We have a regular n-gon $A_1A_2...A_n$. At each vertex, we write one of the numbers $1,2,3,...,n$ and no two vertices have the same number. Let the number written at $A_n$ be called $B_n$.
[color=blue] [b](a)[/b] [/color] Find the maximum of $ \sum^n_{i=1} |B_i-B_{i+1}|$ where $B_{n+1} = B_1$
[color=blue] [b](b)[/b] [/color] For how many arrangements are the maximum in [color=blue] [b](a)[/b] [/color] attained?

Is the solution somewhat related to replacing consecutive nos and finding an invariant???

[quote="blahblahblah"]Rushil, two things

You should stop incessantly asking for hints in threads.
You should also stop saying that 'this is just an application of blah blah blah'.  Occasionally that's alright, most of the time it is not.[/quote]

I agree with blahblahblah  :yoda:  :)


but what do I know?

1. Find all ordered pairs {a, b, c} where a, b, and c are integers that satisfy the equation a^2 + b^2 + c^2 + (ab)^2 + (ac)^2 + (bc)^2 + (abc)^2 = 45654.  Prove that you found all of them.

Hmmm... It looks to me like[hide]

There are no solutions. We add 1 to each side and we get (1+a^2)(1+b^2)(1+c^2)=45655=5*23*397, so 23 divides 1+a^2, 1+b^2 or 1+c^2. However, it's pretty well-known that any prime of the form 4k+3 which divides x^2+y^2 must divide x and y, but this can't be the case if x=1 or y=1 (we have x^2+y^2=a^2+1 or b^2+1 or c^2+1), so any prime which divides a^2+1 must have form 4k+1, so 23 can't divide any of the 3: a^2+1, b^2+1, c^2+1.[/hide]

An other solution :

Looking at parity it is easy to see that a^2 + b^2 + c^2 + (ab)^2 + (ac)^2 + (bc)^2 + (abc)^2 is even iff a,b,c are all even  :)

But then a^2 + b^2 + c^2 + (ab)^2 + (ac)^2 + (bc)^2 + (abc)^2 would be divisible by 4 which is not the case of 45654 = 2*3*7*1087.

So there are no solutions. 45654 may be replaced by 2*(2N+1).

This problem is not too difficult. Once again, parity rules !

actually, i switched to mod 4 and since any n^2 mod 4 is either 0 or 1, i listed out all the n^2 possibilities
000 - not 2 mod 4
001 - not 2 mod 4
010 - etc...
011 -
100 -
101 -
110 -
111 -
none of these were 2  mod4, which the number was.

I prefer "mod 2" because, thanks to symmetry, there are only 3 cases to consider (odd, odd, odd), (even, odd, odd), (even,even,odd)  :D

Btw, where does this problem come from ?? Is-it an AIME ?


I'm using the flashcards package, but I want to produce one flash card per page on my pdf.  And I want my paper to be 3x5 instead of the standard letter paper.  I've tried a handful of things with no luck and I'm can't find anything using google either.  Any ideas?

See [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/flashcards/flashcards.pdf]flashcards documentation[/url] pages 5-6. Or use the flacards package with \pagesetup{1}{1} see [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/flacards/flacards.pdf]flacards documentation[/url] page 2.

I looked at page 5 of the flashcards document and I didn't see anything that would let me create a custom page size.  I was able to change it from letterpaper to things like a4paper, a5paper, etc but none of those helped me.

I tried the flacards package, but when I use \pagesetup{1}{1} it puts one card per page, but the page size is still 8.5 x 11.  I need the page size to be 5 x 3.

Thanks for your help!  I really appreciate any additional ideas you have.

The flashcards package uses the geometry package. The [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/geometry.pdf]documentation for the geometry package[/url] page 9 lets you set the paper size. The flashcards configuration will set the flashcard size.

flacards.cls also loads the geometry package so I expect you can define the page size there as well.

I thought so too, but I can't figure out how to do it.  The only thing I could get to compile didn't work.

[code]\documentclass[frontgrid,papersize=(5in,3in)]{flacards}

\begin{document}
\pagesetup{1}{1}
\card{gehen\\schlafen}{to go/to walk\\to sleep}
\card{hell, klar}{bright}
\card{mit\\ohne}{with\\without}
\card{Lernkarte}{flashcard}
\card{Wasser}{water}
\card{Saft}{juice}
\card{Fenster}{window}
\card{Linie}{line}
\card{und\\oder}{and\\or}
\card{Regen}{rain}
\end{document}[/code]

papersize=(5in,3in) is not an option that can be used with flacards or the article class that it loads.
The easiest way to set paper sizes is to use the geometry package, but as I say, it is already loaded by the flacards class. So edit flacards.cls replacing the line
[code]\RequirePackage[noheadfoot, margin=1cm, outer=1cm, inner=1cm]{geometry}[/code] by [code]\RequirePackage[noheadfoot, margin=1cm, outer=1cm, inner=1cm, paperwidth=5in, paperheight=3in]{geometry}[/code] with other changes to the options if you wish.
Best to change the name of the class to something else then refresh the database so that it's found.
Then use
[code]\documentclass[frontgrid]{your new name for flacards class}[/code]

That worked great!  The only problem is I couldn't get my TeX package (MikTeX 2.7) to see the new cls file.  I had to directly change flacards.cls.  I tried adding a root in the MikTeX settings but go an error.  I also use TeXShop on a mac and couldn't figure out where to copy the flacards.cls file.

Got it working on my Mac, but still can't get it to work in Windows.  Any ideas?

Here is what I've tried:

1. Create a folder called C:\My Styles.  I put a file called tiacards.cls in that folder.  I went to the command prompt and ran

[code]latex --include--directory="C:\My Styles" tiacards.cls[/code]

I get the error: "latex: Invalid command-line option".

2. I also tried adding the folder using the MikTeX settings.  I get the following error.

[img]http://img.skitch.com/20090706-rmb4na9a658d7x7wk7m38r16n9.jpg[/img]

You need a local directory that mirrors the directories that MikTeX uses for all the installed files (TDS-compliant root directory). See [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=tds]TDS[/url]
Create C:\localtexmf\text\latex\tiacards and put tiacards.cls in there. Then go to MikTeX Settings, Roots and add that as a root directory. Then click on Refresh FNDB button in the General tab.

stevem, you are the man!

Now I have my version of flacards working on both my mac and my pc.  Unfortunately I can't get flashcards working for a 5x3 version.  I tried changing the .cls file like we did for flacards, but that didn't work.  I also tried creating my own config file.  Here is my code:

My LaTeX document
[code]\documentclass[tiacard,grid,frame]{flashcards}

\begin{document}

\begin{flashcard}[Title]{Front}
Back
\end{flashcard}

\end{document}[/code]

My config file:
[code]\newcommand{\cardrows}{1}
\newcommand{\cardcolumns}{1}
\newcommand{\paperwidth}{5in}
\newcommnad{\paperheight}{3in}
\setlength{\cardheight}{3.0in}
\setlength{\cardwidth}{5.0in}
\setlength{\topoffset}{0in}
\setlength{\oddoffset}{0in}
\setlength{\evenoffset}{0in}
[/code]

I get an "undefined control sequence \newcommand{\paperheight}{3in}" error.

Here is my version of [url=http://files.getdropbox.com/u/426149/tiaflashcards.cls]flashcards[/url] that I tried to use with the following code:
[code]\documentclass[grid,frame]{tiaflashcards}

\begin{document}

\begin{flashcard}[Title]{Front}
Back
\end{flashcard}

\end{document}[/code]

Sorry for asking so many questions.  As you can tell I'm still a noobie with LaTeX, but this experience is teaching me a lot.  Thanks!

flashcards hasn't been updated after the geometry package was changed. It means that you must swap the lines in the class file
[code]\LoadClass{article}
\RequirePackage{geometry}[/code]
But the real problem is the config file which [b]must[/b] contain lines like
[code]\newcommand{\cardpaper}{letterpaper}
\newcommand{\cardpapermode}{portrait}
\newcommand{\cardrows}{5}
\newcommand{\cardcolumns}{2}
\setlength{\cardheight}{2.0in}
\setlength{\cardwidth}{3.5in}
\setlength{\topoffset}{0.50in}
\setlength{\oddoffset}{0.75in}
\setlength{\evenoffset}{0.75in}[/code]You can't leave out the paper size, so you would have to define a new paper size first but that seems like madness. flashcards is rather old so I'd stick with flacards if I were you, as that works. You could try emailing the author of flashcards to see if he is willing to update it.

Ok I'll stick with flacards, but the problem is I can't get an itemize to work.

Here is my code:
[code]\documentclass[]{tiacards}
\usepackage{lifecont}

\renewcommand{\flfoot}{\tiny CAS6 - \thecardno}
\renewcommand{\blfoot}{}

\renewcommand{\frfoot}{\fboxsep=0pt \tiny \copyright 2009 The Infinite Actuary, LLC}
\renewcommand{\brfoot}{}

\renewcommand{\cardtextstylef}{}
\renewcommand{\cardtextstyleb}{}

\begin{document}
\pagesetup{1}{1}
\fboxsep=0pt

\card{Probabilistic Trend Family (PTF) model}{
\[ y(i,j) = \alpha_{i} + \sum_{k=1}^{j} \gamma_{k} + \sum_{t=1}^{i+j} \ell_{t} + \epsilon_{i,j} \]
\begin{tabular}{ll}
where: & $\alpha_{i}$ is the mean of $\gamma(i,0)$ \\
 & $\gamma_{k}$ is the mean trend between dev yrs $k-1$ and $k$ \\
 & $\ell_{k}$ is the mean of the random inflation between payment \\
 & \hspace{10pt} year $t$ and $t+1$
\end{tabular}
}

\card{Assumptions made by the ELRF model}{
\begin{itemize}
\item Standard errors, $\epsilon(i) / \sigma x(i)^{\delta/2}$
\item $E[y(i) | x(i)]=bx(i)$
\end{itemize}
}

\end{document}[/code]

Here is the [url=http://files.getdropbox.com/u/426149/tiacards.cls]tiacards.cls file[/url] (same as the flacards.cls file with the pageheight and pagewidth items added).

The first card works fine, but when I add the second card I get a "LaTeX Error: There's no line here to end", but I know there is.  I tried an even simpler list and it wouldn't work either.

flacards is trying to add a new line after \end{itemize} when there is nothing on the line. This produces an error and the compilation stops. So all you have to do is put something there, even a space \ will do; the [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=noline]TeX FAQ[/url] recommends \hspace*{\fill} so let's go with that.
Either replace [code]\end{itemize}[/code] by [code]\end{itemize} \hspace*{\fill}[/code][i]or[/i] define your own itemize environment that automatically includes it.
[code]\newenvironment{myitemize}{\begin{itemize}}{\end{itemize}\hspace*{\fill}}[/code]You can of course call it anything you like instead of [i]myitemize[/i]

Perfect!

Is there a way to make the text that is on the front of the card appear on the back left header?  For example, the title of the first card is "Probabilistic Trend Family (PTF) model" and I'd like that to automatically appear in the \blhead position without having to retype that for each card.  I looked around in the .cls file but I'm just not capable of following what is going on.

No need to mess wth the class file. In the document put [code]\newcommand{\mycard}[2]{\renewcommand{\blhead}{#1} \card{#1}{#2}}[/code]then use \mycard in the same way as you would use \card.


This challenge is reserved for Middle School students who have never attended a State or National Competition.  Please DO NOT ANSWER if you have attended a State or National Competition.

The measures of a pair of complementary angles is in the ratio 5:1.  Expressed in degrees, what is the positive difference in the measures of the angles?

Nonie has a correct solution.  Good job.


geometry
geometric transformation
projective geometry

Λύση της 3:

Η ισότητα γίνεται $ (x\plus{}y)^2\equal{}nxy\Longleftrightarrow x^2\plus{}y^2\equal{}(n\minus{}2)xy$.
Αν $ x\equal{}y\Longleftrightarrow n\equal{}4$.
Αν τώρα $ x\neq y$ μπορούμε να θεωρήσουμε ότι $ gcd(x,y)\equal{}1$ γιατί αν ήταν μεγαλύτερος του 1 θα διαιρούσαμε με το τετράγωνο αυτού και θα παίρναμε 2 άλλα $ x_1,y_1$ πρώτα μεταξύ τους για τα οποία ισχύει η ισότητα.
Φανερά όμως αυτό δεν μπορεί να ισχύει γιατί $ x^2\equal{}(n\minus{}2)xy\minus{}y^2$ και άρα το $ y$ διαρεί το $ x^2$ που σημαίνει ότι δεν μπορεί να είναι σχετικά πρώτο με το $ x$.

Αρα η μόνη λύση είναι η $ n\equal{}4$ :)

Άσκηση 1:

$ \frac {1}{x} \minus{} \frac {1}{y} \equal{} \frac {1}{n} < \equal{} > \frac {1}{x} \equal{} \frac {1}{y} \plus{} \frac {1}{n}$
Έστω ορθογώνιο τρίγωνο $ ABC$ με $ y, n$ οι κάθετες πλευρές του τότε για το ύψος u που βαίνει στην υποτείνουσά του είναι:
$ \frac {1}{u} \equal{} \frac {1}{y} \plus{} \frac {1}{n} < \equal{} > \frac {1}{x} \equal{} \frac {1}{u} < \equal{} > x \equal{} u$

Επειδή το τρίγωνο είναι ορθογώνιο ισχύει (από μετρική σχέση):
$ x\sqrt {y^2 \plus{} u^2} \equal{} yu < \equal{} > x^2(y^2 \plus{} u^2) \equal{} y^2u^2 < \equal{} > x^2 \equal{} \frac {y^2u^2}{y^2 \plus{} u^2}$ αντικαθιστώντας στην αρχική εξίσωση και υψώνοντας στο τετράγωνο παίρνουμε:

$ \frac {y^2 \plus{} u^2}{y^2u^2} \equal{} \frac {1}{y^2} \plus{} \frac {1}{u^2} \plus{} \frac {2}{yu} < \equal{} >$
$ y^2 \plus{} u^2 \equal{} u^2 \plus{} y^2 \plus{} 2yu < \equal{} > 2yu \equal{} 0$

Μήπως έχεις κάνει κάπου λάθος. Κάνω μήπως εγώ λάθος (πείτε μου πού);

Αφού στην τρίτη σειρά βαζεις x=u στην εβδομη πως βγάζεις το x συναρτήσει του u ???
Ξαναδέστο και λάβε υπόψη σου όλα τα δεδομένα και θα σου έρθει η λύση.

Για την δεύτερη χρησιμοποιούμε την ταυτότητα $ k^2 \plus{} \frac {1}{k^2} \equal{} \left(k \plus{} \frac {1}{k}\right)^2 \minus{} 2 \ \ (\star)$.

Άρα έχουμε:
$ ab \equal{} \left(x \plus{} \frac {1}{x}\right)\left(y \plus{} \frac {1}{y}\right) \equal{} c \plus{} \frac {x}{y} \plus{} \frac {y}{x}$ και θέτοντας

$ d \equal{} \frac {x}{y} \plus{} \frac {y}{x}$ παίρνουμε: $ \boxed{ab \equal{} c \plus{} d \ \ (1)}$.

Χρησιμοποιώντας την πιο πάνω ταυτότητα $ (\star)$ έχουμε

$ (a^2 \minus{} 2)(b^2 \minus{} 2) \equal{} \left(x^2 \plus{} \frac {1}{x^2}\right)\left(y^2 \plus{} \frac {1}{y^2}\right) \equal{} (xy)^2 \plus{} \frac {1}{(xy)^2} \plus{} \left(\frac {x}{y}\right)^2 \plus{} \left(\frac {y}{x}\right)^2 \equal{} c^2 \minus{} 2 \plus{} d^2 \minus{} 2$

Άρα λοιπόν $ \boxed{(a^2 \minus{} 2)(b^2 \minus{} 2) \equal{} c^2 \plus{} d^2 \minus{} 4 \ \ (2)}$.

Απαλείφοντας το $ d$ από τις $ (1),(2)$ παίρνουμε τη ζητούμενη σχέση

$ \boxed{a^2\plus{}b^2\plus{}c^2\minus{}abc\minus{}4\equal{}0}$.

Αλέξανδρος

Problem 2 reminds me of Japanese University  entrance exam.

[url=http://www.mathlinks.ro/viewtopic.php?t=21371&search_id=234001800]See here![/url]

Could someone problem 2 translate into English?

Thanks in advance.

kunny

Hi kunny. Thanks for your interest in our subforum.

Here is the translation of the problem stated above by kritikos.

"Find the relation between $ a,b,c$ if $ a \equal{} x \plus{} \frac {1}{x}, b \equal{} y \plus{} \frac {1}{y}, c \equal{} xy \plus{} \frac {1}{xy}$. "

The problem above is the "converse" of that given at the entrance exam of the Japanese University.
Alexandros

Ευχαριστω, cretanman :)

I will tackle the problem right now.

Ωραία λύση Αλέξανδρε . Βάλε και εσύ κανένα ωραίο θέμα :)

Σε αυτό που έγραψα έκανα λάθος. Αντί για n έβαλα u μετά την τρίτη σειρά. Πού αλλού κάνω λάθος;

Μανιέ  τα αφήνεις τετράγωνα και αντικαθιστάς στην αρχική χωρίς να πάρεις ρίζα.Νομίζω εκεί είναι το λάθος
Για την άσκηση 1 [hide]Αρκεί κανείς να παρατηρήσει ότι $ y \equal{} \minus{} n \plus{} \dfrac{n^2}{n \minus{} x}$ και στη συνέχεια νομίζω εύκολα καταλαβαίνει κανείς τι γίνετε για n πρώτος και x,y$ \in\mathbb{N}$[/hide]

Υψώνω στο τετράγωνο. (Το αναφέρω)
Πού βρήκες την άσκηση;

[quote="cretanman"]
"Find the relation between $ a,b,c$ if $ a \equal{} x \plus{} \frac {1}{x}, b \equal{} y \plus{} \frac {1}{y}, c \equal{} xy \plus{} \frac {1}{xy}$. "[/quote]

$ ca \equal{} x^2y \plus{} \frac {1}{x^2y} \plus{} y \plus{} \frac {1}{y}\Longleftrightarrow ca \minus{} b \equal{} x^2y \plus{} \frac {1}{x^2y}$

$ \therefore b(ca \minus{} b) \equal{} \left(y \plus{} \frac {1}{y}\right)\left(x^2y \plus{} \frac {1}{x^2y}\right) \equal{} x^2y^2 \plus{} \frac {1}{x^2y^2} \plus{} x^2 \plus{} \frac {1}{x^2}$

$ \Longleftrightarrow b(ca \minus{} b) \equal{} \left(xy \plus{} \frac {1}{xy}\right)^2 \plus{} \left(x \plus{} \frac {1}{x}\right)^2 \minus{} 4 \equal{} c^2 \plus{} a^2 \minus{} 4$, yielding the desired relation for $ a,\ b,\ c$ is

$ \boxed{a^2 \plus{} b^2 \plus{} c^2 \minus{} abc \equal{} 4}$

Alternative Solution:

$ c(c \minus{} ab) \equal{} \left(xy \plus{} \frac {1}{xy}\right)\left\{xy \plus{} \frac {1}{xy} \minus{} \left(x \plus{} \frac {1}{x}\right)\left(y \plus{} \frac {1}{y}\right)\right\}$

$ \equal{} \minus{} \left(xy \plus{} \frac {1}{xy}\right)\left(\frac {y}{x} \plus{} \frac {x}{y}\right)$

$ \equal{} \minus{} \left(x^2 \plus{} y^2 \plus{} \frac {1}{x^2} \plus{} \frac {1}{y^2}\right)$

$ \equal{} \minus{} (a^2 \minus{} 2 \plus{} b^2 \minus{} 2)$

$ \Longleftrightarrow a^2 \plus{} b^2 \plus{} c^2 \minus{} abc \equal{} 4$.

Let $ a,b,c$ be real numbers such that

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

with all denominators non zero. Prove that

$ \frac{1}{a^{2007}}\plus{}\frac{1}{b^{2007}}\plus{}\frac{1}{c^{2007}}\equal{}\frac{1}{(a\plus{}b\plus{}c)^{2007}}$.

Alexandros

Στο Quantum. Να λυθεί η εξίσωση

$ x\plus{}\sqrt{x^2\minus{}1}\equal{}\dfrac{2(x\plus{}3)}{(x\minus{}3)^2}$

[quote="cretanman"]Let $ a,b,c$ be real numbers such that

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

with all denominators non zero. Prove that

$ \frac {1}{a^{2007}} \plus{} \frac {1}{b^{2007}} \plus{} \frac {1}{c^{2007}} \equal{} \frac {1}{(a \plus{} b \plus{} c)^{2007}}$.

Alexandros[/quote]

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

$ \Longleftrightarrow (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}abc\equal{}0$ and $ abc\neq 0$.

$ \Longleftrightarrow (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}0$

$ \Longleftrightarrow a\plus{}b\equal{}0$ or $ b\plus{}c\equal{}0$ or $ c\plus{}a\equal{}0$.

Thus  for $ a\plus{}b\equal{}0$, we have $ \frac {1}{a^{2007}} \plus{} \frac {1}{b^{2007}} \plus{} \frac {1}{c^{2007}}\equal{}\frac{1}{a^{2007}}\minus{}\frac{1}{a^{2007}}\plus{}\frac{1}{c^{2007}}\equal{}\frac{1}{(a\plus{}b\plus{}c)^{2007}}$.

So do for $ b\plus{}c\equal{}0$ or $ c\plus{}a\equal{}0$. Q.E.D.

Needless to say, we can replace $ 2007$ into $ n\in\mathbb{N}.$

[quote="kunny"][quote="cretanman"]Let $ a,b,c$ be real numbers such that

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

with all denominators non zero. Prove that

$ \frac {1}{a^{2007}} \plus{} \frac {1}{b^{2007}} \plus{} \frac {1}{c^{2007}} \equal{} \frac {1}{(a \plus{} b \plus{} c)^{2007}}$.

Alexandros[/quote]

Needless to say, we can replace $ 2007$ into $ n\in\mathbb{N}.$[/quote]

Very nice solution kunny :wink:  but be careful. We can replace $ n$ only with any odd number (not with an even)

[color=red][b]Problem![/b][/color]

 Παιδιά , βάζω αυτό το πρόβλημα στα αγγλικά , που μόλις το πήρα μεταφρασμένο από τα βιετναμέζικα(το έστειλα χθες). Μια και βλέπουν το ελληνικό φορουμ και σε άλλες χώρες, ας βάζουμε που και που και κανένα πρόβλημα , από τις άλλες κοινότητες , στα αγγλικά κυρίως.Αν θέλει κάποιος ας το αποδώσει στα ελληνικά , ώστε να μπορούν όλοι να το καταλάβουν και στο μέλλον.
[u] Μπάμπης[/u]

[i]A friend in Vietnam sent me yesterday a pdf - file with some  tst's in his country. I sent one problem to Vietnamese subforum and Pvthuan traslated it for us. I thank him for this. See the problem and try it ![/i]

  Two circles $ (O_1)$ and $ (O_2)$ intersect at $ A$ and $ B$.Tangents at points $ A$ and $ B$ of the circle $ (O_1)$ meet at $ K$. Point $ M$ (distinct from $ A$ and $ B$) lies on the circumference of $ O_1$. Let $ MA$ intersect the circle $ O_2$ again at $ P$; let $ MK$ intersect the circle $ O_2$ again at $ C$; let $ CA$ meet the circle $ O_2$ again at $ Q$.

1) Prove that the mid-point of $ PQ$ lies o the line $ MC$.

2) Prove that line $ PQ$ has a fixed point as $ M$ varies on the circle $ O_1$

[quote="DIMITRIS T"]Εστω $ n$ σημεία πάνω σε μια περιφέρια κύκλου.Συνδέοντας με όλους τους δυνατούς τρόπους τα σημεία αυτά πόσες χορδές μπορούμε να κατασκευάσουμε;Αν τα σημεία αυτά θεωρηθούν κορυφές,πόσα $ k$-γωνα μπορούμε να κατασκευάσουμε;[/quote]

Ο αριθμός των χορδών είναι ίσος με το πλήθος των τρόπων με τους οποίους επιλέγουμε 2 σημεία δηλαδή $ \binom{n}{2}\equal{}\frac{n(n\minus{}1)}{2}$ και το πλήθος των $ k$-γώνων ($ 3\leq k \leq n$) ταυτίζεταi με το πλήθος των τρόπων με τους οποίους επιλέγουμε $ k$ σημεία από τα συνολικά $ n$ δηλαδή $ \binom{n}{k}$.

Ας συνεχίσουμε στο ίδιο πνεύμα την άσκηση ανεβαίνοντας ακόμη ένα σκαλοπάτι:
[b][size=150][color=red]
ΑΣΚΗΣΗ[/color][/size][/b]

Πάνω σε μία περιφέρεια κύκλου θεωρούμε $ n$ διαφορετικά σημεία ($ n\geq 3$), τα οποία εαν ενωθούν ανα δύο μεταξύ τους χωρίζουν τον κύκλο σε περιοχές (κομμάτια). Να προσδιοριστεί ο μέγιστος αριθμός περιοχών του κυκλικού δίσκου που ορίζονται με τον τρόπο αυτό.

Αλέξανδρος

[quote="kritikos"]Τί λέτε παιδιά ηρθε η ώρα να ασχοληθούμε με θέματα Eυκλείδη ? Λοιπόν ας αρχίσω

1) Αν x,y και n $ \in {N}$ , να αποδειχθεί ότι η εξίσωση : $ \dfrac{1}{x} \minus{} \dfrac{1}{y} \equal{} \dfrac{1}{n}$ (για δεδομένο n)
έχει μοναδική λύση , αν και μόνο αν ο n είναι πρώτος.

2) Να βρεθεί η σχέση που συνδέει τα a,b και c , αν a=$ x \plus{} \dfrac{1}{x}$ , b=$ y \plus{} \dfrac{1}{y}$  και c =$ xy \plus{} \dfrac{1}{xy}$

3) Βρείτε όλους τους ακεραίους n , για τους οποίους η εξίσωση

$ \dfrac{1}{x} \plus{} \dfrac{1}{y} \equal{} \dfrac{n}{x \plus{} y}$

έχει ακέραιες λύσεις x,y ($ x\neq{ \minus{} y}$)[/quote]
Μήπως πρέπει στην άσκηση 1) $ x,y \in Z$ οπότε $ x\equal{}2n,y\equal{}\minus{}2n$? η μοναδική λύση

[quote="r_boris"][quote="kritikos"]Τί λέτε παιδιά ηρθε η ώρα να ασχοληθούμε με θέματα Eυκλείδη ? Λοιπόν ας αρχίσω

1) Αν x,y και n $ \in {N}$ , να αποδειχθεί ότι η εξίσωση : $ \dfrac{1}{x} \minus{} \dfrac{1}{y} \equal{} \dfrac{1}{n}$ (για δεδομένο n)
έχει μοναδική λύση , αν και μόνο αν ο n είναι πρώτος.

2) Να βρεθεί η σχέση που συνδέει τα a,b και c , αν a=$ x \plus{} \dfrac{1}{x}$ , b=$ y \plus{} \dfrac{1}{y}$  και c =$ xy \plus{} \dfrac{1}{xy}$

3) Βρείτε όλους τους ακεραίους n , για τους οποίους η εξίσωση

$ \dfrac{1}{x} \plus{} \dfrac{1}{y} \equal{} \dfrac{n}{x \plus{} y}$

έχει ακέραιες λύσεις x,y ($ x\neq{ \minus{} y}$)[/quote]
Μήπως πρέπει στην άσκηση 1) $ x,y \in Z$ οπότε $ x \equal{} 2n,y \equal{} \minus{} 2n$? η μοναδική λύση[/quote]
oι λύσεις που έκανα είναι στο attachment

[quote="stergiu"] $ O_2$ again at $ P$; let $ MK$ intersect the circle $ O_2$ again at $ C$[/quote]

Το C πρέπει να ανήκει στον κύκλο $ O_1$ και όχι στον $ O_2$ .Αν ισχύει αυτό τότε το πρώτο ερώτημα βγαίνει απευθείαςς

αν δούμε ότι :α) ΜΚ συμμετροδιάμεσος στο τρίγωνο ΜΑΒ .
                β)PQ αντιπαράλληλη στην AB αφού ABQP εγγράψιμο
γ)  αν μια ευθεία είναι συμμετροδιάμεσος τότε είναι διάμεσος στην αντιπαράλληλη  .
Μετά θέλει κάτι ακόμα και βγήκε  :wink:

Το δεύτερο ερώτημα πρέπει να το σκεφτώ λίγο ακόμα. :P

Το σταθερό σημείο είναι το σημείο τομής της παραλλήλου από το Κ στην ΑΒ με την PQ .

[quote="cretanman"]
[b][size=150][color=red]
ΑΣΚΗΣΗ[/color][/size][/b]

Πάνω σε μία περιφέρεια κύκλου θεωρούμε $ n$ διαφορετικά σημεία ($ n\geq 3$), τα οποία εαν ενωθούν ανα δύο μεταξύ τους χωρίζουν τον κύκλο σε περιοχές (κομμάτια). Να προσδιοριστεί ο μέγιστος αριθμός περιοχών του κυκλικού δίσκου που ορίζονται με τον τρόπο αυτό.
[/quote]

Έχουμε $ \binom{n}{2}$ χορδές.
Τώρα αρκεί να παρατηρήσουμε ότι
η πρώτη χορδή χωρίζει τον κύκλο σε $ 2$ χωρία
όταν φέρνουμε την $ k > \equal{} 2$ χορδή, ο αριθμός των περιοχών που υπάρχουν αυξάνει κατά  το πολύ $ k$,
άρα το ζητούμενο βγαίνει αν υπολογίσουμε το άθροισμα

$ 2 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} ... \plus{} \binom{n}{2} \equal{} 1 \plus{} 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} ... \plus{} \binom{n}{2} \equal{} 1 \plus{} \frac {\binom{n}{2}(\binom{n}{2} \plus{} 1)}{2} \equal{}$

$ 1 \plus{} \frac {1}{2}\binom{n}{2} \plus{} \frac {1}{2}\binom{n}{2}^2$

[quote="stergiu"] Two circles $ (O_1)$ and $ (O_2)$ intersect at $ A$ and $ B$. Tangents at points $ A$ and $ B$ of the circle $ (O_1)$ meet at $ K$. Point $ M$ (distinct from $ A$ and $ B$) lies on the circumference of $ O_1$. Let $ MA$ intersect the circle $ O_2$ again at $ P$; let $ MK$ intersect the circle $ O_2$ again at $ C$; let $ CA$ meet the circle $ O_2$ again at $ Q$.

1) Prove that the mid-point of $ PQ$ lies o the line $ MC$.

2) Prove that line $ PQ$ has a fixed point as $ M$ varies on the circle $ O_1$[/quote]

If I'm not wrong, it is from Vietnamese TST 2004. And let me present a solution for this nice problem.

$ 1)$ Let $ N$ be the intersection of $ KM$ and $ PQ$.
We have $ ACBM$ is a harmonic quadrilateral, so that $ \frac{MA}{MB}\equal{}\frac{CA}{CB}$ $ (1)$

On the other hand, we have $ \angle BPQ\equal{}\angle BAQ\equal{}\angle BAC\equal{}\angle BMC$. It implies points $ M$, $ B$, $ N$, $ P$ are cyclic. So, we can easily conclude that $ \triangle BAC\sim\triangle BPN$, which yields $ \frac{CA}{CB}\equal{}\frac{NP}{NB}$ $ (2)$

Combining $ (1)$ and $ (2)$, we deduce $ \frac{MA}{MB}\equal{}\frac{NP}{NB}$. And hence, we have to show that $ \frac{MA}{MB}\equal{}\frac{NQ}{NB}$, i.e. $ \triangle MAB\sim\triangle NQB$, which follows from the cyclic quadrilaterals $ MBNP$ and $ BCNQ$.

$ 2)$ Now, denote points $ L\equal{}KM\cap AB$, $ S\equal{}AK\cap (O_2)$, $ T\equal{}SS\cap BB$, where $ SS$ and $ BB$ are the tangent lines in $ S$ and $ B$ of the circle $ (O_2)$. We will prove that $ PQ$ passes through $ T$, as a fixed point.

Indeed, we have $ (K, L, C, M)$ forms a harmonic division, so the pencil $ A(K, L, C, M)$ is harmonic, too. Intersecting them with the circle $ (O_2)$, we conclude that the quadrilateral $ SBQP$ is harmonic. Hence, $ PQ$ passes through $ T$. Our solution is completed.

[size=150]Merry Christmas, Greek Forum![/size]


Hey Everyone!

I just wanted to know [i] how [/i] you can be elgible for MATHCOUNTS. I mean, my school doesn't have a MATHCOUNTS program, but I still want to be in it.

There are lots of these entries in the forum--use the search function and you'll find many of them.  You can't join mathcounts for this year because the deadline past, but next year you can encourage your school to have a MC program.  You have to be in grades 6-8 to be eligible.  What grade are you in now?  If you are in 6 or 7 you may do it next year, but of you are in 8, you won't be able to do MC.

EDIT: you said you've done mathcounts.  have you moved to a school that doesn't?

you can convince your school to do it
or you can sign up as an individual
go here: http://mathcounts.org/Page.aspx?pid=336
and get the registration form or register online

EDIT: it says you can still register

Thanks for your quick reply!

I'm in seventh grade right now. For your "EDIT" portion, that's correct. I was at a school with MATHCOUNTS but I have moved to one that doesn't.  It's going to be hard to convince my school to start MATHCOUNTS; it's a public school with funding for only sports and 2-3 clubs and [i] no one [/i], I repeat no one, is interested in math at all at public schools. They probably wouldn't start a program just for me  :( .

Sorry for posting a topic already existing  :blush: ...

[quote="xoangieexo"]
EDIT: it says you can still register[/quote]

[quote="Mathcounts Website"]Registration for the 2008-2009 Competition series is now open.  The deadline for MATHCOUNTS Competition registration is [size=200][b]December 12, 2008.[/b][/size]
Be sure to also sign up for the FREE MATHCOUNTS Club program so that you may receive your FREE Club in a Box kit and Vol II of the MATHCOUNTS School Handbook. [/quote]

I thought the school rounds were out already. And isn't the Chapter Round in February or something? It's amazing that you can still register.

(Frankly, I have my doubts. Are you sure about that?

[quote="007math"]Did you read the bold?  It said [b][size=200]December 12, 2008.[/size][/b] We're a year past that ( :P )[/quote]

"Still want to participate? Late registrations for the Competition may still be accepted at the discretion of the National Office.

Download the Registration/Request Form!"

Taken from: http://mathcounts.org/Page.aspx?pid=315

It says it on the Top Right Corner of the sheet:

http://mathcounts.org/Document.Doc?id=229

Eh... Anyway, I'm probably not going to join. I had like NO preparation.


If the arithmetic mean of two numbers is $6$ and thier geometric mean is $10$, then an equation with the given two numbers as roots is:

$\text{(A)} \ x^2+12x+100=0 ~~ \text{(B)} \ x^2+6x+100=0 ~~ \text{(C)} \ x^2-12x-10=0$
$\text{(D)} \ x^2-12x+100=0 \qquad \text{(E)} \ x^2-6x+100=0$

I liked this problem, so I'll post my solution.

[hide]Let the roots be a and b.

We know $a+b=12$ and $ab=100$, so our answer is $x^2-12x+100=0$. (Coefficient rules)[/hide]

Or, just by knowing that $a+b=12$  $ab=100$, you can calculate $a$ and $b$ and then the solution follows...


[i]  Solve the equation  [/i]

   $ x^3 \minus{} 3x^2 \plus{} 2\sqrt{(x\plus{}2)^3} \minus{}6x \equal{} 0$

[i]  Virgil Nicula i think you will be satified with this problem[/i] :lol:

[i]  Love An forever[/i]

If $ y\equal{}\sqrt{x\plus{}2}$, then we have $ x^3\plus{}2y^3\equal{}3xy^2\Rightarrow (x\plus{}2y)(x\minus{}y)^2\equal{}0$. $ x\equal{}y$ gives $ x\geq 0$ and $ 0\equal{}x^2\minus{}x\minus{}2\equal{}(x\minus{}2)(x\plus{}1)$ implying $ x\equal{}2$, and $ x\equal{}\minus{}2y$ gives $ x\leq 0$ and $ 0\equal{}x^2\minus{}4x\minus{}8\Rightarrow x\equal{}2\left(1\minus{}\sqrt{3}\right)$. So the solutions $ \boxed{2,2\left(1\minus{}\sqrt{3}\right)}$.


geometry
geometric transformation
rotation
analytic geometry
congruent triangles

F is the midpoint of BC in triangle ABC. Isosceles triangles ABD and ACE are constructed externally on sides AB and AC with right angles at D and E, respectively. Show that DEF is a right isosceles triangle.

For this one, I read a solution at

http://www.kalva.demon.co.uk/irish/soln/sol964.html

but I didn't understand the last sentence of it (why DF and FE are perpendicular). Does anyone have any idea?

Thanks.

MUCH more straightforward is

B=(0,0), A=(a,b), C=(c,0), F=(c/2,0), E=((1/2)(a+b+c),(1/2)(b+c-a)), D=((1/2)(a-b), (1/2)(a+b)).

From which it is easily checked that DF and FE are perpendicular

hehe

by the way, DF and FE are perpendicular since DMF and FNE are congruent, and DM perpendicular FN, and MF perpendicular NE, hence DF perpendicular FE.

[quote]by the way, DF and FE are perpendicular since DMF and FNE are congruent, and DM perpendicular FN, and MF perpendicular NE, hence DF perpendicular FE.[/quote]

Hmm, I may be missing something obvious, but how does congruency and the perpendicular angles formed by the intersection of two pairs of sides of the congruent triangles imply that the third pair of sides is perpendicular as well?

One triangle can be made to coincide with the other through a rotation followed by a translation. Translation preserves angles of intersection. Rotation alters angles of intersection uniformly.

Hmm, then why do we need to prove that both DM is perpendicular to FN AND MF is perpendicular to NE? Can't we only show one of these, and then rotate by n degrees + translate? After this process, all three pairs of sides are parallel and so we can say that before the n degree rotation, all three pairs met at the same angle, which must be 90 degrees since we know one of the pairs intersected at a right angle.

If both were not perpendicular, one could be a flip rotation of the other.

Actually, now I am becoming confused.

Consider triangle A with coordinates (1,0), (1,2), (0,2)
Triangle B with coordinates (1,0), (1,2), (2,2)
and Triangle C with coordinates (0,2), (0,3), (2,2)

Triangle A and Triangle C have two pairs of perpendicular sides, but the third sides are not perpendicular. So that is not sufficient. Have to somehow show that one can be rotated onto the other, not flip rotated.

Then I think the only reason that happened was because the two triangles are right triangles? So I think 1 pair of perpendicular sides always yields rotation and flip rotation possibility but 2 pairs of perpendicular sides yields rotation possibility only in every case except for when the two congruent triangles are right triangles.

Is it impossible for DMF or FNE to be right triangles? gah i don't know what i'm talking about...

ARGH. This is why one uses analytic geometry.

It might be helpful to look at the Euclidean solution this way:

After you've proven that triangle DMF :cong: triangle FNE, Look at triangle DAE. Since DA = :rt2: DM, AE = :rt2: EN = :rt2: AN = :rt2: MF, and <DAE = 90o + <BAC = <DMF, we have that triangle DMF :sim: triangle DAE. Thus DE = :rt2: DF. Then, the triangle has sides DF, EF = DF, and DE = :rt2: DF, which means that it must be an isosceles right triangle with hypotenuse DE. Thus <DFE = 90o.

Edit: If triangle ABC is obtuse at vertex A, replace

[quote]<DAE = 90o + <BAC = <DMF[/quote]

with

[quote]<DAE = 270o - <BAC = <DMF[/quote]

Cool, that solution works too.

I'd still be interested in seeing why that third side must be perpendicular using John Scholes' argument though. (it seems a lot quicker :D )

Er, H is supposed to be N. On my diagram G is M and H is N. Corrected. Sorry! :)

Re John Scholes' solution:

Extend DM to meet FN at P, and extend EN to meet MF at Q. Let <MDF = <NFE = a, <MFD = <NEF = b, and <MFN = c. Then by either triangle DFP or triangle EFQ, a + b + c = 90o, so <DFE = a + b + c = 90o.

if i may provide an alternate sol. using rotational geometry.

Consider the rotations r1 about point D by 90 degress, r2 about F by 180 degrees, and r3 around E by 90 degrees.

the rotation r3(r2(r1(A))) = A, and since the sum of rotations is 360 degrees, the net rotation is a null translation, or identity.

so lets do r3(r2(r1(D))). r1(D) = D, so this is r3(r2(D)). r2(D) = P where F is the midpoint of DP. r3(P) must then be D. So DPE is a right isosceles triangle, and the median EF must then also be an altitude, with length equal to DF. Therefore DEF is a right isosceles triangle.

By the way, fairyofwind, what method did you use to find E=((1/2)(a+b+c),(1/2)(b+c-a)), D=((1/2)(a-b), (1/2)(a+b)) for yoiur solution using coordinates?


Consider the polynomial $ P \in \mathbb{Z}[X]$ with at least one irrational root. If $ P(1988)\equal{}P(1990)\equal{}1$ then determine all the integer roots of $ P$.

[color=blue]Consider more generally $ P \in \mathbb{Z}[x]$ with at least one irrational root, and such that $ P(k\minus{}1) \equal{} P(k\plus{}1) \equal{} 1$ for some integer $ k$. Determine all integer roots of $ P$.[/color]

[hide]
Use that, for $ P \in \mathbb{Z}[x]$ and $ a, b \in \mathbb{Z}$, one has $ a \minus{} b \mid P(a) \minus{} P(b)$. Let $ r$ be an integer root of $ P$; then $ k \minus{} 1 \minus{} r \mid P(k\minus{}1) \minus{} P(r) \equal{} 1$ and $ k \plus{} 1 \minus{} r \mid P(k\plus{}1) \minus{} P(r) \equal{} 1$, with unique possibility $ r\equal{}k$.
Take $ R(x) \equal{} (x \minus{} k \plus{} 1)(x \minus{} k)(x \minus{} k \minus{} 1) \plus{} 1 \equal{} (x\minus{}k)^3 \minus{} (x\minus{}k) \plus{} 1$; it takes value $ 1$ in $ k\minus{}1$ and $ k\plus{}1$, and has an irrational root in $ (k\minus{}2, k\minus{}1)$.
Now, $ P(x) \equal{} R(x)\cdot(Q(x)(x\minus{}k\plus{}1)(x\minus{}k)(x\minus{}k\minus{}1) \plus{} (x\minus{}k)^2)$, with $ Q \in \mathbb{Z}[x]$, fulfills and has root $ k$, but so does $ P(x) \equal{} R(x)\cdot(Q(x)(x\minus{}k\plus{}1)(x\minus{}k)(x\minus{}k\minus{}1) \plus{} (x\minus{}k)^2 \plus{} (x\minus{}k\plus{}1)(x\minus{}k\minus{}1))$, with $ Q \in \mathbb{Z}[x]$ fulfill, with $ k$ not a root.
Therefore we can only determine that the only integer root of $ P$ could be $ k$.
[/hide]


I just posted in the appropriate competition forum, but I suspect this will be of interest to many who may not frequent that forum ...

I will be running an informal competition for anyone interested which will begin on November 13th, 2009.  I know, the AMC8 is that week, but this may be a good tune-up or post-AMC8 set for many ... answers will be due by the following Friday.  There will be at least 4 additional rounds in the next 2 months and it should be a great way to prepare for MATHCOUNTS.

The format is essentially a MATHCOUNTS Warm-Up in the Warm-Up 15-18+ range for which you will have 20 minutes without a calculator to answer as many questions correctly as possible.   I will post results on my site, http://www.agmath.com on the AGMath Challenge page [url=http://www.agmath.com/57427/223285.html]here[/url].

I hope enough of you will be interested in participating.

Questions?  Post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310667]in this thread[/url].

Problems are posted [url=http://www.agmath.com/57427/223285.html]here[/url] ... enjoy!  Please read and follow the directions on the linked page before attempting the set and strictly follow the time limit of 20 minutes with no calculator.

The forum topic for this discussion is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310667&start=20]here[/url]. Please post there if you have questions, and I will only bump this thread when a new set is posted (if that is alright with the mods) ... thanks!

Round two problems are up [url=http://www.agmath.com/57427/223285.html]here[/url].

Scroll down to see round 1 problems, key, solutions, and results.

Discuss [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310667]here[/url].

Thanks to all who participated in round 1.  You have 2 weeks for round 2 because of the holiday.

Round 4 is up [url=http://www.agmath.com/57427/223285.html]here[/url] along with lots of results and solutions from the first three rounds.

Discuss [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310667]here[/url].


Thank you very much :D . [hide="And my solution"]
[quote="April"]Let be given a triangle $ABC$ with $\angle{A}=45^{\circ}$. Prove that: $a^{4}=\left(b^{2}-a^{2}\right)^{2}+\left(c^{2}-a^{2}\right)^{2}$[/quote]
[b]Solution.[/b]
We have: $a^{2}=b^{2}+c^{2}-\sqrt{2}bc$
$\Longleftrightarrow\left(a^{2}-b^{2}-c^{2}+\sqrt{2}bc\right)\left(a^{2}-b^{2}-c^{2}-\sqrt{2}bc\right)=0$
$\Longleftrightarrow\left(a^{2}-b^{2}-c^{2}\right)^{2}-2b^{2}c^{2}=0$
$\Longleftrightarrow a^{4}+b^{4}+c^{4}-2a^{2}b^{2}-2a^{2}c^{2}=0$
$\Longleftrightarrow\left(b^{2}-a^{2}\right)^{2}+\left(c^{2}-a^{2}\right)^{2}=a^{4}$[/hide]


function
geometry
geometric transformation
reflection

I just have a quick question:

Why is f(-x) the reflection of f(x) in the y axis?  There is probably an obvious answer that I cannot see at the moment.
Thanks,

-interesting_move

In general that's false. It's only true for even functions (such as cosine).

Are you referring to a different function g(x) = f(-x) as being the mirror image of f(x) across the y axis?

If so, then think about the inputs of f and g.  For each x that you plug in to f(x), you have f(x) = g(-x).  The two inputs are additive inverses and will produce the same output.  This will be true for any input you are considering.  Make sense?


This is quite obvious I think.
We know that S(3n) is a multiple of 3, since 3|k <=> 3|S(k).
And in this case obviously 3|3n so indeed 3|S(3n).
Now since 3|S(3n) and S(n) = S(3n), we have 3|S(n).
Hence n is a multiple of 3. Hence 3n is a multiple of 9.
Hence 9|S(3n). But also 9|S(n) (since S(n) = S(3n)) so 9|n.

Yeah, well I didn't say it was a hard problem... :D


function
trigonometry
integration
ratio
calculus
geometric series
real analysis

Let a funtion $ f_n (x) \equal{} \left\{\begin{array}{l}\frac {{\sin nx}}{{\sin x}}|x > 0 \\
n|x \equal{} 0 \\
\end{array} \right.$
Let $ I(n) \equal{} \int\limits_0^{\pi /2} {f_n (x)dx}$, find $ I(2009)$.

First compute $ I(1), I(2), I(3)$.  That may suggest something to do.

it looks like 3rd IMC 1996  first day, problem 2 .  http://imc-math.org/
consider $ I(n) \minus{} I(n \minus{} 2) \equal{} 2\int_{0}^{\pi/2}\cos((n \minus{} 1)x)dx$...

If you've seen a bit of Fourier series/transforms on the circle, this looks familiar. Let's manipulate $ f_n(x) \equal{} \frac {\sin nx}{\sin x}$.

$ f_n(x) \equal{} \frac {e^{inx} \minus{} e^{ \minus{} inx}}{e^{ix} \minus{} e^{ \minus{} ix}} \equal{} \frac {e^{i(n \minus{} 1)x} \minus{} e^{i( \minus{} n \minus{} 1)x}}{1 \minus{} e^{ \minus{} 2inx}}$
The sum of a finite geometric series is (first term-term after last)/(1-ratio); this becomes
$ f_n(x) \equal{} e^{i(n \minus{} 1)x} \plus{} e^{i(n \minus{} 3)x} \plus{} \cdots \plus{} e^{i( \minus{} n \plus{} 1)x}$. If $ n$ is odd, this is $ 2\cos(n \minus{} 1)x \plus{} 2\cos(n \minus{} 3)x \plus{} \cdots \plus{} 2\cos 2x \plus{} 1$

Cosines are easy to integrate, and we do so termwise.
$ \int_0^{\pi/2}f_{2009}(x)\,dx \equal{} 2\left[\frac {\sin(2008x)}{2008} \plus{} \frac {\sin(2006x)}{2006} \plus{} \cdots \plus{} \frac {\cos(2x)}{2} \plus{} x\right]_0^{\pi/2}$
$ \equal{} 0 \plus{} 0 \plus{} \cdots \plus{} 0 \plus{} \frac {\pi}{2} \equal{} \frac {\pi}{2}$

If $ n$ is even, we get $ \frac {2}{1} \minus{} \frac {2}{3} \plus{} \frac25\minus{}\frac27\plus{}\cdots\pm\frac{2}{n\minus{}1}$ instead. That does converge to $ \frac {\pi}{2}$, but it isn't equal.

My solution.

$ \int_{0}^{\frac {\pi}{2}}\frac {\sin nx }{\sin x }$

[u]1st case [/u]

n = even

let n = 2k

therefore  integral becomes $ \int_{0}^{\frac {\pi}{2}}\frac {\sin 2kx }{\sin x }$

Now $ \cos x \plus{} \cos 3x \plus{} \cos 5x \plus{} .......\cos (2k \minus{} 1)x \equal{} \frac {\sin 2kx }{ 2 \sin x }$

therefore  $ \frac {\sin 2kx }{\sin x } \equal{} 2 (\cos x \plus{} \cos 3x \plus{} \cos 5x \plus{} .......\cos (2k \minus{} 1)x )$

therefore  $ \int_{0}^{\frac {\pi}{2}}\frac {\sin 2kx }{\sin x } \equal{} \int _{0}^{\frac {\pi}{2}}2 ( \cos x \plus{} \cos 3x \plus{} \cos 5x \plus{} .......\cos (2k \minus{} 1)x )$

$ \equal{} 2 ( 1 \minus{} \frac {1}{3} \plus{} \frac {1}{5} \minus{} \frac {1}{7} ....... \plus{} \frac {1}{2n \minus{} 1} )$

As  $ n \rightarrow \infty$ this $ \rightarrow \frac {\pi}{2}$

[u]Case 2[/u]

n=odd

therefore $ \int_{0}^{\frac {\pi}{2}}\frac {\sin nx }{\sin x } \equal{} \int_{0}^{\frac {\pi}{2}}\frac {\sin ( (n \minus{} 1)x \plus{} x )}{\sin x }$

$ \equal{} \int_{0}^{\frac {\pi}{2}} \frac {\sin (n \minus{} 1)x \cos x }{\sin x } \plus{} \int_{0}{\frac {\pi}{2}} \frac {\cos (n \minus{} 1)x \sin x}{\sin x}$

$ \equal{} \int_{0}^{\frac {\pi}{2}}{\sin (n \minus{} 1)x \cot x } \plus{} \int_{0}{\frac {\pi}{2}} {\cos (n \minus{} 1)x }$

As n=odd therefore n-1 = even

therefore $ \int_{0}^{\frac {\pi}{2}}{\sin (n \minus{} 1)x \cot x } \equal{} \frac {\pi}{2}$  ( from above ) .

and  $ \int_{0}^{\frac {\pi}{2}} \frac {\cos (n \minus{} 1)x \sin x}{\sin x} \equal{} 0$

therefore $ \int_{0}^{\frac {\pi}{2}}\frac {\sin nx }{\sin x } \equal{} \frac {\pi }{2}$    for n=odd

Therefore I(2009) = $ \frac{ \pi}{2}$


calculus
inequalities unsolved
inequalities

$\displaystyle \frac{ab+c}{ca+cb} = \frac{ab}{c(a+b)} + \frac{1}{a+b} \leq \frac{a+b}{2c} + \frac{1}{4}\left(\frac{1}{a} + \frac{1}{b}\right) = \frac{a+b}{2c} + \frac{1}{4a} + \frac{1}{4b}$.

Similarly,

$\displaystyle \frac{bc+a}{ab+ac} \leq \frac{b+c}{2a} + \frac{1}{4b} + \frac{1}{4c}$.

$\displaystyle \frac{ca+b}{bc+ba} \leq \frac{c+a}{2b} + \frac{1}{4c} + \frac{1}{4a}$.

Adding all three together and we are done.

Using that  :Sigma: ab/c(a+b) <=  :Sigma: a+b/4c we shall prove that  :Sigma: a+b/4c + :Sigma: 1/a+b <=:Sigma: a+b/2c + :Sigma: 1/2c After some calculus wehave to prove that  :Sigma: 1/a+b<= :Sigma: a+b/4c + :Sigma: 1/2c From cauchy we have  :Sigma: 1/2c >=  :Sigma: 1/a+b and  :Sigma: a+b/4c >0 and we are done Also we can easily see that we do nat have equality


MIT's theoretical calculus class uses Apostol, and I like it very much
If you're just starting to learn calculus as a high school student, I recommend Anton.

The calculus book I used to self-teach myself is: Calculus and Analytic Geometry by Thomas and Finney.

Calculus by Spivak is good.  (It proves everything with perfect rigor; explanations are very clear; that may or may not be what you're looking for.)

Spivak also wrote a short book called The Hitchhiker's Guide to Calculus that's pretty good.


algebra
polynomial
conics
parabola
calculus
inequalities
inequalities proposed

Let $ P(x) , Q(x) $ be the polynomials :
  $ P(x) = x^2+ax +b $
  $ Q(x) = x^2+cx +d $
 $ P(x) , Q(x) $ get negative values in two different intervals
 Prove that :
 There exists $ h , k $ such that : $ hP(x) + kQ(x) > 0 $ for all $ x\in R $

The two intervals have no any intersections or they can have common parts.

The intervals on which $P$ and $Q$ are negative would have to be disjoint or the claim would be false. So assume disjoint.

[hide]Let $x_0$ be a point that lies in between the two intervals of negativity. That is, both $P(x_0)$ and $Q(x_0)$ are positive but the interval on which $P$ is negative lies on one side of $x_0$ and the interval on which $Q$ is negative lies on the other side.

The vertex of $P$ is located at $-\frac{a}2$ and lies within the interval on which $P$ is negative. Similarly, $-\frac{c}2$ lies within the interval on which $Q$ is negative. Thus $x_0$ lies in between $-\frac{a}2$ and $-\frac{c}2$ and can be written as a convex combination of those two points.

Find positive $h$ and $k$ with $h+k=1$ such that $x_0=-\frac{ha+kc}2.$ Then the parabola $R(x)=hP(x)+kQ(x)= x^2+(ha+kc)x+hb+kd$ has its vertex, hence minimum, at $x_0.$ But $R(x_0)=hP(x_0)+kQ(x_0)>0$ so $R(x)$ is always positive.[/hide]

[quote="Kent Merryfield"]Now that I've looked at it, it's not clear to me that this is the right forum for it, since it doesn't have to be a calculus problem. Should we leave it here, or does it belong in "Olympiad - Inequalities"?[/quote]

 I don't khow where it should be, but is should be moved :D

I think it is regardless of calculus for Kent Merryfield.allow me to explain it.
first: we can choose X_0 between -a/2 and -c/2,to satisfy P(x),Q(x) acroding  "P(x),Q(x) get negative values in two different intervals "
second :we try to adjust k and h to let Q(x)'s vertex is Q(x_0).
that is all.thank you for Kent Merryfield's nice proof.


An ellipse has foci at $(9,20)$ and $(49,55)$ in the $xy$-plane and is tangent to the $x$-axis. What is the length of its major axis?

[hide="Answer with minimal explanation"]
Just go like WHOA LET'S USE THE DISTANCE FORMULA and do

$d = \sqrt{ (9 - 49)^2 + (20 - (-55))^2 } = \sqrt{40^2 + 75^2} = 5 \sqrt{8^2 + 15^2} = 85$
[/hide]

think about what type of locus the ellipse is


induction
modular arithmetic
group theory
Galois Theory
algebra
binomial theorem
number theory

Prove that if p is a prime then $ (\mathbb{Z}/p^2)^\times \cong C_p$ $ \times$   $ \mathbb{C}_{p\minus{}1} \cong C_{p(p\minus{}1)}$ where $ C_n$ denotes the cyclic group of order n.

My idea was to consider a homomorphism $ \mathbb{Z}/p^2 \rightarrow \mathbb{Z}/p$ and observe that $ (\mathbb{Z}/p)^\times$ is a cyclic group of order $ p\minus{}1$. Then maybe use proof by induction that if $ p$ is an odd prime then $ (\mathbb{Z}/p^n)^\times$ is cyclic for all $ n$.

Next part is to describe the Galois structure of the field of $ p^2$ roots of unity, $ p$ is prime for the cases of a. p=5, b. p=7. How many intermediate fields are there in these cases?

I don't think the induction applies to the problem at hand per se: instead your proof of the problem will hopefully generalize to give the inductive step were one asked to prove the claim for all n.

Like you/murray said: consider the (surjective) map of rings from Z/p^2 to Z/p. It is easy (why?) to see that this induces a group homomorphism on the groups of units. Convince yourself that this is surjective too. Now think about the first isomorphism theorem.

For the Galois group, read Murray's email.

Best,
- Colin

Number-theoretically, it's saying that for every odd prime $ p$ and every natural number $ n$, $ p^n$ has a primitive root. we first prove the following lemma.

Lemma: If $ \gcd(a,p)\equal{}\gcd(b,p)\equal{}1$, then $ a^p\equiv b^p\pmod{p^k}\Rightarrow a\equiv b\pmod{p^{k\minus{}1}}$.

Proof: Let $ b\equal{}a\plus{}rp^m$, where $ \gcd(r,p)\equal{}1$. then, by the binomial theorem, $ p^k|a^p\minus{}b^p\equal{}a^p\minus{}(a\plus{}rp^m)^p\equal{}a^{p\minus{}1}rp^{m\plus{}1}\plus{}\cdots$, where every term after the first is divisible by a power of $ p$ greater than $ p^{m\plus{}1}$. thus the highest power of $ p$ dividing this expression is $ p^{m\plus{}1}$. so it follows that $ m\geq k\minus{}1$, completing the proof.

now to the main result.

Theorem: If $ p$ is an odd prime, $ p^n$ has a primitive root for every natural number $ n$.

Proof: The case $ n\equal{}1$ follows by an easy combinatorial argument or by results from algebra. So let $ g$ be a primitive root modulo $ p$. moreover, assume that $ g^{p\minus{}1}\not\equiv 1\pmod{p^2}$. (if $ g$ is a primitive root modulo $ p$ which doesn't have this property, then it's easy to show that $ g\plus{}p$ is a primitive root that does.)

we're going to show that for an arbitrary natural number $ n$, $ g$ is a primitive root modulo $ p^n$.

let $ d$ be the order of $ g$ modulo $ p^n$. by Euler's extension to Fermat's Little Theorem, $ g^{p^{n\minus{}1}(p\minus{}1)}\equiv 1\pmod{p^n}$. so $ d | p^{n\minus{}1}(p\minus{}1)$. on the other hand, because $ g^d\equiv 1\pmod {p}$, it follows that $ p\minus{}1 | d$ as well. so $ p\minus{}1|d|p^{n\minus{}1}(p\minus{}1)$, and thus $ d\equal{}p^m(p\minus{}1)$ for some $ m\leq n\minus{}1$.

now, repeatedly applying the lemma to $ g^{p^m(p\minus{}1)}\equiv 1\pmod{p^n}$, we get $ g^{p\minus{}1}\equiv 1\pmod{p^{n\minus{}m}}$. since $ g^{p\minus{}1}\equiv 1\pmod{p}$ but $ g^{p\minus{}1}\not\equiv 1\pmod{p^2}$, it follows that $ m\equal{}n\minus{}1$. thus $ d\equal{}p^{n\minus{}1}(p\minus{}1)$ and therefore $ g$ is a primitive root $ \pmod{p^n}$, completing the proof.


inequalities vector geometry circumcircle

$ a,b,c\in R^{+}$, there're an inequality like this,
$ abc\ge\frac{27a^{2}b^{2}c^{2}}{(a+b+c)^{3}}\ge(a+b-c)(b+c-a)(c+a-b)$
How can I prove it?

Well first part is just AM-GM, $ abc \le \left(\frac{a+b+c}{3}\right)^{3}$

Second one is more complicated, I multiplied and did some manipulations, it appears you can group the terms accordingly and use Schur multiple times (eg $ \sum_{\text{cyc}}(a^{6}-a^{4}b^{2}-a^{4}c^{2}+a^{2}b^{2}c^{2})\ge 0$)...though I'm not entirely sure...

Let a = x + y, b = x + z, c = y + z.

At most one of x, y, z is negative, in which case the inequality is immediate. Suppose they are all nonnegative. We could proceed algebraically and it would probably be pretty easy, but I don't feel like it. So, if $ s = \frac{a+b+c}{2}$, we note that the inequality rewrites as

$ 27(abc)^{2}\ge 64s^{3}(s-a)(s-b)(s-c)$, i.e.

$ 27 (4RK)^{2}\ge 64s^{2}K^{2}$, which is equivalent to

$ s \le \frac{3\sqrt{3}}{2}R$.

This is a known inequality (it might be called weitzenbock or something). Anyway, how do we prove it. Well, the vector $ OA+OB+OC$ has nonnegative norm, where $ O$ is the circumcenter, so evaluating the norm yields that $ 9R^{2}\ge a^{2}+b^{2}+c^{2}$, I think, whereupon our desired inequality is immediate.

The following inequality is also true.
Let $a$, $b$ and $c$ are positive numbers. Prove that:
\[4a^2b^2\ge(a^2+b^2)^2\ge (a+b+c)(a+b-c)(b+c-a)(c+a-b).\]


I am now learning about complex numbers. I learnt that $ i^i$ takes infinitely many values. But there are cases that the trick failed.

What is $ i^\pi$ ?

Is the value defined? If not, then when is $ i^x$ defined?

$ i^x\equal{}exp(ix\pi(\frac 12 \plus{}2n))$, we had unique mean only if $ x\in Z$.

[quote="Rust"]$ i^x \equal{} exp(ix\pi(\frac 12 \plus{} 2n))$, we had unique mean only if $ x\in Z$.[/quote]

Oh so $ i^\pi$ can be any complex number with modulus $ 1$! Is this right?


limit
integration
calculus
calculus computations

For each sufficiently small $ \epsilon > 0$, there exists a positive real number $ K$ such that $ |f(x)+f'(x)-1| < \epsilon$ for all $ x \geq K$.

Then

\begin{eqnarray*}&&-\epsilon \; < \; f(x)+f'(x)-1 \; < \; \epsilon\\ & \Leftrightarrow &-\epsilon e^{x}\; < \; \{ (f(x)-1) e^{x}\}' \; < \; \epsilon e^{x}\\ & \Rightarrow &-\epsilon (e^{x}-e^{K}) \; \leq \; (f(x)-1)e^{x}-(f(K)-1)e^{K}\; \leq \; \epsilon (e^{x}-e^{K})\\ & \Rightarrow &-\epsilon \; \leq \; \lim\inf (f(x)-1) \text{ , }\lim\sup (f(x)-1) \; \leq \; \epsilon\\ \end{eqnarray*}

So $ \lim_{x\to\infty}f(x) = 1$ and $ \lim_{x\to\infty}f'(x) = 0.$

[quote="N.T.TUAN"]Let $ f: [0,\infty)\to R$ such that exist $ f'(x)\forall x\in [0,\infty)$ and $ \lim_{x\to\infty}[f(x)+f'(x)]=1$. Find $ \lim_{x\to\infty}f'(x).$[/quote]

By MVT:
$ \lim_{n}f'(n) = 1-f(0)-\sum_{n=1}^{\infty}\frac{f(n)-f(n-1)}{n-(n-1)}= 1-f(0)-\sum_{n=1}^{\infty}f'(c_{n})$ where $ c_{n}\in [n-1,n]$.
Because $ \lim_{n}f'(n)$ exists, a sum in right side converges, so $ f'(c_{n}) \to 0$, as $ n \to \infty$.

[quote="sos440"]

\begin{eqnarray*}&&-\epsilon \; < \; f(x)+f'(x)-1 \; < \; \epsilon\\ & \Leftrightarrow &-\epsilon e^{x}\; < \; \{ (f(x)-1) e^{x}\}' \; < \; \epsilon e^{x}\\ & \Rightarrow &-\epsilon (e^{x}-e^{K}) \; \leq \; (f(x)-1)e^{x}-(f(K)-1)e^{K}\; \leq \; \epsilon (e^{x}-e^{K})\\ & \Rightarrow &-\epsilon \; \leq \; \lim\inf (f(x)-1) \text{ , }\lim\sup (f(x)-1) \; \leq \; \epsilon\\ \end{eqnarray*}

[/quote]
Why we have 1st sign '' $ \Rightarrow$ '', please concrete!

 Hmm, pou's solution is wrong!

Just simple integration. For $ x \geq K$,

\begin{eqnarray*}& &-\epsilon e^{x}\; < \; \{ (f(x)-1) e^{x}\}' \; < \; \epsilon e^{x}\\ & \Rightarrow & \int_{K}^{x}-\epsilon e^{t}\; dt \; \leq \; \int_{K}^{x}\{ (f(t)-1) e^{t}\}' \; dt \; \leq \; \int_{K}^{x}\epsilon e^{t}\; dt\\ & \Rightarrow &-\epsilon (e^{x}-e^{K}) \; \leq \; (f(x)-1)e^{x}-(f(K)-1)e^{K}\; \leq \; \epsilon (e^{x}-e^{K})\\ \end{eqnarray*}

There is a short proof for this I've seen before, but it works only if $ f(x)$ is given to converge to some $ L \neq 0$. It won't work here but, nevertheless, I'm posting it because I found it nice:

$ \lim f(x) = \lim \frac{ e^{x}f(x) }{e^{x}}\left( = \frac{\infty}{\infty}\right) = \lim \frac{ (e^{x}f(x))' }{(e^{x})'}= \lim ( f(x)+f'(x) ) = 1$.


In square $ ABCD$, point $ M$ is the midpoint of side $ AB$ and point
$ N$ is the midpoint of side $ BC$. What is the ratio of the area
of triangle $ AMN$ to the area of square $ ABCD$? Express your
answer as a common fraction.

The area of a triangle is bh/2. The base would be half of the side lenght, and the height would also be half the side length. Thus, the area of the small triangle in terms of s, the side length of the square, would be $ \frac{1}{2}\cdot(\frac{s}{2})^2\equal{}\frac{s^2}{8}$. The area of the square is $ s^2$, so the ratio of the triangle to the square is 1:8.

Except they wanted the answer as a common fraction, so $ \boxed{\frac{1}{8}}$.


Hello, here's a problem that I enjoyed solving, and I hope you do too.  Here it is:

$\text{\"What is the maximum area of a triangle with a perimeter of 16?\"}$

Tru solving this in the most creative fashion. :)

I thought about using this formula:
$S^2=p(p-a)(p-b)(p-c)$, where $p$ is a half of a perimeter.
$S^2=8(8-a)(8-b)(8-c)$, $a+b+c=16$

so we have to evaluate: $(8-a)(8-b)(8-c)$, when it attains maximum value?

[hide="Hint"]
AM-GM, incircle, and area. :D
[/hide]

[/hide]

I had a method involving an inscribed circle, which allowed me to use $A=rs$ with $A=\sqrt{s(s-a)(s-b)(s-c)}$, however, I now have an even simpler method.  Here it is...
We use the formula $A=\sqrt{s(s-a)(s-b)(s-c)}$ with AM-GM as follows.
$s=\frac{a+b+c}{2}=\frac{16}{2}=8$
$\\\frac{8-a+8-b+8-c}{3}=\frac{24-(a+b+c)}{3}=\frac{8}{3}\ge\sqrt[3]{(8-a)(8-b)(8-c)}\\(\frac{8}{3})^3\ge (8-a)(8-b)(8-c)\\8(\frac{8}{3})^3\ge 8(8-a)(8-b)(8-c)\\\sqrt{8(\frac{8}{3})^3}\ge\sqrt{8(8-a)(8-b)(8-c)}$
$\therefore$ The maximum area of a triangle with a perimeter of 16 is $\sqrt{8(\frac{8}{3})^3}$.
$\mathbb{QED}$

That's the same thing as dondigo except he made a mistake in the calculations.

Well, I had already given the AM-GM and area idea in my hint.  And would you please point out my calculation mistake.  Our results are exactly the same.

EDIT: Ohh...he has the mistake...however, his asnwer is right.

Okay, maybe it wasn't a calculation mistake, but I think he put $\left( \frac{8}{3} \right) ^3$ instead of $\left( \frac{8}{3} \right) ^2$

edit: oh, you saw it, okay

Anyway, this was actually a problem given by my father...he had no clue on how to solve it.  Anyway, when I solved it, he somehow did not have interest in the solution, so I poseted it here.:lol: I hope you enjoyed it.

One tirck which makes things easier,  If (a,b,c) are the sides (with 2s=a+b+c) , use the varibles: llet the (s-a)=x, (s-b)=y and (s-c)=z
with $x+y+z=s=8$

We have $(Area)^2 = 8 \cdot x y z$
since $x+y+z = 8$ a constant, max value of xyz is given (AM-GM)  as $(xyz) \le (s/3)^3= (8/3)^3$ etc.

In many IMO type problems it is much easer to work with (x,y,z) instead of (a,b,c) becuase for (x,y,z) all you have to worry about is  that each variable is positive. (If one used a,b,c one has to make sure that  a <=b+c  etc ...)

You're right.  This is better if we're proving something. :)

[quote="boxedexe"][quote="dondigo"]I've done it using my formula.
I'll continue. We have to evaluate when $(8-a)(8-b)(8-c)$ attains maximum value. I'll use AM-GM, because $8-a, 8-b, 8-c$ are positive (traingle inequality).
$(8-a)+(8-b)+(8-c)\geq 3\sqrt{(8-a)(8-b)(8-c)}$
$8\geq 3\sqrt{(8-a)(8-b)(8-c)}$

$(8-a)(8-b)(8-c)\leq \frac{8^3}{3^3}$
So it attains maximum value for $a=\frac{16}{3}=b=c$
Maximum $S=\frac{1}{4}(\frac{16}{3})^2\sqrt{3}$

How about Your solution?[/quote]
Btw, he got this because he knew that an equalateral has the largest area among traingles.  The formula for the area of an equilateral traingle is $A=\frac{\sqrt{3}}{4}s^2$. :)

EDIT:  You could've done this without using AM-GM.  You could just say $\frac{16}{3}$, place it into $A=\frac{\sqrt{3}}{4}s^2=\frac{\sqrt{3}}{4}\left(\frac{16}{3}\right)^2$.  Hope that clarifies things.[/quote]

But notice, that in my solution I actually proved, that among all triangles the biggest area has equilateral triangle.
Look:
I proved, that (8-a)(8-b)(8-c) is always less or equal to something. But the bigger it is the bigger area has the triangle. So it must be equal to this. But
when it's equal? AM-GM tells us that it's equal if and only if $(8-a)=(8-b)=(8-c)$, so $a=b=c$. That's why I concluded it. I didn't assumed it :)

Oh, sorry.  I was giving my explanation to other people that this problem, just for the sake of [i]solving[/i] it, could have been done without AM-GM. :)


integration
logarithms
calculus
calculus computations

we have:
f(x) + (6x+2)f(x)' = 0

Find f(x)

[hide]
For $ x>\minus{}\frac{1}{3}$: The differential equation is of the form $ y^\prime \equal{}\minus{}\frac{y}{6x\plus{}2}$. We have
\[ \int_{\minus{}1/6}^x \minus{}\frac{dt}{6t\plus{}2} \equal{}\frac{\minus{}\ln (6x\plus{}2)}{6}\]
So if we let $ g(x)\equal{}f(x)e^\frac{\ln (6x\plus{}2)}{6}\equal{}(6x\plus{}2)^{1/6}f(x)$, then
\[ g^\prime (x)\equal{}(6x\plus{}2)^{1/6}f^\prime (x)\plus{}(6x\plus{}2)^{\minus{}5/6}f(x)\equal{}0\]
So $ g(x)$ is a constant. If $ g(x)\equal{}A$, then $ f(x)\equal{}\frac{A}{\sqrt[6]{6x\plus{}2}}$.
[/hide]


The problem is that $1-0.99999999...=0$.

The $1$ digit in $1-0.999999...$ only occurs after an infinite number of $0$s (as there are an infinite number of $9$s). Although I don't like to throw the concept of infinity around, as it generally leads to confusing concepts!

[quote="TRAN THAI HUNG"]Can you tell me what's wrong with this problem?
   [b] Prove that "1=0"[/b]
$\frac{1}{9}=0,1111...1$
<=>$\frac{1}{9}*9=0,111...1*9$
<=>$\ 1=0,999...9$
<=>$\ 0,000...1=0$
<=>$\ 1=0$ :)  :D[/quote]

$\frac{1}{9}\neq 0,111...1$, but rather $\frac{1}{9}= \lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{10^{n}}$. You cannot put a final digit in that notation, as it would mean that the decimal expansion is finite, which it isn't.


This topic may have been brought up before (or maybe it was brought up in relation to the AMC 10/12), but for the people here who have made USAMO and/or done well on the AIME: how much time do you spend checking over your answers on the AIME, if you do so at all?

I just took a practice AIME and had solved 10 problems after about 2 hours had passed. I then took the time to check all my answers/solutions, transfer the answers I had written on scratch paper to the official answer sheet with the bubbles, and then check to make sure that I bubbled everything in properly. However, this whole process took about 30 minutes, 1/6 of the 3 hours for the AIME exam, in total -- and this was only with 10 problems answered! I don't know whether I could afford to take 30 minutes on the AIME to do this...

Comments?

If you just want to qualify for USAMO I think that's a fine way to spend half an hour.

If you don't rush and you work neatly, you might not need to check your work at all. Double-check all your arithmetical operations right after you do them to make sure that they are right, and if they are, then what can go wrong? (Ok, lots of things can go wrong, but it's much more likely that you'll get a problem right if you check all the steps as you go.) I'm not going to pretend that I was well-disciplined enough to do things like that on the AIME though. (I wish I had been.)

10 questions should be enough for you to make USAMO.  30 minutes in my opinion is actually less than you actually should spend checking - unless you are absolutely sure.  You don't want to sacrifice USAMO just becuase you were a wee bit impatient.

I think that setting aside time at the end of the test to check answers is a great idea.

Personal anecdotes:

The year I scored my highest AIME score I double-checked every problem and wrote solutions out very neatly on the pages of a legal pad.

My senior year I had my lowest score since my 8th grade year.  I spent a very long time on one problem.  I failed to go back and check my solutions and made three errors.  I felt like I would have caught two or all three of them had I given up on the problem that confounded me for so long and spent 20 minutes reading through my work.

Also, organize your work on different sheets of paper.  It makes it easier to read through each step when there isn't some diagram written over your algebra (something I did for a couple of years when I was young).


The three-dimensional space is coloured with five different colours, all of them occuring at least once. Prove that there is at least one plane that contains four points of pairwise different colour.

Assume the contrary.

Consider a line passing through points with colors $1,2$, say, and another line passing through points with colors $3,4$. We denote the points by the same names as their colors, i.e. $1,2,3,4$ respectively. By assumption, these lines are not coplanar.

If there was a point on $12$ having color $3,4$  or $5$, then one of the planes $(124),(123)$ would contain $\ge 4$ colors. This means that all the points on $12$ have one of the colors $1,2$. Now take any plane $\pi$ passing through three points of colors $3,4,5$. This plane intersects our line $12$ in a point with color $1$ or $2$, so $\pi$ is the plane we're looking for.


I am using TeXnicCenter with MikTex. I'm trying to put theorems in my documents but they don't show up in the output, any ideas? I'm using the following command:

\begin{theorem}
hello
\end{theorem}

See here: http://en.wikibooks.org/wiki/LaTeX/Theorems

I usually just use "\textbf{Theorem 1.1}" and type the theorem in italics.

You will get [i]LaTeX Error: Environment theorem undefined[/i] so ensure you have
[code]\usepackage{amsthm}[/code]in the preamble (before \begin{document}) and then, in the document a line like::
[code]\newtheorem{theorem}{Theorem}[/code]

Hi,

packages are not necessary, the already mentioned line
[code]\newtheorem{theorem}{Theorem}[/code] would fix it, this minimal example is compilable:
[code]\documentclass[a4paper,10pt]{article}
\newtheorem{theorem}{Theorem}
\begin{document}
\begin{theorem}
hello
\end{theorem}
\end{document}[/code]
But it's very recommendable to use either [url=http://ctan.org/pkg/amsthm]amsthm[/url] (mentioned by stevem) or the [url=http://ctan.org/pkg/ntheorem]ntheorem package[/url], their documentations are very useful.

Stefan


geometry
LaTeX
perimeter
function
area of a triangle
Heron\u0027s formula

Third one use right triangles.

14=5+9.

That should do it.

i dont know what happened but your solution isnt hidden...just a heads up

[hide="1"]what is a partition?[/hide]
[hide="2"]$ \frac {1} {n(n \plus{} k)} \equal{} \frac {1} {k} (\frac {1} {n} \minus{} \frac {1} {n \plus{} k})$ i think thats the formula
here k=1 so it becomes 1-1/2+1/2-1/3....+-1/10. everything cancels out to become 1-1/10 which is 9/10[/hide]
[hide="3"]i did the same thing. [hide="question"](is there any other way of doing it? my friend geussed it right on it last year during practice....)
EDIT: the future beat me to the post
[/hide][/hide]

i got a little lazy in the end for #2 to use latex and im short on time

[hide="2"]$ \frac {1}{1*2} \plus{} \frac {1}{2*3} \plus{} ... \plus{} \frac {1}{n(n \plus{} 1)} \equal{} \frac {n}{n \plus{} 1}$.Can be verified through induction. So, it's 9/10.[/hide]

EDIT-[hide]For #3, you draw an altitude to the side with length 14, and some Pythagorean stuff, or noticing 12 satisfies the condiotions needed. So the area is 12*14/2=84.[/hide]

[quote="darkmatter47"]1)What is the number of partitions for the number 7? , given that there are 5 partitions for the number 4.

[hide="answer"] 15   - I got this eventually, but wasted way too much time... More than 3 minutes.[/hide]

2) Express the following sum as a common fraction:

$ \frac {1}{1*2} \plus{} \frac {1}{2*3} \plus{} \frac {1}{3*4} ... \plus{} \frac {1}{9*10}$
[hide="answer"] $ \frac {9}{10}$ [/hide]

3)What is the number of square feet in the area of a triangle with side lengths shown?

The lengths are 14, 13, and 15 ft.
I used Heron's formula.
[hide="my solution for this one. "]
Where s= $ \frac {14 \plus{} 15 \plus{} 13}{2} \equal{} 21$
$ A \equal{} \sqrt {s(s \minus{} 13)(s \minus{} 14)(s \minus{} 15)} \equal{} 84$

But I have a feeling there is a solution without Heron's formula. Is there?[/hide][/quote]
1. What are partitions?
2. The formula for that sum is 1/(1*2)+1/(2*3)+....+1/n(n+1)=n/(n+1). So here n=9 and the sum is 9/10.
3. The perimeter of the triangle is 42 ft. So s=21, and using Heron's formula, we get Area=sqrt(21)(21-14)(21-13)(21-15)=sqrt(21)(7)(8)(6)=sqrt(7056)=84 square feet.

[quote="darkmatter47"]A partition is a way of expresing an integer as a sum of one or more natural numbers.[/quote]
That's what a parition is.There might be a way of carefully using generating functions, but I don't know, since you would be overcounting. So the best way would be to list them.


Find all ordered pairs $\ (x,y)$ $\ 1 \leq x < y \leq 100$ such that $\ i^x + i^y$ is a real number.

there are $100 \cdot 100=10000$ pairs of $(x,y)$ and there are $2450$ pairs where $x \not= y$ where $x$ and $y$ are both even. Half of these are when $x<y$ so there are $1225$ pairs. Now, this condition also works if both $x$ and $y$ are odd and their difference is $\equiv 2( \mod4)$, so there are $2500/2$ pairs of odd $x,y$ where $x\leq y$ and half of that is when $y-x \equiv 2(\mod{4})$ so there are $625$ pairs. So the answer is $1225+625=\boxed{1850}$

More simply...
One possibility is both x and y are even.  There are 50 evens, and they must be distinct, and x < y, so there are $\binom{50}{2} = 1225$ possibilities.
Also, if one is $1$ and the other is $3 \mod 4$, they cancel out.  There are 25 of each, and since x < y, there are $25 \cdot 25 = 625$ possibilities.
Therefore $1850$ total.  Although the question doesn't ask for the total...

Well, i'm assuming it's asking for the number of ordered pairs, not each ordered pair...I MEAN THAT WOULD BE INSANE IF ARML WOULD MAKE PEOPLE WRITE EM ALL OUT..WHOA


integration
real analysis
calculus
algorithm
algebra
polynomial
arithmetic sequence

Let $ N$ be an integer and fix $ N^2$ real numbers $ 0 < a_1, a_2, \ldots, a_{N^2} < 1$. How would you choose $ N$ other real numbers $ 0 < x_1, x_2, \ldots, x_N < 1$ to achieve the infimum:
$ \phi(a_1, \ldots, a_{N^2}) \equal{} \inf_{(x_1, \ldots, x_N)} \sum_{k \equal{} 1}^{N^2} \min_{1 \leq i \leq N} |a_k \minus{} x_i|$
question: find a good upper bound for:
\[ f(N) \equal{} \sup_{(a_1, \ldots, a_{N^2})} \phi(a_1, \ldots, a_{N^2})
\]

Or, more generally, given a measure $ \mu$ on $ [0,1]$, how to choose $ n$ points $ x_j$ such that $ \int_0^1\min_j|t\minus{}x_j|\,d\mu(t)$ is as small as possible? The almost worst case is the case of the Lebesgue measure, where you get total mass $ 1$ and the integral not less than $ \frac 1{4n}$ for every choice with $ \frac 1{4n}$ attained on the arithmetic progression with step $ 1/n$ starting with $ \frac 1{2n}$. The uniform distribution of $ m\gg n$ points on $ [0,1]$ is almost as bad as the Lebesgue measure. On the other hand, for every measure you can consider the progression with step $ 1/{n\minus{}2}$ and random starting point together with 2 endpoints and get $ \frac{\|\mu\|}{4(n\minus{}2)}$ on average.

What seems really interesting is to find an algorithm that allows to get the minimum for any given configuration with an error $ \varepsilon\|\mu\|/n)$ in time polynomial in $ n$ and $ \varepsilon^{\minus{}1}\ll n$. I spent a few hours trying to come up with something but failed :). Some cases are rather trivial: if $ d\mu\equal{}f\,dx$ with continuous positive $ f$, then the almost optimal configuration for large $ n$ is the partition into $ n$ subintervals with equal integrals of $ \sqrt f$ and putting the points at the centers of those intervals but the general case is by no means obvious. Do you need it for something, alekk, or you just were curious?

thank Fedja ! I asked myself this question while thinking about the problem with Steiner Tress etc ..
Btw, do you have an example of non atomic measure such that the strategy you described (put the points according $ \sqrt{f}$) does not work ?


$ 2^x \plus{} 2^(x \minus{} 1) \minus{} 2^(x \plus{} 1) \equal{} \minus{} 32$

I used the exponent law way of solving it and got an answer of 6, which is correct.

However, the log way didn't work out so well.
$ xlog2 \plus{} (x \minus{} 1)log2 \minus{} (x \plus{} 1)log2 \equal{} \minus{} log32$
$ log2 (x \plus{} x \minus{} 1 \minus{} x \minus{} 1) \equal{} \minus{} log32$
$ x \minus{} 2 \equal{} \minus{} log32/log2$
$ x \equal{} \minus{} 5 \plus{} 2$
$ x \equal{} \minus{} 3$

What am I doing wrong?

Thanks.

$ 2^x\plus{}2^y\neq x\log 2\plus{}y\log 2.$

Sorry didn't notice latex sqeezed it on one line :S

[quote]$ 2^{x} \plus{} 2^{y}\neq x\log 2 \plus{} y\log 2.$[/quote]

Yea but isn't...
$ 2^{x} \plus{} 2^{(}x \minus{} 1) \minus{} 2^{(}x \plus{} 1) \equal{} \minus{} 32$
Log both sides:
$ log2^x \plus{} log2(x \minus{} 1) \minus{} log2(x \plus{} 1) \equal{} \minus{} log32$
Rearrange:
$ xlog2 \plus{} (x \minus{} 1)log2 \minus{} (x \plus{} 1)log2 \equal{} \minus{} log32$

Logs aren't distributive.  You would need to write $ log(2^{x} \plus{} 2(x \minus{} 1) \minus{} 2(x \plus{} 1)) \equal{} log( \minus{} 32)$  and stop there.

Also, you can't take the log of a negative number... 10 raised to a power can't equal a negative number.

Er, also, the problem that you wrote doesn't have a solution.  I hope I'm not being dense and not understanding you  :( [/hide]

AHHH forgot the exponent sign :O

sorry...

wait no I did do the exponent but latex didn't read it for some reason, I'll try again.

2^x+2^(x-1)-2^(x+1) =-32

no latex :)

[quote="RoFlLoLcOpT"]
\begin{align*}2^x + 2^{x - 1} - 2^{x + 1} & = 2^x\left(1 + \frac {1}{2} - 2\right) \\
& = 2^{x}\cdot - \frac {1}{2} \\
& = - 2^{x - 1}.\end{align*}
From here, we easily obtain $ x = 5 + 1 = \boxed{6}.$[/quote]

Thanks for that, but that's how I already figured out the answer to the question.  Just wanted to know if there was a way to do it by log.


Not completely sure if it's a true conjecture.  It's a generalization from an AIME problem.

Starting at $ (0,0)$, an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Prove that it is impossible to reach the point $ (n,n)$ where $ n$ is a positive integer and the point $ (n,n)$ must be reached in $ n\plus{}4$ steps if more than one step is taken to the left or downward.

[quote="brady00n"]Starting at $ (0,0)$, an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Prove that it is impossible to reach the point $ (n,n)$ where $ n$ is a positive integer and the point $ (n,n)$ must be reached in $ n\plus{}4$ steps if more than one step is taken to the left or downward.[/quote]

I don't think you've phrased the problem properly.  As it stands, it takes $ 2n$ steps to get to $ (n, n)$ with no restrictions, so I don't see how $ n\plus{}4$ steps is possible for $ n > 4$.

What's the original problem?

The original problem asks for the number of ways to reach (2,2) from the origin in 6 steps (up, left, down, right) or the probability of this happening. I'm not exactly sure which.

I think brady00n would have been better off using 2n+2 steps rather than n+4 steps. This mistake is likely because 6=2+4.
[hide="Proof of this"]
We know it requires at least n steps up and n steps right to reach (n,n). That's 2n steps. Note that if we take a step down, we must take a step up to "offset" this (same goes for left and right). Note that this takes 2 steps and puts us right back where we started. Thus if we take 2 or more steps left or down, it requires a total of 4 steps to take us back to the original position. We must add 2n steps because it takes 2n steps minimum to reach (n,n). Thus if we take 2 steps left or down, we must take at least 2n+4 steps to reach (n,n). Obviously, 2n+4>2n+2, so we are done.[/hide]

A better generalization would be to ask for the number of ways to reach (n,n) from the origin in 2n+2k steps (or the probability of this happening) in terms of n and k. That reminds me of an old IMO problem from the 80's asking the probability of reaching (n,n) from the origin moving only right and up, as directed by flipping a coin, in 2n+k steps (of course, if you reached (n,a) where a<n and flipped a head, you would stay on (n,a)).

[quote="FightTheTide"]
A better generalization would be to ask for the number of ways to reach (n,n) from the origin in 2n+2k steps (or the probability of this happening) in terms of n and k.[/quote]

Well the number of ways can be written as $ \sum_{i\plus{}j\equal{}k}\frac{(2n\plus{}2k)!}{(n\plus{}i)!(n\plus{}j)!i!j!}$...(with $ i, j\in\mathbb{Z}$, $ i, j\ge 0$)


abstract algebra
superior algebra
superior algebra unsolved

[quote="Eneas"]Let $ 0 \to A \to B \to C \to 0$ be an exact sequence of $ R$-modules. Prove that $ \mathrm{Ann}\left(A\right) \plus{} \mathrm{Ann}\left(C\right)$ is in $ \mathrm{Ann}\left(B\right)$. (Note: all the annihilators are over $ R$).[/quote]

If R=Z, A=Z/2Z, B=Z/6Z, and C=Z/3Z, then Ann(A)+Ann(C)=2Z+3Z=Z is not contained in Ann(B)=6Z.

[quote="Nukular"]perhaps you mean Ann(A) $ \cap$ Ann(C)?[/quote]

For R=Z, A=Z/2Z, B=Z/4Z, C=Z/2Z, Ann(A)=Ann(C)=2Z=Ann(A) ∩ Ann(c) is not contained in Ann(B)=4Z.

Ann(A)*Ann(C) ≤ Ann(B) ≤ Ann(A) ∩ Ann(C) if they are left R-modules.

As I like adding fuel onto things:

How about $ \sqrt {\mathrm{Ann}(B)} \supset \mathrm{Ann}(A) \cap \mathrm{Ann}(C)$? (This is what I was actually thinking above...)

[quote="Nukular"]How about $ \sqrt {\mathrm{Ann}(B)} \supset \mathrm{Ann}(A) \cap \mathrm{Ann}(C)$? (This is what I was actually thinking above...)[/quote]

Yes, and even stronger, $ \mathrm{Ann}A\cdot\mathrm{Ann}C\subseteq\mathrm{Ann}B$. Note that we only need the exact sequence $ A\to B\to C$, not the whole $ 0\to A\to B\to C\to 0$.

[color=blue][b]Theorem 1.[/b] Let $ R$ be a commutative ring. Let $ A$, $ B$, $ C$ be three $ R$-modules. Let $ f: A\to B$ and $ g: B\to C$ be two $ R$-module homomorphisms such that $ f\left(A\right)\equal{}\mathrm{Ker}g$. Then, $ \mathrm{Ann}A\cdot\mathrm{Ann}C\subseteq\mathrm{Ann}B$.[/color]

[i]Proof of Theorem 1.[/i] Let $ \left(x,z\right)\in \mathrm{Ann}A\times\mathrm{Ann}C$. Let $ b\in B$. Then, $ x\in \mathrm{Ann}A$ and $ z\in \mathrm{Ann}C$. Now,

$ g\left(zb\right)\equal{}zg\left(b\right)$   (since $ g$ is an $ R$-module homomorphism)
$ \equal{}0$     (since $ g\left(b\right)\in C$ and $ z\in \mathrm{Ann}C$)

and thus $ zb\in\mathrm{Ker}g\equal{}f\left(A\right)$. Thus, there exists some $ a\in A$ such that $ zb\equal{}f\left(a\right)$. Hence,

$ xzb\equal{}xf\left(a\right)\equal{}f\left(xa\right)$   (since $ f$ is an $ R$-module homomorphism)
$ \equal{}f\left(0\right)$      (since $ a\in A$ and $ x\in \mathrm{Ann}A$ yield $ xa\equal{}0$)
$ \equal{}0$   (since $ f$ is an $ R$-module homomorphism).

Thus, we have proven that $ xzb\equal{}0$ for every $ b\in B$. Hence, $ xz\in\mathrm{Ann}B$.
Therefore, we have proven that

[color=green][b](1)[/b][/color] $ xz\in\mathrm{Ann}B$ for every $ \left(x,z\right)\in \mathrm{Ann}A\times\mathrm{Ann}C$.

Now,

$ \mathrm{Ann}A\cdot \mathrm{Ann}C\equal{}\left<xz\mid \left(x,z\right)\in \mathrm{Ann}A\times\mathrm{Ann}C\right>_R$.

In other words, $ \mathrm{Ann}A\cdot \mathrm{Ann}C$ is the smallest ideal of $ R$ that contains $ xz$ for every $ \left(x,z\right)\in \mathrm{Ann}A\times\mathrm{Ann}C$. But $ \mathrm{Ann}B$ is an ideal of $ R$ that contains $ xz$ for every $ \left(x,z\right)\in \mathrm{Ann}A\times\mathrm{Ann}C$ (due to [color=green][b](1)[/b][/color]). Thus, $ \mathrm{Ann}A\cdot\mathrm{Ann}C\subseteq\mathrm{Ann}B$, and Theorem 1 is proven.

  darij


I would like to see your solution(s)? :)

A beginning?

Taking the natural log of both sides, we have

$ \ln{\left(\left (1\plus{}\frac{1}{1^{2}\plus{}1}\right )\left (1\plus{}\frac{1}{2^{2}\plus{}1}\right )...\left (1\plus{}\frac{1}{n^{2}\plus{}1}\right )\right)}$

$ \ln{\left(\left (\frac{1^2\plus{}2}{1^{2}\plus{}1}\right )\left (\frac{2^2\plus{}2}{2^{2}\plus{}1}\right )...\left (\frac{n^2\plus{}2}{n^{2}\plus{}1}\right )\right)}$

$ \ln{(1^2\plus{}2)}\plus{}\ln{(2^2\plus{}2)}\plus{}...\plus{}\ln{(n^2\plus{}2)}\minus{}(\ln{(1^2\plus{}1)}\plus{}\ln{(2^2\plus{}1)}\plus{}...\plus{}\ln{(n^2\plus{}1)})$

Now we must show that that equals two.

This appears to be a weak upper bound... the least upper bound as $ n$ approaches infinity appears to be around $ 2.60243\dots$

[hide="solution"]

Write $ a_n \equal{} \frac {1}{n^2 \plus{} 1}$, and write $ s_n \equal{} \sum_{i \equal{} 1}^n a_i$.  Then we need show that $ \prod_{i \equal{} 1}^n (1 \plus{} a_i) < e^2$.  But we have by AM-GM that $ \prod_{i \equal{} 1}^n (1 \plus{} a_i) \le \left(\frac {n \plus{} s_n}{n}\right)^n \equal{} \left(1 \plus{} \frac {s_n}{n}\right)^n < e^{s_n}$ by the fundamental limiting series for $ e$ - the proof of this is not hard.  As we have $ s_n < \sum_{i \equal{} 1}^{\infty} \frac {1}{n^2} \equal{} \frac {\pi^2}{6} < 2$, we obtain the desired bound.

Indeed, we most likely have $ s_n < 1.1$ for all $ n$ (for example, $ s_{100} \equal{} 1.066...)$ giving a much tighter bound of around 3.  The difference between this and the true limiting value comes from the fact that we obviously don't have equality in AM-GM.
[/hide]


See http://www.mathlinks.ro/Forum/viewtopic.php?p=949616#949616 (I simply copied http://www.mathlinks.ro/Forum/viewtopic.php?p=949609#949609 ).

Can you post the code for what the bug is? For now it just seems you are using additional latex tags inside other latex tags :)

Maybe I misunderstand you, but:
I did nothing except copying identically some posts of mihai miculita (he edited his post a couple of times, but I saw that two as intermediate steps on the way to what the post is now). They are probably intended to be what his post is now. I just feared that it may be harmful, but I know nothing more than that...


Let $ (x_n)$ be a sequence such that: $ x_0 \in (0,1)$, $ x_{n\plus{}1}\equal{}\frac{2\sqrt{x_n}}{1\plus{}x_n}$. The sequence $ (y_n)$ is such that: $ y_n\equal{}n\minus{}(x_0\plus{}x_1\plus{}...\plus{}x_{n\minus{}1})$.

Study the sequences $ (x_n)$ and $ (y_n)$. ([u]if they converge or not[/u])

Both converge.

By AM-GM, $ x_i \le 1$ for $ i > 1$, and obviously $ x_i > 0$ - so $ x_n$ is bounded. $ x_i$ is increasing, because for $ x > 0$: $ \frac {2}{1 \plus{} x^2} \ge x \leftrightarrow x^{0.5} \plus{} x^{1.5} \le 2$
Thus, $ x_n$ converges.
The limit must be a positive solution of $ L \equal{} \frac {2\sqrt {L}}{1 \plus{} L}$. $ L \equal{} 1$ is the only solution (substitute $ L \equal{} t^2$ and it becomes $ (t \minus{} 1)(t^2 \plus{} t \plus{} 2) \equal{} 0$), so $ x_n \to 1$.

$ y_{n} \equal{} \sum_{i \equal{} 0}^{n \minus{} 1} 1 \minus{} x_{i}$
$ 1 \minus{} x_{i \plus{} 1} \equal{} \frac {(1 \minus{} \sqrt {x_i})^2}{1 \plus{} x_i}$
$ \frac {1 \minus{} x_{i \plus{} 1}}{1 \minus{} x_{i}} \equal{} \frac {1 \minus{} \sqrt {x_i}}{1 \plus{} \sqrt {x_i}}\frac {1}{1 \plus{} x_i} < \frac {1}{1 \plus{} x_i} < \frac {1}{1 \plus{} x_0} < 1$
It follows that $ y_n$ converges. I have no clue about the limit...


function
Putnam
number theory
prime factorization

Find all functions from the natural numbers to themselves satisfying

$ f(mn)=f(m)f(n)$ anf f{f(n)}=n

This thread would be more at home in a harder forum, e.g. Preolympiad.

Or, in other words, find all totally multiplicative involutions.

[hide]From the second condition, $ f$ must be one-to-one (it has an inverse).  Put $ m = 1$ into the first equation to get that $ f(1) = 1$ or $ f(n) = 0$; since $ f$ is injective, the latter cannot hold and thus $ f(1) = 1$.  Put $ m = 0$ into the first equation to get that $ f(n) = 1$ or $ f(0) = 0$.  The latter cannot hold, so $ f(0) = 0$.  Thus, $ n > 1$ implies $ f(n) > 1$.

For any integer $ n$ with prime factorization $ p_{1}^{a_{1}}\cdot \cdots\cdot p_{k}^{a_{k}}$, define $ \Omega(n) = a_{1}+\ldots+a_{k}$ to be the number of prime factors of $ n$, counting multiplicity.  Note that $ \Omega(f(n)) = \Omega(f(p_{1})^{a_{1}}\cdots f(p_{k})^{a_{k}}) = a_{1}\Omega(f(p_{1}))+\ldots+a_{k}\Omega(f(p_{k})) \geq a_{1}+\ldots+a_{k}$, with equality if and only if $ f(p)$ is prime for each prime $ p$.  Thus, $ f$ must be an involution on the primes.  In addition, any multiplicative involution on the primes satisfies both of the given equations, so we're done.[/hide]

Follow-up:
Find all functions $ f: \bf N \to N$ such that $ f(f(n)) = n$ for all $ n \in \bf N$ and $ f(mn) = f(m)f(n)$ whenever $ (m, n) = 1$.

It is Putnam,isn't it??? Try to use Pigeonhiole Principal.


inequalities
algebra
polynomial
inequalities open

the ineq at Tuymaada was Let $a,b,c$ be positive reals s.t. $a^2+b^2+c^2=1$. Prove the following inequality \[ \sum \dfrac{a}{a^3+bc} >3 .\]
can we find $A$ such that
$\sum \dfrac{a}{a^3+bc}\geq A$

Why did you post it to the open section ? why not the unsolved one ?
Anyway, let me tell you what i did although I'm not gonna write my solution because it's big, i've not checked, i find it boring to check it, it's just calculations and probably true. You may try it after all.

I set $x=\sqrt{3} \cdot a$, $y= $... Because i find it easier to work with . So $x^2 +y^2 + z^2 = 3$.
I multiply the inequality with the product of the denominators. And as a result i get a polynomial inequality in x,y,z.
Then by picking some terms of the RHS and considering inequalities like $x^2 <3$... I get the LHS. thus the result.
I hope that my calculations are correct and that this method works. I you want more details i'll try to write.

As far as your second question, A=3 does the work. So you asked for the minimun of the expresesion.
Surely it does not look easy to answer.

I hope I helped.

I posted the solution before on http://www.diendantoanhoc.net (Vietnamese site). Now I repost here. This solution submited by my student

+ If all terms of LHS > 1 then OK.

+ If, for example,  $a \leq a^3 + bc$, we continue as follow

$\dfrac{b}{b^3+ac}+\dfrac{c}{c^3+ab} = \dfrac{1}{b^2+ac/b}+\dfrac{1}{c^2+ab/c} \geq \dfrac{4}{b^2+c^2+ac/b+ab/c} > \dfrac{4}{1+ac/b+ab/c}$

+ Put t = ac/b + ab/c then it is easy to show that $\dfrac{a}{a^3+bc} \geq t$
(Substite t = a(1-a)/bc and compute we come exactly to $a - a^3 - bc \leq 0$)

+ Finally  $VT > t + \dfrac{4}{1+t}\geq 3$ (AM-GM)

For the last question, I have no idea

[quote="Namdung"]My hypothese: A = 23/7[/quote]

$A=\dfrac{23}{7}$ is wrong. Take a=b close to $0$ and c close to $1$

OK.

Your example also show that 3 is the best constant. The question is closed.

Namdung


Suppose $r$ is a real number for which

$\left \lfloor {r+ \frac{19}{100}} \right \rfloor + \left \lfloor {r + \frac{20}{100}} \right \rfloor + \left \lfloor {r + \frac{21}{100}} \right \rfloor + \cdots + \left \lfloor {r + \frac{91}{100}} \right \rfloor = 546$

Find $\lfloor {100r} \rfloor$. (For real $x$, $\lfloor {x} \rfloor$ is the greatest integer less than or equal to $x$.)

So basically "steps" are the greater integer function?

[color=cyan]I don't think I've seen this one before, but it's nice.

Standard notation for "step function," as you call it, is [x], or |[size=75]_[/size]x[size=75]_[/size]| (that is, the brackets without the bit at the top).[/color]

See the first problem in the [url=http://www.nsa.gov/programs/mepp/usamts/problems/problems15R1.pdf]most recent USAMTS[/url] to see what it looks like.

mathfanatic wrote:Source: 1991 AIME #6

Suppose r is a real number such that
step(r+19/100) + step(r+20/100) + step(r+21/100) + step(r+22/100) + step(r+23/100) + step(r+24/100) + ... + step(r+91/100) = 546

Find step(100r).

Note: step(x) = the greatest integer that is less than or equal to x.  For instance, step(2) = 2, step(2.5) = 2, step(2.998) = 2, and step(-2.95) = -3.

[hide] each term will give either [r] or [r]+1. Let us say that m of the terms gives ([r]+1). Since there are 73 terms total, the total sum will be 73r+m. 546=73*7+35. Thus [r] is 7, and 35 of the terms gives [r]+1. This means the terms up to (r+56/100) give [r] and the terms after (r+57/100) give [r]+1. Thus r is 7.43... [100r] is thus 743. Correct?[/hide]

[quote="tetrahedr0n"]thats what i got. my solution is slightly different: i divided 546 by 73 (the total number of terms) to find that each term was either 7 or 8. Then i multiplied the remainder of 546/73 by 73 to get the number of terms equal to 8. I found 38 terms are 7, 35 are 8. then the 39th term is exactly 8. so i had r + 57/100=8, from which r = 7 + 43/100 and 100r=743. Looking back, this is far less elegant than the other solution.[/quote]

that's like the same thing. except in reverse sort of. like where i multiplied, you divided. etc.

I've used [x] notation in the past, but I find that the step(x) causes fewer ambiguities.  So in answer to your question, yes.

P.S. Mysticman and tetrahedr0n are right.


Prove that if $ ax^{3}-ax^{2}+9bx-b=0$ has $ 3$ positive roots, then these three roots must be equal.

if the roots are $ m,n$ and $ k$ then from theorem of Viet we get
$ m+n+k=1$
$ mn+mk+nk=9b/a$ and
$ mnk=b/a$
but $ 1=m+n+k\geq 3\sqrt[3]{mnk}=3\sqrt[3]{b/a}\Rightarrow b/a\leq 1/27$
$ 9b/a=mn+mk+nk\geq 3\sqrt[3]{(mnk)^{2}}=3\sqrt[3]{(b/a)^{2}}\Rightarrow b/a\geq 1/27$
So $ b/a=1/27 \Rightarrow (m+n+k)^{2}=3(mn+mk+nk) \Rightarrow m=n=k$
So $ m=n=k$.


In a circle two chords AB and AC have the length 6.

and the radius of the circle is 5.......

Find the length of BC?

What i did----

Join A to O and extend it till it cuts BC at D.
Clearly, /_ADC=/_ODC=90

So, DC² + OD² = 25
and, DC² + (OD+5)²=36

Subtracting we get
OD = -8/5 :o  :o

Tell me where i am wrong.

Oh! You wrong because you draw picture not true. So it wrong at $ AD \equal{} OD \plus{} 5$. The true is $ AD \equal{} 5 \minus{} OD$.

Oh! you known? Ok? :rotfl:  :rotfl:  :rotfl:

[hide="Solution."]
$ AD^2\plus{}DC^2\equal{}36$
$ (5\minus{}AD)^2\plus{}DC^2\equal{}25$

$ \implies 10AD\minus{}25\equal{}11 \implies AD\equal{}\frac{18}{5} \implies DC\equal{}\frac{24}{5} \implies BC\equal{}\boxed{\frac{48}{5}}$
[/hide]


The units digit and the tens digit of every two-digit prime number less than 50 are interchanged. How many of the resulting numbers are prime?

The first digit must be $ 1$ or $ 3$ so it is not even.

$ 13,17,19,31,37$

Reverse digits:

$ 31,71,91,13,73$

$ 91$ is the only one not prime, so the answer is $ \boxed{4}$.

wouldn't the answer be 5?? because 11 is also a prime #...... rite?

yes, it would be [b]5[/b], because technically, 11's digits reversed is 11, a prime number.


hi can anyone help me with these problems link is below post answers here. THANK YOU

http://www.mathhelpforum.com/math-help/urgent-homework-help/49784-help-quick.html#post189876


Find all real solutions to the equation $\log_{x}(2-2x)+\log_{1-x}(2x) = 0$.

${f(x)=\log_{1-x}(2 x)+\log_{x}(2-2 x)}$
${f(x)=0}$

${1-x>0\land 2 x>0\land x>0\land 2-2 x>0}$

Simplifying this

${0<x<1}$

We can easy check

${f\left(x+\frac{1}{2}\right)=f\left(\frac{1}{2}-x\right)}$

Making a graph of this is easy to check that it have only a real root. By Symmetry and newton method.

We test  ${{x = \frac{1}{2}}}$

${f\left(\frac{1}{2}\right)=0}$

That's the right answer, but I don't really understand how you conclude that there aren't any other solutions.


The mean of the integers 9, $ x$, $ 2x\minus{}3$, and 7 is 7. Find the value of the smallest integer.


Soluciones de la semana 2:

Nivel 2:

1.a) (solucion por Franco de Peru) Por el teorema de Cauchy y usando que $p-a$, $p-b$, $p-c>0$, se tieen que:

$(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c})((p-a)+(p-b)+(p-c))\geq 9$ pero usando que $2p=a+b+c$ se concluye que:

$\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\geq \frac{9}{p}$ Como se queria mostrar.

b) (solucion de pascual de colmbia)

para probar $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$ supongamos que $a\leq b\leq c$. entonces $\frac{c}{a+b}< 1$ por desigualdad triangular, y:

$\frac{a}{b+c}+\frac{b}{c+a}\leq \frac{a}{a+b}+\frac{b}{a+b}=1$ Sumando estas dos desigualdades se obtiene la cota superior.

(solucion de Federico castillo de Colombia: Muy elegante esta parte)

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<a/p+b/p+c/p=2$ donde $p$ es el semiperimetro. ($p<b+c$, $a+b$, $c+a$)

Para la cota inferior, tenemos que por Cauchy:

 $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{(a+b+c)^2}{2ab+2bc+2ca}\geq \frac{3}{2}$

porque es equivalente a la popular:

$a^2+b^2+c^2 \geq ab+bc+ca$

Nivel #1 (soluciones Por Manuel de Puerto Rico)

1. Substituimos $ab+bc+ca$ en todos los 1 de la ecuacion, y obtenemos que $bc/a+ca/b+ab/c \geq a+b+c$. Pero esta ultima es cierta por la desigualdad de rearrangement con las secuencias $ab, ca,bc$ y $1/a, 1/b, 1/c$.

2. Si $a+b+c\geq ab+bc+ca$ entonces por Cauchy tenemos que ${\sum\frac{a}[b}=\sum\frac{a^2.c}{abc}\geq a+b+c$.
Por otra parte, $\sum\frac{a}{b}=\sum\frac{a^2.c}{abc}\geq abc$, usando el hecho de que $abc\leq 1$, asi que se sigue lo pedido.

3. Sabemos que $\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}=\frac{a^2}{a(b+2c)}+\frac{b^2}{b(c+2a)}+\frac{c^2}{c(a+2b)}$ y por Cauchy obtenemos que esto es $\geq \frac{(a+b+c)^2}{3(ab+bc+ca)}\geq 1$.

[color=blue][size=150]NIVEL 2: Prob 03[/size][/color]

[b]solucion por Franco de Peru[/b]
Sean las siguientes definiciones:
$S_n$: subconjunto cualquiera de $\{1,2,...,n\}$.
$S_a ^n : {a_i}i \in S_n$
$m_{S_a ^n}: (\sum_{a_k \in S_a ^n} a_k)^2$
$Z_n = \sum_{1 \leq i \leq j \leq n} (a_i+...+a_j)^2$

[b]Lema 1:[/b] Sean $a,b,c$ reales, entonces $2ac \leq (a+b+c)^2+b^2$.
 Demostracion:
Como $(a+b+c)^2 + b^2 = 2ac +(a+b)^2 + (a+c)^2$ y $x^2 \geq 0, \forall x \in R$  el lema esta claramente probado.

Probaremos el problema por inducción: $n=3$

Los $m_{S_a ^3}$ que son terminos de la sumatoria de $Z_3$ son
menores o iguales a este ultimo, ya que los terminos restantes son
no negativos ($x^2 \geq 0$). El unico $m_{S_a^3}$ que no es termino de la sumatoria es $(a_1 + a_3)^2$
pero por el lema 1 tenemos: $(a_1 + a_3)^2 = a_1^2 + a_3^2 + 2a_1a_3 \leq a_1^2 + a_3^2 + (a_1+a_2+a_3)^2 \leq Z_3$ ($x^2 \geq 0$).
Por lo que $(\sum_{k \in S} a_k)^2 \leq \sum_{1 \leq i \leq j \leq n} (a_i+....+a_j)^2$ es verdad para $n=3$.

Supongamos que cumple hasta un $n$ y demostraremos para $n+1$

[b]CASO I:[/b] $\exists l \in \{1,2,\cdots,n \}$ / $a_l,a_{l+1} \in S_a^{n+1}.$
$b_i = a_i, 1\leq i<l.$
$b_l = a_l + a_{l+1}.$
$b_i = a_{l+1}, l < i \leq n.$

El nuevo conjunto es $b_1, b_2,\cdots, b_n$ y $m_{S_a^{n+1}}$ es
igual a $m_{S_b^n},$ pues $(\cdots + a_l + a_{l+1} + \cdots)^2 =(\cdots b_l \cdots)^2.$
Ahora:
$Z_{n+1}= \sum_{1 \leq i \leq j < l} (a_i+...+a_j)^2 + \sum_{1\leq i \leq l < j \leq n+1} (a_i+...+a_j)^2 + \sum_{l+1 < i \leq j \leq n+1} (a_i+...+a_j)^2 + R.$

Donde $R$ es la suma de los terminos faltantes de $\sum_{1 \leq i\leq j \leq n+1} (a_i+...+a_j)^2.$

Luego:
$Z_{n+1}= \sum_{1 \leq i \leq j < l} (b_i+...+b_j)^2 + \sum_{1 \leq i \leq l < j \leq n+1} (b_{i-1}+...+b_{j-1})^2 +\sum_{l+1 < i \leq j \leq n+1} (b_i +...+ b_{j-1})^2 + R.$

Que es equivalente a:
$Z_{n+1}= \sum_{1 \leq i \leq j < l} (b_i+...+b_j)^2 + \sum_{1 \leq i \leq l \leq j \leq n} (b_{i-1}+...+b_{j-1})^2 + \sum_{l < i \leq j \leq n} (b_i +...+ b_{j-1})^2 + R.$

Y juntando las sumatorias:
$Z_{n+1}= \sum_{1 \leq i \leq j < n} (b_i+...+b_j)^2 + R.$

y como lo pedido a probar es $m_{S_a^{n+1}} \leq Z_{n+1}$, lo cual es igual a probar:

$m_{S_b^n} \leq \sum_{1 \leq i \leq j < n} (b_i+...+b_j)^2 + R.$

Como $R$ es la suma de elementos negativos y por hipotesis inductiva $m_{S_b^n} \leq \sum_{1 \leq i \leq j < n} (b_i+...+b_j)^2,$
la desigualdad esta resuelta, probando el paso de $n$ a $n+1$ en el caso I.

[b]CASO II:[/b] Si $\forall l / a_l no \in S_a^{n+1} \Rightarrow a_{l-1}, a_{l+1} \nin S_a^{n+1}, (a_0 = a_{n+1} \wedge a_{n+2} =a_1)$

Para $i \leq j \in S, j-i \geq 2,$ sea entonces:
$(\sum_{k \in S} a_k)^2 = \sum_{k \in S} {a_k}^2 + \sum_{i\leq j\in S} 2a_ia_j \cdots (\alpha)$

$k_{ij} = (a_{i+1} + \cdots + a_{j-1}).$

Luego, escribimos $Z_{n+1} = (\sum_{k \in S} a_k)^2 + \sum_{i\leq j \in S} \{(a_i + k_{ij} + a_j)^2 + k_{ij}^2\} + P \cdots (\beta),$
donde $P$ es la suma de los terminos de $Z_{n+1}$ que no estan
en: $\sum_{k \in S} {a_k}^2 + \sum_{i \leq j \in S} \{(a_i + k_{ij} + a_j)^2 + k_{ij}^2\}.$

Aclaramos que en los terminos de $\sum_{i\leq j \in S} \{(a_i +k_{ij} + a_j)^2 + k_{ij}^2 \}$
no hay terminos repetidos del $Z_{n+1},$ pues para los pares $(i;j) \neq (c;d)$  tenemos que son las sumas del $a_i$ al $a_j$ y del $a_{i+1}$ al $a_{j-1}$ en el primer termino. Por la condicion del caso II tenemos $i \neq c+1, i+1 \neq c,$ y las cuatro sumas son todas terminos distintos del $Z_{n+1}.$

Por el lema 1, tenemos que $2a_ia_j \leq (a_i + k_{ij} + a_j)^2 +k_{ij}^2$, $i\leq j \in S$,
tomando sumatoria tenemos:
$\sum_{i\leq j \in S} {2a_ia_j} \leq \sum_{i\leq j \in S} \{(a_i +k_{ij} + a_j)^2 + k_{ij}^2 \}$,
 luego como $P$ es la suma de elemtos no negativos, de ($\alpha$) y ($\beta$) se concluye:

($\sum_{k\in S} a_k)^2 = \sum_{k \in S} {a_k}^2 + \sum_{i\leq j \in S} {2a_ia_j} \leq \sum_{k \in S} {a_k}^2 + \sum_{i\leq j \in S} \{(a_i + k_{ij} + a_j)^2 + k_{ij}^2\} + P = Z_{n+1}.$

Lo cual es equivalente a: $m_{S_a^{n+1}} \leq Z_{n+1}$ que es lo pedido, probando el paso de
$n$ a $n+1$ en el caso II.

Luego como ambos casos son compplementarios, la induccion estacompleta y:

$(\sum_{k\in S} a_k)^2 \leq \sum_{1 \leq i \leq j \leq n} (a_i + \cdots + a_j)^2$, donde $S$ es un subconjunto cualquiera de
$\{1,2,\cdots,n\}$.

                                                                                [i]Editado por Carlosbr[/i]


I am not sure about this, but I think may be...

If $a^{2}+b^{2}+c^{2}=d^{2}$, then $(a,b,c,d)$ has the form $k(z^{2}-y^{2},2xz-2xy,2x^{2}+2yz,2x^{2}+y^{2}+z^{2})$, for $k,x,y,z\in\mathbb{Z}$

can someone reply this guy?

davron latipov

First, you prove that at least two of $a,b,c$ are even.

Now, assuming that $b$and $c$ are even, and according to Sierpinski, the solutions of the given equation are the quadruples of the form
$a=\frac{x^{2}+y^{2}-z^{2}}z, b = 2x, c=2y, d=\frac{x^{2}+y^{2}+z^{2}}z$

where $x,y$ are arbitrary natural integers, and $z$is a divisor of $x^{2}+y^{2}$ less than $\sqrt{x^{2}+y^{2}}$.
Moreover each solution is obtained exactly once in this way.

Référence :
W.Sierpinski, Elementary theory of numbers, North-Holland, p.66-69.

Pierre.

where is $n$?what happens if $z$ doesn't divides $x^{2}+y^{2}$?


calculus
integration
logarithms
calculus computations

$ \int_0^\infty\;\left(\;\cot^{\minus{}1}\,x\;\right)^3\;\textbf dx$  :D

I think the substitutuion of $ {x}{ \equal{} }{cot t}$ should work...

sorry edited:

\begin{eqnarray*}
\int_0^\infty \mathrm{arc}\cot^3 x dx & = & \int_0^{\left(\frac{\pi}{2}\right)^3} \cot y^{\frac{1}{3}} dy \\
                                                         & = & \int_0^\frac{\pi}{2} 3t^2 \cot t dt \\
                                                         & = & \frac{3}{8}\left(2\pi^2\log 2 -7\zeta(3)\right) \\
                                                         & = & \frac{3}{8}\left(\pi^2\log 4 -7\zeta(3)\right)
\end{eqnarray*}


Can we partition an equilateral triangle to equilateral triangles of different size?

has every of these part-triangles another size or have there only to exist two different triangles?

Naphthalin