LaTeX geometry Asymptote vector articles AoPSwiki integration

[size=150][b]Making Pictures and Inserting them in Latex [/b][/size]
Since a no. of queries have been asked on the topic of adding pictures in Latex, I am posting a topic which lists different programs and methods for inserting different files into Latex. My methods and programs are based on the use of Miktex and Texniccenter in Windows. So I do not know if these will work on your system and software. Please post your queries for your particular system.
Some general documents on inserting images in Latex are available at
1) [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictHow.php]http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictHow.php[/url]
2)[url=http://amath.colorado.edu/documentation/LaTeX/reference/figures.html#pdf]http://amath.colorado.edu/documentation/LaTeX/reference/figures.html#pdf[/url]
3)[url=http://www.haptonstahl.org/latex/use_insertinggraphics.html]http://www.haptonstahl.org/latex/use_insertinggraphics.html[/url]
4)[url=http://people.uncw.edu/hermanr/LaTeX/latex.htm]http://people.uncw.edu/hermanr/LaTeX/latex.htm[/url]

[u][b]MetaFont [/b][/u]

The original creation of the legendary Donald Knuth. It is a command based system rather than a point-and-click program. That is, Metafont is WYWIWYG and NOT WYSIWYG. Through people's recommendations etc, it appears Metafont is very useful for making excellent diagrams though it is a bit tedious. The only problem is that Metafont generates .mf files which are quite difficult to include in Latex. AT least I'm not aware of a simple method.
If you download Miktex, Metafont comes bundled with it. You will find (in the default case) the mf.exe program in C:\texmf\miktex\bin\ A userguide is available at [url=http://metafont.tutorial.free.fr/]metafont.tutorial.free.fr/[/url]. Donald Knuth's book "$MetaFont$" is available in non-compilable Tex format at [url=http://www.ctan.org/tex-archive/systems/knuth/mf/mfbook.tex]www.ctan.org/tex-archive/systems/knuth/mf/mfbook.tex[/url]

[u][b]MetaPost [/b][/u]

The problem of inclusion of the excellent metafont files into Latex is solved by Metapost which has the same functionality and method of input as Metafont ; but it produces Postscript files instead of .mf files. These .eps files can then easily be integrated into Latex.
Download available at [url=http://www.ctan.org/tex-archive/graphics/metapost/]www.ctan.org/tex-archive/graphics/metapost/[/url]
A windows based writer(as compared to a DOS based converter available above) is MPedit available at [url=http://www.ctan.org/tex-archive/graphics/metapost/contrib/tools/MPEdit/]www.ctan.org/tex-archive/graphics/metapost/contrib/tools/MPEdit/[/url]
Documentation is at [url=ftp://ibiblio.org/pub/packages/TeX/graphics/metapost/mpman.pdf]ftp://ibiblio.org/pub/packages/TeX/graphics/metapost/mpman.pdf
[/url]

[u][b]WinGCLC [/b][/u]

This is an excellent program. It is not point-and-click but based on commands. However, it doesn't have the functionality of Metapost or Metafont. Nevertheless, it is useful for beginners and gives excellent results. It has options for analytical geometry as well! It can be easily integrated into latex, This is done by putting the gclc.sty (included in download) in .../texmf/tex/latex/gclc . then just use the \input{filename.pic} command in Latex.
It is available(along with documentation) at [url=http://www.matf.bg.ac.yu/~janicic/gclc/]www.matf.bg.ac.yu/~janicic/gclc/[/url]

[u][b]KSEG [/b][/u]

Another good program for creating simple graphics. Not that many options. Could be useful for maybe Physics related uses. Nevertheless, the ability to manipulate geometry pictures by varying points and lengths of lines by clicking is very useful for understanding. it is WYSIWYG . Has an inbuilt .seg to .png converter that makes files easily added to Latex.
Available at [url=http://www.mit.edu/~ibaran/kseg.html]www.mit.edu/~ibaran/kseg.html[/url]
[b]Advanced Batch converter[/b](trial -- do you know where I can find the full version of a similar program? ) available at [url=http://www.batchconverter.com]www.batchconverter.com[/url]can make .eps files from .png that are even easier to put into LAtex.  Otherwise , you can always use [b]Adobe Photoshop[/b]

[u][b]Asymptote [/b][/u]
This is another command based program that works on the same principle as Metapost. Good. Available at
[url=http://asymptote.sourceforge.net/]asymptote.sourceforge.net/[/url]
Documentation at same link as above.

While some of the above programs will serve your purpose, other such programs are [b][url=http://tug.org/PSTricks/main.cgi/]PStricks[/url], [url=http://tug.org/PSTricks/main.cgi/]TPX[/url],[url=http://jpicedt.sourceforge.net/]Jpicedit[/url], [url=http://mars.netanya.ac.il/~samy/tkpaint.html]TKPaint[/url][/b]....
Most of these are simpler but have less feautures than the above. Of course there are many more and just google to find out. If you have any other recommendations, please post along with the appropriate link. If you think I forgot [b]Geometer's Sketchpad[/b], i did'nt include it for it is expensive, not easily available and it's files are difficult to include into LaTeX.

This looks nice, but can use some improvement. I await suggestions from you folks dealing with these problems ...

There are some other programs like InkSpace(open source scalable vector graphic editor) which can convert your drawings to directly .eps format and many other formats like latex file with PSTricks and also to .ps format.  This program can be downloaded freely from http://www.inkspace.org.  Other programs of this kind but not freeware are Mayura Draw (I think this is Trailware/Shareware), Xara X1(Trailware).

Your link needs to ne corrected . It is [url=http://www.inkscape.org/]www.inkscape.org/[/url]. I think the [b].[/b] at the end caused the trouble.

Also, Inkspace has good conversion capabilities and nice options. However, it is NOT a program fro geometry purposes for it does not have options like $\perp$ etc... Anyways, thanks sumyki!!!

I think you need to fix your link too, Rushil: inkscape $\neq$ inkspace.

I don't understand. My link is working perfectly!

OK- it's actually that the name is wrong in all mentions except that link.

That's Correct Rushil! Sorry for the spelling mistake and for the wrong address i have given.  Your URL  http://www.inkscape.org/ is correct.

This is my latest method:

KSEG to draw the picture and to export its image as a pdf file

Adobe Acrobat to crop the pdf.

Done :D

[quote="Rushil"]Are you sure KSEG has an export to pdf option?? Where is it???[/quote]
I don't think it does, but you can easily print a KSEG picture to a virtual PDF printer (those are free). That's the easiest way to go KSEG-->PDF.

Actually guys the easiest way to go, without installing any additional virtual printers or programs is the following.

PdfTex knows to include .png directly using the graphicx package. So you just put the .png file in the same directory as your .tex file and it will compile nice and easily.

To make a .png file, if you are working in KSEG just draw the picture, print screen, ctrl-v in paint and then save as PNG, and you're done. If you are working with Geometers' Sketchpad then you can actually select the drawing, ctrl-c (to copy it), ctrl-v to dump it into paint, then save as PNG.

I presume you mean, can you use LaTeX commands to create pictures on the forum? Yes, but provided you don't use any special packages and stick to the picture environment. For example:
$\setlength{\unitlength}{0.5cm}\begin{picture}(5,2) \put(5,0){\line(-1,0){5}}\put(0,0){\line(1,1){2}}\put(2,2){\line(3,-2){3}}\put(-0.5,-0.1){\small A}\put(1.9,2.1){\small B}\put(5.1,-0.1){\small C}\put(4,0.1){\tiny 30}\qbezier(3.8,0)(3.9,0.55)(4.1,0.55) \put(0.6,1){\footnotesize 2}\put(3.6,1){\footnotesize 3}\put(2.3,-0.6){\footnotesize 5}\put(-1,-0.2){\tiny \shortstack{i\\m\\p\\o\\s\\s\\i\\b\\l\\e}}\put(6,-0.2){\tiny \shortstack{t\\r\\i\\a\\n\\g\\l\\e}}\end{picture}$ or $\setlength{\unitlength}{2mm}\begin{picture}(30, 20) \linethickness{0.075mm}\multiput(0, 0)(1, 0){31}{\line(0, 1){20}}\multiput(0, 0)(0, 1){21}{\line(1, 0){30}}\linethickness{0.15mm}\multiput(0, 0)(5, 0){7}{\line(0, 1){20}}\multiput(0, 0)(0, 5){5}{\line(1, 0){30}}\linethickness{0.3mm}\multiput(5, 0)(10, 0){3}{\line(0, 1){20}}\multiput(0, 5)(0, 10){2}{\line(1, 0){30}}\end{picture}$

I recommend the geometry program Cinderella ([url]http://cinderella.de/tiki-index.php[/url]), which has a free educational version and can export directly to PNG or PDF.

The first error is that it tells that it can't find asymptote.sty

Sigh! Why won't you give us the details of what you've done? Have you installed asymptote? Did you follow the instructions at [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Getting_Started/Windows/Downloads_and_Installation]Asymptote: Getting Started/Windows/Downloads and Installation[/url] and the following page? Did you move asymptote.sty to the right place as shown in [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Using_Asymptote_in_LaTeX]Using Asymptote in LaTeX[/url]?

I installed asymptote.
I made the asymptote tool in texnic center
I made a folder named "asymptote" and I copied asymptoe.sty and asycolors.sty

But did you do the very important Steps 4 & 5 - refreshing the database so MiKTeX can find the files in the new folder?

Yes it is that. No harm in pressing the Refresh FNDB button again, in case that helps.
Exactly which folder are the asymptote sty files in? What happens if you copy them to the same folder as your tex document?

I put the asymptote.sty in C:\Program Files\MiKTeX 2.9\asymptote.
When I put them in the same folder, it still doesn't give the document.
I think the log file might be helpful.

[edit: posted wrong file]

[edit 2: Also, when I press alt+a, nothing seems to happen]

Not finding asymptote.sty in the same folder is very strange. The log is very helpful but the path where you have saved the document is full of strange characters:
**C:/Documents*and*Settings/ê¹€íƒœí˜•/My*Documents/Latex*Document/Test/Example*(Asymptote)/Example.tex
which shouldn't be there. The MiKTeX paths are fine.
Try saving the document to a folder with a simple name such as C:/tex and see if it compiles there with or without asymptote in the same folder. Try the example document at [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Using_Asymptote_in_LaTeX]Using Asymptote in LaTeX[/url]:
[code]\documentclass{article}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\begin{document}
Hello.
I like to make pics with Asymptote like this one:
\begin{figure}[h]
 \begin{asy}
  include graph;
  size(1inch);
  filldraw(circle((0,0),1),yellow,black);
  fill(circle((-.3,.4),.1),black);
  fill(circle((.3,.4),.1),black);
  draw(arc((0,0),.5,-140,-40));
\end{asy}
\end{figure}
\par It makes me happy,
since I can still type my normal LaTeX stuff around it:
\(\int_0^{\pi}{\sin{x}}\,dx=2\)
\end{document}[/code]

Sorry about the strange characters. It's just that it's. Korean

[quote="Rushil"]
[u][b]Asymptote [/b][/u]
This is another command based program that works on the same principle as Metapost. Good. Available at
[url=http://asymptote.sourceforge.net/]asymptote.sourceforge.net/[/url]
Documentation at same link as above.

While some of the above programs will serve your purpose, other such programs are [b][url=http://tug.org/PSTricks/main.cgi/]PStricks[/url], [url=http://tug.org/PSTricks/main.cgi/]TPX[/url],[url=http://jpicedt.sourceforge.net/]Jpicedit[/url], [url=http://mars.netanya.ac.il/~samy/tkpaint.html]TKPaint[/url][/b]....
Most of these are simpler but have less feautures than the above. Of course there are many more and just google to find out. If you have any other recommendations, please post along with the appropriate link. If you think I forgot [b]Geometer's Sketchpad[/b], i did'nt include it for it is expensive, not easily available and it's files are difficult to include into LaTeX.[/quote]
I have filled out the Asymptote tool as instructed on this page:
[url]http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Getting_Started/Windows/Interactive_Mode[/url]
But how can I use Asymptote within a LaTeX document? Currently I have Asymptote as a tool and when I press Alt-A while having a saved asymptote code, the black command screen pops up and shows me the output on Ghostview. However, when I build a document in TeXnic Center, only the code shows up with no asymptote output. Is there a package I must use?

[url]http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration[/url]

Mine doesn't work though.

I'm re-attempting to try this again.

So far, I have:
[list][*]Installed MikTex
[*]Installed Texnic Center
[*]Installed Asymptote
[*]Copied and pasted in the asymptote styles
[*]Refresh the database
[*]Created a file with [code]\documentclass{article}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\begin{document}
Hello.
I like to make pics with Asymptote like this one:
\begin{figure}[h]
 \begin{asy}
  include graph;
  size(1inch);
  filldraw(circle((0,0),1),yellow,black);
  fill(circle((-.3,.4),.1),black);
  fill(circle((.3,.4),.1),black);
  draw(arc((0,0),.5,-140,-40));
\end{asy}
\end{figure}\par It makes me happy,
since I can still type my normal LaTeX stuff around it:
\(\int_0^{\pi}{\sin{x}}\,dx=2\)
\end{document}[/code]
[*]Build the file[/list]
And then the result is: "Couldn't find asymptote.idx"

So who wants to help me?

Where did you save the asymptote style files? There is a [url=http://tex.stackexchange.com/questions/50911/using-miktex-portable-texmaker-and-asymptote-from-a-usb-drive]tex stackexchange post[/url] that I had found helpful.

I use TeXworks, and when I paste the LaTeX source file from AoPS wiki, it says: "does not recognize file "asymptote.sty"", "does not recognize file "graphix"", etc. [url=http://www.freesmileys.org/smileys.php][img]http://www.freesmileys.org/smileys/smiley-confused004.gif[/img][/url]


function
algebra
polynomial
real analysis
real analysis unsolved

let $f:R->R$ be a continuous fonction such that: $lim_{-\infty}f(x)=lim_{+\infty}f(x)$.
Prove that there exists a sequence of fraction $F_n=\frac{P_n}{Q_n}$ that converges uniformly towards $f$

[quote="alekk"]let $f:R->R$ be a continuous fonction such that: $lim_{-\infty}f(x)=lim_{+\infty}f(x)$.
Prove that there exists a sequence of [b]fraction[/b] $F_n=\frac{P_n}{Q_n}$ that converges uniformly towards $f$[/quote]

Should we assume that you are talking about rational functions? this is, $P_n$ and $Q_n$ should be assumed to be polynomials (probably of degree at most $n$).

Best regards,

oups sorry .. $P_n$ and $Q_n$ are real polynomials of arbitrary degree.  :)


function
limit
integration
calculus
calculus computations

Let $ f$ be a function such that
$ f(1) \equal{} 1$ and $ f'(x) \equal{} \frac {1}{x^2 \plus{} f(x)}$ for $ x\ge1$
Prove that $ f(x)$ has limit for $ x\to\infty$ and $ \lim_{x\to\infty}f(x) < 1 \plus{} \frac {\pi}{4}$

I don't see a typo. The problem is correct.

Note that $ f'(x)>0,$ so $ f(x)$ is increasing. This tells us two things. The first is that $ f(x)> 1$ for all $ x> 1.$ The second is the "monotone function alternative": either $ f(x)$ tends to $ \infty$ or it has a limit as $ x\to \infty$

Since we have $ f(x)> 1,$ we then have $ f'(x)<\frac1{x^2\plus{}1}.$ That means that

$ \lim_{x\to\infty}f(x)<f(1)\plus{}\int_1^{\infty}\frac{dx}{x^2\plus{}1}\equal{}1\plus{}\frac{\pi}4.$

Shouldn't be stated $ f(x)$ is defined for all $ x\geq 1$ in the problem?

This is a differential equation, to which the existence-uniqueness theorem applies. That theorem only guarantees local solutions, but the only way this could fail to have a global solution on $ [1,\infty)$ would be for the solution to tend to infinity in finite time - but the very estimates we have used show that that cannot happen.

Here's an interesting question for numerical analysis: to a reasonable level of accuracy, estimate $ \lim_{x\to\infty}f(x).$

I gave my own numerical question a shot.

Method: Use a Runge-Kutta 4th order method out to a certain point. From that, if we get the approximation $ f(a)=b,$ find the solution to $ f'(x)=\frac1{x^2+b}$ beyond that and thus estimate the limit as $ b+\frac1{\sqrt{b}}(\frac{\pi}2-\arctan(\frac{a}{\sqrt{b}}).$

Do this for several step sizes and intervals, at each step both halving the step size and doubling the length of the interval. In what I did, this ranged from $ a=2,\Delta x=1/8$ to $ a=32,\Delta x=1/128.$

Then run a Richardson extrapolation table from those values.

(I was using a spreadsheet, which is the wrong tool. This really should be done in a more procedural environment such as a programming language or MATLAB.)

It looks to me that I probably got the answer to six decimal digits as $ \lim_{x\to\infty}f(x)\approxeq 1.758598,$ although I invite anyone who wants to check and perhaps improve this value.


search
combinatorics proposed
combinatorics

Count all those permutations $ \sigma$ of $ \{1,\dots,n\}$ that do not preserve neighbourhood.  It means $ |{\sigma(i)\minus{}\sigma(i\plus{}1)|\geq 2}$ for $ i\equal{}1,\dots,n\minus{}1$.

[quote="feniks"]Count all those permutations $ \sigma$ of $ \{1,\dots,n\}$ that do not preserve neighbourhood.  It means $ |{\sigma(i) - \sigma(i + 1)|\geq 2}$ for $ i = 1,\dots,n - 1$.[/quote]

Put $ A_i$, i=1,2,...,n-1: set of all permutations such that $ |{sigma(i) - \sigma(i + 1)|< 2}$

Use Principle of Inclusion and Exclusion we have the number of all permutations that do not preserve neighbourhood is

$ S= n!-|\bigcup_{i=1}^{n-1} A_i|= n!+ \sum_{1\le<i_1<...<i_k\le n-1}(-1)^{k}|A_{i_1}\cap A_{i_1}...\cap A_{i_k}|$

Finally, we should calculate $ |A_{i_1}\cap A_{i_1}...\cap A_{i_k}|$, may be it's easy.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=612482435&t=65890
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=612482435&t=192170
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=612482435&t=72112

I found these by searching for "A002464," the sequence number of this sequence in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].


We have a set $A$ with $n$ elements and $p<n$.
Find the maximum number of subsets of $A$ such that each two of them have in common at most $p$ elements.

I think the number is $\binom{n}{0}+...+\binom{n}{p+1}$. This will surely work (we take the subsets with at most p+1 elements). Now assume we have more. We are only interested in subsets with at least p+1 elements. By the assumtion, we will have more than $\binom{n}{p+1}$ subsets with such a cardinal. We associate each of them to a subset with p+1 elements and in any case 2 will have the same associated subset. Therefore, their intersection has a cardinal greater than p, a contradiction.
 :)


Two fair eight-sided dice have their faces numbered from 1 to 8. What is the expected value of the the sum of the rolls of both dice?

I don't understand the problem...there can be numerous ways how the sum can turn out to be.

Alright, this is how we do it.  We first find the expected value for one die.  That is simply \[ \frac{1}{8}(1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6\plus{}7\plus{}8)\equal{}4.5\]
Since we are rolling two dice, that means that our expected value is doubled, meaning our final answer is $ \boxed{9}$

A simpler way is to multiply the expected value of a single die by 2.

The expected value is the average of the largest and the smallest possible value. so (2+16)/2 = 9?


در یک مسابقه تلویزیونی مجری ۳ جعبه را که تنها یکی از آن‌ها پر از پول است، در برابر شما قرار می دهد. بعد از این‌که شما یکی از آن‌ها را انتخاب کردید، مجری یکی از دو جعبه باقی مانده را باز می‌کند و نشان می‌دهد که خالیست، آن‌گاه از شما می‌پرسد که آیا مایلید جعبه‌ای را که انتخاب کرده‌اید با جعبه باقی مانده عوض کنید؟ چه خواهید کرد؟ برای تصمیمتان دلیل بیاورید.

nabayad jabaro avaz konim
dalilesham ba hesab kardan ehteml dar miad
nemidonam ama fek mikonam age jabamono avaz nakonim 2/3 ehtemal bord darim
 :lol:  ;)

man ke omran avaz nemikonam.akhe hameye mojrea ye rigi bekafsheshoon hast.
 :rotfl:


Given two circle $ (O_1)$ and $ (O_2)$.Find the locus of points M such that $ P_{M/(O_1)}\equal{} \minus{}P_{M/(O_2)}$ which $ P_{M/(O)}$ is power of M wrt circle $ (O)$

For any point $ X$ in the plane, its power wrt $ O_1$ is $ P_{M/O_1} \equal{} XO_1^2 \minus{} R_1^2$,  where $ R_1$ is the radius of $ O_1$.  We therefore have $ XO_1^2 \plus{} XO_2^2 \equal{} R_1^2 \plus{} R_2^2$,  which is the equation of a circle with center in the midpoint of $ O_1O_2$ and radius $ \sqrt {\frac {R_1^2 \plus{} R_2^2}{2} \minus{} \frac {O_1O_2^2}{4}}$.  Note that the locus is empty if $ O_1O_2^2>2(R_1^2\plus{}R_2^2)$.  In particular, if the circles have equal radii, they must either touch or intersect for the locus to be nonempty, the locus reducing to the tangency point in the first case.

Forgot to mention earlier, clearly if both circles intersect or are tangent, the locus must include the intersecion or tangency points, since both powers are zero...  If the circles are tangent but do not have equal radii, then the locus is a circle tangent to the other two at their common tangency point, entirely in the interior of the bigger of the two.

Final note: for those who do not know that the locus of points $ X$ whose sum of squares of distances to two given points $ P,Q$ gives a circle centered at the midpoint of $ PQ$, consider the length of the median $ XM$ in triangle $ PXQ$; it is well known that $ XM^2 \equal{} \frac {XP^2 \plus{} XQ^2}{2} \minus{} \frac {PQ^2}{4}$, and if $ XP^2 \plus{} XQ^2$ is constant, so is $ XM$, hence the center is $ M$, and the radius is $ \sqrt {\frac {XP^2 \plus{} XQ^2}{2} \minus{} \frac {PQ^2}{4}}$.


[quote="myc"]but how'd you know that $24^{10}=24^{4}mod 100$ ?[/quote]

i think you just add 2 to both sides leaving

$24^{10}\equiv24^{4}\pmod{100}$

no i mean how do you come up with $24^{4}$ from $24^{10}$?


analytic geometry
geometry
calculus
trigonometry

[quote="drunner2007"]Without using calculus, find the area of the set of points $(x,y)$ in the cartesian plane that satisfy $x^{2}+y^{2}\leq (\pi)^{2}$ and $y\geq \sin x$.[/quote]

[hide="Erhmm.."]The area of the region is half the area of the circle because you can map the part below the x axis to the part not satisfying $y\geq \sin x$ above the x axis and since $\sin x$ is symmetric about the y axis it works. Then the area of the circle is $\pi^{3}$ and the answer is $\frac{\pi^{3}}{2}$? [/hide]

Edit: Stupid.. I forgot the other side..

The area in question lies above the period of the sinusoid on $[-\pi,\pi]$ and below the semicircle $x^{2}+y^{2}=\pi^{2}, y\geqslant 0$. Since the area cut out of the semicircle by the positive halfperiod of the sinusoid is exactly the same as the area added to it by the negative halfperiod, the area is simply $A={r^{2}\pi\over 2}={\pi^{3}\over 2}$

[hide]Graphing it, the shaded area above the x-axis is equal to the nonshaded area below the x-axis. We see that the area is half the area of the circle w/ radius pi, or $\frac{\pi^{3}}{2}$.[/hide]


projective geometry
trigonometry
geometry proposed
geometry

Let $C_{1}$, $C_{2}$ and $C_{3}$ be three circles that does not intersect and non of them is inside another. Suppose $(L_{1},L_{2})$, $(L_{3},L_{4})$ and $(L_{5},L_{6})$ be internal common tangents of $(C_{1}, C_{2})$, $(C_{1}, C_{3})$, $(C_{2}, C_{3})$. Let $L_{1},L_{2},L_{3},L_{4},L_{5},L_{6}$ be sides of polygon $AC'BA'CB'$. Prove that $AA',BB',CC'$ are concurrent.

First of all observe that perpendiculars at $A$ to $O_{2}O_{3}$,  at $B$ to $O_{1}O_{2}$ and at $C$ to $O_{1}O_{3}$ are concurrent in $R$.
This goes from the fact that these lines are radical axes for our circles and they should meet at the radical center.
Also it is clear that this perpendiculars are bisectors for angles $\angle B'AC', \angle C'BA', \angle A'CB'$.
Thus we can conclude that intersection of this bisectors - $R$ is the center of the circle that is inscribed in our hexagon $ABCA'B'C'$.
It follows from Brianchon Theorem that $AA'$, $BB'$, $CC'$ are concurrent.

Some other thoughts:
From the upper thoughts it can be observed that $O_{1}A', O_{2}C', O_{3}B'$ are concurrent (look the given picture) and

$BC'\cdot AB'\cdot CA' =C'A\cdot B'C\cdot A'B$.

Also $BC'+AB'+CA' =C'A+B'C+A'B$.

It is interesting if $BC'=B'C$ and similarly other pairs are equal.
Also if somebody can prove that $B'C'\parallel BC$, then the result will
follow from Desargues Theorem.
So, the problem can be investigated further.

[quote="prowler"]First of all observe that perpendiculars at $A$ to $O_{2}O_{3}$,  at $B$ to $O_{1}O_{2}$ and at $C$ to $O_{1}O_{3}$ are concurrent in $R$.
This goes from the fact that these lines are radical axes for our circles and they should meet at the radical center.
vestigated further.[/quote]
Can you check it for me please? They are not radical axes of our circles and I think it's not true

No, they are not radical axis, and are not concurrent generally.
Your solution is not correct.

Sorry guys, I was asleep probably. Of course it is not correct and my desire
to prove that there is a circle inscribed in hexagon was false. It is not radical axes - it is just point of similitude.

Denote points $ X=L_1\cap L_6$, $ Y=L_2\cap L_3$ and $ Z=L_4\cap L_5$. We have \begin{eqnarray*}\frac{\sin A'AB}{\sin A'AC}&=&\frac{AC}{AB}\cdot\frac{[AA'B]}{[AA'C]}=\frac{AC}{AB}\cdot\frac{A'B}{A'C}\cdot\frac{\sin AA'B}{\sin AA'C}\\ &=&\frac{AC}{AB}\cdot\frac{A'B}{A'C}\cdot\frac{\sin AA'B}{\sin Z}\cdot\frac{\sin Z}{\sin Y}\cdot\frac{\sin Y}{\sin AA'C}\\ &=& \frac{AC}{AB}\cdot\frac{A'B}{A'C}\cdot\frac{AZ}{AA'}\cdot\frac{\sin Z}{\sin Y}\cdot\frac{AA'}{AY}\\ &=&\frac{AC}{AB}\cdot\frac{A'B}{A'C}\cdot\frac{AZ}{AY}\cdot\frac{\sin Z}{\sin Y}\end{eqnarray*}
Therefore, \begin{eqnarray*} \prod\frac{\sin A'AB}{\sin A'AC}&=&\prod\frac{A'B}{A'C}\cdot\frac{AZ}{AY}=\prod\frac{A'B}{A'C}\cdot\frac{XB}{XC}\\ &=&\prod\frac{[O_1A'B]\cdot[O_1XB]}{[O_1A'C]\cdot [O_1XC]}\\ &=&\prod\left(\frac{O_1B}{O_1C}\right)^2\cdot\prod\left[\frac{\sin A'O_1B}{\sin A'O_1C}\cdot\frac{\sin XO_1B}{\sin XO_1C}\right]\end{eqnarray*}
On the other hand, we can easily to prove that lines $ O_1A'$ and $ O_1X$ are isogonal conjugates with respect to angle $ O_3O_1O_2$. Hence, we conclude that $ \prod\frac{\sin A'AB}{\sin A'AC}=1$, i.e. lines $ AA'$, $ BB'$, $ CC'$ are concurrent, due to the trigonometric form of Ceva's theorem.


function
calculus
integration
limit
trigonometry
real analysis
calculus open

we are given $f,g$ - continuous periodic functions with period 1.
prove that
\[\lim_{n\to \infty}\int_{0}^{1}f(x)g(nx)\: dx\quad=\quad\int_{0}^{1}f(x)\: dx\cdot\int_{0}^{1}g(x)\: dx\]

Here is a method of attach:

Let $x_{n}=\int_{0}^{1}f(x)g(nx)dx$. Then we get, using the substitution $nx=y$, that

$x_{n}=\frac{1}{n}\int_{0}^{n}f(y/n)g(y)dy$. Using the periodicity of both $f$ and $g$ we get after some calculations that

$x_{n}=\int_{0}^{1}\frac{f(\frac{u}{n})+f(\frac{u+1}{n})+\cdots+f(\frac{u+(n-1)}{n})}{n}g(u)du$.

Suffices to show that

$\lim_{n\rightarrow \infty}\frac{f(\frac{u}{n})+\cdots+f(\frac{u+(n-1)}{n})}{n}=\int_{0}^{1}f(x)dx$.

To show this we use the Fourier expansion of f, namely:

$f(x)=\frac{a_{0}}{2}+\sum_{m=1}^{\infty}a_{m}\cos 2\pi mx+b_{m}\sin 2\pi mx$.

A calculation shows that:

$\sum_{k=0}^{n-1}f(\frac{u+k}{n})=n\frac{a_{0}}{2}$, and hence

$\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{u+k}{n})\rightarrow \frac{a_{0}}{2}=\int_{0}^{1}f(x)dx$.

I hope that there are no computational mistakes.

It's a relative of the Riemann-Lebesgue lemma; continuity is quite a bit more than we need, and we don't need to assume that $n$ is an integer or that $f$ is periodic.

First, consider the case where $f$ is $C^{1}$. We can write $G(x)=\int_{0}^{x}g(t)\,dt=Ax+H(x)$, where $A=\int_{0}^{1}g(x)\,dx$ is the average value of $g$ and $H$ is a continuous periodic function. Let $h(x)=g(x)=A$. Now $\int_{0}^{1}f(x)g(nx)=\int_{0}^{1}f(x)(h(nx)+A)\,dx=A\int_{0}^{1}f(x)\,dx+\int_{0}^{1}f(x)h(nx)\,dx$. We want to show that the latter integral tends to zero.

Integrate by parts: $\int_{0}^{1}f(x)h(nx)\,dx=\left[\frac1nf(x)H(nx)\right]-\int_{0}^{1}\frac1nf'(x)H(nx)\,dx$. Now all of $f,f',H$ are bounded, so the $\frac1n$ factors make everything go to zero.

For general $f$, note that we can always find a differentiable$C^{1}$ function $f^{*}$ such that $\int_{0}^{1}|f(x)-f^{*}(x)|\,dx<\epsilon$. If $g$ is bounded by some $M$, we get $\left|\lim_{n\to\infty}\int_{0}^{1}f(x)g(nx)\,dx-\lim_{n\to\infty}\int_{0}^{1}f^{*}(x)g(nx)\,dx\right|\le M\epsilon$. A sequence of $f^{*}$ converging to $f$ gives the desired result.

two posts with proofs in 2 hours, nice :)

thank you very much!!


A non-zero digit is chosen in such a way the the probability of choosing digit d is $log_{10}
(d+1) - log_{10} d.$ The probability of choosing 2 is exactly half that of choosing what set of  digits?

[hide]
The probability of choosing two is log(3/2). If there is some set of digits then the probability of choosing them is 2log(3/2)=log(9/4). We notice that

log9/4= log9/8+log8/7+log7/6+log6/5 + log 5/4, which is the sum of the probability of choosing the digits 8, 7, 6, 5, and 4.

Thus the set of digits is {4, 5, 6, 7, 8}.
[/hide]


inequalities
invariant
inequalities proposed

I wanted to propose this for a competition but then I spoilt it, so it can be posted here:

Let a, b, c be three positive reals. Prove the inequalities

$\frac{a}{2b+a}+\frac{b}{2c+b}+\frac{c}{2a+c}\geq\frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}$

and

$\frac{a^2}{2b+a}+\frac{b^2}{2c+b}+\frac{c^2}{2a+c}\geq\frac{a^2}{2a+b}+\frac{b^2}{2b+c}+\frac{c^2}{2c+a}$.

  Darij

Darij, i'm very very sorry to tell you this :( , but your first inequality it is well-known. In fact, I'll prove that:

                            \[ \frac{a}{2b+c}+\frac{b}{2c+a}+\frac{c}{2a+b}\geq 1\geq\frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}.  \]

   Anyway, first let's prove that: \[ \frac{a}{2b+c}+\frac{b}{2c+a}+\frac{c}{2a+b}\geq 1.  \]

   By Cauchy, we have: \[ [(2ab+ac)+(2bc+ab)+(2ca+bc)]\cdot (\frac{a}{2b+c}+\frac{b}{2c+a}+\frac{c}{2a+b})\geq (a+b+c)^{2}.  \]

  Now, using the well-known inequality $(a+b+c)^{2}\geq 3(ab+bc+ca)$ the inequality follows immediately.

  Let's prove the second one: \[ \frac{a}{2a+b}+\frac{b}{2b+c}+\frac{c}{2c+a}\leq 1.  \]
  Well, this inequality is equivalent with: \[ \frac{2a}{2a+b}+\frac{2b}{2b+c}+\frac{2c}{2c+a}\leq 2,  \]
  or \[ \frac{b}{2a+b}+\frac{c}{2b+c}+\frac{a}{2c+a}\geq 1.  \]
  But, like the first one, this follows from Cauchy. And, by the way, this second inequality was given in Moldova Olympiad in 2002. ;)

Cezar, thanks for the solution. In fact, it was a kind of surprise to me, since I didn't expect the problem to be that simple (in fact, due to a miscalculation I thought 1 couldn't be interpolated between the sides of the inequality). My solution was more difficult; to be brief, I noted that

$\frac{a}{2b+a}-\frac{a}{2a+b}=\frac{a-b}{3\left(2b+a\right)}+\frac{\left(a-b\right)^2}{3\left(2a+b\right)\left(2b+a\right)}\geq\frac{a-b}{3\left(2b+a\right)}$,

so it remained to show that $\sum\frac{a-b}{2b+a}\geq 0$, where the $\sum$ sign stands for cyclic sums, but this is equivalent to $\sum\frac{a}{2b+a}\geq 1$ and can be done by Cauchy-Schwarz.

Note that you slightly misread my problem, proving the inequality $\frac{a}{2b+c}+\frac{b}{2c+a}+\frac{c}{2a+b}\geq 1$ instead of the required inequality $\frac{a}{2b+a}+\frac{b}{2c+b}+\frac{c}{2a+c}\geq 1$, but in fact the required inequality can be proven in the same way (even a bit easier: one doesn't have to apply $\left(a+b+c\right)^2\geq 3\left(bc+ca+ab\right)$ in the final step).

[quote="cezar lupu"]And, by the way, this second inequality was given in Moldova Olympiad in 2002. ;)[/quote]

Here you have probably been speaking about the second inequality from your proof (indeed, now I see it was a problem from Moldova 2002: http://www.mathlinks.ro/Forum/viewtopic.php?t=3672 ). Now, what do you think about the second inequality I proposed, namely

$\frac{a^2}{2b+a}+\frac{b^2}{2c+b}+\frac{c^2}{2a+c}\geq\frac{a^2}{2a+b}+\frac{b^2}{2b+c}+\frac{c^2}{2c+a}$ ?

It's very easy, but I would like to see a different solution.

  Darij

for 1:
I think we have:
$\prod_{i=1}^nx_i=1,,x_i>0$
then $\sum_{i=1}^n \frac{1}{1+(n-1)x_i} \ge 1 \ge \sum_{i=1}^n \frac{1}{n-1+x_i}$

for 2:
I hope you think it is a nice solution.
I proved
$\frac{a^2}{2b+a}-\frac{a^2}{b+2a}=\frac{a^2(a-b)}{(2b+a)(b+2a)} \ge \frac{a-b}{9}$
by this method,
I think we can prove:
$\frac{a^n}{2b+a}+\frac{b^n}{2c+b}+\frac{c^n}{2a+c}\geq\frac{a^n}{2a+b}+\frac{b^n}{2b+c}+\frac{c^n}{2c+a}$ $n \ge 2,n \in R$
also can be general for n variable and general for the denominator.

[quote="Soarer"][quote="zhaobin"]$\frac{a^2(a-b)}{(2b+a)(b+2a)} \ge \frac{a-b}{9}$
[/quote]

How can you think of this??[/quote]
I don't know,just want to find a simple proof :P
And I think it will be a nice problem if we right as following:
$\frac{a^n}{2b+a}+\frac{b^n}{2c+b}+\frac{c^n}{2a+c}\geq\frac{a^n}{2a+b}+\frac{b^n}{2b+c}+\frac{c^n}{2c+a} ,n \ge 1, n \in N$
Don't you think so?we must complete the solution with two sitition.(at least now)

And another problem:
 $\frac{a^k}{2b+a}+\frac{b^k}{2c+b}+\frac{c^k}{2a+c}\geq\frac{a^k}{2a+b}+\frac{b^k}{2b+c}+\frac{c^k}{2c+a} ,1<k <2$

[quote="Soarer"][quote="zhaobin"]$\frac{a^{2}(a-b)}{(2b+a)(b+2a)}\ge \frac{a-b}{9}$
[/quote]

How can you think of this??[/quote]

I find this step quite straightforward, particularly since I had the same solution. One is searching for a constant k such that $\frac{a^{2}\left(a-b\right)}{\left(2b+a\right)\left(2a+b\right)}\geq k\left(a-b\right)$ holds for every two positive reals a and b (since finding such a constant would immediately solve the problem). But requiring that $\frac{a^{2}\left(a-b\right)}{\left(2b+a\right)\left(2a+b\right)}\geq k\left(a-b\right)$ is equivalent to requiring that $a^{2}\geq k\left(2b+a\right)\left(2a+b\right)$ for $a\geq b$ and $a^{2}\leq k\left(2b+a\right)\left(2a+b\right)$ for $a\leq b$. Now, it is clear how to proceed, and the result is $k=\frac19$.

[quote="zhaobin"]$\frac{a^{n}}{2b+a}+\frac{b^{n}}{2c+b}+\frac{c^{n}}{2a+c}\geq\frac{a^{n}}{2a+b}+\frac{b^{n}}{2b+c}+\frac{c^{n}}{2c+a},n \ge 1, n \in N$[/quote]

Thanks! I actually thought about such generalizations, but was too lazy to investigate. As I see now, I should have. In fact, while trying to prove this inequality, I remembered a trick I used in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=204464#p204464]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=149630 post #5[/url] and I will now formulate as a lemma:

[color=blue][b]Lemma 1.[/b] Let a, b, c be three reals; let x, y, z, u, v, w be six nonnegative reals. Assume that the number arrays (a; b; c) and (x; y; z) are equally sorted, and

$u\left(a-b\right)+v\left(b-c\right)+w\left(c-a\right)\geq 0$.

Then,

$xu\left(a-b\right)+yv\left(b-c\right)+zw\left(c-a\right)\geq 0$.[/color]

[i]Proof of Lemma 1.[/i] Since the statement of Lemma 1 is invariant under cyclic permutations (of course, when these are performed for the number arrays (a; b; c), (x; y; z) and (u; v; w) simultaneously), and since the number arrays (a; b; c) and (x; y; z) are equally sorted, we can WLOG assume that
- either $a\geq b\geq c$ and $x\geq y\geq z$,
- or $a\leq b\leq c$ and $x\leq y\leq z$.
In either case, $\left(x-z\right)\left(a-b\right)\geq 0$ (since the numbers x - z and a - b have the same sign: either both $\geq 0$, or both $\leq 0$), and that $\left(y-z\right)\left(b-c\right)\geq 0$ (since the numbers y - z and b - c have the same sign: either both $\geq 0$, or both $\leq 0$). Now,

$xu\left(a-b\right)+yv\left(b-c\right)+zw\left(c-a\right)$
$=u\underbrace{\left(x-z\right)\left(a-b\right)}_{\geq 0}+v\underbrace{\left(y-z\right)\left(b-c\right)}_{\geq 0}+z\underbrace{\left(u\left(a-b\right)+v\left(b-c\right)+w\left(c-a\right)\right)}_{\geq 0}$
$\geq 0$,

and Lemma 1 is proven.

Now, we will show a generalization of Zhaobin's above inequality:

[color=blue][b]Theorem 2.[/b] Let a, b, c be three positive and x, y, z three nonnegative reals such that the number arrays (a; b; c) and (x; y; z) are equally sorted. Then,

$\frac{xa}{2b+a}+\frac{yb}{2c+b}+\frac{zc}{2a+c}\geq\frac{xa}{2a+b}+\frac{yb}{2b+c}+\frac{zc}{2c+a}$.[/color]

[i]Proof of Theorem 2.[/i] We will use the $\sum$ notation for cyclic sums. Then, we have to prove the inequality $\sum\frac{xa}{2b+a}\geq\sum\frac{xa}{2a+b}$. We equivalently transform this inequality:

$\sum\frac{xa}{2b+a}\geq\sum\frac{xa}{2a+b}\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ \sum\left(\frac{xa}{2b+a}-\frac{xa}{2a+b}\right)\geq 0$
$\Longleftrightarrow\ \ \ \ \ \sum x\frac{a}{\left(2a+b\right)\left(2b+a\right)}\left(a-b\right)\geq 0$.

Particularly, applying this to x = y = z = 1, we get

$\sum\frac{a}{2b+a}\geq\sum\frac{a}{2a+b}\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ \sum \frac{a}{\left(2a+b\right)\left(2b+a\right)}\left(a-b\right)\geq 0$.

Since $\sum\frac{a}{2b+a}\geq\sum\frac{a}{2a+b}$ is proven (see e. g. posts #2 and #3), we thus have shown that $\sum \frac{a}{\left(2a+b\right)\left(2b+a\right)}\left(a-b\right)\geq 0$. Consequently, applying Lemma 1 (with $u=\frac{a}{\left(2a+b\right)\left(2b+a\right)}$, $v=\frac{b}{\left(2b+c\right)\left(2c+b\right)}$, $w=\frac{c}{\left(2c+a\right)\left(2a+c\right)}$), we get $\sum x\frac{a}{\left(2a+b\right)\left(2b+a\right)}\left(a-b\right)\geq 0$. As showed above, this is equivalent to $\sum\frac{xa}{2b+a}\geq\sum\frac{xa}{2a+b}$. Thus, Theorem 2 is proven.

  Darij

ok,thanks!Darij. :)
and have think the minimum k,such that for every positive number $a,b,c$ we have:
$\frac{a^k}{2b+a}+\frac{b^k}{2c+b}+\frac{c^k}{2a+c}\geq\frac{a^k}{2a+b}+\frac{b^k}{2b+c}+\frac{c^k}{2c+a}$ :)

For m variables
I thinks it still ture that:m positive numbers $a_1,a_2,\cdots,a_m,m \ge 2$
then we have:
$\frac{a^n_1}{a_1+2a_2}+ \frac{a^n_2}{a_2+2a_3}+\cdots+\frac{a^n_n}{a_n+2a_1} \ge \frac{a^n_1}{2a_1+a_2}+ \frac{a^n_2}{2a_2+a_3}+\cdots+\frac{a^n_n}{2a_n+a_1}$
for $n \ge 2$
it can be proved  by:$\frac{a^n_1}{a_1+2a_2}- \frac{a^n_1}{2a_1+a_2} \ge \frac{a^{n-2}_1(a_1-a_2)} 9$
for $n=1$ it can be proved by what I wrote in #4.
but I can't slove $1<n<2$ I don't think your(Darij's) lemma can work for m variables. :)
what about your idea?

[quote="spanferkel"][quote="zhaobin"]

For m variables
I thinks it still ture that:m positive numbers $a_1,a_2,\cdots,a_m,m \ge 2$
then we have:
$\frac{a^n_1}{a_1+2a_2}+ \frac{a^n_2}{a_2+2a_3}+\cdots+\frac{a^n_n}{a_n+2a_1} \ge \frac{a^n_1}{2a_1+a_2}+ \frac{a^n_2}{2a_2+a_3}+\cdots+\frac{a^n_n}{2a_n+a_1}$
for $n \ge 2$ [/quote]
I think you mean $a_m$ for the last one. it should be
$\frac{a^n_1}{a_1+2a_2}+ \frac{a^n_2}{a_2+2a_3}+\cdots+\frac{a^n_m}{a_m+2a_1} \ge \frac{a^n_1}{2a_1+a_2}+ \frac{a^n_2}{2a_2+a_3}+\cdots+\frac{a^n_m}{2a_m+a_1}$
for $n \ge 2$[/quote]
yes,thanks
and darij,have you think about it?


group theory
abstract algebra
number theory proposed
number theory

Hmm...Nobody interests to solve it? :blush:
Try it!It is very interesting! :)

At least it is an elliptic curve with rank 0 and its torsion subgroup is an abelian group isomorphic to C3 says Sage.


Adults are holding children, waiting their turn. The children are handed (one at a time, usually) to a man, who holds them while a woman shoots them. If the child is crying, the man tries to stop the crying before the child is shot.

Explain why this a realistic situation that happens all the time.

[hide]They are receiving vaccinations (or some other form of injection).[/hide]

wow that was a nice answer lol hmm but if they're crying wouldn't it make the child ignore the pain??? that aside i think that is the correct answer  :lol:

That's what I thought of when I saw this, and I'm pretty sure that that is correct, but would you really call giving someone a shot shooting them? I thought that to shoot someone, it was with a gun, and if you were giving them a shot, you were......giving them a shot.

Think of it this way. A doctor doesn't say "I'm going to shoot you", he/she says "I'm going to give you a shot".

[hide]They are getting their photo taken with some man, the photographer is a woman. The child crying will ruin the picture!

What I thought of were young kids waiting in line with their parents to have their picture taken with "Mickey Mouse", which would fit that description.

The woman is "shooting a picture" of the children.[/hide]

yeah!! i was JUST gonna say that coz i saw the same thing happen...i mean the "MAN" could be a clown, or a politician or someone  and obviously u don't want the baby crying in the pic...but then isn't it shooting a movie and not shooting a picture??


When I'm saving a page, it saves OK but instead of showing the new version, my computer displays the following message:

Database error
From AoPSWiki
Jump to: navigation , search
A database query syntax error has occurred. This may indicate a bug in the software. The last attempted database query was:

    (SQL query hidden)

from within function "SearchMySQL4::update". MySQL returned error "126: Incorrect key file for table './dbaps/aw_searchindex.MYI'; try to repair it (localhost)".

It happens when I'm using Mozilla 1.7 under Windows 95.

As a rather off-topic note, try to update your OS & Browser. They are extremely outdated.  :)


calculus
integration
real analysis
function
Functional Analysis
calculus computations

$ \int_{\minus{}\infty}^{\infty} \frac{\cos(nx)}{(x^2 \plus{} 1)^2} dx$

I'm told that there's a nasty trick behind this one, one that may not seem obvious at first.

The same integral with $ i\sin(nx)$ instead of $ cos$ is zero. So you may add that and get an exponential in the numerator. So it's a fourier transform

Could you elaborate a bit?
I've heard of Fourier Transforms (hence my avatar name :P ) but I'm not entirely clear on the basics.

My calculus experience entails a full year of single variable calculus, and a small bit of multi-variable stuff.

I was told that this requires some complex variable stuff, and I'm trying to learn the basics of that as well.

If it's really too far advanced for me now, I'll come back to this post in about a month or so.

Just let me know roughly how much further I'll need to go in my research in order to get this?

Well, I don't really know and I'm not the best person to ask. I'm not a teacher or good at calculating things.

I was just thinking that the transform of $ 1/(x^2 \plus{} 1)$ is well known (it's something with an exponential), and then the convolution of the transform with itself gives the transform of $ 1/(x^2 \plus{} 1)^2$.

But yes, the convergence of this integral is very good and it is easily evaluated with what is called residue calculus from complex analysis. You could probably learn it easily, although stating a version of the residue theorem formally requires quite a bit of machinery. Similarly, you can learn the mechanics of basic fourier analysis easily, but again it requires integration theory and a little functional analysis to really state things in a satisfactory way.

And on a final note a number of calculus wizards on this forum could probably evaluate this with techniques you know, maybe differentiating under the integral sign or maybe expanding the exponential in a sum or something

The appropriate complex tool is the residue theorem. Since the modified function $ \frac{e^{inx}}{(x^2\plus{}1)^2}$ decays rapidly in the upper half-plane, we can complete the integral on the real line to an integral enclosing a large region in the upper half-plane. That integral can be found by calculating the residue of the function at any poles in the region (in this case, $ i$).

The Fourier transform leads to another approach- if you know some basic transforms, you can use identities to combine them and find the transform of $ \frac1{(x^2\plus{}1)^2}$.

There are also real multivariable tricks, but they require a lot of cleverness. The details here aren't very friendly; I tried to do something, and transformed it into a divergent oscillating integral.

Thank you for all the insightful feedback, Kalle and jmerry :)

I'll check mathworld and wikipedia for more on the residue ideas.

Hey,
[quote="jmerry"]
There are also real multivariable tricks, but they require a lot of cleverness. The details here aren't very friendly; I tried to do something, and transformed it into a divergent oscillating integral.
[/quote]
That sounds very interesting any ideas on how to about approaching this integral from multivariable calculus?

Thanks,

-APSStudent

Took me about forty five minutes but I finally got it...it was all computational mistakes...messing up..forgetting a negative etc.

[hide] Let $ f(n) \equal{} \int_{ \minus{} \infty}^{\infty}\frac {\cos(nx)}{(x^2 \plus{} 1)^2}dx \equal{} 2\int_0^{\infty}\frac {\cos(nx)}{(x^2 \plus{} 1)^2}dx$

Then we would have that

$ f'(n) \equal{} 2\int_0^{\infty}\frac {\partial}{\partial{n}}\frac {\cos(nx)}{(x^2 \plus{} 1)^2}dx$

$ \equal{} \minus{} 2\int_0^{\infty}\frac {\sin(nx)x}{(x^2 \plus{} 1)^2}dx$

Integration by parts gives

$ \equal{} \frac {n\sin(nx)}{x^2 \plus{} 1}\bigg|_{x \equal{} 0}^{x \equal{} \infty} \minus{} \int_0^{\infty}\frac {n\cos(nx)}{1 \plus{} x^2}dx$

Now for a second let

$ f(p) \equal{} \int_0^{\infty}\frac {\cos(px)}{1 \plus{} x^2}dx$

Then

$ f'(p) \equal{} \int_0^{\infty}\frac {\partial}{\partial{p}}\frac {\cos(px)}{1 \plus{} x^2}dx$

$ \equal{} \int_0^{\infty}\frac { \minus{} x\sin(px)}{x^2 \plus{} 1}dx$

Two iterations of integration by parts gives

$ \equal{} \minus{} \int_0^{\infty}\frac {\cos(px)}{1 \plus{} x^2} \equal{} \minus{} f(p)$

Therefore

$ f'(p) \equal{} \minus{} f(p)$

$ \Rightarrow{f(p) \equal{} C_1e^{ \minus{} p}}$

Now using the fact that

$ f(0) \equal{} \int_0^{\infty}\frac {dx}{x^2 \plus{} 1}$

We can ascertain

$ f(p) \equal{} \frac {\pi}{2}e^{ \minus{} p}$

So back to the original problem we have that

$ f'(n) \equal{} \minus{} \int_0^{\infty}\frac {n\cos(nx)}{x^2 \plus{} 1}dx$

$ f'(n) \equal{} \frac { \minus{} ne^{ \minus{} n}\pi}{2}$

Now integrating both sides we get

$ \int_0^{m}f'(n)dn \equal{} f(m) \minus{} f(0) \equal{} \int_{ \minus{} \infty}^{\infty}\frac {\cos(mx)}{(x^2 \plus{} 1)^2} \minus{} \frac {\pi}{2} \equal{} \int_0^{m}\frac { \minus{} ne^{ \minus{} n}\pi}{2}dn$

$ \Rightarrow{f(m) \equal{} \frac {(m \plus{} 1)e^{ \minus{} n}\pi}{2}}$

$ \boxed{\therefore\int_{ \minus{} \infty}^{\infty}\frac {\cos(nx)}{(x^2 \plus{} 1)^2} \equal{} \frac {1}{2}(n \plus{} 1)e^{ \minus{} n}\pi}$

yay! :D [/hide]

I'm sorry for posting twice in a row but no one every verified. Is this the correct answer?

Ok, so I've gone through the solution, and it appears rock solid.

I had to make a few assumptions when you said you integrated by parts (I had to guess what you picked, etc) but other than that it was very well written out.

Thanks for the non-fourier transform solution I was looking for!


I have a spinner that lands on 1 with a probability of $ \frac{1}{10}$, 2 with a probability of $ \frac{2}{10}$, 3 with a probability of $ \frac{3}{10}$, and 4 with a probability of $ \frac{4}{10}$. If Phil and Sarah both spin the spinner, what is the probability that they get the same number?

Probability they get the same number = probability of both getting 1 + probability of both getting 2 + probablity of both getting 3 + probability of both getting 4.  So...

$ (\frac1{10})^2 \plus{} (\frac2{10})^2 \plus{} (\frac3{10})^2 \plus{} (\frac4{10})^2 \equal{} \frac {1 \plus{} 4 \plus{} 9 \plus{} 16}{100} \equal{} \frac3{10}$.

well, there are $ 4$ different ways for this to happen:

Probability that both spins will land on $ 1$: $ \frac {1}{10}\times\frac {1}{10} \equal{} \frac {1}{100}$

Probability that both spins will land on $ 2$: $ \frac {2}{10}\times\frac {2}{10} \equal{} \frac {4}{100}$

Probability that both spins will land on $ 3$: $ \frac {3}{10}\times\frac {3}{10} \equal{} \frac {9}{100}$

Probability that both spins will land on $ 4$: $ \frac {4}{10}\times\frac {4}{10} \equal{} \frac {16}{100}$

So the answer is
$ \frac {1 \plus{} 4 \plus{} 9 \plus{} 26}{100}$

$ \equal{} \frac {30}{100} \equal{} \frac {3}{10}$

EDIT:srry, i didnt see you already posted the solution, iin

 :oops:


Could you find the solution of this equation?
$ e^{|x|}\minus{}e^2\equal{}|x|\minus{}2$

[quote="spinos"]Could you find the solution of this equation?
$ e^{|x|} \minus{} e^2 \equal{} |x| \minus{} 2$[/quote]

I think that:
Determine f(x)= \ e^x - x .easily we have f'(x)>0 for all x >0.so that f(x) is incrise function
and that equation have only one solution.

Assume WLOG that $ x > 0$. At the end, we have to add $ \minus{} x$ as a solution as well.

First, check that $ x$ is not 0. $ e^0 \minus{} e^2 \equal{} 1 \minus{} e^2 \neq \minus{} 2$.

Once we have checked that, we can bring all the $ x$ terms to the left side and everything else to the right side:
\[ e^x \minus{} x \equal{} e^2 \minus{} 2
\]
As math man mentioned, the derivative of $ e^x \minus{} x$ being greater than 0 for all values of $ x$ implies that there is exactly one solution (ignoring the absolute value). Once we have found that solution, we can be sure there are no others.

There is no way to solve this analytically (that I know of at the moment), but a quick glance urges us to check $ x \equal{} 2$, and it turns out that it works.

So the solution set is $ \{ \minus{} 2, 2\}$.


I recently installed Linux on a computer and have installed kile and tetex on it.  I am having some trouble with packages, though.  I need to know the following:
In which folder packages are typically kept?  How I can install a package from the internet? Do I need to do anything special to make LaTeX able to read new packages (e.g., in Miktex, I would need to refresh the file name database)?

MiKTeX follows the same directory conventions as other operating systems, called [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=tds]TDS[/url]. So if a package is installed in a subdirectory in Windows it will be installed in the same subdirectory in Linux. TeX is basically a system that is the same on all operating systems which is helpful.

The Linux command for refreshing the file name database is to type [i]texhash[/i] on the command line.

General instructions for installing packages are at [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=instpackages]Installing a new package[/url] but you'll be pleased to learn that MiKTeX has a package manager for installing packages that works in Linux. It is command line only but makes installation as simple as that on Windows. You can download and read all about it at [url=http://www.miktex.org/unx/]MiKTeX Tools[/url].

Using Miktex definitely would make things easier.
Edit: whoops: I misinterpreted how Miktex is being used.  It should still make life easier, though. :lol:

I installed miktex tools and tried to follow the directions on the web-page, but my computer does not recognise the mpm command.  Is there a specific folder in which miktex tools should be installed, or is there a specific package that will install it correctly (I'm using the debian distribution)?

Did you compile the program first as the Readme file says using ./configure, make, make install? This will then create and install the package manager suitable for your system.


modular arithmetic
number theory
prime numbers
number theory proposed

Let $ S$ be the set of prime divisors of natural numbers formed: $ 2^{n^2 \plus{} 1} \minus{} 3^n , n\in \mathbb N$.Prove that both set $ S$ and set $ \mathbb P \minus{} S$  are infinite.
[u]Note[/u]: $ \mathbb P$;$ \mathbb N$ mean the set of primes and set of positive integers,respectively.

Proving that $ S$ is infinite is easy ;
Let's assume that $ S$ is finite and we have  $ S\equal{}\{p_1,p_2,...,p_k\}$ then u put $ n\equal{}(p_1\minus{}1)...(p_k\minus{}1)$ and u get that $ \gcd(p_i,2^{n^2\plus{}1}\minus{}3^n)\equal{}1$. it means that $ S$ is composed of further primes which leads to what is asked.
For the other part...
let $ n$ be even ,then the set of primes dividing $ 2^{n^2\plus{}1}\minus{}3^n$ must be congruent to $ \pm 1$ modulo $ 8$.
So consider the set of primes congruent to $ \pm 3$ modulo $ 8$.if one of them,say $ q$, divides $ 2^{n^2\plus{}1}\minus{}3^n$ then obviously $ n$ must be odd.
so we obtain that $ (2^{\frac{n^2\plus{}1}{2}})^2 \equiv 3.(3^{\frac{n\minus{}1}{2}})^2 \pmod {q} \implies \left(\frac{3}{q}\right)\equal{}1$.
in this case we obtain that if $ q \equiv \minus{}3 \pmod {8},q \equiv \minus{}1 \pmod{3}$ then $ q$ can't divide such expression.
this kind(!) of $ q$ is infinite and so ,the problem is solved (i hope)!

shocki's solution  also works for the following problem . Let a,b be two non-square numbers and $ ab$ is also not a perfect quare . Then the set of prime numbers that are not divisors of any number in the sequence :$ a^{f(n)}\minus{}b^{g(n)}$ is infinite where $ f(n)\equiv g(n)\plus{}1 (\mod 2 )$ for all $ n$ . But I think this problem is so easy . Maybe ,he wants to find a more elementary solution .


I learned BASIC :oops:  as my first language, and studied Java to pass AP comsci. Now, I am aiming for USACO. Although Java is allowed on USACO, almost all my friends recommended me to switch to C, which is much faster than many other languages. Which C implementation is the best on Windows? Where could I get it?

The two most popular C/C++ compilers for Windows are Microsoft Visual C++ and GNU C/C++. You can download Microsoft Visual Studio Express Edition and GNU C/C++ for free.

http://www.microsoft.com/express/
http://gcc.gnu.org/

But you'll probably want some IDE with your compiler too. This one is widely used and it works really well for USACO and all the other programming competitions:

http://www.bloodshed.net/dev/devcpp.html

(You'll probably want to download the one that already comes with a compiler.)

EDIT: MS Visual Studio may be nice if you want to develop windows applications but I don't think it's really that good for programming contests. Also slightly off topic, use %I64d instead of %lld when printing long longs in windows.

By far, my favorite IDE is Visual Studio. I would use it both for serious software development and for contest programming.

Ah..

It turns out I already have MS Visual Studio at home.

I guess that would be fine, according to Yury.

I've worked in Visual Studio only a couple of times so I don't really know how well suited it is for algorithm contests. Yury seems to have more experience with it so I guess it will be fine.

from what i know visual studio is sloooooooooooooooooooooooooooooooooowwwwwww aaaaaaaaaaaaaaaand willlllllllllllllllllllllll compleeeeeeeeeeeeeete yooooooooooooooouuuuur task next december :P

Yeah... I realized that. I only use it for GUI programming anyways. Some friends I asked recommended VIM and Notepad++ for contest programming. What do those do? Which one is better for C/C++?

Especially if you're just beginning with programming contests, I would strongly recommend that you use GNU C/C++ (particularly, GCC and G++) because these are the versions used in almost all contests, including USACO and the IOI. Visual Studio works as well, but it is somewhat slow, and there are some inconsistencies with GCC/G++ and Visual Studio that could cause headaches in the future.

About Vim and Notepad++: these are just some good coding editors with lots of tools that you can use. While I know several people who use Vim, I use Emacs (another popular editor) for contest programming, but it doesn't matter too much which editor you choose as long as you're comfortable and familiar with it. In particular, if you're just beginning to code, I would recommend starting out with [url=http://www.bloodshed.net/dev/devcpp.html]Dev-C++[/url] (which I used at first), since it uses GCC/G++ and has everything you need for basic C++ coding in one editor.


just out of curiousity...

when does AoPS calculus and trig come out?

trig: supposed to be this year but probably not
calculus: fall 2009 :)

Wow it'll be interesting to see what the AoPS version of calculus looks like, having taken the high school "calculus" last year!


floor function
calculus
integration
induction
Miscellaneous Problems

Determine all pairs $(a, b)$ of real numbers such that $a\lfloor bn\rfloor =b\lfloor an\rfloor$ for all positive integer $n$.

[quote="Peter"]Determine all pairs $ (a, b)$ of real numbers such that $ a\lfloor bn\rfloor \equal{} b\lfloor an\rfloor$ for all positive integer $ n$.[/quote]
http://www.kalva.demon.co.uk/short/soln/sh98n2.html

The above link doesn´t work anymore.. It should be: http://web.archive.org/web/20040821052526/www.kalva.demon.co.uk/short/soln/sh98n2.html

My proof (note that I denote real numbers by Greek letters):

Since it is clear that integral values of $\alpha$ and $\beta$ work and that it is impossible for exactly 1 of the numbers to be integral, I'll asumme $\alpha$ and $\beta$ are non-integral (and by implication non-zero) and show that they must be equal. So, assume $\alpha \left\lfloor \beta n \right\rfloor = \beta \left\lfloor \alpha n \right\rfloor$ for all positive integers n and some non-integral real numbers $\alpha$, $\beta$. Let $\alpha = a + \gamma$ and $\beta = b + \delta$ with $0 < \gamma, \delta < 1$ and $a, b \in Z$. We now have:

$\alpha \left\lfloor \beta n \right\rfloor = \beta \left\lfloor \alpha n \right\rfloor \\
(a + \gamma) \left\lfloor (b + \delta) n \right\rfloor = (b + \delta) \left\lfloor (a + \gamma) n \right\rfloor \\
(a + \gamma)(bn + \left\lfloor \delta n \right\rfloor) = (b + \delta)(an + \left\lfloor \gamma n \right\rfloor) \\
(a + \gamma) \left\lfloor \delta n \right \rfloor + \gamma bn = (b + \delta) \left \lfloor \gamma n \right \rfloor + \delta an \\
\alpha \left\lfloor \delta n \right \rfloor + \gamma bn \&= \beta \left \lfloor \gamma n \right \rfloor + \delta an$      (1)

If $n = 1$, this simplifies to:

$\gamma b = \delta a$

Which implies that (1) simplifies to:

$\alpha \left\lfloor \delta n \right \rfloor = \beta \left \lfloor \gamma n \right \rfloor$      (2)

Now, let $n$ be the smallest positive integer such that $\max(\gamma n, \delta n) \ge 1$. Then $\min(\gamma n, \delta n) \ge 1$ as well, because otherwise we would have $0$ on one side of the equation while the other side is non-zero. For this $n$, (2) simplifies to:

$\alpha = \beta$

So the only real numbers $\alpha$, $\beta$ for which $\alpha \left\lfloor \beta n \right\rfloor$ equals $\beta \left\lfloor \alpha n \right\rfloor$ for all $n \in \aleph$ (and is not always $0$), are those real numbers which are either the same or both integral.

Another, rather cute, proof I found today:

Assume $\alpha$ and $\beta$ are distinct non-integral real numbers and assume $\alpha [\beta n] = \beta [\alpha n]$ for all $n \in \aleph$. Then, after dividing by $\beta[\beta n]$ (which is allowed for large enough $n$) we get that $\dfrac{\alpha}{\beta} = \dfrac{[\alpha n]}{[\beta n]}$ is rational, say $\dfrac{p}{q}$. Since it is clear that $\alpha = -\beta$ is not a solution, we may assume $q \geq 2$. Assume, wlog, furthermore that $p$ and $q$ are coprime. Substituting $\dfrac{\beta p}{q}$ for $\alpha$, dividing by $\beta$ and multiplying by $q$ and we obtain:

$p[\beta n] = q[\beta n \dfrac{p}{q}]$

Since the RHS is divisible by $q$, LHS must be a multiple of $q$ too. And since $\gcd(p, q) = 1$, we have, in the particular case that $n = q$, that $q$ divides $[\beta q]$. Now, put $\beta = b + \delta$ with $b$ integral and $0 < \delta < 1$. Then we have: $\delta \in (0, \dfrac{1}{q})$. Assume otherwise (thus, assume $\delta \in (\dfrac{1}{q}, 1)$):

$[\beta q] = [(b+\delta) q] = [bq + \delta q] = bq + [\delta q] \geq bq + [\dfrac{1}{q} q] = bq + 1$
$[\beta q] = [(b+\delta) q] = [bq + \delta q] = bq + [\delta q] < bq + [q] = bq + q$

And this is clearly not possible if $q$ has to be a divisor of $[\beta q]$.

Now, assume that we have showed $\delta \in (0, \dfrac{1}{mq})$ for some integral $m \geq 1$. Either $\delta \in (0, \dfrac{1}{(m+1)q})$ or $\delta \in (\dfrac{1}{(m+1)q}, \dfrac{1}{mq})$. Assume the latter. Let $n = q(m+1)$. We then have:

$[\beta q(m+1)] = [(b+\delta)q(m+1)] = [bq(m+1) + \delta q(m+1)] = bq(m+1) + [\delta q(m+1)] = bq(m+1) + 1$

And since this is not a multiple of $q$, $\delta \in (\dfrac{1}{(m+1)q}, \dfrac{1}{mq})$ cannot occur. By induction we now have proved: $\delta \in (0, \dfrac{1}{mq})$ for all $m \in \aleph$. This is nonsense. So the assumption that there are distinct, non-integral reals $\alpha$ and $\beta$ for which $\alpha[\beta n] = \beta[\alpha n]$ holds for all positive integers $n$ must be wrong.

Still another variation (where I quote the first part of the last post and then take another road):

Assume $\alpha$ and $\beta$ are distinct non-integral real numbers and assume $\alpha [\beta n] = \beta [\alpha n]$ for all $n \in \aleph$. Then, after dividing by $\beta[\beta n]$ (which is allowed for large enough $n$) we get that $\dfrac{\alpha}{\beta} = \dfrac{[\alpha n]}{[\beta n]}$ is rational, say $\dfrac{p}{q}$. Since it is clear that $\alpha = -\beta$ is not a solution, we may assume $q \geq 2$. Assume, wlog, furthermore that $p$ and $q$ are coprime. Substituting $\dfrac{\beta p}{q}$ for $\alpha$, dividing by $\beta$ and multiplying by $q$ and we obtain:

$p[\beta n] = q[\beta n \dfrac{p}{q}]$

Since the RHS is divisible by $q$, LHS must be a multiple of $q$ too. And since $\gcd(p, q) = 1$, we have, in the particular case that $n = 1$, that $q$ divides $[\beta]$, say $mq = [\beta]$. Assume $kmq = [\beta k] < \beta k < kmq + 1$ for some $m \ge 1$. Then $[\beta (k+1)] < \beta (k+1) < (k+1)mq + \dfrac{k+1}{k}$.
So $[\beta (k+1)]$ is either $(k+1)mq$ or $(k+1)mq + 1$. But since the last one is not divisible by $q$, we have, for all $k \in \aleph$:

$[\beta k] = kmq$

Now, put $\beta = b + \delta$ with $b$ integral and $0 < \delta < 1$ and consider the smallest element $N$ such that $\delta N > 1$. For this $N$ we have:

$Nmq = [\beta N] = [\beta (N -1)] + [\beta] + 1 = Nmq + 1$

A contradiction.


probability
search
geometry
3D geometry
Support
analytic geometry
ratio

[quote="stuck"]Now, a little bit of a disclaimer, this is not the exact problem I saw. However, if someone can point me in the right direction for solving such problems (wikipedia link would be nice), I would be very grateful. I'd search for it myself but I have no idea where to begin.

The problem, roughly, is this. You have three lines, each can be a size from 0 to 5, not necessarily integers. What is the probability that these sides can form a triangle?

This type of problem can be extrapolated to form others (I think). So can someone point me in the right direction?[/quote]

Let $x,y,z$ be the lengths of the sides of the triangle.

Then $x+y>z$ otherwise we have no triangle.

With the condition that $0<x,y,z\leq 5$

So we need to find the probability that $x+y>z$.

[quote="JBL"]As phrased, this is a geometric probability problem: the probability space of all possible values is the cube $0 \leq x, y, z \leq 5$.  Then you need to find the portion of this cube which satisfies $x+y > z$ and $x+z > y$ and $y+z > x$, which is the region above three planes.  Then you have a 3-D geometry problem to find the volume of that region.[/quote]

Thanks a lot! I had no idea the solution was that elegant.

According to my calculations, the answer is 1/2.  Can anyone confirm/deny this?

Yes, that is what I got also. A program that I quickly made also supports this.

Draw in Cartesian Coordinates,the region x+y<z,etc. The probability is the ratio between the shaded area between the unshaded area


linear algebra
matrix
algebra
polynomial
calculus
integration
group theory

Let E be the set of all n*n matrices with integer coefficients. Let F be the set of all ineteger k>=2 such that there exists M in E s.t.
   M^k=In and M^p /= In for p<=k-1
Prove that F is finite.

[quote="alekk"]... M^p /= In for p<=k-1...
[/quote]

Alekk do you mean $M^p\neq I_n$ isn't it ?

I apologise if the following contains major mistakes, but you know, I am quite zero in this field. Anyway, I will try. Let A be an integer nxn matrix and k at least 3 such that $A^k=I$ and k is the least with this property. I want to prove that $k$ is at most $3{}^{n{}^2}$.
  Consider H the set which contains $ I,A,...,A^{k-1}$. I will prove that in H there are no two elements A,B different such that all elements of A-B are multipples of 3. Suppose this is not the case and that $ A^i-A^j =3B$ with $i>j$, where B has integer entries. Since then $ A^{i-j}-I=3A^{k-j}B$. Let $X=A^{j-i}$ and $ X=I+3C$ where C has integer entries. We have $X^k=I$, thus $ (I+3C)^k=I$. Consider now the polynomial $f=(1+3x)^k-1$. All its roots are simple and of modulus strictly smaller than 1. Let $ r_1,...,r_k$ be the eigenvalues of $ C$. Then they have modulus strictly smaller than 1 and moreover for all natural number m the number $ r_1^m+...+r_n^m$ is an integer (it is the trace of ${}^C{}^m$). Even more, it's modulus goes to 0 as m goes to infinity and thus from a certain point M it becomes 0. It follows that from a certain point the trace of $C^m$ becomes 0. it is not difficult at all to deduce that all eigenalues are 0. Next, the relation $ (I+3C)^k-I=0$ can be written $ CD=0$ where D is a polynomial matrix in C such that the last term of the polynomial associated to D is non-zero. Since all eigenvalues of C are 0, it easily follows that detD is nonzero and thus D is invertible. Thus, C=0 and $ A^{j-i}=I$ contradiction with the minimality of k.
   Thus, we proved that in H there are no two elements such that their difference is "divisible" by 3, which shows that H has at most $ 3{}^{n{}^2}$ elements and so $k$ is at most $3{}^{n{}^2}$. Hope that this is correct. Anyway, what is the official solution? And another stupid question: how on earth do they expect such things from us in one hour of oral examination? My future seems to be very unclear from this point of view... :(

wow, incredible solution. Where on earth did you find this idea ?
In fact I don't know the official solution beacause my teacher didn't give me the solution for this one. Here what I did:

T={M s.t. there exist k such that M^k=I}, M with integeral entries.
Suppose that M^k=I. We can consider that M is a matrice with rational entries (Q is a field). So the minimal polynomial of M over Q is a divisor of x^k-1. But (x^k-1)=P1*P2*..*Pq where P_i is a cyclotomic polynomial. Since the cyclotomic polynomials are irreducible over Q, we can conclude that the minimal polynomial Q of M is of the form: Q=F1*F2*..*Fs where F_i is a cyclotomic polynomial of degree d<=n. But Deg(C_m)=phi(m) where C_m is the m-th cyclotomic polynomial and a well know fact state that phi(n)>=ln(n). So there is only a finite number of cyclotomic polynomial of degree d<=n. So F_i=C_j and j<N0 for all i. Hence the eigenvalues z of M are of the form z=exp(i*2*Pi*p/N0) for all matrices in T.
So if M is in T, M^N0=I and E is finite.

I am not very sure of this solution but I'm sure it's not the official one  because at the oral examination we can't use the fact that cyclotomic polynomials are irreducible. So it must exist other solutions like yours Harazi.(but i think there is some more natural ojne   :) )

Well, indeed, it is not the most natural idea for this problem. I also had the approach with ciclotomic polynomials but since I had some doubts (which turned out to be with no reason) and since the method I used here helps us to solve another extremely difficult oral problem posted by me (that one with the subgroup generated by A and B), I prefered this solution. Probably Moubinool will be so kind to delight us with a short and easy solution. Please! :maybe:  :maybe:

[quote="alekk"]

...because at the oral examination we can't use the fact that cyclotomic polynomials are irreducible...[/quote]

alekk,

Can you get more information is it ENS or X ? Wich year ?

-----------------------------------------------------------------------------
Well if a student MP* want get in ENS, I think he should know
thing about cyclotomic polynomials.

I am sure the Examninator will appreciate if you use result
like this appropriatly if

1/  You understand the object you manipulate.
2/  You also know the proof of the result  wich is not in math program
3/  Some application
MPSI/MP*

yes but the proof is quite hard to remember.. especially the day of the exam  :D

in fact, to prove that there is only a finite number of possible eigenvalues we can prove that there is only a finite number of characteristic polynomial. And this is easy because all eigenvalues have modulus 1 and the characteristic polynomial have integral coefficients. The conclusion follows with the relations between roots and coefficients.


geometry
trapezoid
power of a point
radical axis
geometry unsolved

Here is my solution

[hide]

Consider invrsion centered at . Let any point  go to the point .

By the properties of inversion,

,

,

By the properties of inversion, since  and  do not containt center of inversion , they transforn into cyclic quadrilaterals  and .

Due to angle equalities, they are isosceles trapezoids.

Now it is easy to check that  is an isosceles trapezoid too.

Particularly, it is cyclic. So  is also cyclic, and that is what we had to prove.

[/hide]

Hello Sergey, can you post all the problems from Belarus TST?
Thx

Now I see, this problem is on the second Brazilian List for IMO.
I will give my solution:

By radical axis, it's sufficient to prove that $ AD$, $ PQ$ and $ BC$ conccur.
Let $ AD \cap PQ \equal{} X$ and $ BC \cap PQ \equal{} Y$.We will prove that $ X\equal{}Y$.
For this, it's sufficient to prove that $ \frac{XP}{XQ}\equal{}\frac{YP}{YQ}$.
With sine's law on $ XAP$ and $ XAQ$ and using that $ APQD$ is cyclic, we get that:
$ \frac{XP}{XQ}\equal{}\frac{AP}{AQ}.\frac{DP}{DQ}$
In a similar way, we get:
$ \frac{YP}{YQ}\equal{}\frac{BP}{BQ}.\frac{CP}{CQ}$

With sine's law on $ APE$, $ AQE$, $ PDE$ and $ DQE$, we have:

$ \frac{AP}{AQ}.\frac{DP}{DQ} \equal{} \frac{PE^{2}}{QE^{2}}$

In a similar way, $ \frac{BP}{BQ}.\frac{CP}{CQ}\equal{}\frac{PE^{2}}{QE^{2}}.$
Hence, $ \frac{XP}{XQ}\equal{}\frac{YP}{YQ}$, as we wanted.

[quote="Sergey"]Let $ ABCD$ be convex quadrilateral $ P, Q$ - such points in it's interior that $ APQD$ and $ BPQC$ are cyclic. Let $ E$ be a point on the segment $ [PQ]$ such that $ \widehat{EBP} \equal{} \widehat{ECQ}$ and $ \widehat{PAE} \equal{} \widehat{EDQ}$.
Prove that $ ABCD$ is cyclic.[/quote]
Do you have any idea about construting this $ E$ in the problem? :maybe:


Let $a,b,c>0$ and $ab+bc+ca=1$. Find minimum of expreesion:
$A=\frac{a}{(b+2c)^{3}}+\frac{b}{(c+2a)^{3}}+\frac{c}{(a+2b)^{3}}$

$A \sum a(b+2c)\geq (\sum \frac{a}{b+2c})^{2}$

Let $x=b+2c, y=c+2a, z=a+2b$

Then,

$9\sum \frac{a}{b+2c}=\sum \frac{-2x+4y+z}{x}=-6+\sum \frac{4y+z}{x}\geq 9$

Finally, $A\geq \frac{1}{3}$


can someone help me on 19?

i stink at this

wait--- you can skip! but is there any catch?

anyway, can someone help me w/ 24? u can't skip that one

flip thru the colors quickly & yll c a message

i got to 42. so many to choose from
need help w/#40.

[hide="Hint to #40"]Use Alt-Tab.[/hide]
I can't get past question 66.

same-- i need help w/ 42.

another way to do 40 is to not even click over the button and skip it (if you have the option to skip)

I've beaten the game several times
all you need to do is google for answers, or also theres a youtube movie on how to do it
oh yeah and don't use any skips, because u need all of them on the last problem (#110)
on number 42, click the 42nd 42

gotten to 60 something

and did not cheat, except with the help of my class

42's mad easy; it's the 42nd 42.
I'm stuck on 72.  I doubt it's even possible.

Btw lingo: I got stuck on that one for a while.  If you want to know, [hide="Click on"]the exclamation point.[/hide]

So what exactly is this game about? :huh:

It's a quiz with tricky questions.

I attached an example: question 12.
[hide="Answer"]One would think that the answer is the third circle from the left, but it's not.  The correct answer is the dot above the I in "CLICK," which is even smaller.[/hide]

Others are plays on words.  For example: "What is the 7th letter of the alphabet?"  Choices: G, H, I, J
[hide="Answer"]The 7th letter in the phrase "THE ALPHABET" is H, and that is the answer.[/hide]

That's the basic idea.

how do u do 40? i used to skip it, but since we need all our skips... :?:

[hide="How to do #40"]There are two ways of which I know.

The first way involves using Alt-Tab to switch out of the window.  Then move the mouse a little to the right and Alt-Tab back into the window with The Impossible Quiz.  If you do it right, you'll be right in that area.  This was how I first did it, until I discovered an easier way.

The second way is simpler: right-click and a menu will appear.  Use that as a bridge.[/hide]

oh wait! i just got how to do 40!

[hide="answer"] right click and use that as a bridge)[/hide]

[quote]There are two ways of which I know.

The first way involves using Alt-Tab to switch out of the window. Then move the mouse a little to the right and Alt-Tab back into the window with The Impossible Quiz. If you do it right, you'll be right in that area. This was how I first did it, until I discovered an easier way.

The second way is simpler: right-click and a menu will appear. Use that as a bridge.
[/quote]

srry. i didn't see kept this window open and i didn't see that u already posted how to do it.  :oops_sign:

WHAT IS WRONG WITH ME TODAY???

[quote]srry. i didn't see kept this window open and i didn't see that u already posted how to do it.[/quote]
didn't make any sense. i meant that i didn't see that u already posted how to do it. forget the window part

hmmm...sounds like a fun game. is this on newgrounds?

[quote="Identity"]I'm stuck at 56 the invisible cursor help plx[/quote]

[hide="What you do is"] go out of the screen and right click, then you move your cursor on the screen and click the button

[hide="for the second one"]blue red blue yellow[/hide][/hide]

What's the answer to the duck thingy? I forgot what number.

That was #41, and the answer is...
[hide]afro[/hide]

And for #60, just answer it honestly.  But if you hit refresh, the game recycles itself, and it has no previous memory of whether you completed #59 on your first try.

ive gottento 10-something but im stuckon the one where youre on a road...which one's that? anyways, iuse a skip on that one and i get stuck on the next one

[quote="lingomaniac88"]I can't get past question 66.[/quote]

That was fun. Took me three(a lot) tries to figure it out.

I know you got through the quiz, but what the heck:

[hide="66"]click the question mark[/hide]

stuck on 79.

[quote="lingomaniac88"]That's #106.  And don't skip it.

I completed The Impossible Quiz about a month ago.  If anyone has any questions regarding the answers, PM me.

And now I'm working on The Impossible Quiz 2.  I'm up to #108. (I skip the one that says, "If you think long enough the answer is obvious" -- I think that's #82...)[/quote]

82. Click obvious when the bomb starts flashing or gets to 2
108. Follow the green arrow (it's best to remember the path)

I know how to beat #108, I just lost because of the time limit.  As of yesterday, I made it all the way to the end.  So I now completed both quizzes.

[quote="kevinj"]This forum is about the Impossible Quiz.  I am on no. 42.  Any hints or help?[/quote]

42 is the 42ed 42

you guys realize there's already an Impossible Quiz 2, right?

I know about The Impossible Quiz 2.  And I beat that about a week ago.

You can play it [url=http://www.notdoppler.com/theimpossiblequiz2.php]here[/url].

Yah I beat it on October 17 its not to hard

Im stuck on number 15
here's a screenshot:

[img]http://i187.photobucket.com/albums/x30/ssjsongoku/quiz.jpg[/img]

Can anyone please give me a hint?

You can find it in several places.  It's on both AddictingGames and Newgrounds.

[quote="lingomaniac88"][hide="Answer to 15"]HORSE[/hide][/quote]
Thanks!   :)

i got stuck on the number 60 something, where you have to charge the laser. how d'you do that? net hints say mash the mouse. wtf is that?

as for 40, as we later found out, after right clicking, you can take your mouse anywhere only don't left click before getting into the other island.

meanwhile any idea as to the reason behind the werewolf one? why shoe polish?

Repeatedly click the laser (and quickly) until it blows the bombs to bits.


There are some N coins out of which only m are fair, the rest being biased.

Probability of a Head turning up when a biased coin is thrown is 2/3 . A coin is picked up from the lot and thrown twice. The outcome of the event is HT, i.e. the first throw results in a Head and the second results in a Tail.
What is the probability that a fair coin was chosen???

-----------Fair Coin----Biased Coin
HT-----------xxx------------xxx-----
Not HT------xxx------------xxx------

Make a table like this where xxx's are probabilities.  The answer should turn out to be 9/17 if my mental arithmetic is working today.

[quote="se2schul"]Have you tried solving this using conditional probability?

ie finding the probability that the coing is biased [b]given[/b] it flipped a tail then a head.

I haven't worked it out, but this is how I would try to solve it.[/quote]

Applying Bayew's theorem should help ..I solved it using the same..


polynomial
Diophantine equation
number theory proposed
number theory

Prove that there are infinitely many positive integers $ n$ for which all the prime divisors of $ n^{2}\plus{}n\plus{}1$ are not more then $ \sqrt{n}$.

[hide]  Stronger one.
Prove that there are infinitely many positive integers $ n$ for which all the prime divisors of $ n^{3}\minus{}1$ are not more then $ \sqrt{n}$.[/hide]

Obviosly, there are infinetely many solutions $ n^{2}\plus{}n\plus{}1\equal{}3m^{2}$.

[quote="Shishkin"][quote="Rust"]Obviosly, there are infinetely many solutions $ n^{2}\plus{}n\plus{}1 \equal{} 3m^{2}$.[/quote]

Can you show how it helps in this problem?[/quote]
If n satisfyed these equation, the prime divisor of $ n^{2}\plus{}n\plus{}1$ is prime divisor of $ m\equal{}\sqrt{\frac{n^{2}\plus{}n\plus{}1}{3}}$ or 3.

If m is a prime it true but on other case it not true.
$ n^{2}\plus{}n\plus{}1\equal{}3m^{2}$
For example if $ m\equal{}pq$ then $ p|n^{2}\plus{}n\plus{}1$ .

I think you mean the other way around.  In fact, when $ m$ is composite it has a prime factor not greater than $ \sqrt{m}$, and

$ m \equal{}\sqrt{\frac{n^{2}\plus{}n\plus{}1}{3}}\le\sqrt{\frac{n^{2}\plus{}n^{2}\plus{}n^{2}}{3}}\equal{} n$

It remains to show that there exist infinitely many solutions such that $ m$ is composite; this might be overkill, but the Eisenstein integers might be of use in this case.

[quote="TTsphn"][quote="Shishkin"][quote="TTsphn"]If m is a prime it true but on other case it not true.
$ n^{2}\plus{}n\plus{}1 \equal{} 3m^{2}$
For example if $ m \equal{} pq$ then $ p|n^{2}\plus{}n\plus{}1$ .[/quote]
If m is a prime then we have divisor of $ n^{2}\plus{}n\plus{}1$ which is more than $ \sqrt{n}$. It's bad for us.
Also even if $ m$ is not prime it can have a divisor which is more than $ \sqrt{n}$.
I think a problem still alive.[/quote]
But it must be prime divisor.[/quote]
Yes. I wrote about prime divisor.
Every prime divisor of $ n^{2}\plus{}n\plus{}1$ have to be not more than $ \sqrt{n}$.

Sorry, I misread the original problem  :oops:

Let $n=k^m$, then \[n^2+n+1=\frac{n^3-1}{n-1}=\prod_{d|3m,d\not |m}\Phi_d(k).\]
We can choose $m$, suth that $\phi(3m)=deg(\Phi_{3m}(k))<m\epsilon$. It give that any prime divisor of $k^{2m}+k^m+1$ less than $n^{\epsilon}$. For example $m=10010=2*5*7*11*13\to \phi(3m)=1920<5005.$

[quote="horizon"][quote="littletush"]consider $n=l^{20}$.then
$n^2+n+1=(l^2+l+1)(l^2-l+1)*\frac{l^{10}+l^5+1}{l^2+l+1}*\frac{l^{10}-l^5+1}{l^2-l+1}$
when $l$ is sufficiently large,those $4$ polynomials(in respect of $l$) are less than $l^{10}=\sqrt{n}$.[/quote]
Dear littletush, the problem is not so bug.Just let
$n^{2}$+n+1=k$y^{2}$,here k is a fixed number satisfy that
the equation has one solution.(this can be achieved obviously!)
then by Pell equation,the equation has infinitely many solutions.
Hence $n^{2}$+n+1's prime divisors are all less than $\sqrt{n}$[/quote]
horizon's seems wrong......

Construction: Let $k \equiv 1 \pmod {91}$. Then set $n=(2k^2)^4$. I claim that this works.
$n^2+n+1 = (4k^4+2k^2+1)(4k^4-2k^2+1)(4k^4-4k^3+2k^2-2k+1)(4k^4+4k^3+2k^2+2k+1)$
$(4k^4+2k^2+1) \equiv 0 \pmod 7$ and $(4k^4+4k^3+2k^2+2k+1) \equiv 0 \pmod {13}$.
This implies that all prime divisors are less than $\sqrt{n}=4k^4$, so we are done. $\blacksquare$

Rip $\ln(4)>1$. One can show in a non-constructive way the weaker result
[quote=Shishkin]Prove that there are infinitely many positive integers $ n$ for which all the prime divisors of $ n^{2}\plus{}n\plus{}1$ are not more then $n^{0.736}$.[/quote]
It generalizes nicely enough, but I can't see how to use it to get the original problem...

We will show that for any $\varepsilon>0$, there are infinitely many $n$ for which all the prime divisors of $n^2+n+1$ are less than $n^\varepsilon$.

Verify the below estimate:

[color=red][b]Lemma:[/b][/color]      For each \(\delta>0\), there is a \(3\mid m\) with \(\varphi(m)/m<\delta\).

[i]Proof.[/i]      Let the primes be \(p_1<p_2<\cdots\) and set \(m=p_1p_2\cdots p_k\). Then     \[\lim_{k\to\infty}\frac{\varphi(m)}m=\prod_p\left(1-p^{-1}\right)\le\prod_pe^{-1/p}=\exp\left(-\sum_p\frac1p\right)=0.\] \(\blacksquare\)

Let \(\varphi(3k)/k<\varepsilon\) by the lemma, observe that \[n^{2k}+n^k+1=\prod_{\substack{d\mid 3k\\ d\nmid k}}\Phi_d(n).\] If \(p>n^{k\varepsilon}\) divides \(n^{2k}+n^k+1\), then \(p\mid\Phi_d(n)\) for some \(d\mid3k\). However \(\varphi(d)\le\varphi(3k)<k\varepsilon\), so \(\Phi_d(n)\ll n^{k\varepsilon}\), contradiction.


Yeah...
How do you know that
m can be expressed
as t+1 and n
as t-1...

This was posted by accident; this thread should be deleted.

I believe this was intended to be a reply to the thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=247252]found here[/url].


calculus
integration
advanced fields
advanced fields unsolved

This is problem 3.7 from Hartshorne:

Suppose f: X -> Y is a dominant, generically finite morphism of finite type of integral schemes.  I'm suppose to show that there's an open subset U of Y such that the induced map f^(-1)(U) -> U is finite.

I reduced to the part where Y = Spec A, X = Union of Spec (A_i) with each Spec (A_i) -> Spec A finite map.  So we're very close to a finite map but now I can't seem to make X (or some inverse image in X of open in Y) affine.

The user hochs from the channel #math in EFnet told me how to solve this (U_k=Spec A_k):
For $ U_i \cap U_j\neq \emptyset$, $ Z \equal{} f(U_j^c\cap U_i)$ is a proper closed subset of Y, because $ f|U_i$ is finite. For some non-unit g of A we have $ Z\cap D(g) \equal{} \emptyset$ and for $ V_k \equal{} U_k\cap f^{ \minus{} 1}(D(g))$ (which is affine) we have $ V_i\subset V_j$, so $ f^{ \minus{} 1}(D(g))$ can be covered by fewer affine sets.

Hi, thanks for the reply.

That's absolutely correct.

Thank you!


Suppose there are $ N$ cups laid on a table, all either up or down. Define an operation, $ F_m$, which flips over $ m$ cups of your choosing.

If $ F_m$ for a given $ m$ is the only operation allowed, for which $ N$ is it possible to obtain any configuration of Ups and Downs from a finite number of compositions of $ F_m$.

I have a friend who says it is true for $ m$ is odd and $ m<N$.
Im not sure though.


geometry
circumcircle
power of a point
radical axis
geometry proposed

Let E be a point on the median CD of triangle ABC.
Let S1 be the circle passing through E and tangent to line AB at A,
intersecting side AC again at M; let S2 be the circle passing through
E and tangent to line AB at B, intersecting side BC again at N Prove that the circumcircle of triangle CMN is tangent to circles S1 AND S2

Since $ D$ has equal power $ \frac {_1}{^4}AB^2$ with respect to $ (S_1)$ and $ (S_2),$ it follows that $ CD$ is the radical axis of $(S_1),(S_2).$ Inversion with center $ C$ and power equal to the power of $ C$ WRT $ (S_1)$ and $ (S_2)$ takes $ (S_1)$ and $ (S_2)$ into themselves and  $ M,N$ into $A,B,$ respectively. Therefore, circle $ \odot(CMN)$ is taken into the common external tangent $ AB$  of  $ (S_1)$ and $ (S_2).$ Then by conformity, $\odot(CMN)$ is tangent to $ (S_1),(S_2).$


It's enough to show that $x(1+x+\ldots+x^{n-1})+(x^{-n}+x^{-n+1}+\ldots+x^{-1})=\sum^{n}_{i=1} x^i+\sum^n_{i=1} x^{-i}>2n$

But

$\sum^{n}_{i=1} x^i+\sum^n_{i=1} x^{-i}=\sum^n_{i=1} x^i+x^{-i}\geq \sum^n_{i=1} \ 2=2n$

since $x^i+x^{-i}\geq 2$ by AM-GM.  Since equality holds iff $x=1$, we get the strict inequality.

[quote="mitix"]convict[/quote]

is that an interchangeable with "prove"???? If then, I learned something today. I never knew that before.


I've never seen the  notation you used in the first expression but it looks like it's the same thing as the second one.

[quote="WiZ@Rd"]Can someone say me if the following sum are equal or not? Thanks.

(1) $ \sum_{r,s}^{1,2,\ldots,n}a_{rs}$

(2) $ \sum_{r,s \equal{} 1}^{n}a_{rs}$

$ \sum_{r,s}^{1,2,\ldots,n}a_{rs} \equal{} \sum_{r,s \equal{} 1}^{n}a_{rs}$   ???

I think not, but I'm not sure.[/quote]

OK. Thanks.


Each term of a sequence of natural numbers is obtained from the previous term by adding to it its largest digit. What is the maximal number of successive odd terms in such a sequence?

[b]Answer:[/b] 5 terms.
Let $ A$ be the subsequence containing the maximal number of successive odd terms in the sequence. Assume $ A$ contains more than $ 5$ terms. Then, the first six terms of $ A$, $ t_1,t_2,...,t_6$ are odd.
[b]1.[/b] If a term $ k$ ends in $ 9$, the largest digit in this term is $ 9$ so the next term is equal to $ k \plus{} 9$ which will end with $ 8$ and will not be odd. Hence, the term following a term ending in $ 9$ is not odd.
Now, consider four cases.
[b]Case 1[/b] The first term $ t_1$ in $ A$ ends in $ 1$. Because the second term is odd, then we must add an even digit to $ t_1$, so the largest digit of $ t_1$, $ l$, is even, and the following term will end in $ k \plus{} 1 > k$ so the third term will end in $ k \plus{} 1 \plus{} k \plus{} 1\pmod {10}$ (as $ k \plus{} 1$ will be the largest digit of $ t_1$ as no digits besides the last one will change after adding $ l$ as $ 1 \plus{} l\le 9$ if $ l$ is an even digit), which is even, so there are at most two terms in $ A$ which is impossible.
[b]Case 2[/b] The first term $ t_1$ in $ A$ ends in $ 3$. If the largest digit of $ t_1$ is $ l$ such that $ l \plus{} 3\le 9$ then, the largest digit of $ t_2$ is $ l \plus{} 3$ so the last digit of $ t_3$ is $ l \plus{} 3 \plus{} l \plus{} 3\pmod {10}$ which is even. So this case is impossible. Otherwise, $ 3 \plus{} l > 9$ so $ l \equal{} 8$ but then the last digit of $ t_2$ is $ 1$ and using case 1, there is at most $ 1$ term after that, so at most three consecutive odd terms can exist with the first one ending in $ 3$.
[b]Case 3[/b] The first term $ t_1$ in $ A$ ends in $ 5$. If the largest digit of $ t_1$ is $ l$ such that $ l \plus{} 5\le 9$ then, the largest digit of $ t_2$ is $ l \plus{} 5$ so the last digit of $ t_3$ is $ l \plus{} 5 \plus{} l \plus{} 5\pmod {10}$ which is even. So this case is impossible. Otherwise, $ 3 \plus{} l > 9$ so $ l \equal{} 6$ but then the last digit of $ t_2$ is $ 1$ and using case 1, there is at most $ 1$ term after that or $ l \equal{} 8$ but then the last digit of $ t_2$ is $ 3$ and using case 2, there are at most $ 2$ terms after that, so at most four consecutive odd terms can exist with the first one ending in $ 5$.
[b]Case 4[/b] The first term $ t_1$ in $ A$ ends in $ 7$. If the largest digit of $ t_1$ is $ l$ such that $ l \plus{} 7\le 9$ then, the largest digit of $ t_2$ is $ l \plus{} 7$ so the last digit of $ t_3$ is $ l \plus{} 7 \plus{} l \plus{} 7\pmod {10}$ which is even. So this case is impossible. Otherwise, $ 3 \plus{} l > 9$ so $ l \equal{} 4$ but then the last digit of $ t_2$ is $ 1$ and using case 1, there is at most $ 1$ term after that or $ l \equal{} 6$ but then the last digit of $ t_2$ is $ 3$ and using case 2, there are at most $ 2$ terms after that or $ l \equal{} 8$ but then the last digit of $ t_2$ is $ 5$ and using case 3, there are at most $ 3$ terms after that, so at most four consecutive odd terms can exist with the first one ending in $ 7$.
If $ t_1$ ends with $ 9$ the next term is even (see beginning of the solution).
Therefore, there can exist at most $ 5$ consecutive odd terms.
An example when this is the case is the subsequence $ 817, 825,833,831,839$.

A simpler approach:
Step 1:The last digit of each term is odd so that the largest digit of each term is even
Step 2:The largest digit of each term remain the same(each addition at most change the largest digit by $ 1$,but the largest is forever even)
Step 3:After at most $ 5$ additions we will obtain the largest digit of one term is $ 9$,which will become the largest digit(but it is odd)
So our answer is $ 5$


greatest common divisor
number theory
least common multiple
prime factorization
relatively prime

Prove that: $ab= \gcd(a,b) \cdot lcm(a,b)$

[hide="hint"]Suppose $a=p_1^{k_1}p_2^{k_2}\cdots{p}_r^{k_r}u$ and $b=p_1^{l_1}p_2^{l_2}\cdots{p}_r^{l_r}v,$ where $a$ and $b$ share the $p_i$ but not necessarily to the same degree, and the $u$ and $v$ belong solely to $a$ and $b$ respectively. Then we have \[(a,b)=p_1^{\min(k_1,l_1)}p_2^{\min(k_2,l_2)}\cdots{p}_r^{\min(k_r,l_r)}\] and \[[a,b]=uvp_1^{\max(k_1,l_1)}p_2^{\max(k_2,l_2)}\cdots{p}_r^{\max(k_r,l_r)}.\]Thus (as if I have actually proven anything..), $(a,b)\cdot[a,b]=uvp_1^{\min(k_1,l_1)}p_1^{\max(k_1,l_1)}p_2^{\min(k_2,l_2)}p_2^{\max(k_2,l_2)}\cdots{p}_r^{\min(k_r,l_r)}p_r^{\max(k_r,l_r)}$
$=uvp_1^{k_1}p_1^{l_1}p_2^{k_2}p_2^{l_2}\cdots{p}_r^{k_r}p_r^{l_r}$
$=(p_1^{k_1}p_2^{k_2}\cdots{p}_r^{k_r}u)\cdot(p_1^{l_1}p_2^{l_2}\cdots{p}_r^{l_r}v)$
$=ab$[/hide]

My own hint:

Like Antonio's but I guess simplified.
[hide]
Consider any prime $p$, the prime factorization of $m$ has a factor $p^a$, for some $n$, and the prime factorization of $n$ has a factor $p^b$, for some $b$. Say $b \ge a$. The GCF will need $a$ p's and the LCM will need $b$ p's. [/hide]

Answer: Don't look unless you need to.
[hide]
So GCF * LCM has $p^{a+b}$ in the prime factorization, SAME as $mn$. This applies for all primes $p$.[/hide]

I really have no idea what you are talking about...... Isn't it the most elementary number theory formula?? :?
After all, thanks for you help. :D

[hide]
If $a$ and $b$ are relatively prime, then $\gcd(a,b)=1$.  Since they share no prime factors, the $lcm(a,b)=ab$.  Thus, $\gcd(a,b) \cdot lcm(a,b)=1 \cdot ab=ab.$

If $a$ and $b$ are not relatively prime, we can say $gcd(a,b)=d$, and $a=dx, b=dy$ for relatively prime positive integers $x$ and $y$.  In this case, $lcm(a,b)=lcm(dx,dy)=dxy.$  Thus, $\gcd(a,b) \cdot lcm(a,b)=d \cdot dxy=d^2xy=(dx)(dy)=ab.$ [/hide]

That was really good shambles. Very simple proof. I like it.


In doing AIMEs, there are often problems that I don't understand. So, after I take the AIME, what is the best way to review?

Well you could post here and ask us to help. Now the important thing with the AIME isn't the solution, it's how one finds the solution, and make sure you think about how a person would come up with that solution.

Yeah, I have all the solutions, but I am first and foremost looking to improve my knowledge, so I am asking what to do after I check my answers.
-Quickly review the solutions and put the AIME away and do another one?
-Look through each solution thourougly?
-Quickly glance at the solutions, recreate problems.
-etc, etc

Like I said, you should review the solutions fairly thouroughly, but mainly understand how a test-taker would come across the solution, and how they would discover it.

Yeah, and you might want to even change some of the numbers in the problem, put the solution away, and then try the problem again. For some people, looking at the solution alone is good enough, but for others, they have to work a problem to get it. Once, you're done working the problem, go back to the solution and see if you worked the right steps (you won't be able to check your final answer but you can at least check the steps)


combinatorics proposed
combinatorics
Brazilian Math Olympiad
Brazilian Math Olympiad 1990

Show that a convex polyhedron with an odd number of faces has at least one face with an even number of edges.

Just look at the graph having the faces as vertices and in which two vertices are connected iff the corresponding faces share an edge. Since every graph has an even number of vertices of odd degree, every polyhedron must have an even number of faces with an odd number of edges.

Suppose that all faces have an odd number of edges. Let $S_{i}$ be the number of edges of the face $F_{i}$.
Then the sum off all edges is $S_{t} = \frac{S_{1} + S_{2} + \cdot\cdot\cdot + S_{2k+1}}{2}$.
But since all $S_{i}$ are odd then $2$ doesn't divide $S_{t}$. That is impossible.


hello its one of the iranian tests for the students who start learning math....

suppose $ABCD$ is a square.if $AB=a$,find the value of $AC$.

$1)a^ \frac{1}{2}$

$2)2a$

$3)\frac{a}{2}$

$4)2a^2$

$5)a\sqrt[2]{2}$

[quote="Farenhajt"]And what about [hide="this"]$5)\,\text{None of the above}$[/hide]?[/quote]

ure right ...

none of the above ... ;) ...

i will add the fifth ...

$5)a. 2^\frac{1}{2}$

[hide="solution"]AC is a diagonal of a square. Since the length of the diagonal of a square is the lenghth of a side times the square root of two, answer 5 is correct.[/hide]

AC²=AB²+BC²
AC²=a²+a²
AC²=2a²
AC=a√2, so the correct one is (5).

[hide]Rules of a 45-45-90 triangle $\implies a\sqrt{2}$[/hide]

[hide]
$\overline{AC}$ is a diagonal of a square, using the pythagorean theorem:
$\sqrt{2a^2}=a\sqrt{2}$. So $\boxed{5}$[/hide]


Let  $a_n$ be a sequence defined by:
 $a_0 =\alpha$ , $a_1 = \beta$ , $a_{n+2} = \frac{1+a_{n+1}}{a_n}$
 Find $a_{2005}$.

The sequence is periodic (hopefully) ;)

$a_{n+5} = a_n$ so $a_{2005} = a_0 = \alpha$


integration
calculus
derivative
function
absolute value
calculus computations

A particle moves along the x-axis in such a way that its position at time t for t>=0 is given by $ x\equal{}\frac{1}{3}t^3\minus{}3t^2\plus{}8t$.
When t=3, what is the total distance the particle has travelled?

The answer is, in fact, $ \int_0^3\left|t^2\minus{}6t\plus{}8\right|\, dt\equal{}\boxed{\frac{22}{3}}$.

If you graph the equation, you see that the distance travelled is equal to the output UNTIL it reaches a maximum and begins to drop. We see that there is one maximum before we reach $ t\equal{}3$. So if we find the maximum and then find the absolute value of the difference from the maximum and $ t\equal{}3$. So the maximum is when $ t\equal{}2$ so the distance traveled is $ \frac{20}{3}$ and then from that point to when $ t\equal{}3$, the distance traveled  is $ \frac{20}{3} \minus{} 6 \equal{} \frac{2}{3}$ so the total distance traveled is $ \frac{20}{3} \plus{} \frac{2}{3} \equal{} \boxed{\frac{22}{3}}$.

$ x\equal{}\frac{1}{3}t^{3}\minus{}3t^{2}\plus{}8t$

We first find the derivative of this function to find where the velocity is 0.

$ 0 \equal{} t^{2} \minus{} 6t \plus{} 8$
$ t \equal{} 2$ or $ t \equal{} 4$

Since $ t \le 3$, $ t \equal{} 2$.

Then we find $ |x(2) \minus{} x(0)| \plus{} |x(3) \minus{} x(2)|$.

$ x(2) \equal{} \frac{2^{3}}{3} \minus{} 3 * 2^{2} \plus{} 8 * 2 \equal{} \frac{8}{3} \minus{} 12 \plus{} 16 \equal{} \frac{20}{3}$
$ x(0) \equal{} \frac{0^{3}}{3} \minus{} 3 * 0^{2} \plus{} 8 * 0 \equal{} \frac{0}{3} \minus{} 0 \plus{} 0 \equal{} 0$
$ x(3) \equal{} \frac{3^{3}}{3} \minus{} 3 * 3^{2} \plus{} 8 * 3 \equal{} \frac{27}{3} \minus{} 27 \plus{} 24 \equal{} 6$

$ |x(2) \minus{} x(0)| \equal{} \frac{20}{3}$
$ |x(3) \minus{} x(2)| \equal{} \frac{2}{3}$

Therefore, the answer is $ \frac{22}{3}$.

I thought we weren't allowed to use calculus on the forum (saw integration symbol). Just asking for a confirmation...

Well, no, you're not supposed to, but this question is in the wrong forum.