group theory
abstract algebra
geometry
geometric transformation
superior algebra
superior algebra unsolved

Show that an infinite simple group $G$ cannot have a proper simple subgroup $H$ with finite index.

My problem here is that I don't see why we need to assume $H$ to be simple ...

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=normal&t=22796]this[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=normal&t=19724]this[/url], for example. Indeed, if $G$ had a proper subgroup of finite index, it would also have a proper normal subgroup of finite index.

Thanks Grobber ( I did the same using the translations on the left of $G$ over $G/H$, but I did not see where the simplicity of $H$ was intervening. )

[quote="julien_santini"]but I did not see where the simplicity of $H$ was intervening[/quote]

I didn't use it either :?. My guess is that it was somehow supposed to make the problem easier :).


function
group theory
abstract algebra
topology
calculus
vector
induction

[size=100]Hello it's me again The Doctor. Anyway way back when I posted about some college course selection advice. I decided that as is usually the case the AOPS community was right, my proposed schedule would have been at best really stressful and at worst pretty downright unreasonable not to mention I would have little to gain from it even if successfull (thnks again for setting me straight on that btw). So I have overhauled it and made a new (This one actually looks a lot nicer even to me) one and am wondering what you guys think?

Autumn:Credit Hours
LinAlg I:3
Intro to Abstract Math:3
Multivariable Calc:5
Music Appreciation:3 (got to get that half fine art credit)
*English:3
Total 14/17
Winter:Credit Hours
LinAlg II:3
Intro to Analysis:3
AbstractAlg Part I:3
Art Appreciation:3
*English:3
Total:12/15
Spring:Credit Hours
Intro to Analysis II:3
AbstractAlg Part II:3
Differential Equations:3/5
*English:3
Total 9/11/12/14

Issues: There are a few issues with it though for some strange reason the entire LinAlg sequence is a prereq to Intro to Abstract Math (Logic, proofs, set theory, relations, functions, and cardinality) and then that is a prereq for the Analysis Sequence ([size=75]The Real and Rational Number Systems: algebraic, order and completeness properties; Sequences: boundedness, monotonicity, convergence; Limits of Real-valued Functions; Continuous Functions: local and global properties, Intermediate Value Theorem.  [/size] this is for I) and for the Abstract Algebra sequence (Definition and basic properties of groups, subgroups, permutation groups, direct products, isomorphisms, homomorphisms, normal subgroups and factor groups. [/size] again part I) So idk as you can see I didn't follow it a linearly as I maybe should have do you guys think that's okay in this case? Even if you do though I suppose I will have to convince the department/professors teaching the classes. *Not sure yet whether taking English at school or University will be easier

Thanks A Lot
The Doctor

I think you should make your winter and spring course selections when the time comes to.  In other words, there's no real way you can know what you will be taking then.  In my opinion, I would not jump in by taking 3 math courses.  I would personally drop the "Abstract Math" course.  In analysis, you'll probably spend sufficient time covering some basic set theory and cardinality.  Also, topology courses, particularly those using Munkres' book, typically cover some set theory.

You should hold off on "Abstract Math" for now.  I also don't think it's wise to take algebra, analysis, and linear algebra at the same time.  Analysis requires knowledge of linear algebra so you might be behind when you get to metric spaces and certain aspects of linear transformations.  I took 3 math courses in my last semester (analysis, number theory, topology) and although I did well, I would not do it again.

Others may disagree with me but I don't think you rush in.

I'll tell you what I'd say if you were as student at my institution. I'd tell you not to ask a bunch of online semi-strangers, but rather to come into my office so we could talk about this face to face. I have a pretty good idea what's in most of those courses by title - but not a perfect idea, and what, really is the intensity of each? And what's the usual order? And are there some particular local features of some of those courses? Oh, and I'd want to know what your own track record looked like. (Don't tell us! That's not a public question.)

There are people whose business is advising. There are most likely well-connected and student-oriented faculty members in your department, people who instantly know the answers to all the questions I just asked and maybe a couple I didn't think of. So don't ask us - ask a person like that.

I do frequently encounter students who have a chart in their hands that shows exactly what courses they're going to take in what semester for their entire future time as a student. Having some idea of what you're doing, how far along you are, and where certain courses fit in is a good thing, but I've never been that much of a fan of rigidity and over-planning. Sometimes you have to adjust.

[quote="Kent Merryfield"]
I do frequently encounter students who have a chart in their hands that shows exactly what courses they're going to take in what semester for their entire future time as a student. Having some idea of what you're doing, how far along you are, and where certain courses fit in is a good thing, but I've never been that much of a fan of rigidity and over-planning. Sometimes you have to adjust.[/quote]

I did this.  The courses I ended up taking were quite different from the courses I had planned to take, (since, as a high schooler, I really didn't know anything about math!) but I still think it was a very useful exercise.

For one, at my undergraduate institution, 400-level and graduate courses were on a two-year rotation.  This meant that (since I was planning on graduating in three years), I only had one chance at taking these courses.  These scheduling issues are common unless you go to a university with a very large math department.

Scheduling is most important at the beginning of one's math 'career' though, since you want to pick your courses in such a way that you have enough freedom in terms of prerequisites to take any course that you're interested in that comes along.  Once you get 'known' in your department it's generally easy to get the prereqs waived but this isn't really easy for first year students to do in general.

JRav: A first analysis course doesn't really require any linear algebra.  YMMV for number of math courses -- I took four math courses almost every semester of undergrad, and five once -- the latter proved to be a bad idea.

JRav I agree with you about making the winter and spring schedules when the time comes, but I still like to have some idea of what I want to do before hand especially cause a lot of these classes are part of sequences and stuff. I want to sorta avoid a miss mosh of classes that lack cohesion. As for dropping Intro to Abstract Math I need that as a prerequisite for a bunch of stuff in fact my biggest challenge is gonna be to convince them to let me take stuff with it at the same time.

Kent Merryfield: AOPS isn't a bunch of strangers they are AOPS which is completely different lol. After all you are one of the so called semi strangers and your input has been quite use full. What I would like to ask you is who is the best person to try and arrange this kind of meeting with and how should I go about doing it? Is phone better or maybe e-mail or perhaps face to face?What I am essentially afraid of is that I will not be given the opportunity to really try and like drop if it turns out to be to much or something you know. I am afraid of the "NO! these are our little rules and guidelines and we won't let you" Plus track record is something I am trying to fix (idk how good or bad it is all I know is I feel I need to do better) or am I to be bound by the past indefinitely. Idk in my humble opinion I should after discussing it with them be allowed to take w/e I want and then get rid of it if I can't handle it, seems more sensible. Buy anyway ya who should I talk to and how do I arrange to meet them?

I agree with Kent Merryfield completely.  Maybe the people on AoPS aren't perfect strangers, but how many of us have you actually met personally?  We might give you sympathetic advice, but we don't know that much about your background, and it would be impractical and inappropriate to go into it deeply on a message board like this.  Also, most of us are students, and few of us have probably gone through your particular program---in short, most of us probably don't know what we're talking about.  But your math professors have probably watched carefully as dozens of students have passed through the program that many of them probably helped design.

You don't have to be "bound by the past" forever, but you can't disregard it immediately.  Face to face is almost certainly the best way to talk to somebody about this.  (I did this just a week ago!)  It probably isn't a brilliant idea to take 11 credit hours of any subject all at once.  (Do you not have any general education requirements whatsoever?)  These ideas take time to sink in.  I doubt that your math faculty will cling to "little rules"; it's more likely that they'll be happy to have a student who's as interested in math as you seem to be.  And if they are so adamant, you're probably out of luck anyway, because you'll probably need somebody (if not several people) to approve many of the courses that you want to take.

[quote="blahblahblah"]These scheduling issues are common unless you go to a university with a very large math department.[/quote]
I don't think that's necessarily true.  My school does not have a particularly large math department but they tend to offer the same grad courses each year so it's fairly predictable.  I guess some of the upper-upper grad courses are offered depending on interest but I wouldn't think that's true of the first or second-year grad courses.
[quote="blahblahblah"]A first analysis course doesn't really require any linear algebra.[/quote]
Maybe "requires" is the wrong word-but I definitely think it helps in understanding metric spaces and inner products.

Boy Soprano II how did said meeting go for you? Also I agree that a face to face meeting is probably the best but I am not quite sure how to go about requesting one, like what should I say, I guess I am kind of afraid of just being an annoyance that he will not really want to deal with or idk that he will just want ot give me a quick response over the phone and get it over with. I do however have a name now though apparently this guy handles all the scheduling of the math department.

Also yes I agree that deep advice is not to be expected but that's also not what I am looking for. It's more just getting some general thoughts from a community of fellow math lovers. I trust that those who do respond to my queries will do the best that they can. And while deep insights may not be gleaned the comments are often times quite help full?

But anyway what I need to know then I guess is the best way to approach this guy in the math department?

[quote="The Doctor"]
But anyway what I need to know then I guess is the best way to approach this guy in the math department?[/quote]
Just go straight to the director of undergraduate studies (if you do not know who it is, just ask any secretary first) and tell him your story. Then see what he will say. Most likely he'll immediately refer you to somebody else who is directly involved in undergraduate mentoring. If he's out of town, knock at any faculty door and ask whom else you could talk to about this. Then just patiently wait until people stop dragging you up and down the stairs from office to office and talking to each other and give you some definite answer. :)

[quote="fedja"]Just go straight to the director of undergraduate studies (if you do not know who it is, just ask any secretary first) ...[/quote]
Or something like that. Technically, my own title is "Undergraduate Associate Chair" and another faculty member has the title "Undergraduate Advisor." But then if you showed up at our department office window, or called the main department telephone number, and asked a question like that, you'd get directed to one or the other of us rather quickly.

A math department is a human institution and a human community. It's not the same think as its catalog listing. Its courses are not some pure unchanging abstract entity, they're the efforts of humans. (And yes, they have human flaws). fedja has rather vividly described some human interactions.

The one thing that struck me from reading The Doctor's posts on this thread is that he doesn't seem to think of himself as a member of that community, despite his determination to spend a very large amount of time in it. That's what I want to say to him: include yourself in, and think of yourself as belonging. You can even find out what the unofficial student grapevine says (allowing, of course, for the frequent inaccuracies of that grapevine - and bear in mind that the right professor for one student is the wrong professor for another just because the two students are very different).

[quote="JRav"]If you declare as a math major, you should be given a faculty adviser in the math department, who should be a member of the faculty.  Ideally, you'll want someone who has lots of experience, or who knows you well.[/quote]

varies from university to university, and it may not be possible to declare a major until you have approximately two years worth of coursework completed.

The meeting went fine.  We discussed my background a little bit and he asked me some questions; we talked about my situation.  Initially he was rather reluctant to let me take the course I wanted, but in the end he agreed to let me take it and even jumped through several hoops to get me in.  I am indebted to him. :)

Just say that you want advice about what math courses to take, or something like that.  If you just start talking to people, eventually you will get to the person you need.  Think of the math department as your new family (or at least, a new branch of your family).

This may sound sort of stupid, but talking to people and getting to know them is a valuable life skill, and maybe some of us math people tend to neglect it a little.  Sure, you don't want to irritate people.  But it's better to risk irritating a few people than to do something really stupid because you were afraid of irritating people.  Also, I think you'll find that you are more likely to irritate people when you don't talk to people than when you do.  Lack of communication can cause serious problems (look carefully the next time you observe a little spat between two people and I think you'll see what I mean).  It would actually be pretty weird to take all those courses without talking to an advisor or professor.

Ya I know I completely agree with everything you said and its normally not a problem for me at all. It's just in this case I can't really afford to get it wrong and I am a little bit more wary when dealing with idk like administration and stuff. But I mean I e-mailed the guy asked hi if we could meet to talk about some stuff in regards to course selection and am now just waiting for a response.

I completely agree with Professor Merryfield.  The single best reason to speak with a professor at your institution is that none of us really knows what is in those courses you've listed, how hard they are, etc.

On the other hand, I understand The Doctor's apprehension about seeking official departmental advice.  And it is probably true that at some places advising is not very good (or nonexistent).  But you really have nothing to lose.  If you don't like the advice you're given, you don't have to follow it.  And if you are worried that someone will tell you, "Oh, you aren't allowed to do this," keep in mind that if something is not allowed, it's not allowed.  It's hard to imagine a scenario in which you would be prevented from doing something *because* you sought advice.  College is for grownups, so typically you can take whatever courses as you like, as long as you are following the official written rules.  (However, some schools/departments do have an excess of official written rules.)

Hey I am not sure if anyone cares but I thought I would share what happened. I took you guy's advice talked to the right people (although it took awhile so many of them were on vacation). The guy said based on my goals the program outlined for him seemed good (although you guys were right I changed it yet again  :P ). And I was cleared to take everything I wanted. So things are going pretty good for now though I am still wary, stuff never goes the way I would like it to my luck is pretty bad,(lol it's like now that something has I am wondering what the Universe is going to put me through to get back at me), of course despite this I consider myself to be a very optimistic person,wow I am not making any sense now. Anyway thanks to all the AOPser's who so patiently helped me and answered all my annoying ?'s I really appreciate it.

As you've already realized, nothing goes as planned.  You may (or may not) be underestimating the difficulty of these courses, and that could completely change what you do in the future.

AW **** I knew this would happen didn't I say the cosmos would screw me over. The professor I talked to said everything was cool (I liked him very informative). Then the undergrad adviser people caught wind of what was going on and now I am living my worst nightmare of trying to convince them I can handle taking Intro to Abstract Math and Lin Alg at the same time. *ML!!!!!!!

Your worst nightmare is convincing people that you have the capabilities to perform certain tasks? Good luck getting into college or finding a job.

It's not the end of the world to have to talk to people, but if you think you're ready for the classes and you know what is expected of you from each of those classes, it can't be that hard to convince them. Of course, it wouldn't be the end of the world either if they convince you that maybe it's not in your best interests to take all of the classes you want to take. After all, advising is what they do, and if they didn't bother you, they wouldn't be doing their jobs: helping you choose the right courses for you.

Well you know what I mean talking to people is not my worst nightmare but this particular conversation was the worst thing I could have conceived of in this particular situation. With good reason too I just got shot down they are basically like you might even be able to handle it but we don't know and it wouldn't be fair etc etc. But I suppose you are right they are just doing their job which since it is advising may indeed mean that maybe lol they know what is best for me.

I still don't see why you need to overload in your first trimester.  Do well in your classes, and you'll convince the staff that you can handle a higher workload.  Simple as that.

Wow this is so stupid this just keeps getting worse and worse. Now I am seriously up sh** creek without a paddle. This is exactly why I hate stupid rules and regulations and bureaucratic red tape this is so ridiculous.

These rules are there to protect you from the likely possibility that you're overestimating your abilities. You're not about to fall behind any reasonable standard; it's not the end of the world if you don't take everything right away.

You haven't been to college, you don't know what you're in for. I'll list the risks in two categories, with the first being the more likely case.

[u]Suppose you don't do all that hot:[/u]

1. Taking more classes and getting generally lower grades means
- You have lower factors in your GPA
- It's harder to make up for it because these lower grades have a higher weight, since it's a lot of credits

2. Employers, research coordinators, internship application reviewers, etc. will not be impressed if you don't do all that hot because of overloading. It tells them that:
- You're not all that great at planning
- You tend to overestimate your abilities, meaning that the strengths you claim in your interview/cover letter will have a watered down effect

3. You probably invested too much time into your classes, and did not have the chance to settle in to your new environment and explore it. This has short run and long run negative effects on your application

4. People reading your application will think you're boring and useless (unless you're applying for pure math research) for taking just math courses.

[u]Suppose you do well.[/u]

1. You probably invested too much time into your classes, and did not have the chance to settle in to your new environment and explore it. This has short run and long-run and negative effects on your application. See (2).

2. You have a good start to your GPA, but there are upperclassmen that have taken the same courses, and more, and done well in all of them. Meanwhile, you missed out on your chance to get engaged early on, which is what an application reader will be looking for out of a freshman application (if it were just coursework, only juniors and seniors would get any useful offers).

3. People reading your application will think you're boring and useless (unless you're applying for pure math research) for taking just math courses.

I must say that I am a little surprised at how little you guys must think of me to assume and work of of the assumption that I will most likely fail and that I am clearly simply trying to do too much. Well perhaps before you decide to jump to anymore conclusions you should have some information in regards to the actual circumstances.
See here is the thing though you guys might be right except I heard everything you guys said and I knew you were right that's why I trimmed it down made it more reasonable, its not what is in my op of this thread. I actually met with a faculty member and he helped me tweak it a lot but said it was essentially quite a good list of classes to take and very manageable. I was quite please and happily registered it was then that a third party got involved and is basically trying to trash the whole thing. That's when I got upset. All I am saying is this third party is trying to not even give me a chance, add/drop policies exist for a reason do they not? Besides I find it hard to believe that what I am wanting to do is clearly too much when a different adviser helped me make my plan and spoke so highly of it. It seems you guys are super confident in the system and insist that i just don't know what I am getting into and am likely to do poorly, even if I do well I will look bad to emplyers etc etc.  But correct me if I am wrong but in this case there is a complete lack of consistency is there not?

Furthermore many of TZF's points seem irrelevant to me as I am unlikely to use any of this for a cover letter or anything of that nature nor will I be attending this school after  this academic year. In case anyone didn't know I am just a senior taking some classes at the local University. I don't see why anyone would look down on the fact that I decided that after having taken a wide variety of classes from freshman through junior year and taking essentially every math and science class offered by my school I decided for my senior year of high school rather than take a bunch of classes that did not really interest me, I would further pursue classes that were actually of some interest to me.

Ah, k, sorry for jumping on that.

Really, though, that was the most important piece of information you could have provided (apart from the courses themselves). It changes [i]everything[/i]. Just so that you don't think I was trying to be an ---, I took a full load of college courses myself in senior year, and it was a [i]completely [/i]different experience from actually going to college because I had completely different short term goals (and in the case of HS, didn't have a roommate).

Your agenda of classes also makes sense now. I would only say that if you're applying to colleges this year, be ready to justify your selection of courses in the most positive way (which I think you've already done, but it's good to have it at the tip of your tongue -- many admissions counselors don't come from technical backgrounds, and might not be able to infer your motivation right away).

About your actual courses, my only suggestion would be to switch your first intro analysis course with your multivariate calculus course. This way, you'll have a course in linear algebra behind you when you start multivariate calculus -- something that I wish I had done. It's not all that big of a deal though -- many, many people take multivariate calculus before linear algebra, although this is more the result of what math is immediately relevant to non-mathematics majors (like physics or engineering) rather than a result of what makes the most sense mathematically.

I'm curious -- what is this third party? Colleges are full of measures to stop students from screwing up, and this might be even harder because you're a high school student. Don't take no for an answer! Have your academic advisor write a letter.

@ TZF I am trying I am really trying so hard but these people are making me go insane I don't know what to do anymore. The third party is another adviser apparently the guy who was actually gonna remove some of the fail safes that are in place so I could sign up was concerned about one class in particular and so asked that I talk to this person I finally relented etc etc and then I found out that apparently multivariable is called calc IV Calc BC apparently actually dosen't quite prepare you for that it only takes you up to Calc III even then I was like okay fine I can deal with that but now they are trying to make me hold off on taking Linear Algebra I as well and that's when I just plain lost it.

I don't know what to do anymore I talked to the department head who in tern e-mailed my math teacher I think she may have gotten the wrong impression of what I wanted to take you know including the analysis and stuff but I don't want that anymore I just want the Linear Algebra and maybe Differential Equations.

Like I said I have pretty much tried everything except playing the race card. Sorry if I seem incoherent but I am rather distraught at the moment.

I really don't understand what you're saying -- an occasional period might help  :wink:

Anyway, from what I've gathered, you're not in a degree program, so the only rules they can enforce with any justification are prerequisites. And prerequisites can pretty much always be overridden by instructor's permission. Have you tried talking to the professors you hope to have? After all, if the instructor is okay with it AND the college gets to collect your money AND the college doesn't risk inappropriately awarding a degree, then they'd do it!

We don't know what college you're attending, but courses such as abstract algebra and analysis are hard no matter where you take them.  There is no reason to overwhelm yourself with work in your first year.  As TZF said, getting adjusted is extremely important.  If you're not well-adjusted, you will not only have a miserable time socially, but especially academically.  And whatever you may think, orientation is not enough time to get acclimated, especially since there is no work.  In fact, you'll probably need around 3 weeks of classes before you even see what a normal week looks like.

Alright so here is where things stand right now. Apparently Calc BC dosen't quite go up to their Calc IV (multivariable) so I will have to start at III. I highly doubt at this point that any amount of kicking and screaming will get me into the multivariable Calc IV but I think this is acceptable I already know a bunch but not all the material taught in the Calc III so fine. I don't know how feasible it is but you know I am always open to you guys's suggestions so if anyone thinks I should wage WWIII to get into Calc IV I would at least like to hear it. They also after having me jump through like 50 hoops (conservative estimate) and talking to my old teacher decided that Lin Alg I was also fine. Intro to Abstract math is apparently totally out of the question until spring and if I am going to go with TZF's suggestion and not take Analysis in the winter and take Calc IV instead that is fine anyway (although in this case I have like no choice).

Anyway the one other class I may want to ask for in the Fall is Diff Eq. I am afraid however that if I do my "adviser" may just lose it. Correct me if I am wrong but I am guessing that all of you guys advocating the whole ease yourself into it think I should just take it in the winter or spring anyway and let it go.

Ok so anyway as it stands now unless I wage WWIII this would be my schedule any thoughts would be appreciated as to whether or not this is good enough to accept or if those (insert expletive here) have a war on their hands. The one thing that I am worried about is looking lazy to colleges granted this will be my schedule for fall quarter only plz let me know if it seems like it needs more classes added to it to seem full. So I mean ya is it enough in general not just in terms of math?

Calc III&Lab: 6
Lin Alg I: 3
FA Class: 3
We will say eng at my school is equivalent to about 3
Total: 15 Credit Hours

God this is way too much drama for something so simple you would think I was running negotiations in regards to nuclear proliferation with these people.

P.S Ignore the Abstract Algebra and all that other stuff I messed up that stuff turns out isn't even offered in the spring anyway.

We don't know what the difference between Calc III and Calc IV is, or what is taught in either.  What is Calc III Lab, and what is FA Class?  I would guess they're some sort of distribution requirements or something, in which case, it seems fine.

With regards to your second to last sentence, these people have far more experience with this stuff than you do and they know the amount of work that goes into these classes.   Listen to them.

I realize they have far more experience but that still dosen't explain why they were all saying different things.

Sorry about the strange abbreviations FA=Fine Arts class. And um basically Calc IV is multi variable but apparently Calc III has some advanced series and vector calculus stuff that we don't cover in BC. Oh and the lab its just this class that you have to take if you are taking Calc III (i don't see why it is listed seperately it is only once a week, though it is just pass/fail) you use like some computer software stuff to like supplement what you are learning. I don't know exactly what it entails.

Taking 12 credits of college courses alongside a class from your school is definitely not 'lazy'.

Look, ultimately the judgment is [i]yours[/i] as to whether or not you want to settle for Calc III. It's up to you if you think you should try for Calc IV or not. If you feel ready for Calc IV, it is [b]absolutely [/b]worth the effort to get the classes you want -- you're paying for them and investing an entire term of your time! Certainly, even 10-15 hours of negotiating is a small price to pay for taking a worthwhile class instead of a useless class, the latter of which is a waste of money and comes at a high opportunity cost.

If you want to get into Calc IV, I would suggest you look up the instructor for the course and go directly to him/her, and say something like:

- Hey look, I want to take your course
- I've taken Calc BC
- I realize Calc BC doesn't cover such and such that Calc III does
- but I've studied them extensively on my own and am very comfortable with them
- I need your formal permission to override the prerequisite of Calc III, which I have met in terms of knowledge, just not on the books
- can I have your permission to take this class?

The instructor will probably say yes (instructors like proactive, interested students), in which case, ask the instructor to kindly write and sign a letter giving his/her approval (and make sure the letter says that s/he is giving you approval because s/he thinks you are ready). It should be a 5-15 minute process overall.

When you approach your adviser, make sure to sound sure of yourself and confident. You should convey through your tone/mannerisms that you're not seeking advice from your adviser, because you are already well informed and have the information to make a balanced decision -- you only need the signature out of formality. Try not to come across as cocky or rude in the process.

If this letter of approval is not enough for your adviser, then s/he is really incompetent, and you go to the director of advising or to the registrar and ask for a different adviser. After all, it's your money the college is taking, and you're not even bound by the requirements of a degree program!

Of course, if the instructor says no, then even if your adviser lets you in the class, you probably don't want to deal with a class with that instructor, because the instructor will be like, oh, it's the kid I told to not take my class, why is he here???

[quote="TZF"]You can also threaten to publish your difficult and strenuous experience on the internet, which will damage the adviser's and the college's reputation. Why would anyone want to go to a college that doesn't let students take classes that they are prepared for?[/quote]

This is poor advice, as the college in question is a community college and I'm basically 100% sure that they will not care in the slightest, and be less likely to help because of the 'threats'.

I haven't seen the other thread, but now given that it is a community college, listen to [b]blahblahblah[/b] (I'll amend my post).
[i](Even if it's a private college, obviously threats are a measure of last resort, though they are completely valid).[/i]

In general, everything I said is just a suggestion -- they need to be adapted to the specific circumstances (like the particular personalities of the people you're dealing with).

I think some clarification information is in order.

First of all this isn't a community college though I don't think that is too relevant to what I am about to say next. Okay so I already talked to the department head and he seemed that I would be better off starting at Calc III (in general he said he didn't really see any compelling reason to overule my adviser)and my old Math Teacher agreed with him also who the instructor is going to be has not yet been decided apparently idk something about budget cuts and what not. Plus idk if the professors will be more like my adviser or more like the first guy I talked to (btw I am pretty sure that my adviser is like the head adviser for the math department or I think her title might be like undergraduate program director or something like that). I mean I am pretty sure I can learn the stuff before classes start but at the same time if I have to take the class in the winter as opposed to the fall it might not be that big a deal. As far as the costs go I didn't mention this cause I didn't think it was really important before I guess now it has become quite relevant. I am not actually paying for any of this as part of the program through which I gained acceptance to the University my school district foots the bill. So the only cost is time. I am not 100% sure off the top of my head but I am quite certain that just taking the Calc IV in the fall won't change much anyway, to do that I think I would have to skip some more pre-reqs doing some stuff concurrently and then do the same thing again skip some pre-req's and do some stuff concurrently. Or I suppose if I took Multivariable and Diff Eq at the same time then in the winter I would have PDE & Fourier Analysis as an option. Idk weren't you guys the ones recommending easing into it, I remember TZF said that some linear alg before multi variable might really help.

Oh btw I don't know quite how accurate this figure is but apparently even with BC there is about half of Calc III left to learn. Apparently this is a switch they just made this year not letting BC students take Calc IV apparently they used too.

Ugh I don't know the department head seems to think I should take it  a little easy to start off with too, he didn't say anything to me about calc III but he apparently did talk to my old math teacher about what's in it and her advice was to not rush. Idk I am a little confused right now, I am not sure what the right call is and I am afraid that if I don't make the right call it could have serious negative impact on me,although it may not be nearly as big a deal as I think, again I ask do you guys see any strong compelling reason I should try and fight on this?

oh btw I have a much better idea of what is in intro to abstract math now. My adviser says I can't take this cause it has Lin Alg II as a pre-req why I have no idea, but she says I need to wait until the spring quarter? This one might be worth trying to get in to:

Sets, set operations, cartesian products, logic, different types of proofs (contrapositive, contradictions, enumeration of cases), real numbers, equivalence relations, functions, induction, cardinality, Euclidean algorithm, sequences, elements of group theory.

So far in your description, I don't see much evidence of linear algebra.  This seems kind of like a discrete math course, although of course real numbers are not discrete quantities.  Do you have a university-written course description?

Just in case I wasn't clear that is the description of the Intro to Abstract Math class. And you are correct as far as I can tell there is no linear algebra. Which is why I don't understand why Linear Algebra is a pre-requisite for the class that need actually be enforced.

Again, we don't know the courses at your school but usually, a rigorous linear algebra is one's first real experience with higher mathematics and formal proof-writing (I said usually, not always-obviously people who went to summer programs or took rigorous calculus will have some familiarity), so they want you to be comfortable working with abstract concepts before you take the class.  If you're not already proficient in, say, proof by contradiction, it will certainly hurt you later on.


Let $x,y,z,u$ be positive numbers, determine the best constant $k$ for the following inequality.
$x^4+y^4+z^4+u^4+xy^3+yz^3+zu^3+ux^3>=k(x^3y+y^3z+z^3u+u^3x)$.

we can show that $k \leq 2$ by puting $x=y=z=u$
and we can show that $x^4+y^4+z^4+u^4 \geq x^3y+y^3z+z^3u+u^3x$ by using
hardy_littlewood_polya inequality, i was trying to prove that :

$xy^3+yz^3+zu^3+ux^3 \geq x^3y+y^3z+z^3u+u^3x$ considering
$x\geq y\geq z\geq u$ that  depends on some other conditions of hardy_littlewood_polya inequality equality
that i dont know them , any idea?


Farmer Flagg has 400 feet of fencing to enclose a circular garden and a second 400 feet of fencing to enclose a square garden. What is the ratio of the area of the circular garden to the area of the square garden? Express your answer as a common fraction in terms of $ \pi$.

Farmer Flagg has 400 feet of fencing to enclose a circular region
that means that the circumfrence of the region is 400 feet
that means the diameter of the region is $ \frac{400}{\pi}$ (since $ C\equal{}d\pi$) and the radius is $ \frac{200}{\pi}$

the area of a circle is $ \pi r^2$, so we plug the radius into the equation

$ \pi r^2\equal{}A$

$ \pi (\frac{200}{\pi})^2\equal{}A$

$ \pi \frac{40000}{\pi^2}\equal{}A$

$ \frac{40000}{\pi}\equal{}A$

the area of the circle is $ \frac{40000}{\pi}$

now for the area of the square
since it's a square, the four sides are all of equal length
since the perimeter is $ 400$, each side is $ 100$, and the area of the square is $ 10000$

now we want to find the ratio of the area of the circle to the area of the square

$ \frac{\frac{40000}{\pi}}{10000}$

$ \equal{}\frac{40000}{\pi}\times\frac{1}{10000}$

$ \equal{}\frac{4}{\pi}\times1$

$ \equal{}\frac{4}{\pi}$

our answer is $ \frac{4}{\pi}$


[b]In a battle 70% of the combatants lost one eye, 80% an  ear, 75% an arm, 85% a leg, x% lost all the four limbs. Find out the minimum value of x.

Its answer is 10. Please help as soon as possible. Thanks a lot.[/b]

[quote="AndrewTom"]Suppose there are $ 100$ soldiers.

$ 70$ lose an eye and $ 30$ don't.

Those $ 30$ lose an ear only and of the others $ 20$ lose an eye only and $ 50$ lose both.

The $ 30$ and the $ 20$ lose an arm as well and of the $ 50$ others $ 25$ lose an eye and an ear only and $ 25$ lose all three.

Then the $ 75$ who have not lost all three so far lose a leg (and so not all four of eye, ear, arm and leg), $ 10$ lose all four and $ 15$ lose the other three but not a leg.

Therefore the minimum value of $ x$ is $ 10$.

Hope this makes sense (even if it is poorly expressed).[/quote]

Thank you so much for your help. I just understood the first three lines and solved it further myself, you gave a very great hint and a very intellingent solution. Also it made sense for me.Thanks so many times. :-)

See how I solved it after reading your first three lines.

1) $ 70$ out of $ 100$ lost one eye, means $ 30$ didn't lose eye. $ 80$ out of $ 100$ lost ear and $ 20$ didn't. All the $ 70$ soldiers who lost eye may belong to $ 80$ soldiers who lost ear , it nay be true or not, but at least $ 50$ out of $ 70$ belong to $ 80$ soldiers who lost ear. So out of $ 100$, the soldiers who lost eye and ear $ >\equal{} 50$.

2)$ 75$ out of $ 100$ lost an arm, means $ 25$ didn't. So all $ 50$ (who lost eye and ear both) may belong to the $ 75$ (who lost arm) but it may be true or not. But atleast $ 25$ in these $ 50$ (who lost eye and ear both) will be in the $ 75$. So out of $ 100$, the soldiers who lost eye, ear and arm $ >\equal{}25$.

3)$ 85$ out of $ 100$ lost a leg, means $ 15$ didn't. So all the $ 25$(who lost eye, ear and arm) may belong to the $ 85$, but again it may be  true or not, but atleast $ 10$ out of $ 25$ will be in the $ 85$ , otherwise it will not become $ 85$.
So we have reached to our solution that atleast $ 10$ soldiers are there who lost all the four limbs.

Done!

Glad I could help, ninni. Looking at yours, I see I could have explained it better that I did. Cheers.

well al you have to do is draw a venn diagram and try to make the smallest intersection of of different things


conics
ellipse
trigonometry
geometry
analytic geometry
graphing lines
slope

[img]http://img509.imageshack.us/img509/9929/68651651.jpg[/img]

I don't have any good idea :oops:

The circle has standard equation $ x^2 \plus{} (y\minus{}1)^2 \equal{} 1$, and so is the circle of unit radius centered at $ (0,1)$.

Clearly, $ a > 1$ and $ b \ge 2$.  We parametrize the ellipse in the usual fashion and consider the first quadrant:

$ E(t) \equal{} (a \cos t, b \sin t), \quad t \in [0,\pi/2]$.

For a given value of $ a$, the ellipse of minimal area that encloses the circle is the one that is tangent to the circle.  There is at most one point of tangency in the first quadrant, and suppose it occurs for the value $ t \equal{} t_0$.  Then two conditions must be satisfied.  First, the point of tangency is a common point of the ellipse and circle:

$ (a \cos t_0)^2 \plus{} (b \sin t_0 \minus{} 1)^2 \equal{} 1$.

Second, the slope of the tangent line to both curves is equivalent:

$ \frac{b \cos t_0}{\minus{}a \sin t_0} \equal{} \minus{}\frac{a \cos t_0}{b \sin t_0 \minus{} 1}$.

The LHS is the expression for the slope of the ellipse at $ t_0$, and the RHS is the negative reciprocal of the slope of the radius from $ E(t_0)$ to the center $ (0,1)$, and therefore is the slope of the tangent line to the circle.

Solving this system for $ a, b$ and treating $ t_0$ constant yields the relationship

$ b \equal{} \frac{a^2}{\sqrt{a^2\minus{}1}}$.

Since the area of the ellipse is $ \pi a b$, it is now a straightforward exercise to minimize the area as a function of $ a$.


What are you talking about? I don't have this problem. Could it be your browser?

I can't reproduce this problem, so if it comes up again, can you give us some more information, like: your browser, your operating system, and the URL of the page on which it doesn't work.

[quote="math92"]Whenever i try to jump to a forum by using the little scroll down button in the bottom right corner of the page it only lets me jump to forum called "None" which doesn't exist.  Can you fix this problem?[/quote]Delete your cookies. That will solve the problem.


What is the largest integer which must evenly divide all integers of the form n^5 - n? (AHSME 1957)

The answer is [hide]0000005[/hide], I think.   That's wrong, by the way. I just realized.

Is this a challenge or do you need help with the problem?

I think it's probably [hide]10, since the fifth power of any number has the same units digit as the number, so when you subtract the original number the units digit of the result must be 0, therefore it must be divisible by 10.[/hide]

I was being too careless, than answer should be [hide]2*3*5=30.[/hide]

I just noticed that I was wrong before, [hide]10[/hide] can't be right. It must be [hide]30[/hide] instead, but beta, how did you get your answer? I just got it by guessing and checking.

n^5-n=[hide]n(n+1)(n-1)(n :^2: +1). One of n, n-1, n+1 must be divisible by 3. then just as you said, the last digit is always 0. 2*3*5=30. [/hide] For all prime p, n^p-n is divisible by p.

[quote="TripleM"]Yeah, thats about 99% correct. I'll give you 6/7 :) but that would be being nice. Maybe 5/7.[/quote]

I am totally confused. Why do you get a score???

he's saying that the solution is not 100% rigorous. You have to prove that the number can't be bigger than 30. To do this, note that the number must divide 2^5-2=30, so 30 is the maximum possible number.

Yup, thats better. I was just saying if you had submitted that solution in a competition, you wouldn't get full marks at all, since you hadn't proved it was the maximum (even though thats trivial, its still important).


algebra polynomial calculus integration function

Is it possible to find a polynomial, apart from the constant 0 itself, which is
identically equal to 0 (i.e. a polynomial p(t) with some nonzero coefficient such that p(c) = 0 for each number c)? Justify your answer?

This is a problem on a book, there is an answer, but I don't get it...

no.

Assume otherwise, and let it have degree n. Since it has degree n, it has exactly n complex roots, so it has at most n real roots. But the n+1 numbers 1,2,...,n+1 are all roots of the polynomial, meaning it has at least n+1 roots, contradiction.

Zabelman said:
 [quote]But the n+1 numbers 1,2,...,n+1 are all roots of the polynomial, meaning it has at least n+1 roots, contradiction.

[/quote]

could you explain that more thoroughly?

In other words, if P(x) = 0 for all x, then certainly x = 1, 2, ..., n+1 will also produce 0. Thus there are more than n roots, which cannot be.

I see...
So, if a polynomial has zeroes for any possible real value x, then
(x-1)(x-2)(x-3).... = polynomial, which can not be possible...
wow, why couldn't I get that...

Well, does it have to be a finite polynomial?  And does it have to be for any c, or just integral c?  Because if the answers to those are no and yes respectively, you could have:
 :pi:*x -( :pi: x) :^2: /2+( :pi: x) :^4: /24+...( :pi: *x)^(2n)/(2n)!...

A polynomial is defined to be finite.  Theres really no such thing as an "infinite polynomial", thats given another name. And it was meant to be for any c, not just integers.

[quote="Magnara"]Fine.  Ruin my fun :-([/quote]

People do sometimes speak of the Taylor polynomial, which, for certain functions, can be extended to converge everywhere.  So, for example, the function f(x) = (x - x^3/3! + x^5/5! - ...), where the sum on the right side is infinite, is "sort-of" a polynomial -- not really, but sort-of -- and it does exhibit many properties of polynomials (continuity, infinite differentiability, and so on), and it also has infinitely many values at which it is zero.  However, as TripleM said, it isn't really a polynomial -- in fact, it's the sin function.


Hint (to the first problem):
[hide]
Triangle $BDC$ is isosceles hence $BC=BD$.[/hide]


Let $x,y,z$ are real number such that $xyz=-1$
Prove that
$3(x^2-x+1)(y^2-y+1)(z^2-z+1) \geq 1$ :)

There must be a typo :
For all $x$, $x^2-x+1\geq 3/4$, so $LHS \geq 3*(3/4)^3=81/64$.

In fact, since $x^2-x+1 \geq |x|$, $LHS \geq 3|x||y||z|=3$.

And I think it can be proven that $LHS\geq 9$, but I'm not sure.


[quote="hendrata01"]Sorry but I'm a beginner. So can I do this?
It's homogeneous, so we can assume $abc = 1$.

$\frac{a^2}{b+c} \leq \frac{a^2}{2\sqrt{bc}} = \frac{1}{2}a^{\frac{5}{2}}$

Thus now the inequality becomes: $(a^{\frac{5}{2}}+b^{\frac{5}{2}}+c^{\frac{5}{2}})^2 \leq (a^2+b^2+c^2)(a^3+b^3+c^3)$ which is true by Cauchy.[/quote]

Excellent!!  :clap:

My solution involved twice the Cauchy-Schwarz ineq.... :)


In one year, pairs of rabbits grow from babies to adults.  In the next year, they give birth to some amount of pairs of baby rabbits.  The next year they die.  Normally, pairs of rabbits grow according to the fibbanacci sequence, but if an imaginary drug is given to a pair of rabbits, every possible pair of two adults gives birth to four baby rabbits (two pairs).

f(n) = normal fibbanacci growth
          1 - 1
          2 - 1
          3 - 2
          4 - 3
          5 - 5
          6 - 8
          7 - 13
 and so on.

g(n) = drugged rabbit growth

Find:
a. f(n) in terms of n
b. g(n) in terms of n
c. how many more drugged rabbits than normal rabbits there are in any year if they started with one pair of baby rabbits each

$F_n=F_{n-1}+F{n-2}$
$F=1,1,2,3,5,8,13,21,34,55,89,144....$
$G_n=G_{n-1}+G{n-2}$
$G=1,1,3,5,11,21,43,75,161...$

With some research from mathworld ([url]http://mathworld.wolfram.com/FibonacciNumber.html[/url]), we know that the equation for a recursive series $S$ is $\frac{\alpha^n-\beta^n}{\alpha-\beta}$, where $\alpha$ and $\beta$ are the roots of the equation $x^2=Ax+B$, when the function is defined recursively as $S_n=AS_{n-1}+BS_{n-2}$.

For Fibonacci, A and B are 1 and the roots yielded are $\frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}$, commonly known as phi ($\phi$) and psi ($\psi$).  So

$F_n=\frac{\phi^n-\psi^n}{\phi-\psi}=\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n\sqrt{5}}$

For the drugged rabbits, A is 1 and B is 2 and roots yielded are 2 and -1.
$G_n=\frac{2^n-(-1)^n}{2+1}=\frac{2^n+(-1)^{n+1}}{3}$

As for the difference, just subtract the two?

Also, when I plug in for $\phi$ and $\psi$ in the formula for $F_n$, the denominator keeps cancelling to $2\sqrt{5}$, and the $2^n$ from the denominator of the numerator is carried down, which I would think would make the final denominator $2^{n+1}\sqrt{5}$, but mathworld says $2^n\sqrt{5}$ so I put that.  And if anyone can explain [u]why[/u] any of this is true I'd be much more satisfied  :lol:

[quote="jrshoch"]And if anyone can explain [u]why[/u] any of this is true I'd be much more satisfied  :lol:[/quote]

A full constructive answer/proof is complicated and requires the use of generating functions.

Edit:  But here is a nice non-constructive answer.

Let $s_n = a \alpha^n + b \beta^n$.  Let $A = \alpha + \beta, B = -\alpha \beta$.  Then $\alpha, \beta$ are the roots of the quadratic $x^2 = Ax + B$.  Moreover,

$s_{n+1} = A s_n + B s_{n-1}$

Or, in other words,

$a \alpha^{n+1} + b \beta^{n+1} = (\alpha + \beta) \left( a \alpha^n + b \beta^n \right) - \alpha \beta \left( a \alpha^{n-1} + b \beta^{n-1} \right)$

:)

Here's a formula I found during a computer science exercise... I solved it using balls and dividers, getting this formula, and then later the teacher solved it for the class, but using some compsci reasoning, and showed that it was really the fibonacci sequence.  Pretty interesting.

${F_n = \sum_{k=0}^{c \left( \frac{n-2}{2} \right) }} \binom{n-k-1}{k}$ where $c(x)$ is the cieling function (don't know the latex)

jrshoch: your recursion for $g(n)$ looks incorrect.

deej21: http://mathworld.wolfram.com/FibonacciPolynomial.html

[quote="miyomiyo"]every [b]possible[/b] pair of two adults gives birth to four baby rabbits (two pairs).[/quote]

Are you sure this is what you mean?

the proof for the explict formula follows directly from induction...even the general case...but as to why that works, there are a few methods...one way that i found is to assume the truth of newton's sums, and the recursion from newton's sums works the same way as the general linear recursion after a few terms...

look at my post (#7) here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=82478[/url]

$g(n)$ is rather different from $f(n)$ then.  jrshoch, sorry, but your solution is incorrect.

$g(1) = 1$
$g(2) = 1$

At the beginning of year $n+1$ there are $2g(n-1)$ rabbits available to breed.  The recursion is then

$g(n+1) = 2 { 2g(n-1) \choose 2} + g(n)$

 :|

The function must be recursive in some way....

There are 2 baby rabits to begin with. Represent this as (0,2). We always add the second number to the first number for the next first number and then the next second number becomes $4\binom{n}{2}$ where $n$ is the first number. Let me demonstrate:

(0,2)
(2,0)
(2,4)
(6,4)
(10,60)
(70,180)
...

These totals become: (0),2,2,6,10,etc

Let $G_n = g(n)$
Then: $G_n = G_{n-1} + 4\binom{G_{n-2}}{2}$ for $n>1$ if (0,0) is included as a point.

The first values for $G_n$ for $n=1,2,3,\ldots$ are:
2, 2, 6, 10, 70, 250, 9910, 134410, 196530790, 36328358170, etc. I dont' think my calculator will hold much more. Divide by 2 to get the number of pairs.

One more thing. I assumed they were immortal due to the fact that normals don't breed to the fibonacci sequence unless they are immortal ;)
Normals would go like this: (old, adult, new) (pairs)
(0,0,1)
(0,1,0)
(1,0,1)
(0,1,0)
(1,0,1)
etc (assuming they only give birth once.

[quote="miyomiyo"]What about a non-recursive function?[/quote]

Do you have a solution?  I think a non-recursive function would be very complicated.  The function grows with $O( exp(exp(x)) )$.   :huh:


There are $ n$ fleas on an infinite sheet of triangulated paper. Initially the
fleas are in different small triangles, all of which are inside some equilateral triangle consisting of $ n^{2}$ small triangles.
Once a second each flea jumps from its original triangle to one of the three small triangles having a common vertex but no common side with it. For which natural numbers $ n$ does there exist an initial configuration such that after a finite number of jumps all the $ n$ fleas can meet in a single small triangle?


A sequence $\{a_n\}$ satisfies $a_0=0,\ a_1=1,\ a_{n+1}=a_n+a_{n-1}\ (n=1,2,3,\cdots).$
Prove the following equalities.

(1) $\left(\begin{array}{cc} a_{n+1} & a_n \\ a_n & a_{n-1}\end{array}\right)^{-1}=(-1)^n \left(\begin{array}{cc} a_{n-1} & -a_n \\ -a_n & a_{n+1}\end{array}\right)$

(2) $\left(\begin{array}{cc} a_{n+1} & a_n \\ a_n & a_{n-1}\end{array}\right)=\left(\begin{array}{cc} 1 &1 \\ 1 & 0 \end{array}\right)^n$

1) Let $A$ be the matrix on the LHS.  It's enough to prove $\det A=(-1)^n$ - this follows immediately from the formula given in 2).

2) Use Induction, diagonalization, anything you want.

[quote]2) Use Induction, diagonalization, anything you want.[/quote]
That's more that the problem asked for, of course. If you do use diagonalization, you will get the explicit formula for the Fibonacci numbers in terms of powers of $\frac{1+\sqrt{5}}2$ and $\frac{1-\sqrt{5}}2.$


function
floor function
calculus
trigonometry
algebra

How do you do floor function problems? I always find them so confusing...

For how many of the first 200 positive integers $n$ are there no solutions to $\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor = n$?

i think that the main thing to do is focus on the decimal part of x, if x is just below .5 or at .5 (since all x is multiplied by 2) same goes for the decimal place of x is close to 1

[hide]$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor = n$

$x=\lfloor x \rfloor + a$

Two cases-$a < .5$ OR $a \geq .5$

If $a <.5$, then $\lfloor 2x \rfloor = 2(x-a)=2\lfloor x \rfloor$
Then $12 \lfloor x \rfloor = n \Rightarrow \lfloor 200/12 \rfloor =16$ numbers are possible.

If $a \geq .5$, then $\lfloor 2x \rfloor = 2 \lfloor x \rfloor +1$

$\lfloor 4x \rfloor = 4 \lfloor x \rfloor +2$

$\lfloor 6x \rfloor = 6 \lfloor x \rfloor +3$

$12 \lfloor x \rfloor +6 = n \Rightarrow 16$ numbers are possible.

But 200-32 is not the right answer. :(

[/hide]

for $(x| 0 $to$ 0.5 )$ 4 and 5 are skipped $f(.499) = 3$ , $f(.500) = 6$
for $(x| 0.5 $to$ 1 )$ 10 and 11 are skipped $f(0.99) = 9$, $f(1.00) = 12$
pattern: every 6, 2 skipped so like n = 1,2,3,4,5,6 (5,6) skipped
$\lfloor{200/6}\rfloor = 33$ so $33*2 = 66$
the remainder is the first 2 terms of the sequence so they have solutions and do not count on the no solution total

is this the right answer? i guess i was not too clear, i have not gotten to that part in the book yet, but the idea is that there is a cycle

[hide="The correct answer is"]134[/hide]

200-66 = 134 but then the problem should be the number of n that [b]have[/b] solutions, or at least is seems that way

Nobody has answered the original question.

Is there some sort of algorith or approach for solving questions involving greatest integer fn. etc???

I think Altheman is right. :) I got this problem from the Olympiad Problem Solving Pre-test, but I think there's a mistake. It has been brought to the instructors' attention in the Online Classes forum, so I hope they fix it.

Can we start like a Floor function marathon type thing? :) And can everyone post easy-to-understand solutions? Because I really need help with these. Thanks!

[size=150]Problem 1[/size]

Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$.

I know this... we prove the result easily for fraction and then just use this result to prove for all $\mathbb{R} $

since 2 is positive, you can factor that out of the abs expression with out changing the expression so you get
$2|x| + 2|y| >= |x| + |y| + |x + y|$
$|x| + |y| >= |x + y|$
then it is obvious that is true but I dont know exactly how to prove it rigorously

I suppose that you could first do that they are both positive (or 0) and show they are equal

Then if one is negative and the other is positive, one is greater than the other since on the right, the negative and the positive numbers cancel some of their magnitude out so it would be less than their respective magnitudes.

Then if they are both negative or zero, you could factor out the -1 which would become 1 since |-1| = 1 giving you the first case

I think that is all of the cases since a number can either be pos, neg, or 0

woops, i thought those were absolute value signs...woops

Let $x=a + \frac{b}{n} + c$ with $a$ an integer, $b < n$ a positive integer, and $0 < c < \frac{1}{n}$ a real number.  $\lfloor nx \rfloor = \lfloor na + b + nc \rfloor = na+b$.  $\left\lfloor\frac{na+b}{n}\right\rfloor = \lfloor a + b/n \rfloor = a$.  However, $\lfloor x \rfloor$ is also $a$!  QED.

I was gone for three weeks and just found this. I really like floor/ceiling problems, so I'm really happy you did this. I have a nice problem for this too, so if nr1337 doesn't post in awhile, I'll write mine.

Ok, it's nearly been a week, so I hope nobody minds if I post the next one.

[b][size=150]Problem 3[/size][/b]

Prove that, for all positive integers x, $\lfloor \sqrt{4x+3}\rfloor =\lfloor \sqrt{x}+\sqrt{x+1}\rfloor$.

[quote="Rep123max"]We also know that $\displaystyle 4x^2+4>4x^2$, so $\displaystyle 2\sqrt{x^2+1}>2x$. We can add $\displaystyle 2x+1$ to both sides and obtain $\displaystyle 2x+1+2\sqrt{x^2+1}>4x+1$. This means that $\displaystyle \sqrt{x}+\sqrt{x+1}>\sqrt{4x+1}$.[/quote]

Err... $(\sqrt{x}+\sqrt{x+1})^2=2x+1+2\sqrt{x^2+x}$ not $2x+1+2\sqrt{x^2+1}$, but it still works. I didn't expect it to be that easy! I did it completely differently. Anyway...

Let 4/x=y. Then we have $\lfloor y\rfloor +\lfloor 7y/4\rfloor=5$. If y<2, the expression is 1+3 (or less), which is less than 5. But if $y\ge 16/7$, we have a minimum of 2+4=6. But since the second floor is exactly 4, y can be infinitesimally smaller than 16/7. So $\frac{16}7>\frac 4x\ge 2$, $\frac7{16}<\frac x4\le\frac 12$, and finally $\frac 74<x\le 2$. The answer is thus 13.

[size=150]Problem 5[/size] :D

How many of the first 1000 positive integers can be expresssed in the form

\[ \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor ? \]

(ACoPS)

[quote="Rep123max"]Since $\displaystyle x$ is a positive integer, $\displaystyle x>0$ so $\displaystyle 8x>0$. This means that $\displaystyle 4x^2+8x+4>4x^2+4x$ so $\displaystyle 2x+2>2\sqrt{x^2+x}$. Addiding $\displaystyle 2x+1$ to both sides, $\displaystyle 4x+3>2x+1+2\sqrt{x^2+x}$. Thus, $\displaystyle\sqrt{4x+3}>\sqrt{x}+\sqrt{x+1}$.

We also know that $\displaystyle 4x^2+4x>4x^2$, so $\displaystyle 2\sqrt{x^2+x}>2x$. We can add $\displaystyle 2x+1$ to both sides and obtain $\displaystyle 2x+1+2\sqrt{x^2+x}>4x+1$. This means that $\displaystyle \sqrt{x}+\sqrt{x+1}>\sqrt{4x+1}$.

Thus, we have $\displaystyle\sqrt{4x+3}>\sqrt{x}+\sqrt{x+1}>\sqrt{4x+1}$. But $\displaystyle 4x+3$ can never be a perfect square (perfect squares are congruent to $\displaystyle 0$ or $\displaystyle 1\pmod 4$.) Thus, $\displaystyle\left\lfloor\sqrt{4x+3}\right\rfloor=\left\lfloor\sqrt{x}+\sqrt{x+1}\right\rfloor=\left\lfloor\sqrt{4x+1}\right\rfloor$.[/quote]

Woah an application of squeeze theorem.  That has to be the coolest proof in the universe.

[quote="chess64"][size=150]Problem 5[/size] :D

How many of the first 1000 positive integers can be expresssed in the form
\[\lfloor 2x \rfloor+\lfloor 4x \rfloor+\lfloor 6x \rfloor+\lfloor 8x \rfloor ? \]
(ACoPS)[/quote]
[hide]Correct me if I'm wrong, chess64. I'm bad at these. :)
I'm betting there is also a more elegant solution.
We must consider these intervals for $\lfloor x-x$. On the right side, I'll write what the fractional part will transform into under the floor function above:
(I'm letting all the denominators be 24 to make it easy)
$0 \le \frac{3}{24}\rightarrow 0+0+0+0= 0$
$\frac{3}{24}\le x <\frac{4}{24}\rightarrow 0+0+0+1= 1$
$\frac{4}{24}\le x <\frac{6}{24}\rightarrow 0+0+1+1= 2$
$\frac{6}{24}\le x <\frac{8}{24}\rightarrow 0+1+1+2= 4$
$\frac{8}{24}\le x <\frac{9}{24}\rightarrow 0+1+2+2= 5$
$\frac{9}{24}\le x <\frac{12}{24}\rightarrow 0+1+2+3= 6$
$\frac{12}{24}\le x <\frac{15}{24}\rightarrow 1+2+3+4= 10$
$\frac{15}{24}\le x <\frac{16}{24}\rightarrow 1+2+3+5= 11$
$\frac{16}{24}\le x <\frac{18}{24}\rightarrow 1+2+4+5= 12$
$\frac{18}{24}\le x <\frac{21}{24}\rightarrow 1+3+4+6= 14$
$\frac{21}{24}\le x <\frac{24}{24}\rightarrow 1+3+5+7= 16$

So all integers of the form $k \equiv 0,1,2,4,5,6,10,11,12,14,16 \mod 20$ can be written in this form. So a total of $11*50 = \fbox{550}$ intgers?[/hide]

Wow, my hand hurts from Latexing.

[quote="Altheman"]since 2 is positive, you can factor that out of the abs expression with out changing the expression so you get
$|2x| + |2y| = 2|x| + 2|y|$[/quote]

$|2(1.9)| = |3.8| = 3$
$2|1.9| = 2|1| = 2$

Those are absolute value things. He accidentally thought they were abs. value instead of floor value. ;)

[quote="FMako"]
Woah an application of squeeze theorem.  That has to be the coolest proof in the universe.[/quote]

What's squeeze theorem? :huh: 4give me, I'm new to these kind of things

I think it's a calculus theorem sometimes used for finding derivatives... :?

Anyone want to check my solution????
I'd really like the know the answer. :lol:

Is there any other way of checking it (rather than counting one by one?)

Yeah, squeeze theorem is a calculus theorem used for finding limits of things like $\sin \frac 1 x$.  I don't see how it's being used there though...

I was being liberal in the use of the term "squeeze theorem" I was just referring to the way you were able to deduce the value of something by bounding it on both sides.  In a more precise definition you would be taking the limit as the two bounds "squeeze" down on a function such that you can then calculate the limit of the squeezed function..

Prove :
$\lfloor n^\frac{1}{2} \rfloor + \lfloor n^\frac{1}{3} \rfloor + ... + \lfloor n^\frac{1}{n} \rfloor = \lfloor log_2^n \rfloor + \lfloor log_3^n \rfloor + ... + \lfloor log_n^n \rfloor$

[quote="chess64"]Can [i]anyone[/i] verify 1234567890's solution??? :([/quote]
Please? I'd be happy if I knew I was right. :)

1234567890 is wrong.

You guys are making it way too complicated.  Put $f(x)= \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$.  Then $f(x+1/2)=f(x)+10$ for every $x$!.  So the values are uniquely determine by what happens in the interval

$(0,1/2]$

But $f$ jumps by $1$ at $x=1/8$, by $1$ at $x=1/6$, by $2$ at $x=1/4$, by $1$ at $x=1/3$, by $1$ at $x=3/8$, by $4$ at $x=1/2$.  In total, it hits $6$ of the first $10$ positive integers.  Extending using $f(x+1/2)=f(x)+10$, we see that it hits $600$ of the first $1000$ positive integers.

Blah^3
I think that's what I did.... though the last part was unnecessary. I just missed a case, and got 550 instead of 600.

Perhaps.  I saw that you had this big list of equivalence classes $\bmod 20$ and concluded that you were thinking too much  :).  Remember, less casework inevitably leads to less mistakes!

1234567890, you can post the next problem. :)

[quote="1234567890"]Okay, I can't think of a good one, so here's something slightly easier:
$100!_{100!}$ is divisible by what power of 10?[/quote]
You cannot possibly have that:

For example, $2_2$ does not exist in binary.

Also, what does this problem have to do with [b]floor functions[/b] in a [i]non-trivial[/i] sense?

10k, I think he means the digits of 100! as the digits of the base 100! number.  For instance, $12_{12}$ is a number.

Hmm if that's the case:
[hide="Idea"]First find the number of zeroes $N$ in $100!$ and that would yield the minimal exponent $e=N$ of $100!$, i.e. $100!_{100!}=\cdots+d\cdot (100!)^e+0+0+0+\cdots$. Then, the number of zeroes in $100!_{100!}$ is $eN=N^2$[/hide]


[quote="pohoatza"]Fie $ABC$ triunghi si $D,E,F$ punctele de tangenta cu laturile ale cercului inscris. Sa se arate ca daca centrele de greutate ale triunghiurilor $ABC$ si $DEF$ sunt izogonale, atunci triunghiul $ABC$ este echilateral.
[b]Remarca[/b]: Aceeasi problema ramane adevarata si pentru cazul in care centrele de greutate sunt izotomice.[/quote]
[color=darkblue][b][u]Demonstratie.[/u][/b] Vom utiliza calculul vectorial.  $\{\begin{array}{c}AE=AF=p-a\\\ BF=BD=p-b\\\ CD=CE=p-c\end{array}\|$ $\Longrightarrow$

$\{\begin{array}{c}(p-c)\cdot\overline{DB}+(p-b)\cdot\overline{DC}=0\\\ (p-a)\cdot\overline{EC}+(p-c)\cdot\overline{EA}=0\\\ (p-b)\cdot\overline{FA}+(p-a)\cdot\overline{FB}=0\end{array}\|$ $\Longrightarrow$ $\{\begin{array}{c}D=\frac{p-c}{a}\cdot B+\frac{p-b}{a}\cdot C\\\ E=\frac{p-a}{b}\cdot C+\frac{p-c}{b}\cdot A\\\ F=\frac{p-b}{c}\cdot A+\frac{p-a}{c}\cdot B\end{array}$.

Asadar, centrele de greutate ale triunghiurilor $ABC$ si $DEF$ sunt izogonale $\Longleftrightarrow$

$\frac{(p-b)(p-c)}{a^{2}}=\frac{(p-c)(p-a)}{b^{2}}=\frac{(p-a)(p-b)}{c^{2}}=\frac{r(4R+r)}{a^{2}+b^{2}+c^{2}}$ $\Longleftrightarrow$

$a=b=c$, ceea ce inseamna ca triunghiul $ABC$ este echilateral.[/color]

[quote="pohoatza"]Fie $ABC$ triunghi si $D,E,F$ punctele de tangenta cu laturile ale cercului inscris. Sa se arate ca daca centrele de greutate ale triunghiurilor $ABC$ si $DEF$ sunt izogonale, atunci triunghiul $ABC$ este echilateral.
[b]Remarca[/b]: Aceeasi problema ramane adevarata si pentru cazul in care centrele de greutate sunt izotomice.[/quote]
[color=darkblue][b][u]Demonstratie.[/u][/b] Vom utiliza calculul vectorial.  $\{\begin{array}{c}AE=AF=p-a\\\ BF=BD=p-b\\\ CD=CE=p-c\end{array}\|$ $\Longrightarrow$

$\{\begin{array}{c}(p-c)\cdot\overline{DB}+(p-b)\cdot\overline{DC}=0\\\ (p-a)\cdot\overline{EC}+(p-c)\cdot\overline{EA}=0\\\ (p-b)\cdot\overline{FA}+(p-a)\cdot\overline{FB}=0\end{array}\|$ $\Longrightarrow$ $\{\begin{array}{c}D=\frac{p-c}{a}\cdot B+\frac{p-b}{a}\cdot C\\\ E=\frac{p-a}{b}\cdot C+\frac{p-c}{b}\cdot A\\\ F=\frac{p-b}{c}\cdot A+\frac{p-a}{c}\cdot B\end{array}$.

Asadar, centrele de greutate ale triunghiurilor $ABC$ si $DEF$ sunt izogonale $\Longleftrightarrow$

$\frac{(p-b)(p-c)}{a^{2}}=\frac{(p-c)(p-a)}{b^{2}}=\frac{(p-a)(p-b)}{c^{2}}=\frac{r(4R+r)}{a^{2}+b^{2}+c^{2}}$ $\Longleftrightarrow$

$a^{2}(p-a)=b^{2}(p-b)=c^{2}(p-c)$ $\Longleftrightarrow$ $\{\begin{array}{ccc}a(b+c)=b^{2}+c^{2}& \vee & b=c\\\ b(c+a)=c^{2}+a^{2}& \vee & c=a\\\ c(a+b)=a^{2}+b^{2}& \vee & a=b\end{array}\|\ (*)$ $\Longleftrightarrow$

$a=b=c$, ceea ce inseamna ca triunghiul $ABC$ este echilateral.

[b]P.S.[/b] Sistemul $(*)$ exprima faptul ca o solutie sintetica are un grad sporit de dificultate.

[b]Remarca.[/b] Cele doua centroide sunt izotomice $\Longleftrightarrow$

 $\frac{p-a}{a^{2}}=\frac{p-b}{b^{2}}=\frac{p-c}{c^{2}}$ $\Longleftrightarrow$ $a=b=c$.[/color]


tehe i put "thrill and satisfaction of problem solving"......

really i do it to .....

whoa epiphany
i have no idea why i do math.  :huh:

Is there an "All of the above"? Because I would vote for that. :D

I guess all of the above, or it doesn't apply to me yet would apply to my case.

I put 'thrill and satisfaction of problem solving'.

 :D

Hmm... I wonder who put "other questionable sources of motivation"  :rotfl:

i put trophies and prizes...lets see...i have 4 trophies,1 pin, and like 20 certificates.

Everything except proving me and hating math.

Guys why do you hate math?!

I still don't know  :maybe:
maybe I can answer this question later

I do not know why i like math. just have not thought about it yet.  may be I like it because I'm good at it . (at least not bad )

Because of the unbelievable beauty of the most beautiful thing in this world.And the satisfaction of.......And because of the freedom offered here.
I put in the last option.
oh yeah .Should've put in that.Didn't notice.Is there any way I can change it now?

It seems like you should have chosen the 7th option...

by the way if you go into pure mathematics, it doesn't pay very much  :(

Why am I the only one who marked to find a way to prove 1=2? How come 1=2 didn't mark that?

[quote="dragon96"]How come 1=2 didn't mark that?[/quote]

He's already asserted it in his user-name. I suppose he needed a new reason?

I find the "bored" option very funny.

Most people: $ \text{do math} \implies \text{bored}$
Math people: $ \text{bored} \implies \text{do math}$

If people are bored dont they usually do sports?

Haha, that's a very interesting observation Zuton, and pretty much sums us math ppl up  :lol:

let's see... I got 12 trophies and a few medals (most from this year)


MATHCOUNTS Support AMC USA(J)MO USAMO AIME geometry

how many states have actually done it already?

Yesterday was hectic...

Anyways... in Oklahoma.

I started in second before countround with a 41... that's kinda bad.

1st was 42...

Then i got first after countdown!!!

and all of them were 3-2...

that's all the stuff i know as of now...

to make this more organized...

Conghua Ye (Martin Ye)/OK/1st/41.

CONGRATULATIONS Martin (and probably a free AOPS class for you) . too bad i can't see you at nationals this year. i'm too old.

[quote="rrusczyk"]We take the privacy of our members very seriously - please do not discuss the real names of other members unless they have done so themselves.[/quote]

i have disclosed my name. what are the dangers?

Chinaboy, do you think the problems were harder or easier than last year? And what were your scores on the individaul rounds? Remember, don't discuss questions.

40. 26 sprint, 14 target. Third in Connectucut, second was also 40 and I don't know what the winner got. My school was the first place team miraculously because we're all pretty stupid.

yif, do you know what the winners of connecticut got last year?

Sorry. I don't really know anyone else in the top four in connecticut.

[quote="yif man12"]
Is it just me, or was the team round much harder than usual?[/quote]

Our team got 20 at chapter... and only a 14 at state... when in practice we either get 18 or 20...

Sorry, but I had to do some snipping and cleaning to get rid of any talks about the difficulty of the test.

I will say that I did not get into the top 10 in an easy state, and I'm pretty much transitioning towards science now.

20 on sprint...14 on target...34 total...15th place...no nats for me...IN

[u][b]State results:[/b][/u]

[b]State[/b]: Indiana
[b]Score[/b]: 41
[b]Top 4[/b]: 41 (Me! Yay!), 41, 41, 40

Illinois this year:

45, 44, 44, 40

We took a beating last year (4th), but this time our chances are good (hopefully)...

Too bad the Californians steamrolled everyone last year :-(

I probably did the worst here: a 14 on sprint and a 4 on target.  It's a shame - when the Mathcounts people posted the tests online, I got a total score of 40 ( I took it again online). :(

don't worry about it. mathcounts doesn't really test your skills in math that much. for example, some of my friends got really bad scores but are really good at math. then, others who aren't that great do pretty well. i'm sure you're great at math.

yifman's right.  I qualified for the USAMO this year w/ 11 on AIME and I'm still pathetic at National MathCounts contests.

Yeah - I love Mathcounts, but none of the problems really require you to go too in-depth (which is the right thing to do for that level).  However, this leaves it at the point where, for the regional and state competitions at least, the sprint round tests your ability to not make stupid mistakes and to do quick arithmetic by hand/in your head, which, in my opinion, is not really what a math competition should be about.

also, it tests your knowledge of shortcuts, if you know all of them, you do well, if you don't, you're in trouble

California:

1st: not sure (think its 43)
2nd:42
3rd:42
4th: (no idea we have a northern competition and southern and the national people get chosen out of the best 4 from the two contests)

i got 11th: 39 :(

Same is true of USAMO qualification or of cutoffs for National Merit Semifinalists. Its much easier in Alaska or Wyoming, but your education and opportunities, especially in math, and much more limited if you live in such states.

[quote="bob123"]California:

1st: not sure (think its 43)
2nd:42
3rd:42
4th: (no idea we have a northern competition and southern and the national people get chosen out of the best 4 from the two contests)
[/quote]

Wow is this all of California?  Usually California does really really well...

State scores don't well reflect nationals scores though.

[quote="JP"]MathCounts is not really fair for everyone.  If you live in place like California, it will be extremely hard for you to get into national round.  But if you live in place like Alaska or Wyomine, it is much easier.[/quote]

This is true as far as getting to nationals goes.  However, once you're there, the top top people will still be the winners - I doubt there has ever been a time when four people from the same state made countdown.

And if you're in a difficult state (like California or New Jersey) it motivates you to work harder ;)

california's fourth may have gotten better than the first, but i dont know because i went to the northern state contest and 3 people were chosen from northern california and one was chosen from southern california (the person i put as fourth)

i got 11th in the northern california :cry:

#1 at state, with a score of 40  (I got first in written too), 2nd was only 38.  I guess our state didn't do as well as others, considering we got like 37th at nationals.  Anyway it was fun while it lasted, and congrats peeps.  Anyway we got like shirt, air fare, food, hat, calculator, pins, and um, more stuff.  Anyway that was all free so I came out ahead.


Given that $ x,y>0$ and $ x,y$ are integers, find all solutions to $ (3x\minus{}10)(3y\minus{}100)\equal{}40$.

Here's the outline to the solution:

Clearly, $ 3x \minus{} 10$ and $ 3y \minus{} 100$ must be factors of $ 40$. Set them equal to those factors to find all possible values of $ x$ and $ y$. For example, we can have $ 3x \minus{} 10 \equal{} 2$ and $ 3y \minus{} 100 \equal{} 20$ which yields a solution of $ (x,y) \equal{} (4, 40)$.

EDIT: Fixed.  :blush:

Well, we have the bound that $ x>0$, so $ 3x\minus{}10>\minus{}10$, so $ x \neq \minus{}20,\minus{}40$.

[quote="sunehra"]Here's the outline to the solution:

Clearly, $ 3x \minus{} 10$ and $ 3y \minus{} 100$ must be factors of $ 40$. Set them equal to those factors to find all possible values of $ x$ and $ y$. For example, we can have $ 3x \minus{} 10 \equal{} 20$ and $ 3y \minus{} 100 \equal{} 20$ which yields a solution of $ (x,y) \equal{} (10, 40)$.[/quote]

I think there might be something wrong here, $ 20\cdot 20 \equal{} 400$ not $ 40$.

One way is $ 3x\minus{}10\equal{}2$ and $ 3y\minus{}100\equal{}20$ which has a solution in, $ x\equal{}4$ and $ y\equal{}40$.

So $ 3x \minus{} 10$ and $ 3y \minus{} 100$ not only must be factors of $ 40$ but they also have to have a product of $ 40$.

[quote="james4l"]Note that both factors must be 1 mod 3.[/quote]

Don't you mean 2 mod 3?

$ 3x \minus{} 10\equiv (\minus{}1)(10) \equiv 2 \equiv (\minus{}1)(100)\equiv 3x \minus{} 100 \pmod{3}$.

I will try and brute force the solutions for the pairs using casework starting with the negative factors.

[hide]

Case $ 1$: The numbers are both negative.

[hide]Subcase $ a$: One of the numbers, $ n$ is less than $ \minus{}10$
If $ n\le\minus{}10$, then we have that $ x$ or $ y$ is equal to a negative number, which cannot be possible unless the other factor would be greater than $ \minus{}10$, making the product of the factors negative.  Therefore, if there exists a pair of negatives, $ (x,y)$, they must either be greater than or less than (or equal to) $ \minus{}10$

Subcase $ b$: Both of the numbers are less than $ \minus{}10$

$ x,y\le\minus{}10$

Clearly this can never be true, becuase if both of them were less than $ \minus{}10$, the smallest possible product fo these numbers would be $ 100$.

Subcase $ c$: Both of the numbers are greater than $ \minus{}10$

$ x,y\ge\minus{}10$

We see that this case will be true for all the pairs of factors we come up with from here on out.  We also notice that there exists four pairs of negative factors in which both fo the factors are greater than or equal to $ \minus{}10$  These factors are $ (\minus{}5,\minus{}8)$,$ (\minus{}8,\minus{}5)$,$ (\minus{}4,\minus{}10)$,$ (\minus{}10,\minus{}4)$.

When we plug them into the equation we get the following:
$ (\minus{}5,\minus{}8)$
$ 3x\minus{}10\equal{}\minus{}5$
$ x\equal{}\frac{5}{3}$
Which means that this answer cannot be true, and when we use the reverse of this pair it does not work either.

$ (\minus{}4,\minus{}10)$
$ 3x\minus{}10\equal{}\minus{}4$
$ x\equal{}2$
$ 3y\minus{}100\equal{}\minus{}10$
$ y\equal{}30$
So we see that $ (2,30)$ is our first pair.  It is also true that when the coordinates are reversed we get the answer $ (0,32)$[/hide]
Case $ 2$: Both factors are positive

[hide]We have a lot of factors to try for the case above.
$ (1,40),(2,20),(4,10),(5,8),(8,5),(10,4),(20,2),(40,1)$

When we try the first case $ (1,40)$, we notice that we do nto get integer outputs.  We can also eliminate the reverse of this pair for the same reason.

Plugging in $ (2,20)$ yields a solution of $ (4,40)$  When we plug in its reverse, we get an answer of $ (10,34)$

Trying $ (4,10)$ gives non-integer solutions, and its reverse yields no integer solutions either.

Our last case is $ (5,8)$, which gives us $ (5,36)$ as a result.  Its reverse gives $ (6,35)$ as the final solution.[/hide]

Summing up all the possible solutions, we can now finally list all of them out:

$ (0,32)$
$ (2,30)$
$ (4,40)$
$ (5,36)$
$ (6,35)$
$ (10,34)$
[/hide]


calculus
integration
logarithms
limit
complex analysis

Evaluate $\int_0^\infty \frac{x\ln x}{x^3+1}\ dx$

Let $I$ denote the integral's value.
Let $\zeta$ be the third root of unity in the upper halfplane.
Lets also take the main branch of the logarithm.
Take $0<\epsilon <1$ and consider the curve $\zeta \epsilon \to \epsilon \to \frac{1}{\epsilon} \to \frac{\zeta}{\epsilon} \to \zeta \epsilon$, where in the first and third case we take the (shorter) arc around $0$ and straight lines in the other two cases.
Now when $\epsilon$ goes to $0$, the integrals over the arcs vanish and by taking the contour integral over this, the residue theorem gives:
\[ 2 \pi i Res(\frac{x\ln x}{x^3+1} , -\zeta^2) = \int_{0}^\infty \frac{x\ln (x)}{x^3+1} \,dx - \int_{0}^\infty \frac{\zeta x\ln (\zeta x)}{x^3+1}\,dx = (1-\zeta^2)I - \zeta^2 ln(\zeta) \int_{0}^\infty \frac{x}{x^3+1} \,dx \]

So lets start calculating the parts different from $I$:
$ln(\zeta)=\frac{2}{3}\pi i$
$Res(\frac{x\ln x}{x^3+1} , -\zeta^2)=\lim_{x \to -\zeta^2} \frac{(x+\zeta^2)xln(x)}{x^3+1}=\frac{-\zeta^2 \frac{1}{3}\pi i}{(-\zeta^2+1)(-\zeta^2+\zeta)}=\frac{-\zeta \pi i}{9}$
To evaluate $J: = \int_{0}^\infty \frac{x}{x^3+1} \,dx$, use the same contour to get:
\[ 2 \pi i Res(\frac{x}{x^3+1},-\zeta^2)=(1-\zeta^2)J \]
Now, by the same as above, $Res(\frac{x}{x^3+1},-\zeta^2)=\frac{-\zeta}{3}$ and thus $J=\frac{-2 \zeta \pi i }{3(1-\zeta^2)}=\frac{-2 \zeta \pi i (1-\zeta)}{9}=\frac{2\pi i (\zeta^2-\zeta)}{9}$.

Now to finish this, we get
\[ 2 \pi i \frac{-\zeta \pi i}{9} = (1-\zeta^2)I - \zeta^2 \frac{2}{3}\pi i \frac{2\pi i (\zeta^2-\zeta)}{9} \]
which gives then
$I(1-\zeta^2)= 2 \pi i \frac{-\zeta \pi i}{9} + \zeta^2 \frac{2}{3}\pi i \frac{2\pi i (\zeta^2-\zeta)}{9}= 2 \pi^2 (\frac{\zeta}{9} - \frac{2(\zeta-1)}{27}) = \frac{2\pi}{27} (3\zeta - 2 \zeta +2) = \frac{2\pi}{27} (\zeta+2)$
and finally:
\[ I= \frac{2\pi}{27} \frac{2+\zeta}{1-\zeta^2}=\frac{2\pi}{27} \frac{1-\zeta^2}{1-\zeta^2}=\frac{2\pi}{27} \]


Hello!

Let $ C_{1}\left(O_{1}\right)$, $ C_{2}\left(O_{2}\right)$, $ C_{3}\left(O_{3}\right)$ be three circles passing through the point $ O$.
$ C_{1}\cap C_{2}\equal{}\left\{O,P_{3}\right\}$, $ C_{2}\cap C_{3}\equal{}\left\{O,P_{1}\right\}$, $ C_{3}\cap C_{1}\equal{}\left\{O,P_{2}\right\}$.
The points $ A_{2}$, $ A_{3}$ are the midpoints of the arcs $ \widehat{P_{3}P_{1}}$, $ \widehat{P_{1}P_{2}}$ respectively $ \left(A_{2},A_{3}\in ExtP_{1}P_{2}P_{3}\right)$.
The point $ B_{1}$ is the midpoint of the arc $ \widehat{P_{2}P_{3}}$ $ \left(B_{1}\in IntP_{1}P_{2}P_{3}\right)$.
$ \left\{L_{2}\right\}\equal{}A_{2}B_{1}\cap C_{2}$ and $ \left\{L_{3}\right\}\equal{}A_{3}B_{1}\cap C_{3}$.
a) Prove that the points $ P_{1},L_{2},L_{3}$ are collinear (on the line $ d_{1}$).
b) The point $ A_{1}$ is the midpoint of the arc $ \widehat{P_{2}P_{3}}$  $ \left(A_{1}\in ExtP_{1}P_{2}P_{3}\right)$.
The points $ B_{2}$, $ B_{3}$ are the midpoints of the arcs $ \widehat{P_{3}P_{1}}$,  $ \widehat{P_{1}P_{2}}$ respectively $ \left(B_{2},B_{3}\in IntP_{1}P_{2}P_{3}\right)$.
$ \left\{M_{1}\right\}\equal{}A_{1}B_{2}\cap C_{1}$, $ \left\{M_{3}\right\}\equal{}A_{3}B_{2}\cap C_{3}$,
$ \left\{N_{1}\right\}\equal{}A_{1}B_{3}\cap C_{1}$, $ \left\{N_{2}\right\}\equal{}A_{2}B_{3}\cap C_{2}$.
Similarly the points $ P_{2},M_{1},M_{3}$ are collinear (on the line $ d_{2}$) and the points $ P_{3},N_{1},N_{2}$ are collinear
(on the line $ d_{3}$).
Prove that the lines $ d_{1}$, $ d_{2}$ and $ d_{3}$ are concurrent. :)

It's a hard problem! Try to solve it! :)

there are a lot of lines Petry...its difficult!!..but this smells like an inversion problem :?:


Let $ P(x)=c_{n}x^{n}+c_{n-1}x^{n-1}...+c_{1}x+c_{0}$ be a polynomial of degree $ n$ with positive real roots. Prove that $ \frac{c_{n-1}c_{1}}{c_{n}c_{0}}\ge{n^{2}}$


[quote="Vasc"][quote="t_c_lhp"]Prove that for all $a,b,c>0$ then $\sum\frac{a^{2}}{b^{2}-bc+c^{2}}>2$.[/quote]
The stronger inequality $\sum\frac{a^{2}+4bc}{b^{2}-bc+c^{2}}\geq 6$ holds for all $a,b,c\geq 0$.[/quote]
Use the same methods with the above inequality, WLOG, we can assume that $c= \min(a,b,c)$, then
\[LHS \ge \frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}+\frac{4ab}{a^{2}-ab+b^{2}}= \frac{(a-b)^{4}(a^{2}+b^{2}+3ab)}{a^{2}b^{2}(a^{2}-ab+b^{2})}+6 \ge 6 \]

We are done. :)

Nice  :)
Mr Vasc  do you have a solution using Cauchy -Swartz  ? :maybe:


Prove that if $ \mathbf u, \mathbf v$ in inner production space satisfy $ |\langle\mathbf u,\mathbf v\rangle|\equal{} \|\mathbf u\|\|\mathbf v\|,$ then $ \mathbf u$ and $ \mathbf v$ are linearly dependent.

I can prove it now. Suppose that $ \langle\mathbf u,\mathbf v\rangle^2\equal{}\|\mathbf u\|^2\|\mathbf v\|^2$, then

$ 4\langle\mathbf u,\mathbf v\rangle^2\minus{}4\|\mathbf u\|^2\|\mathbf v\|^2\equal{}0.$

Thus the equation $ \|\mathbf u\|^2 x^2\plus{}2\langle\mathbf u,\mathbf v\rangle x\plus{}\|\mathbf v\|^2\equal{}0$, or

$ \langle x\mathbf u\plus{}\mathbf v,x\mathbf u\plus{}\mathbf v\rangle \equal{}0,$

has exactly one root, denoted by $ t$. Then we have $ \langle t\mathbf u\plus{}\mathbf v,t\mathbf u\plus{}\mathbf v\rangle \equal{}0.$

Hence $ t\mathbf u\plus{}\mathbf v\equal{} \mathbf0$. Therefore, $ \mathbf u$ and $ \mathbf v$ are linearly dependent.

p.s.: I had a typing mistake in the post above: "production" -- "product".

Of course, that is simply the proof of the Cauchy-Schwarz inequality in an abstract inner product space.

One specific comment: you have implicitly used that this is a [b]real[/b] inner product space. (For instance, you assumed that $ \langle u,v\rangle\equal{}\langle v,u\rangle$.) While the result remains true in a complex inner product space and the proof is similar, we would need to make some subtle modifications of the details of the proof.

[quote="ifai"]I can prove it now. Suppose that $ \langle\mathbf u,\mathbf v\rangle^2 \equal{} \|\mathbf u\|^2\|\mathbf v\|^2$, then

$ 4\langle\mathbf u,\mathbf v\rangle^2 \minus{} 4\|\mathbf u\|^2\|\mathbf v\|^2 \equal{} 0.$

Thus the equation $ \|\mathbf u\|^2 x^2 \plus{} 2\langle\mathbf u,\mathbf v\rangle x \plus{} \|\mathbf v\|^2 \equal{} 0$, or

$ \langle x\mathbf u \plus{} \mathbf v,x\mathbf u \plus{} \mathbf v\rangle \equal{} 0,$

has exactly one root, denoted by $ t$. Then we have $ \langle t\mathbf u \plus{} \mathbf v,t\mathbf u \plus{} \mathbf v\rangle \equal{} 0.$

Hence $ t\mathbf u \plus{} \mathbf v \equal{} \mathbf0$. Therefore, $ \mathbf u$ and $ \mathbf v$ are linearly dependent.

p.s.: I had a typing mistake in the post above: "production" -- "product".[/quote]
Thank you!  :P


superior algebra
superior algebra solved

Prove that $\mathbb{Q}_{p}$ has no nontrivial automorphisms.


Sami's Candy Store has six bins containing jellybeans. These bins contains 10, 13, 19, 8, 13 and 9 blue jellybeans. What is the mean number of blue jellybeans per bin?

for calculating means in my head, i use a different method, which i consider safer
$ \frac {10(6) \plus{} 0 \plus{} 3 \plus{} 9 \minus{} 2 \plus{} 3 \minus{} 1}{6}$
$ 10 \plus{} \frac {0 \plus{} 3 \plus{} 9 \minus{} 2 \plus{} 3 \minus{} 1}{6} \equal{} 10 \plus{} \frac {12}{6} \equal{} 12$
which is the mean.
this is especially useful with big numbers.


superior algebra
superior algebra unsolved

Whether can one give an example of a finite group $ G$ satisfied  the following conditions:

1.  $ |G| \equal{} p^aq^b$, where $ p,q$ odd primes.
2.  $ exp(G) \equal{} pq$.
3.  $ s_p \equal{} s_{pq}$ and $ s_p\neq s_q$, where $ s_n$ is the number of elements of order $ n$.

What does your expression $ exp(G)$ mean ?

The exponent of a torsion group is the smallest positive integer divisible by the order of every element.


Lately, Pohoata sent me this problem
\[ \sum \frac{a^3\plus{}abc}{b\plus{}c} \ge \sum \frac{a(b^3\plus{}c^3)}{a^2\plus{}bc}\]
for all $ a,b,c \ge 0$.
Then I proved it by using the following identity
\[ \sum \frac{a^3\plus{}abc}{b\plus{}c}\minus{}\sum \frac{a(b^3\plus{}c^3)}{a^2\plus{}bc} \equal{}\frac{abc(a^4\plus{}b^4\plus{}c^4\minus{}a^2b^2\minus{}b^2c^2\minus{}c^2a^2)^2}{(a\plus{}b)(b\plus{}c)(c\plus{}a)(a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)}\]
:)
And here is my question: prove my indentity. :)

[quote="can_hang2007"]Lately, Pohoata sent me this problem
\[ \sum \frac {a^3 \plus{} abc}{b \plus{} c} \ge \sum \frac {a(b^3 \plus{} c^3)}{a^2 \plus{} bc}
\]
for all $ a,b,c \ge 0$.
Then I proved it by using the following identity
\[ \sum \frac {a^3 \plus{} abc}{b \plus{} c} \minus{} \sum \frac {a(b^3 \plus{} c^3)}{a^2 \plus{} bc} \equal{} \frac {abc(a^4 \plus{} b^4 \plus{} c^4 \minus{} a^2b^2 \minus{} b^2c^2 \minus{} c^2a^2)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}
\]
:)
And here is my question: prove my indentity. :)[/quote]
I think, your proof is right, can_hang2007.
$ \sum_{cyc}\left((a^3 \plus{} abc)(a \plus{} b)(a \plus{} c)(a^2 \plus{} bc)(b^2 \plus{} ac)(c^2 \plus{} ab) \minus{} (ab^3 \plus{} ac^3)(b^2 \plus{} ac)(c^2 \plus{} ab)(a \plus{} b)(a \plus{} c)(b \plus{} c)\right) \equal{}$
$ \equal{} \sum_{cyc}(a^5 \plus{} a^4b \plus{} a^4c \plus{} 2a^3bc \plus{} 3a^2b^2c)\sum_{cyc}(a^3b^3 \plus{} a^4bc \plus{} \frac {2}{3}a^2b^2c^2) \minus{}$
$ \minus{} \sum_{cyc}(a^6b^2 \plus{} a^6c^2 \plus{} a^5b^2c \plus{} a^5c^2b \plus{} a^4b^3c \plus{} a^4c^3b \plus{} 2a^3b^3c^2)\sum_{cyc}(a^2b \plus{} a^2c \plus{} \frac {2}{3}abc) \equal{}$
$ \equal{} \sum_{sym}\left(0.5a^9bc \plus{} a^8b^3 \plus{} a^8b^2c \plus{} a^7b^4 \plus{} a^7b^3c \plus{} 2a^7b^2c^2 \plus{} 3a^6b^4c \plus{} 5a^6b^3c^2 \plus{} 2.5a^5b^5c \plus{} 6a^5b^4c^2 \plus{} 4a^5b^3c^3 \plus{} 5a^4b^4c^3\right) \minus{}$
$ \minus{} \sum_{sym}\left(a^8b^3 \plus{} a^8b^2c \plus{} a^7b^4 \plus{} 3a^7b^3c \plus{} 2a^7b^2c^2 \plus{} 3a^6b^4c \plus{} 5a^6b^3c^2 \plus{} a^5b^5c \plus{} 6a^5b^4c^2 \plus{} 4a^5b^3c^3 \plus{} 5a^4b^4c^3\right) \equal{}$
$ \equal{} \sum_{sym}\left(0.5a^9bc \minus{} 2a^7b^3c \plus{} 1.5a^5b^5c\right) \equal{} abc\left(\sum_{cyc}(a^4 \minus{} a^2b^2)\right)^2.$  :)

[quote="can_hang2007"]Lately, Pohoata sent me this problem
\[ \sum \frac {a^3 \plus{} abc}{b \plus{} c} \ge \sum \frac {a(b^3 \plus{} c^3)}{a^2 \plus{} bc}
\]
for all $ a,b,c \ge 0$.
Then I proved it by using the following identity
\[ \sum \frac {a^3 \plus{} abc}{b \plus{} c} \minus{} \sum \frac {a(b^3 \plus{} c^3)}{a^2 \plus{} bc} \equal{} \frac {abc(a^4 \plus{} b^4 \plus{} c^4 \minus{} a^2b^2 \minus{} b^2c^2 \minus{} c^2a^2)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}
\]
:)
And here is my question: prove my indentity. :)[/quote]
Now, I have found an easy proof for this inequality. :)

[quote="can_hang2007"]Lately, Pohoata sent me this problem
\[ \sum \frac {a^3 \plus{} abc}{b \plus{} c} \ge \sum \frac {a(b^3 \plus{} c^3)}{a^2 \plus{} bc}
\]
for all $ a,b,c \ge 0$.
Then I proved it by using the following identity
\[ \sum \frac {a^3 \plus{} abc}{b \plus{} c} \minus{} \sum \frac {a(b^3 \plus{} c^3)}{a^2 \plus{} bc} \equal{} \frac {abc(a^4 \plus{} b^4 \plus{} c^4 \minus{} a^2b^2 \minus{} b^2c^2 \minus{} c^2a^2)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}
\]
:)
And here is my question: prove my indentity. :)[/quote]
I have known a ineq $ \sum \frac {a^3 \plus{} abc}{b \plus{} c} \ge \sum a^2$,it is true with vonicu-schur,but i cant prove your ineq by it ,can you

[-( .If you use this ineq  , you will be wrong ;)  .Please notice when the equality holds ;)

[quote="manlio"]Very nice your proof, can_hang2007. :)

How did you find that nice identity?  By computer, intuition or simple computation or ......!!!!!

Thank you very much.[/quote]
I used pqr technique :)

[quote="manlio"][quote="can_hang2007"]
I used pqr technique :)[/quote]

Does it mean that you made all computation using variables p,q,r instead of a,b,c?

Thank you very much.[/quote]
Yes :)


find pairs $(a,b)$of positive reals such that for every$n\in N$and every real number x satisfying:   $4n^{2} x=log_2(2n^{2}x+1)$
              we have $a^{x}+b^{x}\geq 2+3x$

First, let solve the equation   2x = ln(x+1)/ln(2)!!!


If a,b are pozitive reals then prove that 2a^3b^3+a^3+b^3>=a^2b^2(a+b)+ab(a^2+b^2)

[i]unknown source[/i]

The inequality reduces to a(b - 1)(b + 1) + b(a - 1)(a + 1) \geq 0, which holds.


Sbiecte frumoase , dupa parerea mea , dat la concursul Gh. Vranceanu in decembrie anul trecut.

[quote="rockymarcianocosmin99"]You are welcolme and if you want ore need more i will post them.[/quote]Ideea e să postezi câte un topic pentru fiecare problemă, şi apoi un alt topic în care să aduni link-urile către problemele postate, să pui din nou doar subiectele, şi eventual acolo să ataşezi şi fişierul cu toate problemele.

Uite spre exemplu cum am postat eu subiectele de la barajele de juniori de anul trecut:
http://www.mathlinks.ro/Forum/viewtopic.php?t=5462

(bine în acel topic nu am pus şi .pdf-ul, dar am pus un link către topicul unde se găseşte acesta :) ).

Presupun ca decembrie anul trecut inseamna , decembrie 2004  :D . Ce ai facut la concurs?

PS. Multumesc si eu de subiecte.

Am fost primul si am fost pe locul trei la echipaje . Multumesc de intrebare

There is some typo in this files,

clasa XI

problem III
problem IV

could you complete ?

thank you

Are cineva subiectele din acest an? Stiu ca a fost azi concursul dar nu am facut rost de subiecte...inca... :(


trigonometry
ratio
HMMT
geometry
3D geometry
quadratics
algebra

Compute:

$ \text{arctan} \left( \frac{\cos 40 ^{\circ}}{1\plus{}\sin 40 ^{\circ}} \right)$

where we are operating in degrees.

[hide="Hint"]
$ \tan \frac \theta 2 \equal{} \frac {\sin \theta}{1 \plus{} \cos \theta}$
[/hide]

[hide="Answer"]
Note $ \frac {\sin 2\theta}{1 \plus{} \cos 2\theta} \equal{} \frac {2\sin \theta \cos \theta}{1\plus{}2 \cos^2 \theta \minus{}1} \equal{} \tan \theta$.

Since $ \sin \theta \equal{} \cos (90 \minus{} \theta)$, $ \text{arctan}\left(\frac {\cos 40^{\circ}}{1 \plus{} \sin 40^{\circ}}\right) \equal{} \text{arctan}\left(\frac {\sin 50^{\circ}}{1 \plus{} \cos 50^{\circ}}\right) \equal{} 25^{\circ}$.
[/hide]

Another way to do it(not as short as azips's solution):
Using: cos60=4cos20-3cos^3(20) and solving it with Cardano's formula, we get the value of cos20
Then 1-cos^2(20)=square root(sin)(20) to find out sin 20
After all, substitute all value into the ratio and we will get the result!

Aha, oops, I forgot about that identity. Well anyway, clearly this problem is too easy for intermediate.

It seemed cooler when I made it - it's a modified version of HMMT number 8 Alg from this year.

[quote="ghjk"]After all, substitute all value into the ratio and we will get the result![/quote]

Will we?  I don't know about you, but I can't recognize the value of $ \tan 25^{\circ}$.  Cardano's formula doesn't help, by the way; if you've ever actually tried to calculate the value of $ \cos 20^{\circ}$, the cube root in Cardano's formula is a complex cube root and you need to use de Moivre's to evaluate it, so all you get is $ \cos 20^{\circ} \equal{} \cos 20^{\circ}$.

Well, I will check about that later! Maybe you're right since I haven't learned much about complex roots! So now, I can understand why it is the 8th problem on HMMT but we can still do it without using those formulas above! :)

You misunderstood, it was [b]motivated[/b] by an HMMT problem.

The original problem was to calculate:

$ \text{arctan} \left( \tan 65 ^{\circ} \minus{} 2 \tan 40 ^{\circ}  \right)$

the two are related because the answer to the above is $ 25 ^{\circ}$, which leads to a particular quadratic equation for $ \tan 65 ^{\circ}$ in terms of $ \tan 40 ^{\circ}$

Yup, sorry for misintepreting your post above! But it's easily solved in the original form!
Here's my proof:
tg65-2tg40=tg65-2(tg65-tg25)/[1+tg65.tg25]=-tg65+tg65+2tg25/2(since tg65=cotg25 and cotg25.tg25=1)
Therefore we have a very nice result: tg65-2tg40=tg25
The final answer is: arctg(tg65-2tg40)=arctg(tg25)=25

Nice. My solution to that:

[hide="Solution"]$ x : \equal{} \tan 20$. Then:

$ \tan 65 \minus{} 2 \tan 40 \equal{} \frac {1 \plus{} x}{1 \minus{} x} \minus{} 2 \left( \frac {2x}{1 \minus{} x^2} \right) \equal{} \frac {(1 \plus{} x)^2 \minus{} 4x}{1 \minus{} x^2}$

$ \equal{} \frac {(1 \minus{} x)^2}{(1 \plus{} x)(1 \minus{} x)} \equal{} \frac {1 \minus{} x}{1 \plus{} x} \equal{} \frac {1}{\tan 65} \equal{} \tan 25$[/hide]

Wow thats pretty tricky...Does anybody have a geometric solution?

Not geometric, but it's another approach.  All values are in degrees.

[hide="My Solution"]Loads of trig identities!

Let $ \theta\equal{}\arctan\left(\frac{\cos 40}{1\plus{}\sin 40}\right)$.
Since $ \frac{\cos x}{1\plus{}\sin x}\equal{}\frac{1\minus{}\sin x}{\cos x}$, we make this substitution:
$ \theta\equal{}\arctan\left(\frac{1\minus{}\sin 40}{\cos 40}\right)$
$ \theta\equal{}\arctan(\sec 40\minus{}\tan 40)$
$ \tan\theta\equal{}\sec 40\minus{}\tan 40$
Since $ (\sec x\plus{}\tan x)(\sec x\minus{}\tan x)\equal{}1$ (this rearranges to $ \tan^2 x\plus{}1\equal{}\sec^2 x$), we have:
$ \cot\theta\equal{}\sec40\plus{}\tan 40$
$ \tan\theta\plus{}\cot\theta\equal{}2\sec 40$
We then use $ \tan x\plus{}\cot x\equal{}\frac{2}{\sin 2x}$:
$ \frac{2}{\sin 2\theta}\equal{}2\sec 40$
$ \csc 2\theta\equal{}\sec 40$
Since $ \sec x\equal{}\csc(90\minus{}x)$, we have:
$ 40\equal{}90\minus{}2\theta$
$ \theta\equal{}\boxed{25}$[/hide]

cos40/(1+sin40) = cos90+cos40/sin90+sin40

cosA+cosB = 2cos(A+B/2) cos(A-B/2)
sinA+sinB = 2sin(A+B/2) cos(A-B/2)

Hence cosA+cosB/sinA+sinB = cot(A+B/2)

= cot 65 = tan25

Hence the answer is 25 degrees