MIT
college
Princeton
Harvard
Stanford
geometry
geometric transformation

I'm looking for colleges that have really strong math programs.  I know the obvious schools (MIT CalTech) and I've heard that Princeton, University of Chicago, and UC Berkeley are also really good, but I'm just wondering if there are some less known schools that I don't know about.  I want to major in math so this is my biggest criterion for colleges.  Are there any schools with less harsh acceptance rates that are just as good?  Also are there any other Ivy league schools with particularly good math departments?

Some strong math colleges:

Harvard
Harvey Mudd
U of Mich

Are  Caltech and Stanford good for math (compared to Harvard, MIT, and Princeton)?

By the way, mathkid, how did the hurricanes hit your area? (my area was very closely missed 3 times - Frances, Ivan, Jeanne)

I live in Vero Beach. We got the grunt of the storm during both Frances and Jeanne, so our town was, and still is, in bad shape. .. scary stuff.

Generally, if you want a very strong math department, even as a undergraduate, you should go to a school which has a Ph.D. program in mathematics.  The availability of a large breadth of advanced undergraduate courses, not to mention seminars and colloquia, will generally be much better than a school which doesn't have a graduate program.

Having said that, the American Mathematical Society has a ranking of Ph.D. programs at:

[url]http://www.ams.org/employment/groups_des.html[/url]

As with all "rankings", take this list with a huge grain of salt.  In fact, let me quote from the AMS's disclaimer:

[i]Please keep in mind that these groupings are used for statistical reporting purposes only and do not necessarily accurately reflect current program quality at individual departments.[/i]

But if you're looking for a list of strong math programs, this list is not a bad place to start.  In particular, the list of "Group I Public" are all very strong programs at large state universities.

Hi - I put a post  (very personal opinion)  on this subject before... and anyway  here is some   good  discussion  on this subject . It  took place some time ago in this form. If  you have not seen  it,  please look at

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9170
******

There are many good schools. You  must look at them, (or talk to friends who have gone there). and then make up your mind.

Also remember, among good schools, there is no such thing as a 'bad choice'

On personal level I (with my sons) looked at many schools. Apart from the schools I mentioned  in that post, you may like to check out Washington University (St Louis). ... (Duke and Washingto U are one of  a few schools  that go out of their way to recruit good  students.)

Now both of my son's are in college. One went to   Duke (his other choices were were Caltech and  MIT) and other went to MIT (Other choices Duke and WashU).

(Though I pushed a little, they did not even apply to IIT's  - their mom put the foot down, "her  babies were not going to  go to a school  so far away")

Both are very happy with their decesion (and so am I).

Is any one here thinking of going to  I IT's?

[quote="DPatrick"]Generally, if you want a very strong math department, even as a undergraduate, you should go to a school which has a Ph.D. program in mathematics. The availability of a large breadth of advanced undergraduate courses, not to mention seminars and colloquia, will generally be much better than a school which doesn't have a graduate program.[/quote]
That's a very reasonable point of view. I'd like to put another point of view up here, just as something to think about. I will say up front that I do have an ax to grind, a bias which should be obvious from my signature. I will also say that I am not talking to everyone who reads this board - there are certainly many for whom the above advice does apply.

The only thing is that the business of Ph.D. granting institutions is mathematical research and that Ph.D. program. Teaching loads are low (which means from your point of view relatively few chances to have any particular professor for a class), research pressures are high, and the graduate students may have the first claim on the faculty's time. Those seminars and colloquia? Some of them are wonderfully stimulating, but many of them are highly technical research mathematics, accessible only to those who already have graduate-level knowledge in those fields. The question about those seminars is this: will [i]you[/i] be attending them while you are still an undergraduate student? More broadly, where will you be in your academic development in 3 or 4 years? How much will the availability of graduate courses matter to you? (There are some other possible questions that concern finances.)

Those of you who are reading this are talking about starting an undergraduate program - and while you may be 90% sure you're going to major in math, you do know that students change their minds. If you go to a school without a Ph.D. program, you may see a lot more of the faculty, for whom your undergraduate education is a high priority. You may have more opportunities to spend time with one or two particular professors. Even institutions with less selective admissions have their star students; you shouldn't take for granted that you would be one yourself without working for it.

There are two kinds of colleges without Ph.D programs that I'm talking about here. One is small private liberal arts colleges. There are some, particularly in the Midwest, that have excellent reputations in mathematics. The other is second-tier state universities, such as the one I represent, and as masters-granting institutions, we do have an array of first-year graduate courses available to keep you moving forward should you reach that point. Give us a chance. Weigh the benefits of the personal attention against the fame, and decide what is best for you.

[quote="Gyan"]Is any one here thinking of going to  I IT's?[/quote]

I am definitely curious about the Indian Institute of Technology (IIT) schools. I was just looking up information about the joint entrance examination the other day. We have no Indian ethnic heritage, but that is the whole point--going to IIT would be a great multicultural experience for a half-European-American, half-Taiwanese young man.

What is the process for applying to IIT as a foreign student?

[quote="mathkid"]http://www.jee.iitb.ac.in/Mmaths.htm

There's a sample mathematics entrance exam from IIT.[/quote]

Thank you for letting everybody know what aspiring Indian mathematicians have to do test on to get into the top colleges in India. My question is more directed to the issue of how foreigners can sit for those tests. Does one have to travel to India from abroad to take the preliminary and main JEE exams, or is there a way to take them overseas? Or is there a modified examination procedure for foreigners (as there definitely is for universities in China or in Taiwan)?

Tokenadult:
IIT's require three exams if you are going into a tech field such as Engineering or math: physics, chemistry, and math. To get into medical school, you need to take physics, biology, and chemistry. No writing is needed at all, simply because the primary focus of the IIT's are science/tech/math stuff and medicine.
For internationals, I belive you can substitute each of the Indian tests with a SAT II (2C for math) for whatever field you want to get into....... got this from my relatives (4 in IIT's).
Also, there won't be much culture shock for your son if he chooses to attend, because, aside from the fact nearly everyone is Indian, the professors and students speak English near flawlessly, and English is the only language used in instruction. In dorms, as well, English is spoken around 50% of the time simply because India has hundreds of languages, and English is the one everyone knows. I am confident that your son will not have to know a single word of Hindi to pass the 4 years of college, simply because he won't have to venture outside the campus.
Another thing: Ragging (hazing) is prevalent in these dorms: stick head in toilet, sing, servant to upperclassemen, etc. Your son will be able to escape all that since he is Chinese- American, and people will be intimidated of him for that.
What IIT is he planning to apply to?

Hope that helps.

Sorry if this is long ---
I can talk a little about IIT.  I went to IIT Kanpur a long time  ago  but,  yes, I would still be  very biased  ) .  I found that the quality of students, and faculty was  excellent. There was  diversity, although majority were Indian students,  there  were   a few foreign students (including Americans) . And my time at IIT was simply the best time I had in any school.  Although IITs  are primary  Engineering  schools, the sciences and math programs are also excellent  . (I majored in Physics)

IITs  have kept their standards, and enjoy  a  very good reputation. For example when I applied to graduate schools I got into every school I applied  to (including Caltech, MIT) I did  not even take a GRE.. There was a CBS  segment 60-minutes on IITs and it  said some thing like Put Princeton, Harvard and MIT together .. and you start getting some idea about the prestige it  has  in India.. MIT prof     Keniston   once joked
[url=http://www.kamalsinha.com/iit/news/mafia/article.html]joke is that  MIT is run by IIT Kanpur Miafia ..[/url]

For admissions and other info, you may like to see http://www.iit.org  or the link given above by  mathkid which has more details and the syllabus for the tests etc.

There was a special channel called DASA (Direct admission for students at Abroad) for  American (and other  foreign) students which used SAT (minimum 2100 in SAT IIs combined in  Math, Physics and Chemistry  to be considered  for application)  and other scores  to form comparative merit lists but I think it is no longer used.  The criteria for all (Indians or Foreign students) now is JEE (Joint Entrance Exam  used by IITs and a few other schools) JEE has two parts:  A three hour screening test ( 10th April 2005) (About 15% of about 200,000 students gets to the next stage) and a  six hour  Main-test (3x2hous for Physics, Chemistry , Mathematics)  (15% of above or about  2% of total gets selected) .

Difficulty level  *very rough -  my estimate only*  - If you can qualify for USAMO you  will have no difficulty)

Most of American  High Schools are accredited  but you have to have enough courses in Physics,  Chemistry and Mathematics (see the syllabus in mathkids link).

Just few points which Samus raised -  IIT does not have medical school . Though some  now have a few graduate business related degrees but primary  (as in 99%) focus is math/science/engineering.

Hazing,  I also  heard, could be problem, but honestly I have never encountered it  (or even seen it ) and if you stay away from frat-party  type scene you should  be okay. I believe there was a bad case (some one committed suicide) so I believe that administration is now very strict.

IITs are not very expensive. (Tution+ living expenses+ say one trip per year)   may be around $10,000-12,000.  JEE is for common admissons (to all IIT's)

English is the medium, but it really does not matter, (The JEE exams are offered in many other languages) It is really easy to learn a  new language while you are studding it in a IIT type atmosphere.

One important tip for parents here (in US): Check out the atmosphere in the school ( say while you are in Junior year  summer) to see if  think if you like it.  IIT alumini chapters are  many places in USA and they can be a good resource too...

Hope this information is useful.

 .

[quote]My question is more directed to the issue of how foreigners can sit for those tests. [/quote]

As I said in  the previous post,  at present  only JEE (which is same for all - Foreigners  and Indians) is the criteria. DASA (For foreign students), I belive,  i[b]s no longer (starting from 2003) an option.
[/b]
I  will find out for sure  and the locations  where one can take JEE. (Kids I  knew  (from US) did trave to india but they had family in india and arranged that any way -  Indian Embessy  is/was possibliity  but I am not sure at all so I will post it here when I find out something definitive)

[quote="samus"]Tokenadult:
What IIT is he planning to apply to?
[/quote]

Thank you for the very helpful answer. So far my son is hardly thinking about this at all, but I like to consider lots of possibilities for his education. We have been attending information sessions for United States colleges (the usual "top" math schools) this month. I liked the years I spent in Taiwan after college, during which I was a student part of the time, and I think international experience is a good idea. But my son doesn't know even the names of the IIT campuses yet, although one girl in his car pool to his math program is an abcd.

Yes, I expected higher education in India to be in the medium of English for the reasons you mentioned. When my son was in Ireland this summer he found out that English isn't spoken everywhere the way it is spoken in Minnesota [grin]. Also in Taiwan most of the textbooks are exactly the same English-language textbooks used in the top schools in the United States--although of course some students buy cram books written in Chinese.

We have a lot to think about in the next few years. I first became aware of the international reputation of the IITs when my mother watched a Sixty Minutes report on one of the IIT schools a year or so ago.

Sorry for the confusion. I was thinking of other schools in India that have both engineering and med schools. Also, the general math skills of peope in IIT are of Olympiad (international to national)/pre-Olympiad levels (from my cousin in India, who is a national olympiad qualifier, similar to (better than?) colleges such as MIT CalTech. Finally, a word of caution: the IIT's probably do not offer students a well-rounded education.

Sorry if I'm rambling :( .

can we limit the search just a bit more to great small (about maybe less than 50 students to 1 teacher ratio or more....) colleges with great math departments for undergrad looking forward to pure math and great teachers (teachers there to teach and not for the reseach) to teach them?

Well, at Yale, I'd say we have a pretty good math department. Although I think that a lot of the ivies have very similar math departments, ie. princeton, harvard, yale, cornell. So you can't really go wrong with any of those choices, though I feel that our ratio between faculty and students is relatively low. The only thing that you might have to consider is you might want to look into certain distributional requirements you need to meet, if you are only interested in doing straight math and no humanities or social sciences. But like I said, you can't really go wrong with any of the choices, though i'd strongly suggest yale (guess i'm biased?  :) ).

Yes, it's the US News ranking. But anyways, it may still yield some useful info. BTW, this is for graduate school so liberal arts colleges are out.

*** THE SCIENCES (Ranked in 2002) ***

The Sciences->Applied Mathematics->Top Applied Mathematics Programs

Categories:
Rank.|School|Average assessment score (5 = highest)

1.|Massachusetts Institute of Technology|4.7
  |New York University|4.7
3.|California Institute of Technology|4.6
4.|Stanford University (CA)|4.5
5.|Brown University (RI)|4.4
  |Princeton University (NJ)|4.4
  |University of California-Berkeley|4.4
  |University of California-Los Angeles|4.4
9.|University of Minnesota-Twin Cities|4.3
10.|Cornell University (NY)|4.2
11.|Carnegie Mellon University (PA)|4.0
  |University of Maryland-College Park|4.0
  |University of Texas-Austin|4.0
14.|Northwestern University (IL)|3.9
  |Rice University (TX)|3.9
  |University of Washington|3.9
  |University of Wisconsin-Madison|3.9
18.|Georgia Institute of Technology|3.8
  |University of Chicago|3.8
  |University of Michigan-Ann Arbor|3.8
21.|Harvard University (MA)|3.7
  |Rensselaer Polytechnic Institute (NY)|3.7
  |Rutgers State University-New Brunswick (NJ)|3.7
  |SUNY-Stony Brook|3.7
  |University of Arizona|3.7
  |University of Colorado-Boulder|3.7
27.|Duke University (NC)|3.6
  |Purdue University-West Lafayette (IN)|3.6
  |University of Illinois-Urbana-Champaign|3.6
  |Yale University (CT)|3.6
31.|North Carolina State University|3.5
  |University of California-San Diego|3.5
33.|Columbia University (NY)|3.4
  |Indiana University-Bloomington|3.4
  |Johns Hopkins University (MD)|3.4
  |Penn State University-University Park|3.4
  |University of California-Davis|3.4
  |Virginia Tech|3.4
39.|Ohio State University|3.3
  |Texas A&M University-College Station|3.3
  |University of North Carolina-Chapel Hill|3.3
  |University of Utah|3.3
43.|Boston University|3.2
44.|Arizona State University|3.1
  |University of California-Santa Barbara|3.1
  |University of Delaware|3.1
47.|Claremont Graduate University (CA)|3.0
  |CUNY Graduate School and University Center|3.0
  |Iowa State University|3.0
  |University of California-Irvine|3.0
  |University of Pennsylvania|3.0
  |University of Southern California|3.0
  |Washington University in St. Louis|3.0
54.|Florida State University|2.9
  |Michigan State University|2.9
  |University of Illinois-Chicago|2.9
  |University of Pittsburgh|2.9
  |University of Virginia|2.9
59.|University of Houston|2.8
  |University of Iowa|2.8
61.|University of Florida|2.7
  |University of Massachusetts-Amherst|2.7
  |University of Tennessee-Knoxville|2.7
64.|Brandeis University (MA)|2.6
  |New Jersey Institute of Technology|2.6
  |SUNY-Buffalo|2.6
  |University of Notre Dame (IN)|2.6
  |Vanderbilt University (TN)|2.6
69.|Case Western Reserve University (OH)|2.5
  |Clemson University (SC)|2.5
  |Colorado School of Mines|2.5
  |Colorado State University|2.5
  |Dartmouth College (NH)|2.5
  |Emory University (GA)|2.5
  |Oregon State University|2.5
  |Polytechnic University (NY)|2.5
  |University of Georgia|2.5
  |University of Missouri-Columbia|2.5
  |Washington State University|2.5

The Sciences->Mathematics->Top Mathematics Programs

Categories:
Rank.|School|Average assessment score (5 = highest)

1.|Massachusetts Institute of Technology|5.0
2.|Harvard University (MA)|4.9
  |Princeton University (NJ)|4.9
  |Stanford University (CA)|4.9
  |University of California-Berkeley|4.9
6.|University of Chicago|4.8
7.|Yale University (CT)|4.7
8.|California Institute of Technology|4.6
  |University of Michigan-Ann Arbor|4.6
10.|Cornell University (NY)|4.4
  |New York University|4.4
  |University of California-Los Angeles|4.4
13.|Columbia University (NY)|4.3
  |University of Wisconsin-Madison|4.3
15.|University of Texas-Austin|4.2
16.|Rutgers State University-New Brunswick (NJ)|4.1
  |University of Illinois-Urbana-Champaign|4.1
  |University of Maryland-College Park|4.1
  |University of Minnesota-Twin Cities|4.1
  |University of Pennsylvania|4.1
21.|Brown University (RI)|4.0
  |Northwestern University (IL)|4.0
  |SUNY-Stony Brook|4.0
  |University of California-San Diego|4.0
25.|Duke University (NC)|3.9
26.|Indiana University-Bloomington|3.8
  |Johns Hopkins University (MD)|3.8
  |Pennsylvania State University-University Park|3.8
  |Purdue University-West Lafayette (IN)|3.8
  |Rice University (TX)|3.8
  |University of Washington|3.8
32.|Ohio State University|3.7
  |University of North Carolina-Chapel Hill|3.7
34.|Brandeis University (MA)|3.6
  |Carnegie Mellon University (PA)|3.6
  |CUNY Graduate School and University Center|3.6
37.|Georgia Institute of Technology|3.5
  |Michigan State University|3.5
  |University of Illinois-Chicago|3.5
  |University of Utah|3.5
  |Washington University in St. Louis|3.5
42.|Texas A&M University-College Station|3.4
  |University of Arizona|3.4
  |University of California-Davis|3.4
  |University of Colorado-Boulder|3.4
  |University of Virginia|3.4
47.|University of California-Irvine|3.3
  |University of California-Santa Barbara|3.3
  |University of Notre Dame (IN)|3.3
  |University of Southern California|3.3
51.|Boston University|3.2
  |Dartmouth College (NH)|3.2
  |University of Florida|3.2
  |University of Georgia|3.2
55.|North Carolina State University|3.1
  |University of Iowa|3.1
  |University of Massachusetts-Amherst|3.1
  |University of Oregon|3.1
  |Virginia Tech|3.1
60.|Arizona State University|3.0
  |Iowa State University|3.0
  |Northeastern University (MA)|3.0
  |University of California-Riverside|3.0
  |University of Missouri-Columbia|3.0
  |University of Rochester (NY)|3.0
  |Vanderbilt University (TN)|3.0
67.|Louisiana State University-Baton Rouge|2.9
  |SUNY-Buffalo|2.9
  |University of Kansas|2.9
  |University of Nebraska-Lincoln|2.9
  |University of Pittsburgh|2.9
  |University of Tennessee-Knoxville|2.9
73.|Emory University (GA)|2.8
  |Florida State University|2.8
  |Tulane University (LA)|2.8
  |University of California-Santa Cruz|2.8
  |University of Delaware|2.8
  |University of Kentucky|2.8
  |University of Oklahoma|2.8
80.|Claremont Graduate University (CA)|2.7
  |Colorado State University|2.7
  |Oregon State University|2.7
  |Syracuse University (NY)|2.7
  |Tufts University (MA)|2.7
  |University of Connecticut|2.7
  |University of Houston|2.7
87.|Auburn University (AL)|2.6
  |Case Western Reserve University (OH)|2.6
  |Kansas State University|2.6
  |Oklahoma State University|2.6
  |Polytechnic University (NY)|2.6
  |SUNY-Albany|2.6
  |Temple University (PA)|2.6
  |University of New Mexico|2.6
  |University of South Carolina|2.6
  |Washington State University|2.6
97.|New Mexico State University|2.5
  |SUNY-Binghamton|2.5

[quote="samus"]Also, the general math skills of peope in IIT are of Olympiad (international to national)/pre-Olympiad levels (from my cousin in India, who is a national olympiad qualifier, similar to (better than?) colleges such as MIT CalTech.[/quote]

This is highly unlikely, as the best US math students (who go almost completely to Harvard, Princeton, MIT, Caltech, Duke, Stanford, etc.) are better than the best Indian students, as far as IMO, etc, are concerned.

In fact, India didn't have a single gold medal at the IMO this year, and only one in the past 3 years.

It is quite possible that the best Canadian schools, even, attract better math students than IIT.

[quote="mathkid"]http://www.jee.iitb.ac.in/Mmaths.htm

There's a sample mathematics entrance exam from IIT.[/quote]

mathkid, thanks a lot for posting that.

it's rare that I get to see good calculus/linear algebra questions like that.

You may like to check out NY Times (registration is required) article about
[url=http://www.nytimes.com/2004/10/20/nyregion/20ranking.html?oref=login]Top Colleges, Rated by Those Who Chose Them[/url]

As expected Harvard and Yale  ranked high in general. For Math/Science students  MIT, Caltech, Stanford  and Princeton ranked high.

MIT and Caltech, among science related subjects  had top appeal.

Several "top" schools (as ranked from usual sources) did not do as well.

For example, Washington University in St. Louis,  which I liked when I checked it out for my son,  was   ranked around 10  in this year's U.S. News , was the 62nd rated choice among  students' preferences .  Places like the University of Pennsylvania, the University of Chicago and Duke placed at least eight spots lower on the preference ranking than on the U.S. News list.

Slightly related topic -

IMO, Duke  (and  Wash U to some extent) has another attraction, it does have prestigious programs - like A.B. Duke Scholarship  -  (which opens up lot of priviledged oppurtunities,  not to mention that those programs  provide full tution scholarships too)  for really bright students... So  the  select few "recruited "  Math students,  more often than not, ,  choose Duke over  schools like Harvard... (No wonder that some Ivy college compliain that Duke 'steals' their good potential students by bribing them  :) )

Frankly, I think that's a incredibly pointless rating system.  I have no basis whatsoever to rate my school against other schools, because I've only been to one of them.  I think those schools which drop in the rankings suffer (with the exception of UPenn) from geographic problems -- the places that send the most students to private colleges are California and the north-East, especially the New York City tristate region (that is, New Jersey, New York and Conneticut).  Students from those areas are much less likely to wander out into the midwest or down south.  It doesn't surprise me at all that Duke, Chicago and Washington in St.L ranked lower -- I mean, who wants to go to school in North Carolina, for crying out loud?

And who wants to go to Penn, anyhow?  ;)

[quote="JBL"]I mean, who wants to go to school in North Carolina, for crying out loud?[/quote]

North Carolinians (and South Carolinians) presumably are happy to go to school in the Carolinas, in at least some instances. It is still relevant, for comparing colleges, to find out whether a North Carolinian accepted to Duke and Harvard will prefer Harvard or not, and whether a New Yorker accepted to Harvard, to Yale, and to Princeton will prefer one school over the other two.

The "[url=http://post.economics.harvard.edu/faculty/hoxby/papers/revealedprefranking.pdf]A Revealed Preference Ranking of
U.S. Colleges and Universities[/url]" is pretty interesting and has some good discussion of methodological issues.

[quote="tokenadult"]It is still relevant, for comparing colleges, to find out whether a North Carolinian accepted to Duke and Harvard will prefer Harvard or not, and whether a New Yorker accepted to Harvard, to Yale, and to Princeton will prefer one school over the other two. [/quote]

Interesting, maybe.  I don't see how it is relevant.  Students who have not yet attended college are not in the best position to choose what the best college for them would be, honestly.  What if all my friends at Harvard would really have been happier at Washington University?  There's absolutely no way to include that kind of measure in the ranking.  What it tells you is that, when presented with a choice between two schools, students tended to choose one over the other -- I'm not convinced this would really represent what school would be best for any student.  I think there are much more telling pieces of information, like quality-of-life measurements or perceived enjoyment of already-enrolled students, that could build a much better picture of what students want.  Given the basic types of data they collected, the thing that would most warm me up to it is if they also broke it down according to what reasons students gave for choosing the schools they did -- it would give a picture of why students chose schools which would be much more helpful to me, as a student, picking where I wanted to go.

Don't many of the best students also go to large public institutions? Considering financial and "staying close-to-home" concerns...

It seems like nany great scientists did not all go to the best institutions.

For example, [[Dr. David Goodstein]] only went to state schools. There are plenty of others.

But hmmm...

Do the majority of the best students only go on to state schools? Hmmm... How many people are at age 18? Tkae 1 % of that; and take the addition of the freshman pops of the top 15 universities. This should give an impression...

Simfish -- thousands more students go to large public universities than to selective private colleges or universities, but that wouldn't necessarily move those public schools up in the ranking, because it compared where people chose to go once accepted among their various options.  Relatively few people who apply to both large public universities and selective private universities and get in to both go to the public schools nowadays -- most of the students you are talking about would never have bothered applying to the private colleges in the first place.

Tokenadult -- my entire position can be summed up in the following two sentences:
Having been, relatively recently, among a large group of college-bound high school students, I am not convinced that tracing the patterns of where such students choose to go to school gives sufficient evidence of school quality to declare that the most popular schools are anything other than most popular.  In particular, I feel that the relative rankings of any two colleges is essentially meaningless.

[quote]I mean, who wants to go to school in North Carolina, for crying out loud?[/quote] :)

Reminds me of a T-shirt I saw in Cambridge:

[quote]Harvard: Because every one cant get into MIT ![/quote]
:D
My son is at  Duke. He chose Duke over MIT (and Caltech)  and I  know quite a few kids (My sons friends, and my friends children) who have chosen North Carolina over Harvard and Yale, .and some of them even live closer to Connecticut or Massachusetts !  I think, last year, all (or almost all)    who were offered AB Duke scholarships ( this  I mentioned it the previous post)  did go to Duke  (one can be sure that  they were also accepted in other top colleges in US).. BTW     Duke has done pretty good in Putnam competition, so may be some of those kids who like math wants to go to North Carolina. :)

Seriously, IMO, far from pointless, the survey is has lot of points.  When we were visiting colleges from my sons , one of the most often asked question, (when we were  getting perspective from other students,  and indeed my friends asks me  who are trying to get my perspective) is/was  which other colleges you were seriously considering?  and why you choose x rather than y.  This type of survey gives some statistical perspective to get answer to those types of questions. Sure, you  can have a lot of data points, you  can see rankings, you can talk to your friends, professors, and other students, you can visit colleges and  attend classes there, you can also get perspective on why  a person  would choose college x from college y  specially if your choice also happens to be between x and y. Nothing pointless about this. Just my thoughts.

Blahblahblah -  IITs are mainly engineering schools, and some would say  that   math is not their top field but as I said before, these are, IMO, excellent even  for Math and Physics  etc.  I dont think many typical US  students   go to IIT  (unless  some reason like - their parents went to IIT) but I do know some Indian Kids who chose IIT over top US schools.   In final analysis, its hard do define the best school and in reality  one would not go wrong if  one chose any one of the top schools.

BTW, I may be wrong  as  I dont know how popular AMC/AIME/USAMO type contests are in India today, but when I was there, some of the best math students(and their schools)   did  not even   know  of those (nmo etc)  tests  and   contestants for IMO  did not represent the top math talent  from Indian High schools.

[quote]But no one has shown that students who decide which offer of admission to accept lack access to currently enrolled students, who presumably would tell the newly admitted what they think about each school--plausibly or not..[/quote]
Agree ..IMO they are very well informed

This information about foreign students may be of interesti to some:
[url=http://www.iitb.ac.in:8080/examples/acad/foreignstudents.jsp] link.
 [quote]General Information - Admissions for foreign students

   1. Foreign student is one who holds foreign passport, or has Indian passport but is studying outside India and wants to do partial studies at IIT Bombay

   2. Admission for a full programme with degree from IIT Bombay
          * For B. Tech : Admissions are made through JEE
          * For M. Tech / Ph.D. and other post-graduate degrees
                o Admissions are made based on academic performance
                o Please send your application (brochures / forms are available at our website http://www.iitb.ac.in ) to the department of your interest
                o Offer will be made based on recommendations received from the departments
                o The fees at present are US $ 6000.00 per semester (likely to be reduced). You will be given hostel accommodation on the campus. Your other expenses are expected to be about US $ 200 - 300 pm.
                o You will need to get student visa from the Indian Embassy in your country
                o The schedule for admission is available at our web-site

   3. Admissions for visiting / exchange students
         1. You may wish to study a semester or carry out your project / research work at IIT Bombay for a limited duration (2 months upto 1 year)

         2. If you wish to come as exchange student under memorandum of your Institute with IIT Bombay, the fees / charges will be as per the terms of Memorandum. If you wish to do a semester of course work here as a visiting student, the fee will be US $ 6000.00 per semester (likely to be reduced), and you will have to bear your local expenses. If you are planning to do a short R\&D project here, the fees will be about US $ 500.

         3. The admission procedure will be as follows :
                * If you are an exchange student, send your application through your Dean to the Dy. Registrar (Academic) at IIT Bombay
                * If you are a visiting student, send your application giving full details of your education and academic performance, to the Head of the Department, or the Professor with whom you wish to do the project work.
                * Admission offers will be made based on the recommendation of the department / faculty.
                * You will need to obtain student visa for the duration of your visit.

         4. In case you plan to take transfer of credits for your course work done here, you must plan the courses (in consultation with your faculty advisor at your university) beforehand and inform the Head of the Department here. You should ensure that you have the right background to do the course work here.
[/quote][/url]

[quote]BTW, I may be wrong  as  I dont know how popular AMC/AIME/USAMO type contests are in India today, but when I was there, some of the best math students(and their schools)   did  not even   know  of those (nmo etc)  tests  and   contestants for IMO  did not represent the top math talent  from Indian High schools. [/quote]

By the way Gyan, my brother-in-law, who qualified for the Indian National Olympiad had several other friends do that too at IIT-(Dehli?) when he attended from 1996-2000.

[quote="stargazer"]Thanks for all the info...still trying to decide how much to use the rankings...btw does anyone know how Carnegie Mellon is in more theoretical math (as opposed to applied).  A friend was recommending it and I was just wondering if anyone knew any specifics.  Thanks again for all the replies.[/quote]

I am a proud graduate of Carnegie Mellon, with a B.S. and an M.S. in mathematics.  The program prepared me well enough to get a Ph.D. at MIT, so that says something I hope.

As you note, the department at CMU is heavily tilted towards applied mathematics.  However, they have a very strong undergraduate "pure" mathematics course sequence, called Mathematical Studies, and offer a program to get a master's degree in 4 years (which I did).

See [url]http://www.math.cmu.edu/ug/ungrd/special.html[/url] for more specifics, and feel free to post and/or email me if you have some specific questions.

Isn’t institute of advanced studies good for mathematics?

The Institute for Advanced Studies, as its name implies, is not an institution you can apply to for an undergraduate education.


Find the smallest constant k>0 s.t.

 \frac{ab}{a+b+2c}+\frac{bc}{b+c+2a}+\frac{ca}{c+a+2b} \leq k(a+b+c)

for every positive a,b,c

[quote="maruemu"]Find the smallest constant k>0 s.t.

 $ \frac{ab}{a\plus{}b\plus{}2c}\plus{}\frac{bc}{b\plus{}c\plus{}2a}\plus{}\frac{ca}{c\plus{}a\plus{}2b} \leq k(a\plus{}b\plus{}c)$

for every positive a,b,c[/quote]

  $ LHS\equal{}\sum {\frac{ab}{(a\plus{}c)\plus{}(b\plus{}c)}}\le \sum {\frac{1}{4}(\frac{ab}{a\plus{}c}\plus{}\frac{ab}{b\plus{}c})}\equal{}\frac{a\plus{}b\plus{}c}{4}$

 therefore $ k\equal{}\frac{1}{4}$


A curve has parametric equations

$ x\equal{}2t \plus{} t^{2}$, $ y\equal{} 2t^{2} \plus{} t^{3}$.

(i) Express $ \frac{dy}{dx}$ in terms of $ t$ and find the gradient of the curve at the point $ (3, \minus{}9)$.

(ii) By considering $ \frac{y}{x}$, find a cartesian equatiooj of the curve, giving your answer in a form not involving fractions.

$ \frac {dx}{dt} \equal{} 2 \plus{} 2t$

$ \frac {dy}{dt} \equal{} 4t \plus{} 3t^2$

$ \frac {dy}{dx} \equal{} \frac {dy/dt}{dx/dt} \equal{} \frac {4t \plus{} 3t^2}{2 \plus{} 2t}$

value of t satisfying x = 3 and y = -9 is   $ t \equal{} \minus{} 3$

so the gradient of the curve at the point (3, -9) is  $ \frac { \minus{} 15}{4}$

The second part ...

$ \frac {y}{x} \equal{} \frac {2t^2 \plus{} t^3}{2t \plus{} t^2} \equal{} t$

so, what will be the Cartesian form ? is it a straight line ?

Hi Nora.91.

I have

$ \frac{y}{x} \equal{} t$

$ \Rightarrow\, x\equal{} 2(\frac{y}{x}) \plus{} (\frac{y}{x})^{2}$

$ \Rightarrow\, x^{3} \equal{} 2xy \plus{} y^{2}$.


quadratics
superior algebra
superior algebra unsolved

prove that for a quadratic field K, p is ramified in K if and only if p divides the discriminant of K. (where p is a prime in Z).

This is true for all number fields in general.

For this special case, not much is to do:
Let $ d$ be the discriminant of that extension.
Handle the case $ p\equal{}2$ seperately. Now let $ p$ be odd and let the extension be generated by adjoining the square root of $ a$. Show that $ d\equal{}2^s a$, thus we have to show:
An odd prime $ p$ ramifies iff it divides $ a$.
Now show that if the ideal $ (p)$ factors (into two ideals), then $ x^2 \equiv a \mod p$ has a solution $ x$, and $ (p)\equal{} (p,\sqrt a\plus{}x)(p,\sqrt a\minus{}x)$. Now just show that these factors are equal iff $ p |a$.


From Mathcamp's website:

for houshold income between $\$40,000 and $\$100,000:

discount: $\frac{100,000-x}{50}$

payment: $\frac{91,700+x}{50}$

The sum is $\$3,834 while the full payment is $\$3,195. What is the real value of the discount/fee?

No clue how the person who updated the website arrived at that second formula.  Certainly it should be $(59750+X)/50.


For a given k, what is the minimum number of integers we have to take such that it always possible to choose k of them such that their sum is divisible by k ?

$ 2k\minus{}1$, see http://www.mathlinks.ro/viewtopic.php?p=389030 .


geometry
3D geometry
number theory
prime numbers
prime factorization

Find the largest integer $ n$ such that $ n$ has exactly 4 positive divisors and $ n$ divides $ 100!$.

Even after reading the solution, I couldn't understand this problem.

When two prime numbers multiply together, you get a number with $ 4$ factors.  Call one prime $ a$ and the other $ b$.  The prime factorization is $ a^{1}\cdot b^{1}$.  The number of factors it has is $ (1\plus{}1)(1\plus{}1)\equal{}4$.  The largest $ 2$ prime factors of $ 100!$ are $ 97$ and $ 89$.  Multiplying these together, we get the answer of $ \boxed{8633}$.

@jxl28: However, a prime^3 also yields 4 factors! (1, prime, prime^2,prime^3). The floor of 100/3 is 33, and the greatest prime less than 33 is 31. So the correct answer should be 29791, because that is greater than 8633.

Hope that helped :)

Watermelon99

Hmm. I disagree with these answers. 97*89 may have 4 divisors, but it is not the greatest factor of 100! with 4 divisors. Cubes of prime divisors of 100! will also work, 1, the number being cubed, the number squared, and number cubed will be the divisors. So using this fact we get 31^3 as our answer which is 29791.

yup after ready everybody's posts and i'm pretty sure j_f_c_w is correct and 29,791 is the answer but remember to hide your answers please and thank you!

[quote=Aditya11]Please do not revive old topics.[/quote]

I think he gets the point

what's with the 4 solutions, no hidden ones


Positive reals $a,b,c$ satisfy $abc=1$. Prove that
$\displaystyle 1+ \frac{3}{a+b+c} \ge \frac{6}{ab+bc+ca}$.

old and easy:
http://www.mathlinks.ro/Forum/viewtopic.php?t=14765
http://www.mathlinks.ro/Forum/viewtopic.php?t=32861

  dg

[quote="k2c901_1"]Positive reals $a,b,c$ satisfy $abc=1$. Prove that
$ 1+\frac{3}{a+b+c}\ge \frac{6}{ab+bc+ca}$.[/quote]
We have: \[\left(1-\sqrt{3 \over a+b+c}\right)^{2}\ge 0 \\ \implies 1+{3 \over a+b+c}\ge{6 \over \sqrt{3(a+b+c)}}={6 \over \sqrt{3abc(a+b+c)}}\ge{6 \over ab+bc+ca}.\]

Let $a,$ $b$ and $c$ are positive numbers such that  $abc=1$. Prove that
$1+\frac{8}{a+b+c}\ge \frac{11}{ab+bc+ca}$
Is it true?

The stronger inequality holds:

$ 1+\frac{7}{a+b+c}\ge \frac{10}{ab+bc+ca}$.

[quote="silouan"]
But $xy+yz+zx\leq \sqrt{3(x+y+z)}$
[/quote]
But $xy+yz+zx\geq \sqrt{3(x+y+z)}$ :wink:


inequalities
geometry
area of a triangle
Heron\u0027s formula
inequalities proposed

Prove that in a triangle with sides a, b and c, semiperimeter s and area A,

A^4 <= s^2((a^2+b^2+c^2-s^2)/3)^3

Prove that in a triangle with sides a, b and c, semiperimeter s and area A we have:

$A^{4} \leq s^{2} (\frac{\sum a^{2}-s^{2}}{3} )^{3}$$.$

I have a solution using duality!

$a=y+z$; $b=z+x$ and $c=x+y$ then the semiperimeter $s=x+y+z$ and the area $A=\sqrt{(x+y+z)xyz}$. Thus our inequality will become:

$ (x+y+z)^{2} (\frac{2\sum x^{2}+2\sum xy - \sum x^{2}-2 \sum xy}{3} )^{3}=(x+y+z)^{2}(\frac{\sum x^{2}}{3} )^{3}$ so our inequality is equivalent to:

$(x+y+z)^{2}(\frac{\sum x^{2}}{3} )^{3} \geq (x+y+z)^{2}(xyz)^{2}$ so it will become: $ (\sum x^{2})^{3} \geq 27x^{2}y^{2}z^{2}$ which is true AM- GM.

have fun. what is your solution? does anybody have another approach?

cheers! :D :D

I found a neat identity yesterday, so I thought I'd turn it into an inequality:

if s=(x+y+z)/2, then

x^2+y^2+z^2=s^2+(s-x)^2+(s-y)^2+(s-z)^2    (1)

So by AM-GM, we can get the weak result x^2+y^2+z^2>4A (using Heron's formula), or the stronger one mentioned above if we apply the AM-GM only to the last three terms on the right and then apply Heron's formula.

What's neat is that (1) generalizes:

for B+3A^2-4A=0,
x^2+y^2+z^2=Bs^2+(As-x)^2+(As-y)^2+(As-z)^2

Maybe someone else can make a nicer problem out of that equation...


function
abstract algebra
linear algebra
matrix
integration
calculus
circumcircle

[i]Math Contest-Memorialul "Hegy Lajos" January 2004 Targu-Mures[/i]

[b]9th grade[/b]

[b]1[/b]. Proove that the ecuation x^2-x(n^2+1)/n-(n^2+1)/n has real roots and they are irational for every n \in N*.

[b]2[/b]. Proove that for every x,y,z>0 we have:
sqrt((x+y)/(x+z))+sqrt((x+z)/(x+y)) <=(y+z)/sqrt(yz).

[b]3[/b].In the convex quadangle ABCD we denote with G the weight centre of triangle BCD and with H the ortocenter of triangle ACD. Proove that the points A,B,G,H represent in this order the points of a paralelogram(i don't know the english version for this word:( ) if and only if G is the center of the circumscribed circle of triangle ACD.

[b]4[/b].We have a number with 2 digits. Between their digits we write another digit and we obtain another number. AM of these two numbers is equal with the number with 2 digits obtained if we inverse the order of the digits of the initial number. Find the initial number.

[b]10th grade[/b]

[b]1[/b]. Solve in R the ecuations:
	a) 2^(x+2)+4^(x+1) = 3x^2;
	b) 2^(x+2)-4^(x+1) = x^2+2x+2;

[b]2[/b].The sequence (x_n)n>=1 satisfie the relation
	x_(n+1)+x_n = 2n^2, for every n \in N*, x_1 \in Z.
a)Exists in the sequence 3 consecutive termns to be in geometric progression.
b)If x_1 = -2004, then which is the value of x_2004?

[b]3[/b]. Find x,y,z>1 wich for we have
log{x}(y^2+z^2)+log{y}(z^2+x^2)+log{z}(x^2+y^2)=7 and \sum log{x}2 = 1.
(i denoted with log{p}(q) = ln(q)/ln(p) )

[b]4[/b]. The measurements of the angles A, B, D, C of the convex quadangle ABCD are consecutive terms, in this order, of an increasing  arithmetic progesion. AC and BD made between them a \pi/2 angle. The measurements of A,B,D of the triangle ABD are too in an arithmetic progression(in this order).
Compute AC/BD.

[b]11th grade[/b]

[b]1[/b].Let (a_n)n>=0 be a sequence satisfying the relation
(n+1)a_(n+2)-(n+2)a_(n+1)+a_n = 0, a_0=1, a_1 = 3.
a)Determine the general term of the sequence a_n and proove the convergence of (a_n)n.
b)Compute the limit lim{n->oo}x_n, where x_n=(2+2e-a_n)n!.

[b]2[/b].Compute lim{n->oo}(nln(n))/([n/1]+[n/2]+[n/3]+...[n/n]).
i denote with [x] the integer part of x.

[b]3[/b].Let (a_n)n>0 \in R*. For each n \in N*\{1}, we consider e_n : a_nx+a_(n-1)y=0.
We denote with (b_n)n respectively (c_n)n those (a_n)n for wich e_n and e_(n+1) are perpendicular respectively parallel.
Study the convergence of (b_n)n and (c_n)n.

[b]4[/b]. Find all the matrices X \in M2(R) s.t.
X^2+2X =
(5  5)
(10 10).

[b]12th grade[/b]

[b]1[/b]. Let (G,*) be a group with the property that there exists an injective function f:G->G,s.t. f(x*f(x)) = x²*f(x*y) for every x,y \in G. Proove that (G,*) is an abelian group.

[b]2[/b]. If F is a primitive of f:R->R, f(x) = ex² for evry x \in R, compute lim{x->oo}(xF(x))/f(x).

[b]3[/b].Let A \in Mn(R), n>=2, invertible. Proove that (A*)* = (detA)<sup<n-2A.

[b]4[/b]. Proove that from 20 integer numbers we can choose 2 s.t. their difference is divisible by 19.

[i]What is your impression about this contest? i think it had a very low level... and it was an interdistrict contest :([/i]

Can you give us more precision

12th grade

1. ... f(x*f(x)) = x²*f(x*y) ....

What do you mean by x&2sup

2. ... ex&sup2...

3.... (A*)* = (detA)<sup<n-2A....

A is matrix what is the notation of  (A*)*
 sup<n-2A not clear
------------------------------------------------------------------------------
I think ex&sup2 means exp(x^2)

If F is a primitive of f:R->R, f(x) = ex² for evry x  R,
compute lim{x->oo}(xF(x))/f(x).

F(x)= \int _[1,x]exp(t^2)dt  integration by part gives
F(x)= exp(x^2)/2x - e/2 +  \int _[1,x]exp(t^2)/(2t^2) dt
this leads to
xF(x)/f(x) = 1/2 - x/exp(x^2).e/2 +
x/exp(x^2).1/2. \int _[1,x]exp(t^2)/t^2 dt ---> 1/2 when x-->+oo
 use again integartion by part
\int _[1,x]exp(t^2)/t^2 dt =  \int _[1,x] (2t.exp(t^2))/(2t^3)dt =
exp(x^2)/(2x^3) - e/2 +  3/2.\int _[1,x]exp(t^2)/t^3dt

nice problem for 12th grade, but it could be CE1 grade  :D

maybe it's ^(x^2). this forum cannot display this style.

sorryyy... i'l edit my post... i wrote all the contest in notepad and after that i paste it here and i saw that instead " 2 " is something else... i modified them but i think i missed some :D

please do not reply anymore in this thread.
put one problem / post ! :)


I don't know if this has already made the rounds, but in this clip Art Benjamin makes his case for changing math education in the US.

[url]http://metroplexmathcircle.wordpress.com/2009/10/23/arthur-benjamins-formula-for-changing-math/[/url]

I partially agree with him. What he says is to take the focus off calculus and onto discrete math. I think that the math curriculum should not focus on a particular subject, but rather let the students try different topics in the wide range of mathematics. Every branch of mathematics is important and useful.

Another big point, that I didn't see him mention, is memorization versus understanding, formula versus methods, etc. As a student taking both high school and middle school math classes, as well as the son of two engineering professors, I can say that students in middle and high school are taught insufficient math and lots, including students of many subjects (because math is a foundation for many thing), are not able to take the leap between high school and college/university. This both the teachers' and students' fault, but more so the curriculum and textbooks used, who make math sound like memorization and computation, when it is really deep and creative thinking and problem solving. Mathematicians are not given a worksheet of problems and find theorems and identities, they think.

To fix this, we have to make the idea of "Calculus" not sound hard and cold. We have to make students really understand things rather than memorizing. For example, in one of my classes, we are given a sheet of formulas and theorems to use on tests and homework, while most students don't understand them and haven't seen them proved. To fix the problem, we have to introduce a new curriculum, such as the way AoPS books work, because they emphasize on the understanding, and show the formula once the student has understood it or derived it by himself/herself by answering questions and solving problems.

we know him, he is nice, we went to dinner with him at Davidson in baltimore , and my mom went to school with him :lol:


Hi.

Can someone tell me how to install Ghostscript onto my computer?  Because in the AoPS wiki webpage about Asymptote, it just has a link that says "download GS here".  I tried to download AFPL GhostScript 8.54 onto my computer (I clicked on the AFPL link after clicking on the AoPS wiki link) through sourceforge, but they just gave me a bunch of winRar documents (I had to download each part of AFPL separately).

And now, after "downloading", I still can't get Ghostscript to work. In fact, my computer says that Ghostscript was not downloaded at all.  So, is there any particular site I can go to to download AFPL GS 8.54 all at one time?  And, to make my computer accept GS as downloaded?  Thanks!

Mr.Ocax

[quote="stevem"]Unfortunately you don't tell us
1. which operating system you are using;
2. why you chose the AFPL version* which is the license for older versions. Why not choose the GPL version 8.60 on the same page as Asymptote directed you to http://pages.cs.wisc.edu/~ghost/ ?

So assuming you are happy with GPL** then the links will lead you to [url=https://sourceforge.net/project/showfiles.php?group_id=1897&package_id=108733&release_id=529280]here[/url] and there you can download the relevant file for your system eg Windows 32 bit is [i]gs860w32.exe[/i] which, when run, is the installer.

However, if you are using Linux (which flavour?) then it would be better to use your installer/package manager for your system.

* Aladdin Free Public License
** GNU General Public License[/quote]
The wiki page actually leads you to: http://pages.cs.wisc.edu/~ghost/doc/AFPL/index.htm

Ah, that explains it (the link probably needs updating), though that page leads to http://sourceforge.net/projects/ghostscript/ then http://sourceforge.net/project/showfiles.php?group_id=1897 where there is a choice and the latest version is GPL Ghostscript 8.60. So you still end up on the same pages as going via http://pages.cs.wisc.edu/~ghost/.
[quote="Mr.Ocax"]they just gave me a bunch of winRar documents (I had to download each part of AFPL separately)....So, is there any particular site I can go to to download AFPL GS 8.54 all at one time?[/quote]As far as I'm aware there are no WinRar files and only [i]single[/i] compressed [i]alternative[/i] downloads with tar.bz2, tar.gz, .zip and .exe files, all but the last being marked as source files, so need compiling. Unless Linux users want to compile the program themselves, binary files for Linux are best found using the relevant Linux package manager rather than using the Ghostscript site.

Thank you.  no, I do not have a linux; I have windows, it makes things more straight forward.  Thanks for the updated links and the GPL info,

Mr.Ocax


inequalities
geometry
geometric transformation
reflection

Prove that the cinrcumradius of a triangle is at least twice the inradius.

[hide="One simple way"]Prove $ \sum_{cyc}cosA \equal{} 1\plus{}\frac{r}{R}\leq\frac{3}{2}$[/hide]

[hide="method 1"]Bloundon's[/hide]
[hide="method 2"]Euler's Formula[/hide]
[hide="method 3"]The nine-point circle's radius is at least the incircle's radius.[/hide]
[hide="method 4"]Ptolomey's Inequality on OBMC in triangle ABC with O being the reflection of A across BC.[/hide]


search
combinatorics unsolved
combinatorics

Set $ S\equal{}\{1, 2, 3,\dots 2004\}.$ We denote by $ d_{1}$ the number of subsets of $ S$ such that the sum of elements of the subset has remainder $ 7$ when divided by $ 32$. We denote by $ d_{2}$ the number of subset of $ S$ such that the sum of elements of subset has reaminder $ 14$ when divided by $ 16.$ Compute $ \frac{d_{1}}{d_{2}}.$

Let $ A \equal{} {1, 2, 4, 8, 16}$ and $ B \equal{} S \minus{} A$. For each subset $ X$ of $ B$, there is exactly one subset $ Y$ of $ A$ for which

$ \sum_{x \in A \bigcup B} \equiv 7 mod 32$

In fact, this is equivalent with $ \sum_{x \in A} x \equiv 7 \minus{} \sum_{x \in B}x mod 32$ and this presents a solution, because $ \sum_{x \in A} \equiv t mod 32$ always has a solution. This follows from the case $ 0 \le t < 32$, in which our choice gives the binary representation of $ t$.

Thus, $ d_1 \equal{} 2^{|B|} \equal{} 2^{1999}$. Similarly, we get $ d_2 \equal{} 2^{2000}$.

See also http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1867594669&t=39550.


It gets annoying to be smart sometimes because everybody asks you what you got after every test. They say, "what did you get this time - a 100?" and start to scoff. It is soooooooooo annoying. I was wondering if people in your school ever bothered you solely because you knew more than they did. Also, I can't have a moments peace because they keep asking you questions based on the assumption that you will always be able to provide them with the correct answer. I don't brag, but they just are annoying like that.

P.S. Visit the AMC MATHCOUNTS counting forum!

=)

You know, it's actually nice to be asked questions. Even when you can't give them satisfactory answers (which is often the case with me). But you learn so much more when you try teaching others.

Yay math-team-captains-teaching!

(Sorry for the unrelated nature of this post.)

Cus of my race they automatically assume that I am smart, and plus the people at my school ask for help<cheats>arhm...arhm.. help like I said, but they don't really care. They know it ticks me off so they do it.

I see it ur way I hate being annoyed by them. They ask me for my phone number so they can call me if they ever need help and it drives me up a wall. I politely tell them no Im not a 24\7 homework hotline. I hate it when they bother me when Im trying to get some reading done in homeroom and they assume that I can answer it and I dont mind because Im not busy as they put it. They assume that Im smart because the teachers voted me president of the NJHS and I made the mistake of telling one friend my report card and they brodcasted it all over the school. Now everyone comes up and asks me what did u make on ur report card and then smirk probably a 100 like u always do. I get real tired of it so I go to the library during homeroom now and sit in the comfy chairs and read at least Im not bothered their unless someone wants to check out a book.

Why? You're going to miss that attention when you lose it anyway, fame is like that. People did that to me and I was uncomfortable with it but now that it's over I look back and go *sigh* You'll see, you'll see...

I wouldnt miss the attention for that but if people stopped asking me for advice then I would miss that (I mean emotional advice.) If I dont know the answer I simply say go ask the teacher (sometimes I do that anyways even if I do know the answer hehehe...) I just dont like it when they bother me because it seems as if when I do help them they are ungrateful Ill talk some more tommorow Im in trouble its past my bedtime.

Yeah some are but then you can't win 'em all. I know that most I helped have helped me back in some way whether it was teaching me how to teach better :lol: or how to explain something better. You can always learn something by teaching others (and better understand it yourself).

BS? Does that mean make something up?

People try to annoy me, but never succeed.  Hey, I mean, why tease me, it proves them stupid and makes them look stupider than they actually are (which for some of them is hard to imagine :twisted:)

I'm rather being ignored than being annoyed. I'm just not popular even when I do good in school subjects and in other math contests. I am unpopular everywhere... I'm fated to be lonesome.[/img]

That is not necessarily true, you are lonesome because you believe you are lonesome when in fact if you were to open your eyes you'd see you have a lot of friends...that I figured out on my own :cry:

You don't have to be talented in order to be popular, just kind-hearted and helpful and just be nice.

bleumoose wrote:[hide]BS IT!!! they will never know the difference.[/hide]

im joking, of course

Ah, ha, ha.  Sure, you've never  done anything remotely like that before.

Personally, I say "yes" to half of the question and "no" to the other half.  When someone says "What did you get?  100?  *smirks*" or something like that, I'm irritated.  Also, sometimes people think they're complimenting you (like "you're way too smart"), but they're actually also making you feel "different".  That sort of thing gets annoying.

However, when people ask for help, I generally help them because 1) helping someone can be fun as long as they're not completely inept and 2) I want them to understand and see the beauty of math/science/whatever.  Plus, it gives you a warm feeling like hot cocoa on a snowy day.

No, Tare. They actually ignore me and I have no doubt about that. When I was in Taiwan a few years ago, I was repected. Now that I'm here, I'm feeling like there's a gap between Americans and me. I can seldom find people with commonalities. However, that doesn't mean I don't have friends, I'm just not popular, that's all.

MY problem is that people tease me way to much about it and when I get a grade like a A- thry go crazy and keep rubbing salt in my wounds. Also it's annoying now because people keep calling me "MATH-ew". :cry:

People used to tease me all the time but now they dont because if they do they figure they arent going to pass (I tutor them in math.) Plus it doesnt bother me anymore and even though Im smart Im what they consider I guess "cool" at my school. I dont really care what social status Im in at school. I used to be a zero. I may seem like the type that is goodie goodie but shhhhhh Im not. Im loud, noisy, late to class, and skipping down the hallway all the time outside the classroom that is. Yet I seem to make straight A's so the teachers like me. Thats all that matters is the teachers. The kids arent going to be the ones that enter my essay in a contest or grant me scholarships its the teachers so Im friends with all of the teachers except that of a few coaches who have I have grudges agianst (long story) and do not like me. Oh well just in the end dont worry about what the kids think of you just worry about you teachers and what they think.

Well among normal kids I'm cool, but when it comes to smart kids that I'm smarter than they kill me.

Heh, we were having a discussion in English about whether being the most popular or the smartest was better, I have now voted that the ones who said popular are utmostly stupid, so they have no clue what it means to be smart.

Here's a question:  how do you know that you know what it is to be smart?

I'm not trying to start a debate, but if you're afraid I am, there are a whole bunch of problems to be done in the Intermediate forum and a couple in AMC.

In my experience, a lot of smart people don't want to be popular.  For a long time, I didn't understand that.  In 7th and 8th grade I used to wish I were at least somewhat popular so that I wouldn't have to worry about being made fun of.  Now, I realize that being popular won't make you any less vulnerable - in fact, it'll probably make you more vulnerable.

The only reason guys want to be popular is because of girls, and after [hide]breaking up with my girlfriend I realized how shallow must girls are (no offense to raccoon, aalindsey, rippledance, or any other girls)[/hide]

So a word of advice to anyone: don't worry about popularity at all.

[quote="Tare"]However the second part I disagree, popularity does not necessarily mean a lot of girls it just means it gives you a lot of options to choose from when you want to make new friends.[/quote]

Actually, it almost always does mean "a lot of girls", at least temporarily.  And no, it doesn't give you more friends to choose from, because many of those "friendships" aren't true friendships.  When you're popular, everyone wants to be your friend, but once you become unpopular, you'll be deserted.  You're more likely to have real friends who like you because of who you are if you're not the most popular kid in the crowd.

[quote="aalindsey100"]Well wouldnt u rather have a girl like u for who u are? See I just dont get guys. They are so good with their friends but when it comes to girls they are clueless. Girls genrally dont care about how u look most of the time anyways (at least I dont.) But u seem to think we do. Yall are so complicated I just dont understand. By the way I like one of these people on here but Im going to keep it a secret so u can guess all u want but u aint going to ever know the answer.[/quote]
Yes, I would, and yes, she does. I must contest your statement that guys are so good with their friends. If you would like to be called names, punched in the shoulder, and then would like to go make something explode, all in good fun, then please, let the guys in your life know about that. Most girls would much rather have long talks and walks on the beach.

[quote="white_horse_king88"]Talking about girls... they don't really like smart people who are nerdy like. If you are truly smart, and you hide it to others and act normally around school, they don't annoy you. They don't think of you as different, or aloof from the main group of people. I like someone at my school, but she probably doesn't like me just because im different. That is when it sucks to be different.[/quote]
They don't? This is news.

I will tell them this but thats not what I prefer. What I like to do is something most guys would find interesting themselves. For fun Id rather go to a nascar race, play tackle football, basketball, do math and science, and most of all just hang out. I love it when there is no stress to impress just when the guys are themselves. I think guys have got girls all wrong. Yea sure some girls are romantic but most are rough around the edges like I am. Actually my brothers idea of a romantic date is candle light dinners with a real burning fire and my idea of a romantic date would be to go to an amusment park, bike riding, camping, hikeing, or like rock climbing. Now does that totally change ur perspective on girls. I guess thats a small town girl who doesnt like the big city just likes to get away. I just am not like most girls or what u guys think most girls are like.  By the way this guy he knows I like him or at least I think he does. Well I used to like him alot but now I like him more as a friend.

Ps. some of us like smart guys I know Ive never liked a real dumb guy before theyve all been smart. The person I liked last year looked sort of nerdy but he was cute and I wasnt the only one that thought so either. He ended up likeing me at the end of the year and then I told and he got mad. I told u I didnt understand guys.

[quote="confuted"]If you would like to be called names, punched in the shoulder, and then would like to go make something explode, all in good fun, then please, let the guys in your life know about that. Most girls would much rather have long talks and walks on the beach.[/quote]

Blowing things up is fun!  Hey, you can be a girl and still like boy things.  Besides, I detest one-on-one talks for a long time.  It puts too much pressure on both people and eventually you run out of things to say.  And you can't walk on the beach here anyway.  I'm landlocked.

[quote]I just remebered that lots of girls like guys who can dance really good.. and I'm not one of them. Is that true?
[/quote]

Man I cant even dance Im sure some girls do though.[/quote]

i can't dance. don't even try making me. the only time i did was for PE. it was required.

i like guys for their minds, not how popular they are. Most popular people are stuck-up and think that they are the center of teh universe. I actually start to strongly dislike ( hate, but i feel that word is too strong) them.

[quote]Talking about girls... they don't really like smart people who are nerdy like. If you are truly smart, and you hide it to others and act normally around school, they don't annoy you. They don't think of you as different, or aloof from the main group of people.[/quote]

that was a very sterotypical comment.

Dancing...in PE!? Boy that's even weirder than us (which is pretty weird: fencing :?:)

*sigh* The sad truth is, popularity=beautiful/cute=stuck-up=bad personality yet still you're attracted to them since they look so good...another reason how we have betrayed evolution...

yes, dancing in PE.

sixth: square dancing
seventh: slow dancing or folk dancing/line dancing ( this depends on what teacher you have)
eighth: cha-cha and salsa

my school is very screwed up.

[quote="aalindsey100"]By the way I like one of these people on here but Im going to keep it a secret so u can guess all u want but u aint going to ever know the answer.[/quote]

*checks memberlist for people from texas*

irollupstairs? juggle7? JRBoyd? gmann? brian_31407?  Krenium?  mafps?  Smartjock256?  Rep123max?  wow, there are a lot of people from Texas on AoPS.

It could be Fiery.  Or jojobird.   Or Neal.  Granted, none of them live in Texas, but it's a guess.

Popular people annoy me. when they talk to me I can't figure out whether they are trying to make me look bad, if they are doing it as a joke, or if they are just doing it to annoy me. Yes. Popular people are really stuck-up. They think they are above everybody else, and that annoys me the most.

You know that's contradictory to your signature :lol:

Though I have to agree, when I look back at how I was when I was popular, I cry out in disgust. I was actually saying stuff like "Yeah I am better at math than anybody else in the world" :-P *scrubs tongue with soap*


Find the max value of $K$, such that:

$\lfloor \frac{x^2+6x+14}{x^3+27} \rfloor \geq K$; if $x \in [-2;2]$

[quote="deej21"][hide]
So we pretty much just need the floor of the smallest value of $\frac{x^2+6x+14}{x^3+27}, x \in [-2,2]$
The numerator has no real roots, and it thus always positive.  The denominator is negative for $x < -3$, thus positive on the interval.  Then the expression is always positive.  However, for $x=0$, the expression is $\frac{14}{27}$.  So the minimum value $m$ must satisfy $0 < m \le \frac{14}{27}$, and we have the floor of $m$ is 0.
[/hide][/quote]

we want max value  ;)

[quote="Altheman"]we want max value  ;)[/quote]

You want the maximum of $K$ -- he showed that the given expression is never negative (so that $K \geq 0$) but that $\frac{x^2+6x+14}{x^3+27}$ is sometimes less than 1, so that $\lfloor \frac{x^2+6x+14}{x^3+27} \rfloor$ is sometimes 0, so that $K \leq 0$.

The problem seems odd to me, because it's not particularly a close question: the given function is [i]always[/i] 0 over the interval $[-2, 2]$.

Let ABC be a triangle and the rectangular MNPQ is inscribed ABC(M belong to AB, N belong to AC, P and Q belong BC).Prove that
$\ S_{ABC}\geq\ 2S_{MNPQ}$  ;)

Math is my life :rotfl:

:blush: Oh,sorry I have a mistake, I have pressed the post reply buttom but I have to press new topic.

Math is my life :!:


function
integration
real analysis
real analysis unsolved

Let $ h : \mathbb{R} \rightarrow \mathbb{R}^\plus{}$ be a decreasing function. Prove that $ ( \int_0^{\infty} \sqrt{h(x)}dx)^2 \geq 2 \int_0^{\infty} xh(x) dx .$

Maybe try Riemann sums and use the following: If $ a_1 \geq a_2 \geq ... \geq a_n\geq0$, then $ (\sum a_i)^2 \equal{} \sum a_i^2 \plus{} 2\sum_{i<j} a_ia_j \geq \sum a_i^2 \plus{} 2\sum_{i<j} a_j^2$.


I am wondering if there is a good website for latex use on wordpress. I dont like how you have to put the word \$latex \$ at the start of every dollar sign, for example, and I'm wondering if there is a way to get around this.

Los

Are you using a WordPress.com hosted blog? If so then there's no way round it as that is how the hosts have set it up. But if you have your own WordPress blog then see [url=http://sixthform.info/steve/wordpress/]Using LaTeX in WordPress[/url] or edit the plugin you may be using.


[hide] Clearly, if the two absolute values are both positive or negative, the equation will have no solutions. Therefore, we may confine ourselves to two cases.

Case 1: $ |x\minus{}3|\equal{}\minus{}(x\minus{}3)$
Doing this yields $ 6\minus{}2x\equal{}12 \implies x\equal{}\minus{}3$.

Case 2: $ |3\minus{}x|\equal{}\minus{}(3\minus{}x)$
Doing this yields $ 2x\minus{}6\equal{}12 \implies x\equal{}9$.

Thus, the two solutions are $ x\equal{}\minus{}3$, $ x\equal{}9$[/hide]

[hide="Alternate solution"]
$ |x \minus{} 3| \plus{} |3 \minus{} x| \equal{} |x \minus{} 3| \plus{} |x \minus{} 3| \equal{} 2|x \minus{} 3| \equal{} 12\implies |x \minus{} 3| \equal{} 6\implies x \equal{} \minus{}3,9.$
[/hide]


geometry
calculus
analytic geometry
geometric transformation
trigonometry
calculus computations

Let $A$ be a point on the circle with radius $r$ and $P$ another point on this circle. Choose the point $Q$ on the circle's tangent in $A$ so that the length of $AQ$ is equal to the length of $AP$. $R$ is the point where the line through $P$ and $Q$ intersects the normal to $AQ$ in $A$. Determine the position* of $R$ as $P$ goes to $A$.

I have a very messy solution that shows that $R$ should be at a point $4r$ away from $A$. The answers say $3r$, but I think there is a typo in the answers. I'd like to see how a nice solution to this problem looks. I will refrain from posting mine because it is very ugly.

edit: The question gives the hint to introduce a 'suitable' coordinate system, but that's fairly obvious, because how else would you do it?

*The literal translation of the swedish word would be "limit position", i.e. we want the point which R tends to.

I also got $|AR|=4r$. It is mostly an elementary geometry problem. By looking at the triangles $AOP$ ($O$ is the center of the circle), $AQP$ and $AQR$, it is easy to see that the angle $ARQ$ is $\frac 14$ of the angle $\alpha=AOP$. This allows to write $|AR|=\frac{2r\sin\frac\alpha 2}{\tan \frac\alpha 4}$. To take the limit as $\alpha\to 0$ is now easy.

P.S. I figured out where $3r$ comes from: this is what you get if you take $|AQ|$ equal to the length of the arc $AP$ instead of the length of the segment $AP$. The problem becomes less elementary then...


projective geometry
geometry proposed
geometry

$\Gamma$ is a circle and $O$ a point outside $\Gamma$. The tangents through $O$ touch the circle at $A$ and $B$. The parallel to $(OB)$ through $A$ intersects $\Gamma$ at $A$ and $C$. $[OC]$ intersects $\Gamma$ at $E$. $(AE)$ and $(OB)$ intersect at $K$. Prove that $K$  is the midpoint of $[OB]$.

It looks very familiar. I think it's been posted before.

Anyway, let $T=AB\cap OC$. Since $AB$ is the polar of $O$ wrt the circle, we have $(O,T;E,C)=-1$, so $(AO,AT;AE,AC)=-1$ (those four lines form a harmonic quadruple). Since their intersections with $OB$ must also form a harmonic quadruple, and, since $AC\|OB$, $AC\cap OB$ is the infinity point of $OB$, its harmonic conjugate wrt $O,B$, which is $AE\cap OB$, must be the midpoint of $OB$, Q.E.D.

That's neat, Grobber! A more elementary solution:

We get $AK \cdot EK = BK^2$ by power of a point. On the other hand we have $\angle{EOK} = \angle{ACE} = \angle{OAE}$ and hence triangles $AOK$ and $OEK$ are similar, giving $OK^2 = AK \cdot EK = BK^2$, as required.

Thanks Arne for this pretty solution. I had the same solution as Grobber and thought we could do something else :-)


$ f: R \to R$ satisfies the following for every real number a, b.

$ (a\minus{}b)f(a\plus{}b)\minus{}(a\plus{}b)f(a\minus{}b)\equal{}4ab(a^2\minus{}b^2)$

find $ f(8)\minus{}2f(9)\plus{}f(10)$.

Thanks.

Let $ p\equal{}a\plus{}b,q\equal{}a\minus{}b, g(x)\equal{}\frac{f(x)}{x}$ for $ x \neq 0$, obviously $ p,q$ can take any value.

Now for $ p,q \neq 0$, we divide both sides of the equation by $ a^2\minus{}b^2$, we get
$ g(p)\minus{}g(q)\equal{}4ab\equal{}p^2\minus{}q^2$ for any non-zero $ p,q$,
So $ g(x)\minus{}x^2\equal{}k$, where $ k$ is a constant for $ x \neq 0$, so $ f(x)\equal{}x^3\plus{}kx$ for all non-zero $ x$

$ f(8)\minus{}2f(9)\plus{}f(10)\equal{}(8^3\plus{}8k)\minus{}2(9^3\plus{}9k)\plus{}(10^3\plus{}10k)\equal{}24$

the idea is simpler .Consider three equations in which (a,b)are (5,3);(6,4) and (3/2,1/2) respectively.


rotation
calculus
ARML
projective geometry
geometry
3D geometry
sphere

The circle $(x-2)^{2}+(y+4)^{2}=1$ is rotated about the line $y=1$.  Find the volume of the solid formed.

PS no calculus allowed.

The volume of a torus can't be rigorously established without calculus.   :(

this was given on a chicago arml tryout.  in parenthesis it says theorem of pappus.  i searched a bit for this, but i didn't find anything related to a problem like this.  any ideas?

I think theorem of Pappus says that the volume of a rotational body is equal to the product of the area of the lamina and the lenght of the circle described by the centroid of the lamina. Confer [url]http://mathworld.wolfram.com/PappussCentroidTheorem.html[/url]

Since the centroid of a circle is its center, in this case we would have $(1^{2}\pi)(2\cdot 5\pi)=10\pi^{2}$

And the Theorem of Pappus can't be proven rigorously without calculus, as far as I know.   :mad:

And the formula $V={4\over 3}r^{3}\pi$ for the volume of a sphere [b]can[/b] be proven rigorously without calculus?!  :roll: Yet that doesn't stop us from learning it early on and using it freely.

Yep, it was a cheap trick for the tryout to throw this theorem at ya, but at least you'll probably remember it once you hit Mu.

Basically, area of figure * distance the center (centroid) travels during rotation.


The 24th day of a particular month is a Saturday. On what day of the week was the first day of that month?

[hide] If the $ 24$th day was Saturday, then so was the $ 3$rd. So, the $ 1$st day was a Thursday.[/hide]


algebra
polynomial
calculus
derivative
calculus computations

If $f(\alpha)=f'(\alpha)=f''(\alpha)=0$ and $f(\beta)=f'(\beta)=f''(\beta)=0$where $f(x)$ is a polynomial of degree 6,
proove that all the roots of $f''(x)=0$ are real.

edit:I just came to know about the error I made in the question.I have just edited it.
Really sorry for the inconvinience.Plz. help me with this.

[quote="Anurag Agrawal"]If $f(\alpha)=f'(\alpha)=f''(\alpha)=0$ and $f(\beta)=f'(\beta)=f''(\beta)=0$where $f(x)$ is a polynomial of degree 6,
proove that all the roots of $f''(x)=0$ are real.
[/quote]
I suppose $\alpha \neq \beta$
From $f'(\alpha)=f'(\beta)$ I deduce that there is $\alpha<\gamma<\beta,\ f''(\gamma)=0$
So $f''(x)$ is a real polynomial of degree $4$ with $3$ real roots, so the last root have to be real (q.e.d)
:cool:

Another ideea would be this:

$f(\alpha)=f'(\alpha)=f''(\alpha)=0$ implies that $f(x)=(x-\alpha)^{3}Q(x)$ for some polynomial $Q$ of degree 3.
On the other hand the other condition implies that $f(x)=(x-\beta)^{3}R(x)$ for some polynomial $R$ of degree 3.
Now these imply that $f(x)=A(x-\alpha)^{3}\(x-\beta)^{3}$ where $A$ is a constant.
It is not hard to see that the second derivative of this polynomial has all real roots.
Didi

well you know that $f''(a)=f''(b)=0$ so those are already 2 real roots of $f''(x)$
then from the mean value theorem you have that there is a $c\in[a,b]$ such that
$f''(c)=\frac{f'(b)-f'(a)}{b-a}=0$
and since complex roots come in pairs it is impossible for there to be only one, thus the other root must also be real.
or you could use the mean value theorem in the form $f'(d)=\frac{f(b)-f(a)}{b-a}=0$ to prove there there is another point where $f'(x)$ is 0 and apply the mean value theorem w/ this point and one of the other points.

*note: if $d=c$ then this is a double root since $f'(c)=f''(c)$


I am having a problem with getting a problem with a local dipository while i am downloading the Latex. Does any one know why this is happening?

I think that you're going to have to provide us with more detailed information -- are you using MiKTeX? are you using Windows? what error specifically did you get, and what were you trying to do when you got it? -- in order for anybody to be able to help you.


I try to render the following integrals, but the vertical bar is too short.  Please help.

	\[  \int^{y}_{3}rdr = \int^{t}_{2}\frac{2s}{1+s^{2}}ds. \]

	\[  \frac{r^{2}}{2}|^{y}_{3}= \ln|s^{2}+1\parallel ^{t}_{2}, \]

[quote="1dot6180"]I try to render the following integrals, but the vertical bar is too short.  Please help.
\[ \int^{y}_{3}rdr = \int^{t}_{2}\frac{2s}{1+s^{2}}ds. \]

\[ \frac{r^{2}}{2}|^{y}_{3}= \ln|s^{2}+1\parallel^{t}_{2}, \]
[/quote]

Use the \left and \right parameters.
\[ \left.\frac{r^{2}}{2}\right |^{y}_{3}..\]

Use \left and \right to get |or [ ] or { } or ( ) to be the right height round the text whose height you want to match. That's OK for [...] but in this case you don't want anything on the left so you use\left[size=150][b][color=red].[/color][/b][/size] ...\right|
\[ \left.\frac{r^{2}}{2}\right|^{y}_{3}= \left.\ln|s^{2}+1|\right|^{t}_{2}, \]
or you can insist they are both the same height by moving \left. so \left ... \right surround both sides of the equation
\[ \left.\left.\frac{r^{2}}{2}\right|^{y}_{3}= \ln|s^{2}+1|\right|^{t}_{2}, \]
Personally I prefer the British notation :)
\[ \left[\frac{r^{2}}{2}\right]^{y}_{3}= \left[\ln|s^{2}+1|\right]^{t}_{2}, \]

:D Then stevem edited his post at 3:30 :D

Thanks.  Here is what I got.  The next question is how to make it pretty as the bars on both sides are not the same height.

$  \left.\frac{r^{2}}{2}\right|^{y}_{3}= \left.\ln|s^{2}+1| \right|^{t}_{2},$

As I said in my post move the \left. so there are two at the start
\[ \left.\left.\frac{r^{2}}{2}\right|^{y}_{3}= \ln|s^{2}+1|\right|^{t}_{2}, \]

[quote="stevem"]
Personally I prefer the British notation :)
\[ \left[\frac{r^{2}}{2}\right]^{y}_{3}= \left[\ln|s^{2}+1|\right]^{t}_{2}, \]
[/quote]
I think that is also the french notation, because we use it here in Morocco!


Let ABC be a triangle and M a point in his interior.Prove that \[ MA\plus{}MB\plus{}MC \geq 2(MA'\plus{}MB'\plus{}MC')\], where MA',MB',MC' are the proiections from M to the sides of the triangle.

[quote="balan razvan"]Let ABC be a triangle and M a point in his interior.Prove that
\[ MA \plus{} MB \plus{} MC \geq 2(MA' \plus{} MB' \plus{} MC')
\]
, where MA',MB',MC' are the proiections from M to the sides of the triangle.[/quote]
It's true too for $ MA',$ $ MB'$ and $ MC'$ are bisectors of $ \Delta BMC,$ $ \Delta AMC$ and $ \Delta BMA.$  :wink:

sounds nice...could you post a proof please?:D thanks

Well, if you don;t post the proof, I'll post it...
Thru M we construct an antiparalel with BC. $ <AEF=<ACB=\alpha$ and $ \Delta AEF ~ \Delta ACB$ from where \[ \frac{AE}{AC}=\frac{EF}{CB}=\frac{AF}{AB}=k\] \[ S_AEF = S_AME + S_AMF=\displaystyle\frac{MC' \cdot AE}{2} + \displaystyle\frac{MB' \cdot AF}{2} = \displaystyle\frac{h \cdot EF}{2} <= \displaystyle\frac{AM \cdot EF }{2}\] from where \[ AM \cdot EF \geq MC' \cdot AE+MB' \cdot AF\] \[ AM \cdot CB \geq MC' \cdot AC + MB' \cdot AB\] In the same way we have \[ BM \cdot AC \geq MA' \cdot AB + MC' \cdot BC\] \[ Cm \cdot AB \geq MB' \cdot BC + MA' \cdot AC\] and results that \[ Bm \geq MC' \cdot AC/BC + MB' \cdot AB\BC\]
\[ BM \geq MA' \cdot AB\AC + MC' \cdot BC\AC\]
\[ CM \geq MB' \cdot BC\AB + MA' \cdot AC\AB\]
\[ AM+BM+MC \geq 2(MA'+MB'+MC')\]

Dear Mathlinkers,
you can finf it on
American mathematical Monthly june 1935 Problem 3740 p. 396

I wrote an article on this result in
Reperes, IREM n° 35 avril 1999 p. 59-63.

Sincerely
Jean-Louis


One of my friends sent me this problem.
It is not difficult,but I feel that it's interesting.So I'd like to see some other solution.
For positive reals $x,y,z$ such that $x^2+y^2+z^2 \leq 3$
Find the maximum of $xy+yz+xz-xyz$

This is my solution.
The maximum is 2.
Assume both $x,y$ are no less or no greater than 1.
Then,$z(x-1)(y-1) \geq 0 \iff xz+yz-xyz \leq z$
Thus,we need to prove that $xy+z \leq 2 \leftarrow \frac{x^2+y^2+z^2}{2}-\frac{(z-1)^2}{2} \leq \frac{3}{2}$,
which is trivial.

Thanks.And I'll be most delighted if there are other prooves posted.

I'v solved a problem similar to this one but the condition is$x^2+y^2+z^2=1$
I'll try the same method again,if suceeding,I'll post it

No other people want to have look at this ineq? :?


[hide="If you don’t know what this is"]
Go
http://www.mathleague.com
or
http://www.mathleague.com/reports/understanding.htm[/hide]
If you do, how did you do?

I lost to Kerry  :(
Again.
And I'll probably do so next year as well.


Could somebody explain me and give me the definition of x being the convex combination of points and directions in S?

I have never heard of such a thing. My guess: $x$ is called a convex combination of points $z_{1},\dots,z_{n}$ and directions $\zeta_{1},\dots,\zeta_{m}$, if there exist
1) nonnegative numbers $\lambda_{1},\dots,\lambda_{n}$ whose sum is $1$
and
2) nonnegative numbers $\mu_{1},\dots,\mu_{n}$
such that
$x=\sum_{i=1}^{n}\lambda_{i}z_{i}+\sum_{i=1}^{m}\mu_{i}\zeta_{i}$.


because this was moved to intermediate before i corrected my typo, ill post it again.
Find all x that x^(x^5-13x^3+36x)=1

[hide]first of all, state the obvious. x=1 will work. also, -1 will work because the exponent will always be even for all integers (except 0).

next, we can factor 5x^5-13x^3+36x to 5x^4-13x^2+36, because if we get the exponent to equal 0, then the whole thing will equal 1. if you make x^2=y, then the equation is 5y :^2: -13y=36. the discriminent of this is 169-5*36*4, or 169-720. that comes out to -551. if we're just solving over the real number plane, it'd only be 1 and -1. there would be different answers if we were solving over the complex number plane, but i'm pretty sure that's not getting started. this leaves us with 1 and -1[/hide]

alrighty then

[hide]first of all, the obvious, 1. -1 also works because the exponent will always be even for all integers except for 0.

the expression x^(x^5-13x^3+36x) will =1 if the exponent is equal to 0.

factor x^5-13x^3+36x to x^4-13x^2+36

you can factor that to (x^2-9)(x^2-4)

take (x^2-9) and separate into (x+3)(x-3) and take (x^2-4) and factor to (x+2)(x-2)

the end result is x(x+3)(x-3)(x+2)(x-2)

thus, the terms that will satisfy the equation for x are -3,-2,-1,1,2,3[hide][/hide][/hide]

[hide]x^5-13x^3+36x factors to x(x+2)(x-2)(x+3)(x-3)

when the exponent is 0, the equation works, so x=0, 2, -2, 3, -3

oops, you have to take out 0

also x=1, -1 because 1 to any power is 1...

x=1, -1, 2, -2, 3, -3[/hide]


[b]Thirteen black, twenty green and eighteen purple beads are placed in a bowl. [/b]

	Someone with nothing else to do in their life draws a bead 11,543 times and throws the bead back in after each draw. If you then draw one more  bead, what is the probability that you draw a green bead?

ahahah thats a funny problem.

20 / (13+18+20) = 20/51

[quote="eggwad"]What is the probability of drawing a purple bead on the fourth trial if a black bead was drawn on each of the first three trials and the beads were not put back in?[/quote]

If a black bead was drawn on each of the first three trials then there are 13-3=10 black left
P(purple)= 18 / (10+20+18)= 18/48

=3/8
But yeah, simplification is not needed, you got the idea, i just needed to say something or the other XD


linear algebra
matrix
induction
pigeonhole principle
combinatorics unsolved
combinatorics

Let $P_1,P_2,\dots,P_n$ be $n$ distinct points over a line in the plane ($n\geq2$). Consider all the circumferences with diameters $P_iP_j$ ($1\leq{i,j}\leq{n}$) and they are painted with $k$ given colors. Lets call this configuration a ($n,k$)-cloud.

For each positive integer $k$, find all the positive integers $n$ such that every possible ($n,k$)-cloud has two mutually exterior tangent circumferences of the same color.

[hide="Solution"]We claim that the values of $ n$ such that every $ (n, k)\minus{}\text{cloud}$ has two mutually exterior tangent circumferences of the same color are the naturals that satisfy $ n\geq 2^k\plus{}1$.

Let us identify a $ (n, k)\minus{}\text{cloud}$ with a $ n\times n$ matrix with entries in $ \{1,\cdots,k\}$ (without the diagonal), being the element $ a_{i,j}$ the color of the circle with diameter $ P_iP_j$ (considering WLOG that the set of colors is $ \{1,\cdots,k\}$).

Then a $ (n, k)\minus{}\text{cloud}$ doesn't have two mutually exterior tangent circumferences of the same color [i][b]iff[/b][/i] $ \forall \, i$, $ \{a_{i,1},\cdots,a_{i,i\minus{}1}\}$ e $ \{a_{i,i\plus{}1},\cdots,a_{i,n}\}$ are disjoint sets.

Observation: For us to prove that it is possible to have a $ (n, k)\minus{}\text{cloud}$ without two mutually exterior tangent circumferences of the same color for $ n\leq 2^k$ it is enough to construct such an example for $ n\equal{}2^k$ (because from that example we immediately get such a cloud for smaller values of $ n$). We will prove this by induction on $ k$.

[i][b]Base case[/b][/i] - $ k\equal{}1$

$ \begin{bmatrix} \times & 1  \\ 1 & \times \end{bmatrix}$

[i][b]Induction step[/b][/i]:

Suppose that we have a matrix $ M$ for $ k$, and let $ N\equal{}\begin{bmatrix} k\plus{}1 & k\plus{}1 & \cdots & k\plus{}1 \\ k\plus{}1 & k\plus{}1 & \cdots & k\plus{}1 \\ \vdots & \vdots & \ddots & \vdots \\  k\plus{}1 &  k\plus{}1 & \cdots & k\plus{}1\end{bmatrix}$ be a $ 2^k\times 2^k$ matrix with all entries being $ k\plus{}1$.

Then $ \begin{bmatrix} M & N  \\ N & M \end{bmatrix}$ is a matrix for $ k\plus{}1$, since it is symmetric and if $ i\leq 2^k$, then $ \{a_{i1},\cdots,a_{i,i\minus{}1}\}$ and $ \{a_{i,i\plus{}1},\cdots,a_{i,2^k}\}$ are disjoint by the induction hypothesis, and $ k\plus{}1\not \in \{a_{i,1},\cdots,a_{i,i\minus{}1}\}$ (the case $ i>2^k$ is analogous).

We will prove now that for any such matrix $ n\times n$ with $ n\geq 2^k\plus{}1$ there exists $ i$ such that $ \{a_{i,1},\cdots,a_{i,i\minus{}1}\}$ and $ \{a_{i,i\plus{}1},\cdots,a_{i,n}\}$ are [b]not[/b] disjoint.

Because the matrix is symmetrical, that is equivalent to $ \{a_{i,1},\cdots,a_{i,i\minus{}1}\}$ e $ \{a_{i\plus{}1,i},\cdots,a_{n,i}\}$ having a common element.

Consider the following sets:

$ \{a_{2,1}\}$
$ \{a_{3,1},a_{3,2}\}$
$ \cdots$
$ \{a_{n,1},\cdots,a_{n,n\minus{}1}\}$

Since there are $ n\minus{}1>2^k\minus{}1$ sets and $ \{1,\cdots,k\}$ only has $ 2^k\minus{}1$ non-empty subsets, by the pigeonhole principle we have that two of those sets are equal. Let they be $ \{a_{i1},\cdots,a_{i,i\minus{}1}\}$ e $ \{a_{j1},\cdots,a_{j,j\minus{}1}\}$, $ j>i$. Then $ a_{j,i}$ is in both sets (because they are equal), so $ \{a_{i1},\cdots,a_{i,i\minus{}1}\}$ e $ \{a_{i\plus{}1,i},\cdots,a_{j,i},\cdots,a_{n,i}\}$ are not disjoint, as desired.[/hide]


graph theory
combinatorics proposed
combinatorics

Let $G = (V_1, V_2, E)$ be a connected bipartite graph whose vertices are marked with integers. Thus with $G$ we have a mapping $f: V_1 \cup V_2 \mapsto \mathbb{Z}$ and with $x \in V_1 \cup V_2$ let $f(x)$ be the number assigned to the vertex $x$. Let $S_1 = \sum_{x \in V_1} f\left(x\right)$ and $S_2 = \sum_{x \in V_2} f\left(x\right)$.

Define $B$ as a step in which the numbers/markings $f(x)$ and $f(y)$ of the vertices $x,y$ of an edge $\{x,y\} \in E$ are substituted by the values $f(x) + a$ and $f(y) + a$, respectively, where $a \in \{-1,1\}$.

Define $Z_2$ as the state where all the vertices of the bipartite graph have the number/marking 0.

Under the assumption that we have $S_1 = S_2$ prove that state $Z_2$ can be reached by a finite application of steps $B$.

A related problem goes as follows:  Let $m,n \geq 2$ be integers. To each of the fields of a  $m \times n$ checker board a non-negative integer is assigned. A elementary operation is defined by adding an integer to two neighboring numbers such that in both fields the result is non-negative. Provide a necessary and sufficient condition that by a finite number of elementary operations we obtain a state where 0 is assigned to every field on the checker board. Two fields are called neighbors if they have an edge in common.

[quote="seshadri"]some things seem to be missing....
probably you also need the graph to be connected (for instance in a graph with no edges,no steps are possible).[/quote]

It was obvious that I forgot to add the graph is connected but now the statement should be complete.  :)

It is very simple, isn't it?
By removing some edges from $G$ we can obtain tree. Choose arbitrary vertex $x$ of degree 1 and by means of elementary operations make it 0, now exclude it from consideration. We again have tree and can repeat such procedure, and so on.
Finally, we will obtain tree with only vertex. If $S_1=S_2$ then its marking should be 0.

In order to illustrate the connections between those checker board problems and graphs and the mapping among them I show an example. Depending on the definition of neighborhood we have:

a.) shows the mapping of checker boards to a graph if neighborhood is defined as the property two fields have an edge or a vertex in common

b.) shows the mapping of checker boards to a bipartite graph if neighborhood is defined as the property two fields have an edge in common

c.) shows the mapping of checker boards to a bipartite graph if neighborhood is defined as the property two fields  have a vertex (diagonals) in common

attachment c.) And sorry I was too lazy to prepare a properly drawn diagram. Maybe next year.  :P

In order to see the content of the images better just store it to your hard disk or open the image in a new window (right mouse button -> open in new window).


[quote="i_like_pie"]Solve the following by substitution: $ \frac {1}{x(x \plus{} 2)} \plus{} \frac {1}{(x \plus{} 1)^2} \equal{} \frac {1}{12},\;x\in\mathbb{Z}^ \plus{}$.[/quote]
[hide] Say that $ y \equal{} (x \plus{} 1)^2$.  Thus, we have that
\[ \frac {1}{y \minus{} 1} \plus{} \frac {1}{y} \equal{} \frac {1}{12}
\]

\[ \frac {2y \minus{} 1}{y(y \minus{} 1)} \equal{} \frac {1}{12}
\]

\[ 24y \minus{} 12 \equal{} y^2 \minus{} y
\]

\[ y^2 \minus{} 25y \plus{} 12 \equal{} 0
\]

\[ y \equal{} \frac {25\pm \sqrt {625 \minus{} 48}}{2}
\]

\[ y \equal{} \frac {25\pm \sqrt {577}}{2}
\]

\[ \boxed{x \equal{} \minus{} 1\pm \sqrt {\frac {25\pm \sqrt {577}}{2}}}
\]
[/hide]
EDIT: Ah, okay, that makes life quite a bit easier...
[hide="Revised Solution"]
Again, we substitute $ (x \plus{} 1)^2 \equal{} y$ to obtain that
\[ \frac {1}{y \minus{} 1} \minus{} \frac {1}{y} \equal{} \frac {1}{12}
\]

\[ \frac {1}{y(y \minus{} 1)} \equal{} \frac {1}{12}
\]

\[ y^2 \minus{} y \equal{} 12
\]

\[ (y \minus{} 4)(y \plus{} 3) \equal{} 0
\]

\[ y \equal{} 4, \minus{} 3
\]
y cannot be -3 or else x would be imaginary.  Thus,
\[ (x \plus{} 1)^2 \equal{} 4
\]

\[ x \plus{} 1 \equal{} 2, \minus{} 2
\]

\[ \boxed{x \equal{} 1, \minus{}3}
\]
[/hide]

Sorry, the plus was supposed to be a minus. :ninja:

[quote="i_like_pie"]Solve the following by substitution: $ \frac {1}{x(x \plus{} 2)} \minus{} \frac {1}{(x \plus{} 1)^2} \equal{} \frac {1}{12},\;x\in\mathbb{Z}^ \plus{}$.[/quote]
[hide="solution"]
Let $ s$ = $ x \plus{} 1$
$ \frac {1}{(s \minus{} 1)(s \plus{} 1)} \minus{} \frac {1}{s^2} \equal{} \frac {1}{12}, \;s \ne \pm 1$
$ \frac {1}{s ^ 2 \minus{} 1} \minus{} \frac {1}{s^2} \equal{} \frac {1}{12}$
$ 12(s^2)(s^2 \minus{} 1) \cdot \left[ \frac {1}{s ^ 2 \minus{} 1} \minus{} \frac {1}{s^2} \equal{} \frac {1}{12} \right]$
$ 12s^2 \minus{} 12(s^2 \minus{} 1) \equal{} s^2(s^2 \minus{} 1)$
$ 12s^2 \minus{} 12s^2 \plus{} 12 \equal{} s^4 \minus{} s^2$
$ s^4 \minus{} s^2 \minus{} 12 \equal{} 0$
$ (s^2 \minus{} 4)(s^2 \plus{} 3) \equal{} 0$
The only real roots are in $ (s^2 \minus{} 4) \equal{} 0$
$ s^2 \minus{} 4 \equal{} 0$
$ (s \plus{} 2)(s \minus{} 2) \equal{} 0$
$ s \equal{} \pm2$
$ s \equal{} 2$ $ OR$ $ s \equal{} \minus{} 2$
$ x \plus{} 1 \equal{} 2$ $ OR$ $ x \plus{} 1 \equal{} \minus{} 2$
$ x \equal{} 1$ OR $ x \equal{} \minus{} 3$
$ \boxed{x \equal{} \minus{} 3,1}$
[/hide]

[quote="i_like_pie"]ProtestanT's solution is correct, except for one little mistake...

[hide="Solution"]Given: $ \frac {1}{x(x \plus{} 2)} \minus{} \frac {1}{(x \plus{} 1)^2} \equal{} \frac {1}{12}$.

Expand the denominators to get $ \frac {1}{x^2 \plus{} 2x} \minus{} \frac {1}{x^2 \plus{} 2x \plus{} 1} \equal{} \frac {1}{12}$

Substitute $ u \equal{} x^2 \plus{} 2x$. The equation now becomes $ \frac {1}{u} \minus{} \frac {1}{u \plus{} 1} \equal{} \frac {1}{12}$.

$ 12(u \plus{} 1 \minus{} u) \equal{} u(u \plus{} 1)\quad\to\quad 12 \equal{} u^2 \plus{} u\quad\to\quad (u \minus{} 3)(u \plus{} 4) \equal{} 0$

$ u$ is either $ \minus{} 4$ or $ 3$. The former does not yield real solutions for $ x$, but the latter gives $ x \equal{} \{ \minus{} 3,1\}$.

Remembering that $ x\in\mathbb{Z}^ \plus{}$, we rule out $ x \equal{} \minus{} 3$.

Answer: $ x \equal{} \boxed{1}$.[/hide][/quote]
I didn't get what the uppercase Z meant. Could you explain?

$ \mathbb{Z}$ means the set of integers

$ \mathbb{R}$ means the set of real numbers

$ \mathbb{Z}^\plus{}$ means positive integers

$ \mathbb{Z}^\minus{}$ means negative integers.

etc...

$ \mathbb{Z^\plus{}}\equal{}\mathbb{N}$ isnt it ?

Yes and no. $ \mathbb{N}$ generally does mean the same as $ \mathbb{Z}^\plus{}$, but there are some people who define $ \mathbb{N}$ to be $ \mathbb{Z}^* \equal{} \mathbb{Z}^\plus{}\cup\{0\}$ - the nonnegative integers. Use $ \mathbb{Z}^\plus{}$ and $ \mathbb{Z}^*$ rather than $ \mathbb{N}$ for clarity.

[quote="binomial_4eva"]$ \mathbb{Z^ \plus{} } \equal{} \mathbb{N}$ isnt it ?[/quote]

Somewhere $ \mathbb{N}\equal{}\mathbb{Z}^\plus{}$, and elsewhere $ \mathbb{N}\equal{}\mathbb{Z}^\plus{}_0$

In the system A, the set B is denoted $ \mathbb{N}_0$, and in the system B, the set A is denoted $ \mathbb{N}^*$

EDIT: See above