I need to use a special type of h for a paper I'm writing in LaTeX, and I was wondering if any of you LaTeX whizzes knew a quick and easy way to rig it up.  The character looks like ḫ, an h with a short accent under it (if you're curious, it's used in transcriptions of the Hittite language).    I found one site ([url]http://www.christopherculver.com/en/computing/latex.php[/url]) which says that you can directly enter the Unicode symbol if your LaTeX editor supports unicode.  I don't know if my editor (I'm using TeXShop) can save files in UTF-8 encoding.  I'll have to find a way to use unicode in LaTeX someday, but if that takes some long and involved finagling, today might not be the day.  Is there any easy way to put a short vowel mark under a letter?

You'll be pleased to hear that LaTeX makes it easy to use Unicode characters with inputenc, as the following shows:
[code]\documentclass{article}
\usepackage[utf8x]{inputenc}
\begin{document}
For the Hittite laryngeal type \unichar{"1E2B} and get the symbol that you want.
\unichar{7723} gives the same result.
If you use
\newcommand{\h}{\unichar{"1E2B}}
then you can type \h{} or \h\ to get the symbol. The {} or \ are only required if you want a space afterwards
\end{document}[/code]

\unichar{7722} or \unichar{"1E2A} or \charettesubbreve H. The latter is interesting because it allows you to use the breve under any symbol.


Let $x$ and $y$ be positive real numbers such that $x+y=1.$ Show that $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\ge 9.$

Expand $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$, multiply both sides of the inequality with $xy$ and use the fact that $x+y=1$ to get that \[\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \geq 9\ \Longleftrightarrow\ \left(\frac{x+y}{2}\right)^{2}= \frac{1}{4}\geq xy,\] which is just AM-GM. I guess all inequalities of the 70's are trivial now  :D

[hide="In other words..."]
$(x+1)(y+1) \ge 9xy$
$\Leftrightarrow (2x+y)(2y+x) \ge 9xy$
$\Leftrightarrow 2y^{2}+2x^{2}\ge 4xy$

True by Rearrangement (alternately, AM-GM). [/hide]

Expanding we get: $x+y+1 \geq 8xy$

Replacing $x+y=1$, we have: $xy \leq \frac{1}{4}$, this is true by $AM-GM$

If you don't want to expand, you can just Cauchy it: \[\left( 2+\frac{y}{x}\right) \left( 2+\frac{x}{y}\right) \geq 9 .\]


Compute $ \frac 1{1\cdot 2\cdot 3\cdot 4} \plus{} \frac 1{2\cdot 3\cdot 4\cdot 5} \plus{} \frac 1{3\cdot 4\cdot 5\cdot 6} \plus{} \cdots \plus{} \frac 1{28\cdot 29\cdot 30\cdot 31}$

[hide="hint"]
perhaps partial fractions would help
[/hide]


calculus
integration
calculus computations

[url=http://integrals.wolfram.com/index.jsp]$\text{The Integrator}$[/url] will give you a big nasty answer.


Ειναι η τελευταια φορα που θα μιλησω για εξωμαθηματικο θεμα. αλλα η Μακεδονια ειναι ΕΛΛΗΝΙΚΗ, δεν μπορω να βλεπω να μιλανε για ολυμπιαδα της μακεδονιας στο advance.ελπιζω ολοι να συμφωνουν.παρακαλω οποιος βλεπει τετοια αναφορα να παρεμβαινει με κοσμιο παντα τροπο.

Είδατε οι φίλοι μας οι άγγλοι  στο αγγλία- FYROM που το έγραφαν macedonia ??
E ti περιμένεις μετά........ :mad:  :mad:

Το ειδα και εγω .στενωχωρηθηκα πολυ οταν το ειδα.αλλα επισημα λεγονται ΦΥΡΟΜ.κριμα  :(

Φοβάμαι πως το θέμα με το όνομα έχει λήξει οριστικά.Διεθνώς το όνομα είναι MACEDONIA. Αυτό βλέπω και ακούω να χρησιμποποιούν όλοι , πολιτικοί , επιστήμονες και λοιποί. Μακάρι να βρισκόταν για την Ελλάδα μια πιο καλή λύση.Δε γνωρίζω τι άποψη έχουν για το θέμα οι Ευρωπαίοι.Πάντως και εμένα , χθες που είδα να πανηγυρίζουν κάτι φίλαθλοι με το αστέρι στη σημαία , με πείραξε πολύ.Όμωςπρέπει να παραδεχθώ σιγά σιγά ότι η μάχη με το όνομα έχει χαθεί. Η χώρα μας έχασε τη μάχη πριν αρχίσει , διότι δεν αντέδρασε όταν έπρεπε(πριν το 1960). Τώρα , το μόνο που συζητάμε είναι πώς θα αποκαλούμαστε μεταξύ μας. Αυτό είναι κορο'ι'δία!
  Αλλά .... η ελπίδα πεθαίνει τελευταία.Ίδωμεν !!!
[u]
Μπάμπης[/u]

  [i]Κώστα , συμμερίζομαι την αγωνία  σου. Αν νομίζεις ότι μπορούμε να κάνουμε κάτι στο αγγλικό forum είμαι μαζί σου. Να σου πω όμως τη γνώμη μου: Φοβάμαι τη γελοιοποίηση ,διότι:
   α) ο χώρος δεν είναι κατάλληλος
   β) κανέναν στον κόσμο δεν ενδιαφέρει αυτό το θέμα , εκτός από τους Έλληνες
   γ) Τα επιχειρήματά μας δε μπορούμε να τα διατυπώσουμε με σαφήνεια στην αγγλική και θα μας πολεμίσουν σκληρά και προπαγανδιστικά.Πίσω από το όνομα κρύβεται βαθειά πολιτική και συμφέροντα που εγώ τουλάχιστον δε μπορώ να αντέξω ή να αντιμετωπίσω.Ας είμαστε προσεκτικοί - όπως λες και εσύ![/i].

ΑΥΤΗ Η ΤΣΟΓΛΑΝΟΚΡΑΤΙΑ ΚΙ Η ΜΠΟΥΡΔΟΛΟΓΙΑ ΓΙΑ ΤΗΝ ΜΑΚΕΔΟΝΙΑ ΜΑΣ ΠΡΕΠΕΙ ΝΑ ΤΗΝ ΔΕΧΘΟΥΜΕ ΔΙΟΤΙ ΟΙ ΠΟΛΙΤΙΚΟΙ ΜΑΣ ΜΑΣ ΞΕΠΟΥΛΑΝΕ ΜΕ ΠΟΛΙΤΙΚΗ ΤΟΥΜΠΕΚΙ ΨΙΛΟΚΟΜΜΕΝΟ ΓΙΑ ΑΥΤΟ ΑΣ ΜΗΝ ΠΟΛΥΠΑΡΑΠΟΝΙΟΥΜΑΣΤΕ ΑΝ ΚΙ ΕΓΩ ΠΡΟΣΩΠΙΚΑ ΜΑΧΑΙΡΩΝΟΜΑΙ Α ΤΟ ΒΛΕΠΩ ΑΥΤΟ ΟΠΩΣ ΚΙ ΤΗΝ ΚΥΠΡΟ ΜΑΣ ΔΙΧΟΤΟΜΗΜΕΝΗ ΑΠΟ ΤΟΥΣ ΜΟΓΓΟΛΟΕΙΔΗΣ ΤΟΥΡΚΟΥΣ
ΧΑΙΡΕΤΩ
ΚΙ ΕΛΠΙΖΩ ΝΑ ΜΠΟΡΟΥΜΕ ΝΑ ΤΟΥΣ ΠΟΛΕΜΗΣΟΥΜΕ
ΜΕ ΤΡΟΠΟ ΚΟΣΜΙΟ
ΑΛΛΑ ΜΑΧΗΤΙΚΟ :mad:  :mad:

Perisi stin Valkaniada Newn pou egine sti Veroia, h monadiki apostoli pou dimiourgise provlima stin vathmologia (toulaxiston sto 4o thema sto opoio eimoun vathmologitis) itan i Firom kai oi monadikoi oi opoioi pigan stin epitropi tis Valkaniadas gia na epilisoun to provlima... Oloi dextikan tin vathmologia pou mpike stous mathites giati oloi vathmologithikan apolitws dikaia... [mexri kai o Alvanos arxigos irthe meta stin deksiwsi pou egine, ki edwse sigxaritiria gia tin vathmologisi sto 4o thema (pou simeioteon itan poli anwsto kai me polles periptwseis ara kai diskolo stin vathmologisi tou)]. Kratithikame me ton sinadelfo pou vathmologousame alla telos pantwn...

Distixws ki egw pisteuw oti den mporoume pleon na kanoume tipota! Ante na eksigiseis twra ston kseno oti ton patera tou Alexandrou ton elegan Filippo, tin mitera tou Olimpiada (!!!!) kai oti den katagotan apo Fyrom, alla apo tin Ellada dioti tetoia onomata mallon diskolo na eixan ekei. (Vre mpas kai oi olimpiakoi agwnes eginan prwti fora ekei?)

Alexandros

ΑΓΑΠΗΤΟΙ ΜΟΥ ΦΙΛΟΙ ΟΤΑΝ ΕΙΣΑΙ ΚΟΜΠΛΕΞΑΣ ΔΕΝ ΜΠΟΡΕΙΣ ΝΑ ΚΑΝΕΙΣ ΤΙΠΟΤΑ ΑΚΡΙΒΩΣ ΕΠΕΙΔΗ ΣΥΜΦΩΝΩ ΜΕ ΤΟΝ ΑΛΕΞΑΝΔΡΟ ΕΓΩ ΠΡΟΣΩΠΙΚΑ ΔΕΝ ΘΑ ΤΗΝ ΠΩ ΠΟΤΕ ΜΑΚΕΔΟΝΙΑ ΚΙ ΘΑ ΤΟΥΣ ΤΗΝ ΣΠΑΩ ΔΙΟΤΙ ΠΑΝΤΑ ΜΑ ΠΑΝΤΑ ΜΑΚΕΔΟΝΙΑ ΘΑ ΕΝΝΟΩ ΤΗΝ ΕΛΛΗΝΙΚΗ ΓΗ ΛΟΙΠΟΝ ΧΑΙΡΕΤΩ

Συμφωνω οτι το θεμα αυτο ειναι πληγη ανοιχτη .αλλα δεν μου αρεσει να βλεπω ακομα και στο mathlinks τιτλους macedonian olymbiad.συμωνω σχεδον με ολα οσα ειπαν οι προηγουμενοι (εκτος οτι φυσικα πι ολυμπιακοι αγωνες εγιναν στα σκοπια  :P ) και ναι κ Μπαμπη ειμαι μεσα .προτινω οπου γινεται κατι τετοιο να παρεμβαινουμε με το γνωστο ευγενικο τροπο στο αγγλικο φορουμ.ευχαριστω πραγματικα :)


calculus
vector
LaTeX
linear algebra
calculus computations

Given VectorOP=scalar(K)(one)*vectorOA+ScalaR(K)(two)*VectorOB    and K(one)+k(two)=1
Prove that A,B,P are collinear (on the same line)

In $ \text{\LaTeX}$ so we can all read it:

$ \vec{OP} = k_1\vec{OA} + k_2\vec{OB},$ with $ k_1 + k_2 = 1.$

This is linear algebra, of course, not calculus - but material like this does appear in calculus textbooks. This is also very much the standard way to parameterize a line.

One way to do this is to show that $ \vec{AP}$ and $ \vec{AB}$ are scalar multiples:

$ \vec{AP} = \vec{OP} - \vec{OA} = (k_1\vec{OA} + k_2\vec{OB}) - \vec{OA}$

$ = (k_1 - 1)\vec{OA} - k_2\vec{OB} = - k_2(\vec{OA} - \vec{OB}) = - k_2\vec{AB}.$

And we're done.

Note also that if $ k_1$ and $ k_2$ are both nonnegative (what is called a [i]convex combination[/i] of $ \vec{OA}$ and $ \vec{OB}$),  then $ P$ lies on the line segment $ \overline{AB}.$


AMC
AIME
induction
summer program
PROMYS
geometry
HCSSiM

Hi

I've just joined this forum a few days ago, and I am pretty surprised and impressed by the math skills of people around the forums.

I used to think I was pretty smart, but then, after trying to solve problems posted on the forums, I feel like I have so much to improve if I want to compete with everyone else.

My question is how can I improve at math?  I personally have been studying on my own time to get ahead of the game in school, but it is not apparent that it is still not enough.

I am sure that most students here did not just take standard math classes provided by the school to get to the point that they are at.

So I was hoping if people can help me find better resources to use to improve my math, whether it be through links to websites, recommendations of special books on math, etc.

I am currently in ninth grade, but outside of school, I have spent time studying math at the precalculus level.  My major weakness is proofs as I have never learned it at a higher level than what my school provides.

I thank anyone who is willing to help me, and if you have any questions, please do not hesitate to post.[/code][/list]

I recommend the Art of Problem Solving (AoPS) books sold on this site.  You can find a description of them on the Bookstore link.

After that, you can practice with old contest problems.  For example, if you want to prepare for the AIME contest, you can purchase all the old AIMEs, with detailed solutions, from the AMC website.  (You can also obtain many of the AIME problems free on this website.)

Ok, thanks, I will try those books.

Are there any books/websites for specific topics, instead of practice for tests/contests?

For example, until about a month ago, I was completely unaware of the many kinds of proofs that exist.  i.e. direct proof, indirect proof, proof of the contrapositive, proof by induction, proof by strong induction, etc.

So I want to make sure I have thorough background on all individual topics, and many of the information I want to learn is not provided in school.

There are two books that I highly recommend (though I haven't yet read the AoPS books, and I'm sure they are fantastic).

[url=http://www.amazon.com/Concrete-Mathematics-Foundation-Computer-Science/dp/0201558025/ref=pd_bbs_sr_1/104-9905048-5296752?ie=UTF8&s=books&qid=1176590871&sr=8-1]Concrete Mathematics: A Foundation for Computer Science --- Donald Knuth[/url]

[url=http://www.amazon.com/Art-Craft-Problem-Solving/dp/0471135712/ref=sr_1_3/104-9905048-5296752?ie=UTF8&s=books&qid=1176590921&sr=1-3]The Art and Craft of Problem Solving --- Paul Zeitz
[/url]

Also you may want to check out Ruvim's program, Pushkin Academy:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56198

Good luck on your quest!

Cheers,
Jack

[color=red]Links edited by your favorite moderator[/color]

[quote="JackPo"]There are two books that I highly recommend (though I haven't yet read the AoPS books, and I'm sure they are fantastic).

Concrete Mathematics: A foundation for Computer Science --- Donald Knuth

The Art and Craft of Problem Solving -- Paul Zeitz

Also you may want to check out Ruvim's program, Pushkin Academy:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56198

Good luck on your quest!

Cheers,
Jack[/quote]

Ok thanks a lot...I'll check out all of them

This book is by NO means a book for extremely high level mathematicians.  The guy who wrote that review has no idea what he's talking about.  I assure you that if you get that textbook, you'll have fun with it.

[quote="Metsjetsfan"]I followed the link to Concrete Mathematics: a Foundation for Computer Science, and I read the customer reviews.

It appears to me that this book is for extremely high level mathematicians, and unfortunately, I am only in ninth grade, and the extent to my math level is precalculus.

Do you have any other suggestions?

Thanks[/quote]

First of all, welcome to AoPS!

Well to give you an idea of how I got to where I am from probably below your level when I was in 9th grade (I joined this forum in the spring of my ninth grade year):

I first started by eating whatever problems came my way, and generally doing problems I thought interesting, or bringing the problems I thought of here to the forum.  Around that time I was basically struggling in the Intermediate forum, and trying AIME problems.

And I did more problems.  And more problems.  And more problems.  If problems started becoming at all easy for me, I moved up (for example, when I started trying Olympiad problems, I was maybe able to solve 1 out of 10 with a lot of difficulty) to problems which were challenging.  I didn't expect to be able to solve these problems necessarily: I just found them interesting.  But I learned a lot of things and got a lot of ideas from problems.  I learned lots of stuff from this forum.  And I did more problems.  This wasn't intensive training of any sort, if you're wondering: it was more or less picking up a couple problems in the morning, and thinking about them through the school day, and then coming back and posting my solutions on this forum at the end of the day.

Eventually, I started to see more and more math which you wouldn't learn in high school, and started doing some of my own reading.  Did I mention I still did problems?  I honestly never read a great deal, I mostly just did or thought about problems I found interesting.  Not to mention I ALWAYS looked at other solutions, whether or not I had solved the problems.

Nowadays, when I get back from school, I have a 20-30 minute block of time before The Simpsons starts, and I can't really do much homework in that time.  So what do I do?  I come here and look at a couple problems.  And that's how I got to where I am.

Well you get the picture, here are some books I have seen for problem solving:
Zeitz: The Art and Craft of Problem Solving (well who wouldn't recommend it)
Mathematical Olympiad Challenges

Well, for other topics, most of the books I know are a little higher end graduate level books which might be easier after you've seen some other things, but what I think would benefit you would be a good introduction to number theory.  A great program for that (and for proofs in general) is BU's summer program, PROMYS: if you have time I suggest you look into it.  I did that, and it opened up my mathematical horizons a lot.

You've probably done geometry before, but I don't know if you've seen enough for patterns to really feel natural (for example, the centers of a triangle, cyclic quadrilaterals, etc.).  I don't know a good geometry book atm...

Algebra and group theory are a great grounding as well: the general concept of a "group", the first time you see it, will be a little abstract at first, but once you see more, you'll be filled with happiness.

However, I think that it would be detrimental to start reading and stop doing problems.  You can't simply read stuff and expect to fully understand it if you never use it.  Doing a great deal of problems will strnegthen your fundamentals, and give you a "feel" for higher level math.  It'll also give you confidence.  It's the same as for anything else: the fundamentals are the most important.  To solve a problem, you don't need to know a great deal of math, but you really need to fundamentally and deeply understand what you do know.

Well PROMYS also has a book list for all sorts of topics, so if you want recommendations on specific books on certain topics, I can give you some ideas there.

[quote="K81o7"][quote="Metsjetsfan"]I followed the link to Concrete Mathematics: a Foundation for Computer Science, and I read the customer reviews.

It appears to me that this book is for extremely high level mathematicians, and unfortunately, I am only in ninth grade, and the extent to my math level is precalculus.

Do you have any other suggestions?

Thanks[/quote]

First of all, welcome to AoPS!

Well to give you an idea of how I got to where I am from probably below your level when I was in 9th grade (I joined this forum in the spring of my ninth grade year):

[color=red]<Quote snipped by moderator>[/color]

[/quote]

First of all, thanks for posting.

What you said about trying questions until finding your level was pretty much what I had set out on doing.

But I have one severe problem, and that is when I see the solution to questions I did not understand, I find out that I also do not understand the solution :!:

Thats why I have come to the conclusion that in order to understand the solutions, I first need to have a solid foundation/background.

It has also come to my attention that a large portion of questions posted are focused/involved with proofs, and I personally have zero knowledge on any kind of proof, other than very basic ones taught at school.

Of course, I will continue to browse the forums (pitifully, it appears I belong in the middle school forums) and try out any questions I find.

And Jack, I will probably buy those books...after checking out the index, contents, etc., it may be the right level for me

Thanks

[quote="K81o7"][quote="Metsjetsfan"]I followed the link to Concrete Mathematics: a Foundation for Computer Science, and I read the customer reviews.

It appears to me that this book is for extremely high level mathematicians, and unfortunately, I am only in ninth grade, and the extent to my math level is precalculus.

Do you have any other suggestions?

Thanks[/quote]

First of all, welcome to AoPS!

Well to give you an idea of how I got to where I am from probably below your level when I was in 9th grade (I joined this forum in the spring of my ninth grade year):

[color=red]<Quote snipped by moderator>[/color]

[/quote]

Jesus christ... Krishnu is a machine...

Jack: your urls stretch my screen.  Use url tags or tinyurl.

MetsJetsFan: if you're responding to the post immediately preceding yours, you don't need to quote it.

Krishanu: you should tell your coach to upload your Mandelbrot round 4 scores.

By the way, PROMYS is one of a number of excellent summer math programs: the program at Hampshire College, the Ross program at Ohio State and the Canada-USA mathcamp are the ones I know about off-hand.  All have dedicated forums on AoPS where you can go for more information.  If you haven't made plans for the summer already, I believe it's still early enough to apply.

[quote="JBL"]
Krishanu: you should tell your coach to upload your Mandelbrot round 4 scores.
[/quote]

He uploaded them around the time of the deadline for the last two rounds, but it still took until team tests were scored for the individual scores to come.  But thanks for reminding me...I guess I should remind him.

JackPo: lol, in what way?  If you naturally find math interesting and fun, you can't help spending time on this forum...it gradually came to replace video games for me, in fact  :rotfl:

well, as everyone has said, all the AoPS books are good. the number theory was a bit confusing for me though, at the end.

I've just bought a couple aops books, and I hope they will help me

Does anyone know of books that can teach me how to do proofs?

Thanks

also: could a mod edit the posts that quote k8107 because it is making the thread harder to read and scroll?

In my opinion, very little of what you learn in regular school mathematics will actually help you in contests. What you need to learn is [i]contest math[/i], not high school math. The Art of Problem Solving Volumes 1 and 2 has EVERYTHING you need to hop from a pre-AMC level problem solver to USAMO level. It has all the techniques and thinking strategies you need to do well on these contests.

[quote="Ravi B"]Here are a couple of books on how to do proofs:

Daniel Velleman, [i]How to Prove It: A Structured Approach[/i].

Daniel Solow, [i]How to Read and Do Proofs: An Introduction to Mathematical Thought Processes[/i].[/quote]

Which would you recommend more?  And do you know if I should buy the updated editions or older ones?

And I bought the two aops volumes...it seems like a lot of people recommend it, so I hope it helps me  :lol:

[quote="Ravi B"]Both books are good.  Solow's book was more popular, and was written in a more informal style.  I thought Velleman's book was more systematic.

I used both books as references in a course on discrete mathematics I used to teach (about 8 years ago).  I would start the course with an intro to proofs.  Principles like how to prove a "for all" statement, how to prove a "there exists" statement, how to use a "there exists" hypothesis, how to use a "for all" hypothesis.  Mathematical induction and proofs by contradiction.

I don't know how much the new editions differ from the older ones.  If price is an obstacle, then an older edition is fine.

Since you already own the AoPS books, I would start with volume 1, work through as many problems as you can, and keep going.  If you really get stuck, then you can consider getting one of these proof books.[/quote]

Ok thanks a lot.  I will determine if I need to buy these books when I check out the AoPS books.

Also, when you're new to math, don't spend too much time on this forum. You can easily get discouraged because most of the members here are very advanced in math.

i would like to know helpful methods in becoming a master at math.  I'm surprised with the advance problems that are posted on this site.  I would like to reach the level of competition that most people in this site have.  I would also like to know how I can prepare forproblems on the Columbia university science honors program exam

Welcome to the forum!

There is already some good advice on this thread, but if you could describe your situation a little more, we could give you additional tips.  For example, what grade are you in, where do you go to school, what math class are you taking, have you participated in any math competitions before.

About the exam for the Columbia Science Honors Program, my daughter will probably take it this April, but I don't know much about it.  There is a brief description [url=http://www.columbia.edu/cu/shp/apply.html]here[/url].

Edit: Some students on the College Confidential site discussed the Columbia SHP exam [url=http://talk.collegeconfidential.com/columbia-university/179517-columbia-university-science-honors-program-entrance-examination.html]here[/url].

well its like im the only oddball here on this site due to te fact i've never joined any math competitions.  The level of intellect i've seen on this site was amazing.  I'm a sophmore in high school and math is just my favorite subject ever since from childhood.  Though, I want devote my time since i found this website to become way better at it.  I would like to know where I should start and type sources I would need to achieve this goal.  Im taking Math B so far and as a freshman i took the math b regents and got a 65 which is very disappointing and doesnt befit aops material. My main problem was careliness., and for a person like me, i should be a wiz on the regents.  I just want to harness the abilities of many people on this site and maybe become olympiad material

I didn't participate in my first math competition until sophomore year of high school, so it's never too late.  Here are some specific resources that you may wish to consider:

[b](1) AoPS books:[/b] Take a look through the [url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php]AoPS bookstore[/url].  The Intro series might be a good fit for you now.

[b](2) AoPS classes:[/b]  It's too late for the fall season, but in the spring you can find some [url=http://www.artofproblemsolving.com/Classes/AoPS_C_About.php]AoPS classes[/url] to help you out.  In the meantime, check out [url=http://www.artofproblemsolving.com/Alcumus/Instructions.php]Alcumus[/url], a free series of problems and lessons that for now focuses on Intro to Counting.

[b](3) NYC classes:[/b] Check out [url=http://www.pushkinacademy.com/en/pamm/]Math-M-Addicts[/url] or [url=http://www.nymathcircle.org/]NY Math Circle[/url].  Both are Saturday morning classes that offer two or three levels depending on students' ages and current abilities.


calculus
integration
logarithms
limit
inequalities
calculus computations

For $ n \equal{} 1,\ 2,\ 3,\ \cdots$, let $ (p_n,\ q_n)\ (p_n > 0,\ q_n > 0)$ be the point of intersection of $ y \equal{} \ln (nx)$ and $ \left(x \minus{} \frac {1}{n}\right)^2 \plus{} y^2 \equal{} 1$.

(1) Show that $ 1 \minus{} q_n^2\leq \frac {(e \minus{} 1)^2}{n^2}$ to find $ \lim_{n\to\infty} q_n$.

(2) Find $ \lim_{n\to\infty} n\int_{\frac {1}{n}}^{p_n} \ln (nx)\ dx$.

1) We have $ 1\minus{}q_n^2\equal{}\left(p_n\minus{}\frac{1}{n}\right)^2$

Then we have to prove that $ \left(p_n\minus{}\frac{1}{n}\right)^2\leq\left(\frac{e}{n}\equal{}\frac{1}{n}\right)^2$

But $ q_n\equal{}\ln(np_n)\equal{}\ln n\plus{}\ln p_n\Rightarrow p_n\equal{}\frac{e^{q_n}}{n}$

$ 0\leq q_n\leq 1\Rightarrow\frac{e^{q_n}}{n}\minus{}\frac{1}{n}\leq\frac{e}{n}\minus{}\frac{1}{n}$

Square both sides and we get the inequality.

Applying the limit in the inequality we have $ \lim_{n\to\infty}q_n\equal{}1$

2) Using integration by parts we have

$ \int_{\frac{1}{n}}^{p_n}\ln(nx)dx\equal{}p_n\ln(np_n)\minus{}p_n\plus{}\frac{1}{n}\equal{}\frac{e^{q_n}}{n}\cdot q_n\minus{}\frac{e^{q_n}}{n}\plus{}\frac{1}{n}$

So the limit is $ \lim_{n\to\infty}(q_ne^{q_n}\minus{}e^{q_n}\plus{}1)\equal{}e\minus{}e\plus{}1\equal{}1$

[quote="kunny"]For $ n = 1,\ 2,\ 3,\ \cdots$, let $ (p_n,\ q_n)\ (p_n > 0,\ q_n > 0)$ be the point of intersection of $ y = \ln (nx)$ and $ (x - \frac {1}{n})^2 + y^2 = 1$.

(1) Show that $ 1 - q_n^2\leq \frac {(e - 1)^2}{n^2}$ to find $ \lim_{n\to\infty} q_n$.

(2) Find $ \lim_{n\to\infty} n\int_{\frac {1}{n}}^{p_n} \ln (nx)\ dx$.[/quote]

hello kunny, the problem seems so considerate towards us :roll:

1) $ q_n = \ln ( n p_n ) \; \Leftrightarrow \; p_n = \frac {1}{n} \text{e}^{ q_n}$.

$ ( \frac {1}{n} \text{e}^{ q_n} - \frac {1}{n} )^2 + q_n^2 = 1 \; \Rightarrow \; 1 - q_n ^2 = 1 - \frac {( \text{e}^{q_n } - 1 ) ^2 }{n^2} \le 1 - \frac {( \text{e} - 1 ) ^2 }{n^2}$

therefore, $ ( \text{xxx} )^2 + q^2_n = 1 \; \Rightarrow \; - 1 \le q_n \le 1$.

furthermore, we have $ 1 - \frac {( \text{e} - 1 ) ^2 }{n^2} \le q_n ^2 \le 1 \; \Rightarrow \; 0 \le 1 - q_n \le  \frac {( \text{e} - 1 )^2 }{2n^2} \;\; (1)$

$ (1)$ means: $ 1 - q_n \xrightarrow{n \to \infty} 0$

2) $ I_n = n \int _{1/n}^{q_n} \ln (nx) \text{d} x \;\; \overset{ {\small y : = \ln (nx)}}{ = } \;\; \int _ 0 ^ {q_n} y \text{e}^y \text{d} y$

we consider $ I_n = \int _0 ^1 \underbrace{y \text{e}^y \, \chi_{[0,q_n]} (y)}_{g_n (y)} \text{d} y$

check these properties: $ \{\begin{array}{l} g_n (y) \le g(y) = \text{e}^y \, , \, \forall y \in (0,1) \, , \, \forall n \in \mathbb{N} \\
g \in L^1 ( 0,1) \\
\lim g_n (y) = y \text{e}^y \, , \, \forall y \in (0,1) \end{array}$

those yield: $ \lim I_n = \int_0^1 ( \lim g_n (y) ) \text{d}y = \int _0^1 y \text{e} ^y \text{d} y = 1$.

is there another way, brethrens ?

cheer !  :lol:


linear algebra
matrix
vector
quadratics
topology
induction
linear algebra unsolved

Let $M$ be an $n\times n$ matrix with complex coefficients and trace zero. Prove that there is a non-zero vector $x$ such that $x^*Mx=0$ ($x^*$ is the conjugate transpose of the column vector $x\in\mathbb C^n$).

$M=H+A$ where $A^*=-A$ and $H^*=H$.
$H$ is hermitian, there exists a unitary matrix $U$ such that: $U^*HU=D$ is a diagonal matrix with zero trace.
Let $v=^t(1,1,...,1)$. Then $v^*Dv=0=(Uv)^*H(Uv)$.
Hence $w=Uv$ satisisfies: $w^*Mv=0$

I still don't see it. We've found $w$ such that $w^*Hw=0.$ But we still have to deal with $A.$ All I can see about that is that $\overline{w^*Aw}=-w^*Aw,$ meaning that $\text{Re}(w^*Aw)=0.$ But that makes it pure imaginary, which is not the same as making it zero.

:blush: , really sorry .. I must have been really tired.(and influenced by the real case  :)  )

Since $Tr(M) = 0$, the quadratic form $x \to x^*Mx$ is not defined, and we are done.

[quote="mathmanman"]the quadratic form $x \to x^*Mx$ is not defined[/quote]

what's that supposed to mean?

Well, it is supposed to mean (by definition actually), that there are non zero x's that make it zero.

But, still, [b]grobber[/b] helped me to understand that my proof was flawed.

I hope I will be back with a correct one. :)

Hi,

I have seen a solution to that problem in a book (les maths en tête, algèbre, p273).
Let consider an linear map f of a vector space (E,( | ))
You were to introduce the set H(f)={(f(x)|x)/(x|x) for x<>0} (hansdorffien).
You have to show it is convex then that 0 is in that set.

If you want a more detailed proof, i'll write later.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Toeplitz&t=76732]this[/url]. That result (the convexity of the numerical range) is the Toeplitz-Hausdorff Theorem (as I mentioned in that other thread).

As you can see, I didn't prove it completely in that thread (started by you, Tsss, by the way :)), but that induction argument reduces it to $n=2$, where it can be proven by plain computations.

I remembered having posted the problem in the real case but I did not that you had already answered the complex case.
Anyhow, it is a nice result :).


ratio geometry circumcircle trigonometry

If you construct a regular polygon with side length r, then inscribe regular polygons (or polygrams) inside it, a region is created that is similar to the original polygon. Let it have side length s. Find the ratio of r to s with:

a) a pentagram inside a pentagon

b) 2 triangles inside a hexagon

c) a heptagram inside a heptagon

d) 2 squares inside an octagon

e) 3 triangles inside a nonagon

Some of these may be too hard...

I will leave the 3 equilateral triangles forming a regular nonagon inside a regular nonagon aside, because it is a different problem, and replace it with a nonagram inside a regular nonagon. In general, assume that we have a regular n-gon with the circumcircle $(O)$ and we form a similar regular n-gon either by a regular n-star with step 2 (if n is odd) or by 2 overlaping $\frac n 2$-stars (if n is even). Let $A_1, A_2, A_3, ..., A_n$ be the vertices of the 1st regular n-gon. If n is odd, let $H_1, H_2, H_3, ..., H_n$ be the intersections of the sides of the regular n-star $A_1A_3A_5...A_{n-1}A_2A_4...A_{n-2}A_n$, such that $H_1 \equiv A_1A_3 \cap A_{n-1}A_2$, $H_2 \equiv A_3A_5 \cap A_2A_4$, ... etc. If n is even, let $H_1, H_2, H_3, ..., H_n$ be the intersections of the sides of the 2 regular $\frac n 2$-stars, such that $H_1\equiv A_1A_3 \cap A_{n-1}A_2$, $H_2 \equiv A_3A_5 \cap A_2A_4$, ... etc. In both cases, the points $H_1, H_2, ...$ lie inside the triangles $\triangle OA_1A_2, \triangle OA_2A_3, ...$ etc.

The triangles $\triangle A_nOA_1, \triangle A_1OA_3$ are isosceles and the lines $OA_1, OA_2$ are the bisectors of the angles $\measuredangle A_nOA_1, \measuredangle A_1OA_3$. Hence $OA_1 \perp A_nA_2$ and $OA_2 \perp A_1A_3$. But this means that the lines $A_nA_2, A_1A_3$ are the altitudes of the triangle $\triangle OA_1A_2$ intersecting at its orthocenter $H_1$. Let $a, h$ be the sides of the regular n-gons $A_1A_2A_3...A_n$ and $H_1H_2H_3...H_n$ (these were denoted r and s in the original post). We have to fing the ratio $\frac a h$.

The triangles $\triangle OA_1A_2 \sim \triangle OH_1H_2$ are similar with the sides $a = A_1A_2, h = H_1H_2$ and

$\frac a h = \frac{A_1A_2}{H_1H_2} = \frac{OA_1}{OH_1}$

Let $R = OA_1$ be the circumradius of the regular n-gon $A_1A_2A_3...A_n$ and $r$ the circumradius of the triangle $\triangle OA_1A_2$. We can calculate the smaller circumradius $r$ with the help of two different formulas for the area $\triangle$ of the triangle $\triangle OA_1A_2$:

$\triangle = \frac{abc}{4r} = \frac{aR^2}{4r}$

$\triangle = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{s(s - a)(s - R)^2} = \sqrt{\left(R + \frac a 2\right)\left(R - \frac a 2\right) \left(\frac a 2\right)^2}$

$\left(\frac{aR^2}{4r}\right)^2 = \left(R^2 - \frac{a^2}{4}\right) \left(\frac a 2\right)^2$

$r = \frac{R^2}{\sqrt{4R^2 - a^2}} = \frac{R}{2 \sqrt{1 - \left(\frac{a}{2R}\right)^2}} = \frac{R}{2 \sqrt{1 - \sin^2{\frac \alpha 2}}} = \frac{R}{2 \cos{\frac \alpha 2}}$

where $\alpha = \measuredangle A_1OA_2 = \frac{360^o}{n} = \frac{2 \pi}{n}$ and $\sin{\frac \alpha 2} = \frac{A_1A_2}{2OA_1} = \frac{a}{2R}$

The distance $OH_1$ of the vertex $O$ from the orthocenter $H_1$ of the triangle $\triangle OA_1A_2$ is equal to

$OH_1 = 2r \cos{\widehat{A_1OA_2}} = 2r \cos \alpha$

This is easily verified by putting a line from the vertex $A_1$ through its circumcenter, intersecting the circumcircle at another point $B$ and then showing, with the help of the right angle triangles $\triangle A_1A_2B, \triangle A_1OB$ that the quadrilateral $A_2H_1OB$ is a parallelogram. Consequently, we have

$\frac a h = \frac{OA_1}{OH_1} = \frac{R}{2r \cos \alpha} = \frac{2R \cos{\frac \alpha 2}}{2R \cos \alpha} = \frac{\cos{\frac \alpha 2}}{\cos \alpha} = \frac {\cos{\frac \pi n}}{\cos \frac{2\pi}{n}}$

For a regular pentagon, hexagon and octagon, we obtain ratios, which can be easily verified by other means:

[color=blue][b](a)[/b][/color] pentagon:

$\frac a h = \frac{\cos{\frac \pi 5}}{\cos \frac{2\pi}{5}} = \frac {\cos{36^o}}{\cos{72^o}}$

$\cos{72^o} = \frac{1}{2G},\ \cos{36^o} = \frac{G}{2}$

$\frac a h = G^2 = G + 1$

where $G = \frac{1 + \sqrt 5}{2}$ is the golden ratio.

[color=blue][b](b)[/b][/color] hexagon:

$\frac a h = \frac {\cos{\frac \pi 6}}{\cos{\frac \pi 3}} = \frac {\cos{30^o}}{\cos{60^o}}$

$\cos{60^o} = \frac 1 2, \cos{30^o} = \frac{\sqrt 3}{2}$

$\frac a h = \sqrt 3$

[color=blue][b](c)[/b][/color] octagon:

$\frac a h = \frac {\cos{\frac \pi 8}}{\cos{\frac \pi 4}} = \frac {\cos{22.5^o}}{\cos{45^o}}$

$\cos{45^o} = \frac{\sqrt 2}{2}, \cos{22.5^o} = \frac{\sqrt{2 + \sqrt 2}}{2}$

$\frac a h = \sqrt{1 + \frac{\sqrt 2}{2}}$


group theory
abstract algebra
number theory
prime factorization
superior algebra
superior algebra unsolved

If G is p-nilpotent for every prime divisor p of |G|, then whether or not G is nilpotent?

Let $ |G| \equal{} \prod_{i \equal{} 1}^{k}p_{i}^{\alpha_{i}}$ be prime factorization of $ |G|$, $ P_{i}$ a Sylow $ p_{i}$-subgroup of G, and $ H_{i}$ normal complement of $ P_{i}$. Since $ H_{i}$ is normal in $ G$, $ |H_{i}| \equal{} \frac {|G|}{p_{i}^{\alpha_{i}}}$ for $ i \equal{} 1, \ldots, k$, and all Sylow $ p_{i}$-subgroups of $ G$ are conjugate, then:
$ P_{i} \equal{} \bigcap_{j \neq i}H_{j}\triangleleft G$ - every Sylow subgroup of $ G$ is normal - hence $ G$ is nilpotent.


geometry trapezoid logarithms trigonometry Vieta rotation function

Post your solutions!

More detailed solutions and results are pending. Here are the answers, basic results, and a few solution outlines for now:

General Comments:

Very good test overall, much harder than usual, which is good. One thing I tried to do with the problems is make them have various principles and such that I rarely see on AoPS.

[hide="answers"]1 A
2 A
3 B
4 D
5 D
6 E
7 C
8 E
9 B
10 A
11 D
12 C
13 A
14 D
15 B
16 D
17 B
18 C
19 D
20 B
21 E
22 C
23 E
24 C
25 C[/hide]

Winner: SplashD, 129
Iversonfan2007 and Xantos C. Guin both got 120.

Average: 105.26
Standard Deviation: 13.74
Median: 106.5

[hide="soutions"]
1. answer is E

2. easy

3. It's a trapezoid with sides 7 and 11, $\frac{7+11}{2}$ done!

4. The restriction doesn't matter, you get 35 and 35.

5. Just simplify stuff

6. Telescoping

7.

8

9 The triangle you get has the same height as the original triangle and 1/5 the base! (that was nice, some people used complicated messed)

10. This is an arithmetic series, as it's the logarithm of a geometry one  :lol:

11. Medians divide triangle into 6 regions of equal area, 6 and 7 are superfluous information

12. Express in a+bi form or use conjugates, comes out as circle with radius 3 and center -1

13. $\sqrt{2}$ is irrational so it is not periodic!!!, so 0 periods

14. Just draw a triangle and make an equation, not hard

15. Think of it as a trig value, use trig addition formulas

16. Basic Vieta's

17. Hardest one (should have been switched with 25), you basically get a sum with binomial coefficients and figure it out.

18. Similar Triangles or trig

19. This was really nice, you get geometric series inside a binomial thing, you evaluate those and then use binomial theorem.

20. ALL I DID WAS TAKE A NICE EASY TO FIND POLYGON AND ROTATE IT 30 DEGREES  :rotfl:

This was not messy shoelace, there was a nice solution  :lol:

21. I just rearranged stuff, formed a square, use QM-AM

I told Feiqi to check it

He IMs me back and he's like "wow that's a really nice problem"

Me: huh?

Him: Draw a triangle...

Me: WOAH!

22. Conceptual, this is locus for a parabolloid, if you have four coplanar points on parabolloid and project them then projection lies on a circle

23. Just draw some triangles, this came from WOOT actually

24. Generating functions (this was pretty hard)

25. Easy NT - should have been switched with 17, BUT: very important lesson: Do not be afraid of 25s!

[/hide]

Overall very hard test, based on the Mildorf Principle.

Wow...solution to 25...?...since no "hint" was given... :oops:  :blush:

[hide="25"]
We have $93s\equiv 1\pmod{4093}$. We want to add some multiple of 4093 to the RHS so that it's divisible by 93, i.e. $1+4093t\equiv 0\pmod{93}\iff t\equiv-1\equiv 92\pmod{93}$. Therefore, we can add $92\cdot 4093$ to both sides and divide by 93 to get $s\equiv 4049$ and the sum of the digits is $\boxed{17}$.
[/hide]

Can you post all the scores (without names)?
I did really really bad on this.

i KNEW i had the wrong answer for 17, but i figured 1.5 points wasnt worth not goin balls out every now and then, and sum how i missed #13 :blush: musta been a silly error, but newayz, this test was a demon :diablo: so i dont feel bad about 106.5, although i did turn it in late and doubt tht it counted

Iversonfan2007 wrote:can anyone actually do the solution for 24, i just can't get it

[hide]

The left hand side is each individual person's generating function, and the problem statement demands a generating function as on the right (one way to do all values ).

Now, we want to maximize the number of factors on the right hand side. The right hand side is  Now, we want to maximize the number of factors we can pull out of ; this is done by maximizing the degree of the two term factor that we pull out of it. This is done by factoring it as . Similarly, we do the same to get  and . Putting this altogether, we see that the RHS of our original equation turns into a maximum of six factors, and so the maximum number of people is .

[/hide]

I don't get the triangle thing for #21...elaborate?

[quote="beta"][quote="Elemennop"][quote="Iversonfan2007"]can anyone actually do the solution for 24, i just can't get it  :oops:[/quote]

I don't get the triangle thing for #21...elaborate?[/quote]

Make a triangle with sides of length $a, b$ and let the angle between these sides be 120[/quote]

yea yea i see that but i have absolutely no clue where to go from there

[hide="hint"]Consider angles A and B[/hide]

[hide="hint"]Use law of sines[/hide]

[hide]You get an expression of the form $a \sin{x}+b \cos{x}$ for constants $a$ and $b$, and then find the maximum of that.[/hide]

[quote="Elemennop"]I got a+b = 4(sin A + sin B)=4(sin A + sin (60-A)) = 4(2sin30cos(A-30)) which has a maximum at A=B=30 but then a=b and we have the original given undefinied since b-a = 0[/quote]

You want to maximize $\left(\frac{\sqrt{3}+1}{2}\right) a+b$, not $a+b$.

From your comment about binomial coefficients and from your answer, I think you assumed that each answer was equally likely to be known or unknown by Bob.  In that case the contribution to the expected value made by each possible number of known answers (i.e., 1, 2, 3, ..., 29, 30) would have to be weighted by the number of ways to choose that many answers (to be known).  Knowing 15 would have the largest weight (30C15)

That's a different problem from the one I did; I didn't include any extra weighting beyond the obvious (4/5)^k.  I guess that means that I tacitly assumed that it was equally likely for Bob to know any given total number of answers.  I got the intuitively more appealing (to me) answer of 26.

Your problem statement doesn't hint at either assumption.  My current position is that this is an example of how problem design can get tricky.

In problem 24. you said:"Then sets of some coins have some integers dollars value".Does that mean $ \$$1, $ \$$5,$ \$$20,$ \$$50 and $ \$$100 or it has any values as long as they are integers??Since I proved that if the maximum number >=7 then it always has at least one way to get over $ \$$999. But if the number of people <=6 I also find a way to get over $ \$$999 if I use the assumption as I think!


group theory
abstract algebra
geometry
3D geometry
superior algebra
superior algebra unsolved

Show that the subgroup $ H$ of the group $ G$ is normal in $ G$ iff $ H$ is the union of some conjugacy classes of $ G$.

Show that the symmetric group $ S_4$ has a normal subgroup (usually denoted $ V_4$) of order 4.

To which group of order 6 is the quotient group $ S_4/V_4$ isomorphic?

Find an action of $ S_4$ giving rise to this isomorphism.

$ S_4$ is the group of symmetries of a cube (e.g. by acting on the pairs of antipodal vertices; that is, acting on the pairs of antipodal faces of an octahedron). But it also permutes the pairs of antipodal faces, this giving a (surjective) morphism to $ S_3$.

[quote="ZetaX"]But it also permutes the pairs of antipodal faces, this giving a (surjective) morphism to $ S_3$.[/quote]
Do you mean a surjective [i]homo[/i]morphism? If not, what's the difference?

I found a normal subgroup to be $ \{\textup{id}, (12)(34), (13)(24), (14)(23), \}$. How do you prove this is normal other than by inspection?

There's no difference.
You just need to check that this subgroup is the kernel of the map I gave (kernels are always normal).


calculus
integration
trigonometry
function
calculus computations

[quote="color"]How do you determine
\[\int \cos{(x^{e})}dx \]
[/quote]

You can define a function,

$I(x) = \int_{0}^{x}\cos (t^{e}) dt$ for $x\geq 0$.

Numerically as mlok said.

Then,

$\int \cos x^{e}dx = I(x)+C$


Given that $(x+y)^2=x^2+2xy+y^2$ find the positive square root of ${6+2 \sqrt{5}}$.

Please express your answer in simplest radical form.

[b]No decimals.[/b]  :D

[hide]Lets say $a+b\sqrt5$ is the answer with $a,b\in{\mathbb{Z}}$.  Then $(a+b\sqrt5)^2=6+2\sqrt5$
$a^2+2ab\sqrt5+5b^2=6+2\sqrt5$
$a^2+5b^2=6$
$2ab=2$
$ab=1$
$b=\frac1a$
$a^2+\frac5{a^2}=6$
call $a^2=z$
$z+\frac5z=6$
$z^2+5=6z$
$z^2-6x+5=0$
$(z-5)(z-1)=0,\ a$ is an integer, so $a^2=1$,  $a=1$ since we want the square root to be positive, which makes $b=1$, so the answer is $\sqrt5+1$[/hide]


ABC is a triangle. The bisectors of B,C meet AC,AB at D,E respectively.
BD meets CE at I. P is the intersection of the line DE and the line through I perpendicular to DE.
PI meets BC at Q.
If $IQ=2\cdot IP$ (length)
Proof that <A = 60°

[color=red][Moderator edit: See also http://www.mathlinks.ro/Forum/viewtopic.php?t=3900 for a discussion of this problem.][/color]

[color=red][[b]Moderator edit:[/b] There has been some confusion here, since the problem proposed by wei_wenzhe9170 had some typos. As a consequence, Grobber solved, instead of the problem proposed by wei_wenzhe9170, a very different problem:

Let ABC be a triangle. Let the angle bisectors of its angles B and C meet the sides AC and AB at the points D and E, respectively.

Let the lines BD and CE intersect at a point I. The perpendicular to the line AB through the point I meets the line AB at a point P and the line AC at a point Q.

Assume that $IQ=2\cdot IP$. Prove that < A = 60°.

What follows now is Grobber's solution to this problem:][/color]

We can forget about the triangle for now. We're only concerned with the angle $\angle BAC$ and a circle tangnt to $AB,AC$. $I$ is its center, $IP\perp AB$, and $Q=IP\cap AC$. We want $IQ=2IP\Rightarrow \angle BAC=60^{\circ}$.

First of all, notice that what we want follows from its converse, namely $\angle BAC=60^{\circ}\Rightarrow IQ=2IP$. This is because when we keep line $AB$ and the circle constant and move $A$ along $(PA$, $IP$ remains constant in both direction and length, while $IQ$ changes (when $A$ moves away from $P$, it gets smaller, and when $A$ moves towards $P$, it gets larger). This implies that the only thing we have to prove is the converse of the initial statement: $\angle BAC=60^{\circ}\Rightarrow IQ=2IP\ (*)$.

$(*)$ follows from the fact that we can construct a third tangent $QT$ to the circle with $T\in AB$. It's easy to check that $AQT$ is equilateral, so $I$ is also its centroid and $QB$ its $Q$-median, so $IQ=2IP$, Q.E.D.

I am sorry.
The mistake has been corrected.


What is the answer to this question:

Find the integral of the expression (x^2)(6x+5)^(1/2) within the limits of 1 and 0.

Thank you.

$\int_{0}^{1}x^{2}\sqrt{6x+5}\,dx$

Make the substitution $u=6x+5$ and "force" it through by writing it backwards.

$x=\frac{u-5}6,\ dx=\frac{du}6.$

When $x=0,\ u=5.$ When $x=1,\ u=11.$

$\int_{5}^{11}\left(\frac{u-5}6\right)^{2}\sqrt{u}\,\frac{du}6$

$=\frac1{6^{3}}\int_{5}^{11}(u-5)^{2}u^{1/2}\,du$

$=\frac1{216}\int_{5}^{11}u^{5/2}-10u^{3/2}+25u^{1/2}\,du.$

Now finish it yourself from there (meaning I'm not feeling motivated to do the arithmetic).


vector
geometry
3D geometry
rotation
calculus
geometric transformation
analytic geometry

A cube with a diagonal of (4, 4, 4) rotates about its diagonal at 3 radians per second. What is the veolicity vector of the point (1, 3, 2)? This problem was found in Kaplan's Advanced Calculus Book. I'm really not sure where to start. Should I rotate the cube so that the axis of rotations point to positive k hat? Perhaps there is an easier way. I just need something to get started. Thanks, I'm doing this for self-study, so I really don't have anyone else to ask for help.

Cyclindrical coordinate system seems to be natural. Define z-axis as the axis of rotation. z and r coordinates remains constant. You have only $\theta$ which is $3t$.

What  topic does the question belong, btw?

isn't velocity of a point $p$ of a rotating body $\vec{v}_{p}= \vec{\omega}\times \vec{r}_{p}$, where $\vec{r}_{p}$ is its vector position, referred to any point on the axis of rotation?

if it's like this (and i'm pretty convinced of it :wink: ), then you just have to plug in: $\vec{\omega}= 3\pi\cdot(1,1,1), \vec{r}_{p}= (1,3,2)$ (i suppose endpoints of the diagonal you mentioned are origin and $(4,4,4)$).

Nice!
You'll just have to normalize $\omega$, by adding a $\frac{1}{\sqrt 3}$.

But again, I wonder the original title was? Coordinate systems? Or vector algebra? Or something else?


So I'm told that the Woot classroom is already up, but when I click on the classroom button, nothing happens.
Am I stupid or is something wrong?

maybe if you waited a few minutes, then tryed again, it would come up? i did that once

I've tried numerous times over the past month.


hi. i'm a beginner in astronomy and am in need of links from which i can learn astronomy? please help me
do we get questions in the iao which require physics knowledge?

we do get..speciall "steven boltz" law :D

If you want stargazing materials, I suggest [url]http://www.skymaps.com/downloads.html[/url]. Other places with star maps are [url]http://www.fourmilab.ch/yoursky/[/url] and [url]http://skychart.skytonight.com/observing/skychart.asp[/url]. These are star maps, though, not really lessons.


function
calculus
derivative
induction
limit
algebra unsolved
algebra

Prove that ther isn't existe suche a funtion$f: R^+\to R^+$ that the ineq.
$(f(x))^2\geq f(x+y)(f(x)+y)$ is true for every$x,y\in{R^+}$.

[quote="vietnamesegauss89"]This problem in [b]Bungari[/b] :lol:[/quote]
What mean [b]Bungari[/b] :?  :D

[quote="vietnamesegauss89"]We have:$(f(x))^2\geq f(x+\frac{1}{n}(f(x)+\frac{1}{n})$
$\Rightarrow f(x)\geq f(x+\frac{1}{n})+\frac{f(x+\frac{1}{n})}{nf(x)}$
$f(x+\frac{1}{n})\geq f(x+\frac{2}{n})+\frac{f(x+\frac{2}{n})}{nf(x+\frac{1}{n}}$
............
$f(x+\frac{n-1}{n})\geq f(x+1)+\frac{f(x+1)}{nf(x+\frac{n-1}{n})}$
$\Rightarrow f(x)\geq f(x+1)(1+\frac{1}{nf(x)})...(1+\frac{1}{nf(x+\frac{n-1}{n}})$
$\Rightarrow f(x)\geq lim(f(x+1)(1+\frac{1}{nf(x)})...(1+\frac{1}{nf(x+\frac{n-1}{n}}))$
$\Rightarrow f(x)\geq f(x+1)+1\Rightarrow f(x)\geq f(x+n)+n\Rightarrow f(x)\geq n \forall n$a contration[/quote]
Can you explain,why ther existe $lim(f(x+1)(1+\frac{1}{nf(x)})...(1+\frac{1}{nf(x+\frac{n-1}{n})}))$?,becaus the seqans$a_n=f(x+1)(1+\frac{1}{nf(x)})...(1+\frac{1}{nf(x+\frac{n-1}{n})}))$ maybe havn't $lim$ when $n\to \inf$ :? ,also why if ther is $k=lim{a_n}$ then$k\geq f(x+1)+1$? :?:

Another solution:
We have f is a strictly decreasing function.For$(x,y)=(x_0,f(x_0))$we have:$f(x_0)=a$and:
$\frac{a}{2}=\frac{f(x_0)}{2}\geq f(x_0+f(x_0))=f(x_0+a)$
let $x_1=x_0+a$we have:
$\frac{a}{4}=\frac{f(x_0)}{4}\geq \frac{f(x_1)}{2}\geq f(x_1+f(x_1))\geq f(x_1+\frac{a}{2})=f(x_0+\frac{3a}{2}$
Proceeding in this manner:
$\frac{a}{2^n}>f(x_0+\sum_{i=0}^n)>f(x_0+3a)\forall n$
but $lim \frac{a}{2^n}=0\Rightarrow f(x_0+3a)<0$.a contradition

[quote="Rust"]f(x)=const>0.[/quote], a im sorry it is not contradition

this problem reminds me a functional eq. problem from romania 2001.

[quote="vietnamesegauss89"]Another solution:
We have f is a strictly decreasing function.For$(x,y)=(x_0,f(x_0))$we have:$f(x_0)=a$and:
$\frac{a}{2}=\frac{f(x_0)}{2}\geq f(x_0+f(x_0))=f(x_0+a)$
let $x_1=x_0+a$we have:
$\frac{a}{4}=\frac{f(x_0)}{4}\geq \frac{f(x_1)}{2}\geq f(x_1+f(x_1))\geq f(x_1+\frac{a}{2})=f(x_0+\frac{3a}{2})$
Proceeding in this manner:
$\frac{a}{2^n}>f(x_0+\sum_{i=0}^n)>f(x_0+3a)\forall n$
but $lim \frac{a}{2^n}=0\Rightarrow f(x_0+3a)<0$.a contradition[/quote]

Why[quote]$f(x_1+f(x_1))\geq f(x_1+\frac{a}{2})$[/quote] :? ?
Also why [quote]$f(x_0+\sum_{i=0}^n)>f(x_0+3a)\forall n$[/quote]?

[hide="Hint"]
Assume for a second that $ f$ is smooth. Then differentiating both sides with respect to $ y$ at $ y\equal{}0$, we immediately obtain $ f'(x)f(x)\plus{}f(x)\le 0$, i.e., $ f'<\minus{}1$. This clearly leads to a contradiction because maintaining the derivative below $ \minus{}1$ on $ (0,\plus{}\infty)$ and staying above $ 0$ simultaneously is unfeasible. Of course, our function is not even continuous, so one has to use a similar finite difference argument ;)
[/hide]

Filing in the gaps in [quote="vietnamesegauss89"]Another solution:
...[/quote]
Suppose to the contrary. Then $ f(x)(f(x)\minus{}f(x\plus{}y))\geq yf(x\plus{}y)$ so $ f$ is decreasing. Note that $ f$ must be strictly decreasing, otherwise the contradiction: $ \exists x,y>0$ with $ 0\geq yf(x\plus{}y)$.

Set $ x\equal{}x_0$, $ y\equal{}f(x_0)$ and $ a\equal{}f(x_0)$. Then $ \displaystyle\frac{a}{2}\equal{}\frac{f(x_0)}{2}\geq f(x_0\plus{}f(x_0))\equal{}f(x_0\plus{}a)$. Set $ x_1\equal{}x_0\plus{}a$ and inductively get:
$ \frac{a}{4}\equal{}\frac{f(x_0)}{4}\geq\frac{f(x_1)}{2}\geq f(x_1\plus{}f(x_1))\geq f(x_1\plus{}\frac{a}{2})\equal{}f(\underbrace{x_0\plus{}a\plus{}\frac{a}{2}}_{x_2})$
$ \frac{a}{8}\equal{}\frac{f(x_0)}{8}\geq\frac{f(x_1)}{4}\geq\frac{f(x_2)}{2}\geq f(x_2\plus{}f(x_2))\geq f(x_2\plus{}\frac{a}{4})\equal{}f(\underbrace{x_0\plus{}a\plus{}\frac{a}{2}\plus{}\frac{a}{4}}_{x_3})$
$ \vdots$
$ \frac{a}{2^n}\equal{}\frac{f(x_0)}{2^n}\geq\ldots\geq\frac{f(x_n)}{2}\geq f(x_n\plus{}f(x_n))\geq f(x_n\plus{}\frac{a}{2^{n\minus{}1}})\equal{}f(x_0\plus{}\ldots\plus{}\frac{a}{2^n})$
Thus, by induction $ \displaystyle\frac{a}{2^n}\geq f(x_0\plus{}2a(1\minus{}\frac{1}{2^n}))\equal{}f(x_n)> f(x_0\plus{}2a)$ which means $ \displaystyle0\equal{}\lim_{n\to\infty}\frac{a}{2^n}\geq f(x_0\plus{}2a)$


Anyone can help me with this problem please?

Let U be a diagonalizable linear operator on a finite dimensional inner product space V such that all of the eigenvalues of U are real. Prove that there exists positive definite linear operators T_1 and T_1' and self-adjoint linear operators T_2 and T_2' such that U= T_2 T_1= T_1' T_2'

i need help on the same problem.. anyone can help?


For $ B\equal{}\left(
\begin{array}{cc}
1 & 2  \\
2 & 1
\end{array}
\right)$, find all 2x2 matrices $ A$ such that $ A^2\equal{}B^2$.

we have that $ \det A\equal{}\pm 3$ and since $ A^2\equal{}A \mbox{tr} A\minus{}I \det A$, we also have that $ 10\equal{}\mbox{tr} A^2\equal{}(\mbox{tr} A)^2\minus{}2\det A$...

if $ \det A\equal{}3$, then, $ \mbox{tr} A\equal{}\pm 4$, so $ \left(\begin{array}{cc}5 & 4 \\ 4 & 5  \end{array} \right)\equal{}\pm 4A\minus{} \left(\begin{array}{cc} 3 & 0 \\ 0 & 3 \end{array}\right)$, which implies that $ A\equal{}\left(\begin{array}{cc}\pm 2 & \pm 1 \\ \pm 1 & \pm 2  \end{array} \right)$...

if $ \det A\equal{}\minus{}3$, then $ \mbox{tr} A\equal{}\pm 2$, so $ \left(\begin{array}{cc}5 & 4 \\ 4 & 5  \end{array} \right)\equal{}\pm 2A\plus{} \left(\begin{array}{cc} 3 & 0 \\ 0 & 3 \end{array}\right)$, which implies that $ A\equal{}\left(\begin{array}{cc}\pm 1 & \pm 2 \\ \pm 2 & \pm 1  \end{array} \right)$...

it's easy to note that these matrices satisfy the equation and we're done...  :D


[quote="Virgil Nicula"] [color=darkred][b][u]An easy extension[/u].[/b] Let $ w_1$ and $ w_2$ be two secant circles with $ w_1\cap w_2 \equal{} \{P,Q\}$ . The tangent in $ P$ to $ w_1$

touches again $ w_2$ in $ C$ . The tangent $ t$ to $ w_1$ in a point  $ A\in w_1$ cut $ w_2$ in $ B_1$ , $ B_2$ so that the line $ t \equal{} \overline {AB_1B_2}$

doesn't separate $ P$ , $ Q$ and it is closer to $ P$ . Denote $ R_1\in AP\cap CB_1$ , $ R_2\in AP\cap CB_2$ . Prove that :

$ 1\blacktriangleright\ PB_2\ ,\ R_1B_1$ are tangent to the circumcircle of the triangle $ PQR_1$ .

$ 2\blacktriangleright\ PB_1\ ,\ R_2B_2$ are tangent to the circumcircle of the triangle $ PQR_2$ .

$ 3\blacktriangleright\ \widehat {R_1QR_2}\ \equiv\ \widehat {B_1PB_2}$ and $ \triangle\ B_1QR_1\ \sim\ \triangle\ B_2QR_2\ \sim\ \triangle\ AQP$ .

$ 4\blacktriangleright$ If $ AP$ cut again $ w_2$ in $ S$ , then $ B_1S$ is tangent to the circumcicle of $ B_1AQ$ and $ B_2S$ is tangent to the circumcircle of $ B_2AQ$ .
[/color][/quote]

[url=http://www.mathlinks.ro/viewtopic.php?t=79788][color=red][b]See here[/b][/color][/url] [color=darkblue]a particular case - a problem  from [b]APMO, 1999[/b] posted by [u][b]Shobber[/b][/u]. [/color]

I don't understand  why this problem hasn't at least one reply.

We use directed angles modulo $ \pi$ in order to embrace all the configurations of $ \omega_1,\omega_2$ together with the tangent $ B_1B_2.$

[b]1)[/b] Let $ B$ be an arbitrary point of $ \omega_1$ on the arc $ PQ$ not containing $ A.$ Thus $ \angle(BQ,BP) \equal{} \angle(PQ,\tau) \equal{} \angle(B_1Q,B_1C)$ and $ \angle(BQ,BP) \equal{} \angle(AQ,AP)$ $ \Longrightarrow$ $ \angle(AQ,AP) \equal{} \angle(B_1Q,B_1C_1)$ $ \Longrightarrow$ $ A,Q,B_1,R_1$ are concyclic. Then we deduce that $ \angle(R_1Q,R_1P) \equal{} \angle(B_1Q,B_1A) \equal{} \angle(PQ,PB_2)$ $ \Longrightarrow$ $ PB_2$ is tangent to $ \odot(PQR_1).$ Consider $ M \equiv AQ \cap PB_2$ and $ N \equiv AP \cap QB_1.$ Hence, we get $ \angle(QM,QN) \equal{} \angle(AB_1,AP) \minus{} \angle(B_2B_1,B_2P) \equal{} \angle(PN,PM)$ $ \Longrightarrow$ $ P,Q,M,N$ are concyclic. Then it follows that $ \angle(QA,QB_1) \equal{} (PB_2,PA) \equal{} \angle(QP,QR_1)$ and $ \angle (QA,QB_1) \equal{} \angle(R_1A,R_1B_1)$ $\Longrightarrow$ $ \angle(R_1P,R_1B_1) \equal{} \angle(QP,QR_1)$ $ \Longrightarrow$ $ R_1B_1$ is tangent to $ \odot(PQR_1).$

[b]2)[/b] Since $ \angle(AQ,AP) \equal{} \angle(B_1Q,B_1C) \equal{} \angle(B_2Q,B_2C),$ we deduce that the points $ Q,A,R_2,B_2$ are concyclic $ \Longrightarrow$ $ \angle(R_2P,R_2Q) \equal{} \angle(B_2B_1,B_2Q) \equal{} \angle(PB_1,PQ),$ which implies that $ PB_1$ is tangent to $ \odot(PQR_2)$ and for the same reason that $ Q,A,R_2,B_2$ are concyclic, we get $ \angle(R_2Q,R_2B_2) \equal{} \angle(AQ,AB_2) \equal{} \angle(PQ,PR_2),$ i.e. $ R_2B_2$ is tangent to $ \odot(PQR_2).$

[b]3)[/b] $ \angle(QR_2,QR_1) \equal{} \angle(QA,QR_1) \plus{} \angle(QR_2,QA) \equal{} \angle(PB_2,PB_1)$ and since $ \angle(QP,QR_1) \equal{} \angle(QA,QB_1),$ we have consequently $ \angle(QA,QP) \equal{} \angle(QB_2,QR_2),$ which clearly implies that  $ \triangle B_1QR_1 \sim \triangle B_2QR_2 \sim \triangle AQP.$

[b]4)[/b] $ \angle(B_1B_2,B_1S) \equal{} \angle(PB_2,PA) \equal{} \angle(QA,\angle QB_1)$ $ \Longrightarrow$ $ B_1S$ is tangent to $ \odot(B_1AQ).$ On the other hand, we have $ \angle(B_2S,B_2C) \equal{} \angle(PS,PC) \equal{} \angle(QB_2,QR_2)$ $ \Longrightarrow$ $ B_2S$ is tangent to $ \odot(B_2AQ).$


I'm not sure if this has been posted already (it's been happening for at least a few weeks now), but somehow, attached files often disappear for seemingly no reason at all. Here are some examples:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=625283#625283]Example 1[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=633818#633818]Example 2[/url]

Thanks in advance for any help!

there was also one in the MATHCOUNTS Target Round Marathon (about 2 times) they disappeared, and then 20 minutes later...they reappeared :noo:

2nd. In the Mock MOEMS contest, I posted the questions, and then they disapppeared, I reposted it again in another post and the first one came back  :noo:

there seems to be a problem :(

[quote="Valentin Vornicu"]Is this still happening? :|[/quote]
I haven't seen any disappearing attachments lately, at least not in the forums I go to.

If this is a bug, then could how could it have suddenly stopped occuring? :huh:

I think I might have solved this bug at some point, but I am not sure :)


I can't remember where I got this, if it's on the masters contest then tell me and I'll remove the thread. I don't think it is though.

Where n>1,

1/$\log_{2}N$ + 1/$\log_{4}N$ + 1/$\log_{6}N$ + 1/$\log_{8}N$ = 1/$\log_{x}N$

What is x?

Notice that $\log_a b=\frac{1}{\log_b a}$, so we have

$\log_{n} \left(2\cdot 4\cdot 6\cdot 8\right)=\log_{n} 384 =\log_{n} x\implies x=\boxed{384}$


function
inequalities
inequalities unsolved

Given real positive numbers : $ x_1,...x_n$
Prove that
$ \sum \frac{x_{i}}{\sqrt{x_{i}x_{i\plus{}1}\plus{}x_{i\plus{}1}^{2}}} \geq \frac{n}{\sqrt{2}}$

Hint
[hide]It can be rewritten as $ \sum \frac{1}{\sqrt{y_{i}^2\plus{}y_{i}}} \geq \frac{n}{\sqrt{2}}$ where $ \prod y_i\equal{}1$ but I dont know how to go from there...[/hide]

Thanks  :)

[quote="FOURRIER"]Given real positive numbers : $ x_1,...x_n$
Prove that
$ \sum \frac {x_{i}}{\sqrt {x_{i}x_{i + 1} + x_{i + 1}^{2}}} \geq \frac {n}{\sqrt {2}}$

Hint
[hide]It can be rewritten as $ \sum \frac {1}{\sqrt {y_{i}^2 + y_{i}}} \geq \frac {n}{\sqrt {2}}$ where $ \prod y_i = 1$ but I dont know how to go from there...[/hide]

Thanks  :)[/quote]
It can be rewritten as $ \displaystyle \sum_{i=1}^n{ \frac {y_i}{\sqrt {y_i+1}} \geq \frac {n}{\sqrt {2}}}$ with  $ \prod_{i=1}^n y_i = 1$.
The function $ f(y)=\frac 1{\sqrt{y+1}}>0$ is convex so  $ \displaystyle \sum_{i=1}^ny_if(y_i) \geq (\sum_{i=1}^n y_i) f_{(1)} \geq f(1)n \left(\prod_{i=1}^n y_i\right)^{\frac 1n} = \frac n{\sqrt 2}$
 :cool:

[quote="Diogene"]
The function $ f(y) \equal{} \frac 1{\sqrt {y \plus{} 1}} > 0$ is convex so  $ \displaystyle \sum_{i \equal{} 1}^ny_if(y_i) \geq (\sum_{i \equal{} 1}^n y_i) f_{(1)} \geq f(1)n \left(\prod_{i \equal{} 1}^n y_i\right)^{\frac 1n} \equal{} \frac n{\sqrt 2}$
 :cool:[/quote]

we have $ \prod yi\equal{}1$ and not $ \sum yi \equal{} 1$ so $ yi$ are not weights..

Sorry, Diogene was in an absent mind  :blush: ..  Any way ..
The original inequality is equivalent as $ \sum _{i\equal{}1}^n \frac {e^{t_i}}{\sqrt{e^{t_i} \plus{}1}} \geq \frac n{\sqrt 2}$ or  as $ \sum _{i\equal{}1}^n \frac  1{\sqrt{e^{t_i} \plus{}{e^{2t_i}}}} \geq \frac n{\sqrt 2}$ , with $ \sum _{i\equal{}1}^nt_i\equal{}0$
Is sufficient to see that  $ \frac {e^t}{\sqrt{e^t \plus{}1}}$ or  $ \frac  1{\sqrt{e^t \plus{}{e^{2t}}}}$   are convexe, and here you are  !
 :cool:


[quote="RIP"][hide] \[=n!\sum_{m=2}^{n}\left(\sum_{i+k=m,k\ge2}\frac{(-1)^{k}}{i!k!}\right)=n!\sum_{m=2}^{n}(0-(\frac1{m!}-\frac1{(m-1)!}))=n!(1-\frac1{n!})\] [/hide][/quote]

It's an excelent answer. This result has come up when I calculated the numbers of permutation which has fixed points. And my english is even worse :D


inequalities
logarithms
function
inequalities proposed

Nice solution marshell :D.Now my solution:
[hide]Observe that the inequality is equivalent to : \[ \left ( \frac{x+y}{2}\right )^{2}\left ( \frac{y+z}{2}\right )^{2}\left ( \frac{z+x}{2}\right )^{2}\geq \left ( \frac{x+y+z}{3}\right )^{3}\]
Applying $ \ln$ we obtain : \[ 3\ln \left ( \frac{3}{x+y+z}\right )\geq 2\ln \left ( \frac{2}{x+y}\right )+2\ln\left ( \frac{2}{y+z}\right )+2\ln\left ( \frac{2}{z+x}\right)\]
Denote the function $ f$ defined on $ (0,\infty)$ ; $ f(x)=\ln \frac{1}{x}$.
Important to se that $ f(x)+f(y)+f(z)=0$ because $ xyz=1$.So the inequality is equivalent to :
\[ f(x)+f(y)+f(z)+3f\left( \frac{x+y+z}{3}\right ) \geq 2f\left ( \frac{x+y}{2}\right )+2f\left ( \frac{y+z}{2}\right )+2f\left ( \frac{z+x}{2}\right )\]
wich is actually $ Popoviciu's$ inequality for the convex function $ f(x)=\ln \frac{1}{x}$.[/hide]


function
search
real analysis
real analysis unsolved

Does there exist a continuous function f on the reals such that f(f(x))=x^2-2 for all x real?

Somebody must have figured out that knowledge is power.

Haha thanks for nothing! :)
If anyone actually knows how to do it I would appreciate a solution.

[quote="remnic"]Haha thanks for nothing! :) [/quote]

You're more than welcome; I love helping out people who are completely unappreciative.

@LydianRain
[quote] remnic wrote:
Haha thanks for nothing!

You're more than welcome; I love helping out people who are completely unappreciative.[/quote]
I think he doesn't talk about you because you posted the message at 5:46, and he posted the message at 5:48. Maybe he talks about §outh§tar. That's my opinion. Maybe I'm wrong.

I do appreciate it, of course I do. I thought the smiley at the end of "thanks for nothing" was relevant. In any case, thanks! Meanwhile I have found that  f(x) = 2 cos(sqrt(2) cos^{-1} (x/2)  is a solution, so I guess my question becomes find all continuous solutions. I don't think this is tackled in the linked paper, judging by the abstract at least.

Uhm, there is no such function, it has been discussed before
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=778734284&t=103001

Is the solution wrong or is it not the same problem?

The function I gave is defined on [-1,1] only.


algebra
polynomial
trigonometry
function
real analysis
real analysis unsolved

let $f:\mathbb{R}\rightarrow \mathbb{R}$.if we know that $f.f \in C^{\infty},f.f.f\in C^{\infty}$.does it follows that $f\in C^{\infty}$.

[quote="jmerry"]What do those dots represent?[/quote]
My guess is that they represent muliplication, so the problem is to prove (or disprove) that $f$ is smooth if both $f^2$ and $f^3$ are smooth.

[quote="Moubinool"]yes[/quote]
I would love to see a solution for this one (alas, I spent more than one hour on it and didn't come up with any good idea). Can you post your proof, Moubinool?

Come on, Moubinool, you must be kidding. I am sure not even the most brilliant student from ENS cannot solve this problem. As far as I know, it is a theorem of Joris, solved something like in the 80-s, and the solution is monstruous. They may be mad, but I hope they won't give such problems at oral examinations. Unless they want to do it especially, so that the student is not qualified, which I think it is pretty stupid.  ;)

Here the challenge find a new proof different than Joris proof

Ok, Moubi, I will let you find another one. When you manage, let me know and I will start trying. Till then, I don't feel so strong in analysis...  ;)

[quote="harazi"]Ok, Moubi, I will let you find another one. When you manage, let me know and I will start trying.[/quote]
Well, Harazi, I found another one  :lol:. Will you try now?  ;) Seriously, it is quite useful to think of very difficult problems time to time even when the chances of finding a solution are infinitesimally small (the necessary condition being that the problem be attractive to you, of course).

I am surely asking a very stupid question, but is the following true:

let $f:{\mathbb R}\rightarrow {\mathbb R}$ and for any $x_0\in {\mathbb R}$ and positive integer $n$ there exists a polynomial $P(x)$ of degree at most $n$ such that $|f(x)-P(x)|= o((x-x_0)^n)$ in a neighborhood of $x_0$. Then $f\in C^{\infty} ({\mathbb R})$.
?

I just do not see problems in dividing Taylor series for $f^3$ and $f^2$.

[quote="fedja"]...I found another one...[/quote]

Think for publication ( in AMM for example)

[quote="Fedor Petrov"]
let $f:{\mathbb R}\rightarrow {\mathbb R}$ and for any $x_0\in {\mathbb R}$ and positive integer $n$ there exists a polynomial $P(x)$ of degree at most $n$ such that $|f(x)-P(x)|= o((x-x_0)^n)$ in a neighborhood of $x_0$. Then $f\in C^{\infty} ({\mathbb R})$.
?
[/quote]
Unfortunately, this is false: just take $f(x)=e^{-1/x^2}\sin e^{1/x^2}$ for $x\neq 0$, $f(0)=0$. This function satisfies the condition but isn't even $C^1$. It is a very common mistake, by the way: I even found a web page with the "elementary proof" of Joris' theorem where exactly this approach is  taken. Needless to say, the "proof" is invalid...

[quote="Moubinool"]Think for publication ( in AMM for example)
[/quote] I'll rather just post it here (if I manage to simplify it enough to make it understandable and if nobody finds a better approach in a couple of months: I should confess that my proof uses quite heavy tools from compelx analysis and, in a sense, is even less elementary than that by Joris).


integration
function
Support
probability and stats

Fix a real number $ \alpha >0$. Let $ Z \sim  \mathrm{Exp}(\Lambda)$, where $ \Lambda$ itself is a random variable, satisfying $ \Lambda \sim  \mathrm{Exp}(\alpha)$. Find $ E(\Lambda | Z)$.

you know the density of $ (Z, \Lambda)$:
\[ f(z,\lambda) \equal{} \alpha \lambda \exp(\minus{}(\alpha\plus{}z) \lambda) 1_{z>0, \lambda > 0}\]
Hence:
\[ E[ \Lambda|Z](z_0) \equal{} \frac{ \int_{\lambda } \lambda f(z_0, \lambda) d \lambda }{ \int_{\lambda } f(z_0, \lambda) d \lambda }\]
In short, $ P[ \lambda \in [\lambda_0; \lambda_0\plus{}dx] | Z\equal{}z_0] \equal{} \frac{ 1 }{ \int_{\lambda } f(z_0, \lambda) d \lambda } f(\lambda_0, z_0) d \lambda$

[quote="alekk"]you know the density of $ (Z, \Lambda)$:
\[ f(z,\lambda) \equal{} \alpha \lambda \exp( \minus{} (\alpha \plus{} z) \lambda) 1_{z > 0, \lambda > 0}
\][/quote]

How did you get this?

always the same story:
take a test function $ \phi$ (eb: positive, $ C^{\infty}$, compact support) and compute $ E[\phi(Z, \Lambda)]$:
\[ E[\phi(Z, \Lambda)] \equal{} \int_{\lambda\equal{}0}^{\infty} \alpha \exp(\minus{}\alpha \lambda) \int_{z\equal{}0}^{\infty} \lambda \exp(\minus{}\lambda z) \phi(z, \lambda) dz d \lambda\]
Hence, for every test function $ \phi$:
\[ E[\phi(Z, \Lambda)] \equal{} \iint_{z, \lambda} \phi(z, \lambda) f(z, \lambda) dz d \lambda\]
This gives the answer. Is it clearer ?

Using standard approach with pdf I found $ \frac {2}{\alpha \plus{} Z}$.
Too simple to be the unique possible solution!

It's consistent with the following intuition (this is attempt to provide intuitive explanation): if $ \Lambda$ is on its mean level than it's equal to $ \frac {1}{\alpha}$. Given that $ \Lambda \equal{} \frac {1}{\alpha}$ we expect $ Z$ to equal $ \alpha$. So when $ Z \equal{} \alpha$ we expect $ \Lambda$ to equal its mean level, $ \frac {1}{\alpha}$. And the formula gives the same result $ E(\Lambda|Z \equal{} \alpha) \equal{} \frac {2}{\alpha \plus{} \alpha}$.

Maybe there is intuitive proof? Using some argumentation with limits of iid U[0;1] random variables?


inequalities
trigonometry
geometry
IMO
area of a triangle
Heron\u0027s formula
IMO 1961

[b]If a, b, c are the sidelengths of a triangle and S is its area, then

$ 4 \sqrt3 S \leq a^2 \plus{} b^2 \plus{} c^2$.[/b]

Indeed there are lots of proofs for this inequality. Here is my favorite proof:

Let X be such a point that the triangle BCX is equilateral, and the points A and X lie in the same halfplane with respect to the line BC. Then, after the Cosine Law in triangle ABX, we have

  $ AX^{2} \equal{} AB^{2} \plus{} BX^{2} \minus{} 2\cdot AB\cdot BX\cdot \cos \measuredangle ABX$.

But BX = BC and < XBC = 60°, since the triangle BCX is equilateral; hence < ABX = < ABC - < XBC = < ABC - 60° = B - 60°, and we get

  $ AX^{2}\equal{}AB^{2}\plus{}BC^{2}\minus{}2\cdot AB\cdot BC\cdot \cos \left( B\minus{}60^{\circ }\right)$
  $ \equal{}c^{2}\plus{}a^{2}\minus{}2ca\cos \left( B\minus{}60^{\circ }\right)$
  $ \equal{}c^{2}\plus{}a^{2}\minus{}2ca\left( \cos B\cos 60^{\circ }\plus{}\sin B\sin 60^{\circ }\right)$
  $ \equal{}c^{2}\plus{}a^{2}\minus{}2ca\left( \frac{1}{2}\cos B\plus{}\frac{\sqrt{3}}{2}\sin B\right)$
  $ \equal{}c^{2}\plus{}a^{2}\minus{}ca\cos B\minus{}\sqrt{3}ca\sin B$
  $ \equal{}c^{2}\plus{}a^{2}\minus{}\frac{c^{2}\plus{}a^{2}\minus{}b^{2}}{2}\minus{}\sqrt{3}ca\sin B\text{\ \ \ \ \ \ \ \ \ \ (Cosine Law)}$
  $ \equal{}\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{2}\minus{}\sqrt{3}ca\sin B$
  $ \equal{}\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{2}\minus{}\sqrt{3}\cdot 2S\text{\ \ \ \ \ \ \ \ \ \ (since }S\equal{}\frac{1}{2}ca\sin B\text{)}$;

thus, $ \displaystyle \frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{2}\geq \sqrt {3}\cdot 2S$, so that $ a^{2} \plus{} b^{2} \plus{} c^{2}\geq 4\sqrt {3}S$, what was to be proven. Equality occurs if and only if $ AX^{2} \equal{} 0$, i. e. if the points A and X coincide, i. e. if the triangle ABC is equilateral.

  Darij

[quote="fuzzylogic"]$a^2+b^2+c^2-4\sqrt{3}S = a^2+b^2 + (a^2+b^2-2ab\cos{C})-2\sqrt{3}ab\sin{C} $ $ = 2(a-b)^2 + 4ab(1-\cos(C+\pi/3)) \geq 0$[/quote]

Like darij, I have seen many nice proofs of this inequality, but that is just beautiful!

[b]Weitzenböck's inequality: Prove that if a,b,c are the side lengths of a triangle and Sa it area then: [b]4S√3 ≤ a 2  + b 2  + c 2 [/b][/b]

To prove the inequality of Weitzenbock, we should first prove an aiding inequality.
xy+xz+yz ≥ √3xyz(x+y+z)
(xy+xz+yz) 2  ≥ 3xyz(x+y+z)
(xy+xz+yz) 2  = (xy) 2 +(xz) 2 +(yz) 2 +2xyz(x+y+z) ≥ ⅓(xy+xz+yz) 2 +2xyz(x+y+z)
Therefore
⅔(xy+xz+yz) 2  ≥ 2xyz(x+y+z)
(xy+xz+yz)2 ≥ 3xyz(x+y+z) We now have the required inequality.
Going back to the starting inequality.
S√3 = √3p(p-a)(p-b)(p-c) = √3(p-a+p-b+p-c)(p-a)(p-b)(p-c)
Apply the aiding inequality and get:
S :sqrt: 3<=(p-a)(p-b)+(p-b)(p-c)+(p-a)(p-c)
= ab+bc+ac-p 2
Therefore
4S√3 ≤ 4ab+4bc+4ac-(a+b+c) 2
= a 2 +b 2 +c 2 -(a-b) 2 -(b-c) 2 -(a-c) 2
4s√3 ≤ a2 + b2 + c2
Therefore, Weitzenbock�s inequality is now proved.

1)Aply  am-gm  to  $s-a , s-b , s-c  $  and  find  that  $s\geq3\sqrt{3}r$
2).$ a^2+b^2+c^2\geq\frac{1}{3}(a+b+c)^2=\frac{4}{3}s^2\geq4\sqrt{3}sr=4\sqrt{3}A  $

Consider the squares of each side of inequality we want to prove.Squaring on the left we get
 a^4 + b^4 +c^4 +2(a^2)(b^2 ) +2(b^2)(c^2) +2(c^2)(a^2) and

3.16(S^2) =3 [2(a^2)(b^2)  +2(b^2)(c^2)  +2(c^2)(a^2) -(a^4) -(b^4) -(c^4)]  on the right.
The difference between these two expressions is easily seen to be positive.

This is a very weak inequality because
$a^2+b^2+c^2 \ge (a+b+c)^2/3 \ge ab+bc+ca \ge 3(abc)^{2/3} \ge 4S\sqrt{3}$.

It is not necessary to post meaningless things like 'this is a weak inequality' to threads that have been dead for nearly 5 months.

Come on, this inequality has benn posted many times before.
The following inequalities hold:

      $a^{2}+b^{2}+c^{2}\geq ab+bc+ca\geq a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}\geq 18Rr\geq 4S\sqrt{3}$ ;)

Although here is a stronger one due to Hadwiger-Finsler, 1937 which states that

 $ a^{2}+b^{2}+c^{2}\geq 4S\sqrt{3}+(a-b)^{2}+(b-c)^{2}+(c-a)^{2}.$

[Moderator edit: See http://www.mathlinks.ro/Forum/viewtopic.php?t=31598 for proofs of this inequality.]

Caesar, how show you the inequality $\sum a\sqrt{bc}\ge 18Rr \ ?$

Nice and easy :D. Because $2Rr=\frac{abc}{a+b+c}$ the inequality reduces to the following:

   $\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\geq\frac{9}{a+b+c}$.
 By AM-GM we have,$\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\geq\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}$ and from here it is clear how to proceed. ;)

 P.S. I have found some nice stronger inequalities connected to these ones. :D

[quote="Vasc"]This is a very weak inequality because
$\ldots \geq 3(abc)^{2/3} \geq 4S\sqrt{3}$.[/quote]

How do we prove the last inequality ?   :?

[hide=Solution]
Since $S=\frac{\sqrt{-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2c^2a^2}}4$, it suffices to show $a^2+b^2+c^2\geq\sqrt{-3a^4-3b^4-3c^4+6a^2b^2+6b^2c^2+6c^2a^2}$. Since both sides are positive, we can square both sides, so it suffices to show $4a^4+4b^4+4c^4\geq 4a^2b^2+4b^2c^2+4c^2a^2$, which is equivalent to $2(a^2-b^2)^2+2(b^2-c^2)^2+2(c^2-a^2)^2\geq0$. Equality holds if and only if $a=b=c$.

[quote=Math-wiz][quote=mihaig]In $\Delta ABC$  with area $S,$, prove
$$a^2+b^2+c^2\geq4\sqrt3\cdot S+2\left(a-b\right)^2.$$
When do we have equality?

Roberto Tauraso, 2017.[/quote]

We need to prove $$c^2-a^2-b^2+4ab\geq 4\sqrt{3}S$$
Substituting $c^2-a^2-b^2=-2ab\cos C$ and $ab=\frac{2S}{\sin C}$, we need to prove $2\csc C-\cot C\geq\sqrt{3}$
Let $\csc C=x$
We need to prove $x-\sqrt{x^2-1}\geq\sqrt{3}$ which is equivalent to $$(\sqrt{3}x-2)^2\geq 0$$
Equality when $\angle C=\frac{\pi}{3},\frac{2\pi}{3}$[/quote]

Bravo

By Heron's formula and AM-GM,
$16S^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) \leq (a+b+c)(\frac{a+b+c}{3})^3$
$\implies 4S\leq \frac{(a+b+c)^2}{3\sqrt{3}} =\sqrt{3}(\frac{a+b+c}{3})^2 \leq \sqrt{3}\frac{a^2+b^2+c^2}{3}$
$\implies a^2+b^2+c^2 \geq 4S \sqrt{3}.$
Equality when $a=b=c.$
[Note: I thought the area was $A$!]

Let $ a$, $ b$, $ c$ be the sides of a triangle, and $ S$ its area. Prove:
$$a^2+ b^2+ c^2\geq 4S \sqrt {3}+\frac{( b^2- c^2)^2}{2a^2}$$
SXTB,(3)2022,Q2651

[quote=Iris Aliaj]Let $ a$, $ b$, $ c$ be the sides of a triangle, and $ S$ its area. Prove:
\[ a^{2} \plus{} b^{2} \plus{} c^{2}\geq 4S \sqrt {3}
\]
In what case does equality hold?[/quote]
[url=https://artofproblemsolving.com/community/c1068820h2006025p14040671]North Macedonian 1998[/url]
[quote=sqing][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=193702&hilit=Weitzenb%C3%B6ck]stronger than Weitzenbock, a present for nguyentrang:[/url]\[ a^2\plus{}b^2\plus{}c^2\ge 4\sqrt{4\minus{}sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}}S.\][/quote]
In triangle $ABC,$ [url=https://artofproblemsolving.com/community/c6h1371994p7563396]show that[/url] $$\tan{\frac{A}{2}}+\tan{\frac{B}{2}}+\tan{\frac{C}{2}} \ge \sqrt{4-8\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}}.$$

[quote=sqing][quote=Iris Aliaj]Let $ a$, $ b$, $ c$ be the sides of a triangle, and $ S$ its area. Prove:
\[ a^{2} \plus{} b^{2} \plus{} c^{2}\geq 4S \sqrt {3}
\]
In what case does equality hold?[/quote]
[url=https://artofproblemsolving.com/community/c1068820h2006025p14040671]North Macedonian 1998[/url][/quote]

 :rotfl: Ionescu, early 1900

With using Heron's formula we need to prove $(a^2+b^2+c^2)^2 \ge 48p(p-a)(p-b)(p-c)$ or $(a^2+b^2+c^2)^2 \ge 3(a+b+c)(b+c-a)(c+a-b)(a+b-c)$. Note that by Ravi and simple AM-GM we have $(b+c-a)(c+a-b)(a+b-c) \le abc$ so we need to prove $(a^2+b^2+c^2)^2 \ge 3(a+b+c)abc$ or $(ab+bc+ca)^2 \ge 3(a+b+c)abc$ which is true.

[hide=indeed a good one]from trivial inequality we have that $a^2+b^2+c^2\ge ab+bc+ca =2S(\frac{1}{\sin{\alpha}}+\frac{1}{\sin{\beta}}+\frac{1}{\sin{\gamma}})$, where$\alpha,\beta,\gamma$ are angles of the triangle\\
now we now that $f(x)=\frac{1}{\sin{x}}$ is a convex function. so we get \\
$f(\alpha)+f(\beta)+f(\gamma)\ge 3f( \frac{\alpha+\beta+\gamma}{3})=3f(\frac{\pi}{3})=\frac{3}{\sin{60^{\circ}}}=2\sqrt{3}$\\
which gives $a^2+b^2+c^2\ge 4S\sqrt{3}$ as desired\\
equality holds when $a=b=c$[/hide]

Let $s$ be the semi-perimeter and $\triangle$ be the area of the triangle respectively .\\\\[color=#0f0][b]Lemma:[/b][/color] $s^2\geq 3\sqrt{3}\triangle$\\\\[b][color=#0f0] Proof :[/color][/b] By AM-GM , $\frac{(s-a)+(s-b)+(s-c)}{3}\geq \sqrt[3]{(s-a)(s-b)(s-c)}\implies s\geq 3 \sqrt[3]{\frac{{\triangle}^2}{s}} \implies s^4\geq 27{\triangle}^2$ $\implies  \boxed{s^2\geq 3 \sqrt{3} \triangle}$\\\\ Now , by Titu's Lemma , we have \\\\$a^2+b^2+c^2\geq \frac{(a+b+c)^2}{3}=\frac{4s^2}{3} \geq 4\sqrt{3}\triangle$

$$a^{2} + b^{2} + c^{2}\geq 4S \sqrt {3}$$
Let $a=x+y$,$b=y+z$,$c=z+x$
INEQUALITY becomes $$2(x^2+y^2+z^2+xy+yz+zx) \geq 4\sqrt{3xyz(x+y+z)}$$
well known lemma $xy+yz+zx \geq \sqrt{3xyz(x+y+z)}$
well known lemma $x^2+y^2+z^2 \geq xy+yz+zx$
and rest is easy
Aoxz

$a^2+b^2+c^2-4\sqrt{3}S=2(a^2+b^2)-2ab\cos(C)-2\sqrt{3}ab\sin(C)=2(a-b)^2+4ab(1-\cos(C+\pi/3)) \ge 0$, equality holds when $\cos(C+\pi/3)=0$, which holds when the triangle is equalateral.


floor function
number theory proposed
number theory

Define the sequence of positiev integers $ (x_{n})$ by $ x_{1}=10^{999}$ and for $ n\ge2$ the number $ x_{n}$ is obitained from $ 11x_{n-1}$ by deleting the first digit .Is the sequence bounded ? :wink:

What's wrong with the following? :maybe: Provided $ k \in \mathbb{N}_{0}$ and $ b \ge 2$ is an integer, let $ x_{1}= b^{k}$ and $ x_{n+1}$ be the number obtained by deleting the leftmost digit of $ (b+1)x_{n}$ in its base $ b$ representation, for any $ n \in \mathbb{N}$. Here as in fashion, $ \mathbb{N}= \{1, 2, \ldots\}$ and $ \mathbb{N}_{0}= \mathbb{N}\cup \{0\}$. Then $ x_{1}= x_{2}= \ldots$, so $ \{x_{n}\}_{n \in \mathbb{N}}$ is bounded from above. Indeed, if $ n \in \mathbb{N}$ and $ x_{n}= x_{1}$, then
\[ (b+1)x_{n}= (11\underbrace{00 \ldots 0}_{\tiny{k\mbox{ zeros}}})_{b}\quad\Longrightarrow\quad x_{n+1}= (1\underbrace{00 \ldots 0}_{\tiny{k\mbox{ zeros}}})_{b}= b^{k}= x_{1}. \]

[quote="s.tringali"]What's wrong with the following? $ [...]$
\[ (b+1)x_{n}= (11\underbrace{00 \ldots 0}_{\tiny{k\mbox{ zeros}}})_{b}\quad\Longrightarrow\quad x_{n+1}= (1\underbrace{00 \ldots 0}_{\tiny{k\mbox{ zeros}}})_{b}= b^{k}= x_{1}. \]
[/quote]
hallo $ s.tringali$
here we have for exemple if $ x_{1}=1000$
$ x_{2}=1100$
$ x_{3}=1210$
$ x_{4}=1331$
$ x_{5}=1464$

so here we have the first 1000 terms $ x_{0}=10^{999},x_{n}=11^{n-1}.10^{999-n}$ for $ n\in \{1,2,...,999\}$

[quote="quangpbc"]Define the sequence of positiev integers $ (x_{n})$ by $ x_{1}=10^{999}$ and for $ n\ge2$ the number $ x_{n}$ is obitained from $ 11x_{n-1}$ by deleting [b]the first digit.[/b][/quote]
[quote="s.tringali"]What's wrong with the following? :maybe: Provided $ k \in \mathbb{N}_{0}$ and $ b \ge 2$ is an integer, let $ x_{1}= b^{k}$ and $ x_{n+1}$ be the number obtained by deleting [b]the leftmost digit[/b] of $ (b+1)x_{n}$ in its base $ b$ representation [...][/quote]
Really, I assumed quangpbc meant [i]the first digit[/i] to be the leftmost one. Anyway, because in fact I'm not confident of how to take his words, I suggest we wait for him to disambiguate the problem. :wink:

[quote="aviateurpilot"]
[...] so here we have the first 1000 terms $ x_{0}=10^{999}$, $ x_{n}=11^{n-1}\cdot 10^{999-n}$ for $ n\in \{1,2,\ldots,999\}$[/quote]
Finally, note the right formula would be $ x_{n}= 11^{n-1}\cdot 10^{1000-n}$, whether $ x_{1}= 10^{999}$ and $ n = 1, 2, \ldots, 1000$.

Indeed, quangpbc's clarifications are not really necessary. Here it goes:

[quote]Provided $ b \in \mathbb{N}$, where $ \mathbb{N}= \{1, 2, \ldots\}$, define the sequence $ x_{1}, x_{2}, \ldots$ by taking $ x_{1}$ to be any integer $ \ge b$ and $ x_{n+1}$ equal to the number obtained by deleting the [i]rightmost[/i] digit in the base $ b$ representation of $ (b+1)x_{n}$, for any $ n \in \mathbb{N}$. Is the sequence bounded?[/quote]
By definition $ x_{n+1}= \left\lfloor\left(1+\frac{1}{b}\right)x_{n}\right\rfloor$, and it's clear $ \{x_{n}\}_{n \in \mathbb{N}}\subseteq \mathbb{Z}$. Then $ x_{n+1}= x_{n}+\left\lfloor\frac{x_{n}}{b}\right\rfloor$, and hence inductively $ x_{n+1}\ge x_{1}+n$, for any $ n \in \mathbb{N}$. So $ x_{n}\to+\infty$, as $ n \to \infty$. :maybe:


LHS-RHS = $\sum \frac{a(a-c)(a-b)}{b+c}$
Use Vonicular Schurl.  :D


linear algebra
matrix
linear algebra theorems

How to create Jordan normal form of the matrix using the formula of defects  :(


quadratics
calculus
integration
algebra
quadratic formula

find all pairs of positive integers (a,b) with a,b < 100 such that a + sqrt(2a+b) = b*sqrt (b-a).

does this seem like a real problem? im not sure if the sheet it was on was typed right.

I think it's correct -- it's definitely doable.

I think it's incorrect; there are no solutions.

First, I tried it mathematically.
sqrt(2a+b)=b*sqrt(b-a) = sqrt(b:^3:-ab:^2:)
2a+b=b:^3:-ab:^2:
2a+ab:^2:=b:^3:-b    <-- this step was missing in what I typed before
a(2+b:^2:) = b:^3:-b
a=((b:^3: / (2+b:^2:)) - (b/(2+b:^2:)))
Right there, I took a leap and figured b/(2+b:^2:) would have to be an integer. Letting that =k, rearranging and putting it into the quadratic formula, and then only taking real solutions (positive values of (b^2-4ac)), I got that k^2:le:1/8, which is certainly not possible for integer k.

Then, just to make sure I didn't do anything wrong in the math, I whipped out this program in C++ (took about 2 minutes to write)
[code]
/*
find all pairs of positive integers (a,b) with a,b < 100,
such that a + sqrt(2a+b) = b*sqrt (b-a).
*/
#include <iostream>
#include <math.h>
using namespace std;

int main()
{
	//initialize a and b to 1
	int a=1, b=1;

	//loop through all possible values of a and b
	for (a;a<100;a++)
		for (b;b<100;b++)
			if ( (a+ sqrt(2*a+b)) == (b*sqrt(b-a)) )
				cout<<a<<","<<b<<endl;

	return 0;
}
[/code]
It didn't display anything, so there are no (integer) solutions (in that interval, at least)

Actually, there's a more mathematical way to do it:
:sqrt:(2a - b) will have to be integral, since there's no way the right side can be an integer plus an irrational square root, since its square is an integer.  So the right side is an integer, so b - a is a square.   Thus,
2a + b = m:^2:
b - a = n:^2:

a = (m:^2: - n:^2:)/3
b = (m:^2: + 2n:^2:)/3

Plugging into the original equations, we get

(*) (m:^2: - n:^2:)/3 + m = n * (m:^2: + 2n:^2:)/3

We can solve this as a quadratic in m, which will give us in the end a discriminant of (9 - 4(n - 1)(2n:^3: + n:^2:)), which is only non-negative when n = 0, 1.  (n could also be negative -- however, in that case our equation (*) would be false, because the right side would be negative.)

Then we can go and check and see that it doesn't work.

That doesn't necessarily mean the question is wrong -- I mean, it still yeilds to a solution process.  Could have been mis-typed, though, I suppose.

[quote="confuted"]
a=((b:^3: / (2+b:^2:)) - (b/(2+b:^2:)))
Right there, I took a leap and figured b/(2+b:^2:) would have to be an integer. [/quote]

Thats a pretty big leap. Theres no way you can just assume that b/(2+b:^2:) is an integer, I mean if you wanted to you could have written a as (b:^3:-1)/(2+b:^2:) - (b-1)/(2+b:^2:) and assumed (b-1)/(2+b:^2) is an integer. It can't be broken up like that.

Anyway, what happened to the a on the left hand side? I think you may have missed it..

JBL:
I'm not quite sure where the 2a [b]-[/b] b came from.. But I'm not sure you can assume 2a+b (if that was the bit you were meaning) must be a square, because if you square both sides the right hand bit is an integer and the non-integer right hand bit is 2a:sqrt:(2a+b), couldn't you have something like 2a:sqrt:(k / 4a:^2), so the sqrt bit isn't an integer but the overall bit is?

I'll go try to solve it myself now that I've ruined everyone elses solution :D

Whoops -- that was just a typo.  When I did the math, I did it correctly -- just wrote it down wrong here.  I hope.

Eh, I just skipped a step typing it. I fixed it now. The math was all fine until I took the leap. But I'd already seen at that point that the problem was ugly and wasn't going to be easily simplified... which led me to think that it was probably typed incorrectly.

this problem was supposed to come from one of the mandelbrote individual contests from 95, so clearly its typed wrong. still, the proof that it has no solutions was cool.  :D


I posted in the pre-olympiad section but... I would like to get a solid proof so here it is...
find the minimum of \[ a^2 + 4ab + 8b^2 + 16c^2 \]
if \[ abc=1 \] and $a,b,c>0$

[quote="hydro"][hide]
rewrite as $a^2+2ab+2ab+8b^2+8c^2+8c^2$

by AM-GM, $a^2+2ab+2ab+8b^2+8c^2+8c^2 \geq 6 \sqrt[6]{4\cdot 8^3a^4b^4c^4}=12\sqrt[6]{2^5}$[/hide][/quote]
yes... but the equality is only when \[ a^2=2ab=2ab=8b^2=8c^2=8c^2 \] and this cannot be because \[ a^2=2ab \leftrightarrow a=2b \] and \[ a^2=8b^2 \leftrightarrow a=2\sqrt{2}b \]


[quote="kdpk"]2OH(-) + Cr(OH)3 -> CrO4 (-2) + H20[/quote]
umm im not particularly good at chemistry, but i think you made a balancing error here.. with the number of hydrogen atoms.
i think the following equation will work:
$Cr(OH)_{3}+5OH^{-}\Longrightarrow CrO_{4}^{2-}+4H_{2}O+3e^{-}$

I think you are indeed supposed to have an electron in the answer.  I don't know anything about whether the equation you got in the end is right or not though. :D


Find all polynomials
\[P_{n}(x)=n!x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+(-1)^{n}n(n+1)\]
with integer coefficients such that $P_n$ has $ n$ real roots $ x_{1},x_{2},\ldots,x_{n}$ that have the following condition: for all $ k=1,2,\ldots,n$, we have $ k\leq x_{k}\leq k+1$.

[quote="Rust"]$ x_{1}...x_{n}=\frac{n+1}{(n-1)!}<1$ if $ n\ge 3$. If n=2, then $ P_{2}(x)=2x^{2}+a_{1}x+6=2(x-1)(x-3),a_{1}=-8$.
Therefore it may be if and only if $ n\le 2$.[/quote]

Two little mistakes, Rust :

$ 1).$ For $ n=3$, $ \frac{n+1}{(n-1)!}=2>1$. In fact, we must have $ n!\leq x_{1}...x_{n}\leq (n+1)!$ and [u]this[/u] is not verified for n=3.

$ 2).$ For $ n=2$, two values for $ a_{1}$ fit : $ a_{1}=-7$ and $ a_{1}=-8$ (remember roots need not to be integers).

So exactly three polynomials are solutions :

$ 2x^{2}-7x+6$
$ 2x^{2}-8x+6$
$ x-2$


geometry
circumcircle
perpendicular bisector

Two congruent circles of radius $r$ have their centers on each other. Consider the circle tangent to one of them internally, tangent to the externally, and also tangent to the line containing the two circles' centers. Find the radius of this circle, and find a method to construct it.

[hide="Construction only"]
The invertion w.r.t. the circle $(C,CA)$ transforms
- the given circles to two lines $AD,DE$ and
- the line AB to the circumcircle of $\triangle ABC$. Call this circumcircle $k$

We can prove that $D$ is the center of $k$

Also, it is $\angle ADE = 60^{o}$.

[img]8603[/img]

The circle we are searching for touches the two given circles and the line $AB$. So its inverted image will touch the lines $AD,DE$ and the circle $k$.
There are four such circles. We'll construct the one which is inscribed in  $\angle ADE$

It's easy to see that, if $a$ is the radius of the circle inscribed in the curved triangle $ADE$, then $a=\frac{2}{3}\cdot DE$.

After this, we invert the circle back and we obtain the desired circle.

[b]P.S.[/b]
If we choose another of the four circles we obtain:
- Two circles wich touch the given circles (one of them is symmetric with the above constructed circle, wrt the perpendicular bisector of [b]AB[/b])
- The (upper) external common tangent of the given circles[/hide]


Euler
function
number theory unsolved
number theory

prove that:
$ lim_{n\minus{}>\infty}\frac{n\sqrt3}{\sqrt{\phi(1)\plus{}\phi(2)...\plus{}\phi(n)}}\equal{}\pi$
where $ \phi$ denote the euler function


A convex hexagon $ ABCDEF$ is such that each of the diagonals $ AD,BE,CF$ divides the hexagon into two parts of equal area. Prove that these three diagonals are concurrent.

Assuming that $ AD, BE, CF$ are not concurrent then they make triangle $ XYZ$ as you can see on the figure.
$ S_{ADEF}\equal{}S_{ABEF} \Rightarrow S_{ABX}\equal{}S_{EDX}$
Similarly, $ S_{AFY}\equal{}S_{CYD}, S_{BCZ}\equal{}S_{EZF}$
But $ S_{ABCF}\equal{}S_{AFY}\plus{}S_{ABZX}\plus{}S_{BCZ}\equal{}S_{CYD}\plus{}S_{EFZ}\plus{}S_{EXD}\minus{}S_{XYZ}$
$ \equal{}S_{FCDE}\plus{}S_{XYZ}\equal{}S_{FCDE}$
Thus $ S_{XYZ}\equal{}0$
So $ AD,BE,CF$ are concurrent.


Demonstrati ca planul nu poate fi acoperit cu un numar finit de parabole in pozitie generala ( adik puse cum vrei tu in plan) :D

problema e chiar f draguta, si solutia e f misto. nu se baga nimeni ? :)

[quote="enescu"]Ce înseamnă "a se băga" în acest context?[/quote]

 :rotfl:

o invitatie de rezolvare a problemei adresata celor ce nu o stiu  :blush:

[quote="maky"]f draguta problema asta :)
eu o stiam de ceva vreme, deci nu se pune :)

[hide="hint"]
intersectia unei drepte cu o parabola este un segment daca dreapta nu este axa de simetrie a parabolei.
[/hide]

[hide="solutie"]
se alege o dreapta diferita de toate axele de simetrie ale parabolelor. intersectia acestei drepte cu parabolele va fi o reuniune finita de segmente, deci nu o va acoperi.
[/hide][/quote]

Eu credeam ca intersectia are 0, 1 sau maxim 2 puncte si nu un numar infinit de puncte (caci atatea are un segment, nu?)  ;)

Când zice [i]parabolă[/i], cred că maky se referă şi la punctele din "interiorul" acesteia.


Hey guys for you who are in mit or are familiarized with math in mit I have some questions and comments:
 I've heard that the double major with math as a side major is really popular in mit because  the requisites for a math major are not much. But if you want to be a mathematician I think you cannot just take the required courses for the math major, cuz like you wont know math... So if you want to be a mathematician in the future you shouldnt double major in mit...
is this true?

because i wanted to do a double major in math/music but im think im going just for the music minor because I wont have time to take good math courses...though many people do a double major in math and something else; but few of them are like really into math... and I plan to finish a future PhD in math and i wanna learn like a lot and be involved in research since a young age; so im really into math and wanna keep being.  Also im planning to take stanley's discrete math semminar for freshmen in my first year, also im gonna be in the tennis team, and some current mit freshmen called me to do a urop next year.. do you think with all this excesive stuff i will be able to learn good math? or is it not possible since "mit is too hard because of all of he required classes like physics, chem, bio...and those things" (which by the way im not interested at all, im just itnerested in maths, humanities, arts, music, but mainly.......math). I also want to have a good social life in college, so if all these things are impossible to do in mit i rather go to princeton or harvard (princeton and havard have not accepted me so i still dont know; just hypothetically) that are obviously "easier" to surivive and i will still have a great education in mathematics...

thanks for your comments on this...

MIT lets you start at the level you are ready for in mathematics, so you could take quite a few courses in MIT's math department at a fairly high level while still taking other courses. I wouldn't worry about being a future mathematician if you take a second major--Paul Zeitz majored in history at Harvard and he is now a math professor.

thanks tokenadult, true.

but all my mit friends keep telling me that you have to work extremely hard in mit and there is not much time to do other things, and my harvard and pricneton friends keep telling me that they dont work insanely hard, just in what their area of concentration is.
But anyways, i am like 90% sure i will end in mit becuase of its great combinatorics, and that is what i love.

ok, so it is true that with a double major you don't get to take all that many math classes (unless you're willing to take 5+ classes every term after the freshman credit limit, which is also possible). the math department requires 8 classes past differential equations, which as i found out trying to figure out my schedule, is not that much. a music minor wouldn't be bad though- i think it's the fact that the comp sci department (my second major)  has way too many requirements that would force me to only take the 8 class minimum for math.
getting involved in alot of things is good, and i know a few people with similar situations around MIT- but you have to be able to tell for yourself when you're doing too much and should just stop something or reevaluate your priorities.
apparently harvard doesn't have great combinatorics classes, if any... but you do get cross registration with mit and vice versa, which is cool :)


Actually, you have an error in case II: if $n = 2^{3}\cdot 11^{3}$ then the fifth divisor is $11 = b \neq ab$ (because $a^{3}< b$).  It does indeed seem like this is not a good road to head down.

I think it'll be nicer to go this way:
[hide]We know $k = d_{5}$ and $d_{k}= (d_{2}+d_{4})d_{6}$.  In particular, this tells us that $d_{1}6 \geq d_{k}\geq d_{7}$ so $7 \leq k \leq 16$.  For each $k$ in that range, there is a limited set of possible divisors which make it the fifth largest (for example, when $k = 7$ it's not to hard to show that there are two possibilities, $\{1, 2, 3, 6\}$ and $\{1, 2, 4, 5\}$, and the other cases seem to be of similar difficulty).  You get 10 cases this way instead of 4, maybe 20 total subcases, but each case seems much more tractable to me.  I'm not sure this is the best way, but it seems easier than yours.[/hide]

It has been posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=358021#p358021]here[/url].

[quote="4everwise"][quote="Altheman"][quote="4everwise"]Find all positive integers $n$ such that
I) $n$ has exactly 16 divisors: $1 = d_{1}<d_{2}<\cdots < d_{16}= n$;
II) if $k = d_{5},$ then $d_{k}= (d_{2}+d_{4})d_{6}$.
______
Here is my first attempt, but it seemed too messy for me to continue.

[hide="Dead End?"]Let our number be $n.$ Because it has 16 divisors, $n$ is in the form: \[n=ab^{7}, a^{15}, a^{3}b^{3}, abc^{3},~\text{or}~ abcd.\] In all cases, $d_{k}>6,$ since $d_{k\le 6}\ne d_{6}(d_{2}+d_{4}).$

[hide="Case I: $a^{15}$"]First, $d_{5}=a^{4}.$ We want $d_{a^{4}}=(a+a^{3})a^{5}.$ However, $a^{4}\ge 16$ for all $a\ge 2.$ Hence, no solutions here. [/hide]

[hide="Case II: $a^{3}b^{3}$"]the fifth divisor is always $ab,$ and things start getting messy.[/hide]
[/hide][/hide][/quote]

if you use the divisor formula, then it has 18 divisors[/quote]Huh? The problem states that $n$ has 16 divisors. ($a_{1}=1$ and $a_{16}=n$) Am I missing something?  :maybe:[/quote]

oh sorry...the notation confused me...my fault


probability
conditional probability
probability and stats

You draw a card at random from a standard deck of 52 cards. Find each of the following probabilities:
a. The card is a heart, given that it is red.
b. The card is red, given that it is a heart.
c. The card is an ace, given that it is red.
d. The card is a queen, given that it is a face card.

Do I have to use conditional probability? or what?

yes i think we have to use conditional probability to get the answer.

Just answers, so you sould check your solution:
a) 1/2; b) 1; c) 1/13; d) 1/4 (or 1/3 if ace isn't a face card)

For the last one it should be $ \frac{1}{3}$ only na?? beacuse i dont think ace is a face card .