Prove that there doesn't exist a triple of positive integers $ a,b,c$ satisfying $ \frac{a^2\plus{}b^2\plus{}c^2}{3(ab\plus{}bc\plus{}ca)}\in \mathbb Z$

Is this from a mathlinks contest last year?  :)


Hei alle sammen!
Jeg(Aslan) og Atle tenkte at det er litt for vanskelig å skrive matte formler på msn, så jeg ba om en norsk side på forumet.
Her kan vi skrive i Latex. Latex er en program som skriver fine matte formler  :) For å lære mer om Latex se på meny-feltet til venstre. Der finner dere programmer som trengs for å bruke Latex...

Nordisk 2005
Oppgave 2

$\frac{2a^2}{b+c}+\frac{2b^2}{a+c}+\frac{2c^2}{a+b} \geq a+b+c$

Det er bare for å vise hvor fint det er å bruke Latex

Et lite hint til denne nordisk oppgåva. [hide]Først skriv vi den litt om:

$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}\geq \frac{a+b+c}{2}$
Observer at venstre side kan skrivast som ein funksjon $f(x)=\frac{x^2}{S-x}$, der $S=a+b+c$, og x er enten a, b eller c. Slike oppgåver er typiske Jensen-oppgåver. [/hide]

Her er heile løysinga:[hide]
$f(x)$ er ein konveks funksjon sidan den har positiv andrederivert for positive x. Då gir Jensens ulikskap at:

$\frac{f(a)+f(b)+f(c)}{3}\geq f(\frac{a+b+c}{3})$.
Ved å setje inn $f(a)=\frac{a^2}{b+c}$ etc. får vi dette:

$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}\geq \frac{(a+b+c)^2}{a+b+c-\frac{(a+b+c)}{3}}$ som kan lett omformast til det vi skulle vise.
Det er også mange andre måtar å løyse denne oppgåva på. Jørgen kan sikkert nevne 15[/hide]

Ja, bare at du har en liten feil
[hide]$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}\geq 3 \frac{(\frac{a+b+c}{3})^2}{a+b+c-\frac{(a+b+c)}{3}} =\frac{a+b+c}{2}$[/hide]
Kan løses med Causchy-Schwarz:
[hide]$( (b+c) +(a+c) + (b+a) ) (\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}) \geq (a+b+c)^2$[/hide]


factorial
Ring Theory
superior algebra
superior algebra solved

Prove that the following rings are not factorial:

      $\mathbb{Z}[i\sqrt{5}]$, $\mathbb{Z}[\sqrt{26}]$ $\mathbb{Z}[i\sqrt{3}]$,$\mathbb{Z}[\sqrt{10}]$

first :

$2 . 3 =(1+\sqrt{5} i) (1-\sqrt{5} i )$ where all factors are irreducible and obviously they are not up to associates and order the same

idem for third :

$(1+\sqrt{3} i) (1-\sqrt{3} i )=2 . 2$

second :

$11$ is irreducible and when the ring is UFD this is also a prime
but 11 divides $(2+\sqrt{26})(2-\sqrt{26})$
so it has to divide one of these two factors, which it doesn't

last :
the same but now with $3$

it divides $(2+\sqrt{10})(2-\sqrt{10})$

Sorry if I made a mistake and plz tell me when it is unclear

I love UFD  ;)

In fact when $m>2$  then $\mathbb{Z}[\sqrt{m} i]$ is never a factorial ring


pl send all short cuts of trignometry up to +2 level

there is no short cut

i think you are indian

then go through any of the last time revision package for iitjee
you'll find many in them


if $a\ne0$   $\lim_{t\to t_0} f(at)=L$if and only if $\lim_{x\to at_0}f(x)=L$

You might first try to use the delta-epsilon definition of a limit.

Yes, this is what I am trying but...  :blush:

Suppose $\lim_{x\to at_0}f(x)=L$. Pick $\epsilon>0$. Then there exists $\delta$ such that $|x-at_0|<\delta$ implies $|f(x)-L|<\epsilon$. Thus, $|t-t_0|<\frac{\delta}{|a|}$ implies $|f(at)-L|<\epsilon$. Hence, $\lim_{t\to t_0} f(at)=L$.

Suppose $\lim_{t\to t_0} f(at)=L$. Pick $\epsilon>0$. Then there exists $\delta$ such that $|t-t_0|<\frac{\delta}{|a|}$ implies $|f(at)-L|<\epsilon$. Thus, $|x-at_0|<\delta$ implies $|f(x)-L|<\epsilon$. Hence, $\lim_{x\to at_0}f(x)=L$.

Is that it?

[quote="ICSD"]Can anybody verify tha this is the correct answer? I believe it is.[/quote]

Yes, ICSD , it seems to me that is not only right, but also very well written.  Please be sure you understand the "Thus"  that towersfreak2006 used twice. If you do, it will be evident why delta was chosen that way.


let $ f$ a function  on $ \mathbb{R}$ such that :
$ i)$ $ f$ is a periodic function with period $ T$.$ (T>0)$
$ ii)$ $ f$ is a derivable function.
$ iii)$ $ f'$ is continu on $ \mathbb{R}$
prove that  there is two distancts reals $ x$ and $ y$ in [$ 0,T$[ such that
$ f(x).f'(y)\equal{}f(y).f'(x)$

if $ f(x) > 0, f(y) < 0$ for $ 0\le x < y < T$
then $ \exists a,b: \ f(a) \equal{} f(b) \equal{} 0\ and\ a\in]x,y[,b\in]y,x \plus{} T[$
then $ \exists i\neq j\in [0,T[: \ f(i) \equal{} f(a) \equal{} 0,f(j) \equal{} f(b) \equal{} 0$ then $ f(i)f'(j) \equal{} f(i)f'(j)$

if $ \exists 0 < a < b < T$ such that $ \forall x\in[a,b]: \ f(x) \equal{} 0$ then $ f(a)f'(b) \equal{} f(a)f'(b)$

if there arn't such $ x,y$ then $ f$ is $ > 0$ $ \forall x$ or $ < 0$ $ \forall x$
we suppose that $ f > 0$ and we take $ g \equal{} ln(f)$
we have $ g'$ continous and $ g'(0) \equal{} g'(T)$: then obviously $ g'(x) \equal{} g'(y)$ for some $ x\neq y\in ]0,T[$
gives $ \frac {f'(x)}{f(x)} \equal{} \frac {f'(y)}{f(y)}$


Find the remainder when f(x) = x^4 + 2x^3 - 5x^2 + x + 3 is divided by (x - 1)(x + 2).

This has to be done using the remainder theorum and not by long division.

This is what I have so far.

                f(x) = (d(x))(q(x)) + r(x)
Therefore, f(x) = (x - 1)(x + 2)(q(x)) + r(x)

The degree of the divisor is 2, so the degree of the remainder must be 1. Therefore, let r(x) = ax + b.

                f(x) = (x - 1)(x + 2)(q(x)) + (ax + b)

Using (x - 1) as a factor, sub x = 1.

                f(1) = (1 - 1)(1 + 2)(q(1)) + (a(1) + b)
                f(1) = (0)(3)(q(1)) + a + b
Therefore, f(1) = a + b

Using (x + 2) as a factor, sub x = -2.

                f(-2) = (-2 - 1)(-2 + 2)(q(-2)) + (a(-2) + b)
                f(-2) = (-3)(0)(q(-2)) - 2a + b
Therefore, f(-2) = -2a + b

I don't know what to do from here. I think I have to create some sort of system with the two functions, but I don't know how to do so.

Any help would be greatly appreciated.

....a remainder of 7x - 5!

Thanks so much kunny!  :D


If two or more people have the exact same answers on the individual rounds (e.g., get a perfect score), how is the tiebreaker round supposed to function? I can't find out how on the MATHCOUNTS website.

In our chapter this is what happened.

The two people who had this (1st and 2nd place) went into a separate room and had 2 problems with an alloted time limit (8min I think).  The first to miss a problem lost and the person who won got 1st and lost got 2nd.  In CD though, the second person got 1st.

The Tiebreaker round is somewhat of a cross between the Target and a Countdown.

Like a target round, you get 6 minutes for each pair of questions, and a calculator.  If you finish early though, you can raise your hand or something and they'll collect your paper.

If you get both right and were first, then you win the Tiebreaker.  If you were first, but missed one, and the other person got both, then they win though.

The two people who tied will go into a room and do 2 target-like questions in 6 minutes with a calculator. Whoever gets the most questions done wins. In case of a tie, whoever finished first wins. If one person got #2 right and the other person got #1 right, the person who got #2 would win. If both used up the time and got the same questions right. They repeat the process until they have a winner.

Wow, that's pretty intense and awesome. I could never figure out what exactly happened. I assume this is if only no other tie brakers work. (ie. sprints correct, #30, #29, etc.) I see how most of the time, it's needed only in situations with perfect scores.

Yeah, they only do it if all other things fail.  If the $ x$ people got the exact same questions right, then they would have a tiebreaker.  It doesn't have to be a perfect score though.

so in a tiebreaker, would you want to do question number 2 first?
in case both people get only one right?

Not really...clearly, you're not going to turn in the test with only one problem. You have to finish both problems in any case.

The Mathcounts people should seriously just make the tests harder now... some many chapters would need tiebreakers. For some reason, the three people that tied for first in my chapter didn't have to do a tiebreaker.

[quote="tinytim"]For some reason, the three people that tied for first in my chapter didn't have to do a tiebreaker.[/quote]

I doubt that the coordinators even knew what a TB was... :P


calculus
integration
real analysis
limit
function
real analysis solved

Let f(x)=sin(nx)e^(x^2+1) , g(x)=cos(nx)ln(x^2+1) .
Prove that :  lim{n->+infinity}integral{0<=x<=2Pi}(f(x)dx) = lim{n->+infinity}integral{0<=x<=2Pi}(g(x)dx) .

Riemann Lebesgue theorem:
If k is continuous on the compact [0;2 \pi ] then
 \lim  \int _[0,2 \pi ]e^(int).k(x)dx = 0
The result is also true if k is C^1

Here the proof when k is C^1 just integration  by parts

 \int _[0,2 \pi ]e^(inx)k(x)dx= (k(2pi)-k(0))/(in) +
 1/(in) \int _[0,2pi]e^(int)k'(x) dx--->0 when n->+oo

For k continous less immediat, use the fact that k can be uniformly
approximate by steps functions on the compact [0;2pi]

As for my opinion ,I have a more elementary solution to this problem but I don't know whether it is true or not . Please check and inform me if there is any mistake .
This problem comes from a more general problem and I'll give my solution to the general one : "Given T>0, f is continuous and periodic with period T s.t  \int _[0,T]f(x)dx=0.Let g be a continuous function on [0,T].Prove that  \lim (n->00) \int [0,T]f(nt)g(t)dt=0."
Proof :
"We see that f(R)=f([0,T]) and there exists M  \in  R s.t. : |f(x)| \leq M for every x  \in R.
For every  \varepsilon >0,there exists N  \in N* s.t. : |g(x)-g(y)| \leq (MT+1)^(-1) for every x,y \in [0,T] satisfying |x-y| \leq T/N.
Given n \in N,let a_i=iT/n for every i \in {0,1,...,n}.
| \int[0,T]f(nt)g(t)dt|=| \sum {i=1 to n-1} \int {a[i],a[i+1]}f(nt)g(t)dt| \leq  \sum {i=0 to n-1}| \int {a[i],a[i+1]}f(nt)g(t)dt|.
As | \int{a[i],a[i+1]f(nt)g(t)dt|=|1/n \int {iT,(i+1)T}f(s)g(s/n)ds| \leq |1/n \int {iT,(i+1)T}f(s)(g(s/n)-g(iT/n))ds|+|1/n \int {iT,(i+1)T}f(s)g(iT/n)ds| \leq  \varepsilon /n ( since  \int {iT,(i+1)T}f(s)g(iT/n)ds=0)
Hence | \int [0,T]f(nt)g(t)dt| \leq   \varepsilon for every given  \varepsilon >0.
Therefore  \int [0,T]f(nt)g(t)dt=0."
Apply this problem into the old one by replacing suitable f(x) and g(x) , we'll see that both sides of the expression equal to 0 .
That's enough now.


I tried, and just when it was about to finish, there was an error about fonts or something.  :(

"[i]or something[/i]"? It's hopeless to help with something as vague as that as well as no other details at all. Try reinstalling over the top of the old installation and watch carefully for what's happening.

"Build font information cache files has encountered a problem and needs to close.  We are sorry for the inconvenience."

error concerning "initexmf.exe"

Thanks in advance.

Still very little information to go on. You must say more as there's nothing like this error on the Miktex forums. What version of MiKTeX, which operating system, which folder are you installing MiKTeX to, what stage does this happen etc etc? You cannot expect a diagnosis without [b]full[/b] information.

Also what have you tried to do? Has MiKTeX been installed? What happens if you go to Start, All Programs, MikTeX 2.7, Settings. Can you press the refresh FNDB & Update Formats buttons? If so, can you compile a latex doument?

You should be installing MiKTeX version 2.7 (not an earlier version) on Windows 2000, XP, Vista or Server 2003. Are you installing the basic or complete system, do you have admin permission, are their any access restrictions? Try uninstalling then reinstall to C:\texmf rather than C:\Program Files.

So come back with lots of information.

I am following http://www.artofproblemsolving.com/LaTeX/AoPS_L_Downloads.php.

I downloaded version 2.7 on my XP to the desktop. (It is not in the Start Menu.) The installation under C:\Program Files\MiKTeX 2.7 was very nearly complete when the error occured. A box popped up saying "Build font information cache files has encountered a problem and needs to close. We are sorry for the inconvenience.". Various other error message appeared, saying the error concerned "initexmf.exe". The last message that appeared said that MiKTeX was not successfully installed.

Not knowing what to do, I proceeded to download and install TeXnicCenter according to the instructions on AoPS wiki. The process of creating, compiling, and viewing a file seems perfectly normal.

But I would still like to know...

Are there any problems I should expect? Because of the last little bit of MiKTeX that was not installed, will there be anything missing?

[quote="leoxnlin"] Various other error message appeared, saying the error concerned "initexmf.exe".[/quote]
I don't know how you think "various other messages" conveys any information at all.

[quote="leoxnlin"]Are there any problems I should expect? Because of the last little bit of MiKTeX that was not installed, will there be anything missing[/quote]
Impossible to tell since you don't tell us all the error messages

[quote="leoxnlin"]It is not in the Start Menu[/quote]
What is "[i]It[/i]"?
You will need shortcuts in All Programs to
Browse Packages at C:\Program Files\MiKTeX 2.7\bin\mpm.exe
Settings at C:\Program Files\MiKTeX 2.7\bin\mo.exe
If they aren't there you can create them yourself.

Then go to Settings and press the Refresh FNDB button which starts initexmf.exe http://docs.miktex.org/manual/configuring.html
Then try the Update Formats button.
If these work you should be OK but it is impossible to say without having more information.


how would you do this problem

One gear turns 33 and one-third times in a minute. Another gear turns 45 times a minute. Initially a mark on each gear is pointing due north. After how many seconds will the two gears next have both their marks pointing north?

please include an explanation

 :rotfl:  :gathering:

I dunno how to type mathlike so this is the best I can do:
[hide]
first, find how many seconds it takes for a gear to turn:

60/(33+1/3)=3*60/(3(33+1/3))
so the fractions is 180/100 which is equal to 9/5

60/45 =4/3

Since the denominators 3 and 5 are relative prime or have a greatest common divisor of 1, than means they will only coincide on whole numbers.

9/5*5=9
4/3*3=4

lowest common multiple is 36[/hide]


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

[quote="ZaChoSen1"]Hi!

My name is Kruk Sandor and I'm from Baia Mare, Romania. I participated at AMC12A, too. The results came this week but surprisingly I was't on the list.
On the list there were some unknown papers scoring 37,5 (25*1,5).

After seeing the answers to problems my calculated score is 112,5 (17 good answers, 1 bad an 7 blank). With this score I would have been the first in my region ( the highest score now is 109,5) and I would have qualified to AIME.

What can I do to find my real score? I have already sent an email to amcinfo@maa.org and I'm waiting for their answer.

Thank you very much.

Kruk Sandor[/quote]

based on my knowledge, the contest administrator at your school has all the test scores, but i'm not sure how it is in Romania.

Also, if the test you took was unofficial that may be a reason why you were not qualified as an AIME participant.

The AIME is for US Citizens, or students legally residing in the United States and Canada.

[quote="undefined117"]The *USAMO is for U.S. citizens or legal residents of the U.S. I'm pretty sure anyone can take the AIME if they qualify...[/quote]

i happen to know someone who was not living in the USA who took AIME...  :)

[quote="ZaChoSen1"]As far as I know it is an official test.

The contest administrator at our school told me that he sent all  the papers and he doesn't know why I'm not on the list and he has nothing to do anymore.

But I find it unfair, as I paied for this contest.[/quote]
so you don't even know your score?
tell the administrator to show you the email he received from the MAA and look at the last few pages, where everyone who took the test's grades are

[quote]I have already sent an email to amcinfo@maa.org and I'm waiting for their answer. [/quote]

Then you have made the correct action to analyze and potentially correct this problem.

The AMC office received the email, is aware of this situation,  will investigate it, and will report back to you directly.

Steve Dunbar
MAA Director, American Mathematics Competitions

Thank you very much for analyzing my problem.

I look forward to receiving your answer.

Sandor Kruk


arithmetic sequence
number theory
relatively prime

[quote="digitalboy1"]In an arithmetic sequence $a(n)$ , it is known that $a(1)$ and $a(2004)$ are integers. There is also integer $i$ , $1<i<2004$ , such that $a(i)$ is integer.
Prove that all terms of $a(n)$ are integers ![/quote]

denote the common difference by $ d $.  then, from the given, $ a_1+dk \in \mathbb N $ for some $ k \in \mathbb N $ with $ 0 < k < 2003 $.  since $ a_1+2003d \in \mathbb N $, it follows that $ 2003d - dk = d(2003 - k) \in \mathbb N $.  now, if we let $ d = \frac {p}{q} $, in lowest terms (d must clearly be rational; and we must show it's actually an *integer*), the fact that $ a_1+dk \in \mathbb N $ means that $ k $ is divisible by $ q $. on the other hand, $ d(2003 - k) \in \mathbb N $ means that $ 2003 - k $ must be divisible by $ q $.  since 2003 is prime, it's relatively prime to $ k $, which we know is divisible by q, and hence $ 2003 - k $ can only be divisible by $ q $ if $ q = 1 $.  hence $ d $ is an integer, and the proof is complete.


Hello

I have searched through the forum a dozen times, and I couldn't find the answer ... If you would help me it would be great.
 I am to integrate the following:

4x*(X²+1)*e^(x²+1)

No. I am looking for a step-by-step explanation, from start to finish and then I want to ask questions and get references afterwards, if that's possible? :) That's how I've learned all the other math I know ...

Let $t = x^{2}+1$, as Kalle said, then $4x = 2\cdot 2x = 2\frac{dt}{dx}$, so \[\int 4x(x^{2}+1)e^{x^{2}+1}dx = \int 2\frac{dt}{dx}te^{t}dx = 2\int te^{t}dt\]So we want to find $\int te^{t}dt$. Integrate by parts: \[\int te^{t}dt = te^{t}-\int e^{t}dt = (t-1)e^{t}+C\]So \[\int 4x(x^{2}+1)e^{x^{2}+1}dx = 2(t-1)e^{t}+C = \boxed{2x^{2}e^{x^{2}+1}+C}\]

Thank you very much.
I will try to decipher your answer shortly. :)

By the way, are there alternate ways of solving this?  You see, in the textbook this was in the section BEFORE the one that describes integration in parts, which I by the way have trouble understanding completely.

If integration is an art, I'm not very artistic.

Can you explain how te^t minus the integral of e^t dt is (t - 1) * e^t + C

(OK, I understand the +C part)

Note that $\frac{d}{dt}e^{t}=e^{t}$.  No offense, but I don't think you have learned differentiation to its entirety.

Explain please? Note: If you can't explain it to a seven-year-old, you don't really understand it yourself. (- A. Einstein)

[quote="Flyrev"]Can you explain how te^t minus the integral of e^t dt is (t - 1) * e^t + C [/quote]

Which part don't you understand?  If you're wondering why

$te^{t}-\int e^{t}\, dt = (t-1) e^{t}+C$

Then it is because

$\int e^{t}\, dt = e^{t}$

But I think you might be wondering about the step [b]before[/b] that:

$\int te^{t}\, dt = te^{t}-\int e^{t}\, dt$

Which is an application of the method of [b]integration by parts.[/b]  A derivation:  we begin with the product rule,

$\frac{d}{dx}\left( uv \right) = u \, dv+v \, du$
$u \, dv = \frac{d}{dx}\left( uv \right)-v \, du$

Integrate both sides:

$\int u \, dv = uv-\int v \, du$

Take $u = t, dv = e^{t}\, dt$ to arrive at the result.

Edit:  In any case, I have issues with your Einstein quote.  According to accepted developmental theory, the average seven-year-old would find it very difficult to understand highly abstract concepts, even if they were explained very clearly.

What really puzzles me is that this piece is in the section just before the section about integration by parts, all the information we are given before this piece is the following:

int. of cos x = sin x + C
int. of sin x = - cos x + C
int. of (1+tan^2 x) dx = tan x + C
int. of 1/cos^2 x dx = tan x + C
int. of f(ax + b)dx = (1/a)*F(ax+b) + C

So from that, is there a way to solve it? If you knew ONLY that and no special techniques whatsoever, could you solve it? Or are the textbook writers giving me headaches on purpose here?

EDIT: Just found out I was wrong. The book this piece was taken from wasn't arranged by section. I am sorry. I didn't mean to be lazy and find an easy way out. Thanks for your answers, they will definetly be valuable when I start on the next section. :)


Suppose $f$ is bounded on $[a,b]$ and $f^{2}$ is Riemann-integrable on $[a,b]$.
Does this imply that $f$ is Riemann-integrable on $[a,b]$?
What if we assumed $f^{3}$ is Riemann-integrable on $[a,b]$?

let $f(x)=1$ for $x\in \mathbb{Q}\cap [a,b]$ and $-1$ otherwise, then $f^{2}$ is integrable but $f$ is not.

think about the 2nd question, its not very tough. how is the set of discontinuity pts of $f^{3}$ and those of $f$ related?


algorithm
algebra
polynomial
linear algebra
linear algebra unsolved

I can't figure out a pattern to the following types of problems, although I am sure that they have one. If any of you can help I would appretiate it.

1)Is every ideal of $ \Bbb Z[x]/(x^2 \minus{}1)$ principal?

2)Same for $ \Bbb Z[x]/(x^2)$

3)Same for $ \Bbb Z[x]/(15)$.

First ask yourself the question for $ \mathbb{Z}$, then for $ \mathbb{Q}[x]$, then for $ \mathbb{Z}[x]$.  Then: what is the characterization of ideals of $ R/I$ in terms of ideals of $ R$?

Sorry but I only partially get it. What is the exact relationship between the ideals of $ \Bbb Q[x]$ and $ \Bbb Z[x]$?

You know that $ \mathbb{Q}[x]$ has a division algorithm.  This implies that every ideal of $ \mathbb{Q}[x]$ is principal, hence is generated by a single monic polynomial.  On the other hand, the division algorithm in $ \mathbb{Z}[x]$ only works the same way if the polynomial you're dividing by is already monic.  What happens if it's not?  If you want to work out an example, see if you can figure out what the ideal generated by $ x^2, 2x \plus{} 1$ is.


Hi,

To solve:

.An n-gon has all angles equal and the lengths of consecutive sides satisfy  a1 ≥ a2 ≥ ... ≥ an Prove that all the sides are equal(IMO 63)

using complex numer.

Bye.

Alexis....

Let the vector v be the vector of the first side. Then the vector of the second side is vt(a_2/a_1)  and so on where t is the n-th root of unity. We become the equation:

v+vta_2/a_1+vt^2a_3/a_1+...vt^(n-1)a_n/a_1=0 from which follows:

1+ta_2/a_1+...t^(n-1)a_n/a_1=0.

a_1+a_2t+a_3t^2+...+a_nt^(n-1)=0.

If you consider this equation as a vector sum it is obvious that a_1=...=a_n is the only solution.

Misha


Determine $m \in R$ so that: $\displaystyle \{ x \in R | x^{2}+mx+1 = 0 \}$ $\backslash$ $\displaystyle \{ x \in R | x^{2}+x+m=0\} = \varnothing.$

$ A = \displaystyle \{ x \in R | x^2+mx+1 = 0 \} , B= \displaystyle \{ x \in R | x^2+x+m=0\} , A  \backslash  B = \varnothing  \Longleftrightarrow A\subset B $
$A =  \varnothing   \Longleftrightarrow  m^2 - 4 < 0 $ ;; $ A\neq \varnothing  \Longleftrightarrow  m^2 - 4 \geq 0    \Longleftrightarrow  \forall_{x\in A},x^2+mx+1 = 0=x^2+x+m$ $  \Longleftrightarrow x=1 \Longleftrightarrow m=-2 $
Finally $m \in [-2,2[$
 :cool: Is that for a difficult  PHD ? :cool:

[quote="Diogene"] :cool: Is that for a difficult  PHD ? :cool:[/quote]
What do you mean  ;)  :?


I don't know much about inequalities so maybe somebody can easy see is this right or not :
$ T[6,0,0] \plus{} 6 T[4,2,0] \plus{} 2 T[2,2,2] \geq 9 T[3,3,0]$

[b]zaizai-hoang[/b]
For example T(1,2,3) is
T(1,2,3)=a^1*b^2*c^3+a^1*b^3*c^2+a^2*b^1*c^3+a^2*b^3*c^1+a^3*b^1*c^2+a^3*b^2*c^1

[b]life_is_beautiful[/b]
I think , this inequality is wrong  for a=1,b=1,c=3 too.
One side is 13718 and second 14094


Hello,
I am looking for a (la)tex editor that supports unicode so that I can type in different languages. Do you n\know any editor like that?

The editors might support it (it's in fact the font they are using that must be unicode), however $\LaTeX$ does not (by default). I've heard that it's pretty tricky to make it work. You need some packages from ctan for this.

Can you be more exact? If I install a latex editor that supports unicode what should I do to make it work with unicode? Or does it depend on the editor itself?


trigonometry
limit
abstract algebra
complex numbers
calculus
calculus computations

Let $ r\in\mathbb{R}$ and $ |r|<1$, then find:

$ \text{i)} \sum_{n\equal{}0}^{\infty}{r^{n}\cos(n\phi)}$

$ \text{ii)} \sum_{n\equal{}1}^{\infty}{r^{n}\sin(n\phi)}$

Let us study the sums $ A_{n} \equal{} \sum_{i \equal{} 0}^{n \minus{} 1}{r^{i}\cos(i\phi)}$ and $ B_{n} \equal{} \sum_{i \equal{} 1}^{n \minus{} 1}{r^{i}\sin(i\phi)}$. We use the same approach as in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=253811:
\[ z_{n} \equal{} A_{n} \plus{} i B_{n}
\]
So we have to simplify the real and the imaginary part of $ z_{n}$.

Let $ z$ be $ z \equal{} \cos{\phi} \plus{} i\sin{\phi}$. After using this in $ z_{n}$ we get a geometric sum and after a bit of computation we get:
\[ A_{n} \equal{} \frac {r^{n \plus{} 1}\cos{(n \minus{} 1)\phi} \minus{} r^{n}\cos{n\phi} \minus{} r\cos{\phi} \plus{} 1}{r^{2} \minus{} 2r\cos{\phi} \plus{} 1}
\]

\[ B_{n} \equal{} \frac {r^{n \plus{} 1}\sin{(n \minus{} 1)\phi} \minus{} r^{n}\sin{n\phi} \plus{} r\sin{\phi}}{r^{2} \minus{} 2r\cos{\phi} \plus{} 1}
\]
Therefore for $ |r| < 1$, we have $ \lim_{n \to \infty}{A_{n}} \equal{} \frac {1\minus{}r\cos{\phi}}{r^{2} \minus{} 2r\cos{\phi} \plus{} 1}$ and $ \lim_{n \to \infty}B_{n} \equal{} \frac {r\sin{\phi}}{r^{2} \minus{} 2r\cos{\phi} \plus{} 1}$

Note also that the denominator $ r^{2} \minus{} 2r\cos{\phi} \plus{} 1$ can never be $ 0$, since
\[ r^{2} \minus{} 2r\cos{\phi} \plus{} 1 \equal{} (1 \minus{} r\cos\phi)^{2} \plus{} \sin^{2}{\phi}
\]

The more famous relative of (i) is the [i]Poisson kernel[/i]:

$ P_r(\theta)\equal{}\sum_{n\equal{}\minus{}\infty}^{\infty}r^{|n|}e^{in\theta}\equal{} 1\plus{}2\sum_{n\equal{}1}^{\infty}r^n\cos n\theta.$

We calculate in that case that $ P_r(\theta)\equal{}\frac{1\minus{}r^2}{1\minus{}2r\cos\theta\plus{}r^2}.$

That makes (i) equal to $ \frac12(P_r(\theta)\plus{}1).$

Hmm... shouldn't you have gotten $ \frac{1\minus{}r\cos \theta}{1\minus{}2r\cos\theta\plus{}r^2}\,?$ Re-check the sign of that cosine term in the numerator.

Thank you, Kent Merryfield.

I have corrected the sign.


I hav ebeen told that during this summer africain math olympiad was held is Algeria.
I am from Algeria and didnt have any idea about it.
Could someone tell me some words about thsi olympiad.
Thanks.

yes,like melissa said,

i am an algerian participant (silver medal)

it held in algeria on 30 august
only 8 countries participed :idea:

tunisia was the first
algeria the second
south afriac the third

Does anyone know where PAMO will be held next year?

Hi altusi
Nice to meet you here among the mathlinkers.They are incredible!
I would like to know how algerian participant have been chosen.
And how have you been prepared for that pamo.
Senouy.

i think that it will hold in Mali.

about the tact of choosing the algerian team, it done after 4 steps:
a test at the level of cities then another one who is regional,afer , a national one.

so, we have been only 15 pupils after. here,an training course of 1 month enabled to choose 4 participants.

[quote="Melissa"]It's called PAMO (Pan-African Mathematical Olympiad). Just search for PAMO in Mathlinks, I'm sure you'll find lots of stuff on it.[/quote]To view all the questions (and download the .pdf files associated) please visit: http://www.mathlinks.ro/Forum/resources.php?cid=28 ;)

I just heard that PAMO might be held in Cameroon next year. Anyone else heard anything?

I've just heard that PAMO will be held in Senegal this year :D

[quote]I've just heard that PAMO will be held in Senegal this year Mr. Green[/quote]
Did you go there? I went and on my way had a nice safari trip  http://www.mysafaribuddies.com/general-discussions/a-great-safari-on-senegal.html#381
 :lol:

If you have the problem, please visit my blog http://sinabih.co.cc

At the top of this forum, someone said that PAMO is the Pan-African Math Olympiad.

Most likely there aren't any other users for Djibouti.

[quote="DeejVJ"]Sorry for my ignorance, but what's PAMO?

Am I the only one from Djibouti?[/quote]

Are people the only one from Connecticut?

Woah AkshajK I thought you were Indian and were from Virginia.  I'm just bored and looking through all of the forums.  Obvious I'm not African...

obviously; it says so in my user profile.. I was just saying that since it seemed like there was no one from connecticut either..

AkshajK is from Virginia, not Connecticut.  He is trolling.

It is quite sad that this one middle schooler in the U.S., AkshajK, is more active than all the users from the continent of Africa, which has over a billion people.

Oh well poor third world Africa.  :(

That's a mean thing to say!
And am I the only Floridian here?

Floridans are supposed to just hang out in the beach all day...

South Africa will be home to the 2014 IMO, so perhaps this board will get much more active in the coming days as people begin to prepare.  I would love to hear more about the competition training and educational system in Africa.  I believe this might be the first time the IMO has ever been held in Africa (unless I didn't read the list carefully enough).

[url]http://imo-official.org/organizers.aspx[/url]

Hmmm. . . I never knew that there was a mat Olympiad in Africa. Must be interesting. R)


1. Yes, but they both can't be the same letter. In other words, you can take 10A-12B, 10B-12A, but not 10A-12A or 10B-12B.

2. You don't need to qualify to take the AMC 10/12. You just go and take it.

3. That depends on many factors. How does it matter, anyway?

3. i am pretty confident to say that every year, there will be several perfect scores.

3.  there are MANY perfect scores.

1. YES, I did that last year as 6th grader

2. you can't qualify for AMC/AIME through the AMC 8.

I've been taking the AMC 12 since 10 years old.
If a school requires you to qualify for AMC 10/12, find a different school.
AMC 8 always has many people getting each score(0-25).
You can either take the AMC 10A/B, the AMC 10A/12B, the AMC 12A/10B, or the AMC 12A/B.

[quote="Yankzgodzilla55"]thanks a lot

but the top scorers on the amc 8 are invited to take the amc 10

aime15

but anyway thanks a lot![/quote]

Yes, but you don't need to be a top scorer on AMC 8 to take amc 10.

@pythag: some schools make you qualify. It's unfair that way.

[quote="Yankzgodzilla55"]1. can an eighth grader take both the amc 10 and 12?

im taking algebra 2 on aops

2. if you dont qualify with the amc 8, can you take the amc 10/12?

3. What is the range of high scores on the amc 8?

thanks for answering
 :lol:     :huh:   :rotfl:[/quote]

I think what happens is if you get 20+ on the AMC 8, you are invited to take the amc 10, but not required


for $n\geq5$ it is impossible because 1! + 2! + ... + n! ends in 3.

now check for n=1,2,3,4...


[code]$\arcsin$[/code]
$ \arcsin$

or

[code]$\sin^{-1}$[/code]
$ \sin^{\minus{}1}$

Or pehaps $ \stackrel{\frown}{AB}?$

Or $ \overset{\frown}{AB}?$

(I'm guessing the O.P. really did mean "sign" rather than "sine.")

This has been discussed around here before.

This "frown" is really ugly.
One may usepackage [i]arcs[/i] when writing his own Latex, but I don't think it may be done on this forum.

Yeah, I mean't $ \overset{\frown}{AB}$.
Thanks!


For mathematica users out there,

how do you tell mathematica to expand something like, Sin[(a+b)+(c+d)]?
I tried TrigExpand, TrigFactor, TrigReduce. None of them works.

Could someone kindly direct me to the command?

By expand, do you mean to apply the formula sin(x + y) = sin(x)cos(y) + sin(y)cos(x) ?

Hmm, I am not sure, but I doubt that something so specific would be included as a function in mathematica.

I think most people will agree that Mathematica is the best.

PariGP is very strong for number theoretical issues.

I've heard Maple is good as well.  True Basic is good for smaller problems (easy syntax).

maple is good, particularly because of the large corpus of contributed code, but mathematica is more powerful in general.


group theory
abstract algebra
superior algebra
superior algebra unsolved

How do we construct a subgroup of order $ 180$ from $ \mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{3}\oplus\mathbb{Z}_{3}\oplus\mathbb{Z}_{3}\oplus\mathbb{Z}_{5}$?

Well, $ 180 \equal{} 2^2 3^2 5$, so why not try $ \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5$?  :)


Who would like to enlist in my contest, which will occur shortly. It is very easy to do but it does require full solution.

If you wish to enlist, make a signal like "I'm in", "I'll do it", etc.....

I will record the marks of those who do the contest and post the higher ones (if enough people do them). The higher scorers will move on to the next level, while the rest stays the level. I haven't decided on how many levels there will be, as if it moves too much, I can't handle the difficulty of the tests I'll make.

Again, theses questions are very, very easy. So if you're above this level, please don't laugh and wait for a higher level.

Sample prob?

Or could you supply the date this will happen?

Otherwise, count me in

I'm in.

[quote="sheepwarrior"]do the ppl who take it haveta be in middle skool?[/quote]

Shouldn't you be a middle schooler if you're browsing the Mathcounts forum? Unless you're a coach, in which case you're probably too good at math for this contest.

I'll do it... I thought you posted the exact same thing on the basic forum...  :?

[quote="solafidefarms"]Depends on when it is. If after Saturday a week from now, I'll do it, though that is really word-studying time, and right now is geography-studying time...So hard to study 3 contests at once!

Billy[/quote]

Hey, I have Geography, too.

You do not have to be in the middle school to write this. I'm jst posting it here as the first contest is extremely easy and the problems aren't the best in the world. Yes, I did post it in the basics, as my contest quetsions are most likely easier than mathcounts ones, even though I never wrote it, before I read a suggestion to put it into the mathcountes one. hehe  :blush:

Great.
hurry people, enlistment ends shortly, soon I will post the contest, level 1.

several questions: (sorry I keep on asking questions. Its just me)
How do we take it?
How can we access it?
How do we answer?
When will it be?
How long will it be up?
When will we know how we did?

Prize:  :first:

I am typing up a set of questions as we speak...Hitsme and I are writing questions...He is really the coordinator of this whole thing...I am just helping him out!

That's all?  I say it should be something more, like say $50.  JK.

What time is it.... I'm in it if i can make it :D

It was moved to the Getting Started forum under High School. You have until March 30 to PM him your solutions.


Find all pairs $ (m,n)$ of integer numbers $ m,n > 1$ with property that $ mn \minus{} 1\mid n^3 \minus{} 1$.

can anyone prove that other than general solution (m,n)=(k,k^20,there is no other solution?

$ \frac{n^3\minus{}1}{mn\minus{}1}\equal{}1\plus{}mn\plus{}n^2\frac{n\minus{}m^2}{mn\minus{}1}\equal{}1\plus{}mn\minus{}n^3\plus{}mn^2\frac{n^2\minus{}m}{mn\minus{}1}$.
Because $ (mn\minus{}1,n^2)\equal{}1\equal{}(mn\minus{}1,mn^2)$ must be $ a\equal{}\frac{n\minus{}m^2}{mn\minus{}1}\in Z, b\equal{}\frac{n^2\minus{}m}{mn\minus{}1}\in Z$.
If $ 1<m<\sqrt n$, then $ 0<a<1$ is not integer. If $ \sqrt n <m\le n>1$ then $ \minus{}1<a<0$ is not integer.
If $ n<m<n^2$, then $ 0<b<1$ is not integer. If $ n^2<m$, then $ \minus{}1<b<0$ is not integer.
Therefore $ (m,n)\equal{}(k,k^2),(k^2,k)$.

Notice $0 \equiv n^3-1 \equiv n^3-mn (mod mn-1)$, since $gcd(n, mn-1)=1$, then $mn-1 \mid n^2-m$, now it's easy.


Given that $0 < a < b < c < d$, which of the following is the largest?

$\textbf{(A)}\ \frac{a\plus{}b}{c\plus{}d} \qquad
\textbf{(B)}\ \frac{a\plus{} d}{b\plus{} c} \qquad
\textbf{(C)}\ \frac{b\plus{} c}{a\plus{}d}\qquad
\textbf{(D)}\ \frac{b\plus{} d}{a\plus{} c} \qquad
\textbf{(E)}\ \frac{c\plus{} d}{a\plus{}b}$

[hide]
I would say E.
Biggest numbers for numerator and smallest numbers for denominator.[/hide]

[hide]
My answer is E because c and d are the largest, so you want them on top, and they are in E.[/hide]

[hide]E because when you are finding the largest, you put the bigger ones on top and smaller ones on bottom. It is the opposite when looking for the smallest. If you want the closest together then you do not have sufficent information to solve it. :D [/hide]

[hide]i got E.

         numerator is biggest, denominator is smallest.[/hide]

I vote - *ahem*- pick E. It seems the most logical (plug numbers in).


How many solutions has

$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{1998}$ x,y are positives integers

[hide]We have $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{1998}$.  Multiplying both sides by $1998xy$ gives $1998y+1998x=xy$.  Subtracting $xy$ from both sides and factoring out the $y$ yields $y(1998-x)+1998x=0$.  Add $-1998^{2}$ to both sides and factor, giving $y(1998-x)-1998(1998-x)$.  Grouping $1998-x$ we have $(x-1998)(y-1998)=1998^2$.  Since $1998=2\cdot 3^{3}\cdot 37$, $1998^{2}=2^2\cdot 3^6\cdot 37^2$.  So $1998^2$ has $3\cdot 7 \cdot 3=63$ factors.  So there are $63$ possible values of $(x-1998)$, each of which determines the value of $(y-1998)$.  However, we have not considered the cases where both $x-1998$ and $y-1998$ are both negative.  None of these are possible because if both are negative, then one must be less than or equal to $-1998$.  If either $x-1998$ or $y-1998$ is less than or equal to $-1998$ then it will not be a positive integer.  So the only cases are the positive ones, of which as stated earlier, there are $\boxed{63}$.[/hide]

I think this is right...can someone confirm? :?

yeah i got the same thing.  i suppose you can generalize that the number of solutions in positive integers to $(x-a)(y-b) = n$ is just the number of factors of $n.$ what is the symbol they use to represent the number of factors of a number? i forgot what it was.

Yeah, that's the one! Thanks.  Then what does that lower case sigma function do?

[quote="chess64"]$\sigma$? It can be standard deviation or (as you probably mean) the divisor function. http://mathworld.wolfram.com/DivisorFunction.html[/quote]

Thanks a bunch, you really know your greek letter functions  :surf:   I meant the divisor function of "order" one I suppose - the sum of a number's factors.  Thanks again chess64.


Find a such that equation follow has one root:
$ ax^2 \minus{} 2(a \minus{} 1)x \plus{} 2 \equal{} |ax \minus{} 2|$

This is hard problem,I hope someone can solve it....

[quote="Srhiri"]Let  f(x)=ax^2 - 2(a - 1)x + 2 - |ax - 2|
then f(0)=0
case1: a=0
f(x)=2x
case2: a≠0
f'(x)=2ax + 2 - a  for  ax < 2
f'(x)=2ax + 2 - 3a  for  ax > 2
then
f'(0)=0  implies  a = 2/3  ;   a = 2
conclusion: a=0  ;  a = 2/3  ;   a = 2[/quote]
Are you sure $ a\equal{}\frac{2}{3}$ ?


algorithm
algebra
polynomial
linear algebra
matrix
graph theory

There is a graph G with N vertexes. Each vertex is colored in one color. There may be some vertexes with the same color.
One vertex is marked as vertex A.

John plays a following game.

At each move he:
1) Choose random color C1;
2) Takes maximal connected subgraph S (with maximum number of vertexes) from graph G so that S contains vertex A (colored in color C2) and each vertex from S is colored in color C2;
3) Colors each vertex from S in color C1;

What is minimal number of moves in which he can color original graph in one color?
Is there an algorithm solving that problem in polynomial time (    O(N^k) for some constant k   )?

Sorry for my bad english.

perhaps you should say "a color C1" instead of "random color C1"... if it was random then the problem is unsolvable

Yeap, it's "a color".
I do have a similar game here and I'd appreciate any help. :)

Let an (arbitrary chosen) undirected graph and two colors : blue and red. Initially, all vertices are blue-colored. At each step, John can "hit" any vertex of the graph : the vertex itself along with its neighboring vertices change color (blue -> red, red -> blue). The game's goal is to paint red all the vertices.

1. Show that the order in which John will "hit" the vertices of a given set does not mind at all.

2. Show that John does not have to "hit" any vertex over $ \cancel{\text{two times}}$ [edit] one time.

From the above (1-2) we conclude that, if a solution does exist, it will be enough to find an appropriate subset of the vertices and then John will only have to "hit" these vertices, just once, without having to pay attention to the order we will do so.

3. Given a chain, is it possible to color all vertices red? Given a circle? Given a complete bipartite graph $ K_{n,n}$?

4. It is possible to find a polynomial algorithm which will calculate a set of vertices which, if John "hits" once, without minding the order he will do so, all vertices of the graph are colored red?

Thanks in advance. :)

[quote="fedja"]Ah, modulo, isn't it just an $ N\times N$ linear system in $ \mathbb Z_2$? :roll:[/quote]

Ahm..... I don't get it. :blush:

Silly question, but $ \mathbb Z_2$ = ?? :D
A set of 2 integers?

[quote="JBL"]The set of integers modulo 2.  (Here "red" = 0, "blue" = 1, and "changing colors" = adding 1.)[/quote]

Okey... and where will this help? :(

[quote="fedja"]A linear system can be solved by Gaussian elimination in polynomial time over any field.[/quote]

I see. I still don't get it how this game could be modeled with a linear $ N\times N$ system. :maybe:

By the way, any tips for a more formalistic approach of questions 1-3? Besides common sense. =Ρ

Each variable is 0 or 1 depending on whether you hit each vertex or not. The final colour of a vertex only depends on itself and its neighbours, so you form an equation by adding up the right variables and setting to 0 or 1 depending on what colour you want.

Aha! So if I'm correct, given a chain of n=4 nodes, we have:
[geogebra]71ab36900f7feb2dde3081131f5bcd4649af5e59[/geogebra]

$ \begin{matrix} a: \\
b: \\
c: \\
d: \end{matrix} \begin{Bmatrix} a & \plus{} b & & & \equal{} 1 \\
a & \plus{} b & \plus{} c & & \equal{} 1 \\
& b & \plus{} c & \plus{} d & \equal{} 1 \\
& & c & \plus{} d & \equal{} 1 \end{Bmatrix} \Rightarrow \left\{\begin{matrix} a \equal{} 1 \\
b \equal{} 0 \\
c \equal{} 0 \\
d \equal{} 1 \end{matrix}\right.$

Which means we have to hit nodes A and D... right?

And, given a circle of n=4 nodes:

[geogebra]3903b46340c8b13fb30122e33c5f5de387cf3347[/geogebra]

$ \begin{matrix} a: \\
b: \\
c: \\
d: \end{matrix} \begin{Bmatrix} a & \plus{} b & & \plus{} d & \equal{} 1 \\
a & \plus{} b & \plus{} c & & \equal{} 1 \\
& b & \plus{} c & \plus{} d & \equal{} 1 \\
a & & \plus{} c & \plus{} d & \equal{} 1 \end{Bmatrix} \Rightarrow \left\{\begin{matrix} a \equal{} \frac {1}{3} \\
b \equal{} \frac {1}{3} \\
c \equal{} \frac {1}{3} \\
d \equal{} \frac {1}{3} \end{matrix}\right.$

Em.... meaning that all three nodes should be hit, right? :maybe:
What happens when some variables are not integers?

The answer to both questions to which you said "right?" is "Yes, and you can easily check yourself that this actually works."

[quote="modulo"]What happens when some variables are not integers?[/quote]  You've forgotten that you're working modulo 2.  We have 3 = 1.

Right. :)

Well, thanks everybody for your precious help! :)


Gauss
number theory solved
number theory

Hey, here's something that's been bugging me:
How do you prove that a prime number of the form 4k+1 can be written as a sum of 2 squares? I am aware of Fermat's thm about numbers which can be written as a sum of 2 squares, but how do you actually prove it?

try using the fact that there is a residue a modulo p such that a2=-1 (in fact that residue can be guess, it's ((p-1)/2)! - try to prove that )

further more use Thue's Lemma, which says that for any given residue a there are positive integers x,y smaller in modulus than \sqrt p such that ax+y or ax-y is 0 modulo p.

if you can't figure all of it out from this, ask me again and I will try to post the whole proof ...

10x! Can a similar method be used to prove that (I don't even know if this is true) any prime number of the form 8k+3 or 8k+1 can be written as a^2+2b^2? P is of that specific form so that -2 is a square residue mod p, because (-1/p)=(-1)^((p-1)/2) and (2/p)=(-1)^((p^2-1)/8), and by multiplying them we get that p must be 8k+3 or 8k+1.

I wanted to use this for the following problem, which was given at a selection examination in Romania, in 1996: If p^2 can be written as a^2+2b^2 with b different from 0, then so can p, where p is a prime.

According to many, the most elegant proof that primes of the form 4k+1 can be written as sum of two squares is due to Don Zagier. Check:

http://spicerack.sr.unh.edu/~dvf/532/Zagier

A short and similar proof- if I remember correctly- for the 8k+3 primes
was published in the May 2000 issue of the American Mathematical Montlhy, by  Terence Jackson.

Thanks, the proof is indeed amazingly elegant!

and how can one prove that every n so that
n /= 7 [mod 8] is a sum of three perfect squares ??

Gauss was the first to prove thar every natural number wich is not
of the form 4^s(8k+7), (with s,k positive integers) is the sum of the
squares of three integers

See N.C.Ankeny, sum of three squares, Proc.Amer.Math.Soc.50(1944),
881-882

moubi I must admit that I haven't got a clue as to where to get that book, and probably other people don't either :) if you could spare a web-link or even give away a little clue about the proof it would be very appreciated :)

The book i talked you can find in mathematics research library.

----------------------------------------------------------------------------------
A natural number n can be the sum of three squares only if it is no of the
form 4^k(8s+7), where k,s are integers >=0
----------------------------------------------------------------------------------
Proof

If there is a natural number of the form 4^k(8s+7) that are the sums
of three squares integers.
Consider n the least of them n = 4^k(8s+7)=a^2+b^2+c^2  with
a,b,c integers.

If  among a,b,c there is only one  odd a^2 +b^2+c^2 is of the form
4x+1 =/=n

If among a,b,c there are two odd then a^2 +b^2 +c^2 is of the form
4x+2 =/=n

If a,b,c are odd then a^2 +b^2 +c^2 is of the form 8z+3=/=n

So each a,b,c are even

a=2a_1,  b=2b_1, c=2c_1

and 4^(k-1)(8s+7) = a_1^2 + b_1^2 + c_1^2 wich contradict the

minimality of n

 :D  :D  :D
 :D oubi :D
 :D  :D  :D

Oh OK, but you don't prove the inverse (what Gauss wrote about in his diary as $\delta + \delta +\delta = n$ -- though I don't think he used TeX code).

Anyway, if anyone has a proof of what moubi refers to or the quasi-equivalent statement that every natural can be represented as a sum of three triangular numbers, then please post it. (I know one cannot use the very useful "multiplication" lemma, used in similar problems).

Thanks!

Cheers,
Thanos


Given an integer $ d$.Let $ S \equal{} \{m^2 \plus{} dn^2|m,n \in Z\}$
Let $ p,q \in S$ such that $ q$ is a prime and $ r \equal{} \frac {p}{q}$ is an integer.
Prove that $ r \in S$

well,i've heard a problem like this but it is very old and well-known;
if $ A\equal{}a^2\plus{}nb^2$ where $ n,a,b$ are integers and $ q$ is a prime divisor of $ A$ such that there exists integers $ c,d$ with $ q\equal{}c^2\plus{}nd^2$ then the number $ \frac{A}{q}$ can be written in the form $ x^2\plus{}ny^2$ where $ x,y$ are integers.

[quote="ashwath.rabindranath"]I hope you mean q is a prime?[/quote]
I had edit it  :)


Hello everyone,

Here's my word problem:

• Phil has 48 coins composed of nickels, dimes, and quarters with a total value of $5.10. He has 4 more dimes than nickels and quarters combined. How many coins of each kind does he have?

I tried to set up out these equations but they don't work. Could someone please tell me what the correct system of equations should be?

Let n be the amount of nickels, d be the amount of dimes, q be the number of quarters.

4 + d = n + q -> n - d + q = 4 --- (1)

n + d + q = 48 --- (2)

5n + 10d + 25q = 510 --- (3)

Thank you very much!

Cheers,
Patrick

Your equations look fine.

[hide]
Sub (1) from (2)
$2d = 44 \implies \boxed{d = 22}$

Divide (3) by 5
$n+2d+5q = 102$
Sub (2) from (3)
$d+4q = 54$
Sub d above
$\boxed{q = 8}$
Sub q and d in (2)
$\boxed{n = 18}$
[/hide]

Hello everyone,

Thanks for your help, especially to JavaMan. However, equation #1 is wrong, it's supposed to be:

d = 4 + n + q --- (1)

I forgot that the total of amount of dimes is equal to four more dimes + the nickels + the quarters so d would be isolated.

Thanks though for your help. In any case, the answer would be (d, n, q) = (26, 15, 7)

Patrick


[quote="Klebian"][quote="Klebian"]I will not be accepting very new members to this site, unless they are recommended by other more active members of the site. If you don't think you could at the very least check the game on a daily basis, please don't sign up.
[/quote]

I would like to emphasize both these points made in my post. So, lena, at this point, I will not add you to the list, but I will judge your activity. And those people who cannot access the game [u]everyday[/u] on average, should not sign up. This means, if you only have computers on weekends, or only on weekdays, I'd prefer you not sign up. This just depends on your activity however.[/quote]

Well, that is too bad

I used to play Mafia with my friends live, when we were in an army training course, but since we live in diff cities i dont get to see them much, nor find anyonoe who has the same interest in the game as i do.

Big thanks to Bubka for the recommendation.

Also i'd like to know how many posts i need to make to get rid of the "new member" title

Army training course?
[i]Army training course??[/i]

WHOAA!!

Yes, i spent 2 years in the army from april 2004 till april 2006

And how old are you now?
Army ... omg ... the army ...

Ah so you served from 18 to 20.
I wonder whether you have read Eric Maria Remerque. EMR has soem interesting views on teenage recruits.
But of course you never served at the front.
I will soon write in the Round Table about the book. I finished it three days ago and was deeply affected by it.
That was why the armed forces suddenly struck me in your post.

they sent your sister to the front with wars going on?

[quote="solafidefarms"]But still, what country do you live in?[/quote]

well, lena is a german female name. but, as i said in israel you are obligatory to go to army whether you are male or female.
so, i would say lena is from israel. am i right?

btw, i know that we can't choose what are we going to do in the army.
i am suppposed to go to artillery. army wants to send me there. but after the war, our army isn't so... "safe". especially in the past 7 years. so i tend to do the civil service instead of military one [at least i won't be raped, beaten up without a reason, got shot and be smuggler or steal weapons :P ]

yes i am from israel

for some reason i thought you knew that

Lena is from many countries

it is very common in eastern europe
i dont know how common it is in western europe ... Germany? its the first time i hear it

i usually get "its a russian name, its a greek name, its a whatever name "

I attempted to clear all discussion out of the mafia thread into one topic, but I was unable to move one post. So this post:

[quote="Lena"][quote="deimos"][quote="Lena"]my job in the army was fixing parts of tanks , never been to the front

they sent my sister there however[/quote]

so, your sister joined the army and went to the front as a volunter or, she was forced to go to the front? :|  where was the front? is military service compulsory for girls in your country?  where do you live?

[hide]i heard that in israel males and females must go to army [/hide]

wow, this is offtopic. :roll:[/quote]

You dont actually choose what you will do in the army

Teenagers after finishing school and reaching 18 years old
meaning if you're 19 and havent finished highschool yet, you dont go to the army until you're done with school

Boys go for 3 years, girls go for 2 years

If you got something you wanted to do , then great, if you didnt... tough luck[/quote]

came between deimos and solafidefarm's post.

[quote="Lena"]yes i am from israel

for some reason i thought you knew that[/quote]

well, perhaps (because) i mentioned that possibility in the mafia thread (hidden content)  :)

[quote="Lena"]Lena is from many countries

it is very common in eastern europe
i dont know how common it is in western europe ... Germany? its the first time i hear it

i usually get "its a russian name, its a greek name, its a whatever name "[/quote]

it's the name of one big river in siberian land.

yes it is

this was a very short message, so i came up with something to make it longer, lets see if it works now ...  ready?

[quote="Ars"]deimos it is in Russia as siberia is in Russia

so that is the correct way to say it

also many believe siberia is a sperate country for some reason

Also it is a fairly common Russian name and many German female names are the same as female Russian names because of the marrying of Russian royalty to German princesses[/quote]

first of all i don't think that siberia is a separate country (for the first time in my life i hear that someone thinks that  :huh: )

secondly, are you sure that's the only reason why are some female names in russia and germany similar  :roll:

btw, there wasn't only marrying russian royality with german princesses, there was other way around. and with spaniards, frenchs, italians, dutch, danes, swedes, poles, austrians, hungarians...

[quote="deimos"]i am suppposed to go to artillery. army wants to send me there. but after the war, our army isn't so... "safe". especially in the past 7 years. so i tend to do the civil service instead of military one [at least i won't be raped, beaten up without a reason, got shot and be smuggler or steal weapons  ][/quote]
What about cancelling the army? Serbia will have professional army after 2010. Civil service is pointless here. Most people don't do anything useful-they get job below their qualifications or they are told not to disturb and to go home. This solution is far more acceptable than regular service if you consider number of unsolved murders/suicides in lats few years.


geometry
incenter
circumcircle
power of a point
radical axis
geometry proposed

Let $\gamma$ be the incircle in the triangle $A_0A_1A_2$. For all $i\in\{0,1,2\}$ we make the following constructions (all indices are considered modulo 3): $\gamma_i$ is the circle tangent to $\gamma$ which passes through the points $A_{i+1}$ and $A_{i+2}$; $T_i$ is the point of tangency between $\gamma_i$ and $\gamma$; finally, the common tangent in $T_i$ of $\gamma_i$ and $\gamma$ intersects the line $A_{i+1}A_{i+2}$ in the point $P_i$. Prove that

a) the points $P_0$, $P_1$ and $P_2$ are collinear;

b) the lines $A_0T_0$, $A_1T_1$ and $A_2T_2$ are concurrent.

thnks for telling me. now i can post the solution more comfertabally

for a)

because of the power of $P_{0}$ related to $\gamma_{i}$ we get that $P_{0}T_{0}^{2}=PA_{1}\cdot PA_{2}$, but left side is the power of $P_{0}$ related to $\gamma$ (the incircle) and right side is the power of $P_{0}$ related to the circumcircle. Thus $P_{0}$ is on the radical axis of the two circles, and this is true for $P_{1}$ and $P_{2}$ also
in the exam i realised this AFTER i have shown that the projection of $P_{i}$ on $OI$ is the same independant of $i$.
i am very curios to see the solution at b) because no one i talked to solved it

Isn't part (b) just problem 2 here?
[url=http://www.mathlinks.ro/Forum/contest/4th%20Edition/4th%20edition%20-%20round%207.pdf]www.mathlinks.ro/Forum/contest/4th%20Edition/4th%20edition%20-%20round%207.pdf[/url]

Link to MathLinks Contest discussion: [url]https://artofproblemsolving.com/community/c60h24007p152422[/url]

i can`t find the solution (too many buttons :) )
is it here too ?

edit : i found it
very nice, but very hard to find, especially in the exam

well..but using that steinbart theorem (i hope i write the name ok) it is easy :)
let $S_{i}$ be the intersection between the incircle $\gamma$ and the side $A_{i+1}A_{i+2}$.
Then $S_{i}T_{i}$ is the polar of $P_{i}$ related to $\gamma$. But $P_{i}$ are collinear, and by duality we have that the polars have all three one common point. then, steinbart and we`re done.
i hope i didn`t say a big stupid thing :)

Can anyone give a simple proof for the steinbart theorem, using only projective geometry?

A simple proof for part a follows from the fact that each $P_i$ lies on the radical axis of $\gamma$ and $\gamma_i$ (from $P_iT_i$ is tangent to $\gamma$). From the previous fact, it follows that since $\gamma$ is the incenter, the radical axis of $\gamma$ and the circumcenter contain each of $P_i$ so they are collinear as desired.

Sorry for the revive, but the link posted by nsato is broken

can anyone post a new link?


In a game of endless cricket, the scores $X$ and $Y$ of the two sides are: \[ \text{P}(X = j, Y = k) = e^{-1}\frac{(j + k)\lambda^{j + k}}{j!k!}  \] where $\lambda$ is some positive constant and $j,k \in \mathbb{Z}^+$.

Find $\lambda$ and hence calculate $\text{E}(2^{X+Y})$.

The method is to sum over all $j, k$ and set the result equal to 1 (unless there are properties of the rules of cricket that I don't know about), but it seems to me that there will be no good way to solve for $\lambda$.  In particular, I get
[hide]that $\lambda$ must satisfy $f(\lambda) = 2\lambda e^{\lambda-1}(e^\lambda - 1) = 1$ and that expected value we are looking for is $f(2\lambda)$.[/hide]

Suggestions?

[hide="How To Find Lambda"]
\begin{eqnarray*} \text{P}(X + Y = n) &=& e^{-1} \left ( \frac{n \lambda^n}{j!k!} \right ) \\ &=& \frac{n \lambda^n}{n!e} \sum_{r = 0}^n \binom{n}{r} \\ &=& \frac{(2\lambda)^n}{(n-1)!e} \end{eqnarray*}

We can show that $\text{P}(X + Y = 0) = 0$. So
\begin{eqnarray*} \sum_{r = 1}^{\infty} \text{P}(X + Y = r) &=& 1 \\ \frac{2\lambda}{e} \sum_{r = 1}^{\infty} \frac{(2\lambda)^{n-1}}{(n-1)!} &=& 1 \\ 2\lambda e^{2\lambda - 1} &=& 1 \end{eqnarray*}

Now $f(x) = 2xe^{2x - 1}$ is increasing for $x > 0$ so we have only one real solution to $2\lambda e^{2\lambda - 1} = 1$ which by a little guessing is $\frac{1}{2}$. Thus $\lambda = \frac{1}{2}$.

I'll leave you to show that $\text{E}(2^{X+Y})$ is $2e$.
[/hide]

Ah, I see what I did wrong -- I took $X$ and $Y$ to be always positive while I should have taken them non-negative.


Do you guys know where I can find information regarding the braid lemma of Algebraic Topology?

I'm specifically looking for a precise formulation of the result.

I thank you in advance for your insightful replies.  :wink:

I've never heard of it (and according to Google, not too many people have), but I did find it as an exercise in Munkres' "Elements of Algebraic Topology", p. 148 in my edition which I think is the first.

A braid is the following commutative diagram (of abelian groups and homomorphisms, although I guess this would work in any abelian category):

 [geogebra]84b59562c93710e8d67c0f665d5ec2729568f45c[/geogebra]

We consider the following sequences:

1. $ E \to A \to B \to G \to K$

2. $ E \to I \to J \to G \to C \to D$

3. $ A \to F \to J \to K \to H \to D$

4. $ I \to F \to B \to C \to H$

If these four sequences are all exact, the braid is called an exact braid. The braid lemma states that for this to happen it is enough that 1, 2 and 3 are exact, and that the composition $ I \to F \to B$ is $ 0$.

I didn't try to solve this, but it seems like a fairly routine diagram chasing exercise. Apparently this is useful in proving the exactness of the homology sequence of a triple (the next exercise in Munkres).

[quote="randomgraph"]... but I did find it as an exercise in Munkres' "Elements of Algebraic Topology", p. 148 in my edition which I think is the first.

I didn't try to solve this, but it seems like a fairly routine diagram chasing exercise. Apparently this is useful in proving the exactness of the homology sequence of a triple (the next exercise in Munkres).[/quote]

Thanks for the reference, dear [b]randomgraph[/b]... I've solved the exercise already. The solution of it resorts to anything but diagram chasing, just as you'd first suspected.

Best of luck,
José.

P.S. Wonder how it was that the lemma came into being... More important, if the lemma helps one to prove big results, such as the exactness of the sequence of a triple, how is that it is not that well-known? Im' really curious about it, you know...

[quote="coquitao"]
P.S. Wonder how it was that the lemma came into being...[/quote]
Honestly, I don't think it really came into being as a "Lemma". Munkres seems to be the only person to have called it that, and I'm not sure it really deserves the status of a Lemma.

[quote="coquitao"]
More important, if the lemma helps one to prove big results, such as the exactness of the sequence of a triple
[/quote]
I'm not sure it can be used to prove anything else, honestly. It seems very special-purpose and contrived. Also, I don't know that I would call the exact-sequence-of-a-triple "a big result" - once you have the technical definitions etc., it's also a rather routine and unsurprising result.

[quote="coquitao"]
how is that it is not that well-known? Im' really curious about it, you know...[/quote]
Well, if you believe everything I wrote above, then I don't see why it should be well-known. A lemma used in the proof of just one theorem is not a Lemma, really - not in the [url=http://www.math.rutgers.edu/~zeilberg/Opinion82.html]sense of Zeilberger[/url], anyway. So I wouldn't be too curious as to why it's so obscure  :)

[quote="randomgraph"] I'm not sure it can be used to prove anything else, honestly. It seems very special-purpose and contrived. Also, I don't know that I would call the exact-sequence-of-a-triple "[b]a big result[/b]" - once you have the technical definitions etc., It's also a rather routine and unsurprising result. [/quote]

Right... I guess it's [b][color=red]ubiquitous[/color][/b] the word that I should have used there.

[quote="randomgraph"] A lemma used in the proof of just one theorem is not a Lemma, really - not in the [url=http://www.math.rutgers.edu/~zeilberg/Opinion82.html]sense of Zeilberger[/url], anyway. So I wouldn't be too curious as to why it's so obscure.  :)[/quote]

You actually have a point there, man. Thanks for the link, too. I found it to be very insightful...  :wink:


function
quadratics
inequalities
modular arithmetic
calculus
integration

We have [tex]\frac {z}{e^z-1}=\frac {k_0}{0!}+\frac {k_1}{1!}x+\frac {k_2}{2!}x^2+\dotsm[/tex]

-Show that [tex]k_0=1[/tex] and [tex]k_n=\frac{-\sum_{0\leq i<n} \binom{n+1}{i}k_i}{n+1}[/tex]

-Show that [tex]\frac{\sum_{0\leq i<n} \binom{n+1}{i}x^{n+1-i}k_i}{n+1}=\sum_{j=1}^{x-1}j^n[/tex] (note that this formula is very similar to the previous)

-Show that [tex]k_{2n+1}=0\quad \forall n\in N_0[/tex]

-Knowing that the Bernoulli's numbers are definited such that [tex]B_n=k_n[/tex] show for witch n [tex]B_{2n}^{-1}\in N[/tex]

-Show that [tex]\zeta(2n)=\frac{2^{2n-1}|B_{2n}|\pi^{2n}}{(2n)!}\quad \forall n\in N_0[/tex]

P.S.: the zeta-function [tex]\zeta(-)[/tex] is definite for reals [tex]r>1[/tex] in this way: [tex]\zeta(r)=\sum_{n=1}^\infty \frac{1}{n^r}[/tex]

Probably I've made many mistakes... :blush:

There's just one (technical) mistake: we have $k_{2n+1}=0$ for all $n \in \mathbb{N}$, not for all $n \in \mathbb{N}_0$ (note that $B_1=-\frac{1}{2}$.)

Here's, by the way, an explicit formula for the Bernoulli numbers:
\[
B_n = \sum_{0 \leq m \leq n} \frac{(-1)^{m+n}m!S_2(n,m)}{(m+1)},
\]
for $n \geq 2$, where $S_2(n,m)$ are the Stirling numbers of the second kind, which denote the number of partitions of an $n$-element set into $m$ disjoint non-empty classes, and are given explicitly as
\[
S_2(n,m) = \frac{(-1)^m}{m!} \sum_{l} (-1)^l \binom{m}{l}l^n.
\]

Here is another relationship to the (analytic continuation of the) zeta-function:

\[
\zeta (1-n) = - \frac{B_n}{n}, \forall \, n \in \mathbb{N}.
\]
Thus, the zeta function vanishes at the even negative integers, which are called the "trivial zeros" of $\zeta (z)$. I won't discuss the non-trivial zeros because the topic was not originally posted at the College Playground section.   :lol:

As for the problem of finding all $n$ for which the numerator of $B_{2n}$ is $1$, it is not hard to show that $|B_{2n}|>1$ for large enough $n$.

Here's an additional

Problem 1. Show that there exist infinitely many $n \in \mathbb{N}$ for which $6B_{2n} \in \mathbb{N}$. In other words, the denominator of $B_{2n}$ equals $6$ for infinitely many $n$.

For the generating functions aficionado, here's an important quadratic recurrence relation:

Problem 2. Show that, for any $n \geq 2$, the following identity holds:
\[
\sum_{l \geq 0} \binom{2n}{2l}B_{2l}B_{2n-2l} = (1-2n)B_{2n}.
\]

Enjoy!

--Vesselin

The recurrence relations should be read
\[
\sum \binom{2n}{2l}B_{2l}B_{2n-2l} =(1-2n)B_{2n}.
\]

For this last recurrence, all we have to do is multiply both sides by $\frac {x^{2n}}{2n!}$ and sum it all up. On the left we get $(f(x))^2-1-\frac {x^2}6$ ($f(x)=\frac {g(x)+g(-x)}2,\ g(x)\frac x{e^x-1}$) and on the right we get $f(x)-xf'(x)+\frac {x^2}{12}$. Plain calculations show us that these two are equal, i.e. $(f(x))^2-f(x)+xf'(x)-\frac {x^2}4=0$.

Generating functions also work for the first recurrence relation posted by Simo (the Wolf :)):

It's equlvalent to $\displaystyle \sum_{i=0}^n\binom{n+1}{i}B_i=0,\ n\ge 1\iff \sum_{i=0}^n\binom ni B_i=B_n,\ \forall n\ge 2$. We multiply both sides with $\frac {x^n}{n!}$ and sum it. We get $\frac x{e^x-1}-1+\frac x2$ on the left and $\frac {xe^x}{e^x-1}-1-\frac x2$ on the right, and they're obviously equal.

I suppose we can show the second relation in pretty much the same way, but I didn't do the calculations (too lazy :)).

Sure. It's routine.

Let me add one more problem to the list: the famous result of Jensen on the infinitude of irregular primes.

Problem 3. (Jensen, 1915). Prove that there exist infinitely many primes $p$ with the following property: there exists some even integer $n \leq p-3$ such that the numerator of $B_{n}$ is divisible by $p$.
[The smallest such prime is $p=37$.]

Yes, proving the second relation (Bernoulli's formula fo the sum of the $k$th powers of the first few integers) is *absolutely* straightforward by employing the generating function of the Bernoulli numbers. Proving the explicit formula that I posted is a little more technically complicated, but still a matter of routine with generating functions.

The proof of Euler's formula
\[
\zeta (2n) = \frac{(-1)^{n-1}2^{2n-1}B_{2n}\pi^{2n}}{(2n)!}
\]
is a lot more technical, but also routine :). However, there is a paticularly cute way to derive the general result from the special case $\zeta (2) = \pi^2/6$.

These problems (all mentioned so far) are standard results. Here is a conjecture that I have posed but not proven:

Conjecture. There exist infinitely many primes $p \equiv 1 \pmod{4}$ for which $p$ does not divide the numerator of $B_{\frac{p-1}{2}}$.
[Compare with Problem 3 above.]

I'd welcome any fresh ideas or comments. The reason I'm interested in this conjecture is the following result that I have discovered:

Theorem. (Mine). Let $p \geq 5$ be a prime of the form $4k+1$. Assume that $p$ does not divide the numerator of $B_{\frac{p-1}{2}}$, and let $m$ be a positive integer such that $m \equiv \frac{p-1}{2} \pmod{p-1}$. Then, the following congruence holds:
\[
\sum_{l \geq 1} \binom{m}{l(p-1)}l \equiv 0 \pmod{p}.
\]
There may be a simpler approach of examining this congruence. It is quite possible that the congruence holds, in fact, modulo all primes $p \equiv 1 \pmod{4}$, or even for all primes $p \geq 5$. It is also quite possible that this result can be considerably generalized. If someone's interested and has made some progress on this or an analogous question, contact me! We may write a paper on it.  :D

the riemann hypothesis....the so particular line $x=\frac12$ of the zeta function : this is one of the biggest problems of the moment and many mathematician work on it

Regarding Vess's problem 1 (that one about $6B_{2n}$ being integral), I have recently read about von Staudt's theorem in Hardy and Wright's "Introduction to the Theory of Numbers". It states that for $n\ge 1$ we have $\displaystyle B_{2n}+\sum_{p\mbox{ prime},p-1|2n}\frac 1p\in \mathbb N$. It's now obvious that if $n$ is a prime of the form $3k+1$, then we have $6B_{2n}\in \mathbb N$.


Let $a, b, c>0$ such that $abc=1.$ Prove that:

                              \[ \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}\geq 1  \]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=19469 part 1a.

  darij

I think the following general result holds:
For any positive reals $a,\ b,\ c$ and nonnegative reals $k$, \[ \frac{a^k}{abc+b^k+c^k}+\frac{b^k}{abc+c^k+a^k}+\frac{c^k}{abc+a^k+b^k} \ge \frac{3}{2+(abc)^{1-k/3}}  \]
A similar result: For any positive reals $a,\ b,\ c$ and nonnegative reals $k$, \[ \frac{b^k+c^k}{a^k+abc}+\frac{c^k+a^k}{b^k+abc}+\frac{a^k+b^k}{c^k+abc} \ge \frac{6}{1+(abc)^{1-k/3}}  \]

[quote="zhaobin"]a more directly proof:
$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}=\frac{a^2}{ab+ac+a}+\frac{b^2}{bc+ab+b}+\frac{c^2}{ac+bc+c}\geq \frac{(a+b+c)^2}{2(ab+bc+ca)+a+b+c} \ge 1$ :)
the last one is equal to $a^2+b^2+c^2 \ge a+b+c$ which is obvious :)[/quote]

You used to "svacs" inequality


Hello.  Our high school math team, along with a few teachers, is stuck on a problem that appeared in an old math competition test.  Here is the problem:

 :sqrt:[size=134](15 + 4[/size]:sqrt:[size=134](14))[/size]

The answer is [size=134]2:rt2: +[/size] :sqrt:[size=134](7)[/size]

Thanks for the help.

well you can find this topic in AoPS volume 1, chapter 6, but I'm bored so I'll just do it .
since 14 is a composite number, the roots we are looking for are in the form of (a :sqrt: b+ c :sqrt: d) :^2: . we know that b and d must be 2 and 7 since they are the factors of 14. now we expand to get 2a :^2: +2ac :sqrt: 14+7c :^2: .
Now we get a system of equations
ac=2
7c :^2: +2a :^2: =15
by trial and error we find the solutions c=1, a=2 and thus findind the positive root 2 :sqrt: 2+ :sqrt: 7
I did this so long ago and I'm out of practice...

um, this leads to me to another unrelated question. In AoPS Algebratic Problem Solving Class online, when the topic simplifying such radical expressions was stressed, Mr. McCrawford failed to talk about the case in which composite numbers under the radical were involved. I just don't get it. Just a minor point.

Thanks for the help.  I am sure that my fellow math team students/teachers will appreciate your help.

It's more annoying dealing with composites under the radical since you have to guess a bit. For example, if you have to take :sqrt: (a+b :sqrt: 14), then you have to look at A :rt2: +B :sqrt: 7 as well as A+B :sqrt: 14.

It gets reall tricky when the number is something like 30 under the radical sign. I tried something like this using the method in the AoPS volume, and it took me quite some while.

It's possible for them to be 1 and 14 also.

In fact, the most general approach would be to write it as (a + b :rt2: + c :sqrt:7 + d :sqrt:14) square it and set them equal.

[quote="mathfanatic"][quote="answerseeker95"]we know that b and d must be 2 and 7 since they are the factors of 14. [/quote]

Do you know for a fact that b and d must be 2 and 7?  Why can't they be, say, 1 and 14?  I think you have to try that case and see that it doesn't work before you rule it out...[/quote]
I didn't automaticly rule it out..... I just tried the 2, 7 case first and got the answer, then I decided not to check the other case(my laziness)......


If $ a,b,c$ are odd integers then prove that $ ax^{2}\plus{}bx\plus{}c\equal{}0$ cannot have rational roots.

This is an old CMO problem and has been posted several times.  See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=196301]here[/url].


Give a reason if true or a counterexample if false:

If $ Q$ is orthogonal, then $ Q \plus{} \frac {1}{2} I$ is invertible.

This means that -1/2 cannot be an eigenvalue of Q. Can anyone help me find a good counterexample of an orthogonal matrix with eigenvalue of -1/2? and what is the quickest way to find one?

Suppose $ Q$ is orthogonal and $ Qx\equal{}\lambda x$

Then $ \|Qx\|^2\equal{}\lambda^2\|x\|^2.$ Write that out as matrix multiplication:

$ x^TQ^TQx\equal{}\lambda^2x^Tx$ But $ Q^TQ\equal{}I$ - that's what orthogonal means.

$ x^Tx\equal{}\lambda^2x^Tx.$

Hence $ \lambda^2\equal{}1$ and $ \lambda\equal{}\pm 1.$

A similar argument for complex unitary matrices would show that if $ \lambda$ is an eigenvalue of a unitary matrix, then $ |\lambda|\equal{}1,$ so $ \lambda$ is somewhere on the unit circle.

You're not going to find an orthogonal matrix with $ \minus{}\frac12$ as an eigenvalue.

thanks Kent, that makes a lot of sense now.


AMC
USA(J)MO
USAMO
function
inequalities
algebra
floor function

Let $m>k>l$ be positive integers. Prove that the number $\ds\frac{(ma)!(mb)!}{(ka+lb)!(kb+la)!}$ is integer for every positive integers $a$ and $b$ if and only if $\left[\frac{m-k}{l}\right](m-k)\ge k+l.$

[quote="Peter VDD"]what exactly is [x] in this question? ceil? floor? fractional part?

I thought usually it denotes fractional part, but it seems not correct for fractional part...[/quote]

 $[x]$ means floor function (integer part of $x$)
$\{x\}$ is for fractional part of $x$ :)

[quote="Sung-yoon Kim"]Isn't this a type of number theory?[/quote]
Ok, the problem can be reformutated as
[color=darkblue]Prove that the inequality $[mx]+[my]\ge[kx+ly]+[ky+lx]$ holds for every $x,y\ge0$ if and only if $\left[\frac{m-k}{l}\right](m-k)\ge k+l.$[/color]
and now it is inequality. :)

[i]P.S. USAMO 1975 inequality mentioned in the title is $[5x]+[5y]\ge[3x+y]+[3y+x],\ x,y\ge0.$[/i]

[color=darkblue]Prove that the inequality $[mx]+[my]\ge[kx+ly]+[ky+lx]$ holds for every $x,y\ge0$ if and only if $\left[\frac{m-k}{l}\right](m-k)\ge k+l.$[/color]

Any ideas?


$ f: [0,1] \equal{}\equal{}> R$
 $ f(x)\equal{}x^{4} \minus{} 4x\plus{}1$
$ prove$ $ that$ $ f$ $ is$ $ injective$

Assume that there exist two distinct real numbers $ a,b\in [0,1]$ such that $ f(a) \equal{} f(b)$.
Then, $ a^4 \minus{} 4a \plus{} 1 \equal{} b^4 \minus{} 4b \plus{} 1$
$ \Leftrightarrow (a^4 \minus{} b^4) \plus{} 4(a \minus{} b) \equal{} 0$
$ \Leftrightarrow (a \minus{} b)(a^3 \plus{} a^2b \plus{} b^2a \plus{} b^3 \minus{}4) \equal{} 0$  (*)
However, as $ a$ and $ b$ are distinct, $ a \minus{} b$ cannot be $ 0$, and $ a^3 \plus{} a^2b \plus{} b^2a \plus{} b^3 \minus{} 4$ cannot be $ 0$ as $ a^3\plus{}a^2b\plus{}b^2a\plus{}b^3<4$ as $ 0\le a,b\le 1$ and $ a$ is not equal to $ b$.
Then, equality in (*) cannot be reached, so there do not exist two distinct real numbers $ a,b\in [0,1]$ such that $ f(a) \equal{} f(b)$, and hence $ f(x)$ is injective on $ [0,1]$.

you have  a mistake Mr rem because
 $ (a^{4}\minus{}b^{4}) \minus{} 4 (a\minus{}b) \equal{} 0$  not   $ (a^{4}\minus{}b^{4}) \plus{}  4 (a\minus{}b) \equal{} 0$
 :D

[quote="greatestmaths"]$ f: [0,1] \equal{} \equal{} > R$
 $ f(x) \equal{} x^{4} \minus{} 4x \plus{} 1$
$ prove$ $ that$ $ f$ $ is$ $ injective$[/quote]
$ f'(x)\equal{}4x^3\minus{}4\leq 0$
The equalities hold if and only if $ x\equal{}1$
So $ f(x)$ is decreasing on $ [0,1]$ it mean that $ f(x)$ is injective.


how do i fix this??  this is the screenshot...

and make sure you put the solution in extra-big font

[size=150]I can think of two things:  you can either click on the big A on the index page(up at the upper right below where it has the aops challege but on the right side)  or go up to view... text size and pick another one.[/size]

Or you can do it w/the ball on your mouse.  I think it is CTRL ball but I am not sure.


[quote=arqady][quote="Vasc"]Let $x,y,z$ be positive numbers such that $xyz=1$. Prove that

$x^{2}+y^{2}+z^{2}-3 \geq 18(x+y+z-xy-yz-zx)$.

Determine the equality cases.[/quote]
Let $a,$ $b$ and $c$ are non-negative numbers. Prove that
$a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2}\geq18(a^{2}-bc)(b^{2}-ac)(c^{2}-ab).$
There are trubles with SOS.[/quote]
Because $$\text{LHS}-\text{RHS}=\frac{1}{2}  \left\{a^2+b^2+c^2+3(ab+bc+ca)\right\} \sum\limits_{cyc} (a+b-3c)^2 (a-b)^2$$

[quote=arqady][quote="Vasc"]Let $x,y,z$ be positive numbers such that $xyz=1$. Prove that

$x^{2}+y^{2}+z^{2}-3 \geq 18(x+y+z-xy-yz-zx)$.

Determine the equality cases.[/quote]
Mixing variables kills it. But it's a very interesting inequality:
equality holds for $(1,1,1)$ and $\left(4,\frac{1}{2},\frac{1}{2}\right)$ and cyclic of last.
Nice form for your inequality, Vasc:
Let $a,$ $b$ and $c$ are non-negative numbers. Prove that
$a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2}\geq18(a^{2}-bc)(b^{2}-ac)(c^{2}-ab).$
There are trubles with SOS.[/quote]

how about the method of [b]Lagrange Multipliers[/b] ?!

Due to the homogenous property of the inequality, let us assume that$$\sum_{cyc}(a)=1$$Now, Let us define
$$f(a,b,c) = a^6+b^6+c^6-3a^2b^2c^2 - 18(a^2-bc)(b^2-ac)(c^2-ab)$$and
$$g(a,b,c) = a+b+c=1$$.
Now let $U=(0, 1)^3$ which is open. Now, $\overline{U}=[0, 1]^3$ which is closed.
Also, let us define,$$S={\{(a,b,c)\in U : g(a,b,c) = 1\}}$$which is an open constraint set.
Now, let us define,$$\overline{S}={\{(a,b,c)\in \overline{U} : g(a,b,c) = 1\}}$$which is a a closed constraint set.
We need to prove that $f(a,b,c)$ achieves its min. value over $\overline{S}$ with subject to $ g(a,b,c)$.
Now since $f(a,b,c)$ and $g(a,b,c)$ are polynomial functions, $f$ and $g$ are continuous with continuous partial derivatives on; $\overline{U}=[0, 1]^3 \subseteq R^3$.
Also note that$$\bigtriangledown g = \langle \frac{\partial g(a,b,c)}{\partial a}, \frac{\partial g(a,b,c)}{\partial b}, \frac{\partial g(a,b,c)}{\partial c} \rangle = \langle 1,1,1 \rangle \neq 0$$at all points. Thus, the gradient of $g(a, b, c)$ is non zero.
Now,$$\bigtriangledown f = \langle \frac{\partial f(a,b,c)}{\partial a}, \frac{\partial f(a,b,c)}{\partial b}, \frac{\partial f(a,b,c)}{\partial c} \rangle = \langle 6a^5-6ab^2c^2-18(-a^3c^3 +abc^4 -a^3b^3+cba^4+acb^4),6b^5-6ba^2c^2-18(-b^3c^3 +abc^4 -c^3b^3+cba^4+acb^4), 6c^5-6cb^2a^2-18(-b^3c^3 +abc^4 -a^3c^3+cba^4+acb^4) \rangle $$Now, introducing our multiplier and by the method of Lagrange multipliers, we have,
$$\bigtriangledown f(a, b, c) = \psi \bigtriangledown g(a, b, c)$$or$$\langle \frac{\partial f(a,b,c)}{\partial a}, \frac{\partial f(a,b,c)}{\partial b}, \frac{\partial f(a,b,c)}{\partial c} \rangle = \langle 6a^5-6ab^2c^2-18(-a^3c^3 +abc^4 -a^3b^3+cba^4+acb^4), 6b^5-6ba^2c^2-18(-b^3c^3 +abc^4 -c^3b^3+cba^4+acb^4), 6c^5-6cb^2a^2-18(-b^3c^3 +abc^4 -a^3c^3+cba^4+acb^4) \rangle = \psi \langle \frac{\partial g(a,b,c)}{\partial a}, \frac{\partial g(a,b,c)}{\partial b}, \frac{\partial g(a,b,c)}{\partial c} \rangle $$. On subtracting $6a^5-6ab^2c^2-18(-a^3c^3 +abc^4 -a^3b^3+cba^4+acb^4) =
\psi$ from $6b^5-6ba^2c^2-18(-b^3c^3 +abc^4 -c^3b^3+cba^4+acb^4) = \psi $, we get
$$(a-b)\{(6a^4+6a^3b+6a^2b^2+6ab^3+6b^4)-6abc^2+54c^3a+54c^3b+18c^4+3a^2b^2-c(a^2+b^2)(a+b)+4a^2bc+4ab^2c\}=0$$And other similar terms.

Now we have the following three cases.

[b][u]Case - 1 :[/u][/b]

We get$$a-b=b-c=c-a=0$$or$$a=b=c=\frac{1}{3}$$ (as, $a+b+c=1$) and this satisfies$$f(a,b,c)=f(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})=0$$<br>
Now we consider the case $a=b \neq c$.
We have,
$${(6a^4+6a^3b+6a^2b^2+6ab^3+6b^4)-6abc^2+54c^3a+54c^3b+18c^4+3a^2b^2-c(a^2+b^2)(a+b)+4a^2bc+4ab^2c}= 33a^4 -6a^2c^2 +108c^3a+18c^4+4ca^3=0$$and as a constraint, we have
$a+b+c=+2a+c=1$.
On solving for $a$ and $c$, we get,
$a=0.747$ and $c=-0.494$ which suffices
$f(a, b, c)=f(a, a, c)=f(0.747, 0.747, -0.494 )=0.33 > 0$.

[b][u]Case - 2 :[/u][/b]

Now considering $\psi=0$, we get$$6a^5-6ab^2c^2-18(-a^3c^3 +abc^4 -a^3b^3+cba^4+acb^4)=6b^5-6ba^2c^2-18(-b^3c^3 +abc^4 -c^3b^3+cba^4+acb^4)=6c^5-6cb^2a^2-18(-b^3c^3 +abc^4 -a^3c^3+cba^4+acb^4)=0$$And the only solution we obtain is$$a=b=c=0$$which suffices$$f(a, b, c)=f(0, 0, 0)=0$$.

[b][u]Case - 3 :[/u][/b]

 If we consider $a,b,c$ pairwise wise distinct, from [b]Case - 1[/b], we get, $a=b=c=\frac{1}{3}$, which is a contradiction.

Now, analyzing $(a,b,c)$ at the border of the domain or at the boundary, we consider $c=0$.
We need to prove $$a^6+b^6+18(a^2)(b^2)(ab) = (a^3+b^3)^2+16(a^3+b^3) \geq 0$$, which is obvious.
This ends our proof.
$\blacksquare$

thank you,
best regards,
Aditya Raj
:D


[quote="rrusczyk"][quote="abacadaea"]im too lazy to make a new topic, so again, when will it be out. it says spring 2008 so im going to guess march or april but if the exact date were given, that would be nice. thanks[/quote]

I'm hoping early May.[/quote]

NOOOOOOOO!!!!!
THE WAIT IS KILLING ME!


Let $ z$ be a root of $ x^5\minus{}1\equal{}0$, such that $ z\neq1$. Find the sum of $ z^{15} \plus{} z^{16} \plus{} z^{17} \plus{} ...\plus{} z^{50}.$

[hide]
Since $ z^5 \equal{} 1$ this problem can be reduced into

$ (z^5)^3 \plus{} (z^5)^3 * z \plus{} (z^5)^3 * z^2 \plus{} ... \plus{} (z^5)^10$

$ \equal{} 1 \plus{} z \plus{} z^2 \plus{} ... \plus{} 1$

$ \equal{} 8 \plus{} 7(z \plus{} z^2 \plus{} z^3 \plus{} z^4)$

since $ z \neq 1$, those four powers of $ z$ correspond to the other four fifth roots of unity.

$ 8 \plus{} 7[1cis (72) \plus{} 1cis( \minus{} 72) \plus{} 1cis(144) \plus{} 1cis( \minus{} 144)]$

The imaginary parts nicely cancel, leaving

$ 8 \plus{} 7[ 2\cos(72) \plus{} 2\cos(144)]$

or $ 8 \plus{} 14\cos(72) \plus{} 14\cos(144)$
[/hide]


abstract algebra
superior algebra
superior algebra unsolved

Given $ G\equal{}ab$ with $ \gcd\left(a,b\right)\equal{}1$ and $ G_{a}\equal{}\left\{g\in G\ |\ g^{a}\equal{}e\right\}$ ($ G_{b}$ defined similarly), how can I show that $ G\cong G_{a}\times G_{b}$?

I started by saying that $ G\equal{}G_{\left|G\right|}\equal{}G_{ab}$ and tried to construct an isomorphism $ \varphi : G_{ab}\rightarrow G_{a}\times G_{b}$ but had no luck.

If $ g$'s nonabelian, this result is horribly false: if $ G\equal{}S_3$, then $ G_2$ isn't even a group.

It's true for abelian groups, though: choose $ x,y$ such that $ ax\plus{}by\equal{}1$ and then define $ \varphi$ by
\[ \varphi (g)\equal{}(g^{by},g^{ax})\]
Check that $ \varphi$ is an isomorphism $ G\rightarrow G_a\times G_b$.


[color=darkblue] Find all numbers $ n \equal{} \overline{abcd}$, $ n \equal{} \prod_{i \equal{} 1}^k p_i^{\alpha_i}$, such that $ \sum p_i \equal{} \sum \alpha_i$.
[/color]

Let $ \sum p_i \equal{} \sum \alpha_i \equal{} x$, then
$ 2^x \le n \le 9999$, so $ x \le 13$, $ p_1 \plus{} p_2 \plus{} p_3 \plus{} p_4 \equal{} 2 \plus{} 3 \plus{} 5 \plus{} 7 > 13$
So $ k \le 3$
Case 1:$ k \equal{} 1$, then it's easy to see that there is no solution
Case 2:$ k \equal{} 2$, then $ \alpha_1 \plus{} \alpha_2 \equal{} p_1 \plus{} p_2 \equal{} 5$, $ n \le 3^5 < 1000$, no solution again .
Case 3:$ k \equal{} 3$, then $ \alpha_1 \plus{} \alpha_2 \plus{} \alpha_3 \equal{} 10$, obviously $ \alpha_3 \le 2$ or $ n > 10000$

Case 3(a):$ \alpha_3 \equal{} 2$, then the question $ \Leftrightarrow \alpha_1 \plus{} \alpha_2 \equal{} 8$ and $ 40 \le 2^{\alpha_1}3^{\alpha_2} < 400 \Leftrightarrow \frac {40}{256} \le 1.5^{\alpha_2} < \frac {400}{256} \Leftrightarrow 0 \le \alpha_2 \le 1$,
so $ n \equal{} 2^8 5^2 \equal{} 6400$ and $ n \equal{} 2^7 3^1 5^2 \equal{} 9600$ are the solutions in this case .
Case 3(b):$ \alpha_3 \equal{} 1$, then the question $ \Leftrightarrow \alpha_1 \plus{} \alpha_2 \equal{} 9$ and $ 200 \le 2^{\alpha_1}3^{\alpha_2} < 2000 \Leftrightarrow \frac {200}{512} \le 1.5^{\alpha_2} < \frac {2000}{512} \Leftrightarrow 0 \le \alpha_3 \le 3$
So $ n \equal{} 2^9 5^1 \equal{} 2560$, $ n \equal{} 2^8 3^1 5^1 \equal{} 3840$, $ n \equal{} 2^7 3^2 5^1 \equal{} 5760$, $ n \equal{} 2^6 3^3 5^1 \equal{} 8640$ are the solutions in this case.


Find all integer solutions to $m^3 - n^3 = 2mn + 8$.

[hide]Suppose $m>n$, then $m=n+p$ (with $p\in \mathbb{N}$)
$\Rightarrow (n+p)^3-n^3=2(n+p)n+8$
$\Rightarrow (3p-2)n^2+p(3p-2)n+(p^3-8)=0$
$D\geq 0 \ \ \ \Rightarrow (3p-2)^2-4p(3p-2)(p^3-8)\geq 0 \ \ \ \Rightarrow ... \ \ \ \Rightarrow p\leq 2$
$p=0 \ \ \ \Rightarrow 2n^2=-8$ which gives no solutions...
$p=1 \ \ \ \Rightarrow n^2+n-7=0 \ \ \ \Rightarrow$ no solutions...
$p=2 \ \ \ \Rightarrow 4n^2+8n=0 \ \ \ \Rightarrow n=0$ and $m=2$

maybe i made some mistake on my way... don't have time to check it  :|
[/hide]

Why is $D\geq 0$?
And what if $m\leq n$?

sorry, did you mean positive AND negative integers?  :blush:

[hide]Let $m=n+p$ (with $p\in \mathbb{Z}$)
$\Rightarrow (n+p)^3-n^3=2(n+p)n+8$
$\Rightarrow (3p-2)n^2+p(3p-2)n+(p^3-8)=0$

This is a second-degree equation in $n$, so in order to have (any) solutions, $D\geq 0 \ \ \ \Rightarrow p^2(3p-2)^2-4(3p-2)(p^3-8)\geq 0 \ \ \ \Rightarrow ... \ \ \ \Rightarrow 0<p<3$
$p=1 \ \ \ \Rightarrow n^2+n-7=0 \ \ \ \Rightarrow$ no solutions...
$p=2 \ \ \ \Rightarrow 4n^2+8n=0 \ \ \ \Rightarrow (0,2)$ and $(-2,0)$
it should be right now...
[/hide]

Put m-n=p and m.n=q ,after factorising the equation become p^3-8=-q(3p-2) so 3p-2 divide p^3-8 but we have
27(p^3-8)=(3p-2)(9p²+6p+4)-208,and 208=,we conclud that 3p-2 is one of the divisore of 13*2^4 the reste is to calculate :rotfl:


geometry
circumcircle
perimeter
national olympiad

[b][color=cyan]Here are problems from the 5-th Kyiv mathematical festival (8.04.06)
Any comments are welcome![/color][/b]

[b][size=200][color=orange]The fifth Kyiv mathematical festival[/color][/size][/b]

[b][size=150][color=indigo]8-th form[/color][/size][/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=511094#p511094]1.[/url] Squirrels $A$ and $B$ have $360$ nuts. $A$ divides these nuts into five non-empty heaps and $B$ chooses three heaps. If the total number of nuts in these heaps is divisible by the total number of nuts in other two heaps then $A$ wins. Otherwise $B$ wins. Which of the squirrels has a winning strategy?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=509740#p509740]2.[/url] The number $123456789$ is written on the blackboard. At each step it is allowed to choose its digits $a$ and $b$ of the same parity and to replace each of them by $\frac{a+b}{2}.$ Is it possible to obtain a number larger then a)$800000000$; b)$880000000$ by such replacements?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=509738#p509738]3.[/url] Find all positive integers $a, b, c$ such that $3abc+11(a+b+c)=6(ab+bc+ac)+18.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=509732#p509732]4.[/url] Let $O$ be the circumcenter and $H$ be the intersection point of the altitudes of acute triangle $ABC.$ The straight lines $BH$ and $CH$ intersect the segments $CO$ and $BO$ at points $D$ and $E$ respectively. Prove that if triangles $ODH$ and $OEH$ are isosceles then triangle $ABC$ is isosceles too.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=509739#p509739]5.[/url] Let $a, b, c, d$ be positive integers and $p$ be prime number such that $a^{2}+b^{2}=p$ and $c^{2}+d^{2}$ is divisible by $p.$ Prove that there exist positive integers $e$ and $f$ such that $e^{2}+f^{2}=\frac{c^{2}+d^{2}}{p}.$

[b][size=150][color=indigo]9-th form[/color][/size][/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=509740#p509740]1.[/url] The number $123456789$ is written on the blackboard. At each step it is allowed to choose its digits $a$ and $b$ of the same parity and to replace each of them by $\frac{a+b}{2}.$ Is it possible to obtain a number larger then a)$800000000$; b)$880000000$ by such replacements?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=511102#p511102]2.[/url]   2006 equilateral triangles are located in the square with side 1. The sum of their perimeters is equal to 300. Prove that at least three of them have a common point.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=509735#p509735]3.[/url] Let $x,y>0$ and $xy\ge1.$ Prove that $x^{3}+y^{3}+4xy\ge x^{2}+y^{2}+x+y+2.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=509732#p509732]4.[/url] Let $O$ be the circumcenter and $H$ be the intersection point of the altitudes of acute triangle $ABC.$ The straight lines $BH$ and $CH$ intersect the segments $CO$ and $BO$ at points $D$ and $E$ respectively. Prove that if triangles $ODH$ and $OEH$ are isosceles then triangle $ABC$ is isosceles too.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=509739#p509739]5.[/url] Let $a, b, c, d$ be positive integers and $p$ be prime number such that $a^{2}+b^{2}=p$ and $c^{2}+d^{2}$ is divisible by $p.$ Prove that there exist positive integers $e$ and $f$ such that $e^{2}+f^{2}=\frac{c^{2}+d^{2}}{p}.$

[b][size=150][color=indigo]10-th form[/color][/size][/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=511084#p511084]1.[/url]  Triangle $ABC$ and straight line $l$ are given at the plane. Construct using a compass and a ruler the straight line which is parallel to $l$ and bisects the area of triangle $ABC.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=509735#p509735]2.[/url] Let $x,y>0$ and $xy\ge1.$ Prove that $2(x^{3}+y^{3}+xy+x+y)\ge5(x^{2}+y^{2}).$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=509732#p509732]3.[/url] Let $O$ be the circumcenter and $H$ be the intersection point of the altitudes of acute triangle $ABC.$ The straight lines $BH$ and $CH$ intersect the segments $CO$ and $BO$ at points $D$ and $E$ respectively. Prove that if triangles $ODH$ and $OEH$ are isosceles then triangle $ABC$ is isosceles too.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=509739#p509739]4.[/url] Let $a, b, c, d$ be positive integers and $p$ be prime number such that $a^{2}+b^{2}=p$ and $c^{2}+d^{2}$ is divisible by $p.$ Prove that there exist positive integers $e$ and $f$ such that $e^{2}+f^{2}=\frac{c^{2}+d^{2}}{p}.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=509733#p509733]5.[/url] All the positive integers from 1 till 1000 are written on the blackboard in some order and there is a collection of cards each containing 10 numbers. If there is a card with numbers $1\le a_{1}<a_{2}<\ldots<a_{10}\le1000$ in collection then it is allowed to arrange in increasing order the numbers at places $a_{1}, a_{2}, \ldots, a_{10},$ counting from left to right. What is the smallest amount of cards in the collection which enables us to arrange in increasing order all the numbers for any initial arrangement of them?

[hide="[color=cyan"]authors[/color]]
The problems are [supposed to be] by:
V.Brayman (8.1, 8.2=9.1, 8.3, 8.4=9.4=10.3, 10.2, 10.5),
V.Brayman and M.Karlikova (9.3), A.Dorogovtsev (9.2), O.Kulik (10.1),
O.Rybak (8.5=9.5=10.4)[/hide]

The date of 5-th kyiv mathematical festival was not (8.04.06) but (8.05.06) (it is always in May)
Sorry for mistake [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#506789]here[/url] and [url=http://www.mathlinks.ro/Forum/resources.php?c=180&cid=108&year=2006]here[/url] :(


can someone tell me what is "The Cauchy-Schwartz inequality ?"

http://mathworld.wolfram.com/CauchysInequality.html

MathWorld seems to only mention the form with two pairs of numbers; a more general form is explained at http://planetmath.org/encyclopedia/CauchySchwarzInequality.html

The PlanetMath one is also with only two pairs of numbers. The PlanetMath site generalizes the inequality to other normed spaces.

I like the proof they use on MathWorld. It's the same one we used in my calculus on manifolds class, and I think it's really clever.

I like the proof given in Rudin's Principles of Mathematical Analysis.

[u]Theorem:[/u] If a1, a2,... ,an and b1, b2,... ,bn are real numbers, then (:Sigma: akbk)2 :le: (:Sigma: ak2)(:Sigma: bk2).

[u]Proof:[/u] Put A = :Sigma: ak2, B = :Sigma: bk2, and C = :Sigma: akbk. If B = 0, then b1 = ... = bn = 0, and the inequality is trivial. Assume therefore that B > 0.

Then, :Sigma: (Bak - Cbk)2
= B2 :Sigma: ak2 - 2BC :Sigma: akbk + C2 :Sigma: bk2
= B2A - 2BC2 + BC2
= B(AB - C2).

Since each term of the first sum is non-negative, B(AB - C2) :ge: 0. And since B > 0, it follows that AB - C2 :ge: 0. This is the desired inequality.


search
combinatorics unsolved
combinatorics

Let $ I_{1},I_{2},...,I_{n}$ be $ n$ closed intervals of $ R$ such that among any $ k$ of them there are $ 2$ with nonempty intersection. Prove that one can choose $ k\minus{}1$ points in $ R$ such that any of the intervals contain at least one of the chosen points.

It's on the forum. Here are two links:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=intervals&t=84485[/url]

[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1908245589&t=106694[/url]


Five couples were at a party. If each preson shook hands with everyone else except his/her spouse, how many handshakes were exchanged? (Note: One obviously doesn't shake hands with oneself.)

Here is my reasoning:
There are husbands A B C D E.
And there are wives a b c d e.
For person A, there are 8 possibilities, A-b, A-c, A-d, A-e, A-B, A-C, A-D, A-E
For person B, there are 7 possibilities, B-a, B-c, B-d, B-e, B-C, B-D, B-E (I didn't count B-A because I already counted it in A's possibilities)
And so on until E.
So there are:
A-8 possibilities
B-7 possibilities
C-6 possibilities
D-5 possibilities
E-4 possibilities

Hence, we have 4+5+6+7+8=30 possibilities. But, we must add the wives shaking hands.
So, we have:
a-b, a-c, a-d, a-e
b-c, b-d, b-e
c-d, c-e
d-e
There are 10 possibilities, so we add 30+10 possibilities to get [size=150]40.[/size]

My friend says this is wrong(He is smarter than me), but I don't understand how.
Could someone please explain this?
I don't have the solutions.
Thanks.

This is a basic handshake problem.

Each person shakes hands with $ 8$ other people. So we have $ 10\cdot8$. However we are over counting by $ 2$, we get $ \frac {10\cdot8}{2} \equal{} 40$. You are right, and your "smart" friend is wrong.

BTW, I am $ 100\%$ sure on this.

For further evidence look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2476767&t=275884]this[/url].

I think you are right.

Each person shakes hands with 8 people. Since there are 10 people, there are

$ 8 \times 10\equal{}80$ handshakes

But we counted each handshake twice.
for example, we counted B shake hands with A, and A shake hands with B as 2 different handshakes.
So we divide by 2 for our final answer, 40.

yes your reasoning is correct. i think your friend just wanted to make you feel bad.

well
each person shakes hands with 8 other ppl. So we have 8 x 10. We are uh double counting by 2.
So we divide by 2!. Then you get 10x8/2=40


A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks anh shoes, assuming that, on each leg, the sock must be put on before the shoe?

[hide="hint"]Let's write digits $1,2, ... , 8$ on the spider's legs. The consequense of operations with legs (put sock or shoe on $i$-th leg) ...
[/hide]
[hide="solution"]... would contain each digit twice. The first occurense any digit $i$ in this consequense means putting sock and the second one means putting shoes.
Hence the required number of different orders is the number of all permutations of the letters of the word $1234567812345678$ which  equals to $\frac{16!}{2^{8}}$ [/hide]


Show that for any $ k>1$ there exist $ k$ consecutive positive integers, none of
which is a sum of two squares.

Take $ n \equiv p_i \minus{} i \mod p_i^2$ for $ i \equal{} 1,...,k$ and different primes $ p_i \equiv 3 \mod 4$.

Number of $ S_2(N) \equal{} \#\{n \equal{} x^2 \plus{} y^2|n < N\} \equal{} O(\frac {N}{\sqrt {ln(N)}})$. Therefore exist $ N$, suth that $ \frac{N}{S_2(N)} > k \plus{} 1$. It give exist $ n < N$, suth that $ n \plus{} i\not \in \{n \equal{} x^2 \plus{} y^2|n < N\} \ \forall i \equal{} 1,2,..,k$.


MATHCOUNTS geometry calculus trigonometry

If you want another contest, here's a link:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35184

If you have not done Trig., you may receive help for the second problem.

For those who are not going to Nats., take this as an opportunity to compete at a higher level.  At least one problem will stretch you.

For those who are competitive, know that
pkothari13, Xantos C. Guin, and dk2
have qualified to move up.

I think I saw at least Penguin and pkothari13(?) on the MathCounts forum.

Way to go Mathletes!

[quote="chess64"][quote="Xantos C. Guin"]I'm in middle school and have done basic trig.
aka
sin=o/h
cos=a/h
tan=o/a
sohcahtoa
That's about it.
But Penguin! Math Abilities help.[/quote]

i'm in 8th grade, and have done algebra 2, geometry and about half of pre-calculus[/quote]

I'm in 7th grade and have done half of algebra 2/geom.  and basic trig and dif calc only.

I'm in 8th grade and have started pre-calc. Not as impressive.

i'm in 7th and i am just finishing Geometry


Algebrainic's patience level varies quadratically with the number of minutes that he is continually bothered. If he is bothered for 2 minutes, his patience level will be 297. 4 minutes--290, and 10 minutes--245.

a)What's Algebrainic's original patience level? (has not been bothered yet)

b)Let's say Algebrainic's trying to do his homework and his younger sib decides to crank up the tunes as high as possible for 14 minutes. What percent of his [i]original[/i] patience does Algebrainic have now?

c)How many minutes has Algebrainic been bothered if he has no patience left?

[quote="jan4iceskate"][hide="the great answer Janet came up with! "]i got.... a) 300 patience level

           b)he now has 65% of his original patience level.

           c) when he has no patience left, he has already been bothered for 24 minutes.[/hide][/quote]
[hide=":yup:"]Correct! Finally, an answer to my problem!

btw, do you have the solution?[/hide]
Click me!

do you know the answer to the problem..? or are you just asking others what they think about the problem based on my other half.. :)

Yeah I know the answer to the problem because I wrote it. :)

Jan4iceskate got the correct answers.

[quote="Timecat8"]she did.............. :D lol.....[/quote]
She didn't post the solution. Would you like to post it? :)

yeah i thought she did in the other forum...