated factorizing tricks.
I'll be happy  :)  and u can give me as a problem...

[quote="srulikbd"][quote="surge"]I was wondering what "generalizations" (or clever correspondances) you've encountered when factoring?

I'll start with the basics:

$a^2-b^2=(a+b)(a-b)$
$a^3+b^3=(a+b)(a^2-ab+b^2)$
$a^3-b^3=(a-b)(a^2+ab+b^2)$

$a^n-b^z=b^z(x-1)(x^{n-z-1}+x^{n-z-2}...+1)$
For $a^n+b^z$, it is $(x+1)$ and the signs between $x^{n-z-1}$... alternate. (I guess that's kind of an extension.)
Where $n>z$ and $n$ is an odd integer and $z$ is an even integer...[/quote]
$a^n-b^z=b^z(x-1)(x^{n-z-1}+x^{n-z-2}...+1)$
I don't think this true! and what does X means?
u know how to factor $a^n+-b^n$ right, so now find which of the following $a^x+-b^y$ can u factor?
[hide="hint"]coprime[/hide]
[hide="bigger hint"]u need to use the factorize of  $a^n+-b^n$[/hide][/quote]

My bad... getting a little carried away with the $a$'s and $b$'s there. I mean...

$x^n-x^z$

[quote="srulikbd"]does someone have more complicated factorizing tricks. [/quote]
Here's a bit complicated one (although I don't think there's any "generalizaion" for this):
$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

And here's an interesting one to think about:
$x^4+4y^4=?$ :)

[quote="José"][quote="tz"]All of these and more are in the 1st Art of Problem Solving book.[/quote]

Why every MathLinker study everything with those books??? What do you learn at school then????[/quote]
While students in the U.S. do study math (and other subjects) at school, classroom math is not enough to prepare for math competitions or study math deeply in general. That's why many MathLinkers/AoPSers (in the U.S.) use AoPS books in addition to school textbooks.

ok the first I think u made a mistake cause mathematica(software) coudn'y factorize this.
the second, mathematica factorize it very good, how ever I can't find a way to reach it...does someone have?  :)

Well, my TI-89 can factor it properly... :D

[hide="Hint"]
The first step is to rewrite it as $(\text{something})^2-(\text{something})^2$.

Then what technique can you use from here? :)
[/hide]

Factoring "generalizations"?  Factoring in general is difficult to generalize.  :P  I will then present only a few amusing tricks.

$(ac \pm bd)^2 + (ad \mp bc)^2 = (a^2 + b^2)(c^2 + d^2)$ (incidentally, this is the fastest possible proof for Cauchy in two variables.  It is also equivalent to the statement that the length of the product of two complex numbers is equal to the product of the length of the two.)

$x^4 + 4y^4 = (x^2 + 2xy + y^2)(x^2 - 2xy + y^2)$

[hide="Derivation of the second factorization, which is known as the Sophie-Germaine Identity"]The AoPS books teaches the following "partial" factorization:

$x^2 + y^2 = (x + y)^2 - 2xy$

Using this trick,

$x^4 + 4y^4 = (x^2)^2 + (2y^2)^2 = (x^2 + 2y^2)^2 - 4x^2 y^2 = (x^2 + 2y^2)^2 - (2xy)^2 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)$

:) [/hide]

srulikbd:  frt's first factorization is indeed correct (don't trust software! :P).  In fact, here is a nicer equivalent form:

$a^3 + b^3 + c^3 - 3abc = \frac{1}{2} (a + b + c)( (a-b)^2 + (b-c)^2 + (c-a)^2 )$

Edit:

[quote="xenon_nightmare152"]$a^2 + b^2 = (a+bi)(a-bi)$[/quote]

Let's do the generalization tango, shall we?  :)

$a^n + b^n = \prod_{k=0}^{n-1} \left( a - be^{ \frac{\pi i (2k+1)}{n} } \right)$

[quote="t0rajir0u"][quote="shadysaysurspammed"]
$(b-c)^3+(c-a)^3+(a-b)^3=3(b-c)(c-a)(a-b)$
[/quote]

Indeed, this is a special case of the first factorization frt posted.  :)[/quote]
Yeah, this is a nice way to look at the sum of the cubes...

[hide="Trick"]
Let $x=b-c$, $y=c-a$ and $z=a-b$.

Then $x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz$, but since $x+y+z=(b-c)+(c-a)+(a-b)=0$, we have $x^3+y^3+z^3=3xyz$.

Hence $(b-c)^3+(c-a)^3+(a-b)^3=3(b-c)(c-a)(a-b)$ ;)
[/hide]

[url=http://www.mathpages.com/home/kmath034.htm]here's sth[/url]
[url=http://www.artofproblemsolving.com/Forum/topic-6836.html]and here too[/url]


An equivalent of the expression $ \left ( \frac{x^2 \plus{} 1}{x} \right ) \left ( \frac{y^2 \plus{} 1}{y} \right ) \plus{} \left ( \frac{x^2 \minus{} 1}{y} \right ) \left ( \frac{y^2 \minus{} 1}{x} \right )$, $ xy \neq 0$ is:

$ \textbf{(A)} 1 \qquad \textbf{(B)} 2xy \qquad \textbf{(C)} 2x^2y^2 \plus{} 2 \qquad \textbf{(D)} 2xy \plus{} \frac{2}{xy} \qquad \textbf{(E)} \frac{2x}{y} \plus{} \frac{2y}{x}$

[hide="Click for solution"]

This expression becomes $ \frac{x^2y^2 \plus{} x^2 \plus{} y^2 \plus{} 1 \plus{} x^2y^2 \minus{} x^2 \minus{} y^2 \plus{} 1}{xy} \implies \frac{2x^2y^2 \plus{} 2}{xy} \implies 2xy \plus{} \frac{2}{xy}$, so $ \textbf{(D)}$.
[/hide]

[hide="Click for (sort of) alternate solution"]
The given expression equals:

\[ \left (x\plus{}\frac{1}{x} \right )\left (y\plus{}\frac{1}{y} \right )\plus{}\left (x\minus{}\frac{1}{x} \right )\left (y\minus{}\frac{1}{y} \right )
\\ \equal{} xy\plus{}\frac{y}{x}\plus{}\frac{x}{y}\plus{}\frac{1}{xy}\plus{}xy\minus{}\frac{y}{x}\minus{}\frac{x}{y}\plus{}\frac{1}{xy}
\\ \equal{}2xy\plus{}\frac{2}{xy}
\]

or $ \boxed{\textbf{(D)}}$.
[/hide]

hello, the term can be written in the form $ \frac{2(1\plus{}x^2y^2)}{xy}\equal{}2\left(\frac{1}{xy}\plus{}xy\right)$.
Sonnhard.

$ \left ( \frac{x^2 \plus{} 1}{x} \right ) \left ( \frac{y^2 \plus{} 1}{y} \right ) \plus{} \left ( \frac{x^2 \minus{} 1}{y} \right ) \left ( \frac{y^2 \minus{} 1}{x} \right )\equal{}\frac{2x^2y^2\plus{}2}{xy}\equal{}2xy\plus{}\frac{2}{xy}$ answer $ (D)$


trigonometry
function
geometry
3D geometry
videos
Gauss
Mafia

name two animals and then someone else will say what they think the new animals name should be should be
Ill start...

                          1.Penguin
                          2.dog

Thomas the Tank engine.

Viper.

Manatee

"I got whacked on the head by a ninja!!"

Gyarados, Wario

something halfway between the two.

Aerosmith <-> Led Zeppelin

Led AeroZeppelin Smith.

My Chemical Romance<-> Panic! at the Disco

My Disco Romance at the Chemical Panic!

cherry <-> blueberry

aaaardwolfvark.

zebra
zucchini
zoo
zero
(another random z word)

what was that first one...

pie
cake :huuh:

:wow:  :ddr:  :pilot:

-huh?
-what?
-when?
-why?

Yaay!

Sailingdog
is
having
a
birthday
this
year

Really?

TPAM
TPBM
Cake
Funny Hats
Presents
erabel


Let the function
$f(x)=\begin{cases}0\text{ if } x=\frac{\pi}{2}+k\pi\\\frac{1}{2+tan^2 x} \text{ if } x \neq \frac{\pi}{2}+k\pi \end{cases}$
$k\in\mathbb{Z}$.
Prove that the function $g(x)=f(x)+f(ax)$ is periodic if and only if $a \in \mathbb{Q}$

[quote="vutang"]We can prove that if $a\in \mathbb{Q}$ then $g(x)$ is periodic by choose $T$ such that $g(x+T) = g(x)$.Now we  prove if $g(x)$ is periodic then $a\in \mathbb{Q}$. There is the number$T$:$g(x+T)=g(x)$ or $f(x+T)+f(ax+aT)=f(x)+f(ax)$. Given $x=0$ we have $f(T)+f(aT)=2f(0)=1$.Because $f(x)\le \frac{1}{2}$ implying $\frac{1}{2+\tan^2(T)}+\frac{1}{2+\tan^2(aT)} =1$ Therefore
$\frac{1}{2+\tan^2(T)}$=$\frac{1}{2+\tan^2(aT)}$ =$\frac{1}{2}$. It easy to implying $a\in \mathbb{Q}$   [/quote]
[quote]I think you can read my soluion, i'm not sure.[/quote]

You do not have to quote yourself to be readable. :)


OK I kinda have a problem...

I built the .dvi file, but the only three things saved were the .aux, .tex, and .log file, so I couldn't view the .dvi. I got a thing that says the .dvi didn't exist. Did I do something wrong?

You need to give details of the system you are using and whether it happens with one particular compilation or all of them. If the former you need to show us the contents.

You are probably making some coding error.


parameterization
IMO
algebra
system of equations
IMO 1963

Am I dumb?

Adding all equations we and dividing through $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}$(without watching that $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=0,y$)  we get y=2, so it isn't a parameter!? Afterwards $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}$

I'm sure I made a mistake...

It just means you have to discuss the solvability when $y=2$ and $y\neq 2$.

Sorry, sir. How do you get those solutions, if $y =\frac{-1 \pm \sqrt{5}}{2}$? I don't get it... :noo:
Could somebody help me, please?

My solution is a bit messy but the main idea is to write the equations in Ax = b where b=0. If detA = 0 then there are infinite solutions and the solution space has dim rankA. Otherwise the system has a unique solution.
https://imgur.com/a/873nbon


lol there's a chem one?  there's a physics one, but i've never heard of a chem one, or at least my school doesn't offer any.  not everything has an olympiad i don't think, but correct me if i'm wrong

[quote="8th Army"]Just curious, how do you make the chem olympiad?[/quote]

Try this: http://ftp.cheminst.ca/outreach/hsexam/CICxm.html

I'm not too familiar with the Chem Olympiad though, ask pengshi if you want more details.

hi, thats great link, im practicing for next year chem olympiad if i can survive, i'll try the canadian chem problems!!!
ty for link billzhao

[quote="Lucky707"]pengshi, what was your rank on the Avogadro?

Same question goes to everyone else who took it.

(...I got like ... 98th percentile ... 31.25/40 or 118th in Canada .... I think that's it.)[/quote]

My avogadro is 36.25/40. don't know if it's good or bad. I screwed up 3 questions consecutively.

Also, for the chem olympiad. In ontario, you have to do multiple choice problem sets every month, published in this site: http://www.chem.utoronto.ca/IChO.Ontario/. Moreover, they will give you a take home exam in march, which you mail back. Furthermore, you must write the CIC chemistry exam (parts a and c) in April. Based on the multiple choice, the take home, and the CIC, they will select the top 20 in ontario to a training and selection camp, in which there will be a lab and an exam to complete. (This happens at the same time as the POPTOR training camp, which is the analogous thing for physics.) Based on this, along with the other stuff above, they will invite people to the National Olympiad Finals (NOF) which is in early june. This is a week of TST's, and the top 4 in this gets to be in the Chemistry team.

I hope that this gives you enough information, you can also find information from the site i gave. You can also email the guy from the site. I suspect that the other provinces are similar.

The thing about chemistry is that the material is more or less just university stuff. There is little or no creativity and elegance involved, unlike math. Thus, if you learn grade 1-3 university chemistry, you can basically solve everything. And people like me who don't know anything and don't want to memorize those rules and names, are basically screwed. Physics is a little bit better.

[quote="billzhao"]We're getting off topic ... again ....  Thus I've splitted it.

C'mon people, stay on topic, or create a new one.[/quote]

what do you mean by split it? i did not see you cut it in half :?:

[quote="opple"]
what do you mean by split it? i did not see you cut it in half :?:[/quote]

This topic was originally in a discussion regarding the IMO. For better organization, I've divided the original thread into two, one containing the original discussion, and one (you're looking at right now) containing the off-topic discussion.


If A and B are square matrices of the same size such that ABAB =0, does it follow that BABA = 0?

For 2x2 matrices it probably does, but not in general. For example, $ A=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$ and $ B=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}$. Here $ A$ shifts the coordinates by one: third becomes second, second becomes first (and the new third coordinate is zero). $ B$ is a projection that kills the third coordinate. So, $ ABAB=A^{2}B=0$ but $ BABA=A^{2}\ne 0$.

Here's the proof for $ 2\times 2$ matrices:
$ B(ABAB)A=0$, so $ (BA)^{3}=0$. Any nilpotent $ 2\times 2$ matrix $ X$ has $ X^{2}=0$, so $ (BA)^{2}=0$.


[quote="The soul of rock"]
Why you use $x_{0}=0$? It's not necessary![/quote]

he didn't do it on his will, its a china 1996 probem. the original RHS is $\frac{\pi}{2}$.


Let $ a,b,c>0$, s.t. $ a\plus{}b\plus{}c\geq 3$.Prove that:
$ \frac{1}{a^3\plus{}b\plus{}c}\plus{}\frac{1}{a\plus{}b^3\plus{}c}\plus{}\frac{1}{a\plus{}b\plus{}c^3} \leq 1$.

f(a) =<f(3-b-c) anh then fairly easy (if you strong  :P )

[quote="onlylove_math"]
We have to prove  $ \sum \frac {1}{1 \plus{} 2a} \le 1$
[/quote]

 Try $ a\equal{}b\equal{}0,5$    , $ c\equal{}2$.

[quote="Allnames"][quote="onlylove_math"][quote="Inequalities Master"]Let $ a,b,c > 0$, s.t. $ a \plus{} b \plus{} c\geq 3$.Prove that:
$ \frac {1}{a^3 \plus{} b \plus{} c} \plus{} \frac {1}{a \plus{} b^3 \plus{} c} \plus{} \frac {1}{a \plus{} b \plus{} c^3} \leq 1$.[/quote]
 $ \sum \frac {1}{a^3 \plus{} b \plus{} c} \equal{} \sum \frac {1}{a^3 \plus{} 2 \plus{} b \plus{} c \minus{} 2} \le\ \sum \frac {1}{3a \plus{} b \plus{} c \minus{} 2}$ $ \equal{} \sum \frac {1}{1 \plus{} 2a}$
We have to prove  $ \sum \frac {1}{1 \plus{} 2a} \le 1$
This problem can easy prove by PQR[/quote]
It is  not easy,(im not sure it is  right,) can you detail it,please
May it be solved by AMGM (i remember that can_hang2007 has a similary ineq and he said it can be proved by only AMGM)[/quote]
$ \sum \frac {1}{1 \plus{} 2a} \le 1\Leftrightarrow \sum \frac{a\minus{}1}{1\plus{}2a}\ge 0$

apply Cauchy-Schwarz

$ (\sum (a\minus{}1)(1\plus{}2a))(\sum \frac{a\minus{}1}{1\plus{}2a})\ge (a\plus{}b\plus{}c\minus{}3)^2$

setting $ a\plus{}b\plus{}c\equal{}p, ab\plus{}bc\plus{}ca\equal{}q$ and $ abc\equal{}r$

$ \Leftrightarrow \sum \frac{a\minus{}1}{1\plus{}2a}\ge \frac{(p\minus{}3)^2}{2(p^2\minus{}2q)\minus{}p\minus{}3}$

we need prove that $ 2(p^2\minus{}2q)\minus{}p\minus{}3\ge 0 \Leftrightarrow 2p^2\minus{}p\minus{}4q\minus{}3\ge 0$

we have $ 2p^2\minus{}p\minus{}4q\minus{}3\ge 12\minus{}4q\ge 0$ ($ p\equal{}a\plus{}b\plus{}c\ge 3$)

$ \Leftrightarrow q\ge 3 \Leftrightarrow ab\plus{}bc\plus{}ca\ge 3 \Leftrightarrow (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}
a)^2\ge 0$ which is true :wink:

[quote="bunhiacovski"][/quote][quote="Allnames"][quote="onlylove_math"][quote="Inequalities Master"]Let $ a,b,c > 0$, s.t. $ a + b + c\geq 3$.Prove that:
$ \frac {1}{a^3 + b + c} + \frac {1}{a + b^3 + c} + \frac {1}{a + b + c^3} \leq 1$.[/quote]
 $ \sum \frac {1}{a^3 + b + c} = \sum \frac {1}{a^3 + 2 + b + c - 2} \le\ \sum \frac {1}{3a + b + c - 2}$ $ = \sum \frac {1}{1 + 2a}$
We have to prove  $ \sum \frac {1}{1 + 2a} \le 1$
This problem can easy prove by PQR[/quote]
It is  not easy,(im not sure it is  right,) can you detail it,please
May it be solved by AMGM (i remember that can_hang2007 has a similary ineq and he said it can be proved by only AMGM)[/quote][quote="bunhiacovski"]
$ \sum \frac {1}{1 + 2a} \le 1\Leftrightarrow \sum \frac {a - 1}{1 + 2a}\ge 0$

apply Cauchy-Schwarz

$ (\sum (a - 1)(1 + 2a))(\sum \frac {a - 1}{1 + 2a})\ge (a + b + c - 3)^2$

setting $ a + b + c = p, ab + bc + ca = q$ and $ abc = r$

$ \Leftrightarrow \sum \frac {a - 1}{1 + 2a}\ge \frac {(p - 3)^2}{2(p^2 - 2q) - p - 3}$

we need prove that $ 2(p^2 - 2q) - p - 3\ge 0 \Leftrightarrow 2p^2 - p - 4q - 3\ge 0$

we have $ 2p^2 - p - 4q - 3\ge 12 - 4q\ge 0$ ($ p = a + b + c\ge 3$)

$ \Leftrightarrow q\ge 3 \Leftrightarrow ab + bc + ca\ge 3 \Leftrightarrow (a - b)^2 + (b - c)^2 + (c - a)^2\ge 0$ which is true :wink:[/quote]
it is wrong,you can check it with $ a=b=0,5 c=4$


vector
real analysis
real analysis solved

Suppose in $R^n$ we have some $m$ vectors of euclidean norm 1 and having the sum 0. Prove that for any vector $x$ we have $\sum{||x-v_i||}\geq m$.

We have $||x-v_i||=||x-v_i|| ||v_i||\geq (v_i-x)\times v_i=v_i^2-v_i \times x=1-v_i\times x$.
Summing we get $\sum_{i=1}^m ||x-v_i||\geq \sum_{i=1}^m 1-v_i \times x=m-(\sum_{i=1}^m v_i)\times x$.
Since $\sum_{i=1}^m v_i=0$ this is equal to $m$.
Equality  clearly occurs when $v_i-x,v_i$ have same direction for every $i$ thus $x=0$

I have a question related to this ( sorry if it is too simple ).
We know from the problem that if we have a collection of vectors $
v_{1},v_{2}, \ldots , v_{m} \in \mathbb{R}^{n} $ of the same norm
and with sum equals to $ 0 $, then for every vector $ x \in
\mathbb{R}^{n} $ we have that $ \sum ||x-v_{i}|| \geqslant \sum
||v_{i}|| $. Also this property holds when we can divide our
collection of vectors to some groups, such that in each group its
members have the same norm and sum of all vectors in each group
equals to $ 0 $. My question is - is it true that conversely, if
we have that for every $ x \in \mathbb{R}^{n} $, $ \sum
||x-v_{i}|| \geqslant \sum ||v_{i}|| $, then we can divide the set
of vectors $ \{ v_{1},v_{2}, \ldots , v_{m} \} $ to groups with
this property.

Maybe I'm missing something? :?

We can take $m=2,\ v_1=0$ and $v_2$ anything we want. Then $\|x\|+\|x-v_2\|\ge\|v_2-x+x\|=\|v_2\|,\ \forall x$, and the system $\{0,v_2\}$ cannot be partitioned into systems of the mentioned type.

Sorry, this is me , who missed that  :blush:


algebra
polynomial
linear algebra
matrix
real analysis
real analysis solved

Heya there!!!

Consider the sequences (a_n), (b_n) and (c_b) n>=0 such that a_0=1, b_0=c_0=0 and a_(n+1)=a_n+b_n; b_(n+1)=b_n+c_n; c_(n+1)=c_n+a_n for all n>=0.

a) Prove that the sequence (a_n - 2^n/3);n>=0 is periodical.
b) Compute a_p-2^p/3 where p belongs to {2002,2003}.

cheers!!! :D