Ισόπλευρο τρίγωνο περιστρεφεται περι του κέντρου του περιγεγραμμένου κυκλου κατα γωνία $ a$ του $ (0,\pi)$.
Nα βρεθεί το ελάχιστο του εμβαδου του κοινου μερους των τριγωνων  κατα την περιστροφή οταν η ακτίνα του περιγεγραμμένου κύκλου ειναι $ \sqrt[4]{3}$

[quote="xp"]Ισόπλευρο τρίγωνο περιστρεφεται περι του κέντρου του περιγεγραμμένου κυκλου κατα γωνία $ a$ του $ (0,\pi)$.
Nα βρεθεί το ελάχιστο του εμβαδου του κοινου μερους των τριγωνων  κατα την περιστροφή οταν η ακτίνα του περιγεγραμμένου κύκλου ειναι $ \sqrt [4]{3}$[/quote]

otan mporesete valte tin lusi mias kai dn apantise kaneis

Γιωργο την είχα ξεχασει.....
Δινω μια λυση
Εστω οτι η μια πλευρά του ισοπλεύρου τριγωνου ABC μετα την περιστροφη τεμνει τις αλλες δυο στα σημεια K και L .Αν Ο το κεντρο του κυκλου τότε το ζητούμενο εμβαδό λογω συμμετριας είναι $ E\equal{}6(KOL)\equal{}\frac{3\sqrt3(OK)(OL)}2$ και αντ/ντας τα (OK) kai (OL) απ το νομο ημιτονων στα τριγωνα ΟΚΑ και ΟLΑ παίρνουμε.... $ E\equal{}\frac{3\sqrt3r^2(1\plus{}a^2)}{2(1\plus{}a\sqrt3)}\equal{}\frac{9(1\plus{}a^2)}{2(1\plus{}a\sqrt3)}\geq{(9/2).(2/3)}\equal{}3$ οπου $ a\equal{}tan(t/2)$ και $ t$ η γωνία στρεψης .Aρα $ E_{min}\equal{}3$ και το ισον ισχύει οταν $ t\equal{}60^o$


In how many ways can I choose 3 captains from a team of 11 people?

Just use combinations.

$ C(11,3)\equal{}\frac{11!}{3!8!}\equal{}\frac{11\times 10\times 9}{6}\equal{}11\times 15\equal{}\boxed{165}$ ways.

Further explanation:

Out of the 11 people, you can choose any one of the 11 to be one captain, then any of the leftover 10 to be the second captain, and finally any of the remaining 9 after the first two are picked out for the spot of a third captain. Thus, we have $ 11\times10\times9$. However, the order doesn't matter, as the 3 captains are not differentiable by rank, so we have to consider that by dividing out all $ 3!\equal{}3\times2\times1$ ways to order the 3 captains, thus arriving at the formula $ \frac{n!}{r!(n\minus{}r)!}\equal{}\frac{11!}{8!3!}\equal{}\boxed{165}$.


A farmer has a large lawn with grass that grows at a constant rate. Given that all of his cows eat at the same rate, and that 10 cows can finish the entire lawn of grass in 20 days while 15 cows finish the lawn in 10 days, how long does it take for 25 cows to finish an entire lawn of grass.

(finish = height of grass becomes 0... don't get technical with me here  :wink: )

Let the amount that 1 cow eats in 1 day be r, the amount the grass grows in one day be g, and the volume of the lawn be v.

From the first equation, 200r=v+20g. From the second, 150r=v+10g. We are looking for x so that 25xr=v+xg. we can tell that 5r=g and that 100r=v. So 25xr=100r+5xr, or 20x=100, so x=[b]5[/b].


Mark gave each neighborhood child 6 cookies, and he had 7 cookies remaining. He was going to give another cookie to each child, but he was 1 cookie short. How many cookies had Mark baked?

Each cookie he gives out to each neighbor, he loses 8 cookies, so there are 8 neighbors. So, he baked 8*6+7=55 cookies.


So i know everyone says to study for the USABO, use Campbell 7th edition, and I totally agree Campbell pwns. I was wondering though, there is now an 8th edition, so is that edition not as good as the 7th edition and people avoid it or do people just not know about the 8th edition. Also, if I'm trying to make it to finals, is Campbells all I'll need, or what other books could be useful?

I checked google and I couldn't find a reference to an 8th edition.  Besides, editions don't change much so stick with the 7th.  When I took bio, we used a different textbook (with Campbell as a supplement).  Unfortunately, I can't recall the authors but I can tell you the cover featured a closeup of a duck.  A very good book.

so would having a 6th or 5th edition of campbells be as good as the 7th ed or good enough to make semi-finalist in the USABO?

Yes, those should suffice.  But you have to actually study the material.  You can't just read through it once and say "Okay, I'm making USABO."  You have to actually know what you're doing.

um...would a 2nd edition be sufficient? Because I own the 2nd edition (through my dad) and it's hard to find any edition beyond the 3rd edition in university libraries. Or would 3rd edition be much better than 2nd.

First year biology hasn't changed too much over the years.  That book is probably fine.

i have campbell bioogy 8th edition, but i dont know contents of 7th edition  so i cannot compare them. but in contrast to 6th edition, there is much of new details in 8th edition(e.g.new biotechnological methods,etc.). i found it very nice .

Second might be a bit of a problem.

They gave us 8th at USABO finals, but I studied for Finals using 6th.

They're more or less the same.  If a book has had 8 editions, chances are, the 6-8th editions don't vary by too much.  Once you get below 6, however, then they will differ.

but the updates still make a big difference (especially recent one)
and morever it adds on the fun :oops: of book reading :)

so 8th edition just has more information not revised information cause i also have 6th edition and i dont wanna be studying the wrong stuff since science theories are constantly updated i fu no wut i mean.

If you have the 6th edition, you're more or less fine.  Don't go out and spend $ \$$150 on a new textbook just because you think it will help you make an Olympiad.

i paid 6 dollars for campbell  :rotfl:  :P  :wink:
my library was having a book sale ultra cheap just saw the book and said why not

Does Campbell cover everything on the USABO? Because if the majority of people are using it, shouldnt the cutoff score be significantly higher (along with the max score)?

I had no idea of most of the terminology on the test (my bio book sucks) so i'm not sure if most of the test is reasoning or memorization.

ofc no

the most important thing is to apply ur mind

and its not easy to get cent per cent rite in Biology (unlik physics and chem) :)

Hey, does anybody know where I can buy the 8th edition of Campbell and Reece Biology cheap? Anybody selling :?:

And by the way, does anybody know the approximate difficulty of the Olympiad if you read this textbook? Is it relatively a more involved test or more of a memorization and terminology test?

 :D

Isn't that the newest edition? I doubt you can get it cheap anywhere, just get the 7th or the 6th.

its not that bad online

I got the new ed of the book.
I like it a lot compared to the 4th ed I used to have

Also, check at your school, at our school, the science department let me borrow a 7th edition for the summer. If you check online, used copies are usually around 25 bucks (at least on amazon). Sometimes, I don't like to shop online because the product descriptions can be skewed.

eli140's suggestion is a good one: ask your science department if they will let you borrow a bio textbook.  In my senior year, I was allowed to borrow two physics textbooks in preparation for the science olympiad.  I was during the year, but ask and promise to return it (or give them collateral   :)  ).


function
limit
inequalities
algebra solved
algebra

Let [tex]f : (0,+\infty) \rightarrow R[/tex] be a function such that:
1) [tex]x<y \Rightarrow f(x)<f(y)[/tex]
2) [tex]f(\frac{2xy}{x+y}) \ge \frac{f(x)+f(y)}{2}[/tex] for all [tex]x,y \in (o,+\infty)[/tex]
Show that there is a [tex]x_0 \in (0,+\infty)[/tex] such that [tex]f(x_0) <0[/tex]

There’s something really strange about this problem :?

Since $f$ is strictly increasing, there exists $\displaystyle a=\lim_{x\to 0}f(x)=\inf_{x\in(0,\infty)}f(x)$. Now make $y\to 0$ in the second relation (the inequality). The left part goes to $a$, while the second part goes to $\displaystyle \frac{f(x)+a}2>a$, so we get a contradiction (we can speak about the limits of $f$ because it's strictly increasing, and all monotonic functions have lateral limits, and here we're dealing with the right limit in $0$).

What am I doing wrong?

Why would you be wrong?
It seems ok, even if you would rather study two cases according to $a$ is a real (what you done) and $a = - \infty$ (I agree that in that case the result is obvious).

Pierre.

The answer to that issue is that $\displaystyle \lim_{x \to 0+} f(x) = -\infty$.

Let $g(x)=f(\frac1x)$. We have $\frac12(g(\frac1x)+g(\frac1y)) \le g(\frac12(\frac1x+\frac1y))$. Rewriting for clarity, $g(u)+g(u+2a) \le 2g(u+a)$ for $a>0$. Also, we know that $g$ is decreasing because $f$ is increasing. The key inequality can be rewritten $g(x)-g(x+a) \le g(x+a)-g(x+2a)$, so $g(x)-g(x+na) \ge n(g(x)-g(x+a))$. Since $g(1)-g(2)>0$, this obviously forces $\displaystyle \lim_{x\to \infty} g(x) = -\infty$. In particular, $g(n+1)<0$ if $n > \frac{g(1)-g(2)}{g(1)}$ and $f(\frac1{n+1})<0$ if $n > \frac{f(1)-f(\frac12)}{f(1)}$.

My point is that since we have obtained a contradiction, such a function doesn't exist, so any assertion about it is true, including what we have to prove! :)

I don't see where is the proof that such a function does not exist. It is true if $a$ is a real, but what if $a=-\infty$?

Pierre.

Oh, right... :blush: I forgot about that. Well, then I guess this solves the problem then.

So, you found a solution without knowing it  :D

Pierre.


What is the sum of the first 200 terms of an infinite geometric series that has 5th term 1 and 25th term 3486784401

[hide]

We can find the common ratio $ r$ by taking $ (3486784401)^{\frac{1}{25 \minus{} 5}} \equal{} 3$. We can relate the first term $ a$ to the fifth term by $ ar^{5 \minus{} 1} \equal{} 1 \implies a \equal{} \frac {1}{3^4} \equal{} \frac {1}{81}$.

The sum of the first 200 terms of the sequence is thus:

$ S \equal{} a\left(\frac {r^{200} \minus{} 1}{r \minus{} 1}\right) \equal{} \left(\frac {1}{81}\right)\left(\frac {3^{200} \minus{} 1}{3 \minus{} 1}\right)$
$ S \equal{} \boxed{\frac {3^{200} \minus{} 1}{162}}$
[/hide]

jvenezuela716, I think you mean $ \sqrt[25\minus{}5]{3486784401}$.


geometry inradius inequalities trigonometry

[color=darkblue]Given is a triangle $ABC\ .$ Prove that :

$\{\begin{array}{cccc}1.\blacktriangleright & A=90^{\circ}& \Longleftrightarrow & p^{2}+r^{2}=a^{2}+bc\ .\\\\ 2.\blacktriangleright & A=90^{\circ}& \Longrightarrow & p^{2}+r^{2}\ge 6S\ .\end{array}$

where $2p=a+b+c$ , $r$- inradius and $S$- area of the triangle.[/color]

Assuming 1 is true, [hide="#2"] we want to prove
$a^{2}+bc \geq  6S$
Because ABC is right, we have
$a^{2}= b^{2}+c^{2}$ and
$bc = 2S$
So our inequality to prove simplifies to
$b^{2}+c^{2}\geq  4S$
By AM-GM, $b^{2}+c^{2}\geq  2bc = 4S$
So $p^{2}+r^{2}\geq  6S$
QED[/hide]

[quote="boostage"]Assuming 1 is true, [hide="#2"] we want to prove
$a^{2}+bc \geq 6S$
Because ABC is right, we have
$a^{2}= b^{2}+c^{2}$ and
$bc = 2S$
So our inequality to prove simplifies to
$b^{2}+c^{2}\geq 4S$
[/hide][/quote]
[hide="continue from here.."]

$b^{2}+c^{2}\geq 4S$ is equivalent to

$b^{2}+c^{2}\geq 2bc$, which  is equivalent to

$(b-c)^{2}\geq 0$, which is true
----------------------------------------------------------[/hide]

$1.\blacktriangleright$  the straight $(\Rightarrow )$ case:
$\boxed{A=90^{\circ}\Rightarrow  \& p^{2}+r^{2}=a^{2}+bc}$

[hide]
$A=90^{\circ}\Rightarrow  r=p-a \Rightarrow  p-r=a$

$p^{2}+r^{2}-2pr = a^{2}\Rightarrow $

$p^{2}+r^{2}= a^{2}+2pr \Rightarrow $

$p^{2}+r^{2}= a^{2}+bc$[/hide]

[b]Remarks.[/b] $S=pr\ ,\ S^{2}=p(p-a)(p-b)(p-c)\Longrightarrow \boxed{\ (p-a)(p-b)(p-c)=pr^{2}\ }\ \ ;\ \ \boxed{\ \cos\frac{A}{2}=\sqrt{\frac{p(p-a)}{bc}}\ }\ .$

Therefore, $p^{2}+r^{2}=a^{2}+bc$ $\Longleftrightarrow$ $p^{3}+(p-a)(p-b)(p-c)=p(a^{2}+bc)$ $\Longleftrightarrow$ $p(ab+bc+ca)-abc=p(a^{2}+bc)$ $\Longleftrightarrow$

$p(b+c-a)=bc$ $\Longleftrightarrow$ $[(b+c)+a][(b+c)-a]=2bc$ $\Longleftrightarrow$ $(b+c)^{2}=a^{2}+2bc$ $\Longleftrightarrow$ $b^{2}+c^{2}=a^{2}$ $\Longleftrightarrow$ $A=90^{\circ}\ .$ [b][u]Otherwise[/u][/b] :

 $p(b+c-a)=bc$ $\Longleftrightarrow$ $\frac{p(p-a)}{bc}=\frac{1}{2}$ $\Longleftrightarrow$ $\cos^{2}\frac{A}{2}=\frac{1}{2}$ $\Longleftrightarrow$ $2\cos^{2}\frac{A}{2}=1$ $\Longleftrightarrow$ $1+\cos A=1$ $\Longleftrightarrow$ $\cos A=0$ $\Longleftrightarrow$ $A=90^{\circ}\ .$


So you can have them and cheat because you have yet to take the test?
Just kidding.

But others haven't taken it yet so we can't give answers till Thursday at noon...

[quote="xixi"]could someone please post the answers from the AMC12, given today (Feb. 1st, 2005)? (especially the first 20 problems?) Thanks![/quote]

Discussion is not allowed until Thursday. I'll post the answers then. There'll also be a math jam to discuss solutions to the questions.


Let $P$ be a point inside a triangle $ABC$ and $D,E,F$ be the perpendiculars from $P$ to the sides $AB,BC,CA$ respectively.
Prove that:-
$PA + PB + PC \geq 2(PD + PE + PF)$

Erdos-Mordell inequality.A very famous one,can be find in many books.


I can program stuff that makes you input a variable(s), and it executes a special function(nth Fibonacci number, area of a regular polygon, etc.), but nothing bizarre like a model of a 3*3*3 Rubik's Cube. How about anyone else?

P.S. Does anyone have a Monopoly program for the TI-84? Because I would like it.

I just programmed a program that solves this:

Bob writes the list of numbers from 1 to 99 without breaks, as 12345678910111213…  Aidan writes the list backwards starting with 99, as 99989796959493…  In how many positions are the respective digits of each list the same?

Everybody thought it was 21, but it was really 17. Figures.

I've also programmed not 7, but 8 versions of Risk on my calculator! [size=59]I'm kidding.[/size]

Really, the first Fibonacci with 99 digits is not really a skill, just a fast computer. It's making the program that's the skill.

Sadly, I don't have the skill to enumerate permutations, so I stole it from online articles.

At the beginning of sixth grade I found out how to access program codes hidden by the Mirage OS. Naturally, I didn't understand any of it, and I managed to learn BASIC programming by looking at programs created by others.

I haven't done any serious programming in over 2 years, so my memory is a bit hazy in manipulating certain codes.

I've programmed a clock (analog display that synchronizes to the original calculator clock) and a version of snake that sometimes fails. I'm working on a hangman game right now. And who would want to program 7 version of risk? 1 stable version, or 1 fast version is good enough. Fibonacci is easy. The only trick is to make it display it fast. Using a for loop and 2 variables makes it pretty fast. Same goes for a prime number generator. I made both so that they save into a matrix like an internet cache.

Going from fibonacci to 7 ways of risk is a huge jump.

There is a monopoly game. I used to have it.

I'm pretty good at calculator programming.  I once made a blackjack game on my TI-89 with TI-BASIC, but my memory got corrupted and I lost it. :(

what the heck is calculator programming????? :huh:

You get a calculator at least TI-83 and there is a programming option on it.

@moogra: Could I steal your code for Hangman? Thanks!!! :P

It's still in the process of being made

[quote]I'm working on a hangman game right now.[/quote]

but you can have it when I'm done

Right now all it can do is take in guesses, compare with the original word(s) display guesses, display correct or wrong, but even though the hangman figure shows, I can't really make it die (I've tried pxl-test). So right now it's infinite tries.

How about when you guess a letter, if it's wrong you add 1 to a certain value, and if it gets to a certain point, the guy dies?

Err.. what's Risk

I can program a reasonable amount on the calculator, but in general avoid it since most of the time a proper computer is nearby so I'd rather practice different languages

I made a dictionary for my vocab tests that I update for each vocab test.

Nah, actually I don't even have my own calculator that can program. D'oh.


is there any way to know if the last digit of $n!$ is $n$? $1!,2!,4!$ fulfil it.
edit:lets take the lazst non zero digit!

The last digit is the digit of units? because if it is, for $n \ge 5$ the last digit of $n!$ is obviously $0$

For larger numbers we require a number of the form $n = 10^{k}$, and it's not hard to show that $n!$ will always have more zeroes at the end than $n$.

Did you want the last nonzero digits?

oh you are right!
so lets take the last non zero digit!

[quote="tim1234133"]This doesn't actually answer your question, but the last non-zero digit of $n!$ [b]never becomes periodic[/b]. (This is a question from Engel). I would guess that there's no simple formula for the last digit of $n!$.[/quote]
does the proof is difficult?

From http://mathworld.wolfram.com/Factorial.html:

Let  be the last nonzero digit in , then the first few values are 2, 6, 4, 2, 2, 4, 2, 8, 8, 8, 6, 8, ... (Sloane's A008904). This sequence was studied by Kakutani (1967), who showed that this sequence is "5-automatic," meaning roughly that there exists a finite automaton which, when given the digits of  in base-5, will wind up in a state for which an output mapping specifies .

Further references/links on:
http://www.research.att.com/~njas/sequences/A008904


City A is due north of City B. Find the distance between City A (38 degrees, 21' north latitude) and City B (30 degrees, 20' north latitude). Assume that the radius of the Earth is 3960 miles.

First, find the Earth's circumference.  (2*pi*radius)
The two cities are 8 1/60 (8.01666666) degrees apart. Divide this number by 360 (the total number of degrees in a circle) and multiply by the Earth's circumference to find the distance.


[hide="Solution"]

[b]Part A:[/b]
Arithmetic mean of $ x$ and $ y$ is: $ \dfrac{x + y}{2}$

\begin{eqnarray*} \dfrac{x + y}{2} & = & \dfrac{1 + (\frac {1}{2} - \frac {1}{2}) + (\frac {1}{3} - \frac {1}{3}) + (\frac {1}{4} - \frac {1}{4}) + ... + (\frac {1}{2007} - \frac {1}{2007}) + (\frac {1}{2008} - \frac {1}{2008}) - \frac {1}{2009}}{2} \\
\dfrac{x + y}{2} & = & \dfrac{2008}{2009\cdot 2} \\
\dfrac{x + y}{2} & = & \dfrac{1004}{2009} \end{eqnarray*}

[b]Part B:[/b]

[i]Lemma[/i]: For any two real numbers $ a > b$, we have that $ a > \dfrac{a + b}{2} > b$

[i]Proof[/i]: First we will show that $ a > \dfrac{a + b}{2}$
\begin{eqnarray*} a & > & \dfrac{a + b}{2} \\
2a & > & a + b \\
a & > & b \end{eqnarray*}

Which is true, we prove $ \dfrac{a + b}{2} > b$ in a similar fashion.

Now, all we are missing to show is that $ y > x$ which is true if $ y - x > 0$.

\begin{eqnarray*} y - x & = & 1 - \dfrac{2}{2} + \dfrac{2}{3} - \dfrac{2}{4} + \cdots + \dfrac{2}{2007} - \dfrac{2}{2008} + \dfrac{1}{2009} \\
y - x & = & \dfrac{2}{3} - \dfrac{2}{4} + \cdots + \dfrac{2}{2007} - \dfrac{2}{2008} + \dfrac{1}{2009} \\
y - x & = & 2\left(\dfrac{1}{3} - \dfrac{1}{4} + \cdots + \dfrac{1}{2007} - \dfrac{1}{2008}\right) + \dfrac{1}{2009} \end{eqnarray*}

Note that: $ \dfrac{1}{3} > \dfrac{1}{4}$ and $ \dfrac{1}{5} > \dfrac{1}{6}$ and so on, therefore $ y - x > 0$, which means we can apply the lemma and we are done. :ninja:
[/hide]
Edit: Thank you moldovan


The measures of the interior angles of a particular triangle are in a 5:6:7 ratio. What is the measure, in degrees, of the smallest interior angle?

Let the angles be of measures 5x,6x, and 7x.
Since a triangle has 180 degrees, 5x+6x+7x=18x=180, so x=10
The angle with measure 5x is obviously the smallest, and it has measure $ 5\cdot 10\equal{}\boxed{50}$ degrees.


If a,b ,c are positive reals prove that:

8*(a+b+c)*[ab*sqrt(ab)+bc*sqrt(bc)+ca*sqrt(ca)]>=9(a+b)(b+c)(c+a)

:cool:  It doesn't work ! Take for ex . a=b=c < 1 ..  :cool:

Sorry,I wrote it wrong!!!!!!!!
 the real one is:
4*(a+b+c)*[(a+b)*sqrt(ab)+(b+c)*sqrt(bc)+(c+a)*sqrt(ca)]>=9(a+b)(b+c)(c+a)

Put $a = 1, b = 1, c = 0$ we get $16\leq18$.
Put $a = 4, b = 1, c = 0$ we get $200\geq180$.
So I think we can't say that such inequality exists.

$4(a+b+c)((a+b)\sqrt{ab} + (b+c) \sqrt{bc} + (a+c) \sqrt{ac})) ? 9(a+b)(b+c)(a+c)$


Twenty switches in an office computer network are to be connect so that each switch has a direct connection to exactly three other switches. How many connections will be necessary?

Since each switch is connected to 3 other switches, there are have to be 60 connections to connect all 20 switches. Since each connection is counted twice, the answer is 30.

Because since each connection is a two way street (note that that's different than having 3 connections), because each of the connections connect $2$ switches. So we'll double count at some point and so we need to divide by $2$


[quote="quykhtn-qa1"]Given$ a,b,c > 0$ and$ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {a^2 \plus{} 3b^2c}{b \plus{} c} \plus{} \frac {b^2 \plus{} 3c^2a}{c \plus{} a} \plus{} \frac {c^2 \plus{} 3a^2b}{a \plus{} b} \ge\ 1$
This inequality than stronger the inequality of NGUYEN MANH DUNG in MATHVN.
I hope you will like it  :)[/quote]
Hello, this one is related to a contest problem of Mathvn magazine. So, please dont discuss or post any related result here.
Moderator please lock this topic! Thanks.

@Quykhtn: [b]You should read the regular when joining the forum. Dont help contest cheaters, please![/b]

locking  means  we cant  prove  it?   or  should we not prove  it?

[quote="quykhtn-qa1"][quote="can_hang2007"][quote="quykhtn-qa1"]Given$ a,b,c > 0$ and$ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {a^2 \plus{} 3b^2c}{b \plus{} c} \plus{} \frac {b^2 \plus{} 3c^2a}{c \plus{} a} \plus{} \frac {c^2 \plus{} 3a^2b}{a \plus{} b} \ge\ 1$
This inequality than stronger the inequality of NGUYEN MANH DUNG in MATHVN.
I hope you will like it  :)[/quote]
Hello, this one is related to a contest problem of Mathvn magazine. So, please dont discuss or post any related result here.
Moderator please lock this topic! Thanks.[/quote]
sorry,but my inequality than stronger the inequality in mathvn magazine :([/quote]
It doesn't matter that your inequality is stronger or not. We dont care about it. We just care that it is related to a currently problem of our Magazine. Do you understand?
Regards

[quote="can_hang2007"][/quote][quote="quykhtn-qa1"][quote="can_hang2007"][quote="quykhtn-qa1"]Given$ a,b,c > 0$ and$ a + b + c = 1$.Prove that:
$ \frac {a^2 + 3b^2c}{b + c} + \frac {b^2 + 3c^2a}{c + a} + \frac {c^2 + 3a^2b}{a + b} \ge\ 1$
This inequality than stronger the inequality of NGUYEN MANH DUNG in MATHVN.
I hope you will like it  :)[/quote]
Hello, this one is related to a contest problem of Mathvn magazine. So, please dont discuss or post any related result here.
Moderator please lock this topic! Thanks.[/quote]
sorry,but my inequality than stronger the inequality in mathvn magazine :([/quote][quote="can_hang2007"]
It doesn't matter that your inequality is stronger or not. We dont care about it. We just care that it is related to a currently problem of our Magazine. Do you understand?
Regards[/quote]
I understood,thanks can_hang.
Moderator please lock this topic.
I will post it after Mathvn magazine solved it.

[quote="evarist"][color=purple] Hey,please dont post any problems look like Mathvn Magazine's Problem here,Quy.
 Mod please lock this topic !
[/color][/quote]
Sorry,I will post it after mathvn magazine solved it.
Moderator please lock this topic
@ if you use the inequality in this mathvn magazine,you can't solve it

I proved your inequality, quykhtn-qa1. Thank you!

My inequality is similar to above inequality of quy, and it  was proposed in this issue of Mathvn Magazine.
Let $ a,b,c$ be positive real numbers. Prove that
\[ \sum{\frac{a^2\plus{}b^2c}{b\plus{}c}} \ge \frac{2}{3}
\]
 If you proved it, you can send your solution to us (because my problem is old and I posted it to mathlinks long time ago).
Email: mathvn2008@gmail.com
You can download our magazine here: (Vietnamese version, we are translating it to English now)

[DJVU] (1.35 Mb)

http://www.mediafire.com/file/zxjcgymmmxx/v01n01.djvu

[PDF] (14.8 Mb)

http://www.mediafire.com/file/meitymmiumk/v01n01.pdf

[quote="NguyenDungTN"]My inequality is similar to above inequality of quy, and it  was proposed in this issue of Mathvn Magazine.
Let $ a,b,c$ be positive real numbers. Prove that
\[ \sum{\frac {a^2 \plus{} b^2c}{b \plus{} c}} \ge \frac {2}{3}
\]
If you proved it, you can send your solution to us (because my problem is old and I posted it to mathlinks long time ago).
Email: mathvn2008@gmail.com
You can download our magazine here: (Vietnamese version, we are translating it to English now)

[DJVU] (1.35 Mb)

http://www.mediafire.com/file/zxjcgymmmxx/v01n01.djvu

[PDF] (14.8 Mb)

http://www.mediafire.com/file/meitymmiumk/v01n01.pdf[/quote]
Thanks :)

[quote="nguoivn"][quote="quykhtn-qa1"]Given$ a,b,c > 0$ and$ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {a^2 \plus{} 3b^2c}{b \plus{} c} \plus{} \frac {b^2 \plus{} 3c^2a}{c \plus{} a} \plus{} \frac {c^2 \plus{} 3a^2b}{a \plus{} b} \ge\ 1$
This inequality than stronger the inequality of NGUYEN MANH DUNG in MATHVN.
I hope you will like it  :)[/quote]

Indeed, I don't think it's hard. My proof is only 3 lines  :wink:
[quote="can_hang2007"]Hello, this one is related to a contest problem of Mathvn magazine. So, please dont discuss or post any related result here.
Moderator please lock this topic! Thanks.

@Quykhtn: [b]You should read the regular when joining the forum. Dont help contest cheaters, please![/b][/quote]
@Quykhtn: [hide]Dao nay` anh thay em tien bo. rat nhanh, em co thay the ko? Do chinh la`vi` em ko con` kieu cang nhu trc, anh thay' bay gio` em rat ham hoc hoi?, ton trong. ng` khac nen dc moi ng` quy men' hon. Anh tin rang` trong tuong lai em se~ thanh` cong hon nhieu` neu duy tri` dc phong cach nhu bay gio`. Anh Can? tinh' thang? than', n~ gi` anh y noi thuc ra cung~ muon tot cho em, mong la` em se~ tiep thu de? sua? chua~ n~ loi~ cua? minh`. Anh nghi~ dung` nen so. mac' loi~ ma` chi? nen so. khi minh` ko dam' thua` nhan. loi~ sai cua? minh`  :) [/hide][/quote]
I'm sorry :(
@ nguoivn: [hide] em xin loi, em cam on anh va moi nguoi nhieu lam.
Em se co gang hon nua,mong duoc cac anh chi bao them, BDT em con ga lam .a anh sap viet sach a. Sach ten la gi? bao gio xong a. [/hide] :)

No, you don't necessary to sorry. Sometimes, the another people's proof is shorter than the proof of author or you.
I proved it by Schur ineq which degree 3.
[hide]Em co' the? add nick chat cua? anh la` langtukhongnha1012 va` cua? anh Can? la` babylearmath. Anh nghiu~ chung ta co the? trao doi? nhieu` hon tren YM.  :) [/hide]

[quote=quykhtn-qa1]Given$ a,b,c>0$ and$ a\plus{}b\plus{}c\equal{}1$.Prove that:
$ \frac{a^2\plus{}3b^2c}{b\plus{}c}\plus{}\frac{b^2\plus{}3c^2a}{c\plus{}a}\plus{}\frac{c^2\plus{}3a^2b}{a\plus{}b} \ge\ 1$
This inequality than stronger the inequality of NGUYEN MANH DUNG in MATHVN.
I hope you will like it  :)[/quote]

The following inequality is also true.
Given$ a,b,c>0$ and$ a\plus{}b\plus{}c\equal{}1$.Prove that:
\[\frac{a^2\plus{}3b^2c}{b\plus{}c}\plus{}\frac{b^2\plus{}3c^2a}{c\plus{}a}\plus{}\frac{c^2\plus{}3a^2b}{a\plus{}b} \le\ \frac{a^2+b^2+c^2}{ab+bc+ca}\]

[quote=quykhtn-qa1]Given$ a,b,c>0$ and$ a\plus{}b\plus{}c\equal{}1$.Prove that:
$ \frac{a^2\plus{}3b^2c}{b\plus{}c}\plus{}\frac{b^2\plus{}3c^2a}{c\plus{}a}\plus{}\frac{c^2\plus{}3a^2b}{a\plus{}b} \ge\ 1$
This inequality than stronger the inequality of NGUYEN MANH DUNG in MATHVN.
I hope you will like it  :)[/quote]

The following inequality is stronger:
Given$ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}1$.Prove that:
\[ \frac{a^2\plus{}3b^2c}{b\plus{}c}\plus{}\frac{b^2\plus{}3c^2a}{c\plus{}a}\plus{}\frac{c^2\plus{}3a^2b}{a\plus{}b} \ge\ \frac{3}{2}-\frac{ab+bc+ca}{2(a^2+b^2+c^2)}\]


trigonometry
induction
inequalities proposed
inequalities

Prove that : with n be a possitive integer and [tex] x\in(0,\pi) [/tex] :
   [tex] sinx + \frac{1}{2}sin2x + .. + \frac{1}{n}sinnx > 0 [/tex]

This is called Jackson's inequality. Try induction

I think it is quite difficult inequality.
I will prove it by induction.
For $n=1$ statement is correct.
Consider $n>1$. If some of the sums $\frac{1}{k}\sin kx+\frac{1}{k+1}\sin (k+1)x+...+\frac{1}{n}\sin nx\geq 0$ ($k=2,...,n$) then we are done, since in this case, due to induction proposition, whole sum is
\[S(x)=\left(\sin x+...+\frac{1}{k-1}\sin (k-1)x\right)+\]
\[+\left(\frac{1}{k}\sin kx+...+\frac{1}{n}\sin nx\right)=(>0)+(\geq 0).\]
Now suppose all mentioned sums are negative and whole sum is non-positive. Then sum them up:
\[\sin x+\sin 2x+...+\sin nx<0.\]
Since $S(t)$ is positive in neighbourhood of $\pi$ and $S(x)\leq 0$, we may WLOG assume $S'(x)\geq 0$ (just consider the largest $x$ with condition $S(x)\leq 0$). Thus
\[\cos x+\cos 2x+...+\cos nx\geq 0.\]
Using well known formulas for $\sum \sin kx$ and $\sum \cos kx$ we obtain:
$\cos(x/2)-\cos(n+1/2)x<0$ and $\sin(n+1/2)x-\sin(x/2)\geq 0$.
It immediatelly implies $1=\cos^2(n+1/2)x+\sin^2(n+1/2)x>\cos^2(x/2)+\sin^2(x/2)=1$ -- contradiction.

[b]hello guys :

could anyone tell me what are the answers for following formulas :

$ \sum\limits_{k \equal{} 1}^n {\sin (kx),} \sum\limits_{k \equal{} 1}^n {\cos (kx),}$

and how we find it ?

I need help .

Thanks[/b]


Hi all! :)

Well, let S a set of n people. Each one of them send only one card to any of the (n-1) other people.
(a) In how many different ways can the n letters be sent?
(b) One person P is chosen. Let's denote by $N_{k}$ the number of ways to send the n letters such that P receives k letters. Calculate $N_{k}$.

Good luck!

The choice of sending cards is independent wrt people.  From product rule, it is $(n-1)^n$.

Suppose S is the set of all possible sets of people that send P a card [and remember we are supposing P recieves k cards].  Obviously |S| = (n-1) choose (k).
Then the remaining (n-1-k) people can send to n-1 people, so there are (n-1)^(n-1-k) ways for them to send.

Also, P needs to send (to n-1) choices.  So in total, the product of these, which is (n-1)^(n-k) * (n-1)choose(k).

[quote="Singular"]The choice of sending cards is independent wrt people.  From product rule, it is $(n-1)^n$.[/quote]
Yeah :)

[quote="Singular"]Suppose S is the set of all possible sets of people that send P a card [and remember we are supposing P recieves k cards].  Obviously |S| = (n-1) choose (k).[/quote]
Alright :)

[quote="Singular"]
Then the remaining (n-1-k) people can send to n-1 people, so there are (n-1)^(n-1-k) ways for them to send.[/quote]
Hmm.. I think there's a mistake here, because the (n-1-k) can send to $n-2$ people! :) (yeah, I know, it isn't much different from what you got ;))


Siemens semifinalists(and nothing beyond): how long did you guys work on your projects?
how many revisions for the final paper?  Has anyone gotten mediocre results but had a great final paper and still gotten a semifinalist standing?

thanks

Of course mediocre results and a great final paper can get you a semifinalist!

Spend a while on your paper, and several (intelligent) people read it over for language and style (as well as for logic flow).

A girl from our school had an interesting project topic, but her actual experimental design was somewhat impotent. But she had a good paper, and she got a semifinalist standing. With her permission, I might post it here.


inequalities
induction
inequalities unsolved

Your inequality can be proved with help of induction in the following way:

[b]1.[/b] The statement holds clearly for n=1 in view of the fact that
$ \frac{1}{x_1\plus{}1}<\sqrt{\frac{1}{x_1}}\quad\Leftrightarrow\quad 0<x_1^2\plus{}x_1\plus{}1$.

[b]2.[/b] Suppose that the inequality holds for a fixed positive integer $ n$, i.e. suppose that
$ \frac{1}{1\plus{}y_1}\plus{}\frac{1}{1\plus{}y_1\plus{}y_2}\plus{}...\plus{}\frac{1}{1\plus{}y_1\plus{}y_2\plus{}...\plus{}y_n}< \sqrt{\frac{1}{y_1}\plus{}\frac{1}{y_2}\plus{}...\plus{}\frac{1}{y_n}},$
where $ y_i>0$. Define
$ y_i: \equal{}\left\{
\begin{array}{lll}
x_i&\mathrm{for}&1\le i\le n\minus{}1,\\
 & & \\
\left (\frac{1}{x_n}\plus{}\frac{1}{x_{n\plus{}1}}\right )^{\minus{}1}&\mathrm{for}&i\equal{}n
\end{array}
\right.$
where $ x_i$, $ 1\le i\le n\plus{}1$ be given positive numbers. Then
$ \frac{1}{1\plus{}x_1}\plus{}\frac{1}{1\plus{}x_1\plus{}x_2}\plus{}\cdots \plus{}\frac{1}{1\plus{}x_1\plus{}x_2\plus{}\cdots x_{n\minus{}1}}\plus{}\frac{1}{1\plus{}x_1\plus{}x_2\plus{}\cdots x_{n\minus{}1}\plus{}\left (\frac{1}{x_n}\plus{}\frac{1}{x_{n\plus{}1}}\right )^{\minus{}1}}<\sqrt{\frac{1}{x_1}\plus{}\frac{1}{x_2}\plus{}...\plus{}\frac{1}{x_{n\minus{}1}}\plus{}\frac{1}{x_n}\plus{}\frac{1}{x_{n\plus{}1}}}.$
But we have easy the inequality
$ \frac{1}{1\plus{}x_1\plus{}x_2\plus{}\cdots \plus{}x_{n\minus{}1}\plus{}x_n\plus{}x_{n\plus{}1}}<\frac{1}{1\plus{}x_1\plus{}x_2\plus{}\cdots x_{n\minus{}1}\plus{}\left (\frac{1}{x_n}\plus{}\frac{1}{x_{n\plus{}1}}\right )^{\minus{}1}}.$
Hence, finally,
$ \frac{1}{1\plus{}x_1}\plus{}\frac{1}{1\plus{}x_1\plus{}x_2}\plus{}\cdots \plus{}\frac{1}{1\plus{}x_1\plus{}x_2\plus{}\cdots x_{n\plus{}1}}<\sqrt{\frac{1}{x_1}\plus{}\frac{1}{x_2}\plus{}\cdots \plus{}\frac{1}{x_{n\plus{}1}}}.$

This completes the proof by induction.

[quote="M@ri@n"]Your inequality can be proved with help of induction in the following way:

[b]1.[/b] The statement holds clearly for n=1 in view of the fact that
$ \frac {1}{x_1 \plus{} 1} < \sqrt {\frac {1}{x_1}}\quad\Leftrightarrow\quad 0 < x_1^2 \plus{} x_1 \plus{} 1$.

[b]2.[/b] Suppose that the inequality holds for a fixed positive integer $ n$, i.e. suppose that
$ \frac {1}{1 \plus{} y_1} \plus{} \frac {1}{1 \plus{} y_1 \plus{} y_2} \plus{} ... \plus{} \frac {1}{1 \plus{} y_1 \plus{} y_2 \plus{} ... \plus{} y_n} < \sqrt {\frac {1}{y_1} \plus{} \frac {1}{y_2} \plus{} ... \plus{} \frac {1}{y_n}},$
where $ y_i > 0$. Define
$ y_i: \equal{} \left\{ \begin{array}{lll} x_i & \mathrm{for} & 1\le i\le n \minus{} 1, \\
& & \\
\left (\frac {1}{x_n} \plus{} \frac {1}{x_{n \plus{} 1}}\right )^{ \minus{} 1} & \mathrm{for} & i \equal{} n \end{array} \right.$
where $ x_i$, $ 1\le i\le n \plus{} 1$ be given positive numbers. Then
$ \frac {1}{1 \plus{} x_1} \plus{} \frac {1}{1 \plus{} x_1 \plus{} x_2} \plus{} \cdots \plus{} \frac {1}{1 \plus{} x_1 \plus{} x_2 \plus{} \cdots x_{n \minus{} 1}} \plus{} \frac {1}{1 \plus{} x_1 \plus{} x_2 \plus{} \cdots x_{n \minus{} 1} \plus{} \left (\frac {1}{x_n} \plus{} \frac {1}{x_{n \plus{} 1}}\right )^{ \minus{} 1}} < \sqrt {\frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} ... \plus{} \frac {1}{x_{n \minus{} 1}} \plus{} \frac {1}{x_n} \plus{} \frac {1}{x_{n \plus{} 1}}}.$
But we have easy the inequality
$ \frac {1}{1 \plus{} x_1 \plus{} x_2 \plus{} \cdots \plus{} x_{n \minus{} 1} \plus{} x_n \plus{} x_{n \plus{} 1}} < \frac {1}{1 \plus{} x_1 \plus{} x_2 \plus{} \cdots x_{n \minus{} 1} \plus{} \left (\frac {1}{x_n} \plus{} \frac {1}{x_{n \plus{} 1}}\right )^{ \minus{} 1}}.$
Hence, finally,
$ \frac {1}{1 \plus{} x_1} \plus{} \frac {1}{1 \plus{} x_1 \plus{} x_2} \plus{} \cdots \plus{} \frac {1}{1 \plus{} x_1 \plus{} x_2 \plus{} \cdots x_{n \plus{} 1}} < \sqrt {\frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} \cdots \plus{} \frac {1}{x_{n \plus{} 1}}}.$

This completes the proof by induction.[/quote]
I think your solution is wrong.
Because we must prove :
$ \frac{1}{1\plus{}x_1\plus{}x_2\plus{}...\plus{}x_{n\minus{}1}\plus{}(\frac{1}{x_n}\plus{}\frac{1}{x_{n\plus{}1}})^{\minus{}1}}>\frac{1}{1\plus{}x_1\plus{}x_2\plus{}...\plus{}x_{n\minus{}1}}\plus{}\frac{1}{1\plus{}x_1\plus{}x_2\plus{}...\plus{}x_{n\minus{}1}\plus{}x_n}$
But it is wrong  :(

you post your solution to everybody show

[quote]
I think your solution is wrong.
Because we must prove :
$ \frac{1}{1\plus{}x_1\plus{}x_2\plus{}...\plus{}x_{n\minus{}1}\plus{}(\frac{1}{x_n}\plus{}\frac{1}{x_{n\plus{}1}})^{\minus{}1}}>\frac{1}{1\plus{}x_1\plus{}x_2\plus{}...\plus{}x_{n\minus{}1}}\plus{}\frac{1}{1\plus{}x_1\plus{}x_2\plus{}...\plus{}x_{n\minus{}1}\plus{}x_n}$
But it is wrong.
[/quote]

Sorry! You are right!

[quote="conan_naruto236"]Let $ x_1,x_2,...,x_n > 0$. Prove that :
$ \frac {1}{1 \plus{} x_1} \plus{} \frac {1}{1 \plus{} x_1 \plus{} x_2} \plus{} ... \plus{} \frac {1}{1 \plus{} x_1 \plus{} x_2 \plus{} ... \plus{} x_n} < \sqrt {\frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} ... \plus{} \frac {1}{x_n}}$[/quote]
Nobody can't help me ??? :(

[quote="conan_naruto236"]Let $ x_1,x_2,...,x_n > 0$. Prove that :
$ \frac {1}{1 \plus{} x_1} \plus{} \frac {1}{1 \plus{} x_1 \plus{} x_2} \plus{} ... \plus{} \frac {1}{1 \plus{} x_1 \plus{} x_2 \plus{} ... \plus{} x_n} < \sqrt {\frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} ... \plus{} \frac {1}{x_n}}$[/quote]
By Cauchy Schwarz Inequality, we have that
\[ \left( \frac {1}{1 \plus{} x_1} \plus{} \frac {1}{1 \plus{} x_1 \plus{} x_2} \plus{} ... \plus{} \frac {1}{1 \plus{} x_1 \plus{} x_2 \plus{} ... \plus{} x_n}\right)^2 \le \left( \frac{1}{x_1}\plus{}\frac{1}{x_2}\plus{}\cdots \plus{}\frac{1}{x_n}\right)\left( \frac{x_1}{(1\plus{}x_1)^2}\plus{}\frac{x_2}{(1\plus{}x_1\plus{}x_2)^2}\plus{}\cdots \plus{}\frac{x_n}{(1\plus{}x_1\plus{}\cdots \plus{}x_n)^2}\right)\]
It suffices to prove that
\[ \frac{x_1}{(1\plus{}x_1)^2}\plus{}\frac{x_2}{(1\plus{}x_1\plus{}x_2)^2}\plus{}\cdots \plus{}\frac{x_n}{(1\plus{}x_1\plus{}\cdots \plus{}x_n)^2}<1\]
Let $ x_0\equal{}0$, then we have
\[ LHS\equal{}\sum^{n}_{i\equal{}1}\frac{x_i}{(1\plus{}x_0\plus{}\cdots \plus{}x_i)^2} < \sum^{n}_{i\equal{}1}\frac{x_i}{(1\plus{}x_0\plus{}\cdots \plus{}x_{i\minus{}1})(1\plus{}x_0\plus{}\cdots \plus{}x_i)} \equal{}\sum^{n}_{i\equal{}1}\left( \frac{1}{1\plus{}x_0\plus{}\cdots \plus{}x_{i\minus{}1}} \minus{}\frac{1}{1\plus{}x_0\plus{}\cdots \plus{}x_i}\right)  \equal{}1\minus{}\frac{1}{1\plus{}x_1\plus{}\cdots \plus{}x_n}<1\]
Done! :)


Solve the following equation system as $a,b,c,d,e\in [-2,2]$:
$\begin{cases} a+b+c+d+e=0\\ a^3+b^3+c^3+d^3+e^3=0\\ a^5+b^5+c^5+d^5+e^5=10\\ \end{cases}$
I found one set of solutions but I don't know how to prove its uniqueness.


[quote="Together"]
We have $\sum \frac{a^2}{(2a+b)(2a+c)} \leq \frac{1}{3}(\sum (\frac{a}{2a+b} + \frac{a}{2a+c}))^2$
And we have $\frac{a}{2a+b} + \frac{b}{2b+a} \leq \frac{2}{3}$  <-> $(a-b)^2 \ge 0$[/quote]
OK. This is true. But why $\sum \frac{a^2}{(2a+b)(2a+c)} \leq \frac{1}{3}$? ;)

Guys if you want know,my solution of this problem isn't nice :D  :D

After homogenization, it is the world of Muirhead. :D

[quote="Together"][quote="Tiks"][quote="Together"]I'M sorry .This inenquality is True...

We have $\sum \frac{a^2}{(2a+b)(2a+c)} \leq \frac{1}{3}(\sum (\frac{a}{2a+b} + \frac{a}{2a+c}))^2$

And we have $\frac{a}{2a+b} + \frac{b}{2b+a} \leq \frac{2}{3}$  <-> $(a-b)^2 \ge 0$[/quote]
Why $\sum \frac{a^2}{(2a+b)(2a+c)} \leq \frac{1}{3}(\sum (\frac{a}{2a+b} + \frac{a}{2a+c}))^2$???????????????????????? :roll:
I am not shure that this ineq. is true :ninja:[/quote]

It is : $xy+yz+zx \le \frac{1}{3}(x+y+z)^2$ .... Wrong ???[/quote]

what are your x,y,z?!

[quote="Together"][quote="Tiks"][quote="Together"]I'M sorry .This inenquality is True...

We have $\sum \frac{a^2}{(2a+b)(2a+c)} \leq \frac{1}{3}(\sum (\frac{a}{2a+b} + \frac{a}{2a+c}))^2$

And we have $\frac{a}{2a+b} + \frac{b}{2b+a} \leq \frac{2}{3}$  <-> $(a-b)^2 \ge 0$[/quote]
Why $\sum \frac{a^2}{(2a+b)(2a+c)} \leq \frac{1}{3}(\sum (\frac{a}{2a+b} + \frac{a}{2a+c}))^2$???????????????????????? :roll:
I am not shure that this ineq. is true :ninja:[/quote]

It is : $xy+yz+zx \le \frac{1}{3}(x+y+z)^2$ .... Wrong ???[/quote]
I see you understoud this problem not well,so I writing it again in other form :)
Show that for $a,b,c>0$
$\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+a)(2b+c)}+\frac{c^2}{(2c+b)(2c+a)}\leq \frac{1}{3}$

Tiks, THANK YOU!! Your inequality took me about 9 hours, but that was absolutely not a waste of my time, since finally I obtained something I find more than fantastic:

[color=blue][b]Theorem 1.[/b] Let a, b, c, x, y, z be nonnegative reals such that $a\geq b\geq c$, and such that $ax\geq by$ or $cz\geq by$. Then,

$x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)\geq 0$.[/color]

Do you see the difference to "standard" Vornicu-Schur? The condition $ax\geq by$ or $cz\geq by$ holds for numerous cases which don't fall under the "standard" Vornicu-Schur inequality since they don't satisfy $x\geq y$ or $z\geq y$. I expect Theorem 1 to be of help in some further problems.

[i]Proof of Theorem 1.[/i] First consider the case $ax\geq by$.

Then, we can easily prove that $ax-by\geq c\left(x-y\right)$. In fact, if $x-y\leq 0$, then $ax-by\geq 0$ (since $ax\geq by$) and $c\left(x-y\right)\leq 0$ (since $x-y\leq 0$) immediately yield $ax-by\geq c\left(x-y\right)$, while, if $x-y\geq 0$, we have $b\left(x-y\right)\geq c\left(x-y\right)$ (since $b\geq c$), so that $ax-by\geq bx-by=b\left(x-y\right)\geq c\left(x-y\right)$. So we get $ax-by\geq c\left(x-y\right)$, or, equivalently, $ax-by-c\left(x-y\right)\geq 0$. This rewrites as $\left(a-c\right)x-\left(b-c\right)y\geq 0$. Also, $a-b\geq 0$ since $a\geq b$. Thus,

$x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)=\left(a-b\right)\left(\left(a-c\right)x-\left(b-c\right)y\right)\geq 0$.

On the other hand, $c\leq a$ and $c\leq b$ yield $c-a\leq 0$ and $c-b\leq 0$, so that $z\left(c-a\right)\left(c-b\right)\geq 0$, and thus

$x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)\geq 0+0=0$.

Now, consider the case $cz\geq by$.

In this case, we must have $z\geq y$ (else, we would have z < y, what, together with $c\leq b$, would yield cz < by and thus contradict $cz\geq by$, unless we have b = c = 0 what is a more or less trivial case). Hence, $z-y\geq 0$, so that $b\leq a$ yields $b\left(z-y\right)\leq a\left(z-y\right)$. Thus, $cz-by\leq bz-by=b\left(z-y\right)\leq a\left(z-y\right)$, or, equivalently, $a\left(z-y\right)-\left(cz-by\right)\geq 0$. In other words, $y\left(b-a\right)-z\left(c-a\right)\geq 0$. Also, $b-c\geq 0$ since $b\geq c$. Hence,

$y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)=\left(b-c\right)\left(y\left(b-a\right)-z\left(c-a\right)\right)\geq 0$.

On the other hand, $a\geq b$ and $a\geq c$ lead to $a-b\geq 0$ and $a-c\geq 0$, so that $x\left(a-b\right)\left(a-c\right)\geq 0$. Therefore,

$x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)\geq 0+0=0$.

Hence, Theorem 1 is proved in both cases $ax\geq by$ and $cz\geq by$.

Now, let's prove your inequality:

[color=blue][b]Theorem 2.[/b] If a, b, c are three positive reals, then the inequality

$\sum\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}\leq\frac13$

holds, where $\sum$ is the sign for cyclic sums.[/color]

[i]Proof of Theorem 2.[/i] In fact, this inequality rewrites as

$\frac13-\sum\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}\geq 0$;

but

$\frac13-\sum\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}=\sum\frac{a}{3\left(a+b+c\right)}-\sum\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}$
$=\sum\left(\frac{a}{3\left(a+b+c\right)}-\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}\right)$
$=\sum\frac{1}{3\left(a+b+c\right)}\cdot\frac{a}{\left(2a+b\right)\left(2a+c\right)}\left(a-b\right)\left(a-c\right)$
$=\frac{1}{3\left(a+b+c\right)}\sum\frac{a}{\left(2a+b\right)\left(2a+c\right)}\left(a-b\right)\left(a-c\right)$.

So it remains to prove that $\sum\frac{a}{\left(2a+b\right)\left(2a+c\right)}\left(a-b\right)\left(a-c\right)\geq 0$. This becomes $\sum x\left(a-b\right)\left(a-c\right)\geq 0$ if we denote $x=\frac{a}{\left(2a+b\right)\left(2a+c\right)}$, $y=\frac{b}{\left(2b+c\right)\left(2b+a\right)}$, $z=\frac{c}{\left(2c+a\right)\left(2c+b\right)}$. Now, we can WLOG assume that $a\geq b\geq c$ (since our inequality is symmetric), and then we can easily check that $ax\geq by$ as follows:

$ax\geq by\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ a\frac{a}{\left(2a+b\right)\left(2a+c\right)}\geq b\frac{b}{\left(2b+c\right)\left(2b+a\right)}$
$\Longleftrightarrow\ \ \ \ \ a^2\left(2b+c\right)\left(2b+a\right)\geq b^2\left(2a+b\right)\left(2a+c\right)$
$\Longleftrightarrow\ \ \ \ \ a\left(2b+a\right)\cdot a\left(2b+c\right)\geq b\left(2a+b\right)\cdot b\left(2a+c\right)$;

but this is true because $a\left(2b+a\right)\geq b\left(2a+b\right)$ (since $a\left(2b+a\right)=2ab+a^2\geq 2ab+b^2=b\left(2a+b\right)$) and $a\left(2b+c\right)\geq b\left(2a+c\right)$ (since $a\left(2b+c\right)=2ab+ac\geq 2ab+bc=b\left(2a+c\right)$). Hence, $ax\geq by$ holds, so we can apply Theorem 1 and obtain the required inequality $\sum x\left(a-b\right)\left(a-c\right)\geq 0$. Theorem 2 is proven.

My apologies both for the style and for possible mistakes - it's 2 AM (now when I am writing this post, probably not when I will actually submit it, since internet is temporarily inaccessible here).

  Darij

You're welcom Darij!!!! :D
But my solution is easely.
I only take $\frac{a}{b}=x,\frac{b}{c}=y,\frac{c}{a}=z$
and to the comon denominator :blush: ,and I get to a very easy inequaliti :lol:

Yes, Tiks. I was about to post this solution, although I thought some time at this problem.
Anyway, nice problem. :)  ;)

Guys I think for all of you interesting how I get to this ineq :)
See this
 [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58189]www.mathlinks.ro/Forum/viewtopic.php?t=58189[/url]
Interesting isn't it?:D

[quote="Tiks"][quote="zhaobin"][quote="Soarer"]sign reversed? clear denominator, expand, amgm[/quote]
yes,you are right.
and
$\sum \frac{a^2}{(\frac 4 3 a+b)(\frac 4 3 a+c)}\ge \frac{27}{49}$
ugly...[/quote]
$\sum \frac{a^2}{(\frac 4 3 a+b)(\frac 4 3 a+c)}\ge (a+b+c)^2(\sum(\frac 4 3 a+b)(\frac 4 3 a+c))^{-1}$ and $(a+b+c)^2(\sum(\frac 4 3 a+b)(\frac 4 3 a+c))^{-1}\geq \frac{27}{49}$<=>$a^2+b^2+c^2\geq ab+bc+ac$ :lol:[/quote]

 :P right.

Theorem 1 can be proven by SOS (sum of squart) Because :
        $(a-b)^{2}=(a-b)(a-c)+(b-a)(b-c)$

[quote="Tiks"]Let $ x,y,z > 0$, prove that
$ \sum \frac {x^2}{(2x \plus{} y)(2x \plus{} z)}\leq \frac {1}{3}.$[/quote][quote="leepakhin"]It is the world of Muirhead. :D[/quote]$ 3\prod(2x \plus{} y)(2x \plus{} z)\left[\frac {1}{3} \minus{} \sum \frac {x^2}{(2x \plus{} y)(2x \plus{} z)}\right]$

$ \equal{} 2\left[\sum{x^4\left(y^2 \plus{} z^2\right)} \minus{} 2\sum{y^3z^3}\right]$

$ \plus{} xyz\left[5\sum{x^3} \plus{} 4\sum{x^2(y \plus{} z)} \minus{} 39xyz\right]\geq 0.$

[quote="Tiks"]$ a,b,c > 0$ prove that
$ \sum \frac {a^2}{(2a \plus{} b)(2a \plus{} c)}\le \frac 13$
See this[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58189]www.mathlinks.ro/Forum/viewtopic.php?t=58189[/url][/quote]

This inequality is Problem 2.27, page 76 of new book (where it is presented with a simple Cauchy-Schwarz solution):

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://canhang2007.wordpress.com/2009/12/08/new-book-inequalities-with-beautiful-solutions-v-cirtoaje-v-q-b-can-t-q-anh/][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.

[quote=Tiks]$a,b,c>0$ prove that
$\sum \frac{a^2}{(2a+b)(2a+c)}\le \frac 13$
See this[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58189]www.mathlinks.ro/Forum/viewtopic.php?t=58189[/url][/quote]

\[\frac{1}{3}\,{\frac { \left( z+2\,x+y \right) ^{2}x \left( y+z \right) }{\sum{((z+2x+y)^2x(y+z))}}}\geq{
\frac {{x}^{2}}{ \left( y+2\,x \right)  \left( z+2\,x \right) }}\]

[quote=zhaobin]another simpler one
$\sum \frac{a^2}{(a+b)(a+c)}\le \frac 34$
I guess some one have thought about it :)[/quote]
[url=http://www.artofproblemsolving.com/community/c275h247275p1357373]Moscow[/url]

Dear sqing! Are you sure that this inequality is true?

Sorry.
Thank arqady.
$ (\forall)a;b;c > 0\Rightarrow \frac {a^2}{(a + c)(a + b)} + \frac {b^2}{(a + b)(b + c)} + \frac {c^2}{(a + c)(b + c)}\ge\frac {3}{4}.$
P4)
[url=http://www.artofproblemsolving.com/community/c275h247275p1357373]Moscow[/url]


function
complex analysis
complex analysis unsolved

Let $C$ be the unit disk $|z|\leq 1$. Suppose the function $f$ analytic on $C$ maps $C$ into the open unit disk $|z|<1$ - that is, $|f(z)|<1$ for all $z\in C$. Prove there is exactly one $w\in C$ such that $f(w)=w$

Since on the boundary $|z|=1$ we have that $|-z| = 1 > |f(z)|$, then by Rouche Theorem $f(z)-z$ and $-z$ have the same number of zeros from where the problem assertion follows.

I has just known above result as "Brouwer fixed point theorem"

Of course you could prove by "Brouwer Fixed point Theorem" that $f(z)$ has a fixed point and because of Schwart's Lemma $f'(z)<1$ for each $z\in C$. So $|f(x)-f(y)|<|x-y|$ and $f(z)$ has at most one fixed point, and everything's proved.


Λοιπόν  έχω   μια  ανισότητα  που  την  προσπαθω  καιρο  τωρα  και  ακόμη  λυση  δεν  έχω  βρεί .....

      :lol: Σιλουανέ  και  εσύ  Λουκα  για  ρίχτε  της  μια  ματιά . :lol:  :lol:

Αν  $a>b>c>0$ και $x>y>z>0$  πραγματικοί  αριθμοί τότε  να  δείξετε  ότι
     $\frac{a^2 x^2}{(by+cz)(bz+cy)} + \frac{b^2 y^2}{(ax+cz)(az+cx)} + \frac{c^2 z^2}{(ax+by)(ay +bx)} >\frac{3}{4}$

Γιώργο φοβερή ανισότητα, μπράβο. Νομίζω πως έχω μια λύση.

Από την ανισότητα ΑΜ-ΓΜ αρκεί νδο ότι
$\sum\frac{a^2\cdot x^2}{(b+c)^2\cdot (y+z)^2}\geq\frac{3}{16}$ .Τώρα από τη γνωστή $3(x^2+y^2+z^2)\geq (x+y+z)^2$ αρκεί να δείξουμε ότι  $\sum\frac{ax}{(b+c)(y+z)}\geq\frac{3}{4}$  (1). Οι τριάδες όμως  $(\frac{a}{b+c},\frac{b}{a+c},\frac{c}{a+b})$ και  $(\frac{x}{y+z},\frac{y}{z+x},\frac{z}{x+y})$ είναι ομοιοτρόπως διατεταγμένες άρα από την ανισότητα Τσεμπισεβ και την ανισότητα Nesbitt δηλαδή  $\sum\frac{m}{n+p}\geq\frac{3}{2}$  λαμβανουμε την (1)

Κανένα σχόλιο για τη λύση είμαι κάπου λάθος?

Λοιπόν  Σιλουανε  κατανοώ  την  λύση  σου  μετα  την  χρήση  της  ΑΜ - ΓΜ  ωστόσο  αδυνατώ  να  καταλάβω  το  πως   έφτασες  $\sum\frac{a^2\cdot x^2}{(b+c)^2\cdot (y+z)^2}\geq\frac{3}{16}$ . Για  ρίξε  μια  διευκρίνηση
αν  και  λόγω  έλλειψης  χρόνου  δεν  εχω  πρoλάβει  να  πολυασχοληθώ .

Κοίταξε. Έχουμε από ΑΜ-ΓΜ ότι

$\sqrt{ (by+cz)(bz+cy)}\leq\frac{by+cz+bz+cy}{2}$  ή ισοδύναμα

$(by+cz)(bz+cy)\leq\frac{(by+cz+bz+cy)^2}{4}=\frac{(b+c)^2(y+z)^2}{4}$

Μειώνοντας την ανισότητα και διαιρώντας με το 4 αρκεί να δείξουμε ότι......

Είμαι κατανοητός ?

Ok   Σιλουανέ  αν  και  το  καταλαβα  πριν  δω  το  Reply    μου  φαινεται οτι  η  διευκρινηση  σου  ειναι  αρκετά  κατανοητή  .  Μπράβο  αρκετά  καλή  η  λύση  σου  . :o  :o  :lol: Λοιπόν  δεν  βρίσκω  καμία  άλλη  λυση  γι  αυτό  για  προτεινε  καμία καινούργια  άσκηση " υποψήφια " για  τον  Αρχιμήδη  να  μπούμε  στο  πνεύμα ................


Hello Everyone,

I have been doing a recopilation of some material (math conquest around the world) Where it is included. If someone wants to help me to do it as complete as it may be, please contact me.

Best,

David

As you said, this is "collection v2", do you have v1 for being shared?

Oh, the letter "v" maybe means "version", not "volumn"

Right,

The v2 means version 2. Probably in some months I will have time to release v3.

Best,

David


function
complex analysis
real analysis
real analysis unsolved

$f: \mathbb{N}\to  \mathbb{N}$ a function.
Does there exist a power series $g(x)=a_0+a_1x+..+a_kx^k+..$ with infinite radius of convergence that satisfies:
$f(n)=g(n)$ for every integer ?

This is really a complex analysis question. "power series with infinite radius of convergence" means "entire function".

A sequence of entire functions which converges uniformly on compact sets has an entire limit.

I don't understand what you mean (second sentence). Is it a solution ?

so let's put it on complex analysis section. (but there are very elementary solutions ...)


Prouve that if $n\geq3$ ; then $2^{n}=7x^{2}+y^{2}$

where $x,y$ are odd

ps: MMO stands for (Moscov Mathematical Olympiad)

let's use the fact that $(7a^2+b^2)(7x^2+y^2)=7(ay-bx)^2+(7ax+by)^2$.
and let's note that $2^4=7*1^2+3^2,2^3=7*1^2+1^2,2^2=7*0^2+2^2$ so
$2^{4+2k}$ and $2^{3+2k}$ for $k$ non-negative integer- indeed have the form $7x^2+y^2$,
the only thing needed now is to proove that $ay-bx,7ax+by$ remain odd but it's easy so I skip it.
(q.e.d.)


An $8\times8$ chessboard  is completely tiled by $2\times1$ dominoes. Prove that there exist a king's tour
of that chessboard such that every cell of the board is visited exactly once and such that king goes domino by
domino, i.e. if king moves to the first cell of a domino, it must move to another cell in the next move. (King
doesn't have to come back to the initial cell. King is an usual chess piece.)

If our dominoes will be Horizontal The king can move like $MN$ Horizontal for start from even rows and move like $WN$ Horizontal for start from odd rows If our dominoes will be Vertical only sort of moves will be like $MN$ Vertical for start from even rows and move like $WN$ Vertical for start from odd rows.

I have a weird proof which shows that there is a bijection from the set of king tours to the set of 2x1 domino tilings; it's just enumerative combinatorics now.

I find your claim extremely implausible -- for a 2-by-2 board there are vastly more tours than tilings, and I would expect that to continue for larger boards, too.

Sorry, I realise I was being stupid; that was actually a solution to a different problem (I won't post it here, because the solution has obviously been given away).


It's an easy ineq but the coefficient is not nice because use SS, we are easily to prove:
$ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \plus{} \frac{27(ab\plus{}bc\plus{}ca)}{2(a\plus{}b\plus{}c)2} \geq\ \frac{15}{2}$
 :wink:   :ninja:

[quote="nguoivn"]It's an easy ineq but the coefficient is not nice because use SS, we are easily to prove:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {27(ab \plus{} bc \plus{} ca)}{2(a \plus{} b \plus{} c)2} \geq\ \frac {15}{2}$
 :wink:   :ninja:[/quote]
What is SS an abbreviation for?

SS is an abbreviation for the hint:
$ M(a \minus{} b)^2 \plus{} N(a \minus{} c)(b \minus{} c) \geq\ 0$
with $ c\equal{}$ min $ {a, b, c}$

If you don't understand, you should say "please tell me why?", not "??..."  :)
Using CS, LHS $ \geq\ \frac {(\sum\ \sqrt []{\frac {a}{b}})^2}{13\sum\ a^2 \plus{} 14\sum\ ab}$
Easily to see that: $ 13 \sum\ a^2 \plus{} 14\sum\ ab \leq\ \frac {27(\sum\ a^2 \plus{} \sum\ ab)}{2}$
So, we only need to prove:
$ (\sum\ \sqrt []{\frac {a}{b}})^2 \geq\ \sum\ \frac {27}{2}. \frac {\sum\ a^2 \plus{} \sum\ ab}{(a \plus{} b \plus{} c)^2}$
Because: $ \sum\ \sqrt[]{\frac{a}{c}} \geq\ 3$, so we only need to prove:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {27(ab \plus{} bc \plus{} ca)}{2(a \plus{} b \plus{} c)2} \geq\ \frac {15}{2}$
I think now, you understood  :wink:

oh!no!
$ \sum\(a^2-3ab+2bc-c^2+ca)^2 \ge 0$ is very hard.
I think you try to expand,and with cauchy in engel form.But I can find many inequalities with expand as you.

[quote="quykhtn-qa1"]sorry nguoivn,but my solution very nice.
Use CS: LHS ≥ (∑a)^2 / (∑ab(10a^2+3b^2+14ac))
and (∑a)^4-∑ab(10a^2+3b^2+14ac)
      = (a^2-3ab+2bc-c^2+ca)^2≥0.
 :lol:[/quote]

Yes, it's nice but how will you solve with the harder:
$ \sum\ \frac {a}{b(12a^2 \plus{} b^2 \plus{} 13ca)} \geq\ \frac {3}{26}$
with $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$
Your proof above can't solve it  :)

And with the nicer, my proof still trues:
$ \sum\ \frac {a}{b(a^2 \plus{} bc)} \geq\ \frac {27}{2(a \plus{} b \plus{} c)^2}$
Your proof still can't solve it again.

[quote="nguoivn"]It's an easy ineq but the coefficient is not nice because use SS, we are easily to prove:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {27(ab \plus{} bc \plus{} ca)}{2(a \plus{} b \plus{} c)2} \geq\ \frac {15}{2}$
[/quote]
Here is my proof for it:

But with the stronger following, we have to use the another hint  :)
http://www.mathlinks.ro/Forum/viewtopic.php?t=251503

[quote="quykhtn-qa1"]
(∑a)^4-∑ab(10a^2+3b^2+14ac)
      = (a^2-3ab+2bc-c^2+ca)^2≥0.
 :lol:[/quote]

I checked it and I think it is wrong, instead nguoivn's proof is correct and very nice. :)


So, have any packets or anything actually been mailed out yet?

You don't have to do the induction or the pidgeonhole packets. You were only supposed to fill out the information packet.

I didn't say that I knew anything about packets that may or may not come in the future. I just said that we have not gotten any mathematically related packets that we are supposed to complete.

[quote="Klebian"]oh wait so there's no red/blue/yellow/white/black group thing this year, so we don't have those type of packets this year?[/quote]

I think in the math jam they said they did away with that.


the mean of six positive integers is 5 and the median is 6. wat's te largest the mode could be, giventhat the mode is unique?

[hide]
The numbers are: 1, 1, 4, 8, 8, 8

Dunno if it's right, I guessed and checked.[/hide]

[hide]1, 1, 4, 8, 8, 8

The largest possible mode is $\boxed{8}$.[/hide]

[quote="espark52"]the mean of six positive integers is 5 and the median is 6. wat's te largest the mode could be, given that the mode is unique?[/quote]
The mode cannot be unique... I think

Look at the two posts ahead of you... geez...


abstract algebra
superior algebra
superior algebra unsolved

Let S be non-abelian simple group and Aut(S) be automorphism group of S. Is the Aut(S) a split extension of S?


invariant
linear algebra
matrix
algebra
polynomial
vector
abstract algebra

We have T:R^3 -> R^3 defined by left multiplication by the matrix A

A= 0 1 0
     0 0 1
     1 0 0

- Find T-invariant subspaces W1 and W2 such that dim(W1) =1 and dim (W2) = 2, and R^3= W1 + W2.

- are there T-invariant subspaces U1, U2, U3 of dim 1 such that R^3 = U1 + U2 + U3?[/code]

$ A \equal{} \begin{pmatrix} 0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0 \end{pmatrix}$.

Then $ \chi_T (x) \equal{} x^3 \plus{} 1 \equal{} (x \plus{} 1)(x^2 \minus{} x \plus{} 1)$

So $ W_1 \equal{} \ker(A \plus{} I)$ and $ W_2 \equal{} \ker(A^2 \minus{} A \plus{} I)$.

Since $ x^2 \minus{} x \plus{} 1$ is irreducible in $ \mathbb{R}$, the matrix $ A$ is not diagonalizable. Hence, there does not exists $ U_1, U_2, U_3$ with the property.

The characteristic (and minimal) polynomial of $ A$ is $ x^3\minus{}1,$ which has roots (eigenvalues) of $ 1,\omega,\omega^2,$ where $ \omega\equal{}e^{2\pi i/3}\equal{}\frac{\minus{}1\plus{}i\sqrt{3}}2.$

If we consider $ A$ as acting on $ \mathbb{C}^3.$ Then we can find three one-dimensional subspaces: the spans of the three eigenvectors:

$ \begin{bmatrix}1\\1\\1\end{bmatrix},\ \begin{bmatrix}1\\ \omega\\ \omega^2\end{bmatrix},\ \begin{bmatrix}1\\ \omega^2\\ \omega\end{bmatrix}.$

(Another way to say that: this matrix, as with all circulants, is diagonalized by the discrete Fourier transform.)

But the problem asks us to consider real vectors only. In that case, let's find a real basis for the span of the second and third vectors listed above:

$ W_2\equal{}\text{span}\,\left\{\begin{bmatrix}2\\ \minus{}1\\\minus{}1\end{bmatrix},\begin{bmatrix}0\\ 1\\ \minus{}1\end{bmatrix}\right\}.$

[color=green][\operatorname still not working. Must use \text.][/color]

This would have agreed with what mathangel said, had mathangel gotten the details right. The actual characteristic polynomial is $ x^3\minus{}1\equal{}(x\minus{}1)(x^2\plus{}x\plus{}1)$ and my $ W_2$ is the kernel of $ A^2\plus{}A\plus{}I.$ Which is what mathangel meant to say.


Solve in non-negative integers the equation  $ 2^{a}3^{b} \plus{} 9 \equal{} c^{2}$

I got the same answers as those posted in the forum but how do we know that these are all the answers?

The solutions are based on this $ 2^a3^b\plus{}6\equal{}2^c3^d$, 6 is only one time divisible by 2 and 3, so some of the numbers a,b,c,d are automatic less or equal to 1.

[quote="mavropnevma"]Two solutions have been left out in the previous solution.
These are $ a \equal{} 3, b \equal{} 3, c \equal{} 15$ and $ a \equal{} 5, b \equal{} 4, c\equal{} 51$.[/quote]
 how did u get to this answer

[quote="siddharthanand"][quote="Marius Mainea"]1) $ a\neq 0$ $ b\neq 0$

 then c=3q  q odd ,and $ (q - 1)(q + 1) = 2{}^a3{}^{b - 2}$ so b=2 and a=3 c=9

 Similarly a=4 b=3 c=21;

 2) $ a\neq 0$ $ b = 0$

 Then $ 2^a + 9 = c^2^$ and a=4, b=0 , c=5;

 3) a=0; $ 3^b+9=c^2$

 Then a=0 ,b=3 ,c=6;[/quote]
could u explain me the first step in  detail manner[/quote]

Casework: either $2||q-1 \wedge 2^{a-1}||q+1$ or vice versa, and again $3^{b-2}||q-1$ or $3^{b-2}||q+1$.
[url=http://en.wikipedia.org/wiki/Catalan_conjecture]Mihailescu's theorem[/url] makes it easier in some cases.
Otherwise $0 \in \{a,b \}$.

[quote="cheeseyicecream"][quote="highlander"]
Then we have to solve the ecuation

$ 2^x-3^y=\pm1\\
2^x-3^y=1$ has solutions $ x=1,y=0, x=2,y=1\Rightarrow (a,b,c)\in\{(3,2,9),(4,3,21)\}$

$ 2^x-3^y=-1$ has solutions $ x=1,y=1, x=3,y=2\Rightarrow (a,b,c)\in\{(3,3,15),(4,5,51)\}$

[/quote]

So highlander is using Mihailescu's theorem here to find the only solutions such that x,y > 1 ?[/quote]
what's Mihailescu's theorem ?
I've got exactly the same answer as yours,but the way I achieved it is very complex.

Preda Mihailescu proved Catalan's conjecture that $8$ and $9$ are the only consecutive powers of positive integers.

My solution is a bit long but i didnt miss any cases so here it is:
first look for the cases where a,b are 0; a is 0 and b isnt 0, b is 0 and a isnt 0.These should be easy by factorising.
And for the case where a,b,c dont equal 0 just divide both sides by 4.You will be left with (c-3)/2 and (c+3)/2.Note that both of them cant be even but they can be a power of 3 at the same time.Lets say a>=3 and b=x+y. The cases we have now are:
1)(c-3)/2=2^(a-2) and (c+3)/2=3^y
2)(c-3)/2=3^x and (c+3)/2=3^(y)2^(a-2)
3)(c-3)/2=1 and (c+3)/2=3^(b)2^(a-2)
Then just check the case where a=2 and youre done. :-D

[quote=highlander]$ 2^a\cdot3^b\plus{}3^2\equal{}c^2.\\
i)b\equal{}0 \Rightarrow 2^a\equal{}(c\minus{}3)(c\plus{}3)\Rightarrow 2^x\equal{}c\minus{}3, 2^y\equal{}c\plus{}3,x<y,x\plus{}y\equal{}a \Rightarrow 2^x(2^{y\minus{}x}\minus{}1)\equal{}6\Rightarrow\\
x\equal{}1,y\equal{}3\Rightarrow (a,b,c)\equal{}(4,0,5).\\\\
ii)b\geq1\Rightarrow c\equal{}3k\Rightarrow b\geq2\\
2^{a}\cdot3^{b\minus{}2}\equal{}(k\minus{}1)(k\plus{}1)\\\\
I).a\equal{}0\Rightarrow k\minus{}1\equal{}3^x,k\plus{}1\equal{}3^y,x<y,x\plus{}y\equal{}b\minus{}2\Rightarrow 3^x(3^{y\minus{}x}\minus{}1)\equal{}2\Rightarrow x\equal{}0,y\equal{}1\Rightarrow (a,b,c)\equal{}(0,3,6).
\\II).a\geq1\Rightarrow k\equal{}2p\plus{}1\Rightarrow 2^{a\minus{}2}\cdot3^{b\minus{}2}\equal{}p(p\plus{}1),(p,p\plus{}1)\equal{}1\Rightarrow$
$ p\equal{}2^{a\minus{}2},p\plus{}1\equal{}3^{b\minus{}2}$ or $ p\plus{}1\equal{}2^{a\minus{}2},p\equal{}3^{b\minus{}2}.$
Then we have to solve the ecuation
$ 2^x\minus{}3^y\equal{}\pm1\\
2^x\minus{}3^y\equal{}1$ has solutions $ x\equal{}1,y\equal{}0, x\equal{}2,y\equal{}1\Rightarrow (a,b,c)\in\{(3,2,9),(4,3,21)\}$
$ 2^x\minus{}3^y\equal{}\minus{}1$ has solutions $ x\equal{}1,y\equal{}1, x\equal{}3,y\equal{}2\Rightarrow (a,b,c)\in\{(3,3,15),(4,5,51)\}$
$ \\S: (a,b,c)\in\{(4,0,5), (0,3,6), (3,2,9), (4,3,21), (3,3,15), (4,5,51)\}$[/quote]

Umm, you ordered a and b wrong in the last sol… Cuz I got the last pair was (5, 4, 51), which is correct, not (4, 5, 51) for (a, b, c). Also, I got the obvious solutions which were these, but I wasn’t sure if it was the full solution, because $2^x-3^y=\pm1$ I only got the obvious sols of x=1 y=0 and x=2 y=1, but how do we know there are no larger solutions to x and y? (And analogously for the other cases, because I got the same answer as you but didn’t know if it was all solutions)

Edit: After reading other solutions I see why now. Glad that at least I got the right answers! The results are from Catalan’s conjecture [url=here.] https://en.wikipedia.org/wiki/Catalan%27s_conjecture[/url]

[quote=huashiliao2020]After reading other solutions I see why now. Glad that at least I got the right answers! The results are from Catalan’s conjecture [/quote]To solve the equation $2^x-3^y= \pm 1$ , Catalan's conjecture is not necessary at all.
Try mod 3 and 4


Some circles are in the convex poligon. Circles doens't have common points within it's selfes. Prove that that poligon can be represented as the union of convex poligons such that they doesn't have comon area and every poligon contains exactly one circle.

For each circle $\mathcal C$, consider the region consisting of those points for which the power with respect to $\mathcal C$ is $\le$ than the power with respect to any other circle in our collection. You should get something like a [url=http://mathworld.wolfram.com/VoronoiDiagram.html]Voronoi diagram[/url] (but having circles inside instead of points).


The number 34,459,425 is the product of several consecutive positive odd numbers.  What is the greatest of these numbers?

Try to solve it without a calculator first.

Treething, can you please explain why it is the product of 6 or more numbers?

[quote="Treething"]It's obviously the product of 6 or more numbers.[/quote]

If you think that because 25 divides it, well one of the numbers could be 25...

[quote="G-UNIT"]my first assumption was correct  :lol:

[hide] its the product of all the odd numbers from 1-17 inclusive, the largest of these: $\boxed{17}$[/hide][/quote]

How did you randomly manage to guess that?  You must be good at guessing.


AMC
AIME
USA(J)MO
USAMO
email
AMC 12
AMC 12 A

I convinced my public school teacher this year to let our school have the AMC-12 A and AMC-12 B test.  I took the AMC-12 A test and I got a 112.5, then I took the B test and got a 132.5 (which is good for me).  I put the AMC-12 B test score on the AIME as my AMC score, but today when I asked my teacher about the scores he said that he received the AMC A scores right before the AIME, but he never received the AMC B scores.  I know that I did not qualify for the USAMO, but I was wondering if there is a way for me to check my AMC-12 B score.

Thank You

This is very important to me and a response would be greatly appreciated.  Has this happened to anybody in the past?

I think the best thing would be to wait for your teacher to receive the scores. My school has already received the B scores; you could convince your teacher to email them or something. You could also email them yourself and inquire about your score.

All scores should've been received eons ago.

Email amcinfo@unl.edu with your name, grade, school, state, and the test score you are requesting. Simply note that your teacher did not give it to you, and I'm sure they could find out.

Thank you MithsApprentice for providing an email address.  I sent an email.

It doesn't matter. This was discussed before (topic started by sirorange) and I think the AMC Director said that they look both scores up anyway.

To be safe, if you know both definitely put the AMC12 score.

They sent me a message back saying that my school's AMC B scores were never received.  Has this happened to anyone before?  I got a 132.5, when I checked the answers a few days after the test, but they never received the test.  I am frustrated with my teacher, and I was wondering if anyone knew a way that next year I could make sure that my teacher sends the tests in correctly.  I do not want to have scores good enough to qualify for USAMO next year only to find out that my teacher did not send the tests to the graders correctly.

[quote="pieterminate"][quote="Phelpedo"]You have to put your AMC score on the AIME answer sheet?

I took both the AMC 10 and 12- should I put the higher one (10) or the 12 score, or both?[/quote]

12 definitely, thoug it doesn't actually matter, but.......[/quote]
Yeah but my 10 score is, expectedly, quite a big higher. If they look up both, though, I guess it doesn't matter.

[quote="math88fun"]Has anyone else ever had their AMC or AIME test "lost" in the mail?[/quote] Wow. I kind of know what that feels like. On a local two-round contest, our teacher sponsor forgot about sending in round 1, and we only realized it right before round 2. Fortunately, they were gracious and graded our tests late, but that was a local contest. For us, we have two different sponsors for that contest and for AMC and our AMC sponsor is really on the ball, so he doesn't make mistakes. I would confront your teacher and ask him/her if they really did send it in. If they did, then I don't know what to say, just make sure you're still in contact with AMC just in case they eventually get something...

Yeah I haven't ever had scores lost but I've had certificates "lost" for my AMC 8s and 10s numerous times.... (as in 3: 2 amc 8s and 1 amc 10)


calculus
integration
calculus computations

This problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106770]appeared here[/url] about a month ago.

[quote="maokid7"]Evaluate $\int_{0}^{1}\ln(x)\ln(1-x)dx$

[hide]
i was thinking that you could somehow use $\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx$
[/hide][/quote]
Maokid7, you should use that identity only when $f(a-x)$ differs from $f(x)$ in some manner. If $f(a-x)$ is the same as $f(x),$ then the identity pretty much becomes useless. In that case, you might think about manipulating the integral in some way to use that identity. For this problem, Prof Merryfield's solution is the best one (the only one?) I know of.

[quote="maokid7"]what do you mean by "differs in some manner"?[/quote]
In this case, making that change literally doesn't change anything.  That's a useful switch to make if it will alter the expression somehow and let you do something clever, but in this case you end up with an identical expression and so are equally far from finding the answer.


Shown below is a rectangle that has been cut into eleven squares of various sizes.  The smallest square measures 9 by  9 cm.  Find the dimensions of the large rectangle.

Someone else can do the figuring, but basically you label the next square with side $x$, then the next square with side $9+x$, continue this until you get expressions for each side. Set them equal and solve.

[hide="solution"]small square is 9

square to the right of starter is x

square below starter is 9 + x

square in the right bottom corner is 9 + x + x = 9 + 2x

square to the left of starter is 9 + 9 + x = 18 + x

square above starter is 9 + 18 + x = 27 + x

square in bottom left corner is 27 + x + 18 + x = 45 + 2x

center square along right edge is x + 9 + 2x = 9 + 3x

let S be the side of the small square near the middle. We then have
S + 45 + 2x = 9 + 2 x + 9 + 3x = 18 + 5x
S = 3x - 27

the square in the top right corner is 3x - 27 + 9 + 3x = 6x - 18

the big square is 6x - 18 + 3x - 27 = 9x - 45

finally, the bottom edge of the big square gives an equation we can solve.

9x - 45 + S = 12x - 72 =  45 + 2x + 27 + x = 3x + 72
9x = 144
x = 16

top edge of rectangle is 6x - 18 + 9x - 45 = 15x - 63 = 177
left edge is 9x - 45 + 45 + 2x = 11x = 176

177 x 176[/hide]

[quote="mysmartmouth"]I got [hide]108[/hide].  I think I made a mistake though...[/quote]

the problem asks for the [b]dimensions[/b] ;)

[quote="Altheman"]the problem asks for the [b]dimensions[/b] ;)[/quote]

Sorry.  I meant [hide]108 x 108[/hide].  Once again, however, I am probably wrong.

I saw this problem once in one of the contests for 8th graders

Try to draw the figure and follow lasagnaman's steps

OK...I'm still lost.  Oh well... never mind.


http://www.thepetitionsite.com/1/Rehabilitate-the-slum-dwellers

Please sign! We need your help. This has to go the DM tomorrow so act fast and pass it on to the rest of your friends so that they can also sign the petition.

Thanks,
Gaurav.

P.S. Sorry if this looks like a spam :(

[quote="Pocha"]R u expcting a miracle overnite?
 for  14 to rise to 1000 within 12 hrs? :huh:[/quote]
Hell no!! Even 100 will be enough. That was just the default limit, I didn't see the option to change it.


geometry
circumcircle
incenter
symmetry
geometric transformation
reflection
parallelogram

$ ABC$ is an isosceles triangle.$ CD$ is the bisector of $ <C$.The line that is perpendicular to $ CD$ and pass from the circumcenter of $ ABC$ intersect $ BC$ at $ E$and the line that is parallel to $ BC$ and pass from $ E$ intersect $ AB$ at $ F$.prove that:
$ BE$ $ \equal{}$ $ FD$

Dear behdad,
at which vertex is ABC isosceles?
Is there a typo when you say "the line that is parallel to BC" ?
Sincerely
Jean-Louis

The line through E that is parallel to CD (not to BC) cuts AB at F. Isn't that right ?

yes i think that is parallel to CD not to BC
thanks dear yetti.

$ I$ is incenter of the isosceles $ \triangle ABC$ with $ AB \equal{} BC.$ $ AI, BI, CI$ cut its circumcircle $ (O)$ at $ X, Y, Z.$ By symmetry, $ ZX \parallel CA.$ $ OE \perp CZ$ is perpendicular bisector of $ CZ$ $ \Longrightarrow$ $ \triangle ECZ$ is isosceles with $ EC \equal{} ZE$ and $ \angle EZC \equal{} \angle ZCE \equal{} \angle ZCA$ $ \Longrightarrow$ $ EZ \parallel CA,$ $ X \in EZ.$ $ I$ is orthocenter of isosceles $ \triangle XYZ$ with $ XY \equal{} YZ$  $ \Longrightarrow$ $ B$ is reflection of $ I$ in $ ZX$ $ \Longrightarrow$ $ EB \equal{} IE$ and $ \angle BIE \equal{} \angle EBI \equal{} \angle ABI$ $ \Longrightarrow$ $ EI \parallel AB.$ Then $ EIDF$ is a parallelogram and $ DF \equal{} IE \equal{} EB.$


a+b+c+d+e (is less than or equal to) 100

How many possible answers are there to this problem?

 Source: Took Iversonfan2005's problem and morphed it.

I guess he wants the numbers to be natural.

Oops I'm sorry I wanted to say nonnegative. :blush:

[quote="deej21"]How did you evaluate that sum, shyong?[/quote]

it is Chrismas Stocking Theorem [url]http://mathworld.wolfram.com/ChristmasStockingTheorem.html[/url] :)


It's true if $a,$ $b$ and $c$ are sides lengths  of triangle. :D

[quote="tanpham90"][quote="toanhocmuonmau"]If $a,b,c \ge 0,a+b+c=3$, prove that
\[\sqrt{\frac{a}{1+b+bc}}+\sqrt{\frac{b}{1+c+ca}}+\sqrt{\frac{c}{1+a+ab}}\ge \sqrt{3}\]
:)[/quote]

Very easy , toanhocmuonmau :D ! Thank a Million !
We need to prove that :

$\sum \sqrt{\frac{a}{a^{2}+4b^{2}+c^{2}+5ab+2ac+14bc}}\geq\ \sqrt{\frac{1}{a+b+c}}$

Apply Holder we have :
$[ \sum \sqrt{\frac{a}{a^{2}+4b^{2}+c^{2}+5ab+2ac+14bc}}]^{2}\geq\ \frac{[ \sum a(2a+b+3c) ]^{3}}{ \sum (2a+b+3c)^{3}a^{2}(a^{2}+4b^{2}+c^{2}+5ab+2ac+14bc)}$

Finally , it suffices to prove that :

$[ \sum a(2a+b+3c) ]^{3}(a+b+c) \geq\ \sum (2a+b+3c)^{3}a^{2}(a^{2}+4b^{2}+c^{2}+5ab+2ac+14bc)$
Assume that $c=min(a,b,c)$ , after expanding all we are done !
 :lol:  :D  :P  :rotfl:[/quote]
Yes, it's also my solution. But we don't need to expand here since
\[[ \sum a(2a+b+3c) ]^{3}(a+b+c) \geq\ \sum (2a+b+3c)^{3}a^{2}(a^{2}+4b^{2}+c^{2}+5ab+2ac+14bc) =(a+b+c)(4\sum ab(a^{4}+b^{4})+26\sum a^{2}b^{4}+39\sum a^{4}b^{2}+54\sum a^{3}b^{3}+261x^{2}y^{2}z^{2}-24\sum a^{4}bc-93\sum a^{3}b^{2}c-97\sum a^{2}b^{3}c) \]
which is obviously nonnegative by AM-GM. :)

[quote="tanpham90"]
Apply Holder we have :
$[ \sum \sqrt{\frac{a}{a^{2}+4b^{2}+c^{2}+5ab+2ac+14bc}}]^{2}\geq\ \frac{[ \sum a(2a+b+3c) ]^{3}}{ \sum (2a+b+3c)^{3}a^{2}(a^{2}+4b^{2}+c^{2}+5ab+2ac+14bc)}$
[/quote]

Can anyone give more explanation?
Thx..


graph theory
number theory proposed
number theory

Let $ m,n$ be two integers such that $ \varphi(m) \equal{}\varphi(n) \equal{} c$. Prove that there exist natural numbers $ b_{1},b_{2},\dots,b_{c}$ such that $ \{b_{1},b_{2},\dots,b_{c}\}$ is a reduced residue system with both $ m$ and $ n$.

you see the system $ x (mod m) \equal{} r,x(mod n) \equal{} s$has a solution if $ (m,n) \equal{} d$devides $ r\minus{}s$.
if you make a bipartite graph with respect to  two reduced residue system for $ m$ and $ n$,and join $ r$ from the first part to $ s$ from second part if $ r\minus{}s$ is devisible by $ d$.by inclusion principle we conclude that every part is regular,so we have a matching,using hall theorem.

can we make some bounds for the number of matchings?

in fact we can compute the exact number of matchings.

Also a generalization for this for $ m,n,k$ which $ \varphi(m)\equal{}\varphi(n)\equal{}\varphi(k)\equal{}c$ there exist $ m,n,k$ such that $ a_{1},a_{2},\dots,a_{c}$ such that is reduced residue system for $ m,n,k$.
As I know the problem is not correct for more than three numbers.

[quote="dashmiz"]you see the system $ x (mod m) \equal{} r,x(mod n) \equal{} s$has a solution if $ (m,n) \equal{} d$devides $ r \minus{} s$.
if you make a bipartite graph with respect to  two reduced residue system for $ m$ and $ n$,and join $ r$ from the first part to $ s$ from second part if $ r \minus{} s$ is devisible by $ d$.by inclusion principle we conclude that every part is regular,so we have a matching,using hall theorem.[/quote]

I think this graph will be union of $ \phi (d)$ complete bipartite graphs, in each component there will be $ \frac{\phi (m)}{\phi (d)}$ =$ \frac{\phi (n)}{\phi (d)}$=$ v$ vertices in each vertice class. Even without Kőnig-Hall thm we can find $ v!^{\phi(d)}$ matchings.

Correct me please if I was wrong... :blush:

Can we solve the problem without using graph ?


given quadratic equation  $f(x)=ax^2 +bx+c$ such that $f(-3)<-10 , f(-1)>0$and $f(1)<-1$ determine the sign of the coefficient   of    $a$

We cannot determine the sign of a.

Take b = -2,  a + c = 0 then f(-1) = 2 > 0, f(1) = -2 < -1 and the condition f(-3) < 10 is equivalent to  9a - 3b + c < 10 <=> a < 1/2. So, we can take a = 1/3 > 0 and a = -1 < 0 which both satisfy.

Maybe I'm wrong somewhere again?

Namdung


Two numbers, a and b, has product of 384. a is 8 more than b and both a and b are natural numbers.

Find a+b.

[hide="PRIZE"] :first: [/hide]

[hide]ab=384
a+8=b
a(a+8)=384
a 2 +8a-384=0
(a+24)(a-16)=0
since they are natural numbers a is 16 and then b is 24.
soooooooo..16+24=[b]40[/b][/hide]

do i get a cool trophy?

[quote="Silverfalcon"]To dinoman:

[hide="DINOMAN_ONLY"] :first: [/hide]

Tada![/quote]

lol.

[hide="to dinoman"]oh dinoman,

it is a=b+8....i think[/hide]
[color=darkred]
It's dinoman >=o[/color]

That was just for fun. You can just draw your prize up.

jli, it doesn't matter whether it's a=b+8 or b=a+8.


Let $ P(x)$ be a polynomial which gives remainder $ 4$ when divided by $ (x\minus{}3)$ , and remainder $ 3$ when divided by $ (x\minus{}4)$ . The remainder when $ P(x)$ is divided by $ (x\minus{}3)(x\minus{}4)$ is

$ \textbf{(A)}\ 7 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 7\minus{}x \qquad \textbf{(D)}\ 4\minus{}x \qquad \textbf{(E)}\ 3\minus{}x$

[hide]
We have $ P(3)\equal{}4, \ P(4)\equal{}3$

$ P(x)\equal{}(x\minus{}3)(x\minus{}4)Q(x)\plus{}ax\plus{}b$

$ P(3)\equal{}3a\plus{}b\equal{}4, \ P(4)\equal{}4a\plus{}b\equal{}3$

Now solve the system.
[/hide]

[quote="hasan4444"][quote="red_dog"]
$ P(x) \equal{} (x \minus{} 3)(x \minus{} 4)Q(x) \plus{} ax \plus{} b$
[/quote]
Sorry but this is a little confusing is $ Q$ means quotient?[/quote]

Yeah, $ Q(x)$ is the quotient function, basically multiplying this by the divisor and adding the remainder will give you the original polynomial.

Also, this problem is incredibly guessable (which admittedly, is how I did it). One polynomial that satisfies this is $ x^2 \minus{} 8x \plus{} 19$ and finding the remainder when that is divided by $ (x\minus{}3)(x\minus{}4)$ isn't tough.

Red dog's solution assumes a remainder in the form of ax + b. This is because the degree of the remainder is always less than the quotient, which I just figured out in about 10 minutes.

Using the remainder theorem eliminates all of the wrong choices at once. Remember that the remainder of $ P(x)$ upon division by $ x\minus{}a$ is $ P(a)$. [hide="rest"]Then we have $ P(3)\equal{}4$ and $ P(4)\equal{}3$. If $ P(x)\equal{}Q(x)(x\minus{}3)(x\minus{}4)\plus{}R(x)$, where $ Q(x)$ is the quotient and $ R(x)$ is the remainder, then $ R(3)\equal{}4$ and $ R(4)\equal{}3$. Every choice except C fails both of these conditions.[/hide]


calculus derivative Gauss MATHCOUNTS articles

[quote="aasdfg1234"]...that depression[/quote]
I don't think depression has anything to do with appearance for most people here- but it's pretty common. The same personality types correlate to both mathematical aptitude and depression.
If you're falling into depression- seek help. Don't think you'll beat it entirely by yourself.

Check out my album, and you'll get out of the depression

Well, I dont think you will be depressed for not being handsome or pretty, bare in mind the possibility that someone is less handsome or pretty than you is very high.  :wink:

"Think angry thoughts, like milk....no...not milk...kittens..." -RvB.

lol

talk to guidance.

lol uhh well

i've heard suicide doesn't work too great

and you don't need to be handsome/pretty/naked in order to have a good social life, trust me

Uh hello, implants?

Seriously though, who cares? People usually don't have any friends because they look like a beast. Get some social skills and maybe come out of the shell once in a while and you can get a social life. If you want a girlfriend, impress her with math knowledge. Do the derivative tangent to curves thing that's in someone's signature (I forget whose). But seriously, to get a life, you don't need that.

Alter your genes. :rotfl:

Seriously, beauty is not the most important thing in life.

[quote="Treething"][quote="1=2"]I am the worst runner in my math circle, but I managed to tie with six other people in my school at NYSML(yes, I live in NY) with 7's.(easy year, I'm 11 years old...)[/quote]

Okay, what the...[/quote]

uuhh... your point being?

Don't act like a jerk. I for one hate jerks. I think that sensible people you should hang out with like to be around nice people and shouldn't expect you to look anything.

[quote="1=2"]I am the worst runner in my math circle, but I managed to tie with six other people in my school at NYSML(yes, I live in NY) with 7's.(easy year, I'm 11 years old...)[/quote]

What the - are you on?

Why are people asking me these questions? It was just an example.

off topic discussion ends....

NOW!!!!

[quote="life=tennis+math"]Do the derivative tangent to curves thing that's in someone's signature (I forget whose).[/quote]

"I wish to be your derivitive so I could lie tangent to your curves!"

is that what you mean  :D

[quote="noneoftheabove"]Do math! math>beauty[/quote]

No, dude, math=beauty, just not necessarily physical beauty.

[quote="Oscar Wilde"]But beauty, real beauty, ends where an intellectual expression begins. Intellect is in itself a mode of exaggeration, and destroys the harmony of any face. The moment one sits down to think, one becomes all nose, or all forehead, or something horrid. Look at the successful men in any of the learned professions. How perfectly hideous they are![/quote]
Don't say that you were not warned.

[quote="life=tennis+math"][quote="1=2"]I am the worst [b]runner[/b] in my math circle, but I managed to tie with six other people in my school at NYSML(yes, I live in NY) with 7's.(easy year, I'm 11 years old...)[/quote]

What the - are you on?[/quote]

implants

[quote="Oscar Wilde"]But beauty, real beauty, ends where an intellectual expression begins. Intellect is in itself a mode of exaggeration, and destroys the harmony of any face. The moment one sits down to think, one becomes all nose, or all forehead, or something horrid. Look at the successful men in any of the learned professions. How perfectly hideous they are![/quote]

Genius.

[quote="mlok"][quote="aasdfg1234"]what happen if you are not handoosome or pretty?[/quote]
You post on AoPS.  :ninja:[/quote]

 :clap: Don't fuss too much about it because I am probably uglier. Post on AoPS makes everything better. :D

The age-old quote "Beauty is in the eyes of the beholder."

Anyways, I don't care much about my looks. And I don't care much about other people's looks unless I happen to be attracted to them.(I am a straight guy, after all)

implants, pianoforte? uh huh.... :maybe: Well, I'm a guy...i don't think about it, but I doubt high school/middle school girls are getting implants.

[quote="Fabiomr1"]Try using a spherical mirror instead of a planar one

or maybe, don't use Gauss referencial

maybe the $AgNO_{3}$ in the mirror is old. try changing it[/quote]

 :rotfl:  :rotfl:

[quote="ashpotter"]About inpressing a girl with math: I am good enough in math to get fourth o(but same score as second) in a mathcounts contest. Just because of that i have lost half of my popularity. The thing with me is that i like two girls. One lives on the other coast and the other is always in the middle of her click. I can never find her alone to impress her with math anyway[/quote]

haven't u found out yet that it's near impossible to impress a girl with math to the point where she likes u? (trust me....ive had experiance =P)

[quote="trigonometry1.01"]guys guys

I don't think girls are impressed by math.[/quote]

LOL, what a sad fact. Maybe they will be impressed if they like math as well.

[quote="trigonometry1.01"]guys guys

I don't think girls are impressed by math.[/quote]

dude dude what are you serious

DAMMIT

THIS SUCKS

SUCKS

oh well I'm still manly enough to get all the girls

@Miss: is your avatar a real picture of you?

 :wink:

huh how is it mean?

but whoever that's a picture of lemme get her number or something

[quote="Fabiomr1"]hey

don't give up

I got an theory:

1Girl + 1Guy = 1 Couple
1 + 1 = 1
2 = 1

what is wrong. love isn't math, math isnt love and vice-versa. they are distinct things

er... but I still think that spherical mirrors and brand-news $AgNO_{3}$ in no gaussian referencial are better[/quote]

Good theory. But what happens if I change the girl to some new variable.... I don't know... How about Hot Girls? :roll:

To all, I really agree with the thing about trying to impress girls to the extent that they like you with math.... NOT GOING TO HAPPEN!

uhhh if my personal experience applies everywhere
girls like six packs
biceps are secondary

This is the most off topic post i have ever seen...
@ the.Demon.117
THAT IS WHAT I WAS TRYING TO SAY!! you can't get a girl to like you by impressing her with math. The only time girls are impressed at me is when i get another boy back for somthing he did to the girl.

girls are impressed with YOU, dancing banana? :rotfl:  :rotfl:  :rotfl:

[quote="aasdfg1234"]what happen if you are not handoosome or pretty? well, the first thing is that you seem to lose a lot of confident, you're not going to be the most active guy in ur school. how can you get out of that depression?[/quote]

Step 1: Read the Wikipedia article on [hide="Alex Rodriguez"]Not really A-Rod, Ron Jeremy[/hide]
Step 2: Laugh at the picture of [hide="Alex Rodriguez"]Not really A-Rod, Ron Jeremy[/hide]
Step 3: Realize that you look better than he

Step 4: Feel better about yourself

lol, They arn't inpressed with old dancing bannana all the time
:(


search
graph theory
combinatorics open
combinatorics

i want some beautiful graph problems,if you have a problem please send.and if you know where can i find it please tell me.
thank you very much.
 :D

come on boys,sent your nice graph problem here if you want help me and i know you want,so dont waist the time.thank you. :D

****** What do you means by nice problem :
  I suppose you want question rather simply stated , still do you want :
  A) A problem Mostly unsolved  or partially solved ?
  B) How simple in terms of backgound ?

 ****** Yet here is one  ***********
 The Gossip problem:
 - N persons (N >= 4) exchange informations on the phone.
 - At start each person possess it's own information ( all infos are different).
 - Any phonecall binds 2 persons who will exchange all the informations they learned so far.
    that includes past information they had from previous phone calls.
 - The AIM of the game is that everybody  ( N persons) must end with all N informations.
 Q 1 : Show that you can do this in (2.N - 4) phones calls !
 Q 2 : Show that you can do better than this!

 WARNING : To show minimality as in asked in Q2 is hard ( nasty and tricky ) ,
 For a solution see Douglas West graph theory.

 ****** Yet here is another one  ***********
 Looking for BLACK OR WHITE GROUP of size 10 :
 Show that in a graph of 2 millions vertices or more :
 You are guaranteed to find an induced subgraph of size 10 that is either edgeless or complete!
 PERSPECTIVE :
  	- Be sure you known that a subgraph is not necessarily induced.
 	- The title is justified because any graph can be thought as a complete graph (Kn)
 	   whose edges have been colored black (edge is kept) or white (edge is removed).
 	- This is a specific example of the General problem  :
 		If you require a "group" of size p , a graph if big enough (depends on p)
 		will always "possess the group" as an induced subgraph.
 REFERENCE: This is Ramsey theory starting point.

thanks jerjer,if you you have more plz send. :D

HERE IS ONE MORE

Here is an unsolved one name " The second neighboorhood conjecture" 	 of Paul Seymour

Let G be a finite digraph with NO loop and NO "digons" that is no a-->b AND b-->a.
Otherwise said the digraph G has a finite set of vertex V , and for any pair
of different vertices {a,b} either there no edge or one oriented edge in between.
You may interpret the edge as a domination relation, but note that there is
no requirement of transitivity that is we may have :
	A may dominate B who dominates C who itself dominates A .
Define the POWER of someone as its outdegree ( the number of persons it dominates).
Define the INFLUENCE of someone as the number of people he dominates either directly or
with one ( only 1 ! ) indirection ( A-->X-->B : B is "influenced" ) .
Show for any G there is someone whose INFLUENCE is twice or more its POWER.
NOTE : This problem is solved only (not easy) for tournament , that is when G is complete ( more precisely
when the underlying simple graph of G is complete.
Remark: that if someone has power zero his influence is zero also and twice zero
is zero so this person is a good candidate.

Also in the "beautiful, but very very hard and still unsolved" corner:

Let $G$ be a graph so that every vertex is contained in at least three edges.  Prove that $G$ contains a cycle whose length is a power of 2.

Some problems i like:

For every natural $r$ there exists another natural $n$ such that every graph of order at least $n$ contains either a complete graph in $r$ vertices or an empy graph n $r$ vertices as a induced subgraph (ramsey's theorem)

Let G be a connected graph. If G is neither connected nor an odd cycle then $cr(G) \leq d(G)$. where $cr(G)$ denotes the minimum number of colors needed to color $G$ properly and $d(G)$ denotes the maximum degree in $G$.

Every 4-connected planar graph has a hamiltonian cycle.

I've posted some interesting graph problems in the forums, so you can search my posts.


Hi,

i recently came across this problem. Its a Nim game. But this is misere play approach.(player who takes the lest stone(s) loses)

Here is the game description!

There are four stone piles with n1, n2, n3, n4 stones. One is permitted to take any number of stones from one pile for a move. The one who takes the last stone(s) LOOSES.

Ex:
 1 0 0 0 ---------- Second Wins.
 3 2 1 0 ---------- Second wins.
 2 2 1 0 ---------- First Wins.

Question is how can we determine who will win given these four number?
If it is normal play(the player who takes the last stone(s) WINS), the method is very simple!! jus take the XOR of the all the numbers and see if it is equal to zero. if it is then second wins, else first wins.

But in misere play i am stuck with this approach :( . but the method for misere play is also similar.
Please if anyone knows, help me.

[quote="tjhance"]Sorry, but what is "misere play"?  :blush:[/quote]
Possibly he uses this term to denote that it's different from the classic game of Nim

We discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=118422]here[/url] (in the Greek forum) about the "misere" version of the game

Each possible situation of the game can be represented by a quadraple $(n_{1},n_{2},n_{3},n_{4})$

[color=blue][b][u]The classic Nim Game[/u][/b]
We define the sets:
$L = \{(n_{1},n_{2},n_{3},n_{4})\ |\ n_{1}\otimes n_{2}\otimes n_{3}\otimes n_{4}=0\}$
$W = \{(n_{1},n_{2},n_{3},n_{4})\ |\ n_{1}\otimes n_{2}\otimes n_{3}\otimes n_{4}\neq 0\},$ where $\otimes$ denotes the XOR operation

The Nim Game strategy says:
The player who starts playing from an [b]L[/b]-state will definetely loose
The player who starts playing from a [b]W[/b]-state will definetely win

This is because
$L\to W:$ An [b]L[/b]-state can lead only to [b]W[/b]-states

$W\to L,W:$ A [b]W[/b]-state can lead to at least one [b]L[/b]-state (maybe to some [b]W[/b]-states but you won't do that) [/color]

[b][u]The misere Nim Game[/u][/b]

Now we define the set $F = \{(1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)\}$

This is the set of the final situations. Every non-zero pile consists of exactly [b]1[/b] stone

The misere game has two stages:
- At the first stage, we play in the same way with the classic Nim game
- At the second stage we change strategy

The second stage starts when we reach any situation of the final set $F$
Clearly $F\to F$

It's easy to see that the loosing ($F_{L}$) and the winning ($F_{W}$) positions of $F$ are
$F_{L}= \{(1,0,0,0),(1,1,1,0)\}$
$F_{W}= \{(1,1,0,0),(1,1,1,1)\}$

Notice that $F_{W}\subset L$. To avoid this confusion (winning=loosing) we define $L_{0}=L-F_{W}$, and $W_{0}=W-F_{L}$
If you follow the strategy of the classic Nim game, you'll keep your opponent always on $L_{0}$

We should ensure that there is not a move $L_{0}\to F_{L}$

[hide="Indeed"]Suppose that there is a quadraple $X\in L_{0}$ which can lead to some element of $F_{L}$.

Then the XOR operation in $X$ gives zero. But $1\otimes 0\otimes 0\otimes 0 = 1\otimes 1\otimes 1\otimes 0 = 1$

This means that we removed a single stone.
So $X=(1,1,0,0)$ or $X=(1,1,1,1)\Rightarrow X\in F_{W}$, contradiction, because we supposed that $X\in L-F_{W}$
$\hline$[/hide]

The only think that you should have in your mind is that you shouldn't make a move $W_{0}\to F_{W}$

For example, let's check which is the previous $W_{0}$ situation of $(1,1,0,0)$

$(1,1,k,0)$, with $k\geq 2$. Solution:$(1,1,1,0)$
$(1,m,0,0)$, with $m\geq 2$. Solution: $(1,0,0,0)$
$(1,1,1,0)$. Not a $W_{0}$ element

Similarly, the previous situation of $(1,1,1,1)$ could be only $(1,1,1,m)$, where $m\geq 2$ and the solution is $(1,1,1,0)$.

[b][u]Answer[/u]:[/b] If the XOR of all the numbers is equal to zero then second wins, else first wins. Just like the original Nim Game

[color=darkblue][u]Note[/u]:
We assume that both te players know the strategy [/color]

hi Pontios,

Your answer was:
[b]If the XOR of all the numbers is equal to zero then second wins, else first wins. Just like the original Nim Game [/b]

But , please can u explain for the below cases

3 2 1 0  - Second wins
1 0 0 0  - Second wins
1 0 1 1  - Second wins

For the 1st case XOR of all nos is ZERO, for the remaining 2 cases the XOR of all nos is ONE. But still Second wins in all these cases.

Actually I think i have narrowed down the solution!
If the XOR of all the numbers is ZERO then its definitely SECOND WINS.
But if the XOR value is NON ZERO, we have to decide on some basis who will win!!

because for the below cases:
1 1 1 0 ---- XOR VALUE is 1 , but second wins.
2 2 1 0 ---- XOR VALUE is 1,  but First wins.

 :(  :(

[quote="pontios"]I'm sorry, I should clear it

I assumed that we start with large numbers.
I mean that, in the initial quadraple $(N_{1},N_{2},N_{3},N_{4})$ there is at least one $N_{k}\geq 2$

I explain it in my previous post, but I didn't clear it in the conclusion

[b][u]Corrected answer[/u]:[/b]

Let $N_{i}$ be the initial number of stones in the pile-$i$.

If $N_{i}\leq 1, \ \forall i$, then $\{\begin{array}{cc}\text{if XOR}(N_{i})=0 \text{ then winner is 1}\\ \text{if XOR}(N_{i})=1 \text{ then winner is 2}\end{array}$

If $\exists k: N_{k}\geq 2$, then $\{\begin{array}{cc}\text{if XOR}(N_{i})=0 \text{ then winner is 2}\\ \text{if XOR}(N_{i})=1 \text{ then winner is 1}\end{array}$[/quote]

Hi pontios,

Thanks a million!! your solution is perfect!! i am much relieved now!! i'v been breaking my head slightly for 20 days. Thanks again.

By the way i came up with another version of the Nim game called "Stone Game".
And i hav posted that question as a seperate post in this forum itself. Do look into that, and help me if u can!! The solution to this problem is also small and with xor operaions.

Santosh.

When I first faced the game of Nim I found it difficult; I didn't know how to start.
But, when I found a solution in [url=http://www.cut-the-knot.org/nim_theory.shtml]cut-the-knot[/url], I saw that this kind of games is easy, but also excellent!
(the hard part is to find a [u]formula[/u] to express the [b]W[/b] and [b]L[/b] states)

I'll try the [url=http://www.mathlinks.ro/viewtopic.php?t=152824]Stone Game[/url]


guys!!!!Can you believe this!? Virginia was first last year at team... this year they were 33rd!yes 33rd!...and i thought we did bad.....

The inappropriate posts in this thread have been deleted. [b]Do not[/b] talk about a student or his performance on the test if he is not posting in the conversation also. It is an invasion of his privacy.

Please do not make posts like this in the future.


Let A, B ∈ Mat{nxn} (ℝ)

Proof that Det(AB)=Det(A)Det(B)

You can read the proof of this theorem in many books about Linear Agebra. To prove this theorem i realize that you should know about the define of determinant.

Good luck!!!!!!!


search
number theory unsolved
number theory

Find all $a,b,c,d,k\in Z^+$ such that
$a+b+c+d=k\sqrt{abcd}$
Please help me! I need a right and clear solution for it  :(

It was already posted. Search in the solved problems area.

Sorry I could't find it . Can you help me with giving me the link  :D
Thanks

I could only find this topic http://www.mathlinks.ro/Forum/viewtopic.php?t=21090
I remember harazi also posted his solution, but it was actually the same.