So I was doing this devil problem :diablo: , and I seemed to be getting the answer wrong. The problem was
Twelve people, three of whom are in the Mafia and one of whom is a police inspector, randomly sit around a
circular table. What is the probability that the inspector ends up sitting next to at least one of the Mafia?

The website says the answer is 28/55. Could someone please give me a jist of the solution.

Thanks in advance.

It seems to me like it would be easiest to find the probability that the inspector [i]doesn't[/i] sit next to mafia.  If we fix the inspector's seat, then the mafia occupy 3 of the 11 other seats.  But only 9 of these seats are [i]not[/i] next to the inspector.  Thus the probability we want is $ 1 \minus{} \dfrac{\binom{9}{3}}{\binom{11}{3}}$, which is actually 27/55.  So I think the website is wrong.  (The complementary probability is 28/55.)

Most tournaments make occasional errors.  There was even a mistake at ARML last year---admittedly on the second or third tiebreaker question, which ultimately didn't make any difference, if I recall correctly.

Thanks for the explanation Boy Soprano. I seem to have unnecessarily complicated the problem by thinking in terms of circular permutations. As for the ARML, you may be right but I personally don't know.


Hameye Adade Tabiyi mesle $(x,y,z,n)$ ra peida konid be tori ke : $x^{3}+y^{3}+z^{3}=nx^{2}y^{2}z^{2}$

soale ghashangie,bazam ba namosavi hal mishe
[hide]aval az hame tavajoh dashte bashid ke chon moadelamoon nesbat be $x,y,z$ moteghaiere mishe farz kard $x\geq y\geq z$...

pas darim:

$x^{3}+y^{3}+z^{3}\leq 3x^{3}$

$\Rightarrow nx^{2}y^{2}z^{2}\leq 3x^{3}$

$\Rightarrow ny^{2}z^{2}\leq 3x$

$\Rightarrow \frac{ny^{2}z^{2}}{3}\leq x \Rightarrow x^{2}\geq \frac{n^{2}y^{4}z^{4}}{9}$

khob hamchenin midoonim $x^{2}$ samte raste moadelaro ad mikone pas bayad samte chapesho ham ad kone yeni darim $x^{2}\mid y^{3}+z^{3}$

$\Rightarrow x^{2}\leq y^{3}+z^{3}$

$\Rightarrow \frac{n^{2}y^{4}z^{4}}{9}\leq y^{3}+z^{3}$ $(I)$

vali chon $y\geq z$ pas darim $y^{3}+z^{3}\leq 2y^{3}$

pas natije mishe ke:

$\frac{n^{2}y^{4}z^{4}}{9}\leq 2y^{3}\Rightarrow y\leq \frac{18}{n^{2}z^{4}}$

khob hala azin namosavi natije mishe ke $z=1$ chera ke age dashte bashim $z>1$ oonvagh natije mishe ke $y\leq \frac{18}{n^{2}z^{4}}\leq \frac{18}{16n^{2}}$

pas natije mishe $y\leq 1$ amma dashtim $y\geq z$ ke inam tanaghoze...

pas $z=1$...

hala dota halat darim:

------------------------------------------

$i)y=z=1$ ke ba jagozari too soorate masale natije mishe $x^{3}+2=nx^{2}$ pas $x^{2}\mid 2$ yeni $x=1$ pas $3=n$
pas daste javabe $(x,y,z,n)=(1,1,1,3)$ too moadele sedgh mikone...

------------------------------------------

$ii)y>1$ khob hala age $z$ ro tooye namosavie $(I)$ jagozari konim natije mishe:

$\frac{n^{2}y^{4}}{9}\leq y^{3}+1 \Rightarrow n^{2}y^{4}\leq 9y^{3}+9 \Rightarrow n^{2}y\leq 9+\frac{9}{y^{3}}$

khob hala chon $y>1$ pas $y^{3}\geq 8$ pas namosavie akhar be shekle zir mishe:

$n^{2}y\leq 9+1=10$

pas $n^{2}\leq 10$ yeni $n=1,2,3$ pas kafie in $3$taro check konim...  :wink:

[/hide]

[size=150]ordere samt rast ziady bozorgtar az samte chape[/size]

hameye setayi ha mesle $(p,q,r)$ az adade avval ro peyda konid tori ke $p,q,r$ do be do motemayez bashand va :
$p|q+r$ , $q|r+2p$ , $r|p+3q$

ye rahnamayi mikonam bazam az namosavi estefade konin va masale ro dar se halate $p>q,r$ ,$q>p,r$,$r>p,q$ hal konin :)

[quote="sunset"](p,q,r)=(5,3,2)[/quote]
in tahna javabesh nist masalan $(p,q,r)=(2,11,7)$ ham javabe masalast :) . mitooni rahe haleto befresti ??

[quote="saeed123"][quote="BaBaK Ghalebi"]hagh ba toe chon ordere samte rast $n^{6}$e vali male chap $n^{3}$e...
vali inike orderesho baresi konim faghat bemoon mige ke az ye jai be bad namosavi mikhoreo dge javab nadare,yeni javabash mahdoodeo karandareo az ye jai be bad dge javab nadare,pas bemoon mige ke ba idea namosavi masalaro pish bebarim...
hamin na chize dgei,pas bazam bayad hame karai ke too poste ghabli anjam shod ro anjam bedim,gheir az ine?  :maybe:[/quote]
Na mosavy khasi nemikhad zira kafist tarafein ro taghsim bar moraba [b]XYZ[/b] konim oonvaght seta kasre koochektar az 1 darim va halate mahdood ra mitavan hesab kard[/quote]
saeed harfeto ghabool nadaram to migi ke : $\frac{x}{y^{2}z^{2}}\leq 1$ amma in harf dorost nist chon momkene $x>y^{2}z^{2}$. albatte ghabool daram hatman 2 ta az kasra kamtar ya mosavie 1 hastan vali in komaki nemikone :)

age mikhayn mitoonam rahe halle soale nazarie adado bezaram !


Diophantine equation
algebra
binomial theorem
number theory proposed
number theory

Let $\ n$ be a natural number and $\ x=[(2+\sqrt 3)^{n}]$($\ [a]$ denotes the integer part of the real number $\ a$).Prove that the number $\frac{(x-1)(x+3)}{12}$ is a perfect square.

Let
$s=\frac{(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}}{2}$,
$d=\frac{(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}}{2\sqrt{3}}$.
We can show that $s$ and $d$ are integers (by Newton's Binomial Theorem, for instance).

Since $0<(2-\sqrt{3})^{n}<1$,
by definition of $x$, we have $x=2s-1$.

We need to show that there's an integer $z$ with
$(x-1)(x+3)=12z^{2}$, that is $(2s-2)(2s+2)=(2\sqrt{3}z)^{2}$.

We can easily see that $(2s)^{2}-(2\sqrt{3}d)^{2}=4$, so we've found $z=d$, an integer indeed QED.

[i][color=red]Theorem[/color][/i]
The equation $\ x^{2}-dy^{2}=1$, $\ x,y,d\in\mathbb{R}$, is named [b]Pell's equation[/b]and it has the property having a infinity of solutions in $\mathbb{R}$.
 [color=blue]Application[/color]
  Prove that the equation $\ x^{2}+2y^{3}+3z^{4}=t^{5}$ has an infinity of solutions in $\mathbb{N}$(Marius Ghergu, Slatina)

lordWings:

While d is always positive for n > 1,

z can be either positive or negative d to be an integer.

z = d or z = -d.

@pleaseread: so what¿

@moderators: please move this to number theory.

There is nobody here named "pleaseread."

If you're commenting on the last post, don't do it with "So what?"  It doesn't work that way.

All you have to do is follow the last post (before yours) and what it referenced to the other poster.  That's all you'll need.
[Hint:  The other poster mentioned only one value for z.]

[quote="perfect_radio"][quote="stancioiu sorin"][i][color=red]Theorem[/color][/i]
The equation $\ x^{2}-dy^{2}=1$, $\ x,y,d\in\mathbb{R}$, is named [b]Pell's equation[/b]and it has the property having a infinity of solutions in $\mathbb{R}$.[/quote]

Err.... I think it is like this:

[i][color=red]Theorem[/color][/i]
The equation $x^{2}-d y^{2}= 1$, where $d \in \mathbb N$, $d$ not a perfect square, is named [b]Pell's equation[/b] and has an infinity of solutions $(x,y) \in \mathbb Z^{2}$.[/quote]
 :oops: yes, I didn't specified some things...

[quote="perfect_radio"][quote="stancioiu sorin"][i][color=red]Theorem[/color][/i]
The equation $\ x^{2}-dy^{2}=1$, $\ x,y,d\in\mathbb{R}$, is named [b]Pell's equation[/b]and it has the property having a infinity of solutions in $\mathbb{R}$.[/quote]

Err.... I think it is like this:

[i][color=red]Theorem[/color][/i]
The equation $x^{2}-d y^{2}= 1$, where $d \in \mathbb N$, $d$ not a perfect square, is named [b]Pell's equation[/b] and has an infinity of solutions $(x,y) \in \mathbb Z^{2}$.[/quote]
I see... Thanks. But here's another question, the [b]Pell equation[/b] has the property having a infinity of solutions for $(x,y)$, but it does not mean that for any given natural number $x, d$, exists a natural number $y$ which satisfies the equation. So why could Solar Plexsus say in his answer to the problem that $m$ is a solution of the equation $m^{2}-3 n^{2}= 1$ ? Sorry I just can't understand it.

I hope this's right.(I change $m,n$ into $A,B$)
$A^{2}-3B^{2}= 1$ Pell's solution is in this form :
Let $(A_{1},B_{1})$ is the fundamental solution which is $(2,1)$, and $(A_{k},B_{k})$ are the other solutions where $A_{k}< A_{k+1}, B_{k}< B_{k+1}$ for $k \in \mathbb N$

Then, $A_{k}+\sqrt3B_{k}= (A_{1}+\sqrt3B_{1})^{k}= (2+\sqrt3)^{k}$
It's well known that $A_{k}-\sqrt3B_{k}= (A_{1}-\sqrt3B_{1})^{k}= (2-\sqrt3)^{k}$

Hence, $A_{k}= \frac{(2+\sqrt3)^{k}+(2-\sqrt3)^{k})}{2}$

In Solar Plexsus' answer
$A_{n}= \frac{(2+\sqrt3)^{n}+(2-\sqrt3)^{n})}{2}$
And $x = [(2+\sqrt3)^{n}]$
From $(2-\sqrt3)^{n}< 1$ and $(2+\sqrt3)^{n}+(2-\sqrt3)^{n}$ is integer, we get $x+1= (2+\sqrt3)^{n}+(2-\sqrt3)^{n}= 2A_{n}$ and the result follows...

And I think this is NT not Algebra...  :maybe:

this one is easy, just take $x_{n}=[(2+\sqrt{3})^{n}]+1$ then we can prove that $y_{n+2}=4y_{n+1}-y_{n}$ the rest follows

I have tried and reach that equation but I don't know how to go on after that


I know how to use the internet and microsoft word and that's it.  I feel like an idiot when people I know talk about computers and computer programming and computer science and all that... AND I have to take 2 computer classes in the fall!  Can someone explain to me what exactly are all the basics like computer programming languages and stuff like that?  Sorry this is so vague, but I don't know what I am talking about.  I don't want any info about actually how to implement the stuff...just what it is.  Thanks and please forgive my ignorance.

A computer program is a set of instructions for a computer to complete a specific task.

A computer program is written in one or more computer languages. A computer language is one of:

1) a set of instructions that a computer can understand. For example: MOV A 1 sets the value of variable A to 1. (Assembly)

This kind of computer language is called a low-level language. Its grammar and semantics are harder for humans to understand, but easier for a computer to. Example: the Assembly language (which you will not learn in detail in introductory computer science.)

2) a set of phrases that, through an interpreter or a compiler, can be translated into a low-level language. For example: int A = 1; sets the value of variable A to 1. (C++) Think of the interpreter or the compiler as a translator between you and the computer.

This kind of computer language is called a high-level language. Its grammar and semantics are easier for humans to discern and understand. It is easier to program in a high-level language.

The following two sets of codes do the same thing. They all compute the sum of all natural numbers from 1 to 100.

C++:

int i = 1;
int s = 0;
while (i <= 100) {
  s = s + i;
  i ++;
}

I don't exactly know how to write Assembly (except ACSL-asm which isn't really assembly), so I'll just outline the number of lines needed in assembly:

put value 1 into I
put value 0 into S
put value 1 into T
put value 100 into E
Flag1: move the value of I into AX
move the value of S into BX
move the value of T into CX
sum the values in AX and BX and put it in BX
sum the values in AX and CX and put it in AX
move the value of AX into I
move the value of BX into S
move the value of E into CX
subtract the value of CX from the value of AX and put it in AX
if the value in AX is less than zero, jump to Flag2
jump to Flag1
Flag2: the rest of the program

As you can see, C++ is a lot easier to understand (and write) than Assembly.

A program written in a high-level programming language roughly consists of variables, operators, keywords, and functions. Consider the following statement:

int [b]a[/b] = 1;

This line initialises a block of memory space, calls it [b]a[/b], and defines it to be an [b]int[/b]eger. It also assigns the numerical value 1 to this block of memory space. In a C++ program, each line must end with a semicolon.

Consider the following statement:

b = max (c, d);

This line compares the value in variables c and d, and returns the bigger one and puts it in b.

I'll stop here and let you absorb all this; don't hesitate to reply with any questions. I'm not that good a teacher.  :P

[quote="JS1527"]So wait, what are the uses of computer languages?  Everything we have on the computer, like AIM and Microsoft Office and all that, were all written in computer languages?

What are the different types of each type (high level and lower level)?

I understood everything you said.  And I do know how to make programs for the TI, so I guess that would be a "high level" language (well there is TI assembly too - for games, but I don't know that).

What stuff is there other than computer languages (to do with computers that I am not familar with...)?[/quote]

1. Yup.

2. High level: BASIC, Visual Basic, C, C++, Java, Fortran, Pascal, Delphi, etc...

Low level: Assembly languages differ from processor to processor.

3. Yes, TI-Basic would be considered a high-level language. TI-Asm, on the other hand, would be a low-level language.

4. Every piece of computer software is written by one or more computer languages.

High level languages can be compiled by a compiler program into machine code so they can be processed by the computer. It is almost like a stream of inputs and outputs.

Example of JAVA:

final int MAX = 10;
for(int count = 0; count < MAX; count++)
    System.out.println(count+1);

Notice since JAVA is a high level programming language, the grammar must be perfect for the program to compile. I don't have a driver program or the class, so this wouldn't run, but what this particular program actually does is print out the integers from 1 to 10 each on a separate line. JAVA is an object oriented programming language, which means objects are the fundamental pieces of the language structure. C++ and other languages don't enforce objects, so the structure isn't as solid and program designs can get confusing.

The forms of programming languages machine, assemply, high-level, and fourth-generation languages.

Overall, programming is really fun.

[quote="JS1527"]Can someone explain to me what exactly are all the basics like computer programming languages and stuff like that?[/quote]

There are a lot of popular books to explain computer science concepts to educated members of the general public such as you. You can find and read some of these before your classes start, and that should help give you a conceptual framework into which to put the facts you learn in class.

[url=http://www.amazon.com/exec/obidos/tg/detail/-/0735611319/learninfreed]Code: The Hidden Language of Computer Hardware and Software[/url], by Charles Petzold (Microsoft Press, 2000).

[url=http://www.amazon.com/exec/obidos/tg/detail/-/046502596X/learninfreed]Pattern on the Stone[/url], by Daniel Hillis (Perseus Books, 1999).

[url=http://www.amazon.com/exec/obidos/tg/detail/-/1893115232/learninfreed]How Computer Programming Works[/url], by Dan Appleman (APRS, 2000).

[url=http://www.amazon.com/exec/obidos/tg/detail/-/0789730332/learninfreed]How Computers Work[/url], by Ron White and Timothy Edward Downs (Que, 7th ed. 2003).

[url=http://www.amazon.com/exec/obidos/tg/detail/-/0764535242/learninfreed]Computers Simplified[/url], by Ruth Maran (Wiley, 5th ed., 2000).

There are other books like this that you can find by browsing around Amazon for the names of related books.

Hope this helps!

[quote="white_horse_king88"]High level languages can be compiled by a compiler program into machine code so they can be processed by the computer. It is almost like a stream of inputs and outputs.[/quote]
A technical note: there are many languages (such as all variants of BASIC) which are interpretted, rather than compiled, and I believe there are some languages which simply [i]cannot[/i] be compiled, only interpretted. (That last statement may seem far fetched, but all it would require was a language with a radically different flow structure from what is normally found in a program - such as extensively and dynamically modifying the code based on user input)

The difference between interpretting and compiling is fairly simple, really. I'll use an analogy: there's some book (a computer program) written in ancient greek (a programming language). If I hire someone who interprets it, they'll just go through one word or sentence at a time and tell me what it means - but I still won't be able to read the thing. Compiling is equivalent to translating it into English (or for the computer, into machine code (machine code and ASM are equivalent - it's just a simple substitution for each command)).

[quote]The forms of programming languages machine, assemply, high-level, and fourth-generation languages.
[/quote]

Can you tell me what these are?

Thanks for the suggestions, tokenadult.  I need to pay off my 30 dollar library debt, but I'll try to get a simple book out of the library.

Okay, so you guys know all these programming languages and all this stuff about computers.  What exactly do you do with this knowledge?

[quote="takuanitromars36"]Most programs are written in C and C++, if you are using Microsoft Windows. In fact, Windows is written in C. [/quote]

Seriously!?!? OK, that explains a lot.  ;)

I'd recommend learning Java and C as your first two programming languages. Many of the commands between the two are similar, so it's not all that hard.

Java is also helppful if you want to pass the AP Computer Science exam.

[quote="takuanitromars36"]
You use the knowledge to write programs, so that users can be satisfied, and you can gain money.  :D[/quote]

Yea okay, but what sort of programs do you guys write?

Everything that you've used, are using, and will use on a computer. Me myself, I've worked on some games (for my C++ and Java final projects), and I've been coding a utility application in VB. I've also written programs for ACSL (a computing competition), which are more theoretical than the programs that you use every day.

Thanks guys.

What exactly does a computer scientist learn about in college?  What about a computer programmer?  Is computer programming useful in other math/science/engineering stuff?  How?

[quote="JS1527"]Is computer programming useful in other math/science/engineering stuff?  How?[/quote]

Tedious calculations you could do easily by writing a program.  For example, the sum of all the primes less than 2004 is 279053.

[quote name="HamstPan38825" url="/community/p19216365"]
It's interesting looking at 16-year-old topics thinking about how the world has evolved.
[/quote]

And it's disruptive to bump these topics without any contributions to the discussion.


$ x,y\ge\frac{1}{2}$
the first equation is equivalent to
$ (x\minus{}y)(\sqrt{x\plus{}3}\plus{}\sqrt{y\plus{}3}\minus{}\sqrt{2x\minus{}1}\minus{}\sqrt{2y\minus{}1})\equal{}0$
with x=y substitute to the second equation we get
$ \sqrt{2x\minus{}1}\plus{}\sqrt{x\plus{}3}\equal{}\frac{3x\plus{}7}{2}$
this equation is easy to solve notice that x=1satisfy the equation we can reduce it to a cubic equation by squaring 2 sides of the equation
with $ \sqrt{x\plus{}3}\plus{}\sqrt{y\plus{}3}\minus{}\sqrt{2x\minus{}1}\minus{}\sqrt{2y\minus{}1}\equal{}0$ (3)
from the system and (3) we have
$ 3(\sqrt{x\plus{}3}\plus{}\sqrt{y\plus{}3})\equal{}3x\plus{}7$ and from the first and (3)we have
$ 2\sqrt{2x\minus{}1}\plus{}\sqrt{y\plus{}3}\equal{}3\sqrt{x\plus{}3}$
which is equivalent to
$ 6(2\sqrt{x\plus{}3}\minus{}\sqrt{2x\minus{}1})\equal{}3x\plus{}7$
hence it is more simple


Let n is a possitive integer , a is a real number , a doesn't equal to 0 . Find f:R--->R satisfying :
  $ f(x-f(y)) = f(x+y^{2n}) + f(f(y)+y^{2n})+a $

Something tells me that I have solved this forum in the past. Grobber, do you remember anything very similar?

No, but I have a very bad memory, so you shouldn't be discouraged by my failure to remember it :D.

Oh, boy, Alzheimer is coming to take me!

You should replace VVT by Mr.Th, Nttu. We must pay some respect on our teacher.
" TT loves TT"  :D  :D  :D  :D


The Sum of 40 consecutive integers is 100. What is the samllest of these integers.

Let the smallest integer be $ x$. Then you can write $ x \plus{} (x \plus{} 1) \plus{} (x \plus{} 2) \plus{} \dots \plus{} (x \plus{} 39) \equal{} 100$. This simplifies to $ 40x \plus{} 780 \equal{} 100 \implies x \equal{} \boxed{\minus{}17}$.

Or: average is 100/40=2.5, and we must subtract 19.5 since the first number and the 41/2 = 20.5th number are 19.5 apart. Thus the answer is 2.5-19.5=[b]-17[/b]


Welcome to my riddle tourney. due to my dumbness  :rotfl:  i will have "helpers" to make riddles
to become a helper pm me a riddle and an answer the 5 riddles i like the best will be my "helpers" and all riddles will be posted in first round!!!

pm me your answers for first round of elimination and after that i will choose a helper to post a riddle. the first to post the correct ans wins a point (added to the points of the first round) at the end of 10 rounds whoever has the most points wins!!!!!!!  :first:

i no this is confusing if u have any qs just post and ill answer them. FOR THOSE OF U WHO WANT TO POST RIDDLES YOUR PM IS EXCPECTED THIS FRIDAY AT MIDNIGHT FORUM TIME!!!!!! JOIN NOW!!!!!!!!  :D

This might belong in the 'Puzzles & Brainteasers' section.

I think this would be much better if the rules were altered so that thee was a deadline, rather than the first to answer. Because it might just be pure luck that one person is there when the problems are posted, e.g. some one was awake, others were sleeping.

Anyway, I just think it would be more fair to the people that don't sit at their computer and hit "refresh" until the problems are posted.

wher are the puzzles n brain teaser section


geometry
geometric transformation
reflection
inequalities
inequalities proposed

One more problem proposed by me for Mathematical Reflections, which I like very much.

Let $a,b,c$ be positive numbers, prove the following inequality

 $(a^{3}+b^{3}+c^{3})^{2}\geq (a^{4}+b^{4}+c^{4})(ab+bc+ac)$

It is also interesting for me to find the maximum value of $n$ such that the following inequlaity is always true for $a,b,c \geq 0$:

$ \left(\frac{a^{n}+b^{n}+c^{n}}{3}\right)^{2}\geq \sqrt[3]{abc}\left(\frac{a^{2n-1}+b^{2n-1}+c^{2n-1}}{3}\right)$

I think $n \in (\frac{3}{2},2)$

According to EV-Theorem, it suffices to set $b=c=1$.

Ok, I cannot disagree with EV-Theorem, but I suggest to try to solve it another way.


Spurs vs. Suns series

Discuss the series here

I think its going to be the Spurs in Game 7

Note:
Spurs won today 111-106 on Phoenix's court
Spurs lead the series 1-0

[quote="ccy"]blah blah blah blah blah (random incoherent stuff that no one on this forum cares about)
Happy cheeseday to you. :)[/quote]

Happy cheeseday to you too.
Spurs in 6

Why are you comparing elson to stoudamire?  elson just starts but he doenst really play that much (like 10 mins), same goes with jjones for the suns.  Tis really diaw and finley for their respective teams that you should be comparing.

Someone in 7.

Both teams really good.

Suns might actually be deeper... and more talented.

Spurs are amazing every year though.

I think it'll come down to the big man. So Duncan will win it for the Spurs, probably in 6 or 7.

[quote="Aznwithabrain"]Breakdown:

Parker vs. Nash
Steve Nash

Ginobli vs. Bell
Manu Ginobli

Bowen vs. Jones
Bruce Bowen

Duncan vs. Marion
Tim Duncan

Elson vs. Stoudemire
Amare Stoudemire

San Antonio's Bench vs. Phoenix's Bench
Phoenix's Bench

Copied and pasted this from the "NBA Playoffs" topic
Post your own analysis of the series please[/quote]

What is this again?  Comparing players instead of "teams"..
 :D
Nash will finally win, Bowen is overrated defender, Horry wont make anymore big shots, Marion is going to be the best player.

Mhmm.... I'll call the Suns in 7. What about the Warriors vs. Jazz?

Id take the warriors in 6.

[quote="usaha"][quote="Aznwithabrain"]Ok I'll try my hand at comparing teams
but the reason i compare players is that a basketball game CAN be taken over by one player
Players like Michael Jordan, Kobe Bryant, Tracy McGrady, Oscar Robertson, Wilt Chamberlain, Magic Johnson, Kareem Abdul-Jabbar, Larry Bird, and Dr.J have proven that to us
but here it goes:

Phoenix is a fast paced fast break style of basketball
They can also slow down the pace and go to a low post player like Amare Stoudamire late in the game with a lead
They can go to three point shooters like Raja Bell, James Jones, and Shawn Marion
They have probably the best passing point guard in the league. He is definitely a playmaker.
6th Man of the Year Leandro Barbosa brings energy and speed to the Suns
Role-players like Boris Diaw, James Jones, and Kurt Thomas will be important
Someone you might want to watch for is Jalen Rose
I know Jalen Rose has not fit well in the Phoenix offense
and
he plays absolutely no defense
but
who knows what a veteran could do to help his team in the playoffs
though I highly doubt Rose will make an impact
The team chemistry is definitely there

San Antonio is more of a half-court style of basketball
Tony Parker runs the offense like a playmaker
Michael Finley/Manu Ginobli provide the team with nice outside offensive weapons
Bruce Bowen has a deadly three pointer
Bruce Bowen is NOT  a overrated defender. Kobe Bryant has praised Bowen as being the best outside defender in the league. Most players don't like playing Bowen. He maybe considered overrated because he plays a bit dirty.
Their inside game is pretty good with Tim Duncan
Fred Elson and Oberto are make great hustle plays
The team chemistry is also there[/quote]

You seem to know alot about basketball?  Do you jsut go to ESPN or something, or can you actually play?[/quote]

I can play point guard pretty well
I do watch Sportscenter like everyday :rotfl:
so yeah
both

[quote="usaha"]Lol asians (excpet yao) can only play point guard because there so tiny :rotfl:[/quote]

Thats somewhat true I guess :rotfl:
but
Sun Yue is coming out of China and entering the draft
He's 6-9 and plays point guard
so who knows
but yeah i guess thats kinda true :rotfl:

[quote="ccy"][quote="Aznwithabrain"][quote="ccy"]Seriously, we don't need another stupid sports thread.  Sports are like the devil's cheese with which he eats bagels (cream cheese, that is).  Secondly, since sports are obviously wastes of time and money, sports suck just like you would suck on a lollipop, therefore you suck, so stop wasting valuable forum space on lollipops that make everyone sad.  Thirdly, basketball is for basket weavers, none of which we have on AoPS, so your topic is both irrelevant to the forum and stupid, so you should take your baskets to the market, or better yet, the compost pile, assuming the material you made the baskets with is organic, or if it is not, to your nearest recycling station.  Last but not least, basketweaving is neither a game, nor fun, and although is made frequently in factories, does not deserve respect in this atmosphere of fun and games, so I suggest you take your materials and supplies elsewhere.

Happy cheeseday to you. :)[/quote]

Uh no
Your opinion is plain spam
We've had 4 sports thread so far by my count
YOU take your opinions elsewhere[/quote]
[img]http://www.roflcat.com/images/cats/emo.jpg[/img][/quote]

hilarious  :rotfl:

cmon lets actually have people that "play" basketball discuss this.

[quote="ra5249"]duncan>stoud
ginob>marion
parker<nash
bowen>bell
finley/horry about equal to barbosa/diaw in clutch time (i.e. this series)
SPURS IN 6[/quote]

no marion better than ginobili, bell is a lot better on offense than bowen, adn they both were all nba 1st team defense, but I gotta admit horry is way more experiecne especially in crunch time but he isnt goign to play and barbosa is reguarly better than finley and horry together so suns in 6 or 7, and duncan=foul trouble.

im talking potentially, ginobilli > marion
he can cut to the hoop
he can make clutch shots
bowen > bell in defense (bell barely made it to first team)
bell > bowen in offense, but when both teams already have pwnage offense in their stars, defense is what counts
so spurs in 6/7
spurs are deeper, more experienced
barbosa cant play in crunch time, as u saw in game 1
duncan isnt in as much foul trouble as stoud

btw, spurs are up 2-1. they still have two home games. :P
thats all i have to say phoenix fans.
oh, and 1 more thing, spurs CAN win on the road (see game 1), so there.
spurs in 6 or 7

as far as dominant players, you didnt mention shaq.  he took over games.  even if he did suck at freethrows.

[quote="Aznwithabrain"]
They can also slow down the pace and go to a low post player like Amare Stoudamire late in the game with a lead
[/quote]

when was the last time they slowed the game down because they were winning?
amare's post game is actually not that great.
he relies on the offense to get him buckets.  dunks, freethrows, open midrange jumpers, etc.  I mean, not to take anything away from him, but it's not like you can just slow it down, and dump it into him like you can (err... could) with shaq.
or at least the suns dont do it.

[quote="Aznwithabrain"]
Michael Finley/Manu Ginobli provide the team with nice outside offensive weapons
Bruce Bowen has a deadly three pointer
[/quote]

i wouldnt say bowen is quite at the "deadly" status.  he has an okay three point shot, and good midrange jumper, and is useful because when duncan turns and has nowhere to go, he likes to chuck it to bowen in his favorite spot - the left corner
brent barry?  ROBERT HORRY = INSTANT WIN!?!?
oh wait, he got suspended.

[quote]
Bruce Bowen is NOT  a overrated defender. Kobe Bryant has praised Bowen as being the best outside defender in the league. Most players don't like playing Bowen. He maybe considered overrated because he plays a bit dirty.
Their inside game is pretty good with Tim Duncan
Fred Elson and Oberto are make great hustle plays
[/quote]

bowen is not overrated - agreed
pretty good?  are you trying to make tim duncan sound bad?
Do they really call that guy fred?  don't all of the spurs make great hustle plays?

and yeah i used to play basketball... now i just watch lol.

[quote="bloodysloth"]
pretty good?  are you trying to make tim duncan sound bad?
[/quote]

 :rotfl:
haha no
I just couldn't think of a better word without sounding redundant
Tim Duncan is a legend.

*crap*
spurs down SO friggin bad in the first quarter -_-


Given $ 2n-1 $ integers , prove that one can chose exactly $ n $ of them whose sum is divisible by $ n $

This is Erdös-Ginzburg-Ziv theorem, and it appears many times here.
Have alook at :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=nullstellensatz&t=19496#

Pierre.


I already competed for as long as I can.

I'm still posting here because I'm in middle school.

I have competed for three years. Last year (or rather this year) was my last year, and my first and last year at Nationals.

[quote="isabella2296"]Last year was my first year, and this will be my second(I'm in 7th grade).[/quote]

Me too. I have only competed once, and that was when I was a sixth grader.

[quote="ajain"][quote="gauss1181"][quote="ernie"]As I am in 8th grade, this will be my final year.[/quote]
Same here! :D[/quote]

Same Here! :D[/quote]
Hey, you copied my post! =P (except for a capital H) :rotfl:

And why would someone not want to compete (or can't)?

I've been with MC since 6th grade.

I'm too old now though  :yoda:

[quote="AIME15"]This will be my first year, in 7th grade.  My school didn't have mathcounts last yeer[/quote]

My school still doesn't have it and I'm in 7th grade. Is there a way to get individually?

[quote="gauss1181"][/quote][quote="ajain"][quote="gauss1181"][quote="ernie"]As I am in 8th grade, this will be my final year.[/quote]
Same here! :D[/quote]

Same Here! :D[/quote][quote="gauss1181"]
Hey, you copied my post! =P (except for a capital H) :rotfl:

And why would someone not want to compete (or can't)?[/quote]

Well maybe they weren't one of the eight best in their school, or their school doesn't do it, or they just didn't want to compete.

[quote="FantasyLover"]this is my first and last year  :(

blah i didnt even know what mathcounts was last year  :rotfl:[/quote]
awww....

But I competed in both 6th and 7th grade. In 6th grade I only got 6th in chapter, and in 7th grade i took 4th chapter and 6th state. I guess I phailed in both years... :(

Now I'm only having one more shot at Nationals :| . And I'm in 8th grade, competing in Illinois.

Last year I did not make it on the team to compete (7th grade). So this is my first and last year to compete. There is no Mathcounts until middle school (7th grade) here so I had no chance in 6th. Which was lame.

i suck. i made like 20th or something chapter in 5th. Now, in 6th, i doubt i'll do any better :( i just can't do any of those tests fast :mad:

I thought you can't do mathcounts in 5th grade?

[quote="gauss1181"]I highly doubt that an AoPSer would be worse than the 8th best in his/her school.
If that person was, I'd feel bad for him. :([/quote]

You never know. Would if there was a sixth grader at a school who was really good at math, but the team for their school was composed of a majority of very intelligent eighth graders, a few seventh graders, and maybe one sixth grader that is smarter than that other sixth grader. It could happen at really advanced schoools.

[quote="gauss1181"]I highly doubt that an AoPSer would be worse than the 8th best in his/her school.
If that person was, I'd feel bad for him. :([/quote]

Jason Li from Lexington was 13th in his school for the school round (but he still made the team because of other trials), and he still made nationals :D

[quote="gauss1181"][quote="Quickster94"]I officially took the test three yrs (school-chapter-state-nats) when I was in 5th grade, me and 2 of my friends unofficially competed at chapter, according to other ppl, I like made the cd round or something, but couldn't do it (my chapter sucks)[/quote]
How can u do mathcounts in 5th grade if it's limited to 6th-8th grade? :maybe:[/quote]

THBAPSO gauss1181, TY.

he said unofficially competed.

I think that there was a girl from Nevada that took the test unofficially in fifth grade, and then competed in 6th, 7th, and 8th grade, and I think she ranked 13th at Nat's.


[color=green][size=167]Mock ARML Individual III for 2008 [/size][/color]

Written by Alex Anderson (Altheman) [Thanks to everybody who has given me feedback about the test]

Attached is the .pdf with the problems. The rules are written on the cover sheet. Everybody pretty much knows the rules...no calcluators, etc. The time limit is 10 minutes per round. I found this website helpful: http://www.online-stopwatch.com/

Submit answers to me, Altheman, through a private message. I will be accepting answers for the next week or so.

[b]A short description of how you did the problem is required.[/b] I want to be able to gauge the types of solutions that people are giving; that helps me understand how to better write the tests. Furthermore, I like to discuss problems with other people.

[b]Assume that the problems are correct before you send a complaint. While I have been quite careful to make sure that everything is correct, mistakes can happen. Send all complaints to me by using an PM. It causes confusion when somebody posts a complaint in the actual thread. [/b]

Lastly, I hope you enjoy the test. I loved writing it. Good luck and have fun!

--Alex

all problems have been posted in separate threads in HSM  , I just created the related post collectionw

[url=https://artofproblemsolving.com/community/c3731805_2008_mock_arml_3_individual]here[/url]

enjoy  / start solving

more USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more ARML mocks [url=https://artofproblemsolving.com/community/c2442143_arml_mocks]here[/url],
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


find all solutions $(a, b)$

where $a^2+b^2=2005$

$a, b \in {Z+}$

What is the largest number of different squares that when summed give $2005$?

[quote="shadysaysurspammed"]$(a^2+b^2)(c^2+d^2)=(ac \pm bd)^2 + (ad \mp bc)^2$

In the above,why do we write $\mp$ and not $\pm$ in the second bracket.[/quote]

This is meant to imply that when the $+$ sign is taken in the first square, the $-$ sign must be taken in the second, and vice versa.

[quote="chess64"][quote="puuhikki"]An another way is to factorize the expression in $\mathbb{Z}[i]$ and read the decomposition from the factorization.[/quote]

What?[/quote]

Instead of considering the expression as merely an integer, he suggests to consider it as a complex number, and factor it as a complex number instead, and to deduce some facts from the factorization.  This works because

$(a + bi)(a - bi) = a^2 + b^2$

:)

[quote="ArcticMonkey"][hide="Solution"]
We will use formula $(a^2+b^2)(c^2+d^2)=(ac \pm bd)^2 + (ad \mp bc)^2$.

$2005=401 \cdot 5=(20^2+1^2)(2^2+1^2)=(40 \pm 1)^2+(20 \mp 2)^2$

$(a,b) \in \{ (41;18), (18;41), (39;22), (22;39) \}$
[/hide][/quote]
I think this is not a full proof.
Because you're asked to find all solutions.
Imdeed, here this method finds all of them. But the mistake is that you can't be sure these are ALL the solutions (there may be other).
So after finding (41;18), (18;41), (39;22), (22;39), you have to prove that they are the only solutions.

[quote="krassi_holmz"]
So after finding (41;18), (18;41), (39;22), (22;39), you have to prove that they are the only solutions.[/quote]

That's true.  But it's easy;

$2005 = (2 + i)(2 - i)(20 + i)(20 - i) = (a + bi)(a - bi)$

:)

See the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76610]Number Theory Formulary[/url], section "On sums of squares", for a formula for the number of representations of $n$ as sum of two squares:

A positive integer $n$ can be represented as sum of two perfect squares iff all prime factors $p \equiv 3 \mod 4$ of $n$ occur an even number of times in the factorisation of $n$. $n$ can be written as sum of squares $\neq 0$ iff the previous condition holds and it has at least one prime factor $\equiv 1 \mod 4$ or $v_2(n)$ is odd.
There are exactly
\[ \mathrm{r}_2 (n) = 4 \cdot \sum\limits_{\substack{ d \in \mathbb{N} \\ d|n \\ d \equiv 1 \mod 2}} (-1)^{\frac{d-1}{2}} = 4 \cdot \prod_{\substack{p \in \mathbb{P} \\ p\equiv 1 \mod 4}} (v_p(n)+1)  \]
different solutions $(a,b) \in \mathbb{Z}^2$ to $n= a^2+b^2$.


Solve for XY+1=Z where X, Y, Z are prime numbers.

First we see that either  X or Y must be 2 . Or else if both are odd prime , then XY+1 will be even and Z=2 which gives us XY=1 .Impposible .

Suppose X=Y=2 , then Z=5
If X=2 and Y=odd prime we have 2Y+1 = Z . But there are infinity many prime satisfy Y, Z .....

The smallest values are X=2, Y=3, and Z=7.
There are unlimited bigger values.


Yann and Camille go to a restaurant.  If there are 10 items on the menu, and each orders one dish, how many different combinations of meals can Yann and Camille order (note that they are allowed to both order the same dish, and that two combinations are considered different if either Yann or Camille orders a different meal)?

If my interpretation is correct...
Yann can order 10 meals, and Camille can also order 10 meals.
Thus the total number of possible meals is $ 10 * 10 \equal{} \boxed{100}$


How do I get the result found in the picture attached?

I can get to something close, but not quite the same:

$\frac{\left( -1 \right) ^{n-1}n!}{1\cdot 3 \cdot 5 \cdot \dots \cdot \left( 2n -1 \right) }$

As you can see, I can't get the dots to be slightly different in size.

Thanks

$\frac{(-1)^{n-1} n!}{1\bullet 3\bullet 5\bullet \cdots\bullet (2n-1)}$ is close. Why do you want that though? Doesn't $\frac{(-1)^{n-1}n!}{1\cdot 3\cdot 5\cdots(2n-1)}$ (with only three dots) look much better?

Well, it's not exactly what I need. The dots should be a little smaller than that.

The reason why I wish to obtain this final result is that it was published that way in a book I have. Thanks, anyway.  :)

How about

$\frac{(-1)^{(n-1)} \cdot n!}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}$

Use the \boldsymbol command under the amsmath package:

\[
\frac{(-1)^{n-1}n!}{1\boldsymbol{\cdot}3\boldsymbol{\cdot}5\boldsymbol{\cdot}\cdots\boldsymbol{\cdot}(2n-1)}
\]


algebra
polynomial
complex numbers
greatest common divisor

An extension of http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169026

I'd like to know more about them.
So in wikipedia or mathworld, they had really heavy notations that I don't understand.

All I picked up was:
$ z^1 \minus{} 1 \equal{} z \minus{} 1$
$ z^2 \minus{} 1 \equal{} (z \minus{} 1)(z \plus{} 1)$
$ z^3 \minus{} 1 \equal{} (z \minus{} 1)(z^2 \plus{} z \plus{} 1)$
$ z^5 \minus{} 1 \equal{} (z \minus{} 1)(z^4 \plus{} z^3 \plus{} z^2 \plus{} z \plus{} 1)$
$ z^7 \minus{} 1 \equal{} (z \minus{} 1)(z^6 \plus{} z^5 \plus{} z^4 \plus{} z^3 \plus{} z^2 \plus{} z \plus{} 1)$

and then they talked about the first few cyclotomic polynomials...

Can someone please explain what this is all about?

I probably should've warned you that cyclotomic polynomials are a fairly advanced concept.  :oops:

You know that if $ d | n$ then $ x^d - 1 | x^n - 1$.  When $ n$ doesn't have many factors (especially if $ n$ is prime), then $ x^n - 1$ doesn't, either.  Similarly, when $ n$ has many factors then so does $ x^n - 1$.  We can encapsulate this statement in a well-known lemma.

[b]Lemma:[/b]  $ \gcd(x^a - 1, x^b - 1) = x^{gcd(a, b)} - 1$

Cyclotomic polynomials arise when we try to answer the question of how we can [b]completely factor[/b] $ x^n - 1$ into polynomials with rational coefficients.  When a polynomial with rational coefficients cannot be factored further, it is said to be [i]irreducible[/i] over the rationals.  It turns out that there is a set of polynomials $ \Phi_d(x)$, known as the [i]cyclotomic polynomials[/i], with the property that

$ x^n - 1 = \prod_{d | n} \Phi_d(x)$.

[b]Example 1:[/b]  \begin{eqnarray*}
x^6 - 1 &=& (x^3 - 1)(x^3 + 1) \\
 &=& (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1) \\
 &=& \Phi_1(x) \Phi_2(x) \Phi_3(x) \Phi_6(x)
\end{eqnarray*}

[b]Example 2:[/b]  \begin{eqnarray*}
x^8 - 1 &=& (x^4 - 1)(x^4 + 1) \\
 &=& (x^2 - 1)(x^2 + 1)(x^4 + 1) \\
 &=& (x - 1)(x + 1)(x^2 + 1)(x^4 + 1) \\
 &=& \Phi_1(x) \Phi_2(x) \Phi_4(x) \Phi_8(x)
\end{eqnarray*}

Moreover, each $ \Phi_d(x)$ is irreducible, so this represents the complete factorization of $ x^n - 1$.  This factorization provides an immediate proof of the above lemma.

Note that each factor of $ n$ corresponds precisely to a factor of $ x^n - 1$; that's why for prime $ p$ the polynomial $ x^p - 1$ only has two factors (in particular, the second factor $ \frac{x^p - 1}{x - 1}$ is irreducible).

Now, you may have seen the factorization

$ x^n - 1 = \prod_{k=1}^{n} (x - e^{i \frac{2 \pi k}{n} } )$.

This represents a complete factorization of $ x^n - 1$ over the complex numbers; the roots of $ x^n - 1$ are called the [i]$ n^{th}$ roots of unity[/i].  Now, from this we would expect that we can represent cyclotomic polynomials as a product of these terms with roots of unity in them, and in fact

$ \Phi_d(x) - 1 = \prod_{1 \le k \le d, \gcd(k, d) = 1} (x - e^{i \frac{2 \pi k}{d} } )$.

The roots $ e^{i \frac{2\pi k}{d} }$ for $ gcd(k, d) = 1$ are called the [i]primitive[/i] $ d^{th}$ roots of unity.  From this we deduce that the degree of $ \Phi_d(x)$ is $ \varphi(d)$ (one motivation for the notation).  This provides an interesting way of thinking about the divisors of $ n$ and consequently the $ n^{th}$ roots of unity, which I will illustrate by considering the $ 12^{th}$ roots of unity.

For notational convenience, let $ \zeta_k = e^{i \frac{2 \pi k}{12} }, k = 1, 2, 3, ... 12$.  Now, first we have the primitive $ 12^{th}$ roots of unity

$ \zeta_1, \zeta_5, \zeta_7, \zeta_{11}$.

The polynomial with these roots is $ \Phi_{12}(x)$.  Once we remove these roots of unity, we are left with the roots $ \zeta_k$ such that $ gcd(k, 12) > 1$.  The next possibility is that $ gcd(k, 12) = 2$.  This corresponds to the primitive sixth roots of unity

$ \zeta_2, \zeta_{10}$.

Next, we have the case $ gcd(k, 12) = 3$, which correspond to the primitive fourth roots of unity

$ \zeta_3, \zeta_9 = i, -i$.

Next, $ gcd(k, 12) = 4$, which corresponds to the primitive third roots of unity

$ \zeta_4, \zeta_8 = \omega, \omega^2$.

Next, $ gcd(k, 12) = 6$, which corresponds to the primitive second root of unity

$ \zeta_6 = -1$.

Finally, $ gcd(k, 12) = 12$, which corresponds to $ \zeta_{12} = 1$.  So now we have divided up the positive integers less than or equal to $ 12$ according to their greatest common factor with $ 12$, an interesting concept which, when generally applied, provides a proof of the identity

$ n = \sum_{d | n} \varphi(d)$.

(Consider the degree of $ x^n - 1$.)

What is $ \varphi(d)$?  At first I thought it was the same thing as $ \phi(x)$ but that doesn't make sense in context...

And also, are you saying that $ \forall d \in \mathbb{N}$,  $ \Phi_d(x)$ is a unique (cyclotomic) polynomial?  So $ \Phi_6(x)$ is always $ x^2\minus{}x\plus{}1$?  If so, how do we find any arbitrary $ \Phi_d(x)$, such as $ \Phi_d(8)$ for example?

Let $ z_n\equal{}e^{\frac{2\pi i}{n}}$. If we start with the definition:

$ \Phi_n(x)\equal{}\prod_{(d,n)\equal{}1, n>d>0}(x\minus{}(z_n)^d)$,

we can establish all the standard properties. As you can see, $ deg(\Phi_n(x))\equal{}\phi(n)$.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Prove that if $ S$ is a linear operator on $ R^2$ such that $ S^2\equal{}S$, then $ S\equal{}0$, or $ S\equal{}I$, or there is an ordered basis $ B$ for $ R^2$ such that $ [S]_B\equal{}
\begin{pmatrix}
1 & 0  \\
0 & 0  \\
\end{pmatrix}$

$ S \equal{} 0$ and $ S \equal{} I$ satisfy our conditions.
We can now, assume that $ S \notin \{0,I\}$.

Let $ A$ be the matrix of the operator w.r.t. some basis.
$ A^2 \equal{} A \Rightarrow (det A)^2 \equal{} det A$
$ \Rightarrow det A \in \{0,1\}$

[u]Case I:[/u] $ det A \equal{} 0$
$ A^2 \minus{} (tr A) A \plus{} (det A) I \equal{} 0$
$ \Rightarrow A \equal{} (tr A) A$
As $ A \ne 0$, $ tr A \equal{} 1$.
The characteristic polynomial is $ t^2 \minus{} t$ has two distinct roots i.e. $ 0,1$.
So with a suitable change of basis, we get the matrix of the operator as $ \begin{pmatrix} 1 & 0 \\
0 & 0 \\
\end{pmatrix}$

[u]Case II:[/u] $ det A \equal{} 1$
$ \Rightarrow A \textrm{ is invertible}$

$ (det A) I \equal{} A ((tr A) I \minus{} I)$
$ \Rightarrow (adj A) A \equal{} ((tr A) I \minus{} I) A$
$ \Rightarrow adj A \equal{} (tr A) I \minus{} I$.

If $ A \equal{} \begin{pmatrix} a & b \\
c & d \\
\end{pmatrix}$
we have $ a \plus{} d \minus{} 1 \equal{} a \equal{} d$ and
$ \minus{} b \equal{} \minus{} c \equal{} 0$
So $ A$ is the identity matrix.

[quote="Nukular"]Using $ S$ as a projection mapping $ \mathbb{R}^2 \to \mathrm{im}(S)$,

$ \mathbb{R}^2 \equal{} \mathrm{im}(S) \oplus \ker(S)$

the condition says that $ S|_{\mathrm{im}(S)} \equal{} I$, and clearly
$ S|_{\mathrm{ker}(S)} \equal{} 0$. Take a basis $ B$ respecting
this splitting and you're done. :)[/quote]

Can you explain in more details? I am new in Linear Algebra so I could not fıgure out what do you mean.


inequalities
calculus
inequalities proposed

Let $ x,y,z$ are positive real number such that $ xyz\equal{}1$ Prove that :
1) $ \frac{xy}{x^8\plus{}xy\plus{}y^8}\plus{}\frac{yz}{y^8\plus{}yz\plus{}z^8}\plus{}\frac{zx}{z^8\plus{}zx\plus{}x^8} \leq 1$
2) $ \frac{(xy)^7}{x^8\plus{}(xy)^7\plus{}y^8}\plus{}\frac{(yz)^7}{y^8\plus{}(yz)^7\plus{}z^8}\plus{}\frac{(zx)^7}{z^8\plus{}(zx)^7\plus{}x^8} \leq 1$

Hello,
If i didn't make calculus mistakes, 1) is equivalent to :
$ [16,8,0]\geq[8,1,0]\equal{}[13,6,5]$ because $ xyz\equal{}1$
which is true by muirhead.

[quote="QuyBac"]Let $ x,y,z$ are positive real number such that $ xyz \equal{} 1$ Prove that :
1) $ \frac {xy}{x^8 \plus{} xy \plus{} y^8} \plus{} \frac {yz}{y^8 \plus{} yz \plus{} z^8} \plus{} \frac {zx}{z^8 \plus{} zx \plus{} x^8} \leq 1$
[/quote]

Using Am-Gm, $ LHS \leq\ \sum\ \frac {1}{x^6 \plus{} y^6 \plus{} 1}$
Let $ a \equal{} x^2; b \equal{} y^2; c \equal{} z^2$ we obtain an old and easy problem  :)

[quote]@QuyBac: Oi anh Chi' nha` em day a` =)). Dao. nay` bo? half life than` cong, quay tro? lai con duong` luong thien sao   :rotfl:
Em xin phep' chem' tam. 1 bai` trc nha'   :roll:
Using Am-Gm, $ LHS \leq\ \sum\ \frac {1}{x^6 \plus{} y^6 \plus{} 1}$
Let $ a \equal{} x^2; b \equal{} y^2; c \equal{} z^2$ we obtain an old problem
PS: Che' cung~ kheo' day' nhi? :P [/quote]
  Đây là một câu trả lời thú vị tôi sẽ suy nghĩ !!!  :P :P  :P Not spam  :mad:

[quote="QuyBac"]  Đây là một câu trả lời thú vị tôi sẽ suy nghĩ !!!  :P :P  :P Not spam  :mad:[/quote]
@QuyBac: [hide]Sorry ban. rat nhieu`. Minh` nham` ban. voi' 1 ng` ban. cua? minh` ten la` Chi', nick la` Silver, cung~ hoc bach khoa  :oops: . Ko sao, tui minh` lam` quen nhe'. Minh` la` Quoc Anh, hoc khoa toan-tin K52.  :)  [/hide]
Using Am-Gm, we have:
$ LHS (2) \leq\ \sum\ \frac {1}{z^6(x^6 \plus{} y^6) \plus{} 1}$
or we can rewrite: $ \sum\ \frac {1}{a^3 \plus{} b^3 \plus{} abc} \leq\ 1$ with $ a^3 \equal{} y^6z^6...$
It's an old problem and very easy  :)

[quote="Mathias_DK"]1) $ \sum_{cyc} \frac {xy}{x^8 \plus{} xy \plus{} y^8} \le \sum_{cyc} \frac {xy}{x^3y^3(x^2 \plus{} z^2 \plus{} y^2)} \equal{} 1$. :)

2) $ \sum_{cyc} \frac {(xy)^7}{x^8 \plus{} (xy)^7 \plus{} y^8} \equal{}$ $ \sum_{cyc} \frac {x^5y^5}{x^8z^2 \plus{} x^5y^5 \plus{} y^8z^2} \le \sum_{cyc} \frac {x^5y^5}{x^5y^3z^2 \plus{} x^5y^5 \plus{} z^2y^5x^3} \equal{} 1$ :)[/quote]
 Solution of you same of me  :)


Here are good 3 problems that can develop your problem skills. They're not very difficult but all of them require some work.

1. (Algebra) If you remove the first digit of $n$, you get a new number which is 1/57 the size of $n$. Find $n$. (For example, removing the first digit of 284 results in the number 84 but 284/57 $\neq 84$ so 284 is not the correct answer).

2. (Number Theory) For how many different positive integers $n$ is $n^4+125$ divisible by $n+5$?

3. (Geometry) A right triangle has one leg of length 20 cm. If the circle inscribed in the triangle has a radius of 6 cm, find the length of the other leg.

[hide="3"]
Let the unknown leg be x. We can denote the hypotenuse as $\sqrt{x^2+400}$.

We know that the inradius is 6.

The area of the triangle can be found out in two helpful ways: $\frac{1}{2}bh=10x$ and $rs$, with r=6 (the inradius) and $s=\frac{1}{2}\left( 20+x+\sqrt{x^2+400}\right)$ (the semiperimeter). Setting these equal to each other:

$10x=3\left( 20+x+\sqrt{x^2+400}\right)\\ 10x=60+3x+3\sqrt{x^2+400}\\ 7x-60=3\sqrt{x^2+400}\\ 49x^2-840x+3600=9x^2+3600)\\ 40x^2-840x=0\\ x=\frac{840}{40}=\boxed{21}$
[/hide]

-BLt

[hide="2"]
$n^4 + 125 \equiv 0 \bmod n+5$

Rewrite in terms of $m = n + 5$.  We have

$(m-5)^4 + 125 \equiv 0 \bmod m$
$625 + 125 \equiv 0 \bmod m$
$750 \equiv 0 \bmod m$
$750 = 2 \times 3 \times 5^3$

So there are $2 \times 2 \times 4 = 16$ factors, but remembering that $m = n + 5$ so $m > 5$ we subtract out the factors $2, 3, 5$, so our answer is $\boxed{13}$ numbers.
[/hide]

[hide="1"]
We seek a number $m = \frac{n}{57}$ such that $57m - m \equiv 0 \bmod 10^k$ where $k$ is the number of digits of $m$.

In order that $56m \equiv 0 \bmod 10^k$ we need $m$ to be divisible by some factors of five - if $m = 5$ we get $n = 285$, which is too large.  $m = 25$ gives us $n = 1425$, which is also too large.  Finally, $m = 125$ gives $n = \boxed{7125}$.
[/hide]

[hide="Faster method for 3"]
The inradii tangent to the triangle divide the sides $a, b, c$ into lengths $x+y, y+z, x+z$.  We know $x+y = 20$ and because $\angle C = 90^{\circ}$ we can easily deduce $y = 6, x = 14$.

Next, Pythagoras gives us $(x+y)^2 + (y+z)^2 = (x+z)^2$, so we plug in to get $20^2 + (6+z)^2 = (14+z)^2$ which, after simplification, becomes $16z = 240, z = 15$, so $b = y+z = \boxed{21}$.
[/hide]

[hide="1"]Notice that $57$ is one greater than a multiple of $8$. So, we can have the last three digits as $125$, so that both numbers will end in $125$. The smallest such number is then $n=57\cdot 125=\boxed{7125}$.[/hide]

[hide="3"]$\frac{1}{2}(20)x=6\left(\frac{1}{2}\right)(20+x+\sqrt{20^2+x^2})$, where $x$ is the other leg of the triangle.
$7x=60+3\sqrt{20^2+x^2}\Rightarrow (7x-60)^2=9x^2+3600\Rightarrow 2x(x-21)=0$
$x=\boxed{21}$[/hide]

Number Three

[hide] r=2A/P.  Let missing leg be x and hypotenuse be sqrt{x^2+400}.  Plug in, solve, you should get [b]21[/b].[/hide]

An easier way for 3 is to note the inradius of a right triangle is $\frac{a + b - c}{2}$


What's the difference between Mathlinks and AoPS?

They were originally separate sites. Mathlinks was a forum focused on olympiad mathematics, with users worldwide but centered in Europe. AoPS was focused on the American contests and the books of the same name. They merged in July 2004, with the founder Valentin Vornicu of Mathlinks later coming to San Diego.

Ever since, the site has maintained both urls and both forum styles.

So it doesn't really matter which style you use? Does one have an advantage over the other?

Your preference. The AoPS style has the side bar for better navigation and the animation, but the MathLinks style has more space. I prefer the MathLinks style.

Also, some places like Portal are accessible from mathlinks skin only.


MATHCOUNTS
rotation
geometry
geometric transformation
complex numbers

Post any words or phrases in a MathCounts context whose meaning you find ambiguous. My first:

When they say "rotated about the x-axis" do they mean 180 or 360 degrees?

It's about time for this post!

Is 0 a whole number? Apparently it is, but that sounds to me as if a bucket that holds no water is completely full when it's empty. The term 'whole number' is rarely used in higher mathematics, and is rarely clearly defined. JUST SAY NONNEGATIVE!

Argh.  This stuff used to trouble me to no end.  I have long forgotten what the definition of "whole number" was, but I *think* it means non-negative integer.

[quote="JesusFreak197"]Natural numbers (a.k.a. counting numbers) are all positive integers.
Whole numbers are natural numbers and zero.
Integers are all natural numbers, their negatives, and zero.
Rational numbers are numbers that can be expressed algebraically (i.e., non-repeating decimals).
Real numbers are all non-imaginary numbers.

That just about sums it up.[/quote]No rational numbers can be expressed as $ \frac ab$ where $ a $ and $ b $ are integers. And Irrationals cant. Reals are the set of Rationals and Irrationals.

[quote="JesusFreak197"]Natural numbers (a.k.a. counting numbers) are all positive integers.
Whole numbers are natural numbers and zero.
Integers are all natural numbers, their negatives, and zero.
Rational numbers are numbers that can be expressed algebraically (i.e., non-repeating decimals).
Real numbers are all non-imaginary numbers.

That just about sums it up.[/quote]

I'd like to add to that:

Complex Numbers: Numbers of the form $a+bi$, where $i=\sqrt{-1}$ and $a$ and $b$ are real numbers
Imaginary Numbers: Complex numbers where $b\neq 0$

Pure Imaginary Numbers: Imaginary numbers where $a=0$
Real Numbers: Complex numbers where $b=0$

Natural Numbers: $1,2,3,4,...$
Whole Numbers: Natural numbers + zero
Integers: Numbers of the form $\pm a$, where $a$ is a whole number
Rational Numbers: Real numbers that can be expressed as $\frac{a}{b}$, where $a$ and $b$ are integers

Irrational Numbers: Real numbers that cannot be expressed as $\frac{a}{b}$, where $a$ and $b$ are integers (i.e. $\sqrt{2}$)
Transcendental Numbers: Irrational numbers that cannot be expressed as $\sqrt{a}$, where $a$ is a positive, real number (i.e. $\pi$ and $e$)

And I derived all those definitions by myself, just right now.  :lol:

[quote="NoSoupForYou"]And I derived all those definitions by myself, just right now.  :lol:[/quote]

Go Alex?

I don't have time to correct them now, but some of the definitions provided by JF137, B&D, and NSFY are correct.  Some are imprecise.  Some are just plain wrong.  I suggest you use the definitions at your own risk.

[quote="probability1.01"]If it's to generate a solid, it's 360 degrees always.  If it's an actual 2-D rotation, it should be specified.  Otherwise, probably assume 180.  However, given that it's about the x-axis, it ought to be a 3-D rotation.[/quote]

Yeah, it was a 360 degree 3-d rotation (from a problem on the 2004 national sprint) . However, a 180 degree rotation would also produce a solid, so I was confused and ended up with exactly half the correct answer.

I think I remember another mathcounts in which they used the same term but meant a 180 degree rotation- can anyone clarify?

[quote="bob123"][quote="236factorial"]What's a deltoid? Is it a four sided figure? It wasn't in my dictionary...[/quote]

Go here: http://dictionary.reference.com/search?q=deltoid[/quote]

Here is a much better site: http://mathworld.wolfram.com/Deltoid.html

[quote="rcv"]Some are just plain wrong.  I suggest you use the definitions at your own risk.[/quote]

Yeah like the transcendental thing.  mathworld.wolfram.com is your friend... :)

I know most of the terms. It's just whole numbers that were kind of confusing. And I didn't think that 1+1[i]i[/i] was considered imaginary. I thought that imaginary was of the form n[i]i[/i]. Also, transcendental numbers can not be the root of any polynomial. Through some hefty work, you can prove that pi and e are transcendental. Anyway, whole numbers were my only confusion.

back to the MathCounts terms. Just wanted to make sure: 0 is not positive, but it is nonnegative, correct?

[quote="09husbri"]1+i isn't pure imaginary, but I believe it is still imaginary.[/quote]

Yeah, imaginary is pure imaginary and complex numbers.

I think that the 3-d geometry involving painting wierd shapes are hard.

Is 0 divisible by everything? Is anything divisible by 0? Just checking...

You can find a lot of math vocab definitions at http://www.mathwords.com

Nothing is divisible by zero, technically. $\frac{0}{0}$ is indeterminate, and $\frac{a}{0},$ where $a$ is any real number, is infinity.


geometry
rotation
analytic geometry
graphing lines
slope

Find the area of the largest triangle $ ABC$ [and prove this is the maximum] whose interior is entirely within the region bounded by $ y \equal{} \sqrt {3}x \minus{} 1$ and $ y \equal{} 3x^2 \minus{} 12x \plus{} 1$.
[hide="Comments"]
I think it's a little easy for PO, it's just ugly, because [hide="hint"] we just rotate the graph by 60 degrees clockwise and find the minimum of the second graph then. That's the height, so it's then trivial.[/hide]

EDIT: Now that Altheman's pointed it out, wow, I like doing things the hard way and this does not belong in PO.[/hide]

$ 6x\minus{}12\equal{}\sqrt{3}$ is the vertex [tangent line with slope of intersection line], the other two points are on the intersection of the line and the parabola.


geometry
geometric transformation
reflection

Hard problem:

We have a glass sheet with a thin cover of water on it. A laser is pointed towards the sheet, with an angle of incidence $\alpha$. The laser ray is reflected by the water (partially) and by the glass. We can see that there is destructive interference in the reflected ray for two values of the lenght of wave of the laser. The thickness of the water cover is $h$.

We know that: $\lambda_1= 385 nm ; \lambda_2= 693 nm ; 3,5 \mu m < h < 5,5 \mu m ; n_{water}= 1,33 ; n_{glass}>n_{water} ; \alpha \rightarrow 0$

Find the exact value of $h$.

I want to say that $\alpha$ is 0,0000000000000000.....1°. In other words, the laser is almost perpendicular to the glass sheet.

I apologize for my horrible english; if you have doubts about the text, please ask me.

We may assume $n_{\mathrm{air}} = 1$ (or at least $n_{\mathrm{air}} < n_{\mathrm{water}}$; I think that's all we need). Anyway, as $n_{\mathrm{air}} < n_{\mathrm{water}}$, there will be a phase shift $\phi_1 = \frac{\lambda}{2}$ at the first reflection; since $n_{\mathrm{water}} < n_{\mathrm{glass}}$ as well, the phase shift $\phi_2$ at the second reflection will also be $\frac{\lambda}{2}$. Thus, if we are to have destructive interference, $\exists\,l \in \mathbb{Z}, \sum \phi + \Delta d = (l + \frac{1}{2}) \lambda \Rightarrow \exists\,m \in \mathbb{Z}, \Delta d = (m + \frac{1}{2}) \lambda$. Now, if $\alpha_2$ is the refracted angle in the water, we see that $\Delta d = 2h \sec \alpha_2$, but $\alpha \to 0 \Rightarrow \alpha_2 \to 0$ by Snell's law $\Rightarrow \Delta d \to 2h$, so $\exists m \in \mathbb{Z}, 2h = (m + \frac{1}{2}) \lambda$. Note that this wavelength $\lambda$ is in the water and is thus equal to $\frac{v}{\nu} = \frac{\frac{c}{n}}{\nu} = \frac{\frac{c}{\nu}}{n_{\mathrm{water}}} = \frac{\lambda_0}{n_{\mathrm{water}}}$, so $2hn_{\mathrm{water}} = (m_i + \frac{1}{2}) \lambda_i, i \in \{1, 2\} \Rightarrow \frac{2m_1 + 1}{2m_2 + 1} = \frac{\lambda_2}{\lambda_1} = \frac{9}{5}$; thus, $(m_1, m_2)$ has the form $(9k + 4, 5k + 2)$; this yields $h = 651(2k+1) \mathrm{ nm}$, so $5. 37 = \frac{3500}{651} < 2k+1 < \frac{5000}{651} = 7.68$, so we have $k = 3$, yielding $h = 4.56 \mu\mathrm{m}$.

Very good.... I used another tecnique for the last part of the problem, using the Chinese Remainder Theorem after writing $m_1 = f(m_2)$. I haven't fully understood what you've done; however i got the one solution $h = 4,56 \mu m$ ; actually, if $h = 3,91 \mu m$ you would have $m_1 + 0,5 = 27$ but it's impossible because $m_1$ has to be integer, otherwise you wouldn't have destructive interference.

I noted 2m_1 + 1 / 2m_2 + 1 = 9/5 means that 2m_1 + 1 = 9(2k+1), 2m_2 + 1 = 5(2k+1).

Apologies for the error (I think switched a 9 for a 4.5 somewhere and confused myself); I think it is fixed.


geometry
floor function
number theory
prime factorization
algebra
partial fractions
system of equations

4) Find the constant term in the expansion of $ (x^8 \minus{} \frac {2}{x^6})^7$

13. Stephen has a really,-really,-superball (RRSB) which rebounds to 5/3  of its original height.  If Stephen drops the RRSB from a height of 4 feet, after how many bounces will the RRSB first be higher than the roof of Hoover High School, which has a height of 55 feet?

14) . Find the value of  (1+i / 1-i )^2007.

16) If a solution to $ x^6 \minus{} 64i$ may be written as $ a \plus{} ai$, where $ a$ is a positive real number, then $ a \equal{}$?

17) What's a harmonic mean?: Let X= the geometry mean of 2 and 12. and Y= the arithmetic mean of 9 and 13. Find the harmonic mean of X and Y.

20) The number 2007! has a prime factorization written in the form $ 2^{a1}3^{a2}5^{a3}$....b^(a base n), where the prime factors are written in increasing order and b is the last prime factor in the prime factorization. Find b++a5+a13.

21) In the equation $ \sqrt {a \plus{} 4\sqrt {5}} \plus{} \sqrt {a \minus{} 4\sqrt {5}} \equal{} \sqrt {b \plus{} 4\sqrt {5}}$, where $ a$ and $ b$ are positive integers, $ b$ must be equal to what number?

22) If Joyce decomposed $ \frac {x^2 \plus{} 29}{(x \plus{} 1)^2 (x^2 \plus{} 4)}$ into partial fractions, what did she get for the numerator of the term with $ (x^2 \plus{} 4)$ as its denominator?

I don't understand from reading what partial fractions are. Harmonic means seem confusing.

With partial fractions, you decompose a fraction that you already have into a sum of its components.  So in your example,
$ \frac{x^{2}\plus{}29}{(x\plus{}1)^{2}(x^{2}\plus{}4)}\equal{}\frac{A}{(x\plus{}1)^{2}}\plus{}\frac{B}{(x^{2}\plus{}4)}$
Now handle this as you would handle any other sum:
$ x^{2}\plus{}29\equal{}A(x^{2}\plus{}4)\plus{}B(x\plus{}1)^{2}$
However, it appears that the only roots of these terms are complex.

4

Multiply the expression by $ (x^6)^7$ and we will obtain $ (x^{14} \minus{} 2)^7$. Since every term in the expression in the expansion of the original expression of $ (x^8 \minus{} \frac {2}{x^6})^7$ is multiplied by $ x^{42}$, our problem is now equivalent to that of finding the coefficient of $ x^{42}$.

Notice that this is also equivalent to finding the coefficient of $ x^3$ in the expansion of $ (x \minus{} 2)^7$.

By considering the binomial expansion, the answer is $ \binom {7}{4} \cdot (\minus{}2)^{4} \equal{} \boxed {560}$.

13

Suppose there were $ n$ bounces before it exceeded the height of the roof. Then $ 4 \cdot ({\frac {5}{3}})^n \geq 55$. Take logs on both sides and rearrange. The greatest $ n$ is $ \boxed{5}$.

17) The [[Harmonic mean]] is the reciprocal of the average of the reciprocals.

[quote="JRav"]With partial fractions, you decompose a fraction that you already have into a sum of its components.  So in your example,
$ \frac {x^{2} + 29}{(x + 1)^{2}(x^{2} + 4)} = \frac {A}{(x + 1)^{2}} + \frac {B}{(x^{2} + 4)}$
Now handle this as you would handle any other sum:
$ x^{2} + 29 = A(x^{2} + 4) + B(x + 1)^{2}$
However, it appears that the only roots of these terms are complex.[/quote]
Actually, you have to decompose $ \frac {x^{2} + 29}{(x + 1)^{2}(x^{2} + 4)}$ as $ \frac {A}{x+1} + \frac {B}{(x+1)^2} + \frac {Cx+D}{x^2+4}.$ From this, we have:
$ x^{2} + 29 = A(x^{2} + 4)(x+1) + B(x^2+4) + (Cx+D)(x + 1)^{2}$
Either multiply out each term and collect different powers of x (A, B, C, and D are all constants) to make a big system of equations, or plug in some nice values for x. Plug in $ x=-1$ to get:
\begin{eqnarray*}(-1)^2 + 29&=&B((-1)^2+4)\\
B&=&6\end{eqnarray*}
Plug in the value for B, then make two separate equations for 2 easy values of x. (In this case, 0 and 1.)
\begin{eqnarray*}
29&=&A(4)(1) + 6(4) + (Cx + D)(1)\\
29&=&4A + 24 + Cx + D\\
5&=&4A + D + Cx\\
30&=&A(5)(2) + 6(5) + (Cx + D)(4)\\
0&=&10A + 4D + 4Cx\\
\end{eqnarray*}
Since A, B, and C are constants, we have C=0. So our system of equations reduces to:
\begin{eqnarray*}
5=4A+D
0=10A+4D
\end{eqnarray*}
Solve it to get D=-25/3. This is the term going above $ x^2+4$ in the partial fraction decomposition, and thus is your answer.

14

[hide]$ \frac {1 \plus{} i}{1 \minus{} i} \equal{} i$, after multiplying the top and bottom by $ (1 \plus{} i)$ to make the denominator real. So $ i^{2007} \equal{} i^{2004} \cdot i^3 \equal{} (i^4)^{501} \cdot (\minus{}i) \equal{} 1 \cdot (\minus{}i) \equal{} \boxed {\minus{}i}$.[/hide]

21

[hide]Squaring both sides, we have $ 2a \plus{} 2\sqrt {a^2 \minus{} 80} \equal{} b \plus{} 4\sqrt {5}$. Comparing the surd and surd-free term, we must have $ 2a \equal{} b$ and $ 2\sqrt {a^2 \minus{} 80}  \equal{} 4\sqrt {5}$. So $ a^2 \minus{} 80 \equal{} 20$, $ a^2 \equal{} 100$, $ a \equal{} 10$, ($ a$ is positive) and therefore $ \boxed {b \equal{} 20}$. [/hide]

[hide="Number 20 Solution"]First, we find $ b$. The largest prime number less than $ 2007$ is $ 2003$, so $ b\equal{}2003$.

Then, we need to find the exponent of the third prime number, which is $ 11$. A factor of $ 11$ will appear once for every multiple of $ 11$ less than $ 2007$.
There are $ \lfloor \frac{2007}{11} \rfloor \equal{}182$ of these.
Then, for every multiple of $ 11^2$, there is another multiple of $ 11$.
There are $ \lfloor \frac{2007}{11^2} \rfloor \equal{}16$ of these.
Finally, for every multiple of $ 11^3$, there is another multiple of $ 11$.
There is $ \lfloor \frac{2007}{11^3} \rfloor \equal{}1$ of these.
Totaling them up, we have $ a_5\equal{}182\plus{}16\plus{}1\equal{}199$.

Now, we need to find $ a_13$. Similarly to finding $ a_5$, we just find the thirteenth prime number and proceed in the same manner. The thirteenth prime number is $ 41$.
$ \lfloor \frac{2007}{41} \rfloor\plus{}\lfloor \frac{2007}{41^2} \rfloor\equal{}48\plus{}1\equal{}49\equal{}a_13$

In conclusion, we have $ b\plus{}a_5\plus{}a_13\equal{}2003\plus{}199\plus{}49\equal{}\boxed{2251}$.[/hide]

[quote="Bicameral"]16) If a solution to $ x^6 \minus{} 64i$ may be written as $ a \plus{} ai$, where $ a$ is a positive real number, then $ a \equal{}$?[/quote]

[hide="solution"]$ (a\plus{}ai)^6\equal{}a^6(1\plus{}i)^6\equal{}a^6(2i)^3\equal{}a^6(\minus{}8i)$

$ \minus{}8a^6i\equal{}64i$

$ a^6\equal{}\minus{}8$

$ a^2\equal{}\minus{}2$

UH-OH. Did I do something wrong?[/hide]


1. Can the beginner/intermediate/advanced option be changed mid-course. for example if i start out at beginner and some time during the program, i become a lot smarter, could i change to intermediate on the next olympiad test?

2. For the beginner/intermediate/advanced option, what is the USAMO score range for each. for example beginner: 0-9, interm. 10-19, advanced: 20+ or something.

3. If i choose say, beginner, and i submit my solutions to those questions, but i still want to try the advanced/interm problems, will the solutions to those questions be available to everyone?

4. Are the assignments different for the different levels of people?

5. Is there a schedule for when the assignments are given out? If so, where can it be accessed?  :)

6. If come up with more questions later on, am i free to post them on this topic??

1.  Absolutely - you choose what you'd like to do on each test.

2.  Tough to say, because the difficulties of USAMOs vary so much.  I might go 0-4 beginner, 5 - 15 middle, 16+ advanced.  It's not unusual for only 20-40 people to clear 15 on the USAMO.

3.  Yes, all problems and solutions are available to everyone.

4.  The extra problems on the message board are the same for everyone, and will typically include problems with a wide range of difficulty.

5.  You should be able to find that on the WOOT homepage (in the My Classes section).  If not now, then very soon.

6. You can post them in the WOOT forum - you should have access to it now (just below this forum when you are logged in).

i was wondering how many people take WOOT per year?


induction
geometry
algebra
polynomial
integration
3D geometry
rotation

hi there.

I guess all of you guys know that 1³+2³+...+n³ = (1+2+...+n)²

so that the sum of all i³ = (sum of all i)²

and I guess all of you can prove it with induction, but can you find another way how you can [b]get to[/b] this idea, so getting the formula in some way? [i][u](without induction thus)[/u][/i]

Let's keep it as original as possible here, mine is pretty short and even uses a little geometry in it :D

(i'll post mine if you guys like it, but I guess you'll bring some original solutions as well)

You can do it even using calculations only (no induction):

(1 + 2 + ... + n)²
= (1 + 2 + ... + (n - 1))² + 2n(1 + 2 + ... + (n - 1)) + n²
= (1 + 2 + ... + (n - 1))² + n²(n - 1) + n²
= (1 + 2 + ... + (n - 1))² + n³

Repeating this step n times we get the result, of course.
However, the idea of induction is not far away here...
I suppose you wanna have something really surprising :)

ya like I said: some way you can get it without knowing the thing in advance (ok it will look kind of artificial any way but that one is very close to induction indeed :?)

[quote="Peter VDD"][quote="liyi"]1^3 + 2^3 + ... + n^3 = n^2 (n+1)^2 / 4.
1 + 2 + ... + n = n(n+1)/2.
i think most of us know these two equalities.[/quote]

did you know that      1^3 + 2^3 + ... + n^3 = n^2 (n+1)^2 / 4.     ??

then you're smarter than me, I had to prove it first :?[/quote]

(n+1)^4-n^4=4n^3+6n^2+4n+1
...
2^4-1^4=4*1^3+6*1^2+4*1+1
Sum them up
Using the known fact 1^2+...+n^2=(1/6)n(n+1)(2n+1) and 1+...+n=n(n+1)/2 you can finish it.

and there we get down to induction-like stuff again :)

To find S_k=1^k+..+n^k more easy, we have the following reccurence:

(n+1)^(k+1)-(n+1)=S_1*C(1,k+1)+...+S_k*C(k,k+1);

where C(k,n)=n!/(k!(n-k)!).

yes indeed, but unfortunetely by not using induction VDD I think that ment not using any higher-than-induction mathematical argumentation :D:D

anyway what's your idea VDD? :)

[quote="TripleM"]I find the easiest way to calculate is instead do this:

If S_k is a polynomial with variable n, then S_(k+1) = k (\int S_k  dn) + cn, where you choose c so the sum of the coefficients is 1.[/quote]

well, TripleM perhaps it is easier but I don't know integrals....

ya well I'm not even on college yet, so I don't know all mysterious math-solving tricks :P but anyway, mine was as follows (y is vertical axis)

[i]the sum of 1³ and 2³ and so on, means we gotta take an nxnxn cube, place an (n-1)x(n-1)x(n-1) cube on top of it, etc, until you reach the 1x1x1 cube, and count the volume taken by it.[/i]

Let's first suppose it were cilinders, we can move little parts of the volumes so we'd get a threedimensional parabolus (thus a parabolus, rotated around the y-axis) (which could easily be proven)

So we count the volume of that, and supposing the vertical axis is y (as said), we get y=n(n+1)/4 (with y being the place where the parabolic thing cuts the cilinders)

Thus an integral gives us V = \pi/4 . \int(0,n) (n³+3n²/2+n/2), which is \pi.n²(n+1)²/16 for the cilinders
Now go back to cubes (*4/\pi) and you get n²(n+1)²/4
In which we all see sum of all i...


I want to know if this problem is elementary(easy) or not:
Prove that if $k=\frac{a^2+b^2}{ab+1}$ is an integer then k is a perfect square.
(I DONT WANT THE SOLUTION.)

If this is in fact  the IMO problem proposed from Germany , then it is one of the most difficult problems. I have read this comment in << Englel - Problem solving Strategies , Springer >>. Otherwise I can not decide , because I have not tried to  solve it.

 Babis

http://www.mathlinks.ro/Forum/viewtopic.php?t=33781
http://www.mathlinks.ro/Forum/viewtopic.php?t=26374

:)


Let [b]A[/b] be a set of all infinite subsets of [b]N[/b], such that for every infinite sequence of natural numbers [b]Bn[/b] there is an [b]a[/b] from [b]A[/b] such that [b]a[/b] is subset of [b]Bn[/b]. Prove that [b]A[/b] is uncountable.

First, when you say "infinite sequence," I am assuming that you mean "infinite set".

Suppose that $ S\equal{}\{A_1,A_2,...\}$ is countable.  Pick some $ b_1\in A_1$, $ b_2\in A_2$ with $ b_2 > b_1 \plus{}1$, $ b_3\in A_3$ with $ b_3> b_2 \plus{}1$, etc. (we can do this because each $ A_i$ is infinite)

Let $ B\equal{}\{b_1,b_2,...\}$, and let $ C$ be the complement of $ B$.  Since $ C$ is missing an element from each $ A_i$, it cannot have any $ A_i$ as a subset.  But for every $ i$, $ b_i\plus{}1\in C$, so $ C$ is infinite, contradiction.


$cosxcos2x=sin(\frac{\pi}{4}+4x)sin(\frac{\pi}{4}+x)+sin(\frac{3\pi}{4}+4x)cos(\frac{7\pi}{4}-5x)$
Solve for x?

[quote="midorya"]sorry for reviving thread
[hide="answer"]
x=pn/2
x=p/14+pn/7
[/hide][/quote]
Solution please?
[hide] $x=\frac{n\pi}{2}$, $x=\frac{\pi}{14}+\frac{n\pi}{7}$ [/hide]


Putnam Princeton college Harvard MIT Columbia Duke

Results are out!

Winning teams:

1. Princeton
2. Harvard
3. MIT (picked the wrong team again!)
4. Toronto
5. Chicago

Honorable Mention:

U. of British Columbia
Brown
Duke
Harvey Mudd
Stanford
Wash U. (St. Louis)

Putnam Fellows:

Hansheng Diao, MIT
Daniel Kane, MIT
Tiankai Liu, Harvard
Po-Ru Loh, Caltech
Yufei Zhao, MIT

(The city of Madison, WI, is doing pretty well there)

N1:

Timothy Abbott, MIT
Ralph Furmaniak, Waterloo
Anders Kaseorg, MIT
Sungyoon Kim, MIT
Yuncheng Lin, MIT
Alison Miller, Harvard (Elizabeth L. Putnam prize)
Kevin Modzelewski, MIT
Andrei Negut, Princeton
Aaron Pixton, Princeton
Eric Price, MIT

N2:

Saran Ahuja, Brown
Park Doo Sung, Caltech
Shinn-Yih Huang, MIT
Matthew Ince, MIT
Theodore Johnson-Freyd, Stanford
Cedric Lin, British Columbia
Thomas Mildorf, MIT
Xuancheng Shao, MIT
Andrei Ungureanu, Princeton
Yeoil Yoon, Caltech
Rumen Ivanov Zarev, Caltech

--

My school, Cal State Long Beach, had a team rank of 78, which is the best we've ever done.

Some minimum scores:

To be a Putnam fellow required a score of 92. To be listed in the top 200 required a score of 31, with some tiebreaking. To be listed in the top 500 required a score of 14.

Big congrats to forum member billzhao on being a Putnam Fellow! :)

Indeed!  And to, um, everyone else I know on that list!  (Also, everyone laugh at MIT again!  Hehe.  Okay, all done.)  :)

Who is billzhao?

OK, this year I got 11 points, which is significantly lower than 29, which was my score in 2005. Progress! Woohoo!  :lol:

By the way, how did the other forum members do? What about Xevarion, JBL and blahblahblah?  :)

Kent tells me that I was Honorable Mention this year.  Our math department hasn't told me anything yet, so I don't know my score; I also don't remember what score I was hoping for based on my solutions  :oops:.

How was the honorable mentions list?  I'm particularly interested in how fellow waterlooians did.

Thanks,
Ralph

Ralph: Besides you as an N1, Waterloo also had Tor Gunnar Myklebust and Xiao Heng Wang as Honorable Mention. Also four other people in the 100-200 range and eight in the 200-500 range.

[quote="QQQ"]Ah, thanks for this. Who was on UChicago's team?[/quote]
David Coley, Junehyuk Jung, Zhiwei Calvin Lin. Lin was Honorable Mention, the other two were in the 100-200 range. Chicago picked the right team.

Oops - I see one place with the name listed as David Coley and another place with it listed as David Caley. Maybe you can tell us which one is right.

This was my second test. I got 20, down from 22. Ranked 390.5, up from 391. That's somewhat an improvement!
Brown placed surprisingly high this year, I think.

Prof Merryfield:
Princeton's on spring break right now so we're not going to hear details of the results for a while yet. I am wondering 2 things:
1) what was the minimum score for HM?
2) who at Princeton got HM? (there might be a good number so if you don't want to type them all just for me I won't be offended)

[quote="NTIM"]Go Brown!! We did so well this year - has Brown ever been in the top 10 before?[/quote]I believe so. There's a plaque in the math department from a previous Honorable Mention finish. Among the team members that year was Joseph H. Silverman...

[quote="MysticTerminator"]
also if you have an chinese IMO perfect scorer, why is he not on the team...[/quote]

Cause we had 3 (returning) fellows last year?

Everyone seems to comment on how MIT sucks at choosing its team. But really, what else can we do?

I don't blame our organizers at all. The number of people at MIT taking the Putnam is simple enormous. If we were to hold some sort of a preliminary selection exam, then the magnitude of that would be on the same scale as that of a national olympiad! Surely you can't expect the Putnam organizers at MIT to grade such a thing! I doubt we have the resource.

That being said, I think picking the highest ranking individual from the previous year seems like a fair strategy. It has certainly worked in the past :P

That strategy usually seems to work for Princeton, although Princeton also tends not to pick freshmen and this has cost us in the past.

One option would be to participate in the Virginia Tech contest, but that still might not work because I could see easily envision significantly more than 3 MIT students dominating that exam. Also the results for that might come too late for Putnam team selection.

Yufei, you better step up next year! Not to put pressure on you but I know where you live...  :P

In all seriousness (and in conjunction with Yufei's comment), there really aren't too many other good ways to pick the team. If you think about it, if a more qualitative selection method is used, then the school is opening itself to even more criticism in the event that the team members screw up.

Also, in MIT's defense, I believe we did put Reid on the team as a freshman, which turned out pretty nicely.

Personally, Professor Richard Stanley is my advisor and I've spoken to him a few times about team selection. It's definitely not like Professor Stanley and Professor Rogers (the other primary person involved in selecting the team) don't think hard about how to improve our standing.

But when you have 3 returning fellows what (else) are you supposed to do?

I've even witnessed Professor Stanley at the receiving end of a little good natured ribbing about our team's ineptitude (relative to our overall performance) at faculty functions and the like.

It's not easy for me to pick a team, and I have 6-8 people taking the exam, and my main goal in picking the team is to try to find three people who will get positive scores. And I did pick the right team this year, even though it didn't quite work out, as one of the three, a little bit sick at the time, had a bad day and got a zero. He's had a positive score before and will likely have a positive score in the future - stuff happens.

I can't imagine trying to pick the MIT team.

They should really do away with the preliminary team selection method and take the top three scorers from each school. The way the team score is computed now is unreasonable to me.

I disagree that it should just be the top three scorers, because that gives an enormous advantage to big schools. Big schools already do have an advantage in that they have a deeper pool from which to pick the team; the fact that everyone's scoring team is the same size at least ensures that the risk of bad test days impacts all schools equally (to the extent that it affects all students equally relative to their demonstrated ability).

That said, I think a very reasonably solution might be to have school nominate a team of six or seven instead of three, and choose the top three ranks among those six or seven. This would keep the test-day variance equal at all schools, but reduce it overall.

Besides selecting the team in advance, another factor affecting team ranks is that the team score is the sum of the individual ranks, not the sum of the individual scores (I think).  Sum of the ranks means that the team rank is most influenced by the weakest team member.  It's better to have team members ranked 33rd, 34th, and 35th than 1st, 2nd, and 100th, even though the second team will probably have more points.

I don't have a position on whether the current method of team ranking should be changed.

That is correct, about how teams are ranked.

What I like about summing the ranks is that it removes any effect from the shape of the score distribution by considering only relative position. This makes sense to me in a lot of ways.

Of course, there are other funny results of it. E.g. in MIT's case, all of their non-team high scorers are literally harming the team's score by pushing their ranks down. Obviously, they are pushing all the other schools' ranks down, too, but it's a funny thing to think about.

Yay! My school placed quite well for being a really local state school with only four people taking the test. We got 63rd, which we were told there were over 400 teams. I'm not sure if that is accurate.

There were 402 teams. There were students from 508 colleges, but only 402 formed teams. And your 63rd beats our 78th. (They only rank teams out to about 150; beyond that, teams are unranked.) Doing that well is largely about avoiding zeros on your team - each person with a zero was worth 2501 rank points.

Is Yufei Zhao a Chinese?
It seems that he studied in Canada.

5 Chinese names appear in the list.

Hansheng Diao,Yuncheng Lin,Xuancheng Shao are all Shanghaiese and are all IMO gold owners.
I think there will be more Chinese names on the list.....Because in the past years only Shanghaiese knew the chance to go to study in U.S(especially MIT).
But this year I know two IMO gold owners who are not Shanghaiesehave transfered to MIT from Peking University and should have attended putnam 2007.
They are Qingchun Ren from Tianjin and Long Jin from Jilin.

I think Qingchun Ren will be a awesome scorer in putnam.Putnam fellow Diaohan Sheng ranked 3rd(or 2nd?I don't remember clearly) in 2005 Chinese team selection(include 7 exams,the former 6exams and the last exam either holds half weight of final selection).But Qingchun Ren was somewhat more amazing.He ranked 1st in the Chinese team selection in both 2005 and 2006 and got IMO gold medals in both IMO2005(full marks) and IMO2006.Long Jin also has a IMO2006 medal.

Yufei posts here under the name billzhao. He's definitely Canadian; I'd guess he was either born in Canada or moved with his family at a very young age.

Of course, there are far more than 5 Chinese names here by language; you're just looking for the minority that grew up in China.

[quote="galoisj"]
I think there will be more Chinese names on the list.....Because in the past years only Shanghaiese knew the chance to go to study in U.S(especially MIT).
But this year I know two IMO gold owners who are not Shanghaiesehave transfered to MIT from Peking University and should have attended putnam 2007.
They are Qingchun Ren from Tianjin and Long Jin from Jilin.

I think Qingchun Ren will be a awesome scorer in putnam.[/quote]

Just as I expected,Qingchun Ren have been a putnam fellow in putnam 2007.......
He is tooooo awesome.


A farmer spent 3 equal sums of money on buying hens, pigeons and parrots;each hen cost $1 more than a pigeon and $2 more than a parrot.Altogether he bought 47 birds and the number of pigeons exceeded that of the hens by as many parrots as he could have bought for $ 9.Find the number of birds of each kind.Also find the total amount of money spent.

No. It's sufficient. I have a really ugly solution which I will post later but I know that it's possible to find all the prices, then find the numbers of each bird. (Then in turn the total amount spent.)

solved it at last!Sorry for troubling you.

  [hide]Let the each  parrot cost x,hen cost x+2 and pigeon cost x+1(Allow me to exclude the units of currency).
Let amount spent in buying the parrots ,hens, pigeons etc be A each.

So find the number of parrots, pigeons etc.

So, A= $ \frac {9(x \plus{} 1)(x \plus{} 2)}{x}$.

Again, $ \frac {A}{x}$+$ \frac {A}{x \plus{} 1}$ + $ \frac {A}{x \plus{} 2}$=47

Putting the value of A in this equation,we get $ 10x^{2} \minus{} 27x \minus{} 9\equal{}0$

So, x=3.
Then we can find the rest.[/hide]

You are right.The process isn't nice.


Copy the argument used for the determinant here: http://www.mathlinks.ro/Forum/viewtopic.php?t=205595 to calculate det A explicitly, the limit is then easy to calculate.


At Jefferson Summer Camp, 60% of the children play soccer, 30% of the children swim, and 40% of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer?
(A) 30% (B) 40% (C) 49% (D) 51% (E) 70%

[hide="Solution"]Let's assume that there are $ 100$ people at the camp.

We know this:
$ 60$ people play soccer.
$ 30$ people swim.
$ 24$ of the soccer players swim.

There are $ 70$ nonswimmers - as $ 100\minus{}30\equal{}70$.

Because $ 24$ of the soccer players swim, $ 36$ of the soccer players are non swimmers.  $ \frac{36}{70}\equal{} \frac{18}{35} \approx 51$ percent.

Therefore, the answer is $ \boxed{\textbf{(D)}}$.[/hide]

This has already been posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1404705#1404705]here[/url]. Lots of problems' links can be found by going to the resources section or just searching.


Given an $ n x n$ table and $ k<n^{2}$ a natural number, what is the number of different ways to colour $ k$ of the $ n^{2}$ unit squares of the table?  (Two colorings are considered different if one can not be obtained from the other by rotating the sqare)

No...It's about [i]different[/i] colorings with resprect to the rotation...In your solution some colorings are count twice...some four times...In fact it is enough to count the colorings which are counted twice or four times and substract :P

We can go for a inclusion-exclusion attack as you suggest. But we donot get a neat closed form.

1. We count a coloring twice if the coloring is anti symmetric about one of the diagonals.
2. We count a coloring four times if it is symmetric with respect to both diagonals.

For even $ n$ we get,

$ {n^2\choose k_1} \minus{} {{\frac{n^2\minus{}n}{2}}\choose k_2} {{\frac{n}{2}}\choose k_3} {{\frac{n}{2}}\choose k_4} \plus{} \frac{n^2 \minus{}n\minus{}(n/2)}{4} {{\frac{n}{2}}\choose k_5}$

I would suggest to try this problem with $ k\equal{}n$, maybe the result is nicer and could help finding the general one...


All right this may sound like a silly question but how does one extend lines in Asymptote (e.g. I want to extend sides AB and CD of a quadrilateral until they meet in a point).

Figured it out: it's the command extension:

[code]pair extension(pair P, pair Q, pair p, pair q);[/code]


Prove that any function $f$ defined over thel real numbers can be uniquely expressed as $f=E+O$, where $E$ is an even function and $O$ is an odd function.

This is pretty easy, but it's interesting to visualize.

[hide]

Identity is trivial, and one can clearly see that the two terms are an even and odd function, respectively.[/hide]

Substituting $f(x)=e^x$ we get our respective hyperbolic sine and cosine parts.

That has some pretty awesome properties.


From elsewhere on this forum:
USA 4 6 7 3 7 3 7 33 (Miller)
USA 5 7 7 2 7 7 7 37 (Pixton)

What works differently/better in your heads from the rest of us that the two of you are able to pull of things like this?

Old, but I just saw this.

Aaron Pixton goes to my area's high school! Surprising to see an Olympian from this quiet dead place...

[quote="demji"]did you know that pixton is also an awsome chess player.  I think he was NYS Junior Champion at one time.[/quote]

He was actually NYS Champion... No Junior... He was amazing... He's an FM with like a 2400+ USCF rating and I'm not sure if he has even played at all since going to college...

i thought he got recruited by harvard for chess


function
Support
integration
complex analysis
real analysis
real analysis theorems

Suppose $\phi$ is of compact support. What are the restrictions on its Fourier transform? That is, what kind of power do I have to choose a function $\widehat{\phi}$ while still making sure that it is the Fourier transform of a function of compact support?

If $\varphi$ has compact support, then $\widehat{\varphi}$ is entire.

Let's be more specific: suppose $\text{supp}(\varphi)\subset[-A,A]$ and $\varphi\in L^1.$

Then $\widehat{\varphi}(z)=\int_{-A}^A\varphi(x)e^{-2\pi ixz}\,dx$

$\left|\widehat{\varphi}(z)\right|\le \int_{-A}^A|\varphi(x)|e^{2\pi A|\text{Im}(z)|}\,dx \le\|\varphi\|_1e^{2\pi A|\text{Im}(z)|}$

$\widehat{\varphi}$ is an entire function that grows no faster then a particular exponential.

Want more? Check your sources - textbooks, reference books, online - for the topic of [b]Paley-Wiener Theorems.[/b]