MIT
college
Princeton
Harvard
calculus
Stanford
University of Chicago

i have been looking at many different colleges and am interested in majoring in mathematics.  however, not many colleges that i have fould talk about their math departments. does anyone have any sugesstions or names of colleges with great math departments.

1. MIT
1. Princeton
2. Harvard
3. Berkeley
4. U Michigan
5. Rutgers

I am applying to those because I believe they have the best math departments, but I bet that If we go to Russia, Romania, or someplace in eastern europe we would learn more math, by less money, but I am not as crazy to go to a place where I don't speak the main language, LOL.

[quote="manuel"]
I am applying to those because I believe they have the best math departments, but I bet that If we go to Russia, Romania, or someplace in eastern europe we would learn more math, by less money, but I am not as crazy to go to a place where I don't speak the main language, LOL.[/quote]

Certainly less money, but the 'more math' is not so clear.  There's definitely a dissonance between US high schools and top high schools abroad, but I haven't seen any evidence that this is the case in undergraduate or graduate education.

true, but a lot of the professors in the US's universities come from eastern europe. i think that in this countries mathematics is part of the culture. what is cultural for the US? business?
but anyways, i think we can create our proper small culture by communicating the mathematical and artistical curiosity and apprecation to the ones that are near to us

wow most of the schools that manuel listed earlier i dont think that i am capable of getting accepted into. but it cant hurt to try. :?

maokid7, don't worry. Rutgers, UMichigan, and Berkeley have great mathematics departments, and its not hard to get into those.

This thread would be very welcome on the [url=http://www.artofproblemsolving.com/Forum/index.php?f=143]College Forum[/url], if the moderator here would kindly care to move it.

How are we defining "great math department" for undergraduates?  (Really what I'm asking is, what qualities are you looking for in a department?  There are lots of great departments, but they're all great for different reasons.)  I'd definitely add St. Olaf and Swarthmore to the list, for numbers of math majors and depth of program respectively.

Reed College
Harvey Mudd College
California Institute of Technology
Rose-Hulman Institute of Technology

Cal Tech and Reed College both have very strong math and science programs and are among the top 10 all-around schools in the nation. Reed is actually probably the best college in the USA, but most people have not heard of it because they don't let reviews rank their school. Harvey-Mudd College is an excellent college with a strong math department and is part of a consortium of colleges in southern cali. Rose-Hulman also has an awesome math department and has had the #1 undergrad engineering program in the nation for 7 consecutive years.

The drawback is that all of these school are very expensive . . . though definitely worth it if you can dish out the dough! Cal Tech also has one of the best financial-aid programs in the nation.

[quote="tokenadult"][quote]Reed is actually probably the best college in the USA, but most people have not heard of it because they don't let reviews rank their school.[/quote]

What do the alumni of other colleges have to say about this?

BTW, there is a very detailed write-up about Reed in the current year's U.S. News College guide, which I think is still available in most places where magazines are sold.[/quote]
Thanks for pointing that out. I didn't see the write-up.
I'm sure alumni of some other colleges would disagree, but ehy, who wouldn't if they liked their school? ;)

Hi,

Yale's math department is pretty good  :D

But ya, any of the math departments listed in this thread will be good... I'm pretty sure that Cornell's math department, if it wasn't listed, is good as well.

This is useful stuff.

I have no knowledge to offer.... but I'd like to ask what everyone thinks the best places are for theoretical math, as opposed to applied stuff....... I don't wanna be an engineer (yet)

For 95% of all students, there's really no difference between universities that have a focus on applied math and pure math at the undergraduate level.  There will be a few that have courses better tailored to advanced students (harvard's math 25/55 spring to mind, as does princeton's 215/217/218 sequence) but you can take most math courses anywhere.

If you want a short list, some of the best undergraduate math universities are Harvard, Princeton, MIT, Stanford, UChicago, Caltech, UC Berkley, Michigan, and Duke.

[quote="tokenadult"]BTW, there is a very detailed write-up about Reed in the current year's U.S. News College guide, which I think is still available in most places where magazines are sold.[/quote]Did you mean to say The Atlantic magazine rather than U.S. News, because the current(?) issue of The Atlantic has an article about Reed's refusal to participate in the U.S. News rankings.  This issue also has several other thought-provoking articles about the college admission process.

Here the [url=http://www.ams.org/employment/group_i.html]American Math Society's list of the top Ph.D.-granting math departments[/url].  This list doesn't even mention the many, many excellent places that don't have Ph.D. programs but are nonetheless excellent places to go for undergraduate work.  Several have been mentioned in this thread; I would specifically add Williams College, and there are many, many others.

The point is that there are dozens of places that you can get an excellent education in the United States.  I say this over and over again: my advice is to pick places from these 50-or-so schools that you like, and not worry too much about whether your school is #1 or #2 or #19 or whatever.  (I'm also aware that not everybody agrees with this advice.)

[quote="DPatrick"]Did you mean to say The Atlantic magazine rather than U.S. News, because the current(?) issue of The Atlantic has an article about Reed's refusal to participate in the U.S. News rankings.[/quote]

In this thread I was referring to this year's "magazine" version of the U. S. News annual guide to colleges, which may still be on newstands. It's not a regular issue of U. S. News, but a separate publication about the size of a thick magazine issue.

But, yes, I'm glad you mentioned the Atlantic Monthly issue, with the November cover date, because it is also quite interesting. The issue of "enrollment management" illustrates that applicants have to be valuable to colleges to have the best chance of admission, financially or academically.

[quote="ankur87"][quote="ZennyK"][quote="ankur87"]I go to Berkeley, I find it extremely cool that two of the Math 1A (Calculus I) teachers are Field Medalists!

One of them is my teacher...I find that to be quite the honor....[/quote]
Wow! Anyone we've heard of before?[/quote]

My teacher is Vaughan Jones, he is on Wikipedia and received a Field's Medal in 1990 for his work in knot theory. He has an equation named after him!
http://en.wikipedia.org/wiki/Vaughan_Jones

The other teacher is Richard Borcherds, received the Fields Medal in 98, wikipedia says he invented the notoin of vertex algebra...
http://en.wikipedia.org/wiki/Richard_Borcherds

wow![/quote]

heh, my step dad studied with jones in grad
supposedly, he loves beer :D

Whenever someone asks about the best colleges for math, they have to be more specific.

Do they mean the school with the smartest math students?

Or the school with the smartest average math student?

Perhaps the school with the best math instruction?

The school with the toughest, most challenging, and advanced courses?

The school with the best math research opportunities?

Or do they instead want to know about the school with the smartest professors?

Based upon which of these is more important in one's criteria, the ranking could be diametrically opposite.  Also, keep in mind that the name of the university is only part of one's educational experience; the other part is how well a student takes advantage of them.

With that caveat in mind, all the following US schools have at least decent math departments;

Harvard
MIT
Caltech
Princeton
Chicago
Utah
Northwestern
UC Berkeley
UC Davis
Stanford (for practical applications)

[quote="RoyceGracie"]With that caveat in mind, all the following US schools have at least decent math departments;

Harvard
MIT
Caltech
Princeton
Chicago
Utah
Northwestern
UC Berkeley
UC Davis
Stanford (for practical applications)[/quote]
[quote="tokenadult"]Based on what criteria did Utah get on that list?[/quote]
If the criteria is "at least [a] decent math department", then certainly Utah belongs on the list.  Along with probably at least 3 dozen schools that were omitted.

If the claim is the Utah is one of the top 10 math departments in the country (by whatever criteria), then I would be very skeptical of a list that contains Utah and Northwestern but not Michigan or Wisconsin or Washington (to cite just three examples).  This is not to argue that Utah is in any meaningful way "worse" than those other three (or Penn State or Illinois or UCSD or USC or Oregon or several others I could list), just that it's somewhat pointless (in my opinion) to compare them in this manner.

Missing  also above is Duke,  It sure has good math students  (Consistantly ranked among the  top 3 teams in Putnam for last 5-6 years) .
.

Yeah, I missed a few; Duke, Michigan, Washington, UPenn, UCSB, and Wisconsin are all decent places too.

And yes, the criteria is simply universities that offer good [b]opportunities[/b] (they're up to the student to take advantage of) in undergraduate mathematics.

On the other hand, I'd love to know the casually noted "few dozen" places that were missing from my last post.

Oh, okay, DPatrick picked up the detail that I missed. Yeah, to the best of my knowledge, the flagship state university in most states has a decent mathematics department, at which a MOTIVATED student who goes beyond what is merely assigned could probably get a pretty good mathematics education. But, not to disparage Utah, I would recommend attending the University of Washington if you want to study math at a place with gorgeous mountain views.

[quote="RoyceGracie"]On the other hand, I'd love to know the casually noted "few dozen" places that were missing from my last post.[/quote]Here are 48 of them: [url]http://www.ams.org/employment/group_i.html[/url]

And this doesn't even include places without Ph.D. programs, like Harvey Mudd, Rose-Hulman, Williams, Reed, etc.

[quote="Isura"]But the difference in quality between schools 6 to about 25 is very small. Do you guys tend to agree with this?[/quote]I would go farther -- I would say the difference in quality between 1 and 25 is very small.  (It largely depends on what you mean by "quality".  For some definitions of "quality", the difference may not be so small.)

[quote]I would go farther -- I would say the difference in quality between 1 and 25 is very small. (It largely depends on what you mean by "quality". For some definitions of "quality", the difference may not be so small.)[/quote]

I was originally referring to research strength, specifically about studying for a PhD at these schools. The gap is probably even smaller when comparing undergraduate program difficulty/quality.

[quote="RoyceGracie"][quote="DPatrick"]Here are 48 of them: [url]http://www.ams.org/employment/group_i.html[/url]

[/quote]

Let's be honest; many of the colleges listed there are far, far below the level of Harvard/MIT/Princeton/Caltech etc.

Ohio State?  Brandeis?  SUNY?

While these universities may be decent in math and great in a number of other departments, anyone wondering about the best colleges for math should realize that they're very far below a Harvard, for instance.[/quote]I never claimed they were at the same level as Harvard.  But I also contend that for an [i]undergraduate[/i] education they are not [i]very far[/i] below.

Some anecdotes with your examples:

- two of my classmates at MIT, who received Ph.D.'s in math from MIT, are now professors at SUNY Binghamton;

- MIT has joint colloquia with Brandeis, and perhaps the best conference I ever attended as a grad student was at Brandeis;

- (alright, I don't have an example for Ohio State :) )

My point is not to quibble about individual schools.  My points are: there are lots of places that you can get an excellent undergraduate mathematics education; the phrase "best colleges in math" is pretty vague, especially at the undergrad level; and the difference in quality between the "top" schools and the "decent" schools is not that great.

This is not to say that a degree from Harvard will not have more cachet on the job market than a degree from SUNY or Ohio State, because that's nonsense.  But that's a different issue.

After all, Ohio State runs the Ross Program, and I think there are some strong mathematicians who teach there/used to teach there/did their undergraduate or graduate degrees there. DPatrick is quite correct that one issue of ranking schools by "quality" is determining which dimension of differences among schools has the most to do with "quality." I personally think (based on my own undergraduate program, in completely different subject, at a state flagship university) that general smartness of classmates, which is approximated but not completely predicted by SAT scores, makes a substantial difference, but I also look at library size as one gauge of a school's quality. And I throw up my hands in despair at trying to rank schools on quality of faculty--how am I supposed to know whether or not last year's Fields medalist being on the faculty really makes all the undergraduate classes at a school better classes? How am I supposed to know who will win the next Fields medal? When I want to get into highly detailed looks at particular classes, I download their syllabuses from the Web.

[quote="DPatrick"]I never claimed they were at the same level as Harvard.  But I also contend that for an [i]undergraduate[/i] education they are not [i]very far[/i] below.
[/quote]

Depends what you mean by "very far".  I certainly agree that one can get a poor education at Harvard and an excellent one at say, Brandeis University, based on how much the individual student studies.  However, the opportunities at the former vastly exceed the latter, and any claims to the contrary are wrong.

[quote]
Some anecdotes with your examples:

- two of my classmates at MIT, who received Ph.D.'s in math from MIT, are now professors at SUNY Binghamton;
[/quote]

So?  What is that supposed to show?

Just because two MIT PhDs teach at a university, doesn't mean the math department is good, nor even that those two PhDs are good at teaching OR research.

We had a Stanford PhD and an MIT PhD teaching math at my high school.  Does that mean my former high school's math department is equatable to Harvard's?

It's a nice anecdote, but it doesn't illustrate anything.

[quote]
- MIT has joint colloquia with Brandeis, and perhaps the best conference I ever attended as a grad student was at Brandeis;
[/quote]

Again; what's your point?  There are great conferences at any semi-respectable college or university, regardless of whether they have a good math department or not.

[quote]
My point is not to quibble about individual schools.  My points are: there are lots of places that you can get an excellent undergraduate mathematics education;[/quote]

I would even go as far as to say "any place", if a student is motivated enough.

[quote]the phrase "best colleges in math" is pretty vague, especially at the undergrad level; and the difference in quality between the "top" schools and the "decent" schools is not that great.[/quote]

I'm not really sure what "quality" means, but educational opportunities at "top" schools and "decent" schools differ quite tremendously in my opinion.

[quote]
This is not to say that a degree from Harvard will not have more cachet on the job market than a degree from SUNY or Ohio State, because that's nonsense.  But that's a different issue.[/quote]

Funny you mention this, as I think the effect of a Harvard/MIT/Princeton/etc. (top school) degree relative to a degree from a lesser school has gone down tremendously in recent years, and will continue to do so.

For instance, a friend recently interviewed three guys for a job.  Two Stanford PhDs were rejected in favor of a PhD from the University of Minnesota.

On the job market, actual skills vastly trump diplomas.

Why is CUNY so great?

I've seen the work of many professors there... and it is incredibly beautiful, for example Janos Pach a close friend of Erdos and you can see the wuality of his ideas in his work (search for his site...)

[quote="Isura"]I prof told me that Harvard, MIT, Princeton, Berkeley, and Stanford have been the top mathematics departments for about 30 years. But the difference in quality between schools 6 to about 25 is very small. Do you guys tend to agree with this?[/quote]

This sounds like a comment about graduate departments rather than undergraduate departments to me.  I'd say the graduate breakdown is roughly the top 5 or 6 (Chicago should be on that list) and then the next category goes through the 10th-12th, and after that (fuzzy) line the difference in quality is small through the 25th-ish school, and there's another drop after that (again fuzzy) line.

The variability is much larger for undergraduate institutions, if only because the institutional cultures vary so much in their attitudes toward undergraduate education.


Does anyone know how to force a line break after \begin{proof}?

Use a space (\ ) and a blank line after \begin{proof} so you have

\begin{proof}\

First line of proof
Second line
\end{proof}


Note that $ 16 \equal{} 2^4.$

So we have $ 2^{16^x} \equal{} 2^{4 \cdot 2^x} \implies 2^{16^x} \equal{} 2^{2^{x \plus{} 2}}.$

Since they have the same bases, their exponents must be the same as well. So we have $ 16^x \equal{} 2^{x \plus{} 2} \implies 2^{4x} \equal{} 2^{x \plus{} 2}.$ Therefore, $ 4x \equal{} x \plus{} 2,$ so $ x \equal{} 2/3.$

EDIT: Yay! :)

Simply convert $ 16$ into $ 2^4$, and use laws of exponents to manipulate the algebra and stuff.

So $ 2^{16^x}\equal{}2^{2^{4x}}$ and $ 16^{2^x}\equal{}2^{4*2^x}\equal{}2^{2^2*2^x}\equal{}2^{2^{x\plus{}2}}$.

Equating both expressions gives

$ 2^{2^{4x}}\equal{}2^{2^{x\plus{}2}}$

$ 4x\equal{}x\plus{}2$

$ 3x\equal{}2$

$ x\equal{}\boxed{\frac{2}{3}}$.

And that's our final answer.

EDIT: Beaten, darn.


I own a special digital clock. It beeps evrytime the hour number, minute number, and the sum of the digits of the hour and minute numbers are all prime. For example, my clock will beep at 2:23. IF Zach's clock is a military clock (1:00pm is 1300 and 11:59pm is 23:59), how many times does my clock beep everyday?


geometry AMC AIME calculus USA(J)MO USAMO MATHCOUNTS

My basic idea was to write my story about a reality show called [i]Of girls and math[/i] where the contestants would compete for a slot in the Canadian equivalent of the MOP. And that the main character would audition for the show.

All that I know about the format of the reality show in my story is that they are in a MOP-style math camp. The weekly event that would decide of a contestant's fate would be CMO-level and, later, IMO-shortlist-level (unless they are the same), albeit with only 1 question each and every time, with the notable exception of the final four-way match, which will have 6 shortlist questions of previous years all condensed into one test, for which they will have eight hours to complete in a single day.

I ask for your aid because I know very little about the life in advanced math camps.

[quote="CEM_College_Champ"]

I ask for your aid because I know very little about the life in advanced math camps.

[/quote]

Well, if you consider last year's IMO training camp as an advanced math camp, here is what happened in the end:

Cynthia turned into a monster;
Alex became an amp;
Jon became a doll;
I was a perve;
Jarno was a skeleton;
Chen took off his mask - he confessed that he is a moose.

Hopefully this helps. :)

I know that IMO training camps are advanced math camps. I wish to know if CMO problems are any different, in terms of difficulty, from IMO shortlists. Danny, I know you have the experience of this, you should tell us. This is also for people out there all over Canada that would strive for an IMO participation.

Just by looking at CMO problems, I can tell they are quite easier than IMO.

Your brain will crash if you do SL problems for 8 hrs straight.

At least do the 6 IMO SL questions on two days, otherwise everyone dies and the show ends sadly.

Please don't use one question to decide a contestant's fate, as it may result in serious consequences. If this happens in our IMO training, when you put a geometry question someone will press the button for killing you. But one good thing is that your arms are safe. For a week at least.

All right, 3 questions for the weekly event. However, the counselors and instructors decide whether a contestant is considered "endangered" and it's up to the audience to submit ideas for the questions. For the first three weeks of the show, it's CMO-level and from the 4th week onwards, IMO-level, in which case it will boil down to two questions. No more than four contestants can be endangered in any single given week and one will be spared, through the whims of the audience.

The audition process begins with questions taken at random from the COMC (and, ironically, has the same format as the COMC) so the first audition round is basically a mock COMC test. The "Short-listed" then go for a second round of auditions with a medley of questions from both past CMOs and AIMEs, following the CMO format. Any math is authorized at any stage, even complex analysis, multi-variable calculus, and so on, so forth.

All right, but then, there will be a checkpoint after the first 3 questions. The worst contestant after 3 questions will be eliminated, leaving three contestants for the final 3 questions. The time limits are the same as on IMO regarding the final six questions: 4.5 hours to do 3 problems in the first day and the same for the second.

You have just outlined pretty much the Canadian IMO selection and Training Camp and IMO itself system, just a bit more problems and shorter overall span of time. :wink:
[ie COMC, AMC, AIME -> APMO, CMO, USAMO -> Training -> IMO]
 :lol:

comment: Canadians are the most amusing people I have ever met. :P

question: What format would the contest be? MathCounts CountDown style(intense)? Test taking conditions(boring)? Cross-country-running-while-doing-problems format(silly)?
question 2: Why would the audience sit for more than three hours to watch math students writing on paper???

The show itself would last 30 minutes (or 60, depending on the weeks). A weekly show begins with the announcement of who is endangered, the best moments for the last week in camp for every contestant, the questions submitted by the audience during the week that were used for the weekly contest, along with the answers of the endangered contestants. (The audience never gets to see the contestants actually writing their papers; a MC-style countdown wouldn't be very practical with CMO-level or IMO shortlist-level questions)

Tie-breakers are a lot more intense than the contest in itself. The rules of the tie-breaker are that of the [i]old[/i] MathCounts Countdown, i.e. a best of five, but the "sudden death" in case of a tie has nothing to do with solving math problems whose difficulty can be compared with the second third of either AMC12 or COMC, as it was in the countdown phase. If two or more contestants get through the countdown phase with a tie to break, they throw darts until either one scored a bull's-eye. And the tie-breakers would be shown on TV, unlike the test that led to said tie-breaker.

Hehe.  Interesting idea...I love creative writing, fun stuff!

I have a suggestion...the 'after competition', i found, is a great part of camps and competitions.  See how the contestants interact with each other, play games, tensions, and of the occassional romances (lol) that can develop at camps.  If you can throw some of that in there, might be fun.  Just an idea though =)

Even in such an advanced math camp, it's not all math and calculations: too much math can ruin a person. There are interactions and games at every step of the process, but my main character will find romance only in the IMO.

Will the character find romance only [i]in[/i] the IMO or [i]at[/i] the IMO? For many folks, math qualifies as a love; for some, they find a romantic somebody at <X competition here>. I believe finding love at the IMO would be astonishingly cool, if you could make it sappy but still mathy. Merely being in love with math is also interesting, but putting a guy in it might be enjoyable.

My .02 dollars.

I never thought there was a significant difference between the two wordings. Now that I'm aware of it, I decided that the main character found love [i]at[/i] the IMO.

Here is a sample portion of what I have. This is located about 15-20% into the story:

[quote]"It has come to this. I was lucky to have made it this far. If they accepted the two people who got 24/35, there would be one competitor too many. The cutoff was 65/80 to qualify for the 2nd round of auditions and I got 68" Janick says.

"From what I heard, you defeated the defending champion of the Canadian Open Mathematics Challenge for your college because the first audition round was basically a mock COMC test and the score of the defending champion in the COMC was only 50/80. And you may as well decide your fate on the roll of dice at this stage" Aqqar says.

"Aqqar, even though you're my brother, you don't even know who this defending champion is. I spent the last three weeks preparing for these auditions and I don't want to spend Christmas bickering over a mere tie-breaker! I know that the tie-breaker is, in part, a contest of speed and I have to answer four questions correctly. I have 45 seconds to answer" Janick says.

"It sucks!" Cartman screams. "I hate math; you're telling me that you're learning advanced math only so that you could try to qualify for some prize!"[/quote]

If Cartman (one of Janick's brothers) has such an attitude towards countdowns, then one can easily imagine his attitude towards math in general...

What are you guys talking about?!  :?:

And Jarno...I really admire your humor and wish sincerely that you would not get your arm ripped off.

btw, who is CEM_College_Champ?

Why would any bookmaker make a bet on a math contest in the first place?

Here's the thing...

The world of competitive mathematics is full of gangs, drugs, lies, deceit, conspiracy, philosophy, money, high-tech escapades, gambling, and secret clubs.

But you CAN'T TELL ANYONE. It's a secret known only to to a few selected individuals called...Mathons.

We are a secret fraternal society who practices the philosophy of free-mathonry. We developed out of a small group of people called the stone-mathons who would meet in secret, discussing great mathematical ideas, and carving their great formulas into stones, despite having pencils and paper to write their ideas down on. From these humble beginnings, we have spread our influence across Europe, and across the world. Our symbol since ancient times is the "compass and square", for it's great significance in the art of Euclidean Geometry, which despite the beliefs of J.S., is still a valid area of mathematics. As a member of the group, we are required to declare our belief to a supreme math god, who controls all of our mathematical destinies, and controls how large or small our individual Erdos number ends up being. We have survived as a group for many millennia, and would not want any publicity tearing us apart.

If your story ever became widely known, our society would be exposed, and internal struggles would make our fraternal organization too hard to hold together, and we would inevitably implode as a group.

CEM_College_Champ, you should not meddle in affairs that are far greater than yourself, and anything you can even imagine. Someday, the brotherhood may accept you into their society, but for now, just live your life in the pursuit of knowledge.

Stay Frosty Yvan, and stay on the path of enlightenment.

Sincerely,

Rofler

Edit: Lol, this was my insane moment. Everybody gets 1.

lol actually I think that would make a GREAT story for college champ... although it seems so familiar... where have I seen it before... hmm... certainly not from all those RCMExams...

[quote="Rofler"]As a member of the group, we are required to declare our belief to a supreme math god, who controls all of our mathematical destinies. [/quote]

Yes Rofler?

Since when does a persons AOPS username reflect their mathematical ability Danny?

Also, the real MathGod wouldn't tell really innapropriate jokes, and then when asked about his masculinity respond with disturbing answers in a really high voice.

Just Saying...:ninja:

[quote="U§ername"]now it makes sense...

as I've said though, you had a good idea, but then you went in the wrong direction. All that crazy maths repels people. And why did they suddenly start skateboarding anyway? seems kind of random to me... perhaps there's a story behind it or something? Maybe the dad never wanted his child to become a nerd?[/quote]

On the surface, it may seem as though they wanted to start skateboarding because they may have to face a tie-breaker at the final round. And also that the format of the final round, which has been explicitly mentioned, included a team relay race with the skateboard as a baton.

Deep down, though, here's the passage pertaining to the more profound reasons that motivated them to take skateboarding up:

[quote]The next day, the Canadian team receives their skateboard for use in the team round. However, it isn't long before any of the six state their reasons as to why they really took up skateboarding.

Zhou: Because I have to take care of my health.

A long moment of silence ensues. Then Janick steps up to the plate and gives her answer.

Janick: Some of you may be familiar with how cardiovascular endurance influences our brains... if I told you that if you had a greater circulatory endurance you may be more efficient with your intellectual capabilities. Plus you can use it longer in your life!

The other five decided that they have enough of the medical hullabaloo, so they have reported some pieces of news regarding the betting war that opposes them to the Swedish IMO team:

Yvan: The Swedish IMO team has modified its betting orders. They kept to a 4,000 dual wager on Priscille Demoly (W-FR) and PR China for team gold, as far as Vans M.C is concerned. Only they wagered on the IMO in a peculiar way; they wagered only on if they would win a gold or not.

Daniel: What are our betting orders?

William: Here are the women's odds: 1-2 on Janick Beauchamp (CA), 2-1 on Priscille Demoly (FR) and 9-1 on anyone else, regardless of competitor.

Yan: Odds, in the bookmaking world, are noted a-b and represents the probability of an event to happen, calculated as b/(a+b).

Yvan: Give me the men's odds and the team's odds.

Zhou: Men? Han Hsichuan (CN): 1-3, Yvan Ung (CA): 1-4, Dieter Brugger (DE): 2-11, the list is too long to go on...

Daniel: Teams? PR China: 3-7, Canada: 1-3, Finland: 1-4, Japan: 1-7, France: 1-9, Germany: 1-9 and for anyone else (regardless of country), 1-14

Yvan: From now on, I am your betting overseer.

The Canadian IMO team has also wagered on if they won golds or not. This left the Vans M.C. team with about 10,000 euros to wager from all the sources they had. Since Janick was the surest of values for the Canadian team, they decided to wager 1,000 on her, 2,000 on Yvan, 2,000 on the team as a whole and 5,000 on all three happening.

Janick: We are no longer compelled to bet the farm on how we will fare in the days to come.[/quote]

Sorry, my brother got the odds all wrong. Only the women's odds were right in the first place, but I'll give the full list anyway.

Women's odds: 1-2, Janick Beauchamp (CA), 2-1, Priscille Demoly (FR), 9-1 for anyone else

Men's odds: 3-1, Han Hsichuan (CN), 4-1, Yvan Ung (CA), 11-2, Dieter Brugger (DE), and many more...

Team's odds: PR China: 7-3, Canada: 3-1, Finland: 4-1, Japan: 7-1, France: 9-1, Germany: 9-1 and for anyone else (regardless of country), 14-1

The day the competition officially begins... just one warm-up question for us.

[quote]Their betting orders are frozen, and they realize that the probability of all three happening, according to the bookmaker, is 1/39. Basically, they stand to earn a fortune.

Maria: Yvan, you guys borrowed 4,000 euros from me. I just want a compensation: the prize money you would get yourself.

Yvan: Prize money... We have to work in euros since it's the responsibility of the host country to award the prizes. They say: 5,000 for an individual gold, 3,000 for an individual silver, 2,000 for an individual bronze, 30,000 for a team gold, 20,000 for a team silver and 10,000 for a team bronze.

The Canadian team has a tradition of merging all the prize money and distribute the share of it equally among the team members in the event of a team medal. So far, it only happened twice, once for silver and another for bronze.

Daniel: Now... to warm up! In the diagram, the triangle ABC is equilateral and the radius of its incircle is 1. A larger circle is drawn in the vertices of the rectangle ABDE. What's the diameter of the larger circle?

It's completely different from what they trained in the past few months. Only Zhou could recognize it was a question from a past COMC, so he alone could raise the issue.

Zhou: It's completely unrelated to what the contest is!

The other four, of a single voice: $ \sqrt {21}$

Janick: It's completely inappropriate... give us a Lagrange multiplier question instead.

The first day of the competition begins early in the morning. They changed the number of questions to 30 for an hour in the sprint round and then they could remain on the premises for 15 minutes until they find out that the target round has been upgraded to eight questions in a 45-minute timeframe. After all first-day questions have been answered...

Janick: I know this contest puts a lot of emphasis on speed, but maybe that's why this bookmaker initially accepted wagers on the Vans M.C. only rather than all three of these contests: Vans M.C., CGMO and IMO.

Yvan: Rumor has it that a Las Vegas gambler wanted to wager on both CGMO and IMO but found no one in Nevada to take this wager. Result: it had to turn to this Dutch bookmaker.

Janick: What about... an ice cream cone?

Yvan: Don't talk about ice cream cones... I had enough of these ice cream cones since I had a question in my multi-variable calculus exam which got transposed here again. The exact same question, I swear!

Janick: Me too... A sundae then!

Yvan: That's better, as long as it is not served into a cone.[/quote]

The great moment... as requested.

[quote]Yvan: Jane, do you really want to go through this? You're aware that you'd be fined or worse, be thrown in jail, if we're caught!

Janick: Let's do it. As you drove past that lighthouse to get here (albeit at 25 m/s) I remember someone saying the lighthouse ran at 1 radian per minute. We're in a two-seater rowing boat, how fast are we to row in order not to be caught?

Yvan: We'd better keep this for when we come back.

Janick wants to help out the CGMO team, out of a favor she made by resigning two weeks into the show. The captain of the CGMO team has requested a copy of [i]Vasc's Classical Inequalities[/i] for her team. She didn't find any place in North America where to find a copy of this book. But, upon hearing that the Swedish IMO team has two copies of this book, she finds it tempting to go out and try to obtain it.

Yvan: 3, 2, 1, go!

The couple rows at a brisk pace, not suspecting that the occupants of the lighthouse are planning their moves in the betting war that is about to sweep the entire world of competitive mathematics. That's why they are completely oblivious as to what happens outside the lighthouse.

Swedish team member #1: I just received the standings from the Vans M.C. as would every other national team in play.

Swedish team captain: How are other countries responding to these?

Swedish team member #2: The only foes that did not change anything are Canadian.

The couple opens the door, taking all the occupants by surprise.

Janick: True, we didn't change anything, knowing that the odds of everyone else would somehow favor us as they changed.

Yvan: The U.S. IMO team is forced to take out a loan so it could outwager everyone.

Swedish team member #3: How much did the U.S. IMO team borrow?

Yvan: $ \$CAN 30,000$ since they took the loan from TD Canada Trust.

The bookmaker's website indicated the latest development in the odds of Vans M.C. competitors. All the wagers were registered in their specially configured message boards. Here is a (partial!) list of moves made by their enemies:

- The French team wagered 10,000 euros on Priscille, knowing none of their male players can win individually or as a team
- The Italian team wagered 5,000 euros on Chiara Micheletti (W-IT), 5,000 on Harald Thunberg (M-NO)
- The Russian team wagered 15,000 on Svetlana Stigova (W-RU), 15,000 on Nikolai Beriev (M-RU) and 7,000 on the Russian team
- The Dutch team wagered 6,000 on Carolina Holstrom (W-SE), 10,000 on Vladimir Medvedev (M-BG) and 9,000 on the Romanian team
- The Indonesian team wagered 7,000 on Hanna Kolehmainen (W-FI), 7,000 on Tara Whitten (W-GB) for silver, 7,000 on Balbir Singh (M-IN), 9,000 on the Indonesian team
- A joint American-Malaysian betting consortium wagered 15,000 on Janick, 15,000 on Elaine Yusoff (W-MY) for silver, 15,000 on Yvan, 15,000 on Manolo Sotormayor (M-MX) for silver and 40,000 on the United States

As far as the odds go, here they are:

Women: 2-1 Janick Beauchamp (CA), 3-1 Priscille Demoly (FR), 7-2 Svetlana Stigova (RU), 5-1 Elaine Yusoff (MY), 6-1 Hanna Kolehmainen (FI), 6-1 Tara Whitten (GB), 7-1 Carolina Holmstrom (SE), 8-1 Chiara Micheletti (IT), 24-1 for anyone else

Men: 3-1 Yvan Ung (CA), 4-1 Han Hsichuan (CN), 4-1 Manolo Sotormayor (MX), 4-1 Nikolai Beriev (RU), 5-1 Vladimir Medvedev (BG), 6-1 Balbir Singh (IN), 9-1 Harald Thunberg (NO), 11-1 Dieter Brugger (DE), 37-1 anyone else

Teams: PR China: 4-1, Canada 4-1, United States 4-1, Romania 7-1, Russia 8-1, Finland 8-1, Indonesia 9-1, Japan 10-1, France 23-3, Germany 23-2, 14-1 for any other country

Janick: To help fund your betting warfare operations, we would like to purchase one copy of Vasc's Classical Inequalities.

Swedish team captain: 500.

Yvan: 100 now and the other 400 once the wagers' payouts are paid. If you're victorious, you could refund us the 100 euros in deposit.

With a copy of the book secured, they hid it under Janick's seat to then go back to their team's HQ, where they are discussing the standings.

William: We are in the lead as a team. However, it is odd that neither of you (or us) lead individually.

Zhou: Have you seen how close the scoring is? Janick is a very CLOSE 6th with only 3 hundredths of a point behind the leader, Priscille.

Daniel: What about Yvan? He is 4th and it's every bit as close as Janick. He is 2 hundredths of a point behind the leader, Singh.

Yan: We are to brace ourselves if we are to seep the honors tomorrow and win this betting war!

The couple arrived, with the others not suspecting that they purchased a very expensive book.

Yan: What took you so long?

Yvan: It's to spy on other teams, and we decided not to wager any more money. They are driving up the odds on our choices; we stand to win more if we do well tomorrow.

Janick: Driving up? The bookmaker does not know that we could win every single penny from our enemies in battle if we win tomorrow. Well, not exactly. The American team will earn a mere 20 grand, everyone else will lose everything.

William: If we won everything, how much profit we will earn?

Zhou: 200 grand.[/quote]

The last time odds will ever be mentioned here.

[quote]Hunter: Guys, what are the odds we win a gold medal in the IMO?

Janick: To the other Canadian IMOers, I'm the doll, you're the moose.

Yvan: Here's how it works: if you win one gold, you get the nominal payout, according to the odds currently in effect. If you win more than one, you get $ (odds)^{\minus{}N}$ times the original amount you guys wagered where N is the number of gold medals.

Zhou: The odds are currently 13-7 (or 35%.)

The whole team warms up in preparation to the skateboard race that is about to take place in one of Bremerhaven's parks.

Janick: Remember, even if we are first as a team, we have many opponents closing on us, by as little as one-tenth of a point.

Daniel: One-tenth of a point? Who?

William: Germany. Of course, they have the home advantage...

Their heat is scheduled two heats after Germany, who is in the lineup of the opening heat. They'd get the result of their heat a few minutes after the next heat starts, along with the points they'd get for their performance. As far as Germany is concerned, they won 46.3 I + 60.5 J. Which prompted the Canadian team to convene a few minutes before the end of the second heat.

Janick: It's clear that Germany has lost a shot for a medal. If we want to remain in contention, we have to skate faster than them.

Yan: Jane, for your information, the new leader clocked in at 33 minutes 20 seconds. If we are to win, we have to be flawless in our problem solving as well as quick on our skateboard.

Yvan: I'd go first and Janick will be the anchor leg.

All goes well for the Canadian team until the anchor leg comes. Even with the Canadian team lapping some other teams (and their answers to their allotted questions seem correct as well) she lost her balance and she fell off the skateboard. By her clumsy act, the Canadian team lost part of its momentum and she had to rush some bit through her allocated problem.

Yvan: Jane, are you all right?

Janick: I don't feel... very... well. I want to go... to... bed. (She lies flat on the ground)

Yvan drove the team back to their headquarters, where Yan and Yvan carried Janick back to her bed.

Janick: I fear... that we... lost.

Yvan: Why?

Janick: Because... I... fell off... the... skateboard.

But, as Janick lies flat on the bed, the 5th heat was underway. Yan comes knocking at her door for some heartening news.

Yan: Jane, Now, we're the ones whom the enemies must defeat in order to take the team gold away. You're in the lead, with Chiara in tow, ready to race away. Priscille is running in the 6th heat, so we'd have to wait the end of the heats to say who won this event.

Three hours pass and all heats are finished... the team eagerly awaits the moment of the final results while Janick is not done with her recovery.

Yvan: Janick! All our wagers are (drum roll) VICTORIOUS!

Janick: By how much?

Yvan: What aspect, you, me or the whole team?

Janick: Start by me first.

Zhou: Jane, you won by three-tenths of a point over Tara Whitten with Priscille in third.

Janick: Three-thirds of a point? That's a lot to me!

Zhou: Yvan! Your margin of victory is tighter than that of your love interest, you won by 0.07 point over Nikolai Beriev and Dieter Brugger, who will have to run their way out.

The other three got into Janick's room. They'd got to hear whatever result there were for the team.

Zhou: It's as though the vast majority of our foes neglected the skateboarding and the closest to us is like five points behind.

Janick: I consider my oath fulfilled.

Yvan: Hunter foresaw we'd win... but he expected we win [i]by default[/i].

Meanwhile, in Hunter's room...

Jonathan: I resigned from the $ \sqrt{ANS} M.C.$ but look at what I'm missing! Our national representatives in that contest swept all golds, I could have been a part of it...

Hunter: Try skateboarding: you'll see whether you could have won the gold as well. I heard we're so... sluggish compared to them.

Jonathan: I feel deeply guilty about me surrendering the Vans M.C. for the IMO.

Danny: Look at the bright side, though. We won TWO gold medals! So we didn't wager in vain!

Once everyone has received their money, Yvan goes out to pay the remaining 400 to the Swedish team. And she hands 10,000 euros to Maria, so she actually made 6,000 in profit, as his deal with her was only to give her, in the event of a victory in the gambling war, his share of the prize money. And the IMO team would share 16,500 euros between themselves, since their wager worked and they got their service fees reimbursed.[/quote]

To be continued...

[quote="CEM_College_Champ"]
To be continued...[/quote]

Noob. Stop this nonsense.

[quote="CEM_College_Champ"]To be continued...[/quote]

n00b

:ninja:

Dude your last post was unrelated to math and totally unrelated to math competitions; this definitely does not sound like a math story anymore :| just a lot of random numbers

My error. The story is completed but I waited until the IMO results came out to complete it (since there was an event in the story that was dependent on IMO results)

I admit the last bit is a bit confusing. At some point of the story, you see Jonathan, Hunter and Zhen resigning from the Vans M.C. to pursue their CMO/IMO paths.

And it was the day of the team event. Again, I made no mention of whatever problems each person had to solve (it would be too boring anyway) but it was a relay race where each person runs for a distance (I think it's 400 m) and then solves a problem before passing the skateboard on. One component was the speed of the team, another is based on the answers.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1594511[/url]

Tsk Tsk, we weren't being harsh on mathcounts, we were being harsh on mathcounts fanfiction, lol :P

:ninja:

Edit: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=290228[/url] and your past posts on the various fun factory/round table subforums makes it more clear where your story is coming from :P

my next-door neighbor has pointed out that this is one of the best stories of all time

I agree

it may be on the same level as "A Boy and His Game"

thanks for the entertainment guys


let $r$ be a root of the equation $ax^{2}+bx+c=0$
and $s$ be a root of $-ax^{2}+bx+c=0$
show that there exist a root $t$ of $-\frac{a}{2}x^{2}+bx+c=0$
where $r<t<s$ or $s<t<r$. :)


algebra
polynomial
search
quadratics
difference of squares
special factorizations
quadratic formula

Source: Pomona-Wisconsin Math Talent Search

Find all positive integers n such that n:^2: + 25n + 19 is a perfect square.

Wow, this took me. A LONG. LONG. TIME.

[hide]

n:^2:+25n+19 = (n+a):^2:

25n+19 = (n+a):^2:-n:^2:

25n+19 = 2na+a:^2:

a:^2:+n(2a-25) = 19

Now we check a from 0 to 12 (too lazy to explain why.)

And we n = 5, 34, 125

[/hide]

Wow, that problem has some pretty big numbers...

Yes, TheSouthPaw is right.

Here's another way:

[hide]n:^2: + 25n + 19 = m:^2:

4n:^2: + 100n + 76 = 4m:^2:

4n:^2: + 100n + 625 - 549 = 4m:^2:

(2n + 25):^2: - 549 = 4m:^2:

(2n + 25):^2: - 4m:^2: = 549

Factor the difference of squares (a:^2: - b:^2: = (a+b)(a-b)):

(2n + 25 + 2m)(2n + 25 - 2m) = 549

since 549 = 3:^2: * 61, the two factors can be (1,549), (3,183), (9,61), (61,9), (183,3) or (549,1) or their negatives.

But since n is positive, at least one of the two factors is positive.  The other factor must also be positive, since positive * negative ≠ positive.  So both factors must be positive.

Now, adding the two factors together, we get:

4n + 50 = (1+549)

or

4n + 50 = (3+183)

or

4n + 50 = (9+61)

Solving these three equations individually for n, we find that:

n = 125, 34, or 5.

Answer.

[/hide]

This method avoids having to test cases, instead, you just find factors and then add the equations together to cancel.

I think he wanted to somehow get n^2 + 25n into a perfect square. It just took quite a bit of manipulation.

One other way to do it is to bound it by (n+12)^2 and (n+13)^2 sort of. The first will come out to n^2+24n+144, and the second will be n^2+26n+169. Eventually the terms with the n will be way larger than the constant term, and you can figure out when that happens. Note also that those are consecutive squares, so there won't be any in between.

TheSouthpaw wrote:One question.

[hide]

HOW did you know to do that????

[/hide]

Well, we are given an equation n:^2: + 25n + 19 = m:^2:, which isn't very nice since there are four terms.  If we can convert this into something x something = something, we know that the two somethings must be factors of the third one, so there are fewer possibilites that we have to test.

So we factor the left side into a perfect square, leaving a constant behind, and then use the difference of squares property to turn a subtraction into a product.

One more thing: since the middle term (25x) has an odd coefficient, trying to complete the square will be ugly, since (x:^2: + 25x + 625/4) doesn't leave us with all integers.  To fix this, we multiplied by four before completing the square.

I'll give you another similar problem to try out:

For what values of x is x:^2: + 9x + 44 a perfect square?

Let x^2+9x+44=k^2. Thus, x^2+9x+44-k^2=0. Solve for x using the quadratic formula, knowing that the discriminant, or
81-4(44-k^2), must be a perfect square. Thus, let 4k^2-95=m^2.
4k^2-m^2=95.
(2k+m)(2k-m)=95.

Now, set 2k+m, 2k-m to be the factors of 95, and then solve. Remember to plug them back in to see which ones will actaully generate integers for x.

I'm really busy, so I'll leave the rest to you.

And as always, someone check my work?


calculus
integration
calculus computations

According to Kaplan's (test company) Calculus AB/BC book, I need to know how to use calculator well in AP Calculus test. But the book compares between TI-83 and TI-89 and tend to suggest that TI-83 is not a good calculator. My TI-84 seems to be closer to TI-83. Does that mean that I'll need a new calculator in AP Calculus test?

If this belongs to other forum, please move it but I wasn't sure since this is calculus stuff (just not problems but yeah).

I used TI-89 and it was really convenient. But I pretty much haven't used it since then (well I guess I did some in physics, but I don't plan to do physics ever again, except maybe quantum, and I'm sure I won't need a calculator for that).

My opinion is that you would be fine on the AP test with just a scientific calculator if you are good enough at calculus. You don't [i]need[/i] a new calculator. TI-89 will probably make it easier. But you'll be better at doing stuff yourself if you learn with the 84. :D

the difference between the 83 and 84 is that the 84 has more memory and is faster.
however, for the AP test having an 89 will help only with the 2-3 questions that ask for an indefinite integral (one that is not too easy) and a $n$ degree taylor polynomial. both of which can be done by hand, and relatively quick.

I used and 84 and 89 on the exam and realized that the capabilities of the 89 were made obsolite by the test writers.


abstract algebra
algebra unsolved
algebra

[quote="RDeepMath91"]$7^{x}-3^{x}=4(7^{x-1}+\cdots+3^{x-1})$
But $7^{k+1}-3^{k+1}> 7^{k}-3^{k}\geq 7^{1}-3^{1}=4$ for any $k \in %Error. "bbmath" is a bad command.
N$
So the only $X$ is $1$...

Isn't this NT ??[/quote]
sorry
find x,y is integer
 $4^{x}+3 = 7^{y}$

$4^{x}$+$3$=$7^{y}$
$4^{x}$-$4$=$7^{y}$-$7$
Evidentially, $x\geq 1$ and so is $y$.
Lets suppose $x\geq 2$, $y\geq 2$. Then $4$ exactly divides left side. So, $4$ exactly divides 7^(y-1)-1, and that gives the same reminder as (-1)^(y-1)-1 when its divided by $4$, so we conclude that $y$ is odd. Now,  as $y$ is odd
$A$=7^(y-1)-1 =$49^{z}$-$1$ where $y$-$1$=$2z$. $A$ is congruent to $(1)^{z}$-$1$ by module $4^{2}$, so $16|A$. So, if right side is divisible by $4$, then its also with $16$, but left is divisible only with $4$. Contradiction :lol:
Now, its easy to conclude that $x=1$, $y=1$ is the only solution.


there must be 2 answers since x is raised to the 2nd power.

[quote="Hamster1800"]you sure? doesn't the equation cross teh x axis in two places if it were graphed?[/quote]
$a<0$ and the parabola intersect the y-axis at 196. So it must have two intersections with the x-axis.

[hide]im probably wrong but i got -3.5[/hide]

x is equal to the square root 49 so x is 7 or -7

Exactly. the step you take from $X^2-49$  to $(x+7)(x-7)$ is called the difference of squares


inequalities
geometry
geometric transformation
reflection
complex numbers
algebra solved
algebra

Prove the following inequality:
sqrt(x^2-x+1)+sqrt(x^2-xy+y^2)+sqrt(y^2-y+1) >2 for every positive numbers x,y

sqrt(x^2-x+1)=|x-w|, sqrt(y^2-y+1)=|y-w| and sqrt(x^2-xy+y^2)=|x-wy|, where w=cos(60)+isin(60).

Let's consider the following points (the stuff in brackets are the complex numbers representing the points):

O(0)=the origin of the complex plane. W(w), A(y), D(x), B(wy), C(-w^2y). What we must prove is AW+WD+DB>2. It's obvious that BD=DC, so WD+DB=WD+DC>WC, so it would be enough to prove AW+WC>2. If we reflect A in the line OB we get A'. We can see that angle WOA'=angle WOA = 60 deg and angle WOC=120, so points A', O, C are collinear, WA'=WA so
WA+WC=WC+WA', so we have to prove WA'+WC>2=2WO=2*(the median in the triangle WA'C), but this is a well-known problem: in any triangle the sum of 2 sides > 2*(the mdian corresponding to the other side); it can be easily proved with triangle ineq. I think this ends the proof.


trigonometry
complex numbers
calculus
calculus computations

is there a formula for the indefinite trig sum?
like:
sin(1)+sin(2)+...+sin(n)
in terms of n

It's the imaginary part of $\sum_{k=0}^{n}e^{ik}=\frac{e^{i(n+1)}-1}{e^{i}-1}$. Since $\frac{e^{i(n+1)}-1}{e^{i}-1}=\frac{e^{i(n+1/2)}-e^{-i/2}}{e^{i/2}-e^{-i/2}}=\frac{e^{i(n+1/2)}-e^{-i/2}}{(2i)\sin(1/2)}$, the imaginary part is $\frac{\cos(1/2)-\cos(n+1/2)}{2\sin(1/2)}$ unless I am mistaken. Another way to get this is to multiply the sum by $\sin(1/2)$ and telescope it.

is there anyway to do this with more complex things like tangents and secants etc.

There is no general method to express partial sums of a series in a closed form.


If nonnegative reals $x_1, x_2, \ldots, x_n$ satisfy
\[
\sum_{i=1}^n x_i^2 + 2\sum_{1 \leq k < j  \leq n} \sqrt{\frac{k}{j}}x_kx_j = 1
\]
what are the minimum and maximum values of $\sum_{i=1}^n x_i$?

yehy.it is a very difficult problem.I rememeber that there no students in zhejiang province had waken out it in the competition.but I have the offical key.if you want I can post it. :)

Yes, thank you very much  :)

I am always curious to study a hard proof.

we know that $\sum_{i=1}^n x_i^2 + 2\sum_{1 \leq k < j  \leq n} \sqrt{\frac{k}{j}}x_kx_j = 1$
let $x_{k}=\sqrt{k}y_{k}$  we get $\sum_{i=1}^n ky_i^2 + 2\sum_{1 \leq k < j  \leq n}ky_ky_j = 1$
formulate it we can know   $(\sum_{i=1}^n ky_i)^2+\sum_{i=2}^n ky_i^2+(y_{n-1}+y_{n})^{2}+y_{n}^{2}=1$
let $(\sum_{i=k}^n ky_i)=a_{k}$ and the problem is converse to $(\sum_{i=1}^n a_i^{2})=1$
let $u=\sum_{i=1}^n x_i=sum_{i=1}^n \sqrt{k}y_i=\sum_{i=1}^n \sqrt{i}(a_i-a_{i+1})=\sum_{i=1}^n (\sqrt{i}- \sqrt{i-1})a_i$
with Caughy inequality we can get  $\sum_{i=1}^n (\sqrt{i}- \sqrt{i-1})a_i\leq[\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}]^{\frac{1}{2}}$
clearly it can be hold.    :)

Thank you very much, Zhaobin. It is very nice.

I understand the max, but which is solution for the min?

Thank you very much.

The min is easy
\[
1=\sum_{i=1}^n x_i^2 + 2\sum_{1 \leq k < j  \leq n} \sqrt{\frac{k}{j}}x_kx_j \leq \left( \sum_{i=1}^n x_i\right)^2
\]
So the min is 1, with equality when one variable is 1 and all the others are 0.


What is the reciprocal of the common fraction form of $ .\overline{2}$?

$ .\overline{2} \equal{} \frac{2}{9}$

Reciprocal = $ \frac{9}{2}$

It is well known that $ .\overline{n}$ is the decimal representation of $ \frac{n}{9}$*. Thus, $ .\overline{2}$ is $ \frac{2}{9}$. The reciprocal of $ \frac{2}{9}$ is $ \frac{9}{2}$, or $ \boxed{4.5}$.

*Derived this way:
$ x\equal{}.\overline{n}$
$ 10x\equal{}n.\overline{n}$
$ 10x\minus{}x\equal{}n$
$ 9x\equal{}n$
$ x\equal{}\frac{n}{9}$


geometry
3D geometry
tetrahedron
symmetry

Mike starts at one corner of a tetrahedron. In a move, he walks along any edge to a different corner. In how many ways can he end up where he started after 6 moves?

lemme try

[hide]I got 18 9 on one side and I doubled it[/hide]

I think you'll find its a lot more than that. (More than you could count by hand.)

This looks like a pretty classical recursion problem, unless I'm mistaken.

he walks along any edge to a different corner... is it a web or just a regular tetrahedron?  Anyway the only edge I know of are the sides of the figure.

[quote="WarpedKlown1335"]I get 183.  You can use a net to visualize block walking for this problem I think.[/quote]

Theres a very nice way to solve this without having to think much about counting like that at all. It also easily generalises to (an explicit formula for) the number of ways he can return in n steps.  I wonder if anyone can find it..

Well I used recursion (I didn't use the net when I solved it by the way), but I don't know if this is what you were talking about before.

A_n:
The number of ways to reach point A (home) after n moves on a tetrahedron with points A, B, C, D.  B_n is defined similarly.

A_0 = 1
B_0 = 0

A_1 = 0
B_1 = 1
For C and D, due to symmetry, the number of ways to reach B is also the number of ways to reach C and D also.

A_2 = 3*B_1 = 3
B_2 = 2*B_1 + A_1 = 2

A_n = 3*B_(n-1)
B_n = 2*B_(n-1) + A_(n-1)

A_n - B_n = B_(n-1) - A_(n-1)
A_n + A_(n-1) = B_n + B_(n-1)

A_n + 3*B_n = 3^n because for each move, there are 3 possible ways to do so.
3*B_(n-1) + 3*B_n = 3^n
B_(n-1) + B_n = 3^(n-1)
A_n + A_(n-1) = 3^(n-1)
A_(n-1) + A_(n-2) = 3^(n-2)
A_(n-2) + A_(n-3) = 3^(n-3)
...
A_1 + A_0 = 3^0 = 1

If n is odd,
(A_n + A_(n-1)) - (A_(n-1) + A_(n-2)) + (A_(n-2) + A_(n-3)) - ... + (A_1 + A_0)) =
A_n + A_0 = A_n + 1 = 3^(n-1) - 3^(n-2) + 3^(n-3) - 3^(n-4) ... + 1 = (3^n + 1)/4
A_n + 1 = (3^n + 1)/4
A_n = (3^n - 3)/4.

If n is even,
(A_n + A_(n-1)) - (A_(n-1) + A_(n-2)) + (A_(n-2) + A_(n-3)) - ... - (A_1 + A_0))  =
A_n - A_0 = A_n - 1 = 3^(n-1) - 3^(n-2) + 3^(n-3) - 3^(n-4) ... - 1 = (3^n - 1)/4
A_n = (3^n + 3)/4.

There's a shorter way...there must be.

There is a much shorter way :) I could probably write it up in 2 or 3 lines. (The general case even). I might give a hint a bit later, if anyone wants it.

(edit; OK, so its basically the same thing, just less calculation involved.)


Suppose that $n$ is a positive integer and $p$ a prime number such that
$n | p-1$ and $p | n^{3}-1$. Show that $4p-3$ is a perfect square.

[hide="A start..."] $n^{3}\equiv 1 \bmod p$

If $n | (p-1)$ then let $p-1 = nk$.  Then

$(nk)^{2}\equiv 1 \bmod p \implies$
$n^{2}(k^{2}-n) \equiv 0 \bmod p$

Clearly $gcd(n, p) = 1$ so it follows that $k^{2}\equiv n \bmod p$.  Then

$p-1 = nk \equiv k^{3}\equiv-1 \bmod p \implies$
$n^{2}k^{2}+k^{3}\equiv 0 \bmod p \implies$
$k^{2}(n^{2}+k) \equiv 0 \bmod p$

Again, $gcd(k, p) = 1$ so it follows that $n^{2}\equiv-k \bmod p$.

Now, $4p-3$ is a square iff $p-1$ can be written in the form $t^{2}+t$.  [/hide]


1st one-
y = -x + 5

2nd one-
2x + y -3 = 0

This should probably be in the classroom forums in middle school, but I'll help you a little.

4x + 4y = 20.

You want to get y on one side first (usually the left). To do that, subtract both sides by 4x. that leaves
4y = -4x + 20.

The standard slope intercept form follows y = mx + b. There's nothing in front of the y. In our equation, we see a 4 in front of the y. How do we get rid of it? Let's divide both sides by 4. That leaves

y = -x + 5. Now it's in slope intercept form.

2x + 4y + 2 = 3y + 5

First, let's get the y on one side. Subtract both sides of the equation by 2x to get it on the right side. We have

4y + 2 = 3y + 5 - 2x.

Next, we want the two to move out of the left also. To do that, subtract both sides of the equation by 2.

4y = 3y -2 + 5 - 2x.
Simplify.
4y = 3y + 3 - 2x.

We see y on both sides of the equation. We want it on the left side, so subtract both sides of the equation by 3y to cancel it on the right, and get,

4y - 3y = 3 - 2x.
Simplify and rearrange.

y = -2x + 3.

luckily, there wasn't anything in front of the y this time, so that's our equation.

Hope this help a bit.

[quote="mathegurl"]Rewrite the equation 4x+ 4y=20 in slope intercept form.

and

Rewrite the equation 2x+4y+2=3y+5 in general form.

can anyone explain how to do this?[/quote]

[hide]$4x+4y=20$
$x+y=5$
$y=-x+5$

$2x+4y+2=3y+5$
$2x+y+2=5$
$2x+y=3$[/hide]


function
topology
real analysis
real analysis unsolved

1. If K is a compact, connected subset of $\mathbb{R}^{p}$ and $f : K \rightarrow \mathbb{R}$ is a continuous function, what can you say about $f(K)$?

2. If $F : \mathbb{R}^{p}\rightarrow \mathbb{R}^{p}$ is continuous and A is a bounded subset of $\mathbb{R}^{p}$, prove that $\overline{F(A)}= F(\overline{A})$. Is this necessarily true if A is not bounded?

3. Does uniform continuity make sense for a function from one metric space to another? If so, how would you define it?

i dont have much time so here are some hints

for the first one- what can you say about the continuous image of a compact set? likewise for a connected set? can you characterize the connected and compact sets of $\mathbb{R}$? from here you should be able to get the answer for this one

for the last one- there is nothing much different in defining a uniformly continuous fn from one metric space to another from the way you define in the reals. given $\epsilon >0$ $\exists \delta >0$ such that we have

$d(x,y)<\delta \Rightarrow \rho(f(x),f(y))<\epsilon$ where $d$ and $\rho$ are the metrics in each metric space

for the second one i will come back later and reply, thats also quite easy but i really have to go now and i could do only the above 2 now, sorry for that


When is a LaTeX solutions template for round 2 supposed to be out?  I clicked on "solutions template", but it said "Round 1" and I thought that there was going to be some code to help typeset the grid problem (#1)

It's the same solution template.  Just change the line:

\usamtsround{2}

so that it's round 2 (as above).  We've also changed the default template so that it's set to round 2.

Regarding LaTeX code for the grid problem: you can download the LaTeX source file for the problems from the Problems page at

[url]http://www.usamts.org/AoPS_U_Problems.php[/url]


Is every $ \lambda$ -system a $ \sigma$ -algebra? It seems to be true.
we know that every $ \sigma$ -algebra is a  $ \lambda$ -system. But then are they one and the same?!!!!

Yeh sorry I ve realized that not every $ \lambda$ -system is a $ \sigma$ -algebra because we can't get the closure under countable unions.

Can anyone give an example of an $ \lambda$ -system which is not a $ \sigma$ -algebra.


inequalities
algorithm
algebra
polynomial
inequalities proposed

I am posting it here only to revive it it was in one another topic but was left un-answered
For $ a,b,c \ge 0$ , we have,
$ \frac{3a\plus{}b}{b\plus{}c} \plus{}\frac{3b\plus{}c}{c\plus{}a}\plus{}\frac{3c\plus{}a}{a\plus{}b} \ge 6\frac{\sqrt{3(a^2\plus{}b^2\plus{}c^2)}}{a\plus{}b\plus{}c}$
i haven't been succesful for so far.. :(

$ \displaystyle\sum_{cyc}{ \frac {3a \plus{} b}{b \plus{} c}}\ge  \displaystyle\sum_{cyc} {\frac {4a}{b \plus{} c}}$
now by nesbitts $ \sum_{cyc}{\frac {4a}{b \plus{} c}} \ge 4*\frac {3}{2} \equal{} 6$.
now we need to prove that $ \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} > \equal{} a \plus{} b \plus{} c$.which is equivalent to $ a^2 \plus{} b^2 \plus{} c^2 > \equal{} ab \plus{} bc \plus{} ca$which is well known

[quote="aadil"]$ \displaystyle\sum_{cyc}{\frac {3a \plus{} b}{b \plus{} c}}\ge \displaystyle\sum_{cyc} {\frac {4a}{b \plus{} c}}$
now by nesbitts $ \sum_{cyc}{\frac {4a}{b \plus{} c}} \ge 4*\frac {3}{2} \equal{} 6$.
now we need to prove that $ \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} > \equal{} a \plus{} b \plus{} c$.which is equivalent to $ a^2 \plus{} b^2 \plus{} c^2 > \equal{} ab \plus{} bc \plus{} ca$which is well known[/quote]

You are wrong.

[quote="aadil"]$ \displaystyle\sum_{cyc}{\frac {3a \plus{} b}{b \plus{} c}}\ge \displaystyle\sum_{cyc} {\frac {4a}{b \plus{} c}}$[/quote]
this is wrong actually the reverse holds and even if this were right,
[quote]
now we need to prove that $ \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} > \equal{} a \plus{} b \plus{} c$.which is equivalent to $ a^2 \plus{} b^2 \plus{} c^2 > \equal{} ab \plus{} bc \plus{} ca$which is well known[/quote]
you actually have to prove the reverse which is wrong obviously..
i think the inequality is tough and not solved by normal methods i guess there is a certain trick involved..

$ \frac {3a \plus{} b}{b \plus{} c} \plus{} \frac {3b \plus{} c}{c \plus{} a} \plus{} \frac {3c \plus{} a}{a \plus{} b} \ge 6\frac {\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}{a \plus{} b \plus{} c}$
$ \Leftrightarrow \sum_{cyc} (\frac {3a \plus{} b}{b \plus{} c} \minus{} 2) \geq 6(\frac {\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \minus{} (a \plus{} b \plus{} c)}{(a \plus{} b \plus{} c)}) \minus{} 6$
For LHS
$ \sum_{cyc} (\frac {3a \plus{} b}{b \plus{} c} \minus{} 2) \equal{} \sum_{cyc} \frac {(a \minus{} b)}{b \plus{} c} \plus{} \frac {2(a \minus{} c)}{b \plus{} c} \equal{} \sum_{cyc} \frac {(a \minus{} b)}{b \plus{} c} \plus{} \frac {2(a \minus{} b)}{c \plus{} a} \equal{} \sum_{cyc} \frac {(a \minus{} b)^2}{(c \plus{} a)(c \plus{} b)} \minus{} \sum_{cyc} \frac {(a \minus{} b)(b \plus{} c)}{(c \plus{} a)(c \plus{} b)}$
However since $ \minus{} \sum_{cyc} \frac {(a \minus{} b)(b \plus{} c)}{(c \plus{} a)(c \plus{} b)} \equal{} \frac {\sum_{cyc}a^3 \minus{} \sum_{cyc} ab^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \equal{} \frac {\sum_{cyc} (a \minus{} b)^2(a \plus{} 2b)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
we can rewrite LHS as
$ \sum_{cyc} \frac {(2a \plus{} 3b)(a \minus{} b)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
For RHS
$ 6(\frac {\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \minus{} (a \plus{} b \plus{} c)}{(a \plus{} b \plus{} c)}) \minus{} 6 \equal{} 6\frac {\sum_{cyc} (a \minus{} b)^2}{(a \plus{} b \plus{} c)(\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \plus{} (a \plus{} b \plus{} c))}$
So the problem can be rewritten as
$ S_a(a \minus{} b)^2 \plus{} S_b(b \minus{} c)^2 \plus{} S_c(c \minus{} a)^2\geq 0$
where
$ S_a \equal{} \frac {(2a \plus{} 3b)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \minus{} \frac {6}{(a \plus{} b \plus{} c)(\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \plus{} (a \plus{} b \plus{} c))}$
$ S_b \equal{} \frac {(2b \plus{} 3c)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \minus{} \frac {6}{(a \plus{} b \plus{} c)(\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \plus{} (a \plus{} b \plus{} c))}$
$ S_c \equal{} \frac {(2c \plus{} 3a)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \minus{} \frac {6}{(a \plus{} b \plus{} c)(\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \plus{} (a \plus{} b \plus{} c))}$

Now I will show that
1. $ S_a \plus{} S_b \plus{} S_c > 0$ which is equivalent to
$ \frac {5\sum_{cyc} a}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} > \frac {18}{(a \plus{} b \plus{} c)(\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \plus{} (a \plus{} b \plus{} c))}$
or
$ \frac {5p}{pq \minus{} r} > \frac {18}{p(\sqrt {3p^2 \minus{} 6q} \plus{} p)}$ but since $ \frac {18}{p(p \plus{} p)} > \frac {18}{p(\sqrt {3p^2 \minus{} 6q} \plus{} p)}$ (from $ 3\sum_{cyc} a^2\geq (\sum_{cyc} a)^2$)
it suffices to show that
$ \frac {5p}{pq \minus{} r} > \frac {18}{2p^2}$
which is equivalent to
$ 5p^3 \plus{} 9r > 9pq$ which is obviously true from $ p^3\geq 3pq$
2. $ S_aS_b \plus{} S_bS_c \plus{} S_cS_a > 0$ which is equivalent to
$ \sum_{cyc} \frac {(2a \plus{} 3b)(2b \plus{} 3c)}{((a \plus{} b)(b \plus{} c)(c \plus{} a))^2} \minus{} \frac {6(2a \plus{} 5b \plus{} 3c)}{(a \plus{} b \plus{} c)(\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \plus{} (a \plus{} b \plus{} c))(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {36}{(a \plus{} b \plus{} c)^2)(\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \plus{} (a \plus{} b \plus{} c))^2} > 0$
or
$ \frac {6\sum_{cyc} a^2 \plus{} 19\sum_{cyc} ab}{((a \plus{} b)(b \plus{} c)(c \plus{} a))^2} \minus{} \frac {60\sum_{cyc} a}{(a \plus{} b \plus{} c)(a \plus{} b)(b \plus{} c)(c \plus{} a)(\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \plus{} (a \plus{} b \plus{} c))} \plus{} \frac {108}{(a \plus{} b \plus{} c)^2)(\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)} \plus{} (a \plus{} b \plus{} c))^2} > 0$
which can be rewritten as
$ \frac {6p^2 \plus{} 7q}{(pq \minus{} r)^2} \plus{} \frac {108}{p^2(\sqrt {3p^2 \minus{} 6q} \plus{} p)^2} > \frac {60}{(pq \minus{} r)(\sqrt {3p^2 \minus{} 6q} \plus{} p)}$
But from AM-GM
$ \frac {64p^2}{27(pq \minus{} r)^2} \plus{} \frac {108}{p^2(\sqrt {3p^2 \minus{} 6q} \plus{} p)^2}\geq \frac {32}{(pq \minus{} r)(\sqrt {3p^2 \minus{} 6q} \plus{} p)}$
it remains to show that
$ \frac {(6 \minus{} \frac {64}{27})p^2 \plus{} 7q}{(pq \minus{} r)^2} > \frac {28}{(pq \minus{} r)(\sqrt {3p^2 \minus{} 6q} \plus{} p)}$
but from $ 3\sum_{cyc} a^2\geq (\sum_{cyc} a)^2$ it suffices to show that
$ \frac {(6 \minus{} \frac {64}{27})p^2 \plus{} 7q}{(pq \minus{} r)^2} > \frac {28}{(pq \minus{} r)(p \plus{} p)}$
or
$ 14r \plus{} \frac {98p^3}{27} > 7pq$ which is obviously true...
Thus,
$ S_a(a \minus{} b)^2 \plus{} S_b(b \minus{} c)^2 \plus{} S_c(c \minus{} a)^2\geq 0$ is true...

 :)

Generally, we can prove that: if $ a,b,c>{0}$,then

$ \frac{ka+b}{b+c}+\frac{kb+c}{c+a}+\frac{kc+a}{a+b}\ge {\frac{3(k+1)}{2}\frac{\sqrt{3(a^{2}+b^{2}+c^{2})}}{a+b+c}}$

holds if and only if ${ k\ge{\frac{5}{2}}}.$

[quote="fjwxcsl"]Generally, we [b]can prove[/b] that: if $ a,b,c > {0}$,then

$ \frac {ka + b}{b + c} + \frac {kb + c}{c + a} + \frac {kc + a}{a + b}\ge {\frac {3(k + 1)}{2}\frac {\sqrt {3(a^{2} + b^{2} + c^{2})}}{a + b + c}}$

holds if and only if ${ k\ge{\frac {5}{2}}}.$[/quote]
[b]can prove[/b] how?? i am interested in seeing your solution for this..

Sorry, we have established  a new  mechanical proving algorithm   which  can be  read  for  a  kind of polynomial inequality , but still not published...

$ a^2(2b \plus{} c) \plus{} b^2(2c \plus{} a) \plus{} c^2(2a \plus{} b) \plus{} 2abc \ge a^3 \plus{} b^3 \plus{} c^3$ i seemed to have solve the problem(with simpler methods) if this is true..(but not if and only if!)
and if $ a,b,c$ are the sides of a triangle then i believe i have more than one way to solve it of which this is one..
when an inequality is homogenous is it valid to assume that the variables are side-lengths of a triangle??(i kno i am horribly wrong but jus in case.. )

OH! so sorry actually the conditions state that $ a,b,c$ are [b]positive reals[/b] and [b]not[/b] non-negative
sry abt that loophole !
and actually even if it were non-negative then u wud have the added condition that no two are 0 for eg .$ b\equal{}c\equal{}0$, this would mean one of the fractions in the LHS $ \frac{3a\plus{}b}{b\plus{}c}$ has denominator zero!!

[quote="geniusbliss"]OH! so sorry actually the conditions state that $ a,b,c$ are [b]positive reals[/b] and [b]not[/b] non-negative
sry abt that loophole !
and actually even if it were non-negative then u wud have the added condition that no two are 0 for eg .$ b \equal{} c \equal{} 0$, this would mean one of the fractions in the LHS $ \frac {3a \plus{} b}{b \plus{} c}$ has denominator zero!![/quote]
Actually it does not really matter, just let $ b \equal{} c$ both get infinitely close to $ 0$, and $ a \to \infty$ then your inequality is contradicted. Also, note that it is homogeneous so there is nothing wrong in Felix's assumption.

[quote="geniusbliss"]
when an inequality is homogenous is it valid to assume that the variables are side-lengths of a triangle??(i kno i am horribly wrong but jus in case.. )[/quote]
I don't think so. For instance, the inequality $ 2(ab \plus{} bc \plus{} ca)\geq a^2 \plus{} b^2 \plus{} c^2$ holds only for $ a,b,c$ as the sides of a triangle, but it is homogeneous. You cannot just say, " since this is homogeneous, we may assume $ a,b,c$ to be sides of a triangle" because this is illogical. :)


Παιδιά ας τα πούμε από εδω για το Μεσογειακό κι την προφορική
οκ??
ό,τι κρίσεις απόψεις κι σχόλια ας τα πούμε εδώ
περιμένουμε όλοι όλους  :wink:

Το προβλέψεις έγραψε  :D

Καλή επιτυχία σε όλους !  :)

Μπορεί να μου πει κάποιος με σιγουριά τι ώρα αρχίζει ο διαγωνισμός και πού ακριβώς γίνεται (στο χημείο ή στη μαθηματική εταιρεία) :?:

Τα γραπτά στο Χημείο, τα προφορικά στη Μαθηματική.

Καλή μας επιτυχία. :)

Πώς τα πήγατε παιδιά Μεσογειάδα?

Πολύ δύσκολα θα έλεγα εγώ προσωπικά. :mad:

Συγκεκριμένα:

Το 1ο αρκετά δύσκολο ως προς το δεύτερο ερώτημα το ψιλοέβγαλα απ' ότι είδα και τη λύση.

Το 2ο βατότατο και προσεγγίσιμο από όλους πιστεύω. Ήθελε λίγη προσοχή στις πράξεις και σε ένα τέχνασμα για να μη βγουν μεγάλα τα νούμερα. Το έβγαλα πρώτο πρώτο. :wink:

Το 3ο πολύ δύσκολο. Έφτασα σε ένα ικανοποιητικό σημείο και το παράτησα γιατί λόγω χρόνου αλλά και λόγω περιπλοκότητας. :(

Το 4ο μέτριας δυσκολίας απ' ότι άκουσα, γιατί εγώ θεώρησα πως είναι το πιο δύσκολο. Το έφτασα μέχρι κάπου σωστά αλλά μετά για κάποιο ανεξήγητο λόγο κόλλησα και έτσι δε νομίζω ότι θα πάρω πάνω από ένα-δύο βαθμούς γι' αυτό. :(

Αναμενόμενο σκορ: 11-12 πόντοι αν έχω κάνει σωστή κρίση. :|

protonie apo pou eides tis lyseis... :?:
exoun anarti8ei kapou?

Είχαν βγει οι πρώτες εκεί και τις είχαν κάποια παιδιά με τους δασκάλους τους και πήγα και ρώτησα. Μην ανησυχείτε. Αργά ή γρήγορα θα κυκλοφορήσουν. Εγώ μόνο τη λύση του πρώτου είδα.

Για το πρώτο θέμα το αρχικό ερώτημα ήταν αρκετά προσβάσιμο
, το δεύτερο λίγο περίεργο αλλά υπάρχουν τεχνικές για να αντιμετωπιστούν τέτοιου είδους θέματα με βελτιώσεις σταθερών.
Το δεύτερο εκτός από κάποιες πράξεις που είχε δεν ήταν δύσκολο εγώ το έλυσα με τρία όμοια και μια φορά Πτολεμαίο.
Το τρίτο θέμα ήταν υπολογιστικό .
Η κύρια ιδέα που χρησιμοποίησα ήταν ότι $Rr=\frac{abc}{4t}$
kai ότι $tan\ \frac{w}{2}=\frac{r}{t-a}$
Χρησιμοποιώντας και τη δοσμένη καταλήγεις σε μια δευτεροβάθμια που σου δίνει τα $b,\ c$ . (Από την πρώτη έχεις το γινόμενο και από τη δεύτερη το άθροισμα  :wink: )

Τέλος για το τελευταίο εκφράζουμε το $x$ ως συνάρτηση των
$\left[ x\right]$  και $\left\{x\right\}$ από την σχέση

$x= \left\{x\right\}+\left[ x\right]$  . Οπότε η ανισότητα γίνεται δύο μεταβλητών και θέτωντας την μία $a$ και την άλλη $b$ και κάνοντας απαλοιφή παρονομαστών φτάνουμε σε κάτι προφανές .

Η εκφώνηση του 4ου;

Silouan ('h opoiosdipote allos pou pire meros sti Mesogeiada), mporeite sas parakalw na scannarete ta themata kai na ta steilete gia na ta katevasoume kai na ta ektipwsoume?

Alexandros

Εγώ έχω σκανάρει τα θέματα αλλά δεν ξέρω πως να τα αναρτήσω στο forum...καμιά βοήθεια :?:
Εκτός κι αν θέλει κάποιος να του τα στείλω προσωπικά (e-mail)
καο να τα αναρτήσει αυτός...(βασικά το πρόβλημα είναι ότι ,για κάποιο λόγο, το attachment είναι πολύ μεγάλο σε χωρητικότητα :huh: )

Εντάξει τελικά, το πρόβλημα ήταν το μέγεθος τηε εικόνας. Κατάφερα να την μικρύνω χωρίς ν αχάσουμε σε ποιότητα. Πολλά ευχαριστώ στον bilot4

Υποχρέωση μου :)  και μπράβο που κατάφερες να το αναρτήσεις ώστε να δουν τα θέματα κι οι υπόλοιποι :wink:

Gia tin 4h me mia grigori matia pou tis eriksa, mou vgike to fragma ligo kalitero kai iso me $5$ (anti gia $\frac{9}{2}$). Vrike kaneis antistoixo apotelesma? Mporw na steilw tin lisi mou afou tin koitaksw ligo kalitera mipws iparxei kapoio lathos.

Alexandros

Αν και έμαθα ότι ή ανισότητα πιθανώς βελτιωνόταν αμφιβάλλω λίγο αν η σταθερά μπορούσε να γίνει 5...Ανυπομονώ να δω τη λύση σας :wink:

Ti symainoyn ta symbola $\{x\}$ και $[x]$ kai ti diafora exoyn ?  :blush:

mailo ta $[x]$ kai $\{x\}$ simvolizoun to akeraio meros kai to klasmatiko meros tou pragmatikou arithmou $x$. Akeraio meros enos arithmou $x$ einai o megaliteros akeraios pou den ipervainei to $x$. Ara gia paradeigma $[3.21]=3, [2]=2, [-3.21]=-4$ ktl...  kai antistoixa gia to klasmatiko meros exoume:

$\{3.21\}=0.21, \{2\}=0, \{-3.21\}=-0.21$

Opote profanws an $x\geq 0$ isxiei $x=[x]+\{x\}$ enw an $x<0$ isxiei $x=[x]+\{x\}+1$. (Dokimase merika paradeigmata gia na peisteis).

Alexandros

cretanman το καταλαβα  :wink:
Αυτοι οι συμβολισμοι ειναι standart? Δηλαδη θα μπορουσαν να μην πουν τι παριστανουν στη εκφωνιση της ασκησης?

Akoma kai na min einai kathierwmenoi, sigoura einai apo tous simvolismous pou einai poli sinithismenoi. Sinithws omws meta apo ton simvolismo, eksigoun ti simasia tou gia na ginei diakrisi apo to anwtero akeraio meros enos arithmou $x$ pou orizetai ws o mikroteros akeraios pou ipervenei ton $x$ kai simvolizetai me paromoio tropo. Gia to klasmatiko meros tou arithmou $x$ den exw dei allo simvolismo gia na pw tin alitheia.

Alexandros

[color=red]Y.G. [/color] Mipws tixainei kapoios apo edw na exei ta themata twn mikrwn? An nai kante sas parakalw ena kopo na ta scannarete kai na ta valete stin selida.

poli xalari i anisotita sto tesera mu fenete... :huh:
8etontas $r=\frac{[x]}{\{x\}}$
kataligo oti $\frac{8 r^{4}+6 r^{3}-13 r^{2}+6r+8}{2r(2r^{2}+5r+2)}>0$
pu ine profanestato afu  $r=\frac{[x]}{\{x\}}>\frac{1}{\{x\}}>1$

 :D

Αλέξανδρε με το $LHS$ τι εννοείς?

[quote="giannis"]Αλέξανδρε με το $LHS$ τι εννοείς?[/quote]

[color=red]LHS = Αριστερό μέλος( Left Ηand Side). [/color]

    Συνηθίζεται στην αγγλική ορολογία.


inequalities
geometry
trigonometry
geometry solved

Let $\triangle ABC$ be a triangle and denote I its incenter. Lines AI, BI, CI cut the incircle (I) for the first time in D, E, F respectively. Prove that

\[
AD + BE + CF \ge \frac{{DE + EF + FD}}{{\sqrt 3 }}
\]

It seems to me that this is correct.  We have to prove $AI+BI+CI\geq3r+1/\sqrt3(DE+EF+FD)$, or after dividing by r:
\[\sum\frac{1}{\sin\alpha/2}\geq3+(2/\sqrt3)\sum\sin(45^\circ+\alpha/4).\] And now just notice that $1/\sin(x/2)$ is convex and $\sin(45^\circ+x/4)$ is concave and apply Jensen two times.


Square ABCD has area of 400. Points E and F are the midpoints of AB and AD respectively. What's the area of the largest circle that can be inscribed in EBCDF?

I can't remember if i've already seen this somewhere, oh well.

Does inscribed mean the circle has to touch all sides of the pentagon? That's sorta impossible...

the problem basically means the biggest circle that can fit inside the pentagon. The circle doesn't have to touch all sides of the pentagon, in fact it can't.

[quote="pianoforte"][hide]r is the radius. $5\sqrt{2}+r+r\sqrt{2}=20\sqrt{2}$
$r(1+\sqrt{2})=15\sqrt{2}$
$r=\frac{15\sqrt{2}}{\sqrt{2}+1}$
$r=30-15\sqrt{2}$
$Area=(1350-900\sqrt{2})\pi$[/hide][/quote]
How do you know the first step?

[hide]green line is $5\sqrt{2}$, blue line is $r$, red line is $r\sqrt{2}$, and the total diagonal is $20\sqrt{2}$.
$5\sqrt{2}+r+r\sqrt{2}=20\sqrt{2}$
[/hide]

Isn't the green line [hide]$10\sqrt2$?[/hide]So the area is[hide]$(20-10\sqrt2)^2*\pi$[/hide]

Yes it is $5\sqrt{2}$
It is half of FE
FE=$10\sqrt{2}$
so $5\sqrt{2}$


23. Each face of a cube is painted either red or blue, each with probability 1/2. The
color of each face is determined independently. What is the probability that the
painted cube can be placed on a horizontal surface so that the four vertical faces
are all the same color?

What are vertical faces?

The faces that are perpendicular to the horizontal surface.

Oh.

How would we have figured that out? Is there some vocabulary that I don't know?

ver·ti·cal
1. being in a position or direction perpendicular to the plane of the horizon; upright; plumb.

How could I have realized which sides they meant?

Well, the definition of vertical, it seems that they are the faces that are standing "straight", perpendicular to the ground.

Well, if you place a cube on a surface, 2 of the 6 faces are parallel to the surface and the other 4 are perpendicular. Isn't it kind of obvious which ones the problem is referring to?


calculus
integration
function
algebra
floor function

Does anyone know how to solve the following integration problem?

[i]For any integer k, integrate the greatest integer function, f(x) = [[x]], on the interval [0,k].[/i]

Thanks.

Brett

Well...  You know that for the interval [n, n+1), the height is n.  Thus, it would, on that interval from 0 to k, be $\frac{k(k-1)}{2}$.  That's how I would do it...


If you want to propose [i]creative[/i] names so that we can be something other than SC A and SC B, do it here!

I don't really have any good ideas...but I was hoping someone else might :D

Maybe fire and ice? Oh wait, that's probably already taken....hmmm....maybe Gold and Silver...??? Wind and Earth??? Why do I keep thinking of elements??? I don't know... :|

Bacon & Grits!  It's as original South Carolina as you can get.  Plus, it'll make them underestimate us!

[quote="mysmartmouth"]Bacon & Grits! [/quote]
haha  :rotfl:
that would so awesome and funny.

bacon and grits...lol

i wouldnt mind having that on our shirts.

Look, you can't name us "Bacon and Grits." Half the people north of the Mason-Dixon Line haven't even [i]heard[/i] of grits.

[quote="furious"]ok, i've got it. these two names really represent SC as a whole.

the first will represent the upstate with its forests and stuff and will be called GREEN

the second team will represent the coastline and will be called WAVES

so together it will be the South Carolina GreenWaves  :D[/quote]

umm...David....think of a better one....please

How about we draw on our southern heritage and go with....never mind.  I think a poll is in order.  I will add one soon, with my moderating powers.  But first, what are the choices?  We'll go with anything anyone wants to submit, as long as it's appropriate.  Keep in mind, [b]THIS IS NOT AN OFFICIAL POLL, IT'S UP TO MR R. TO CHOOSE TEAM NAMES.[/b]  With that, let the fun begin!

[u]Suggestions:[/u]
Bacon & Grits
Green & Waves
Gold & Silver
Wind & Earth
"A" & "B"

As soon as we gather a few suggestions, I'll add the poll.  If you were not serious about your suggestions, tell me and I will take them off this list.

maybe you can start a poll after we get some decent suggestions. all the current ones need to be thrown out.
bacon and grits is too stupid to be taken seriously
greenwave was a joke. that would never make it
colors are stupid.
the elements are already taken. one team from last year was fire and ice
and A and B are what we are trying to get rid of.
now maybe Z and X......only kidding joe.....lol

[quote="DummAddict"]did we ever decide on shirt colors?[/quote]

Mr. R. said blue and white, i believe.

I actually like the bacon and grits idea :D

[quote="furious"][quote="joml88"]Mr. R. said blue and white, i believe.

I actually like the bacon and grits idea :D[/quote]

like the SC navy blue or a light blue or what?
and yes, the bacon and grits is kinda funny, but it might make us stand out like the spinny hubcap guys at furman, and i don't want that. :|[/quote]

no, we're not doing bacon and grits, it's just some stupid idea me and sean thought of (although pretty funny), add this to the poll:

The Phantoms and the Dark Knights

That should be our team names, it's also very intimidating for other teams :lol:

my favorite by far is the bacon and grits one at this point. itd be really funny esp if we placed..."And South Carolina Grits Team Places Xth!"  :D

Shouldn't it be biscuits and grits?  Anyways, the colors will be SC blue (navy blue) and white.  Unless the t-shirt shop cannot do this.  Also,  how about the following names (most are just for laughs):

fruit / cake

yin / yang

luke warm / room temperature  (take that Lehigh Valley)

genuine / imitation

stars / bars

Stars and bars is definitely inappropriate. Google it if you're a Yankee and don't understand the reference.

[u]New Suggestions:[/u]
Bacon & Grits
Green & Waves
Gold & Silver
Wind & Earth
Luke Warm & Room Temperature
Genuine & Imitation
Phantoms & Dark Knights

[quote="mysmartmouth"]Stars and bars is definitely inappropriate. Google it if you're a Yankee and don't understand the reference.
[/quote]

lol, my Northern mind didn't know what stars and bars was, i just thought it sounded cool :rotfl:  :rotfl:

If we go up there with Stars & Bars, people will think we are bigoted Southerners.  If we go up there with Bacon & Grits, they'll laugh at our ingenuity.  Or we could go with:  [b]Grits (A-Team) and Gits (B-Team).[/b]  Yes, I'm aware that I'll probably be on the B-Team!   :D

i like the genuine and imiatation idea, and the shirts sound good

luke warm and room temperature is now my favorite. btw, what is the connected to lehigh valley (were they fire and ice?)

I'm about to edit Joe's post and add the poll.  Actually, Joe, you wanna edit your own post and add the poll?

You could do the Fast and the Furious.

The Fast would be the A team (for obvious reasons), and the Furious would be the B team (they're angry because they didn't make the A team, maybe?). Just an idea.

[quote="mysmartmouth"]How about the Fast (everyone) and then furious (David).[/quote]
haha, very funny.
wow, i just noticed i have a smiley named after me ": furious :" = :furious: lol

Actually, I think the Fast and Furious idea is the best so far :coolspeak:

[quote="furious"]*bump* :ninja:
we need to get serious on the team names again, since i don't think we want to be the generic A and B teams. and please, keep all suggestions that you know are stupid to yourself, or possibly create another topic called "stupid team names not to be considered"[/quote]

You're the one who came up with green and waves...
edit:  btw, if you didn't understand david's attempt at humor with the greenwave thing...summerville high school's mascot or whatever is the greenwave

Anyway, the poll has been added to the first post of the thread...so let's vote!!!

I like Dumb and Dumber!  That's a good one.

[quote="osama"]:starwars:    :bruce:      why not good team and crappy team[/quote]

[quote="rs9999"]you should put dumb and dumber :rotfl:[/quote]

Who are you people? Please introduce yourselves in the Introduction Thread. Thank you :)

I voted for fruit and cakes.
Certainly describes a bunch of people on the team...
 :rotfl:

It is official.  The A team for South Carolina will be Grits and the B team will be Bacon.  I hope that those using this site appreciate that these are the names.  I don't know how the other kids or sponsors are going to feel about it.  It won't really matter, unless we place.

I think that's great.  It will enlighten other teams about our Southern culture... :P   Also, it's an incentive to place: we all want them to announce the "Grits" team on the mic!

Also, I made a South Carolina Blog.  See if you can find it.

[b]Should have gone with my idea [/b]:ewpu:  :wallbash:

[quote="joml88"][quote="rs9999"]you should put dumb and dumber :rotfl:[/quote]

..........[/quote]

Ummm.....what Joe?


Let m,n>2 and let p>2 be a prime divisor of both m and n. Consider all products of the form x_1*x_2*...*x_m, where x_i belong to {1,2,...,n} and their sum is a multiple of p. Prove that the sum of all these products is a multiple of (n/p)^m.

Maybe I didn't express what I thought:
  Prove that (n/p)^m is a divisor of  \sum x_1*x_2*...*x_m, where the sumation is taken for all systems (x_1,...,x_m) of numbers from {1,2,...,n} having their sum a multiple of p. So, in this sum omly systems which have the sum of their components divisible by p appear.


Is there any formula to derive  an nth term of the fibbonacci sequence if you don't know the previous terms? Just wondering... :?:

You probably wanted that whole thing over the denominator.  In other words, it is

$F_{n}= \frac{1}{ \sqrt{5}}\left( \left( \frac{1+\sqrt{5}}{2}\right)^{n}-\left( \frac{1-\sqrt{5}}{2}\right)^{n}\right)$

[quote="i_like_pie"]No, it's correct.

From: http://www.research.att.com/~njas/sequences/A000045

Formulas are near the bottom of the page.[/quote]
t0rajir0u's version is the more common form though.

[quote="1=2"]ok. Here's mine:

Where $\theta=\frac{1+\sqrt{5}}{2}$,

$\frac{\sqrt{5}\left(\theta^{n}-(-\theta)^{-n}\right)}{5}$

Gives you the nth term. I've tried it, and it works.[/quote]
No, that formula is wrong because $-\theta=\frac{-1-\sqrt{5}}{2}$
Instead consider where $\alpha$ and $\beta$ are the roots to the quadratic $x^{2}-x-1$.
$\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}}$

[quote="i_like_pie"]No, it's correct.[/quote]

It would be correct if the denominator in your second term were also under your first term, but it is not.  (Please read your posts when I tell you you have made an error before you assume that you are right.  You did this for three posts in another topic and it was quite a waste of time.)

[quote="Ubemaya"]Have you people actually memorized this formula?

Where would you need to use it?[/quote]

The formula is not useful for calculations, but it offers an immediate proof of several identities, including the lemma $F_{n}| F_{mn}, m, n \in \mathbb{N}$.

[quote="bpms"][quote="1=2"]Where $\theta=\frac{1+\sqrt{5}}{2}$,

$\frac{\sqrt{5}\left(\theta^{n}-(-\theta)^{-n}\right)}{5}$

Gives you the nth term. I've tried it, and it works.[/quote]
No, that formula is wrong because $-\theta=\frac{-1-\sqrt{5}}{2}$[/quote]

Notice the negative exponent on the second term.  In fact, $(-\theta)^{-1}= \frac{1-\sqrt{5}}{2}$, which is correct.

Oh, sorry, I should've looked more carefully :blush:

[quote="t0rajir0u"][quote="i_like_pie"]No, it's correct.[/quote]

It would be correct if the denominator in your second term were also under your first term, but it is not.  (Please read your posts when I tell you you have made an error before you assume that you are right.  You did this for three posts in another topic and it was quite a waste of time.)[/quote]
I didn't assume anything. I tried it and it works.

Also, you're not obligated to waste your time on correcting me. :dry:

Eh, this is getting a bit out of hand.  Either Wiki "Binet's Formula" or look at sequences and series in book 2 (AoPS of course!).  This can be proved using recursion.  locked.


$ x^{6} \minus{}6x^{5} \plus{}15x^{4} \minus{} 20x^{3}\plus{}16x^{2}\minus{}10x\plus{}5$

$ \equal{} x^{6}\minus{}6x^{5}\plus{}15x^{4}\minus{}20x^{3}\plus{}15x^{2} \minus{}6x \plus{} 1 \plus{}x^{2}\minus{}4x\plus{}4$

$ \equal{}(x\minus{}1)^{6} \plus{}(x\minus{}2)^{2} > 0$.

Its least value occurs when $ x\equal{}1$ and this value is $ 1$, so the curve is always above the $ x$-axis and the equation has no real roots.

Just as a thought, I'd substitute $ x\equal{}y\plus{}1$ because that would seem to simplify the equation a lot. After that, the problem is pretty obvious as you'd get $ y^6\plus{}(y\minus{}1)^2$.


trigonometry
logarithms
function
real analysis
real analysis unsolved

continuous $ f: (0,\infty)\to(0,\infty)$ satisfies $ f(t)\equal{}pf(qt)$ for all $ t>0$, where $ p,q$ - positive const.

prove that $ f$ is of the form $ ct^d$, where $ c,d$ - const. dependent on $ p,q$

The assertion is false.  Counterexample:
\[ f(t) \equal{} \sin^2\left(\pi \log_q t\right) t^{ \minus{} \log _q p}
\]
More generally, let $ g: \mathbb{R} \to (0,\infty)$ be any continuous function with period $ \log q$, then
\[ f(t) \equal{} g(\log t)t^{ \minus{} \log_q p}
\]
is continuous, positive, and solves the functional equation.  The particular solution you found is equivalent to choosing $ g(t)\equal{}c$.


algebra
polynomial
function
complex numbers
superior algebra
superior algebra unsolved

Given any $n$ complex numbers $x_{i}$, does there exist a polynomial $f \in \mathbb{Q}[X]$ whose roots approximate the $x_{i}$ arbitrarily good?

a) If $f$ is allowed to have more than $n$ roots, so one may choose a subset of them.

b) If $\deg{f}= n$

Yes either way, by approximating the coefficients of $\prod(x-x_{i})$ with rationals.

Edit: Of course, this only works if the set $(x_{i})$ is closed under complex conjugation. If you don't assume this, then the answer is (a) yes, (b) no.

Lets directly address b):

Assume the roots as real.
Let $\sigma_{k}(x_{1},...,x_{n})$ be the $k$-th elementary symmetrical polynomial in $n$ variables (and give the coefficients up to sign here). Then $f_{k}(y)=\sigma_{k}(x_{1}+y_{1},...,x_{n}+y_{n})$ is a polynomial in $y=(y_{1},...,y_{n})$ (note that $x_{i}$ are the roots), especially $f(y)$ is continous, thus also $y \mapsto (1,f_{1}(y),..., f_{n}(y))$ is a continuous map.
Note that $x_{i}+y_{i}$ is always a root of the polynomial given by $x^{n}+f_{1}(y)x^{n-1}+...+f_{n}(y)$. Now we just have to take small enough $y_{i}$ such that $x_{i}+y_{i}$ is rational and the result follows.
For complex roots (conjugated pairs assumed), the argument still holds, but we have to change two variables simultanously, still not really restricting us.

And what if we demand that $f$ be irreducible?

[quote="ZetaX"]Nothing except having conjugated roots was used.[/quote]I don't understand. You use that the $x_{i}$ are closed under complex conjugation. But why is your $x^{n}+f_{1}(y)x^{n-1}+...+f_{n}(y)$ irreducible?

i) The roots of a complex polynomial are continuous (but non differentiable except in the case of single roots)  functions of the polynomial's coefficients.
ii) The irreducible polynomials are dense everywhere; thus you can take f as an irreducible polynomial.
Example: assume that x1,x2,x3 are the roots of
sqrt(2)*x^3+e*x+pi; then the polynomials
1.414*x^3+2.718*x+3.142,
1.4142*x^3+2.7183+3.1416,
1.41421*x^3+2.71828*x+3.14159
are irreducible over Q (use Maple).
It is necessary an extraordinary bad luck so that does not function (especially when n is large).

Just a small remark why they are dense:
You could always change every coefficient $a_{k}$ of $a_{n}x^{n}+...+a_{1}x+a_{0}$ to some (very) close rational $\hat a_{k}$ with $p| \hat a_{k}$ for $k=1,2,...,n-1$, $p \nmid \hat a_{n}$, $p^{2}\nmid \hat a_{0}$.
Eisenstein then shows that it is irreducible.