I've decided to make this thread for discussion of the Olympics. I don't know how many of you acctually watch the Olympics but I watch it quite alot.

To Mods: Move to Fun Factory if necessary. Or Delete if their is already one.

This morning I watched the US vs China basketball game. My mom was rooting for China and I was rooting for the US. She called me a traitor for doing that. Anyway, does anyone think the "Redeem Team" will not get gold. As for me I think they probably will since they have the most elite basketball players in the world and looked quite sync'd during their first game.

dad's watching China vs. US women's basketball in the living room
making it hard for me to do math!
personally i think BASEBALL is the sport for math people :D

Unfortunately, if you recall Athens '04, their "Dream Team" had the elite players in the world but lost 3 times.  This year, I think they have a far greater chance.  In '04, they practiced for the first time only 20 days before the game so there was no chance that they would gel together and play as a team.  This year, they've spent a lot of time together, and you can tell that those from the '04 team, such as Kobe, LeBron, and Carmelo, are all very driven to get gold this year.  Expect tough competition from Spain, Lithuania, and Argentina.

The 4x100 Freestyle was the event of the night, though.  Unbelievable race.

[quote]personally i think BASEBALL is the sport for math people [/quote]

Of course.  Baseball = godliness.  Unfortunately, the IOC is stopping baseball and softball after this year.

Thats the plan but their is a chance that it won't get taken off by the IOC

[quote="JRav"]

The 4x100 Freestyle was the event of the night, though.  Unbelievable race.
[/quote]

AMAZING RACE.  Now that's what makes the olympics exciting....down to the wire finishes like the US did there.  Who knows, maybe that swimmer will get his 8 gold medals...

I've never played baseball.  (I know, shame).  Actually, the first time I put on a glove and played catch was when I was 18....I have no idea why it was just about the only sport I never tried growing up.

While it is unlikely that baseball will see the 2012 games in London, don't forget that the World Baseball Classic will be taking place again (either next year 0r 2010, I don't remember).

On a different note, does anybody actually believe that China's women gymnists are under 16?

They probably are knowing China, they probably did....cheat

W00t US picked up another 7 medals in 30 minutes

I don't think that the US "Redeem Team" will get gold, because thats what Lithuania is going to get.

 :rotfl:  :rotfl:

GO CHINA! they got the gold in artistic gymnastics! (It's always fun to see the gold medalists bite their medals =])

if baseball is going to be an Olympic sport, it sucks how the US will always send minor leaguers not major leaguers to the Olympics, so the best US team will not be out there.

It coincides with the regular season.  They would have to do what the NHL does and suspend the season for 3 weeks, and Selig would never do that.  And teams would never agree to it.

Did anyone see Micheal Phelps BEASTLY out touch of Covic by .01 seconds. That was amazing it looked like Micheal was going to lose and then the scoreboard said he outtouched by a hundreth of a second.

So far, that has been the top moment of the Olympics.  Even Phelps was excited and in his individual races, he's pretty stoic.

can't even spell michael right crazychinesemaniac :P

he's getting gold #8 tonight

U.S. basketball team will win gold
and Michael Phelps got gold medal #8!!! :D  :first:  :D

Speaking of that, I find it sort of annoying how they keep saying he's the top [i]Olympic athlete[/i], in terms of his medals. At one point, they said something like 10,000 athletes have competed, but none had ever gotten 8 golds in a single Olympics. How many sports are there in which that is even possible? It's very plausible that if other sports had more medal oppurtunities, there would have been others who could have gotten 8 or more. I'm not knocking Phelps, just stating my opinion.

hmm..I think its more of the fashion in which he received his medals. Most people cant win 8 swims in 8 days due to tiredness. You'd probably experience that in every sport that it's possible to win 8 medals. Also Phelps over came alot to become an Olympic athlete. From an early age he had ADHD and was bullied at school because of how he looked and his behavior. (According to his mom)

Well, like I said, I do acknowledge that Phelps is an excellent (perhaps the best ever) swimmer. It's just that very few sports have the oppurtunity that swimming does to receive so many golds.

This was a really good article on the gymnastic's controversy. [url]http://sports.yahoo.com/olympics/beijing/blog/fourth_place_medal/post/The-REAL-Olympic-medal-count?urn=oly,101537[/url]

Indeed.  The last part of the article was very telling: in almost all of the sports decided by pure competition (races, games, non-subjective competitions), the US is far superior to China.

A little out dated, but nice US sweep in the mens 200m. I think it was the 200m but it was a nice sweep anyway.

US definitely did not sweep the 200m since Usain Bolt set another WR.  However, Crawford and Dix took silver and bronze.  US swept 400 m hurdles.

According to my method of picking the best all around country. (Gold=3pts Silver=2pts and Bronze=1pt.) The US loses to China by the score of 220 to 223. However these scores aren't final. If the IOC proves that even just one of the Chinese gymnast are under age the US will win by the score of at least 220-221. This is because if even one gymanst is underage China gets stripped of its Team gold and it goes to the US. So US gains another pt and china loses 3. Even though I am chinese I am a heavily biased towards the US.

[quote="Bugi"]Maybe you should include number of participants into your method. China would then lose![/quote]

Do you mean ratio of medals won over number of participants? Or something else.

I think he's trying to say that the US won gold in sports such as indoor volleyball, men and women's basketball, and women's soccer, which has many more athletes than say weightlifting.

But many of the events China got gold in are judged events, or events that no other country cares about.  The US, as I mentioned, earned gold in soccer, basketball, volleyball, track and field, swimming, while Chinese earned in events such as diving, gymnastics, shooting, weightlifting, and ping pong.

No they have world championships for many events that can be found at the summer Olympic games (held in odd numbered years).

[quote="crazychinesemaniac"]Now I can't wait till 2012. Thats my graduation year too, instead of getting a car I might ask my parents for tickets to the Olympics. Anyone think we should hold the Olympics every year instead of waiting 2 years? So its like this year is the summer olympics. Next year is the winter olympics and so on.[/quote]

I'm sorry to say this...but that would be incredibly hectic

Besides, they have to determine who the favorites are, through the use of world championships and other meets.  You can't have countries scrambling together every year to create teams.  The money factor is big.  In addition, it would be beyond chaotic.

[quote="Sunny"][quote="crazychinesemaniac"]Now I can't wait till 2012. Thats my graduation year too, instead of getting a car I might ask my parents for tickets to the Olympics. Anyone think we should hold the Olympics every year instead of waiting 2 years? So its like this year is the summer olympics. Next year is the winter olympics and so on.[/quote]

Graduate from University or High School?  If it's high school, lol ask for the Olympic tickets.  If you have a car in university, you will be the chauffeur for all your friends, and the designated driver (since you're the only one) when you go to parties or even dinner.  It just sucks.[/quote]

Highschool. I don't want a car in college for the reason you stated.  :P

[quote="crazychinesemaniac"]

Highschool. I don't want a car in college for the reason you stated.  :P[/quote]

  I just saved you tons of gas money and the prevention of people who are interested in your car and not you  :lol:

[quote="JRav"]Actually, he swam something like 17 races in 8 days, not 8.  Just 8 finals.  It is possible to win multiple gold medals in track and field, although not 8, since you'd be hard-pressed to find anyone that can do the 100 and the 400, let alone the 100, 200, 400, 800, 1500, 1600, 3000, 3200, 5000, 10000, and marathon (I know that's 11 but you get the idea).  But expect Usain Bolt to earn some more medals.[/quote]

that's not a very good comparison...

running the 100 backwards, forwards, sideways, etc. would be more like it...

phelps doesn't swim anything over 400/800 last time I checked...

[quote="JRav"]Actually, he swam something like 17 races in 8 days, not 8. Just 8 finals. It is possible to win multiple gold medals in track and field, although not 8, since you'd be hard-pressed to find anyone that can do the 100 and the 400, let alone the 100, 200, 400, 800, 1500, 1600, 3000, 3200, 5000, 10000, and marathon (I know that's 11 but you get the idea). But expect Usain Bolt to earn some more medals.[/quote]

Well there are the hurdles... but we get your point

[quote]that's not a very good comparison...

running the 100 backwards, forwards, sideways, etc. would be more like it...

phelps doesn't swim anything over 400/800 last time I checked...
[/quote]

My point isn't the race itself.  It's that there's so many of them.

Although Vietnam takes 1 Silver Medal but I love Vietnam!!!  :)  :)

Two highlights in this year's olympics (at least for me):

1. Michael Phelps slaughtering other swimmers, except one of events, of course. (He DID win that one, OK?!!!)

2. KOREA wins gold in BASEBALL :lol:

The problem I have with baseball is that the US, and many of the Latin countries do not send their best players since they are in MLB.  Countries like Cuba, Japan, and Korea can send their best players because not too many of their players are in other leagues, and in the case of Cuba, none.


Solve for x

$ 4x^3 \minus{} 4x^2 \minus{} 7x \plus{} 4 \equal{} 0$

hello, $ x\equal{}\frac{1}{2}$ is a solution. Now you can use polynomial long division.
Sonnhard.


salam
tooye yeki as topic ha rajebe masaleye jabr zir bahs shode
a,b,c >0, a+b+c >= abc
sabet konid agar a^2 + b^2 + c^2 >= abc
hala man migam basharayete gofte shode hokme ghavitare zir ra sabet konid
a^2 + b^2 + c^2 >=3^ 0/5 * abc
bebakhshid agar asoone vali ghashange . as negareshe badamam mazerat .bedrood.

$b,a,c >0$&$a+b+c\geq abc$
prove that $a^2+b^2+c^2\geq\sqrt3abc$ ;)
oon bala ro nega koni latex help dare :roll:

salam mamnoon hosein joon
faghat eshtebah neveshti soaalo
soorate soaal ine :
 $a,b,c>0      a+b+c>=abc   prove that    a^2 + b^2 + c^2 >= \sqrt{3}abc $
albate baazam ghalate !

editesh kardam.manzooret az bazam ghalate chiye?

salam aval mamnoon as lotfet  badam begam ke manzooram as bazam ghalate  neveshteye khodame
man natoonestam mesle to benevisam :P
$\2^2\$ :P

Latex help ke khoob tozih dade shode baraye bayad samte chap va samte raste ebarat
sign dolar bezari .

man alan daghighan oon chizi ro ke gofte mizanam
$2^2$

$a+b+c \le abc$ pas $3*{a^2+b^2+c^2} \le (a+b+c)^2 \le a^2b^2c^2$(1)
bana bar vaseteha $\frac{(a^2+b^2+c^2)^3}{27} \le (abc)^2$(2)
age 1 ,2 dar ham zarb beshan darim $(a+b+c)^4 \le 9(abc)^4$
$(a+b+c) \le \sqrt3(abc)^4$


PM me if you want to join, space for 20 signups. If replacements are needed, it will be posted in the replacements sticky.

In the middle of the Dark Ages, there was a time of knights.

You work for a Lord who owns a fortress, but THE DARK FORCES are attacking and you must defeat them!

You are a Knight hired by the lord. Your quest is to get to the other outpost and warn them of attacks, while fighting on the way. You have a day to reach each outpost.

Once you join, you will be PMed a role and you must play as one.

There can be 3 Lords, 3 Dark-Lords, and 7 knights and 7 Shadow Knights.

And remember, don't let this topic be all spammed up, since spam is bad.

Oh, and if you're bored, you can help find the ninja in this post.

[hide="."]Nope[/hide][hide="."]:ninja:[/hide]

EDIT: Lords can fight too, and they can hire knights, BUT, unlike knights, they cannot defect to the enemy.

KNIGHTS PRIVELAGES:

Can fight

Can defect to enemy

I sign up

The message is too small. Please make the message longer before submitting

you're supposed to sign up through PM (private message)

This prevents spammy posting in this topic and makes AOPS' site have more memory :D

GAME UPDATE:

This game is about activeness and traveling to other bases. To travel to an allied outpost, post:

I traveled to an allied out post in (land given)

To defect to the enemy, say "I defected to the enemy"

To travel to an enemy base say "Enemy base/outpost detected"

Note: all those commands have to be in [b]Bold Text[/b]

You can also play as a dragon, however, you must complete 10 trips to other bases first.[/b]

EDIT: GAME ISSUE RESOLVED:

Now you must [b]sign up by posting, not pm![/b]

Game dead yet? If not, I join I'm bored

whats defect?

It's not dead, it's waiting until Epic Space War ends.

I guess I'll join this game. A little confused :maybe:

I'll Join...I'll be the Lord.

Isn't the Dark Lord Voldermort? :P


The second set of instructions says:

[b] The first page should contain at the very beginning:
  	a) the title and authors (written in sentence-case, with no
 	   affiliation data - see item 6.j) which should be followed by
 	b) short abstract, MSC2000 and Keywords (written in normal font). [/b]

and respectively

[b] at the very end of the paper, leftside, should be inserted the
  		coordinates of the author(s), as follows:
 	   First Author,  address and email,
 	   Second Author, ddress and email,... .[/b]

1. What is sentence-case?
2. For the title and author I used the commands [quote] \title and \author and \maketitle [/quote] My name appeared bigger than the normal font. What can I do?
3. For the abstract I used [quote] \begin{abstract} and \end{abstract}[/quote]. Which made the word "Abstract" appear centered, above the text of the abstract. Which is not their case (see http://www.mathem.pub.ro/bjga  ). How can I make it to appear as in their papers?
4. For the MSC2000 (subject classification according to AMS) and keywords I didn't use any command . I separated them by [quote] \\[/quote]. Are there any commands?
5. Which bothers me more is how to add at the end of the paper the author and address, and leftside, with which commands?

Thanks
Jean

I would ask the publishers for a sample approved tex file then you could use that without worrying about setting yours up.
1. http://en.wikipedia.org/wiki/Sentence_case
2. http://www.tex.ac.uk/cgi-bin/texfaq2html?label=titlsty
3. Try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=abstract]abstract package[/url]
4. If \\ works what's wrong with that?
5. Just type the author and address. You don't need commands for them.


A integer x is [i]nice[/i] if it can be represented as sum of five nonzero squares.
The set $ L(x,n)\equal{}\{x\plus{}1,x\plus{}2,...,x\plus{}n\}$ is called [i]beautiful[/i] if $ x\plus{}i$ is nice $ \forall 1 \le i \le n$.

Show that there are infinitely many [i]beautiful[/i] set $ L(x,23)$.
 :D

Every large enough integer can be written as a sum of five nonzero squares. Proof was posted here before (using the expressions of $ 169$ as sums of squares).

really? can you give me a link?
thanks  :lol:

My (now not useful) solution:
Let $ (m,n)$ be coprime odd integers and define $ b\equal{}m^2\minus{}n^2$ and $ c\equal{}2mn$ and $ d\equal{}m^2\plus{}n^2$. Now we can note that the equation $ e^2\equal{}2b\equal{}2(m\minus{}n)(m\plus{}n)$ has infinite solution since a class is $ (m,n,e)\equal{}(2k^2\plus{}h^2,2k^2\minus{}h^2,4hk)$ where h,k are coprime odd integers. Now we have:
$ a\equal{}4b^2\plus{}c^2$
$ a\plus{}1\equal{}3b^2\plus{}d^2\plus{}1$
$ a\plus{}2\equal{}2b^2\plus{}(b\minus{}1)^2\plus{}(b\plus{}1)^2\plus{}c^2$
$ a\plus{}3\equal{}b^2\plus{}(b\minus{}1)^2\plus{}(b\plus{}1)^2\plus{}d^2\plus{}1$
$ a\plus{}4\equal{}2(b\minus{}1)^2\plus{}2(b\plus{}1)^2\plus{}c^2$
$ a\plus{}5\equal{}(b\minus{}2)^2\plus{}b^2\plus{}(b\plus{}1)^2\plus{}c^2\plus{}e^2$
$ a\plus{}6\equal{}(b\minus{}2)^2\plus{}b^2\plus{}2(b\plus{}1)^2\plus{}c^2$
$ a\plus{}7\equal{}(b\minus{}2)^2\plus{}2(b\plus{}1)^2\plus{}d^2\plus{}1$
$ a\plus{}8\equal{}(b\minus{}2)^2\plus{}2b^2\plus{}(b\plus{}2)^2\plus{}c^2$
$ a\plus{}9\equal{}(b\minus{}2)^2\plus{}b^2\plus{}(b\plus{}2)^2\plus{}d^2\plus{}1$
$ a\plus{}10\equal{}(b\minus{}2)^2\plus{}2(b\plus{}1)^2\plus{}d^2\plus{}2^2$
$ a\plus{}11\equal{}(b\minus{}1)^2\plus{}b^2\plus{}(b\plus{}1)^2\plus{}d^2\plus{}3^2$
$ a\plus{}12\equal{}(b\minus{}2)^2\plus{}b^2\plus{}(b\plus{}2)^2\plus{}d^2\plus{}2^2$
$ a\plus{}13\equal{}(b\minus{}3)^2\plus{}(b\plus{}2)^2\plus{}b^2\plus{}c^2\plus{}e^2$
$ a\plus{}14\equal{}(b\minus{}3)^2\plus{}(b\plus{}2)^2\plus{}e^2\plus{}d^2\plus{}1$
$ a\plus{}15\equal{}(b\minus{}2)^2\plus{}2(b\plus{}1)^2\plus{}d^2\plus{}3^2$
$ a\plus{}16\equal{}2(b\minus{}2)^2\plus{}2(b\plus{}2)^2\plus{}c^2$
$ a\plus{}17\equal{}2(b\minus{}2)^2\plus{}(b\plus{}3)^2\plus{}d^2\plus{}e^2$
$ a\plus{}18\equal{}(b\minus{}3)^2\plus{}2b^2\plus{}(b\plus{}3)^2\plus{}c^2$
$ a\plus{}19\equal{}(b\minus{}3)^2\plus{}(b\plus{}3)^2\plus{}b^2\plus{}d^2\plus{}1$
$ a\plus{}20\equal{}(b\minus{}3)^2\plus{}(b\plus{}3)^2\plus{}(b\minus{}1)^2\plus{}(b\plus{}1)^2\plus{}c^2$
$ a\plus{}21\equal{}(b\minus{}2)^2\plus{}(b\plus{}1)^2\plus{}d^2\plus{}e^2\plus{}4^2$
$ a\plus{}22\equal{}(b\minus{}3)^2\plus{}(b\plus{}3)^2\plus{}b^2\plus{}d^2\plus{}2^2$


Ring Theory
algebra
function
domain
factorial

Comment traduire Unique Factorization Domain en francais ?

Domaine de factorisation unique?  :P

Toi aussi tu commences à te rendre compte qu'il y a des expressions mathématiques que tu connais en anglais.. mais pas en français! :)


There are no calculators allowed. There are a specific number of questions for each round. This round is worth 3 points. You must get at least 2 to continue. You get 1 point for a correct answer and 1 bonus point for each problem you show your work. Also, there will be enough rounds to determine a first placee winner. This round is due August 10, 2005.

\question{1} I choose 3 letters from the word MATHCOUNTS. What is the probability that at least 1 is a vowel?
\question{2} Simplify:

\[\displaystyle (\frac{\sqrt{9x}-9}{(x+1)^2})^2\]
\question{3} A right triangle in which a and b are the legs, and c is the hypotenuse, $a=b$. If c=6, what is the area of the triangle.

Why doesn't it work? I tried the following code:

[code]\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{array}
\usepackage{xy}
\usepackage[left=2cm,top=1cm,right=2cm,nohead,nofoot]{geometry}

\pdfpagewidth 8.5in
\pdfpageheight 11in

\newcommand{\question}[1]{\paragraph{Question #1.}}
\title{Round 1 Version 3}
\author{anirudh}
% Remove command to get current date
\date{\today}
\begin{document}
\maketitle
\paragraph{Directions} There are no calculators allowed. There are a specific number of questions for each round. This round is worth 3 points. You must get at least 2 to continue. You get 1 point for a correct answer and 1 bonus point for each problem you show your work. Also, there will be enough rounds to determine a first placee winner. This round is due August 10, 2005.

\question{1} I choose 3 letters from the word MATHCOUNTS. What is the probability that at least 1 is a vowel?
\question{2} Simplify:

\[\displaystyle (\frac{\sqrt{9x}-9}{(x+1)^2})^2\]
\question{3} A right triangle in which a and b are the legs, and c is the hypotenuse, $a=b$. If c=6, what is the area of the triangle.
\end{document}[/code]

I don't see your problem.
I copied the text run tex and everything was perfect.

I've attached the result (It looks quite normal)

I want to use that in the forums.  :(  :(  :(  :(

Oh, then I'm sorry: I've nothing to do with the latex in the forum, so I can't help :(
I only thougt by the title you where using texniccenter.

    M

The use of LaTeX in the forums is designed for [i]short[/i] pieces of LaTeX code, not entire documents.  You cannot use commands like \documentstyle, \usepackage, etc. in the forum.


My children, I'm ashamed of you.  Not in the AMC forum!  And not anywhere, really... but this seems most appropriate for other problem solving topics.

Moved from AMC --> Other Problem Solving Topics

[quote="pilot"]Does anyone have a set of solutions to the Art and Craft of Problem Solving?[/quote]

Yes, I have it.  :)

orl -  could you send me a copy of the solutions you have? btw, where did you get it?

The author provides solutions manuals to problem-solving instructors, I'm pretty sure. Most math books have instructor's solution manuals that are only supposed to be for instructors.

Well I'm not using this book as part of a class. Just working through some of the problems, and having a set of solutions would be a lot of help.


superior algebra
superior algebra solved


James and Jasmine decide to play a game of Illusion: The Gathering. James owns five decks, while Jasmine owns four. If they each pick a deck to play with, how many different match-ups are possible?

This really making fun of Magic: the Gathering

James has 5 choices; Jasmine has 4.  Thus the answer is $ 5 \cdot 4\equal{}\boxed{20}$.

Jasmine has four choices for each of James' five. $ (4)(5)\equal{}\boxed{20}$

Annoying problem:
Couldn't this be interpreted as... there are nine total decks:call them ABCDEFGH and I.
Jasmine picks one, James picks one... that makes 9x8=72 matchups.
or... A vs B is equivalent to B vs A, so 9x8/2 = 36 matchups.
I don't know how Magic is played, but I assume you can only use one of your own decks.


linear algebra
matrix
linear algebra unsolved

Let $ A$ be an $ nxn$ complex matrix with $ Tr(A)\equal{}0$.Show that $ A$ is similar to a matrix with all $ 0'$s along the mail diagonal.
------------- :( -------------

Standard result.
More difficult: $ A$ is unitarily similar to a matrix with all 0's along the mail diagonal.


Λοιπόν το βιβλίο της πέμπτης δημοτικού (ναι καλά διαβάσατε )
έχει ένα πολύ ενδιαφέρον πρόβλημα .Παίρνουμε έναν τριψήφιο  αριθμό που να έχει ψηφίο μονάδων $\geq 2$ έστω ο 527.
Παίρνουμε δύο μοναδες και τις βάζουμε στην αρχή δηλαδή ο αριθμός γίνεται 2525 .
ι) Να γράψετε τον 2525 ως άθροισμα 5 τριψήφιων αριθών .
ιι) Γενικεύσετε το αποτέλεσμα
ιιι) Κάντε μαθηματική απόδειξη .

Καλή διασκέδαση  :D

Εγραψες:Παίρνουμε έναν τριψήφιο αριθμό που να έχει αριθμό μονάδων  έστω ο 527.Τι εννοεις?

Γράφω που να έχει αριθμό μονάδων  >=2  σε  latex .
De φαίνεται ?

Αριθμό μονάδων; Μήπως εννοείς αριθμό ψηφίων; Ο αριθμός μονάδων πάντα είναι ένας σε κάθε αριθμό. Π.χ. για τον 527 το 7 είναι το ψηφίο των μονάδων, το 2 των δεκάδων και το 5 των εκατοντάδων.

Μπορείς να αναδιατυπώσεις την άσκηση;

Ναι το ψηφίο των μονάδων εννοώ (νομίζω ότι από το παράδειγμα φαινόταν)

Α,ρε siluan,με εκανες να ψαξω στο παταρι για τα βιβλια της πεμπτης δημοτικου.... :P
Βασικα με την ασκηση,δεν πολυκαταλαβαινο τι παιζει....
για το α)αρκει νααποδειξω οτι μετα απο αυτη την μεταφορα μοναδων σε ολουσ τους τριψηφιους,προκυπτει αριθμος μεγαλυτερος η ισος του 555 για να μπορει να γραφει σαν αθροισμα πεντε τριψηφιψν αριθμων (του 111 πεντε φορεσ)   [δεν διευκρινιζεις αν ειναι διαφορετικων τριψιφιων...τοτε θα πρεπει νδο >=565=111+112+113+114+115)
ομως ο ελαχιστος τριψηφιος με ψηφιο μοναδω μεγαλυτερο η ισο του 2 ειναι ο 112.ο οποιοσ μτα απο αυτη την <πραξη>γινεται 2110(αν καταλαβα καλα,αφερεις 2 απο το ψηφιο των μοναδων+τισ βαζεις μπροστα απο τον αριθμο...)το οποιο ειναι κατα πολυ μεγαλυτερο του 555(η 565,αντιστοιχα)και αρα σιγουρα μπορει να γραφει σαν αθροισμα 5 τριψηφιων...
για το β)..οταν λεσ γενικευση???ν-ψηφιοσ αροθμοσ με ψηφιο μοναδων >+2??? το ζητουμενο ισχυει παντα....

πολυ πιθανον να μην εχω καταλαβει πως παιρνουμε τον καινουργιο αριθμο...,δηλαδη να μην εχω καταλαβει το προβλημα...
α,+κατι αλλο...το παραδοξο ποιο ειναι τελικα???....
 :D

Εγώ δεν κατάλαβα το "παίρνουμε [b]2[/b] μονάδες" και τις βάζουμε στην αρχή... Μπορείς να εξηγήσεις τι ακριβώς εννοείς; Αφού το ψηφίο των μονάδων σε κάθε αριθμό είναι 1, πώς παίρνεις 2 "μονάδες" και τις τοποθετείς στην αρχή;

Σόρρυ αν φαίνεται χαζή η ερώτηση, μπορεί να υπάρχει κάτι που όντως δεν το έχω προσέξει, αλλά ξανά δες τη λίγο..

PS: Ίσως εννοείς ότι αφαιρούμε το ψηφίο των μονάδων κατά 2,
τοποθετούμε το 2 μπροστά από τον υπάρχων αριθμό και αυτό που βρίσκουμε από την αφαίρεση το τοποθετούμε σαν ψηφίο μονάδων στο νεό αριθμό κρατώντας όλα τα υπόλοιπα του ψηφία ως έχουν; Π.χ. για τον 348 θα έχουμε 8-2 = 6 =>
=> 2346

μοστελ νομιζω οτι εννοει : περνουμε 2 απο το αριθμο  των μοναδων που εδω στον αριθμο 527 ειναι το 7.στο νεο αριθμο (τετραψηφιο) γραφουμε 252-και τεταρτο ψηφιο οτι μενει δηλ 5. αρα 2525

Kostas: αν διαβάσεις προσεχτικά τι έγραψα, αυτό ακριβώς λέω και εγώ αλλά με διαφορετικά λόγια  :P

δεν το ειδα σορυ. :lol: αρα αυτο ειναι που εννοει ο Σιλουαν

i)$2525=503+504+505+506+507$  :P

ii)Κάθε φυσικός $x$ με $500\leq x\leq 4995$ γράφεται ως άθροισμα πέντε τριψήφιων.  :lol:

iii)Ο αριθμός που προκύπτει αφού "πάρουμε" τις δυο μονάδες είναι της μορφής $x=2000+100a+10b+c$ με $a,b,c$ ψηφία και $c\leq 7$. Επειδή τώρα ισχύει $500\leq x\leq 4995$ σύμφωνα με το (ii) προκύπτει το ζητούμενο...  :rotfl: !

LοOOOoOOOoooooooooOoooool!

P.S. Την άσκηση την έλυσε η αδερφή μου που πάει δημοτικό κι εγώ απλά έγραψα τη λύση της...!  :P

Και το παράδοξο ξέρετε ποιο είναι; Ότι η συγκεκριμένη άσκηση έχει κινήσει το ενδιαφέρον πολλών μέσα από το forum (συμπεριλαμβανομένου και εμού... :lol: ) και το παρών topic πάει να γίνει ένα από τα πιο δημοφιλή έχοντας ήδη στην πρώτη ημέρα δημιουργίας του 10 posts!  :rotfl:

Kostas: Αν είναι έτσι όντως η σωστή εκφώνηση και ο Silouan έχει αντιγράψει την άσκηση από το βιβλίο της 5ης δημοτικού κατά γράμμα... Τότε μπορούμε να μιλήσουμε για εκφώνηση που δεν ορίζεται στο R  :rotfl:

οπως την βλεπω μαλλον δεν οριζεται στο R αλλα ποτε δεν ξερεις κατι μπορει να μην εχουμε δει στην ασκηση. παντως ο Σιλουαν μας ξεγελασε ειτε με το  θαλη ειτε με την συμπαθητικη ασκησουλα  :P

Αυτό είναι σίγουρο  :lol:

Λοιπόν θα πρέπει αν παραδεχτώ οτι ο Σιλουάν μας αναστατώνει ποικιλοτρόπως  κι φαντάζομαι ότι η επόμενη άσκηση θα είναι από το βιβλίο της Α δημοτικού ίσωσ κι του νηποιαγωγείου :rotfl:  :rotfl:  :rotfl:

Λοιπόν παιδιά τα διαδοχικά πόστς σε αυτό το τόπικ μου κίνησαν τρομερά το ενδιαφέρον κι παραθέτω την λύση μου προς έλεγχό σας  Λοιπόν
Ο ελάχιστος αποδεκτός αριθμός είναι ο 102
Ο μέγιστος  αποδεκτός  αριθμός είναι ο 999
(Αδιαφορούμε παντελώς για την ύπαρξη μη αποδεκτών αριθμών ανάμεσα στους 2 αυτός αριθμούς)
Άρα ο χ του προβλήματος κινείται
2100=<χ=<2997 (1)
Επίσης ο χ ισούται με
Χ=2000+100α+10β+γ με γ=<7 (2)
Κι επειδή 2000= 5 * 400
Έχουμε μόνο να δείξουμε ότι ο 100α+10β+γ είναι πάντα τριψήφιος   αν συμβαίνει αυτό απλά θα προστίθεται σε ένα 400αρι κι θα αποδεικνύεται το ζητούμενο αλλά
Από (1) κι (2)
100=< 100α+10β+γ=<997 με γ=<7 που ισχύει πάντοτε
Άρα αποδείχτηκε το ζητούμενο
Ομοίως κάποιος μπορεί να σκεφτεί κι για την  περίπτωση που το ψηφίο των μονάδων είναι 1 κι να καταλήξει στο ίδιο συμπέρασμα ( η περίπτωση που το ψηφίο των μονάδων είναι 0 είναι αδύνατη )
Οφείλω να ομολογήσω ότι η άσκηση αυτή είναι ευχάριστο να μαθαίνει κανείς ότι μπαίνει σε παιδιά 5ης δημοτικού διότι δείχνει την προσπάθεια αναβάθμισης του επιπέδου της μαθηματικής παιδείας στην Ελλάδα μας Τέλος θα ομολογήσω ότι πραγματικά είναι μια πολύ ευχάριστη άσκηση

Λοιπόν ωραία όλοι σας . :wink: Εγώ είχα το εξής στο μυαλό μου .
Για το πρώτο :
2525=527+472+527+472+527 .Όπου παρατηρούμε ότι 527+472=999  :lol:
Δηλαδή κάθε αριθμός δημιουργείται από αυτή τη διαδικασία γράφεται ως άθροισμα 3 φορές του αρχικού αριθμού + 2 φορές τον αριθμό που προκύπτει ώστε το άθροισμα του αρχικού και αυτού να είναι 999 .Απλά είναι ένα ωραίο τρικάκι που μου άρεσε τίποτα παραπάνω.  :wink:
Ελπίζω να σας άρεσε και εσάς .


[quote="daermon"]Two students attempted to solve a quadratic equation x^2+bx+c=0.  Although both students did the work correctly, the first miscopied the middle term and obtained the solution set (-6,1).  The second student miscopied the constant term and obtained the solution set (2,3).  What are the correct solutions?  (MAO 1992)

[hide="hint"]what equations did the students write down?[/hide][/quote]

I actually used a different method from the book. [hide] Since the first student copied down the last term correctly, and $c$ is equal to the product of the roots. $c=-6$. The second student copied the second term correctly, and $b$ is the sum of the roots multiplyed by $-1$. So $b=-1\cdot5=-5$. Solving $x^{2}-5x-6=0$, we get $x=6,-1$ [/hide]


In order to direct obsevers to a school exhibit, the students are building a road of paper sheets that measure $ 18^{\prime\prime}$ by $ 21^{\prime\prime}$ each. The path is $ 18^{\prime\prime}$ wide. They need to tape around the outside of each sheet of paper completely, either to join the sheet to another or to stick the paper to the floor. (When two sheets meet at an edge, only one piece of tape is used to hold the sheets together.) What is the least number of feet of tape they need for a 100 foot long path? Express your answer as a decimal to the nearest tenth.

[quote="1996 Chapter Sprint #28"]In order to direct obsevers to a school exhibit, the students are building a road of paper sheets that measure $ 18^{\prime\prime}$ by $ \boxed{24^{\prime\prime}}$ each. The path is $ 18^{\prime\prime}$ wide. They need to tape around the outside of each sheet of paper completely, either to join the sheet to another or to stick the paper to the floor. (When two sheets meet at an edge, only one piece of tape is used to hold the sheets together.) What is the least number of feet of tape they need for a 100 foot long path? Express your answer as a decimal to the nearest tenth.[/quote]

This is what it should be. I don't know what I should do if I see some problem like this, so mods do whatever you need to do.

Solution to the corrected problem:

The perimeter will be $ 100\cdot12\cdot2 \plus{} 18\cdot2 \equal{} 2436$ inches. Since $ \frac {100\cdot12}{24} \equal{} 50$, there will be exactly $ 50$ sheets in this path, for $ 49$ junctions, each of which takes $ 18$ inches of tape. Thus, we need $ 2436 \plus{} 49\cdot18 \equal{} 3318$ inches of tape at least, or $ \boxed{276.5}$ feet.


[quote="Labiliau"]Give a method to find an left inverse to $A$ (if it exists of course).[/quote]

u mean $BA$=$I$ ? For square nonsingular matrices it's usual inversion, of course. Methods observed near, at topic.  If it's exists, a solution with minimal norm is $B$=$A^{+}$ (pseudoinversion), B=pinv(A) in matlab. Among methods there are enough standard column or row algorithmes. Other solutions could be impressed through $A^{+}$: $B$=$A^{+}$+$C$($I$-$AA^{+}$), $C$ is matrix of any coefficients (corresponding dimensions). It follows from $AA^{+}A$=$A$.


function
trigonometry
limit
real analysis
real analysis unsolved

Consider $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that $ |f(x)\minus{}x \cdot \sin {x}| \le |x^3|$, $ \forall x \in \mathbb{R}$. Find $ \lim_{x\to 0}\frac{f(x)}{x^2}$.

answer is $ 1$
here is the hint.
we need to prove :
$ \lim_{x \to 0}\frac {f(x)}{x^{2}} \equal{} 1$ this means :
 for all $ \epsilon$ there exist $ \delta$ such as
$ |\frac {f(x)}{x^{2}} \minus{} 1| < \epsilon$ provided $ x < \delta$
let's fix some $ \epsilon$
but we have
$ |f(x) \minus{} x^{2}| \minus{} |x^{2} \minus{} x \sin x| < |f(x) \minus{} x\sin x| < |x^{3}|$
thus
$ |f(x) \minus{} x^{2}| < |x^{3}| \plus{} |x^{2} \minus{} x  \sin x|$
show that if $ |x| < 1$ then there exist $ C > 0$ such as $ |x^{2} \minus{} x\sin x| < C|x^{3}|$
hence
$ |x^{3}| \plus{} |x^{2} \minus{} x \sin x| < (C \plus{} 1)|x^{3}|$
and if we choose $ \delta \equal{} \frac {\epsilon}{1 \plus{} C}$ then
$ (C \plus{} 1)|x^{3}| < \epsilon x^{2}$
hence :
$ |f(x) \minus{} x^{2}| < \epsilon x^{2}$
or
$ |\frac {f(x)}{x^{2}} \minus{} 1| < \epsilon$ provided $ x < \delta$

$ |f(x)\minus{}x \cdot \sin {x}| \le |x^3| \Longrightarrow  \minus{}|x| \plus{}\frac {sin(x)}x \leq \frac {f (x)}{x^2}\leq \frac {sin (x)}x \plus{}|x|$. If you know that $ \lim_{x\rightarrow 0}\frac {sin(x)}x\equal{}1$, then $ \lim_{x\rightarrow 0}\frac {f(x)}{x^2}\equal{}1$
 :cool:


Soppose that $p,q,r$ are prime and distinct numbers and \[ A={\{p^aq^br^c \mid 0\leq a,b,c\leq 5}\}.  \] Find the least of $n$ such that every subset with $n$ element like $B$ that $B\subset A$, has two elements $x,y$ such that $x$ be one of the divisor of $y$.

there are 216 elements. (no point in listing them).
if we have
p^5
q^5
r^5
p^a*q^(5-a) for 0<a<5
p^b*r^(5-b) for 0<b<5
q^c*r^(5-c) for 0<c<5
p^d*q^(e)*r^(5-d-e)
for 0<d+e<5, 0<d, 0<e,

we have a subset of 21 non-dividing elements which <i>appears</i> to be maximal, so my conjecture is 22.

It has been on a Czech competition. I didn't manage to solve it, butthe right answer is 28. If you don't believe, see http://mks.mff.cuni.cz/naboj/s_kratkym_riesenim.ps , task 34. Hope it|s not neccesary to translate. The set of numbers, that no divides the other can contain 27 elements: consider all numbers
$p^a.p^b.p^c$ where $a+b+c=7$, $a,b,c\geq 0$. there are ${9 \choose 2} =36$ such triplets of a,b,c, 6 of which are permutations of (6,1,0) and 3 are permutations of (7,0,0), so 27 of them fulfill all conditions.


Let be $x$ a rational number.
Prove: There are only finitely many triples $(a,b,c)$ of integers with $a<0$ and $b^2-4ac=5$ such that $ax^2+bx+c$ is positive.

Some more sharper statement (follows from Hurwitz' theorem):
For any irrational $x$, there are $\infty$ many such triples $(a,b,c)$.

Another sharpening is given by the following (thanks to Naphthalin for noticing):
Let $\{ (a_n , b_n , c_n) \}$ be the tripels that work for $x=\frac{k}{l}$, $\gcd(k,l)=1$.
Then $\sum_n a_n = -2 l^2$ , $\sum_n b_n = 4kl$  and $\sum_n c_n = -2 k^2 +2$.


inequalities
calculus
trigonometry
geometry
integration
function
calculus computations

using calculus (or otherwise, e.g. trigonometry, series expansion(s), etc.),
prove the following 3 inequalities:

${\bf (i)}\;\;\;3\;<\;\pi\;<\;3\frac{1}{2}$

${\bf (ii)}\;\;\;2\frac{1}{2}\;<\;e\;<\;3$

${\bf (iii)}\;\;\;\frac{1}{2}\;<\;\gamma\;<\;1$

[hide="another way to do (i)"]
We have
\[\sin \theta <\theta <\tan \theta\]
for $0<\theta <\frac{\pi}{2}$ and $\theta$ in radians, which can be shown using a sector of radius 1 and angle $\theta$. If we let $O$ be the the center of the circle, $A,B$ be the endpoints of the chord, and $C$ be the intersection of $\overline{OB}$ and the tangent to the circle at $A$, then:

$\theta$ is twice the area of the sector

$\sin \theta$ is twice the area of $\triangle OAB$ which is inside sector

$\tan \theta$ is twice the area of $\triangle OAC$ which covers sector

In particular, setting $\theta=\frac{\pi}{6}$ gives
\[3<\pi <2\sqrt{3}<\frac{7}{2}\]
[/hide]

#1 we can do better $\pi < 3\frac{1}{7}= \frac{22}{7}$.

Consider,

$\int_{0}^{1}\frac{(1-x)^{4}x^{4}}{1+x^{2}}dx = \frac{22}{7}-\pi$

Note, the function just mentioned is non-negative. So, $\frac{22}{7}-\pi >0$.

[quote="misan"]interesting... hmmm, i wonder if we can find a similar function that would show $\pi\;<\;3\frac{16}{113}\;=\;\frac{355}{113}\;\;\;$ ?[/quote]

Actually what you can do is use continued fractions. Express $\pi = [3;1,...]$ as a continued fraction. Then the odd covergents always exceede the limit value and the limit value always exceeds the even convergents. (This is a well-known theorem).

But I think such an approach is messy.


I think I posted this problem in ML a long time ago in Combinatorics section, but there was no reply. Here is the problem :
    Let $m, p$ be two natural numbers and $m \geq p$. Denote $(m,p)=\frac{m!}{p!(m-p)!}$. Let $f(m,p)=\sum (-1)^{m-i}(m,i)(ip,p)$ for $i=1$ to $m$.
    Prove that if $p$ is a prime then $p^m|f(m,p)$ and there exists $m$ such that $f(m,p)=p^m$. :)

There 's something I want to correct here. The condition $m \geq p$ here just serves for the defintion of the denotion $(m,p)$. In the problem, it is not neccessary that $m \geq p$. Sorry!  ;)

Consider a table consisting of $p$ rows and $m$ columns and denote the number of ways one can choose $p$ squares of the table such that there are at most $i$ columns containing selected squares by $g(i)={m\choose i} {pi \choose p}$.

By principle of implosion and explosion $f(m, p)=\sum_{i=1}^m (-1)^{m-i} g(i)$ equals the number of ways of choosing $p$ squares such that there are exactly $m$ columns containing squares.

For $m>p$ we therefore have $f(m, p)=0$.
Now assume $m \le p$. Some calculation shows that $f(m, p)=m^{p-m} p^m$.
Setting $m=p$ yields $f(m, p)=p^m$.


superior algebra
superior algebra unsolved

Let $ R$ be a ring with 1. An element $ a\in R$ is said to have a left inverse if $ ba\equal{}1$ for some $ b\in R$. Show that if the left inverse $ b$ of $ a$ is unique, then $ ab\equal{}1$ (so $ b$ is also a right inverse of $ a$).

I know how to do this if we have no zero divisors. This is just by doing $ bab\equal{}b\rightarrow b(1\minus{}ab)\equal{}0\rightarrow ab\equal{}1$ since $ b\neq 0$. But I can't seem to get this result if $ R$ has zero divisors.

You have to use uniqueness, otherwise the result is false.

We have $ (b \plus{} ab \minus{} 1)a \equal{} 1$, so $ b\plus{}ab\minus{}1$ is a left inverse for $ a$.  It follows that $ b\plus{}ab\minus{}1\equal{}b$, so $ ab \equal{}1$.


sorry!!i forgot to check one case...i will try it again..

I can proove that no solution exists if $ a,c > 1$ and $ b > 0$. But i can't kill the case $ c \equal{} 1$ and $ b > 0$. Mybe someone can find all solutions to the equation $ 3^a \equal{} 5^b \plus{} 2$ and finish it. (I was thinking about using the ring $ \mathbb{Z}[\sqrt {2} i]$ but didn't quite finish it)

the proof:
[hide]
If $ a,c > 1$ we have $ 5^b \equal{} 1$ mod $ 9$ which yields that $ b$ is of the form $ b \equal{} 6\beta$
By mod $ 4$ we have $ 3^a \equal{} 3^c$ therefore $ a \equal{} c$ mod $ 2$. By mod $ 5$ ($ b > 1$) we have $ 1 \plus{} 3^a \equal{} 3^c$. If we combine these last two facts we can conclude that $ a \equal{} 4\alpha \plus{} 3$ and $ c \equal{} 4\gamma \plus{} 1$.
If $ b$ was devisable by $ 4$ than we can rearange the equation into $ 3^{4\gamma \plus{} 1}(3^{4(\alpha \minus{} \gamma) \plus{} 2}) \equal{} 5^{12\beta} \minus{} 1$ where RHS is devisable by $ 16$ (because $ 16|5^4 \minus{} 1$) and LHS is devisable by $ 8$, but not by $ 16$ which is a contradiction. Therefore $ b$ is of the form $ b \equal{} 12\beta \plus{} 6$. So now the equation looks like this:
\[ 1 \plus{} 3^{4\alpha \plus{} 3} \equal{} 5^{12\beta \plus{} 6} \plus{} 3^{4\gamma \plus{} 1}\]
Now consider this mod $ 13$. Note that $ 5^2 \equal{} \minus{} 1$, $ 5^4 \equal{} 1$ and $ 3^3 \equal{} 1$ mod $ 13$. We get $ 3^a \minus{} 3^b \equal{} 11$ mod $ 13$. Analysing all posibilities we could conclude that the only posibility is that that $ a \equal{} 0$ and $ b \equal{} 1$ mod $ 3$. By chinese reminder theorem we can now conclude that $ a$ and $ b$ are of the form $ a \equal{} 12\alpha \plus{} 3$ and $ b \equal{} 12\gamma \plus{} 1$. The equation looks like this:
\[ 1 \plus{} 3^{12\alpha \plus{} 3} \equal{} 5^{12\beta \plus{} 6} \plus{} 3^{12\gamma \plus{} 1}\]
Finally consider this last equation by mod $ 31$. Note that $ 5^3 \minus{} 1 \equal{} 124 \equal{} 4\cdot 31$ or in other words $ 5^3 \equal{} 1$ mod $ 31$. Therefore we have $ 3^{12\alpha \plus{} 3} \equal{} 3^{12\gamma \plus{} 1}$ mod $ 31$. But the smallest integer $ k$ such that $ 3^k \equal{} 1$ mod $ 31$ is $ 30$. Therefore $ 12\alpha \plus{} 3 \equal{} 12\gamma \plus{} 1$ mod $ 30$ implying $ 12\alpha \plus{} 3 \equal{} 12\gamma \plus{} 1$ mod $ 6$ which is equivalent to $ 3 \equal{} 1$ mod $ 6$ obviously a contradiction.

Therefore since obviously $ a\geq c$ we must have $ c \equal{} 0$ or $ c \equal{} 1$. Meaning all solutions are the trivial $ (a,b,c) \equal{} (k,0,k)$ for any natural $ k$ and the nontrivial solutions of the equation $ 3^a \equal{} 5^b \plus{} 2$ of which unleast $ (3,2,1)$ exists...
[/hide]

ok i have solved the last case too. $ 3^a \equal{} 5^b \plus{} 2$
if we have $ a \equal{} 1$ we get $ b \equal{} 0$. now assume $ a > 1$. by mod $ 9$ we have $ 5^b \equal{} 7$ or in other words $ b$ is of the form $ b \equal{} 6\beta \plus{} 2$.
Now consider the equation mod $ 31$. Like mentioned before $ 5^3 \equal{} 1$ mod $ 31$. We therefore have $ 3^a \equal{} 27$ mod $ 31$. Because $ 30$ is the smallest integer $ k$ such that $ 3^k \equal{} 1$ mod $ 31$ we have $ a$ of the form $ a \equal{} 30\alpha \plus{} 3$.

We see that $ a$ is divisable by $ 3$ and $ b$ is divisable by $ 2$. We may therefore rewrite the equation into:
\[ x^3 \equal{} y^2 \plus{} 2\]
where $ x \equal{} 3^{10\alpha \plus{} 1}$ and $ y \equal{} 5^{3\beta \plus{} 1}$.

this is a known equation of Fermat with unique solution $ (x,y) \equal{} (3,5)$ (fact which we proove using the ring $ \mathbb[\sqrt {2}i]$) leading to $ (a,b) \equal{} (3,2)$ be the only solution left.
Q.E.D.

Can you solve with elementary Number Theory? :roll:

The equation: $ x^{3}\equal{} y^{2}\plus{}2$ can not be solved by elementary methods.
To resolve this problem your need to know some algebraic number theory.

Actually you could, but that gets really ugly (look at Euler's elementary proof of Fermat for exponent $ 3$, or his proof for $ x^3\plus{}1\equal{}y^2$). Just using arithmetic in $ \mathbb Z[\sqrt{\minus{}2}]$ is just so much easier.

ZetaX do u know where i could find such an elementary proof? i know how to use algebraic number theory, but im interested in the method of an elementary proof.

as for the the sub-case of the equation... i'm not sure, but i think it cannot be solved using just congruences..

[quote="Jure the frEEEk"]ok i have solved the last case too. $ 3^a \equal{} 5^b \plus{} 2$[/quote]
Considering this equation modulo $ 8829$, it is easy to obtain that $ (a,b)\equal{}(1,0)$ and $ (3,2)$ are the only solutions.
See http://www.mathlinks.ro/viewtopic.php?t=48431 for details.


Let $ a,b,c$ be positive real numbers from the interval $ \left[ \frac {1}{\sqrt {2}},2\right]$ such that $ abc\equal{}1$. Prove that
\[ 0 \ge \frac {a \minus{} 1}{2a^2 \minus{} 2a \plus{} 1} \plus{} \frac {b \minus{} 1}{2b^2 \minus{} 2b \plus{} 1} \plus{} \frac {c \minus{} 1}{2c^2 \minus{} 2c \plus{} 1} \ge \minus{} \frac {4}{5}.
\]
:)

Good luck in your exams, my younger brothers. Also, please take care for yourselves to work more with me at MIC. :)

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers from the interval $ \left[ \frac {1}{\sqrt {2}},2\right]$ such that $ abc \equal{} 1$. Prove that
\[ 0 \ge \frac {a \minus{} 1}{2a^2 \minus{} 2a \plus{} 1} \plus{} \frac {b \minus{} 1}{2b^2 \minus{} 2b \plus{} 1} \plus{} \frac {c \minus{} 1}{2c^2 \minus{} 2c \plus{} 1} \ge \minus{} \frac {4}{5}.
\]
:)

Good luck in your exams, my younger brothers. Also, please take care for yourselves to work more with me at MIC. :)[/quote]

[quote="nguoivn"]Here is my gift for my younger brothers: Honey_S (Loc) and tranquoc (Luat).
Given $ a, b, c > 0$. Prove that:
$ \sqrt {\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc}} \geq\ \frac {4}{3} (\sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b}})$

PS: Think about Am-Gm, my brothers  :)[/quote]

I, and Luat also, want to say "Thank you very much both of you, my brothers." :)

We will try our best on working with the MIC. Indeed, your inequalities is very nice. :)

I will show that
\[ \frac{{a \minus{} 1}}{{2a^2  \minus{} 2a \plus{} 1}} \plus{} \frac{{b \minus{} 1}}{{2b^2  \minus{} 2b \plus{} 1}} \plus{} \frac{{c \minus{} 1}}{{2c^2  \minus{} 2c \plus{} 1}} \le 0
\]
The rest is for Luat. :)
Indeed, without loss of generallity we may assume that $ a \ge b \ge c$. Then we consider 2 cases.
Case 1. $ a \ge b \ge 1 \ge c$ then we have
\[ \frac{{a \minus{} 1}}{{2a^2  \minus{} 2a \plus{} 1}} \plus{} \frac{{b \minus{} 1}}{{2b^2  \minus{} 2b \plus{} 1}} \equal{} \frac{{a \minus{} 1}}{{a^2  \plus{} \left( {a \minus{} 1} \right)^2 }} \plus{} \frac{{b \minus{} 1}}{{b^2  \plus{} \left( {b \minus{} 1} \right)^2 }} \le \frac{{a \minus{} 1}}{{a^2 }} \plus{} \frac{{b \minus{} 1}}{{b^2 }} \equal{} \left( {\frac{1}{a} \plus{} \frac{1}{b}} \right) \minus{} \left( {\frac{1}{{a^2 }} \plus{} \frac{1}{{b^2 }}} \right) \le \left( {\frac{1}{a} \plus{} \frac{1}{b}} \right) \minus{} \frac{2}{{ab}}
\]
And
\[ \frac{{c \minus{} 1}}{{2c^2  \minus{} 2c \plus{} 1}} \equal{} \frac{{c \minus{} 1}}{{2c\left( {c \minus{} 1} \right) \plus{} 1}} \le c \minus{} 1 \equal{} \frac{1}{{ab}} \minus{} 1
\]
It suffices to prove
\[ \left( {\frac{1}{{ab}} \minus{} 1} \right) \plus{} \left( {\frac{1}{a} \plus{} \frac{1}{b}} \right) \minus{} \frac{2}{{ab}} \le 0
\]
\[ 1 \plus{} \frac{1}{{ab}} \ge \frac{1}{a} \plus{} \frac{1}{b}
\]
\[ \left( {a \minus{} 1} \right)\left( {b \minus{} 1} \right) \ge 0
\]
Case 2. $ a \ge 1 \ge b \ge c$. We have to prove
\[ \frac{{1 \minus{} b}}{{2b^2  \minus{} 2b \plus{} 1}} \plus{} \frac{{1 \minus{} c}}{{2c^2  \minus{} 2c \plus{} 1}} \ge \frac{{a \minus{} 1}}{{2a^2  \minus{} 2a \plus{} 1}}
\]
In fact
\[ \frac{{1 \minus{} b}}{{2b^2  \minus{} 2b \plus{} 1}} \plus{} \frac{{1 \minus{} c}}{{2c^2  \minus{} 2c \plus{} 1}} \ge 2 \minus{} b \minus{} c
\]
and
\[ \frac{{a \minus{} 1}}{{2a^2  \minus{} 2a \plus{} 1}} \le \frac{1}{a} \minus{} \frac{1}{{a^2 }} \le 1 \minus{} \frac{1}{a} \equal{} 1 \minus{} bc
\]
It suffices to prove

\[ 2 \minus{} b \minus{} c \ge 1 \minus{} bc
\]

\[ \left( {b \minus{} 1} \right)\left( {c \minus{} 1} \right) \ge 0
\]
which is obvious. We have done. :)

[quote="Honey_S"]I will show that
\[ \frac {{a \minus{} 1}}{{2a^2 \minus{} 2a \plus{} 1}} \plus{} \frac {{b \minus{} 1}}{{2b^2 \minus{} 2b \plus{} 1}} \plus{} \frac {{c \minus{} 1}}{{2c^2 \minus{} 2c \plus{} 1}} \le 0
\]
The rest is for Luat. :)
Indeed, without loss of generallity we may assume that $ a \ge b \ge c$. Then we consider 2 cases.
Case 1. $ a \ge b \ge 1 \ge c$ then we have
\[ \frac {{a \minus{} 1}}{{2a^2 \minus{} 2a \plus{} 1}} \plus{} \frac {{b \minus{} 1}}{{2b^2 \minus{} 2b \plus{} 1}} \equal{} \frac {{a \minus{} 1}}{{a^2 \plus{} \left( {a \minus{} 1} \right)^2 }} \plus{} \frac {{b \minus{} 1}}{{b^2 \plus{} \left( {b \minus{} 1} \right)^2 }} \le \frac {{a \minus{} 1}}{{a^2 }} \plus{} \frac {{b \minus{} 1}}{{b^2 }} \equal{} \left( {\frac {1}{a} \plus{} \frac {1}{b}} \right) \minus{} \left( {\frac {1}{{a^2 }} \plus{} \frac {1}{{b^2 }}} \right) \le \left( {\frac {1}{a} \plus{} \frac {1}{b}} \right) \minus{} \frac {2}{{ab}}
\]
And
\[ \frac {{c \minus{} 1}}{{2c^2 \minus{} 2c \plus{} 1}} \equal{} \frac {{c \minus{} 1}}{{2c\left( {c \minus{} 1} \right) \plus{} 1}} \le c \minus{} 1 \equal{} \frac {1}{{ab}} \minus{} 1
\]
It suffices to prove
\[ \left( {\frac {1}{{ab}} \minus{} 1} \right) \plus{} \left( {\frac {1}{a} \plus{} \frac {1}{b}} \right) \minus{} \frac {2}{{ab}} \le 0
\]

\[ 1 \plus{} \frac {1}{{ab}} \ge \frac {1}{a} \plus{} \frac {1}{b}
\]

\[ \left( {a \minus{} 1} \right)\left( {b \minus{} 1} \right) \ge 0
\]
Case 2. $ a \ge 1 \ge b \ge c$. We have to prove
\[ \frac {{1 \minus{} b}}{{2b^2 \minus{} 2b \plus{} 1}} \plus{} \frac {{1 \minus{} c}}{{2c^2 \minus{} 2c \plus{} 1}} \ge \frac {{a \minus{} 1}}{{2a^2 \minus{} 2a \plus{} 1}}
\]
In fact
\[ \frac {{1 \minus{} b}}{{2b^2 \minus{} 2b \plus{} 1}} \plus{} \frac {{1 \minus{} c}}{{2c^2 \minus{} 2c \plus{} 1}} \ge 2 \minus{} b \minus{} c
\]
and
\[ \frac {{a \minus{} 1}}{{2a^2 \minus{} 2a \plus{} 1}} \le \frac {1}{a} \minus{} \frac {1}{{a^2 }} \le 1 \minus{} \frac {1}{a} \equal{} 1 \minus{} bc
\]
It suffices to prove
\[ 2 \minus{} b \minus{} c \ge 1 \minus{} bc
\]

\[ \left( {b \minus{} 1} \right)\left( {c \minus{} 1} \right) \ge 0
\]
which is obvious. We have done. :)[/quote]
Very nice your proof, Loc. My proof for the left part is as follow:
Setting $ a: \equal{}\frac{1}{a},b: \equal{}\frac{1}{b},c: \equal{}\frac{1}{c}$, the inequality becomes
\[ \sum \frac{a(1\minus{}a)}{a^2\minus{}2a\plus{}2} \le 0\]
As
\[ 1\minus{}\frac{4a(1\minus{}a)}{a^2\minus{}2a\plus{}2} \equal{}\frac{(a\minus{}1)^2\plus{}(2a\minus{}1)^2}{a^2\minus{}2a\plus{}2}\]
and by Cauchy Schwarz,
\[ \sum \frac{(a\minus{}1)^2}{a^2\minus{}2a\plus{}2} \ge \frac{(a\plus{}b\plus{}c\minus{}3)^2}{a^2\plus{}b^2\plus{}c^2\minus{}2(a\plus{}b\plus{}c)\plus{}6}\]
\[ \sum \frac{(2a\minus{}1)^2}{a^2\minus{}2a\plus{}2} \ge \frac{(2a\plus{}2b\plus{}2c\minus{}3)^2}{a^2\plus{}b^2\plus{}c^2\minus{}2(a\plus{}b\plus{}c)\plus{}6}\]
the inequality reduces to
\[ (a\plus{}b\plus{}c\minus{}3)^2 \plus{}(2a\plus{}2b\plus{}2c\minus{}3)^2 \ge 3(a^2\plus{}b^2\plus{}c^2)\minus{}6(a\plus{}b\plus{}c)\plus{}18\]
\[ (a^2\plus{}b^2\plus{}c^2)\plus{}5(ab\plus{}bc\plus{}ca)\minus{}6(a\plus{}b\plus{}c) \ge 0\]
\[ (a\plus{}b\plus{}c\minus{}3)^2 \plus{}2(ab\plus{}bc\plus{}ca\minus{}3) \ge 0\]
which is true by AM-GM. :)


Show that for any positive $x,y,z$ not all equal, we have:

\[(x+y)(y+z)(z+x) > 8xyz\]

[hide="Hint"]
AM-GM...[/hide]

[quote="amirhtlusa"][quote="Rep123max"]That should be a $\geq$ sign.[/quote]
the problem says NOT ALL EQUAL... ;)[/quote]

Right and this is what they have in Kalva (although sometimes they're wrong but I was pretty sure this one is right).

 ;)

Another solution.

$(x+y)(y+z)(z+x)-8xyz=x(y-z)^2+y(z-x)^2+z(x-y)^2>0$, because $x,y,z$ are not equal.

[hide="An ugly solution"]
Multiplying the LHS, we have $(x+y)(y+z)(z+x)=xyz+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+xyz$
Applying AM-GM to this sum, we have
$\frac{xyz+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+xyz}{8} \geq \sqrt[8]{(xyz)(x^2y)(x^2z)(y^2x)(y^2z)(z^2x)(z^2y)(xyz)}$
$\Rightarrow \frac{(x+y)(y+z)(z+x)}{8} \geq \sqrt[8]{x^8 y^8 z^8}$
$\Rightarrow (x+y)(y+z)(z+x)  \geq 8xyz$
The AM-GM inequality says that equality only occurs when all the terms are equal, so with the additional restriction that the $x,y,z$ not all equal, we can add strictness to the inequality.
[/hide]

i think if you scaled x, y,z, so that xyz=1 that might make the ugly proof above a little nicer.

[quote="blahblahblah"][quote="kunny"]Using $A.M\geqq G.M.$, $x+y>2\sqrt{xy}>0,y+z>2\sqrt{yz}>0,z+x>2\sqrt{zx}>0$, because$ x,y,z$ are not equal.
Therefore Multiplying these inequalities $(x+y)(y+z)(z+x)>8xyz$, we are done.[/quote]

That's not true.  It says not ALL equal; one of the inequalities can be an equality.[/quote]

Oh, you are right, thank you, blahblahblah.

By the way, the original problem is as follows.

[b]If $a,b$ and $c$ are any three positive real numbers, prove that $(a+b)(b+c)(c+a)\geqq 8abc$.
Show that equality holds only when $a=b=c$.

1st British Mathematical Olympiad 1965 #9[/b]

x + y >= 2sqrt(xy)
y+z >= 2sqrt(yz)
x+z >= 2sqrt(xy)

However, one of the inequality signs must be > and not >=.  If a>b and c >= d, then we (kind of obviously) have ac > bd, so we can multiply the above inequalities to get

(x+y)(y+z)(x+z) > 8xyz


[quote="quangtrung"]prove that for any irrational $\alpha$ there exists an integer number n so that
             {$\alpha^{n}$}$\in [\frac{1}{3}: \frac{2}{3}]$[/quote]
   don't you have any idea .just only idea  please!
  thanks

idea: search ;)
http://www.mathlinks.ro/Forum/viewtopic.php?t=48461


So I'm playing DDR occasionally to try to slim down for Meggy-Meg but I suck too much. I can't even do three footers and get perfects on all of them. I *always* get a ton of greats. Any tips on hitting notes perfectly?

And my feet are really slow and clumsy (big surprise there) so I can't really hit multiple notes fast with one foot... and my brain is too slow to think a few notes ahead so that I can alternate feet on all of them.

I'm a DDR god.
I do regular 9 and 10 footers.
The key is practice.
Sight reading DDR is like sight reading music.

Also, get a computer version like DWI or Stepmania, and practice with that (with your fingers).
It will help your step timing. Don't forget that you will need more practice with your feet, but I found this to be a solid way to train.

Eh, I'm too lazy to download Stepmania. I'm having trouble navigating the website... there's so much confusing stuff you have to downloads, and I can't tell the difference between Stepmania and Stepmania Online and etc.

I'm pathetic. I tried doing Breakdown (from MAX2) on training mode on speed 2 (out of 5, which is normal speed) and I failed miserably with like a B or something. The highest thing I've AA'd is a 3 footer.

can meggy-meg even dance, or does she need to get better too

 :|  :!:

Well I know for a fact that she takes dance classes... not sure what type though. And she's like in Jazz Dance and Colorguard and every dance and music related thing in the school.

And she's hott.

I rock at DDR.  :wink: I can AA 10-footers.
But I like playing ITG better. Harder songs, more of a challenge, existence of songs that I actually still can't beat, etc. etc.

im ok at ddr but i'm quite good at ffr...
i guess those are kind of related.

...sorta.
I find that FFR < SM, though. At least you can adjust the syncing in SM.

Who is the Meggy-Meg I keep seeing posts/sigs about?

FFR -> http://www.flashflashrevolution.com

I personally think it's a bit lame, but whatever.
YES! Since I'm going home today, I have DDR time!

Shame...  :lol: . So who/what is Meggy-Meg?

i SUCK at DDR, guitar hero, and anything of the sort.  i tried DDR on like speed 1 and i failed like a D or something. Ubemaya, your still better than me, not much to be proud of though.

[quote="moogra"][quote="CheeseIsGood"][quote="Ubemaya"]What's FFR, and who the heck has sigs about Meggy-Meg?[/quote]

Well, pianoforte does, and I think you did before.[/quote]

he has it to "publicly shame" Ubemaya[/quote]

it's a parody of PenguinIntegrals sig, only making fun of Ubemaya instead of me

guitar hero=MAJOR PWN>>>>>>>>>>>ddr=semi pwn

[quote="veryfriendly"]DDR is an awesome game I love it. I play all the time!![/quote]

you say: DDR is an awesome game I love it. I play all the time!!
i think: DDR is a not so awesome game. i hate it. i dont play all the time and i SUCK at it.

i know this is the second time im posting something like this, but you guys get the point

@junggi
They made a game trying to mimic GH its called frets on fire, but it sucks.

Ubemaya make sure you have a shock device nearby, most likely a cattle prod just in case you pass out, and when there are four arrows, you have to go down on all fours.

And now a song from GH2

[youtube]VY_DOeCCbCg[/youtube]

In The Groove has triples and quads too, i.e. three or four arrows at the same time.

[quote="hunter34"]@junggi
They made a game trying to mimic GH its called frets on fire, but it sucks.[/quote]
yes i know
awesomemath counselors(zac cox and oleg to be specific) were playing it in the office
zac said that the audio is late and such so it sucks

Yes, And now for another GH2 music.

[youtube]FDxbTNwQyvg[/youtube]

foo fighters=crap
post some megadeth please

Done and Done

[youtube]MZmPD6-vNVY[/youtube]

[youtube]-YczYC2yHTM[/youtube]

[quote="hunter34"]Done and Done[/quote]
wtf
iirc it was hangar 18 not holy wars
hangar is a better song too
but
they need tornado of souls
brilliant song

@albert: you are a noob

[quote="Ubemaya"]And you have poor taste in music.[/quote]
hmm?
how so?
[url]http://www.allmusic.com/cg/amg.dll?p=amg&sql=10:oekqoaeaqij9~T1[/url]
[url]http://www.allmusic.com/cg/amg.dll?p=amg&sql=10:necibkj9aakm~T1[/url]
[url]http://www.allmusic.com/cg/amg.dll?p=amg&sql=10:bifnzfahehpk[/url]
[url]http://wc03.allmusic.com/cg/amg.dll?p=amg&sql=10:2k9sa9yge23d[/url]
[url]http://www.allmusic.com/cg/amg.dll?p=amg&sql=10:1mzyxdabjola[/url]
[url]http://www.allmusic.com/cg/amg.dll?p=amg&sql=10:4yhqoayabijx[/url]
[url]http://www.allmusic.com/cg/amg.dll?p=amg&sql=10:avfixqlald0e[/url]
unlike some people, i actually listen to music for what music is supposed to be:creative/complex songwriting, not catchyness of the vocals

Uh oops here is hanger 18

[youtube]rtsD2tBPZgo[/youtube]

Here is Tornado of souls, but it is GH2 hack,

[youtube]ND_sf5zWoos[/youtube]

misirlous ftw!
except all the youtube vids of it are wayy off

[quote="hunter34"]Maybe to make Ubemaya happy, I will post a electronic song that may fit DDR.
[youtube]xqKielID-mc[/youtube][/quote]
prettay cool song actually
just needs better drums and bass
keyboards sounds awesome

Did anyone hear about guitar hero 3 and the PSP guitar hero? it's pretty old news actually

Uh, yeah. I just want to know the music.


[quote="Beat"]Let $x, y$ and $z$ be positive real numbers satisfying $^{2}+y^{2}+z^{2}= 1$. Prove that $\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}\ge \frac{3\sqrt{3}}{2}$.[/quote]
Let $x=\frac{a}{\sqrt3},$ $y=\frac{b}{\sqrt3}$ and $z=\frac{c}{\sqrt3}.$ Hence, $a^{2}+b^{2}+c^{2}=3$ and
$\sum_{cyc}\frac{x}{1-x^{2}}\geq\frac{3\sqrt{3}}{2}\Leftrightarrow\sum_{cyc}\frac{a}{3-a^{2}}\geq\frac{3}{2}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(\frac{a}{3-a^{2}}-\frac{1}{2}-\frac{1}{2}(a^{2}-1)\right)\geq0\Leftrightarrow\sum_{cyc}\frac{a(a+2)(a-1)^{2}}{3-a^{2}}\geq0.$ :)


algebra
polynomial
modular arithmetic
ratio
number theory
relatively prime
IMC

Let $ n, k$ be positive integers and suppose that the polynomial $ x^{2k}\minus{}x^k\plus{}1$ divides $ x^{2n}\plus{}x^n\plus{}1$. Prove that $ x^{2k}\plus{}x^k\plus{}1$ divides $ x^{2n}\plus{}x^n\plus{}1$.

we can find roots of the first polynom x^2k-x^k+1 and it alows as to find a relation between n and k.
we check easily that roots of x^2k+x^k+1 are roots of x^2n+x^n+1.

[quote="t0rajir0u"]Er... I may have miscalculated, but are you sure you don't mean $ x^{2n} \minus{} x^n \plus{} 1$ for the first condition?[/quote]I think the stated problem is correct as well.

We can calculate the roots of $ x^{2k} \minus{} x^{k} \plus{} 1$: $ cis (\frac {\pm60}{k} \plus{} \frac {360i}{k}), 0 \le i \le k \minus{} 1$
They are also roots of $ x^{2n} \plus{} x^{n} \plus{} 1$, hence: $ \frac {\pm 60n}{k} \plus{} \frac {360in}{k} \equal{} 120$ or $ \minus{} 120 \pmod {360}, 0 \le i \le k \minus{} 1$ (1)

It is sufficient and necessary to prove that k divides n, and $ (\frac {n}{k},3) \equal{} 1$ (sufficiency can be seen considering the polynomial $ P(x) \equal{} x^{2n/k} \plus{} x^{n/k} \plus{} 1 : x^2 \plus{} x \plus{} 1|P(x)$ because $ {2,1,0} \equal{} {\frac {2n}{k},\frac {n}{k},1} \pmod 3$, and we choose $ x \equal{} y^k$)

The equation (1) is equivalent to:$ \frac {\pm n \plus{} 6in}{k} \equal{} 2$ or $ \minus{} 2 \pmod {6}$ for every $ 0 \le i \le k \minus{} 1$.
for i=0 it implies: $ \frac {n}{k} \equal{} \pm2 \pmod 6$, and than k divides n and also the ratio is relatively prime to 3, and we are done.

EDIT: we only used the fact that $ cis \frac{60}{k}$ is a root of the polynomial. Can anyone validate the solution\present a more elegant one?

Try the following

[b]Lemma:[/b]  Given a prime $ p$ and a positive integer $ n$, we have

- $ p | n \implies \Phi_n(x^p) \equal{} \Phi_{np}(x)$
- $ p \not | n \implies \Phi_n(x^p) \equal{} \Phi_{np}(x) \Phi_n(x)$.

This can be proven by root-counting or by Mobius inversion.


Determine the number of solutions to the equation:
$ 3x\plus{}4y\plus{}5z \equal{} 2008$, where $ x, y, z \in N$.

[quote="luimichael"]Determine the number of solutions to the equation:
$ 3x \plus{} 4y \plus{} 5z \equal{} 2008$, where $ x, y, z \in N$.[/quote]
We can use generating functions I think:
We want the coefficient of $ x^{2008}$ in
$ (1\plus{}x^{3}\plus{}x^{6}\plus{}...)(1\plus{}x^{4}\plus{}x^{8}\plus{}...)(1\plus{}x^{5}\plus{}x^{10}\plus{}...)$
$ \frac{1}{(1\minus{}x^{3})(1\minus{}x^{4})(1\minus{}x^{5})}$, which I have no idea how to get.

Consider the equation modulo $ 5$, that is $ 3x\plus{}4y\equiv 3\pmod 5\iff 3x\plus{}4y\equal{}3\plus{}5s$.  Notice that $ (x,y)\equal{}(1\minus{}s,2s)$ is a solution, so $ (x,y,z)\equal{}(1\minus{}s\plus{}4t,2s\minus{}3t,401\minus{}s)$.  Then limit $ s,t$.

[quote="luimichael"]Determine the number of solutions to the equation:
$ 3x \plus{} 4y \plus{} 5z \equal{} 2008$, where $ x, y, z \in N$.[/quote]

$ 4y \equal{} 2008\minus{}3x\minus{}5z$

$ y \equal{} \frac{2008\minus{}3x\minus{}5z}{4}$

$ y \equal{} 502\minus{}z\minus{}\frac{3x\plus{}z}{4}$

So $ \frac{3x\plus{}z}{4}$ must be an integer $ t$.

$ t\equal{}\frac{3x\plus{}z}{4}$

$ \Rightarrow 4t \equal{} 3x\plus{}z$

$ \Rightarrow x \equal{} \frac{4t\minus{}z}{3} \equal{} t\plus{}\frac{t\minus{}z}{3}$

So $ \frac{t\minus{}z}{3}$ must be an integer $ u$.

$ u \equal{} \frac{t\minus{}z}{3}$

$ 3u \equal{} t\minus{}z$

$ z \equal{} t\minus{}3u$

Substituting back, we get:

$ x \equal{} \frac{4t\minus{}(t\minus{}3u)}{3} \equal{} \frac{3t\plus{}3u}{3}\equal{}t\plus{}u$

$ y \equal{} 502\minus{}(t\minus{}3u)\minus{}\frac{3(t\plus{}u)\plus{}(t\minus{}3u)}{4}$

$ y \equal{} 502\plus{}3u\minus{}t \minus{} \frac{3t\plus{}3u\plus{}t\minus{}3u}{4}$

$ y \equal{} 502\plus{}3u\minus{}t\minus{}\frac{4t}{4}$

$ y \equal{} 502\plus{}3u\minus{}2t$

So the solution set is:

$ \lbrace{ (x,y,z): (t\plus{}u,502\plus{}3u\minus{}2t,t\minus{}3u) \rbrace}$


Prove that for any positive real $ t$, there are integers $ a,b,c,d$ such that:
$ a^3b\plus{}b^3c\plus{}c^3d < tabcd$

Take $ a\equal{}c\equal{}\minus{}1,b\equal{}d\equal{}1$, the inequality becomes $ t>\minus{}3$, which is true for any positive real $ t$.

[quote="hendrata01"]Prove that for any positive real $ t$, there are integers $ a,b,c,d$ such that:
$ a^3b \plus{} b^3c \plus{} c^3d < tabcd$[/quote]

with $ (a,b,c,d)\equal{}(\minus{}1,1,0,0)$, the inequality becomes  $ \minus{}1<0$, true for any real $ t$, and not only for positive reals ...  :)

[quote="The Enigma"]the problem seems interesting is there a variation or a typo??[/quote]

I think the real problem likely was :

[quote="pco modifying hendrata01"]Prove that for any positive real $ t$, there are [u][b]positive [/b][/u]integers $ a,b,c,d$ such that:
$ a^3b \plus{} b^3c \plus{} c^3d < tabcd$[/quote]

And then, just choose $ (a,b,c,d)\equal{}(n,1,1,n^3)$ and the inequality becomes $ t>\frac{2n^3\plus{}1}{n^4}$ and you just have to choose $ n$ great enough :)


How many license plates consist of 2 letters followed by 3 numbers?

There are 26 possible letters and 10 possible one-digit numbers. Since repititions are allowed, the total number of ways is $ 26\times{26}\times10^3\equal{}676\times{1000}\equal{}\boxed{676000}$


this is permutations and combination problem.
 problem is as follows.
  if 3 distinct integers are chosen at random . show that we can choose 2 of them
 lets say 'a' and 'b'. then prove that (a^3)b-a(b^3) is divisible by 30.
please provide answer by using permutations and combinations.

My proof doesn't use combinatorics.
[hide]For every two integers $ a^{3}b \minus{} b^ab^{3} \equal{} ab(a \minus{} b)(a \plus{} b)$ is divisible by 2, because
it's trivial if one of the integers is even, and then 2 integers are odd their sum is divisible by 2.
Similarly this expression is  divisible by 3. If the integers (not divisible by 3) are equal mod 3, then it's difference is divisible by 3, if not they are 1 and 2 mod 3 so their sum is divisible by 3.
To check if we can choose 2 integers from 3, so that expression above is divisible by 5, we can assume that none 2 of 3 integers don't equal mod 5 and none of them is divisible by 5. So they can equal 1,2,3,4 mod 5. If one the integer equals x mod 5 and the other equals 5-x, the its sum is divisible by 5, if not the remainders of 3 integers are x and 2 numbers from 1,2,3,4 without 5-x, the sum of 5-x+5=5, so the sum of this two remainders is 10-5=5.[/hide]


Chris's running schedule dictates that the never run more than two days in a row, nor does he go more than two days in a row without running. In two weeks what is the positive difference between the fewest number of days and the greatest number of days that he could have run?

(_ indicates no run and X indicates run)

Least: _ _ X _ _ X _ _ X _ _ X _ _ (10 days)
Greatest: X X _ X X _ X X _ X X _ X X (4 days)
10-4=6

EDIT: Oh, in 2 weeks!

6

10 is the most days he can run, and 4 is the least days he can run.


[quote="radio"]Given is a sphere of radius $ 1$. Two parallel planes cut the sphere. Look at the total surface $ S$ of the sphere cut by the planes.

Find the maximum of $ S$.[/quote]

[hide]
Assume the planes are each (insert infinitesimal here) from the center.  Then when you cut it, you get a circle with radius 1 + 2 hemispheres of radius 1
[/hide]

seems too simple, maybe i misinterpreted it?

Well, actually I want to count the surface only if it lies on the outside of the sphere... you look at the surface "between" the two planes.

Sorry for being unclear


Which of the following numbers is the greatest?

$ \textbf{(A)}\ 2^{10} \plus{} 2^{ \minus{} 10} \qquad \textbf{(B)}\ 2^{10} \minus{} 2^{ \minus{} 10} \qquad \textbf{(C)}\ 2^{10} \plus{} 10^{ \minus{} 3}$
$ \textbf{(D)}\ 10^3 \plus{} 2^{ \minus{} 10} \qquad \textbf{(E)}\ 10^3 \plus{} 10^{ \minus{} 3}$

[hide]$ 2^{10} \equal{} 1024$ and $ 10^3 \equal{} 1000$ . Since the second term of each choice is less than 24, any answer that starts with $ 2^{10}$ will be greater than one that starts with $ 10^3$. So the answer is either A, B, or C. Since choice B is subtracting, it won't be greater than A or C. Also, since $ 2^{10} > 10^3$ , $ 2^{ \minus{} 10} < 10^{ \minus{} 3}$ . Therefore, the answer is $ C$ , $ 2^{10} \plus{} 10^{ \minus{} 3}$ . Or you could just enter each one into your calculator and see which one has the greatest value.[/hide]

Edit: I didn't realize this was a challenge question, so I edited my answer to hide it, sorry.

I want younger people to try to understand how this works. So, answer this question first (the solution above is correct).

Let's say we have two positive integers such that $ a > b$. What relations do we have for $ \frac{1}{a}$ and $ \frac{1}{b}$? Which is greater?

Now, does your answer change if $ a,b$ are negative integers?


Let $ k\geq2$ be a positive integer and $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}3$.Prove that
$ \sum_{cyc}\sqrt{a\plus{}k^{2}\minus{}1} \geq k(ab\plus{}bc\plus{}ac)$.

[quote="arqady"][quote="stephen38"]Let $ k\geq2$ be a positive integer and $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$.Prove that
$ \sum_{cyc}\sqrt {a \plus{} k^{2} \minus{} 1} \geq k(ab \plus{} bc \plus{} ac)$.[/quote]
I think, it's true for all $ k\geq1.$ :wink:[/quote]

Dear arqady,have you a proof for it? :)

[quote="stephen38"]Let $ k\geq2$ be a positive integer and $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$.Prove that
$ \sum_{cyc}\sqrt {a \plus{} k^{2} \minus{} 1} \geq k(ab \plus{} bc \plus{} ac)$.[/quote]
$ \sum_{cyc}\sqrt {a \plus{} k^{2} \minus{} 1} \geq k(ab \plus{} bc \plus{} ac)\Leftrightarrow$
$ \Leftrightarrow3k^2\plus{}2\sum_{cyc}\sqrt{(a\plus{}k^2\minus{}1)(b\plus{}k^2\minus{}1)}\geq k^2(ab\plus{}ac\plus{}bc)^2.$
For $ k\geq1$ we obtain: $ \sqrt{(a\plus{}k^2\minus{}1)(b\plus{}k^2\minus{}1)}\geq\sqrt{ab}\plus{}k^2\minus{}1.$
Id est, it remains to prove that
\[ 9k^2\minus{}6\plus{}2\left(\sqrt{ab}\plus{}\sqrt{ac}\plus{}\sqrt{bc}\right)\geq k^2(ab\plus{}ac\plus{}bc)^2,\]
which equivalent to the following inequality:
\[ (k^2\minus{}1)\left(9\minus{}(ab\plus{}ac\plus{}bc)^2\right)\plus{}3\plus{}2\left(\sqrt{ab}\plus{}\sqrt{ac}\plus{}\sqrt{bc}\right)\geq (ab\plus{}ac\plus{}bc)^2.\]
Thus, it remains to prove that $ 3\plus{}2\left(\sqrt{ab}\plus{}\sqrt{ac}\plus{}\sqrt{bc}\right)\geq (ab\plus{}ac\plus{}bc)^2.$
Let $ a\equal{}x^2,$ $ b\equal{}y^2$ and $ c\equal{}z^2,$ where $ x,$ $ y$ and $ z$ are non-negative numbers.
Hence, $ 3\plus{}2\left(\sqrt{ab}\plus{}\sqrt{ac}\plus{}\sqrt{bc}\right)\geq (ab\plus{}ac\plus{}bc)^2\Leftrightarrow$
$ \Leftrightarrow(x^2\plus{}y^2\plus{}z^2)^4\plus{}2(xy\plus{}xz\plus{}yz)(x^2\plus{}y^2\plus{}z^2)^3\geq27(x^2y^2\plus{}x^2z^2\plus{}y^2z^2)^2,$
which  true ( after expanding ) by AM-GM and Muirhead.


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

hi can you help me ?
every r_module flat is projective?

Well, what is $r$¿ The reals¿ Any ring (some doubts)¿

[b]Theorem.[/b][i] Every projective module is flat.[/i]

[b]Theorem (Bass).[/b] [i]The following conditions are equivalent:
(а) Every right flat $R$-module is projective.
(b) $R/rad(R)$ is semisimple and for any sequence $a_{1},a_{2},\ldots$ from $rad(R)$ there exists $n\in\mathbb N$ such that $a_{n}a_{n-1}\ldots a_{1}=0$.
(c) $R$ satisfies the descending chain condition for principal left ideals.[/i]

The latter theorem is extremely hard to prove.

@sepidetaghizade: No ... take $R = \mathbb{Z}, M = \mathbb{Q}$. Then $M$ is a flat $R$-module. But every projective $R$-module is free (since $R$ is a PID, submodules of free $R$-modules are also free), and clearly, $M$ is not.


algorithm
superior algebra
superior algebra unsolved

Let $ F$ a free group.Prove that, if $ ab\equal{}ba$ then $ \exists z$ such that $ a\equal{}z^k,b\equal{}z^l$ for uniq $ k,l \in \mathbb{Z}$

Suppose the free generators of $ F$ are $ (x_i)_{i\in I}$ where $ I$ is a set.
Write $ a\equal{}ya'x$ and $ b\equal{}x^{\minus{}1}a'y^{\minus{}1}$ so that $ l(ab)\equal{}l(y)\plus{}l(a')\plus{}l(b')\plus{}l(x)\equal{}l(ba)$ where $ l(t)$ is the length of $ t$ in terms of the free generators (i.e. if $ t\equal{}x_1x_2....x_k$ with $ x_i\neq x_{i\plus{}1}^{\minus{}1},\forall i\equal{}1,2,..,k\minus{}1$ then $ l(t)\equal{}k$ ).

(I'm working with elements written in the shortest form, i.e. when two generators are neighbors in the writing of some element they are not inverse one to each other to avoid things like $ ...x_i\cdot x_i^{\minus{}1}...$)

You can write $ a$ and $ b$ in this form because writing $ a$ and $ b$ in terms of the free generators first you look at the begining of $ a$ and at the ending of $ b$ and use the following algorithm:
0. set $ x\equal{}1$.
1. if $ a$ ends in $ x_i$ and $ b$ begins with $ x_i^{\minus{}1}$ then you set $ x\equal{}x\cdot x_i$.
1'. else algorithm stops and you have your $ x$.
2. replace $ a$ with $ a\cdot x_i^{\minus{}1}$ and $ b$ with $ x_i\cdot b$
3. go to step 1.

similarly you find $ y$.

Now you have from $ ab\equal{}ba$ that $ ya'b'y^{\minus{}1}\equal{}x^{\minus{}1}b'a'x$. Since all the elements (and the products) are written in the shortest form we get that  $ l(x)\equal{}l(y)$ and hence $ y\equal{}x^{\minus{}1}$.

Now we have to prove that if $ a',b'$ are in this free group so that $ l(a'b')\equal{}l(b'a')\equal{}l(a')\plus{}l(b')$ and $ a'b'\equal{}b'a'$ then $ a'\equal{}z^k$ and $ b'\equal{}z^l$ for some $ z$.
Find $ z$ with the shortest length so that $ a'\equal{}z^k$ for some $ k>0$ with the following algorithm:
0. set $ g\equal{}1$
1. Take the first letter of $ a'$ and find the next position $ >g$ on which this appears, let's say $ l$.
2. Consider the word $ z$ formed by the letters from the position 1 to position $ l\minus{}1$ and find if $ z^{l(a')/(l\minus{}1)}\equal{}a'$.
    If yes then this is our $ z$, else set $ g\equal{}l$ and go to step 1.

You find a $ z$ so that $ z^r\equal{}a'$ for some $ r$.
Similarly you find $ w$ so that $ w^m\equal{}b'$ for some $ m$.
Suppose $ A_2\equal{}l(z)<l(w)\equal{}A_1$.
Partition $ w$ in bunches of $ A_2$ letters. If $ A_1$ mod $ A_2$ is not zero then the last bunch of letters will have $ A_3\equal{}(A_1$ mod $ A_2)$ letters.
From $ z^rw^m\equal{}w^mz^r$ and from the fact that the group is free and this writing is with minimal length we get that $ w\equal{}z^{r_1}u_1$ where $ l(u_1)\equal{}A_3$.
Now $ z^{r}u_1w^{m\minus{}1}\equal{}u_1w^{m\minus{}1}z^r$
Partition $ z$ in bunches of $ A_3$ letters. If $ A_2$ mod $ A_3$ is not zero then the last bunch of letters will have $ A_4\equal{}(A_2$ mod $ A_3)$ letters. In the same manner we get an element $ u_2$ with $ l(u_2)\equal{}A_4$ and $ z\equal{}u^{r_2}_1u_2$.
In this process at some time we will have $ A_{i\plus{}1}|A_i$ (from Euclid's algorithm) which will lead to a some contradiction with the minimality of lengths of $ z$ and $ w$ (in this situation $ z$ (or $ w$)  will be a power of some $ u_{i\minus{}1}$ with the length of $ u_{i\minus{}1}$ strictly smaller than that of $ z$ (or $ w$) ).
Similarly you treat the case $ A_2>A_1$. So $ A_1\equal{}A_2$ and the conclusion follows.


[quote="Virgil Nicula"]$\boxed{\ \left\{\begin{array}{c}\{x,y,z\}\subset (0,\infty )\\\\ xy+yz+zx+xyz=4\end{array}\right\|\Longrightarrow\frac{(x-1)(x+4)}{x+1}+\frac{(y-1)(y+4)}{y+1}+\frac{(z-1)(z+4)}{z+1}\ge 0\ }$[/quote]
My solution:
The condition $xy+yz+zx+xyz=4$ can be rewritten by $\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=1$
Setting $a=\frac{1}{x+2}$ and $b,c$ is similar, then we have $a+b+c=1$ and from this, we have $a,b,c$ be three length sides of an triangle.
The inequality becomes
\[\sum\frac{(1-3a)(1+2a)}{a(1-a)}\ge 0 \]
\[\Leftrightarrow \sum\frac{(b+c-2a)(1+2a)}{a(1-a)}\ge 0 \]
\[\Leftrightarrow \sum(a-b)^{2}(c-2ab)c(1-c) \ge 0 \]
\[\Leftrightarrow \sum(a-b)^{2}S_{c}\ge 0 \]
WLOG, we can assume that $b+c \ge a \ge b \ge c \ge 0$. Then we have $S_{a}, \ge 0$ and
\[S_{b}=(b-2ac)b(1-b)=(b(a+b+c)-2ac)b(1-b) \ge 0 \\ S_{b}+S_{c}\ge c(1-c)(b+c)(1-2a)=c(1-c)(b+c)(b+c-a) \ge 0 (\text{since }b(1-b) \ge c(1-c)) \]

We are done. :)

[quote="toanhocmuonmau"]
\[\sum\frac{(1-3a)(1+2a)}{a(1-a)}\ge 0 \]
[/quote]

This is equivalent to
$\frac{(3a-1)^{2}(2-5a)}{2(1-a)a}\geq 0$

.Could anyone prove it in that form ?

[quote="Virgil Nicula"]$\boxed{\ \left\{\begin{array}{c}\{x,y,z\}\subset (0,\infty )\\\\ xy+yz+zx+xyz=4\end{array}\right\|\Longrightarrow\frac{(x-1)(x+4)}{x+1}+\frac{(y-1)(y+4)}{y+1}+\frac{(z-1)(z+4)}{z+1}\ge 0\ }$[/quote]

What about your solution, virgil?

 :)

i think a very nice solution is with the following substitutions:

$x+1=a, y+1=b, z+1=c$ and then our condition becames $a+b+c+2=abc$ and now we make the following ones:

$a=\frac{m+n}{p}, b=\frac{n+p}{m},c=\frac{p+m}{n}$ , with $a,b,c > 1$ therefore $m,n,p$ sides of a triangle.

and the rest is now history  :blush:


Prove that there exists $x,y,z \in N$ such that $n=x^{2}+y^{2}+z^{2}$ $\leftrightarrow$ $n \neq 4^{m}(8l+7)$

It was posted before:
http://www.mathlinks.ro/Forum/viewtopic.php?t=5269
http://www.mathlinks.ro/Forum/viewtopic.php?t=112973

Let us be clear: from now on, absolutely any post whose title is "help me", "selling shoes", "huoc tran dung...", "have merci", "god save us", "nice", "hard"  and any other such inventions will be deleted with absolutely no worries and without regarding the content.   :mad:


Ross Mathematics Program
calculus
integration
probability
limit
probability and stats

consider the following SDE:
$ dX \equal{} (1\minus{}X) dt \plus{} \sqrt{|X|} dW$, $ X_{t\equal{}0} \equal{} 1$
where $ W$ is a brownian motion. It is quite intuitive that $ X$ stays positive for all time because when $ X$ goes near $ 0$, the term $ \sqrt{|X|} dW$ becomes very small, while the term $ (1\minus{}X) dt$ make $ X$ goes up again. How would you properly prove that ? Can anyone compute the law of $ X_t$? (I was told that there is no closed form for the solution of this SDE)

I don't know how to prove it, but this is related to the Cox-Ingersoll-Ross model of interest rates; you may want to look there.

yes, and this is why it is interesting that it stays positive (interest rates are positive). A friend told me today that it was distributed like a chi square ..

Here's one suggestion as to why it may be true.  It's not a rigorous proof, but it may be the skeleton of one.  Then again, I'm not very comfortable with stochastic integration at all, so this argument may be complete garbage...

Let $ t_0 < t_1 < ...$ be the times at which $ X$ is a power of $ \frac {1}{2}$.  Let $ X(t_j) \equal{} 2^{ \minus{} Y_j}$.  The idea here is to view $ Y_j$ as a random walk on the nonnegative integers, and that if $ Y_j$ is large, that walk should be biased heavily towards 0.

For each step, we view it as a separate SDE process where we start at $ 2^{ \minus{} k}$ and stop when we hit either $ 2^{ \minus{} (k \minus{} 1)}$ or $ 2^{ \minus{} (k \plus{} 1)}$.   Let $ Y$ be an auxillary variable satisfying the SDE
\[ dY \equal{} (1 \minus{} 2^{ \minus{} k \plus{} 1})dt \plus{} \sqrt {(2^{ \minus{} k \plus{} 1})}dW, Y_{t \equal{} 0} \equal{} 2^{ \minus{} k}
\]
$ Y$ has less positive bias than $ X$, and a larger Brownian component.  It therefore seems obvious (and maybe not too hard to prove) that if we run the same stopping time process on $ Y$, then $ Y$ will be more likely to reach $ 2^{ \minus{} (k \plus{} 1)}$ than $ X$.  However, the probably that $ Y$ reaches $ 2^{ \minus{} (k \minus{} 1)}$ before $ 2^{ \minus{} (k \plus{} 1)}$ can be computed explicitly by treating the system
\[ E(Y \minus{} Y_0) \equal{} (1 \minus{} 2^{ \minus{} k \plus{} 1})E(t_{stop}), Var(Y \minus{} Y_0) \equal{} 2^{ \minus{} k \plus{} 1} E(t_{stop})
\]
as two equations in the two unknowns $ E(t_{stop}), P(Y_{stop} \equal{} 2^{ \minus{} (k \minus{} 1)})$.
When I did so, I got that $ Y$ stopped at $ 2^{ \minus{} (k \minus{} 1)}$ with probability  approaching $ \frac {5}{9}$ for sufficiently large $ k$.  It follows that there is some $ k_0$ for which $ P(Y_{k \plus{} 1} \equal{} Y_k \minus{} 1 | Y_k \geq k_0) \geq \frac {1}{2} \plus{} \frac {1}{20}$.  But a random walk with a positive bias towards 0 remains bounded almost surely, so we should be done.

There's probably a very way to go to make this argument rigorous.  My main concern would be translating back from a statement about the $ Y_j$ to a statement about the behavior of $ X(t)$.  What's missing is some sort of argument along the lines of "$ Y_j$ remains bounded, therefore $ \lim_{t \rightarrow \infty} t_j \equal{} \infty$ with high probability", in order to be sure that we've actually shown something for all $ t$ as opposed to just a finite interval.


inequalities
function
geometry unsolved
geometry

Given are $\triangle ABC$and Fermat point $F$ (suppose to exist $F$) prove that $\sqrt{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge \frac{9}{FA+FB+FC}$ :)

This inequality is not true in general.

Your inequality is equivalent to showing for every point $P$ we have $(PA+PB+PC)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq 3\sqrt{3}$.

Suppose $ABC$ is an isosceles triangle at $A$ and $\frac{1}{\sqrt{3}}\leq \frac{b}{a}\leq \frac{3\sqrt{3}-4}{2}$. Notice that at that time $\measuredangle{A}\leq 120^{0}$. Let $P$ be $A$. Then your inequality becomes $2b(\frac{1}{a}+\frac{2}{b}) \geq 3\sqrt{3}$ which is not true according to what we have assumed above.

Maybe, you miss some conditions such as $ABC$ is acute?

ok treegoner youa are right but If I suppose all angle of $\triangle ABC <120^{0}$ then is it true, I hope it true because I will have very nice consequence! :)  :)

To gemath,

Probably you didn't read my post thoroughly. In my previous post, I showed a counter example of your inequality. The triangle in my counter example is obtuse and has all angles $\leq 120^{0}$.

However, your inequality is true for any acute triangle. Here is my long proof

[i]Proof[/i]

[b]Lemma 1[/b] $p^{2}\geq \frac{27Rr}{2}\geq 27r^{2}$.

This lemma is quite obvious since $(a+b+c)^{3}\geq 27abc$ and $R \geq 2r$.

[b]Lemma 2 [/b] If $F$ is the Fermat point, then $FA+FB+FC = \sqrt{{\frac{a^{2}+b^{2}+c^{2}+4\sqrt{3}{S}}{2}}}$

This lemma is easy to prove based on the definition of $F$ and the law of cosine.

[b] Lemma 3 [/b]

$tg{\frac{A}{2}}+tg{\frac{B}{2}}+tg{\frac{C}{2}}= \frac{4R+r}{p}$

$a^{2}+b^{2}+c^{2}= 2(p^{2}-r^{2}-4Rr)$.

$cotgA+cotgB+cotgC = \frac{a^{2}+b^{2}+c^{2}}{4S}$

This lemma is quite classical.

[b] Lemma 4 [/b] Let $f$ be a convex function on the interval $[a, b]$ (i.e. $\frac{f(x)+f(y)}{2}\geq f(\frac{x+y}{2})$ for every $x, y$ on the interval $[a, b]$). Then

$f(x)+f(y)+f(z)+3f(\frac{x+y+z}{3}) \geq 2(f(\frac{x+y}{2})+f(\frac{y+z}{2})+f(\frac{z+x}{2}))$

From this lemma, the following corollary immediately follows

[b] Corollary [/b]

If $ABC$ is an acute triangle, then

$cotgA+cotgB+cotgC+\sqrt{3}\geq 2(tg{\frac{A}{2}}+tg{\frac{B}{2}}+tg{\frac{C}{2}})$

[b] Lemma 5 [/b] If $ABC$ is a triangle, then for every real numbers $x, y, z$ we have

$(tg{\frac{A}{2}}+tg{\frac{B}{2}}+tg{\frac{C}{2}})(yzsinA+zxsinB+xysinC) \leq \frac{(x+y+z)^{2}}{2}$

[hide="Hint"] This inequality is equivalent to the well-known inequality

$yzsin^{2}M+zxsin^{2}N+xysin^{2}P \leq \frac{(x+y+z)^{2}}{4}$

here $MNP$ is a triangle such that $sin^{2}M = \frac{1}{2}(tg{\frac{A}{2}}+tg{\frac{B}{2}}+tg{\frac{C}{2}})sinA, ...$

[/hide]

Those lemmas are enough for solving this problem  :lol:.

From lemma 2, it follows that we need to show $\sqrt{\frac{a^{2}+b^{2}+c^{2}+4\sqrt{3}{S}}{2}}(bc+ca+ab) \geq 3\sqrt{3}abc$.

or equivalently,

$(cotgA+cotgB+cotgC+\sqrt{3})(sinBsinC+sinCsinA+sinAsinB)^{2}\geq 27sinAsinBsinC$.

Applying the corollary, it follows

$LHS \geq 2(tg{\frac{A}{2}}+tg{\frac{B}{2}}+tg{\frac{C}{2}})((sinBsinC+sinCsinA+sinAsinB)^{2}$

Here is the hardest part. Write

$(sinBsinC+sinCsinA+sinAsinB)^{2}$

$= sin^{2}Asin^{2}Bsin^{2}C(\frac{1}{sinA}+\frac{1}{sinB}+\frac{1}{sinC})^{2}$

$= 2sin^{2}Asin^{2}Bsin^{2}C(\sum \frac{sinA}{sinBsinC})(tg{\frac{A}{2}}+tg{\frac{B}{2}}+tg{\frac{C}{2}})$

$= 4sin^{2}Asin^{2}Bsin^{2}C(tg{\frac{A}{2}}+tg{\frac{B}{2}}+tg{\frac{C}{2}})(cotgA+cotgB+cotgC)$

Thus

$LHS \geq 8sin^{2}Asin^{2}Bsin^{2}C(tg{\frac{A}{2}}+tg{\frac{B}{2}}+tg{\frac{C}{2}})^{2}(cotgA+cotgB+cotgC)$.

It is now sufficient to show that $8sinAsinBsinC(tg{\frac{A}{2}}+tg{\frac{B}{2}}+tg{\frac{C}{2}})^{2}(cotgA+cotgB+cotgC) \geq 27$

Now, substitute from lemma 3 and 4, the inequality is equivalent to $2(p^{2}-r^{2}-4Rr)(4R+r)^{2}\geq 27R^{2}p^{2}$. From lemma 1 and well-known inequality $R \geq 2r$, this inequality becomes obvious. And the problem is completely proved.  :)

Hopefully, everything is correct.

I have just checked again my computation and found out that I made some mistakes in the last inequality. In fact, that inequality is not right. One can show a counter- example $A = 80^{0}, B = C= 50^{0}$. I will think about it again.

Right now, I can prove that the initial inequality is true when $p \geq \sqrt{3}(R+r)$. But I don't know about other cases.

To gemath, can you post the sources? Or did you create it by yourself?

Thank treegoner for your trying to solve my problem! Really my problem begin when I try to solve this problem
http://www.mathlinks.ro/Forum/viewtopic.php?t=17228
If in this inequality you let FA=x,FB=y,FC=z then we have
$\sum\frac{x+y+z}{\sqrt{y^{2}+yz+z^{2}}}\ge 3\sqrt{3}\forall x,y,z>0$ It is the thing that I hope It true but as your idea It is false,it bring about the problem given are equaliteral triangle $ABC$ with any $P$ in side find min $\frac{1}{PA}+\frac{1}{PB}+\frac{1}{PC}$ won't give a solution $P\equiv O$(center of $ABC$).

In your solution I see a very interesting Inequality:
[quote="treegoner"]
Let $f$ be a convex function on the interval $[a, b]$ (i.e. $\frac{f(x)+f(y)}{2}\geq f(\frac{x+y}{2})$ for every $x, y$ on the interval $[a, b]$) Then
$f(x)+f(y)+f(z)+3f(\frac{x+y+z}{3}) \geq 2(f(\frac{x+y}{2})+f(\frac{y+z}{2})+f(\frac{z+x}{2}))$
[/quote]

Can you give me a poof of this nice inequality
and I think about extension of it:
Let $f$ be a convex function on the interval $[a, b]$ for every $x, y$ (on the interval $[a, b]$) and $\alpha,\beta,\gamma>0$ we will have

$\alpha f(x)+\beta f(y)+\gamma f(z)+(\alpha+\beta+\gamma)f(\frac{\alpha x+\beta y+\gamma z}{\alpha+\beta+\gamma}) \geq (\alpha+\beta)f(\frac{\alpha x+\beta y}{\alpha+\beta})+(\beta+\gamma)f(\frac{\beta y+\gamma z}{\beta+\gamma})+(\gamma+\alpha)f(\frac{\gamma z+\alpha x}{\gamma+\alpha})$
I hope it true!
Thank you again treegoner!  :)  :)

To gemath,

I didn't say that this inequality is wrong with acute triangle. What I said in my previous message is my solution isn't right.

Your inequality is equivalent to the following which looks easier but is still difficult to prove:

Given $F$ the Fermat point of an acute triangle $ABC$. Then

$\frac{1}{a.FA}+\frac{1}{b.FB}+\frac{1}{c.FC}\geq \frac{9}{4S}$

Too difficult!!!!  :blush: (I still can't find a counter-example for your inequality in the acute case. But, I really hope that this inequality is right, or else, I would be mad that I had spent so much time trying to solve it.)

To Treegoner: really I have checked the problem find min of $\frac{1}{PA}+\frac{1}{PB}+\frac{1}{PC}$ in equaliteral triangle, the result $P\equiv O$ is false, really reasult is $P\equiv$ three midpoint of three side, this brings about my inequality is false  :blush: , but the inequality you write $\frac{1}{aFA}+\frac{1}{bFB}+\frac{1}{cFC}\ge \frac{9}{4S}$ seem true (maybe with any triangle when $F$ outside I only try by computer)
Thank you treegoner !!!

There is a weaker inequality that I believe it is true but it is quite ugly:
$(FA+FB+FC)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+2\frac{\sqrt{(FB-FC)^{2}+(FC-FA)^{2}+(FA-FB)^{2}}}{FA+FB+FB}\ge 3\sqrt{3}$
It begin from quite nice problem:
Given equaliteral $ABC$ triangle center $O$find min value
$\frac{1}{PA}+\frac{1}{PB}+\frac{1}{PC}+\frac{3PO}{R^{2}}$ with any $P$ inside triangle result is $P\equiv O$  :)  :)


When a line cuts the area of a polygon in half, does it always go through the center of that polygon?  If not, when will it have to go through the center of a polygon (which type)?

I think that for regular polygons it is intuitive that the line must cross the center of the polygon.

but doesnt it work for all quadrilaterals?

The question is too vague for non-regular polygons. What does "cutting down the middle" mean for an a-symmetrical concave polygon?

Define "center of a polygon."

Also, there are lots of lines that cut a polygon in half area-wise.

[quote="cincodemayo5590"]I think he was wondering if a [b]straight[/b] line had to go through the [b]centroid[/b] of the polygon.[/quote]

yea thanks for rephrasing the question.

So is this always true?  I think its always true for a quadrilater but Im not sure about anything else.

It is not true...just think about taking a square, then adding like 100 verticies to one side [a degenerate 104-gon] this shifts the centroid, but not the line that bisects the area.

Here is a nicer problem:

For any polygon, $P$, does there exist a line through the centroid of $P$ that bisects the area of $P$?

[quote="Altheman"]It is not true...just think about taking a square, then adding like 100 verticies to one side [a degenerate 104-gon] this shifts the centroid, but not the line that bisects the area.[/quote]

The centroid of the convex set enclosed by the square, not the centroid of its vertices.

http://en.wikipedia.org/wiki/Centroid

Okay I was wrong to think about a polygon as a set of verticies. I suppose that it follows directly by definition then...


Suppose $ \psi(abc) \equal{} ab^c$. Compute $ \psi(642) \plus{} \psi(933) \plus{} \psi(824)$.

We have $ \psi(abc)\equal{}ab^c$.  Then $ \psi(642)\equal{}6 \times 4^2\equal{}6 \times 16\equal{}96$, $ \psi(933)\equal{}9 \times 3^3\equal{}9 \times 27\equal{}243$, and $ \psi(824)\equal{}8 \times 2^4\equal{}8 \times 16\equal{}128$.  Adding these yields $ 96\plus{}243\plus{}128\equal{}\boxed{467}$.


So what's the most complex calculator program you have ever written during a math competition to solve a problem?  :mrgreen:

i forgot, but i did it for a math circle competition once.

On a Mathleague contest probability question concerning flipping 3 coins or something, I created a program to do the Monte Carlo method for it.  (It was a really easy problem, I just wanted to try it out for kicks.)  After running it all school day, I think I had 10,000 trials and was accurate to 1 decimal place.  Then I went home and ran it on Mathematica for a few million trials and still got pretty awful accuracy.

the greatest calculator program is the chess program for the ti-89. i haven't figured out how to defeat it when it is at level 3 (the hardest because if it is higher the time takes too long). there are puzzle packages with 1000 puzzles. you can also watch the calculator face each other. or play 2 on 2.

during a team round i made a pretty awesome program involving the fibbonaci sequence and its respective sums for any term [i]n[/i].

the most difficult though was a closed form base converting program that works for all fractions, repeating decimals, whole numbers, et cetera. BASIC is fun, lol.

Coolest program I ever made, although not for a math competition, was a fractal reiterater. You simply inserted a complex number in and bam, out came a fractal. Although the resolution was crappy and it took forever to graph :)

I had a student who wrote a program to calculate product (i = 2 to 100) (1 + 1/i).  Fortunately, he also saw the trick right after.


geometry
rectangle
vector
trigonometry
combinatorics proposed
combinatorics

On a $49\times 69$ rectangle formed by a grid of lattice squares, all $50\cdot 70$ lattice points are colored blue. Two persons play the following game: In each step, a player colors two blue points red, and draws a segment between these two points. (Different segments can intersect in their interior.) Segments are drawn this way until all formerly blue points are colored red. At this moment, the first player directs all segments drawn - i. e., he takes every segment AB, and replaces it either by the vector $\overrightarrow{AB}$, or by the vector $\overrightarrow{BA}$. If the first player succeeds to direct all the segments drawn in such a way that the sum of the resulting vectors is $\overrightarrow{0}$, then he wins; else, the second player wins.
Which player has a winning strategy?

Got little time to explain, so I'll be brief

look at the picture. Consider those edges + new segments, then we get a graph that has even number of edges coming out from every vertex, so we can find a circuit that go through every edge exectly once. We can easily cancel these out.

Remaining ones are all horizontal, same length, and there are even number of them, so we are done.

lightrhee, your post is a mess, your hints mean nothing and the picture looks more like abstract art to me.

Could you please detail your proof?
Thank you

Or can anybody explain lightrhee's proof? (I've tried hard to understand, but...)

Well, I think lightrhee misunderstood the problem totally.

Does anybody know the solution?

the first wins.
lemma if two segment have an arc between  0 to 60 degree we can sum or minus them such that the result segment have length under the max length of two other segment .this lemma is very easy and proved by cosine theorem.
the prove have four steps:
1-at first the first one try to choose the adjacent cells horizontally and vertically at least 5*70 of any of them (prove of this step is a little long and hard two write)
2-according to  lemma after the play end first one do this Strategy choose two segment (except segment in step 1)which the arc between them is between 0 to 60 degree and erase them and draw the segment result in the lemma after doing this at last remind at most 5 segment the and sum of them become at most 5*70 in vertical and  5*70 in horizontal
3-sum the segment how introduce in step 1  in the way that the sum become 0
4-prove of that the strategy work depend on the sum of two step 1 and 2 have similar binary(even or odd) that proof by coloring the cells by row and another one coloring them by column in two color white and black.

[hide="I hope this works"]I claim that the first player wins. Tile the lattice points with horizontal 1 by 2 dominoes. On his first move, the first player connects two lattice points within the same domino. Then after each of the second player's moves, the first player does the following:

[b]Case 1:[/b] The second player connects lattice points from two different dominoes.
The the first player connects the corresponding squares in the same two dominoes.

[b]Case 2:[/b] The second player connects lattice points within a domino.
The first player then chooses any domino and then connects the two lattice points within it.

This strategy always works because after the first player's move, there is no domino with exactly one connected lattice point.

There are 1750 dominoes in total. There are two categories of dominoes: those with lattice points connected within themselves, and those with lattice points connected to some other domino. Since there is an even total of dominoes and an even number of paired dominoes, there must be an even number of dominoes with lattice points connected within themselves. Furthermore, the first move the game produces a self-connected domino, so there are at least 2 self-connected dominoes at the end.

Consider all the domino pairs. In general, for each pair, we can either produce the zero vector or a horizontal vector of length two pointing either direction. Thus, player 1 can direct all the paired dominoes so that the total group of paired dominoes produces either exactly one horizontal vector of length 2, or a resultant sum of 0. Since there is an even number of self-connected dominoes, player 1 can choose to produce either the zero vector with them or produce exactly one horizontal vector of length 2, so depending on the vector he creates with the paired dominoes, player 1 can always cancel it out. [/hide]

[hide = Solution]

We claim that player $1$ has a winning strategy. His strategy is as follows. At the beginning of the game, he will create $840$ pairs of horizontally adjacent lattice points and $910$ pairs of vertically adjacent lattice points. To see that he can do this, simply partition the top $24 \times 70$ into horizontal pairs and the bottom $26 \times 70$ into vertical pairs. Now, in the first part of the game, observe that player $2$ can "mess up" at most $2$ pairs at a time, by picking two lattice points which aren't paired with each other. With this in mind, it's easy to see that $A$ can always ensure that he ends up drawing at least $200$ horizontal pairs and $200$ vertical pairs of lattice points. After the first stage of the game, let the vectors which have length $>1$ be $v_1, v_2, \cdots, v_k$. We will prove the following key lemma.

$\mathbf{Lemma.}$ $|v_1 \pm v_2 \cdots \pm v_{k-1} \pm v_k| < 200$ for some choice of signs.

$\mathbf{Proof.}$ Simply observe that whenever we have three vectors $a, b, c$, one of $a+b, b+c, c+a, a-b, b-c, c-a$ has a magnitude of at most $\max(|a|, |b|, |c|).$ To see this, notice that two of them must have either an angle of $\le 60$ between them or $\ge 120$, and so now simply adding them if $\le 60$ and subtracting them if $\ge 120$ finishes. Now, repeatedly apply the operation of selecting three random $v_i$'s and then merging two of them such that the magnitude of their sum is minimal. By the above observation, we end up with $2$ vectors, each with a magnitude of $\le \sqrt{70^2 + 50^2} < 100.$ Hence, adding these two proves the lemma.

$\blacksquare$

Let $v_1 \pm v_2  \pm \cdots \pm v_k = \left < a, b \right >$ with $-200 < a, b < 200$, where a choice of signs exists by the lemma. Now, simply direct the horizontal and vertical pairs accordingly, by parity and because there are $\ge 200$ of each, there must exist a way.

$\square$

[/hide]


A problem that solves [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105991]this one[/url].
Suppose $X_{1},X_{2},\dots,X_{t}\in\mathbb R^{n}$ that for each $i\neq j\neq k\neq i$ we know that $\angle X_{i}X_{j}X_{k}\leq 90^\circ$. Prove that $t\leq2^{n}$

I know the solution,but it is too long.The idea is that  any ensemble with no angle greater than 90 has this property that :

  For any two points $s_{i},s_{j}$ consider two (parallel)plane which are perpendicular to segment $s_{i}$ $s_{j}$at $s_{i}$ and $s_{j}$.Then all the other points are between these two planes.
  Let $p$:=$conv${$s_{1},s_{2},...,s_{n}$} then make it symmetric to origin with minkowski transform.Among this transform it's property is reversed.
  Exact proof is available upon request :)

Also You might find it in the favorite book of erdos(Some aspects of that book ) \[\text{Proofs from the BOOK }\] By Martin Aigner and Gunter M.Zeigler


analytic geometry
algebra unsolved
algebra

In a planar coordinate system we got four pieces on positions with coordinates. You can make a move according to the following rule: You can move a piece to a new position if there is one of the other pieces in the middle of the old and new position. Initially the four pieces have positions $ \{(0,0),(0,1),(1,0),(1,1)\}$. Given a finite number of moves can you yield the configuration $ \{(0,0), (1,1), (3,0), (2, \minus{} 1)\}$ ?

solution
[hide]
if we can get from the begining to the end we can do the opposite too
suppose so. notice that for all 4 ending points we have $ 3|x_i\minus{}y_i$.
if we move point $ (a,b)$ over $ (x,y)$ we get the point $ (2x\minus{}a,2y\minus{}b)$ for which we again have $ 3|x\minus{}y$
since the begining points aren't all like that the statement is false.[/hide]


algebra polynomial AMC AIME parameterization

There exists a polynomial \[P(x) = x^{3}+(4k+9) x^{2}+(-438-17k) x+(13k-1118) \] for some integer $k$ such that all of its roots are integers, and no other polynomials of this type have a root that is larger than one of the roots of $P(x)$.  Find the remainder when this largest root is divided by $1000$.

it would be nice if x=1 was a root

but it is not

for some reason i feel that there is a wrong coefficient

i might be wrong...

[quote="Altheman"]for some reason i feel that there is a wrong coefficient

i might be wrong...[/quote]

Nope. I've checked.  :P

[quote="Altheman"][hide="Rawr"]well...if you find k, then you partial frac it, that limits your values of x significantly to try for k, then you can be like blerg casework...or you can be lazy and just assume that the problem works out relatively nicely[/hide][/quote]

But there's a much better solution than that.   :P

(And hide your answer. Rawr.  :D )

I don't know if this helps but I noticed that the only time $P$ has a double root is when $k=\frac{18138}{2884}=\frac{3*3023}{14*103}$ and for this $k$, the third root is an integer !!! $29$

I have no idea how to approach this. Can someone post a solution/hint?

sam, that is the nature of aime-like problems

but....umm...we can be like woa we need real roots...and if we consider k as a parameter, then we have like we are more or less shifting the cubic up as k increases...we want to shift the cubic down as much as possible, because the end behavior of the cubic is to positive infinity, thus increasing our largest root...but of course, we are limited in this movement; we need 3 real roots, this obviously happens when the roots of the left are double roots; note the k=integer thing is not really important, rather, it just so happens the answer was made to work out as an integer; anyway, we have a double root, now the problem becomes mere computation...something that i don't feel like doing...

[quote="shyong"][hide="Solution"]let the roots be $a,b,c$ then we have

$a+b+c=-(4k+9)$
$ab+ac+bc=-438-17k$
$abc=1118-13k$

Hence $a+b+c-ab-ac-bc+abc=1547$

so $(a-1)(b-1)(c-1)=1546=2\cdot 773$[/hide][/quote]

That's it.  :D

Completely counter to what altheman was doing in his last post... very nice and tricky impossible problem

[quote="Altheman"]i would use the word: contrived ...lol...the fundamental observation was: 4+14=17...[/quote]

I hope nobody successfully made that fundamental observation.

you know waht i mean  :lol:

that would be pretty awesome though...the secret of the problem is to realize 0=1