geometry
CEMC
calculus
geometric transformation
reflection
Putnam
ratio

Heh, well it's about 3 AM here and I've spent the past 1 or 2 hours kicking around on google looking at math stuff... eventually I stumbled upon this place via the name Yufei Zhao.

Anyways, from what I saw I'm the only Saskatchewan person here... and it's sad because you all seem so far ahead.  I've been really angry about the education system here, especially math wise.  My high school always scores well and all that nationally but academically we aren't aloud to excel.  They only let me skip one grade (I completely crumbled when I was looking through the pascal results and saw a grade 8 that is 11!) and they only let me skip 2 math classes.  Skipping 2 was pretty good, but the math I'm taking is still too easy, it's June and we're doing permutations and combinations... Ugh!

Anyways, yeah... Anyone else have the same problems with the system? :(

I've talked with the school board so much and they refuse to let me do extra work and stuff... I got first in the province this year on Cayley, well tied for first.  I made a stupid counting mistake on the box moving question.  One question dropped me from 4th nationally to 91st  :P But ugh! They still want me to do all this boring math... it's so easy! >_<

Shout out to Saskatchewan!  I lived for 5 years there in Saskatoon. :)

I'm in the same situation, because I'm living in Nova Scotia, which has an even smaller population than Sask.  Frankly, I don't think schools anywhere do the stuff on the harder math contests (like proving inequalities...I looked at that, and I don't think any high school teaches that stuff).  However, there's a reason many of the people here are ahead of you (and me): almost all the best math students post on this site.  You have Euclid winners, CMO winners, Fermat winners, and all that jazz.  Don't be discouraged!  Try to get invited to UWaterloo Math Contest Seminar next year (if you weren't invited this year).  It won't be too hard from Saskatchewan; you just need to win the Fermat there.  That's how I got invited. :D

There all slow. Its not just your area of the world. Believe it or not, but most class rooms in the western world are HORRIBLE places to learn things. You learn alot more by self discovery, and with gentel guidance. Hang around here, order some books that really make you think your stupid, and then you'll be learning the math you should be.

Bah :( It's too bad that the math programs everywhere are so slow... we have enough kids in our school to run a special class or something, and a teacher that would do it, but I don't think the school board would approve.  Our school got 7th in Canada on the Euclid with a team of all grade 11s, including 7th in Canada individually.  I'll try to win the Fermat, I havn't received any kind of invitation anywhere as of yet, but I'd love to get one.  I know that all the competition for the Fermat is inside my school, so as long as I stay above the 5 people biting at my heels I should be able to win it.  I did all the Fermats on the CEMC site last year and I got around the 120-134 range on all of them, and since then I've completed 4 math credits, my grade 10, 11 and 2 grade 12s, so I might be able to do well :)

I'll try to score well in the COMC and Euclid, I hear those are the most important for scholarships and camps.  Actually, the CEMC site has decent workshops that outline the Euclid, so for self discovery it might not be that hard to find.

At least I might be able to take some university classes once I finish grade 11, in grade 11 I'm taking my last 2 high school math courses, one of which is Calculus, so I'm going to ask my friend to talk to his math professor father and see if I can get something going.

:o Surprising to see someone else from Saskatoon :P

well same thing happens in BC too.
Everyone thinks BC has a comparatively large population, thus the math education would be sufficient. BC hasn't had an IMO member for 5 years already(except for this year, Farzin made it). I'm taking math 10 now, I don't see any point to take such a dumb math course, but the school board always says 'you have the provincial exam this year, you have to take the course to write it.' Also, I can't get any help from the teachers with math problems beyond AMC.
Yesterday I talked about it with the math department head in our school. He suggested that I could go to SFU(which is only 3 miles from my school) for further math (calculus year II and linear algebra and stuff). Would that be a good idea? I would have much more schoolwork when I enter higher grades. Also, I'm not planning to take math as my major in university. If anyone has tried this way to make progress in math, please help me out.

Yeah, I'm surprised that you say that.  I'd say go for it with SFU, but at the same time mattering what kind of person you are it could backfire.  It'd be more efficient then just working on your own I'd say, because then you at least receive some recognition for it.

Personally I don't really intend on majoring in math in university, but after looking at some IMO stuff, the competitive part of me really desires to learn more... I regret not pushing forward more in elementary school, but partially I wasn't really aloud to, and then the other part was that I didn't know how to.  Plus, primary school kids shouldn't be expected to have to be ambitious to excellerate their pace, I'd say that's more on the shoulders of the teacher.  I had all my multiplication tables down and could divide and do long division when I was in kindergarten, so I mean if I didn't have to formally go through grade 1-5 here doing all that stuff I could have been way ahead, not to mention I could do algebra in grade 1 and stuff...

But I read about people outside of Canada that finish university when they're 20 ish, or I hear about these phenoms and it's frustrating because I'd say the school system here doesn't really allow for kids like that to attain what they could.  I mean, they wouldn't let my friend take calculus this year, so he isn't taking a math because he's in grade 11 and done all his except calculus.  He scored 7th in Canada on the Euclid.  He's beyond all the math we're doing in school, and actually I think a surprisingly large amount of students are.  Definately I'd say the curriculum doesn't reflect students ability.  My friend was looking through some math textbooks from China, and some of the grade 3 textbooks from China teach our grade 8 curriculum.  It's ridiculous.  I wish they'd push us more here.

What I'm trying to get at, I guess, is that if you can do it, you should.  Don't let little technical things hold you back.  Personally I wish I started looking into this stuff earlier  :( I feel so behind now that I'm in grade 10, I doubt I could even do the first IMO question  :blush: But this summer I hope I can change some of that :)

If you think that you won't have enough time to do your high school coursework plus the SFU coursework as well, you could always look into doing a program over the summer.  But bleh! *Pokes math B30 textbook* This is supposed to be grade 12 curriculum? Give me a break x_x

actually do not over-expect education in other countries which seems spendid for you now. I've been in a so-called high-school-gifted-kid program for two years before, but the education I received can't guarantee me a place in IMO at all! the stuff I learnt was surely ahead of most people of my age, however I felt that it's somehow futile.

The goal of education is not to stuff as much knowledge in your brain as possible. Heurism is proven to be a more successful way. Math is not just knowing how to do calculations, it requires more logic and creativity(especially when it comes to olympiads questions), so a better ideation would be more likely to lead you to success in math. And that's also why I hate school math so much----everything with a calculator, then who's thinking?the dumb machine or my brain?
For contest math, I think the best way to practise is not to do tons of problems, but to have a general idea about each topic it would be testing. I try to analyze every question I do, and that does help me a lot do similiar questions

I got so much things to say about education.

When I found out that our family is going to immigrate to Canada, I was actually very happy(opposed to my little brother, who hated losing friends). The main reason was that the education system in Korea was really disappointing. Students were pressured too much, always had so much to work on, yet the gifted students had no way to learn more in the system.

When I came here, it first looked good. Now I actually had free time to do other things. However, after a while I found out that this is even worse. Every course was going too slow(I had hard time in English though, because of the obvious reason). I graduate in a month, and I feel that I gained very little from the school. I think the whole high school curriculum could be learned within a year.

I really hope that the system is modified so that there is an option of going subjects over faster, for some people who can learn certain subject faster. Students shouldn't suffer through the course and learn less stuff, just because other students learn slower. :(

Yeah, the school system here is very... well, it tailors to the average, or slower students.  I think in Asia the schools are more made so that the brightness and genius rise up, almost like sifting out the bad ones?  But I've never been in Asian schools, that's just what I've heard.  My dad skipped 2 grades in China when he was growing up though (I hear that's impossible? :P) but they barely let me skip 1 even though my test scores and comprehension proved that well... I could finish all of elementary school in probably a year (so instead, in grade 1 I just always played sick and then went in the last week and did all my work), and I really got discouraged academically :/

It's almost like they chain weights to your legs that you must drag behind yourself, and then you eventually just get tired and bored and collapse.  If it wasn't for math contests, I'd probably have demanded that my parents send me to school in Ontario at least, preferrably some private school in the US that would let me be smart instead of dulling me to blend with the norm :/

i heard system in asia is changing to accomodate weaker students too... the truth is, no matter how easy the system is, there will always be ppl struggling, so no point making it easier. with the ontario government wanting to cut out gr 12 calc, which is :mad:. in all my courses, i've never had one that didn't use calculus!

when i moved to canada in gr 3, i was learning the same stuff here that i was learning in gr 1 back in asia. in gr 5, we started learning stuff that i learnt in gr 3. basically gr 3,4,5 was learning pretty much the same stuff. if they want to cut the high school curriculum because it's too hard, why not change the curriculum at the elementry level?

canadian kids are not dumber, if u take psych courses, u'll see that there is virtually no difference in intelligence between races. just that the curriculum needs to be changed if canadians wish to remian competitive internationally. <---- quoted from a prof

also, avg canadians r not competitive in academics. i think the reason why is, there is really no top canadian university, plus there r so many schools, why worry which one u go into? even the top schools in the country are not that hard to get in at all, so why try?

Well I think a lot of the people here get lazy since their life is fine.  In say China, there's WAY more people, WAY more competition, and the average person struggles a lot growing up.  This stress and pressure pushes them to perform in my opinion. :/ I think that elementary school curriculum could be covered in maybe a bit more than 2 years for most "gifted" children.  Kindergarten and Preschool are really just like daycare.

Rant alert:
[rant]
I wouldn't know fully, because I've never been to Korea or China, but I know that in Canada we spend most of our elementary education reading children's books.  Since the English language has such an abundance of literature, it's mandatory for multiple years in high school too.  Frankly, studying literature is a waste of time.  Writing it, or reading it, is different, but studying literature does not increase appreciation of literature nor does it teach us life skills.
Another problem I see is that much of the "social studies" we learn are redundant from year to year.  Mostly, they focus on local or Canadian history up to Grade 10, which is an absolute waste of time.  Canada has been around for 139 years; the world has been around for billions.  I don't know if Asian schools waste so much time on teaching this crap.
When I think back, I realize I learned about 5 things in math up to grade 5:
a) addition;
b subtraction;
c) multiplication;
d) division; and
e) that "guess and check" is an appropriate way to do math.
Tell me, what is the difference between adding a two-digit number to a three-digit number and adding a two-digit number to a four digit number?  The textbooks I had in those grades were full of crap like this.   In the Nova Scotia public system, there is absolutely NO streaming of students who find the work too easy.  The homework was mostly art projects, and since my art skills are terrible, my grades were never very good.  I don't know if my being in French Immersion slowed me down.
I think what it comes down to is a culture of entitlement.  All of the people at my school are going to unversity (it is a private school, but there are still dummies).  I know people who have never read a book in their life who plan on going to university  People say they want to major in English or Psychology, which seem to be a waste of time.  Also, regardless of what you hear students whine about, post-secondary education is not overwhelmingly expensive.  It [b]is[/b] a worthwhile investment if you are not going to party all day, skip most of your courses, or major in Comparitive Literature.
[/rant]
:diablo: 's day today!

I'm honestly happy I made it out of highschool with ambition and curriosity. There were several points where it actualy did KILL them. They grew back in new forms, but I was never again interested in some of the things I was when I went into highschool.

I noticed something recently: If I hadn't found math in the last 12 months. I would have ended up as one of them.....

Not that math is the 'only' path to 'salvation'   :D , but I honestly had no interest in anything besieds getting out of highschool.

I don't think I'm being dramatic when I say "Highschool destroys children"/

I agree that kids should challenge themselves, hell, my life became [b]fun[/b] when I relized I can a) choose my own constructive activities I can do, and b) work to better myself, reguardless of whats going on around me.

But I didn't figure this out untill begginning of grade 11. We all think the advice is obvious, but no one tells kids to 'work hard at something important to them'. Thats why things are so messed up, kids have no idea how to do better, or be constructive.

To be fair, its pretty easy to complain about the ontario cirriculum, becuase it resists [i]everything[/i] you do. It doesn't let kids move forward 'within' the cirriculum. If you want to advance faster, you HAVE to have your parents pay for private schools ( at least that was the only option left for my family).

Its not that people aren't trying to fallow all of this easy, good advice, its the fact that the system in place is just so hard to work with.

example: when my brother got his gifted designation, my parents spent almost a month going to meetings with his grade 8 principle. They [/i] refused[i] to give him an IEP ( the Individual education plan ), that he would need to be let into the gifted program in grade 9. Why? becuase they had never done it before. They make IEPs all the time to get kids shoved into the sp-ed programs, but when a kid came in, trying to get an IEP, to help him excel, they actualy thought the system shouldn't be used this way, and resisted for as long as possible.

example 2: I love ultimate frisbee. since there was no team running for my school on my senior year, I went on a mission to start it up. They made this impossible. They told me I couldn't even play on school grounds witouth being a 'team' and having a 'teacher supervisor'. I asked every teacher in the school to be an advisor, but the apathy was  too thick. after a few weeks of all this running in circles, trying to get some help, we just left, and play in a local shitty park on weekend.  =(

in conclusion: I agree kids should complain less, do more. But remember: there are some elaborate shut down systems in place, some people feel fine complaining after so much crap.[/i]

i think the issue of defining "gifted" is a big problem itself. one of my friends, he was invited to participate in 3 olympiads, would he be "gifted" by the standards discussed?
he isn't particularly smart, but he works really hard, he's got bronze in ioi and silver in ibo (only participated in 2 olympiads because third one conflicted)
he wasn't accepted to the typical gifted program we have here in ontario because his iq score wasn't high enough.
things didn't always come easy to him, but he definitly tried hard. would he be denied to such program?
how do u measure one's "giftness"?
the current iq test is quite flawed actually, so do we do it by academic achievements? if so, wouldn't we run into the same problems about gifted kids being not sociable?
if a school will take kids who r simply "gifted" not by academic achievements, then why isn't university acceptance based on "giftedness?"
u don't need to be gifted to excel in something, all u need is a certain standard of intelligence (not too high, i dunno what the level of measurement is), perserverance, interest, and hard work. for ex. the person who discovered the double helix of dna has iq of 104, he won nobel prize. thomas edision once said "99% hard work, 1% intelligence"

i'm always interested about subjects of measurement of "smarts", let me know if u have any comments on that

To make J-I-M-B-O's point of view even more punching to young mathematicians, I'll add that most (if not all) of widely known mathematicians were "stakhanovist's". From Poisson, to Erdos, all of them worked much and hard ! Being gifted is not revelant. If you are "gifted" (whatever that means) you have a better starting point, but if you work harder than the "gifted", you learn more, hence you surpass him unevitably.

As to the IQ discussion it is to be noted that it heavily depends on your standard of living. Given the same kid, psychologist's measured an IQ of 100 in poor living conditions, and 115 in considerably better conditions. Having this in mind, don't you find IQ measurement in schools an awful perspective ? More generally, has the notion of "gifted child" any sense if it does oscillate in the limits of 15 IQ points  (since with 105 you are normal, and with 120 that you possibly might have you are "gifted") ?

[quote="lightrhee"]I think the whole IQ thing is complete nonsense.[/quote]

it is... it was proved to be flawed...

the point of iq test is to test a person's potential
while the point of academic test is to test a person's knowledge

however, one's iq score changes from year to year. it was shown that a person's iq drop over the summer when not in school. also, an interesting note was that avg person's iq increased from 70s to 100 since 1930s, which coincidently was when education was made widely available to the public. either human beings evolved intellectually over the past 70 years or iq is very muchly influenced by education or knowledge.

so iq doesn't accurately test a person's potential but rather than test one's current knowledge.

so the question is... how do we measure intelligence? current iq test gives us an IDEA, but not accurate but it does give us an idea of how we stand to the avg joe.

I think a comparison I first heard from singular is relavent:

" the discussion of nature vs nurture is interesting, but not incredibly important. Nature sets the stage and determines the small game, but nurture seems to win out every time in the big game"

Some of us have distinct advantages in the womb, but most people here would agree, its what you do with it, not what your born with.

Well, it's hard to say what effect schooling actually has on IQ.  There could be the fact you learn things in school that make oneself aware of certain things.  For instance, I remember on my IQ test in grade 3 I never could figure out how to get 4 litres of water given a 5 litre and 3 litre bucket.  Of course, after seeing the solution to one of these questions, I was able to do them all.  At the same time, even if you do not "learn" anything new from the schooling, it keeps your mind active.  Over the summer, lots of people get lazy and their minds relax more, the synapses don't fire as quickly and what not, so getting a lower IQ would be very permissable due to inactivity.

Of course, I think one can learn to score well on an IQ test, but I think that an IQ test is pretty accurate when someone is a kid and hasn't been exposed to questions that would be on an IQ test.  I think it's useful to test a child's logical potential, but of course, that's just potential.  But really, I don't think it can be that heavily weighted.

Even tests such as the SATs are underfire for not being an accurate assessment.  I think one of the best ways to reach someone's intellegence is to talk to them and observe them.  I think most reasonably educated or smart people can differentiate between someone that is very smart, and someone that just says a bunch of BS.

I also find it odd that the school system is more willing to cater towards children that have difficulty learning due to a disease, disability, or just the fact they aren't all that "smart" compared to say a gifted child that has difficulty learning because the curriculum doesn't teach all that much.  I mean, I'm taking grade 11 chemistry this year and for me it's a joke.  It's just basic math all over again.  Stoiciometry, moles, gas laws, it's all pretty basic once you wrap your head around it.  But I know kids that have a LOT of trouble wrapping their heads around it, and of course the teacher is more willing to give them extra work to practice, but none for me to further my knowledge base.

It's really hard to title someone gifted though, and I'll agree to that.  But I think that some people at least have a sense for who is gifted and who isn't.

And also, hard work I feel can only get you so far.  I think that everyone needs a certain level of "intellegence" of giftedness to meld with their hard work.  I mean, I know guys that try so hard in classes but still get 60s or 70s because they just don't have the head for that class.  And I mean, that's normal.  I think the whole thing is having a balance.  I watched a show once... W-Five or the Fifth Estate or something that was about people with very high IQs that ended up being like.. bouncers and stuff, or just sit around and do nothing.  It's a certain balance between intellegence and how far you want to take it.  So I mean, someone with a 110 IQ compared to someone with a 150 IQ (assuming IQ is accurate, just for a hypothetical example) could score way better on tests, and get further academically and what not just because they push their intellegence to it's limit, and the 150 IQ person is just content doing whatever.

I just think though all in all that the school system should cater to gifted children to some extent.  I mean... in grade 5-8 my teachers just got me to mark other kids work and help other kids with their assignments when I finished early... and I was in a gifted program that is supposed to have a 130 IQ minimum... It was ridiculous.

[quote="J-I-M-B-O"]i'd like to add that from the textbook i have, for students who succeed in post-secondary (i dunno what they mean by success) has 0.44 correlation with their iq tested in gr 3, while SAT score has correlation of 0.78.
it's very true we can tell if the person is smart by talking to them, but there r a lot of exceptions, like einstein, who was thought to be slightly slow in his early teens.
again, some of the top students have trouble communicating ideas, thus sound/look stupid.
i think it's much better to stick with the system we have right now and just tighten it up slightly. truly gifted students can simply study on their own.

no institution will take anyone based on their intelligence score, they at most take combination of academic and intelligence.

btw, since u need certain intelligence to excel in something and especially when mentioned that ppl getting 60s and 70s because they're not smart enough to get the subject, wouldn't we just pick ppl based on their accomplishment??? just like that we r doing right now?? since ppl who accomplish big r ppl who ACTUALLY cares so they try and is smart as well.  contests r also another good indicator!! thats what waterloo does. if u excel in contests and have sufficient marks, they'll take u.

this world is not just about nature, it's nature AND nurture.
why take ppl who won't work hard but ridiculously smart, just because they r born lucky and put them in class with other smart ppl?
how is this feasible?
it's like saying, there's a football star in my school. instead, since playing locally is impeding his progress and us normal ppl is slowing his progress, so he gets go to training camp with the pros. so far that hasn't happened yet, not to my knowledge anyways. even if it did happen, it's private company that pays for it. why would the government pay for that?
i am sure that we're not the first ones to think of this, if it was a feasible plan, any country would of done it by now.[/quote]

Actually I think some parts of China do it :P *pokes their IMO score*  :|

And I think that sure, gifted kids can study on their own but they shouldn't have to.  The government is willing to title kids retarded and offer them funding, special programs, and TAs (one student in my school was given a laptop, a TA to himself and special sessions with a tutor just because he has aspergers, he isn't even stupid, and thats what.... probably 70,000 of funding right to him, and I always beat him on contests, so consider that 70,000 of taxpayers money down the drain).  They do this, but they aren't willing to title a child GIFTED and give him the same funding?  I think that is skewed.  I really do believe the school system is made for the lower hinge of kids, and I don't think that's fair... I'm just saying it would seem more logical to put all that power where the potential is.

Oh, and anyone know what kind of test scores and marks in high school classes one would need to go somewhere like Stanford or Harvard?  I hear that it's mostly about extra curricular activities.

[quote="{x}"]To make J-I-M-B-O's point of view even more punching to young mathematicians, I'll add that most (if not all) of widely known mathematicians were "stakhanovist's". From Poisson, to Erdos, all of them worked much and hard ! Being gifted is not revelant. If you are "gifted" (whatever that means) you have a better starting point, but if you work harder than the "gifted", you learn more, hence you surpass him unevitably. [/quote]

Of course being gifted is relevant.  Probably some 'famous' mathematicians are more innately intelligent than others, but I'm pretty sure they would all rank in the 99.9th percentile of any standardized intelligence test.

[quote="CaucasianAsian"] And I think that sure, gifted kids can study on their own but they shouldn't have to.  The government is willing to title kids retarded and offer them funding, special programs, and TAs (one student in my school was given a laptop, a TA to himself and special sessions with a tutor just because he has aspergers, he isn't even stupid, and thats what.... probably 70,000 of funding right to him, and I always beat him on contests, so consider that 70,000 of taxpayers money down the drain).  They do this, but they aren't willing to title a child GIFTED and give him the same funding?  I think that is skewed.  I really do believe the school system is made for the lower hinge of kids, and I don't think that's fair... I'm just saying it would seem more logical to put all that power where the potential is.
[/quote]
yo man always remember that 144 or w/e on Cayley is not the best in Canada and someone else should get more funding than you bcz of the contest. is that fair?
and also, potential is not only in math contests. the contests in high school are only as piddling in your life as kindergarten games in your education.

[quote="Sunny"][quote="CaucasianAsian"]
And I think that sure, gifted kids can study on their own but they shouldn't have to.  The government is willing to title kids retarded and offer them funding, special programs, and TAs (one student in my school was given a laptop, a TA to himself and special sessions with a tutor just because he has aspergers, he isn't even stupid, and thats what.... probably 70,000 of funding right to him, and I always beat him on contests, so consider that 70,000 of taxpayers money down the drain).  They do this, but they aren't willing to title a child GIFTED and give him the same funding?  [/quote]

Little bit off topic.  Kids with Aspergers AREN'T stupid, they tend to especially excel at one topic.  And they actually need that TA, cuz they just don't function with large groups of people.  I've worked with a couple kids like that, its crazy what they do.  But agreed, the government and individual schools need to have some way of encouraging talent in other students.  I suppose they just can't afford it in general, not now at least.  Besides, those kids falling behind need the help more than kids who can potentially help themselves.[/quote]

Yeah, my point for that is just that it is a perfect example of everyone's tax money going towards one person's [well, I guess two people's] child.

[quote="nneonneo"]Chris, are you going to quit Advanced?
I don't think that's a good idea--I personally still think it is a great experience. However--your crop of Advancies seems a bit stranger and more arrogant than ours...maybe that's just me.

I think Advanced will do you good, especially the IRP (I recommend it despite the stress; you learn a lot more about how to learn and yourself from the project than regular will ever give you).

*sigh* I wish we had IB.[/quote]

Yeah I am Bobby, sorry to disappoint, but I hate the group of children that I am with for it.  Nothing personal with them, but I'm not a big fan of comparing marks for the most part, especially in a class like english.  IRP isn't going to be a big loss for me, I had to teach myself math 10 and 20 over last summer, so I have a decent idea of how I learn, and I mean... I know I won't be able to do the same with IRP, but I finished my MIRP essentially in one night over the course of 3 hours, or at least I did all the research, and all the writing except maybe a couple proof reading things.  Not just that, but I ended up with one of the best marks in our class.... apparently though, our classes marks were bad  :P I got a 90 I think.  I also prepared for my presentation over the course of 5 minutes, before that I had no idea what I wanted to do for it, and I had no notes for reference, and Mr Davis said it was one of the best ones.

A perfect example of why I'm leaving advanced though, is after he said that a couple kids of course tried to say how bad my presentation was, etc etc because of how competitive the advanced class is  :blush: not to say I can't compete, but I'd rather enjoy my time in school and be relaxed than be all stressed over marks and comparing myself to others.  We both know I'm a pretty....... confident person.  :P


here are a few exam archives from australia:

4 unit maths:

http://www.angelfire.com/ab7/fourunit

http://www.geocities.com/fourunitmaths

http://www.geocities.com/ext2papers

maybe some other australians can post up their papers here on mathlinks. ok?

That isn't really problem solving is it? :?

It's just the typical schoolish calculation...

Yeah, these above links are just "4 unit maths" papers which is just the end-of-school level maths here in Australia --- no problem solving. In terms of olympiad problem solving here we have a two-day contest, the Australian MO (run by the Australian Maths Trust --- http://www.amt.edu.au), in february each year. All the papers used to be on http://www.kalva.demon.co.uk until John Scholes suddely took them all off. I think he thought they were too easy, and they certainly are easy compared to APMO and IMO which we of course have in common with everyone else.

[quote="Dionysius"]All the papers used to be on http://www.kalva.demon.co.uk until John Scholes suddely took them all off. I think he thought they were too easy, and they certainly are easy compared to APMO and IMO which we of course have in common with everyone else.[/quote]

No, he was forced to take them off. The AMT left him te choice: taking it off voluntarily, or ask a court what they think of it.

They did the same thing with the Chinese olympiad. I really wonder what the AMT has to say on the CMO but anyway, it's happened: http://www.mathlinks.ro/Forum/viewtopic.php?t=5304


A fair coin is tossed 1997 times. Find the probability that at no point during the
tossing are two heads flipped consecutively.

Uh..how is the answer expressed?

As a sum of combinations..?

beucase its too big..

[quote="JavaMan"][hide="Solution"]
I will consider 1 to be heads, and 0 to be tails.

[b]Count number of binary strings that do not contain two consecutive 1s:[/b]

Let $ f_{a}(n)$ denote the number of binary strings of length n, with last digit $ a$ (0 or 1).

$ f_{0}(1) = 1\; f_{1}(1) = 1$ (There's one binary string that ends in 0, and one that ends in 1).

The number of binary strings of length n ending with 0, is the sum of the number of binary strings of length n-1 ending in 0 or 1.
$ f_{0}(n) = f_{0}(n-1)+f_{1}(n-1)$
Also the number of binary strings of length n ending with 1, is the sum of the number of binary strings of length n-1 ending in 0.
$ f_{1}(n) = f_{0}(n-1)$

Adding the previous two equations together;
$ f_{0}(n)+f_{1}(n) = 2f_{0}(n-1)+f_{1}(n-1)$
$ = [f_{0}(n-1)+f_{1}(n-1)]+[f_{0}(n-2)+f_{1}(n-2)]$

The LHS represents the total number of binary strings of length n without two consecutive heads. This is the fibonacci sequence since $ F_{n}= F_{n-1}+F_{n-2}$

So the answer is $ \frac{F_{1997}}{2^{1997}}$
[/hide][/quote]

Um if its $ \frac{F_{1997}}{2^{1997}}$, then you have to note that $ F_{1}=2$ (then fibionacci), not $ 1$.Otherwise if $ F_{1}=1$and$ F_{2}=1$(then 2,3..so on), then its $ F_{1999}$ over $ 2^{1997}$

[quote="JavaMan"]I should've been more explicit there, though I did mention $ f_{0}(1) = 1,\; f_{1}(1) = 1 \implies F(1) = 2$[/quote]
and i shouldve read your full solution..I just went straight to your answer and compared it to mine..because I was thinking about putting the answer as a summation of combinations, which results in that fibionacci pattern.


function
calculus
derivative
integration
real analysis
inequalities
real analysis unsolved

i cant seem to pick the test functions right? the question asks if $ \exists p \in[1,\infty)$

s.t. $ u \in W^{1,p}([0,1])$ then show that the weak derivative $ u'$ exists and is in $ L^p(0,1)$ and that u is equal a.e. to an absolutely continous function.

i think by definition of the sobolev space we get the u' exists and is in L^p, but im having trouble isolating something to look like

$ \sum_k |u(x_k) \minus{} u(y_k)|$ without being able to use all the normal calculus tools..

picking $ \phi_m \to \sum_k 1_{[x_k,y_k]}(x)$ made sense to me, but then i use the FTC which im not sure is legal on a weak derivative.. help!

part 2: asks us that if $ p\in (1,\infty)$ then show

$ |u(x)\minus{}u(y)| \le |x\minus{}y|^{1\minus{}\frac{1}{p}} \left( \int_0^1 |u'(t)|^pdt\right)^{\frac{1}{p}}$

Let $ v(x)\equal{}u(0)\plus{}\int_0^x u'(t)\,dt$. This is an absolutely continuous function because $ u'$ is Lebesgue integrable. It is also in $ W^{1,p}$, since $ u'$ is its weak derivative. Finally, prove that $ u\minus{}v\equal{}0$ a.e. Hint: if $ \int (u\minus{}v)\phi \equal{}0$ for any test function, then $ u\minus{}v\equal{}0$ a.e..

The second part is Holder's inequality applied to $ \int_x^y 1\cdot u'(t)\,dt$.


Find $ \boxed{S\equal{}\left( \frac{C_n^0}{1}\right)^2\plus{}\left( \frac{C_n^1}{2}\right)^2\plus{}\left( \frac{C_n^2}{3}\right)^2\plus{}...\plus{}\left( \frac{C_n^n}{n\plus{}1}\right)^2}$

Yes. It's true and it's a nice solution. Thank you very much.


Harvard college MATHCOUNTS Gauss AMC AIME USA(J)MO

Congrats to Alex Zhai on his perfect score on the IMO!  :lol:

yes. he's a senior from downstate who (as you can see) is pretty beastly.

[quote="RussianRocket"]Oh this guy...
http://www.unl.edu/amc/e-exams/e9-imo/e9-1-imoarchive/2008-ia/2008imoteamannounce.shtml

he's amazing[/quote]
Yep. And he's going to Harvard this fall.

He tied for 1st along with two members of the China team. :first:

Is china usually the best or are there other countries that are usually in the top?
I wouldn't usually expect for a USA person to win.

China and Russia are always very good; US is usually in the top 5.
I'm pretty sure the records for teams are online, but I can't find them right now.

Yeah my dad tied for first at the Soviet Union Olympiad and made the IMO in 1985 but he did not do as good as he wanted to there.
That year the Union of the Soviet socialist republics aka Russia got 6th. I am not sure what my dad got but I think he was second with a score of 28.

Here is the website
http://www.imo-official.org/year_country_r.aspx?year=1985

Wow...your dad is so much better than my dad in math...

 :(  :(  :(

Yah...  :rotfl:  right now im not living up to his expectations   :wallbash_red:
But im getting better  :football:

I challenged my dad to a MATHCOUNTS countdown and nuked him :D  :D  :D

The one game i can beat my dad at is 24. (you know when you take 4 numbers and you have to make 24 out of it).
We had 9 rounds to see woould figure out the answers faster and i beat him
5-4. LOL!   :D
He can destroy me at any other game.  :rotfl:

[quote="Aryth"]dude, who is this stephen character? why would he ever sing the teletubbies song? geez. what a moron he must be.  :roll:[/quote]

i don't know, but if I ever meet him I am turning and running in the opposite direction. He sounds really really really really scary/silly.

[quote="gauss1181"]i think he's stupidityismygam.[/quote]

oh what a lame screen name, this furthers my utter fright and disrespect for this stephen character. I hope I never meet him at a math tournament or something  :oops:

[quote="gauss1181"]i think he's stupidityismygam.[/quote]

uh, i was being sarcastic (again). i lived with him for 3 weeks once.

[quote="gauss1181"][quote="avec_une_h"]Do I seem like a guy?  Because I'm definitely not.[/quote]
I'm not saying you...they're saying ME :D[/quote]

I thought you were a guy for a minute.  Then I realize who you were.

The AMC's are so easy...

It doesn't take much to get a perfect on those...

how can you easily get a perfect AMC 12? :huh:

Have you ever taken Amcs before? what scores do u get? (and on aime too?)

[quote="avec_une_h"][quote="gauss1181"]:rotfl:  :rotfl:  :rotfl:

by the way, avec_une_h are you gonna take AMCs for 2009?

I AM![/quote]

Definitely.  There's not much else in high school.  (I'm still trying to work out how existent my high school's math team is.[/quote]

yeah. go amc. is there anything else in high school to do 4 math? actually i'm not sure if my school has a math team...

um

ARML, local contests (ICTM, NSML, IML, ...), Mandelbrot...
just try to make it, and then do well, on the usamo. that's enough of a field to keep you busy for a while (forever probably)

china is usually #1 or #2, and this year there was an american person in the top ;)


Eight lamps, each with an on-off switch, are arranged in a circle. A lamplighter can flip (in
other words, change the state of) the switches, but he is not allowed to flip just one at a time.
When he switches lamps on or off, he is required to flip the switches of three consecutive
lamps simultaneously. (For example, he can flip the switches of lamps 7, 8 and 1 at the same
time.) Prove that no matter what set of lamps was turned on at the start, the lamplighter
can turn all the lamps on.

did u get these from pomona?

http://www.math.pomona.edu/ps1_06.pdf

[hide]It suffices to show that there is a combination that, overall, will switch one light and keep the others the same.

Suppose you want to change light number 1 alone. You can switch {2,3,4}, {3,4,5}, {5,6,7}, {6,7,8}, {8,1,2}, which in effect switches all switches but 1 twice, and 1 once.[/hide]

[b]Generalization:[/b] If there are $ n$ lights, and you can switch $ m$ consecutive lights at a time, for which ordered pairs $ (m,n)$ is this doable?

[hide="Solution"] This is doable precisely when $ (m, 2n) \equal{} 1$.

First, we rephrase the problem in terms of linear algebra.  We are given the $ n$ vectors in $ \mathbb{F}_2^n$ consisting of $ m$ consecutive entries of $ 1$s with the rest zero and we wish to prove it spans $ \mathbb{F}_2^n$.  A rather general result in linear algebra tells us this is possible if and only if the $ n$ vectors are linearly independent.  (This is actually easier to prove over finite fields than infinite ones.)

- If $ (m, n) > 1$, then there are $ \frac {2n}{(m,n)}$ vectors that sum to the zero vector.  For example, $ (1, 1, 1, 0, 0, 0) \plus{} (0, 0, 0, 1, 1, 1) \plus{} (0, 1, 1, 1, 0, 0) \plus{} (1, 0, 0, 0, 1, 1) \equal{} \mathbf{0}$.

- If $ (m, 2) \equal{} 2$, then all $ n$ vectors sum to the zero vector.

- Finally, if $ m$ is odd and $ (m, n) \equal{} 1$, the sum of $ (1, 1, 1, ...) \plus{} (0, 1, 1, ...)$ is a vector that switches $ 1$ and $ m \plus{} 1$.  Generally, we can switch $ 1$ and $ am \plus{} 1 \bmod n$ simultaneously.  Since $ am$ goes through all residue classes $ \bmod n$, we can switch any two lamps simultaneously.  Then starting from $ (1, 1, 1, ... )$ we can remove every $ 1$ except the first in pairs to obtain $ (1, 0, 0, ...)$, and since the same logic holds cyclically we can obtain every element of the standard basis of $ \mathbb{F}_2^n$.  (There is in all likelihood a much slicker way to handle this case. See [url=http://en.wikipedia.org/wiki/Circulant_matrix]circulant matrix[/url].) [/hide]


[quote="Klasz"]Hi, i have a problem understanding a thing in the joined text.
all is clear until the author introduces this "t" (see the picture attached).

Could someone explain me what it means ? where does this t com from ?
is it a known notation i don't know ?

Thank you very much[/quote]

It means [url=http://mathworld.wolfram.com/Transpose.html]Transpose[/url], go there, they have a nice definition and explanation.

Best,

[quote="Klasz"]Great Thanks   :blush:

in france we put this at the upper left of the matrix name....[/quote]

Yeah, not long ago Valentin mentioned that in Romania that was the standard notation too. But such is life. Here people use to write the t on the upper right  of the matrix name, and most of the time they write a capital T instead of a lower case t.

Actually, there are tons of things that we use to assume they are standard notation, and they are not. For example, when you have an interval that is half open you write a round parenthesis towards the number, and in most of Latin America you see a bracket toward the other side. For example, you see $[0,1[$ instead of $[0,1)$, and $]1,2]$ instead of $(1,2]$ :? .

Best,


Does anybody have any idea how to get a solution manual for Spivak's second edition Calculus book??

I doubt they still make the solution book for the second edition, since there is now a 4th edition out.  Some university bookstores may have old copies of the third edition but finding the second edition would be difficult.


What does this sign mean?

        U (pretend it's upside-down)

And is there a formula for A U(upside-down) B?

You might be interested in [url=http://www.artofproblemsolving.com/Wiki/index.php/Set]this.[/url] It explains everything about sets and the notation you were wondering about.

That notation means "intersection", which is the set of numbers that two or more sets have in common. For example, if you have sets $ (1,2,3,4,5)$, $ (2,3,4,5,6)$ and $ (3,4,5,6,7)$ The intersection of these three sets are $ (3,4,5)$


[quote="jmerry"]They're not similar, since their traces are different.[/quote]

I'm sorry. I corrected the problem.
Thank you.


[color=red]Do not post the answers.(If you find the answer,pm me.)[/color]
#1.
[color=green]The students in Mr.Beaty’s[/color] [color=red]Math[/color]Counts [color=green]designed the decorations that are hung on the city’s street-light poles.  Each decoration is either in the shape of an equilateral triangle or a regular hexagon.  The perimeter of each decoration is 24 feet.  What is the ratio of the area of the triangle to the area of the hexagon? Express your answer as a common fraction.[/color]
--------------------------------------------------------------------------------
#2.
[color=red]In December, 1990, a holiday tree in the shape of a right circular cone, had a height of 8 feet and a radius of 2 feet.  By December 1991, the tree’s height and radius each increased by 10%, and the tree’s measurements continued to increase each year at a rate of 10% of its previous year’s size.  In December, 2006, the tree will be decorated with lights.  The lights will be placed along the curved lateral surface of the tree at the rate of one light per square foot.  No lights will be placed within the branches or on the underside of the tree. The formula for the lateral area of a cone is LA = π × r × l where r is the radius and l is the slant height of the cone.  How many lights are needed to decorate the tree in December, 2007? Express your answer to the nearest ten. [/color]
--------------------------------------------------------------------------------
#3.
[color=green]Each year[/color] [color=red]Mario :lol: [/color] [color=green]gives [/color][color=olive]Luigi  [/color]:lol: [color=green]a gift (mushroom) that is wrapped in five boxes that are five similar rectangular prisms.  After he unwraps the first box, there is another box inside to be unwrapped, and then another and another and another. He does not get to his gift until he has unwrapped all five boxes.  The length, width, and height of the smallest box are 4 inches, 2 inches, and 1 inch respectively.  The dimensions of each larger box are 20% greater than the dimensions of the previous smaller box.  By what factor does the volume of the smallest box need to be multiplied to obtain the volume of the largest box?   Express your answer as a decimal to the nearest tenth.[/color]

im guessing these r the "holiday" problems that u talked about.....

I'm going to post the answers to thesewhen 5 people send me a pm with the right answer.

Or I'll just post the answers when somebody asks me to.

Okay I'll post the answers now.
1. [hide]$ \frac{2}{3}$[/hide]

2.[hide]$ 1090$[/hide]

3. [hide]$ 5.9$[/hide]


Can anyone give me a code (what to write in $\LaTeX$) for 4x4 square in Problem 1?

I can't reveal my reason for it but I just want it. Thanks in advance for anyone who can give me.  :D

You could just make the square in Excel, copy it and save it as an image (in Paint or whatever), and implement it into $\LaTeX$ using the \includegraphics command.

[quote="joml88"]\[ \begin{tabular}{|c|c|c|c|}\hline x & y & z & w\\ \hline &&&\\ \hline &&&\\ \hline &&&\\ \hline \end{tabular}  \]

[code]\begin{tabular}{|c|c|c|c|}\hline
x & y & z & w\\ \hline
&&&\\ \hline
&&&\\ \hline
&&&\\ \hline
\end{tabular}[/code][/quote]
That's exactly what I did - also, if you want to put several tables across one row, use the align command with something like \quad between tables

[quote="chess64"]Anyone know how to make a thick border?[/quote]
same question
why do they not give out the latex code and images for the questions  in each round?
that would help so much... we could just edit the images and copy-paste it into our solutions

Unfortunately, there's not much to glean from the LaTeX files.  This year, we did our graphics externally (even the square) with METAPOST, so looking at the LaTeX files won't really show you anything.

[quote="chess64"]Oh, I was going to use metapost but I wanted to know if it could be done in LaTeX first. (I can't find any good tutorials for Metapost :()[/quote]

Oh, there's a number of them:
[url=http://remote.science.uva.nl/~heck/Courses/mptut.pdf]remote.science.uva.nl/~heck/Courses/mptut.pdf[/url]
[url=http://cm.bell-labs.com/who/hobby/MetaPost.html]cm.bell-labs.com/who/hobby/MetaPost.html[/url]
[url=http://www.tug.org/metapost.html]www.tug.org/metapost.html[/url]

[quote="Silverfalcon"]Thanks.

I didn't really need the thick border line anyway. I just needed a way to make tables and this is one thing I should learn in $\LaTeX$ (along with eqnarray..).[/quote]
lol eqnarray is my favorite command. :lol:

[quote="nsato"][quote="ZennyK"]lol eqnarray is my favorite command. :lol:[/quote]

Actually, if you're using $\LaTeXe$ (you can tell by whether the document begins with documentclass, instead of documentstyle), I recommend using the align environment, or the split environment, instead of eqnarray.  I think it looks nicer.[/quote]
I've switched over to almost exclusively using the align environment - it's just as good as eqnarray but is more powerful.

I don't mean to change the topic or anything, but how do you an align environment?

Here's an example:
[code]
\begin{align*}
a^2 + b^2 &= c^2, \\

x + y &= z.
\end{align*}
[/code]

[quote="nsato"][quote="ZennyK"]lol eqnarray is my favorite command. :lol:[/quote]

Actually, if you're using $\LaTeXe$ (you can tell by whether the document begins with documentclass, instead of documentstyle), I recommend using the align environment, or the split environment, instead of eqnarray.  I think it looks nicer.[/quote]
Thanks! I'll try using that in my solutions for round 2.

[quote="nsato"][quote="chess64"]Okay, but I don't see any difference.[/quote]

You might have to put two examples next to each other.  In the align/split environments, there's not as much space around the equal signs, or whatever you're aligning around.[/quote]
Oh, good. That's one thing I never liked about the eqnarray command; it always seemed kind of weird that there was so much space. :?

Oh, I see it. Thanks! (It's also less to type... :D)


MATHCOUNTS
number theory
Diophantine equation

I had this question during mathcounts, the answer is 10, but I can't think of a way to get to that besides guess and check.

A number, if divided by 2, has a remainder of 0

if divided by 3, remainder of 1

divided by 4, remainder of 2

I assume the question asks for the smallest positive integer? If, so

[hide="Solution"] Call the integer $ y$. Then from the second equation we get that $ y \equal{} 3k \plus{} 1$, for some positive integer $ k$. From the third condition, we get that $ y \equal{} 4x \plus{} 2$, for some positive integer $ x$ (we can ignore the first condition). Then, we have

$ y \equal{} 3k \plus{} 1$

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

This implies that

$ 3k \plus{} 1 \equal{} 4x \plus{} 2 \implies 4x \plus{} 1 \equal{} 3k$

This is a Diophantine Equation with smallest solution at $ x \equal{} 2$, $ k \equal{} 3$. Thus, the smallest positive integer is

$ y \equal{} 4x \plus{} 2 \implies y \equal{} 4(2) \plus{} 2 \implies y \equal{} \boxed{10}$.[/hide]

Another way to think of the question...

[hide="Another Solution"]Let's consider this: If we add 2 to the number, and then divide it by 2, 3, and 4, we would get a remainder of 0 for each answer.

Therefore, what we are now looking for is the $ LCM(2,3,4)$ subtract 2.

$ LCM(2,3,4) \equal{} 12$

$ 12 \minus{} 2 \equal{} \boxed{10}$[/hide]

I will have to remember that in the future,

and oh, yes, it was for the smallest integer

you will see a lot of these problems in mathcounts where the difference is constant.

For example State Sprint a couple of years ago was:

A rectangular band has a certain number of members between 100 and 200. When the band is divided into 8 rows there are 2 spots left unoccupied. When it is divided into 9 rows 3 spots are left unoccupied..how many people does the band have?

Well,what can i do for you? How to get the answer?


algorithm
number theory
prime numbers
number theory unsolved

problem : $d,n \in N$ so that $d|n$ and d prime.proove that $d\leq \sqrt{n}$.

(i think i just wasted eight pages on this and got nowhere,i just
thought i got two solutions,but then looking back i always found
an error in my proof. I remember i used this to make allot of
algorithms,but i never had to proove it,it seemed natural,
in Knuth they say that one should consider the fact
that $n=d\cdot \frac{n}{d}$,by this i thought about a solution
with a friend ,we made it but i feel that its...too complicated)

The way it is stated, it's wrong. Counterexample: let $n=100$, $d=20$. $20|100$ but $20>\sqrt{100}=10$. I think you meant there exists a $d$ that divides $n$ such that $d\le\sqrt{n}$

[EDIT]Well, my wording is too obvious by letting $d=1$. Hmm, maybe this: let $a,b\in\mathbb{Z}^+$ so that $n=ab$. Then prove that $\min\{a,b\}\le\sqrt{n}$. :)

The corrected version doesn't make sense either. Let $d=5$ and $n=10$. $d=5$ is prime but $5>\sqrt{10}$.

[EDIT]If you read the statement I put in my previous post under EDIT, you can realize that's a true statement.

Let $a,b\in\mathbb{Z}^+$ so that $ab=n$. Prove that $\min\{a,b\}\le\sqrt{n}$.

Proof: Without loss of generality (WLOG) let $a\le b$. Now, we have to prove that $a\le\sqrt{n}$. We use a simple proof by contradiction. Assume that $a>\sqrt{n}$. Then,
\[ n=ab\ge a^2>(\sqrt{n})^2=n \]
a clear and easy contradiction. Thus, $a\le\sqrt{n}$.

10000th User i am really sure that when trying to find if a number n is prime or not,
you would try to see if there is a number d smaller than the square root of n so that
d divides n.
i'm really sure of that.
about your last post,for sufficiently big n the property works

I think I understand what you are trying to do. You want an algorithm to test whether $n$ is prime by testing all primes $p\le\sqrt{n}$.

Let's get our ideas organized. From my previous post, the proof I gave was essentially that if a number is composite, then one of its positive factors is less than or equal to $\sqrt{n}$. The statement I proved works for any positive integer $n$.

Now, take the contrapositive of my statement. If none of the positive integers less than or equal to $\sqrt{n}$ divide $n$, then $n$ is prime.

So it suffices to only test prime numbers less than or equal to $\sqrt{n}$. If none of these divide $n$, then $n$ is prime.

Example: let $n=101$. OK we know 101 is a prime number but for this example let's assume we don't know if it is prime or not. Now, $10<\sqrt{101}<11$, so we only need to test the primes 2,3,5,7. After you test, you realize none of these prime numbers divide 101, so therefore 101 is prime.

[quote="spx2"]so it follows that all prime divisors of n are in the interval $[1,[\sqrt{n}]]$ ?[/quote]No, this is something else different, check again the example I gave before, i.e. $n=10$. $2\in [1,\sqrt{n}]$ and $5\in [\sqrt{n},n]$. Nowhere in my previous posts says that all prime divisors of $n$ are in that interval, this is a false statement. Perhaps you want to know what happens to prime numbers larger than $\sqrt{n}$ like 5|10 ?

Look at the statement I wrote before. Now I challenge you this:

Let $n$ be a product of two primes $p,q$ and given that $p<q$. Prove that $q>\sqrt{n}$.

Here is a second related problem to this discussion:

Prove that if all the prime numbers in $[1,\sqrt{n}]$ do not divide $n$ evenly, then so does all the prime numbers in
$[\sqrt{n},n]$. Then show that in this case $n$ is prime.

If you can do the above problems, you are good to go.

so the algorithm is based on this:
if a number n is composite then it must have a factor in $[1,\sqrt{n}]$.

about the first problem :
$n=pq ,\ p<q \ | \cdot q \Rightarrow  q^2 >n \Rightarrow  q>\sqrt{n}$

about the second problem what do you mean by "divide n evenly " ?

That's probably an American slang, sorry. It means "divide $n$ exactly". So if all primes $p\le\sqrt{n}$ do not divide $n$, then all primes $q>\sqrt{n}$ do not divide $n$.

ok i understood..now about the 2nd problem,
if there is a p prime number in $[\sqrt{n},n]$ that divides n,
then $n/p$ must be in $[1,\sqrt{n}]$ and it must also divide n.
but we started from the hipothesis that none of the numbers in
$[1,\sqrt{n}]$  divide n. so :) there it is.

Excellent job, and thus from 2nd problem, we conclude that $n$ is prime. :)

If there are any questions, post here, I'll check back.


Of the following sets of data the only one that does not determine the shape of a triangle is:

$ \textbf{(A)}\ \text{the ratio of two sides and the included angle} \\
\qquad\textbf{(B)}\ \text{the ratios of the three altitudes} \\
\qquad\textbf{(C)}\ \text{the ratios of the three medians} \\
\qquad\textbf{(D)}\ \text{the ratio of the altitude to the corresponding base} \\
\qquad\textbf{(E)}\ \text{two angles}$

[hide="Click for solution"]
$ \boxed{\textbf{(D)}}$ is incorrect because we can "shift" the altitude over, thus changing the shape of the triangle.
[/hide]

It can't be E, because then the triangles would be similar by AA.


combinatorics unsolved
combinatorics
Combinatorial games
IMO Shortlist
game strategy
game

Two players play alternatively on an infinite square grid. The first player puts an $X$ in an empty cell and the second player puts an $O$ in an empty cell. The first player wins if he gets $11$ adjacent $X$'s in a line - horizontally, vertically or diagonally. Show that the second player can always prevent the first player from winning.

It is known that a "$k$-in-a-row" game is a draw for $k\geq 8$, with a simple pairing strategy for $k\geq 9$. See [url]http://www.weijima.com/index.php?option=com_content&view=article&id=11&Itemid=15[/url]. The $5$-in-a-row is also called Gomoku, and its outcome is not known. $4$-in-a-row (and less) are a win for first player.

Divide the infinite grid into staggered columns of 3x3 grids, and treat them as separate tic-tac-toe games where you can force a draw. To get 11-in-a-row, it's clear that should it occur in a column or row, one must have 3-in-row in one of those 3x3 grids. After drawing it out, we can see that at most ten can be placed in a diagonal so that none of the 3x3 grids has 3 diagonally in a row.

[hide = Solution] Our intuition leads us to first consider a finite grid. Most easy to work with being a square. Playing around with various squares, we begin to develop a strategy to tie in an odd square board. (Boards of the form $(2k + 1) \times (2k + 1)$ for some integer $k$).

[b][u]Lemma[/u]: [/b] We have a winning strategy for $(2k + 1) \times (2k + 1)$ boards described as follows. If the first player $A$ puts an $X$ in a corner of a square board, then reflect over the line $"y = x"$. If $A$ marks an $X$ on an edge that is on the top or bottom of a square board, $B$ places an $O$ to the reflection of $X$ over the $y$-axis. If $A$ marks an $X$ on an edge that is on the left or right of a square board, $B$ places and $O$ to the reflection of $X$ over the $x$-axis.

[hide = Proof of lemma] We induct on $k$. For $k = 1$, we actually have a slight variation in the figure below. If $k = 2$, then this strategy holds true. Now assume this is true for $k - 1$. We show it is true for $k$. If $A$ marks and $X$ anywhere that isn't an edge then this is simply the case for $k - 1$ or smaller. Now if $A$ marks an $X$ in a corner, $(\pm 2k + 1) \times (\pm 2k + 1)$, then we just swap the signs and now it is guaranteed $A$ cannot make any diagonal of length $2k + 1$.

If $A$ marks an $X$ on an edge from the $x$-axis, $(x, \pm 2k + 1), x \neq 0$, then mirorr about the $y - axis$. This guarantees he cannot make a row of $2k + 1$ adjacent $x$'s along that row. If $x = 0$, then mirror about the $x$-axis.

Similarly if $A$ marks an $X$ on an edge from the $y$-axis, $(\pm 2k + 1, y), y \neq 0$, then mirror about the $x$-axis. If $y = 0$, mirror about the $y$-axis.

This always guarantees that $A$ can never fully make a row consisting of $2k + 1$ adjacent $X$'s. Thus we have that this strategy holds for the case of the $(2k + 1) \times (2k + 1)$ board. $\blacksquare$ [/hide]

With this lemma we can now implement the following strategy in the case of infinite boards: If $A$ places $X$ somewhere, $B$ should imagine the smallest board of the form $(2k + 1) \times (2k + 1)$ centered at the origin such that $X$ lies on its edges or corners. We then have using the above lemma that we have a winning strategy. $\blacksquare$ [/hide]

Sorry my solution is wrong. Don't read it.

Consider the following 10 x 10 square. There is one pair of numbered squares in every row, column and diagonal of length 5 or more. Repeat this square indefinitely. Then any line of 11 adjacent squares must include a numbered pair. Whenever the first player plays on a numbered square, the second player plays on the other number of the pair.

 n   1     2     3    4     5     6     6     7   n
 8   8     1     2    3     4     5     9     7  10
11  12  12   n   13   14   15   9    10  16
11  17  18  18  13   14   15   n    16  19
17  20   n   21   22   23   24  24  19  25
20  26   n   21   23   22   27  27  25  28
26  29   n    n     n     n     30  30  28  31
29  32  33   n     n     n     n    34  34  31
32  35  33  36   37  38    39  40  41  41
 n  35  42   42   36  37    38  39  40   n

(put this numbers respectively in a 10*10 grid and follow the strategy that I have represented earlier.)
(if you check the grid you'll see that it doesn't matter what you want to put instead of {n}.)
[Note that to prove the result for 11 in this way the square can be at most 10 x 10 (not 11 x 11).]


[u][i]    Problem: [/i][/u]
  [i]
   Solve the equation

   $ 4^{a} \plus{} 9^{a} \equal{} 1 \plus{} 4 .6^{a \minus{} 1 }$[/i]

 [i]Wait  for a  Nice - short and  creative  solution [/i] :wink:

This is the very interesting problem
[hide]
Let  us take $ m\equal{} 2^a$ and $ n\equal{}3^a$
then our equation becomes $ m^2\plus{}n^2\equal{}1\plus{}\frac{2}{3}mn$
we can rewrite above as $ (m\plus{}1)(m\minus{}1)\equal{}n(\frac{2}{3}m\minus{}n)$
now for $ a >\equal{} 0$ R.H.S is always negative and L.H.S is positive,
and for $ a < 0$ L.H.S is negative and R.H.S is positive.
Hence for no value of a the equation is satisfied.
 [/hide]

It is easy to prove, that equation had unique real solution $ \minus{}0.5<a<0$.


inequalities
function
real analysis
inequalities proposed

Prove the following inequality for all positive real numbers $ a,b,c$ and $ k\equal{}3\sqrt[3]{4}\minus{}2$:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}k\frac{ab\plus{}bc\plus{}ca}{a^{2}\plus{}b^{2}\plus{}c^{2}}\ge 3\plus{}k$

[quote="evarist"][color=purple]Hi 10 maths  :D ! Can you show me a expression of A,B,C denote it is M such that :
     $ M\ge\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}$
[hide] M doi' xung' em nhe' cam on em nhieu`[/hide][/color][/quote]

Sorry you, but I have no inequality such that except $ M \equal{}\frac{a^{3}\plus{}b^{3}\plus{}c^{3}}{abc}$. I think when we let $ a\to 0$ then $ VT\to\plus{}\infty$, so it's hard to find M satisfying the condition you've given. But I will think about it.

[hide] xin lỗi anh nhé nhưng em không có bài nào cả, em chưa nghĩ đến TH này, nếy M ở vế kia thì có khá nhiều nhưng M như trên thì em chưa có, em sẽ suy nghĩ về nó.[/hide]

[quote="john_terry"]this ineq's so hard .... Do you use " SS " ? I want to see your solution  :)  and.....sorry but .... in your solution , Do you use " chia de tri " ??.... I think you must do it because this ineq so hard...  :D[/quote]
I think that you mean to use SOS,not SS.What does "chia de tri" mean? :maybe:

Sorry... " S-S " means " S.O.S + Shur " ... "

[quote="Erken"]I think that you mean to use SOS,not SS.What does "chia de tri" mean? :maybe:[/quote]

$ \text{SS}$-method :

We must prove that : $ \text{A}_{cyc}(a,b,c)\ge\text{B}_{cyc}(a,b,c)$ then we rewrite it in the form:

$ M(a\minus{}b)^{2}\plus{}N(a\minus{}c)(b\minus{}c)\ge 0$ which $ M,N\ge 0$ and suppose $ c \equal{}\min(a,b,c)$  :wink:

And "chia de tri" is Vietnamese . It is a strong method, created by Kimluan- member of VMF and MNF ( two sites about maths in Vietnam )

Thank you very much :lol: But how could this problem be solved using $ SS$?

Well, I think it is not so good to mention too much about the name of method, without a solution, except its author gives an official announcement. It may annoy some people maybe. Here we have tens posts without any real analysis :(

Personally, I know this problem of 10maths, since he sent this to my book before. I don't solve it  :blush: I just read the solution he did, and my feeling is that: It is so so nice and difficult ;)

That is why I want to say "Congratulation, 10maths..."

My solution, example 5.
You can do some hard problem by using the similar way.

=====
Sorry for my bad English, if you don't understand, please PM to me.

Your solution so nice ...
Can your way solve the problem? :
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca})^k$ With $ k\equal{}\frac{4}{5}$

Please help me !!  :D

TO Doviettuan: I think it's not hard, I'll try to do it.
TO everyone: Sorry, I had a mistake in my file attached here. That's the notation $ X_{ct}$, I don't remember that it's still in Vietnamese. For explaining:
Consider the function $ f(X) = AX^2 + BX + C$ then we have $ f^' (X) = 2AX + B$ thus the minimum value of the above function occurs when $ X = \frac { - B}{2A} = X_{ct}$
 :oops: $ ct$ abbreviates for "cuc tieu" in Vietnamese, and it means [i]minimum value[/i] in English :oops:

:D  wow!! I thinks it hard and I have 2 solution for that problem ... I want to see your solution

I think the inequality you gave is wrong, try that:
$ a \equal{} 5.644734398,b \equal{} 0.153118687,c \equal{} 0.023098808$ then $ LHS \minus{} RHS \equal{} \minus{} 4.43390211$
The inequality is still true for $ k \equal{} \frac {2}{3}$, it was posted by toanhocmuonmau, I have solved the inequality in that case, I will post my solution next Thursday.

Sorry for later!
I have fixed some mistakes in the file I posted before, and the above problem is the example 17 now.

[quote="Altheman"]I like the packet and the technique a lot. I enjoyed studying it.

Hello, I think this would be a nice identity to add [seeing as how you use it almost every step]:
Let
$ A \equal{} a^2b \plus{} b^2c \plus{} c^2a$; $ B \equal{} ab^2 \plus{} bc^2 \plus{} ca^2$

Then we notice
\[ B \minus{} A \equal{} (a \minus{} b)(b \minus{} c)(c \minus{} a)
\]

\[ A \plus{} B \equal{} (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \minus{} a^3 \minus{} b^3 \minus{} c^3
\]
Hence
\[ \boxed{2A \equal{} \left[ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \minus{} a^3 \minus{} b^3 \minus{} c^3\right] \minus{} \left[ (a \minus{} b)(b \minus{} c)(c \minus{} a)\right]}
\]

\[ 2B \equal{} \left[((a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \minus{} a^3 \minus{} b^3 \minus{} c^3\right] \plus{} \left[ (a \minus{} b)(b \minus{} c)(c \minus{} a) \right]
\]
Another comment: we can also treat certain equality cases easily. If $ (a \minus{} b)(b \minus{} c)(c \minus{} a)\le 0$, then we still retain all equality cases from before because equality must occur when $ (a \minus{} b)(b \minus{} c)(c \minus{} a) \equal{} 0$ for that case.[/quote]

Thank you, the idea you given is also mine, note that we can always tranform cyclic expression as the following form:
$ f(a,b,c) \plus{} g(a,b,c)(a \minus{} b)(b \minus{} c)(c \minus{} a)$ in which $ f(a,b,c)$ and $ g(a,b,c)$ are two symmetric expressions.
for example:
$ a^3b \plus{} b^3c \plus{} c^3a \equal{} \frac {1}{2}\left(\sum\limits_{sym}a^3b\right) \plus{} (a \plus{} b \plus{} c)(a \minus{} b)(b \minus{} c)(a \minus{} c)$
...
Sorry for my bad English, but I don't understand this:
[quote="Altheman"]
Another comment: we can also treat certain equality cases easily. If $ (a \minus{} b)(b \minus{} c)(c \minus{} a)\le 0$, then we still retain all equality cases from before because equality must occur when $ (a \minus{} b)(b \minus{} c)(c \minus{} a) \equal{} 0$ for that case.[/quote]
For example, consider the first example in my file I posted here. Equality occur when $ a \equal{} 2,b \equal{} 1,c \equal{} 0$ and their permutations. For the last one, equality holds for $ a \approx 0.483998187, b \approx 0.301572589,c \approx 0.214429222$ which is not $ (a \minus{} b)(b \minus{} c)(c \minus{} a) \equal{} 0$ ( I means that the equality doesn't occur only when $ (a \minus{} b)(b \minus{} c)(c \minus{} a) \equal{} 0$, and of course the last example equality holds also for $ a \equal{} b \equal{} c$). In the last example, for $ k\equal{}k_{max}$, if $ (a\minus{}b)(b\minus{}c)(a\minus{}c) \le 0$ equality holds only when $ a\equal{}b\equal{}c$ and we don't retain all equality cases.
Thank you.

P/s: sorry for mistake, but in the last example in my file, $ t \equal{} \sqrt {3 \minus{} 6q}$, not $ t \equal{} 3 \minus{} 6q$

sorry again, but  there's still a little important mistake in the uploaded file, in the beginning of page 2, it's $ \delta \le 0$ instead of $ \delta \ge 0$ as shown in the file.

Another problem.
Find the greatest constant $ k$:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}k\left(\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\right)^{\frac{2}{3}} \ge 3\plus{}k$


logarithms
trigonometry
LaTeX
function
calculus
calculus computations

Bita=Beautiful :pilot:

$lim_{x\rightarrow0}\frac{(2^{sinx}-1)ln(1+sin2x)}{xArctanx}=?$

The limit will be $2\ln 2.$

For $x$ near zero, $2^{\sin x}-1$ looks like $x\ln 2$ and $\ln(1+\sin2x)$ looks like $2x,$ while $\arctan x$ looks like $x.$

$\LaTeX$ tip: for "named" functions like sin, ln, or arctan, put a slash before them: \sin, \ln, \arctan. Just be careful to put in a space if you need to: \sin x works, but \sinx doesn't.


geometry
3D geometry
pyramid
AMC
AIME
tetrahedron
Pythagorean Theorem

Consider the paper triangle whose vertices are (0,0), (34,0), and (16,24). The vertices of its midpoint triangle are the midpoints of its sides. A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid?

[color=cyan]This is a fun question.  AIME #15 from 1999.  And I know that without looking it up -- pretty cool, huh?[/color]

This "pyramid" as you call it is in fact a tetrahedron, rite?

A pyramid is any figure consisting of a polygonal base and triangular faces drawn from the base to a vertex. A tetrahedron is a pyramid.

[color=cyan]Which means the answer to your question is "yes."[/color]

Well, the following is a hint, but not a fabulously helpful one:

[hide]We know what the area formula for a pyramid is.  We can figure out what the area of the base is.  So all we need is the height.  So you have to figure out where that top vertex will be in order to find the height of the tetrahedron is.  More hints will follow, maybe.[/hide]

Well I got as far as figuring out the midpoints, but there I'm stuck. I'm pretty sure you use the Pythagorean Theorem to figure out the height of the base triangle, then use it again to figure out the height of the whole pyramid...never mind, Joel finished his post before me. The vertex is obviously [hide]the apices on the original triangle, probably  [/hide]

School starts tomorrow, so I have to go to bed now. I'll post more about this later, if no one posts them before then.

Yes, that's a good start.  But you need to figure out where that top vertex lies if you intend to apply Pythagorean Theorem the second time.  Following is a much bigger hint, if you really want to do this on your own, don't read it: So, you need to [hide]trace the paths that the vertices as you fold the tetrahedron up, and see where they all connect and such.  That wasn't phrased well, but I hope you know what I mean.[/hide]

Once you've solved this question, read this for another problem.  But if you haven't solved the problem, don't read, because it basically contains the solution:
[hide]This problem was doable because as you folded the tetrahedron up, the vertex ended up lying over one of the edges.  Describe the set of triangles for which this would happen.[/hide]

Question (spoiler):

[hide]But how could I possibly trace the paths of the vertices?[/hide]

i believe u need to trace them in ur head....

Spoiler answer to fiery's spoiler question:
[hide]Just consider one of the three that gets folded up for a moment.  "Folding" is not some random procedure -- it happens in a special way, in this particular case with respect to an edge of the triangle that will become the base of out tetrahedron.  So, you need to figure out what this path looks like.  You could even try folding something a few times and seeing what happens to the corner.  (You do, after all, get paper on the AIME).[/hide]

In spoiler:

[hide]Well, it appears that folding the three triangles in this question involves moving their vertices (the three that meet at the top) along the altitudes to the sides of the base triangle. Thus, finding where they intersect would be a matter of finding the intersection of the altitudes on the base triangle. Not sure how to do this though given only the sides lengths...[/hide]

[color=cyan](1) Given only the side lengths, the triangle is uniquely defined and therefor the point you want is uniquely defined.  I never said you wouldn't have to do any work!

(2) The problem-writers were kind enough to slap the triangle down into the Cartesian plane, so that makes the job easier.[/color]

Solution:

[hide]Ok, so I found the equations for the lines that for the base triangle. Using those, I then found the equations of the altitudes, and then the intersection of the altitudes, which was at (16, 12). I found the area of the base triangle (using Heron's) to be 102, and the altitude of the side triangle with vertices (8,12),(16,24), and (25, 12) to be 12 (that is, the altitude from the point (16,24)). Using V=(1/3)bh, I found the volume to be 408.[/hide]

Thanks for the help Joel.

You're missing an explanatory step:  why does Quote:[hide]the intersection of the altitudes, which was at (16, 12)[/hide] lead to the conclsion that you can [hide]simply plug in the altitude of the top triangle as the altitude of the tetrahedron?[/hide]

Response:

[hide]Because, as you had stated, that intersection is along one of the sides of the base triangle. Thus, because the top vertex of the tetrahedron is also the top vertex of the triangle with vertices (8,12),(16,24), and (25, 12), the height of that triangle is the height of the tetrahedron.

By the way, if the vertex of the tetrahedron was not above one of the segments of the base triangle, would this problem have been realistically solvable (by that, I mean without having to memorize obscure tetrahedron formulas)?

[/hide]

[color=cyan]Well, maybe.  It would definitely be messy.  But I bet you could do it.  Give it a try!  (I will too, just be sure to remind me if I haven't said anything more in a week or so :))[/color]


Let $ n$ belong to the set of positive integers.

Prove that:

$ 1 \minus{} n^2 \plus{} \frac {2n^3}{n \plus{} 1} \minus{} \frac {n^4}{(n \plus{} 1)^2} > 0$

This is the correction.  Thank you for showing an error.

$ n\equal{}1 \Rightarrow 1\minus{}1^{2}\minus{}\frac{2(1)^{3}}{1\plus{}1}\minus{}\frac{1^{4}}{(1\plus{}1)^{2}}\equal{}\minus{}1.25< 0$

Is there a typo with the inequality sign?

[quote="Arrange your tan"]Let $ n$ belong to the set of positive integers.

Prove that:

$ 1 \minus{} n^2 \plus{} \frac {2n^3}{n \plus{} 1} \minus{} \frac {n^4}{(n \plus{} 1)^2} > 0$

[/quote]

This is the correction.  Thank you for showing that there's an error.

[hide="Proof"]
Multiply through by $ (n\plus{}1)^{2}$ (a positive number) to get $ (n\plus{}1)^{2}\minus{}n^{2}(n\plus{}1)^{2}\plus{}2n^{3}(n\plus{}1)\minus{}n^{4}\equal{}2n\plus{}1>0$.
Since $ 2n\plus{}1>0$ when $ n>0$, we are done.[/hide]


Does anybody know how to use OpenGL in C# (under Windows) ? I've heard about / tried CsGL but couldn't make it work, do you guys have any experience with it or know about some other way to have OpenGL functionality in C#?

For [b]C++[/b], you have to download the header files "gl.h", "glu.h", "glaux.h" and link the libraries "opengl32.lib", "glu32.lib", "glaux.lib". These are found everywhere on the web.

I don't know how to do it in C#, but I'm guessing it's pretty similar. You might want to post this on
[url=http://www.gamedev.net/]www.gamedev.net/[/url]
This is a very good game programming forum and they'll probably be able to answer your question much better and faster.

.NET provides some stuff for openGL

http://sourceforge.net/project/showfiles.php?group_id=33241

And gamedev is also good as philip suggested.

Yes, I've played with OpenGL in C++ for some time now but the libraries in C# seem to be a little different. Doo posted a link to CsGL and some examples - that's pretty much the only resource on C# OpenGL that I know about too. I'll have a look on gamedev.net then... thanks to both of you.


calculus
integration
induction
trigonometry
calculus computations

You don't need to use power sreies, amirhtulsa :D

You are right, erdos. :)

Let $x=\cos \theta$, we have $\int_0^1 (1-x^2)^n dx=\int_0^{\frac{\pi}{2}} \sin ^{2n+1} d\theta\ (n=0,1,2,\cdots)$

which is like Wallis' integral.


[quote="Vasc"]If $a,b,c$ are the side lengths of a triangle, then

$\frac 1{5a+b+c}+\frac 1{5b+c+a}+\frac 1{5c+a+b}\geq \frac 2{7}(\frac 1{b+c}+\frac 1{c+a}+\frac 1{a+b})$.[/quote]
Let $a=y+z,$ $b=x+z$ and $c=x+y,$ where $x,$ $y$ and $z$ are positive numbers. Then
$\frac 1{5a+b+c}+\frac 1{5b+c+a}+\frac 1{5c+a+b}\geq \frac 2{7}(\frac 1{b+c}+\frac 1{c+a}+\frac 1{a+b})\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{1}{6y+6z+2x}\geq\frac{2}{7}\cdot\sum_{cyc}\frac{1}{2x+y+z}\Leftrightarrow$
$\Leftrightarrow(7x+7y+7z)\cdot\sum_{cyc}\frac{1}{3y+3z+x}\geq(4x+4y+4z)\cdot\sum_{cyc}\frac{1}{2x+y+z}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(3y+3z+x)\cdot\sum_{cyc}\frac{1}{3y+3z+x}\geq\sum_{cyc}(2x+y+z)\cdot\sum_{cyc}\frac{1}{2x+y+z}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(\frac{3x+y+3z}{x+3y+3z}+\frac{x+3y+3z}{3x+y+3z}-2\right)\geq\sum_{cyc}\left(\frac{x+2y+z}{2x+y+z}+\frac{2x+y+z}{x+2y+z}-2\right)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{4(x-y)^{2}}{(3x+y+3z)(x+3y+3z)}\geq\sum_{cyc}\frac{(x-y)^{2}}{(2x+y+z)(x+2y+z)}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x-y)^{2}(x+y-z)(3x+3y+z)(x+y+2z)\geq0.$
Let $x\geq y\geq z.$ Then $S_{z}\geq0,$ $S_{y}\geq0$ and $S_{x}+S_{y}=$
$=(x+y)(x-y)^{2}+6z^{3}+13(x+y)z^{2}+4(x^{2}+4xy+y^{2})z\geq0.$
Id est, it's proved. :)

Nice solution Arqady.  :lol:
We can also work directly with $a,b,c$ in SOS method. :)


otro problemita : sea $n$ nuemeros donde hay $p$ de estos repetidos ¿de cuantas formas se pueden permutar $q$ elementos?

Creo que es asi: primero permuto los n numeros con p elementos repetidos, numero de formas es $n!$/$p!$, de los cuales es necesario solo escoger q de estos lluego el numero de formas es ${n!/p!\choose q}$


advanced fields
advanced fields theorems

The following is probably trivial... You may assume choice.

Let S be a chain. Suppose that for every x, y in S the set T={u | x < u < y} is countable. Can S be uncountable?

Cheers.

The chain of all countable ordinals is uncountable.


20. A gum ball machine contains 9 red, 7 white and 8 blue gum balls. What is the least number of gum balls a person must buy to ensure getting four balls of each color?

Can you guys post solutions with your answer?

[quote="Quickster94"]For these kind of problems, u always need to think of the worst case scenario.  So, the worst case scenario, is that u get all the 9 red balls first, then the 8 blue, and then finally 4 white.  9+8+4=21 gumballs[/quote]

Oh... Okay, I got it.

[hide="My Solution"]
So in order to make sure you get 4 of each color, you have to max out the 2 largest and get 4 in the smallest.
So $ 9\plus{}8\plus{}4\equal{}21$.
Therefore, the answer would be 21.
[/hide]

[quote="BlackSwordsman"][hide]Consider the worst possible case:

You get all the reds. After that you get all the blue's. after that all you need is 4 whites, so the maximum number is

$ 9 \plus{} 8 \plus{} 4 \equal{} \boxed{21}$[/hide]
edit: o.o..i was only typing for a second  :|[/quote]

I know.. They are really fast typers, aren't they?

[quote="tpulak"]Hi there:
I agree with what Quickster94 said: to always think of  the situation that makes you swerve the most, in which this case, you have to add all the gumballs, so its 9+7+8, which come up with your final answer:21[/quote]

You mean 9+8+4=21..right?

Whoops, i'll make sure that doens't happen again :(

[quote="ZhangPeijin"][quote="DonkeyKong"][quote="tpulak"]Yeah, sorry for my mistake :( , and thanks for the correction:so 9+8+4=21[/quote]
No problem.[/quote]
These threads aren't a chat fourm, please don't write responses such as "Ok" "Sure" "No Problem"... etc.
 :lol:[/quote]

Actually, they kinda are, but yeah, it isn't a good idea to use up one response to say one of those


Three pencils and a jumbo eraser cost $ \$1.24$. Five pencils and a jumbo eraser cost $ \$1.82$. No prices include tax. In cents, what is the cost of a pencil?

Subtracting the first equation from the second gives 2 pencils cost 58 cents. Thus one pencil costs $ \boxed{29}$ cents.

[hide]The answer is 29  Use elimination to solve this problem[/hide]

[hide=More graphic solution]
[b]Step 1: Assign Variables[/b]
Have $p$ be the cost of a pencil and $j$ be the cost of a jumbo eraser.
[b]Step 2: Set up Equations[/b]
We have $2$ equations:
$$3p+j=1.24$$
$$5p+j=1.82$$
[b]Step 3: Elimination[/b]
$$3p+j=1.24$$
$$5p+j=1.82$$
$$\implies (3p+j)-(5p+j)=1.24-1.82\implies -2p= -.58\implies \frac{\cancel{-2}p}{\cancel{-2}}=\frac{-.58}{-2}\implies \boxed{p=29}$$
[/hide]


What is the ratio of the area of a square that is inscribed in a regular octagon to that of the area of the regular octagon?? Express the final answer in simplified radicals.

If this is what you mean by inscribed, refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1398535#1398535]this[/url].


linear algebra
matrix
function
calculus
integration
linear algebra unsolved

So I'm really stumped with this:

So I have some matrix $ A$ in $ S^m$ where $ A$ is negative semidefinite and has rank $ r < m$.
Now consider matrices $ X$ in $ S^m$ in a neighborhood of $ A$ (not necessarily neg. sdf.)
By $ S(X)$ denote the eigenspace of $ X$ corresponding to its m-r largest eigenvalues.

Let $ P(X)$ denote the orthogonal projection matrix of $ X$ onto $ S(X)$.
($ P(X)$ can be given explicitly by $ P(X) = E(X)E(X)^T$, where $ E(X) = [e_p_ - _r_ + _1(X),...,e_p(X)]$, where $ e_i(X)$ is the eigenvector corresponding to the eigenvalue $ lambda_i(X)$)

What I want to understand is why $ P(X)$ is supposedly a twice cont. differentiable function of $ X$ in a suff. small neighborhood of $ A$ (in fact even analytic, but I don't need this). Does this follow generally from $ P(X)$ being a symmetric operator? Can I see this directly from my explicit form for $ P$?

I'm not sure about how to prove $ C^2$ directly, but analyticity immediately follows from the fact that the projection in question is just the Cauchy integral of the resolvent over a small contour containing $ 0$. In general this gives you the spectral projection (i.e., the operator that is the identity on some eigenspace and $ 0$ on the complementary eigenspace, but, since all your matrices are self-adjoint, it is the same as the orthogonal projection you asked about).

Okay, thanks a lot. That was a big help.

I looked a bit into your suggestion and it seems this is really the way this is done in the literature. Sadly, I couldn't find anything on a more elementary approach either.

I know this is old, but I still have a question on this actually:

So I understand that given some fixed matrix $ A$ I can express the orth. projection onto some eigenspace of $ A$ as a Riesz-integral (which is the Cauchy integral over a contour containing only the eigenvalues in question). I can also see easily that the resolvent is analytic in terms of the variable in the resolvent set of $ A$.

Now, if for some operator $ T$ I formulate my perturbation as $ T(x) \equal{} T \plus{} xT' \plus{} x^2T'' \plus{} ...$, then for its resolvent I can show analycity in terms of $ x$, and use some kind of power series representation of this resolvent to prove analycity of $ P(T(x))$

However, what I don't know is if this works for my actual problem, since my problem features arbitrary perturbations of $ A$. Any ideas?

Also, I would be interested in calculating the differential of such a projection $ P$.

Thank you for any help.


I got this forwarded to me from someone I know, with the line "Why you don't tick off an Engineer who calls trying to straighten out his phone bill. " In it was a scanned copy of a check, payable to a phone company for this amount:

\[\$\ .002+e^{i\pi}+\sum_{n=1}^{\infty}\frac1{2^{n}}\]

LOL!

message is too small . please make the message longer before submitting.

[quote="tjhance"]geez that's such a small phone bill  :rotfl:[/quote]

I think that was the disparity between the actual phone bill and the bill according to the engineer?


Let $G$ be the centroid of a triangle $ABC$ and let $d$ be a line that intersects $AB$ and $AC$ at $B_{1}$ and $C_{1}$, respectively, such that the points $A$ and $G$ are not separated by $d$.
Prove that: $[BB_{1}GC_{1}]+[CC_{1}GB_{1}] \geq  \frac{4}{9}[ABC]$.

We can transform the inequality $ [B B_{1} G C_{1}]$ + $ [C C_{1} G B_{1}]$ $ \geq$ $ \frac {4}{9} [A B C]$
$ LHS$ = $ 2 [G B_{1} C_{1}]$  + $ [B B_{1} C_{1}] \plus{} [C C_{1} B_{1}]$ = $ 2[G B_{1} C_{1}]$ + $ 2[B_{1} M C_{1}]$ = $ 2 [G B_{1} M C_{1}]$ = $ \frac {2}{3}[A B_{1} M C_{1}]$
We put $ \delta$ = $ [B B_{1}M]$+ $ [C C_{1} M]$ M is the middle of $ BC$
The inequality becomes $ \frac {2}{3}([A B C] \minus{} \delta)$ $ \geq$ $ \frac {4}{9}[A B C]$ $ \Leftrightarrow$ $ [ABC] \geq 3\delta$
 taking the projections of $ B_{1} C_{1} on BC$  to be $ B_{2} C_{2}$   computing $ \delta$ we will get $ \delta \equal{} \frac {B_{1}B_{2} * \frac {BC}{2}}{2}$  + $ \frac {C_{1}C_{2} * \frac {BC}{2}}{2}$ after this we replace $ B_{1} B_{2} \plus{} C_{1} C_{2}$ with $ 2*D D'$ where $ D$ and $ D'$are the middle of $ B_{1} C_{1}$ and his projection on $ BC$ the only thing left to prove is that $ DD' \leq \frac {1}{3}h_{A} \equal{} h_{G} , h_{A} height from A$ on  $ BC , h_{G}$ similar not dificult,have in mind here that $ d$ passes under $ G$ , $ D$ will always be under the paralel to  $ BC$ through $ G$


1. Several sets of prime numbers, such as ${7,83,421,659}$, use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?
(A) 193
(B) 207
(C) 225
(D) 252
(E) 477

2. A base-10 three-digit number $n$ is elected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of n are both three-digit numerals?
(A) 0.3
(B) 0.4
(C) 0.5
(D) 0.6
(E) 0.7

3. Let $n$ be a 5-digit number, and let $q$ and $r$ be the quotient and remainder, respectively, when $n$ is divided by $100$. For how many values of $n$ is $q+r$ divisible by 11.
(A) 8180
(B) 8181
(C) 8182
(D) 9000
(E) 9090

[hide="Answer to 1"]The hardest digits to use will be the non-prime digits (1, 4, 6, 8, 9).  The even numbers will have to be 10's digits to create prime numbers.  In the 40's, the first prime is 41, which also gets rid of the 1.  In the 60's, the only prime is 67, which eliminates 6 and 7.  In the 80's, there are 2 primes to consider:  83 and 89.  Because the 3 can be used by itself, you want to make it 89.  The remaining digits are all prime (2, 3, 5).  That makes the sum of the numbers $2+3+5+41+67+89=207$.  Therefore, the answer is $\boxed{B}$.[/hide]

[hide="Answer to 2"]Find the lower limit, which will be $100_{11}$ in base 10.  100 is the lowest 3-digit number in base 11.  $100_{11}=121_{10}$  Then find the upper limit, which is $888_9$.  888 is the largest 3-digit number in base 9.  $888_9=728_{10}$.  There are $728-(121-1)=608$ 3-digit numbers that work.  There are a total of 900 3-digit numbers.  $\frac{608}{900}\approx0.66$, which is closest to $0.7$.  Therefore, the answer is $\boxed{E}$.[/hide]

[hide="Answer to 3"]To find if a number is divisible by 11, add the alternating digits and see if the difference is divisible by 11.  When you divide by 100 and take the remainder, you are asking for the last 2 digits.  The first 3 digits are the quotient.  The problem is really asking how many 3-digit numbers are divisible by 11.  The lowest is 10,010.  The highest is 99,990.  Find the number of multiples of 11 in that range.  $\frac{99,990-(10,010-11)}{11}=\frac{89991}{11}=8181$.  That makes the answer $\boxed{B}$.[/hide]

On Number 1, you messed up.  I agree with you rlist of primes, but $2+3+5+41+67+89=207$ so the answer is B.  You said that they summed to 225.


[quote="themonster"]How huge a disadvantage does a late application put you at?  Due to problems with my teachers e-mail (and my laxadaisicalness in dealing with it), mine was just barely submitted.  I solved 1-5,6b,7, and 9.  I also had some insights on 8.  Do I have any chance?[/quote]

Here's how we handle late applications.   From the pool of on-time applicants, we will admit enough students to fill the camp.  Inevitably, there are students who decide not to attend (Mathcamp
is not the only program to which they're applying), and late applicants will be considered for these spots alongside the remainining on-time applicants.

[quote="JGeneson"]I could not find out this answer in other topics I looked at, so here is question: Is it too late to apply late? Furthermore, is Monday too late to apply late? Thanks for the help.[/quote]

Any time in May is generally not too late to apply late.  However, I suspect we will take very very few late applicants this year.

Sorry for being clueless, but I have another late application question:

It seems the site does not want to upload my solutions file (at least it has not been happy the last couple days when I have tried it). My solutions are images, and I have compressed the file so that it is around 18 MB. Does anyone have any advice for how I could shrink this more or get the site to upload it? Is there a fax number that I have not been able to spot? Any help would be greatly appreciated. Thanks.

There's a fax number somewhere on this forum. I'm not quite sure where it is, but there is one. And you can always snail mail it.

You could try to fit your file on an e-mail and send it that way.

its only late after the dealine of April 26

[quote="Hellstar"]When you say late applications, do you mean applying after April 26th, or just applying very late, but still before April 26th?[/quote]

Note, you are replying to a thread from 2005. In fact, if I remember correctly, they couldnt even come close to accomodating all the on time applications last year, and no one got in who applied late.

[quote="lfm"]In fact, if I remember correctly, they couldnt even come close to accomodating all the on time applications last year, and no one got in who applied late.[/quote]

We had to interpret the deadline somewhat generously, because of problems with the application site that prevented many people from submitting their applications by the precise deadline, but this is basically accurate.  We've got our fingers crossed that the web application problems are fixed for this year, so one should probably take the deadline quite seriously this year.


Solve the system of equations:
$\begin{cases}\tan x-\tan y-\tan x.\tan y=1\\ \cos 2y+\sqrt{3}\cos 2x=-1\end{cases}$

Hint: first equation rewrite: $sin{x}.cos{y}-sin{y}.cos{x}=cos{x}.cos{y}+sin{x}.sin{y}$ or $sin(x-y)=sin[\frac{\pi}{2}-(x-y)]$,...


geometry
3D geometry
tetrahedron
geometry unsolved

Let $SABC$ be a tetrahedron with $SA\perp SB\perp SC$ satisfying the condition $SA+SB+SC=k$, $k$ is constant. Find the locus of the centroid $G$ of the triangle $ABC$.

[quote="kunny"]The answer is the plane $x+y+z=\frac{k}{3}.$[/quote]

Did you use analytic? :|  Please post a full solution


What is the minimum value of $x \sin x$ where $0 \leq x < 2 \pi$?
Maximum?

Dude... this is beastly...

I can't even get an exact answer using calculus...

Same, the answer doesn't come out nicely, more like a random question.

We have to solve the equation $x\cos x+\sin x=0$ and we get two roots $x_{1},x_{2}$. Playing around with trigonometry, the equality $x=-\tan x$ yields
the max/min:
$f(x_{i})=x_{i}\sin x_{i}=\frac{(-1)^{i+1}x_{i}^{2}}{\sqrt{1+x_{i}^{2}}}$, with approximate values $x_{1}\approx2.0287578381$ and $x_{2}\approx4.9131804394$


Is there a book or a site where i can find methods and/or lists to beyond Alg 1 factoring?
e.g.
(a^2)(c)+(b^2)(a)+(c^2)(b)-(a^2)(b)-(c^2)(a)-(b^2)(c)
   =(a-b)(b-c)(c-a)

$=a^{2}c+b^{2}a+c^{2}b-a^{2}b-c^{2}a-b^{2}c$
$=a^{2}c-a^{2}b+ab^{2}-abc+bc^{2}-b^{2}c-c^{2}a+abc$
$=a(ac-ab+b^{2}-bc)-c(b^{2}-cb+ac-ab)$
$=a(b-a)(b-c)-c(a-b)(c-b)$
$=a(b-a)(b-c)-c(b-a)(b-c)$
$=(a-c)(b-a)(b-c)$
$=(a-b)(b-c)(c-a)$

hmm lol i expanded it first and saw that the term "abc" cancels out..but anyway the trick with this is to split up the terms and usually add and subtract another key term

[quote="Gyan"][hide]One can observe, the expression is zero if $a=b$ so $a-b$ is a factor, similarly $(b-c)$ and $(c-a)$ are also factors. Since it's a third degree, we have
$LHS=k(a-b)(b-c)(c-a)$ one can easily deduce that k=1 (For example, put a=0, b=1, c=-1, and you have 2=2k or k=1)[/hide][/quote]

Factor Theorem? :clap: Quite elegantly done...

It's also a symmetric polynomial, isn't it?

[quote="mathnerd314"]It's also a symmetric polynomial, isn't it?[/quote]

No. Intuitively, a polynomial that is symmetric shouldn't change when you permute the variables. So if you switch $a$ and $c$, you should not get another polynomial. That's not the case here.

In fact, it's anti-symmetric: transposing any two variables changes the sign of the polynomial.

Interesting fact: if you take [i]any[/i] anti-symmetric polynomial in the variables $a, b, c$, it is equal to the product of this polynomial with some symmetric polynomial.

What if you are asked to factor a symmetric polynomial? is there any tricks or theorems that deal with that?


Given positive integer $n (n \geq 2)$, find the largest positive integer $\lambda$ satisfying :
For $n$ bags, if every bag contains some balls whose weights are all integer powers of $2$ (the weights of balls in a bag may not be distinct), and the total weights of balls in every bag are equal, then there exists a weight among these balls such that the total number of balls with this weight is at least $\lambda$.

\[\lambda = \lfloor \frac{n}{2} \rfloor + 1\]

[hide = Solution]
We claim that the answer is $\lambda = \left \lfloor \frac{n}{2} \right \rfloor + 1$. First, we will establish that this is a lower bound. Firstly, we observe that any sum of $m$ as a sum of powers of two can be obtained from $m$'s binary representation by applying the operation $2^i \rightarrow 2^{i-1} + 2^{i-1}$ repeatedly. For example, $7 = 2^1 + 2^1 + 2^1 + 2^0$ can be obtained from $7$'s binary representation $2^2 + 2^1 + 2^0$ by simply replace $2^2$ with $2^1 + 2^1$. The proof of this is easy by induction. Notice that this claim allows us to assume, WLOG, that the sum of the weights of the balls in every bag is a power of $2$. The reason we can do this is because if not, we can just take the balls which "split off" of one of the powers of two in the binary representation (as in the result above) of the common sum of the bags. So suppose that the sum of the balls in every bag has weight $2^k$. Then, we know that $\lambda ( 2^k + 2^{k-1} + \cdots + 2^1 + 2^0) \ge n * 2^k \Rightarrow \lambda \ge \frac{2^kn}{2^k + 2^{k-1} + \cdots + 2^1 + 2^0} > \frac{n}{2}$, hence establishing that $\lambda \ge \left \lfloor \frac{n}{2} \right \rfloor + 1$.

Now, all that remains is to construct an example where $\lambda = \left \lfloor \frac{n}{2} \right \rfloor + 1$ is achieved for all $n \in \mathbb{N}$. Write $2^k \le n < 2^{k+1}$, and let $\lambda' = \left \lfloor \frac{n}{2} \right \rfloor + 1$. Then, consider writing $2^k$ $\lambda'$ times in a row, and then $2^{k-1}$ $\lambda'$ times in a row, $\cdots$, and finally writing $2^0$ $\lambda'$ times in a row. Then, let's draw a divider so that between any two dividers the sum is zero. It's easy to show that this "works" since $2^i \lambda' > n$ for $i \in \mathbb{N}$ and $\lambda' (2^0 + 2^1 + \cdots + 2^k) = \lambda' (2^{k+1} - 1)  \ge \frac{n+1}{2} (2^{k+1} - 1) \ge n * 2^k$. The last part follows since $\frac{n+1}{2n} = \frac12 + \frac{1}{2n} \ge \frac12 + \frac{1}{2(2^{k+1}-1)} = \frac{2^k}{2^{k+1}-1}$.

Example:

$$8|8|8|8|8|44|44|422|22211|111$$
$\square$
[/hide]


linear algebra
matrix
probability
LaTeX
vector
probability and stats

Given the transition matrix below, find all communicating classes, classify all states, and discuss the
long run behaviour.

(1/4) (1/8)  (0) (5/8)
(1/3)  (0)    (0) (2/3)
(1)     (0)    (0) (0)
(1/2)  (0)   (1/2) (0)

States 1,2 and 4 communicate as well as they can. (In fact, each one can go to each of the others in one step.) But state 3 is an absorbing state. And since there is a pathway (specifically from state 4) from the set {1,2,4} to state 3, the result will be that in the long run, the system will wind up in state 3 with probability 1.

Useful exercise: treat each positive value in that matrix as if it were 1 and draw a picture of the resulting directed graph.

So, the classes are {1,2,4} and {3}. Are all of these states ergodic?

Kent, it looks to me like you've misread the transition matrix.  (Either that, or I'm misreading it.)  In particular, the way I read it, state 3 immediately empties into state 1, and it's not possible to get from state 4 to state 2 in only one step.

BrainMan, it would be very helpful if you learned a little bit of [[LaTeX]] (see also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 ).

Oops, yes I misread the matrix. In the long run, all four states will be populated. The long-term behavior is to approach the vector

$ \begin{pmatrix}\frac{48}{105}&\frac{6}{105}&\frac{17}{105}&\frac{34}{105}\end{pmatrix}.$


Is there a certificate of participation for those who participate in the USAMO?

Yes, the AMC office creates and sends a Certificate of Participation with the USAMO results and the solutions.  This material will be sent sometime in late May, after the grading is complete and all results have been compiled.

Steve Dunbar
MAA Director, American Mathematics Competitions