number theory
relatively prime
number theory unsolved

prove that there exist infinitely many pairs (a,b) of relatively prime poistive integer such that $\frac{a^{2}-5}{b}$and $\frac{b^{2}-5}{a}$are both integers

its a nice problem. it was in hojoo lee notes.my solution is :
we can prove basing in the fact that $a$ and $b$ are relativly prime that
$\frac{a^{2}-5}{b}$ and $\frac{b^{2}-5}{a}$ are both integers <=>$\frac{a^{2}+b^{2}-5}{ab}$ is integer. and we are done :)

From here [quote="fermat3"]$\frac{a^{2}+b^{2}-5}{ab}$ is integer[/quote]
I think you will continuous with method of [url=http://www.kalva.demon.co.uk/imo/isoln/isoln886.html]IMO 1998/B3[/url], Am I true?  :maybe:


logarithms
limit
algebra
polynomial
induction
calculus
calculus computations

What is the value of $ \lim_{x \to 0^\plus{}} x \cdot \ln x \cdot (1 \plus{} \ln x)$ ?

Value is $ 0$.

[hide="solution"]
Let $ \ln x\equal{}\minus{}t$, if $ x\to 0^\plus{}$, then $ t\to \plus{}\infty$ and we can calculate easier limit:

$ \lim_{t\to\infty}\frac{\minus{}t(\minus{}t\plus{}1)}{e^t}\equal{}0$

Becouse:

$ \lim_{x\to\infty}\frac{W_n(x)}{e^x}\equal{}0$ for every $ n\in\mathbb{N}$, where $ W_n$ is polynomial of degree $ n$. It's easy to show by induction and de l'Hospital rule.
[/hide]


Cant you do a scan? (on my comp, you can scan for viruses before opening an attachment)

That new virus is really nasty. I think it checks places like ebay for credit card information. I've been getting tons of those, but fortunately my antivirus software catches them.

Yeah so does mine.  I heard that it was the ones with Hi or something that you weren't supposed to open.  Is this a new one?

this thread sure is popular... 4 posts in 1 minute!!

By the way, how did you find out about it?

i was wondering if AOL at all scans attachments, i know MSN does because it has a scan checker, and a cool download manager I gave up on AOL because it used to be very crappy (did'nt even had junk mail fileter i think it does now) i'm staying with msn because my email won't change, they have free programs such as Microsoft Money, Microsoft Encarta, and Microsoft Photo Editer, and MSN Groups, and MSN Music (probably AOL is better). AOL Broadband also costs like $10 more than MSN Broadband over here in Utah.

[quote="confuted"]...and neither AOL nor MSN are any good. (final similarity)[/quote]

I agree...that's why i have Roadrunner...

hmm.. this is a pretty silly post really.. who on earth would open email attachments in the first place unless you are 100% expecting it.. its just common sense. in fact, I can't believe how many people end up opening things like that, its quite funny.

That's true...normally people don't do that unless it's something that a realli good friend sent them or they were expecting it from someone...but then again, I think one of the viruses going around has the capability (and I forgot what this is formally called) of automatically sending it out to everyone in your address book or something like that...

There was a virus a few decades ago that did it (it was a christmas greeting) and it took out a lot of computers b/c it would send itself to everyone and then duplicate itself on your computer until it crashed...

In any case, rep, your computer isn't going to "be okay" until you get rid of it...especially if it's a time bomb...

Even so, its pretty simple to tell the difference between a real attachment and a fake one..

that's true...yet u'd be surprised how many people dont know what a processor does...

if you get an error message with an attachment dont open it.  Also, if security@microsoft.com sends you a message with an attachment, don't open it.  Microsoft doesn't send attachments, it only gives warnings and links to sites.

Also if anyone else sends u an attachment don't open it. unless you asked for it. and even then, scan it with an antivirus program first.

These virus people really enjoy spamming. I got 11 messages in my bulk mail all having the title "hi" and "hey". That was annoying - just like it is when webmaster sends me about 40 e-mails every two days after I visited a forum.

And people dont be stupid please and open an email if you dont recognize the address. Also, scan all your newsletters or things that you've signed up for, those can have a virus on it too.

Hmm. I wonder what happens if you open a mail from your account at school on their computers... I need to try that. <jk>

I heard there is like this new really wierd virus, it messes up your computer just when you open the email, not the attachment... scary.


[quote="mdk"]Prove that there is no positive integer n for which 2^n +1 is divisible by 7.[/quote]

[hide]Note that the powers of two residue $ \pmod{7}$ are $ 2,4,1$. This is because
$ 2^{1}\equiv 2\pmod{7}$
$ 2^{2}\equiv 4\pmod{7}$
$ 2^{3}\equiv 1\pmod{7}$,

and any higher power can be formed by raising the last equality some number of times. So, there doesn't exist $ n\in\mathbb{N}$ such that
$ 2^{n}\equiv 6\equiv-1\pmod{7}$
$ 2^{n}+1\equiv 0\pmod{7}$,

as desired.[/hide]


function
topology
real analysis
real analysis unsolved

Prove that if $f(x)$ and $g(x)$ are semi-continuous functions then function $f(x)+g(x)$ is semi-continuous one, too

I'm sory for my typing.I have edited it right

Doesn't this follow from the definition of [url=http://en.wikipedia.org/wiki/Semicontinuity]semi-continuity[/url], hence it's "Computations" material?

[quote="Wikipedia"]Suppose $X$ is a topological space, $x_{0}$ is a point in $X$ and $f : X \to \mathbb R$ is a real-valued function. We say that $f$ is [b]upper semi-continuous[/b] at $x_{0}$ if for every $\varepsilon > 0$ there exists a neighborhood $U$ of $x_{0}$ such that $f \left( x \right) < f \left( x_{0}\right)+\varepsilon$ for all $x$ in $U$. Equivalently, this can be expressed as \[\limsup_{x \to x_{0}}f(x) \leq f \left( x_{0}\right) .\] [/quote]

There are some interesting things about semi-continuity on that webpage, although they're pretty easy to prove.

$1.$ If $f$ is upper semi-continuous on $X$, then $\left\{ x \in X | f(x) < a \right\}$ is an open set for all $a \in \mathbb R$;

[i]Proof:[/i] We'll prove that its complement, $\left\{ x \in X | f(x) \geq a \right\}$, is closed. Take a sequence $\left\{ x_{n}\right\}_{n \geq 1}\to k$ such that $f \left( x_{n}\right) \geq a$. If $f(k)$ were $<a$, then $f$ would take values in $<a$ in a neighbourhood of $k$, which is contradiction. So $f(k) \geq a$.

$2.$ If $f$ is upper semi-continuous on a compact space $C$, then $f$ has a maximum.

[i]Proof:[/i] If it doesn't have a maximum, then we'd find a sequence $\left\{ x_{n}\right\}_{n \geq 1}$ such that $f \left( x_{n}\right) \geq n$. It will have a subsequence converging to some $t$. But in $t$ the definition doesn't hold, so we have contradiction.

Now suppose that it doesn't reach its maximum. Again take a sequence, then some subsequence of it converging to some $t$. Let $\epsilon > 0$. We know that $\left\{ x \in C \left| x \geq \sup_{x \in C}f(x)-\epsilon \right. \right\}$ is closed, so $f(t) \geq f(x)-\epsilon, \, \forall \epsilon > 0$. Hence, $f(t) \geq \sup_{x \in C}f(x)$, so $f(t) = \sup_{x \in C}f(x)$.

Suppose $\{f_{\alpha}\}_{\alpha\in A}$ is a collection of continuous functions on the same domain. Let $g(x)=\sup_{\alpha\in A}f_{\alpha}(x).$

Is

*- $g$ necessarily upper semicontinuous?

*- $g$ necessarily lower semicontinuous?

*- Both?

*- Neither?

1. [color=blue]Let $\left\{ f_\alpha \right\}_{\alpha \in A}$ be a collection of lower semi-continuous functions. Let $g(x) = \sup_{\alpha \in A}f_\alpha (x) < \infty$.

Then $g$ is lower semi-continuous.[/color]

[i]Proof.[/i] Let's suppose the opposite, i.e. there is $x_{0}\in X$ such that $g$ is not lower semi-continuous, i.e. $\exists \, \varepsilon > 0, \, \forall U \in \mathcal V \left( x_{0}\right), \, \exists x \in U \cap X$ s.t. $f(x) \leq f \left( x_{0}\right)-\varepsilon$. Clearly, there is $\alpha \in A$ s.t. $f_\alpha \left( x_{0}\right) > f \left( x_{0}\right)-\varepsilon$. But in the vicinity of $x_{0}$ there are numbers $t$ s.t. $f_\alpha (t) \leq f \left( x_{0}\right)-\varepsilon$, which is a contradiction.

[i]Note.[/i]On the Wikipedia page, the condition $g(x) < \infty, \, \forall x \in X$ was imposed, but I think we can drop it. I hope I'm not wrong.

2. Let $f_{n}: \mathbb R \to \mathbb R$, $f_{n}(x)=0, \, \forall x \leq 0$, $f_{n}(x)=1, \, \forall x \geq \frac1{n}$, then join $(0,0)$ and $\left( \frac1{n},1 \right)$ with a line. $\sup_{n \in \mathbb N^\ast}f_{n}(x)$ is $0$ if $x \leq 0$ and $1$ if $x > 0$, a function that is not upper semi-continuous.

We have a result:
" A necessary and sufficient condition that $f$ be l.s.s(u.s.c) in $X$ is that, for each $\lambda\in R$, the set {${x\in X|f(x)>\lambda}$}
$[{x\in X|f(x)<\lambda}]$ be open


How do you use the math signs in the forum?

it's called latex... just click [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5261]here[/url] :)

Put dollar signs around the following expressions to experiment with $\LaTeX$, the markup language used on this site:

\LaTeX
2^5
2^{10}
a_1
a_{10}
\sqrt{n}
\frac{a}{b}
\mathbb{N}
\alpha\beta\gamma\Gamma
\sum_{i=1}^\infty\frac{1}{i^2}=\frac{\pi^2}{6}
\grave{x}\bar{x}\acute{x}
\widehat{a+b}
\lceil{x}\lfloor{x}\langle{x}
\frac{-b\pm\sqrt{b^2-4ac}}{2a}

***SPAM DETECTED, spams by Ricardo***
 :spam:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=241708#p241708]www.mathlinks.ro/Forum/viewtopic.php?p=241708#p241708[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=241706#p241706]www.mathlinks.ro/Forum/viewtopic.php?p=241706#p241706[/url]

If they were in one same forum ok, maybe he clicked twice. This is purposely SPAM :spam: since they were posted in different forums. Please consider removing the others.

[quote="blahblahblah"]In the future, why don't you just PM a moderator about it?[/quote]
Certainly...  :dry:

Thanks.  :)  I just hope
latex s system isnt so difficult to learn!  ;)

$\LaTeX$ shouldn't take you more than a week to master....Provided you spend around 10 minutes on it everyday...
 :)

Look at the links in my signature. It took me 30 minutes to master the basic writing in $\LaTeX$, so it shouldn't take you longer ;)


induction
algorithm
number theory
Euclidean algorithm

[quote="Inspired By Nature"][quote="fireemblem13"]Prove: Two consecutive Fibonacci numbers are relatively prime.[/quote]
[hide="proof"]
We'll proceed by contradiction.
Assume that there are some two consecutive Fibonacci numbers say $F_{n}$ and $F_{n+1}$ that have a common divisor d, where d is greater than 1. So, their difference $F_{n+1}-F_{n}= F_{n-1}$ will also be divisible by d. So, by induction, all lower fibonacci numbers must also be divisible by d. But, we know that $F_{1}=1$ which is clearly not divisible by d. Thus, we have reached a contradiction; and so consecutive Fibonacci Numbers are Relatively Prime.[/hide][/quote]

Hmm, or just by plain induction?

[hide]Let $F_{n}$ be the Fibonacci sequence. Clearly, $\gcd\left(F_{1},F_{2}\right) = 1$, so suppose $\gcd\left(F_{k},F_{k+1}\right) = 1$, for a positive integer $k$, then $\gcd\left(F_{k+1},F_{k+2}\right) = \gcd\left(F_{k+1},F_{k+2}-F_{k+1}\right) = \gcd\left(F_{k+1},F_{k}\right) = 1$.[/hide]

This is from the first AoPS book, yes?

[hide]Assume that $F_{k}$ and $F_{k+1}$ are not retatively prime. Then $F_{k}+F_{k+1}= F_{k+2}$, the next Fibonacci number. Therefore, $F_{k+2}$ cannot be reletively prime to $F_{k+1}$ because $F_{k}$ and $F_{k+1}$ share a common multiple, and in adding them together, we get the next Fibonacci number, $F_{k+2}$, that is a multiple of the same number. Thus, we are forced to conclude that no Fibonacci number is relativily prime to another, because all are a ultiple of this number. Therefore, if we can give a counterexample to this statement, we have proved that because those two numbers in our counterexample are reletively prime, the numbers before and after the pair are reletively prime, until every Fibonacci number is reletively prime to its neighbor. 8 and 13, two adjacent Fibonacci numbers, are reletively prime, becasue the only factor they share is one. Thus, we have proved that becase 8 and 13 are reletively prime, no two consecutive Fibonacci numbers are reletively prime. $QED$[/hide]

its very confusing, but i think its a basic proof.

[quote="Totally Zealous"]Actually the Fibonacci sequence proof in that book is different.
pg. 285 AoPS volume one.[/quote]

The proof I gave is one I derived. Not the one from the book.

[quote="t0rajir0u"][quote="buzzer11"]This is from the first AoPS book, yes?[/quote]

buzzer11 refers to the question, not to his proof.

Your solution is unnecessarily long.  Kurt Godel gives the nice proof  :)[/quote]

I know... sigh. I'm no good at proofs.  :lol:


In the diagram, AM=10, MN=20, NT=9, AB is parallel to NT, and TN is perpendicular to AN. If P is the point on AB so that MP+PT is minimal, then AP=?

A. 1.25
B. 1.5
C. 1.75
D. 2.25

[hide]
Let $ C$ be the foot of the perpendicular from $ P$ to $ NT$. Let $ TC\equal{}x$. Then $ AP\equal{}9\plus{}x$

Then

$ PM \plus{} PT \equal{} \sqrt{x^2\plus{}18x\plus{}181} \plus{} \sqrt{x^2\plus{}900}$

This value is minimized when $ x\equal{}0$, so $ \text{min}\{PM\plus{}PT\}\equal{} \boxed{30 \plus{} \sqrt{181}}$

[/hide]

Can $ AP$ be less than $ 9$ in length?  The diagram would suggest otherwise, which might be confusing.  Assuming it can be:

[hide="Answer."]
$ \frac{9}{4}.$
[/hide]


geometry
probability
algorithm
calculus
linear algebra

I am an undergraduate student. I have to do a research project for my homework assignment. This is my first maths research so I have no idea what topic should I select and how to start....

I want to do a research which related to Linear algebra because i have just completed this course. Can anyone suggest me the possible and interesting research topics in linear algebra?

==========================
Below is the guideline of the research areas which given by my professor:

The areas of research must be mathematics and its applications, such as:
Basic research: algebra, analysis, geometry, probability, statistics, etc.
Engineering applications: computing, internet, communications, information and digital technology, etc.
Commercial applications: economics, finance, logistics, management, decision science, operations research, transportation, etc.
Scientific applications: medicine, physics, chemistry, biology, environment and health problems, etc.
Innovation designs: image and visualization, games, puzzles, algorithms, etc.
Thank you very much !

It depends on your current knowledge on Linear Algebra. At undergraduate level, research on LA's applications should be easier and a good point to start.

Do you think they expect something new from me or just a conclusion of research at my current level?

I can also do researches related to Calculus or Multivariable Calculus, do you have any topics to reconmend in those subjects?

Thank you!


If anybody here likes fractals, I suggest you take a look at this:

http://www.fractalus.com/cgi-bin/glist


logarithms
integration
real analysis
real analysis unsolved

$ a_{n}\approx C \plus{} B\frac {1}{n\ln n}$ e.t.c.

[quote="Svejk"]Can you give some details ?How did you find this result ?[/quote]
Sure  :) .
1) $ \ln_{n}(n \plus{} 1) \equal{} \frac {\ln(n \plus{} 1)}{\ln n}$
2) Let's try to investigate sum $ \sum_{j \equal{} 2}^{n}\frac {\ln (n \plus{} 1)}{\ln n}$
Use $ \ln (1 \plus{} n) \equal{} \ln(n) \plus{} \ln(1 \plus{} \frac {1}{n}) \equal{} \ln n \plus{} \frac {1}{n} \minus{} \frac {1}{2n^{2}} \plus{} O(\frac {1}{n^{3}})$
Thus we have :
3)$ \sum_{j \equal{} 2}^{n}\frac {\ln (n \plus{} 1)}{\ln n} \equal{} \sum_{j \equal{} 2}^{n} \frac {\ln n \plus{} \frac {1}{n} \minus{} \frac {1}{2n^{2}} \plus{} O(\frac {1}{n^{3}}) }{\ln n} \equal{}$
$ \equal{} \sum_{j \equal{} 2}^{n} 1 \plus{} \frac {1}{n\ln n } \minus{} \frac {1}{2n^{2}\ln n} \plus{} O(\frac {1}{n^{3}\ln n})$
4)$ \sum_{j \equal{} 2}^{n} 1 \equal{} n \minus{} 1$,
5) $ \sum_{j \equal{} 2}^{n} \frac {1}{n\ln n } \equal{} \int_{2}^{n}\frac {1}{x\ln x}dx \plus{} o(1) \to \ln \ln n \plus{} c_{1} \plus{} o(1)$
6) $ \sum_{j \equal{} 2}^{n} \minus{} \frac {1}{2n^{2}\ln n} \to \minus{} \int_{2}^{n}\frac {1}{2x^{2}\ln x}dx \approx c_{2} \plus{} \frac {1}{2\ln n}\int_{2}^{n}\frac {1}{x^{2}} \plus{} o(.)\approx c_{3} \plus{} c_{4}\frac {1}{n\ln n }$
7) $ \sum O(\frac {1}{n^{3} }{\ln n}) \approx c_{5} \plus{} c_{6}\frac {1}{n^{2}\ln n }$
Hence :
$ \sum_{j \equal{} 2}^{n}\frac {\ln (n \plus{} 1)}{\ln n}\approx n \plus{} \ln \ln n \plus{} b \plus{} c_{4}\frac {1}{n\ln n } \plus{} O(\frac {1}{n^{2}\ln n })$
Therefore :
$ a_{n} \equal{} \sum_{j \equal{} 2}^{n}\frac {\ln (n \plus{} 1)}{\ln n} \minus{} (n \plus{} \ln \ln n ) \approx b \plus{} c_{4}\frac {1}{n\ln n } \plus{} O(\frac {1}{n^{2}\ln n })$
Simple problem for you  :
try to find $ c_{4}$  :)

[quote=Svejk]Study the convergence of the following sequence :

\[ a_n\equal{}\log_2 3\plus{}\log_3 4\plus{}...\plus{}\log_n(n\plus{}1)\minus{}(n\plus{}\ln \ln(n))\][/quote]
[url]https://artofproblemsolving.com/community/c7h402665p2245265:[/url]
$\left[\sum_{k=2}^n \log_{k}(k+1) \right] - (n+ \ln \ln n)$

$=\left[\sum_{k=2}^n \log_{k}(k+1) -1\right] - \ln \ln n$

$= \left[ \sum_{k=2}^n \log_k\left(1+ \frac{1}{k} \right)\right]- \ln \ln n$

$=\left[ \sum_{k=2}^n \frac{\ln (1+k^{-1})}{\ln k}\right] - \ln \ln n $

$=\left[ \sum_{k=2}^n \frac{ k^{-1} +O(k^{-2})}{\ln k}\right] - \ln \ln n$

$=\left(\left[ \sum_{k=2}^n \frac{1}{k\ln k}\right] - \ln \ln n\right) + \sum_{k=2}^n O\left(\frac{1}{k^2\ln k}\right)$

thus, we see that $a_n \to C$ some constant?


No, there are 3 cases:
1) $ x\leq\minus{}6$
2) $ \minus{}6<x<\minus{}2$
3) $ x\geq\minus{}2$

[quote="SteveB."]Are there 4 cases?
f(x) = 2|x+6| - 3|x+2|[/quote]
[hide]The critical points are -6 and -2 where the graph has breaks. So the graph will consist of 3 straight lines as given below.

When $ x\le\minus{}6$, $ f(x) \equal{}\minus{}2(x\plus{}6)\plus{}3(x\plus{}2) \equal{} x\minus{}6$.
When $ \minus{}6\le x\le\minus{}3$, $ f(x) \equal{} 2(x\plus{}6)\plus{}3(x\plus{}2) \equal{} 5x\plus{}18$.
When $ x\ge\minus{}2$, $ f(x) \equal{} 2(x\plus{}6)\minus{}3(x\plus{}2) \equal{}\minus{}x\plus{}6$.[/hide]


Today I was having an argument with a chemistry major about electrons changing $ n$ levels in an atom. I was convinced (from quantum) that every element has a sequence of visible frequencies that could be emitted (the Balmer series changing from $ n\equal{}3,4,5\ldots$ to $ n\equal{}2$).

He was convinced (from chemistry classes), that every element has only one visible frequency that is emitted, and this can be seen in some sort of burning experiment? I'm not a chemist, so I didn't really get the gist of the experiment.

I figured that it was probably some sort of simplification to not confuse chemistry majors about quantum mechanics, but he claims that chemists learn this because the outer shell is the only place in which shifts can naturally occur. The shells lower are stable and full. You have to force it unnaturally to get other emissions, so chemists don't care about the others.

I'm pretty sure this is not true. (You determine what element a planet is made of not be examining a single frequency, but by looking at the entire spectrum of emissions. This must therefore occur naturally).

I don't really know how to explain this, though, since I don't know enough about it. Does anyone have any suggestions for explanations? Or know what he is talking about? Are chemists actually taught this on a regular basis?

What you are talking about is called a flame test. http://en.wikipedia.org/wiki/Flame_test

And the emission spectrum of a planet or star is due to emission from a bunch of elements on the planet. Although yes, there is an emission spectrum for hydrogen for all the series. I'm not sure if that is true for all elements, but theoretically it should be yes.

In practice, though, I suspect that emission may be limited to the more reactive valence electrons? Not too sure.

Thanks. That is helpful...kind of. I'm going to ask my prof tomorrow. I just wanted to make sure the question wasn't going to sound overly stupid (like I wasn't paying attention all semester).


Prove that the equation
log2(1-x^2)=log|x-1|(2) has at most one solution
(the bases of the logs are 2 and |x-1|)

clearly x must lie in (-1,0).
so the equation becomes
ln (1-x^2) / ln 2 = ln 2 / ln(1-x)
which is
ln(1-x^2)ln(1-x) = (ln2)^2
if we can proof that
f(x)=ln(1-x^2)ln(1-x) is monotonic in (-1,0),
then f(x)=(ln2)^2 has a unique solution.

in fact, f(x) is increasing.


Can you people tell me the artist of the song The late George Wallno worked on High Street and lived on manners. I have heard it at my friend’s place but not been able to find its artist.


Positive integers $ a$ , $ b$ , and $ c$ are chosen so that $ a$ is less than $ b$ is less than $ c$, and the system of equations \[ 2x\plus{}y\equal{}2003\text{ and }y\equal{}|x\minus{}a|\plus{}|x\minus{}b|\plus{}|x\minus{}c|\] has exactly one solution. What is the minimum value of $ c$?

I keep getting that the minimum value is $ 669$, but that's not the correct answer.

I got the following values to get the minimum of $ c$:
$ a \equal{} 667 , b\equal{}668 , c\equal{}669 , x \equal{} 1 , y \equal{} 2001$

Could somebody help me on this?

See the resources section:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1366815#1366815[/url]


limit
real analysis
real analysis unsolved

If not why?

http://www.physicsforums.com/showpost.php?p=1083012&postcount=1

This is not a proof straight from the definition of the limit -- it uses the additivity of $\lim$ and the continuity of two functions. I would write something like:
if $\sqrt{(x-3)^{2}+(y+1)^{2}}<\delta$, then $|x-3|<\delta$ and $|y+1|<\delta$, hence $|x-2y-5|<3\delta$. Thus, the definition of the limit holds with $\delta=\epsilon/3$.


Is there any math tricks that can be used to win Chess?

Well there is probability, but to use it in chess is absurd. There are billions of ways that the board could be arranged. For chess I suggest studying good openings to build up a good defense. Its also good to do puzzles cause they help your end game. Just think out your moves.

Wrong forum by the way.

I don't know if there is any math trick to win chess, but I know there are  some problems how with use of the chess board and figures. for example one problem is to arrange  8 "queens"(I don't know how this figure is called in english) on the board so that none of them was able to kill another. and there are many problems like that...

P.s if I'll find some math trick to win at chess, I'll tell you ;)

Are you trying to improve your chess skills?  :P

If there was a mathematical approach to chess, it wouldn't be practical in a real life situation. Calculating everything under a time frame (I doubt they would let you have a calculator, pencil, or paper) is hardly a benefit. It's better just to, well, play.

It's not as simple as what's the probability of this random move being a winner, or simple combinatorics. Mathematical chess analysis consists of complex algorithms (to suit for individual piece movement).

Moved to Math and Strat.

[quote="Elio (n)"]The problem you talked about is discovered  (and solved sure) by Frederich GAUS.[/quote]
 I know it.  There are 92 ways to do that. ;)

More possible positions than particles in the universe, the longest chess game can be about 5700 moves!

How many ways can you complete the "Knight game" in which a knight meets every square of the board?

1. Prove impossible if you omit a square.
2. Possibilities lessen if the knight can meet each square only once. How many now?

You can use math when dealing with opposition.

mathelete, DON"T REVIVE!

Staying on topic, being good at math will hlep in chess, as it requires logic, but I'm not sure of any tricks that have a direct correlation with math....except for probability.


function
real analysis
real analysis unsolved

Let $ f: (a,b)\to{\mathbb{R}}$,$ a,b>0$ a function with the derivate increasing. Prove that $ g: (a,b)\to{\mathbb{R}}$,$ g(x)\equal{}xf'(x)\minus{}f(x)$ is increasing on $ (a,b)$

with the derivetive increasing means that : $ f''>0$
it's a simplest question ( :read: ) just write $ g'(x)\equal{}xf''>0$ on $ (a,b)$ $ a>0$

[quote="tarta"]Sorry but I don't understand something. If $ f'$ is increasing we know about the existence of $ f''$?
I think we should know that $ f'$ is derivable[/quote]
OOK,  :lol:  just consider $ \frac{g(x_{1}) \minus{} g(x_{2})}{x_{1}\minus{}x_{2}}$ when $ x_{1} > x_{2}$ and use same argument. I mean don't write $ f''$ , instead $ f''$ write  $ \frac {f'(x_{1}) \minus{} f'(x_{2})}{x_{1} \minus{} x_{2}}$
you will get :
$ \frac {g(x_{1}) \minus{} g(x_{2})}{x_{1} \minus{} x_{2}} \approx \frac {f'(x_{1}) \minus{} f'(x_{2})}{x_{1} \minus{} x_{2}} > 0$
I mean : how we prove that : $ (fg)' \equal{} f'g \plus{} g'f$ ?
use same argument with this $ xf'(x)$ or I mean :
 $ \frac {x_{1}f'(x_{1}) \minus{} x_{2}f'(x_{2})}{x_{1} \minus{} x_{2}} \equal{} \frac {x_{1}f'(x_{1}) \minus{} x_{2}f'(x_{2})}{x_{1} \minus{} x_{2}} \plus{} \frac {x_{1}f'(x_{2})}{x_{1} \minus{} x_{2}} \minus{} \frac {x_{1}f'(x_{2})}{x_{1} \minus{} x_{2}}$ and e.t.c


* In intermediate section you can insert the problem source when editting your topic only, not while initially posting it.

* How are we supposed to use the other problems section? Only moderators can post new topics there :?

I think you'll want to get these things fixed, thx! :)

*ahem*

*AHEM*

(desperately tries to draw attention :))

For you second question, I think that's been fixed, I posted a test topic today

[quote="Peter VDD"]yes, the second one is solved.

but I tried some hours ago and it didn't work so it's only recently been done.

and for the second you'll need source code access - I don't know who has it on AoPS? Valentin and who from AoPS?[/quote]Richard also manages the code. However there is no need for that, because that is how it is intended to be. As in the old beginner's corner on ML, when first posting a problem you need not input the source field (because many problems do not have a source, people are not used with using that field there, etc.) If one however wants to specifically put the source, he can edit the message afterwards and add it.

seems an awkward logics :D

(if people don't want it they don't use it alright? :?)

Oh anyway they you'll just have to accept I put it in the topic title :D


I've only taken Physics honors. Is there any chance of me making semifinalist?

Although I am not aware of the selection procedure in your country(US???), I believe taking or not taking a course is not that much a 'deciding' factor. Just know the basic all concepts and think!!!


1) CH3COCH2CH2CH2Cl + aq NaOH ;then H2O --> Product

2) (C2H5COO)2 + C6H5CH2COOC2H5 + C2H5ONa + HEAT --> Product

[hide="Answer for 2"]Assuming that "(C2H5COO)2" is EtOCOCOOEt, then the product would be PhCH(CO2Et)COCO2Et.[/hide]

yes, but my book says on heating, PhCH(CO2Et)COCO2Et loses carbon monoxide to form, diethyl phenylmalonate. I didnt understand why it should though. Any ideas?

How about the first one

I think the product is [b][i][u]CYCLOPENTANONE[/u][/i][/b]
The alpha hydrogen next to the carbonyl carbon is abstracted and then inramolecular Ei takes place due to anchimeric assistance to give cyclopentanone

[quote="Zimrock"]yes, but my book says on heating, PhCH(CO2Et)COCO2Et loses carbon monoxide to form, diethyl phenylmalonate. I didnt understand why it should though.[/quote]

I have never heard of something like that. However, we are always learning. Therefore, provide the mechanism of that transformation (such a claim on a reasonable book must by acompanied by a mechanism).

[quote="chemrock"]I think the product is [b][i][u]CYCLOPENTANONE[/u][/i][/b]
[/quote]
i was thinking of the same but why is the treatment with water req.

[quote="zimrock"]
yes, but my book says on heating, PhCH(CO2Et)COCO2Et loses carbon monoxide to form, diethyl phenylmalonate. I didnt understand why it should though. Any ideas?
[/quote]
The compound is unstable as it has two carbonyl groups adjacent to each other, so one of the two groups has to be eliminated on heating. My guess is that the keto group is attacked by the base and cooc2h5 is eliminated to give the ester. I know -cooc2h5 is not a very good leaving group but probably the relief from electronic repulsion favors the forward reaction. And –cooc2h5 eliminates –oc2h5 to give CO. for all I know, there  could be an intramolecular attack by oc2h5 of the ester on the keto foll by elimination of CO. but that would probably not be the case here bcoz the rigid structure of ester(unless the elimination is concerted.)
P.S I am only guessing and the above proposed could be and probably is wrong

In organic chemistry, reactions that occur with extrusion of carbon monoxide are quite rare, and almost always with harch conditions (an exception is alpha-hydroxiacids). And I certainly have never heard of such a reaction for alpha-keto esthers. But I am always prepared to learn something new. If somebody can provide a reference, that would be nice.

Zimrock, can you give the first answer and chemrock what role does anchimeric assistance play here.

Sorry it is not anchimeric assistance :blush:
I just wanted to say that it followed a Ei mechanism. My guess about water being included is probably that they want to indicate a polar solvent so that the Cl- is eliminated easily (as it is effectively solvated).

[quote="chemrock"]
Sorry it is not anchimeric assistance
I just wanted to say that it followed a Ei mechanism. My guess about water being included is probably that they want to indicate a polar solvent so that the Cl- is eliminated easily (as it is effectively solvated).
[/quote]
the way the problem is stated h2o seems to be more than a solvent. why is it aq NaOH ;then H2O??

yes the answer is cyclopentanone.
but i was thinking on the lines of SN2 followed by cyclic hemiacetal formation

That cannot be, because you would obtain a ring with an oxygen atom. Enolate formation followed by intramolecular SN2 seems a better pathway.


I don't know about the picture, but you assign letters to the circles with the top one being $ A_1$, the next three being $ B_1, B_2, B_3$, etc. You get the point.

I know it doesn't really matter and that everyone will be doing that to represent it, but still remember that the rules say you shouldn't be posting up stuff on how you're solving it.

[quote="crazypianist1116"]Is there any way to make the picture associated with Problem 1 in Latex? Is it possible to fill in the circles with numbers? If so, how?[/quote]

You can just draw the picture in Microsoft Paint or whatnot, which allows you to label/fill in the circles to your specifications. From there, you can use the /includegraphics command.


A scarf is on sale at 20 percent off the list price.  The following week, the price of the scarf is reduced 25 percent of the sale price.  Rita called a friend and told her that the price of the scarf is now 45 percent off the original price.

(a) Do you agree with Rita?

(b) If you don't agree with Rita, what is the ACTUAL percent of the discount off the original price?

MY WORK:

For part (a), I disagree with Rita because list price is not the same as the sale price, right?  Rita added 20 percent and 25 percent and got 45 percent.  She added correctly but one thing does NOT go with the other.  List price DOES NOT = sale price.

For part (b), I said the answer is 20 percent or 80 percent.

The book's answer is 40 percent for part (b).

How do I get 40 percent for part (b)?

ALSO:

Is the list price the same as the original price?

(a)No, you cannot simply add them
(b)For simplicity's sake, let's say the scarf was originally 100 dollars. First it is marked 20% off, so you will pay 80%. $100*(.8)=80$ dollars. Now this 80 dollar sale price is marked down 25%, which means you will pay 75% $80*(.75)=60$ dollars Thus, the price has gone from 100 dollars down to 60, which is a 40% decrease.

Yes, the list price is the original price.

Understood except when you said that "Now this 80 dollar sale price is marked down 25 percent."

What do yo mean by MARKED DOWN?


function
integration
inequalities
floor function
algebra unsolved
algebra

Calculate $ \lfloor1\plus{}\cfrac{1}{\sqrt{2}}\plus{}\cfrac{1}{\sqrt{3}}\plus{}...\plus{}\cfrac{1}{\sqrt{10^6}}\rfloor$

I think it's 1998.

You 're right. I wrote a small Java program to test it.

[code]
public class Summation_Of_Invert_SQRT{
	public static void main (String[] args){
		double r = 0;
		for (int i=1; i<1000001; i++){
			r += ((double) 1) / (Math.sqrt(i));
		}
		System.out.println(r);
	}
}
[/code]

That's the computer's stuff. I 've thought of a new kind of approach such a problem which is making use of the Taylor expansion of the function $ f(x) \equal{} \frac{1}{\sqrt{1\plus{}x}}$ in order to estimate the sum. But it seems that this does not help.

Define $ f(x) \equal{} \frac {1}{\sqrt {x}}$ on the positive reals and call $ S \equal{} \sum_{n \equal{} 1}^{10^6}\frac {1}{\sqrt {n}}$.  Clearly $ f(x)$ is strictly decreasing, or $ \frac {1}{\sqrt {n}} > \int_n^{n \plus{} 1}f(x)dx > \frac {1}{\sqrt {n \plus{} 1}}$ for all positive integers $ n$.  Therefore, adding the previous inequality for $ n$ from $ 1$ to $ 10^6 \minus{} 1$, we obtain
\[ S \minus{} \frac {1}{1000} > \int_1^{10^6}\frac {dx}{\sqrt {x}} \equal{} 2\sqrt {10^6} \minus{} 2\sqrt {1} \equal{} 1998 > S \minus{} 1,
\]
and hence
\[ 1999> S > 1998 \plus{} \frac {1}{1000},
\]
yielding trivially $ \lfloor S\rfloor \equal{} 1998$.


Game 3 is starting! Just a few rules.(the same as 1 and 2)
1. Anyone can hurt and heal.
2. To go quicker, hurt twice and heal once.
3. After a post you must wait ONE hour before hurting and healing again.
4. No double posting
5. No healing yourself.
6. You start off with 25 points.

To start the game off, hurt ccy twice, heal now a ranger
i_like_pie-25
BanishedTraitor-25
worthawholebean-25
ckck-25
now a ranger-26
ccy-23
junggi-25
anirudh-25

i_like_pie-25
BanishedTraitor-25
worthawholebean-25
ckck-23
now a ranger-26
ccy-24
junggi-25
anirudh-25

hurt ckck heal ccy

Hurt anirudh, heal ccy

i_like_pie-25
BanishedTraitor-25
worthawholebean-25
ckck-23
now a ranger-26
ccy-25
junggi-25
anirudh-23

i_like_pie-25
BanishedTraitor-25
worthawholebean-25
ckck-21
now a ranger-26
ccy-25
junggi-25
anirudh-25

hurt ckck heal ccy

It should be:

i_like_pie-25
BanishedTraitor-25
worthawholebean-25
ckck-21
now a ranger-26
ccy-26
junggi-25
anirudh-23

Hurt now a ranger, heal i_like_pie

i_like_pie-26
BanishedTraitor-25
worthawholebean-25
ckck-21
now a ranger-25
ccy-26
junggi-25
anirudh-23

hurt ccy heal ckck

i_like_pie-26
BanishedTraitor-25
worthawholebean-25
ckck-22
now a ranger-25
ccy-25
junggi-25
anirudh-23

hurt anirudh twice, heal ckck

i_like_pie-26
BanishedTraitor-25
worthawholebean-25
ckck-23
now a ranger-25
ccy-25
junggi-25
anirudh-21

hurt I-like-pie *2 heal anirudh

i_like_pie-24
BanishedTraitor-25
worthawholebean-25
ckck-23
now a ranger-25
ccy-25
junggi-25
anirudh-22

hurt anirudh twice, heal ckck.

i_like_pie-24
BanishedTraitor-25
worthawholebean-25
ckck-24
now a ranger-25
ccy-25
junggi-25
anirudh-20

Hurt anirudh twice, heal now a ranger

i_like_pie-24
BanishedTraitor-25
worthawholebean-25
ckck-24
now a ranger-26
ccy-25
junggi-25
anirudh-18

BanishedTraitor forgot to take a point off NAR.

Hurt ckucku, heal anirudh

_like_pie-24
BanishedTraitor-25
worthawholebean-25
ckck-22
now a ranger-25
ccy-25
junggi-25
anirudh-19

[u]Proposition: The Anirudh-God plan[/u] [i]Anirudh gains 500 extra points for no reason whatsoever. [/i]

If you agree, vote[color=red] AYE[/color] along with your hurt/heal. If not, then screw you

hurt ckck heal anirudh
_like_pie-24
BanishedTraitor-25
worthawholebean-25
ckck-20
now a ranger-25
ccy-25
junggi-25
anirudh-20

Hurt junggi twice.

i_like_pie-24
BanishedTraitor-25
worthawholebean-25
ckck-20
now a ranger-25
ccy-25
junggi-23
anirudh-20

Hurt i_like_pie twice, heal now a ranger

i_like_pie-22
BanishedTraitor-25
worthawholebean-25
ckck-20
now a ranger-26
ccy-25
junggi-23
anirudh-20

i_like_pie - 6
ckck - 33
NAR - 3

hurt i like pie twice , heal nar

hurt NAR heal pie

i_like_pie - 7
ckck - 33
NAR - 1

hurt i like pie twice heal nar

i_like_pie - 5
ckck - 33
NAR - 2

hurt i like pie twice heal nar

i_like_pie - 3
ckck - 33
NAR - 3

hurt nar hurt i like pie heal ckck

i_like_pie - 3
ckck - 34

nar is now dead.

The game will end VERY soon...

Hurt 1lp*2, heal ckck

i_like_pie - 1
ckck - 35

[quote="now a ranger"]hurt i like pie

[size=150][b]ckck - 35 IS THE WINNER!!!  :lol: [/b][/size][/quote]
Make that 36-you are going to heal him that last turn, right?

[quote="junggi"]gosh alts should be outlawed?
its very unfair to have alts
i mean come on, who could be so lifeless to cheat in a pointless onliine forum game?
i_like_pie - 4
ckck - 33
NAR - 1
heal pie hurt nar[/quote]

As i had to explain in H/H 2, no one was using any alternates.

ckck goes to my school.  his name is chris.
nar lives in alabama
asianpower goes to my school.  his name is patrick. or daniel, they have similar names, i forget.
im me, obviously.
ragnarok goes to my school.  idk if he wants me to tell u his name, but it starts with a p, and ends with a ratik.

i think thats everyone.
quit accusing people of cheating.
its more like, ganging up.
 :lol:

[quote="junggi"][quote="Walk Around The River"]
its more like, ganging up.
 :lol:[/quote]
which is still unfair, you guys have like 2000 people in the alliance, its guaranteed you guys will win
i think someone proposed this earlier, [b]You can't heal or hurt the same person twice in a row[/b]
that should make it more fair[/quote]
People outside the alliance could form a much bigger alliance and give them a taste of their own medicine.  :)

caricature was an alt. 100% sure of this


This question is again from AoPS book and I have trouble on understanding it's wording on my answer. I hope I can get better answer from here :).

Here it is:
(#35 page 32, chapter 3 Linear Equation)

Mike and Joey bought identical loaves of bread and packages of bologna.  Mike made sandwiches with 5 slices of bologna and had 4 slices of bread left when he ran out of meat. Joey made sandwiches with 4 slices of bologna and had 4 slices of meat when he ran out of bread. How mnay slices of bread were in each loaf?
(MA :theta:  1992)

Call the number of slices of bread in a package x and the number of slices of bologna in a package y.

So we have

y/5=x/2-2

y/4-1=x/2

From the first equation we get

y/5+2=x/2=y/4-1

y/5+3=y/4

4y+60=5y

60=y

28=x

I'm assuming your average sandwich looks like this:
----------
|Bread|
----------
|Meat|
----------
|Bread|
----------

Or when we got the equations we could have graphed them

Hello, Silverfalcon!

The set-up was certainly challenging.

I approached it differently . . .

Quote:Mike and Joey bought identical loaves of bread and packages of bologna.
Mike made sandwiches with 5 slices of bologna each
and had 4 slices of bread left when he ran out of meat.
Joey made sandwiches with 4 slices of bologna each
and had 4 slices of meat when he ran out of bread.
How many slices of bread were in each loaf?

[hide]

Let M = number of sandwiches Mike made.

Mike used: 5M slices of bologna and 2M slices of bread.

He had 4 slices of bread left over and 0 slices of meat.

His package of bologna contained: 5M slices.

His loaf of bread contained: 2M + 4 slices.

Let J = number of sandwiches Joey made.

Joey used: 4J slices of bologna and 2J slices of bread.

He had 4 slices of bologna left over and 0 slices of bread.

His  package of bologna contained: 4J + 4 slices.

His loaf of bread contained: 2J slices.

Since the packages of bologna and loaves of bread were identical

we have: 5M = 4J + 4, and 2M + 4 = 2J

Solving the system: M = 12, J = 14

Therefore, there were 28 slices of bread in each loaf

 and 60 slices of bologna in each package.

[/hide]

Thanks to both of the users for helping me comprehending this question.

Soroban, an excellent way you gave me. It's so EASY to understand. Thanks.

RC-7th, thanks too. Except I didn't quite get what you meant.

But it's okay..

Thanks!!

\[a_{0} = 1\]
\[a_{1} = 3\]
\[a_{n}^2 - a_{n-1}a_{n+1} = (-1)^n\]


Find all values of x which satisfy: \[\displaystyle 4^{x^3+5x^2-6x}=1\]

Billy

$4^{x^3+5x^2-6x}=1\Longleftrightarrow x^3+5x^2-6x=0\Longleftrightarrow x(x-1)(x+6)=0$.
The answer is $x=-6,0,1$

HOw did you get that. Next time, please post the explanation

Oh. Whoops. I forgot the 4^0 th rule. Well, here's my way:

$4^{x^2-5x^2+6x}=0
\\
\\
x^3-5x^2+6x=0
\\
\\
x(x^2-5x+6)=0
\\
\\
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
\\
\\
a=1, b=-5, c=6
\\
\\
x=\frac{5\pm \sqrt{25-4(1)(6)}}{2}
\\
\\
\frac{5\pm1}{2}
\\
\\
(3, -2, 0)$ except I think I made a mistake somewhere but where. Can someone find out where I made a mistake?

It would be ideal next time to copy the problem down correctly  :rotfl:

:rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl:
:rotfl: :rotfl: Oh i copyied it wrong. :rotfl: :rotfl:
:rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl::rotfl:

$\displaystyle 4^{x^3+5x^2-6x}=1$

$x^3+5x^2-6x=0$

$x(x^2+5x-6)=0$

$x(x+6)(x-1)$

which is 6, -1, 0

[hide]
Since anyting to the power 0 is 1, the answer is 0.
Factoring x gives you
$\displaystyle x(x^2 + 5x - 6) = 0$
$\displaystyle x(x - 1)(x + 6) = 0$
So the answers are: $ x = \framebox{-6,1,0}$[/hide]

[quote="anirudh"]why -6 and +1, when in the factoring it was 6 and -1 :?[/quote]

Many people fall for this trick.

You set each factor to zero.

x+6=0
x-1=0

You solve the equations. x is always the opposite sign as it is in factoring  :)

ok thanks. Also, was this moved. here. I thought i saw this at the getting started forum.  :)

No, it is not copied from another forum.


Let $ a;b;c > 0;abc \equal{} 1$
Find max value of
$ T \equal{} \sum_{cyc} \frac {bc}{a\plus{}1}$

It doesn't have a maximum value.

Consider just one term, $ \frac {bc}{a \plus{} 1}$. This can be simplified as $ \frac {1}{a^2 \plus{} a}$. Note that as a approaches 0, this fraction grows larger and larger without bound, and so does the sum.

(This holds because all terms of the cyclic sum are positive.)

$ \frac{bc}{a\plus{}1}\equal{}\frac{1}{a^2\plus{}a}$
If we let $ a\minus{}>0$ then it will be ∞....

I think may be $ a\plus{}b\plus{}c\equal{}1$,not $ abc\equal{}1$


Let $S(x)$ be sum of digits of the natural number $x$. If $S(1+2+...+n)$ has all equal digits $a$, find all posiblle values of $a$.


geometry
circumcircle
geometric transformation

Let $CP\cap{AB}=\{X\}$, then $AX=XP$ $\Longleftrightarrow$ $\angle{BRA}=\angle{AQB}$.
Consequently, $AQRB$ is cyclic $\Longleftrightarrow$ $\angle{BPR}=\angle{PRB}=\angle{PQR}$.

i think that's wrong... or, at least, it's not obvious and there's a considerable number of steps missing. explain.

Dear pleurestique
There is nothing wrong.

1. $x=\angle{AQP}=\angle{PAB}=\angle{APX}=\angle{CPR}$.
2. $y=\angle{PBA}=\angle{BQP}=\angle{BCP}$.
3. $\angle{AQB}=x+y=\angle{ARB}=\angle{RPB}$, thus $AQRB$ is cyclic.
4. It remains to see that $\angle{RQB}=x$.

I hope this time I presented the data clearly enough for you.

I suppose I'll flesh out Sailor's solution, as it took me a bit to work through it and see why it was true.  It's a very slick solution, however.

1. We have $ \angle AQP \equal{} \angle PAB$ since both are half the measure of minor arc $ AP$ (since $ AB$ is tangent to $ \Gamma_1$).  Then since $ APX$ is isosceles with $ AX \equal{} PX$, we have $ \angle PAB \equal{} \angle PAX \equal{} \angle APX$.  Finally, since $ A, P, R$ are collinear and $ X, P, C$ are collinear, we have $ \angle APX \equal{} \angle CPR$.

2. We have $ \angle PBA \equal{} \angle BQP$ since both are half the measure of minor arc $ BP$.  Then $ \angle BQP \equal{} \angle BCP$ since $ BPQC$ is cyclic by definition.

3. Then $ \angle AQP \plus{} \angle PQB \equal{} \angle AQB \equal{} x\plus{}y$.  But also $ \angle ARB \equal{} 180^{\circ}\minus{}\angle CRP \equal{} \angle CPR \plus{} \angle PCB \equal{} x\plus{}y$.  Similarly, $ \angle RPB \equal{} \angle RPB \equal{} 180^{\circ}\minus{}\angle BPA \equal{} \angle PBA \plus{} \angle PAB \equal{} x\plus{}y$.  Since $ \angle AQB \equal{} \angle ARB$, $ AQRB$ is cyclic.

4. Since $ AQRB$ is cyclic, $ \angle RQB \equal{} \angle RAB \equal{} \angle PAB \equal{} x$.  Then we have $ \angle PQR \equal{} \angle RQB \plus{} \angle BQP \equal{} x\plus{}y$.

5. (to finish) Let $ O$ be the circumcenter of $ PQR$.  Then $ \angle POR \equal{} 2(x\plus{}y)$.  We have also $ \angle PBR \equal{} 180^{\circ}\minus{}(\angle RPB \plus{} \angle PRB$).  But $ \angle PRB \equal{} \angle ARB \equal{} x\plus{}y$ and $ \angle RPB \equal{} x\plus{}y$ as above, hence $ \angle PBR \equal{} 180^{\circ}\minus{}2(x\plus{}y)$.  Thus $ PBRO$ is cyclic.  Since $ POR$ is isosceles, we have $ \angle OPR \equal{} \angle ORP$.  But also $ \angle BPR \equal{} \angle BRP$, hence $ \angle BPO \equal{} \angle BRO$.  But $ \angle BPO \plus{} \angle BRO \equal{} 180^{\circ}$ since $ PBRO$ is cyclic, hence $ \angle BPO \equal{} \angle BRO \equal{} 90^{\circ}$.  It follows that $ BP$ and $ BR$ are tangent to the circumcircle of $ PQR$.

I think my solution is quicker and it is purely angle chasing.
[hide="Solution"]
Use the facts about all the tangents and let $\angle{PAB} = \angle{PQA} = \alpha$ and $\angle{PQB} = \angle{PBA} = \beta$ and $\angle{QAP} = \angle{QPC} = \angle{QBC}= \gamma$. Note that $ABRQ$ is cyclic because $\angle{QAR} = \angle{QBR}=\gamma$. This means that $\angle{RAB} = \angle{RQB} = \alpha$. Furthermore, $\angle{PQR} = \alpha+\beta$. Therefore, it suffices to show that $\angle{BPR} = \angle{BRP} = \alpha+\beta$. $\angle{BPR} = \angle{BAP} + \angle{ABP} = \alpha+\beta$. Additionally, because $ABRQ$ is cyclic, $\angle{BRP} = \angle{AQB} = \alpha+\beta$. This completes the proof.
[/hide]

Edit: I noticed that $Q$ was the center of a spiral similarity that took $APQ$ to $BRQ$. If anyone has a nice solution using spiral similarity, could they post it.

Join AQ and BQ. RBQ = QPC = PAQ (alternate segment theorem). So, ABRQ is cyclic.
Again, BQR = BAR = AQP.
So, AQB = PQR
or, ARB = PQR.
Performing similar angle chasing, we get BPR = PQR. This completes the proof.

Assume $| \Gamma_1 | > | \Gamma_2 |$,
$$\angle CPQ=\angle PAQ=180^{\circ}-\angle AQR-\angle ARQ=180^{\circ}-\angle PQR-\angle AQP-\angle ARQ=\angle QPR-\angle PAB$$
$$\angle CPR=\angle PAB \implies \Delta APB \sim \Delta PRC \implies \Delta BPR \text{ is isosceles}$$
$$\angle QAR=\angle QPC=\angle QBR \implies AQRB \text{ is cyclic} \implies \angle BQR=\angle BAR=\angle AQP \implies \angle PQR=\angle AQB=\angle PRB $$

Pretty easy. Three steps:

1) Show that $BP=BR.$ (Motivation: This is true by problem statement and probably will make life a lot easier.)

2) Show that $ABRQ$ is cyclic. (Motivation: We have no idea what the heck $R$ is, and messing around with angles long enough/having a good diagram will make this [i]feel[/i] correct.)

3) Reflect $P$ about the midpoint of $AB$ to finish. (Motivation: Cyclic quadrilaterals are cool, $P'$ lies on $(ABRQ),$ and $\triangle ABP'$ is really easy to handle.

Okay, full solution.

Note that
\[\angle BPR=\angle BAP+\angle ABP=\angle AQP+\angle PBQ=\angle AQB\]
and that
\[\angle BRP=\angle RPC+\angle RCP=180^{\circ}-\angle APC+\angle BCP=\angle AQP+\angle BQP=\angle AQB,\]
so $\angle BPR=\angle BRP.$

Now note
\[\angle AQB=\angle BAP+\angle ABP=180^{\circ}-\angle APB=\angle BPR=\angle BRP,\]
so $ABRQ$ is cyclic.

Now reflect $P$ about the midpoint of $AB$ to get $P'.$ Then note
\[\angle PQR=\angle P'QR=\angle P'AR=\angle P'AB+\angle BAR=\angle ABR+\angle BAP=\angle BPR.\]

[quote=shobber]Let $\Gamma_1$ and $\Gamma_2$ be two circles intersecting at $P$ and $Q$. The common tangent, closer to $P$, of $\Gamma_1$ and $\Gamma_2$ touches $\Gamma_1$ at $A$ and $\Gamma_2$ at $B$. The tangent of $\Gamma_1$ at $P$ meets $\Gamma_2$ at $C$, which is different from $P$, and the extension of $AP$ meets $BC$ at $R$.
Prove that the circumcircle of triangle $PQR$ is tangent to $BP$ and $BR$.[/quote]
Could you send an appropriate visual for the problem if I ask?

a kind of different angle-chasing solution

[hide = Solution]

Notice that proving the tangency of $\overline{BP}$ and $\overline{BR}$ to the circumcircle of $PQR$ is equivalent to showing that $\measuredangle RQP = \measuredangle RPB = \measuredangle BRP$ thus meaning that we can prove the desired statement by showing that $\triangle BRP$ is isosceles with $\overline{BP} = \overline{BR}$ and then that $\angle BRP = \angle RPQ$. Let $BQ \cup QR = F$.

Angle chasing for $\angle BPR$ with cyclic quadrilateral $BPQC$, $\angle LAP = x, \angle CBQ = z$ and $\angle LBP = y$ yields $\angle LPA =x, \angle PLB = 180 - 2x, \angle LPB = 180 - (2x + y)$ and $\angle BPR = x+y$.

Similarly, for $\angle BRP$ we find that $\angle BFR = 180 - (x+y+z)$ and thus $\angle BRF = \angle BRP = x+y$.

Thus, $\angle BPR = \angle BRP = x+y$ which implies that $\triangle BPR$ is isosceles.

By the properties of tangents to a triangle and cyclic quadrilaterals we find that $\angle BQP = \angle BCP = y$ and $\angle PQA = x$ thus meaning that $ABRQ$ is cyclic since $\angle BRA = \angle BQA = x+y$. Then, $\angle ABR + \angle AQR = 180 - (2x + y) + x + y + \angle BQR = 180 \Rightarrow \angle BQR = x$.

Thus, $\angle BRP = \angle PQR = \angle PQB + \angle BQR = x + y$ so $\overline{BP}$ and $\overline{BR}$ are tangents to the circumcircle of $\triangle PQR$ and we are done.
[/hide]

Good practice for angle-chasing.

In the diagram attached, note that $\angle ABP=\angle BCP=\angle MQB,$ the first congruence coming from the fact that $AB$ is tangent to $\Gamma_2$ and the second coming from the fact that $P,B,C,Q$ are concyclic(these are the red angles in the diagram). Next, note that $\angle QBC=\angle QPC=\angle QAP,$ the first coming from concyclic and the second coming from the fact that $PC$ is tangent to $\Gamma_1$. From this, it is easy to see that $A,B,R,Q$ are concylic. Notice that we are trying to prove that $\angle BRA=\angle BPR=\angle PQR.$ Let's first try proving that $\angle BRA=\angle PQR,$ and then prove that $\angle BRA=\angle BPR$(equivalent to proving that $BR=BP$). Notice that $\angle BRA=\angle AQB,$ so the first statement is equivalent to proving that $QP$ and $QB$ are isogonal with respect to $Q$. This is easy to see because $\angle AQM=\angle BAR$ from $AB$ being tangent to $\Gamma_1$, and then $\angle BAR=\angle BQR$. Now note that $\angle BQR=\angle BAR$ and $\angle ABP=\angle MQB$(from tangents), so by exterior angle sum, we have that $\angle BPR=\angle BRP$. We are done!!

Claim: $BP = PT$.
Proof : $\angle BTP = \angle BEP + \angle EPT = \angle ABP +\angle BAP = \angle BPT$.

Claim: $ABTQ$ is cyclic.
Proof : $\angle BQA = \angle BQP + \angle PQA = \angle PAB + \angle PBA = \angle BPT = \angle BTP = \angle BTA$.

Now we have $\angle PQT + \angle PQB + \angle BQT = \angle PEB + \angle TAB = \angle BTP$ so $BT$ is tangent to $PQT$. we have $BP = BT$ so $BP$ is tangent too.

Another finish after proving that $ABPR$ is cyclic and $BP=BR$ is to consider the midpoint of $PR$, let it be $X$. Then $M$ is the circumcenter of $(ABX)$ since $BX\perp PR$. Note that $$\angle MXA=\angle MAX=\angle AQP.$$
So $A,M,X,Q$ are concyclic. This gives $$\angle MQX=\angle MAX=\angle BQR.$$
Hence $BQ$ is $Q$-symmedian of $PQR$. This with the fact that $BP=BQ$ gives the desired result.

(using @2above diagram)


geometry
cyclic quadrilateral
national olympiad

British Mathematical Olympiad 2005-2006 round 1 was held in Nov 30, 2005.

Here are the problems.

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=63971
Let $n$ be an integer greater than $6$. Prove that if $n-1$ and $n+1$ are both prime, then $n^{2}(n^{2}+16)$ is divisible by $720$. Is the converse true?

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=63965
Adrian teaches a class of six pairs of twins. He wishes to set up teams for a quiz, but wants to avoid putting any pair of twins into the same team. Subject to this condition:
i) In how many ways can he split them into two teams of six?
ii) In how many ways can he split them into three teams of four?

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=63969
In the cyclic quadrilateral $ABCD$, the diagonal $AC$ bisects the angle $DAB$. The side $AD$ is extended beyond $D$ to a point $E$. Show that $CE=CA$ if and only if $DE=AB$.

4. http://www.mathlinks.ro/Forum/viewtopic.php?t=63966
The equilateral triangle $ABC$ has sides of integer length $N$. The triangle is completely divided (by drawing lines parallel to the sides of the triangle) into equilateral triangular cells of side length $1$. A continuous route is chosen, starting inside the cell with vertex A and always crossing from one cell to another through an edge shared by the two cells. No cell is visited more than once. Find, with proof, the greatest number of cells which can be visited.

5. http://www.mathlinks.ro/Forum/viewtopic.php?t=63970
Let $G$ be a convex quadrilateral. Show that there is a point $X$ in the plane of $G$ with the property that every straight line through $X$ divides $G$ into two regions of equal area if and only if $G$ is a parallelogram.

6. http://www.mathlinks.ro/Forum/viewtopic.php?t=63967
Let $T$ be a set of $2005$ coplanar points with no three collinear. Show that, for any point of the $2005$ points, the number of triangles it lies strictly within, whose vertices are points in $T$, is even.

Here is the PDF files:

Thank you zhaoli. when the second round takes place ?Thanks

The second round takes place on 31 Jan.

In other news, I did so badly on this one that it's not even funny.


trigonometry
geometry
quadratics
algebra
system of equations
trig identities
Law of Cosines

$ x,y,z \in \mathbb{R}_\plus{}$, satisfying that:

$ x\plus{}y\equal{}\sqrt{14}$
$ x^2\plus{}z^2\minus{}\sqrt{2\plus{}\sqrt{3}} xz\equal{}7\minus{}4\sqrt{3}$
$ y^2\plus{}z^2\plus{}\sqrt{2\plus{}\sqrt{3}} yz\equal{}7\plus{}4\sqrt{3}$

Then, find $ z$ :oops:

hello, from your first equation we have $ y\equal{}\sqrt{14}\minus{}x$.By substituting this
in your second and third equation we have
(4) $ x^2\plus{}z^2\minus{}\sqrt{2\plus{}\sqrt{3}}xz\equal{}7\minus{}4\sqrt{3}$.
(5)$ (\sqrt{14}\minus{}x)^2\plus{}z^2\plus{}\sqrt{2\plus{}\sqrt{3}}(\sqrt{14}\minus{}x)z\equal{}7\plus{}4\sqrt{3}$
Solving equation (3) with respect to $ x$ we have
$ x_1\equal{}\frac{1}{4}z\sqrt{6}\plus{}\frac{1}{4}z\sqrt{2}\plus{}\frac{1}{2}\sqrt{\minus{}2z^2\plus{}z^2\sqrt{3}\minus{}16\sqrt{3}\plus{}28}$
$ x_2\equal{}\frac{1}{4}z\sqrt{6}\plus{}\frac{1}{4}z\sqrt{2}\minus{}\frac{1}{2}\sqrt{\minus{}2z^2\plus{}z^2\sqrt{3}\minus{}16\sqrt{3}\plus{}28}$
By substituting $ x_1$ in (5) and after simplifying we have
$ \minus{}8\sqrt{3}\plus{}14\minus{}\sqrt{\minus{}(2\minus{}\sqrt{13})(z^2\plus{}4\sqrt{3}\minus{}8)}\sqrt{14}\equal{}0$
with the solutions
$ z_1\equal{}\sqrt{1}{7}(2\plus{}\sqrt{3})(\minus{}\sqrt{21}\plus{}\sqrt{7})$
$ z_2\equal{}\minus{}\sqrt{1}{7}(2\plus{}\sqrt{3})(\minus{}\sqrt{21}\plus{}\sqrt{7})$
By substituting $ x_2$ in (5) we get the following equation
$ \minus{}8\sqrt{3}\plus{}14\plus{}\sqrt{\minus{}(2\minus{}\sqrt{13})(z^2\plus{}4\sqrt{3}\minus{}8)}\sqrt{14}\equal{}0$
which has no real solutions.
Sonnhard.

Dr SG you are wrong (I think so):
We find with Maple: $ 49z^4\minus{}56z^2\plus{}4\equal{}0,(49x^4\minus{}700x^2\plus{}625)(49x^4\minus{}756x^2\plus{}729)\equal{}0$.
Let $ a\equal{}\dfrac{9\minus{}3\sqrt{3}}{\sqrt{14}},b\equal{}\dfrac{\sqrt{3}\plus{}1}{\sqrt{7}},c\equal{}\dfrac{5\minus{}5\sqrt{3}}{\sqrt{14}}$.
Then $ 2$ solutions: $ x\equal{}a,z\equal{}b$ and $ x\equal{}c,z\equal{}\minus{}b$.

hello, sorry,it was a typo from me, it must be
$ z_1\equal{}\frac{1}{7}(2\plus{}\sqrt{3})(\minus{}\sqrt{21}\plus{}\sqrt{7})$
$ z_2\equal{}\minus{}\frac{1}{7}(2\plus{}\sqrt{3})(\minus{}\sqrt{21}\plus{}\sqrt{7})$.
Sonnhard.

Algebra?  Really, now...

Recall that $ \sqrt {2 \plus{} \sqrt {3}} \equal{} 2 \cos \frac {\pi}{12}$.  Draw a line segment $ XY$ through the origin $ O$ such that $ OX \equal{} x, OY \equal{} y$ and $ XY$ is at an angle of $ \frac {\pi}{12}$ clockwise from the positive $ x$-axis, and let $ Z \equal{} (z, 0)$.  In other words, we now have $ \angle ZOX \equal{} \frac {\pi}{12}$ and $ \angle ZOY \equal{} \frac {11\pi}{12}$.

We are given the side lengths of $ XYZ$ and asked to find $ z$.  Note that $ XZ^2 \plus{} YZ^2 \equal{} XY^2$, so our triangle is in fact a right triangle with area $ \frac {1}{2}$.  It follows that the length of the altitude from $ Z$ to $ XY$ is $ \frac {1}{\sqrt {14} }$ and from there

$ z \equal{} \frac {1}{\sqrt {14} \sin \frac {\pi}{12} } \equal{} \frac {\sqrt {3} \plus{} 1}{\sqrt {7} }$

as before.

Edit:  I should mention that this type of problem (where a quadratic system of equations is solved by a geometric interpretation using the Law of Cosines) is strangely common; it is discussed by Engel, for example.  Another example is given [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=144598]here[/url].


function
ratio
real analysis
real analysis unsolved

Hi,
I've got some interesting math problems and no idea for it :)

here is the first one:

find all continuous functions $ f: \mathbb{R}\to\mathbb{R}$ such that
there exists continuous $ g: \mathbb{R}^2\to\mathbb{R}$ satisfying  $ f(xy)\equal{}g(x,f(y))$.

On each semi-axis the function must be either strictly monotone or constant. Indeed, the immediate corollary of the condition is that if $ f(y_1)\equal{}f(y_2)$, then $ f(xy_1)\equal{}f(xy_2)$ for all $ x$. If $ f$ is not strictly monotone on $ [0,\plus{}\infty)$, then there exist $ 0<y_1<y_2$ with $ f(y_1)\equal{}f(y_2)$ and the ratio $ x\equal{}y_2/y_1$ arbitrarily close to $ 1$. But then all the values $ f(x^jy_1)$ ($ j\in \mathbb Z$) must be the same, from where it is not hard to see that $ f$ must be constant on $ [0,\plus{}\infty)$. I leave it to you to finish the problem from here.

could you write something more; I don't see how to continue it :(


i thinks it is a famous problem but :

can you find 3 ratinal number x,y,z that
xyz=x+y+z=8

I reduce it to this... find positive integers $ a,b$ so that $ a (a^3\minus{}16 a^2 b\plus{}64 a b^2\minus{}32 b^3)$ is a perfect square


An arc of $ 55$ degrees on circle $ A$ has the same length as an arc of $ 40$ degress on circle $ B$.  What is the ratio of the area of circle $ A$ to the area of circle $ B$? Express your answer as a common fraction.

This means that

If D=Diameter,
$ 55/360 (D_A) \equal{} 40/360 (D_B)$
$ 55(D_A) \equal{} 40(D_B)$
$ \frac {D_A}{D_B} \equal{} \frac {8}{11}$

Since the ratio of the areas is

$ \frac {D_A}{D_B})^2$

We know that it is

$ \frac {8^2}{11^2} \equal{} \boxed{\frac {64}{121}}$


vector
linear algebra
linear algebra unsolved

given vector space $ V$ and $ V_1,V_2 \subseteq V$ such that dims of quotien spaces $ V/V_1$ and $ V/V_2$ are finite.Prove that dim of quotient space $ V/(V_1 \cap V_2)$ is finite

There is a natural injective map from $ V/(V_1\cap V_2)$ to $ (V/V_1)\oplus (V/V_2)$:
$ x\plus{}V_1\cap V_2\mapsto (x\plus{}V_1)\plus{}(x\plus{}V_2)$.

The two spaces aren't naturally subspaces of some larger object, so the direct sum is the only meaningful kind of sum. We could also write it as a Cartesian product.
It's just notation.

so how to prove that map is injective ? :(

I probably should have written it as a Cartesian product- so I'll switch.
If $ (x\plus{}V_1,x\plus{}V_2)\equal{}(y\plus{}V_1,y\plus{}V_2)$, $ x\minus{}y\in V_1$ and $ x\minus{}y\in V_2$- so $ x\minus{}y\in V_1\cap V_2$ and $ x\plus{}V_1\cap V_2\equal{}y\plus{}V_1\cap V_2$.

This is very general stuff; it works for lots of algebraic structures, not just vector spaces.


One digit of the decimal representation of $ \frac{5}{7}$ is randomly selected. What is the probability that the digit is a 4? Express your answer as a common fraction.

When you divide 5 by 7, you get .714285
There is one four and six digits includong four.
The probability of choosing four is 1/6.

An important property of sevenths:

$ \frac17 \equal{} 0.\overline{142857}$

$ \frac27 \equal{} 0.\overline{285714}$

$ \frac37 \equal{} 0.\overline{428571}$

$ \frac47 \equal{} 0.\overline{571428}$

$ \frac57 \equal{} 0.\overline{714285}$

$ \frac67 \equal{} 0.\overline{857142}$

$ \frac77 \equal{} 0.\overline{999999} \equal{} 1$


Let $S$ be the subset of $N$($N$ is the set of all natural numbers) satisfying:
             i)Among each $2003$ consecutive natural numbers there exist at least one contained in $S$;
             ii)If $n \in S$ and $n>1$ then $[\frac{n}{2}] \in S$
Prove that:$S=N$
I hope it hasn't posted before. :lol:  :lol:

Assume $n \in S$ is binary represented, i.e.

$n \;=\; \sum_{i=1}^{k}c_{k}\cdot 2^{k}\;=\; c_{k}c_{k-1}\ldots c_{0}$

where $c_{i}\in \{0,1\}$ and $c_{k}= 1$. Then

$[n/2] \;=\; c_{k}c_{k-1}\ldots c_{1}\in S.$

From this it follows inductively that

$(1) \;\; [n/2^{i}] \;=\; c_{k}\ldots c_{i}\in S$

for every $i \in \{0,1,2, \ldots,k\}.$

Next let $N$ be a natural binary number. Now there exists an integer $M \in [2^{11}N, \: 2^{11}N+2002]$ such that $M \in S.$ Hence $M = 2^{11}N+R$ where $0 \leq R \leq 2002 < 2^{11},$ giving

$(2) \;\; [M/2^{11}] \;=\; [N+{\textstyle \frac{R}{2^{11}}}] \;=\; N+[{\textstyle \frac{R}{2^{11}}}] \;=\; N \in S$

according to (1). Seeing that $N$ is an arbitrary natural number, (2) implies $S ={\bf N}.$

here, consider a abitrary $n\in N*$ . $n\in S$ if we found $2n$ or $2n+1$ in $S$. Similarly if we found any one of $4n,4n+1,4n+2,4n+3$in $S$ then also $n$.... Similarly , if we found any of $2^{2003}n,2^{2003}n+1,....,2^{2003}n+2002$ in $S$ then also $n$. but ${2^{2003}n,2^{2003}n+1,....,2^{2003}n+2002}$ are $2003$ consecutive natural numbers  ,so at least on of them in $S$, hence the result follows,..


limit
advanced fields
advanced fields unsolved

i am not 100% sure if this is in the right section but here is the question.

does the following expression diverge or converge and if it converges to what value does it converge.

$\displaystyle \lim_{n\to\infty}\sqrt{1+\sqrt{2+\sqrt{3+\dots\sqrt{n}}}}$

i have no clue what to do, any help would be greatly appreciated.

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58927[/url] for the proof of convergence. As to the value of the limit, I doubt it is a nice number you can express in terms of standard mathematical constants. On the other hand, it is not hard to find its approximate numerical value with any prescribed accuracy.


trigonometry
geometry
circumcircle
ratio
geometry proposed

Let $\triangle ABC$ be an isosceles triangle $(AB=AC)$. And, $P$ is an interior point such that:
$\angle BPC=147, \angle BAP=57$ and $\angle CAP=39$.
What is the measure of $\angle PBC$ in degrees?

For x=m(<ACP), I obtained a relation in x:
 :rotfl: \[\cos{48}.\cos{72}=\cos{6}.\cos{(30+2x)}\] :rotfl:
I ask you: the correct answer verify this relation or this problem is a in bad taste gist?

all the six angle at the vertices performed by P are integers.

:rotfl: I like this (gist)problem. A hard hoax !:lol:

${m=\overline{\underline{\underline{|\ m(<BPC)\ |}}}}=147(rad)=\overline{\underline{\underline{|\ \frac{147}{\pi}\cdot 180^\circ\ |}}}\ (degree)$ ! It is incorrectly, to my mind ! You must needs do another explanations because m=147 at no time belongs to $(0,\pi)$, where $\pi$=3,14159... !.

Let (O) be the circumcircle of the triangle $\triangle BCP$ and R its circumradius. The central angle $\angle COB$ is equal to

$\angle COB = 2(180^\circ - \angle CPB) = 2(180^\circ - 147^\circ) = 2 \cdot 33^\circ = 66^\circ$

and its base angles are equal to

$\angle CBO = \angle BCO = \frac{180^\circ - \angle COB}{2} = \frac{180^\circ - 66^\circ}{2} = 57^\circ$

The triangles $\triangle ABC, \triangle OBC$ with the common base BC are both isosceles, hence, we can calculate the internal angles of the triangle $\triangle ABO$.

$\angle OAB = \frac{\angle CAB}{2} = \frac{96^\circ}{2} = 48^\circ$

$\angle AOB = \frac{\angle COB}{2} = \frac{66^\circ}{2} = 33^\circ$

$\angle ABO = 180^\circ - (\angle BAO + \angle BOA) = 180^\circ - (48^\circ + 33^\circ) = 99^\circ$

Using the sine theorem for this triangle,

$\frac{OA}{R} = \frac{OA}{OB} = \frac{\sin \widehat{ABO}}{\sin \widehat{OAB}} = \frac{\sin 99^\circ}{\sin 48^\circ} = \frac{\sin 81^\circ}{\sin 48^\circ}$

Now we turn to the triangle $\triangle PAO$. The angle $\angle OAP$ is equal to

$\angle OAP = \angle BAP - \frac{\angle CAB}{2} = 57^\circ - \frac{96^\circ}{2} = 9^\circ$

Using the sine theorem for this triangle,

$\frac{OA}{R} = \frac{OA}{OP} = \frac{\sin \widehat{APO}}{\sin \widehat{OAP}} = \frac{\sin x}{\sin 9^\circ}$

Combining the 2 sine theorem formulas,

$\frac{OA}{R} = \frac{\sin 81^\circ}{\sin 48^\circ} = \frac{\sin x}{\sin 9^\circ}$

${{\sin 81^\circ \cdot \sin 9^\circ} = \sin x \cdot \sin 48^\circ}$

${\cos(81^\circ - 9^\circ) - \cos(81^\circ + 9^\circ) = 2 \sin x \cdot \sin 48^\circ}$

${\cos 72^\circ = 2 \sin x \cdot \sin 48^\circ}$

From a regular pentagon, we have $2 \cos 36^\circ = G,\  2\cos 72^\circ = \frac 1 G = G - 1$, where $G = \frac{1 + \sqrt 5}{2}$ is the golden ratio. Hence,

$\cos 72^\circ = \frac G 2 - \frac 1 2 = \cos 36^\circ - \cos 60^\circ =$

$= \cos(48^\circ - 12^\circ) - \cos(48^\circ + 12^\circ) = 2 \sin 12^\circ \sin 48^\circ$

It follows that

${\cos 72^\circ = 2 \sin x \cdot \sin 48^\circ} = 2 \sin 12^\circ \sin 48^\circ$

$\sin x = \sin 12^\circ$

Since $90^\circ < x < 180^\circ$, ${x = \angle APO = 180^\circ - 12^\circ = 168^\circ}$ and the remaining angle of the triangle $\triangle APO$ is

$\angle AOP = 180^\circ - ( \angle APO + \angle OAP) =  180^\circ - (168^\circ + 9^\circ) = 3^\circ$

As a result, the central angles of the isosceles triangles $\triangle BPO, \triangle CPO$ are equal to

$\angle POB = \angle AOB + \angle AOP = 33^\circ + 3^\circ = 36^\circ$

$\angle POC = \angle AOB - \angle AOP = 33^\circ - 3^\circ = 30^\circ$

and their base angles to

$\angle PBO = \angle BPO = \frac{180^\circ - \angle POB}{2} = \frac{180^\circ - 36^\circ}{2} = 72^\circ$

$\angle PCO = \angle CPO = \frac{180^\circ - \angle POC}{2} = \frac{180^\circ - 30^\circ}{2} = 75^\circ$

The angles in question are then equal to

$\angle PBC = \angle PBO - \angle CBO = 72^\circ - 57^\circ = 15^\circ$

$\angle PCB = \angle PCO - \angle BCO = 75^\circ - 57^\circ = 18^\circ$

It is a nice problem. The only thing I do not get is why is levi laughing so much - there is nothing funny in his replies, but that's me.

:spider:
Yetti

I am sorry, Xeroxia specified that "the angle is integer". But a integer is a real number which, with respect to a angle and in my opinion, means "radian". From here set out the misunderstanding. And I like very much this problem. Xeroxia, beg you to forgive me for my rash words !

Fair enough. Thanks for your reply.

Yetti

[b]Respectfully and with friendship[/b] for you and Xeroxia, Levi !

In the my first message  found the relation: [b]cos48.cos72=cos6.cos(30+2x)[/b]    [b](*)[/b]
This ecuation has only one solution. From the Yety's message rezults [b]x=24 [/b]  (**)(This text isn't a solution of the propose problem !). Indeed, cos48.cos72=cos6.cos78 iff cos120+cos24=cos84+cos72 iff cos84+cos60=cos24-cos72 iff cos12.cos72=sin24.sin48 iff cos72.cos12=2sin48sin12cos12 iff cos72=2sin48sin12 iff sin18=cos36-cos60. I multiply with "2sin36" and I obtain cos18-cos54=sin72-sin36. It is true.
[u]I don't manage to do from (*) to (**) naturally. [/u] [b]Help me ![/b] In fact [b]I don't [/b][b]know to solve the ecuation (*) ![/b]

A alike own problem is the following:
 :arrow:   Let ABC be a triangle in which $B=A+60^\circ.$ I note $N\in (AC),\ M\in (AB)$ such that $m(\widehat {ACM})=\frac 14\cdot C,$
$m (\widehat {ABN})=30^\circ.$ Ascertain $\ \ m(\widehat {CMN}).$

 :?: It is perhaps to transfer this last problem in the section of geometry (propose and own) of FORUM  without to write directly ? It is able ? Thank.

[i]Dear my friends[/i], Xeroxia and Yeti, I am lucky !.
I managed to solve the ecuation from the preceding message :
**************************************************
 :?: $cos48\cdot cos72\cdot =cos6\cdot cos(30+2x)\ \ (*)$ :?:

I note [u]A=cos48.cos72[/u], [u]B=cos6.cos(30+2x)[/u]. I multiply by "4sin36" both A, B:

 :o [b]4Asin36[/b]=2cos48(sin72-sin36)=sin60+sin24-sin84+sin12=
....................=sin12+sin60-2sin30cos54=sin12+sin60-sin36=
....................=sin12+2sin12cos48=sin12(1+2cos48)=([u]1+2sin42[/u])[b]cos78[/b];
 :o [b]4Bsin36[/b]=2(sin42+sin30)cos(30+2x)=([u]1+2sin42[/u])[b]cos(30+2x).[/b]
*****************************************************
Thus,  (*)  iff  A=B  iff  (1+2sin42)cos78=(1+2sin42)cos(30+2x)
...............iff cos(30+2x)=cos78  iff  30+2x=78  iff  $\underline{\overline{| x=24 |}}$. :rotfl:  :rotfl:  :rotfl:

 :roll: ${\underline{\overline{|\ Good\ luck\ for\ Math\ and\ besides\ a.s.o.\ !\ |}}}$ :roll:   [i]Levi[/i]

[b]P.S.[/b] You are young men, I am a retired teacher ! There is a great difference,
       but at the same time and a huge common passion for Math.

Hello levi,

It was a nice solution of your trigonometric equation. Nevertheless, if you are willing to use $\sin 30^\circ = \cos 60^\circ = \frac 1 2$ (and you have to), I do not see anything unnatural in using other values, such as $\sin 54^\circ = \cos 36^\circ = \frac G 2$ and $\sin 18^\circ = \cos 72^\circ = \frac{G - 1}{2}$, where $G = \frac{1 + \sqrt 5}{2}$ is the golden ratio, etc.

Yetti


[quote="bodan"][quote="José"][quote="bodan"]Who's going to JBMO 2006 Moldova?[/quote]

When is it?[/quote]

25th--30th June.[/quote]

OK. Are you going?

[quote="José"][/quote][quote="bodan"][quote="José"][quote="bodan"]Who's going to JBMO 2006 Moldova?[/quote]

When is it?[/quote]

25th--30th June.[/quote][quote="José"]

OK. Are you going?[/quote]

Yes (Why would I write this post if I hadn't? :) ) and most probably will go next year too.

I see, good luck!! any other MathLinker???

Does  anyone know the official  site  of  JBMO  2006?

Could anyone post the problems ??
I think today is the day of the competition

Here are the problems:

1.
If $n>4$ is a composite number, then $2n$ divides $(n-1)!$.

2.
The triangle $ABC$ is isosceles with $AB=AC$, and $\angle{BAC}<60$.
The points $D$ and $E$ are chosen on the side $AC$ such that, $EB=ED$, and $\angle{ABD}\equiv\angle{CBE}$. Denote by $O$ the intersection point between the internal bisectors of the angles $\angle{BDC}$ and $\angle{ACB}$. Compute $\angle{COD}$.

3.
We call a number $\it{perfect}$ if the sum of its positive integer divisors(including 1 and $n$) equals $2n$. Determine all $\it{perfect}$ numbers $n$ for which $n-1$ and $n+1$ are prime numbers.

4.
Consider a $2n$ x $2n$ board.  From the $i$th row we remove the central $2(i-1)$ unit squares, if $i\leq n$ and $2(n-i)$ central squares if $i>n$.  We obtain a symmetrycal figure with respect to the line that separates $n$th and $n+1$th row.
What is the maximal number of rectangles $2$x$1$ and $1$x$2$ that can be placed on the obtained figure without overlapping or getting outside the board?

Someone post the problems in the adequate forum.
Try not to post solutions here.

Thanks freemind .Did you solve it all ?? :D

I took 36. It's not so bad, but i lost 3 points because of a particular case at the geometry... try the last one, the others are easy.

[quote="freemind"]Here are the problems:

3.
We call a number $\it{perfect}$ if the sum of its positive integer (including 1 and $n$) equals $2n$. [/quote]

I think you mean
We call a number $\it{perfect}$ if the sum of its positive integer [b]divisors [/b] (including 1 and $n$) equals $2n$.
[/b]

[quote="freemind"]The points $D$ and $E$ are chosen on the side $AC$ such that, $EB=ED$, and $\angle{ABD}\equiv\angle{CAE}$[/quote]

Please see it again . If $E$ is on $AC$  then $\angle{CAE}=0$

sure you can share the problems so early after the competition?

1: http://www.mathlinks.ro/Forum/viewtopic.php?t=99348

2: http://www.mathlinks.ro/Forum/viewtopic.php?t=99349

3: http://www.mathlinks.ro/Forum/viewtopic.php?t=99350

4: http://www.mathlinks.ro/Forum/viewtopic.php?t=99351

Will somebody post an understandable version of JBMO problem 4?

what were the cutoffs? and which countries took the gold medals?

well, I only know that Turkey got the 3rd place (by country) and a student from Turkey got a perfect score, maybe someone will continue.

davron

Romania got the first place! :lol: A boy from Romania took 38 points, because at the last problem, whith the chessboard, he lost 2 points. :( Anywhay, it's a very good result.

1. Romania - 206 points.
2. Serbia and Montenegro - 194 points.
3. Turkey - 193 points.
4. Moldova (1st team) - 168 points.
5. Bulgaria - 165 points.
6. Kazahstan (unofficial) - 151 points.
7. Macedonia - 109 points.
8. Moldova (2nd team, unofficial) - 98 points
9. Greece - 93 points.
10. Bosnia & Herzegovina - 90 points.
11. Cyprus - 58 points.
12. Albania - 53 points.

what about the cutoffs and the origin of the proposed problems...

What is the "Zaimi--Marku" inequality, Albanian Eagle?


Yay!! i am the first one to start :D
How did you all do :?:

dunno :| but think it was tougher than 07.might be at par with 06!


Sam and Janet each have a whole number of dollars, and $ \frac13$ of Sam's money equals $ \frac12$ of Janet's money. Together, they have more than $ \$10$. What is the least number of dollars they could have combined?

Let $ S=\text{The number of dollars Sam has}, J=\text{The number of dollars Janet has}$. We set up two equations.
\[ J+S>10\newline\frac{1}{3}S=\frac{1}{2}J\]

Substituting \[ S=\frac{3}{2}J\] into the first equation gives \[ J+\frac{3}{2}J>10\implies\frac{5}{2}J>10\implies J>4\]. Since $ J$ must be even (because if $ J$ was odd, then $ \frac{1}{2}J$ would contain decimals, and $ 3(\text{any decimal number})\neq\text{a whole number}$)
Now all that's left is to test.
Testing \[ J=6\] gives \[ \frac{1}{3}S=\frac{1}{2}^\implies\frac{1}{3}S=3\implies S=9\], and this also satisfies the first inequality, so $ J+S=\boxed{15}$

Alternatively, you could have noticed that $ LCM(2,3)=6$, so one of them could have $ 6$ to minimize the sum. Testing, we get the same result.


MATHCOUNTS
quadratics
Pythagorean Theorem
geometry
algebra
quadratic formula

didnt number 1 only have a,b,c, and d in the progression earlier?

and you could still get an answer, sorta, with that

??? you mean the 1st contest???

anyways, like you said, its doable with the given info (just w/o the sorta, lol)

i know its doable, but the version without the 10 was doable too

no... if it was

{a, b, n, c, d,}, it wouldn't have... it would've been either a+b+c+d

no, i was saying for (a,b,c,d) as an arithmetic progression, that was how  it used to be until he added a 10

[hide]ah, adding a 10, its much easier[/hide]

will you be posting the scoreboard soon? before the due date?

[quote="Tare"]I haven't actually studied series seriously so for number 11 is n supposed to be integers greater or equal to 1?[/quote]
How could it possibly be anything else?

I don't know, negative numbers, 0, that sort of stuff.

I'm almost done Max...hold on...

Can we start a new thread now for solutions and stuff?

This is almost done, so i de-stickeyed it

Ok this will probably just be another exhibition of my stupidity, but can someone please post a solution to number 10: Find all real x which satisfy the equation (x - 3)(x^2 + 4x + 4) = (x - 3).

Here's what I did:
Divided both sides by (x-3) and got x^2 + 4x +4 = 1.
x^2 + 4x + 3 = 0.
(x+3)(x+1) = 0
x = -3, -1
but this is wrong, maybe I'm forgetting some solutions.

Who said you were allowed to divide by x-3?  :)

*hits self*
ack, once again, I am an idiot.

tripleM: lol, well I'm fairly sure those others that I got work, or at least they did the last time I checked....

why can't we divide by x-3? i plugged -1 and -3 for x and got the right answers.

Yes, they are answers, but they are not the only answers. You can divide each side by x-3, but only when x-3 is not 0..

Right. At least I won't miss that again...I mean I like these contests because they're good for practice.

most of these are straight from mathcounts and amc tests

I'm bored, so here are my solutions:

1. Call A=10-2x, B=10-x, (10), C=10+x, D=10+2x...adding all these, the x's cancel and the answer is 40.

2. ....didn't answer this one, someone else can post their solution

3. a=1, d=3, a+(n-1)d=193, from which n=65. The sum is then         N(a1 + an)/2= 65(194)/2 = 6305.

4. 6/7 = .857 142857 142857 ... every 6 digits after the 857, the digits repeat, so the after 16 cycles of 142857 (96 digits) + the .857, that's 99 digits, so the 100th would be the first term of the .142857 = 1.

5. 2^(3x+2) = (2^3x)(x^2) = (7^3)(4) = 1372.

6.
system:
Q     +   D = 63
25Q+10D=1035
Solving for Q, answer is 27.

7. I didn't get this one correct, so...

8. The third term is 59, and this term repeats over and over so that a1999=59.

9. Factor out an 8! to get 8!(1+9+9*10+9*10*11)=8!(1090)=8!(10)(109), and 109 is prime, so it's the answer.

10. Grr...I missed this one, but here's a solution:
Factor out the x-3 to get x^2+4x+3=0, x= -3, -1, but I forgot to note that if x=3, then both sides equal zero, so the solutions are
-3, 3 -1.

11. I factored 5814 and got it down to 17(18)(19), and got k=17.

12. 23b = 3 + 2b....plugging in small numbers, b=11.

13. P/Q= :rt2: , so P=Q:rt2: ...
(3Q - Q:rt2:)/(Q:rt2: - Q)
Q(3 - :rt2:)/Q(:rt2: - 1)
= 1 + 2:rt2:

14. Divide everything by 2 to get x:^2: + 5x +y:^2: - 3y = 9...complete the square, r:^2:=35/2,  :alpha: = 35 :pi: /2

15. We can call the sides of the triangle 2, 2r, and 2r :^2: , and use the pythagorean theorem to get
4+4r :^2: =4r :^4:
Divide by 4...1+r:^2:=r:^4:
Calling r:^2: A,
1+A=A:^2:
Solving with the quadratic formula, A= (1+:rt5:)/2, and since A = r :^2:  and the hypotenuse is 2r :^2: , the answer is 2(1+:rt5:)/2 = 1+ :rt5:.

[quote="MagnusRex"]
10. Grr...I missed this one, but here's a solution:
Factor out the x-3 to get x^2+4x+3=0, x= -3, -1, but I forgot to note that if x=3, then both sides equal zero, so the solutions are
-3, 3 -1.
[/quote]

Ok, so this is the correct solution.. it just sounds a little funny. It sounds as if you just had to 'notice' that x=3 was also a solution.. where of course it should have been part of your argument.. the best solution to this problem would be:

(x-3)(x^2 + 4x + 3) = (x-3)
Either
x-3 = 0 or x^2 + 4x + 4 = 1
x = 3 or (x+3)(x+1)=0
x = 3 or -3 or -1.

Lol, yes....correction: TripleM told me the answer ;). Thanks for a nicer solution.


probability
counting
derangement
geometry
3D geometry
combinatorics unsolved
combinatorics

1. A random number generator randomly generates the integers 1,2,...,9 with equal probability. Find the probability that after $ \mbox{n}$ numbers are generated, the product is a multiple of 10.

2. Let $ a_1,a_2,...a_n$ be an ordered sequence of n distinct objects. A derangement of this sequence is a permutation that leaves no object in its original place. For example, if the original sequence is 1,2,3,4 then 2,4,3,1 is not a derangement, but 2,1,4,3 is. Let $ D_n$ denote the number of derangements of an $ \mbox{n}$-element sequence. Show that $ D_n \equal{} n!\left(1 \minus{} \frac{1}{1!} \plus{} \frac{1}{2!} \minus{} ... \plus{} (\minus{}1)^n\frac{1}{n!}\right)$

3. Imagine you are going to give $ \mbox{n}$ kids ice-cream cones, one cone per kid, and there are k different flavours available. Assuming that no flavours get mixed, show that the number of ways we can give out the cones using all k flavours is $ k^n \minus{} \binom{k}{1}(k\minus{}1)^n \plus{}  \binom{k}{2}(k\minus{}2)^n\minus{} \binom{k}{3}(k\minus{}3)^n\plus{}...\plus{}(\minus{}1)^k \binom{k}{k}0^n$

Thanks :)

Problem 1 was already posted here,  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=316155

Thanks pleurestique, you have helped sooo much!! I really appreciated it :)

Solution to problem 2: let $ f\equal{}\sum\limits_{n\equal{}0}^{\infty}\frac{D_{n}x^{n}}{n!}$ and $ g\equal{}\sum\limits_{n\equal{}0}^{\infty}\frac{x^{n}}{n!}$. It is easy to see that for each $ n$ we have $ \sum\limits_{i\equal{}0}^{n}D_{i}\binom{n}{i}\equal{}n!$ or equivalently $ \sum\limits_{i\equal{}0}^{n}\frac{D_{i}}{i!}\cdot \frac{1}{\left(n\minus{}i\right)!}\equal{}1$. Thus $ fg\equal{}\sum\limits_{n\equal{}0}^{\infty}x^{n}\equal{}\left(1\minus{}x\right)^{\minus{}1}$ or $ f\equal{}g^{\minus{}1}\left(1\minus{}x\right)^{\minus{}1}\equal{} \mathrm{e}^{\minus{}x}\left(1\minus{}x\right)^{\minus{}1}$. Therefore
$ D_{n}\equal{} f^{\left(n\right)}\left(0\right)\equal{} \sum\limits_{i\equal{}0}^{n}\binom{n}{i} \left(\mathrm{e}^{\minus{}x}\right)^{\left(i\right)}\left(0\right) \left(\left(1\minus{}x\right)^{\minus{}1}\right)^{\left(n\minus{}i\right)}\left(0\right)\equal{}$
$ \equal{}\sum\limits_{i\equal{}0}^{n}\binom{n}{i}\left(\minus{}1\right)^{i}\left(n\minus{}i\right)!\equal{} n!\sum_{i\equal{}0}^{n}\frac{\left(\minus{}1\right)^{i}}{i!}$.


Consider the equation $ ax^3\minus{}x^2\plus{}bx\minus{}1\equal{}0$
where $ a,b$ are real numbers, $ a \neq b, a \neq 0$ such that all of its roots are positive real numbers.
Find the smallest value of $ P\equal{}\frac{5a^2\minus{}3ab\plus{}2}{a^2(b\minus{}a)}$

[quote="thesunflower"]Consider the equation $ ax^3 \minus{} x^2 \plus{} bx \minus{} 1 \equal{} 0$
where $ a,b$ are real numbers, $ a \neq b, a \neq 0$ such that all of its roots are positive real numbers.
Find the smallest value of $ P \equal{} \frac {5a^2 \minus{} 3ab \plus{} 2}{a^2(b \minus{} a)}$[/quote]
To Mods: Please lock this topic, this is a problem from M&Y, issue 10, and we can't discuss it now
Thanks!

What a poor cheater. I locked all your attempts to cheat in Superior Algebra now.


We are very pleased to announce that the first Metroplex Math Circle of the Fall 2008 semester will feature Richard Rusczyk.  Please go to http://mmc.awesomemath.org for more details as the date approaches.

Metroplex Math Circle now has a new web site at http://www.metroplexmathcircle.org


The volume of a cone is given by the formula $ V \equal{} \frac{1}{3}Bh$, where $ B$ is the area of the base and $ h$ is the height. The area of the base of a cone is 30 square units, and its height is 6.5 units. What is the number of cubic units in its volume?

In the equation, substitute $ B$ for $ 30$ and $ h$ for $ 6.5$ to get:
\[ \frac{1}{3}\cdot30\cdot6.5\equal{}10\cdot6.5\equal{}\boxed{65}
\]


LaTeX
limit
logarithms
real analysis
real analysis unsolved

Prove that

[size=150] [i]lim Π(n=1 to m) (1+1/n)^(n/m) = e, as m -> ∞. [/i][/size]

I try to write it on latex:
\[ \lim_{m\to \infty} \prod\limits_{n \equal{} 1}^{m}\left( 1 \plus{} \frac {1}{n} \right) ^{n/m} \equal{} e
\]
hint willl be consider
[hide="here"]
\[ \sum\limits_{n \equal{} 1}^{m}\frac {n}{m}\ln \left( 1 \plus{} \frac {1}{n} \right) \to \sum\limits_{n \equal{} 1}^{m}\frac {n}{m} \frac {1}{n} \to \sum\limits_{n \equal{} 1}^{m} \frac {1}{m} \equal{} 1
\]
as $ m \to \infty$
[/hide]
there is also another hint:
[hide="here #2"]
consider:
\[ \lim_{m\to \infty} \prod\limits_{n \equal{} 1}^{m}\left( 1 \plus{} \frac {1}{n} \right) ^{n/m} \equal{} \lim_{m\to \infty} \prod\limits_{n \equal{} 1}^{m}\left( \frac {n \plus{} 1}{n} \right) ^{n/m} \equal{} \lim_{m\to \infty}\left( \frac {2^{1}3^{2}4^{3}...m^{m \minus{} 1}(m \plus{} 1)^{m}}{1^{1}2^{2}3^{3}...m^{m}} \right) ^{1/m} \equal{} \lim_{m\to \infty}\left( \frac {(m \plus{} 1)^{m}}{(m!)} \right) ^{1/m} \equal{} e
\]
by stirling
[/hide]


linear algebra
matrix
algebra
polynomial
invariant

Prove that the sum of eigenvalues is equal to the trace of the matrix.

[hide="Method 1, sketch"] Follows from the fact that the roots of the characteristic polynomial are the eigenvalues. [/hide]
[hide="Method 2, sketch"] Show that the trace satisfies $ \text{tr}(ABC) \equal{} \text{tr}(CAB)$.  Conclude that traces are invariant under similarity; triangularize. [/hide]

In order for the question to make sense, the characteristic polynomial must split over the the field we're using - in other words, we must have a full set of eigenvalues.

Here's one pathway:

Similar matrices have the same trace.

Similar matrices have the same eigenvalues.

A matrix which (as discussed above) has a full set of eigenvalues can be shown to be similar to an upper triangular matrix. (This is Schur's Lemma without all its features; the proof is elementary and inductive on dimension.)

For an upper triangular matrix, the eigenvalues are the elements on the main diagonal.

Note that this proof also shows that the product of the eigenvalues is the determinant.

[Edit: essentially the same as t0rajir0u's method 2, just a little more filled out.]

[quote="Kent Merryfield"].

A matrix which (as discussed above) has a full set of eigenvalues can be shown to be similar to an upper triangular matrix. (This is Schur's Lemma without all its features; the proof is elementary and inductive on dimension.)

For an upper triangular matrix, the eigenvalues are the elements on the main diagonal.

Note that this proof also shows that the product of the eigenvalues is the determinant.

[Edit: essentially the same as t0rajir0u's method 2, just a little more filled out.][/quote]
 To see it every $ A \in M_n{(\mathbb{C})}$ have eigenvalues are $ \lambda_1,\lambda_2,..,\lambda_n \in \mathbb{C}$  the exist $ T \in GL_n{(\mathbb{C})}$
 Such that $ A \equal{} T\begin{bmatrix}\lambda_1 & * & * & * & ... & * & * \\
0 & \lambda_2 & * & * & ... & * & * \\
0 & 0 & \lambda_3 & * & ... & * & * \\
... & ... & ... & ... & ... & ... \\
0 & 0 & 0 & 0 & ... & 0 & \lambda_n\end{bmatrix}T^{ \minus{} 1}$
 Use $ Tr(BC) \equal{} Tr(CB) , \forall B,C \in M_n{\mathbb{(C)}}$
  $ Tr(A) \equal{} Tr(T\begin{bmatrix}\lambda_1 & * & * & * & ... & * & * \\
0 & \lambda_2 & * & * & ... & * & * \\
0 & 0 & \lambda_3 & * & ... & * & * \\
... & ... & ... & ... & ... & ... \\
0 & 0 & 0 & 0 & ... & 0 & \lambda_n\end{bmatrix}T^{ \minus{} 1}) \equal{} Tr(T^{ \minus{} 1}T\begin{bmatrix}\lambda_1 & * & * & * & ... & * & * \\
0 & \lambda_2 & * & * & ... & * & * \\
0 & 0 & \lambda_3 & * & ... & * & * \\
... & ... & ... & ... & ... & ... \\
0 & 0 & 0 & 0 & ... & 0 & \lambda_n\end{bmatrix}) \equal{} Tr(\begin{bmatrix}\lambda_1 & * & * & * & ... & * & * \\
0 & \lambda_2 & * & * & ... & * & * \\
0 & 0 & \lambda_3 & * & ... & * & * \\
... & ... & ... & ... & ... & ... \\
0 & 0 & 0 & 0 & ... & 0 & \lambda_n\end{bmatrix}) \equal{} \sum_{k \equal{} 1}^{n}{\lambda_k}$
 More you can see that :
$ \begin{bmatrix}\lambda_1 & * & * & * & ... & * & * \\
0 & \lambda_2 & * & * & ... & * & * \\
0 & 0 & \lambda_3 & * & ... & * & * \\
... & ... & ... & ... & ... & ... \\
0 & 0 & 0 & 0 & ... & 0 & \lambda_n\end{bmatrix}^m \equal{} \begin{bmatrix}\lambda_1^m & * & * & * & ... & * & * \\
0 & \lambda_2^m & * & * & ... & * & * \\
0 & 0 & \lambda_3^m & * & ... & * & * \\
... & ... & ... & ... & ... & ... \\
0 & 0 & 0 & 0 & ... & 0 & \lambda_n^m\end{bmatrix}$
 then $ Tr(A^m) \equal{} \sum_{k \equal{} 1}^{n}{\lambda^m_k} ,\forall m \in \mathbb{N}$ ....


PLEASE POST A WARNING WHEN POSTING SITES THAT HAVE NUDITY ON THEM. THANKS.

Warning given, then. Please, avoid these types of hosting sites.

Yea, sorrrry.  The ads change every time, so I didn't realize how bad it was.  I'll repost it on Imageshack and then link here.

EDITED: OK it's rehosted, so we shouldn't have a problem.

[quote="mysmartmouth"]I try to avoid the spam factory, but this is worse then when I missed #11 at USC:

http://img179.imageshack.us/img179/2632/333imagedm0.jpg[/quote]

lol ive seen that b4

NOW THAT IS SAD.  :D Probably didn't read it carefully. No one is that dumb. Bet that was $\$$100 question.

WAHAHAHAHAHA! WAHAHAHAHAHAHAHA! That is [b][i]pitiful![/i][/b] ABSOLUTELY PITIFUL!  :rotfl:  :rotfl:  :rotfl:

HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA.............. to ∞.

elephant is bigger than moon. :P

lol, it looks like koda just copy/pasted what nutz said.

Captain Obvious....To the rescue!!!!

<====


modular arithmetic
number theory
Diophantine Equations
pen

Determine all pairs $(x, y)$ of integers such that \[(19a+b)^{18}+(a+b)^{18}+(19b+a)^{18}\] is a nonzero perfect square.

I guess the $ (x,y)$ should be $ (a,b)$ or vice versa...

By Fermat's little theorem, $ (19a \plus{} b)^{18},(a \plus{} b)^{18},(19b \plus{} a)^{18}\in\{0,1\}\pmod{19}$. Since $ 2$ and $ 3$ aren't squares modulo $ 19$, we need the sum it to be $ 0$ or $ 1$.

If $ 19\not|\ a \plus{} b$ then we need $ 19|18a$ and $ 19|18b$ by the above, contradiction. So $ 19|a \plus{} b$.
If $ 19\not|\ a$ and $ 19\not|\ b$ then the sum is $ 2$, contradiction. So at least one divides $ 19$, and by the above, the other does as well.

So we need $ 19|a,b$. Now take a nonzero solution where $ |a| \plus{} |b|$ is minimal. By the above, $ \left(\frac {a}{19},\frac {b}{19}\right)$ is an integer solution, and it has $ \left|\frac {a}{19}\right| \plus{} \left|\frac {b}{19}\right| < |a| \plus{} |b|$, contradiction.
So we only have the zero solution: $ a \equal{} b \equal{} 0$.

sketch:

1. if all a,b,c are not divisible by 19 ,then the term is equivalent to 3 mod 19 (by fermat little theorem)

but 3 is not a quadratic residue modulo 19 (since $3^9 = -1 (mod 19)$  ) ,contradiction!

2.if one of a,b (wlog ,a)  is divisible by 19 so the term is equivalent to 2 mod 19, again , 2 is not a quadratic residue modulo 19

contradiction!

3.let $a=19a_1 , b=19b_2$  then the term will be   $19^{18}((19a_1+b_1)^{18}+(a_1+b_1)^{18}+(19b_1+a_1)^{18})$

but$ (19a_1+b_1)^{18}+(a_1+b_1)^{18}+(19b_1+a_1)^{18}$ is not perfect squares as proved above, contradiction.
so, no solution.

(a=0,b=0 gives 0 which is not satisfies the problem condition)