MIT college Harvard Support email Duke Princeton

Soon I will have to start applying to colleges as will many students.  Can anyone offer me some suggestions for good colleges that I might be interested in based on the following criteria:

-on the east coast (or close)
-affordable or has good merit scholarships
-good math/science/engineering departments (I don't care much about humanities, etc.)

That's very broad, but any help is appreciated.  Also, below are some colleges that I've found and sound decent and I'm thinking about, please give some input on them if possible:

-MIT
-Princeton
-University of Chicago
-Johns Hopkins
-Duke
-UNC Chapel Hill
-University of Pennsylvania
-Boston University
-Furman
-University of Florida

well the first few on your list are obvious, so I won't comment on those.

If money is a problem, and your "need' isn't sufficient enough to get you a reduced price, then they might not be the best idea, especially if you want a merit scholarship.

I don't know about a lot of those, but I do know about the Universty of Florida because I'm from Florida.

Are you a national merit finalist? If so, UF will treat you like a God... no joke. You get to select classes first before everyone else regardless of credits (a big deal if you attend a larger university such as UF). Of course it's completely free. They pay for your classes, your room and board, your books, and I forget the exact number but I somehow remember like an additional $5,500 per year in cash to spend on whatever. They'll give you the best advisors and cater to your every little need. I've heard from people that's it truly ridiculous.

On another note, UF heavily recruits National Merit (for some reason, dunno why), as last year they were ranked #2 in the nation in national merit freshman behind only Harvard. Now, obviously the percentage is less because the student body is much larger at UF, but hey, the recruiters seem to be proud of it. (I personally am not too big a fan of national merit scholars as I don't think it means anything, and, as you might have guessed, I wasn't one ;); i missed it by 1 point.)

I also know that UF is making a big effort and spending a lot of money trying to break themselves from the stereotypical view that the nation has toward Florida education (even though K-12 definitely is such crap). UF easily towers over every other institution in Florida by a long shot.

There Engineering department is top notch. Math and Physics are good considering it's a public school. Obviously they don't compare to MIT :). I think I heard once that there journalism program was good or something, who knows. But for engineering in particular, it's a good choice. Team CIMAR (UF's Darpa Team) finished a very impressive 8th last year.

Another thing to consider when choosing a college is weather. Let me tell you, I live in South Florida and I hate it. I hate the heat lol. I can't say that I love the frigid cold either. Gainesville has probably some of the nicest weather in the nation overall. It doesn't get too cold and UF isn't really far enough south to get super hot.

Coming from Florida and hearing people talk about the poor education that Florida has, it makes me really happy to know that at least one institution is trying to change Florida's reputation :).

Oh another fact I just found that might interest you. UF's freshman class last year (c/o 2008) had an average SAT score of 1300. When you consider that the class has 6,000+ students, it's public, and it's not one of the "greatly desired" schools, that's pretty damn impressive. It impressed me when I heard it.

Spoon, you go to MIT right? What are your thoughts on it? Obviously it's a very tough school to go to, with excellent departments and students, but how does it fare compared to your expectations?

The University of Georgia has a good math program and offers good merit based scholarships. Also, though they're not really east coast, Rice University in Houston and Washington University in St. Louis are schools you might want to look into. They're good schools that offer a few very good merit based scholarships. I considered both of them when I was looking into colleges.

Go to Duke Joe, that's where i'm planning to go...It's not very far away, it's great if you're planning to become a Doctor, and it has a great basketball team :P

[quote="joml88"]Can anyone offer me some suggestions for good colleges that I might be interested in based on the following criteria:

. . . .
-affordable or has good merit scholarships
. . . .
[/quote]

Hi, joml88, I know you have been following this forum fairly closely for a while, so you already have a conceptual base for thinking about these issues. I'll just comment very specifically about the "affordable" issue. In my day, I was determined to go to college without borrowing any money (for a lot of reasons). As it happened, my parents didn't pay out the full calculated EFC (estimated family contribution) expected for families of our income level, although they did provide financial support for my studies up to my eighteenth birthday (which happened in my freshman year spring quarter). So I had a VERY limited college budget, with neither parents' money nor loans to work with, and my idea of "affordable" was my local state university, which I worked my way through.

In online and private email comments I have seen from Richard Rusczyk, it appears that his family devoted somewhat more resources to his education, and he thought big about where he applied, and the place he was admitted was "affordable" partly in view of taking on a fair amount of debt. (The financial aid policies there have changed to a no-loan policy, which is even more affordable.) But the end of that story is that the debts have long since been paid back. Because a good college education has economic value, gauging how affordable it is should probably include the idea that borrowing money to obtain it is an investment. I have no idea how much your parents can/will contribute to your education--and I'm not asking--but I would offer that thought to keep in mind--list price isn't the whole story about how much college costs. If the college is affordable in the BROADEST sense (that is, you can meet deadlines to pay its bills with your family's maximum creditworthiness applied to that task), then you may as well consider going there if you can get in and if the college has a truly stellar reputation, as some of the colleges you mentioned do.

That's the advice I give my children, and it's the advice I give to parents who email me privately (including one parent of an AoPSer who has been CONSPICUOUSLY successful in math contests this year, and who has gained admission to a very expensive, very famous college that is on your list). Only your family can decide what's worthwhile for them, but that's my friendly advice.

Staying on the east coast is to stay near home mostly, right? That is very much what I did (I usually took a city bus to go to college, living in various places nearby, sometimes at home), but on another thread I might explore rationales for going farther away for college.

Good luck in your application process, and in deciding what to do once you know where you are admitted. Thanks for your many contributions to the forums.

hey joe, you forgot about clemson honors.  :D

and i'm almost 100% sure that i will be a national merit commended scholar (even though they don't release the cutoffs till this fall). does that get anything special or do you need to be a finalist to get the scholarships and whatnot?

203 for last year?

It takes them a year to figure it out?

[quote="tetrahedr0n"]The cutoff for National Merit Commended has been released already, it is 202 (3?). You're probably refering to National Merit Semifinalist.[/quote]

yea. i got a 206. i'm just saying that i'm not gonna be a semifinalist. do i get anything special for being a commended scholar or is it a "you did good but not good enough" award.

Ok, the topic has gone off topic a little.  Start a new thread if you want to talk about National Merit.

Spoon, that info on UF is very much appreciated.  I lived there for 5 years and my parents invested in a college prepaid program (they didn't expect to move to SC).  My sister went in state (in SC, that is) for college and thus if I decided to go to a Florida school I would have a bunch of money to pay for it already.  UF is looking like a very nice option right now for me.

Marcus (or whoever else has an opinion on this), which school do you think is better: University of Georgia or Georgia Tech?  Also, Rice and Washington University in St. Louis sound good also.  I need to remember to look into them.

tokenadult, I'm not really sure how much my parents would help me with college.  I have talked to them about going to a place like MIT which would be expensive.  If I went somewhere like that they would probably work with me on the financial matters but I know that they would rather me not go to such an expensive institution.

Thanks for all the advice so far!

Georgia Tech and UGA are completely different types of schools. It would be hard to compare them. GT is a very popular school for the best math/science students from the south. I think more seniors from the SC All-State Math Team went to GT than any other school last year. The only reason I didn't mention it is because I don't know if they have many/any really good merit based scholarships. I actually applied to GT when I was in high school and closely considered going there.

I know UGA, Rice, and Wash U. all have merit based scholarships that range up to full tuition and board. In some cases you have to specifically apply for them, and in some cases they use the regular applications to decide who gets them.

For right now though, I think you should keep your options open about schools. If you think you might want to go to GT or MIT then you should apply and see what happens. You can make decisions about cost after you see the final financial aid packages from each school.

[quote="tokenadult"]Because a good college education has economic value, gauging how affordable it is should probably include the idea that borrowing money to obtain it is an investment.[/quote]

I think it is important to remember that the idea that "If you graduate from X school you will make more money than if you graduate from Y school because of X's reputation," has been shown to be wrong. Studies have shown that students who are accepted to the most prestigious institutions yet decide to attend a less prestigious school make just as much money coming out as the student who deciced to attend the more prestigious school. Basically the money you make coming out of college depends on you and what you do while in college rather than the college itself. The more I observe in and around college students from all different universities, the more I realize this is true.

[quote]Spoon, that info on UF is very much appreciated. I lived there for 5 years and my parents invested in a college prepaid program (they didn't expect to move to SC). My sister went in state (in SC, that is) for college and thus if I decided to go to a Florida school I would have a bunch of money to pay for it already. UF is looking like a very nice option right now for me.
[/quote]
I also had Florida Pre-Paid, but my parents transfered it to my brother who is entering high school this up coming school year. If you have pre-paid AND national merit, then at UF your education would be free and you'd walk out after 4 years with like $30,000 in your pocket. I wish I was in that position :(

[quote]Studies have shown that students who are accepted to the most prestigious institutions yet decide to attend a less prestigious school make just as much money coming out as the student who deciced to attend the more prestigious school.[/quote]

This is an issue that is extensively discussed elsewhere online among parents. I think the correct factual statement is that ONE study in the United States that sampled a limited universe of schools appeared to show that for cross-admits. BUT, that would mean that the study results can only be something to take to the bank for cross-admits in that universe of schools. (Even at that, the study might just plain be flat wrong; I have some methodological quibbles with that study that I won't go into here.)

Of course "make just as much money" isn't the only gauge of success of a college graduate. For someone who desires an academic career, pursuing fulfillment of personal goals will almost always mean making less money in nominal salary than than someone who pursues a career in private industry. But I find it VERY hard to believe, for instance, that a cross-admit to the University of Minnesota and to Harvard who chooses to go to MN as an undergrad will find it as easy, years later, to become a professor at some second-tier school as the person who did the Harvard undergrad degree. It's not impossible, not at all, to have great career fulfillment after attending any of the schools that have been mentioned in this thread, but it's more difficult, and that is part of the economic value of going to the more famous school--the "foot in the door" it can provide to a person who has competitive personal goals.

But, yes, the many AoPSers who will, alas, NOT make it into their first choice schools next year shouldn't feel demoralized but should maximize their initiative at whatever school they are admitted to. My point was more directed to the person who applies to and obtains admission to an expensive but famous school. The expense will be a lot more temporary than the fame, I think, and may very well be worth paying a premium for.

joml88, you made USAMO, right? So that means that you are a supergenius. Maybe you should consider the West Coast--specifically, CalTech. The admissions officer told me they had lots of merit money (Axline). I also asked him about USAMO quals, and he told me that there were not a significant amount. So you would probably also get money from them, from full tuition up.

[quote="sran"]Maybe you should consider the West Coast--specifically, CalTech.[/quote]

*gags* *grabs throat*

[quote="Spoon"]
I think it is important to remember that the idea that "If you graduate from X school you will make more money than if you graduate from Y school because of X's reputation," has been shown to be wrong. [/quote]

I suspect this is very highly dependent upon profession.  Besides, money isn't the only reason to aim high in terms of undergraduate school.

[quote="Spoon"]Yes, I was assuming a private industry job (e.g. business owner, consultant etc.), definitely not one that is academically oriented. And, I mentioned the aspect of money because the topic that I quoted was strictly about going into debt. Of course there are always benefits to attending one college over another.[/quote]

I think it varies wildly within private industry careers, as well.  I would actually guess that the variance is smaller in academia and far more based on *graduate* institution than undergrad.

Also, you'd have to look very closely at how the study is done.  If, for example, it argues that 'Ivy leaguers in profession X make the same as non-Ivies in profession X', this doesn't really say anything if profession X is highly selective.  If the Ivy leaguers are twice as likely to get the breakthrough starting position in the profession than the non-Ivy leaguers (which is probably an underestimate for a profession like hedge-fund trader), then clearly this study misses something important.  (Note: I'm intending 'Ivy league' to be a catch-all for all 'top-tier' schools.)

[quote="tokenadult"][quote]Studies have shown that students who are accepted to the most prestigious institutions yet decide to attend a less prestigious school make just as much money coming out as the student who deciced to attend the more prestigious school.[/quote] [/quote]
These studies are averages of large (or sometimes, shockingly small) numbers of people.   At the upper outlier end of income, jobs, and schools, it can make a large, definite and hard-to-substitute difference where you go to university.   Closer to the middle, the student's abilities and exploitation thereof count more than the particulars of the certificate, so one doesn't expect studies to detect much on the average except some minor "added value" of a top 10 school over a top 50,100, or 500 school (which may anyway be overbalanced by higher fees at the fancy schools).

[quote]But I find it VERY hard to believe, for instance, that a cross-admit to the University of Minnesota and to Harvard who chooses to go to MN as an undergrad will find it as easy, years later, to become a professor at some second-tier school as the person who did the Harvard undergrad degree.[/quote]
Believe it.  What matters for academic employment happens at (and past) the Ph.D level, not undergraduate.  The Harvard graduate would have an easier time getting to any given graduate school, all other things being equal, but once there there is no disadvantage to the UMN alumnus.   The Harvard alum will have more college friends who are professors, but the hiring process is not so simple that this alone can compensate for (e.g.) an inferior publication history or Ph.D thesis compared to the UMN graduate.  Also, the playing field at the first- and second-tier university professor level is international and highly competitive, so the Harvard fellow's college friends who are professors will mostly be at third-tier institutions and not necessarily in a position to help with employment at the types of schools you're talking about.
In other words, a degree from Harvard, Cambridge, etc  is nothing special within the professoriate, though it doesn't hurt.

Joml88,

Among the schools you listed here, I would put Duke as the first choice for your situation. Duke is a very fine school, probably better than some of the Ivy leagues. Duke does provide several merit scholarships. One of them, The Angier B. Duke Memorial Scholarships, is nationally based, which provides full tuition for 4 years (total today around $160,000 for 4 years) to about 15 new students. Since this is a nationally based merit scholarship, it is extremely difficult to get. However, Duke provides additional merit scholarships for similar amount specifically for students from NC and SC. These scholarships provide about additional 15-20 students from NC and SC. Since you are from SC and you have superb qualification (as a lone USAMO qualifier from SC; I assume you have excellent SAT score and GPA), your chance of getting one of those merit scholarships is great. By going to Duke, you do not lose any future opportunities (regardless of academic career or ability to make money in the future) as long as you continue to keep up with your academic achievements.

For most people, making significant more money would require going to professional schools. In that regard, what ever the undergraduate school a person attends will not make too much of a difference. In the end, it is really not the specific school, but the specific career choice will make the real difference in ones earning. For example, doctors with MDs from state medical schools will generally make significant more than math professors with PhDs from Harvard/Princeton. Doctors with MDs from state medical schools will generally not make any less than doctors with MDs from Harvard medical school. A person’s earning does not correlate that well with his math/ science IQ, but involving multiple other factors, such as verbal ability, wiliness to work hard, personality, considerate for others, presentable, etc. (Anyway, this paragraph is off topic).

Duke also has very strong professional schools. Duke has combined medicine/law (6 years), medicine/MBA (5 years) and law/MBA. For joml88, I believe Duke is the most practical choice for its quality and your probability of getting merit scholorship. Good luck.

sran, yes I did qualify for USAMO this year.  Do you really think that I could get a full ride from Cal Tech?  I mean, is it reasonably likely to happen?  If so I might look into that option.

Thanks qvc122.  Duke [i]is[/i] near the top of the list (though I don't really have a number 1 school right now).

Also, does anyone have any input on The University of Michigan?  Their math department has sent me letters because of my AIME performance and I don't believe I've gotten anything from other schools because of AIME.  In fact, they just sent me a letter offering to pay for a trip for a trip with one of my parents to their campus for a day or two with hotel, food, and transportation paid by them.

This to me shows that they have a pretty ambitious math department.  So any comments?

Without going on anything but reputation and course selection, I would say that UMich is an excellent math university.  Of course, you can't go wrong by visiting it.

As for Caltech, you're probably best off assessing your chances by seeing if you can find the achievements of some of the other Axline winners.  I will note that Caltech gives out a lot less financial aid than other elite schools, at least as far as I've heard.

[quote="blahblahblah"]Caltech gives out a lot less financial aid than other elite schools, at least as far as I've heard.[/quote]

I could have gone there free.
so I guess I have to disagree on that.

I know two girls who got into Caltech and Princeton and MIT and a few other schools.  One of them ended up going to Caltech, the other to Princeton, but in both cases, Caltech was ~$10000 CDN per year more after financial aid.

They have a very good (Top- 10) math department with a tough course sequence for honors math majors (MTH 295- 396), and I'd like to go there (I'm in-state), but I don't think it compares to Harvard/ Princeton/ MIT/Caltech/ Duke/ Chicago...etc.  The fee is high ($40,000ish non-state), though.

University of Michigan is one of my top six schools I'd like to go to for graduate school. (I haven't ranked them in any way, so I can't give you anything more precise than that.) So at least I think they have a good graduate program. And one of my advisors got his PhD there.

[quote="blahblahblah"]Without going on anything but reputation and course selection, I would say that UMich is an excellent math university.  Of course, you can't go wrong by visiting it.[/quote]
I have a close friend who turned down MIT, CalTech, and Havard to go to UMich, but he also had a athletic scholarship (he does pole vault). He's going into his 4th year doing a double M.S. in math and physics. I hear from him that UMich is a great school, the only downfall being the extreme cold (even colder than Boston, ouch). He, like me (and like you!) is from Florida, so it's a hard switch. You definitely can't go wrong there.

$blah^3$ As some one has mentioned before, Caltech  (and Duke) offers meriit scholarships which covers full tution  (and  , some times, more)..Of course  only a  very small number of students get selected for the scholarships.

I have a whole bunch of friends who are UM math majors. They seem to really enjoy the program, and the honors math sequence is supposedly fairly challenging. Plus, UM has merit scholarships, some of them specifically for math students.

[quote="Spoon"] hear from him that UMich is a great school, the only downfall being the extreme cold (even colder than Boston, ouch). [/quote]

It really isn't that bad. There's only a few days a year it goes below 0, and most of the winter its 20-30 degrees. I actually hate the summer more, since it's humid and the temperature has been in the 90s for a while now.


geometry
3D geometry
calculus
real analysis

Hey there,

I'm looking for a proof for the surface area of a cone:

Surface Area = Pi * Radius * Slantheight

Does anyone know where I can find such a proof?
Any hints?

By the way, I am currently learning about antidervatives, so the proof will need to be accomplished w/ antiderivs.

Thanks!

Riemann sum the changing circumference as you slide down the cone.

(But the non-calculus proof is simple.  It requires only the insight that the lateral surface area of a cone can be "rolled out" into 2-D. )


I posted about five or six questions on getting started level and I found it very fun and learning thing to do..
But, I wonder like how am I going to post the difficult or more advanced question, and I want to see opinions..

Especially for authors of this site and the book Art of Problem Solving, I want to know what made you to publish the book.. I think it's really great but very difficult thing to do....

I LOVE MATH PROBLEM AND TO CREATE ONESSSSS  :mrgreen:

Sandor and I wrote and published the book as a result of our experience in the first 3 years of the Mandelbrot competition.  The first 3 years were way too hard, and the scores were very low.  We realized that there was a pretty large body of ideas that most students hadn't seen, which Sandor and I had picked up over years of solving and writing problems.  So we decided to write a book - the book we wish we'd had when we were 13-16.  It became 2 books, then 2 more solution manuals.  It was a lot bigger task than we had imagined.  When we finally finished, we started thinking about software.  But that was 1994, and the tools weren't really there for two guys who knew little about computers to do something like, well, this site.  Now it's 2004, and the tools are out there.

As for writing problems, I've easily learned as much math trying to write problems as trying to solve them.  Even better for learning is trying to teach people how to solve problems.  I've learned more math in the last year than any year before in my life - because I know the students will ask exactly whatever it is that I don't completely understand.  Give it a try.

Interesting.. If your text became like textbook for advanced math in school, do you hope to publish the third edition of the art of problem solving?
I'm very interested in people who done something interesting, and I'm very curious on mathematics...

And for creating problems, you said teaching is more efficient way to learn the the math, right?  I definitely agree that teaching is good way too, but like is there any other way I can produce better quality of question?

[quote="Silverfalcon"]And for creating problems, you said teaching is more efficient way to learn the the math, right?  I definitely agree that teaching is good way too, but like is there any other way I can produce better quality of question?[/quote]

I don't think he meant that teaching is an efficent way to learn more math.  In my experience, it's the other way around - it often takes a lot of work to effectively teach math.  However, teaching is a [b]thorough[/b] way of learning math, because you need to be prepared for any questions that your students ask of you.

[quote="mathfanatic"]it often takes a lot of work to effectively teach math. [/quote]

This is exactly why I say teaching is such an effective way to learn.  You're forced to learn it yourself much more deeply - you can't just duck the tough questions.

[quote="Silverfalcon"]do you hope to publish the third edition of the art of problem solving?
[/quote]

I consider this site the third volume (and fourth, and fifth, and 0th, and expansions of 1st and 2nd) of the books.

[quote="rrusczyk"] Give it a try.[/quote]

Hmm... Since you've done it before, is there a tip for making good problems because I know better the quality of problems, more I must learn to produce them...

[quote="masamune2387"]Want to start an inequality problem creating chain?[/quote]

what's that~~~


Regions I, II and III are bounded by squares. The perimeter of region I is 12 units and the perimeter of region II is 24 units. What is the ratio of the area of region I to the area of region III? Express your answer as a common fraction.

[asy]draw((0,0)--(9,0)--(9,9)--(0,9)--(0,0)--cycle,linewidth(2));
draw((9,0)--(15,0)--(15,6)--(9,6)--(9,0)--cycle,linewidth(2));
draw((9,6)--(12,6)--(12,9)--(9,9)--cycle,linewidth(2));
label("III",(4.5,4.5));
label("II",(12,3));
label("I",(10.5,7.5));[/asy]

The perimeter of region 1 is $ 12$ so you know the side length of the square enclosing region 1 is $ 3$. If we do the same to region 2, we find that the square enclosing region 2 has a side length of $ 6$. By adding these two side lengths together, we find the side length of the square enclosing region 3 has a side length of $ 9$. Thus, we find that the area of region 3 is $ 9$ and the area of region 3 is $ 81$, so the ratio of the area of region 1 to region 3 is $ 1/9$.


\[ \begin{array}{l}
 find\_\min imum\_\& \_\max imum\_of \\
 \frac{{x \plus{} 1}}{{xy \plus{} x \plus{} 1}} \plus{} \frac{{y \plus{} 1}}{{yz \plus{} y \plus{} 1}} \plus{} \frac{{z \plus{} 1}}{{xz \plus{} z \plus{} 1}} \\
 \end{array}
\]

[quote="anorithdialga"]
\[ \begin{array}{l} find\_\min imum\_\& \_\max imum\_of \\
\frac {{x \plus{} 1}}{{xy \plus{} x \plus{} 1}} \plus{} \frac {{y \plus{} 1}}{{yz \plus{} y \plus{} 1}} \plus{} \frac {{z \plus{} 1}}{{xz \plus{} z \plus{} 1}} \\
\end{array}
\]
[/quote]
It not exists!
$ \sup\left(\frac {{x \plus{} 1}}{{xy \plus{} x \plus{} 1}} \plus{} \frac {{y \plus{} 1}}{{yz \plus{} y \plus{} 1}} \plus{} \frac {{z \plus{} 1}}{{xz \plus{} z \plus{} 1}}\right)\equal{}\plus{}\infty,$
$ \inf\left(\frac {{x \plus{} 1}}{{xy \plus{} x \plus{} 1}} \plus{} \frac {{y \plus{} 1}}{{yz \plus{} y \plus{} 1}} \plus{} \frac {{z \plus{} 1}}{{xz \plus{} z \plus{} 1}}\right)\equal{}\minus{}\infty.$ :wink:

[quote="anorithdialga"]

$ x,y,z$
are positive reals[/quote]
It still does not exist.

$ \sup\left(\frac {{x \plus{} 1}}{{xy \plus{} x \plus{} 1}} \plus{} \frac {{y \plus{} 1}}{{yz \plus{} y \plus{} 1}} \plus{} \frac {{z \plus{} 1}}{{xz \plus{} z \plus{} 1}}\right)\equal{}3$ ( $ x\equal{}y\equal{}z\rightarrow0^\plus{}$ ) and
$ \inf\left(\frac {{x \plus{} 1}}{{xy \plus{} x \plus{} 1}} \plus{} \frac {{y \plus{} 1}}{{yz \plus{} y \plus{} 1}} \plus{} \frac {{z \plus{} 1}}{{xz \plus{} z \plus{} 1}}\right)\equal{}0$ ( $ x\equal{}y\equal{}z\rightarrow\plus{}\infty.$ )


calculus
function
integration
real analysis
real analysis unsolved

If f(x) = (1 if x is rational) and (0 if x is irrational) on the interval [0,12]... how do you find upper and lower partitions?

Ah, the classic example of a function that's not Riemann integrable. Making the interval $[0,12]$ sounds like whoever posed this problem was bored with the usual - not that it makes any real difference.

The key observation: any interval of positive length contains both rational and irrational numbers. Hence the supremum of $f$ over that interval is 1 and the infinum is 0.

Let $P$ be any partition of $[0,12].$ Then

$U(f,P)=\sum_{j}1\cdot\Delta x_{j}=12$ and

$L(f,P)=\sum_{j}0\cdot\Delta x_{j}=0$

The upper integral is 12, the lower integral is zero, and the function is not Darboux integrable (hence not Riemann integrable).


calculus
integration
real analysis
real analysis unsolved

hello, the integral is $ \frac{x^2\plus{}2x\plus{}1\plus{}xe^x}{e^x\plus{}x\plus{}1}\minus{}\ln(e^x\plus{}x\plus{}1)$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, the integral is $ \frac {x^2 \plus{} 2x \plus{} 1 \plus{} xe^x}{e^x \plus{} x \plus{} 1} \minus{} \ln(e^x \plus{} x \plus{} 1)$.
Sonnhard.[/quote]
could you plz show the working . Thx

Let's start by noting that

$ \left(\frac{e^x}{e^x \plus{} x \plus{} 1}\right)' \equal{} \frac{xe^x}{(e^x \plus{} x \plus{} 1)^2}$

and $ [(x\plus{}1)e^{\minus{}x}]' \equal{} \minus{}xe^{\minus{}x}$

and so

$ P \equal{} \int\frac{x(x\plus{}1)}{(e^x \plus{} x \plus{} 1)^2}\,dx \equal{} \int\frac{xe^x \cdot (x\plus{}1)e^{\minus{}x}}{(e^x \plus{} x \plus{} 1)^2}\,dx \equal{} \frac{e^x \cdot (x\plus{}1)e^{\minus{}x}}{e^x \plus{} x \plus{} 1} \minus{} \int\frac{e^x \cdot (\minus{}xe^{\minus{}x})}{e^x \plus{} x \plus{} 1}\,dx$

$ \equal{} \underbrace{\frac{x\plus{}1}{e^x \plus{} x \plus{} 1}}_{\equal{} A} \plus{} \int\frac{x}{e^x \plus{} x \plus{} 1}\,dx \equal{} A \plus{} \int\frac{x \plus{} (e^x \plus{} 1) \minus{} (e^x \plus{} 1)}{e^x \plus{} x \plus{} 1}\,dx$
$ \equal{} A  \plus{} \int\frac{e^x \plus{} x \plus{} 1}{e^x \plus{} x \plus{} 1}\,dx \minus{} \int\frac{e^x \plus{} 1}{e^x \plus{} x \plus{} 1}\,dx \equal{} A \plus{} x \minus{} \ln(e^x \plus{} x \plus{} 1) \plus{} C$

and so

$ P \equal{} \frac{x\plus{}1}{e^x \plus{} x \plus{} 1} \plus{} x \minus{} \ln(e^x \plus{} x \plus{} 1) \plus{} C$.

Could you explain the following step in a bit more detail   :blush:  Thx.
[quote="Carcul"] $ \int\frac {xe^x \cdot (x \plus{} 1)e^{ \minus{} x}}{(e^x \plus{} x \plus{} 1)^2}\,dx \equal{} \frac {e^x \cdot (x \plus{} 1)e^{ \minus{} x}}{e^x \plus{} x \plus{} 1} \minus{} \int\frac {e^x \cdot ( \minus{} xe^{ \minus{} x})}{e^x \plus{} x \plus{} 1}\,dx$ [/quote]

I have just used integration by parts:

$ \int\frac {xe^x \cdot (x \plus{} 1)e^{ \minus{} x}}{(e^x \plus{} x \plus{} 1)^2}\,dx \equal{} \left(\int\frac {xe^x}{(e^x \plus{} x \plus{} 1)^2}\right) \cdot (x \plus{} 1)e^{ \minus{} x} \minus{}$

$ \minus{} \int\left(\int\frac {xe^x}{(e^x \plus{} x \plus{} 1)^2}\right) \cdot [(x \plus{} 1)e^{ \minus{} x}]'\,dx$.


Let the equation $\sqrt{x}+\sqrt{x-4}-\sqrt{x-1}-\sqrt{x-3}=a\sqrt{\frac{x-3}{x}}$.($a$ is a parameter).
How many roots does it have?

hello, let $ f_1(x)\equal{}\sqrt{x}\plus{}\sqrt{x\minus{}4}\minus{}\sqrt{x\minus{}1}\minus{}\sqrt{x\minus{}3}$ and $ f_2(x)\equal{}a\sqrt{\frac{x\minus{}3}{x}}$, then we get
$ f_1(x)$ is for all $ x \in \mathbb{R}$ with $ x\geq4$ strictly monotonoulsy increasing and negative, hence we have for $ a\geq0$ no real solution  of our equation $ f_1(x)\equal{}f_2(x)$. Since $ f_1(0)\equal{}1\minus{}\sqrt{3}$ and $ f_2(x)$ is for $ a<0$ strictly monotonously decreasing we have exactly one real solution if $ 2(1\minus{}\sqrt{3}) \le a<0$ (Hint: we have $ f_2(4)\equal{}\frac{a}{2}$). In the case $ \minus{}\infty<a<2(1\minus{}\sqrt{3})$ the equation has also no real solution.
Sonnhard.


Which integers can be written as a difference of two squares?

[quote="Rust"]$4k=(k+1)^{2}-(k-1)^{2}$ and $2k+1=(k+1)^{2}-k^{2}.$
It give $n=p^{2}-q^{2}$ if and only if $n\not =2(mod 4)$.[/quote]

How to prove the "only if" part?

If $n=p^{2}-q^{2}$ and p, q are even, then $n=0(mod 4)$. If one of them even, another odd, then n is odd. If p,q are odd, then $n=0(mod 8)$. For all cases $n\not =2(mod 4)$.


Dla każdego skończonego zbioru $Z$ liczb natruralnych dodatnich definiujemy $\mathrm{NWD}(Z)$ jako największy wspólny dzielnik elementów tego zbioru, przy czym przyjmujemy $\mathrm{NWD}(\{ a\})=a$, $\mathrm{NWD}(\emptyset)=1$.
Niech $n$ będzie liczbą naturalną dodatnią. Udowodnić, że dla dowolnych liczb $N_{0},\ldots,N_{n-1}\in\mathbb{N}^{+}$ zachodzi nierówność \[\prod_{Z\subset\{0,\ldots,n-1\}\atop |Z|\equiv 1\pmod{2}}\mathrm{NWD}\left(\left\{N_{i}: i\in Z\right\}\right)\geqslant \prod_{Z\subset\{0,\ldots,n-1\}\atop |Z|\equiv 0\pmod{2}}\mathrm{NWD}\left(\left\{N_{i}: i\in Z\right\}\right).\] Podobna nierówność (różniąca się jedynie tym, że oba $\mathrm{NWD}$ są podniesione do $|Z|$) jest [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=106173]tutaj[/url]. Jak na razie nierozwiązana.

To dość proste w sumie, nazywa się zasadą włączeń i wyłączeń z tego co wiem.
A polega na tym, że jak podzielimy lewą stronę przez prawą i sprawdzimy ile razy w ilorazie występuje jakaś arbitralnie wybrana liczba pierwsza $ p$ to okaże się, że tyle, ile w największej potędze $ p$ mniejszej od n. Dlaczego? Otóż:
Najpierw dodajemy ile razy występuje w każdej liczbie, ale okazuje się, że dodaliśmy o wiele za dużo, zatem odejmujemy to ile razy występuje w każdej możliwej parze, ale okazuje się że odjęliśmy za dużo itd, nie chce mi się tego formalnie dowodzić, zasada włączeń i wyłączeń jest dość intuicyjna mam nadzieje.
Ta nierówność zachodzi jeśli sumujemy po zbiorach na dowolnych liczbach, nie tylko na kolejnych mniejszych od n.


function
trigonometry
analytic geometry
graphing lines
slope
LaTeX
real analysis

If $ f$ and $ g$ are uniformly continuous on $ \mathbb{R}$ to $ \mathbb{R}$ show that $ fg$ need not be uniformly continuous on $ \mathbb{R}$ even when one of $ f$ and $ g$ are bounded.

If $ f$ and $ g$ are uniformly continuous on a closed bounded interval $ [a,b],$ then $ fg$ is uniformly continuous on that interval - this follows from the fact that merely being continuous on such a set implies uniform continuity. So our example will have to involve something going bad as $ x\to \infty.$

Let's see: something going bad at $ \infty,$ but one of the functions is bounded. We need a bounded uniformly continuous function that isn't too "nice" at infinity - having a limit at $ \infty$ would be too nice. All right. try $ f(x)\equal{}\sin x.$ But they didn't say both bounded - hmm, maybe there's a theorem there. So an unbounded uniformly continuous function. Might as well be a function of constant slope. All right $ g(x)\equal{}x.$

So is $ x\sin x$ uniformly continuous on $ \mathbb{R}?$ Check it out; pay particular attention to the slope near the places where it crosses the axis.

---

And maybe there is a theorem.

Claim: if $ f$ and $ g$ are both uniformly continuous and both bounded on a set $ S,$ then $ fg$ is uniformly continuous on $ S.$

Can you prove it?

then for every $ \epsilon > 0$,

$ |f(x)g(x) \minus{} f(a)g(a)|\equal{}|f(x)g(x) \minus{}f(x)g(a)\plus{}f(x)g(a)\minus{} f(a)g(a)|$
$ \le|f(x)g(x) \minus{}f(x)g(a)|\plus{}|f(x)g(a)\minus{} f(a)g(a)|
\equal{}|f(x)\parallel{}g(x)\minus{}g(a)|\plus{}|g(a)\parallel{}f(x)\minus{}f(a)|\equal{}M|g(x)\minus{}g(a)|\plus{}M|f(x)\minus{}f(a)|$.

and since both are bounded then there exists $ \delta_{1} >0$ s.t $ |x\minus{}a|<\delta$ and we can pick $ |f(x)\minus{}f(a)|<\epsilon/2M$ and similarly with$ |g(x)\minus{}g(a)|$.

then we should get
$ M|g(x)\minus{}g(a)|\plus{}M|f(x)\minus{}f(a)|<M(\epsilon/2M\plus{}\epsilon/2M)$. which shows that $ fg$ when both are bounded, is uniformly continuous.

So should I use a counterexample to say something about $ fg$ when only one of the $ f,g$ is bounded or should I cite the proof above?

You need a counterexample. The proof may sometimes guide you in how to construct a counterexample when some of the hypotheses fail, but it doesn't actually say anything about those cases.

(Note- long $ \text{\LaTeX}$ expressions like that look better when you break them up. In a document, that would likely run off the edge of the page.)

I decided I like this example instead:

$ f(x)\equal{}\frac1x\sin(x^2),\ g(x)\equal{}x.$

$ f'(x)\equal{}2\cos(x^2)\minus{}\frac{\sin(x^2)}{x^2}.$ This is bounded, so $ f(x)$ is Lipschitz, hence uniformly convergent. (Yes, it's even OK at $ 0$ if we make the natural extension, not that that's the focus of our attention.)

Let $ h(x)\equal{}f(x)g(x)\equal{}\sin(x^2).$

Let $ x_n\equal{}\sqrt{\left(n\plus{}\frac12\right)\pi}.$ Then $ h(x_n)\equal{}(\minus{}1)^n$ and $ |h(x_n)\minus{}h(x_{n\plus{}1}|\equal{}2.$

But $ |x_n\minus{}x_{n\plus{}1}|\equal{}\sqrt{\pi}\left|\sqrt{n\plus{}\frac12}\minus{}\sqrt{n\plus{}\frac12}\right|\equal{}O(n^{\minus{}1/2}).$

This tends to zero as $ n\to\infty.$ Finding points $ x_n$ such that $ |x_n\minus{}x_{n\plus{}1}|$ tends to zero while $ |h(x_n\minus{}h(x_{n\plus{}1}|$ remains bounded away from zero contradicts uniform convergence.

Hey Kent,

I have a question out of curiosity. You previously said that "being continuous on such a set implies uniform continuity"--in which the set in question was a closed set. However, $ f(x)\equal{}x^2$ is not uniformly continuous, so that does that mean that on some bounded interval $ [a,b]$ that it is uniform continuous?

Would you mind explaining?

Thanks

A function that is continuous on a [b]compact[/b] set is necessarily uniformly continuous. In $ \mathbb{R}$ (or for that matter, $ \mathbb{R}^n$), the compact sets are the sets that are closed and bounded.

$ f(x)\equal{}x^2$ is uniformly continuous on every bounded subset of $ \mathbb{R},$ but it is not uniformly continuous on all of $ \mathbb{R}.$


There are exactly twelve children in Ryan’s class. Only four of the children are boys. The following questions refer to a time when all the children are present in the class. There are no visitors in the class. There might be more than one correct anwer to a question.

1. Which of the lettered statements must be true?

2. Which of the lettered statements cannot be true?

3. Which of the lettered statements could be true or not true?

a. There are twice as many girls as boys in Ryan’s class.

b. There are eight more girls than boys in Ryan’s class.

c. There are four more girls than boys in Ryan’s class.

d. If Ryan is sitting at a table with all the girls, there are exactly nine children at that table.

e. If only three of the boys are standing on their heads, one of the boys is not standing on his head.

This is really for elementary school, not for pre-olympiad.


LaTeX
trigonometry
geometry
trig identities
Law of Sines

Equilateral triangle ABC has sides of length 6. Squares ABMN and BCXY are drawn on the triangle's exterior. Find the distance between M and Y. I got [hide="this."]$ \boxed{6\sqrt{3}}$. I drew triangle BYM, with sides 6, 6, and MY. Because angle MBY = $ \frac{2\pi}{3}$, I saw that the triangle was one third of a larger equilateral triangle, and I solved accordingly.[/hide]  Please confirm, or tell me if there's an easier/different method. Also, what is the LaTeX code for the degree symbol. (e.g., 120 degrees  :) )

To do degrees:
[code]
$120^{\circ}$
[/code]

This comes out as $ 120^{\circ}$

Draw perpendicular from $ B$ to $ MY$. Call intersection $ E$. The perpendicular from $ B$ is the bisector of angle $ MBY$ and it also bisects line $ MY$.Then angle $ MBE \equal{} 60$ and triangle $ MBE$ is a special triangle 30-60-90, then since $ MB\equal{} 6$ It follows that $ ME\equal{}3 \sqrt{3}$ So $ MY \equal{}2*3\sqrt{3} \equal{} 6\sqrt{3}$. I think this is easier..

This is my solution;

If you connect M and Y you form triangle BMY. We know that angle MBY is $ 120^{\circ}$ by subtracted the sixty degree angle and the two ninety degree angles from 360 degrees. We know that the other two angles are the same, because the triangle is isoceles. Using algebra, you find that one of the other two angles is 30 degrees, so the third angles is as well. So the triangle is a 120-30-30 obtuse triangle. We know three angles and two sides, so we can{x} use the law of sines to find the third side length;

$ \frac{sin30^{\circ}}{6} \equal{} \frac{sin120^{\circ}}{x}$

$ sin30^{\circ} \equal{} \frac{1}{2}$

and

$ sin120^{\circ} \equal{} \frac{1}{2}$

... But that means $ x \equal{} 6$

Okay what did I do wrong?  :(

Oh!...

So we get;

$ \frac{\frac{1}{2}}{6} \equal{} \frac{\frac{\sqrt{3}}{2}}{x}$

Cross multiplying yields;

$ \frac{x}{2} \equal{} 3\sqrt{3}$

$ x \equal{} 6\sqrt{3}$

Yes!!!

How does $ \sin120^{\circ} \equal{} \frac {\sqrt {3}}{2}$

Since $ \sin 180^{\circ} \minus{} \alpha \equal{} \sin \alpha$

We have that $ \sin 120^{\circ}\equal{}\sin 180^{\circ}\minus{}60^{\circ}\equal{}\sin 60^{\circ}$

You could also draw the unitary circle and then draw the angle and see that it is definitely not $ 30^{\circ}$

Well, to be honest, the unit circle was how I reached the conclusion that sin120 was 1/2 in the first place. Thanks though! :)


geometry
incenter
circumcircle
geometry unsolved

[quote="Vorelin Tancu"]Let circle $ O_1,O_2,O_3$ have a common point $ O$. Let $ ABC$ triangle such that sides $ AB$ tangent with $ O_1$ and $ O_2$, $ BC$ tangent with $ O_2$ and $ O_3$, $ CA$ tangent with $ O_3$ and $ O_1$
Prove that points $ O$, incenter of $ ABC$ and circumcenter of $ ABC$ are collinear[/quote]

You should require an additional condition, e. g. that the circles $ O_1$, $ O_2$, $ O_3$ are congruent. See http://www.mathlinks.ro/Forum/viewtopic.php?t=179930 for this case. Did you mean a different condition?

  darij


complex analysis
complex analysis unsolved

I tried to compute the residues for cot(piZ) over the entire complex plane and I got an answer of (1/pi) but the other guy I work with  says the answer is that there is no residue at infinity. Who's right?

$\infty$ is not an isolated singularity of the cotangent. Residues only make sense at isolated singularities.


$ a,b,c \in (0,1]$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$
Prove $ \frac {1 \minus{} b^2}{a} \plus{} \frac{1 \minus{} c^2}{b} \plus{} \frac{1 \minus{} a^2}{c} \le \frac {5}{4}$
equality holds when $ a \equal{} 1,b \equal{} \frac {1}{2},c \equal{} \frac {\sqrt {3}}{2}$ ...

[quote="hxy09"]$ a,b,c \in (0,1]$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$
Prove $ \frac {1 \minus{} b^2}{a} \plus{} \frac {1 \minus{} c^2}{b} \plus{} \frac {1 \minus{} a^2}{c} \le \frac {5}{4}$
equality holds when $ a \equal{} 1,b \equal{} \frac {1}{2},c \equal{} \frac {\sqrt {3}}{2}$ ...[/quote]
Sorry but if $ b \equal{} c \equal{} 10^{ \minus{} 9}$ so the $ LHS$ will be greater than $ 10^9$.
May be I wrong ?.But where? :)
P/s;A stupid question,I missed that condition,Sorry for all.It is realy hard

[quote="Allnames"][quote="hxy09"]$ a,b,c \in (0,1]$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$
Prove $ \frac {1 \minus{} b^2}{a} \plus{} \frac {1 \minus{} c^2}{b} \plus{} \frac {1 \minus{} a^2}{c} \le \frac {5}{4}$
equality holds when $ a \equal{} 1,b \equal{} \frac {1}{2},c \equal{} \frac {\sqrt {3}}{2}$ ...[/quote]
Sorry but if $ b \equal{} c \equal{} 10^{ \minus{} 9}$ so the $ LHS$ will be greater than $ 10^9$.
May be I wrong ?.But where? :)[/quote]

substituting $ b$ and $ c$ in $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$ you'll get that $ a>1$ which contradicts with $ a,b,c \in (0,1]$. So we can't take $ b\equal{}c\equal{}10^{\minus{}9}$

In my opinion ,this is the hardest 3-variable inequalities in the history of China team selection.Most of students had no idea towards this ineq(my teacher told me about it) I deem that it is really a challenging one.I present it to all inequalities-lovers.And hope we will get nice solutions here :)

[quote="hxy09"]$ a,b,c \in (0,1]$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$
Prove $ \frac {1 \minus{} b^2}{a} \plus{} \frac {1 \minus{} c^2}{b} \plus{} \frac {1 \minus{} a^2}{c} \le \frac {5}{4}$
equality holds when $ a \equal{} 1,b \equal{} \frac {1}{2},c \equal{} \frac {\sqrt {3}}{2}$ ...[/quote]
equality holds when $ a \equal{} 1,b \equal{} \frac {1}{2},c \equal{} \frac {\sqrt {3}}{2}$,so I think we should use trigonometrical method to solve this(1=cos45,$ \frac {1}{2}$=cos60,$ \frac {\sqrt {3}}{2}$=cos30)

Strange one  :roll: The first we call:
\[ f(a,b,c) \equal{} \frac {5}{4} \minus{} \frac {1 \minus{} b^2}{a} \minus{} \frac {1 \minus{} c^2}{b} \minus{} \frac {1 \minus{} a^2}{c}
\]
I have tried to prove that $ f(a,b,c)$ have the minimum value at ponit is $ 1$. But this one is fail ... Because we have:
\[ f(a,b,c) \minus{} f(1,b,c) \equal{} \frac {(1 \minus{} a)[a^2 \plus{} a \plus{} c(1 \minus{} b^2)]}{ac}\le 0,\forall a,b,c\in (0;1]
\]
But we need $ f(a,b,c) \minus{} f(1,b,c)\ge 0$  :o

Try another way, we have:
\[ f(a,b,c) \minus{} f(a,1/2,c)\ge 0 \forall b\ge 1/2
\]
There are a lot of ideas but really hard to find out a nice way for the moment. I will try later  :)

Another one:
Find the maximum value of $ f(a,b,c),\forall a,b,c\in (0,1]$.

[quote="BG Yoda"][b]zaizai-hoang[/b], notice that for 2 fixed variables, then the third is fixed, too. Therefore, you can't change only one of the variables.[/quote]
I misunderstand your opinion  :maybe: Some problems can be solve easily by the same way. For example $ f(a,b,c)\ge f(a,b,0)\ge 0,etc...$

If you substitute $ c \equal{} 0$, then $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$ wouldn't be true anymore for the initial $ a$ and $ b$ and the new $ c \equal{} 0$.  :!: Therefore, if you change $ c$, you have to change the other variables, too.

Possible solution:

 Let $ b \equal{} \min\{a,b,c\}$.
 If $ a \equal{} 1$ $ \Rightarrow \frac {1 \minus{} c^2}{b} \equal{} b$ and the inequality will be reduced to $ 1 \minus{} b^2 \plus{} b\le \frac {5}{4}\Leftrightarrow 0\le \left(b \minus{} \frac {1}{2}\right)^2$.

 So we may conjecture that $ \frac {1 \minus{} b^2}{a} \plus{} \frac {1 \minus{} c^2}{b} \plus{} \frac {1 \minus{} a^2}{c} \le 1 \plus{} b \minus{} b^2\le \frac {5}{4}$.

 $ A \equal{} \frac {1 \minus{} b^2}{a} \plus{} \frac {1 \minus{} c^2}{b} \plus{} \frac {1 \minus{} a^2}{c} \minus{} 1 \minus{} b \plus{} b^2 \equal{}$ $ b^2\frac {a \minus{} 1}{a} \plus{} \frac {a^2 \plus{} b^2 \minus{} 1}{b} \minus{} b \plus{} \frac {(1 \plus{} a)(1 \minus{} a)}{c} \plus{} \frac {1}{a} \minus{} 1 \equal{}$ $ (a \minus{} 1)\left[\frac {b^2}{a} \plus{} \frac {a \plus{} 1}{b} \minus{} \frac {a \plus{} 1}{c} \minus{} \frac {1}{a}\right] \equal{} (a \minus{} 1)\frac {cb^2 \minus{} b(a^2 \plus{} a \plus{} c) \plus{} ac(a \plus{} 1)}{abc}$

$ f(b) \equal{} cb^3 \minus{} b(c \plus{} a \plus{} a^2) \plus{} ac(a \plus{} 1) \equal{}$ $ \minus{} \sqrt {2 \minus{} a^2 \minus{} c^2}(2\minus{}a^2\minus{}c^2)(c \plus{} a \plus{} a^2) \plus{} c(2 \minus{} a^2 \minus{} c^2) \plus{} ac(a \plus{} 1) > 0$ (Perhaps, I didn't check this form. The derivate is too ugly to do it.  :oops: ) for the given conditions so $ A < 0$ and we are done.

Sorry. I forgot that the inequality is not homonegeous with the condition  :) The subsitution $ b\equal{}a\plus{}q,c\equal{}a\plus{}p$ also works well for this kind. But it is so ugly  :) I will try to prove this one again, but not now. See all later  :wink:

[quote="BG Yoda"]If you substitute $ c \equal{} 0$, then $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$ wouldn't be true anymore for the initial $ a$ and $ b$ and the new $ c \equal{} 0$.  :!: Therefore, if you change $ c$, you have to change the other variables, too.

Possible solution:

 Let $ b \equal{} \min\{a,b,c\}$.
 If $ a \equal{} 1$ $ \Rightarrow \frac {1 \minus{} c^2}{b} \equal{} b$ and the inequality will be reduced to $ 1 \minus{} b^2 \plus{} b\le \frac {5}{4}\Leftrightarrow 0\le \left(b \minus{} \frac {1}{2}\right)^2$.

 So we may conjecture that $ \frac {1 \minus{} b^2}{a} \plus{} \frac {1 \minus{} c^2}{b} \plus{} \frac {1 \minus{} a^2}{c} \le 1 \plus{} b \minus{} b^2\le \frac {5}{4}$.

 $ A \equal{} \frac {1 \minus{} b^2}{a} \plus{} \frac {1 \minus{} c^2}{b} \plus{} \frac {1 \minus{} a^2}{c} \minus{} 1 \minus{} b \plus{} b^2 \equal{}$ $ b^2\frac {a \minus{} 1}{a} \plus{} \frac {a^2 \plus{} b^2 \minus{} 1}{b} \minus{} b \plus{} \frac {(1 \plus{} a)(1 \minus{} a)}{c} \plus{} \frac {1}{a} \minus{} 1 \equal{}$ $ (a \minus{} 1)\left[\frac {b^2}{a} \plus{} \frac {a \plus{} 1}{b} \minus{} \frac {a \plus{} 1}{c} \minus{} \frac {1}{a}\right] \equal{} (a \minus{} 1)\frac {cb^2 \minus{} b(a^2 \plus{} a \plus{} c) \plus{} ac(a \plus{} 1)}{abc}$

$ f(b) \equal{} cb^2 \minus{} b(c \plus{} a \plus{} a^2) \plus{} ac(a \plus{} 1) \equal{}$ $ \minus{} \sqrt {2 \minus{} a^2 \minus{} c^2}(c \plus{} a \plus{} a^2) \plus{} c(2 \minus{} a^2 \minus{} c^2) \plus{} ac(a \plus{} 1) > 0$ for the given conditions so $ A < 0$ and we are done.[/quote]

$ f(b)$ should be:
\[ f(b) \equal{} cb^3 \minus{} b(c \plus{} a \plus{} a^2) \plus{} ac(a \plus{} 1)\]
And why $ f(b)>0$  :wink:

[quote="arqady"][quote="hxy09"]$ a,b,c \in (0,1]$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$
Prove $ \frac {1 \minus{} b^2}{a} \plus{} \frac {1 \minus{} c^2}{b} \plus{} \frac {1 \minus{} a^2}{c} \le \frac {5}{4}$
equality holds when $ a \equal{} 1,b \equal{} \frac {1}{2},c \equal{} \frac {\sqrt {3}}{2}$ ...[/quote]
It's a known Walther Janous's inequality.  :lol:
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1459[/quote]

Wow,thank you so much arqady :lol:
I knew the famous Walther Janous's inequality,which can be solved by Mixing variable.But I didn't come up with the idea that this one can be transformed into a previous ineq!So the 2007 China IMO seletion used an old problem.What a pity!

[quote="hxy09"]$ a,b,c \in (0,1]$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$
Prove $ \frac {1 \minus{} b^2}{a} \plus{} \frac {1 \minus{} c^2}{b} \plus{} \frac {1 \minus{} a^2}{c} \le \frac {5}{4}$
equality holds when $ a \equal{} 1,b \equal{} \frac {1}{2},c \equal{} \frac {\sqrt {3}}{2}$ ...[/quote]
This is equivalent to the following famous inequality of Jack Garfunkel:
\[ \frac{x}{\sqrt{x\plus{}y}}\plus{}\frac{y}{\sqrt{y\plus{}z}}\plus{}\frac{z}{\sqrt{z\plus{}x}} \le \frac{5}{4} \sqrt{x\plus{}y\plus{}z}.\]
This inequality is very well known and up to now, I have found 7 proofs (my own) for it in total. ;) :)

[quote="can_hang2007"][quote="hxy09"]$ a,b,c \in (0,1]$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$
Prove $ \frac {1 \minus{} b^2}{a} \plus{} \frac {1 \minus{} c^2}{b} \plus{} \frac {1 \minus{} a^2}{c} \le \frac {5}{4}$
equality holds when $ a \equal{} 1,b \equal{} \frac {1}{2},c \equal{} \frac {\sqrt {3}}{2}$ ...[/quote]
This is equivalent to the following famous inequality of Jack Garfunkel:
\[ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}} \le \frac {5}{4} \sqrt {x \plus{} y \plus{} z}.
\]
This inequality is very well known and up to now, I have found 7 proofs (my own) for it in total. ;) :)[/quote]

Oh that's nice.Can you post it to us :lol: I am eager to see it.(I knew two proofs,one is Mixing Variable,the other is the official solution from China Team Selection which use lots of calculation)


Let V be a vector space and S, T subspaces of V such that direct sum of S and T is V.
Let U be a subspace of V. Determine whether U = direct sum of A and B, where A is intersection of S and U
and B is intersection of T and U.

The key for a direct sum is that $ W_{i}\cap\sum_{j\neq i}^{n}W_{j}\equal{}\{0\}$ so you have to check that $ A\cap B\equal{}\{0\}$.

[quote="JRav"]The key for a direct sum is that $ W_{i}\cap\sum_{j\neq i}^{n}W_{j} \equal{} \{0\}$ so you have to check that $ A\cap B \equal{} \{0\}$.[/quote]
@ JRav That $ A\cap B\equal{}\{0\}$ is quite clear in the above question, according to the choice of $ A$ and $ B$, but it does not suffice as this simple example shows:

Let $ V\equal{}\mathbb R^2, \ S\equal{}<i>,\ T\equal{}<j>,\ U\equal{}<i\plus{}j>$.
(So the answer is negative.)

Generally, to prove that something is an inner product, you seek to show that its representation is unique, which is usually accomplished through contradiction.  And why did you use inner product notation?

I haven't used Dirac's notation which is used to denote inner product between [u]two[/u] vectors. This is a standard notation for the genrated subspace of a set. Here the set is a singleton and it should be clear from the context that $ i\equal{}(1,0),\ j\equal{}(0,1)$.

Using The Captain's example, if we let U = c(2,1) where c is a scalar
then is the question is false?

Your statement does not hold in that case either. What I said should be quite easy for you to visualize in the real vector space $ \mathbb R^2$; Note that $ S,T,U$ are three different lines passing through the origin.


A 1.12 kg block is attached to a horizontal spring with spring constant 2750 N/m. The block is at rest on a frictionless surface. A 9.60 g  bullet is fired into the block, in the face opposite the spring, and sticks.

What was the bullet's speed if the subsequent oscillations have an amplitude of 12.1 cm?

Consider the potential energy in the spring at its maximum extension.

Find the max potential energy and use that to find the velocity at equilibrium.  Then use conservation of momentum to find the velocity of the bullet.

[quote="Immanuel Bonfils"]The mechanical energy is not conserved.[/quote]
Yes, but did I say you use the mechanical energy in the block to find the mechanical energy of the bullet?
The energy can find the momentum transferred to the block. I think that was quite self-evident.

At the moment the bullet, $ m$ strikes the block, $ M$, its velocity was $ v_f$ while before the collision it is $ v_i$.

[b]After[/b] collision, the mechanical energy is conserved.

$ \therefore \frac{1}{2}v_f^2(m\plus{}M) \equal{} 0 \plus{} \frac{1}{2} kx^2$

Gives $ v_f \approx 6\;m/s$

By Conservation of Momentum,
$ mv_i \equal{} (m\plus{}M)v_f \\ \\
v_i \equal{} \frac{m\plus{}M}{m}v_f \\ \\
v_i \approx 702.5 m/s$


real analysis
algebra
polynomial
modular arithmetic
number theory unsolved
number theory

Let $ a(x)\in(\mathbb{Z}/q\mathbb{Z})$, where $ q$ is a prime.

(i) Prove that $ a(1)\equiv 0\pmod{q}$ iff $ (x\minus{}1)|a(x)$ in $ (\mathbb{Z}/q\mathbb{Z})$.

(ii) Prove that if $ a(1)\equiv 0\pmod{q}$, then $ a(x)$ is not invertible in $ R_q$

(i) follows from the division algorithm.
As for the second, define $ R_q$ again?  Remember $ a(x)$ is a polynomial, and therefore belongs to a polynomial ring...

(ii)  Suppose $ a(x)$ has an inverse $ b(x)$ (in what I assume is $ \mathbb{Z}_q[x]$).  Then $ a(x) b(x) \equiv 1 \bmod q$.  But $ a(1) b(1) \equiv 0 \bmod q$.  Contradiction.


This should have gone to the Chemistry/Science thread.

Yes there are some real chemistry geniuses in that forum

but for a problem as simple as this one, just consult your textbook! (I don't see anything out of the ordinary... :maybe: )

Hint: Are you sure about the "$K_{2}SO_{3}(aq)$" term?

1. Do your own homework.
2. Read the forum descriptions and post threads where they belong.

I assume that this is a redox reaction in which the sulfite anions $(\text{SO}_{3}^{2-})$ are oxidized to sulfate $(\text{SO}_{4}^{2-}).$ The only problem: what is nitrate reduced to? There are several possibilities. Nitrogen has some pretty complicated chemistry, with oxidation states including $+5$ (nitrate), $+4\ (\text{N}_{2}\text{O}_{4}),$ $+2\ (\text{NO})$, $+1$ (nitrous oxide), $0\ (\text{N}_{2}),$ and $-3$ (ammonia), and I left out some possibilities.


How many edges dose a hexagonal prism have?

A hexagonal prism has 6 edges on each base and 6 running straight across to the other base, so $ \boxed{18}$


[b]Prove that:[/b] $ \boxed{\boxed{\sqrt{A \pm \sqrt{B}}\equal{}\sqrt{\frac{A\plus{}C}{2}} \pm \sqrt{\frac{A\minus{}C}{2}}}}$, [b]where[/b] $ C^2\equal{}A^2\minus{}B.$

First square both sides.

$ A \pm \sqrt {B} \equal{} \frac {A \plus{} C}{2} \plus{} \frac {A \minus{} C}{2} \pm 2 \cdot \frac {\sqrt {(A \plus{} C)(A \minus{} C)}}{2}$

$ A \pm \sqrt {B} \equal{} A \pm \sqrt {A^2 \minus{} C^2}$

$ \pm \sqrt{B} \equal{} \pm \sqrt {A^2 \minus{} C^2}$

$ B \equal{} A^2 \minus{} C^2$

$ C^2 \equal{} A^2 \minus{} B$ which is given.


Prove that each two numbers in the sequence
$2+1, 2^{2}+1,....,2^{2^{n}}+1$  are relative prime numbers


number theory
greatest common divisor
function
inequalities
linear algebra
matrix
number theory proposed

Let $ p,q$ be two fixed primes such that $ \text{gcd}(q,p^2 \minus{} p) \equal{} 1$. Define now the function $ f(x): \mathbb{N}_0 \to \mathbb{N}$ the integer such that $ q^{f(x)} \mid x$ but $ q^{f(x) \plus{} 1} \nmid x$. Define also for all positive integer $ j$ the function $ w_j(x): \equal{} 1 \plus{} x \plus{} x^2 \plus{} \ldots \plus{} x^{j \minus{} 1}$.

Prove or disprove [i]without any use of p-adic valuation[/i] if the inequality $ f(w_1(p) \cdot w_2(p) \cdot \ldots w_n(p) ) \ge f(n!)$ holds for all positive integer $ n$.

[i](Carlo Pagano)[/i]

[quote="bboypa"]...[i]without any use of p-adic valuation[/i]...[/quote]
what does this mean?
can someone explain plz? :blush:
anyway, it's very easy by using $ v_q(x^n\minus{}y^n)\equal{}v_q(x\minus{}y)\plus{}v_q(n)$.
maybe p-adic valuation means this!?

Exactly, without it (i.e. also without De Polignac and all formula involving counts of the exact number of primes..)  :D

[quote="bboypa"]Define now the function $ f(x): \mathbb{N}_0 \to \mathbb{N}$ the integer such that $ q^{f(x)} \mid x$ but $ q^{f(x) \plus{} 1} \nmid x$

[...]

[i]without any use of p-adic valuation[/i][/quote]

Sorry, but that's already ridiculous in itself.

I'm sorry for the misundersting, in a better way i should say that we are interested in a combinatorical solution or something that is not closed with number theory.. in that way it can be seen as a very nice one exercise  :wink:

It is passed enough time..

It is enough to show that for every positive integer n and prime p we have that n! divides $ \prod_{i\equal{}0}^{n\minus{}1}{(p^n\minus{}p^i)}$. But it is true since rhs is the number of non sigular matrices in with elements in $ \mathbb{F}_p$, that is clearly divisble by n!: in fact we fix a non singulat matrix A then we can obtain always other (n!-1) non singular matrices exchanging the columns of A.


inequalities
logarithms
inequalities unsolved

If $ x,y,z>1$ then prove that : $ x^{\log_y z}\plus{}y^{\log_z x}\plus{}z^{\log_x y}\ge x\plus{}y\plus{}z$ .

[quote="Mateescu Constantin"]If $ x,y,z > 1$ then prove that : $ x^{\log_y z} + y^{\log_z x} + z^{\log_x y}\ge x + y + z$ .[/quote]
very easy by AM-GM. Putting $ a=\ln x$ etc., we have $ a,b,c>0$ and

$ LHS=\sum_{cyc} e^\frac{ab}c=\sum_{cyc} \frac{e^\frac{ab}c+e^\frac{bc}a}2$ ${ \ge\sum_{cyc} \sqrt{e^\frac{ab}ce^\frac{bc}a}}=\sum_{cyc} \sqrt{e^{b(\frac{a}c+\frac{c}a)}}\ge\sum e^b=\sum x=RHS$. :)


analytic geometry
geometry
rectangle
area of a triangle
Heron\u0027s formula

A point inside a square is positioned so that the distances to the four vertices are 27, 21, 6, and x units. If x is a whole number, what is the value of x? :D

would solve if I could draw the picture. I know the solution but artpad.art.com isn't working.
[url]http://artpad.art.com/?iyo3zgler2g[/url] is where I put a hint if it works.

Solution:

[hide]
Picture a square with our point in the upper-leftish corner. Let the lower left corner be at the origin, and let the length of a side of the square be $a$. Then the other three points are $(0,a)$; $(a,a)$; and $(a,0)$. We have visualized the point as being closest to $(0,a)$.

Let the coordinates of our point be $(c,d)$. By the distance formula, we have four equations:
1) $c^2+(d-a)^2=6^2$ (distance from point to upper left corner)
2) $(c-a)^2+(d-a)^2=21^2$ (distance from point to upper right corner)
3) $(c-a)^2+d^2=27^2$ (distance from point to lower right corner)
$c^2+d^2= x^2 = ?$ (distance from point to the origin)

To achieve the last equation, which describes $x$, add equations 1 and 3 and subtract equation 2; then $c^2+d^2=x^2=6^2+27^2-21^2=324=18^2$.

Thus $x^2=18^2$, and the answer is $x=18$.
[/hide]

[hide]Theorem: for rectangle ABCD, and a point P within the same plane,
$(AP)^2+(CP)^2=(BP)^2+(DP)^2$.

$27^2=729, 21^2=441, 6^2=36$.

We can test the following for a whole number x:
$\sqrt{729+441-36}=x$
$\sqrt{729+36-441}=x$
The second one works, with $x=18$.

EDIT: ok, I searched- British Flag Theorem[/hide]

[hide="hint without some weird theorem"]
Without using some strange theorem we have never heard of, the best way I can think of is using Heron's formula.

Think of how you can find the area of the square, if the side is $n$[/hide]

Gah! Just accidentally saw a question on '05, wasn't supposed to do '05 until May 9th :(

[quote="slamchess"][quote="daermon"][hide="solution"]its pretty easy if u know brittish flag theorem.
using BF theorem we get 765=441+x^2
x^2=324
x=[color=red]18[/color]  :D [/hide][/quote]
could you please elaborate on the brittish flag theorem??[/quote]
"Brittish" only has one t.  :P I googled it and there are only 2 [hide] ONLY TWO!!! [/hide] websites that include it, other than these math forums. It's not a very popular theorem.
And pianoforte basically explained it in his solution.

Lmao, I was doing this during practice and I used a series of four equations and had to spend about five minutes using math-0 and extensive graphing on my calculator. Then my teammate pointed out the obvious solution that pianoforte posted.

It is possible with Heron's formula, it's just extremely long, and you'll probably make a mistake or two.

[hide="Hint to Derivation of the theorem"] Take the point and drop perpendiculars to each of the sides, call those $u,v,x,y$.

Then, call the segments that are 27,21,6, and x $a,b,c,d$.  Write out the pythagorean theorems that exist (there should be 4), and try to figure it out from there.[/hide]

[hide="Now that we know"] Knowing the theorem, we check the cases, and find x = 18.[/hide]


function
advanced fields
advanced fields unsolved

using  the axiom of choice , construct a function $ f: R\to R\setminus\{0\}$ such that $ f(P)\equal{}R\setminus\{0\}$  for every non degenerate interval $ P\subset R$.

There are equivalent constructions [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=31493]here[/url]. All it takes is a bijection between $ \mathbb{R}/\mathbb{Q}$ and the desired range $ \mathbb{R}\setminus\{0\}$.

The axiom of choice is required for that sort of cardinality arithmetic.


group theory
abstract algebra
linear algebra
linear algebra unsolved

A finite subgroup in $GL_2(C)$ has an abelian subgroup of index at most 12.

Next: prove that there exists a finite subgroup of $GL_2(C)$ such that any normal abelian subgroup has index at least 60. Can we replace 60 by 61?  :P


Determine last three digits of $2^{2^{2003}}$
I think this problem is too short and too easy for you!

It is a standart technical problem using Fermat theorem.

That's problem in the first day.
This problem in the second day.
5.(Tmo) Give $P$ is prime and $p^{2}+2543$ has less than 16 differrent divisor.Find all value of $P$.

It is also a problem for the old well known idea. Indeed, we are to check $P=2,3$, and all other primes have the form $6k\pm 1$, then $P^2+2543=24\cdot \left(\frac{3k^2\pm k}{2}+106\right)$. It is quite easy to show that $P^2+2543$ has at least 16 divisors.

To Myth:
I spend a long time to learn your solution.Now I understand it.
Thank you very much!

You can tell me about my wrong gramma in English language.


Is it possible to find integers m,n,x such that

$11^m+3^n-2^n=x^4-1$?

[hide]Look at it mod 8.
Since any square mod 8 is 0,1, or 4.  Any fourth power is 0 or 1.  Therefore, minus one of a fourth power is 7 or 0.
Now, let's assume  x m n is a whole number.  Then $3^n-2^n$ can be 0 1 3 or 5 mod 8 where 5 can only exist when n=2.  11=3 mod 8 so it alternates between 3 and 1.  Therefore, 3+5=0 is the only possible way it works.  Therefore, n=2.  In other words, $x^4-6=11^m$.  Now look at this mod 12.  $x^4$ can be 0,1,4, or 9 mod 12.  Minus 6 is 6, 7, 10, 3.  $11^m$ is 11 or 1 mod 12.  Therefore, it can never be true for any whole number.
Now, assume these three are negative.  $x^4$ is positive regardless of x.  Since x is an integer, $x^4-1$ is an integer.  Since 11, 3, and 2 are relatively prime and only int. powers are allowed, there's no way the numerator of the resulting fraction on the left side could ever be an integer multiple of the denominator.[/hide]


geometry articles LaTeX algebra function domain

my idea is to include only theorems without any problems and also to include a proof for each theorem. If someone else has a good idea - share it ! :)

I'm not sure the project (especially the "include the proofs" part of it) is feasible unless by "theorems" you mean just very elementary and standard stuff that can be found quite easily on the web even now. Otherwise it is equivalent to writing textbooks in all subjects simultaneously, each textbook covering everything from the very basics to the latest achievements in the subject. This is not impossible in principle (see Bourbaki, etc.) but it requires one to devote almost his entire life to it and to have an almost encyclopedic knowledge of mathematics. I'm certainly not qualified for such task, are you? :? Looking at what you wrote in your other post, namely at
[quote]
Hello,
I am a freshman in high school and I really enjoy math. I would like to start solving problems that are given in competitions but I don't know where to start.
[/quote]
I have some doubts about it. If you really want to try (just to try, not to make anything real yet), pick your favorite subject, sit down, and make the list (just statements, or even names; no proofs or anything) of the theorems in just that subject that are worth including. See whether you can make a reasonably comprehensive list and how much time it'll take. If you are still enthusiastic about your idea after that, post your list here. We'll see if it makes sense and how to go best about it :).

[quote="hsiljak"][hide="Offtopic"]
When talking about this kind of sites: has anyone of you ever contributed to MathWorld? I've used their online form for contributions on two articles but never received any response whatsoever. Is that standard procedure (not answering people's mails) or is there a technical issue?[/hide][/quote]

I have commented one or two times on Mathworld, but not recently -- I recall that I received some sort of response at least once, but maybe not until many weeks later.  That's definitely something to be said in favor of the wiki model.

I like fedja's suggestion, too.

Um... this math website you are talking about, I'm guessing you are somehow going to have to get a server or access to a server and install latex and asymptote. Also how large is this site going to be and are you going to program it from scratch or use an editor.

[quote="JgFi"]Hello guys,
Does someone want to help me making a website, which will be about math theorems? We will try to include as many as possible.
Thank you,
JgFi[/quote]
Try to make the website open to a large audience.

I like [b]fejda[/b]'s suggestion. I would add to it that you use an electronic file, and link any proofs of the theorems that you find. That way, even if you don't end up refining it into a website of your own, it will still be a great fast-reference for you. And if you wanted, you could share it, too  :wink:

[quote="myyellowducky82"]Try to make the website open to a large audience.[/quote]
What do you mean? Obviously he's not going to password-protect it...

Do you need help with the website designing or what to put in the website?

[quote="splatyango"]um... [color=#FF0000]are you willing to pay anything?[/color]

If you are, there are several good domain providers that charge pretty low (i think), but you still have to pay...

As for the "open to a large audience" thing, if you don't want to pay, you have to do something like freewebs.com, which do not have a very high attraction level...

EDIT: sorry accidentally pressed submit.[/quote]
Actually freewebs is quite popular.

freewebs is weird
yo just like get mediawiki and get a nice theme
and a domain
and host it yourself

[quote="hsiljak"]What will differ your site form the ones already existing, such as MathWorld, PlanetMath, CutTheKnot etc (especially the first two sites, concerning their encyclopedic model)?
Is it going to be a general mathematics project, or specialized for certain areas of mathematics?

[hide="Offtopic"]
When talking about this kind of sites: has anyone of you ever contributed to MathWorld? I've used their online form for contributions on two articles but never received any response whatsoever. Is that standard procedure (not answering people's mails) or is there a technical issue?[/hide][/quote]

I've made several contributions which they never responded to.

Me too.  I guess they don't have time to respond to so many.

[quote="myyellowducky82"]Actually freewebs is quite popular.[/quote]

Yeah, if you don't want to use programming, use freewebs. My school has 2 or 3 really good gossip websites made using freewebs.

however, do you need help with making the website or finding stuff to put on the website?

I'm quite good at the programming, at least enough to get a decent site. The only problem is, does anyone know how to get it online for free?

[color=green]Moderator says: there is never a good reason to quote the entire post immediately preceding yours.[/color]

freewebs.com ? you would guess by the name

also, domains are pretty cheap. i am sure that you can get your parents to foot the bill. if you can't, whatever, everyone has friends with credit cards

[color=green]Moderator says: there is never a good reason to quote the entire post immediately preceding yours.[/color]

Let me put it this way:

I don't want it on freewebs or else its not unique in any sense
My parents probably won't pay for me to have a domain
My friends don't have credit cards (yet). My friends and I are still young...

Yes... I have a freewebs site and I never use it.... Blogger is pretty good and so is wikispaces... both which i have used. freewebs is annoying IMO (not the math thing in my opinion)


Let $x_1,x_2$ be the roots of the quadratic equation: $x^2+px+q=0$.
Let $y$ satisfy:
\[ x_1 \le \frac{y^2-q}{2y+p} \le x_2  \]
Find $\min y, \max y$.

This would be even more exciting if you change your post into LaTeX.

LaTeX'ed and corrected :) (I know, too much free time in my hands... :P)

[quote="Lovasz"]Let $x_1, x_2$ be the roots of the quadratic equation:
                                           $x^2+px+q=0$.
                Let y satisfy:
                                     $x_1 \leq \frac{y^2-q}{2y+p} \leq x_2$
                Find  min y, max y.[/quote]

did you find any solutions ? is not this problem interesting?


Who else is going?  I just found out I'm going today. :D

all I know is that I'm going to regionals...

Well my school does something similar called science fair, which is from the IJAS.

I had regionals on March 14. (PI DAY YAY) and I took 1st place in my category, Computer Science. My project was about computer fonts and how they affect file sizes of documents.

Well so that meant I qualified for State Science Fair on May 8-9, 2009, at the University of Illinois in Champaign. But sadly, I'm not going there because of National Mathcounts. :( (well, nats is a lot more fun but oh well)

EDIT: FIXED.

lol you mean May 8-9?

You haven't had regionals yet appilai?  When do you have regionals?

[quote="suma_milli"]Bleh no one? :(

Usually mathy ppl are also sciency?[/quote]

Not always. Then again, it depends on what science. I have a math genius friend, and she loves chem and physics, hates bio with a passion. I'm OK at all three.
Yeah, we had regionals. I was on JV. :(
4 medals in my 4 events :)... so I should have been an alternate for state. I'm not, but at least it makes my life less busy.
Do you do JETS?

[quote="suma_milli"]Oh err.. I'm in 8th grade :P

This sounds fun, I want to do it next year now! lol ^^[/quote]

Oops... right, cuz ur in mathcounts. Fun stuff.
Try out next year. It's fun!

[quote="Aryth"][quote="appillai"][quote="suma_milli"]Aww :(  Yeah I was also disappointed. There were a lot of good people on our JV team that should have made the state team but didn't.

Nope,  what's JETS?[/quote]

Junior Engineering and Technical Society. They have tests in sciences, engineering, math, and english. I did bio and math... failed in both... Kevin Hu rocked math at the state competition (as always).[/quote]

That would be WYSE (Worldwide Youth in Science and Engineering), not JETS.
And I got rocked by math at state, actually...[/quote]

OK but technically (at least at our school) the major organization is called JETS. Yeah, TEAMS is the real JETS competition, but they're practically synonyms. Did you get 3rd? (not sure if I remember) Either way, I failed.

Yeah I did. Largely because I am horrible at calculus =\


CEMC
Gauss
AMC
USA(J)MO
USAMO
AoPS Books

Do you think "Art of Problem Solving" books really work? I have tried textbooks, contest-related books; spending an incredible amount of money. I need a fine resource that gets to the point and teachs how to solve problems and do well in Canadian contests. Fellow Canadians, give me some responses!  :)

Which Canadian contests in specific ?

Also which books have you from studied from in the past.
Some of us may be able to make a comparison with Aops.

There are few books available in North America that have the breadth of Aops, which covers most of the precalculus topics you would find any regional or national contest, along with its wonderful collection of simple to intermediate problems.
The only other all in one handbook I'm familiar with is The Thrill of Pre-College Mathematics, though this is published only in India. With these expceptions, contested related material is scattered across dozens of texts, which you'll notice if you check the references in some of the MAA IMO problem books. I can't imagine Aops doing anything but improving your problem solving skills. You will be exposed
to some very interesting math,problems, and solutions.

Good point. I generally use the Math Power textbooks for basic knowledge. Using numerous CEMC practice books and Math League books, I practice my problem-solving skills. What do you think?

[b]Still I need to know; is it worth paying $68 for the Art of Problem Solving books?[/b] :?

That is good to know. Do you have a specific reason why you think so?

The reality is that any set of textbooks are only as good as the amount of effort you're willing to put into it.  Certainly the AoPS textbooks are going to give you the most out of whatever effort you're willing to put in, but you can probably achieve the same results if you're diligent enough by just doing past contests or doing problems in these forums.  I will say, however, that 68 bucks is pretty small change considering that you'll be spending thousands of dollars on texts once you get to university.  I tend to think its a good investment.

Good point stated by everyone.

[b]Please Canadians, give more responses![/b]  :lol:

[quote="TheDreamer"]Are there lots of Problems/Solutions?  How many exercises are there after teaching each new concept?[/quote]Each volume has over 500 problems.

[quote="d.lam.86"]That is good to know. Do you believe that these volumes can be read by beginning high school students?[/quote]Certainly Volume 1 can -- it is primarily intended for students who not do have a lot of experience with the type of mathematics found on the more challenging math contests.  Volume 2 is, naturally, for students who have mastered the material in Volume 1.

Although we (obviously) recommend our own Art of Problem Solving books, we have a list of several other books that we also strongly recommend:
[url]http://www.artofproblemsolving.com/Resources/AoPS_R_Books.php[/url]

Since you are Canadian, let me especially draw your attention to the book [url=http://www.amazon.com/exec/obidos/ASIN/1895997046/artofproblems-20]A Mathematical Mosaic[/url] by Canadian IMO veteran Ravi Vakil.

Yeah, I read A Mathematical Mosaic; it is really good.  It doesn't teach problem solving in the sense that the AoPS books do.  Instead, it is a really fun math book to read that explores cool math problem solving concepts.  Each chapter is followed up by a "food for thought" section of problems, but the solutions are omitted.

that brings up a question i have----- are the books with solutions helpful????

Some people say solutions might make you peak at it without making much progress on your own. BUt some say that looking at solution gives you different perspectives of the problem which you can employ in other problems with similar types of questions.

Some say books without solution force you to get the answer by yourself, but you are unable to see the patterns and the steps that you go through, and furthermore, you dont see any other ways that you could approach the problem, which certainly helps if you are stuck on a problem doing it one-way. Of course, AoPS is why it exist (to discuss problems), but everyone cannot come here all the time

Which would you like? Solutions or No Solutions?

I would say get the solutions! When I first ordered the AoPS books, I made sure I got Volume I and II with solutions just because I knew I wouldn't be able to solve every problem. Solutions are useful when you can't go to the forums, and they also provide an alternate way to solve the problem. Who knows, maybe the solution that you came up with is better than the solution provided, or maybe its much worse. Either way they can only help...

You have raised an important issue whether to obtain solutions or not. I think it depends on your problem-solving capabilities. If you are capable of solving problems and being able to understand how and why you know this is the answer, you don't necessarily need the solutions. But for beginners (like me), I believe that it is crucial to have the solutions.

So after all of this talk, I think it depends on the style of learning a person has.

[b]Does anyone know where I can purchase "Art of Problem Solving" books in Toronto, Ontario (Canada)?[/b]

If you want to buy the books, go to the site below:

http://www.artofproblemsolving.com/Books/AoPS_B_Order.php

Hmm,

That's the first time I heard about how getting solutions makes you want to peek and give up.  However I've always thought that it's reading the solution of a problem you can't do that makes you learn; the process of doing the problem is only for practice.  I've always gone right to the solutions if I had no clue on how to proceed with the problem at first glance.  Am I doing something wrong?  Because honestly I don't see a point of getting problems without solutions since you can neither check your work nor learn new concepts to solve problems you couldn't solve the first time through.

[quote="d.lam.86"]still I am a beginner; inadequate of attacking difficult problems at a regular basis[/quote]
Then, attack problems with medium difficulty, rather than difficult problems. I think the problems, which a person choose to solve, need to be somewhat challenging, but I also think they should be reasonably solvable. For example, I wouldn't expect my grade 7 younger brother to solve IMO shortlist(he doesn't even like math). I also wouldn't work on extremely easy problem.
Finding problems at your level is hard to do. Solving them is also hard and time consuming. But if you work hard, you definitely would become a good problem solver. :)

Thank for your encouragement.  :lol:  I will continue to strive for strong problem-solving skills starting with the AoPS books.  ;)

[quote="d.lam.86"]Are you David Rhee?  :)[/quote]
Yes

After being totally rejected by University of Toronto Schools, I should've went to Don Mills Collegiate. It's only a bit farther than North Toronto Collegiate. I regret not working harder for the U of T schools..... *once a failure... probably always going to be a failure*  :( :'(

[quote="lightrhee"][quote="blahblahblah"]I wouldn't buy the solutions if I were you.  But that's just me...[/quote]
I agree. Having solutions can be dangerous.[/quote]

Lol k. :)

But like what if there is a problem that one can't solve after so long?  Do they just have to hang forever?

[quote="d.lam.86"]Are you David Rhee?  :)[/quote]

hahaha lightrhee you're a legend.

find a teacher or someone just as adequate  ;)

[quote="d.lam.86"]Well it's nice to meet you, Mr. Rhee :D ...... a legend? (according to Sunny?)  ;)  Yeah, I've seen your name a lot—mostly from the CEMC website...... Good luck to everyone with this year's IMO! Make CANADA PROUD!  :D  :D  :D  Anyway, I'm probably the youngest in this forum. I'm in eighth grade; going to ninth..... Anyone surprised?  :?  LOL! I have a cloning essay to finish... CIAO (for now).[/quote]
I'm not really a legend. I'm a good problem solver, but not the best. There are some high school students in Canada who are much better than me. Last IMO, I made a horrible mistake and could win any medal :( , while half of the competitors get a medal :( .

I'm surprised that you are in grade 8. Didn't sound like one. :) BTW I hate essays. I am always get 50's for essays that I write. If multiple choice test didn't exist in English, I would be failing it :lol: .

Wow, so close to a medal....  :blush: I haven't even come across a medal... let alone a math medal  :oops:

I got 97/150 in Gauss 2005. I am really poor at problem-solving!  :(

Yeah that happens to me sometimes. I usually end up cutting out the solutions and hiding them. LOL!  :blush:

d.lam.86, you know that you can edit posts right? There is really no need to post three consecutive short messages one after another 5 minutes apart. This isn't a chat room, it's a discussion forum. Anyways, that was a just me playing my role as the moderator.

On the question of solutions, I highly recommand that your training books have solutions to the problems. In fact, solutions are often the crux to problems books. Without solutions, you may go into great frustration and possiblity be lost for a long time; you probably won't learn very much either. In my opinion, the second most important thing in  contest preparation is reading other people's solutions (the first is of course solving the problems oneself). Honestly though, if you don't have the ability to control yourself from constantly peaking at the solutions, you probably don't have the ability to find your way out of darkness without a solution book either.

[quote="d.lam.86"]After being totally rejected by University of Toronto Schools, I should've went to Don Mills Collegiate.[/quote]
You should know that, contrary to popular belief, Don Mills doesn't have any special programs in math (unlike Vincent Massey). As well, it is a lot more difficult to stand out here - for example, I know a number of students from my school who would have been invited to Waterloo Seminar if they were at any other school. For reasons unknown to me (I personally think it's very coincidental), we have had the most number of IMOers in (recent?) history. In any case, I have great pride in my school.

[quote="d.lam.86"]I am dearly sorry but I am new and was unaware of the "edit" function. Don Mills Collegiate has had recently the most IMO students; a very spectacular coincidence. Anyway it's nice to meet the infamous Yufei Zhao.  :) I agree with your strong opinion on the solutions issue but I tend to struggle and ask peers about questions until resorting to the solution.

How come there is another website [url=http://www.mathlinks.ro/]www.mathlinks.ro/[/url]? I don't understand; how come I was directed back to the AoPS forum? This is really confusing... Is MathLinks another website that I must register with?

Please reply back soon, Mr. Zhao.[/quote]
I'm sure that Yufei would know better, but I think MathLinks and AoPS is essentially same place. Difference seems to be how the forums are viewed.

Right, last year in June, the two math contest forums, MathLinks and AoPS, merged. Thus they are now exactly the same forum, just with different skins.

You don't need to register for both of them. One is enough.

[quote="d.lam.86"]Well it's nice to meet you, Mr. Rhee :D ...... a legend? (according to Sunny?)  ;)  Yeah, I've seen your name a lot—mostly from the CEMC website...... Good luck to everyone with this year's IMO! Make CANADA PROUD!  :D  :D  :D  Anyway, I'm probably the youngest in this forum. I'm in eighth grade; going to ninth..... Anyone surprised?  :?  LOL! I have a cloning essay to finish... CIAO (for now).[/quote]

  No, I think im the youngest at gr.7. :harhar:

I highly recomend the aops books, I just got them, and I am finding them my invaluable. Also, a friend of mine ( who Practices for USAMO regularly ) still likes to go through vol 2, and won't shut up about how he wish he had them in grade nine.

Wow G-UNIT, you are only in grade 7? Lucky... :(
I wish I did more math in my middle school years...


If arctan [2mn/(m^2 - n^2)] + arctan [2pq/(p^2 - q^2)] = arctan [2xy/(x^2 - y^2)],then

((m + n)p + (m - n)q)/((m + n)q - (m - n)p) =

A)(x + y)/(y - x)
B)(x - y)/(x + y)
C)x/y
D)(x + y)/(x - y)

Only one option is correct.

The given expression can be simplified into [2mn/(m^2 - n^2) + 2pq/(p^2 - q^2)]/( 1 - [2mn/(m^2 - n^2).2pq/(p^2 - q^2)] = 2xy/(x^2 - y^2)
But any amount of manipulation by me didn't yield the required result.

Recall that $ \tan 2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$.  What does that tell you about these arctans?

[hide="It tells you..."] that $ \arctan \frac{2mn}{m^2 - n^2} = 2 \arctan \frac{n}{m}$. [/hide]
[hide="With complex numbers"] This is the argument of $ m^2 - n^2 + 2mni = (m + ni)^2$, which is clearly the square of the argument of $ m + ni$. [/hide]


function
real analysis
real analysis solved

Prove that there doesn't exist functions wich are continuous on Q and discontinuous on R-Q.
What about functions continuous on R-Q and discontinuous on Q?

I think the first problem might have been solved before here. Here's an answer for the second one:

f(p/q)=1/q for integer p and natural q, different from 0, s.t. (p,q)=1, and f(x)=0 for x irrational (I think this was first built by Riemann). It can be showed easily that it's continuous in every irrational point and it's obviously not continuous in any rational pt.

The best thing I could do to the first problem was to reduce it to the following:

Show that there are no functions g:R->R<sub>+</sub> s.t. g(x)=0 iff x is rational and g is continuous in every rational point. However, this is a particular case of the initial problem, so it might not be easier to deal with. I couldn't find it, but I'm sure it was solved here.

Useless post :)

There is no function f : |R - > |R which is continuous at each rational point and discontinuous at each irrational point since it can be shown that the set of point of continuity C(f) of a real function f is a countable intersection of open sets.

Suppose, for a contradiction, that a real function f such that C(f) = Q does exist :
Then, Q is a countable intersection of open sets, and then R-Q is a countable union of closed sets, with empty interiors (since R-Q has empty interior). But, in another hand, since Q is countable, it is a countable union of closed sets (the singletons) with empty interiors.
It follows that |R = Q \cup (|R-Q) is a countable union of closed sets with empty interiors. Thus, from Baire's theorem, |R has empty interior. A contradiction.

Pierre.

[quote="pbornsztein"] ...from Baire's theorem...[/quote]
I haven't seen for a long time someone using Baire's Theorem. Good job Pierre.

Never mind, I found it in a book (and I could have seen what it meant from the last sentence in pborn's post :)). However, the fact that the set of points of continuity of a real function is a countable intersection of open sets is new to me.

And conversely, for each set S which is a countable intersection of open sets, you may construct a real function f such that S is the set of points of continuity of f.

Pierre.


geometry
parallelogram
trapezoid
similar triangles

Isosceles, similar triangles $QPA$ and $SPB$ are constructed (outwards) on the sides of parallelogram $PQRS$ (where $PQ = AQ$ and $PS = BS$). Prove that triangles $RAB$, $QPA$ and $SPB$ are similar.

$\Delta QPA \sim \Delta PBS$.
$QR = PS = BS$. $QA = PQ = RS$. $\measuredangle AQR = \measuredangle BSR$. Hence $\Delta AQR \equiv \Delta BSR$.
$180^{\circ} - (\measuredangle QAR + \measuredangle QRA) = \measuredangle AQP + \measuredangle PQR$.
$180^{\circ} - \measuredangle QPS = \measuredangle PQR$.
Subtracting these $\measuredangle QPS - (\measuredangle QAR + \measuredangle QRA) = \measuredangle AQP = \measuredangle QRS - (\measuredangle QRA + \measuredangle BRS) = \measuredangle ARB$.
Also $AR = RB$. Hence $\Delta QPA \sim \Delta PBS \sim \Delta ARB$.

[quote="vanu1996"]Note that $AQ=PQ=RS$ so $QASR$ is isosceles trapezoid,similarly $QRSB$ also.hence pentagon $AQRSB$ is cyclic,also $AQ=RS,QR=SB$ and $\angle AQR=\angle RSB$ because $\angle AQP=\angle BSP$,hence $AQR$ and $RSB$ are congruent. so $RA=RB$,also $\angle ABQ=\angle RBS$ ($AQ=RS$),so $\angle ABR=\angle PBS$,hence we are done.[/quote]

To revive such an old thread with a wrong solution... how did you get isosceles trapezoids??

Best regards,
sunken rock

Let $AR \cap PQ=L$,we can easily prove that $\angle QAL=\angle SRB$ and $AR=RB$,Now $\angle QAL+\angle AQL+\angle ALQ=180=\angle RLP+\angle LRB+\angle SRB$ hence $\angle AQP=\angle ARB$.
@sunken,I am sorry.

Let $\angle QAP=x$ and $\angle PSB=y$. As we have $\Delta QPA\sim \Delta SPB$ then,
\[\frac{PA}{PB}=\frac{PQ}{BS} ...(i)\]
We have $\Delta BSR\cong \Delta RQA$, hence we get $RA=RB$.

Let us assume $\angle RBS=b$ and $\angle RAQ=a$, then, we get $\angle RSB=\angle RQA=180-(a+b)=z$. Let the intersection of $RB$ and $PS$ be named as $K$, then let $\angle RKS=k$. Hence we get $\angle SPQ=k+a$ and assume

Assume,
\[360-(k+a)-2x=z=\angle RSP+y\]
\[180=y+2x\]
which is true, so we get $\angle APB=z$ and $\Delta RSB\sim \Delta APB$, from (i), therefore we have $\angle RAB=x-a+a=x$.

Hence,
\[\Delta QPA\sim\Delta SPB\sim\Delta RAB\]


I'm going to try something new: a multi problem marathon. Every problem you solve, post one. Every problem must be from the CanMO.
1)S is any set of n distinct points in the plane. The shortest distance between two points of S is d. Show that there is a subset of at least n/7 points such that each pair is at least a distance d(sqrt3) apart.

2)Show that we can find infinitely many positive integers n such that 2^n - n is a multiple of any given prime p.

3)  Tranform a number by taking the sum of its digits. Start with $1989^{1989}$ and make four transformations. What is the result?

4)  There are five ladders. There are also some ropes. Each rope attaches a rung of one ladder to a rung of another ladder. No ladder has two ropes attached to the same rung. A monkey starts at the bottom of each ladder and climbs. Each time it reaches a rope, it traverses the rope to the other ladder and continues climbing up the other ladder. Show that each monkey eventually reaches the top of a different ladder.

5)What is the largest possible number of elements in a subset of {1, 2, 3, ... , 9} such that the sum of every pair (of distinct elements) in the subset is different?

6)Show that ($1 + 2 + ... + n$) divides ($1^k + 2^k + ... + n^k$) for k odd.

7)Show that the sum of 1984 consecutive positive integers cannot be a square.

8)Show that there are infinitely many integers which have no zeros and which are divisible by the sum of their digits.

For 2, you probably want $2^n - n$, or else it's extremely trivial.

[quote="PenguinIntegral"]7)Show that the sum of 1984 consecutive positive integers cannot be a square.[/quote]

[hide]
Let the first integer be $a$. Then \[ \frac{1984}{2} \cdot (2a+1983) = 1984a+1967136. \] Modulo 64, we get \[ 1984a + 1967136 \equiv 32 \pmod {64}. \] 32 is not a quadratic residue $\bmod 64$, so it cannot be a perfect square. $\boxed {}$
[/hide]

Yeah, cheap, I know. :(

[quote="joml88"]Why are you competing with 4everwise so much?  I don't believe he really wants to make this a competitive thing (nor did he ever want to).  It's getting a bit obnoxious.[/quote]
I'm being sarcastic. This was never serious.

[quote="chess64"][quote="PenguinIntegral"]3)  Tranform a number by taking the sum of its digits. Start with $1989^{1989}$ and make four transformations. What is the result?[/quote]

This reminds me of that IMO problem where it was $4444^{4444}$. The answer is easy but a rigorous proof is difficult (for me). Are we supposed to post proofs for these problems?[/quote]
Proofs are fine, but outlines will suffice.


linear algebra
matrix
geometry
geometric transformation
rotation

Let M be an 2x2 matrix with entries

a11 = -9
a12= 6
a21= -6
a22 = -9

Find general formulas for the entries the the matrix M^n , where n is an integer.

Hint: This matrix is in "complex number" form; the product of a rotation and a scalar.


Determine all natural number solutions $a, b, c$ of the system of equations

$a+b=(a,b)^2$
$b+c=(b,c)^2$
$c+a=(c,a)^2$,

I assume $(a,b)$ stands for $gcd(a,b)$ according to the tittle.

Let $n = (a,b,c)$.
Statements proves that $n^2 \mid (a+b) + (b+c) - (c+a) = 2b$ and similarly for $a$ and $c$

But then $n^2 \mid (2a,2b,2c) = 2n$ : $n = 1$ or $n = 2$

If $n=1$ then $lcm((a,b),(a,c)) = (a,b) \times (a,c)$ which must divides $a$ which implies $a \geq (a,b) \times (a,c)$
Same thing for $a$ and $b$

$(a,b)^2 = a+b \geq (a,b)((a,c)+(b,c))$ or $(a,b) \geq (a,c) + (b,c)$ and similar relations.
Adding every gives : $(a,b)+(a,c)+(b,c) \geq 2((a,b)+(a,c)+(b,c))$ which is impossible.

So $n=2$.
Again, $lcm((a,b),(a,c)) = (a,b) \times (a,c)/2$, ...., we end with $(a,b)+(a,c)+(b,c) \geq .... \geq ((a,b)+(a,c)+(b,c))$ which is possible only if there is equality everywhere between, i.e
$(a,b)=(a,c)=(b,c)=(a,b,c)=2$

Finally : $a=b=c=2$