geometry algebra polynomial ARML email AMC AIME

[quote="bugzpodder"]on the contrary, i believe toronto have all those math circles (including the one w/ Barbeau and Rice that dished out the last question in Euclid before the contest!!)
and  all those toronto area competitions as well as tourney of towns which are olympiad trainings that massey doesnt have.  massey only focuses on pre-olympiad stuff (up to AIME/Euclid, not so much USAMO/CMO/APMO).  Personally, I was not aware of olympiads at all until I somehow qualified for CMO in grade 10.  As for Olympiad team, Michael lipnowski claimed that he was placed in the top 6 in terms of olympiad score, but he did not get into the team simply because he was grade 12 while most other guys on the team are grade 11 (w/ except of jacob+janos?)[/quote]

I totally agree.

I think Michael was not selected, for the reason that the canadian team leaders wanted to place younger students to gain experience for this years IMO.  It is so sad(if that is the truth).

Another thing I was so dissapointed, is the fact that no student born in the year 1986 (and maybe 1985) never made it on the canadian IMO team.  Why?

Secondly, Toronto and Calgary (canada's second largest english-speaking city) and most large cities have so much resourses in math olympiads that other cities don't.  Just look at the last three IMO teams...where are these students from?   The only way we are introduced to olympiad math is through the Internet (Olymon).  In fact the city I live in... the word Olympiad doesn't exsist... I don't think I even knew about it until my senior year.  For example, the chances of a student growing up or living in a small city and making the team is so low if not impossible, unless they are Immigrants who brought olympiad resourses with them from home.

This is something I hope will change soon...espicially with websites like this one, that have so much resources.

I totally agree with you... everyone should be offered the same opportunities to develop their problem solving.  What upsets me is that cheating on the 2003 TLE eliminated the 2004 TLE, so some students may have been invited to the Winter Camp, completely based on their success in the Olymon program... which is outrageous since on the average, I can easily find half of the solutions on the Internet, and the other half in books I have at home. (Which is why I quit Olymon back in 2003.)

What happened to original problems and mathematical ingenuity?  It's not fair that people with resources get even more opportunities to improve while others stay in the dark... I'm lucky enough to attend an excellent math school so I shouldn't complain... but how about others across Canada?

So I feel that the math competition committees should try to propose the most creative and original problems that they can.  That way, no student with access to math books and previous contests should have a substantial advantage.

Can you believe that Open2002#4 was taken from a Singapore contest, Open2003#4 was taken from Barbeau's Polynomials Book, and even Euclid2004#10 was taken from the 102 Combinatorial Problems Book (question#42/51)? How is this fair? Hopefully, this year's CMO and Euclid will consist of totally original problems... every student across the country should have the same chance of succeeding, even if they can't attend an amazing school or afford amazing math books...

I hope someone agrees with me... lol... gotta go watch football...

This is getting off-topic, so I've split the thread.

As for original problems, no one can deny the fact that problems get recycled. It is extremely difficult to create good and original problems. Heck, even IMO 2004 #3 was known for a long time (but then again, how many contestants solved it? Not many).

If any contestent had actually carefully gone through every single problem in the history of math Olympiads, he might as well deserve a gold medal anyway.

The sad thing is, and this doesn't just apply to math olympiad, there are rules for everything, and we have to play by them. It may not up to us to decide what rules follow. And yes, I am talking about Olymon. Shaun isn't the only person that was disadvantaged because he chose to quit Olymon. As trivial as the program is to us more experienced competitors, we still have to play it.

And just for the record, I've never gone to the Toronto math circle (though I might just go the week before Euclid this year :lol:)

By the way, mak, who are you?

I'm not too sure of the winter camp invitation thing.  Did anyone at all got invited this year because of Olymon only?  With the absence of TLE, most people who got invited were those ones who did good on last year's CMO.  (people who got 3 questions and more?)  Barbeau emailed me that Olymon's only subsidiary while the real contests (COMC) mean a lot more. :?

As for the Toronto math circle, I've been there lots of times this year.  Just before COMC we started talking about recurrences and stuff.  Barbeau even talked about the fibonacci sequence, which might have been helpful on COMC last question.  :P   But then again, he personally didn't know it would be on the contest himself.

dam bill you split the topics so it looks like i was the one who's complaining by starting this topic.  it is NOT my intention so please at the least change the topic, otherwise you are forcing me to delete my original post.
lotus, the point is that in TLE, the cheaters gets the chance to do the problem set just like OLYMON (while you dont), which means they have resources such as books, internet (if they are smart enough, the'd post the whole problem set on artofproblemsolving waiting for someone to solve it for them :D), or even writing computer programs to find all the patterns!  for example, 2004 TLE day 2 question 1 is question 2 in "problems with elementary solutions", day 1 question 2 I saw it on the net...

[quote="anipoh"]I'm not too sure of the winter camp invitation thing.  Did anyone at all got invited this year because of Olymon only?  With the absence of TLE, most people who got invited were those ones who did good on last year's CMO.  (people who got 3 questions and more?)  Barbeau emailed me that Olymon's only subsidiary while the real contests (COMC) mean a lot more. :?[/quote]

From what I gather, this year's winter camp invitations are based on three factors: Olymon, CMO and COMC.

I was invited for Olymon + CMO, but certain people got invited just for CMO.. (notice the plural form!)

It seems like I should know all of you, but I really have no clue who you guys are.. other than Yufei..
Mind telling me who you are??

And what's this with Toronto math circle discussing contest problems before the actual contest?

Hey you don't know me lol? Shaun? Ever since I quit Olymon two years ago, people don't seem to know me anymore... it's ok though... hey Jack, don't forget TLE 2003 Day 1 Problem 3!

I'm not sure about the guy giving away the answer to Euclid2004 #10 but if that story is true, he should be fired from the contest committee... there is enough bias and cheating going around already... we don't need anymore.

Here at Massey, we prepare for the open as soon as October arrives, and we go through hundreds of different problems... and even though our teacher is on the Open Committee, he would never ever consider giving us a hint to solve one of the problems, or even a similar one! (It might just be because he can't remember them lol...)

I really enjoy solving math problems and competing in math contests and I know I shouldn't complain about the bias and corruption in canadian math contests but... last year at my school, two people... Jack and John, who were both amazing problem solvers who could have made the imo team... ended up in group 3 for the cmo, when I'm sure they solved 4+ problems each (including a mistake here and there)... and I met Michael at the Waterloo Seminars... he definitely should have been on the imo team...

I'll leave this for you guys to think about... does it really help our country to send grade tens to the imo so that they gain experience, when there are senior students who are more deserving to go?

P.S. I'm not talking about Yufei... he's one of the best!

i thought number_cruncher was shauns...  ??  my bad

> I'm not sure about the guy giving away the answer to Euclid2004 #10

as I know Barbeau did it unintentionally and he is not on the committee (or maybe he is but he doesnt know about it)... just a really big coincidence i guess.  What he did was gave out an old CMO, which contained that question.  Larry Rice was there, he knew about it, but didnt say anything.  maybe he should've said something but cant really blame him much either.

> last year at my school, two people... ... ended up in group 3 for the cmo, when I'm sure they solved 4+ problems each (including a mistake here and there)...

well, i am definitely not nearly as qualified as michael as far as the IMO team goes.  As for my score on CMO... I could never figure it out.  my best take is that I got
1 on question one (for a pitful mistake near the end that produced an incorrect solution as a result of a typo) and 0 on the last question (three/four people in our school got the same (wrong) answer and all of us got 0/1).  so I probably deserve group 3 anyways  :D
I cant complain much since the report for Q1 said that wrong answers only got 0 or 1...  maybe they are changing marking style  :(

Hmm.. Shaun, I don't think people not knowing who you are is [i]because[/i] you're not doing olymon..

As for Barbeau + Rice giving out some similar problems, I don't believe it was intentional. Dr. Rice also teaches at Waterloo math circle, and many times he has clearly stated that he did not want to steer anyone towards or away from the problem, so he prefers to teach thngs totally unrelated to the contest when we're near the contest.. I mean, he has no reason to favor Toronto over Waterloo, and if he was going to give out problems, he would have done that for us in Waterloo too. As I recall, we studied circle geometry and polynomials for the five weeks before the contest.. I think.. :blush:

And it really is sad for the IMO, I'm in grade 11, and I'm already worrying about me being too old..  :D
Hopefully they want some new grade 11s there too.. I also agree with all of you deserving to be on IMO! you guys are soooooo good!!   :)

[quote="mak"]
Secondly, Toronto and Calgary (canada's second largest english-speaking city)
[/quote]

Pshh, Vancouver is bigger and better than Calgary.

I can't believe that Barbeau and Rice would give away the problem intentionally... I'm sure it was just a coincidence... I think Larry Rice is a really cool guy.

I think we shouldn't worry and just enjoy this year's contests... plus, at massey we're writing twice as many american contests than canadian ones this year...

Hey, does everybody know about the canadian version of ARML? It's being held in London... in June I believe... I hope a lot of you are planning on attending and representing your high schools... it's the first year so it should be interesting!  It's pricey though so prepare to get yourselves some sponsors!  Spread the word!

[quote="mak"]

Secondly, Toronto and Calgary (canada's second largest english-speaking city) and most large cities have so much resourses in math olympiads that other cities don't.
[/quote]

Hmm, I know this is a bit biased, but believe it or not, Calgary doesn't have nearly as many resources as Toronto.  NOT NEARLY as much.  I should know, lol.  If it did have a bit more maybe, I might not have dropped out of the Olympiad streams...

I agree with Shaun, smaller cities towns don't have a lot unless you try to make the trip to other places.  And living in the middle of no where, our school doesn't have anything in terms of math prep, I do it on my own, and that didn't get me far.  hehehe

[quote="lotus87"]Hey, does everybody know about the canadian version of ARML? [/quote]

Guess what? Waterloo's planning on going!

Btw, mak, who are you???

a ha ha…. guys, guys… come on…. relax! :lol:  :lol:  :lol:

[quote="blahblahblah"] Pshh, Vancouver is bigger and better than Calgary.[/quote]
First of all Calgary is the second-largest English-speaking city in Canada…go to http://www.mongabay.com/igapo/Canada.htm and check it out for your self!  Vancouver is something like 7th…ha ha ha!

Secondly,
[quote="lotus87"]I totally agree with you... everyone should be offered the same opportunities to develop their problem solving.  What upsets me is that cheating on the 2003 TLE eliminated the 2004 TLE, so some students may have been invited to the Winter Camp, completely based on their success in the Olymon program... which is outrageous since on the average, I can easily find half of the solutions on the Internet, and the other half in books I have at home. (Which is why I quit Olymon back in 2003.)[/quote] “There is an Honour System in higher-level education(this includes Olympiads) here in Canada, and it is expected that every student would respect and honor it!?…this is quote from one of Canada’s greatest math profs… please pass it on.

Those that are cheating are just screwing themselves…and the chances of making the IMO or getting a descent score on Olympiads is like impossible for them and further they are likely to do the same in University where they really get screwed!  I can think of many instances in the newpaper where academic dishonesty has come back and haunted them…one that comes to mind is the “child genius? Justin Chapman’s Mother, check out   http://cnnstudentnews.cnn.com/2002/fyi/teachers.ednews/03/26/boy.genius.ap/ .

Olymon was created to help us prepare for Olympiads, and to raise the standards for Canadian Math Olympiads:
1.  Olymon will help give feedback on writing solutions and learning new math for beginners.
2.  For those of us who have started to do well on Olymon…you wouldn’t get much feedback…the purpose for you to submit your solutions is to give a signal to CMS and tell them that you are active in math-olympiad problem solving and would like to be considered as a candidate for the IMO.

[quote="lotus87"]
What happened to original problems and mathematical ingenuity?  It's not fair that people with resources get even more opportunities to improve while others stay in the dark... I'm lucky enough to attend an excellent math school so I shouldn't complain... but how about others across Canada?

So I feel that the math competition committees should try to propose the most creative and original problems that they can.  That way, no student with access to math books and previous contests should have a substantial advantage.[/quote]
As Yufei mentioned it already, it is so hard to come up with new and original problems.  Believe me, I am trying to do so for some contests and it is very difficult.  Therefore it is almost impossible for barbeau to do so seven times every single month!  As yufei mentioned earlier, just play by the rules and do the Olymon...think of it as an assignment.

About the selection process in general for Canada:
The thing is that 10-years from now, we are the ones running the selection process and believe me…a lot of changes are going to happen!  For winter camp selection, if the organizers are planning to bring back the TLE, I suggest they should just instantly email the exam and after the time is over, the students should then fax in their solutions.  Secondly, I am predicting that some time in the future the COMC will be broken into two exams just like the US (AMC then AIME->USAMO) and UK(UKMTC then BMO1->BMO2) , so selection to the winter camp and the CMO will be more accurate.

[quote="Sunny"]...And living in the middle of no where, our school doesn't have anything in terms of math prep, I do it on my own, and that didn't get me far.  hehehe[/quote]
[quote="jpark"]And it really is sad for the IMO, I'm in grade 11, and I'm already worrying about me being too old...[/quote]
You two are two of the smartest people I ever met.  Just keep up the good work and you'll be on the IMO. :)  There is still one and a half years left... Who knows?  ;)  Next years IMO team will be the first Canadian team with more than two females.

[quote="billzhao"]By the way, mak, who are you?[/quote]
Some of you know me some of you don’t.  Lets leave it at that!  I like to remain Mysterious!...ha ha ha ha! :lol:
Here is a hint Jenn… we met at the Wat. Sem. 2003 and I was Andrea(flute guy)’s roommate. :)

Anyways guys...good luck on this years Olympiads.  Hope you will ALL do well!!!  Just keep doing lots and lots of problems! Take Care.;)

Hmm... I think it's weird that you choose to remain mysterious... but ok...

I agree with your idea of instantly sending in solutions to the TLE... that's exactly how I wrote the USAMO last year... if the USAMO committee does not receive the solutions within 15 minutes after the contest has ended, you're disqualified, simple as that (I think...).

But I really don't believe it's that hard to create original olympiad problems... especially with the great amount of brainpower Barbeau and the CMO committee has... but I can see your point.  I guess it's because even though I don't do particularly well on math contests, I dedicate tons of my time to creating problems and it's really a lot of fun, and not as hard as it seems.

If anyone would like to try mine, I can send you some or post some here (let me know)... they're really cool... especially problems involving prime numbers, inequalities and diophantine equations... they're my favs lol...

But anyways, compared to other countries, I think Canada does a pretty good job of selecting the right students to write olympiads and attend imo training camps... I just have one question... if the IMO is a 2-day, 9-hour, 6-problem exam, why does the USAMO have the same format, while the CMO is a 1-day, 3-hour, 5-problem exam?  Since usually the top 3 scoring students on the CMO qualify for the IMO, wouldn't it make sense for the CMO to parallel the IMO?  Does anybody know why the two competitions are so different?  I could never figure that one out...

[quote="lotus87"]Hmm... I think it's weird that you choose to remain mysterious... but ok...

I agree with your idea of instantly sending in solutions to the TLE... that's exactly how I wrote the USAMO last year... if the USAMO committee does not receive the solutions within 15 minutes after the contest has ended, you're disqualified, simple as that (I think...).

But I really don't believe it's that hard to create original olympiad problems... especially with the great amount of brainpower Barbeau and the CMO committee has... but I can see your point.  I guess it's because even though I don't do particularly well on math contests, I dedicate tons of my time to creating problems and it's really a lot of fun, and not as hard as it seems.

If anyone would like to try mine, I can send you some or post some here (let me know)... they're really cool... especially problems involving prime numbers, inequalities and diophantine equations... they're my favs lol...[/quote]

ha ha ha  :)  I like to be mysterious, it is sooo cool!

By the way, yeah I love how they organized the USAMO for us, and I hope they could do the same for the TLE!
Second, I love to see some of your Number Theory problems!

[quote="lotus87"]But anyways, compared to other countries, I think Canada does a pretty good job of selecting the right students to write olympiads and attend imo training camps... I just have one question... if the IMO is a 2-day, 9-hour, 6-problem exam, why does the USAMO have the same format, while the CMO is a 1-day, 3-hour, 5-problem exam?  Since usually the top 3 scoring students on the CMO qualify for the IMO, wouldn't it make sense for the CMO to parallel the IMO?  Does anybody know why the two competitions are so different?  I could never figure that one out...[/quote]

As pathetic as it may sound, the CMO must appeal to the general population as well (by that I mean the 80 who qualify for it). I assume that that most of those people do not practice specifically for Olympiads. That's probably they don't intend the CMO to be too difficult.

If you look at other national Olympiads, you see that the 2-day-9-hour-6-problem format is usually reserved for the stronger countries (e.g. China, USA, Vietnam) and team selection tests (which we don't have).

As for USAMO, did you know that before 1995, the exam used to be 1-day-3.5-hour-5-problems? The format has underwent changes at least two time since then.

[quote="lotus87"]if the USAMO committee does not receive the solutions within 15 minutes after the contest has ended, you're disqualified, simple as that (I think...).[/quote]

actually not as simple.  when they didnt receive a submission, they actually call the school to see what happened and then decide on that.  there is some leniency involved, believe it or not.

[quote]
But anyways, compared to other countries, I think Canada does a pretty good job of selecting the right students to write olympiads and attend imo training camps... [/quote]
Lets just say that what they are doing now is better than they were doing in the past two or three years.

as for "re-use" of problems... I've seen enough to understand its unavoidable.  But it would be desirable if they dont do it intentionally (which i believe they dont) and possibly reduce the chance of that.

[quote]Hey, does everybody know about the canadian version of ARML? It's being held in London... in June I believe... I hope a lot of you are planning on attending and representing your high schools... it's the first year so it should be interesting! It's pricey though so prepare to get yourselves some sponsors! Spread the word![/quote]
The date is May 20-21; the name is (potentially) Mc... um, something.  I forget.  I'm not quite sure how pricey it is - Mr. Griffiths hasn't told us - but it involves staying in residence at UWO overnight.
There will be individual, team, relay, and power questions, along with mental math individual questions, where you're not allowed to write anything on any paper except for an answer.  It promises to be quite fun - and a good time for Windsor/London/Waterloo/Toronto? math students to get together, especially those who don't get invited to Waterloo Seminar or IMO Training Camp.

Also, from what I've heard about Winter Training Camp invitations (as mentioned in previous posts), only two people were invited after COMC - myself, and Viktoriya.  Francis Chung, who came 4th in Canada, was not invited.  This seems to be in keeping with the theme of "Invite the grade tens and grade elevens" (alluded to in other previous posts).  Rather unfortunate, as Francis is an outstanding mathematician.

By the way, mak, do I know you?

[quote]and a good time for Windsor/London/Waterloo/Toronto? math students to get together, especially those who don't get invited to Waterloo Seminar or IMO Training Camp. [/quote]

I'm not getting it...  :?  :blush:

Muwhahaha I finally know who mak is, lol.  Took me a long long time.

Alberta needs an ARML team, I wanna do it!

Francis wasn't at Winter Camp?  I'm surprised.

its quiet funny... if you dont get to go, then you really want to go.  but when you actually do get to go, its really not that great cept for the free food (i feel really down last year at W.C. for some reason i cant remember) unless you like doing math in caf  ;)  no pun intended bill... just had to say it :P :D.  one exception, Andy Liu is the funniest guy - awsome story teller, and he actually taught stuff (no idea how this year's went -- but last years was uncomparable to the year before, they didnt even give free breakfast :P, so...)!!

:D

Yes, I remember talking about math problems over... all meals..
I really enjoyed the camp though.

Unfortunately, Andy Liu wasn't there, or Ian VanderBurgh.

However Dr. Barbeau stayed with us for the entire week, and it was amazing.. I've finally learned inversion  :blush:

[quote="jpark"]:D

Unfortunately, Andy Liu wasn't there, or Ian VanderBurgh.

 [/quote]

Aww, that's too bad.  They're two of my favorite lecturers, along with Dr. Rice, hehehe

[quote]
[quote]Quote
and a good time for Windsor/London/Waterloo/Toronto? math students to get together, especially those who don't get invited to Waterloo Seminar or IMO Training Camp.  [/quote]
I'm not getting it...  [/quote]
er... well, it's always sort of a gamble whether or not you'll get invited to Waterloo Seminar or an IMO training camp, isn't it?  So if you want to see your math friends in Windsor, London, Waterloo, and Toronto, there's more of a guarantee that you'll actually see them if you choose to go to the London math contest, rather than hoping for an invitation to something.  Oh - and the question-mark in my original post was intended to indicate that I was unsure whether or not Toronto was planning on sending any teams.  Did I explain that well enough this time?

By the way, does anyone know anything about some Friar Arthur Holmes Senior Invitational Mathematics Contest?  My school's been invited to write it... I've never heard of it.  Supposedly it's curriculum-based.[/quote]

[quote="geoffreynham"]
er... well, it's always sort of a gamble whether or not you'll get invited to Waterloo Seminar or an IMO training camp, isn't it?  So if you want to see your math friends in Windsor, London, Waterloo, and Toronto, there's more of a guarantee that you'll actually see them if you choose to go to the London math contest, rather than hoping for an invitation to something.  [/quote]
But when you're a non-Ontario person, those camps are the only hope for the year to see your math friends again.

[quote="geoffreynham"]
By the way, does anyone know anything about some Friar Arthur Holmes Senior Invitational Mathematics Contest?  My school's been invited to write it... I've never heard of it.  Supposedly it's curriculum-based.
[/quote]
A quick search on google gave me this. The actual name is "The Father Arthur Holmes Ontario Mathematics Competition" and it seems to be only popular with catholic schools. This website has some sample papers from the last couple of years: http://www.stmichaelscollegeschool.com/hm/inside.php?id=713.

wow, the canadian forum on AoP is much more fun to read than a year ago   :lol:

Just want to know, how does a school register for that Canadian Version of ARML? although I'm in university, I think I can still let my school know.

By the way, I do agree with that people in Toronto and Calgary has more resources to work with than other cities, but maybe people in Toronto does not has much advantage as others imagined... I'm from Toronto and I didn't even know there is CMO/IMO until grade 11. And for someone like Peng Shi, I think he didn't even do Tournament of Towns in Grade 10, and only started to go to Math Circle in last April. Resource he used before is just his Chinese Olympic book  ;)  doesn't seem he took advantage of all resources in Toronto, does it?

Also, there is a painful part being in Toronto, if you are not good enough (like me :) ). I never got invited to Waterloo Seminar, and I believe I should make it at least twice if I'm in an easier region. This is also the case for something like Physics National Camp, only 5 people from Ontario were selected for the national camp, and 3 of them made the IPHO team. (i'm a bit off topic now)

Edmonton is THE resource in Alberta.  Trust me on that one.

Good point. Toronto (also known as Ontario Metro) is always the most competition region in the Waterloo contests. One always needs a top-notch score to receive regional recognition in Toronto.

[quote="bugzpodder"]i believe toronto have all those math circles (including the one w/ Barbeau and Rice that dished out the last question in Euclid before the contest!!)[/quote]

Just to clarify things a bit, this was the day before CMO, and Barbeau was explaining an  CMO problem which is exactly like Euclid #10. So it was not really his fault that the Euclid contest makers chose to make a problem exactly like that CMO question (It was #5 in one of the CMOs in the 1990's).  :-)


For those in ARML, don't forget that the first online meeting is tomorrow: Sunday, 10 September 2006. Hope to see you there! Bring yourself and problems  :D

For those who will not be able to make the meetings, we shall record the conversation (at least, I will). And we can give you a copy of the "transcript."

The meeting will start circa 1:00 PM,

won't be attending this one because of scheduling conflict, but let me know when the next one is.

Hey Karth, let me get the transcript. I'll try to make the next one. When is it?


I need a reply fast--- does even on a non integer mean that its floor is even???

well what i would do is just test some numbers such as 3.14, -1.2, and 2.34.
First 3.14 the integer directly below it is 3  so unless i did it wrong, your statement is already false.

Second, -1.2 the integer directly below it is -2 so that works

Third, 2.34 the integer below it is 2 so that works

If i messed up somehow on the first one, tell me please!

[quote="joml88"]Well, this is really random and far off topic...  But his statement holds true for nonnegative numbers.  However, that's basically the definition of the floor...so I'm not sure what he's asking.[/quote]

As suggested above 3.12 returns 3 and thus is not even.
Or didn't I get the original question? ^^
=)

Non-integers aren't even. You can divide any non-integer by 2, and for all non-integers you will get a second non-integer factor. Your question doesn't really make sense (or I'm misreading something).

Even is a subset of integers that are divisible by 2 and odd is the subset of noneven integers...

[quote="mysmartmouth"]All integers are even unless they are odd.[/quote]

 :roll:

[quote="Apprentice"]What about 0??
Isn't zero not even and not odd? (double negative!)[/quote]
I think 0 is even.

It's even because it differs by $2$ from an even number (namely from $2$).

Are non-integers considered either even or odd?? :?:

No. The property of "oddness/evenness" is defined only for integers.

So, the question doesn't make much sense, does it?


geometry
incenter
geometric transformation
geometry proposed

Let $ ABCD$ be a square and $ P$ a point on $ \overline{CD}$. Let $ X,Y,Z$ be the incenters of $ PDA$, $ PAB$, $ PBC$. Prove that $ P, X, Y, Z$ lie on a circle.

[quote="immodestius"]Let $ ABCD$ be a square and $ P$ a point on $ \overline{BC}$. Let $ X,Y,Z$ be the incenters of $ PDA$, $ PAB$, $ PBC$. Prove that $ P, X, Y, Z$ lie on a circle.[/quote]
How come the $ \triangle PBC$?? Maybe you made a typo.

Dear Mathlinkers,
I think we have to read PDC and not PBC.
I have mentioned recently a developpment of this problem in
http://www.mathlinks.ro/Forum/viewtopic.php?t=282316

In this : http://www.mathlinks.ro/Forum/viewtopic.php?t=218636, I mentioned that the three circles have a common tangent.

For the cocyclicity that I have proved in a next paper on my website :
1. I consider the common tangent of PBA and PDC
2. the quadrilateral that we obtain has an incircle according Pitot's theorem
3. then we are done with a simple angle chasing.

Sincerely
Jean-Louis

The proposition is true for any quadrilateral tangential to a circle:
square at [url]http://www.mathlinks.ro/viewtopic.php?t=273939[/url]
kite at [url]http://www.mathlinks.ro/viewtopic.php?t=218636[/url]

[quote="jayme"]Dear Yetti,
If my memory is well, I think you have worked on the common tangent of the three circles...can you mention the reference?
Sincerely
Jean-Louis[/quote]

Dear Jayme:
The underlying theorem about the common directed tangents of 4 directed circles tangent to a single directed circle is at
[url]http://www.mathlinks.ro/viewtopic.php?t=247411[/url].
The theorem comes from Yaglom's Geometric Transformations, last chapter: "Circular Transformations" of Russian original.
English translation of the last chapter was not published.


For which $ n\in \mathbb{N}$ do there exist rational numbers $ a,b$ which are not integers such that both $ a \plus{} b$ and $ a^n \plus{} b^n$ are integers?

Here is my solution .We only need consider $ n>1$
The first we consider :
$ a\equal{}\frac{x}{p},y\equal{}\frac{y}{q}$
where $ \gcd(x,p)\equal{}\gcd(y,q)\equal{}1$ and $ p,q>0$
The condition equivalent
i)$ pq|xq\plus{}py$
From 1 we have $ q|p,p|q$ so $ p\equal{}q$
So we only need to consider when $ \gcd(x,y)\equal{}1$
Then condition equivalent :
i)$ p|x\plus{}y$
$ p^n|x^n\plus{}y^n$
[u]Case 1[/u] $ n\equiv 0 (\mod 2).$
Because $ \gcd(x\plus{}y,x^n\plus{}y^n)\in\{1,2\}$ do so $ p\equal{}2$
But $ 2^n|x^n\plus{}y^n$ then $ 2|x,2|y$ illogic.
[u]Case 2[/u] $ n\equiv 1 (\mod 2)$
Call p is a prime divisor of n . Then
$ v_p (x^n\plus{}y^n)\equal{}v_{p}(x\plus{}y)\plus{}v_{p}(n)$
If we chose $ v_p(x\plus{}y)\geq v_{n}(n\minus{}1)$ then it satisfies conditon .
So answer is $ n\equiv 1 (\mod 2)$

What is your function $ v$? Is it the power of $ p$ that divides $ n$? Why $ v_p (x^n\plus{}y^n)\equal{}v_{p}(x\plus{}y)\plus{}v_{p}(n)$?

To eay you only chose $ p^n|x\plus{}y$
This lemma posted many time on Mathlinks  :wink:

[quote="TTsphn"]
This lemma posted many time on Mathlinks  :wink:[/quote]

Which lemma are you referring to?

Hi. I was just wondering where you got that problem from? It was in Croatia TST a week ago, but I'm sure it's from some older competition. I'd just like to know which. I've solved it (Wee!) so I'd like to know where it's from (I actually develop a sort of an emotional bond with the not-so-easy problems I solve on important competitions :blush: ). Thanks a lot!

Well, mornik, I took it from Croatia TST  :D  It's in the imo.org.yu website.

Anyway, for the $ n$ odd case, I guess we can just choose $ a\equal{}\minus{}b$ not being integers. But what if we want positive $ a,b$?

Can you help me?

Yes chose $ p^n|x\plus{}y$ and $ \gcd(x,y)\equal{}1$
There exist infinite such pair $ (x,y)$

Simply choose $ a \equal{} \frac {2^n \plus{} 1}{2}$ and $ b \equal{} \frac {2^n \minus{} 1}{2}$ and you got yourself a solution for odd $ n$.


A number is randomly selected from the set $ \{0,3,5,8,10,13,15,18,20\}$. What is the probability that the number is divisible by 3 or 5? Express your answer as a common fraction.

By 3: 4 numbers
By 5: 5 numbers
By 15: 2 numbers
All: 9 numbers
Probability: $ \frac{4\plus{}5\minus{}2}{9}\equal{}\frac{7}{9}$

[quote="JulieLeung"]Why 0?
Please help![/quote]

By definition, $0$ is divisible by any number (but definitely not the other way around).

Then Bugi's solution is an example of the [b]P[/b]rinciple of [b]I[/b]nclusion and [b]E[/b]xclusion, or PIE for short.

On the other hand, Mewto55555's solution is probably easier to understand.

The thing doesn't show the numbers after 13.. You have to scroll to the right


calculus
integration
function
real analysis
real analysis unsolved

if $ f: [0,2]\rightarrow\mathbb{R}$ is defined by $ f(x)\equal{}1$ if $ x\neq 1$ and $ f(1)\equal{}0$.

How do you prove that $ f$ is integrable on $ [0, 2]$ and how do you calculate its integral?

To prove that this function is Riemann integrable, you should fix an $ \epsilon > 0$ and find a partition $ 0 \equal{} x_0 < x_1 < \cdots < x_n \equal{} 2$ of $ [0,2]$ such that the upper Riemann sum and the lower Riemann sum for this partition are within $ \epsilon$ of each other. For this problem, the upper Riemann sum is always equals to $ 2$. You should convince yourself that you can make the lower Riemann sum as close to $ 2$ as you want by chose a partition where $ x_i < 1 < x_{i\plus{}1}$ and $ x_{i\plus{}1} \minus{} x_i$ is very small.

For integrability use the  theorem that states: If the bounded function $ f: [a,b] \rightarrow \mathbb{R}$ has a finite number of discontinuities, then $ f$ is integrable on $ [a,b]$.
For calculation, a hint: $ \int_{0}^{2}f(x)dx\equal{}\int_{0}^{1}f(x)dx\plus{}\int_{1}^{2}f(x)dx$.


Sophia has 16 plants and one window. Only five plants can be placed in the window at any given time. If each of the 16 plants spends the same amount of time in the window during an eight-hour period of sunlight, what is the greatest number of minutes in the sun that is possible per plant?

8 hours is 480 mins, so if you divide that by 16 plants, each plant gets 30 mins of sunlight. Since 5 plants can be placed in the window at a time, each time gets the 30 min period 5 times. So our answer is 30*5, or 150 mins.

Can you offer a timetable that shows how?  If you simply put 4 sets of plants up, five-five-five-one, it's clear that not each plant can have 150 minutes.. that'd be at least 600 minutes.

Certainly 120 minutes is achievable using the naive method of 5-5-1, that's just 2 hour shifts for those four groups.

For these kinds of problems, you do not need a timetable.

The formula is, answer = $ \frac{\text{\# of spots}}{\text{\# of total things}} \times \text{time}$

[size=75][color=red]Used LaTeX to clarify. -Isabella2296[/color][/size]

Ok, well you've shown that you can blindly apply a mysterious formula... but should I conclude you can't tell someone how to do the task physically?

Better yet:  3 spots, 2 things, 1 hour
Answer: obviously each plant can get 1 hour
Formula: 1.5 hours

No idea... maybe 150!  Any way to look it up?

I know for a fact the answer is $ \boxed{150}$.

Everyone can relax!  Yesterday I figured out how to achieve the maximal solution of every problem like this type.

It's probably too far off topic to post though, so PM me if you really want to know how!


Shift the coordinate system from $ (0,0)$ to $ (1,0)$...so let $ z_{1} \rightarrow z_{1}^{'}$ clearly $ z_{1}^{'}\equal{}z_{1}\minus{}1$ and similarly
now $ z_{i}^{'}$ all lie on a circle with centre at origin.So for them $ \sum z_{i}^{'}\equal{}0$ implies the centroid coincides with the circumcentre...so the triangle is equilateral.
so the sides must be of same length so all $ |z_{i}\minus{}z_{j}|$ must be same for $ i \neq j$ and $ i,j \in [1,2,3]$

[quote="pardesi"]Shift the coordinate system from $ (0,0)$ to $ (1,0)$...so let $ z_{1} \rightarrow z_{1}^{'}$ clearly $ z_{1}^{'} \equal{} z_{1} \minus{} 1$ and similarly
now $ z_{i}^{'}$ all lie on a circle with centre at origin.So for them $ \sum z_{i}^{'} \equal{} 0$ implies the centroid coincides with the circumcentre...so the triangle is equilateral.
so the sides must be of same length so all $ |z_{i} \minus{} z_{j}|$ must be same for $ i \neq j$ and $ i,j \in [1,2,3]$[/quote]

Yes vertices of equilateral triangle. Right you are harry.

BTW, why did you shift and consider a circle and all???? Just say that (1,0) is circumcentre and z1+z2+z3/3 (centroid) = 1 and done!


linear algebra
matrix
algebra
polynomial

Use the Spectral Decomposition theorem to prove that the trace of a symmetric real matrix is equal to the sum of its eigenvalues.

let A be any real symmetric matrix.
by the spectral decomposition theorem,
A=PDP' , where P is orthogonal and D is diagonal.
so tr(A)=tr(PDP')=tr(P'PD) (please check this relation!!)
=tr(D)(since P is orthogonal)
=sum of eigenvalues of A.
 :)

You can actually prove this result for every square matrix, i.e., the trace of a matrix equals the sum of its eigenvalues. For this you can use, for example, Schur's lemma, of simple examination of the coefficients of the characteristic polynomial of a matrix.


Can someone help me solve this problem, it is urgent..thanking you in advance..

Find all triangles (determined by the lengths of their sides) with the
following properties:
the lengths of sides are consecutive integer numbers, and
one of the angles is twice as large as another angle.

Assume, $ \angle A \equal{} 2*\angle B$ , in $ \triangle ABC$
$ \Rightarrow a^2 \equal{} b(b\plus{}c)$
Clearly, $ a > b$

Assuming, $ a > b > c$
$ \Rightarrow (x\plus{}1)^2 \equal{} x(x\plus{}x\minus{}1)$
$ \Rightarrow x^2\plus{}1\plus{}2x \equal{} 2x^2\minus{}x$
$ \Rightarrow x^2\minus{}3x\minus{}1\equal{}0$
$ \Rightarrow$ No integer solution.

Assuming, $ a > c > b$
$ \Rightarrow (x\plus{}1)^2 \equal{} (x\minus{}1)(x\minus{}1\plus{}x)$
$ \Rightarrow x^2\plus{}1\plus{}2x \equal{} 2x^2\minus{}3x\plus{}1$
$ \Rightarrow x\equal{}5$

Assuming, $ c > a > b$
$ \Rightarrow x^2 \equal{} (x\minus{}1)(x\minus{}1\plus{}x\plus{}1)$
$ \Rightarrow x^2 \equal{} x^2\minus{}x$
$ \Rightarrow x\equal{}0$

So  Ans : [b] Required sides are a:6 , b:4 , c:5[/b]

guru_bhai,

the work of your last case is wrong.

Using your work from your last case:

Instead, it should be (more steps
included for emphasis):

Assuming c > a > b
=> x^2 = (x - 1)(x - 1 + x + 1)
=> x^2 = (x - 1)(2x)
=> x^2 = 2x^2 - 2x
=> ... 0 =  x^2 - 2x
=> ... 0 =  x(x - 2)
=> x = 2  (This is the value different from yours.)
=> It's a "collapsed triangle" with sides of 3, 2, & 1
    units, so this case gives no triangle.


Let $ a, b, c$ be possitive numbers such that $ a\plus{}b\plus{}c\equal{}3$.
Prove that \[ abc\plus{}\frac{12}{ab\plus{}bc\plus{}ca}\geq 5.\]

third degree Schur's inequality
$ (a\plus{}b\plus{}c)^{3}\plus{}9abc\geq4(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ac)$ and $ a\plus{}b\plus{}c\equal{}3$ --->
$ 27\plus{}9abc\geq12(ab\plus{}bc\plus{}ac)$------>  $ abc\geq\frac{12(ab\plus{}bc\plus{}ac)\minus{}27}{9}\equal{}\frac{1}{3}(4(ab\plus{}bc\plus{}ac)\minus{}9)$ ---->
 $ abc\plus{}\frac{12}{ab\plus{}bc\plus{}ac}\geq\frac{1}{3}(4(ab\plus{}bc\plus{}ac)\minus{}9)\plus{}\frac{12}{ab\plus{}bc\plus{}ac}\equal{}\frac{4}{3}(ab\plus{}bc\plus{}ac)\plus{}\frac{12}{ab\plus{}bc\plus{}ac}\minus{}3\geq2\sqrt{\frac{4(ab\plus{}bc\plus{}ac)}{3}\frac{12}{ab\plus{}bc\plus{}ac}}\minus{}3\equal{}8\minus{}3\equal{}5$

$ a\equal{}b\equal{}c\equal{}1$

[quote="Cyberjunkie"]Can please someone explain to me which is shur's inequality?[/quote]
Let $ a,$ $ b$ are $ c$  are non-negative numbers and $ n\geq1.$  Then
$ a^{n}(a\minus{}b)(a\minus{}c)\plus{}b^{n}(b\minus{}a)(b\minus{}c)\plus{}c^{n}(c\minus{}a)(c\minus{}b)\geq0.$

[quote="arqady"]
[...]$ n\geq1.$  Then[...][/quote]
actually this inequality is true for $ n\in\mathbb{R}$,for example $ n\equal{}\frac{1}{2}$,or even $ n$ could be negative...

Certainly, you are right, BaBaK Ghalebi.
But I think, for the Schur's inequality $ n\geq1$ and $ n\in\mathbb N$ even.
However, it how to agree. :wink:

The problem is equivalent with this one
If $ a\plus{}b\plus{}c\equal{}1$,then $ 27abc\plus{}\frac{4}{3(ab\plus{}bc\plus{}ac)}\geq 5$
By Vo Quoc Ba Can method,let $ ab\plus{}bc\plus{}ac\equal{}\frac{1\minus{}q^{2}}{3}$
and $ abc\geq\frac{(1\plus{}q)^{2}(1\minus{}2q)}{27}$,after expanding we must prove
$ 2q^{3}\plus{}3q^{2}\minus{}2q\plus{}1\geq 0$,while we have $ 3q^{2}\minus{}2q\plus{}1\geq 0$,thus the proof follows. :)


Hoi,

Ik wil via matlab wat trekkingen doen uit een bepaalde kansverdeling. Allemaal leuk natuurlijk, maar om om aan die trekkingen te komen moet ik een aantal inverses functies berekenen. Ik heb het via maple geprobeerd maar dat lukt niet.

Een voorbeeldje:

functie zie plaatje. De functie is op u uit [0 1] en parameter b uit [0 inf) gedefinieerd en het bereik is [0 1]. Ik zou graag een gesloten uitdrukking willen voor de inverse...  kan iemand me helpen hoe ik dat in maple doe??

Alvast heeeell erg bedankt!!!

haha,

ik twijfel of ik dat nog kan na al die tijd...

als ik het 'gewoon' in maple gooi krijg ik een onredelijke functie terug omdat maple denkt dat de functie op de reele rechte is gedefinieerd... Kan iemand pleasse even helpen hoe ik dat doe??

Ik moet er namelijk een stuk of 10 doen en ik ben heel slordig met de hand....

Die functie heeft gewoon geen gesloten uitdrukking voor haar inverse... je kunt er een numerieke benadering voor zoeken, maar daar houdt het ook op.

okay bedankt....
ik dacht dat er mischien nog wel een gesloten uitdrukking zou zijn voor het [0 1] interval met b>0.


calculus
analytic geometry
graphing lines
slope
quadratics
derivative
vector

Find the min value of $\sqrt{x^{2}+y^{2}}$ if $5x+12y=60$.

Are you in anyway restricting $x$ and $y$?  If you don't, then you could easily have imaginary numbers, and if you are discussing if imaginary numbers are less than real numbers, ect, then it is not Getting Started difficulty.

No, find the real value, no complex values.

[hide="Easy one"]Just substitute $y=\frac{60-5x}{12}$ into $\sqrt{x^{2}+y^{2}}$ and use the formulae for the minimum of a quadratic function. If I'm not mistaken, the result is $\frac{60}{13}$[/hide]

[quote="vishalarul"]IF we are restricted to real numbers, here is a really nice way without calculus:
Graph $5x+12y=60$. If we take the circle $x^{2}+y^{2}=n^{2}$, and notice that we want the line $5x+12y=60$ to be tangent to this circle, we can easily find $n^{2}$, and then $n$. Also, the center of this circle is (0,0). We want a perpendicular line through (0,0) perpendicular to $5x+12y=60$. The intersections of these lines will be the desired point:
$m=-\frac{5}{12}$
The line perpendicular to $m$ has slope $\frac{12}{5}$. So the equation of this line is given by: $y=\frac{12x}{5}$. Substituting,
$5x+\frac{144x}{5}=60$
$\frac{169x}{5}=60$
$x=\frac{300}{169}$
$y=\frac{720}{169}$
The calculation is very messy, so I'll let you figure it out.[/quote]
Your solution is very similar to mine, but like you said, yours is a bit messy. We end up with $n=\frac{60}{13}$.
[b]Is there a solution that gives an exact answer?[/b] (your solution is very nice too vishalarul :) )
[quote="Farenhajt"][hide="Easy one"]Just substitute $y=\frac{60-5x}{12}$ into $\sqrt{x^{2}+y^{2}}$ and use the formulae for the minimum of a quadratic function. If I'm not mistaken, the result is $\frac{60\sqrt{2}}{13}$[/hide][/quote]
It is an easy one, but i think you made a mistake Farenhajt.

First derivative, taking into consideration that the minimum of x is at the same x-value as the minimum of $\sqrt{x}$.

 :wink:

Equivalent to Farenhajt's solution.  (Cauchy is neater anyway  :D )  Here is [hide="another"] There is a vector technique that allows you to find the distance from an arbitrary point to an arbitrary line in $n$ dimensions.

Here, our point is the origin.  Let's pick an arbitrary point, say $(12, 0)$, on our line and consider the vector from the origin to that point.  This is $\vec{V}= \left< 12, 0 \right>$.

Let's consider next a vector perpendicular to the line.  We take $\vec{N}= \left< 5, 12 \right>$ (can you see why this vector is perpendicular?)

Now, consider the projection of $\vec{V}$ onto $\vec{N}$.  The vector produced is perpendicular to our given line and has length equal to the perpendicular distance from our point to our line - precisely our desired distance!

The length of this projection is given by

$\frac{ | \vec{V}\cdot \vec{N}| }{ | \vec{N}| }= \frac{12 \cdot 5+0 \cdot 12}{ \sqrt{5^{2}+12^{2}}}= \frac{60}{13}$. [/hide]

My method still works when there are 3 variables, while Farenhajt's way won't work when there are 3 variables (without using calculus :D ). Suppose we have this question:

Find the minimum value of $\sqrt{x^{2}+y^{2}+z^{2}}$ if $ax+by+cz=d$, where $a$, $b$, $c$, and $d$ are constants, and everything else is a variable.

Both the Cauchy and vector methods work for $n$ variables  :wink:

I was referring to the method where he expressed $y$ in terms of $x$, and substituted. The Cauchy and vector methods were really nice.  :winner_second:

Well, all methods lead to the same solution, so they are all fine.
P.S. no more ':wink:'!


Solve the equation arctan |x| - arctan |x - 1| = arctan |2|

Take the tangent of both sides and use the tangent of a sum/difference formula on the left side. It will become an algebra problem.

For solving it, we first need to ensure that Range of L.H.S = Range of R.H.S. So we have to develop a restriction on x.


FTW
Gauss
geometry
3D geometry
MATHCOUNTS

1) Mark Twain and Harry Trumann were born here

2) The population center of the U.S. is in Missouri

3) Brad Pitt is from Missouri

4) Bob Costas lives in Missouri

5) Missouri was the northernmost slave state

6) Missouri is the only state to have a street named Barack Obama Blvd

7) Missouri has the spammiest forum

8) Missouri has the most combined multiple accounts, FTW mutes, and pisses off the admins the most

9) If you are reading this then there is a 90% chance that you dont live in Missouri because that's how popular this forum is

10) You are tempted to post something bad about Missouri right now but the awesomeness of Missouri prevents you from doing so.

And this:

11) tinytim, math154, and Mewto55555 are funny.

12) Because gauss1181 says so.

lol I like #8.

ok I was trying to think of a reason MO pwns just now..but I can't do it.

13>  Because only people from MO can think of good reasons of why MO pwns.  You have to be there!
14>  If you add an o to MO you get moo!

 :|

15) Because suma_milli used to live there when she was little.

16) Because Ladue Middle School pwns. (go tinytim)

Anheuser Busch used to be located here.

First ever Olympics hosted in US was hosted in MO.

Jimmy Connors

Yogi Berra

Scott Joplin

Jackie Joyner-Kersee

Stan Musial

Ozzie Smith

Daniel Boone

OMG, it's Max Schindler!!!!!

and gauss, i'm not funny??  :(

[quote="suma_milli"]If you add an o to MO you get moo!

 :|[/quote]

18. AND, if you add an I to MO, you get... IMO!

Missouri is Emo!!!! :P  :rotfl:  :rotfl:

[quote="math154"]We save our worst roads for Kansas bonding. :P[/quote]

Haha. :P

omg mewto is making more multiple accounts?!

19>  The ice cream cone was invented in Missouri.

Pizza Hut was invented in Kansas. PWNAGE

1. First pancake mix ever was made in Missouri by "Aunt Jeminah."

2. Missouri held the 1904's World's Fair.

3. Starting point of the Oregon Trail.

4. All non-Missourians are jealous

5. Kansas has worse roads! (for pianoforte..)

Kansas and worst roads??

oh yeah, do you guys have the NATIONS ONLY WWI MEMORIAL??

or SPRINTS HEADQUARTERS??

all you guys have is an arch made of steel which sways up to 16 feet in the wind.

oh yeah, do you guys have an boeing, cessna, airbus, etc. plant?

And does your capitol building have the 2nd (or 4th) tallest dome?

AND do you have a person who can do a nationals sprint round and get a 30 in less than 10 minutes?

/me wonders if me should join debate.

and if you are thinking why I am posting in the Muissuori forum, here is why:  I live 1 mile away from Missouri.

Kansas:

* It is illegal for restaurants to sell cherry pie a la mode on Sundays.
* In Wichita, a man's mistreatment of his mother-in-law may not be used as grounds for divorce.
* In Wichita, it is illegal to carry a concealed bean snapper.
* In Lang, it is illegal to ride a mule down Main Street in August, unless the animal is wearing a straw hat.
* In Natoma, it is illegal to throw a knife at anyone wearing a striped shirt.

Missouri

* In Saco, women are forbidden from wearing hats that "might frighten timid persons, children or animals."
* In St. Louis, it is illegal for an on-duty firefighter to rescue a woman wearing a nightgown; in order to be rescued, a woman must be fully dressed.
* While children may purchase shotguns in Kansas City, they are not allowed to buy toy cap guns.
* Missouri considers drunkenness an "inalienable right."
* Bars are allowed to give beer to go.

I think that makes Missouri so much better.

007 math, you stated that the St.Louis Arch is 16 feet tall?  It is actually 630.  And it doesn't sway. :P

1. Missouri invented iced tea.  :P
2. Missouri is one of the most neighbourly states in the Union. That's why everyone loves it and it's considered a pwnage state.  :P
3. Our state contains the 2nd ranked medical school in the nation: Washington University (does that compare with KU?) - ( I don't think so.. :P )
4. The St. Louis Cardinals are better than the Kansas City Royals.  The Rams are also better than the Chiefs.  :P

That's why Missouri isn't nationally considered a [b]boring[/b] state like Kansas.

[/b]

Missouri beat Kansas at 2008 Mathcounts and ARML.

Missouri will beat Kansas at 2009 Mathcounts and ARML.

STL Cardinals won World Series in 2006.

I don't even remember the name of the KS baseball team so they must fail.

Who cares about WWI? It was just a stupid war that resulted from this random unimportant person being shot.

MO has more CIVIL WAR memorials than KS.

MO joined the Union earlier than KS.

Obama and Mccain didn't bother to campaign in KS but it has like 5 electoral votes and is full of n00b farmers that always vote republican.

and much, much more...

p.s.  [quote]do you have a person who can do a nationals sprint round and get a 30 in less than 10 minutes? [/quote]

and who might that be?  :P

It's an adult sadly.  The adult is in Mensa, has probably over 100 trophies from various competitions (probably 60 from MathCOunts), and his living is to teach kids how to pwn missuori.

I think we are going to own you team in this year in nats with our secret weapon!!!(sadly, without me)

The arch [b]does[/b] sway up to 16 feet.  It sways up to 8 feet each direction.

Missuori, on the other hand, made the KKK who came over to Kansas and killed inccoent little people.

Plus, Obama has a house here.  Also, Ex-President Abraham Lincold came here teh day before he was shot.

[size=59][color=white]its all bs for the last 2 facts[/color][/size]

You know that getting four trophies in every MATHCOUNTS competition in your life will only give you $ 36$, right?

AH...now I know what he's talking about... Charles McCalla, mathcounts coach, retired from Pembroke, teaches a mathcounts class in the summer, state coach multiple years, including 1993 National Individual and Team Champions from KS (and 2nd individual I believe)--really has like 60 Mathcounts tropies. Or as my middle schooler from LKMS said, "the fat mathcounts guy who goes to all the competitions".

Trick question... (and sad that I know so much considering I never heard of him until after qualifying for nats in 8th grade, despite my own classmates having been in his class--that actually might be sadder.)

Who is this "secret weapon" you speak of? I have a guess.

007 math, before you contradict the obviously better state of Missouri, please check your spelling and grammer.

I will say it again: Kansas is extremely [b]boring[/b]

@007math

By secret weapon, do you mean the midget indian kid who's like beast at spelling bees, geography bees, mathcounts, and every other academic competition in existence. I bet that he will get top 30 at nats in 6th grade.

Lakewood DOMINATION! He's good, he'll be our third-fifth nats qualifier yet, and I'd be surprised if he did not do very well at nats. LIKE <OMG>.

we'll see, but i'll see to it that if he is not, then he will be next year. and the next.

[quote="math154"][color=red][size=200][u][b]DARN THIS GUY SEEMS LIKE A BEAST[/b][/u][/size][/color][/quote]

Wrong.  [b]Uber[/b] Beast.

lol I could see him fighting with the judges because he missed one question in cd but still got first. Who knows, maybe he will throw a tantrum for missing one question on the sprint.  :rotfl:

[edit] Missouri is pwnage because we're not Guam or South Dakota or West Virginia and we have 4 people from the same school on the nats team.

That means Missouri sucks since it is the only state with only 1 good school. :rotfl:

Or it means that that one school is too beastly...

You're definitely not the only team to have done that...

I'm pretty sure we are... counterexample?

Guam '06
South Dakota '06
West Virginia '06

There are three from the only year I've checked...


modular arithmetic
function
number theory
totient function
relatively prime

There's a pattern:

2,4,8,6,2, etc...

now, 2003 is 3 mod 4, meaning that it will match up with 8.

LaTeXed:

$ {2003} \equiv 3 \pmod{4}$

So 8..

Sorry, I had messed up.

[quote="007math"]LaTeXed:

$ 2^{2003} \equiv 2^3 \equiv8 (\mod{4})$[/quote]

Little correction

$ 2^{2003} \equiv 2^{3} \equal{} 8 \pmod{10} \ne 2^{2003} \equiv 0 \pmod 4$

That [i]is[/i] what modular arithmetic is.

There is a pattern to the units digit of powers of 2.

2^1=[color=red]2[/color]
2^2=[color=red]4[/color]
2^3=[color=red]8[/color]
2^4=1[color=red]6[/color]
2^5=3[color=red]2[/color]
2^6=6[color=red]4[/color]
2^7=12[color=red]8[/color]
2^8=25[color=red]6[/color]
2^9=51[color=red]2[/color]
2^10=102[color=red]4[/color]
2^11=204[color=red]8[/color]
2^12=409[color=red]6[/color]

With powers of 2, the pattern repeats itself every 4 powers.

So, to find the units digit of 2^2003

2003/4= 500 remainder 3

The units digit of 2^2003 will be the 3rd number in the repeating pattern so the units digit of 2^2003 is 8.

Here's another way.
For reference:
- http://en.wikipedia.org/wiki/Euler%27s_totient_function
- http://en.wikipedia.org/wiki/Euler%27s_theorem

$ 2^{2003}\equiv x\pmod{10}$. It's pretty clear that $ x\equiv 0\pmod 2$, so let $ x\equal{}2z$ for $ z\in\mathbb{Z}$. Then, $ 2^{2002}\equiv z\pmod{5}$.

Noting that $ \varphi(2002)\equal{}\varphi(2\cdot7\cdot11\cdot13)\equal{}2002(1/2)(6/7)(10/11)(12/13)\equal{}720$, we have, by Euler's theorem, $ 2^{720}\equiv 1\pmod 5$.

Thus, $ 2^{2002}\equiv (2^{720})^2\cdot 2^{562}\equiv 2^{562}\pmod 5$.
Again, $ \varphi(562)\equal{}280$, so $ 2^{562}\equiv (2^{280})^2\cdot 2^2\equiv 2^2\equiv 4\pmod 5$.

So $ z\equal{}4$ and $ x\equal{}2z\equal{}\boxed{8}$.

In general, [url=http://www.artofproblemsolving.com/Wiki/index.php/Fermat%27s_Little_Theorem]Fermat's Little Theorem[/url], [url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_Totient_Theorem]Euler's Totient Theorem[/url], and the [url=http://www.artofproblemsolving.com/Wiki/index.php/Carmichael_function]Carmichael Function[/url] (listed from least to most powerful) are useful in these sorts of exponentiation questions.

Also useful is the fact that if we want to find $ x\pmod{A_1\cdot A_2\cdots A_{n \minus{} 1}\cdot A_n}$, where $ A_1,A_2,\ldots,A_{n \minus{} 1},A_n$ are relatively prime, it suffices to find

$ x\pmod{A_1},x\pmod{A_2},\ldots,x\pmod{A_{n \minus{} 1}},x\pmod{A_n}$, which follows directly from the [url=http://www.artofproblemsolving.com/Wiki/index.php/Chinese_Remainder_Theorem]Chinese Remainder Theorem[/url].

This technique was used by grn_trtle, in noting that $ 2^{2003}\pmod{10}$ could be found by separately finding $ 2^{2003}\pmod2$ and $ 2^{2003}\pmod5$.

Actually, I'd quite forgotten about Fermat's Little Theorem  :blush: Bit out of practice with modular arithmetic.
Once we have $ 2^{2002}\equiv z\pmod 5$, we can note that $ 2^{4}\equiv 1\pmod 5$ by Fermat's Little Theorem. We have $ 2002\equiv 2\pmod 4$, so then $ 2^{2002}\equiv 2^2\equiv 4\pmod 5$.
That saves you the trouble of using Euler's totient theorem twice.

Also, $ \lambda(5)\equal{}4$, so $ 2^{\lambda(5)}\equal{}2^4\equiv 1\pmod 5$ by Carmichael's theorem, and we arrive at the same result.

[quote="DaReaper"]how do you find the units digit of 2^2003?[/quote]

Look at the pattern: 2^n= 2, 4, 8, 16, 32, ... The units digit repeats 2, 4, 8, 6, 2, 4, 8, 6...etc. So, since $ 2^1\equal{}2, 2^2\equal{}4, 2^3\equal{}8, 2^4\equal{}16$ , and so on, $ 2^{2000}$ ends with a units digit of 6. $ 2^{2003}$ ends with a units digit of 8. Answer = $ 8$

[quote="grn_trtle"]

Noting that $ \varphi(2002) \equal{} \varphi(2\cdot7\cdot11\cdot13) \equal{} 2002(1/2)(6/7)(10/11)(12/13) \equal{} 720$, we have, by Euler's theorem, $ 2^{720}\equiv 1\pmod 5$.

[/quote]

Euler's theorem states that $ 2^{720} \equiv 1\pmod{2002}$. The fact that $ 2^{720} \equiv 1\pmod 5$ is probably a coincidence, as 720 is a multiple of $ \varphi(5) \equal{} 4$.


LaTeX integration Alcumus AoPSwiki geometry

how do you take notes involving math signs wheter in maths or physics or whatever ?

I started using Word to save on paper, for now it's taking a lot more time than it would take to do this on paper, I'm just wondering if it's worth it or not.

I know several people who use a tablet PC, which basically allows you to write on the screen with a little stylus, and the computer saves the material, in your handwriting, digitally.  I've seen some people do exactly what you're doing and just bullet notes on Word and in one case, I saw someone who's computer could listen to the lecture and take notes for him.  I have no idea how it works or what program it is but it's pretty cool.

I still take notes the old-fashioned way because I think it's more convenient.  Notes on Word are probably fine for non-math/science courses but for math/science, unless you have a tablet PC or that cool program, I would stick to pen and paper.

You could try taking notes in latex, although that might be just as bad if not worse than Microsoft Word.

I completely agree with everything TZF stated.  Scrap paper is great, especially for homework problems.  My desk is covered in old essays and handouts I no longer have use for, and they're now covered in scrap work.

Word's new equation maker is now quite strong, but it still takes some time to input and format everything.

[quote="TZF"]If you want to save paper, just use scrap paper. Go to the library of some town or university, they generate so much scrap paper that's only printed on one side. If you don't care about a graded essay or something, you can write on the back of that, too. Re-using paper is much more efficient than recycling paper, anyway.

Writing on blank paper gives you an enormous amount of freedom in terms of how you lay out your notes, and you can exercise it very quickly, instead of having to set up all these formatting things. Computers won't be able to match that until they can read minds.

Meanwhile, Word 2007 has a [i]great[/i] equation editor (IMO) now. It admits a lot of LaTeX commands for symbols and will auto-format as you type. For example, [i]"\int _0^\infty (x^2+7 \alpha)/\sqrt(2-x^2) dx"[/i] produces $ \int_0^{\infty} \frac {x^2 \plus{} 7 \alpha}{\sqrt {2 \minus{} x^2}} dx$. If any symbols are not covered, you can include them using "Math AutoCorrect".

You can also set some keyboard shortcut to initate a new equation. So if you're used to writing in LaTeX, you will be able to naturally add in equations without having to take your fingers off the keyboard.[/quote]

cool, I'll try writing equations down that way, I think it'd much simpler to learn lines of codes instead of using the object stuff although it's really good as you pointed out but there might be too much pointing and clicking at some point

is there a tutorial available somewhere for this latex stuff ?

Look at the left of this website(if you use the AoPS skin.) There should be a navigation bar that looks like this:

Bookstore
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Community
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Click LaTeX.

[quote="leoxnlin"][hide="irrelevant"][quote="TZF"]Meanwhile, Word 2007 has a great equation editor (IMO) now.[/quote]

There are some serious typographical issues with Word (if I'm not mistaken), but I won't go into them...[/hide]

Notes? I'm surprised everyone suggesting fancy stuff like Word and LaTeX. Notes are not supposed to be serious or well-formatted, which is what stuff like Word and LaTeX are for. They have tons of junk and distractions on them. Use Notepad.

EDIT:... Oh, [i]math[/i] stuff. For what purpose and when are you taking these notes?[/quote]

these days I'm reading about 10 books at once, 1 on maxwell's equations, maxwell's geometry of space, dirac ... books on rings, differential geometry ... music ... people's history of the US and several others. Basically what I'd like to do is take notes on my computer so I could make resumes of each book. Instead of carrying several books at once I could keep this all clean on my computer and I'd like to add to these readings later on. So for example, let's say I'm reading on rings, I'd start with the basic information and build from other books later on. I'd think it would be fairly messy to do this all on paper.

If I'm just working with notes or the like, I'll use paper. If it's on the computer, it's for communication to someone else. That comes in two modes:
- plain text pseudo-markup, which will probably go in the body of an email.
- $ \text{\LaTeX}$, which will probably be sent as an attached pdf or pdf and source.

Instead of Microsoft Word, use Microsoft OneNote.  (It's included in the student editions of Microsoft Office).
It's much much better for taking notes, basically it's a sheet of paper you can write on.  For instance, you can write anywhere on the page, and move thoughts and bullets around really nicely.

I agree with Sunny. MS OneNote is not for serious and well formatted text. It is also easy to use - just click anywhere on the page and the cursor will show up.

However, I don't think you can write math stuff. Maybe its possible, but I don't use OneNote a lot, so I really dont' know

onenote sucks.

Well, maybe for notes for like English II or something but not for math.  There's a freeform pen tool that is very hard to control if you don't have a tablet PC.

Or go buy a [url=http://www.wacom.com]Wacom tablet[/url]... which doesn't cost as much and plugs in via USB

[quote="Nick31"]
these days I'm reading about 10 books at once, 1 on maxwell's equations, maxwell's geometry of space, dirac ... books on rings, differential geometry ... music ... people's history of the US and several others. Basically what I'd like to do is take notes on my computer so I could make resumes of each book. Instead of carrying several books at once I could keep this all clean on my computer and I'd like to add to these readings later on. So for example, let's say I'm reading on rings, I'd start with the basic information and build from other books later on. I'd think it would be fairly messy to do this all on paper.[/quote]

It seems to me you want to keep nice, neat references for yourself over the material. Then you [i]do[/i] want well-typeset and well-formatted documents, and $ \text{\LaTeX}$ is clearly the way to go.

$ \text{\LaTeX}$ is of course really nice, but it takes sort of a long time to type everything out.  I wouldn't recommend using it to take notes in class.

With what it sounds like you're trying to do, I think the best approach would be to make a first round through the books, taking notes as you go along with all the information, and then a second round through your first notes re-ordering/refining them and compiling them into a single reference (re-writing notes in a different way is actually a good learning strategy).

On the first round through (notes you take while reading the books), you can keep your notes really neat and number your pages/sections. Then, if you read about the same topic later, include a reference at the top of the page to the relevant previous pages, and update the references on the previous relevant pages to include the new page. You could also directly add to previous pages. Of course, if you want to do this, you would have to space out your notes and leave some amount of white-space in them for potential future use.

[quote="JRav"]$ \text{\LaTeX}$ is of course really nice, but it takes sort of a long time to type everything out.[/quote]

I haven't personal experience (since I haven't done really large documents yet), but I believe $ \text{\LaTeX}$ing should be fast once all formatting issues are set: one of the major purposes of $ \text{\LaTeX}$ was to separate content from form, so that one can focus all attention on the actual content at the same time and the formatting at a different time (as opposed to being distracted by the formatting while working on content (which happens to me a lot with MS Word), and vice versa). If you are creating a large document, or a large number of documents, the time saved should accumulate into a net time-save.


trigonometry
inequalities
geometry
geometric transformation
rotation
combinatorial geometry
inequalities unsolved

Would it be of any use considering an equilateral triangle $ABC$ of side length $\frac{2}{\sqrt{3}}$ and a point $P$ in its interior such that calling $P_{a}$,$P_{b}$,$P_{b}$ the foots of $P$ on the sides of the triangle we have  $P_{a}=x$,  $P_{b}=y$,  $P_{c}=z$  and the inequality to prove becomes $\frac{2}{PA}+\frac{2}{PB}+\frac{2}{PC}\geq 3\sqrt{3}$?

No. $P$ is torricelli's point of $\Delta ABC$ and $PA=x,$ $PB=y$ and $PC=z,$
$\measuredangle APB=\measuredangle APC=\measuredangle BPC=120^{\circ}.$

[quote="shuchang"]x,y,z>0  x+y+z=1
prove
$\sum\frac{1}{\sqrt{x^{2}+xy+y^{2}}}$>$3\sqrt3$[/quote]
It is not true! :lol:
Try $x\rightarrow0^{+}$ and $y=z\rightarrow\frac{1}{2}^{-}.$ :wink:
Id est, in $\Delta ABC$ $\measuredangle BAC=120^{\circ}$ and $AC=AB=\frac{1}{2}.$

[quote="zhaobin"]Yes,I think it is has been posted that the Minimum of $LHS$ is $4+\frac{2}{\sqrt 3}$[/quote]

yeah , and the proeblem is very hard . I remember xtar posted a solution with Langrange Multipliers and even so I wonder if there is a solution elementary possible........
Maybe Arqady has an answer   :lol:

Ok, I have a solution for this inequality :

Let's denote by $S$ the sum to minimize and let's consider in the complex plane the points $O(0), \ A(1)$ and $B(-j)$, where $j =-\frac{1}{2}+i \frac{\sqrt{3}}2$.
Also, let's denote by $T$ the convex hull of the points $O, \ A$ and $B$ (full equilateral triangle), $z = a-bj, \ z' = b-cj, \ z'' = c-aj$, $M(z), \ M'(z')$ and $M''(z'')$.
It is then easy to check that $M, \ M'$ and $M''$ are points of $T$, and that $S = \frac 1{|z|}+\frac 1{|z'|}+\frac 1{|z''|}= \frac 1{OM}+\frac 1{OM'}+\frac 1{OM''}$.
Besides, the relations $z+jz'+j^{2}z'' = 0$ and $z+z'+z'' = 1-j$ show that the triangle $AM'M''$ (no, it has no link with the famous newspaper :P)   is equilateral and its centroid is $G \left(\frac{1-j}3 \right)$ which is also that of the equilateral triangle $OAB$.
Using the rotation $r$ of center $G$ and angle $\frac{2 \pi}3$ and rearranging the points $A, \ M'$ and $M''$ so that we have $M' = r(M)$ and $M'' = r^{2}(M)$ we check easily that $OM' = AM$ and $OM'' = BM$ and we see that the problem comes down to minimizing the sum $S(M) = \frac 1{MO}+\frac 1{MA}+\frac 1{MB}$ when $M$ describes $T$.
Hence it remains to show that $S(M)$ is minimum when $M$ is the midpoint of one of the sides of the equilateral triangle $OAB$ which will yield the desired inequality.

See here for the initial post and solution......
http://www.mathlinks.ro/Forum/viewtopic.php?t=52376&start=20


[hide]Let's see... 1-9999 and we want to see how many numbers do not have 3 or 4 different digits.
The probability all the digits are different are 10*9*8*7
The probability that 3 digits are different and 1 is the same is what we need to find next.  Without a loss of generality, let's set the digit to be considered the same last.  We have 10*9*8*3 since the digit to be copied has 3 choices.  This adds up to 100*72=7200 so 9999-7200=2799.[/hide]
Is this right?

It has to be inbetween 0 and 999, right?

I think you can divide up the problem into cases for different digit-numbers, which I just did, but I screwed up horribly on the cases. :mad:

Let's use my figuring out of cases that don't work idea but let's use some common sense.  If a number has 4 different digits, it has 4 digits not 3 or 2 or...
Therefore, the number of possibilities for 4 different digits is 9*9*8*7 or 4536.
Now for the 3 numbers case.
The simplest case is when there's three digits.  9*9*8=648.
With four digits, let's consider the case where the first digit is anything and the second digit must be the same as the first.  There's 9*1*9*8 possibilities.  This case is identical to the cases where the 1st and 3rd or 4th digits are the same.  This results in 9*3*9*8=1944 possibilities.  For the other three equivalent cases, let's investigate the case when the two similar xs don't equal 0.  We have 9*3*8*8=1728 possibilities.  Now, let's consider the cases where it equals 0.  We have 1*3*8*8=192 so we have 10*3*8*8=1920 possibilities.  10000-(1944+1920+648+4536)=952.  Is this right?

[quote="joml88"]How many positive integers less than 10,000 have at most two different digits?[/quote]

[hide]
Case 1(all digits the same): $9 + 9 + 9 = 27$
Case 2(all digits same except one):
with 3 digits you have $aab = \displaystyle{3 \choose 2}$ and $9\cdot 8$ choices for $a$ and $b$
with 4 digits you have $aaab = \displaystyle{4 \choose 3}$ and $9\cdot 8$ choices for $a$ and $b$
or you can have $aabb = \displaystyle{4 \choose 2}$
so if you add it all up you hve: $(4 + 6 + 3) \cdot 72 + 27 = 936 + 27 = 963$

$\fbox{963}$[/hide]

I think you guys are missing two things:
1. Restrictions with the number 0
2. 1,2,3-digit numbers.
The toughest part of this problem is the cases; I don't think its that hard, but its tricky. I keep getting different answers.

[quote="joml88"]How many positive integers less than 10,000 have at most two different digits?[/quote]

I think people require to read the question well

Anyone get an answer? I'm getting 1936 for some reason :(

[hide="Solution"]
Case 0:  One-and-two-digit numbers.  They all work, and there are $99$ of them.  Each of the other cases deal with three- and four-digit numbers.

Case 1:  All digits the same.  $9 + 9 = 18$ numbers.

Case 2:  One digit off that isn't the first digit.  Take each of the numbers above and change each digit (except the first) to any of the $9$ other digits.  This gives $2 \times 9 \times 9 + 3 \times 9 \times 9 = 405$ numbers.

Case 3:  One digit off that is the first digit.  Take each of the numbers above and change the first digit to any of the $8$ other nonzero digits.  This gives $9 \times 8 + 9 \times 8 = 144$ numbers.

Case 4:  Numbers that end in a string of zeroes.  $9 + 9 = 18$ numbers.

Adding up, we have $99 + 18 + 405 + 144 +18 = \boxed{684}$ numbers.
[/hide]

Gosh, no two people have agreed yet  :huh:  I think my method is right, though - you can't beat casework!

I checked the AMC site, the answer is actually [hide]927[/hide].

[hide="yet another different answer (!)"]When there are less than 3 digits, all work [b](99)[/b].

When there are 3 or 4 digits which are all the same, 9 each [b](18)[/b].

When there are 3 digits, exactly 2 of which are the same, then either:

the repeated digit is zero, so nonzero digit leads [b](9)[/b]

the single digit is zero, so 9 digits to choose from and 2 places to put 0 [b](18)[/b],

or there is not a zero, so 9 digits to choose for single, 8 for repeated, 3 places to put single [b](216)[/b].

When there are 4 digits, exactly 3 of which are the same, then either:

the repeated digit is zero [b](9)[/b],

the single digit is zero, so 3 places to put it [b](27)[/b],

or there is no zero, making 9 choices for single, 8 for repeated, 4 places to put single [b](288)[/b].

Finally, if the digits come in pairs, then:

9 choices for leading, 9 for other (including zero), 3 places to put copy of leading digit [b](243)[/b].

for a total of $\boxed{927}$; wow, nobody is agreeing on this!!![/hide]

edit: i got it! yay!

[quote="joml88"]How many positive integers less than 10,000 have at most two different digits?[/quote]

[hide="a more clever approach"]how about take number of 3 different digits, 4 different digits, subtract from 9999

-4 diff digits, 9*9*8*7

-3 diff digits

--3 digit
9*9*8

--4 digit
---abcc,acbc,accb
3*9*9*8
---cabc,cacb,ccab
3*9*9*8

9999-9*9*8*(7+1+3+3)=9999-14*9*9*8=$\boxed{927}$[/hide]

[hide]All positive integers 1 digit are ok. So 9 total.

All integers 2 digits are ok. SO 90 total.

For 3 digits, subtract ones with all digits different.

So $900 - 9 \cdot 9 \cdot 8 = 252$.

For 4 digits, it's a little casework.

Case: All digits same. There are 9.

Case: Digits contain '0'.

It can only be of combinations of xxx0, xx00, x000, with no 0's as first digit.

So there are xxx0, xx0x, x0xx, xx00, x00x, x0x0, and x000, for 7. Now, $x$ can be chosen from 9 digits, so $9 \cdot 7 = 63$.

Case: Digits don't contain '0', and not all same.

Of the forms xxxy, xxyy, and xyyy.

For xxxy, there are 4 combinations. For xxyy, there are $\binom{4}{2} = 6$ combinations, and for xyyy, there are 4. So $4+6+4 = 14$.

And there are $\binom{9}{2}$ ways to choose x and y.

So $\binom{9}{2} * 14 = 504$.

I'm getting $927$.[/hide]

BTW, Scorpius, this a problem from this week's AIME B problem set. :)

[quote=hmoonster][hide="slightly different (less casework for the 0 case)"]
Considering $0$ as a digit,
Case 1: 1 distinct digit
1111, 2222, 3333, ... 9999
$9$

Case 2: 2 distinct digits
We have $\binom{10}{2}$ ways to choose the two numbers.
we have $2^4-2$ possible arrangements for $a$'s and $b$'s (since $a$ and $b$ are distinct we ignore $aaaa$ and $bbbb$)
So $(2^4-2)\binom{10}{2}=720$

Case 3: 3 distinct digits (e.g. 0035, 0355)
2 digit-number: they have to be distinct digits so $9\cdot 8=72$
3 digit-number: we can permutate the above in 3 ways so $9\cdot8\cdot3=216$

$9+720+72+216=927$
[/hide][/quote]

But in your second case you said $(2^4-2)\binom{10}{2}=720$, but that's not 720, that's 630.....

So I was looking at one subcase, which is if two digits are the same (and hence three are distinct) and one of the two same digits is the first digit. Then we can have the 1st&2nd, 1st&3rd, 1st&4th be the two same digits, so $3$ ways for that. Now $9$ ways to pick what the same digit actually is, and $9 \cdot 8$ ways to pick the other two digits. But that in itself is greater than $1000$. What happened?

There are at most two different digits according to the problem. An integer of the form $ABAC$ does not work.

Case 1: one digit

$\Rightarrow\frac{9}{10}\binom{10}{1}=9$ numbers.

Case 2: two digits

$\Rightarrow\frac{9}{10}\binom{10}{2}\left(2^2-2\right)+9=90$ numbers.

Case 3: three digits

$\Rightarrow\frac{9}{10}\binom{10}{2}\left(2^3-2\right)+9=252$ numbers.

Case 4: four digits

$\Rightarrow\frac{9}{10}\binom{10}{2}\left(2^4-2\right)+9=576$ numbers.

The answer is $9+90+252+576=927$ numbers.

[quote=OlympusHero]Okay, now what is the issue?

If the number is one or two digit, all work, so $99$ ways. For three digit, the only not working way is if all digits are different, so $900-9 \cdot 8 \cdot 8 = 252$ ways. For four digit, we split into two cases. Case 1: All digits are the same. Obviously $9$ ways. Case 2: 3 digits are the same. Subcase 1: One of these three same digits is the first digit. We have $3$ ways to pick the other two same digits and then $9$ ways to pick the value of the same digit and $9$ ways to pick the final digit. That's $3 \cdot 9 \cdot 9 = 243$ ways. Summing gives $693$.[/quote]

You suddenly stopped your casework. I'd suggest getting through the remainder of the cases.

Okay - but the only other case I can see is if the three same digits are the second, third, fourth digits for 10*9=90 ways? That still doesn’t make 927. If two digits are the same then there are three different digits so that does not work.

Why can't we have something like $AABB$?

Ok, so we have 6 arrangements of AABB. There are 9 ways for the first digit and the digit which is the same as that has 1 way (it’s already determined) and we can pick the other digit in 9 ways (can be zero but cannot be equal to A), so 6*81 = 486 but if I add this to what I have the result is too big… sorry for all these questions, I’m really trying to get better at combinatorics.

You're overcounting by $2$. $AABB$ is equivalent to $BBAA$.

Okay; I see now, thanks! Haha, I have just one more question.

[quote=OlympusHero]Okay, now what is the issue?

If the number is one or two digit, all work, so $99$ ways. For three digit, the only not working way is if all digits are different, so $900-9 \cdot 8 \cdot 8 = 252$ ways. For four digit, we split into two cases. Case 1: All digits are the same. Obviously $9$ ways. Case 2: 3 digits are the same & one is different. Subcase 1: One of these three same digits is the first digit. We have $3$ ways to pick the other two same digits and then $9$ ways to pick the value of the same digit and $9$ ways to pick the final digit. That's $3 \cdot 9 \cdot 9 = 243$ ways. Subcase 2: These three digits are the last three. Then there are $1 \cdot 10 \cdot 9 = 90$ ways.[/quote]

If I add $243$ to this, I get $936$ ways. So this seems like I counted the case where all digits where the same twice, but I don’t see where I did this. Thanks so much for your help and sorry for bugging you :(

I think the overcounting is in Case 2, Subcase 2.

You claim that there are $10\cdot 9$ ways, but in fact there are $9\cdot 9$ because $0$ can't be the first digit.

Final solution:

If the number is one or two digit, all work, so $99$ ways. For three digit, the only not working way is if all digits are different, so $900-9 \cdot 8 \cdot 8 = 252$ ways. For four digit, we split into two cases. Case 1: All digits are the same. Obviously $9$ ways. Case 2: 3 digits are the same & one is different. Subcase 1: One of these three same digits is the first digit. We have $3$ ways to pick the other two same digits and then $9$ ways to pick the value of the same digit and $9$ ways to pick the final digit. That's $3 \cdot 9 \cdot 9 = 243$ ways. Subcase 2: These three digits are the last three. Then there are $1 \cdot 9 \cdot 9 = 81$ ways. Case 3: Two pairs of same digits. There are $\frac{\binom{4}{2}}{2}=3$ position possibilities, and $9 \cdot 9$ ways to pick the two digits, for a total of $243$ ways. Summing gives $99+252+9+243+81+243 = \boxed{927}$.


$x$ acute angle is degree. If $\frac{1}{sin^{2}x}-\frac{sin^{4}x}{cos^{2}x}=-\frac{11}{12}$  then find $x$ .


What is the formula to compute a fixed percentage of profit, with a
beginning number, over time (20 years)?

It depends on what you want really.
A general formula is:
$P=I\left(1+\frac{p}{n}\right)^{nt}$ .
Where
$P=$profit from percentage compounded on the intial maount
$I=$the initial amount
$p=$percentage, interest rate
$n=$number of compoundings per time, $t$
$t=$time observing


AMC USA(J)MO USAMO algebra polynomial factorial

What is $1^{a}+2^{a}+3^{a}+...+x^{a}$ In terms of x and a?
GIve a proof. (Is this a USAMO level problem?)

What? There's no closed form for that really...

I can find a summation formula if I know a and I'm not the sharpest tool in the shed.

[quote="JSteinhardt"]There's a method for finding it though, think Hockeystick.[/quote]

True, and there's also an inductive way (on a) of finding it. (And even a "fake" way where you assume it's a polynomial and solve for the induction. Although it's probably not hard to prove it's a polynomial based on finite differences...)

the question asked for in terms of a and x

[quote="Andrew Kim"]the sum would be infinite unless the absolute value of x is less than 1[/quote]

No it wouldn't. It is only infinite if either $x$ or $a$ was infinite.

So, is anyone going to try deriving a formula.

There is no reasonable closed-form, only a method for finding it for a particular a, as i said earlier. Try googling 'summation housepainting numbers' or something like that.

If the Bernoulli numbers are allowed, then we can find a closed-form solution.  See [url]http://mathworld.wolfram.com/BernoulliNumber.html[/url] (Equation 43).

But they (Bernoulli numbers) don't have an explicit formula. =(

The problem is a pretty interesting problem though, one of the major achievements in algebraic combinatorics.

What if you know what A is?
I noticed that $1^{a}+2^{a}+3^{a}+...+x^{a}$ can always be written as $bx^{a+2}+cx^{a+1}+dx^{a}+...+$some constant. I can find any summation when I know a this way.

[quote="bpms"]What if you know what A is?
I noticed that $1^{a}+2^{a}+3^{a}+...+x^{a}$ can always be written as $bx^{a+2}+cx^{a+1}+dx^{a}+...+$some constant I can find any summation this way.[/quote]
Yes, if you know what $a$ is you can find an explicit formula (AoPS volume 2 outlines a method in its combinatoric section) - however, the question is how to find a formula written out in terms of $x$ and $a$, rather than just in terms of $x$ for a given $a$.

I think you've already established that there's no simple closed form solution. But here's a quick routine way to find it (my favorite way):

If $x_{n}=x(x-1)(x-2)...(x-(n-1))$ is a falling factorial, then $\sum_{i=0}^{k}i_{n}=\frac{(k+1)_{n+1}}{n+1}$, which is easily proven by induction/telescoping sum. Also notice that $x_{n}$ is an nth degree polynomial, so we can write x^n as a "polynomial" in falling factorials and then go from there. For instance, $\sum_{i=0}^{n}i^{3}=\sum_{i=0}^{n}i_{3}+3i_{2}+i=\frac{(n+1)_{4}}4+(n+1)_{3}+\frac{(n+1)_{2}}2=(\frac{n(n+1)}2)^{2}$.

I realize it's already been brought up that this isn't what you're looking for, but you'll have to settle for that.

So if we all agree that there isn't any way to write it should this be locked?


[quote="236factorial"]Bob and Ray are describing the same number. Bob say, "the number is a positive even integer less than or equal to 20. Ray says, "the number is divisible by 4." If Bob's statement is true and Ray's is false, what are the possible numbers they're talking about?[/quote]
[hide]
According to Bob's true statement, the only possibilities right now are:

2, 4, 6, 8, 10, 12, 14, 16, 18, and 20.

According to Ray's falso statement, all multiples of 4 are eliminated, leaving:

2, 6, 10, 14, and 18.[/hide]

[hide] just eliminate all the even positive integers under 20 that are multiples of 4. Or eliminate every other number starting from 4.
answer: 18, 14, 10, 6, 2.  [/hide][/hide]


What is the sum of the three numbers less than 1000 that have exactly five positive integer divisors?

When you write a number x in it's canonical for, like this: $ x \equal{} a_1^{i_1}\cdot a_2^{i_2}\cdot a_3^{i_3}\cdot \ldots \cdot a_n^{i_n}$, then every number in form $ y \equal{} a_1^{j_1}\cdot a_2^{j_2}\cdot a_3^{j_3}\cdot \ldots \cdot a_n^{j_n} \mid a_1^{i_1}\cdot a_2^{i_2}\cdot a_3^{i_3}\cdot \ldots \cdot a_n^{i_n}$ iff $ \forall  z : j_z \le i_z$

In other words, to find divisors of x, you can just substitute lower exponents instead of $ i_c$ in x.

So how many different divisors of x there are? Well, let's find one $ y | x$ to answer that. To determine exponent of $ a_1$, we have choices 0,1,2,..., $ i_1$. Thus $ i_1\plus{}1$ choices. So, independently of that, we have $ i_2\plus{}1$ choices to determine exponent of $ a_2$. Continuing like that, we get that we have $ (i_1\plus{}1)\cdot (i_2\plus{}1)\cdot (i_3\plus{}1)\cdot \ldots (i_n\plus{}1)$ choices to pick a divisor of x.

Put as 5=5*1 is a prime number, it follows that x has only one factor $ i_1$ and it is to the power of 4.

So, the searched powers of primes are 2^4, 3^4, 5^4. Sum of those is 16+81+625=$ \boxed {722}$


estas seguro de que asi es el problema, a mi me sale que si ABC es escaleno, entonces la recta que dices se corta con BC, y si es isosceles(equilatero) entonces I coincide con O(no estaria definida la recta IO)

si eso pasa tendrias que demostrarlo, si ya lo demostraste postealo porfavor!! :D

[quote="Tony2006"]a qui posteo este problema , espero que haya otra solucion aparte de la que busque,
buscar una razon necesaria y suficiente para que si el angulo en A de un triangulo ABC es 60º entonces la recta que une el circuncentro y el incentro es paralelo a BC.[/quote]

Sea $M_{b}$ el punto medio de $\ovl{BC}$. Desde que $OI$ y $BC$ son paralelos se sigue que la distancia de $O$ a $BC$ es $r$ con lo que $OM_{b}=r$. En el triangulo $OBM_{b}$ tenemos que $\frac{BC}{2r}=tgA$ con lo que $tgA=\frac{a}{2r}=\frac{2RsenA}{2r}=\frac{RsenA}{r}$ y de ahi que $cosA=\frac{r}{R}$. Si $A=60$ se sigue que $\frac{r}{R}=\frac{1}{2}$ con lo que el triangulo es equilatero y $I=O$ y la recta $IO$ degenera en un punto.

Podemos sin embargo caracterizar los triangulos tales que $\overline{OI}\parallel \overline{BC}$ (omitiendo que $A=60$).

De la identidad $cosA+cosB+cosC=1+\frac{r}{R}$ se sigue que $cosB+cosC=1$.

Facilmente obtenemos que $AI\perp IH$.

José Carlos Chávez Sandoval

[quote="Tony2006"]y que es H??[/quote]

El ortocentro del triangulo ABC

[color=darkblue][b]An interesting remark[/b] (from the my friend [i]Marcel Chirita[/i]) : $A=60^{\circ}\Longleftrightarrow OI=2R\sin\frac{|B-C|}{4}\ .$
Generally, I proved that for any $\lambda\in (0,1)\ ,\ A=2\arcsin\lambda\Longleftrightarrow OI^{2}=R^{2}\cdot\left[(2\lambda-1)^{2}+8\lambda\sin^{2}\frac{|B-C|}{4}\right]\ .$[/color]


number theory
prime numbers
number theory proposed

$ a\equal{}1,b\equal{}1,c\equal{}n\minus{}2$
and if u want them to be distinct then
$ a\equal{}1,b\equal{}p,c\equal{}n\minus{}p$
where $ p$ is a prime for which $ p \nmid n$ which can easily be found.
am i missing sth?

[quote="hophinhan"]Prove that for every integer $ n \ge 18$,  we can find three integer numbers $ a,b,c > 1$ are pairwise relative prime and $ n \equal{} a \plus{} b \plus{} c$

@: I got a solution which use the lemma :there is a prime between $ n$ and $ 2n$ ... but it's not very nice :)[/quote]
Can u post ur solution?

[quote="hophinhan"]Prove that for every integer $ n \ge 18$,  we can find three integer numbers $ a,b,c > 1$ are pairwise relative prime and $ n \equal{} a \plus{} b \plus{} c$
[/quote]
[color=blue]
As I said, my solution used the lemma : there is a prime between $ n$ and $ 2n$ (*)

For $ n \ge 18$:

Let $ P\equal{}\{p_1,p_2,...,p_k \mid p_i<p_j \forall i<j\}$ is the set of all primes $ \le n$

There is a prime $ p_i$ of set $ P$ such as  $ 2p_i > n$ and $ 2p_{i\minus{}1} <n$ . We write :$ n\equal{}p_i\plus{} m$  $ (n> p_i > \left[\frac{n}{2}\right] > m)$

Similarly, We can write $ m\equal{}q_j\plus{}y \  (  m > q_j > y)$

We got the result : $ n\equal{}p_i\plus{}q_j\plus{}y$ ( $ p_i > q_j > y$ and $ p_i$ ; $ q_j$ are two prime numbers) $ \Longrightarrow (p_i \ ; \ q_j )\equal{}(p_i \ ; \ q_j)\equal{}( q_j \ ; \ y)\equal{}1$

@Case $ n\equal{}17$: $ y\equal{}1$ :) . $ 3\plus{}5\plus{}7 < 17 < 5\plus{}7\plus{}9$ . The existence of $ p_i$ was proved by lemma (*).[/color]


limit
logarithms
calculus
calculus computations

let sequence:  $ x_1 \equal{} 2$ and $ x_{n} \equal{} 2 \plus{} \frac {1}{x_{n \minus{} 1}}$
  find $ \lim_{n \to \infty} x_n$

It obvious that $ x_n \in \left[ 2, \frac{5}{2}\right] \forall n \in\mathbb{N^*}$.Now let $ f: \left[ 2, \frac{5}{2}\right]\to\mathbb{R}$ $ f(x)\equal{}2\plus{}\frac{1}{x}$.It's easy to verify that $ |f(x)\minus{}f(y)|\leq \frac{1}{4}|x\minus{}y|$ ,therefore $ f$ is a contraction hence the sequence $ x_{n\plus{}1}\equal{}f(x_n)$ , $ x_1 \in \left[ 2, \frac{5}{2}\right]$ is convergent to the unique fixed point of $ f$ , which in this case is $ 1\plus{}\sqrt{2}$

find lim
 $ \lim_{n \to \infty }xarctan(n cotan x)$
  and set up graph
 last problem:
$ \lim_{n \to \infty} \frac{1}{n} \plus{}\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}\plus{}...\plus{}\frac{1}{2n}$

[b](*)[/b]$ \lim_{n\to \infty}{x\arctan(n\cot x)}\equal{}\frac{\pi x}{2} \newcommand{\sign}{\mathop{\mathrm{sign}}}\sign{x}$
[b](**)[/b]$ \lim_{n\to \infty}{\sum_{k\equal{}n}^{2n}{\frac{1}{k}}}\equal{}\lim_{n\to \infty}{\left(\ln 2n \minus{} \ln (n\minus{}1)\right)}\equal{}\ln 2$


I am looking for an [b]elegant[/b] solution to the following geometry problem. I'm just not seeing it. (I have a very ugly solution using coordinates.)

(From Kiran Kedlaya's Notes. Also on MOP 1995)
Let $BB', CC'$ be altitudes of triangle $ABC$, and assume $AB \neq AC$.  Let $M$ be the midpoint of $BC$, $H$ the orthocentre of $ABC$ and $D$ the intersection of $BC$ and $B'C'$. Show that $DH$ is perpendicular to $AM$.

Any help will be greatly appreciated.

See the following posts:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110887[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23319[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=552[/url]


$a,b,c,d\ge 0$
Prove or disprove:
\[ (a+b+c)(b+c+d)(c+d+a)(d+a+b) \ge 4(a+b)(b+c)(c+d)(d+a)  \]

Correct, of course - how can such a nice inequality be wrong?

First, we have

$\left(a+b+c\right)\left(b+c+d\right)\geq\left(a+b+c+d\right)\left(b+c\right)$,

since $\left(a+b+c\right)\left(b+c+d\right)-\left(a+b+c+d\right)\left(b+c\right)=ad\geq 0$. Similarly,

$\left(c+d+a\right)\left(d+a+b\right)\geq\left(a+b+c+d\right)\left(d+a\right)$.

Multiplying these two inequalities, we get

$\left(a+b+c\right)\left(b+c+d\right)\left(c+d+a\right)\left(d+a+b\right)$
$\geq\left(a+b+c+d\right)\left(b+c\right)\left(a+b+c+d\right)\left(d+a\right)=\left(a+b+c+d\right)^2\left(b+c\right)\left(d+a\right)$.

Now, $\left(a+b+c+d\right)^2=\left(\left(a+b\right)+\left(c+d\right)\right)^2\geq 4\left(a+b\right)\left(c+d\right)$ (since $\left(u+v\right)^2\geq 4uv$ for any two reals u and v), so we get

$\left(a+b+c\right)\left(b+c+d\right)\left(c+d+a\right)\left(d+a+b\right)\geq 4\left(a+b\right)\left(c+d\right)\left(b+c\right)\left(d+a\right)$
$=4\left(a+b\right)\left(b+c\right)\left(c+d\right)\left(d+a\right)$,

so we are done.

  Darij

That is great, Zhaobin and Darij. I like it.


The sum of four consecutive odd integers is 112. What is the greatest of the four integers?

$ x\plus{}x\plus{}2\plus{}x\plus{}4\plus{}x\plus{}6\equal{}4x\plus{}12\equal{}112\Rightarrow x\equal{}25$, which is odd.  So the largest number is $ x\plus{}6\equal{}31$.

This requires a little explanation -

JRav let $ x$ be the smallest of these four odd integers. So, in terms of $ x$, the integers are $ x, x\plus{}2, x\plus{}4, x\plus{}6$. Then you can set up your equation and solve, as JRav demonstrated.


This one is well-known here (it appeared in an usamo).
See here for example :
http://www.mathlinks.ro/Forum/viewtopic.php?t=5726


inequalities
geometry
probability
number theory
\/closed

Is AOPS going to bring back the Olympiad Problem Solving, Olympiad Combinatorics, Olympiad Number Theory and Olympiad Inequalities Classes?
Olympiad Geometry is really great but what about the others? If not are there any other Olympiad level books that focus on how to solve these kinds of problems and not just a collection of math facts or a collection of problems with solutions but something that teaches you how to think about Olympiad problems. I know there is WOOT but that doesn't cover very much material class time or have that many problems to gain a deep understanding of the topics and know how to think about and approach the problems. Is AOPS going to sell the transcripts or make a book out of these classes soon? That would also be really awesome.

Most of the Olympiad Problem Solving course has been included in WOOT and the Intermediate Counting & Probability course.  We never had Olympiad classes in the other areas you mention.  I'll get around to writing a geometry class someday.

(Also, I'm not sure I agree with your critique of WOOT -- true, it's not as in-depth as the Olympiad Geometry class on that topic, but we've had more complaints about too many problems than we've had complaints about too few problems!)

Okay sorry about not knowing actually which classes were offered. I just read some of that stuff on the forum. Also I don't see why anyone would complain about too many problems or too many classes. If you don't have enough time to do all the problems thats fine too. I mean people should feel bad if they aren't able to solve all the problems or go to all the lectures. This is a training program so you should just try to have fun learning the material and not stressing that you didn't solve most of the problems like you would feel on a contest. But really are you releasing any Olympiad level books? By that I mean not interm counting or algebra but something at an olympiad level but still written in the AOPS style that helps us discover things and understand how to think.

It's on the todo list, but it's a ways in the future.


inequalities
search
algebra
polynomial
inequalities proposed

Let $ a,b,c$ be positive real numbers. Let $ p=a+b+c,q=ab+bc+ca,r=abc$. Prove that:
\[ (*)\frac{64p^{3}}{27}-8pq+16r\le 8 \]
when $ 16p^{2}-36q=36$
\[ (**)\frac{4p}{3}\ge \frac{64p^{3}}{27}-8pq+12r+2q \]
when $ 16p^{2}-36q=9$

Comment: the first inequaity, it is easy to prove with $ p\le \frac{3}{2}$, and For the second inequality, it is easy to prove with $ p\le \frac{3}{4}$.
How to solve it, argady? Can you show me your proof using Schur's inequality? :) Thank you very much  :wink:

Thank you, zaizai-hoang, for your gift. I'll try it.

yeah :) I created it to solve some inequatilies of 4-variable. ;) it is really interesting :)

Argady, where are you? Can you post yours?
I created it to solve the following inequalities (in my notebook):
\[ \bigtriangledown\]
$ \bigstar$[b] Prolem 24[/b][color=green][b] [Pham Van Thuan][/b][/color]
 Let $ a,b,c,d$ be non-negative real numbers such that $ a^{2}+b^{2}+c^{2}+d^{2}=1$. Prove that:
\[ a+b+c+d\ge a^{3}+b^{3}+c^{3}+d^{3}+ab+bc+cd+da+bd+ac \]

Solution:
By Roll's theorem we will prove the following inequality:
\[ \frac{4p}{3}\ge \frac{64p^{3}}{27}-8pq+12r+2q \]
when $ 16p^{2}-36q=9\to q=\frac{16p^{-}9}{36}$
Using Schur's inequality:
\[ r\le \frac{pq}{9}=\frac{p\left(\frac{16p^{-}9}{36}\right)}{9}=\frac{16p^{3}-9p}{324}\]
Let:
\[ f(p,q,r)=\frac{4p}{3}-\frac{64p^{3}}{27}+8pq-12r-2q \]
We will prove that:
\[ f(p,\frac{16p^{-}9}{36},\frac{16p^{3}-9p}{324})\ge 0 \]

\[ \Leftrightarrow \frac{(4p+3)(6p+1)(3-4p)}{18}\ge 0 \]

If $ p\le \frac{3}{4}$ we are done. If $ p\ge \frac{3}{4}$, it is not hard to solve. But my solution is not nice for remain case. Argady, Can you help me?
\[ \bigtriangledown\]
And it is also the key of the following inequality

$ \bigstar$ [b]Prolem 23 [color=green][Nguyen Thuc Vu Hoang][/color][/b]
 Let $ a,b,c,d$ be non-negative real numbers such that $ a^{2}+b^{2}+c^{2}+d^{2}=4$. Prove that:
\[ a^{3}+b^{3}+c^{3}+d^{3}+abc+bcd+cda+dab\le 8 \]
\[ \bigtriangledown\]
$ \bigstar$ [b]Prolem 21 [color=green][Pham Kim Hung][/color][/b]
 Let $ a,b,c,d$ be positive real numbers such that $ a+b+c+d=1$. Prove that:
\[ a^{3}+b^{3}+c^{3}+d^{3}+10(ab+ac+ad+bc+bd+cd)\le 64\]
\[ \bigtriangledown\]
$ \bigstar$ [b]Prolem 19[color=green] [Nguyen Thuc Vu Hoang][/color][/b]
Let $ a,b,c,d$ be positive real numbers such that $ abc+bcd+cda+dab=4$. Prove that:
\[ 16(a^{2}+b^{2}+c^{2}+d^{2})+48\ge 7(a+b+c+d)^{2}\]
\[ \bigtriangledown\]

[quote="zaizai-hoang"]
$ \bigstar$[b] Prolem 24[/b][color=green][b] [Pham Van Thuan][/b][/color]
 Let $ a,b,c,d$ be non-negative real numbers such that $ a^{2}+b^{2}+c^{2}+d^{2}=1$. Prove that:
\[ a+b+c+d\ge a^{3}+b^{3}+c^{3}+d^{3}+ab+bc+cd+da+bd+ac \]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135903
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158933

[quote="pvthuan"]For problem 23, look at

http://www.mathlinks.ro/viewtopic.php?search_id=2076432006&t=99927

Up to now I can claim that all three variable inequalities, four variable inequalities (polynomial and with fractions, without square roots) can be solved using the symmetric elementary expressions $ p,q,r$. A large section of my book demonstrates the ideas and techniques needed to solve such types.[/quote]
I am really curious about that. All four variable inequalities can be solved using $ p,q,r$. Can you solve the following inequality by $ p,q,r$?
Let $ a,b,c\ge 0$ such that $ a+b+c+d=4$. Prove that:
\[ abc+bcd+cda+dab+(abc)^{2}+(bcd)^{2}+(cda)^{2}+(dab)^{2}\le 8\]

For problem 24, argady, can you post solution for remain case? It means: $ p\ge \frac{3}{4}$