geometry
rhombus
circumcircle
geometry proposed

Rhombus $ CKLN$ is inscribed into triangle $ ABC$ in such way that point $ L$ lies on side $ AB$, point $ N$ lies on side $ AC$, point $ K$ lies on side $ BC$. $ O_1$, $ O_2$ and $ O$ are the circumcenters of triangles $ ACL$, $ BCL$ and $ ABC$ respectively. Let $ P$ be the common point of circles $ ANL$ and $ BKL$, distinct from $ L$. Prove that points $ O_1$, $ O_2$, $ O$ and $ P$ are concyclic.

[i]Author: D.Prokopenko[/i]

∠NPK=180-∠ACB=180-∠O1CO2=∠O1OO2
notice ∠NPO1=∠O2PK  =>   ∠O1PO2=∠O1OO2


If your favorite color did not get on the list, I am sorry. It would not allow me to have more than 10 poll options. You may post your color here.
[hide="lots of aliens"] :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien::alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: :alien: [/hide]

[color=darkblue]Dark Blue totally! [/color]

Wow I can't believe so many people like the color that I like too.

Hi Everyone,

I dont know about some of you, but I find it hard to determine things like favorite color or favorite food.  I do have preferences, but they are not very clear cut as there is such variety in the Universe.  I like many colors.  Often times its the context in which the colors appear that imprints them into my memory.  As another example of the sort of problem of determining a definite favorite, would you absolutely be able to say what's your favorite theorem youve encountered in mathematics?  For me, the Fundamental Theorem of Galois Theory is just plain beautiful, but I couldnt say that it was my favorite, though it probably is among my set of favorite theorems.  I guess the set of my favorite colors is defined as the set of sets of particular favorite colors, these sets having a finite cardinality greater than 1.

Happy New Year Everyone!

M.E.B.

[color=darkblue] yay for dark blue[/color]

i think blue is the universally preferred color  :D

Not that pink's my preferred color but because someone asked:

I'm not sure if this is pink, as it's not on the drop-down list...

[quote="KBabe"][color=#FF6EC7][size=200]PINK!!!!!!!!!!!!!!!!!!!!!!!!!![/size][/color][/quote]

[quote="h_s_potter2002"]How could you leave out pink?[/quote]

 I started the list out with basically the AoPS drop down list which doesn't include pink but it wouldn't let me have that many poll options so I deleted colors from there.

 Why is Valentin Vornicu's so obvious?

[color=red] RED [/color] and [color=black] BLACK [/color]

[quote="yif man12"]Hot pink. Definitely.

[color=white]heh, just kidding. probably blue or red for me.[/color][/quote]

lol. at first, i didn't see the white part and i was like HOT PINK?!?!?!?!?!?

[color=white]Invisible :P .[/color]

I like black a lot :P .

Green, they say genius' choose green (thats what movies teach me).  Regardless I like it the best... dark blue is ok

[quote="yif man12"]Hot pink. Definitely. [/quote]

my thoughts exactly  :)
[hide]actually i'm more into red.[/hide]

[quote="Super Naterul"]Why is Valentin Vornicu's so obvious?[/quote]

I'm pretty sure its blue since the Mathlinks skin of this site is all blue and white.  That might also explain why it is so obvious.

I dunno.......
on Valentin's original post when he said that his favorite color is obvious the post subject was a Mr. Green.   :D   Sooooooo....maybe green might be his favorite color..... How about him just telling us?

btw my favorite color is [color=darkblue]DARK BLUE!!!!!!!!!!!![/color]

Hm.. I'm the only one who likes orange... interesting :?  ;)


inequalities
logarithms
inequalities theorems

Please help me  :(
How can I prove the Bernoulli inequality by Jensen inequality?
Thanks.

I don't think that is possible since Bernoulli inequality has been used to prove very basic facts about the real numbers. But, on the other hand, I have never seen a proof of Jensen inequality starting with axioms of real numbers.

We'll prove:

\[ [x>-1\wedge a>1]\Leftrightarrow [(1+x)^a\geqq1+ax].  \]

Take ln, we have:

\[ a\ln(1+x)\geqq \ln(1+ax).  \]

It's Jensen for f(x)=-lnx. Oh, $0=\left(1-\frac{1}{a}\right)\ln 1$ :)

I think that lotharek want to write that $0=(1-a)ln1$ (not $(1-\frac{1}{a})$)


Can we email the solutions tonight and snail mail the entry form tomorrow or does it need to be mailed by tonight as well?

You can mail the entry form tomorrow.

Basically, as long as we have it by the time we release the Round 1 scores (which won't be until mid-November), you're fine.  But your scores won't show up unless we have the form on file.


calculus
derivative
function
real analysis
real analysis unsolved

Find n-th derivative for following functions:
$ y \equal{} \frac {1}{a^{2} \minus{} b^{2}x^{2}}$
$ y \equal{} \frac {1}{a^{2} \plus{} b^{2}x^{2}}$

hello, rewriting your first term in the form
${\frac{1}{2a}(\frac{1}{a+bx}-\frac{1}{-a+bx}})=f(x)$
we get
$f^{(n)}(x)=\frac{1}{2a}((-1)^nn!(a+bx)^{-(n+1)}b^n-(-1)^nn!(-a+bx)^{-(n+1)}b^n)$.
Sonnhard.


'Εστω $\pi(n)$ το πλήθος των πρώτων που είναι μικρότεροι ή ίσοι του $n$ και $p(n)$ ο $n$-ιοστός πρώτος αριθμός. Να δειχθεί ότι:

$1) \pi(n)=-1+\sum_{i=3}^{n}{\left((i-2)!-i\cdot\left[\frac{(i-2)!}{i}\right]\right)}$ για κάθε φυσικό $n\geq 4$ ,

$2)p(n)=2+\sum_{i=2}^{[2+2n\cdot ln(n)]}{\left(1-\left[\frac{\pi(i)}{n}\right]\right)}$   για κάθε φυσικό $n\geq 2$

(με [...] συμβολίζω το ακέραιο μέρος).

Στέλιος

gia to proto....
me epagogi...
gia n=4....  2=-1+(1+2) isxii
esto oti isxii gia n
8do isxii gia n+1
π(n+1)=π(n)+(n-1)!-(n+1)[(n-1)!/(n+1)]    (1)

8do to akolu8o <limataki>...to opio an to apodiksume 8a sinepagete tin isxi tis(1) kai ara 8a oloklironete i epagogi...

esto x fisikos...

i parastasi x!-(x+2)[x!/(x+2)] ine isi  me miden an o x+2 ine sin8etos i me monada an o x+2 ine protos...

an o x+2 ine sin8etos tote oli i proti paragontes tu stin kanoniki morfi  8a ine mikroteri i isi tu x, (o x+1 den mpori na ine protos paragontas tu x+2...) kai ara 8a iparxun san paragontes sto x!....epomenos to [x!/(x+2)] ine akereos kai ara x!-(x+2)[x!/(x+2)]=0

 gia to deytero melos (an o x+2)ine protos  ..den to exo apodiksi akoma...

pantos mon ayto meni gia na apodix8i to 1)...otan to apodikso..8a to stilo...

 :D

to apediksa...
8eorontas tin eyklidia dieresi tu x! me to x+2 pernume x!=k(x+2)+u
opote  i parastasi ginete x!-(x+2)[x!/(x+2)]=k(x+2)+u-(x+2)k=u diladi isute me to ipolipo tis eyklidias dieresis tu x!me to x+2 ....
omos o x+2 ine protos...apo to 8eorima wilson lipon pernume...
(x+1)! ine isoipolipo -1 mod(x+2) kai epidi x+1 isoipolipo -1 mod(x+2) ,
ine x! isoipolipo 1 mod (x+2) ,diladi to ipolipo tis eyklidias dieresis tu x! me to x+2 ine 1.......kai i apodiksi olokliro8ike...

@stedes
[hide]pou tus vrikes tus tipus?[/hide]

 :D

[quote]esto x fisikos...

i parastasi x!-(x+2)[x!/(x+2)] ine isi me miden an o x+2 ine sin8etos i me monada an o x+2 ine protos... [/quote]
Με μια διευκρίνιση. Αν x=2 ο x+2 είναι σύνθετος αλλά η παράσταση ισούται με  2. :wink:
Κατά τα άλλα ok. Επίσης βγαίνει απευθείας χρησιμοποιώντας ένα γνωστό θεώρημα (ποιό; ).
Στο δεύτερο ερώτημα δυσκολεύτηκα πιο πολύ πάντως.

[hide="@RAKAR"]Τους βρήκα στο internet, χωρίς απόδειξη όμως και γι'αυτό προσπάθησα να τους αποδείξω. Κι έχω κι άλλους αν θες.  :wink: [/hide]

[quote="stedes"]
Με μια διευκρίνιση. Αν x=2 ο x+2 είναι σύνθετος αλλά η παράσταση ισούται με  2. :wink:
[/quote]

 :huh:  :huh: .........??
afu i arxiki isxii gia $n\geq 4$ o $x=n-1$ 8a ine $\geq 3$..... :wink:

vsk..... i ikasia tu rieman ti lei [b]akrivos[/b]....??

 :D

edit:to provlimataki ayto emfanizete otan ston paranomasti iparxi..dinami kai ara prepi na eksetasume an iparxun oli i paragontes ayti sto x!...
pantos mono gia x=2 exume to prvlm....(apodikniete eykola)+gia ayto malon o tipos lei $n\geq 4$....
diladi iparxi tipos pu na dini ton n-osto proto ari8mo+to pli8os olon ton proton mexri enos do8entos n..... :o  :o

 :D

[quote="RAKAR"]afu i arxiki isxii gia $n\geq 4$ o $x=n-1$ 8a ine $\geq 3$..... Wink[/quote]
Μα αφού στην απόδειξή σου λες x γενικά φυσικός και δε λες x=n-1... . Anyway, άσ'το.

[quote]diladi iparxi tipos pu na dini ton n-osto proto ari8mo+to pli8os olon ton proton mexri enos do8entos n..... Shocked Shocked[/quote]

 :yup: Καλό ε; :P

[quote="RAKAR"]
diladi iparxi tipos pu na dini ton n-osto proto ari8mo+to pli8os olon ton proton mexri enos do8entos n..... :o  :o
[/quote]

Fisika, apla texnika einai sxetika "axristos" mias kai gia na vreis px ton $25o$ sti seira prwto arithmo, o ipologistis prepei na douleuei kamposi writsa (gia enan psilokalo algorithmo) kai isws merika deuterolepta [(kai pali polla einai) gia enan kalo algorithmo]. Fantasteite na thelete na vreite ton 15657o prwto arithmo ti tha ginei!!!

Autoi oi algorithmoi den exoun opws leme kali poliplokotita (complexity sta agglika) kai argoun na ektelestoun apo enan ipologisti. Genika grigoroi algorithmoi thewrountai ekeinoi pou exoun poliwnimiki poliplokotita (dioti enas ipologistis tous epeksergazetai poli poli grigora).

Sas episinaptw kai ena diaforetiko tipo apo ekeino tou Steliou pou vriskei ton n-osto prwto, den exei kamia texniki aksia, apla einai enas omorfos tipos.

(Sta Ellinika sxoleia iparxei o mithos oti den iparxei tipos pou na dinei ton n-osto prwto arithmo gi'auto kai katalavainw tin ekpliksi tou RAKAR (ki egw tin idia aisthisi eixa). Apla den iparxei "grigoros" tipos gia na ton vreis... An omws exeis oreksi gia prakseis tote iparxoun mpolikoi :) ).

Alexandros

[quote="RAKAR"][color=blue][b]Λήμμα[/b]
Έστω x φυσικός...

Η παράσταση $x!-(x+2)\frac{x!}{x+2}$ είναι ίση με μηδέν αν ο $x+2$ είναι σύνθετος ή με μονάδα αν ο $x+2$ είναι πρώτος... [/color]

Αν ο $x+2$ είναι σύνθετος τότε όλοι οι πρώτοι παράγοντές του στην κανονική μορφή θα είναι μικρότεροι ή ίσοι του $x$,
(o $x+1$ δεν μπορεί να είναι πρώτος παράγοντας του $x+2$...) και όλα θα υπάρχουν σαν παράγοντες στο $x!$ .... επομένως το $\frac{x!}{x+2}$ είναι ακέραιος και άρα $x!-(x+2)\frac{x!}{x+2}=0$
[/quote]
[quote="RAKAR"]edit: Το προβληματάκι αυτό εμφανίζεται όταν  στον παρονομαστή υπάρχει..δύναμη και άρα πρέπει να εξετάσουμε αν υπάρχουν όλοι οι παράγοντες αυτοί στο $x!$
Πάντως μόνο για $x=2$ έχουμε το πρόβλημα ....([u][color=red]αποδεικνύεται εύκολα[/color][/u]  :huh: ) και γι αυτό μάλλον ο τύπος λέει $n\geq 4$....
[/quote]

@RAKAR
Μπορείς σε παρακαλώ να γράψεις πώς αποδεικνύεται ότι για $x> 2$ ο $\frac{x!}{x+2}$ είναι ακέραιος;
Γιατί το απέδειξα αλλά μάλλον διάλεξα το δύσκολο δρόμο...  :(

@cretanman
Αυτός ο αλγόριθμος έχει μεγαλύτερη πολυπλοκότητα πχ από το κόσκινο του Έρατοσθένη;

Λουκάς

Λουκά το ότι ο $\frac{x!}{x+2}$ είναι ακέραιος αν ο $x+2$ είναι σύνθετος και $x>2$ προκύπτει απευθείας από το θεώρημα Wilson (μια απόδειξη υπάρχει [url=http://planetmath.org/?op=getobj&from=objects&id=1343]εδώ[/url]).

Όσον αφορά τώρα τη δεύτερη ερώτηση, εμπειρικά και μόνο απαντάω ότι το κόσκινο του Ερατοσθένη έχει μικρότερη πολυπλοκότητα. Άλλωστε θεωρείται από τους πιο γρήγορους αλγόριθμους για την εύρεση πρώτων και το μόνο μειονέκτημα του είναι ότι απαιτεί πολύ μνήμη αφού δουλεύουμε με πίνακες.
Όπως είπε και ο Αλέξανδρος, οι παραπάνω τύποι δεν έχουν καμία απολύτως πρακτική αξία για τον υπολογισμό πρώτων (...αφού για μεγάλα $n$ περιέχουν πολύ μεγάλους αριθμούς που οι γλώσσες προγραμμματισμού γενικής χρήσης δεν υποστηρίζουν). Απλώς είναι όμορφοι.  :)

Το κόσκινο του Ερατοσθένη έχει πολυπλοκότητα $\Theta(n \ln \ln n)$, σαφώς καλύτερη από την παραπάνω (αλλά ναι, χρησιμοποιεί πολύ από τη μνήμη)


Let $(u_{n})$be a complexe sequence; such that
$\forall k$ non negative integer we have , $\sum_{n=0}^{\infty}u_{n}^{k}=0$
 and $\sum_{n=0}^{\infty}|u_{n}|$converge
proof that : $\forall n$ we have $u_{n}=0$

it's been posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=entire&t=101156]here[/url] (3rd post)


I'm new to AOPS.  What are these ratings that everybody has?  How is one rated?  And why are the millenium problems (like the riemann hypothesis, poincare conjecture, hodge conjecture)underneath eveyones user names?

This would go in the questions forum...

But those ratings are based entirely on post count...except of course for administrators.

[u][b]Post count [/b][/u] -  [u][b]Rank[/b][/u] - [u][b]Number of stars[/b][/u]
below 20 -      New member - $0$
20 to 49 -      P vs NP - $\frac1{2}$
50 to 99 -      Hodge Conjecture - $1$
100 to 249 -   Poincare Conjecture - $2$
250 to 499 -   Reimann Hypothesis - $2\frac1{2}$
500 to 999 -   Yang Mills Theory - $3$
1000 to 1999 - Navier Stokes Equation - $4$
2000 or above - Birch and Swinnerton Dyer - $5$

[quote="bubka"]and with that double post, you move closer to P vs NP, see?[/quote]

(chan) just clicked twice...

It doesn't matter, since G&FF's post count is about to be nuked anyway.

You need 2500 for Birch and Swinnerton Dyer, not 2000.

But this question occured at least 5 times before...

If I have any chance of getting B&SD, it will be washed away if VV does what he threatened to! :(

Why do you lot care what your post count and rank is? If you get 2500 posts by posting non-maths posts in G&FF, it doesn't mean anything.

I don't see how the next post rank is anything to aim for.

[quote="mbabbitt2003"]It makes you more distinguished.[/quote]
I doubt.
Especially having as good as all posts here is more or less shameful (think about: you are on a forum about math!) ;)

[quote="mbabbitt2003"]well, if you almost never posted, you would look like a new member when you've been here for two years![/quote]That means nothing, the content of the posts is all that matters. There are certain people in here with lotsa posts, but after browsing their content, more than 80% of their posts I would classify as 'SPAM'. Not pointing fingers to anybody, but really guys, you know it's true. I spam sometimes too (and with a reason :P )

The major upsetting issue with most users here that I see is that most of us (yes that's including me) feel compelled to post something. It is always that 'unnerving' hand that wants to type up something or that math problem fiddling with your brain and you want to post your hint/solution, and so on.

What are the kind of AoPSers that usually get my attention when they post something? Those that post helpful contents like t0rajir0u, frt, kunny, or guys from the Olympiads section like grobber, Arne, blahblahblah, or admins and professors like jmerry and Kent Merryfield, just to name a few (sorry if I missed you :) ). I would rather view a post from a user with as little as 50 post counts but containing beautiful solutions by him/her than a magnificent guy with 3000 posts but all that he/she posts is 'Hi guys, wanna dance?' or 'Answer is 1' type of posts.

[quote="bubka"]Oh! I was not rebuking him in the very least! I love double posts, especially when they look accidental, and nobody can scold you! And the best part is this ...

[hide="A Secret"]When you double post, and delete the latter, your post count goes up by 2  [i]and stays up by 2!![/i][/hide][/quote]

 :|  And by doing so, you could get your post number up to astronomical heights without anyone knowing you are spamming?  I doubt our Überlord hasn't thought of that?

[quote="10000th User"][quote="filletwho"][quote="10000th User"]

Those that post helpful contents like t0rajir0u, frt, kunny, or guys from the Olympiads section like grobber, Arne, blahblahblah, or admins and professors like jmerry and Kent Merryfield, just to name a few (sorry if I missed you :) ). [/quote]

I think you're forgetting Arnav.[/quote]Ah yes, MysticTerminator, towersfreak2006, tetrahedr0n, and the list goes on...[/quote]

 :rotfl:  Mysitc?You gotta be kidding me....

How does it matter??? There's really no point.

only 75 more till Yang-Mills Theory. my avg post count used to be so high
[u]COUNTDOWN[/u]
75
74
73

mine said that it was three right after the day that I posted 100 times WITHOUT spamming in ONE DAY.

i was so excited the day i saw this site :lol: ...March 15... but it was 3 days before state mathcounts. if only i had discovered this before.... in those days, my avg post count was above 10 for a week or so.

my post per day count used to be near 16.   :o

[quote="nutz_for2.718281828"]I have never seen him post something helpful in a maths forum yet.
(I've been here for almost 5 months, so that figures why.)(no offense, Mystic) (oh! He also makes it VERY difficult to post in the quoting competition. Can't he quote the SMALLL posts?[/quote]

lol, it's cool ... I only post in f&g nowadays (I used to post in the olympiad forums)

I had crossed twenty before MithsApprentice did all those horrible things. I am struggling to keep a steady eleven now. But just reached NSE though.

last summer, i posted a good 6-7 posts per day, but every one of them was in either intermediate or pre-olympiad, with a few in puzzles and brainteasers/math and strategy.  Now, I post only occasionally, and that too in mostly G&FF and olympiad section.

[quote="bubka"]Why do people post in the olympiad section?[/quote]
Wow that's a toughie..... :P  :huh:

[quote="bubka"]Noooooooooooooooooooo!
It isn't, it isn't, it isn't!

NOOOOOOOOOOOOOOOOOOOOOOOOOO! NEVER! NEVER EVER![/quote]

*cough* All of your "GAMES" were nuked especially your Tug of Wars *cough* and that dropped everyones who really participatedpost count by like a 100.

[quote="bubka"]Why do people post in the olympiad section?[/quote]

There are those who have competed in Math Olympiad and still enjoy doing the problems or just want to help out their felllow Mathlete.

maybe ... i just don't like it.

By the way I am into the hundred top posters, with this post leaving me at number 97, one post behind ch1n353ch3s54a1l.

well you are right, but what are you doing right now?


articles Asymptote LaTeX trigonometry search

Ok so I reinstalled Asymptote.
And I test it out using a few simple lines.
[code]
\documentclass{article}
\usepackage{asymptote}
\begin{document}
\begin{asy}
draw((0,0)--(1,1));
\end{asy}
\end{document}
[/code]
I named it "Example"
So I build the file, and this is what I get:
[code]
! Package Asymptote Warning: File Example_1.pdf does not exist on input line 6.
  (Example.Aux) )
  No pages of output.
[/code]
Thus, I went to line 6, and line 6 simply says "\end{asy}"
Ok I have no idea what's wrong here, maybe I installed asymptote incorrectly?

Did you compile the .tex file with pdflatex or something before compiling the asymptote?

meh I get the same errors after trying to compile the example given [url=http://asymptote.sourceforge.net/doc/LaTeX-usage.html]here[/url]:

[code]Package asymptote Warning: file Untitled_1.pdf does not exist on input line 67.

Package asymptote Warning: file Untitled_2.pdf does not exist on input line 83.

Package asymptote Warning: file Untitled_3.pdf does not exist on input line 125
[/code]

[hide="heres the full Untitled.log file if anyone can make sense of it.."][code]This is pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6) (format=pdflatex 2008.4.20)  27 APR 2008 09:08
entering extended mode
\write18 enabled.
 %&-line parsing enabled.
**Untitled.tex
(./Untitled.tex
LaTeX2e <2005/12/01>
Babel <v3.8h> and hyphenation patterns for english, usenglishmax, dumylang, noh
yphenation, arabic, basque, bulgarian, coptic, welsh, czech, slovak, german, ng
erman, danish, esperanto, spanish, catalan, galician, estonian, farsi, finnish,
 french, greek, monogreek, ancientgreek, croatian, hungarian, interlingua, ibyc
us, indonesian, icelandic, italian, latin, mongolian, dutch, norsk, polish, por
tuguese, pinyin, romanian, russian, slovenian, uppersorbian, serbian, swedish,
turkish, ukenglish, ukrainian, loaded.
(/usr/local/texlive/2007/texmf-dist/tex/latex/base/article.cls
Document Class: article 2005/09/16 v1.4f Standard LaTeX document class
(/usr/local/texlive/2007/texmf-dist/tex/latex/base/size12.clo
File: size12.clo 2005/09/16 v1.4f Standard LaTeX file (size option)
)
\c@part=\count79
\c@section=\count80
\c@subsection=\count81
\c@subsubsection=\count82
\c@paragraph=\count83
\c@subparagraph=\count84
\c@figure=\count85
\c@table=\count86
\abovecaptionskip=\skip41
\belowcaptionskip=\skip42
\bibindent=\dimen102
)
(/Users/yoos/Library/texmf/tex/latex/misc/asymptote.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/graphics.sty
Package: graphics 2006/02/20 v1.0o Standard LaTeX Graphics (DPC,SPQR)

(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/trig.sty
Package: trig 1999/03/16 v1.09 sin cos tan (DPC)
)
(/usr/local/texlive/2007/texmf/tex/latex/config/graphics.cfg
File: graphics.cfg 2007/01/18 v1.5 graphics configuration of teTeX/TeXLive
)
Package graphics Info: Driver file: pdftex.def on input line 90.

(/usr/local/texlive/2007/texmf-dist/tex/latex/pdftex-def/pdftex.def
File: pdftex.def 2007/01/08 v0.04d Graphics/color for pdfTeX
\Gread@gobject=\count87
))
\AsyStream=\write3
\AsyTestStream=\read1
\openout3 = `Untitled.asy'.

\c@asy=\count88
)
(./Untitled.aux)
\openout1 = `Untitled.aux'.

LaTeX Font Info:    Checking defaults for OML/cmm/m/it on input line 15.
LaTeX Font Info:    ... okay on input line 15.
LaTeX Font Info:    Checking defaults for T1/cmr/m/n on input line 15.
LaTeX Font Info:    ... okay on input line 15.
LaTeX Font Info:    Checking defaults for OT1/cmr/m/n on input line 15.
LaTeX Font Info:    ... okay on input line 15.
LaTeX Font Info:    Checking defaults for OMS/cmsy/m/n on input line 15.
LaTeX Font Info:    ... okay on input line 15.
LaTeX Font Info:    Checking defaults for OMX/cmex/m/n on input line 15.
LaTeX Font Info:    ... okay on input line 15.
LaTeX Font Info:    Checking defaults for U/cmr/m/n on input line 15.
LaTeX Font Info:    ... okay on input line 15.

(/usr/local/texlive/2007/texmf-dist/tex/context/base/supp-pdf.tex
[Loading MPS to PDF converter (version 2006.09.02).]
\scratchcounter=\count89
\scratchdimen=\dimen103
\scratchbox=\box26
\nofMPsegments=\count90
\nofMParguments=\count91
\MPscratchCnt=\count92
\MPscratchDim=\dimen104
\MPnumerator=\count93
\everyMPtoPDFconversion=\toks14
)

Package asymptote Warning: file Untitled_1.pdf does not exist on input line 67.

LaTeX Font Info:    External font `cmex10' loaded for size
(Font)              <12> on input line 73.
LaTeX Font Info:    External font `cmex10' loaded for size
(Font)              <8> on input line 73.
LaTeX Font Info:    External font `cmex10' loaded for size
(Font)              <6> on input line 73.

Package asymptote Warning: file Untitled_2.pdf does not exist on input line 83.

Package asymptote Warning: file Untitled_3.pdf does not exist on input line 125
.

[1

{/usr/local/texlive/2007/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]
(./Untitled.aux) )
Here is how much of TeX's memory you used:
 836 strings out of 94073
 10334 string characters out of 1162984
 56824 words of memory out of 1500000
 4172 multiletter control sequences out of 10000+50000
 6376 words of font info for 23 fonts, out of 1200000 for 2000
 645 hyphenation exceptions out of 8191
 25i,4n,19p,168b,110s stack positions out of 5000i,500n,6000p,200000b,5000s
</usr/local/texlive/2007/texmf-dist/fonts/type1/bluesky/cm/cm
r12.pfb></usr/local/texlive/2007/texmf-dist/fonts/type1/bluesky/cm/cmr8.pfb>
Output written on Untitled.pdf (1 page, 11745 bytes).
PDF statistics:
 14 PDF objects out of 1000 (max. 8388607)
 0 named destinations out of 1000 (max. 131072)
 1 words of extra memory for PDF output out of 10000 (max. 10000000)

 [/code][/hide]

There's probably an error when you ran Asymptote itself so no picture was created - what error was it? See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148557]Reading asymptote error messages[/url] or [[Asymptote:_Advanced_Configuration#Showing_Asymptote_error_messages_in_TeXnicCenter]] if the DOS window closes too quickly to read the error.

Do you know of a similar program for TeXShop on a Mac, or at least the filename of any log files that are produced?

Ugh.. looking forward to digging through the mess of root files again :P

Just run Asymptote from the command line and it will tell you the error (that's all the Windows version does in reality).

Hmm okay this might be a bit off-topic but here's what I get in asy mode in Terminal:
[code]> size(100);
> pair A,B,C,O,I;
> A=(0,0); B=(10,0); C=(5,15);
> draw(A--B--C--cycle);
> Cannot execute gv
Please put in config.asy:

import settings;
psviewer="PATH";

where PATH denotes the correct path to your PostScript viewer.

Alternatively, set the environment variable ASYMPTOTE_PSVIEWER
or use the command line option -psviewer="PATH"

> draw(circumcircle(A,B,C));
-: 1.6: no matching variable 'circumcircle'
[/code]

I guess that last part is behaving as it's supposed to, but where might I find the psviewer? A google search returned with no answers and Spotlight can't find it either..

You need to install [url=http://www.macwrite.com/critical-mass/ghostscript-mac-os-x]Ghostscript[/url]

[hide="ignore this.."]I compiled ghostscript from the source release.. I found gs in usr/local/bin but I still can't find gv..[/hide]

EDIT: wait I'm an idiot I didn't compile ghostview.. :blush:

Thanks for the help so far :)

EDIT: actually, it would be good to ask exactly how to make a config.asy file - I just used a normal text editor to input the content, saved/exited, then just changed the extension.. is that the wrong way to do it?

EDIT2: maybe installing windows via boot camp and using TeXNicCenter will be easier..

[quote="undefined117"]actually, it would be good to ask exactly how to make a config.asy file - I just used a normal text editor to input the content, saved/exited, then just changed the extension.. is that the wrong way to do it?[/quote]
That's fine; config.asy is just an ordinary text file.

Guys, I get the error before I run Asy, so it can't be an asymptote error can it?

Delete all the files except the tex one. This means any errors will be new. Then
1. compile to pdf
2. run asymptote
3. compile to pdf
[b]but[/b] stop as soon as you get an error. That will tell you where it's occurring.

I see what happening. It's not really an error at step 1. The purpose of step 1 is just to create an asy file, not to create a pdf. Did step 1 create Example1.asy? If it did then you can do steps 2 (which will create Example1_1.pdf) and step 3 and get Example1.pdf

Let [b]RandomName.tex[/b] be some .tex file with Asymptote code in it.

There are 3 steps to compiling [b]RandomName.tex[/b] to a complete .pdf file:

[list]1. [color=darkred][b]pdflatex[/b][/color] - a main [b]RandomName.pdf[/b] file, which contains the entire output of [b]RandomName.tex[/b], is created. This file does not yet contain the Asymptote images you coded.

2. [color=darkred][b]asy[/b][/color] - 1 [b]RandomName.asy[/b] file is created, which contains all the Asymptote code found in [b]RandomName.tex[/b]. Also, [b]RandomName_1.pdf[/b], [b]RandomName_2.pdf[/b], etc. files are created, one for each Asymptote code in [b]RandomName.asy[/b].

3. [color=darkred][b]pdflatex[/b][/color] - [b]RandomName.pdf[/b] incorporates the now-existing .pdf files that contain the Asymptote images.[/list]
If you're still having trouble, delete everything except for the .tex file itself, then [b]pdflatex[/b], [b]asy[/b], and [b]pdflatex[/b] again.

[quote="ZhangPeijin"]Yeah It created Example.asy, not Example_1.asy though, and the warning says the file Example_1.pdf doesn't exist, while I named my file Example.tex, not Example_1.tex.
Also, when I do name my file Example_1.tex, the warning says Example_1_1.pdf doesn't exist.[/quote]
I don't think you understood what I said. If it created Example.asy from Example.tex then that is [b]correct[/b] and you can [b]ignore[/b] the error. That is only step 1. Now you must use Asymptote to create Example_1.pdf then compile again to create Example.tex.

I will repeat the instructions again, which i_like_pie has also done. If you start with Example.tex then
1. run TeX to create Example.asy. A pdf will only be created if there's some text. In your example there's none, just a picture so no pdf will be created;
2. run Asymptote on Example.asy which will create Example_1.pdf;
3. run TeX on Example.tex and it will create Example.pdf with the picture from Example_1.pdf in it.

Does running asymptote create Example_1.pdf? If not, what error message do you get from Asymptote?

asy shows no errors.
Maybe I should go install GS...

If there are no errors then a file will have been created. What file is created? Are you using pdflatex or latex? If you only use latex then you won't be getting a pdf. See [[Asymptote:_Advanced_Configuration#PDF_output_configuration]].

You must also give us much more information, not very short postings. How about answering i_like_pie's questions?

[quote="i_like_pie"]The code you provided in post #1 does not produce any errors when I run it; are you sure it's identical to what you're actually compiling?

Also, what editor are you using, and on what OS?

I'm not sure, but maybe your editor wants to include the non-existing .pdf files on the first compilation. However, those can't be made until you have a .asy file, which doesn't happen until you compile the .tex file itself, and that brings us back to the original problem. If this is the case, install the latest version of the editor (maybe the problem has been fixed)/get a different editor.[/quote]
Yes I'm sure.
Umm, I'm using Texnic Center and Foxit Reader
Microsoft Windows XP Home
Yes, an asy file was created when I ran asy.

[quote="ZhangPeijin"]Yes, an asy file was created when I ran asy.[/quote]No it doesn't! When you compile using TexnicCenter's LaTeX=>PDF you get the asy file. When you run asy you get a [i]different[/i] file - a ps or pdf depending how you run asy.
Thanks for answering i_like_pie's questions so how about answering mine?
[quote="stevem"]If there are no errors then a file will have been created. What file is created? Are you using pdflatex or latex? If you only use latex then you won't be getting a pdf. See [[Asymptote:_Advanced_Configuration#PDF_output_configuration]].
[/quote]We need to know how you run asymptote? What commands do you use? Do you use pdflatex in the command line - see [[Asymptote:_Advanced_Configuration#PDF_output_configuration]]?
It's impossible to diagnose the problem without all the information.

Yeah ok, umm, I use pdflatex.
Command is Alt-A
EDIT: Asy says "C:\Program Files\Asymptote\redir.exe" is not recognized as an internal or external command, operable command, or batch file.

Aha! Each time a little more info. So you are using [[Asymptote:_Advanced_Configuration#Showing_Asymptote_error_messages_in_TeXnicCenter]].
Don't be shy - aways give as much information as possible.
[quote="ZhangPeijin"]EDIT: Asy says "C:\Program Files\Asymptote\redir.exe" is not recognized as an internal or external command, operable command, or batch file.[/quote]That probably means that you haven't saved redir.exe to C:\Program Files\Asymptote. The instructions say:
[quote="Showing_Asymptote_error_messages_in_TeXnicCenter"]Download [redir.zip], extract redir.exe, virus check it and save in C:\Program Files\Asymptote[/quote]

Eh, ok I did that, but the command now says nothing.
Just says Asymptote is working ... Please wait.
And yeah, nothing happens.

Looks like a problem with Asymptote. Do you know how to run asymptote from the command line? If so, run it and see what errors you get. If not, follow the instructions in the first post at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148557 and see what happens. Report back what you see. A screenshot would be even better.
You could also try reinstalling Asymptote in case that's causing the problem.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1122464#1122464]I've got it![/url] :lol:

[quote="ZhangPeijin"]Yes, an asy file was created when I ran asy.[/quote]

If you have an .asy file, try dragging that file onto the Asymptote icon you should get on your desktop once you install the windows binary.. I'm thinking it'll output a filenamehere_1.pdf onto the Desktop, after which you can move that file into the directory that has all your .tex .aux .log files and compile the .tex file again.


modular arithmetic
number theory
prime numbers

I saw an interesting question today that asked to find the remainder of $ 1*2*3\cdots*22$ when divided by $ 23$. In other words $ 1*2*3*\cdots*22\equiv x\pmod{23}$ Find $ x$.

After some tries for the first couple of prime numbers and multiplying the modular inverse of each number I found that it was $ 22$ or in general $ \minus{}1$.

$ 1*2 \equiv \minus{}1\pmod{3}$
$ 1*2*3*4\equiv \minus{}1\pmod{5}$
$ 1*2*3*4*5*6\equiv \minus{}1\pmod{7}$
etc...

Im just wondering if the following holds for all prime numbers:
If $ p$ is prime, then
$ 1*2*3\cdots *(p\minus{}1)\equiv \minus{}1\pmod{p}$

If so can you prove it?

[url=http://www.artofproblemsolving.com/Wiki/index.php/Wilson%27s_Theorem]Wilson's Theorem[/url] states that $ (p \minus{} 1)!\equiv{ \minus{} 1}\pmod{p}$, where p is a prime.

There's an extension to Wilson's theorem which is easy to prove.

If $ n\neq4$ is composite, then $ (n\minus{}1)!\equiv 0\mod n$.  If $ n\equal{}4$, then obviously $ (n\minus{}1)!\equiv 2\mod n$.
The only case not covered is if $ n\equal{}0$ or $ 1$, which are easy to deal with.

Thanks to the both of you, I wish I would have known it earlier to answer my problem tons quicker :)


Let $ X$ be a Hausdorff topological space, and let $ A,B\subset X$ be disjoint compact sets in $ X$. Show that there exist sets $ U,V\subset X$ s.t. $ U$ and $ V$ are open, $ A\subset U$, $ B\subset V$, and $ U\cap V \equal{} \O$.

Doesn't this only apply if $ A\cap{B}\equal{}\emptyset$? Unless you mean find open sets $ U,V$ such that $ U\subset{A}$ and $ V\subset{B}$

But this isn't necessarily true either since finite sets are compact and open sets could be inifinite.

Sorry, yes, $ A\cap B \equal{} \O$ as well. That's all you need, I believe, as I have the proof right here. When I first did it I wrote "...disjoint compact sets $ A,B\subset X$..." But the rest of the proof looks good. (I accidentally used $ \in$ instead of $ \subset$ in my post, but it's fixed now).

Lemma 1: Let $ x \in B$. Then there are open sets $ U_x, V_x$ such that $ A \subset U_x$, $ x \in V_x$, and $ U_x \cap V_x \equal{} \emptyset$.

Proof: For any $ y \in A$, there are open sets $ R_y,S_y$ such that $ (y,x) \in R_y \times S_y$ and $ R_y \cap S_y \equal{} \emptyset$. There are finitely many $ y$ such that $ \bigcup_y R_y$ contains $ A$. Then our desired sets are $ \bigcup_y R_y \equal{} U_x$, $ \bigcap_y S_y \equal{} V_x$. (union and intersection over these finitely many $ y$)
==========================

From here, finitely many $ V_x$ cover $ B$. Then our desired open sets are $ U \equal{} \bigcap_x U_x$, $ V \equal{} \bigcup_x V_x$. (union and intersection over finitely many $ x$)

Exactly what I did. The problem that follows from this one is the (admittedly easy) corollary: If $ X$ is a compact Hausdorff topological space, $ X$ is normal.


Support
AMC 10
AMC 10 B
analytic geometry

2004 AMC 10 B Problem 20

In $ \triangle ABC$ points $ D$ and $ E$ lie on $ BC$ and $ AC$, respectively. If $ AD$ and $ BE$ intersect at $ T$ so that $ \frac{AT}{DT}\equal{}3$ and $ \frac{BT}{ET}\equal{}4$, what is $ \frac{CD}{BD}$?

Can someone help me with this problem? Sorry I couldn't make a diagram.

you may want to look up mass point on AoPS wiki or somewhere else
... ahh it seems that mass point does not have a decent explanation... guess ill do it myself

mass point concerns placing weights on vertices and uses the principle of first class levers
the fulcrum must support the sum of the weights on the ends of the lever
the lever must balance, so length of 1 side of lever * mass on corresponding vertex = length of other side * mass on other vertex
if a point is on 2 levers, consider the point as if it was only on one of those levers (it doesnt matter which, you get the same answer)
this rule applies to all existing levers

yeah... im horrible at explaining mass point
try to look at the picture

Ummm. lol? I searched it and got "point mass"...

Can someone show me how to do the problem?

ok here is a last ditch thing you can do
assume it is right and plot the triangle on the cartesian plane :D

but if anyone else knows how to explain mass point please do
textangle, you may want to ask someone else... im afraid im really bad at masspoint

Thanks anyways.

For everyone else: The diagram wasn't a right triangle.


inequalities
logarithms
inequalities proposed

Show that for all positive reals $a,b,c$ for which $a+b+c=1$ the following inequality is always true:
[list]$\sqrt{a}  + \sqrt{b} + \sqrt{c} \geq ab+bc+ca   $[/list]

what's the equality case? It seems that the inequality is strict.
\[
\begin{gathered}
  \sqrt a  + \sqrt b  + \sqrt c  \geqslant \frac{{1 - (a^2  + b^2  + c^2 )}}
{2} \hfill \\
   \Leftrightarrow \sum {a^2  + 2\sqrt a }  \geqslant 1 \hfill \\
\end{gathered}
\]
while we have
\[
a^2  + 2\sqrt a  \geqslant 3a
\]
So L.H.S. is larger or equal to 3 indeed, qed.

[quote="jaka"]Equality would be when $a^2= \sqrt{a}$, that is when $a^3=1$, so $a=1$, or when $a=0$. When $a=1$, we can have $b=c=0$ and the equality holds. When $a=0$, we can have $b=0, c=1$.

So it isn't a strict inequality. Or am I wrong?

Does anyone have any other solutions?[/quote]

You are wrong.

1. My solution also used $b^2 + 2 \sqrt{b} \ge 3b$ and so on, so to make my arguments true, you need to have all $b^2=\sqrt{b}$ and $c^2=\sqrt{c}$.

2. When two variables equal 0, R.H.S. = 0, L.H.S.=1, equality case cannot be attained.

[quote="Enrique Lesseps"]Yes, the inequality must be strict. I guess there is a mistake in the book i got it from  ;)[/quote]

If a+b+c=3, then everything's okay i think.

The problem is weaker then one of my old and nice inequalities (in fact, not that old, since we discussed its generalization today :D ):
   For any positive real numbers we have $ (x^2+y^2)(y^2+z^2)(z^2+x^2)(1/x+1/y+1/z)^2\geq 8 (x^2+y^2+z^2)^2$

and the problem  is equal to find the min of $k$,    $a,b,c$ are nonegetive numbers,$a+b+c=3$,then:
$2\sum a^k+\sum a^2 \geq 9$

i think it can be proved when $k  \geq  1$,$k \leq 0$,or $k(1-k)3^{(k-1)(k-2)} \leq 2$,by largrange multiply

It is Russian problem:
For all nonnegative $a,b,c$  for which $a+b+c=3$ prove that the following inequality is always true:
$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq ab+ac+bc$
Solution:
$\sum(a^2+2\sqrt{a})=\sum(a^2+\sqrt{a}+\sqrt{a})\geq \sum 3a=(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$

Oh yeah.Russia problem is easy.
Zhaobin,can you clearly your solution?

note $L(a,b,c, \lambda)=2\sum a^k+\sum a^2+\lambda(a+b+c-3)$
we let $L_a=2ka^{(k-1)}+2a+\lambda=0$
$L_b=2kb^{(k-1)}+2b+\lambda=0$
$L_c=2kc^{(k-1)}+2c+\lambda=0$
$L_{\lambda}=a+b+c-3=0$
if $2ka^{(k-1)}+2a+\lambda=0$,has one solution we will have $a=b=c$,
let $h(a)=2ka^{(k-1)}+2a$,we can prove h(a) is always increasing when $k(1-k)3^{(k-1)(k-2)} \leq 2 $ or $k \geq 1$ or $ k \leq 0$
so we have $L(a,b,c, \lambda)$ get extremum when one of $a,b,c$=0,or $a=b=c$,
but I don't know the other case.
indeed I'm not sure of my solution :(

[quote="Enrique Lesseps"]Show that for all positive reals $a,b,c$ for which $a+b+c=1$ the following inequality is always true:
[list]$\sqrt{a} + \sqrt{b} + \sqrt{c} \geq ab+bc+ca$[/list][/quote]
a and b and c are positifs then a<1 and b<1 and c<1 then $\sqrt{a}+\sqrt{b}+\sqrt{c}>a^2+b^2+c^2>ab +ac +bc$

[quote="hungkhtn"]We can general this problem.It will become more difficult alot:

Find the best constant $k$ such that:

If $a,b,c$ are nonegative real and $a+b+c=3$ then:

$a^k+b^k+c^k \ge ab+bc+ca$[/quote]

putting $a=b=\frac{3}{2}$ we get $k\geq 2-\frac{\ln 2}{\ln 3 -\ln 2}=k_0$
and we need only check the case when $a=b\geq k_0 , c\le k_0$ Im gonna do that later...
and I think akhrif's solution is ok! arqady. $\sqrt{\frac{1}{3}}>\frac{1}{9}$  :lol:

[quote="arqady"][quote="akhrif"]
a and b and c are positifs then a<1 and b<1 and c<1 then $\sqrt{a}+\sqrt{b}+\sqrt{c}>a^2+b^2+c^2>ab +ac +bc$[/quote]
You are mistaked.
Try $a=b=c=\frac{1}{3}.$ ;)[/quote]
[quote="Albanian Eagle"]
and I think akhrif's solution is ok! arqady. $\sqrt{\frac{1}{3}}>\frac{1}{9}$  :lol:[/quote]
I think you are mistaked, Albanian Eagle: $\sqrt{\frac{1}{3}}>\frac{1}{9}$ is true,
but  $3\cdot\sqrt{\frac{1}{3}}>\frac{1}{3}>\frac{1}{3}$ is wrong.  ;)
And where akhrif's solution? :?

:rotfl: I guess the only mistake in akhrif's solution is that he should have written $\ge$ instead of $>$. That is forgiveable. ;)

[quote="akhrif"]!!!!! anyone?[/quote]
What do you want $akhrif$? :?  :)
Mybe prove of this question
[quote="akhrif"]$x^5-x^3+x\geq p \Longleftrightarrow x^6\le 2p-1$[/quote]
then tell us what mean $(x;p)>0$.
If it just mean $x>0,p>0$,then in your question you have written somthing wrong.

[quote="akhrif"]yes $x>0,p>0$
where is the mistake $Tiks$ :?[/quote]
You think there aren't any mistake :)
Look! for example take $x=1$ then $x^5-x^3+x=1\geq p$
you want to prove that $1\geq p$<=>$1\leq 2p-1$<=>$p\geq 1$ :blush:
if you can do this,then I am Merelin Monro :rotfl:  :blush: .
Sorry :blush: ,
I think you must writed $x^6\geq 2p-1$ and not $x^6\leq 2p-1$ ;) .

[quote="akhrif"]thanks.
now  $(x,p)>0$  prove $x^5-x^3+x\geq p \Longleftrightarrow x^6\le 2p-1$[/quote]
OK,here is the solution for this one

$x;p>0$  prove $x^5-x^3+x\geq p \Longleftrightarrow x^6\geq  2p-1$

We just need to prove that $\frac{x^6+1}{2}\geq x^5-x^3+x$<=>

$(x-1)^2((x^2-1)^2+x^2)\geq 0$

so $\frac{x^6+1}{2}\geq p$=>$x^6\geq  2p-1$(and not$x^6\le 2p-1$ :lol: )
 ;)

[quote="Enrique Lesseps"]Show that for all positive reals $ a,b,c$ for which $ a \plus{} b \plus{} c \equal{} 3$ the following inequality is always true:
[list]$ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \geq ab \plus{} bc \plus{} ca$[/list][/quote]
The following stronger inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a\plus{}b\plus{}c\equal{}3.$ Prove that:
\[ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \ge \sqrt [7] {9(ab \plus{} bc \plus{} ca)^5}\]

[quote="arqady"][quote="Enrique Lesseps"]Show that for all positive reals $ a,b,c$ for which $ a \plus{} b \plus{} c \equal{} 3$ the following inequality is always true:
[list]$ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \geq ab \plus{} bc \plus{} ca$[/list][/quote]
The following stronger inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \ge \sqrt [7] {9(ab \plus{} bc \plus{} ca)^5}\]
[/quote]
It's a result from a can_hang's problem, see it here:
http://mathvn.org/forum/viewthread.php?thread_id=1180

[quote="nguoivn"][quote="arqady"]
The following stronger inequality is also true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \ge \sqrt [7] {9(ab \plus{} bc \plus{} ca)^5}\]
[/quote]
It's a result from a can_hang's problem, see it here:
http://mathvn.org/forum/viewthread.php?thread_id=1180[/quote]
What do you mean?
For example, $ 4(\sqrt a \plus{} \sqrt b \plus{} \sqrt c)\geq3(ab \plus{} ac \plus{} bc \plus{} 1)$ does not help here. :wink:

[quote="arqady"]
What do you mean?
For example, $ 4(\sqrt a \plus{} \sqrt b \plus{} \sqrt c)\geq3(ab \plus{} ac \plus{} bc \plus{} 1)$ does not help here. :wink:[/quote]
Sorry, I was wrong  :oops: .
But your ineq can be solved by this ineq:
Given $ a, b, c \geq\ 0$ and $ a\plus{}b\plus{}c\equal{}3$. Prove that: $ (ab)^{\frac{27}{20}}\plus{}(bc)^{\frac{27}{20}}\plus{}(ca)^{\frac{27}{20}} \leq\ 3$
The general problem for this ineq was solved by Pham Kim Hung and Vo Quoc Ba Can  :)


geometry
3D geometry
calculus
integration
modular arithmetic

Find all perfect cubes that can be written as the sum of two consecutive perfect squares.

[hide="Hint for one approach"]
$ k^2\plus{}(k\plus{}1)^2 \equal{} n^3 \iff (2k\plus{}1)^2 \equal{} 2n^3\minus{}1$

It suffices to find solutions to $ m^2 \plus{} 1 \equal{} 2 n^3$[/hide]

[quote="The Zuton Force"][hide]It suffices to find solutions to $ m^2 \plus{} 1 \equal{} 2 n^3$[/hide][/quote]

This is actually where I started. Is this form easier to solve?

I don't know, but I think I've seen treatment of that form of the equation around somewhere before. If I recall correctly, the solution set is:
[hide="Solutions"]$ (m,n) \equal{} (1,1)$, which gives $ 1^3 \equal{} 0^2 \plus{} 1^2$[/hide]

I'm blanking out, but I find a good proof (better than trying all integral m,n in [1, 10 000 000]), I'll post it  :)

Playing with mods a little bit,

For cubes, the recursive pattern in mod 9 is

$ 1,8,0,1,8,0,.....$
for $ 3m\plus{}1,3m\plus{}2,3m\plus{}3,..$

For Squares, it is

$ 1,4,0,7,7,0,4,1,0$
for $ 9n,9n\plus{}1,9n\plus{}2,9n\plus{}3,9n\plus{}4,9n\plus{}5,9n\plus{}6,9n\plus{}7,9n\plus{}8,...$

Summing consecutive square mods, we find that the only pairs that work are

$ 1 \plus{} 0 \equal{} 1$, $ 0 \plus{} 1 \equal{} 1$.

Thus we can rewrite 2 equations:

$ (9n)^{2} \plus{} (9n\plus{}1)^{2} \equal{} (3m\plus{}1)^{3}$
or
$ (9n)^{2} \plus{} (9n\minus{}1)^{2} \equal{} (3m\plus{}1)^{3}$

I don't know where to go from there

Maybe we can try considering the sum of two perfect squares to equal a perfect cube and go from there?

$ k^2\plus{}(k\plus{}1)^2\equal{}n^3\Longleftrightarrow (2k\plus{}1)^2\equal{}2n^3\minus{}1$
$ (2k\plus{}1)^2\minus{}1\equal{}2n^3\minus{}1\Longrightarrow k(k\plus{}1)\equal{}\frac{(n\minus{}1)(n^2\plus{}n\plus{}1)}{2}$

Observing the parity of the right side we see that for all even $ n$ the RHS is odd and therefor not divisible by two. Because $ n$ may not be even, the cube of $ n$ must also not be even.

From our first line we have the identity $ (2k\plus{}1)^2\equal{}2n^3\minus{}1$. Working in modulo ten, we can show that $ u\equal{}2k\plus{}1\Rightarrow u^2\equiv\{1,5,9\}\pmod{10}$ which implies that $ n^3\equiv \{\text{zero},1,3,5\}\pmod{10}$. Zero may be discarded because $ a^2\plus{}b^2\equal{}0$ has no real roots. We are limited to $ n^3\equiv\{1,3,5\}\pmod{10}$ and so to $ n\equiv\{1,5,7\}\pmod{10}$.

$ (n\minus{}1)(n^2\plus{}n\plus{}1)\equal{}2k(k\plus{}1)$
Then letting $ n$ be congruent to 1,5,7 modulo ten we have
$ \{0,4,6\}\cdot\{3,1,7\}\equal{}\{0,4,2\}\pmod{10}\Rightarrow k^2\plus{}k\equiv\{1,2,5\}\pmod{10}$
But because $ k(k\plus{}1)$ must be even we have $ k^2\plus{}k\equiv 2\pmod{10}$ implying $ k\equiv\{1,3\}\pmod{5}$. It follows that $ n^3,n\equiv0\pmod{5}$. Testing $ k$ of the form $ 5z\plus{}1,5z\plus{}3$ we notice that they always have a units digit of five when their squares are summed, and there are a good number of values under three-thousand of this form that are within $ 1/1000$ of a perfect cube. However, I am as of yet unable to prove that there exist no solutions of this form...

but I tried. :(

Anyone got a lead from there?

[hide="Discussion"] The equation $ y^2 \equal{} 2x^3 \minus{} 1$ is an [url=http://mathworld.wolfram.com/EllipticCurve.html]elliptic curve[/url], and well-known techniques exist for finding its set of integer points.  In particular, the substitution $ y \to \frac{y}{2}, x \to \frac{x}{2}$ gives the [url=http://mathworld.wolfram.com/MordellCurve.html]Mordell curve[/url] $ y^2 \equal{} x^3 \minus{} 4$; according to Mathworld, there is an elementary solution for this case, but I like talking about theory.  :) [/hide]


geometry
trigonometry
Pythagorean Theorem

These are all from the Texas UIL 2007 regional math test.

1. ACD is a triangle where m∠DAC=25° and m∠DCA=90°. Point B lies on segment AC and is 5 cm from point C and m∠DBC=50°. How far is point B from point A? (nearest tenth)

(A)7.8 cm (B)2.5 cm (C)10 cm (D) 1.0 cm (E) 2.2 cm
Answer: [hide]A[/hide]

2. A triangle is drawn as shown: [url]http://www.iaza.com/work/090425C/pic75466.bmp[/url] Find h if XZ=10", XY=9", and YZ=15". (nearest tenth)

(A)8.7" (B)8.4" (C)8.1" (D)7.7" (E)7.5"
Answer: [hide]A[/hide]

3. Points A, B, and C are on circle O such that segment PA is a tangent to circle O and segment PC is a secant to circle O through point B. If PB=3cm and PC=12 cm then PA=?

(A)9 cm (B)6 cm (C)5 cm (D)4 cm (E)3 cm
Answer: [hide]B[/hide]

#1-  $ \tan{40}\equal{}\frac{5}{CD}$ and $ \tan{65}\equal{}\frac{AB\plus{}5}{CD}$. Solving for $ CD$ in each and substituting, you get $ AB\equal{}\frac{5\tan{65}}{\tan{40}}\minus{}5\equal{}\boxed{1.6}$

#2- Let the segment from X to where the altitude drops. Use Pythagorean theorem:

$ 100\plus{}20a\plus{}a^2\plus{}h^2\equal{}225$
$ a^2\plus{}h^2\equal{}81$                           Then subtract them:
_____________________(-)

$ 100\plus{}20a\equal{}144$
$ 20a\equal{}44$
$ a\equal{}2.2$

$ 2.2^2\plus{}h^2\equal{}81$
$ h^2\equal{}76.16$
$ h\equal{}8.7"$  $ \boxed{A}$

#3 - $ PA^2\equal{}PB*PC$

$ PA^2\equal{}3*12$
$ PA^2\equal{}36$
$ PA\equal{}6$  $ \boxed{B}$

In number 2, where does the 20a come from?

Thank you!

From the first equation $ (10\plus{}a)^2\plus{}h^2\equal{}225$.


this is crazy  ur going to let superllamel, Fantasy Lover, mewto5555, AND RomanianGenius be on a team?????!!!!! THATS MADNESS!!! btw ill join. gimme some time to find a group

no i'm not....but um...why did ustart a differnt thread?

[quote="ash-Pokemon"]about the thread................. um........... it was an accident[/quote]

btw, u dont need a subject 4 every post...

 :rotfl:


geometry
inequalities
function
geometry unsolved

Let AA1,BB1,CC1 be the inner angle bisectors of triangle ABC and A2,B2,C2 be the touching points oh the incircle of triangle ABC with the sides BC,Ca,AB respectively.Let S,S1,S2 be the areas of triangles ABC,A1B1C1,A2B2C2 respectively.Prove that
      3/S1 - 2/S2 <=  4/S

Man what a solution.Here it goes>We have to prove the inequality (after we express $S_{1},S_{2}$ as a function of S):$\frac{3\sum_{cyc}bc(b+c)}{2abc}\leq\frac{8abc}{\prod_{cyc}(a+b-c)}+1$.After we make $a=y+z,b=x+z,c=x+y$ we have to prove $\frac{9}{2}+\frac{3(\sum_{cyc}x^{3}+xyz)}{\sum_{cyc}xy(x+y)+2xyz}\leq\frac{\sum_{cyc}xy(x+y)}{xyz}$.Now we bring them to the common denominator and after we simplify we must prove $6x^{2}y^{2}z^{2}+\frac{xyz\sum_{cyc}xy(x+y)}{2}+xyz\sum_{cyc}x^{3}\leq 2\sum x^{3}y^{3}+\sum_{cyc}x^{4}(y^{2}+z^{2})$.Now we have $2\sum x^{3}y^{3}\geq 6x^{2}y^{2}z^{2}$ and we have $\sum_{cyc}x^{4}(y^{2}+z^{2})\geq 2xyz\sum_{cyc}x^{3}$ so we have to prove just $\sum_{cyc}xy(x+y)\leq 2\sum_{cyc}x^{3}$,which comes from $xy(x+y)\leq x^{3}+y^{3}$.thats all :)


conics
ellipse
geometry
3D geometry
sphere
Pascal\u0027s Triangle

What are the formulas for these?

These always mean MOST whatever produced

1. number of regions of the plane produced by n straight lines
2. number of regions of the plane produced by n intersecting circles
3. number of intersection points produced by n straight lines
4. number of intersection points produced by n circles
5. number of regions of a crescent moon produced by n straight cuts
6. number of regions of the plane produced by n intersecting ellipses
7. regions of space produced by n intersecting spheres

There are more, but these are the ones I can think of at the moment.

EDIT: 8. number of pieces of pie that can be made with n straight cuts

you mean the MAX amount of regions/intersections produced, right?

(3) is the triangular numbers, or $ \binom{n\plus{}1}{2}$

That's all *I* can think of at the moment.

[hide="Number 1, Number 8, and Generalization"]
Generalization: in the $ d^{th}$ dimension, the maximum number of regions that $ n$ things of $ (d \minus{} 1)^{th}$ dimension can cut it into is the sum of the first $ d \plus{} 1$ numbers in the $ n^{th}$ row of Pascal's Triangle, where the zeroth row is $ 1$, the first row is $ 1,1$, etc., i.e. $ R_d \equal{} \left\{\binom dj\right\}_{j \equal{} 0}^d$, or equivalently, $ \sum_{i \equal{} 0}^d\binom ni$, where if $ n < i$, we have $ \binom ni \equal{} 0$. This is because we start with $ 1$ region, and after drawing $ n \minus{} 1$ objects, when we draw the $ n^{th}$ object, it will divide each object into $ \sum_{i \equal{} 0}^{d \minus{} 1}\binom{n \minus{} 1}i$ $ (d \minus{} 1)^{th}$ dimension objects, each giving one more region.

Specifically for number $ 1$, this is $ \binom n0 \plus{} \binom n1 \plus{} \binom n2 \equal{} \frac {(n^2 \plus{} n \plus{} 2)}2$, and for number $ 8$, this is $ \binom n0 \plus{} \binom n1 \plus{} \binom n2 \plus{} \binom n3 \equal{} \frac {(n \plus{} 1)(n^2 \minus{} n \plus{} 6)}6$.
[/hide]
[hide="Number 2, Number 7, and Generalization"]
Number $ 2$: After the first circle is placed, we have $ 2$ regions. For each $ n^{th}$ circle placed after that, we intersect each of the $ n \minus{} 1$ other circles at two places, forming $ 2(n \minus{} 1)$ arcs on the $ n^{th}$ circle. We add one region for each of these arcs since they divide a previously drawn region in half, so the number of regions after the $ n^{th}$ circle is drawn is $ 2 \plus{} 2(1 \plus{} 2 \plus{} \cdots \plus{} (n \minus{} 1)) \equal{} 2\binom n0 \plus{} 2\binom n2 \equal{} n^2 \minus{} n \plus{} 2$.

Number $ 7$: After the first sphere is placed, we have $ 2$ regions. For each $ n^{th}$ sphere placed after that, we form $ n \minus{} 1$ new circular regions, and each of these makes $ (n \minus{} 1)^2 \minus{} (n \minus{} 1) \plus{} 2 \equal{} n^2 \minus{} 3n \plus{} 4$ new regions, so we have $ \sum_{k \equal{} 1}^nn^2 \minus{} 3n \plus{} 4 \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}6 \minus{} \frac {3n(n \plus{} 1)}2 \plus{} 4n \equal{} \frac {n(n^2 \minus{} 3n \plus{} 8)}3$
$ \equal{} 2\binom n3 \plus{} 2\binom n1$.

Generalization: number $ 2$ is for $ 1$-spheres and number $ 7$ is for $ 2$-spheres. For $ 3$-spheres, we note that it is similarly $ \sum_{k \equal{} 1}^n\frac {(n \minus{} 1)((n \minus{} 1)^2 \minus{} 3(n \minus{} 1) \plus{} 8)}3$, and we can continue in this manner. In fact, for $ n$ $ d$-spheres in the $ (d \plus{} 1)^{th}$ dimension, there are at most $ 2\binom n{d \plus{} 1} \plus{} 2\binom n{d \minus{} 1}$ regions.
[/hide]
[hide="Number 3"]
The optimal case is when each line intersects each other line at one distinct point, so the maximum is $ \binom n2 \equal{} \frac {n(n \minus{} 1)}2$.
[/hide]
[hide="Number 4"]
The optimal case is when each circle intersects each other circle at two distinct points, so the maximum is $ 2\binom n2 \equal{} n(n \minus{} 1)$.
[/hide]
[hide="Number 5"]
The first line divides it into $ 3$ regions. Then, for every subsequent line, it intersects the previous $ n \minus{} 1$ lines, making one new region for each of these intersections, and $ 2$ other regions as a result of the concave part of the crescent, so this is $ 3 \plus{} (3 \plus{} 4 \plus{} 5 \plus{} \cdots \plus{} (n \plus{} 1)) \equal{} \binom{n \plus{} 2}2 \equal{} \frac {(n \plus{} 1)(n \plus{} 2)}2$.
[/hide]
[hide="Number 6, Generalization"]
After the first ellipse is placed, we have $ 2$ regions. For each $ n^{th}$ circle placed after that, we intersect each of the $ n \minus{} 1$ other ellipses at four places, forming $ 4(n \minus{} 1)$ arcs on the $ n^{th}$ ellipse. We add one region for each of these arcs since they divide a previously drawn region in half, so the number of regions after the $ n^{th}$ circle is drawn is $ 2 \plus{} 4(1 \plus{} 2 \plus{} \cdots \plus{} (n \minus{} 1)) \equal{} 2\binom n0 \plus{} 4\binom n2 \equal{} 2(n^2 \minus{} n \plus{} 1)$.

Note that this can be generalized in a similar fashion to that of the first hide tag and the second. In the $ (d \plus{} 1)^{th}$ dimension where we have $ n$ $ d$-ellipsoids (hopefully that's what it's called; the above case is the $ 2^{nd}$ dimension with a $ 1$-ellipsoid and higher analogs follow accordingly), there are at most $ 4\binom n{d \plus{} 1} \plus{} 2\binom n{d \minus{} 1}$ regions.
[/hide]
[hide="Given a circle and its inscribed regular $ n$-gon, the number of regions formed by the diagonals is..."]
First, we have $ n \plus{} 1$ regions, the polygon itself, and the surrounding $ n$ gaps. Then, each time we draw a diagonal, we add a region, giving us $ n \plus{} 1 \plus{} \left(\binom n2 \minus{} n\right) \equal{} \binom n2 \plus{} 1$ so far. However, if we drew a diagonal that intersected another diagonal, we would have to add another region. In every quadrilateral, there are two unique intersecting diagonals, so we finally have $ \binom n0 \plus{} \binom n2 \plus{} \binom n4 \equal{} \frac {n^4 \minus{} 6n^3 \plus{} 23n^2 \minus{} 18n \plus{} 24}{24}$ regions.
[/hide]

Tags: [i]regions, lines, planes, spaces[/i]

[quote="leewongjao"]

1. number of regions of the plane produced by n straight lines
[/quote]

$ \frac{n(n\plus{}1)}{2}\plus{}1$

[quote]3. number of intersection points produced by n straight lines  [/quote]

$ \frac{n(n\minus{}1)}{2}$

Thats all I'm gonna post now.  I know some more, but it's more helpful if [i]you[/i] figure the formula for each, it just some induction.

When trying to find out the most sections n numbers of lines can form:
Where n is the number of lines there are, then nC0 + nC1 + nC2
                                                               n Choose 0 + n Choose 1 + n Choose 2


function
real analysis
algebra unsolved
algebra

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be the differentiable function and for every $n\in \mathbb{N}, x\in \mathbb{R}, f'(x)\leq f'(x+\frac 1n)$. Prove that $f^'$ is the continuous function.

I'm pretty sure this has been discussed before, and you could probably find it in the "Real Analysis" section. That's where these problems belong anyway, you know :).

[quote][b]grobber[/b] wrote: you could probably find it in the "Real Analysis" section[/quote]

 :?:   :|


Can you find the next two terms in each of the following sequences?

THESE ARE PRETTY CHALLENGING

A)
2, 12, 1112, 3112, 132112, ..., ...

B) 1, 1, 1, 2, 3, 7, 23, ..., ...

C) 1, 1, 1, 2, 3, 5, 11, ..., ...

$ \text{HINT:}$ patterns B and C are very similar

No , he tells you that A is 2, 12, 1112, 3112, 132112, ..., ... You need to find the next two terms

your A isn't correct by the way

[hint] don't look at the number, look at the digits

you don't need to post this in multiple forums

[hide]A) 1113122112, 311311222112
The other two are too difficult for me  :oops:

EDIT: errrrr I've just seen you told someone else with the same answer as me that his answer was wrong, but I've checked it several times and didn't see anything wrong  :? [/hide]


[quote="Karth"]What do you think the most useful AoPS class is, and why?[/quote]

This is really hard to answer depending on how good you are on what area and etc.

For me, I found AMC-12 Problem Series to be the BEST class I ever took on AoPS. The reason is that it taught enough skills for me to qualify to AIME this year. But then, I did a lot of problems during class so some of the problems were review. Still, it was an excellent choice.


I do not know whether this problem has been discussed on the forum , however I think it is an interesting geometry problem.

Let $ABC$ be a triangle and let $P$ be a point in the interior of $ABC$ such that $<PBC=30^{0}$, $PM=\frac{AC}{2\sqrt{3}}$, where $M$ is the midpoint of $AC$, and $PN=\frac{AB}{2\sqrt{3}}$ where $N$ is the projection of $P$ on $AB$. Prove or disprove that $ABC$ is an equilateral triangle and $P$ is its centroid.

sincepb bisects angle-PBC....so..it lies on angle-bisector..,median,altitude of triangle-ABC..now this is when triangle is equilateral....
see...if we take that..ABC is equi. then...PM=PN...as AB=AC...equal distance from arms of angle shows it lies on angle bisector..etc. of angle-BAC.....hence proved that p is meeeting point of angle bisectors...i.e. medians..therefore..P is the centroid...
proving.. ABC equi.
use trigo...u get PM/CM = tan30...or PM/AM=tan30...therefore angles concerned are 30....if u take N  as mid-point..then both statements are proved.....from  proof at the top we can make this assumption as true..


quadratics
inequalities
inequalities unsolved

For $a,b,c >0 $ and $x,y,z \geq 0$ prove

$(y+z)\frac{yc+zb}{a}+(z+x)\frac{za+xc}{b}+(x+y)\frac{xb+ya}{c}\geq  (x+y)(x+z)+(y+z)(y+x)+(z+x)(z+y) $

I would like to see a solution that doesn't use Sylvester theorem for quadratic form.

Sorry , but the what is the Sylvester theorem ?

Using AM-GM
\begin{eqnarray*}
&& (y+z)\frac{yc+zb}{a}+(z+x)\frac{za+xc}{b}+(x+y)\frac{xb+ya}{c} \\
&=& \sum \frac{b}{a}z(z+y) + \frac{a}{b}z(z+x) \\
&\geq& \sum 2 z\sqrt{(z+x)(z+y)}
\end{eqnarray*}
So it suffices to prove that
\[
2\sum_{\mathrm{sym}} x\sqrt{(x+y)(x+z)} \geq \sum_{\mathrm{sym}} (x+y)(x+z)
\]
Set $p^2 = y+z$, $q^2 = x+z$, $r^2 = x+y$, the above ineq turns into
\[
\sum_{\mathrm{sym}} (q^2+r^2-p^2)qr \geq \sum_{\mathrm{sym}} q^2r^2
\]
rearranging
\[
2\sum_{\mathrm{sym}} p^3q \geq \sum_{\mathrm{sym}} p^2q^2 + p^2qr
\]
which follows immediately from Muirhead.

And I have the same question, what is the Sylvester theorem? Does it trivialize this ineq?

And just out of curiousity, why is the source Ford Motor Company Scientific Laboratory? And what does that have to do with Klamkin?

[quote="billzhao"]
$
\[
2\sum_{\mathrm{sym}} p^3q \geq \sum_{\mathrm{sym}} p^2q^2 + p^2qr
\]$
[/quote]

I think it's $\sum_{\mathrm{sym}} p^3q$ on the left hand side?

Nice solution, Billzhao.

This inequality can be regarded as the validation of the positiveness of a quadratic form and Sylvester's theorem is a very powerful criterium to test that in a simple way.
However I enclose this post a document with the complete theory of quadratic forms and a very elegant proof of Sylvester theorem, so
if you are interested, you can study it in detail.

About your second question, Klamkin worked in 1970 year period at Ford Scientific Laboratory and many inequalities, proposed by him in that period, were so referenced.

Here is another interesting document with the proof of Sylvester's Theorem

[quote="siuhochung"][quote="billzhao"]No, it's $2\sum_{\mathrm{sym}} p^3q$. If you write everything out, you'll see.[/quote]

Sorry, can you multiply it out once for me?[/quote]

$\sum_{\mathrm{sym}} p^3q= p^3q + p^3r + q^3p + q^3r + r^3p + r^3q$

Just remember that the sym sum always runs through all (6) permutation of the varibles. In particular, $\sum_{\mathrm{sym}} pq = 2pq + 2pr + 2qr$ NOT $pq + qr + pr$


Show that associativity does not imply communativity and that communativity does not imply associativity.

Isn't that more like common sense kid of?

Like say heat and ice-cream sales are correlated, but that does not mean ice-cream sales cause heat.

I think what he really means is to give functions that don't satisfy the properties (called counterexamples).
For the second one (commutativity to associativity), we can use the following function as our counter example:
[hide]Define $a \ @ \ b = \frac{a+b}{2}$. Then we have $a \ @ \ b = b \ @ \ a$, but we don't have $(a \ @ \ b) \ @ \ c = a \ @ \ (b \ @ \ c)$.[/hide]

For the 1st case, there are many examples available.

However, most of these are operations on other sets, such as matrices.

Here is an example for an operation defined on the real numbers (or rationals, or integers).

Let $?$ be defined by $a ? b = a |b|$.

Then, $a ? (b ? c) = a ? (b |c|) = a |b |c| | = a |bc|$.
And, $(a ? b) ? c = (a |b|) ? c = (a |b|) |c| = a |bc| = a ? (b ? c)$.

Thus, $?$ is associative.

Now, let $a =-1, b = 2$.

Then, $a ? b =-1 (|2|) =-2$,
while $b ? a = 2 (|-1|) = 2 \neq a ? b$.

Thus, $?$ is not commutative.

QED.


Let V be an infinite dimensional metric space that has norm metric. W is a finite dimensional subspace of V. Prove that W is closed.

Hint: Every finite dimensional normed vector space over $ \mathbb R$ or $ \mathbb C$ is complete, by the norm equivalence theorem.

I think the OP should restate the question more carefully.
For example, it seems he does not mean "finite-dimensional metric space".


As a passionate chess player, I had to post this topic...
True genius has died in age of 64 (symbolically, 64 squares!). Brilliant chessplayer, rebel with a cause...

RIP.

I heard about that last night too.  :( . My chess instructor at my old middle school is going to be really sad.

I also heard that he said 9/11 was a good thing?

I saw it on the news as well, and I thought he was already dead because when I first learned about him, he was in a black and white photo  :blush:.

RIP Fischer.

Oh my god!!! I had no idea about that  :(  :(  :( From the same point of view as [b]hsiljak[/b] I must , we will most probably not have such beautiful chess games like he produced. I know that he was not as good as he had been, but it's simply very sad :(

As far as Fischer's statements about 9/11 are concerned, they're affectuous words of a man angry of his fellow Americans and Jews-keep in mind, B. Fischer was a Jew (even religious one in the first days, maybe someone remembers the story about Bobby in Sus on Sabath). After the "Belgrade-Sveti Stefan" match and the storm it caused, he felt not as a betrayer, bat as betrayed...

But I hope he will be remembered as a true hero of the ancient game... Not as an eccentric dissident ;)

Well, we should remember, that he was not so ancient :wink: In my opinion he was the most talented chess player of all time (before the famous match for the world champion title he proved his domination over many IM's and GM's getting few times an incredible score (at that level) of 6-0). And my favourite game all-time was played then vs Bent Larsen. For me he was, simply put, the greatest  :)

It must be also said that he has done a great deal in making chess popular in the USA, where a sport discipline not including a large number of spectators (and great amount of money) couldn't get much appreciation. Fischer's unique style on the board, and in real life (the "Sus" case, his comments about other players...) made the difference.

Maybe he wasn't the all time best, maybe there is no way to determine who was no.1, but he is definitely one of  the greatest geniuses of this game.

Yeah, according to the ICC newsletter, he had strange opinions about a lot of things. Still, his death is rather sad.  :(

Here are a [url=http://www.chessgames.com/perl/chessplayer?pid=19233]list of games[/url] he played if anyone's interested.

Fischer, RIP.

He was the first famous chess player I learned about. Can't believe he.... :(

This is really sad.

A good part of my life in 8th and 9th grade was spent on my floor playing over his games and reading stories about him before he left chess.

Even though Bobby Fischer added controversial comments to his speech, he still will be remembered as a great chess player.

RIP

FISCHER :noo:

I love chess (rating 1455) but I don't really care Fisher's dead. Sure he was a good chess player. But he was a racist againist Jews. Also he was a radicalist before he was declared insane. So really hes dead and thats just a fact. I really have no feelings about it. My morals-Since he descriminated againist race hes not that great. Part of an article I found on his racial issue below.

In 1961 Fischer "made his first public statements despising Jews."[101] In recent years, Fischer's primary means of communicating with the public was via sometimes-outrageous radio interviews. Fischer participated in at least 33 such broadcasts between 1999 and 2005, mostly with radio stations in the Philippines, but also with stations in Iceland, Colombia, and Russia.

In 1999, he gave a call-in interview to a radio station in Budapest, Hungary, during which he described himself as the "victim of an international Jewish conspiracy." Fischer's sudden re-emergence was apparently triggered when some of his belongings, which had been stored in a Pasadena, California storage unit, were sold by the landlord, who claimed it was in response to nonpayment of rent.[102]Fischer interpreted this as further evidence of a worldwide Jewish conspiracy perpetrated by "the Jew-controlled US Government" to defame and destroy him. In 2005, some of Fischer's belongings were auctioned on eBay. In 2006, Fischer claimed that his belongings in the storage unit were worth millions.[103][104] Fischer, whose mother was Jewish,[105][106] made occasional hostile comments toward Jews from at least the early 1960s.[107][108] From the 1980s and thereafter, however, his hatred for Jews was a major theme of his public and private remarks.[109] He denied the "Holocaust of the Jews," announced his desire to make "expos[ing] the Jews for the criminals they are [...] the murderers they are" his lifework, and argued that the United States is "a farce controlled by dirty, hook-nosed, circumcised Jew bastards."[110] In one of his radio interviews, Fischer said that it became clear to him in 1977, after reading The Secret World Government by Count Cherep-Spiridovich, that the Jews were targeting him.[111]

Fischer also made anti-Zionistic statements in connection with anti-American statements; see below.

 Anti-American statements
Hours after the September 11, 2001, attacks Fischer was interviewed live by Pablo Mercado on the Baguio City station of the Bombo Radyo network, shortly after midnight September 12, 2001 Philippines local time (or shortly after noon on September 11, 2001, New York time). Fischer commented on U.S and Israeli foreign policy that "nobody cares ... [that] the US and Israel have been slaughtering the Palestinians for years".[112][113][114] Informed that "the White House and Pentagon have been attacked", Bobby Fischer proclaimed "This is all wonderful news."[112][113] Fischer stated "What goes around comes around even for the United States."[112][113] and said that if the US fails to change its foreign policy, it "has to be destroyed." After calling for President Bush's death, Fischer also stated he hoped for a coup d'état in the US, and that the military government would then execute "hundreds of thousands of American Jewish ring-leaders", "arrest all the Jews", and "close all synagogues". [112][113]Subsequent to that interview, Fischer's "right to membership in the United States Chess Federation [was] canceled" by a unanimous 7-0[115] of the USCF,[116] taken on October 28, 2001.

Chess columnist Shelby Lyman, who in 1972 had hosted the PBS broadcast of that year's Championship, said after Fischer's death that "the anti-American stuff is explained by the fact that ... he spent the rest of his life [after the game in Yugoslavia] fleeing from the US, because he was afraid of being extradited".[117]

You said that he was possibly insane: I agree that he said a lot of bad things, and I think that honestly, something probably cracked in him. Maybe it was the pressure of competing in chess, maybe it was several tragic events we don't know about. You're right, I think some of the things he has said are simply inexcusable.

But we also know that he was once a great man, a great chessplayer, and very polite and a gentleman as well. We know that Kasparov has gone to do great things in politics after his chess career, while Fischer has strayed off the deep end. What makes this difference, nobody knows. But I think that it was clearly something that happened to him that made him lose everything he had.

I haven't played chess in who-knows how long but I still remember, Fischer is even today a chess icon for young kids playing chess, and it makes a big difference. And in that way, he should be remembered.

[quote="crazychinesemaniac"][quote="distracted523"]You're sorta plagiarizing...not to mention posting an excessively long post...not to mention reviving an old topic.[/quote]

not really its Wikipedia so its a mixture of a whole bunch of people no one knows[/quote]

your ignorance of how wikipedia actually works is pretty amusing given that you're pretending to know about it. the content in wikipedia is licensed under the GFDL, which indeed allows for reproduction under the same license. however, claiming that the writing is "a whole bunch of people no one knows" is pretty ridiculous. the majority of blocks of content of wikipedia are written by one sole editor and then spellchecked and proofread by a few others, so a clear author can be identified. that's not even mentioned your claim that "no one knows" the author, which is just... wow.

about the content of the post, I have no comment, really. you can remember someone for their positive or their negative attributes, it's your choice.

Plagiarism is the sin of claiming or passing off someone else's work as your own.  It doesn't matter if you know who that person is, or if it's actually many people -- to present someone else's work without attribution is plagiarism.  The license issue is relevant for questions of copyright (which Wikipedia also holds), not plagiarism.


[quote="7h3.D3m0n.117"][quote="vishalarul"]How do I get the source code for a TI-89 [b]assembly[/b] program?

If I already have the source code (in text), how do I sent it to my TI-89?[/quote]

to send it you need the USB port thingy...lol

source code for an ASM prog? umm you have to make one by urself and here is a good place:

[url]http://www.technoplaza.net/assembly/[/url][/quote]

I already have the program that I downloaded. I want the source code for it.

Assembly source code can not be regenerated from the machine code it is converted to after it has been compiled, because all custom things, such as labels, variable names, etc are stripped of their meaning.

The only way to get it would be to find the original author and ask them for it.

Moved to Other Problem Solving Topics.

In the manual, it says something about the Exec command.

What does that do?

I'm pretty sure that the Exec command runs actual machine code (probably Hex form of input).

I would NOT recommend messing with it to find out - you could mess up your calculator.


function
algorithm
logarithms
algebra
functional equation
number theory unsolved
number theory

Prove that there is infinitely many different functions $ f: N\rightarrow N$ such that $ f(f(n))\equal{}n^3\plus{}1997$.

Can the function be $ 1\minus{}1$?

Let $ g(n) \equal{} n^3\plus{}1997$.
Let $ g^p$ denotes the $ p^{th}$ iterate of $ g$ (with $ g^0 \equal{} Id$).

Let's construct some sets, according to the following algorithm :
$ S_0 \equal{} \{g^p(0)/p \in \mathbb{N} \}$.

If $ S_k$ is defined, then let $ a$ be the least integer which belongs to non of the $ S_i$ for $ i \leq k$ (if any).
Then $ S_{k\plus{}1} \equal{} \{g^p(a)/p \in \mathbb{N} \}$.

Then :

- For each $ i \neq j$, we have $ S_i \cap S_j \equal{} \emptyset$ :
Otherwise, there would exist $ n,m,p,q$ such that $ p \leq q$ and $ n \neq m$ such that $ g^p(n) \equal{} g^q(m)$.
Since $ g$ is clearly injective it follows that $ g^{p\minus{}q}(n)\equal{}m > n$.
Thus $ m$ has been chosen after $ n$ in the above procedure, which forces $ m$ to not belong to any of the preceding $ S_i$, so that $ m \neq g^{p\minus{}q}(n)$. A contradiction.

- The procedure ensures that any integer belongs to some $ S_k$.
Since these sets are pairwise disjoints, they form a partition of $ \mathbb{N}$.

- The sequence $ (S_k)$ is infinite :
It is easy to verify that, for all $ a>0$ and $ p\geq 0$, we have $ g^p(a) \geq 1997^{3^{p}}$ thus the number of such terms which are not greater than $ n$ is at most $ c \cdot \log(\log (n))$ for some appropriate constant $ c>0$.
The same kind of evaluation is valid for $ a\equal{}0$.
Thus, if there are only $ m$ sets $ S_k$ they contains at most $ c' \cdot \log(\log (n))$ (for some appropriate constant $ c'>0$) integers not greater than $ n$. Then, for sufficently large $ n$, the set $ \{1, \cdots, n \}$ is not contain in $ S_0 \cup \cdots \cdots S_m$, a contradiction.

- Now, divides the non-negative integers into pairwise disjoint pairs $ (a,b)$.
Let $ m,n$ be the least elements of $ S_a$ and $ S_b$ respectively.
Thus, using all the pairs, define $ f$ by $ f(m)\equal{}n$ and $ f(n)\equal{}m^3\plus{}1997$ (this consists to intricates $ S_a$ and $ S_b$).
Then $ f$ satisfies the given functional equation and is bijective.

Since there are an infinite number of pairing, this gives an infinite number of such functions.

Pierre.


Which ones are on the waitting list?
These kind of laws are real culture shocks, what's the reason to make such ridiculous(for me, at least) laws?
Some jokes said this is the aftereffect of the drugs taken by the young generation in 50's~60's, what's your opinion?

Im sorry, Im not aware of what you mean by 'shoot first'.

Is this referring to new police protocols where they shoot first and ask questions later?

Here is the related old news, the problem is, more and more states became the same as Florida since then.
[url]http://www.washingtonpost.com/wp-dyn/content/article/2005/04/30/AR2005043000735.html[/url]
Roughly speaking, a "shoot first" law allow you to take ANY action when you feel you're in danger, ANYWHERE, not only in your home. You won't be sued for that, you don't need to pay any for the damage, unless the victim is a cop.
Good law for the killers.

[quote="Buffalo"]what's the reason to make such ridiculous(for me, at least) laws?[/quote]
It's not just you who thinks that kind of thing is ridiculous. And to answer your question, [b]Jeb Bush[/b]. Well actually he might not be behind this one but he's a moron and he should have pwned it (any of them should have killed that bill).

My mom and I always joked that you have to be careful not to cut anyone off when you're driving in Texas because the person might be some hick in his huge pick-up truck (like 50% of vehicles in Texas are these) and he'll pull out his hunting rifle and shoot you.

With that kind of law, it might not be a joke any more.

You know what, I've found a good way to legally kill some people. It's simple, you spit him(better not let the others see this), definitely he will be enraged and walk towards you with his powerful fists, now you get a perfect reason in your hand to shoot him. :rotfl:

[quote="solafidefarms"]Hmmm... I take a bit of offense to saying it's "Shoot First" and that it's shooting for any reason at all. (Disclaimer: Proud to be in the NRA.) Your justification for shooting the person must stand up in court or you will be treated like an ordinary murderer, so you still aren't going to shoot somebody except in extraordinary circumstances. No, it's not going to help the killers. You've still got to have a incredibly good reason to feel threatened. In fact, iirc there have been *fewer* murders in FL than there were before this law.[/quote]

What you're implying is bull and a logical fallacy. The fewer murders in Florida are not probably tied to the law ESPECIALLY as under that law, the killing wouldn't even COUNT as murder. It wouldn't be counted in statistics at all for crime becasue it's not considered a crime! Also, the murder rate is probably correlated to the general drop in murder rate in the United States, there are many other factors.

And yeah, I was going to say Jeb Bush is a hick but Xevarion said it for me. *Gives cookie to Xevarion*

[quote="G-UNIT"][quote="solafidearms"]"You've still got to have a incredibly good reason to feel threatened." [/quote]

  Spotting an intruder on the premises is enough to take his life.  Apparently, the potential possibility of him remotely aspiring to consider whether or not he should be prompted to seek the motivation required to analyze the position in which he might very well take to contemplating the hypothetical scenario in which he might choose to induce or bring harm upon you - is sufficiently threatening.[/quote]

OK, do you know what the law in FL was *before* this law? Before what you call "shoot first" you had to run away and do everything you could to flee the would-be assailant before you could demonstrate that you had a gun (even if you didn't shoot himer). Now you may legally use the gun without going to extreme lengths not to use the gun (which gives the assailant, if he has a gun, ample opportunity to shoot _you_). Studies show that higher gun ownership reduces crime, but only if the citizens are free to use said guns. This law makes it a little easier for Floridians to use their rights.

@Buffalo: Haven't you ever heard of forensic evidence?

[b]solafidearms[/b], I would be much happier with the law in question if I were more confident of the competence of our justice system and of the competence of the average citizen of the USA. As it stands, I am not at all confident of either; thus, such a law worries me. This law is certainly appropriate for someone like [b]you[/b], who has a gun (I presume), is competent enough to use it, and has good enough judgment to use it only when it is absolutely necessary.

But as PenguinIntegral (I think) said in the death penalty thread, isn't one wrongful death already one too many? And imagine this situation:
Some dude is innocently sitting in his living room late at night watching The Man Show (does this show exist, or did I make it up?). He hears the sound of breaking glass from the bedroom -- someone just broke in his window. He grabs his gun off the rack, and when a shadowy figure emerges from the bedroom with one hand in his jacket pocket, he fears the guy may have a knife or gun in there and shoots. Seems reasonable under this law, right?
In effect, this institutes the death penalty without trial for crimes that usually carry relatively light sentences, such as breaking and entering.
At my high school, students who fought were punished equally, no matter who started the fight. I believe that fighting fire with fire is not the answer.
And I would rather die with a clear conscience than kill another man, under any circumstances. So I have no problem with the old law.

Well then, when kids come to your door trick or treating, and you blow their heads off, is that legal?  It seems like this law could turn halloween into a bloodbath in some places  :D

Well you could then just kill anyone and say the person was scaring you.  I mean, fear is a natural response to many situations.  It is something we have to deal with internally, and just because think we are going to die does not mean we should kill...

There's really no way to get evidence for many murder with this new law...

Important question, I just talked about this with my uncle, who was born as a Belgian but now he lives in Georgie and has obtained the USA nationality.

Feeling threatened, does that mean I feel threatened, or "I felt one of my close ones was in danger".
In other words, such a law, does it allow me to shoot whoever points a knife at me, but it doesn't allow me to shoot someone strangling my child at the moment? :maybe:

[quote="funcia"]I feel that children are their parents' property so then you could feel threatened if it were your child-if not-shoot anyway. There's got to be some legal way to do it...[/quote]  And what about my wife (is she my property) or my father, aunt, grandmother and sister?

YES! Okay I'm not trying to be crude but I have that opinion and no other people probably don't agree but if you're the one with he gun in the family you own EVERYONE!(in the family of course..)

Okay  :wink: , but what does the state law say?  To what extent does it allow protection of other people?  Where does it stop?  Only your children, your spouse too, perhaps your parents too if they live in the house as well, .... your friends if they are guest in your house too....  :maybe:

What if there are several guns in the house, everyone owns each other.  And what if your father threatens your mother, can you shoot him with another weapon?
Trust me, it happens, famous (at a national level, that is..) Belgian lawyer Jef Vermassen always repeats this claim :"The gun in the house is practically always used against other inhabitants, not an intruder."

MOVE SOMEWHERE ELSE NOW. AND GET YOUR KIDS AND WIFE AWAY FROM GUNS(and your household.) Or become a rebel...

but if you wanted to be law-abiding you'd have to check the laws of your state which is tedious so you should always be prepared and move to the place you can be most law-abiding. I have no idea about my place of dwelling(haven't lived there whole life) so I'd assume what I have said...I'd shoot and if you have to go to prison so be it...at least a member of your family or a friend is alive...of course you do have a point...if that were the case where everyone had a gun I'd assume the "man of the family" to be the one who owns everyone.

[quote="funcia"]MOVE SOMEWHERE ELSE NOW. AND GET YOUR KIDS AND WIFE AWAY FROM GUNS(and your household.) Or become a rebel...
 [/quote]
Are you talking to me? :o   I don't live in a "state"??

No..I meant if you were in those conditions...

[quote="fredbel6"]Feeling threatened, does that mean I feel threatened, or "I felt one of my close ones was in danger".[/quote]

Well, if you could extend it to that, I could say "oh, I feel really close to this rock.  Oh no you're stepping on it.  Bang.  Headshot."

For example, if you were Jain, or just in general believed that things most people consider inanimate actually have a soul and suffer...and live...rocks might very well live for all we know...

Okay, I know I might be going overboard, but I think this law is just ridiculous (as I'm sure most of you do).  The whole point of defending yourself in a situation is that you feel you'd rather face the consequences of defending yourself than not defend yourself.  If one would face no consequences from such an action, then how can we define such situations?  This in combination with the second amendment makes for chaos: when you see someone with weaponry, that makes the situation much more tense and makes people much more likely to react on impulse...

Uhm, what you quote was actually a question  :blush:  not a request for opinions (which is welcome).
But seriously, these "Shoot First" states or whatever their official name is, what do they allow, officially.

Yeah, I know it was a question...but I'm saying that if it [i]were[/i] that way, then you could do things like that...so it's probably not.  Well, I guess these laws are kinda fuzzy...

call me a redneck and verbally hang me but i think the new law's a good idea, now you can't get vivisected by the intruder's lawyer if he really is posing a threat and you shoot him.


calculus
integration
absolute value
calculus computations

hi,

1- (U(n)) is a sequence  such that :  $ U(n) = \sum_{k=1}^{n}\frac{1}{k\times2^{k}}$

find the limit of U(n).

2- we have $ S(n) = \sum_{k=0}^{n}\frac{(-1)^{k}}{1+k}$ and L is its limit.
find a sequence V(n) such that : $ | S(n)-L | \le V(n)$.

For (1), it helps to know your power series.

For $ -1<x\le 1,$ we have $ \ln(1+x)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}x^{k}}k.$

Replace $ x$ by $ -x$ and we get $ -\ln(1-x)=\sum_{k=1}^{\infty}\frac{x^{k}}k$ for $ -1\le x<1.$

(You should be able to figure out what that has to do with this problem.)

For (2): one of the conclusions of the Alternating Series Test is that for a series satisfying the conditions of the test, the remainder (the difference between a partial sum and the limit) is, in absolute value, less than the next term.

thanks for your answer :lol: .i guess there is also another solution using the integral of $ f(x) = t^{k}$  :wink: .


i) Solve the system
$ m^{3}\minus{}n^{3}\minus{}q^{3}\equal{} 3mnq$
$ m^{2}\equal{} 2(n\plus{}q)$
in nonnegative integers.

ii) Find all $ m,n$ positive ingers such that $ ^{60}\sqrt{m^{n^{5}\minus{}n}}$ is an integer. For $ n \equal{} 60k$ it's true. Then I've put $ n^{4}\equiv1\mod 60$, but couldn't find them all.

[hide="i"]
Use the identity
$ a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}bc\minus{}ca)\equal{}\frac{1}{2}(a\plus{}b\plus{}c)((a\minus{}b)^{2}\plus{}(b\minus{}c)^{2}\plus{}(c\minus{}a)^{2})$

Rearranging the first equation gives us
$ m^{3}\plus{}(\minus{}n)^{3}\plus{}(\minus{}q)^{3}\minus{}3m(\minus{}n)(\minus{}q)\equal{}0$
$ (m\minus{}n\minus{}q)((m\plus{}n)^{2}\plus{}(m\plus{}q)^{2}\plus{}(n\minus{}q)^{2})\equal{}0$

So we have either $ m\minus{}n\minus{}q\equal{}0$ or $ (m\plus{}n)^{2}\plus{}(m\plus{}q)^{2}\plus{}(n\minus{}q)^{2}\equal{}0$. In the first case,

$ m\equal{}n\plus{}q\equal{}\frac{m^{2}}{2}$
$ 2m\minus{}m^{2}\equal{}0$
$ m(2\minus{}m)\equal{}0$
$ m\equal{}0$ or $ m\equal{}2$.
For $ m\equal{}0$, we have $ n\plus{}q\equal{}0$, and since both are nonnegative, $ n\equal{}q\equal{}m\equal{}0$
For $ m\equal{}2$, we have either $ n\equal{}q\equal{}1$ or $ n\equal{}2,q\equal{}0$ or $ n\equal{}0,q\equal{}2$.

For the next case,

$ (m\plus{}n)^{2}\plus{}(m\plus{}q)^{2}\plus{}(n\minus{}q)^{2}\equal{}0$
Each term is nonnegative, so they must all be 0. Then $ m\plus{}n\equal{}0$ but m and n are nonnegative so they are both 0. Then $ q\equal{}0$.

Thus the only solutions in nonegative integers are $ (m,n,q)\equal{}(0,0,0);(2,1,1);(2,2,0);(2,0,2)$
[/hide]

wow! how do you come up with a solution like that?! do you just know those factorizations?

The factorization is well-known and quite useful. I wouldn't have gotten it if I hadn't already known it.

Heh.

It's Adam Hesterberg's factorization, and it has come in handy for me a LOT. Learn it well :p.

uh, no it's not his, and your the first person that i've ever heard refer to it as adam hesberg's factorization

Adam Hesterberg's Factorization was the name popularized by Daesun after the second red team contest at MOP07. He cited it as Adam Hesterberg's Factorization Trick because Adam taught it to him, and the judges were like, "What is Adam Hesterberg's Factorization Trick?"

The factorization can be given any name you want, it's not named as rigidly as say difference of squares.

In a cosmic way, this is rather in accordance with the mathematical habit of naming concepts after mathematicians to whom they are distantly related, if at all  :P

I'm gonna call it larry the cable guy's factorization

[quote="t0rajir0u"]I've heard a lot of people on this forum refer to it as Adam Hesterberg's factorization, but I don't really know why.  I believe [i]Mathematical Olympiad Treasures[/i] by Titu Andreescu is the first place most people who didn't learn it on the forums learned it.

[hide="Large hint on ii"] Fermat's Little Theorem ;) [/hide][/quote]

Or [i]USSR Olympiad Problem Solving Book[/i]

Urm, excuse me, but I don't really think this is that uncommon of an identity.
I saw it in an Algebra textbook once. (not a problem solving book)

Im going to try #2...
[hide]
$ m^{\frac{n^{5}\minus{}n}{60}}\in\mathbb{Z}$
Assuming $ m$ is raised to the first degree, $ 60 \equal{} 4\cdot3\cdot5 | n^{5}\minus{}n$

Using the hint (Fermats little theorem), $ n^{5}\minus{}n\equiv n\minus{}n\equiv 0(mod5)$ and $ n^{5}\minus{}n\equiv n^{3}\minus{}n\equiv n\minus{}n\equiv 0(mod3)$, so we need to show $ 4|n^{5}\minus{}n$.

$ n^{5}\minus{}n \equal{} n(n^{2}\plus{}1)(n\plus{}1)(n\minus{}1)$.
Note that $ (n\plus{}1)(n\minus{}1)$ have the same parity, so if $ n$ is odd then $ n^{5}\minus{}n$ is disivible by 4. If n is even, let $ n \equal{} 2(n')$ so that:
$ \frac{n^{5}\minus{}n}{2}\equal{}\frac{32(n')^{5}\minus{}2n'}{2}\equal{} 16(n')^{5}\minus{}n'$ and we want this congruent modulo 2.
$ 16(n')^{5}\minus{}n'\equiv\minus{}n'\equiv 0(mod2)$, which means that $ n'$ is even and thus $ n$ is divisible by 4.

So assuming $ m$ is raised to the first degree, $ n\equiv 0, 1, 3 (mod4)$

Raising $ m$ to any degree would be helpful only if it is divisible by 60. Because $ n^{5}\minus{}n$ is always divisible by 3 and 5, only raising $ m$ to an even degree with help. If it is raised by 2, then $ 30|n^{5}\minus{}n$ which is true for all $ n$, so the answer is:

$ (m, n); n\equiv 0, 1, 3 (mod4)$
and
$ (m^{2i}, n); i\ge 1$
[/hide]


In a triangle $ABC$, there is an arbitrarily chosen circle $O_{1}$ tangent to the segments $AB$ and $AC$. Let $O_{2}$ be the circle tangent to the circle $O_{1}$ and to the segments $BA$ and $BC$. Let $O_{3}$ be the circle tangent to the circle $O_{2}$ and to the segments $CB$ and $CA$. Continue the construction in a similar way to obtain the circles $O_{4}$, $O_{5}$, $O_{6}$ and $O_{7}$.

Prove that $O_{1}= O_{7}$. In other words, prove that the circle $O_{6}$ is tangent to the circle $O_{1}$ again.

Also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=45972 (note: the problem was posted in "Unsolved Problems", so the author is looking out for a solution).

  darij

I have a solution but I don't remember it all. But I think that with my hint someone will finish this problem. If nobody will do that, I will ask my teacher (he show me solution but a long time ago), but now I have holidays and I will see him for 2 weeks.

$x1,x2,..$ determine radius of circles. $r1=tg\frac{A}{2}x1$
$KL=\sqrt{r1r2}$
$x1+x2+\sqrt{r1r2}=c$
$x1+x2+\sqrt{tg\frac{A}{2}tg\frac{B}{2}x1x2}=c$
We can make 7 analogical equations. Now we must to make trygonometryczny substitution, but I don't remember what a substitution. After that we will have easy linear equations.

Here is the prove :
[url=http://24img.com/768211430][img]http://24img.com/image_min.php?id=768211430[/img][/url]

[url=http://24img.com/839454874][img]http://24img.com/image_min.php?id=839454874[/img][/url]


Does there exist a finite set A made up of straight lines in the three-dimensional space,all lines in A passed
through a mixed point,and each line is perpendicular with at least 100 other lines?