geometry Harvard college Princeton Stanford Yale Gauss

Just about every college in the United States offers math as a major, and there are lots and lots of colleges with good math programs. The question is more of which one is the right fit *for you.*

Do you like big programs, that often offer more diversity in classes? Or smaller programs, that usually offer more intimate classroom settings? Do you like colleges that focus a lot of their resources and attention on the top few students (like in an honors college), or ones that are more ecumenical?

Are you interested in double majoring with someone else, or integrating your math education with some other areas of knowledge? Do you want to live close to home? Does money affect your decision? Do you like large cities, small towns, college towns, athletic schools, ethnicly diverse schools?

These are just a few of the questions that you have to ask yourself to find the right program for you. The worst thing you can do is pick a college based on what *other* people think is the best one. Luckily there are enough great math programs at a diversity of types of schools to fit just about anyone's preferences. It just takes figuring out what system suits you best, and then doing your research to find them.

Incidentally, since I see you're from Michigan, you will get a great deal on your education if you go to UMich, which has a quite good honors program in mathematics.  Although the massive-state-school format is not for everyone.

[quote="hikari"]Does anyone know of colleges that have a high emphasis on math?[/quote]

What do you mean by "a high emphasis on math"?

I could say a lot more now, because I have been researching this issue for quite a while for my family, but please tell me more about what you are looking for.

I've taken out the discussion of affirmative action and moved it to the Round Table forum, as it's more appropriate to be discussed there.

When I say "a high emphasis on math" I mean that I want to find a place that will help me further develope my mathematical talents. I would like the school to have a strong math program that is open mostly to people who are good at math. The reason for that is I usually get stuck in math classes where I have to listen to a review of the most basic concepts, and I think it would be beter for me to find a place that has a fast paced program that is not necessarily an honors program, though I am open to them if they turn out to be right.

Hikari

[quote="hikari"]When I say "a high emphasis on math" I mean that . . .  I would like the school to have a strong math program that is open mostly to people who are good at math.[/quote]

Okay, I kind of figured that that was what you meant.

The first thing you would want to check for is whether that college expects ALL applicants to take an SAT II subject test in math (the IC or the IIC) and whether it tests the ability of students before placing them into math courses. You definitely do NOT want a college in which many entering students guess their own level and start math classes that they can't finish.

At my alma mater, the University of Minnesota, you could find a track of math courses (they would mostly be labeled "honors" math courses) that would develop that higher level of understanding, and give you a chance to meet other talented individuals. I wouldn't advise you to apply to any college any worse than the the U of MN, as that would be a waste of opportunity for you. I expect my son, who is basically an automatic admit at the U of MN for as long as I remain a Minnesota resident, to apply to some of the small number of colleges that are arguably "better" than the U of MN. I'm still figuring out which schools those are. Names of colleges I hear from a friend whose daughter is applying to college this year include Harvard, Princeton, U of Chicago, Stanford, M.I.T., Caltech, and a few others. (These are not listed in any particular order.) Even at those schools, there are multiple tracks through the math courses--read the course catalogs carefully to figure out which courses are the courses for the best-prepared students.

How many years do you have to prepare to meet this goal? What preparations are you making now?

I just applied to college last year (being that I'm a freshman now) and I was looking at roughly the same thing.  The fact of the matter is that you will have a better experience in mathematics at just about any college that has an honors track than you have at high school, because the degree of self-selection is much greater.  That is, people taking higher-level or more advanced college math classes are doing it because they are much more math-oriented.  People taking higher-level high school math classes are doing it because, well, high school classes aren't really that hard for very capable people no matter what, so they [i]can[/i].
I would add Harvey Mudd College on to tokenadult's list.  I also know that Carleton College (which I may have mispelled) in Minnesota, which is a small liberal arts school, has a quite good math program.  Nearly every big state school has some sort of honors math program or degree.  I happen to know that UMich's is quite good.  I generally feel that Amherst can be relied upon to be pretty good at just about anything, including math.  Yale is trying to increase the number of math-kids that it gets, which generally has been small compared to Harvard and M.I.T.  I don't really know how that's going.  They do have Benoit Mandelbrot somewhere there, which is kind of neat.  You would probably not end up unhappy at any of these schools, at least not because of mathematics.

Oh, and I'm sure Gauss (the resident Southerner) can tell you about schools from the part of the country I've left out.  Duke is very good -- they win the Putnam competition about every third or fourth year.  UNC, UVA.  I think Rice isn't bad, either.  But I don't know very much about any of these.

I'm from the south, but I did my undergrad at Princeton. I know a little bit about some other school's departments though. The thing is, that just about any top tier school has a solid math department - and the basic classes and curricula don't vary all that much from school to school. You would be surprised to find that most of them offer more or less the same classes, and use the same textbooks as each other. But that doesn't mean that your experience would be the same at each school!!

Some schools have bigger math departments and some have smaller. And some have fundamentally different approaches to education. That can affect everything from the number of diversity requirements that you have to take; to the sizes and structure of the classes; to the amount of work that you have to do.

I have spent some time in the math departments at Princeton, Harvard, MIT, Washington University, Rice, and the University of South Carolina - and they all are fairly different. Which one is better? Well it really depends on the individual person!!

The best advice I can give you is to go and visit as many schools as you can. Take a look around ,and talk to people who go there. See what kind of feel you get from the place, read the course cataloge to see what classes they offer each semester, maybe sit in on a few classes. Find out if they have an active undergradate math club or if the students in the department are close knit. Are they the kind of people you could see yourself "hanging out" with for four years. Do they have a good math departmental library? And this is not even to mention all of the things you should look at outside of the math department - which are equally as important!!

If the prestige factor is most important to you (and it can be useful, especially in certain professions like business or finance), then of course "big name" schools like Harvard, Princeton, Yale, Duke, Stanford, etc will have some benefit to you. But for getting into grad school, it's most important to go somewhere where you can do well, get to know some of the faculty, and enjoy your experience there.

Thanks to everyone who has posted advice for this topic. If anyone has more, feel free.
To answer tokenadult's questions, I am hoping to start college in the fall of 2005. Right now I am preparing by finding out the branches of math I like most. I am just starting to research colleges. I started about a month ago and have found a few that sound really great. I started this topic to see if anyone knew of good ones and I am looking into some of the ones that people have listed.

Hikari

This is probably too far forward for you (it certainly is for me, and I'm working two years ahead of you), but if you get your heart set on a [i]graduate[/i] school, it's probably a good idea [i]not[/i] to go there for undergrad -- most schools don't accept very many of their own students as graduate students.  This is probably not something that should be of terrible importance to you at this point, but if you do happen to decide that you are going to do your PhD thesis with John Conway as your advisor, you might not want to go to Princeton undergrad.

Honestly, I don't think reputation for undergraduates is particularly important. Undergraduates don't usually get to work with world-famous professors anyway -- they have better things to do. As a graduate student, everything is different. I, for one, am not going to a prestigious college, but I can almost guarantee I'll get more math classes in than just about anyone at a prestigious school. Whether I'll get any sort of a decent education outside of mathematics is up to me (or, more precisely, my desire to take way more classes at once than any sane person would dream of taking), but with at least four math classes per quarter, I'll be sure to learn plenty of math here.

In the  hope that info is useful the message may be a little 'politically incorrect'  but  with the disclaimer that this is my personal opinion (gotten from my son's friends and other friends) here I am going to name some schools..

Duke - This has excellent undergradute math (and physics etc) department and the "other coach K" recruits  good math students.    I know many stars from mathcounts.IMO/USAMO have been activly recruited there and were/are very happy there. Check that out. For last few years, this school have been doing extremely good in Putnam contest. One of the very few top school which even gives academic merit scholorships.

Caltech -  Very good but atmosphere can be a little tense.
MIT       -  Excellent.
(IMO - These may pay less attention to undergraduate than Duke)

Less known but excellent schools (I went to graduate school here) are  Indian Institue of Technology. There are a few (about 7) campuses in India- These are primarly for Engineering but do have excellent Math program too. ..

The admisson, here, is purely based on merit, and selection criteria is much higher than any of the top schools in USA. (2% compared to 10% in harvard)  (It is not uncommon for students who had Ivy league school as a backup school ) but if you are really very good you may  like to check that out.  They have web pages (serch for IIT)....As quality of students and Faculty goes IIT's are  second to none..if you are willing to go far.. check that out...

Regards

Hope this helps.

yeah, my dad's gone on sprees about how IIT's are awesome nation and super hard to get into

[quote="Gyan"]Less known but excellent schools (I went to graduate school here) are  Indian Institue of Technology. There are a few (about 7) campuses in India- These are primarly for Engineering but do have excellent Math program too. ..

The admisson, here, is purely based on merit, and selection criteria is much higher than any of the top schools in USA. (2% compared to 10% in harvard)  (It is not uncommon for students who had Ivy league school as a backup school ) but if you are really very good you may  like to check that out.  They have web pages (serch for IIT).[/quote]

I appreciate you mentioning the various IIT campuses in India. I'd like to know a lot more about those. I am not too familiar with the situation in India. (Taiwan and other east Asian countries are the places I have lived and visited outside the United States.) There are some very impressive students whose parents are first-generation immigrants from India in the [url=http://www.itcep.umn.edu/umtymp/]accelerated math program[/url] my son now attends at the U of MN. I'd like to figure out how some of that Indian math-loving culture can come over here to Minnesota--the schools here could sure use it.

Here's an interesting list that I found of the mathematicians who are members of the National Acadamy of Sciences: http://www4.nationalacademies.org/nas/naspub.nsf/urllinks/$$Section11?OpenDocument&Count=50000

All of the people on this list are not only great mathematicians, but also tend to be great expositors and teachers. Going to schools where you can learn from some of these people could be a major advantage. Beware though, that a few of them are getting up in age and don't teach much anymore. Also remember that this is by no means a complete list of great mathematicians or teachers in America.

Note: You can also do a search for acadamy members by state or institution.

[quote="JBL"]Hmm . . . this is only very tenously related, but Paul Salley at the University of Chicago is the best teacher I've ever seen, ever.   Ever.[/quote]

Well, I think this relates well to the overall subject of the thread, which is criteria an applicant might apply to deciding where to go to college to study math. I happen to know locally two U. of Chicago Ph.D.s in math (husband and wife, who met during their graduate program) and at least one of them was a student of Paul Sally's. I'm curious about what setting you saw Sally teaching in--during a college visit, to help you decide where to apply, or during a summer program, or on Internet streaming video, or where? I get the impression that the teaching at U. of Chicago in math has been pretty high quality for a long time, yet nonetheless people continue to apply to other schools and accept their offers of admission.

Part of the reason I, who was interested in EVERYTHING when I was in high school, never considered majoring in math in college was that I mistakenly believed that my high school teachers were at least somewhat like what college math teachers would be like--in my high school, the foreign language teachers were sometimes star teachers while the math teachers were mostly sports coaches who just taught school as their "day job." Today opportunities abound (compared to the 1970s) for high school students to see REAL math teachers in action, and I think it would help the original poster to discuss how a young person in high school could get a glimpse at how mathematics is taught at a university level.

Is he the guy with the eye patch?  If so, I think I've heard of him from a friend of mine who goes there.  I'd be interested in hearing your impression of U of Chicago -- although I'm seriously interested in it (and was accepted EA :-) I've yet to visit.

[quote="JBL"]Hmm . . . this is only very tenously related, but Paul Salley at the University of Chicago is the best teacher I've ever seen, ever.   Ever.[/quote]

Yep, he's the one with the eye-patch.  I sat in on a class of his when I visited while deciding where to apply/attend.  I spent a very long time choosing between Harvard and Chicago, and his class was part of the reason.  (I think the math department there specifically points visiting students interested in math to his class because, well, he's just so good.)  I think Chicago does have quite a good reputation over-all (not just Salley) for teaching math.

As for my overall opinion, and why I think fewer people accepted go there:  I just saw a friend of mine who goes there and we were talking about schools.  Basically, Chicago has a reputation as a grind school, and it's not inaccurate.  By her third night of classes, my friend (who is not the kind of person who struggles academically, ever) was up until 2 AM doing homework.  They have a quarter system, which has the advantage of allowing you to only take 3 courses many quaters but the disadvantage of meaning that everything is squeezed into 5 fewer weeks -- classes last only 10 weeks, not 15.  This creates a much higher-pressure atmosphere.  It has a very intellectual atmosphere -- there is not very much party culture at all, which I thought was great, but I am in the minority of 18-year-olds in that respect.
I think that combo of intellectual atmosphere and hard work turn many students off.  Also, it's not in prime territory geographically, since California and the East Coast generate a proportionally larger body of college students than their population dictates, and they have bigger-name schools nearer by.

Other general points of knowledge -- due to location, it is less diverse (and possibly slightly more conservative) than the East Coast schools.  (For the political point, I'm just guessing -- both my friend and I are far enough left that nearly any group on either of our campuses is more conservative than we are, and we didn't really get to do a comparative-politics study.  Maybe Harvard Socialist Alternative is more left-wing.  But I digress.)  It's also very, very cold and windy in Chicago.  ("It's nice being in Boston, because the wind doesn't prevent you from walking in a straight line.")  If you're going to visit, Alison, you've probably picked the worst time of year.

One last thing -- the Chicago curriculum is more structured than many other schools.  Their Core offers fewer options for fufilling requirements, since each covers a smaller intellectual area.  So there are a lot of courses whose curricula are along the lines of "Dead White Male Authors, 1800-1850."  Obviously, that is an exaggeration, but not a very large exaggeration.  Probably the thing that chased me away from Chicago in the end was the prospect, presented to me by a sophomore math major there, that I might have to take biology.  (The decision, in the end, was based on rather trivial things -- I decided to go to Harvard leaving the house one morning in order to make my parents stop bothering me to decide.  Not that I think I would have chosen differently given more time, necessarily -- I probably just wouldn't have come to a decision.  Except for that biology . . .)

That was rather long, and I hope it flowed somewhat and didn't just come out as a random-association-type excersize.  Alison, I really liked Chicago, far more than 'mathier' schools like Caltech or MIT.  But I'm not by any means the same person you are, so I don't have any idea if my feelings carry over.  But I do think you should give it a serious consideration, take a visit (wear warm clothing).

Oh, one last thing I forgot that doesn't fit in anywhere else:  the cafeteria food at Chicago was better than any other school I visited.  I mean, it was quite good food.  Yeah.

I've been to the University of Chicago a couple of times. My girlfriend used to live practically right next door, and I have a cousin who lives very near there now. Their math and physics departments both have very good reputations - especially for undergraduate teaching.

The school has, as JBL described, somewhat of a reputation of being a "grind" school. I hear that the students there generally have more work than students at some other top schools. Part of that may be because of the quarter system, and some of it may just be the educational philosophy there. I think they once had some kind of "Great Books" philosophy, where students were encouraged to read the "great books" of their discipline. I'm not sure if this is still true, but I generally found that they had a lot of reading to do.

The school as a whole has a very intellectual atmosphere. They don't have a major sports program, and I don't think the party scene is very strong on campus (the city though is another matter); and the undergraduate community tends to be pretty closely knit. They also have some great bookstores nearby.

I disagree with Joel about the diversity though. I found the school to be pretty diverse academically, culturally, and politically (compared to most other elite universities). It's in a fairly diverse area of Chicago, which is itself a fairly diverse city. So I think you would have an exposure to a lot of different people and cultures there. It's also one of the premier Universities in the midwest, so it draws a lot of students from all over the midwest and the nation as a whole.

The math department I would characterize as fairly large for a school of it's size (closer to the sizes of Harvard or MIT's, than to Princeton or Yale's). Maybe the reason we don't hear about it as often on the East Coast is because they attract most of their students from the midwest. Here's an article on the size and popularity of their math department: http://www-news.uchicago.edu/releases/03/030407.math.shtml

They also offer two different courses of study in Mathematics - a BA and a BS degree. That adds a certain degree of flexibility in your options there. You can read about the difference between these two degree here: http://zaphod.uchicago.edu/undergrad/concentration.html

I found that the students seem to be very happy and take a lot of pride in their school. Here are some student reviews of the math department: http://www.studentsreview.com/IL/UC_c.html

And I can confirm that the food there is very good :D As is the food in the surrounding area of Chicago. Chicago is a beautiful city with lots of intellectual, cultural, and fun things to do. You just have to beware of the weather there!! The kind of cold that I felt when I was there is unlike any other kind of cold I have ever felt before! (though note I'm from South Carolina!)

The people that live in Chicago though, seem to have a lot of pride in their city and don't seem to mind the weather at all. In fact they are some of the most "ourdoorsy" people I have ever seen in my life. They have a tremendous number of parks there.

I also considered Chicago when I was applying to colleges (back in the mid 90s). The weather was a big reason why I chose not to go there though. Being from the south (and having basically never seen snow before), I think that would have been too much of a climate shock for me! I chose the moderately warmer Princeton instead. :D

Thanks, guys, for the info about the University of Chicago.  Joel, what were the qualities that made Prof. Salley such a good teacher?

Wow... I didn't expect such detailed responses.  Thanks a lot!


Can you please help me with the following problem:

Two of the roots for x of a polynomial equation with intergal coefficients are sqrt(5+sqrt(3)) and sqrt(7-sqrt(2)). If the equationis 209x^4+x^a-24x^k-px^2+m=0, find the value of a+k+p+m.

pardon my ignorance, but couldnt you also just plug the roots in for x and solve?

There are too many variables to just plug in and solve.  Is there the stipulation that $a, k \le 4$?

i think it's $x^8-24x^6+209x^4-778x^2+1034$=0

since we wanted integers, and we knew sqrt(5+sqrt(3)) was a root, i made -sqrt(5+sqrt(3)) a root, etc, since this would help get integer coefficients.  so then the polynomial can be written like $(x-\sqrt{5+\sqrt{3}})(x+\sqrt{5+\sqrt{3}})(x-\sqrt{7-\sqrt{2}})(x+\sqrt{7-\sqrt{2}})\cdot(stuff)$, and that simplifies to $(x^2+\sqrt{2}-7)(x^2-\sqrt{3}-5)(stuff)$.  you know you want integers, so we make $stuff$ =$(x^2-7-\sqrt{2})(x^2-5+\sqrt{3})$, and you get the polynomial i gave at the top.

That seems to work, but how would you show that it's the only solution, and how would you generalize?  I don't think there's an easy way to solve this problem without a partly guess and check solution though.

wouldn't the conjugates have to be the roots in order for us to get integral coefficients?

Yeah, but you just have to "trust" that it turns out as something compatible with the 209.  Actually, I think there might have been a solution that dealt with this somewhere, but I don't know where...


Let a,b,c such that $ a^4\plus{}b^4\plus{}c^4\equal{}3$.Prove that $ 3(a^2\plus{}b^2\plus{}c^2)\minus{}2(ab\plus{}bc\plus{}ca)\geq 3$

[quote="natural"]Let a,b,c such that $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3$.Prove that $ 3(a^2 \plus{} b^2 \plus{} c^2) \minus{} 2(ab \plus{} bc \plus{} ca)\geq 3$[/quote]
Since $ a^4  \plus{} b^4  \plus{} c^4  \equal{} 3 \Rightarrow a^2  \plus{} b^2  \plus{} c^2  \equal{} \sqrt {3 \plus{} 2\left( {a^2 b^2  \plus{} b^2 c^2  \plus{} c^2 a^2 } \right)}$
Hence the inequality is equivalents with
\[ 3\sqrt {3 \plus{} 2\left( {a^2 b^2  \plus{} b^2 c^2  \plus{} c^2 a^2 } \right)}  \minus{} 2\left( {ab \plus{} bc \plus{} ca} \right) \ge 3
\]
\[ \Leftrightarrow 18 \plus{} 14\left( {a^2 b^2  \plus{} b^2 c^2  \plus{} c^2 a^2 } \right) \ge 8abc\left( {a \plus{} b \plus{} c} \right) \plus{} 12\left( {ab \plus{} bc \plus{} ca} \right)
\]
But it's true obviously since
\[ 8\left( {a^2 b^2  \plus{} b^2 c^2  \plus{} c^2 a^2 } \right) \ge 8abc\left( {a \plus{} b \plus{} c} \right)
\]
And
\[ 6\left( {a^2 b^2  \plus{} b^2 c^2  \plus{} c^2 a^2 } \right) \plus{} 18 \equal{} 6\left( {a^2 b^2  \plus{} 1} \right) \plus{} 6\left( {b^2 c^2  \plus{} 1} \right) \plus{} 6\left( {c^2 a^2  \plus{} 1} \right) \ge 12\left( {ab \plus{} bc \plus{} ca} \right)
\]

[quote="natural"]Let a,b,c such that $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3$.Prove that $ 3(a^2 \plus{} b^2 \plus{} c^2) \minus{} 2(ab \plus{} bc \plus{} ca)\geq 3$[/quote]
The following statement is stronger.

If $ a,b,c$ are real numbers such that $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3$, then

$ (1 \plus{} \sqrt 2)(a^2 \plus{} b^2 \plus{} c^2)\ge \sqrt 2 (ab \plus{} bc \plus{} ca) \plus{} 3$,

with equality for $ a\equal{}b\equal{}c$, and again for $ a\equal{}b\equal{}\frac 1{c}$ or any cyclic permutations.

The following is nice.

Let $ a,b,c,d,e$ be real numbers such that $ a^4\plus{}b^4\plus{}c^4\plus{}d^4\plus{}e^4\equal{}5$. Then,

$ 7(a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2)\ge (a\plus{}b\plus{}c\plus{}d\plus{}e)^2\plus{}10$,

with equality for $ a\equal{}b\equal{}c\equal{}d\equal{}e$, and again for $ a\equal{}b\equal{}c\equal{}d\equal{}\frac 1{e}$ or any cyclic permutation.


geometry
geometric transformation
reflection
circumcircle
trigonometry
perpendicular bisector
geometry unsolved

[quote="shobber"]Let $ABCD$ be a convex quadrilateral with $\angle CBD = 2 \angle ADB$, $\angle ABD = 2 \angle CDB$ and  $AB = CB$.

Prove that $AD = CD$.[/quote]

let $m\angle ABD=2m\angle BDC=2x$ y sean $m\angle CBD = 2 m\angle ADB =2y$, it's easy to discover that $m\angle CAD=90-x$ and $m\angle ACD=90-y$ so, aplyyn the sinus law we have:

$sin y. cos y. sin 2y. cos (x+y)=sin x. cos(x+y).sin2x.cosx$

so, [i]multiplicando[/i] by 2 we have:

$sin^2(2x)=sin^2 (2y)$

so, as $0<x+y<120$, $x=y$ and $90-x=90-y$ so $m\angle CAD=m\angle ACD$ and $AB=CB$

PS. I think when you post this sort of problems you're looking for a sinthetycal solution, just using pure geometry, :P  but all the solvers of your problems use trigo  :P I'm so sorry....

[img]http://img431.imageshack.us/img431/8826/belgian7wr.png[/img]

$BK$ is the bisector of $\angle DBC$
$BH$ is the bisector of $\angle DBA$

$\left( \frac{AB}{BD}=\frac{AH}{DH} \land \frac{BC}{BD}=\frac{CK}{DK}\right) \Rightarrow \frac{DH}{DK}=\frac{AH}{CK}$

$\triangle BKD= \triangle DHB$ so $\frac{DH}{DK}=\frac{BK}{BH}$

$\frac{BK}{BH}=\frac{AH}{CK}$

so triangles $BKC$ and $BHA$ are similiar, but $AC=BC$, so $\triangle BKC=\triangle BKA$ and $\angle HBA=\angle KBA$

$\angle BDA=\angle DBH$ and $\angle BDC=\angle DBK$ so $DK=DH$

We have proved that $HA=KC$ and $DK=DH$, hence $CK+KD=AH+HD$ q.e.d.

Reflect B over AC to B', and assume B' is not on BD, that is, CA is not perpendicular to BD. Note that B'A=B'C and also <AB'C=2<ADC. Thus, B' is the circumcenter of ADC. Let the parallel to BD through B' meet AC at X, and BD meet the circumcircle of ADC again at Y. Then, <CB'X=<ABD=2<CDB=2<CDY=<CB'Y (angles directed). Thus, B', X, Y are collinear, but B'X is parallel to DB which contains Y, contradiction. Thus, B' is on BD, CA is perpendicular to BD, so D is also on the perpendicular bisector of AC and DA=DC.

[hide="Solution"]Let $\angle BDC=y$ so that $\angle ADB=2y$ and let $\angle ADB=x$ so that $\angle DBC=2x$.Now , in $\triangle ADB$, by the Sine rule we have \[\frac{AB}{\sin x}=\frac{DB}{\sin(\pi-(2y+x))}\]Again, in ${\triangle DBC}$, we have \[\frac{DB}{sin(\pi-(2x+y))}=\frac{BC}{\sin y}\].Combining and utilising the equality of $AB$ and $BC$, we get $\sin (2y+x)\cdot \sin x=\sin(2y+x)\cdot \sin y$, which gives $\cos (3x+y)=\cos (3y+x)$.Now, $(2x+2y)=\angle ABC<\pi$, so, $(3x+y)+(3y+x)<2\pi$, which yields $x=y$.In $\triangle ADB$ and $\triangle CDB$, $BD$ is common side, $AB=BC$ and $\angle ABD=\angle DBC\implies AB=BC$ as the two triangles are congruent.[/hide]

$EB,FB$ bisect $\angle CBD,\angle ABD$ respectively, then $FBED$ is a parallelgram, $G$ is the mid-point of $EF$

By sin law, $\frac{AB}{sin(\theta+\alpha)}=\frac{AF}{sin\alpha},\frac{BD}{sin(\alpha+\theta)}=\frac{BE}{sin\alpha},$hence $\frac{AB}{BD}=\frac{AF}{BE}=\frac{AF}{DF},$ similarly $\frac{BC}{BD}=\frac{EC}{DE},$ thus $\frac{AF}{DF}=\frac{EC}{DE} \rightarrow EF \parallel AC \rightarrow H$ is the mid-point of $AC \rightarrow BD \perp AC \rightarrow AD=CD.$

Let Μ the intersections of AD and CB  extensions  and   N the intersections of  AB and CD extensions  .
Then we have:

\[\begin{array}{l}
\left. \begin{array}{l}
BM = BD = BN\\
\\
BC = BA
\end{array} \right\}\mathop {}\limits^{}  \Rightarrow \mathop {}\limits^{} tr.ABM = tr.CBN\mathop {}\limits^{}  \Rightarrow \\
\\
N\hat CB = M\hat AB\mathop {}\limits^{}  \Rightarrow \mathop {}\limits^{} tr.ABD = tr.CBD\mathop {}\limits^{}  \Rightarrow \mathop {}\limits^{} DA = DC
\end{array}\]

[quote=sayantanchakraborty]Appluing sine rule and simple calculation leads to <ADB=<CDB and so triangles ADB and CDB are congruent.[/quote]

A possible intermediate pathway:

Draw angle bisectors of CBD and ABD (call them BE and BF)=> show that [ABE] = [CBF] => congruent angles via sine rule

*showing that the areas are equal involves some basic length/ratio manipulation

less clever stewart solution
-------------
Let $P,Q$ be points on $AD,CD$ such that $BP,BQ$ bisect $\angle ABD,\angle CBD$ respectively. Let $AD=p,$ $CD=q,$ $AB=CB=m,$ $BD=n,$ and WLOG let $m+n=1.$

Note that $BPDQ$ is a parallelogram. From angle bisector theorem, we have $AP=mp,$ $PD=BQ=np,$ $CQ=mq,$ $QD=BP=nq.$ Now by Stewart,
\begin{align*}
mnp^3+n^2pq^2&=m^2np+mn^2p &\implies mp^2+nq^2=m(m+n)\\
mnq^3+n^2p^2q&=m^2nq+mn^2q &\implies mq^2+np^2=m(m+n).
\end{align*}
Then we have
$$(m-n)(p^2-q^2)=0.$$

Suppose that $m=n,$ then $BP\perp PD,$ and
$$90^{\circ}=\angle PBD+\angle BDA=\angle PBD+\angle DBQ=\angle PBQ=\frac12\angle ABC,$$
so $A,B,C$ are collinear, but $ABCD$ is convex, contradiction.

Then $m\ne n,$ so $p=q,$ as desired.

Let $AB \cap CD = E$ and $AD \cap BC = F$. Notice $$\angle DFB = \angle CBD - \angle FDB = 2 \cdot ADB - \angle ADB = \angle ADB = \angle FDB$$ so $BF = BD$. Similarly, we conclude $BE = BD$. Thus, $BAF \cong BCE$ by SAS, and $\angle AFB = \angle CEB$. The rest is trivial. $\blacksquare$

[b]Remark:[/b] The motivation to construct isosceles triangles should be fairly obvious, as $2 \cdot ADB = \angle CBD$ is screaming for an application of exterior angles.

Let $X=\overline{AB}\cap\overline{CD},$ $Y=\overline{BC}\cap\overline{AD},$ $\angle ABD=\alpha,$ and $\angle CDB=\beta.$ We see that $$\measuredangle CXB=180-(\beta+2\alpha)-(180-2\alpha-2\beta)=\beta$$ and similarly $\measuredangle BYA=\alpha.$ Hence, $BX=BD=BY$ and $\triangle BCX\cong\triangle BAY,$ implying that $\alpha=\beta.$ We are done as $\triangle CBD\cong\triangle ABD.$ $\square$

Note that problem is equivalent to the following:
[quote]Consider triangle $ABC$ with $BE,CF$ as its angle bisectors ($E,F$ lie on $AC,BC$, respectively). Prove that if $BF=CE$, then $CF=BE$. [/quote]
Indeed, by angle bisector theorem, we get that $\triangle ABE\sim\triangle ACF$, thus $BFEC$ is cyclic. This means that $AB=AC$ and hence $BE=CF$.


I don't understand how you got from  (a+b)(a-b) = 100+99+98+97...+1 could you please explain?  :maybe:

Consider the terms consecutively. For example, $ 100^2\minus{}99^2\equal{}(100\minus{}99)(100\plus{}99)\equal{}100\plus{}99$. The same is true for all the pairs, and then summing them is not that hard.


function
induction
real analysis
real analysis unsolved

Let f from (a,b) to R  be a cont. function satisfying f(x+y/2)<=1/2*(f(x)+f(y)) (i.e. mid pt. convex) for all x,y in (a,b).show f is convex on (a,b).

By induction you have that $ af(x)+(1-a)f(y) \geq f(ax+(1-a)y)$ for all $ a$ of the form $ m/2^k$.

To see the above, note that, for example

\begin{eqnarray*}
\frac{f(x_1)+f(x_2)+f(x_3)+f(x_4)}{4} &\geq& \frac{f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_3+x_4}{2}\right)}{2} \\
&\geq& f\left(\frac{x_1+x_2+x_3+x_4}{4}\right)\\
\end{eqnarray*}

then use induction to get

\[ \sum^{2^n}_{i=1} \frac{f(x_i)}{2^n} \geq f\left(\sum^{2^n}_{i=1} \frac{x_i}{2^n}\right)
\]

and then set $ m$ of the $ x_i$'s equal to $ x$ and the other ones equal to $ y$.

Numbers of the form $ m/2^k$ are dense in $ (0,1)$; now use continuity.


If $a,b,c$ are sides of a triangle prove or disprove this:
$a+b-c \leq \sqrt{(b+c-a)(c+a-b)}$


From some website...

Prof Matherjee had a high school maths teacher as fellow passenger on his flight from Kolkata to Delhi. "What's your family like?" asked the teacher. "Well, there's my wife Theeta and our two children Alfa and Beta." The teacher then asked: "That's great. How old are the kids?" The Professor replied: "Well to tell you mathematically, if $ a$ and $ b$ are integer values of their ages and if I say $ 10a \plus{} 2ab \minus{} 4b \minus{} a^2 \minus{} 2b^2$ is a positive integer, there is no reason why you'll not be able to deduce their ages." The teacher smiled and said: "Yes Professor, now I know their ages." How did he know?

Perhaps he was psychic, as there are several possible solutions to that equation.

a few thoughts
[hide]
well since we know that a and b each represent a child's age, we know that
a>0 and b>0 ( These > are greater than or equal signs, forgive my nonuse of LaTex)
it is assumed a child cannot be of negative age

simplifying a bit, we have the following

a(10-a) - 2b(b+2)  + 2ab > 0

(thanks to Begoner for catching my careless mistakes.)
so a must be less than 10, leading to a maximum value for (10-a)a is 25 (when a=5)
and  2b (b+2) +10b <25, leading to a maximum value the expression being 16 ) (b=1)

[/hide]

[quote="cipos2"]simplifying a bit, we have the following

10a(1-a) - 2b(2-b) + 2ab > 0[/quote]

It's $ a(10\minus{}a) \minus{} 2b(2\plus{}b) \plus{} 2ab$.

wow whoops, i didn't catch these two careless mistakes....
my apologies

I'm sure this is clearly a "simplest form" question. You should more likely be trying to simplify it to the multiplication of two values to get EXACT results, as the teacher in question did.


I'm typing a report for AP US History using LaTeX. Is there a way for me to double space it?

Use:

\linespread{2}

Or use the setspace package and the \doublespacing command.


If three coins are tossed at the same time, what is the probability of getting two tails and one head? Express your answer as a common fraction.

The probability is $ 3(\frac{1}{2})^2\frac{1}{2}\equal{}\boxed{\frac{3}{8}}$. Hopefully.... :maybe:

Of 3 tosses, there are 8 possible outcomes. You need 2 of them to be heads. We can choose this in $ \binom{3}{2}\equal{}3$ ways, for $ P\equal{}\frac{3}{8}$

[quote=AwesomeToad]Of 3 tosses, there are 8 possible outcomes. [b]You need 2 of them to be heads[/b]. We can choose this in $ \binom{3}{2}\equal{}3$ ways, for $ P\equal{}\frac{3}{8}$[/quote]

You need 2 to be tails. But the answer wont change either way

 ;) Sorry for reviving


The polynomial $P(x)=x^{3}+ax^{2}+bx+c$ has the property that the mean of its zeros, the product of its zeros, and the sum of its coefficients are all equal. If the $y$-intercept of the graph of $y=P(x)$ is $2$, then what is $b$?

A. $-11$
B. $-10$
C. $-9$
D. $1$
E. $5$

[hide="uhhhh"]ok so the sum of the roots is -a, so -a/3 is the average. the product of the roots is c. the y intercept is found by inputting 0 for x, so since P(0)=2, c=2. so -a/3=2 --> a=-6. the sum of the coefficients is 2 so -6+1+2+b=2 --> b=5, e  [/hide]


inequalities
geometry
inradius
trigonometry
inequalities unsolved

[hide]Denote $x+y=a,y+z=b,z+x=c$, we have $a,b,c$ are length sides of a triangle, When , your problem from a acquainted inequalities .  :wink: [/hide]

:D OK if let as you please prove your "acquainted" inequality
$a^{2}+b^{2}+c^{2}\le 4\sqrt{3}S\frac{R}{2r}$
For all triangle $ABC$, side $a,b,c$ area $S$, cricumradius $R$ , inradius $r$
Is it clear?  :D

No, it shall is $\frac{m_{a}m_{b}m_{c}}{m_{a}^{2}+m_{b}^{2}+m_{c}^{2}}\geq r$. Where $m_{a}$ is length of segment joining $A$ with midpoint of $BC$, etc.

You should check your calculation  :D or you should explain clearly how can you get your problem from my inequality ? :D

Exists triangle with length of sides are $\frac{2}{3}m_{a},\frac{2}{3}m_{b},\frac{2}{3}m_{c}$ :D

And you think inradius of your triangle (median triangle) "equal" inradius of triangle $ABC$
In my inequality we need prove $\frac{\sqrt{3}abc}{a^{2}+b^{2}+c^{2}}\ge 2r$
If you want to change into median triangle you need to prove
$\frac{4\sqrt{3}m_{a}m_{b}m_{c}}{3(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})}\ge 2r_{m}$ and you shoudn't think $\sqrt{3}r_{m}= 2r$ or between them exixt an inequality.
That's a funny thing!  :D  :D :D

[quote="gemath"]prove $\frac{\sqrt{3}abc}{a^{2}+b^{2}+c^{2}}\ge 2r$
[/quote]
Use $a=2R\sin{A},...,r=4R\prod\sin{\frac{A}{2}}$ and $\sum \sin^{2}{A}=2+2\prod \cos{A},\sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}$. We need prove $\sqrt{3}\prod \cos{\frac{A}{2}}\geq 1+\prod \cos{A}$. This is easy problem, are you agree?  :o

:D Really I don't have any solution therefore I have to post it in Unsolve box , if you think it is easy why don't you finish your post ?

Ok now stop this game here because my inequality is false, and it isn't easy as you thought !  :D  :D Anyway thanks :P

[quote="gemath"] and it isn't easy as you thought !  [/quote]

 :D I am sorry  :P


I looked at the IMO photos at http://www.imo2004.gr and I must admit I mostly cannot map any of the faces to people. That is why I really would appreciate it if people could post the link to the image and say "who is who".  :)

I go first:

http://www.imo2004.gr/fimo/foto_4/023_athens/ac.JPG

The guy with the blue T-shirt is Eric Mueller and the guy right to him (mixed coloured shirt, mostly red) is Dieter Gronau.

http://www.imo2004.gr/fimo/foto_5/029_athens/P1010025.JPG

The guy with the blue shirt in the front is Titu Andreescu and right to him is Mircea Becheanu (blue/white T-Shirt, glasses).

In the second Romanian picture orl described, the person to the right of Mr. Becheanu (from our perspective) is Bogdan Enescu (deputy of Romanian team this year). The hair that can be seen just below Titu Andreescu most probably belongs to yours truly :D

I see you all want to remain anonymous or don't you really know anybody on the IMO pictures. Anyway here is the picture of the [url=http://www.mathematik-olympiaden.de/IMOs/imo04pic.htm]German IMO team.[/url]  :)


Points $A$ and $C$ lie on a circle centered at $O$ with radius $\sqrt{50}$.  There exists a point $B$ inside the circle such that $\angle ABC = 90^\circ$, $AB=6$, and $BC=2$.  Find $OB$.

If BP=x then BQ=3x

Thus MP=(6+x)/2 and BM=|(6-x)/2|
Also  NQ=(2+3x)/2 and BN=|(2-3x)/2|

Hence by Power of a point \[ \begin{array}{l} AB.BQ = r^2 - OB^2 \\ \Rightarrow OB^2 = 50 - 6x \\ \end{array}  \]

Also\[ \begin{array}{l} OB^2 = BN^2 + BM^2 \\ \Rightarrow OB^2 = (3 - \frac{x}{2})^2 + (1 - \frac{{3x}}{2})^2 \\ \Rightarrow 50 - 6x = 9 - 3x + \frac{{x^2 }}{4} + 1 - 3x + \frac{{9x^2 }}{4} \\ \Rightarrow x^2 = 16 \\ \Rightarrow x = 4 \\ \end{array}  \]

So \[ OB = \sqrt{50 - 24}  \]
Thus \[ OB = \sqrt{26}  \]


inequalities
function
inequalities proposed

Given that $0<a,b,c<3,$ and $a^n+b^n+c^n=3$, prove the following inequality for any positive integer $n.$

[tex]\displaystyle \frac{1}{a^n(b+c)}+\frac{1}{b^n(c+a)}+\frac{1}{c^n(a+b)} \geq \frac{3}{2}[/tex].

Let $S$ be the LHS of the desired inequality.
From Cauchy :
$6S = S \cdot [(b+c) + (c+a) + (a+b)] \geq ( \frac 1 {a^{ \frac n 2}} + \frac 1 {b^{ \frac n 2}} + \frac 1 {c^{ \frac n 2}})^2 $
Now, from Jensen's inequality for the convex function $f(x) = \frac 1 {x^{ \frac n 2}} $, we have :
$6S \geq ( \frac 3 { (\frac {a+b+c} 3 )^{\frac n 2}} )^2 = 9$, hence result.

Note that equality occurs if and only if $a=b=c = 1$.

Pierre.

[quote="pbornsztein"]Let $S$ be the LHS of the desired inequality.
From Cauchy :
$6S = S \cdot [(b+c) + (c+a) + (a+b)]$ [/quote]
Why this?

I see the condition is $a^n+b^n+c^n=3$, not $a+b+c=3$.

Ooops... sorry I read the other one :blush:

Pierre.

Ok, let's try again. Hope this will be good this time...

Using the convexity of $f(x) = \frac 1 x$, we have :
$\frac 1 {a^n(b+c)} + \frac 1 {b^n(c+a)} + \frac 1 {c^n(a+b)} \geq \frac 9 {a^n(b+c)+b^n(c+a) +c^n(a+b)}$

In another hand :
$a^n(b+c)+b^n(c+a) +c^n(a+b) = a(3-a^n)+b(3-b^n) + c(3-c^n)$
$=3(a+b+c) - (a^{n+1} + b^{n+1} + c^{n+1})$.

From the inequality between means of order 1 and $n$, we have :
$a+b+c \leq 3(\frac {a^n+b^n+c^n} 3 )^{\frac 1 n} = 3$

From the inequality between means of order $n$ and $n+1$, we have :
$a^{n+1}+b^{n+1}+c^{n+1} \geq 3(\frac {a^n+b^n+c^n} 3 )^{\frac {n+1} n} = 3$

Thus $a^n(b+c)+b^n(c+a) +c^n(a+b) \leq 6$ hence result.

Pierre.

[quote="pbornsztein"]Let $S$ be the LHS of the desired inequality.
From Cauchy :
$6S = S \cdot [(b+c) + (c+a) + (a+b)] \geq ( \frac 1 {a^{ \frac n 2}} + \frac 1 {b^{ \frac n 2}} + \frac 1 {c^{ \frac n 2}})^2 $
[/quote]

Your original proof almost works. Note that from the given condition, we have $a+b+c\le3$ (by Power Mean) and $abc\le1$.

So $\displaystyle 6S \geq S \cdot [(b+c) + (c+a) + (a+b)] \geq \left( \frac 1 {a^{ \frac n 2}} + \frac 1 {b^{ \frac n 2}} + \frac 1 {c^{ \frac n 2}}\right)^2 \geq 9$, the last "$\geq$" is by AM-GM.

From this proof, it seems that the condition $a^n+b^n+c^n=3$ is too strong. It can be changed to $a+b+c\le 3$.

Here's my solution...

From Holder,

$\displaystyle \left( \frac{1}{a^n(b+c)}+\frac{1}{b^n(c+a)}+\frac{1}{c^n(a+b)} \right)^{1/3}
 \left( (b+c)+(c+a)+(a+b) \right)^{1/3}
 \left( a^n+b^n+c^n \right)^{1/3}$

$\geq (1+1+1) = 3$.

Thus,

$\displaystyle \frac{1}{a^n(b+c)}+\frac{1}{b^n(c+a)}+\frac{1}{c^n(a+b)} \geq \frac{27}{2(a+b+c)(a^n+b^n+c^n)} = \frac{9}{2(a+b+c)}$ $\geq \displaystyle\frac{3}{2}$,

since from Power-Mean, $a+b+c \leq 3$.

[quote="ThAzN1"]Here's my solution...
[/quote]
Your proof fails if we change the condition from $a^n+b^n+c^n=3$ to a weaker one $a+b+c\le 3$.  :D  :D

On the other thread, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5875  we were talking about one line proof of a French TST problem.

The IMO 95 problem can be proved in one line too.
[quote]Let $a,b,c>0$ with $abc=1$. Prove that
\[
\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}.
\][/quote]

$LHS = \frac{1/{a^2}}{a(b+c)} + \frac{1/{b^2}}{b(c+a)} + \frac{1/{c^2}}{c(a+b)} \geq \frac{(1/a + 1/b+1/c)^2}{2(ab+bc+ca)} = \frac{ab+bc+ca}{2}\geq \frac{3}{2}$. The 1st "$\geq$" is Cauchy-Schwarz, the 2nd "$\geq$" is AM-GM.

Haha, it does fit in one line.  :)

Lol.  :lol:

Well even if it's $a+b+c \leq 3$, Holder still does the trick  :D

$(LHS)^{\frac{1}{n+2}} ((b+c)+(c+a)+(a+b))^{\frac{1}{n+2}} (a+b+c)^{\frac{1}{n+2}} (a+b+c)^{\frac{1}{n+2}} \cdots (a+b+c)^{\frac{1}{n+2}}$

$\geq (1+1+1) = 3$,

where there are n $(a+b+c)^{\frac{1}{n+2}}$'s.

It follows that

$LHS \geq \displaystyle \frac{3^{n+2}}{2(a+b+c)^{n+1}} \geq \frac{3}{2}$.  :D  :D  :D

More general version:

[quote]Given that $a,b,c > 0$ and $a+b+c \leq 3$, prove that the following inequality holds for any [b]real[/b] $s\geq0$:
[tex]\displaystyle \frac{1}{a^s(b+c)}+\frac{1}{b^s(c+a)}+\frac{1}{c^s(a+b)} \geq \frac{3}{2}[/tex].[/quote]

[b]ThAzN1[/b], nice proof with Holder. I think you might want to refine a little bit for real power $s\geq0$ as above instead of positive integer $n$.

we can prove the result with the condition abc=1

yes we can assume[tex]abc=1[/tex] and prove the result[/tex]

why can we use the condition [tex]abc=1[/tex]
did i miss something guys!

ok! i see man :)
u talk about another problem :)


Can you help me with testing of numerical method to solve following type of equation :

$ A \frac {\partial^2 f} {\partial x^2} \plus{} B \frac {\partial f} {\partial x} \plus{} C f \minus{} \frac {\partial f} {\partial y} \equal{} 0$          ,  diff eq. (1)

[i]A,B,C[/i] can by any - also constant. Maeby one of them should be function of [i]y[/i] (to not have trivial solution).

I need [i]A,B,C [/i]and [i]boundary condition[/i] for which diff. equation [i](1)[/i] has analytical solution - to use it for testing precision of numerical solution.

Can you tell me any combination of [i]A,B,C+boundary.conditions[/i] with analytical  solution?

Ok, no answear at all.

Let say that  f = e^(-xy)   how will look A, B, and C ?

Or, let say that f = e^(xy) , after A, B, C = ?


function
vector
algebra
polynomial
geometry
3D geometry
sphere

Given a real variable $x \in \mathbb{R}^n$, and a generic vector  $v \in \mathbb{R}^n$, I define a unidirectional function a function that can be expressed as a function of only the standard scalar product between x and v:
$\begin{array}{l} g: \mathbb{R}^n \mapsto \mathbb{R} \\ g(\langle {x,v} \rangle ) \\ \end{array}$,
where $\langle {x,v} \rangle$ is the standard scalar product between x and v.
I wonder, which classes of functions can be expressed as a sum of unidirectional functions?
You can prove that functions that are developable with Taylor can be expressed this way: i begin by showing this in steps:
the generic term of a polynomial in n variables can be expressed as a sum of another kind of terms:
$\begin{array}{l} p \in \mathbb{N}^n \\ x \in \mathbb{R}^n \\ t(x,p) = \prod\limits_{i = 1}^n {x_i ^{p_i } } \\ v_k \in \mathbb{R}^n \\ t(x,p) = \sum\limits_k {a_k ( {\langle {x,v_k } \rangle } )} ^{\sum\limits_{i = 1}^n {p_i } } \\ \end{array}$

You can use linear algebra to prove this thesis. Here are a few examples:
$\begin{array}{l} xy = \frac{1}{4}[ {( {x + y} )^2 - ( {x - y} )^2 } ] \\ x^2 y = \frac{1}{6}[ {(x + y)^3 - (x - y)^3 - 2y^3 } ] \\ \end{array}$

So if a f(x) can be developed in Taylor series, then it can also be developed in a series of unidirectional functions:
$\begin{array}{l} i \in \mathbb{N} \\ j \in \mathbb{N} \\ v_i (j) \in \mathbb{R}^n \\ x \in \mathbb{R}^n \\ b_i (j) \in \mathbb{R} \\ f(x) = \sum\limits_i {\sum\limits_j {b_i (j)( {\langle {x,v_i (j)} \rangle } )^i } } \end{array}$

So which classes of functions can in general be expressed as:
$f(x) = \sum\limits_i {g_i ( {\langle {x,v_i } \rangle } )}$
?

Can we further extend the class of functions if we consider also the possibility that the set of the v_i is not countable, such as the set of all the possible directions, D? D can be a subset of the unit sphere S, but is not the whole sphere because the two vectors v and -v define the same direction.

${\begin{array}{l} S = \{ s \in \mathbb{R}^n | {\langle {s,s} \rangle = 1\} }  \\ d = ( \begin{array}{l} d_1 \\ d_2 \\ ... \\ d_n \\ \end{array} ) \\ D = \{ {d \in S| d_n \ge 0} } \} \\ \end{array}$

For instance, for n=3, it can be the upper semisphere with z>=0.

So the most generic expression I can think of is:

$f(x) = \sum\limits_i {g_i ( {\langle {x,d_i } \rangle } )} + \int\limits_D {g( {d,\langle {x,d} \rangle } )}$

I give you finally these questions:

1-which classes of functions can be expressed in this way?
2-given a f(x) that can be expressed in this way, how can you deduce the unidirectional functions that compose it?

$T[ {f(x)} ] = g( {d,\langle {x,d} \rangle } )$

Sorry, a small correction. When i defined the unidirectional function $g(\left\langle {x,v} \right\rangle )$, i said that : $g: \mathbb{R}^n \mapsto \mathbb{R}$. Since $\left\langle {x,v} \right\rangle$ is a scalar, i should have said that $g: \mathbb{R} \mapsto \mathbb{R}$


linear algebra
matrix
linear algebra unsolved

given $ n, p  \in N (p\leq n)$. consider all square matrix $ A\equal{}[a_{ij}]_{n\times n}$( we consider in R) satifying : $ rank(A)\equal{}p, A^2\equal{}A$. find all possibile value of $ S\equal{} a_{11}\plus{}a_{22}\plus{}...\plus{}a_{nn}$

$ A$ has only eigenvalues $ 0, 1$; in Jordan normal form, its trace is precisely its rank $ p$.


Este vorba de:

Fie $(x_{n})_{n\geq 1}$ un sir de numere reale , convergent, cu proprietatea ca $x^{2}_{n}-x_{n}x_{n+1}-\frac{1}{n}\geq 0$, pt orice n≥1. Sa se arate ca $\lim_{n\to\infty}\ x_{n}\geq 0$

-Am ajuns si eu la o solutie dar e putin cam urata.
-Astept raspunsuri :)

Era $\lim_{n\to\infty}\ x_{n}\leq 0$ , ma scuzati. :P

uh never mind postasem o prostie mai devreme  :blush:

eu m`am gandit asa
[hide="click"]
voi presupune prin absurd ca $\lim x_{n}>0$. atunci, de la un rang incolo, toti termenii vor fi strict pozitivi. fie $p$ acest rang. pentru $n\geq p$, am ca $x_{n}^{2}>x_{n}x_{n+1}$, deci sirul $\big(x_{n}\big)$ este strict descrescator (de la rangul $p$). apoi, $x_{n}(x_{n}-x_{n+1})\geq \frac{1}{n}$. insa $x_{n}<x_{p}$, si deci $x_{n}-x_{n+1}\geq \frac{1}{nx_{p}}$. insumand dupa $n$ (de la $p$ la $m$) aceste relatii voi obtine :
$x_{p}-x_{m+1}\geq \frac{1}{x_{p}}\left(\frac{1}{p}+\frac{1}{p+1}+\ldots+\frac{1}{m}\right)$.
de aici se vede cam clar ca $RHS\rightarrow\infty$ si e gata.
[/hide]


Primer nivel, problema 2:

Sean $X=a1b9$ y $Y=51ab$ dos enteros positivos, donde $a$ y $b$ representan dígitos. Sabemos que $X$ es múltiplo de un entero positivo $n$ de dos cifras, y que $Y$ es el múltiplo siguiente de $n$. Hallar $n$ y los dígitos $a$ y $b$.

[color=blue]Parece bonito para una olimpiada de mayo.
¿En qué olimpiada apareció?[/color]

pues en esta olimpiada de mayo!!! :roll: hasta donde yo se los problemas son originales

[quote="tipe"]Pues sí, los problemas son originales. Cada país envía sus problemas a un Comité Central, en Argentina, ahí se escogen los problemas.
En este año dos problemas los propuse yo.  :D

$Tipe$.[/quote]

en serio?? cuales?? :D

El 3 del nivel 1 (que lleva mi nombre  :D ) y el 1 del nivel 2.

$Tipe$

Tenemos:

X=1000a+10b+109
Y=5100+10a+b

Como Y es el multiplo seguinte de n apos X, tenemos que Y-X=n => n=4991-990a-9b

Haciendo acotacciones para a y b, vemos trivialmente que el unico valor possible de a es 5.
Entonces, n=41-9b.Como n es um numero de 2 digitos, los valores para b son: b=1,2,3.
Se b es impar, n es par, una contradiccion, pues X es impar.Entonces, el unico valor para b es 2.
Com isso, obtemos n=23.

La unica soluccion es portanto: n=23, a=5 y b=2.

Saludos!
Perdone por mi "Portunhol" xD


I    -> A and B are 1000x1000 matrixes.
II   -> A is different from B.
III  -> A²B=AB²
IV  -> A³=B³

[b]Check if A²+B² is invertible.[/b]

PS: Please, DON'T USE DETERMINANTS CONCEPTS. Just matrix operations.

[quote="brazil"]
III  -> A²B=AB²
[/quote]

Could you check this statement, I wonder if it is $A^{2}B=B^{2}A$

[quote="me@home"]$AB^2\not =(AB)^2$
$AB^2=A(B^2)=B^2A$ :)[/quote]

Sorry, but I haven't understood what you mean. Could you explain again, plaese?

Thanks a lot. ^_^

[quote="me@home"] By the way, can this matrix problem be answered with only a high school knowledge (we hardly learned matrices in Alg. 2)[/quote]

Yes, it can be. ^^

[quote="me@home"]I was trying to clarify to Moubinool that his "correction" is the same as what was stated in your problem.
By the way, can this matrix problem be answered with only a high school knowledge (we hardly learned matrices in Alg. 2)[/quote]

Obviously not; matrices don't necessarily commute!

Anyways, I suspect that Moubinool is correct and it should be $A^2B=B^2A$; then we get $(A^2+B^2)A=(A^2+B^2)B$.  If $A^2+B^2$ were invertible, then we would have $A=B$, an obvious contradiction.


calculus
integration
logarithms
calculus computations

\[\int_{0}^{1}dx \frac{\log(1-x)}{x}\]

Hi guys, I'm a new member of the forum so I don't hope you have've seen this integral already a thousand times.

Not one thousand, just 999 :).  Probably, the easiest way to compute it is to expand $\log(1-x)$ into a power series, integrate term by term, and then use the well-known formula for the sum of the inverse squares.

Anyway, welcome to the forum! :welcomeani:

Let $-\log (1-x)=t$
So,
$\int_{0}^{1}dx \frac{\log(1-x)}{x}=\int_{0}^{\infty}\left(-\frac{t}{e^{t}-1}\right) dt=-\frac{\pi^{2}}{6}$
(by Laplace Transform)


geometry
trapezoid
geometric transformation
reflection
circumcircle
symmetry
parallelogram

Consider $ ABCD$ a cyclic quadrilateral.
(a) The Simson lines of $ A$ wrt. triangle $ BCD$, of $ B$ wrt. $ CDA$, of $ C$ wrt. $ DAB$, and of $ D$ wrt. $ ABC$ concur at the [i]anticenter[/i] of $ ABCD$, the intersection point of the nine-point circles of triangles $ BCD$, $ CDA$, $ DAB$, $ ABC$.
(b) If $ X$, $ Y$, $ Z$, $ T$ are the projections of $ P$ on the sides $ AB$, $ BC$, $ CD$, $ DA$, respectively, and if $ M$, $ N$ are midpoints of $ XZ$, and $ YT$, the point $ P$ lies on the line $ MN$.
(c) The quadrilateral $ XYZT$ is circumscribed, if and only if, $ ABCD$ is an isosceles trapezium or $ AC \perp BD$.

Note: Problem (c) follows from a result due to Darij: [url]http://www.mathlinks.ro/viewtopic.php?t=38712[/url]. (scroll down)

PS: Thanks to yetti for poiting out the initial errors of the original statement.

[color=blue]The 9-point circles $ (N_d),\ (N_a),\ (N_b),\ (N_c)$ of the $ \triangle ABC,\ \triangle BCD,\ \triangle CDA,\ \triangle DAB$ meet at the anticenter $ P$ of the cyclic quadrilateral $ ABCD.$[/color]

A', B', C', D' are midpoints of AB, BC, CD, DA, U, V are midpoints of AC, BD. Let $ O_a,\ O_u$ be reflections of the circumcenter O in AB, AC. $ A'U \parallel BC$ is midline of the $ \triangle ABC,\ \triangle OO_aO_u \Longrightarrow$ $ O_aO_u \parallel BC.$  $ N_c, N_b$ are midpoints of $ DO_a,\ DO_u,$ $ N_cN_b \parallel O_aO_u$ is midline of the $ \triangle OO_aO_u\ \Longrightarrow N_cN_b \parallel BC.$ Similarly, $ N_dN_c \parallel CD,\ N_aN_d \parallel DA,\ N_bN_a \parallel AB.$ In exactly the same way, we could obtain $ N_cN_a \parallel AC,\ N_dN_b \parallel BD$ (but it is not necessary). In any quadrilateral ABCD, the 9-point circles $ (N_a),\ (N_b),\ (N_c),\ (N_d)$ concur at a point P. Since our ABCD is cyclic, the $ \triangle ABC,\ \triangle BCD,\ \triangle CDA,\ \triangle DAB$ have common circumcircle (O) with circumradius R, their 9-point circles are all congruent with radius $ R/2.$ Thus $ N_aN_bN_cN_d$ is cyclic with circumcenter P and centrally similar to the cyclic ABCD (with corresponding sides parallel) with coefficient $ - 1/2.$ PA' is radical axis of $ (N_d), (N_c),$ $ PA' \perp (N_dN_c \parallel CD),$ A' is midpoint of AB $ \Longrightarrow$ PA' is maltitude of ABCD and similarly, PB', PC', PD' are maltitudes concurrent at the anticenter P.

[color=blue][b](a)[/b] Simson lines of the $ s_d, s_a, s_b, s_c$ of the $ \triangle ABC,\ \triangle BCD,\ \triangle CDA,\ \triangle DAB$ with poles $ A, B, C, D$ concur at the anticenter $ P.$[/color]

Let $ A_b, A_d$ be feet of normals from A to CB, CD and let $ C_b,\ C_d$ be feet of normals from C to AB, AD. $ s_a \equiv A_bA_d$ is Simson line of the $ \triangle BCD$ with the pole A and $ s_c \equiv C_bC_d$ is Simson line of the $ \triangle DAB$ with the pole C. $ C_b, A_b, A_d, C_d$ all lie on a circle (U) with diameter AC. $ A_bC_b$ is radical axis of $ (U),\ (N_d),$ $ A_dC_d$ is radical axis of $ (U),\ (N_b),$ they meet at the radical center R of $ (U),\ (N_d),\ (N_b),$ $ RA_b \cdot RC_b = RA_d \cdot RC_d.$ But $ A_bC_b = A_dC_d$ as sides of orthic triangles of the $ \triangle ABC,\ \triangle CDA$ with the common side AC and $ \mod 180^\circ$ equal angles $ \measuredangle ABC = \measuredangle ADC.$ Thus $ RA_b = RA_d,\ RC_b = RC_d$ and $ A_bC_bC_dA_d$ is isosceles trapezoid with $ A_bA_d \parallel C_bC_d \parallel N_dN_c$ and with symmetry axis RPU. If $ B_u, D_u$ are feet of the B- and D-altitudes of the $ \triangle ABC,\ \triangle CDA,$ then

$ \measuredangle A_bPA_d = \measuredangle A_bPU + \measuredangle UPC_d + \measuredangle C_dPA_d =$

$ = \measuredangle A_bB_uU + \measuredangle UD_uC_d + \measuredangle C_dD_uA_d =$

$ = \measuredangle CBA + \measuredangle CBA + 2 \measuredangle ABC = 0$

and the points $ A_b, P, A_d$ are collinear, the Simson line $ s_a \equiv A_bA_d$ of the $ \triangle BCD$ with pole A goes through P. Similarly, the remaining Simson lines $ s_b, s_c, s_d$ of the $ \triangle CDA,\ \triangle DAB,\ \triangle ABC$ with poles B, C, D go through P.

[color=blue][b](b)[/b] Points $ P, M, N$ are collinear.[/color]

A', B', C', D' are midpoints of AB, BC, CD, DA as before, A'B'C'D is a parallelogram, its diagonals cutting each other in half at the centroid G of ABCD. The anticenter P is reflection of the circumcenter O in the centroid G. A'OC'P and B'OD'P are also parallelograms, their diagonals cutting each other in half at G. For angles $ \mod 180^\circ,$

$ 0 = \measuredangle A'OG + \measuredangle OGB' + \measuredangle B'OA' = \measuredangle A'OG + \measuredangle OGB' - \measuredangle B'BA'$

A'XZC' and B'YTD' are both cyclic because of the right angles $ PX \perp AB \equiv A'X,\ PZ \perp CD \equiv C'ZPY \perp BC \equiv B'Y,\ PT \perp DA \equiv D'T,$ respectively. The $ \triangle OA'C' \cong \triangle PC'A' \sim \triangle PZX$ with medians OG, PG, PM are similar and likewise, the $ \triangle OB'D' \cong \triangle PD'B' \sim \triangle PTY$ with medians OG, PG, PN are also similar. Consequently,

$ \measuredangle MPN = \measuredangle MPT + \measuredangle TPZ + \measuredangle ZPN =$

$ = (\measuredangle MPT + \measuredangle TPZ) + (\measuredangle TPZ + \measuredangle ZPN) - \measuredangle TPZ =$

$ = \measuredangle MPZ + \measuredangle TPN - \measuredangle TPZ = \measuredangle A'OG + \measuredangle OGB' - \measuredangle B'BA' = 0$

and the points M, P, N are collinear.

[color=blue][b](c)[/b] Quadrilateral $ XYZT$ is circumscribed, iff $ ABCD$ is an isosceles trapezoid or $ AC \perp BD.$[/color]

If $ Q \equiv AC \cap BD$ is the quadrilateral diagonal intersection and X', Y', Z', T' feet of normals from Q (not from the anticenter P) to AB, BC, CD, DA, the quadrilateral X'Y'Z'Y' is always circumscribed. AT'QX', BX'QY', CY'QZ', DZ'QT' are all cyclic because of the right angles, AQ, BQ, CQ, DQ their circumcircle diameters.

$ \frac {BD}{DA} = \frac {\sin A}{\sin \widehat{ABD}},\ \ \frac {AC}{DA} = \frac {\sin D}{\sin \widehat{ACD}}$

$ T'X' + Y'Z' = AQ \sin A + CQ \sin C = AC \sin A =$
$ = BD \sin D = BQ \sin B + DQ \sin D = X'Y' + Z'T'$

Alternately, let  $ (A_q), (B_q)$ be circumcircles of AT'QX', BX'QY' with diameters AQ, BQ, let QX' meet CD at R, and let E, F be feet of normals from Q to T'X', X'Y'. From similar $ \triangle QAD \sim \triangle QBC$ with Q-altitudes QT', QY', $ \angle DQT' = \angle CQY'.$ From isogonal altitudes and circumradii $ QE,\ QA_q$ amd $ QF,\ QB_q,$ $ \angle T'QE = \angle X'QA_q = \angle RQC,$ $ \angle Y'QF = \angle X'QB_q = \angle RQD.$ It follows that $ \angle RQE = \angle RQF,$ RQX' bisects the $ \angle EQF,$ the right $ \triangle X'QE \cong \triangle X'QF$ are congruent, X'Q bisects the $ \angle T'X'Y'.$ Similarly, Y'Q, Z'Q, T'Q bisect the $ \angle X'Y'Z',\ \angle Y'Z'T',\ \angle Z'T'X',$ Q is incenter of the circumscribed X'Y'Z'T'.

When ABCD is a cyclic (isosceles) trapezoid, XYZT is a kite, which is circumscribed. When ABCD has perpendicular diagonals, the anticenter P is identical with the diagonal intersection Q, $ P \equiv Q,$ and $ XYZT \equiv X'Y'Z'T'$ is also circumscribed with incenter P (even bicentric, centrally similar to the tangential quadrilateral of the contact cyclic ABCD with perpendicular diagonals).  But the problem is that for other points (besides the diagonal intersection Q), X'Y'Z'T' is circumscribed (the locus is a curve passing through Q). It appears separated from the anticenter P for $ AC \not\perp BD,$ but I do not know for sure and cannot prove it.  :(

What Pohoatza calls "Simson point" is indeed called Euler point or anticenter and, I think, also Lemoine point (please don't use this name, since it can be mistaken for other things). Besides, it is a particular case of the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=109112]Poncelet point[/url] of an arbitrary quadrilateral ABCD (see Theorem 4: if ABCD is a cyclic quadrilateral, then the pedal circles degenerate to Simson lines).

  darij


Find the total number of integers $ 1000 \leq \overline{abcd} \leq 9999$ such that $ \overline{dcba} \minus{} \overline{abcd} \equal{} 2088$.

Find the total number of triangles such that the lengths of all three sides of each triangle are natural numbers and the length of the longest side of each triangle is 15.

Each time the two hands a certain standard 12-hour clock form a $ 180^{\circ}$ angle, a bell chimes twice. How many chimes will be heard in a day (24 hours)?

(Refer to the attached PDF file for diagram) Seven identical dominoes of size $ 1 \times 2$ and with identical faces on both sides are arranged to cover a rectangle of size $ 2 \times 7$. One possible arrangement is shown in the diagram. Find the total number of ways in which the rectangle can be covered by the seven dominoes.

Find the total number of triangles such that the lengths of all three sides of each triangle are natural numbers and the length of the longest side of each triangle is 15.

[hide="Solution"]
I really don't know if i am missing something. Could just use the property of a+ b > c and think of triples and got 605 :(
[/hide][/hide]

[quote="analytic"][hide]ans to domino question.)
observe that the vertical dominoes always come in pairs. so, they are three combinations of vertical dominoes 2,4,6. so, let the units of 2two,s be one unit. so, we have three cases,. let units be U and others be B
combinations are : UBBBBB
                          UUBBB
                          UUUB
CALCULATING COMBINATIONS FOR EACH AND ADDING WE GET (6+10+4)=20[/hide][/quote]

According to my math olympiad instructor the answer is 21.

[quote="solveitall"][hide="Solution"]
I really don't know if i am missing something. Could just use the property of a+ b > c and think of triples and got 605 :(
[/hide][/quote]

The answer is much smaller than that. Remember that the longest side is 15. (Note that another side can be 15 too, but not more than 15)

absolutely right !theta pi. [b]the answer is 21 as i didnt count the case when there are no vertical dominoes or U as given notation above.so, 20+1=21[/b]

[u]answer to first question[/u]
expand the numbers in final equation i.e.
dcba-abcd=2088
$ \implies1000d\plus{}100c\plus{}10b\plus{}a\minus{}1000a\minus{}100b\minus{}10c\minus{}d\equal{}2088
\implies999d\plus{}90c\minus{}90b\minus{}999a\equal{}2088
\implies111d\plus{}10c\minus{}10b\minus{}111a\equal{}232
\implies111(d\minus{}a)\plus{}10(c\minus{}b)\equal{}232$
from here u get $ (d\minus{}a)\equal{}2 and (c\minus{}b)\equal{}1$
a and b are dependnt on d and c. total combinations for d are 7 and for c are 9 taking othere also in consideration.
so, total pairs are 9*7=63


What is the probability of rolling a sum of 7 or 11 by tossing
two fair dice? Express your answer as a common fraction.

i would just like to add the following:

with two dice, it is possible to roll a number with the following ways:

2: 1
3: 2
4: 3
5: 4
6: 5
7: 6
8: 5
9: 4
10: 3
11: 2
12: 1


calculus
integration
trigonometry
logarithms
calculus computations

Evaluate $ \int_0^{\frac{\pi}{4}} \frac{\sin ^ 2 x}{\cos ^ 3 x}\ dx$.

[quote="kunny"]Evaluate $ \int_0^{\frac {\pi}{4}} \frac {\sin ^ 2 x}{\cos ^ 3 x}\ dx$.[/quote]

 given integral $ =\int_0^{\frac {\pi}{4}} \frac  {\sin ^ 2 x}{\cos ^ 4 x}\ d\sin x$,then make a variable substitution,let $ t=\sin x$,given integral can be translated into the following integral

${{ \int_{0}^{\frac{\sqrt{2}}{2}}\frac{t^2}{(t^2-1)^2} dt=-\frac{1}{2}\int_{0}}^{\frac{\sqrt{2}}{2}}t d(t^2-1)^{-1}dt=-\frac{1}{2}t(t^2-1)^{-1}|_{0}^{\frac{\sqrt{2}}{2}}+\frac{1}{2}\int_{0}}^{\frac{\sqrt{2}}{2}}\frac{1}{t^2-1}dt=\frac{\sqrt{2}}{2}+\frac{1}{4}\ln 2-\frac{1}{2}\ln(2+\sqrt{2})$

That's correct answer.
You can write it as $ \frac 12 \{\sqrt {2} \minus{} \ln (1 \plus{} \sqrt {2})\}$. :)


MATHCOUNTS
AMC
AIME
geometry
USA(J)MO
USAMO
AMC 12

Recently, I just started AMCs and AIMEs and my Mathcounts "career" is over.. Does anyone have any tips/suggestions/ideas for study for someone making the transition from Mathcounts to things such as the AMCs and AIME?

haha thanks, I'm probably going to have to study that thing hard, incredibly hard. I have to overcome the geometry....  :ninja:

Yeah, I got it last summer or so.. I pretty much got all of it, except skipped all the geometry.. I didn't feel like dealing with all the thereoms and stuff like that... But besides that, i've got that book down and i have AoPS 2 sitting on my shelf somewhere and i'm probably going to start it a week or two..

The one major transition you will have to make is from working very fast to working at a bit slower tempo, as to avoid making careless errors, which hurt more (especially in the AIME) now.

If you did all of AOPS 1 except geo, I recommend you do all of that geo first before going to aops2

I found the geometry in Vol. I to be extremely useful.
Working speed:
Mathcounts -> AMC -> AIME -> USAMO:
Very fast -> fast -> pretty slow -> slow.

One thing:  Think reallly realllly hard for high AMC12, AIME, and USAMO problems.

AMC10's should be relatively comfortable, so study for AMC12 first.  And yes, DO the geometry, its very helpful (and the intro to geometry book challenge problems are good for AMC10/12).  I guess sections of AoPS II such as vectors and matrices and limits aren't [i]that[/i] important (I haven't learned them and I got a decent score on AMC12B, luck...) but will probably still help.

also you might want to learn a couple of the more advanced topics such as trig and complex numbers and such

Here's my take on the matter:

Math books like AoPS 1/2 are important resources and you should definitely read them through. But I would refrain from anything you label "studying". The best way to read them is to [i]scan[/i] through them and work sample problems. Make sure you really understand everything, and I don't just mean read through the proof and think, "that makes sense." You should try (for most of the ideas) to really instincively understand them when you look at them later (this way you will remember them better). But don't focus your time/energy on memorizing anything. Just quickly go through the books once, then do old tests, and read through the solutions. You will learn much better if you allow yourself to auto-memorize and learn important problem-solving techniques along the way rather than spend time "studying". At least for me, working problems was ten times faster than studying facts.

I don't know how helpful that is, but I always try to provide non-cliché answers to this question. Hopefully it is helpful in some way.

Oh, and don't overextend yourself. Focus on the extremely important topics first, and once you get those down, go to the harder and/or less common things like complex numbers, Euler's totient theorem, etc. That should be more of a phase-in to the AIME.

Yes, the other major difference is that there will be trigonometry and complex numbers and logs etc. (more advanced materials).

[quote="davidyko"]
Mathcounts -> AMC -> AIME -> USAMO:
Very fast -> fast -> pretty slow -> slow.[/quote]

oh.my.god.you.are.a.genius.no.one.would.ever.be.able.to.figure.that.out.
sometimes I love how you post the most obvious things that anyone with some amount of common sense could deduce on their own.

So anyway like there are some topics that are vital if you want to be successful on the AMC 12, and especially the AIME, the big ones are logs and trig/complex, as mentioned earlier. But you really can't shy away from the geometry, it may be easier for you to do introduction to geometry instead of AoPS 1 for geometry, it's a little more elaborate, contains a little more problems, has more up-to-date contest problems (AoPS 1 was first published in 1993 if I'm not mistaken), and is a little bit easier to read, because AoPS refined their writing style between the first two books and the Introduction series.

To elaborate on kyyuan's comment about thinking hard, I think that's really important. If you see a difficult problem that first looks beyond your ability, try it. Obviously, if it involves a particularly obscure theorem that is a little difficult to derive from nothing and hard to solve without it, you're a little out of luck, but if there's an interesting problem you find and you feel that you have the tools to solve it, by all means, work on it, for an hour, three, 24, whatever. And doing math in English, History, or Spanish class helps.

But above all, you shouldn't be doing the problems only to do well on contests. Yes, it's good to do well on contests, but you should be doing this because you find it interesting and area actually having fun with it.

Well, some people have no common sense, and besides, you can't say no to an admirably short summary.
Besides, that's my job. Remind others of obvious facts.  :P

thanks a lottt

Hopefully these tips will help me


I try to compute the Napier Constant $e : =\lim\limits_{n\to \infty}\left(1+\frac{1}{n}\right)^{n}$ with many  decimals.

 A very good algorithm is the following one:
[b]Consider the sequence $\left(X(n)\right) \; ,\; X(n)=X(n;a,b) ,$ defined  by means of the recurrence
     \[X(j+1)=(4j+2)X(j)+X(j-1) \; \; ,\; \; j\in \{1,2,...\}\; , \; X(0)=a , X(1)=b . \]
Let $A(n): =X(n,1,3) ,$  i.e.  $(a,b)=(1,3)$ and $B(n): =X(n,1,1).$
If one defines   $\begin{array}{|c|}\hline \\ E(n): =\frac{A(n)}{B(n)}\\ \\ \hline \end{array}\; ,\;$      it may be shown that $\lim_{n\to \infty}E(n)=e$ and moreover $E(2m)< e<E(2k+1) .$
For instance $E(1)=3 , E(2)=19/7 ,E(3)=193/71 , ... .$  More precisely, it may be shown that $|e-E(n)|< 4 (4n)^{-2n-1}.$
 I am interested to compute $E(n),$ for $n$ large , and in this way  to find many decimals  of  number ,, $e$" .[u][color=blue]
It is possible to find on computer $A(10000)$ and $B(10000)$  [/color] ? [/u]
[/b]
[size=75]
[b]Remark :[/b] On my PC such simple computations are not possible .  [/size]

This computation is very easy for any modern computer. What kind of problem do you have when you compute A(10000) and B(10000)?

Update: I did compute the numbers. But I thought they were too long for posting here...

BigIntegers in Java are quite handy :).

You may also want to check out a program called "bc". You can find it in most Linux distributions and is also ported into Windows as part of UnxUtils. It's free and it can work with arbitrary precision numbers. Calculating A[10000] in bc took 10 seconds on my computer.

[quote="bus"]BigIntegers in Java are quite handy :----Calculating A[10000] in bc took 10 seconds on my computer.[/quote]
OK ! But please inform us about $\frac{A[10000]}{B[10000]}.\;$ Thanking you in advance/the proposer.

Bc handled it pretty fast, I compared the result to http://www.gutenberg.org/dirs/etext94/ee710.txt and it's accurate to first [b]83359[/b] digits.

EDIT: Seems like the forum won't let me post such a long message here so that's all I have to say, if you want to see the actual output look at the first 83359 digits of e in the above link :). This is what the end of the computed value looks like:

34884415610695529562209602272988034902440030940866574392815996102205564011534302

and what it should really be:

34917684119102024662747957956434961501265743384575863883473583224253532814204782

By the way, I solved many of the other problems you posted here quite fast and easily using Bc, you should really get hold of it if you're interested in this kind of calculations...

Yes, bc is quite amazing. For bc, though, it takes longer to output for this question than it does to calculate.

Try printing the output to file instead of the screen :).


Let $ b$ a positive integer.Determine all 2002-tuples of nonegative integer $ (a_1,a_2,..,a_{2002})$ such that:
  $ \sum_{j\equal{}1}^{2002}a_j^{a_j}\equal{}2002b^b$

It's really. Can you post your solution for this? I have a solution, but it's very complicated and maybe wrong. I hope there's a "elementary" solution for this.

Locked (bad title and posted before).

See http://www.mathlinks.ro/viewtopic.php?t=135914 .


There are a number of ways of getting at this: strict diagonal dominance, the Gershgorin circle theorem ...

Let's do this: $ A=2006(I+B).$ $ B$ is a matrix with row sum equal to $ \frac1{2006}.$ That means that the $ \ell^{\infty}$ operator norm of $ B$ is $ \frac1{2006}.$ (And, incidentally, the spectral radius of $ B$ is less than or equal to $ \frac1{2006}.$)

Then the series $ C=\frac1{2006}\sum_{k=0}^{\infty}(-1)^{k}B^{k}$ converges, and, by direct computation, equals $ A^{-1}.$

[quote="N.T.TUAN"]Let $ A\in M_{n}(\mathbb{R})$ such that every its row contain exactly two elements unequal to zero, where element on mean diagonal is equal to $ 2006$, other element is equal to $ 1$. Prove that $ A$ is invertible.[/quote]

Another solution?  :P


So you have a circle, and a line going through that circle. You also have a point X on the line, but it's not inside of the circle.
How would construct a circle tangent to the line at point X and hitting the circle on the side of the line that's opposite the center of the circle?

Let the line be $ l$ and the first circle be $ C$ with center $ O$. Construct a circle $ C_2$ with the radius as $ C$ and center $ X$, and the line $ l^\prime$ perpendicular to the given one at $ X$. Where these two intersect on the side of the line with the center, call that $ P$. Construct the perpendicular bisector of $ PO$. Where this intersects $ l^\prime$ is the center of the desired circle, $ O^\prime$. The proof that this works is basically that $ POO^\prime$ is both isosceles and similar to $ XYO^\prime$ where $ Y$ is the intersection of $ C$ and $ OO^\prime$.


[quote="woofunky"]1.In the angle BAC, there are P,Q =>BP=CP, BQ=CQ, angle ABP + angle ACQ=180

  Then, show angle BAP = angle CAQ

2. There are three spots A,B,M in a row, a secant MDC cut an half cirle( diameter is AB)

     Is, intersection points of circumcircle of AOC and BOD is O,and K, show angle MKO=90

PLZ HELP ME!![/quote]
[hide="1"]
Consider $ P'$, the isogonal conjugate of $ P$ with respect to $ \triangle ABC$.  Let $ CP'$ and $ BP'$ meet $ AB$ and $ AC$ at $ M$ and $ N$ respectively.  Notice that $ \angle ABN \equal{} \angle PBC \equal{} \angle PCB \equal{} \angle ACM$, so $ MNCB$ is cyclic.  Now, notice that\[ \angle QBC \equal{} \angle ABQ \minus{} \angle ABC \equal{} 180 \minus{} \angle ACP \minus{} \angle ABC \equal{} 180 \minus{} \angle MBC \minus{} \angle MCB\]which means that $ \angle QBC \equal{} \angle BMC$, so $ QB$ is tangent to the circumcircle of $ MNCB$.  Similarly is $ CQ$.  It follows that by Pascal on $ BBMCCN$, we have that $ A$, $ P'$, and $ Q$ are collinear, so $ \angle QAB \equal{} \angle P'AB \equal{} \angle PAC$.
[/hide]

[hide="2"]
Invert around the semicircle, mapping any point $ X$ to $ X'$.  Then, $ K$ inverts to the intersection between lines $ AC$ and $ BD$ and $ M$ inverts to the intersection between the circumcircle of $ \triangle OCB$ and $ AB$.  Notice that $ \angle ACB \equal{} 90 \equal{} \angle ADB$, so $ C$ and $ D$ are the respective feet of the altitudes from $ B$ and $ A$ of $ \triangle K'BA$.  Furthermore, $ O$ is the midpoint of $ AB$, so the circumcircle of $ \triangle DOC$ is the nine-point circle of $ \triangle AK'B$, so it intersects $ AB$ at $ M'$ so that $ K'M'\perp AB$.  Inverting back, we see that this is equivalent to $ OK\perp KM$. [/hide]


1.Name a compound of boron that is optically active....
2.GIve the molecular formula of talc and Muscovite
3.What is 'GHost light'?

[hide="answer"]
1.Boron salicylate

2.Talc-$ \ce{Mg(Si2O5)2.Mg(OH)2}$

Muscovite-$ \ce{KAl_2 (AlSi_3 O_10) (F / OH)_2}$

3.$ \ce{PH_3}$
[/hide]

$ Mg(Si_2O_5)_2.Mg(OH)_2$

Just made your 2nd ans clear. :D

That was straight from one of the excercise questions of OPT

ur third answer is wrong chemrock.. think abt it...

It's either $ \ce{P2H4}$ or $ \ce{C2H2}$ :maybe:

I think its the former, cant remember though....

Btw, nice avatar Rak...

One mokkai:
The third answer is the light coming from a ghost(Seen casper :P ) :rotfl:

@Rakesh: Lol. That was brilliant albeit a little mokkai..  :rotfl:  :rotfl:  :rotfl:

The answer is white $ \ce{P_4}$ or $ \ce{P_4 O_10}$ For info see swern oxidation topic where I showed a moving picture of phosphorous moon. :D


algebra
function
domain
limit
logarithms
integration
real analysis

Show

$ \sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k}{\sqrt{k^2\plus{}a^2}}\equal{}\frac{1}{2a} \quad \forall a>1$

Well, the series converges uniformly on compact sets in the domain $ \Re a>0, |\Im a|<\Re a$, so it must represent an analytic function of $ a$ in that domain. If it is $ \frac 1{2a}$ for $ a>1$, then it is $ \frac 1{2a}$ for all $ a>0$. But the sum is $ \ge \frac 1a\minus{}1$ by the alternating series property, which is larger that $ \frac 1{2a}$ for small $ a$. So, the formula must be false as written. What may be true is that the series is [i]asymptotically[/i] equivalent to something like the right hand side for $ a\to\plus{}\infty$. ;)

Also, if

$ \sum_{k \equal{} 0}^{ \plus{} \infty} \frac {( \minus{} 1)^k}{\sqrt {k^2 \plus{} a^2}} \equal{} f(a)$

then we must have

$ \lim_{a \to 0} f(a) \equal{} \ln 2$.

On the other hand, provided I have made no mistake,

$ \sum_{k \equal{} 0}^{ \plus{} \infty} \frac {( \minus{} 1)^k}{\sqrt {k^2 \plus{} a^2}} \equal{} \int_0^{\plus{} \infty} \frac{J_0(at)}{1 \plus{} e^{\minus{}t}}\,dt$.

[quote="Carcul"]Also, if

$ \sum_{k \equal{} 0}^{ \plus{} \infty} \frac {( \minus{} 1)^k}{\sqrt {k^2 \plus{} a^2}} \equal{} f(a)$

then we must have

$ \lim_{a \to 0} f(a) \equal{} \ln 2$.[/quote]  No: the first (zeroth) term of the series dominates in this case, so $ \lim_{a \to 0} f(a) \equal{} \infty$.

Could anyone show more elementary solution please? :blush:


x1, x2, x3, and x4 are selected from the set {1,2,3,4,5,6}.  Determine the number of selections so that x1<x2<x3<x4

Given four distinct numbers in the set, there is only one way to arrange them in increasing. There are $ \binom{6}{4} \equal{} 15$ ways to choose four numbers. Therefore, there are $ 15$ ways to choose four numbers in increasing order.

Oh, I was thinking $ \binom{6}{4}$ would overcount.  That makes sense, thanks :)


A triangle ABC,A,B,C are the midpoints of BC,CA,AB.X is outside the triangle
ABC.Prove the area of one of the triangle XAM,XBN,XCP is the sum of the others.

um... canton?

What exactly does that mean?

And how can A,B,C be midpoints of BC,CA,AB??????

 :o

I think this should be the correct statement:

In $\triangle ABC$ the midpoints of $BC, CA, AB$ are $M, N, P$ respectively, and $X$ is a point outside the triangle. Prove that the area of one of $\triangle XAM,\triangle XBN, \triangle XCP$ is the sum of the other two.

[hide="Hint"]Easy using vector products.[/hide]


For the indians here..
Have you ever heard of the book "Challenge and thrill of pre-college mathematics" by ranganathan
Its the training book for the indian imo team.
The coach of the team recommended it for me.
is it good for usamo practice?

Theres this chinese book called "Gao Zhong Shu Xue Bi Shai Cheng."  It's what China and Taiwan use to train for IMO.


[color=blue][size=150]Theorem[/size]
Let $ ABCD$ be a convex quadrilateral with $ \angle ABC = 150^\circ$. Let $ E$ be the intersection point between the diagonals.
If $ \angle CED = 120^\circ$ and $ \angle DCA = 2\angle BCA$ then $ AB=AD$
[/color]

[size=150]Proof[/size]
Let $ I$ be the incenter of $ \triangle CDE$. Then

$ \angle ICE = \frac{1}{2}\angle DCE = \angle BCE$

$ \angle IDE = \frac{1}{2}\angle CDE = \frac{1}{2}(60^\circ-\angle DCE) = 30^\circ-\angle BCE = \angle BAE \ (1)$

[img]9702[/img]

$ \underline{\underline{ \triangle CBE, \triangle CIE}}:$
$ \begin{array}{rcl}\angle BCE & = & \angle ICE\\ CE & = & CE\\ \angle CEB & = & \angle CEI = 60^\circ \\ \end{array}\}(a.s.a.)\Rightarrow \triangle CBE \equiv \triangle CIE$ and consequently $ IE=BE$

$ \underline{\underline{ \triangle ABE, \triangle AIE}}:$
$ \begin{array}{rcl}BE & = & IE\\ \angle BEA & = & \angle IEA\\ EA & = & EA\\ \end{array}\}(s.a.s.)\Rightarrow \triangle ABE \equiv \triangle AIE$

So $ \angle IAE = \angle BAE \ \ (2)$ and also $ \angle EIA = \angle EBA \ \ (3)$

From $ (1),(2)$ we get that the quadrilateral $ EADI$ is cyclic.
So $ \angle EDA = \angle EIA$ and now from $ (3)$ we get
$ \angle EDA = \angle EBA \Rightarrow \angle ABD = \angle ADB \Rightarrow AB=AD$

So $ \triangle ABD$ is isosceles $ \Rightarrow \frac{AB}{AD}=1$


geometry
circumcircle
trigonometry
perpendicular bisector
geometry unsolved

Let a triangle $ABC$, with $\angle B=90$, $AB=BC$, and $P$ an interior point such that $\angle ABP=5\alpha$, $\angle PAC=2\alpha$ and $\angle BCP=3\alpha$. Find $\alpha$

Are you sure that there is a condition $AB=BC$?
I got a contradiction.

Sorry, I made a mistake. Your question is fine.

Because $\angle BPC=90^\circ+2\alpha$, the circumcenter of $\triangle PAC$ must lie on $BP$. Also, the circumcenter of $\triangle PAC$ must lie on the perpendicular bisector of $AC$ which is the bisector of $\angle B$. If $P$ is not on the bisector of $\angle B$, then $B$ is the intersection of those $2$ lines i.e. $B$ is the circemcenter of $\triangle APC$. Contradiction! So $P$ lies on the bisector of $\angle B$. Therefore, $5\alpha=\angle ABP=45^\circ\Rightarrow \alpha=9^\circ$. We get $\angle PAC=2\alpha=18^\circ$ and $\angle BCP=3\alpha=27^\circ$. There is such a point $P$ that satisfies the conditions.

[color=darkblue][b][u]Nice and very easy problem ![/u][/b] There is no need to lose times ... thinking !

Apply the trigonometric form of the Menelaus' theorem $(\alpha : =x)$ :

 $\sin 5x\sin 2x\sin 3x=$ $\cos 5x\sin (45^{\circ}-2x)\sin (45^{\circ}-3x)$ $\Longleftrightarrow$

$\sin 5x(\cos x-\cos 5x)=$ $\cos 5x(\cos x-\sin 5x)$ $\Longleftrightarrow$ $\tan 5x=1$ $\Longleftrightarrow$ $\boxed{\ x=9^{\circ}\ }\ .$[/color]


The number of square inches in the area of a rectangle is equal to the number of inches in its perimeter. Each dimension is a whole number, and the rectangle is not a square. What is the number of inches in the perimeter of the rectangle?

If l = length and w = width, than lw = 2l + 2w. Try some different numbers for l and w. The answer is length = 6 and width = 3, or other way around. The perimeter is 6 x 2 + 3 x 2 = 18, so that is answer.

PS: Sorry for poor English. It is my second language.


I seem to remember a Math Jam scheduled for the day before the AHSIMC, with difficulty 5, but now I can't find it anywhere. Has it been canceled? Or is this just a mistake?

I couldn't find any better place to put this than here; feel free to move it.

The reason for the cancellation is because I realized much, much too late that it is inappropriate for me to be running the Math Jams as the director of the competition.

My belief is that students need to be teaching students in our Math Jam.  After the 2005 competition, I will approach the winning team about hosting dual Math Jams next year (one before and one after the 2006 competition), which will become an honorary (and paid) position.

I understand that Mathew approached you, Joseph, to run the Math Jams, but I certainly understand not being able to do it with such late notice.  I appreciate you considering it.

We want to continue our relationship with AoPS and support its innovative approach to math education.  Part of this will be making sure these Math Jams happen next year - if you do well on the 2005 iTest, you may be hearing from me on this!

Bradley

Wait ---- so Joseph was offered with the math jam moderator and he rejected it??? (Certainly I could have helped, but I am not as frequent to the fora as Joseph is ;) )

I had a scheduling conflict.  Otherwise I would definitely be doing it.


analytic geometry
geometry
AMC
USA(J)MO
USAMO
Purple Comet

PURPLE COMET MATH MEET

(Formerly the University of Wisconsin – Whitewater/
University of Texas – Dallas Mathematics Competition)

Monday April 24 – Thursday April 27, 2006

FREE ON-LINE TEAM MATHEMATICS COMPETITION

<<<< TWO LEVELS >>>>
•High School (25 problems in 90 minutes)
•Middle School (15 problems in 60 minutes)

<<<< THREE CATEGORIES OF COMPETITION >>>>
•All team members from the same large school (over 1200 students)
•All team members from the same small school (not more than 1200 students)
•Team members from more than one school

• Teams of up to six students each
• Each team must have an adult supervisor
• Problems from previous meets available
• Open to teams from around the world

Registration and rules at http://math.uww.edu/mathmeet

Well what does the supervisor actually do?And what time is it(U.S.A time?)

Does the supervisor have to be present while we do t the questions(how are they going to know whether he is or not)  and why do we have to give our school phone nos. ?

The supervisor helps ensure that the rules of the contest are enforced, and can help coordinate the use of the computer in cases when there are several teams competing at the same time, and there is only one computer available that the teams must share to submit their problem answers.  The supervisor is given the information required to begin each team's competition.  Once the teams begin the competition, they will be able to see the contest problems on the screen, but even before the competition begins, the supervisor will have access to printable copies of the problems which the supervisor can have ready to hand out to team members when the competition begins.  The supervisor also gives the contest administrators a contact person in case there are any technical or administrative problems with the contest.  That is also why we need contact information.

No, there is no way for us to ensure that the supervisor is actually present during the competition just like there is no way we can assure that teams are following the rules of the competition.  But this is a mathematics contest being offered for fun, excitement, and education.  It is not like there are any great prizes at stake.  What is there to be gained by a team cheating?  If a team gets three Ph.D. mathematicians to work the problems for them, that would just rob the team of everything it could have gained from participating.  The only enforcement tool we have is that if we discovered a rule violation, we could ban the participating school from future competitions.

The PURPLE COMET MATH MEET is being offered over an 84 hour period.  High School teams need 90 minutes to compete, and Middle School teams need 60 minutes to compete.  Teams can choose the exact time that they begin the contest during that 84 hour period.  Once they begin, their clock starts and they have 90 or 60 minutes to complete the contest.

O.K,so will they call the supervisor before or during the competition?Because no one in our city has ever accessed matjh contests through internet(or otherwise :roll: ),so the sir won't be willing to be with us for 90 mins.We can follow the rules(we aren't bogus people,we know what the beauty of math is).So we can just give the mobile number of our sir and start away?And i think they'll call incase we win ,right?

thanks

There is generally no need for us to contact the supervisor directly.  Contest winners will be posted on our web site on May 1.  Have fun competing.

I took the Middle School one with a friend and we got 12/15 leaving 15 seconds  :D

what did you get?

Well I took high school and I fudged up the contest.my friend and I did not attempt a simple geometry uestion(thjopugh i guessed some  :D ) and out of the others we just got like 15,though i'm sure our score will be much less because our calculations are stupid,we didn't have calculators,were we allowed to use them?

I think so, but we didn't use them at all

[quote="chess64"]you're allowed to use calculators

they definitely used MS frontpage to make that website

{chess64, 1234567890, h_s_potter2002} $\in$ hungry usamo rejects[/quote]

O.k,first about the calculator thing---> :wallbash:

*Seeing your team members--->  :wallbash_red:   :wallbash_red:

I participated in this contest with 5 more people from my school.
Unfortunately, I don't have our answers with me, so I don't know how our team did. (Minimum=0, Maximum=24 :P )

I almost got #25, but I made a dumb mistake at the end... Oh well.

I took the test with only one friend, but we took the middle school one

but do you know when the actual result is coming out?

According to the web site...
[quote]the announcement of winners for the Spring 2006 contest
has been delayed until noon CDT on May 2. [/quote]

lol

i just saw the school results and found out my team got fourth :P

Our team didn't do well... we got 14/25.
Congrats to everyone who participated :)

I liked the contest overall. The problems were quite interesting (especially #25!) :lol:

Only[size=59] $9$[/size] bleh.We  could have got more if we used calculators.Not bad for a team of 2(including a dumb one)

I THOGUHT I NAILED 1-10.AARRRGGHHHh

Hungry U rejects got an honorable mention :D

HM on highschool (we didn't do that great though and most of smart (relative to others) ppl on my team were gone, so yeah i was the only one doing questions 15+  :P )

i got #25 right :D

8/15 on MS division, bleh. To note, we had a 6 person team...

well....
this is the thing...

on team contests like this, problems are really difficult. So it doesn't really matter how many people are in a team.......

my team was interesting.... we had only two people and we did pretty good :D

[quote="shinwoo"]...So it doesn't really matter how many people are in a team.......[/quote]

ummm... yeah it does....

but anyone we only had three people and we did pretty bad

My team got an Honorable Mention for schools with less than 1200 students. :)

We had a 6 person team, but that's about 40% of the high schoolers in our "school". :P

[quote]ummm... yeah it does....

but anyone we only had three people and we did pretty bad[/quote]

No it doesn't

It's really rare to have many skilled people in one school

also, when some horrible kid attempts to solve a problem, and screws it, you guys will just waist time and lose a point

Gays,what were your passwords?Mine were really nasty :rotfl:  Even my supervisor got a cheap one  :rotfl:

I remember it being very odd...
It had something to do with elephants I think...

Some of m friends and i combined our passwords to get-->"his hot arrow","her rapid preach" and "My bad curve" :rotfl:  :rotfl:

i want to learn lapplace theroem. how to learn that ?


It's disappeared; http://www.artofproblemsolving.com/Forum/ is now no different from http://www.artofproblemsolving.com/Forum/index.php .

mathlinks
.
ro

Hehe... It automatically converts to AoPS.com...


function
algebra
polynomial
inequalities
complex analysis
superior algebra
superior algebra solved

If an entire function $ f: \mathbb{C} \rightarrow \mathbb{C} $ commutes with
a polynomial (you can take it with complex coefficients) of degree at least
$ 2 $, then $ f $ is a polynomial.

Here is a sketch of how I did it. Let $M_f(R)=\max_{|z|=R}|f(z)|$. Assume that $f$ commutes with some polynomial $p$ of degree $n\geq 2$. Let $|z|=R^n$. If $R$ is large enough, then the roots of the equation $p(\zeta)=z$ are not greater than $CR$ where $C>0$ is some constant depending on $p$.  This results in the inequality $|f(z)|=|f(p(\zeta)|=|p(f(\zeta))|\leq C \max\{1,M_f(CR)^n\}$. Since $z$ was arbitrary, we conclude that $M_f(R^n)\leq C \max\{1,M_f(CR)^n\}$.  From here it is possible to derive that $M_f(R)$ grows only as some power of $R$ as $R\to+\infty$, so $f$ is a polynomial.
Is your proof the same, Leva?

Yes, my proof is exactly the same  ;)  .

What about other problems?


algebra
polynomial
complex numbers
algebra unsolved

Find all polynomials $p$ with real coefficients such that for all reals $a, b ,c$ such that $ab+bc+ca =1$ we have the relation \[p(a)^{2}+p(b)^{2}+p(c)^{2}=p(a+b+c)^{2}.\]

$ (a+b+c) ^2 = a^2 + b^2 + c^2 +2 (ab +bc +ca) $

Thus since $ (ab +bc +ca) = 1$ , $ (a+b+c) ^2 = a^2 + b^2 + c^2 +2 $
Let us assume that the polynomial is $ p(x) = a_n x^n + a_{n-1} x^{n-1} + ..... + a_0 $

Thus we get,

$a_n ( a^{2n} + b^{2n} + c^{2n} ) + a_{n-1} (a ^ {2n-2} + b^{2n-2} + c^{2n-2} ) +.... + 3 a_0 = a_n ( a^2 + b^2 + c^2 +2 )^n +.... a_0 $

Comparing coefficient of different powers of $x$ , we get,
$a_0 = 0 \\
\\ a_1(a^2 + b^2 + c^2 ) = a_1 (a^2 + b^2 + c^2 +2 ) \Longrightarrow a_1 = 0 \\
\\ a_2 ( a^4 + b^4 + c^4 ) = a_2 (a^2 + b^2 + c^2 +2 )^2  $ which on simplification again gives that $ a_2 = 0 $ for we get $ a_2 \times positive\ = 0 $

Thus , going on in this way, we get that all coeeficients are zero. Hence the rquired polynomial is identically $0$ !!!

Something is definitely wrong!!!! :(  :huh:

confused, is it $p(a^2)$ or $[p(a)]^2$?  i thought it was the second version.

interesting. I never paid heed to that . I tooh $ p(a^2) $..... :blush:

rushil, even so, i think you can probablly use the same method to solve the problem, and i wouldnt be too surprised if it had the same answer.

edit: by using your notation just look at the $a_n^2$ terms, you get $a_n^2(a^{2n}+b^{2n}+c^{2n})=a_n^2(a+b+c)^{2n}=a_n^2(a^2+b^2+c^2+2)^n$, but for positive integer values of $n$ RHS>LHS so no positive $n$ works ---> $n=0$ is only solution, i.e it's constant, and u can show the constant is 0.

So you think my solution is correct(in either case..) :)

[quote="cats..."]i didnt read the whole thing but the idea is right.[/quote]
Actually, it's wrong.
Answer: $p(x)=m(x\pm 1)$.
Hint: if the given equality holds for all real $a,b,c$ such that $ab+bc+ca=1$, then it also holds for all complex $a,b,c$ such that $ab+bc+ca=1$.

[quote="enescu"]
Actually, it's wrong.
Answer: $p(x)=m(x\pm 1)$.
Hint: if the given equality holds for all real $a,b,c$ such that $ab+bc+ca=1$, then it also holds for all complex $a,b,c$ such that $ab+bc+ca=1$.[/quote]

Interesting exercise!! :)

Enescu could you explain in detail the proof??  :?:

But arent $ a,b,c \in \mathbb{R} $ Am i missing sth???

Consider a slightly more general problem: find $f\in R[X]$ such that
\[
f\left(  a\right)  +f\left(  b\right)  +f\left(  c\right)  =f\left(
a+b+c\right)  ,
\]
for all real $a,b,c$ satisfying $ab+bc+ca=1.$

We can rewrite the latter as
\[
f\left(  a\right)  +f\left(  b\right)  +f\left(  \frac{1-ab}{a+b}\right)
=f\left(  \frac{a^{2}+b^{2}+ab+1}{a+b}\right)  ,
\]
for all real $a,b$ (if $a+b=0,$ it follows that $-a^{2}=1,$ a contradiction).
If $\deg f=n,$ we obtain
\[
\left(  a+b\right)  ^{n}\left(  f\left(  a\right)  +f\left(  b\right)
+f\left(  \frac{1-ab}{a+b}\right)  \right)  =\left(  a+b\right)  ^{n}f\left(
\frac{a^{2}+b^{2}+ab+1}{a+b}\right)  ,
\]
which is a polynomial equality, valid for all $a,b\in R,$ and hence also valid
for all $a,b\in C.$

It follows that $f$ verifies
\[
f\left(  a\right)  +f\left(  b\right)  +f\left(  c\right)  =f\left(
a+b+c\right)  ,
\]
for all complex $a,b,c$ such that $ab+bc+ca=1.$

Now, let $a=-2x+1,b=3x+\frac{i-3}{2},c=6x-3-2i.$ A short computation shows
that $ab+bc+ca=1,$ and hence
\[
f\left(  -2x+1\right)  +f\left(  3x+\frac{i-3}{2}\right)  +f\left(
6x-3-2i\right)  =f\left(  7x-\frac{7}{2}-\frac{3}{2}i\right)  ,
\]
for all $x.$ Checking the leading coefficients yields
\[
\left(  -2\right)  ^{n}+3^{n}+6^{n}=7^{n},
\]
which is possible only for $n=1$ or $n=2.$ Since $f=p^{2},$ we must choose
$n=2.$ Set $f\left(  x\right)  =qx^{2}+rx+s;$ the above equality gives $q=s.$
But $f$ is also a square, therefore $r^{2}=4q^{2},$ that is, $r=\pm2q$ and
$q>0,$ say $q=m^{2}.$ Finally, $f\left(  x\right)  =m^{2}\left(  x\pm1\right)
^{2}$ and $p\left(  x\right)  =m\left(  x\pm1\right)  .$

It is not difficult to check that these polynomials indeed satisfy the
requested condition.

[quote="enescu"]
\for all real $a,b$ (if $a+b=0,$ it follows that $-a^{2}=1,$ a contradiction).
If $\deg f=n,$ we obtain
\[
\left(  a+b\right)  ^{n}\left(  f\left(  a\right)  +f\left(  b\right)
+f\left(  \frac{1-ab}{a+b}\right)  \right)  =\left(  a+b\right)  ^{n}f\left(
\frac{a^{2}+b^{2}+ab+1}{a+b}\right)  ,
\]
which is a polynomial equality, valid for all $a,b\in R,$ and hence also valid
for all $a,b\in C.$

[/quote]

I didn't quite follow this and the steps after.. Can you plz explain??? :starwars:  :spider:

Also, can you tell what's wrong eith my solution. I still think complex nos. should not be considered!!!

[quote="Rushil"]

Also, can you tell what's wrong eith my solution. I still think complex nos. should not be considered!!![/quote]
Well, try the polynomial $p(x)=x+1$. Does it work? Is it true that $p(a)^2+p(b)^2+p(c)^2=p(a+b+c)^2$ whenever $ab+bc+ca=1$? Then your "solution" implying that $p=0$ is not correct. Replace in your computations $p$ with $x+1$. Eventually you will find the reason why it doesn't work. And yes, do not consider complex numbers if you don't like it. Find another proof.