algebra
polynomial
quadratics
quadratic formula

[quote="TheBB"]Well, which equations, if you don't mind me asking?[/quote]

That one equation. Then go and solve it...  :)

The only solution i could find was

$x= \frac{127^2}{129}$

Am i right?

[quote="orl"]That one equation. Then go and solve it...  :)[/quote]

There is no equal sign, so there is no equation. Do you want the roots, is that it?

Now you have your equal sign. Oops. Sorry for all the confusion.  :)

And if that $=$ sign was a $+$, and we were asked to get the real roots of that poynomial, we would have gotten my answer!

[quote="TheBB"]No problem. Now let's see. :)

[hide="I get..."]
$x_1=\frac{1+\left(\frac{25}{2}\right)^\frac{7}{3}}{1-\left(\frac{25}{2}\right)^\frac{7}{3}}127, \text{    } x_2=\frac{1-\left(\frac{11}{2}\right)^\frac{7}{3}}{1+\left(\frac{11}{2}\right)^\frac{7}{3}}127$
[/hide][/quote]

That is not correct. Please be so kind to provide a solution not just answers.  ;)

[hide=""answers""]

Solutions are $x = \{0,129\}$

[/hide]

[quote="The Original Pi Guy"]um... i don't think 129 works back in the equation...[/quote]

129 satisfies the equation. My solution is below, but I think a TI-89 might be able to verify the two solutions as well ;)

[quote="orl"]Find all real solutions of the following equation:

\[\sqrt[7]{(x + 127)^6} - 8 \cdot \sqrt[7]{(x - 127)^6} = 7 \cdot \sqrt[7]{(x^2 - 16129)^3}.\][/quote]

Here is my solution...

[hide]We divide both sides through by $\sqrt[7]{(x - 127)^6}$ to obtain:

\begin{eqnarray*}
\[\sqrt[7]{\left(\frac{x+127}{x-127}\right)^6} - 8 &=& 7 \cdot \sqrt[7]{\left(\frac{x^2-127^2}{(x-127)^2}\right)^3}\\ &=& 7 \cdot \[\sqrt[7]{\left(\frac{(x+127) \cdot (x-127)}{(x-127) \cdot (x-127)}\right)^3}\\ &=& 7 \cdot \[\sqrt[7]{\left(\frac{x+127}{x-127}\right)^3}\\
\end{eqnarray*}

We now set $\[\sqrt[7]{\left(\frac{x+127}{x-127}\right)^3} = y$  to obtain $y^2-8 = 7y$, and we get $y \in \{-1, 8\}$

For $y=-1$, we obtain the following:

$\[\sqrt[7]{\left(\frac{x+127}{x-127}\right)^3} = -1  \Longleftrightarrow \left(\frac{x+127}{x-127}\right)^3 = -1
\Longleftrightarrow   \left(\frac{x+127}{x-127}\right) = -1 \Longleftrightarrow x+127 = -(x-127)  \Longleftrightarrow x = 0$

For $y = 8$, we obtain the following in a similar fashion:

$\frac{x+127}{x-127} = 2^7 = 128 \Longleftrightarrow  x = 129$

Thus, the only two solutions to this problem are $\boxed{x = 0}$ and $\boxed{x = 129}$

[b]Note:[/b] We can divide through by $\sqrt[7]{(x - 127)^6}$ only because $x=127$ is not a solution to the problem.[/hide]

[quote="fanaticsm"]

Here is my solution...

[hide]We divide both sides through by $\sqrt[7]{(x - 127)^6}$ to obtain:

\begin{eqnarray*}
\[\sqrt[7]{\left(\frac{x+127}{x-127}\right)^6} - 8 &=& 7 \cdot \sqrt[7]{\left(\frac{x^2-127^2}{(x-127)^2}\right)^3}\\ &=& 7 \cdot \[\sqrt[7]{\left(\frac{(x+127) \cdot (x-127)}{(x-127) \cdot (x-127)}\right)^3}\\ &=& 7 \cdot \[\sqrt[7]{\left(\frac{x+127}{x-127}\right)^3}\\
\end{eqnarray*}

We now set $\[\sqrt[7]{\left(\frac{x+127}{x-127}\right)^3} = y$  to obtain $y^2-8 = 7y$, and we get $y \in \{-1, 8\}$

For $y=-1$, we obtain the following:

$\[\sqrt[7]{\left(\frac{x+127}{x-127}\right)^3} = -1  \Longleftrightarrow \left(\frac{x+127}{x-127}\right)^3 = -1
\Longleftrightarrow   \left(\frac{x+127}{x-127}\right) = -1 \Longleftrightarrow x+127 = -(x-127)  \Longleftrightarrow x = 0$

For $y = 8$, we obtain the following in a similar fashion:

$\frac{x+127}{x-127} = 2^7 = 128 \Longleftrightarrow  x = 129$

Thus, the only two solutions to this problem are $\boxed{x = 0}$ and $\boxed{x = 129}$

[b]Note:[/b] We can divide through by $\sqrt[7]{(x - 127)^6}$ only because $x=127$ is not a solution to the problem.[/hide][/quote]

Perfect Solution. Well done and written.  :first:

[quote="joml88"]Unrelated to the problem but I looked up DeMO and apparently it is the German Math Olympiad.  I must be overlooking something obvious....but where does "De" in DeMO come from?[/quote]

DeMO stands for Deutsche Mathematik-Olympiade (German Mathematical Olympiad). This is a competition as the USAMO. There is also a homework competition called BWM which stands for Bundeswettbewerb Mathematik, Federal Mathematical Competition. First one is originated in the former German Democratic Republic. Latter was established in Western Germany. Both competitions can help you to qualify for the German IMO Team Pre-Selection Tests.  :)


USAMTS
AMC
AIME
USA(J)MO
USAMO
AMC 12
ARML

Has anyone qualified for the AIME via the USAMTS? If so, how high did you have to score and how hard was it in relation to the AMC tests. Any thoughts on this topic would be appreciated.

I never qualified based only on my USAMTS score (that is, I qualified by AMC also). If you aim for USAMO, qualifying only by USAMTS is not the best idea in the world. I think they give you a score of 100 on the AMC12 if you qualify by USAMTS, which makes qualifying for USAMO extremely difficult.

as long as we're on the topic, what do you guys think of the USAMTS? Does it help prep for olympiad type probs? Do you take it? Do you score high? (I know I don't)  Just wondering.

I think the problems on USAMTS are really too easy to be of much use for preparation for challenging olympiads such as the USAMO. But one of my friends said that USAMTS and ARML alone made him as good as he is. (He was in the C team for SFBA ARML in his sophomore year and in the A team his junior year.) I got a 99 on USAMTS in my junior year (I missed a point in the last round because I was writing my solutions too quickly), and I was just too lazy to do it my senior year (which is a shame because they gave as a prize a book that I could use).

What book would that be Zeta? And can you guys explain how the USAMTS works for those who don't know?

The book is [u]Diophantine Equations[/u] by Titu Andreescu and Dorin Andrico. I don't know enough about Diophantine Equations.

USAMTS consists of four rounds. In each round, you are given five problems and roughly a month to work on them. It is run on the honor code; that is, you aren't supposed to get help from other people. But you can use books, and you can use any calculators or computers you want. You have to write up solutions to all the problems you solved at the end of the round and then mail them to USAMTS. If you just give the right answer, you will probably only get a points or two (out of five) for each question.

what does USAMTS stand for anywho. i know in the time i have taken to type this out i could have already found out. especially taking the amount of time for someone to post the answer. oh well. im writing a story so im busy. ajods

USAMTS stands for USA Mathematical Talent Search.

how would u compare these problems to AMC12 and AIME problems?
i am able to do, or have a slight idea of where to begin, on most of the problems that are not geometry related.
o and what is a good book to learn geometry from, i have already finished the AoPS vol1 and chapters 2,3, and 4 in vol2. and i feel that my ability to solve geometry problems is severly lacking for some reason.  :(

Geometry usually just takes more practice.  Knowing theorems doesn't matter as much as being able to "see" the big picture.  And that will only come from either innate intuition or practice.


Well, i used to be addicted to chess :blush:.

But ever since the school's chess club went down (the teacher who organized it had to get back surgery :(),

i never was into chess anymore for some reason.

[quote="jjfun1"]I don't like it that much, probably because I suck at it.[/quote]

Cmon jjfun.....
Chess is AWESOME.. Beastly game of skill and strategy...

I love chess, but i don't havve anyone to play it with!

I play my brother, and I've drastically improved. The first time I played him, I didn't beat him for like 100 games, but now I win like 75% of the time.  :D

i use to like chess but now not so much.

[quote="jjfun1"]I don't like it that much, probably because I suck at it.[/quote]

And yet the better I get, the less I like chess :what?:

Chess is alright and I'm pretty good, but it's REALLY overrated.

I think that's because too many people like it :)

Chess is great. Although I stopped for a while since math. :P

Love chess.

Love chess so much, that I'm playing in a chess tournament tomorrow. Literally.

LOLZ!!!!!! chess is awesome. alright, if you really want a good online multiplayer chess server, go to funorb.com and click full list of games. you will see chess as one of the games. you need a runescape account though. doesnt matter if you are a member or nonmember. who wants to organize a chess game there? if you do, my usernames hamberger670. ADD ME!!!

lolz.
Do ne of u guys play on ICC?
I love chess. I play bullet, blitz, wild, standard, etc.

[quote="AwesomeToad"]Well, usually you're good at visualization if good at Blindfold chess[/quote]

Definitely doesn't describe me.  :)  :lol: I [u][i][b]would[/b][/i][/u] fail at blindfold chess.......[/u][/i]

heh, not good enough to beat kasparov :P or topolov or howeva that dpelled

What? Jail!? WHO's in jail?! Whay in jail?! I'm confused!!!

[quote="AwesomeToad"]Garry Kasparov, former pwn World Champion is in Russia jail, right?[/quote]

I don't think so.... Where did you hear this?

I read it and my friend told me about half a year ago.

Please do not revive way long dead threads.

PS: Go Chess! I luv chess!

[quote="Joe10112"]
PS: Go Chess! I luv chess![/quote]

Seconded.

Isn't this a revive...

BTW FICS and ICC pwn all other chess websites.  Seriously PWN.

(except for ICC dasher :P)

Now no one else post after me :P

chess is good...at least at the fact that i usually beat a lot of people and i still get beaten. each time you get beaten you become a little stronger.
that's probably cuz after winning 13th place in a chess tournament it made me feel better about chess
i mean it is a good game and it helps your brain which makes you smart and that helps u in math and other subjects. it seems boring to some but it is actually really awesome if u actually understand it and get down to the core. lets relate this to the hot topic of the FIFA world cup which is awesome if u actually know how to play an know how it works. you make up a strategy and use it to beat the others. it really is interesting if u see the point.
CHESS PWNS

[quote="andrew99"]I think chess is a good game but i would rather prefer go.[/quote]

Go sounds quite fun. I haven't really gotten a chance to try it out.


today the smileys celebrate 25yrs.
 :D  :)  :wink:

WTH????

Well, to keep up with common fasion:
 :roll:  :P  :wink:

Really? Had no idea :)  :blush:   :heli:

[quote="mz94"]:censored:  :diablo:  :cursing:[/quote] :D  :D  :D

:a smiley:  :another smiley:  :yet another smiley:

:winner_first:  :winner_second:  :winner_third:

[quote="mz94"]:idea:    :!:    :arrow:[/quote]
 :huuh:  :wow:  :gathering:

I post here late with glee:

 :)  :|  :dry:

[quote=1=2]I post here late with glee:

 :)  :|  :dry:[/quote]

you call that late?

Oh I missed it it was last month
I didn't know there was a world smiley day
Well there's an everything day
Is there an Everything Day?  :P


Let $ABC$ be an equilateral triangle and $D$ an internal point of the side $BC$. A circle, tangent to $BC$ at $D$, cuts $AB$ internally at $M$ and $N$, and $AC$ internally at $P$ and $Q$.

Show that $BD + AM + AN = CD + AP + AQ$.

Let $R,S$ be the midpoints of $MN,PQ$.
Then the three perpendiculars pass through a point: Perpendicular from $D$ to $BC$, Perpendicular from $R$ to $AB$ and the Perpendicular from $S$ to $AC$. (Let them pass through $O$ the center of circle)
Now we have $(AR^2-BR^2)+(BD^2-CD^2)+(CS^2-AS^2)=0$
Now note that $(AR^2-BR^2)=(AR-BR)(AR+BR)=-(AB-2AR)(AB)$
And similarly we have $(BD^2-CD^2)=(BD-CD)(BC)$ and $(CS^2-AS^2)=(AC-2AS)(AC)$
Now putting these into our equality and factoring $AB=AC=BC$ we have
\[ -AB+2AR+BD-CD+AC-2AS=0 \]
which is equivalent to the equality we wanted.


calculus
integration
trigonometry
function
factorial
complex numbers
calculus computations

[quote="tmrfea"]Find the volume of the region enclosed by $ (\frac {x}{a})^{\frac {2}{3}} \plus{} (\frac {y}{b})^{\frac {2}{3}} \plus{} (\frac {z}{c})^{\frac {2}{3}} \equal{} 1, a,b,c > 0$.[/quote]

If you don't know the [url=http://mathworld.wolfram.com/GammaFunction.html]Gamma function[/url] yet (it interpolates the factorial, extending it to all complex numbers except negative integers), now is a good time to go meet it.

One kind of Dirichlet Integral is this: if $ V^{n} \equal{} \left\{ (x_{1},x_{2},\ldots ,x_{n})\in\mathbb{R}^{n}|\ x_{1},x_{2},\ldots ,x_{n}\geq 0, \sum_{k \equal{} 1}^{n}\left( \frac {x_{k}}{\alpha_{k}}\right)^{\beta_{k}}\leq 1\right\}$, then
\[ \iint\cdots\int_{V^{n}}x_{1}^{\gamma_{1} \minus{} 1}x_{2}^{\gamma_{2} \minus{} 1}\cdots x_{n}^{\gamma_{n} \minus{} 1}\ dx_{1}dx_{2}\cdots dx_{n} \equal{} \frac {\prod_{k \equal{} 1}^{n}\left[\frac {\alpha_{k}^{\gamma_{k}}}{\beta_{k}}\Gamma \left( \frac {\gamma_{k}}{\beta_{k}}\right)\right]}{\Gamma \left( 1 \plus{} \sum_{k \equal{} 1}^{n}\frac {\gamma_{k}}{\beta_{k}}\right)}\]

[quote]$ \mathrm{Jacobian: } 9abcr\cos^{2}{\phi}\sin^{2}{\phi}\sin^{2}{\theta}\cos^{5}{\theta}$[/quote]

Jacobian is in fact $ 9abcr^2\cos^{2}{\phi}\sin^{2}{\phi}\sin^{2}{\theta}\cos^{5}{\theta}$. Therefore I have to multiply my answer with $ \frac{2}{3}$ and get $ \frac{4\pi}{35}$. And since we observe the curve just in the first octant, I have to divide also by $ 8$ and the answer really becomes $ \frac{\pi}{70}$, which can be checked also with benorin`s suggestion:

$ \int\int\int_G dxdydz\equal{}\frac{(\frac{3}{2})^3 abc \Gamma^3(\frac{3}{2})}{\Gamma(1\plus{}\frac{9}{2})}\equal{}\frac{(\frac{3}{2})^3 abc \Gamma^3(\frac{1}{2})(\frac{1}{2})^3}{\frac{9}{2}\frac{7}{2}\frac{5}{2}\frac{3}{2}\frac{1}{2}\Gamma(\frac{1}{2})}\equal{}\\
\frac{1}{70}\frac{\pi}{\sin{\frac{\pi}{2}}}\equal{}\frac{\pi}{70}$

why am I wrong when calculating Jacobian?


will someone help me solve this problem?

For which digits "a" do exist integers n>3 such that every digit of n(n+1)/2 equals a ?


I am quite OK with taking both classes as i'm not too strong at either but would like to make a decision as to which would be appropriate for me. Mr Sato knows what I'm like.
   Ashwath

You seem confident in Olympiad Problem Solving, so I think you may be ready for these classes.  How do you did on the "Are you Ready" tests?

Pretty Good. I actually haven't taken Olympiad Geometry but I comfortably crossed the Inequalities benchmark.
   Ashwath

I also have a question about these two particular classes.  I think I'll be taking one and only one, but I'm having trouble deciding which.

For Olympiad Inequalities, I was wondering which of {Jensen, Weighted Power Mean, Holder, Rearrangement, Chebyshev, Schur, Newton, Maclaurin, Majorization, Popoviciu, and Bernoulli} this class goes through, and how light or extensive of a treatment they would be.  Or could I see a copy of the Lesson Plan schedule (Lesson I:, Lesson II:, etc)?

Whoa, you already know all those inequalities! While you can still learn a lot of stuff from the inequalities course (which I thought wasn't going to be held anymore!?), you might benefit more from the geometry course, which a lot of people seem to like.

In deciding between these two courses, two main criteria should be the subject matter and how much time you have.  The geometry class is our hardest and most time-consuming class.  Furthermore, the geometry classes themselves are longer and at a somewhat faster pace than the inequalities classes.  So, if you're a student who is looking to just get started with olympiad-type problems, the inequality class is probably a better fit.  If you have solid geometry skills and time to invest in mastering approaches to very advanced geometry problems, the geometry class may be the better solution for you.


geometry
ARML
trigonometry
ratio
geometric transformation
reflection

it looks to me like it should be $\frac{AC}{PQ} = \frac72$, not $\frac27$...

Let $H$ be the intersection of $AN$ and $CM$. Notice that angle $B$ is congruent to angle $MHA$ (they're both complements of angle $BAN$), that $QM = MH$, and that triangles $QHP$ and $AHC$ are similar, so

$\sin B = \sin \angle MHA = \frac{MH}{AH} = \frac12 \cdot \frac{QH}{AH} = \frac12 \cdot \frac27 = \frac17$

(if $\frac{AC}{PQ}$ were $\frac27$ then we'd get $\sin B = \frac12 \cdot \frac72 = \frac74$, which is impossible)

Hmmm, interesting, first of all you calculated $\cos{B}$, secondly the problem actually said the ratio as I wrote it, thirdly, I checked your solution and it seems correct (It was basically what I had been doing except $QM = MH$ eluded me somehow.  However, the solutions and answer say the answer is $\frac{1}{4}$.  I will look into that some more,
but it is very strange.

yes, you're right, i calculated cos though i was writing sin...my mistake. still doesn't come out to 1/4 though, according to my solution. that's weird, maybe i made an error somewhere, though i don't see it

[quote="pdiao"]Tonight, my school held an ARML practice.   In the individual round we had a question that nobody actually solved.  Can anybody post a solution?

Inscribe acute triangle ABC inside a circle.  Extend altitude AN to P on the circle and extend altitude CM to Q on the circle.  If we are given that AC:PQ is 2:7, calculate sin B.

Have fun with it.  I will probably work on it some more too and maybe post a solution if I can find one.[/quote]

Well, Menyoung and Kevin guessed it right >_>

What a cute problem! (How come they were never like that in my day?) Note that the orthocenter reflects over AB to Q and over BC to P (This is well known; it follows from <HBC=90-<BCA=<CAN=<CAP=<CBP.) Therefore <PAQ = 2<ABC. Then AC/PQ = sin(B)/sin(2B) = 1/(2cos(B)). So, if AC/PQ=2/7 then the problem had a typo, since cosine is not allowed to be 7/4. If instead AC/PQ = 7/2, which is more likely, then cos(B) = 1/7 and so sin(B)=4sqrt(3)/7.

yeah, that agrees with my answer. definitely not 1/4


calculus
integration
trigonometry
induction
calculus computations

In fact I think the result is the same even if $n$ is not integer. I am not sure how to show it though.  :blush:

[quote="kunny"]Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}n\theta}{\sin^{2}\theta}\ d\theta\ (n=1,\ 2,\ \cdots).$[/quote]
Let $I_{n}= \int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}n\theta}{\sin^{2}\theta}\ d\theta$

$\Rightarrow I_{n+1}-I_{n}= \int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}(n+1)\theta-\sin^{2}n\theta}{\sin^{2}\theta}\, d\theta$

Now, note that
$\sin^{2}(n+1)\theta-\sin^{2}n\theta$
$= \left( \sin (n+1)\theta-\sin n\theta \right]\left[ \sin (n+1)\theta+\sin n\theta \right]$

$= 2\cos (\frac{2n+1}{2})\theta \sin (\frac{\theta}{2}) \cdot 2\sin (\frac{2n+1}{2})\theta \cos (\frac{\theta}{2})$
$= \sin \theta \sin (2n+1)\theta$

$\therefore I_{n+1}-I_{n}= \int_{0}^{\frac{\pi}{2}}\frac{\sin (2n+1)\theta}{\sin \theta}\, d\theta = B_{n}$ (say)

Again, $B_{n+1}-B_{n}= \int_{0}^{\frac{\pi}{2}}\frac{\sin (2n+3)\theta-\sin (2n+1)\theta}{\sin \theta}\, d\theta$

$\Rightarrow B_{n+1}-B_{n}= \int_{0}^{\frac{\pi}{2}}\frac{2\cos (2n+2)\theta \sin\theta}{\sin \theta}\, d\theta = \frac{1}{n+1}\sin 2(n+1)\theta \Big|_{0}^{\frac{\pi}{2}}= 0$

$\therefore B_{n+1}= B_{n}\ \forall n \ge 1$

Also, $B_{1}= \int_{0}^{\frac{\pi}{2}}\frac{\sin 3\theta}{\sin\theta}\, d\theta = \int_{0}^{\frac{\pi}{2}}(3-4\sin^{2}\theta)\, d\theta$

$\Rightarrow B_{1}= \int_{0}^{\frac{\pi}{2}}3\, d\theta-2\int_{0}^{\frac{\pi}{2}}(1-\cos 2\theta)\, d\theta = \int_{0}^{\frac{\pi}{2}}\, d\theta+\sin 2\theta \Big|_{0}^{\frac{\pi}{2}}={\frac{\pi}{2}}$

Thus, $B_{n}= \frac{\pi}{2}\ \forall n\ge 1.$

So, we finally have $I_{n+1}-I_{n}= \frac{\pi}{2}\ \forall n\ge 1.$

A simple induction on $n$ shows that $I_{n+1}= n(\frac{\pi}{2})+I_{1}.$
But, $I_{1}= \int_{0}^{\frac{\pi}{2}}\, d\theta = \frac{\pi}{2}\Rightarrow I_{n+1}= (n+1)\frac{\pi}{2}$

$\Rightarrow I_{n}= \boxed{\frac{n\pi}{2}}.$

[[b]N.B.[/b] Evaluating first $(I_{n+1}-I_{n})$ instead of $I_{n}$ is a useful technique in such problems.]

the substitution: $\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ solves the problem assuming you allow $i$ as a constant (i have not seen rigirious definitions of complex integrals so i don't know if that can be assumed to work or not)

Of course you can use complex numbers. But I don't see how the problem solves itself after this  :maybe:

[color=darkblue][b]Very nice exercise, Kunny ![/b]
Shortly, the [u]FieryHydra's proof :[/u] $I_{1}=\frac{\pi}{2}\ ;\ I_{2}=\pi \ ;\ I_{n+2}-I_{n+1}+I_{n}=0$ $\Longrightarrow$ $I_{n}=(n-1)I_{2}-(n-2)I_{1}$, i.e. $I_{n}=\frac{n\pi}{2}\ .$[/color]

Thank you, Virgil Nicula. Your solution reminds me of another problems.

Thanks [b]FieryHydra![/b]
Your solution is very nice!
I must confess that I was having trouble when calculated it.
However, when I viewing your solution, it's a surprise, and it is very interesting.


Find the 49th term of the sequence
3,6,9,15,24...
(add the previous 2 terms to get the next term)

ugh.

thats 50th term of fibonacci sequence *3.

...does somebody know the 50th term of fibonacci sequence?

The 50th term of the Fibonacci sequence is 12586269025, so three times that is $ \boxed {37758807075}$.

Umm, well, [url=http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibFormula.html]this[/url] may help you.  :lol:

is there a proof for Binet's formula?  :maybe:

Yes, there are many... one of them uses generating functions and the fact that the generating function for $ F(n) \equal{} F(n \minus{} 1) \plus{} F(n \minus{} 2)$ is $ G(x) \equal{} \frac 1{1 \minus{} x \minus{} x^2}$.

Note that $ G(x)$ is NOT the Fibonacci function, just that its Taylor expansion has its coefficients in the order $ 1, 1, 2, 3, 5, 8, \dots$ In fact, $ G(x) \equal{} 1 \plus{} x \plus{} 2x^2 \plus{} 3x^3 \plus{} 5x^4 \plus{} 8x^5 \plus{} \dots$

If this is confusing, ignore it. This is... higher-level stuff... :D

i don't really see why G(x) is not the fibonacci function but the taylor expansion does give the Fibonacci sequence!

Well $ G(x)$ is not the Fibonacci function because
\begin{align*}G(1) & = \frac 1{1 - 1 - 1} = - 1 \\
G(2) & = \frac 1{1 - 2 - 4} = - \frac 15 \\
G(3) & = \frac 1{1 - 3 - 9} = - \frac 1{11}\end{align*}
And those are definitely not Fibonacci numbers :D
It's just the generating function that "encodes" the sequence in the coefficients of its Taylor Polynomial.

Theres a whole chapter on the fibbonacci numbers(and it contains the proof of Binets formula) in Intermediate Counting and Probability. Ironically, im reading that chapter right now(this moment).

For those of you that know the golden ratio, the higher n gets, the closer $ \frac {F_n} {F_{n \minus{} 1}}$ gets to $ 1.618$, or the golden ratio. In shorthand notation, this is $ \lim_{x\rightarrow \infty} \frac {F_n} {F_{n \minus{} 1}} \equal{} \phi\sim 1.618$

the rectangle diagram thing for the golden ratio makes it really very clear...

Well actually, the real golden ration is $ \frac {1 \plus{} \sqrt {5}}{2}$, which is about $ 1.618034$

wow
this drifted from it's topic a bit...
especially the part where someone asked what grade yongyi was in...[hide="cough"]Bachukas[/hide]
anyway, in the future, plz remember to mostly stay on topic
if you don't understand something, try pming the person to explain instead of posting it here for all to see


function
real analysis
real analysis unsolved

Let $A$,$B$ be two sets.
Prove that $A$~$B$ if and only if there is two functins $f,g$ one to one such that :
$f: A--->B$
$g: B--->A$

It's a definition, if there exists a 1-1 mapping of $A$ onto $B$, then $A \sim B$.  It's clear that this is onto because of the inverse function.

He was referring to the Cantor-Bernstein Theorem:

If there is an injection from $A$ to $B$ and another injection from $B$ to $A$, then there is a bijection from $A$ to $B$.


induction
number theory unsolved
number theory

(a) Do there exist 2009 distinct positive integers such that their sum is divisible by each of the given numbers?

(b) Do there exist 2009 distinct positive integers such that their sum is divisible by the sum of any two of the given numbers?

[quote="April"](a) Do there exist 2009 distinct positive integers such that their sum is divisible by each of the given numbers?[/quote]
Suppose it exists $ n$ distinct positive integers $ a_i$ such that $ \sum\frac{1}{a_i}\equal{}1$, then let $ b_k\equal{}\frac{\prod a_i}{a_k}$ :

$ \{b_k\}$ are $ n$ distinct positive integers whose sum is $ S\equal{}\sum b_k\equal{}\prod a_i\sum\frac{1}{a_k}\equal{}\prod a_i$ and each $ b_k$ divides $ S$.

In order to end the demonstration, it's easy to show with induction that it is always possible, when $ n>2$, to find $ n$ distinct positive integers $ a_i$ such that $ \sum\frac{1}{a_i}\equal{}1$ :
It's true for $ n\equal{}3$ : $ 1\equal{}\frac 12\plus{}\frac 13\plus{}\frac 16$
If it's true for $ n$, just replace $ \frac 1a$, for the greatest $ a$ with $ \frac 1{a\plus{}1}\plus{}\frac 1{a(a\plus{}1)}$ and so it's true for $ n\plus{}1$

Hence the result.

b) The answer is no . We can solve problem with m numbers instead of 2009 .Suppose the  existence of  such m numbers .We can assume that $ x_1<x_2<...<x_m$  .Let $ s\equal{}x_1\plus{}...\plus{}x_m$ ,by the condition we have $ x_n\plus{}x_j|s,\forall s\equal{}1,..,m\minus{}1$ . We also have :
$ m>\frac{s}{x_m\plus{}x_1}>\frac{s}{x_m\plus{}x_2}>...>\frac{s}{x_m\plus{}x_{m\minus{}1}}>1$ ,but it gives contradiction to the fact that all number are integers .

[quote="TTsphn"]b) The answer is no . We can solve problem with m numbers instead of 2009 .Suppose the  existence of  such m numbers .We can assume that $ x_1 < x_2 < ... < x_m$  .Let $ s \equal{} x_1 \plus{} ... \plus{} x_m$ ,by the condition we have $ x_n \plus{} x_j|s,\forall s \equal{} 1,..,m \minus{} 1$ . We also have :
$ m > \frac {s}{x_m \plus{} x_1} > \frac {s}{x_m \plus{} x_2} > ... > \frac {s}{x_m \plus{} x_{m \minus{} 1}} > 1$ ,but it gives contradiction to the fact that all number are integers .[/quote]

Very nice !!
Just a very little precision, you need $ m>2$ in order to have $ \frac {s}{x_m \plus{} x_{m \minus{} 1}} > 1$

See topic $ \textbf{hard about set}$ by mathnice1 at http://www.mathlinks.ro/viewtopic.php?t=295676.

(a) It is a well known result that for any $n$, one can choose $1/a_1+...+1/a_n=1$ distinct integers. This is called "Egyptian fractions".

Now let the numbers be $(\prod a_i)/a_k$ for $k=1,...,n$.

(b) No. Take the 2008 greatest sums $s_1,...,s_{2008}$ and let $S$ be total sum. then $S/s_i$ are 2008 distinct integers strictly between 1 and 2008, impossible!


Συγχαρητηρια σ ολους για τα μετάλλια στη Β.Μ.Ο.
Σιλουανε η προσπαθεια και το μερακι σου πιασαν τοπο.
Επειδή οι μιγαδικοι ειναι χρήσιμοι (Sil συμφωνείς φανταζομαι) ας μην τους ξεχναμε
Αν οι εικόνες των μιγαδικων $ w\equal{}x\plus{}ci, z_1\equal{}a\plus{}bi, z_2\equal{}b\plus{}gi, z_3\equal{}g\plus{}ai$  βρισκονται στο πρώτο τεταρτημοριο ν.δ.ο. $ \frac{|wz_2|^2}{Re(w\overline{z_1})Im(wz_1)}\plus{}\frac{|wz_3|^2}{Re(w\overline{z_2})Im(wz_2)}\plus{}\frac{|wz_1|^2}{Re(w\overline{z_3})Im(wz_3)}\geq3$

Πολυ απλή.Θεωρούμε $ w \equal{} (r_1)*(cosk \plus{} i*sink)$, $ z_1 \equal{} (r_2)*(cosm \plus{} i*sinm)$,  $ z_2 \equal{} (r_3)*(cosn \plus{} i*sinn)$, $ z_3 \equal{} (r_4)*(cost \plus{} i*sint)$ με $ 0\le k,m,n,t\le\frac {\pi}{2}$
Τότε $ \sum\frac {(r_1*r_3)^2}{(r_1*r_2)^2[cos(k \minus{} m)*sin(k \plus{} m)]}\ge\sum\frac {r_3^2}{r_2^2}\ge3$ με $ AM \minus{} GM$
Η ισότητα ισχύει για $ k \equal{} m \equal{} n \equal{} t \equal{} \frac {\pi}{4}$ και  $ r_2 \equal{} r_3 \equal{} r_4$
Το λάθος στο typing διορθώθηκε.

Kαι χωρις τρ. μορφες επισης είναι (οπως αναφερω στο topic) ευκολη
(Ειδικα για Βαλκανιoνικες)
$ \sum{\frac{(x^2\plus{}c^2)(g^2\plus{}b^2)}{(xa\plus{}bc)(ac\plus{}bx)}}\geq\sum{\frac{2(x^2\plus{}c^2)(g^2\plus{}b^2)}{(x\plus{}c)^2(a^2\plus{}b^2)}}\geq\sum{\frac{g^2\plus{}b^2}{a^2\plus{}b^2}}\geq3$
Για $ a,b,c$ προφανως ενοεις τα $ z_1,z_2,z_3$


Hi. I am actually a member of the CTC and i just want to ask some people:
Is CTC necessary?

[b][u][i]I am not against the CTC[/i][/u][/b]. I just want to know some of your opinions.

[quote="jjx1"]I THINK THAT THE ELECTIONS ARE NOT USEFUL.  ALL THE VOTES ARE JUST POPULARITY AND NOT HOW PEOPLE ARE ACTUALLY QUALIFIED[/quote]

In saying that you imply that I, the only one who created a finishing tournament, am not qualified to head a tournament group...

Wait-just asking-I havn't bothered to read that entire CTC post by Mewto, but is there going to be a organized tournament?
If there is, then I think that CTC is a good idea, because it cuts the spam and unfinished tournaments. However, the only problem is that, for instance, jain_harshi could not create a tournament of his own, due to the CTC. So there has to be some trade off.
However, I think it is a good idea.
Just my 2 cents.  :D

i dont think its that necessary but it does help a bit i guess

Yeah, CTC is trying to stop the tourney mania and prevent more tourney mania.

*Note: Mania was a word used by mewto55555.

Since CTC hasn't made any tournaments yet, I guess we'll have to wait to see how well it works.

JUst so everyone knows, we are expected to start signup in mid-August and startat the beginning of September. We cannot start until board member election is over, so please vote ASAP!

if there are gonna be tourneys, they should be more frequent. like bi monthly :idea:

i agree with jain.

we signed up for CTC because we didn't want like hundreds of tourneys to happen.
we don't want too few tourneys, either.

Don't worry, we just needed to get everyone elected and stuff. Within a week of the election, plans for a new tourney will have begun.

[quote="superllamel"]Do you really need everyone to vote?

Can you just set a deadline and then begin tourney planning immediately?[/quote]

We need to see the likings of our board members before we continue.  RG is under 75%.

If you set a deadline, then you won't have to wait for every single person to vote. Some people hardly ever come on anyway so it will take forever if you want everyone to vote.

Deadlinis 6 pm forum time on Friday. In the meantime we are getting everyone's votes.

ctcs a waste of time....its not gonna work, and when school comes, no ones gonna have time to organize/participate in a tourny anyways..

Weekends too. We'll have just as much time during school. However, I do think CTC will die down soon, though Tourney Mania will not come back.

[quote]its not gonna work, and when school comes, no ones gonna have time to organize/participate in a tourny anyways..
[/quote]

that's actually a valid point
the only reason i gave up being VP is because i wouldn't have as much time as i would need once school starts

Wow... maybe I shoulda done that.  Not gonna be on AoPS much when school starts. :P

I TOTALLY AGREE WITH DING. WHEN SCHOOL STARTS NOT MANY KIDS WILL COME ON. SOME WILL, BUT NOT AS MUCH. ANYWAYS EVEN IN THE SUMMER THE MOST WE HAVE IS LIKE 20.Also what happens if you won't be on the computer when the tournament. How can we be exactly on at lets say 7:30. its hard with different timezones

That's why you pm your opponent, so you can set up a time of your choice.

And you can meet your opponent within a week.  But if you have communication problems, we can extend the deadline. (we as in mewto and i)

[quote="ernie"][quote="myyellowducky82"][quote="jjx1"]I THINK THAT THE ELECTIONS ARE NOT USEFUL.  ALL THE VOTES ARE JUST POPULARITY AND NOT HOW PEOPLE ARE ACTUALLY QUALIFIED[/quote]
Yes, I totally agree.[/quote]

Actually, some votes are for qualification, for example, mewto55555 voted as president.  And ducky, you only voted izzy and angie without telling me why, I asked you.[/quote]
Well I thought they'd be a good choice.
SOME. The key word is SOME!

[quote="myyellowducky82"][/quote][quote="ernie"][quote="myyellowducky82"][quote="jjx1"]I THINK THAT THE ELECTIONS ARE NOT USEFUL.  ALL THE VOTES ARE JUST POPULARITY AND NOT HOW PEOPLE ARE ACTUALLY QUALIFIED[/quote]
Yes, I totally agree.[/quote]

Actually, some votes are for qualification, for example, mewto55555 voted as president.  And ducky, you only voted izzy and angie without telling me why, I asked you.[/quote][quote="myyellowducky82"]
Well I thought they'd be a good choice.
SOME. The key word is SOME![/quote]

So you thought only those two were deserving?

i kinda know what ctc is and i don't think it will be useful because school starts soon

I never came on to FTW from January till June because of school. I also live in the Cupertino school district which means my school opens earlier and i get more work from school. I dont think a lot of people will log on on weekdays. CTC might work but there is a chance it may totally fail.

even if nobody goes on, CTC has still served it purpose: to stop all the tournies being made
as you can see, there is currently only one going right now, which is the first CTC tourney

even if CTC fails later on, it has still worked for some time and that's all that matters


[quote="ramanujam"]can someone plz tell me how to go abt solving 17^-1 mod 100?
plz give a detailed solution as i have just started with number theory..

thanks[/quote]
[hide]So you have that
$17k \equiv 1 \mod 100$
That means that $100n+1 \equiv 0 \mod 17 \Rightarrow 15n \equiv 16 \mod 17$.
So $17a+16 \Rightarrow 2a+1 \equiv 0 \mod 15$. We see that a=7, which means that n=9, so our answer is $\frac{901}{17}=53$.[/hide]

thanks diophantient that was very comprehensive n nice..


geometry
angle bisector
geometry unsolved

Let $O_{A}$ be the center of the excircle of triangle $ABC$ opposite $A$. Let $M$ be the midpoint of $AC$, and let $P$ be the intersection of lines $MO_{A}$ and $BC$. Prove that if $\angle BAC = 2\angle ACB$, then $AB=BP$.

I'll use $O$ instead of $O_{A}$. Let $Q$ be the intersection of $BC$ and $AO$.

It's easy to see that $AB=BP\iff \angle PAQ=\angle PAC$.

Since $AQ$ bisects $\angle BAC$, $\frac{AC}{QC}=\frac{AB}{BQ}$.

Since $BO$ bisects external $\angle ABQ$, $\frac{AB}{BQ}=\frac{AO}{QO}$.

Applying Menelaus' with $\triangle AQC$ and the line $OM$, $\frac{AO}{QO}=\frac{PC}{PQ}$.

Those three equations give $\frac{AC}{QC}=\frac{PC}{PQ}$. Because $AO$ bisects $\angle BAC$ and $\angle A=2\angle C$, $\angle QAC=\angle QCA$ i.e. $QA=QC$.

Therefore, $\frac{AC}{QA}=\frac{PC}{PQ}$ i.e. $\angle PAQ=\angle PAC$.

Let be D, intersection point of $BC$ and $AO$.By menelaus: $\frac{OD}{OA}\cdot \frac{AM}{MC}\cdot \frac{CP}{PD}=1$.Thus, $\frac{CP}{PD}=\frac{OA}{OD}...(1)$.
In $\triangle ADQ,CO$ is external bisector of $\angle ACD$.So, $\frac{AC}{CD}=\frac{OA}{OD}...(2)$.
By (1) and (2), $\frac{CP}{PD}=\frac{AC}{CD}$.Since $AD=DC$, we find that $\frac{CP}{PD}=\frac{AC}{AD}$. Therefore, $AP$ must be interior angle bisector of $\triangle DAC$.Clearly, $\angle BAP=\angle BPA$ and $BA=BP$
Lokman GÖKÇE


AMC
USA(J)MO
USAMO
logarithms
national olympiad

USAMO 2005 just took place these days - The official site is http://www.unl.edu/amc/
[url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/questions/Day1/05USAMOday1.pdf]PDF Version DAY 1[/url] [offsite link]
[url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/questions/Day2/05USAMOday2.pdf]PDF Version DAY 2[/url] [offsite link]
Official solutions set [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/Solutions/05SOL.pdf][PDF][/url]

On site-pdf version is below the questions. Please discuss the problems in each problems thread by following the link attached to the problem number.
\[ \textrm{Day I}  \]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34314]Problem 1[/url]
Determine all composite positive integers $n$ for which it is possible to arrange all divisors of $n$ that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34315]Problem 2[/url]
Prove that the system

$x^6+x^3+x^3y+y \& = 147^{157}$
$x^3+x^3y+y^2+y+z^9 \& = 157^{147}$

has no solutions in integers $x$, $y$, and $z$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34316]Problem 3[/url]
Let $ABC$ be an acute-angled triangle, and let $P$ and $Q$ be two points on side $BC$. Construct point $C_1$ in such a way that convex quadrilateral $APBC_1$ is cyclic, $QC_1 \parallel CA$, and $C_1$ and $Q$ lie on opposite sides of line $AB$. Construct point $B_1$ in such a way that convex
quadrilateral $APCB_1$ is cyclic, $QB_1 \parallel BA$, and $B_1$ and $Q$  lie on opposite sides of line $AC$.

Prove that points $B_1, C_1,P$, and $Q$ lie on a circle.
\[ \textrm{Day 2}  \]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34317]Problem 4[/url]
Legs $L_1, L_2, L_3, L_4$ of a square table each have  length $n$, where $n$ is a positive integer.  For how many ordered  4-tuples $(k_1, k_2, k_3, k_4)$ of nonnegative integers can we cut  a piece of length $k_i$ from the end of leg $L_i \; (i=1,2,3,4)$ and still have a stable table?

(The table is [i]stable[/i] if it can be placed so that  all four of the leg ends touch the floor. Note that a cut leg of length 0 is permitted.)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34319]Problem 5[/url]
Let $n$ be an integer greater than 1. Suppose $2n$ points are given in the plane, no three of which are collinear. Suppose $n$ of the given $2n$ points are colored blue and the other $n$ colored red. A line in the plane is called a [i]balancing line[/i] if it passes through one blue and one red point and, for each side of the line, the number of blue points on that side is equal to the number of red points on the same side.
Prove that there exist at least two balancing lines.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34318]Problem 6[/url]
For $m$ a positive integer, let $s(m)$ be the sum of the digits of $m$. For $n\ge 2$, let $f(n)$ be the minimal $k$ for which there  exists a set $S$ of $n$ positive integers such that  $s\left(\sum_{x\in X} x\right)=k$ for any nonempty subset $X\subset S$.  Prove that there are constants $0<C_1<C_2$ with
\[ C_1 \log_{10} n \le f(n) \le C_2 \log_{10} n. \]

[u][b]Please do not discuss in this topic about the solutions, but follow each problem's link to discuss each problem separately. Thanks![/b][/u]

Two questions:

1.What score do you imagine will qualify for IMO/Honorable Mention? (My guess: 30 for IMO/ 25 for honorable mention)

2.When will they have results?

Thx.

[quote="DCo"]2.When will they have results?[/quote]

i believe in the teachers manual it said April 29th, but i think it will come out sooner. Obviously they will make phone calls to the MOPpers over the weekend, but I think the whole results will be out sometime early next week.

hey,
i've got a little query about q4. i know mayb this isn't the right forum to post it in, but i didn't want to go to the solutions thread because i didn't want any hints/spoilers for this problem.
Do the legs have to be perpendicular to the ground? And does the plane which the square table lies in be parellel to the ground?

vinoth: no, they don't have to be perpindicular to the ground... on an unrelated note: the table is, much to the chagrin of one of my friends, of infinite width, so that it balances even with leg lengths of 0, 1000000, 2000000, 1000000, which wouldn't balance in real life if the table were, say, 3x3. :)

Edit, that's not quite right... something like 999999, 1000000, 1000001, 1000000 instead, is what I mean...

[quote="samath"]Just a question; I know that MOSP takes 30 students 9th grade and below.  What score is usually needed to be in this top 30?[/quote]

It's top 30 students in 9th grade... NOT below. So 8th graders can't get the benefit of this. And check out the AMC forum for more details (many threads on this).

I know an 8th grader that went last year.  Does that mean that he must ahve been in the top group overall since he couldn't have been taken because he was younger?

i believe you are refering to Eric Larson.
the Red MOP only applies to 9th graders, he might have scored high enough for Black MOP. (not sure though).
anyways, this is defenatly not the right forum to discuss this, for more info go to AMC forum. ;)


Ready for The school, an exciting mystery game?

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.

Join and you'll see. Anway, please update the list while you post, instead of just writing your name.

[quote="r15s11z55y89w21"]1. Joe10112
2. r15s11z55y89w21
3. dragon96
4.
5.
6.
7.
8.
9.
10.[/quote]

I'll try not to be lazy and inactive, like my previous games.
And try keeping up a quote stack!

[quote="dragon96"][quote="r15s11z55y89w21"]1. Joe10112
2. r15s11z55y89w21
3. dragon96
4. BOGTRO
5.
6.
7.
8.
9.
10.[/quote]

I'll try not to be lazy and inactive, like my previous games.
And try keeping up a quote stack![/quote]

This better be a good game.

Quote stacks are BAD.

1. Joe10112
2. r15s11z55y89w21
3. dragon96
4. BOGTRO
5. vahalla
6.
7.
8.
9.
10.

It's a trait of the ol' Misoorah forum, so I guess it ain't (:)) good.

1. Joe10112
2. r15s11z55y89w21
3. dragon96
4. BOGTRO
5. vahalla
6. ernie
7.
8.
9.
10.

I hope this is cool

1. Joe10112
2. r15s11z55y89w21
3. dragon96
4. BOGTRO
5. vahalla
6. ernie
7. auroraak
8.
9.
10.

1. Joe10112
2. r15s11z55y89w21
3. dragon96
4. BOGTRO
5. vahalla
6. ernie
7. auroraak
8. bluecarneal
9.
10.

sounds fun, I'm in!

yeah, lets go Dojo

1. Joe10112
2. r15s11z55y89w21
3. dragon96
4. BOGTRO
5. vahalla
6. ernie
7. auroraak
8. bluecarneal
9. Bugi
10.mz94

[quote="ernie"]It's a trait of the ol' Misoorah forum, so I guess it ain't (:)) good.[/quote]

Yes, if this happens to actually be Missouri-related, I would voluntarily lend my spot to you.

egglv, the sign-up sheet is filled

and when does this gaem start?


3 < N/12 < 4;that is N/12 is between 3 and 4  if  N/12 is a proper reduced fraction, what is the sum of all possible values of N?

A.84	B.168	C.184	D.252	E.462

[hide=" click to reveal hidden content"]

Unless I've misunderstood the question, I make this $ 168$, the proper reduced fractions arising from $ N\equal{} 37, 41, 43, 47$ as the other numbers between $ 37$ and $ 47$, inclusive, have common factors with $ 12$.

[/hide]

This is essentially asking us for values of N, given that 36 < N < 48 and that N is relatively prime to 12.

37+41+43+47 = 78+90 = 168.

The answer is B


Solve the equation $\frac{1}{\sin x}+\frac{1}{\cos x}=\frac{1}{p}, $ where $p$ is a real parameter. Investigate for which values of $p$ solutions exist and how many solutions exist.

(Of course, the last question ''how many solutions exist'' should be understood as ''how many solutions exists modulo $2\pi $''.)

Obviously multiples of :pi:/2 and :pi: are reulted out for a solution to exist.

Then the expresssion is equivalent to 2:rt2:p sin(x+:pi:/4)= sin2x.
Let sin(x+:pi:/4)= y.
We have 2 y2-2:rt2:p y -1=0.

Solving for y we have y= [p :pm:  :sqrt: (p2+1)]/ :rt2:. We can solve for the correct values of p by using the criteria:
-1<=y<-1/:rt2:
-1/:rt2:<y<1/:rt2: &
1/:rt2:<y<=1.


Μπορεί κάποιος να μου αποδείξει χωρίς την χρήση παραγώγων ότι η $f(x)=4x+e^{sinx}$ είναι γνησίως αύξουσα??

Έχει κάποια σχέση το 4? Γιατί, άμα ήταν 3χ θα άλλαζε κάτι? :?:

Μαλλον δεν ειναι γν. αυξουσα στο $R$

Η συναρτηση $sinx$ δεν ειναι γνησιως μονοτονη στο $R$ αρα δεν μπορει να ειναι και η $f$ η οποια περιλαμβανει το $sinx$ στο τυπο της...

Η f σαφώς και είναι γνησίως αύξουσα και βγαίνει πάρα πολύ εύκολα με την παραγώγιση
$f'(x)=4+e^{sin(x)}cos(x)>0$

Χωρίς τη χρήση παραγώγων γίνεται πολύ περίπλοκο αν προσπαθησεις να πάρεις $x_{1}< x_{2}$ κοκ και δεν ξέρω αν αξίζει να το προσπαθήσεις

[quote="mailo"]Ναι εχεις δικιο alexandros_plus, εκανα λαθος. Ομως δεν ειναι τελικα δυσκολο να το αποδειξεις με $x_{1}<x_{2}$

Για καθε $x_{1},x_{2}\in R$ με $x_{1}<x_{2}$ θα δειξουμε οτι $f(x_{1})<f(x_{2})$

$f(x_{1})-f(x_{2})=4x_{1}+e^{sin(x_{1})}-4x_{2}-e^{sin(x_{2})}=4(x_{1}-x_{2})+\left(e^{sin(x_{1})}-e^{sin(x_{2})}\right)$

Ισχυει
$-1\leq sin(x) \leq 1$

$\frac{1}{e}\leq e^{sin(x)}\leq e$

και επομενως $f(x_{1})-f(x_{2})>0\Leftrightarrow f(x_{1})>f(x_{2})$

Αρα η $f$ ειναι γνησιως αυξουσα[/quote]

Κάτι εδώ πάει στραβά :huh:
Αρχισες με την υπόθεση $x1<x2$ και κατέληξες στο $f(x1)>f(x2)$ και το συμπέρασμα σου ήταν ότι είναι γνησίως αύξουσα...
Μάλλον η υπόθεση ήταν $x1>x2$ :wink:( δεν έχω μάλλον αρκετές γνώσεις για να καταλάβω "τι πήγε στραβά" :blush: )
Εξάλλου δεν καταλαβαίνω πώς απ' τη σχέση "$\frac{1}{e}\leq e^{sin(x)}\leq e$ φτάνεις στο συμπέρασμα...

[quote="mailo"]Ναι εχεις δικιο alexandros_plus, εκανα λαθος. Ομως δεν ειναι τελικα δυσκολο να το αποδειξεις με $x_{1}<x_{2}$

Για καθε $x_{1},x_{2}\in R$ με $x_{1}<x_{2}$ θα δειξουμε οτι $f(x_{1})<f(x_{2})$

$f(x_{1})-f(x_{2})=4x_{1}+e^{sin(x_{1})}-4x_{2}-e^{sin(x_{2})}=4(x_{1}-x_{2})+\left(e^{sin(x_{1})}-e^{sin(x_{2})}\right)$

Ισχυει
$-1\leq sin(x) \leq 1$

$\frac{1}{e}\leq e^{sin(x)}\leq e$

και επομενως $f(x_{1})-f(x_{2})>0\Leftrightarrow f(x_{1})>f(x_{2})$

Αρα η $f$ ειναι γνησιως αυξουσα[/quote]

 Ποιος το έδωσε αυτό το πρόβλημα και σε ποια τάξη; Αν είναι για τη Γ΄Λυκείου , χάνετε το χρόνο σας. Θέλει παράγωγο.Χωρίς τη χρήση ανισοτήτων που βγαίνουν με παράγωγο , δε  βλέπω να βγαίνει το θέμα. Παλαίψτε το όμως γιατί ποιος ξέρει !
 Πάντως η παραπάνω προσπάθεια είναι ελλειπής.Έχει και λογικά κενά.
 Μπάμπης

Ναι sorry, ειναι $x_{1}>x_{2}$

Λοιπον, ειναι $4(x_{1}-x_{2})>4$ και

$\frac{1}{e}-e \leq e^{sin(x_{1})}-e^{sin(x_{2})}\leq e-\frac{1}{e}\Leftrightarrow-2,3 \leq e^{sin(x_{1})}-e^{sin(x_{2})}\leq 2,3$

Αρα χωρις δευτερη σκεψη $f(x_{1})>f(x_{2})$  :wink:

Σωστός :wink:  κατάλαβα...

[quote="mailo"]Ναι sorry, ειναι $x_{1}>x_{2}$

Λοιπον, ειναι $4(x_{1}-x_{2})>4$ και

[/quote]

Βσκ...αυτό δεν ισχύει....γιατί δεν ξέρεις αν  $x_{1}>x_{2}+1$

Βάλε πχ $x_{1}=1$ και $x_{2}=1/2$ να δεις οτι δεν ισχύει....

Πράγματι Φώτη έχεις δίκιο :oops:
Το μόνο που ξέρουμε είναι ότι $4(x1-x2)>0$ :wink:
Ισως θα έπρεπε να πάρουμε περιπτώσεις για τα πρόσημα του χ1, χ2 και να προχωρήσουμε έτσι...

Την άσκηση αυτή την είχα βρει σε ένα βιβλίο ανάλυσης και ακολουθώντας την σειρά του βιβλίου έπρεπε να το αποδείξω χωρίς την χρήση παραγώγων..Μάλλον όμως είναι πιο δύσκολο τελικά και χωρίς μεγάλη σημασία.. :wink:

Thanks σε όλους που αφιέρωσαν χρόνο και κόπο....

Σε ποιο βιβλίο ανάλυσης; Του Michael Spivak μήπως;

Γιατί έχει κάτι τέτοιες που ζωγραφίζουν  :)

Όχι δεν είναι στου Spivak,που παρεπιπτόντως είναι και πολύ καλό βιβλίο,αλλά σε αυτό του Σωτήρη Ντούγια..

Δυστυχώς δεν είχε καμιά υπόδειξη για την λύση της,εγώ έφαγα καιρό χωρίς αποτέλεσμα οπότε είπα να την βάλω μήπως κάποιος έχει καμιά καλύερη ιδέα....

Η βασική ιδέα έχει ως εξης Θέτω a=sinx , b=siny
Στον λογο μεταβολής εμφανίζεται 4+κλάσμα. Πολλαπλασιάζω και διαιρώ το κλάσμα  με a-b. Προκύπτουν δυο νεα κλασματα. Χρησιμοποιώ την 1+a-b<e^(a-b) και δείχνω ότι το πρώτο κλάσμα μεταβάλλεται από 1/e ως e είτε a>b είτε a<b αφού -1<a,b<1. Το δεύτερο που είναι το a-b/x-y είναι γνωστό ότι βρίσκεται ανάμεσα στα -1 , 1 (χορδή<τόξου) . Άρα το αρχικό κλάσμα κυμαίνεται σπό το -e ως το e Οι ανισότητες δεν είναι γνήσιες,το υπόλοιπο είναι απλό. Η 1+a-b<e^(a-b) προκύπτει από Bernoulli στην ακολουθία που ορίζει το e^x

Βασικά νομίζω οτι την λύσατε σχεδόν και οτι το μόνο που χρειάζεται να κάνετε ειναι μερικοί έλεγχοι-περιορισμοί.Η ελάχιστη τιμή που παίρνει η e^ημχ1 - e^ημχ2 ειναι περίπου το -2,3.Αρα πρέπει να  τσεκάρετε εάν και πότε η 4(χ1-χ2)<=2,3(αφού και ο να γινει μας χαλάει αφού τη θελουμε γνησίως).Αυτο γινεται οταν:
4(x1-x2)<=2,3=>x1-x2<=2,3/4=1/2+0,075.Aρα εξεταζουμε χ1=χ2(1/2+0,075) που ικανοποιει και χ1>χ2.Αρα e^sinχ1-e^sinx1(1/2+0,075).θεσε χ1=π/2(οποια τιμη θες βασικα απλα για ευκολια επειδη ειναι γνωστο ημιτονο) τοτε χ2=2π/2,3=>e^sinπ/2-e^sin2π/2,3=>e-e^sin(π(2/2,3))
το sin(π(2/2,3)) αφορα την γωνια 0.87π(περιπου)=78.3 μοιρες=>1ο τεταρτημοριο αρα μικροτερο της μοναδας το ημιτονο αφου η μεγιστη τιμη του ειναι το 1 και ημχ γνησιως αυξουσα.Αρα το e^sin(π(2/2,3))<e και αρα και το δευτερο μελος βγαινει>Ο.
Απο την αλλη οταν η σχεση παιρνει το-2,3=e-1/e πρεπει τα ημιτονα χ1,χ2 να δινουν e κ 1/e αντιστοιχα.αυτο συμβαινει οταν χ1=3π/2 και χ2=π/2 και μονο τοτε μιας και πρεπει χ1>χ2.τοτε ομως και το πρωτο μελος βγαινει >4.Αρα σε καθε περιπτωση γνησιως αυξουσα.Πιστευω οτι ειναι σωστη η διευκρινηση που εδωσα,ελπιζω να σε βοηθησει. :)

Οταν λεω οτι ημχ γνησιως αυξουσα εννοω στο [ο,π/2].

Νομιζω εχω κανει ενα τραγικο λαθος!θα την πιασω απ'την αρχη.αφου χ1>χ2 τοτε χ1-χ2>0 για καθε χ1,χ2εΔf,πρεπει να ελενξω αν e^ημχ1-e^ημχ2<0
αυτο ισχυει οταν ημχ2>ημχ1=> στα πεδια που ειναι δηλ γνησιως φθινουσα στα [π/2,π] κ [3π/2,2π].η ελαχιστη τιμη εινα5 περιπου το -2,3 οπου
αυτο γινεται εαν χ1=π κ χ2=π/2.τοτε ομως χ1-χ2=π/2=>4(χ1-χ2)=2π=6,28>-2,3=>αυξουσα


Let L(0)=2 ;L(1)=1 and L(n+2)=L(n+1)+L(n)  be the famous Lucas’s sequence. Then the only n>1 such that  L(n) perfect square is n=3

This problem can be posed as the following 4th degree Pell-like equation:
\[ x^4 \minus{} 5 y^2 \equal{} \pm 4\]

$ L_k^2\equal{}L_{2k}\plus{}2(\minus{}1)^k$, therefore $ L_{2k}$ is not square for any k.
From $ L_n\equal{}q_1^n\plus{}q_2^n,q_1\equal{}\frac{1\plus{}\sqrt 5}{2},q_2\equal{}\frac{1\minus{}\sqrt 5}{2},q_1q_2\equal{}\minus{}1$ we get:
(1) $ L_{mk}\equal{}L_{(m\minus{}1)k}L_k\minus{}(\minus{}1)^kL_{(m\minus{}2)k}, L_k^2\minus{}5F_k^2\equal{}4(\minus{}1)^k$.
If n is odd and $ L_n\equal{}x^2,F_n\equal{}y$, then $ x^4\minus{}5y^2\equal{}4$ or
(2) $ (x^2\minus{}2x\plus{}2)(x^2\plus{}2x\plus{}2)\equal{}5y^2$.
If x odd $ (x^2\minus{}2x\plus{}2,x^2\plus{}2x\plus{}2)\equal{}1$, if x even $ (x^2\minus{}2x\plus{}2,x^2x\plus{}2)\equal{}2$.
If $ x$ odd we get $ x^2\minus{}2x\plus{}2\equal{}5z^2,x^2\plus{}2x\plus{}2\equal{}(x\plus{}1)^2\plus{}1\equal{}t^2\to (t\plus{}x\plus{}1)(t\minus{}x\minus{}1)\equal{}1\to x\equal{}\minus{}1$ or
$ (x\minus{}1)^2\plus{}1\equal{}z^2,x^2\plus{}5x\plus{}2\equal{}5t^2\to x\equal{}1$ - solution with $ n\equal{}1$.
If $ x\equal{}2s$ is even, then $ n\equal{}3k$ and $ x^2\equal{}L_{3k}\equal{}L_k(L_{2k}\minus{}(\minus{}1)^k)\equal{}L_k(L_k^2\minus{}3(\minus{}1)^k)\equal{}y(y^2\minus{}3(\minus{}1)^k)$.
If $ 3|y$, then k and n are even (not solutions). If $ 3\not y$ we get $ L_k$ is square (because $ (L_k,L_k^2\pm 3)\equal{}1$).
Because $ L_9$ is not square, $ L_1\equal{}1^2,L_3\equal{}2^2$ are unique solutions.

See also [url=http://www.mathlinks.ro/viewtopic.php?t=290317]4.21[/url] :)


Show that, for each positive integer $n$, it is possible to write $\frac{4}{n}$ as a sum of three reciprocals of positive integers.

What's the point of posting open questions in number theory in this section?  :lol:

Mathlink Contest 5th edition round 6 problem 2 has discussed a similar one.

As far as I know this problem (posted in this topic) is an open problem due to Erdos.


[size=75]arashk hamidi[/size]
$a,b \in \mathbb{Z}$,$a\neq b$,$p$ is  prime ,$p\nmid ab$,$(n \in \mathbb{N})$

A)$p\neq 2$,$p\mid a-b \[ \Longrightarrow e_{p}(a^{n}-b^{n})=e_{p}(n)+e_{p}(a-b)$

B)$n$ is odd,$a+b\neq 0$,$p\mid a+b \Longrightarrow e_{p}(a^{n}+b^{n})=e_{p}(n)+e_{p}(a+b)$

($e_{p}(x)=k$ if and only if $p^{k}\parallel x$)

Let$p$be a prime number and $a$,$n$positive integers. Prove that if$2^{p}+3^{p}=a^{n}$then $n = 1$.(Ireland )
you can use that formula


Να λυθει στους πραγματικους αριθμους το συστημα

$x+y+z=3$
$x^{3}+y^{3}+z^{3}=15$
$x^{5}+y^{5}+z^{5}=83$

 :D  :D  :D

Ωραιος...
Ολογος που την εβαλα ηταν για αυτες τις ταυτοτητες...
Εγω προσωπικα δεν την ηξερα εκεινη με την 5η δυναμη και με εφαγε αρκετο χρονο να κανω την παραγοντοποιηση εκεινη την στιγμη....

 :D  :D  :D

Α,Ramanujan η καποιος αλλος,μηπως ξερεις καμια παρομοια τατοτητα για την τεταρτη δυναμη,δηλαδη πως παραγοντοποιειτε το
$(x+y+z)^{4}-x^{4}-y^{4}-z^{4}$  ??
To χρειαζομαι σε ενα αλλο συστημα(σε λιγες μερες θα το βαλω στο foroum...)

 :D  :D  :D

Δεν ξέρω κάτι...

Εικάζω ότι παραγοντοποιείται "καλύτερα" όταν έχουμε περιττές δυνάμεις .

 :D

Basika den paragontopoieitai katholou otan exoume arties dinameis. To dokimasa kai me mathimatika paketa alla den dinei apotelesma...

Yparxei kai i thewria twn simmetrikwn poliwnimwn kai to mono pou katafera einai na to grapsw se auti ti morfi...

Auto pou exoume diladi einai:

$(x+y+z)^{4}-(x^{4}+y^{4}+z^{4})=4(x+y+z)^{2}(xy+yz+xz)-4xyz(x+y+z)-2(xy+yz+xz)^{2}$

Den kserw ean auto voithaei katholou... Otan exoume na lisoume sistimata opou mas dinoun dinameis $x+y+z=... , x^{2}+y^{2}+z^{2}=.... ,x^{3}+y^{3}+z^{3}=...,$ 'h kai anwteris taksis, kalo einai na tis grafoume sinartisei twn simmetrikwn poliwnimwn tous $x+y+z, xy+yz+xz, xyz$ etsi wste na vriskoume ta teleutaia, ta opoia ftiaxnoun tin $3$-vathmia eksiswsi (mesw twn tipwn tou Vieta) $t^{3}-(x+y+z)t^{2}+(xy+yz+xz)t-xyz=0$ tin opoia kai linoume kai vriskoume tis rizes oi opoies profanws einai oi $x,y,z$...

Elpizw na voithisa ligaki! Estw nomizw oti ksekatharise to provlima me tetoiou eidous sistimata...

Gia efarmogi tou parapanw, na lithei to sistima (stous pragmatikous 'h migadikous, den diaferei o tropos lisis).

$x+y+z=3$
$x^{2}+y^{2}+z^{2}=3$
$x^{3}+y^{3}+z^{3}=6$

Alexandros

Y. G. San deuteri askisi prospatheiste na lisete tin parapanw askisi tou RAKAR me ton parapanw tropo. Oxi giati pisteuw oti einai kaliteros apo ton tropo pou parethese o Ramanujan, alla giati prosferei mia sigouria xwris na pigaineis sta tifla alla kai didaktika einai pio katanoitos. Sigoura pio omorfos tropos omws einai tou Ramanujan.
Paratiriste episis oti mporeite na genikeusete to apotelesma me mia tautotita me tin opoia tha mporeite na vreite kathe athroisma $x^{n}+y^{n}+z^{n}$ se sinartisi me ta simmetrika poliwnima!

συμμετρικες συναρτησεις εννοεις τα γινομενα ανα 2 ορους ,ανα 3 κτλ των ν μεταβλητων του συστηματος?

Akrivws Kwsta. Ean exoume gia paradeigma $4$ metavlites $x,y,z,w$ tote ta simmetrika poliwnima pou prokiptoun apo autes einai

$E_{1}=x+y+z+w$
$E_{2}=xy+xz+xw+yz+yw+zw$
$E_{3}=xyz+xyw+xzw+yzw$
$E_{4}=xyzw$

kai genika an exw $n$ metavlites $x_{1},\ldots,x_{n}$ tote to simmetriko poliwnimo $E_{i}$ einai to athroisma olwn twn diaforetikwn ginomenwn me $i$ paragontes twn parapanw metavlitwn. Profanws apo tin sindiastiki to kathe $E_{i}$ periexei ${n \choose i}$ orous.

Koitakste kai to proigoumeno post to opoio exw kanei edit kai zitaw akomi mia askisi gia eksaskisi ston tropo pou proteinw...

Alexandros

[quote="cretanman"]$x+y+z=3$
$x^{2}+y^{2}+z^{2}=3$
$x^{3}+y^{3}+z^{3}=6$
[/quote]

Θα κάνω λίγο χαλάστρα  :P
Είναι $3(x^{2}+y^{2}+z^{2})\geq (x+y+z)^{2}$ δηλαδή $9\geq 9$ που ισχύει μόνο ως ισότητα . Άρα x=y=z
Sto  R δεν έχει λύσεις ......
Στο C έχει ..... :wink:

[quote="silouan"][quote="cretanman"]$x+y+z=3$
$x^{2}+y^{2}+z^{2}=3$
$x^{3}+y^{3}+z^{3}=6$
[/quote]

Θα κάνω λίγο χαλάστρα  :P
Είναι $3(x^{2}+y^{2}+z^{2})\geq (x+y+z)^{2}$ δηλαδή $9\geq 9$ που ισχύει μόνο ως ισότητα . Άρα x=y=z[/quote]

Είχε βάλει και η ΕΜΕ στην άσκηση του μήνα μια τέτοια  :)

Edit: άρα η εξίσωση δεν έχει λύσεις   :!:

Tin parapanw askisi den tin evala gia na lithei me ena "kolpatzidiko" tropo opws polles olimpiakes askiseis (opws kai i lisi pou edwse o Rananujan etsi kai i diki sou Silouan einai poli kali). Tin evala gia provlimatismo gia na deiksw tin geniki methodo pou epiliei SIGOURA tetoia provlimata kai i opoia odigei se SIGOURO apotelesma... Epeidi polles fores rwtate gia tin methodologia tetoiwn askisewn kai pws skeftike ekeinos pou tin eftiakse...

Meletiste tin periptwsi auti mi epireazomenoi apo to pws tha lithei "eukola" alla pws tha lithei me ta simmetrika poliwnima kai prospathiste tin genikeusi pou anaferw... Nomizw oti tha sas voithisei na ton gnwrizete ton tropo auto!

Alexandros

Συμφωνώ για τον τρόπο λύσης και τα λοιπά .
Διαφωνώ στο ότι η ανισότητά μου ισχύει μόνο για θετικούς αριθμούς  :wink: αφού είναι ισοδύναμη με
$(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$  που ισχύει για κάθε a,b,c sto R
Αν τη μεθοδοποιήσουμε θα πρέπει να τη λύσουμε ως εξής .
Από Euler έχυμε
$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$ από όπου βρίσκουμε τη σχέση του αβγ με το αβ+βγ+γα
Όμως το αβ+βγ+γα μπορούμε να το βρούμε από την
$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$
και μετά συνεχίζουμε όπως ο ramajuan stin pproti lisi

Exeis apolito dikio gi'auto to esvisa amesws (prin prolaveis na kaneis post tin apantisi) kathws otan sou apantousa koitaza to parapanw sistima tou RAKAR (to opoio exei pragmatikes rizes) kai oxi ekeino pou arxika proteina.

Geniki methodos lisis tetoiwn provlimatwn kai efarmogi sto parapanw sistima tou RAKAR:

Gia $3$ metavlites (opws to provlima tou RAKAR)
Isxiei i tautotita tou Newton

$x^{n}+y^{n}+z^{n}=(x+y+z)(x^{n-1}+y^{n-1}+z^{n-1})-(xy+yz+xz)(x^{n-2}+y^{n-2}+z^{n-2})+xyz(x^{n-3}+y^{n-3}+z^{n-3})$

gia $n\geq 3$ (Gia $n=2$ kai $n=1$ einai tetrimenes periptwseis). Etsi loipon xrisimopoiwntas diadoxika tin parapanw tautotita grafoume px to $x^{5}+y^{5}+z^{5}$ se sinartisi me ta simmetrika poliwnima $x+y+z$, $xy+yz+xz,xyz$ linoume tin antistoixi 3-vathmia eksiswsi kai exoume ta $x,y,z$.

Ean omws oi metavlites einai parapanw apo $3$ (ligo spanio alla oxi anefikto) as poume $x_{1},x_{2},\ldots,x_{n}$ kai $S_{k}=\sum_{i=1}^{n}x_{i}^{k}$ kai $E_{j}$ einai ta simmetrika poliwnima tote isxioun oi sxeseis:

$S_{1}=E_{1}$
$S_{2}=S_{1}E_{1}-2E_{2}$
$.$
$.$
$.$
$S_{k}=-\Big(\sum_{j=1}^{k-1}(-1)^{j}S_{k-j}E_{j}\Big)-(-1)^{k}kE_{k}$

Nomizw oti einai arketa endiaferon tropos pou linei ola ta provlimata tetoiou eidous. Ti lete ki eseis?

Alexandros

Ωραία η ταυτότητα  :) .

ωραια ! τυποι που εγραψες το γενικευει για ν μεταβλητες το συστημα που εδωσες Αλεξαντρε.ειναι απο τα ωραιοτερα θεματα που εχω δει ποτε( με την προυποθεση οτι στο δευτερο μελος καθε εξισωσης ειναι αθροισμα k-οστων δυναμεων για την k-οστη εξισωση. :)

Xαιρομαι πραγματικα που η ασκηση αυτη ηταν αφορμη για τετοια συζητηση και σημειωτεον μια ταυτοτητα του creatman με βοηθησε να λυσω το ακολουθο προβλημα....
Λοιπον το α συστημα που ειναι στην αρχη του post ηταν στα longlist μιας χρονιας
και επειδη με τα θεματα που υπηρχαν εκεινη την χρονια δεν υπηρχε περιπτωση να ειναι ενα απο τα 6 θεματα της imo,του αλλαξαν εκφωνηση και τελικα περασε και στα shortlist....Αλλα μεχρι εκειι...Αυτα για την ιστορια του θεματος....
Γνωριζοντας οτι το συστημα
$x+y+z=3$
$x^{3}+y^{3}+z^{3}=15$
$x^{4}+y^{4}+z^{4}=35$(για αυτο ειχα ζητησει την ταυτοτητα 4 βαθμου... :wink: )
εχει μια πραγματικη λυση (x,y,z) με $x^{2}+y^{2}+z^{2}<10$, βρειτε την τιμη της $x^{5}+y^{5}+z^{5}$ αυτης της λυσης

Creatman πολυ σωστες και οντως διδακτικες οι παρατηρησεις σου!!Το α συστημα +εγω το ειχα λυσει με συμμετρικα πολυωνυμα,αλλα αυτο τωρα χρειαζεται μια ταυτοτητα(Νewton) απο αυτες που εχεις γραψει....

 :D  :D  :D

Afou loipon xrisimeusan kapou ta parapanw, simainei oti epiteuxthike o skopos gia ton opoio mpikan! Xarika pou se voithisan estw kai ligo RAKAR kathws kai opoiondipote pou mporei na ekane xrisi twn parapanw idiotitwn...
 Molis eida oti xreiazosoun kapoies sxeseis gia 4-vathmia poliwnima gia na liseis kapoio sistima, katalava oti isws na se ksemperdepsoun oi sxeseis pou anefera.

Filika,
Alexandros

Y.G. To nick mou einai cretanman (apo to creta=Kriti) kai oxi creatman (apo to kreas) heheh


Prove that if |U|>5, then the number of all 2-element subsets of U is less than the number of its 3-element subsets.

Let $ n$ be the cardinality of $ U$.
The number of two-element subsets of $ U$ is $ \binom{n}{2} \equal{} \frac {n(n \minus{} 1)}{2}$
The number of three-element subsets of $ U$ is $ \binom{n}{3} \equal{} \frac {n(n \minus{} 1)(n \minus{} 2)}{6}$
Then, the result of the problem is equivalent to proving that:
$ \frac {n(n \minus{} 1)(n \minus{} 2)}{6} \ge \frac {n(n \minus{} 1)}{2}$ for $ n \ge5$
$ \Leftrightarrow \frac {n \minus{} 2}{3} \ge 1$
$ \Leftrightarrow n\minus{}2\ge 3\Leftrightarrow n \ge 5$ which is true if $ n \ge 5$. The result follows.
Edited because of arithmetic error :oops: Sorry I was too careless

How did you get:

$ \frac{n(n\minus{}1)(n\minus{}2)}{3} \ \equal{} \ \frac{(n\minus{}3)}{3}$

he didn't...
according to what he said:
we must prove that:

$ \binom{n}{3}\geq \binom{n}{2}$

$ \iff \frac{n(n\minus{}1)(n\minus{}2)}{6}\geq\frac{n(n\minus{}1)}{2}$

now devide both sides by $ \frac{n(n\minus{}1)}{2}$,to get that:

$ \Rightarrow \frac{n\minus{}2}{3}\geq 1$

$ \iff n\minus{}2\geq 3$

$ \iff n\geq 5$


This was from our Trigonometry test today - The problem askd to simplify this quantity:

[tex]\cos \frac{\pi}{33} \cos \frac{2\pi}{33} \cos \frac{4\pi}{33} \cos \frac{8\pi}{33} \cos \frac{16\pi}{33}[/tex]

After a rather large amount of products to sums procedures, I had it down to this form

[tex]\displaystyle\left(\frac{1}{16}\right)\displaystyle\left(\cos \frac{\pi}{33} + \cos \frac{3\pi}{33} + \cos \frac{5\pi}{33} + \cos \frac{7\pi}{33} + . . . + \cos \frac{31\pi}{33}\right)[/tex]

I couldn't simplify it down past this point... I stared at this for a while after the test had ended, and then plugged it into the calculator, and the huge sum of cosines there was equal to [tex]\frac{1}{2}[/tex], making the answer [tex]\frac{1}{32}[/tex].

My question is, how does one determine that that large summation of cosines is equal to [tex]\frac{1}{2}[/tex]?

The key here is probably to recognize how to apply the identity $\sin(2\theta)=2\sin\theta\cos\theta$. Once you do that, the product collapses.

Let $\displaystyle P=\cos\frac{\pi}{33} \cos \frac{2\pi}{33} \cos \frac{4\pi}{33} \cos \frac{8\pi}{33} \cos \frac{16\pi}{33}$.

$\displaystyle P\sin\frac{\pi}{33}=\sin\frac{\pi}{33}\cos \frac{\pi}{33} \cos \frac{2\pi}{33} \cos \frac{4\pi}{33} \cos \frac{8\pi}{33} \cos \frac{16\pi}{33}$

$\displaystyle P\sin\frac{\pi}{33}=\frac{1}{2}\sin \frac{2\pi}{33} \cos \frac{2\pi}{33}\cos \frac{4\pi}{33} \cos \frac{8\pi}{33} \cos \frac{16\pi}{33}$

$\displaystyle P\sin\frac{\pi}{33}=\frac{1}{4}\sin\frac{4\pi}{33}\cos \frac{4\pi}{33} \cos \frac{8\pi}{33} \cos \frac{16\pi}{33}$

$\displaystyle P\sin\frac{\pi}{33}=\frac{1}{8}\sin\frac{8\pi}{33} \cos \frac{8\pi}{33}\cos \frac{16\pi}{33}$

$\displaystyle P\sin\frac{\pi}{33}=\frac{1}{16}\sin \frac{16\pi}{33}\cos \frac{16\pi}{33}$

$\displaystyle P\sin\frac{\pi}{33}=\frac{1}{32}\sin\frac{32\pi}{33}$

Recognizing that $\sin\theta=\sin(\pi-\theta)$,
$\displaystyle P\sin\frac{\pi}{33}=\frac{1}{32}\sin\frac{\pi}{33}$

$\displaystyle P=\frac{1}{32}$

Wow. I totally took the wrong path on that.

Thanks.


geometry
geometric transformation
reflection
circumcircle
analytic geometry
ratio
search

Let AEC be an acute triangle but not an issoceles triangle. Let B,D,F are points such that AC,EC,AE are respectively the midperpendiculars of OB,OD,OF (O is the circumcircle center of ABE). A',C',E' are projections of A,C,E in CE, AE, AC. Let M, N, P are intersections of A'E' and BD, A'C' and FD, E'C' and FD. Prove that M,N,P are collinear.

Just a quick observation:

It's equivalent to proving that DA', BE' and FC' are comcurrent, because both Desargue's theorem and its converse are true.

We could continue computationally, but I'm looking for a nice way.

[quote="treegoner"]Let AEC be an acute triangle but not an issoceles triangle. Let B,D,F are points such that AC,EC,AE are respectively the midperpendiculars of OB,OD,OF (O is the circumcircle center of ABE).[/quote]

I guess you mean O is the circumcenter of ACE.

Well, of course, B, D, F are just the reflections of O in AC, EC, AE.

[quote="treegoner"]A',C',E' are projections of A,C,E in CE, AE, AC. Let M, N, P are intersections of A'E' and BD, A'C' and FD, E'C' and FD. Prove that M,N,P are collinear.[/quote]

WOW ! This is even more fascinating in the form stating that the lines DA', BE' and FC' are concurrent. In fact they concur at the X(155) center of triangle ACE, i. e., at the orthocenter of the tangential triangle of triangle ACE. Curses - I have no idea how to prove this!

With changed notations:

If O is the circumcenter of a triangle ABC, and D, E, F are the reflections of O in the lines BC, CA, AB, and A'B'C' is the orthic triangle of triangle ABC, then the lines DA', EB', FC' concur at the X(155) center of triangle ABC, i. e. at the orthocenter of the tangential triangle of triangle ABC.

Well, for any P, if D', E', F' are the reflections of P in the lines BC, CA, AB, then the lines D'A', E'B', F'C' concur. But why is the point of concurrence X(155) for P = O ?

  Darij

You're absolutely right, but I'm corious: how on Earth did you spot this? Do you use geometry software, or do you have so much experience that when you see a triangle, you automatically know the "whereabouts" of X(155)? :D:D. I drew a picture with Euklides and realized that you were right.

Let's look at this from the point of view of triangle ABC (this isn't a solution, just an outline). In Darij' last notation DEF is the symmetric of ABC with respect to the nine-pt circle (X(5)) of ABC (which is also X(5) of DEF). The concurrence can be looked-at from the point of view of triangle ABC:

Let A" be the projection of O on EF, B" the projection of O on FD, C" the projection of O on DE. We must prove that AA", BB", CC" are concurrent in the reflection of X(155) about X(5). This point is X(68) and it has barycentric coordinates tan2A:tan2B:tan2C.

It's not that hard to compute the ratios in which the cevians AA" and the others divide the corresponding sides, thus showing that the concurrence point of these 3 is exactly X(68).

However, I'm sure there's a nice, non-computational solution out there somewhere...:D.

Grobber, thanks for the kind words!

[quote="grobber"]You're absolutely right, but I'm corious: how on Earth did you spot this? Do you use geometry software, or do you have so much experience that when you see a triangle, you automatically know the "whereabouts" of X(155)?[/quote]

[b]Step 1.[/b] Draw a dynamic geometry sketch for an arbitrary triangle ABC. Your lines concur at a point P. Make a macro which constructs the point P from an arbitrary triangle ABC.

[b]Step 2.[/b] Open the 6-9-13 sketch file. This is a sketch file containing a triangle ABC having the sidelengths 6, 9, 13. Construct, using the macro, the point P of this triangle ABC. Measure (the program should have a good measuring precision) the distance d from P to BC. If the points P and A lie on the same side of the line BC, then call x = d; if the points P and A lie on different sides of BC, then call x = -d.

[b]Step 3.[/b] This x is the actual (normalized) trilinear a-coordinate of the point P with respect to triangle ABC. Now go to [url=http://faculty.evansville.edu/ck6/encyclopedia/search.html]the ETC search site[/url] and find the triangle centers whose actual trilinear a-coordinate is approximately equal to x. If the point P is a triangle center in the ETC, then it should be among these ones.

[b]Step 4.[/b] In most cases, you either get one possible point; then, with a good probablity, you have got the right one, but it is still useful to check the description of this center in the ETC, look at the list of prominent lines passing through this point and check whether these lines actually pass through your point P for an arbitrary triangle ABC.

If you get no points, then your point P is not in the ETC. (Or, in some seldom cases, the search index was calculated incorrectly.)

If you get two or more possible points (e. g., your dynamic geometry program has shown that x is approximately 0.021 and you wonder whether it is 0.021787866127 (X(2160)) or 0.020959929318 (X(1403))), then it is often helpful to check out prominent lines through these points, or some important properties; finally, you can always construct a point if its trilinear coordinates are given, so in the worst case, you can just construct the possible points from the ETC for an arbitrary triangle and look which of these points coincides with your P. (If any of them does.)

This method looks difficult but it is quite quick when you have a 6-9-13 sketch file at hand (but it is also very easy to make such a file yourself: it's just the construction of a triangle with its sidelengths given) and the ETC together with its Search page downloaded for offline use.

This was the way I found out that the lines DA', EB', FC' concur at X(155). In some cases, I "see" the point of concurrence at first sight, but here it was not as easy.

[quote="grobber"]Let's look at this from the point of view of triangle ABC (this isn't a solution, just an outline). In Darij' last notation DEF is the symmetric of ABC with respect to the nine-pt circle (X(5)) of ABC (which is also X(5) of DEF). The concurrence can be looked-at from the point of view of triangle ABC:

Let A" be the projection of O on EF, B" the projection of O on FD, C" the projection of O on DE. We must prove that AA", BB", CC" are concurrent in the reflection of X(155) about X(5). This point is X(68) and it has barycentric coordinates tan2A:tan2B:tan2C.[/quote]

This is [b]wonderful[/b]: I do have a proof that the lines AA", BB", CC" meet at X(68); in fact, the points A", B", C" are the points diametrically opposite to A', B', C' on the nine-point circle, and there is a synthetic proof that the lines joining the vertices A, B, C to these diametrically opposite points A", B", C" meet at one point. This point is X(68), the so-called [b]Prasolov point[/b] of triangle ABC.

But I have no proof that X(68) is the reflection of X(155) in X(5) !

  Friendly,
  Darij Grinberg

I don't have a proof for that either, but the ETC site says so..:). I'm sure we can use barycentrics to show that.

In fact, the computational proof isn't that bad. It's really easy to show that if AA" cuts BC at A1 then A1B/A1C=-tan2C/tan2B. Then it's even easier to continue the solution (we must prove that the intersection of these cevians has barycaentrics tan2A:etc.)

[quote="grobber"]I don't have a proof for that either, but the ETC site says so..:). I'm sure we can use barycentrics to show that.[/quote]

Okay, but it shouldn't be really simple :?
I'll continue searching for a synthetic proof. In the meantime, let me comment on what I wrote before:

[quote="darij grinberg"]Well, for any P, if D', E', F' are the reflections of P in the lines BC, CA, AB, then the lines D'A', E'B', F'C' concur.[/quote]

In fact, they concur at the isogonal conjugate of P with respect to triangle A'B'C'. (This follows from the simple fact that the lines D'A' and PA' are reflections of each other in the line BC, hence also in the altitude AA', which is perpendicular to BC, i. e. in the angle bisector of C'A'B'.)

Hence, the result to be proven can be rewritten as follows: The points O and X(155) are isogonal conjugates of each other with respect to triangle A'B'C'.

  Greetings,
  Darij Grinberg

Nice! With so many relations between X(i) with i from 1 to 400, you'd think that some problems are simply impossible to solve without consulting the ETC site :D.

[quote="darij grinberg"]If O is the circumcenter of a triangle ABC, and D, E, F are the reflections of O in the lines BC, CA, AB, and A'B'C' is the orthic triangle of triangle ABC, then the lines DA', EB', FC' concur at the X(155) center of triangle ABC, i. e. at the orthocenter of the tangential triangle of triangle ABC.[/quote]

Now I have a synthetic proof of this. At first, rewrite the theorem "from the viewpoint of the tangential triangle":

Let the incircle of a triangle ABC have the center I and touch the sides BC, CA, AB at the points X, Y, Z. Let X', Y', Z' be the feet of the altitudes of triangle XYZ, and let A', B', C' be the reflections of the point I in the lines YZ, ZX, XY, respectively. Then, the lines A'X', B'Y', C'Z' concur at the orthocenter H of triangle ABC.

[i]Proof.[/i] We are going to prove that the line B'Y' passes through H; then, all the rest will follow by analogy.

Since B' is the reflection of I in ZX, we have ZB' = ZI and XB' = XI; but trivially ZI = XI. Hence, the quadrilateral ZB'XI is a rhombus; therefore, ZB' || XI, i. e., the line ZB' is perpendicular to BC. Similarly, the line XB' is perpendicular to AB. Hence, B' is the orthocenter of triangle BZX.

Let the line ZX meet the line CA at Y". The well-known Steiner theorem states that the orthocenters of the four triangles formed by four given lines are collinear; hence, the orthocenters of triangles ABC, AY"Z, BZX and CY"X are collinear. But the orthocenter of triangle ABC is H, and the orthocenter of triangle BZX is B'. Let the orthocenters of triangles AY"Z and CY"X be D and F, respectively. Then, the points H, D, B' and F are collinear; hence, instead of proving that the line B'Y' passes through H - i. e., that the point Y' lies on the line HB' -, we can just prove that the point Y' lies on the line FD.

At first, we will prove ZY' / XY' = DZ / FX. Of course, this is quite a straightforward trigonometry exercise; if one wants to avoid trigonometry, one can proceed as follows:

Let D' and F' be the feet of the perpendiculars from A and C to the line ZX, respectively. Then, obviously D' lies on AD and F' on CF. Now, it is easily seen that the triangles DZD' and FZF' are similar; thus,

  DZ / FX = D'Z / F'X.

Triangles AD'Z and CF'X are similar, too; hence,

  D'Z / F'X = AZ / CX,

implying DZ / FX = AZ / CX.

Now, some primitive angle chase shows that triangles AIZ and YXY' are similar, and triangles CIX and YZY' are similar. Hence, AZ / IZ = YY' / XY' and IX / CX = ZY' / YY'; thus,

  DZ / FX = AZ / CX = AZ / IZ * IX / CX     (since IZ = IX)
          = YY' / XY' * ZY' / YY' = ZY' / XY'.

Hence, we have proven ZY' / XY' = DZ / FX. Now, it is very easy to see that Y' lies on FD: In fact, the lines ZD and XF are parallel. If the line FD meets the line ZX at a point T, then ZT / XT = DZ / FX. Hence, ZT / XT = ZY' / XY', so that T = Y', and consequently, the point Y' lies on the line FD. Proof complete!

  Darij

Here is my solution to this one:
     It is very easy to prove that A, C,E are respectively the exincenter of A'C'E'.
     We denote (A,a),(C,c),(E,e) those circles.
     Let E1 be the projection of  E on E'C' and A1 be the projection of A on A'C'.
     Call F' the intersection of AA1 and EE1.
     We have to prove that F'=F. That is equivalant to proving EF'=AF'=R(AEC).
     By a few computations, we have EF'=AF =2*R[A'C'E']=R(AEC) (the Euler circle's ratio = R/2).
      =>F=F'.
    So sin(FC'A1)/sin(FC'E1) = FA1/FE1=(R(AEC)-a)/(R(AEC)-e).
     Similarly and by applying  Ceva's theorem, we have what we want. :P

You can also seen,
http://perso.orange.fr/jl.ayme  vol. 1 le point de Prasolov
Sincerely
Jean-Louis


My physics teacher isn't the most excited about signing up for it, because he doesn't have any details, such as dates and registration materials. I tried looking on the usapho website but couldn't find anything. Am I just not looking in the right spot, or do people not know this yet.

I want to know this so I can at least try the test.

http://www.aapt.org/olympiad2006/pdf/application_2006.pdf
Also see
http://www.aapt.org/olympiad2006/


integration
calculus
derivative
real analysis
real analysis unsolved

Let $ k$ be a positive integer.

Proove That:

$ \int_0^1 \left(\sum _{n\equal{}\minus{}x}^x n^{2 k}\right) \, dx\equal{}\frac{2}{2 k\plus{}1}$

That doesn't make sense. The indices and limits of a sum should be integers.

do you ever have seen something about fractional calculus?

in normal definition, a half derivative dont make sense, but it exist.

let's a small example in sums

$ \sum _{i\equal{}1}^x i^2\equal{}\frac{1}{6} x (x\plus{}1) (2 x\plus{}1)$

the formula in the rigth is the generalization of the formula in the left.
after computed, the result could be real, complex or someting of high dimension.

is something totally abstract. a generalization.

if we thing just in your sense of thougth, the integer numbers is the only kind of number that exist.

but mathematicians have more number than common persons.

so in this abstract poin, the formula have sense and is correct.

Lordkelvin Explain What does it mean :
$ \sum_{i \equal{} 1}^{x}a_{n}$?? or  what is definition of this expression?
I have no definition.
as I know
when we wrote
$ \sum _{n}$ we mean that $ n\in Z$
but when you wrote $ \sum_{n\equal{}\minus{}x}^{...}$
this $ n\equal{}\minus{}x$  ? you mean $ x\in Z$? or what ?

$ \sum_{k\equal{}0}^{n} k^p \equal{} \frac{\varphi_{p\plus{}1}(n\plus{}1)\minus{}\varphi_{p\plus{}1}(0)}{p\plus{}1},$

with that $ n$ dont more need to be integer and we have a generalization.


(a) Let [b]x[/b] and [b]y[/b] be vectors with ||[b]x[/b]|| = ||[b]y[/b]||.  Prove that the vector [b]x[/b] + [b]y[/b] bisects the angle between [b]x[/b] and [b]y[/b].

(b) More generally, if [b]x[/b] and [b]y[/b] are arbitrary nonzero vectors, let a = ||[b]x[/b]|| and b = ||[b]y[/b]||.  Prove that the vector b[b]x[/b] + a[b]y[/b] bisects the angle between [b]x[/b] and [b]y[/b].

(a) follows from the definition of the angle between two vectors by means of inner products and (b) is a straightforward consequence of (a).

Let $ \theta_1$ be the angle between $ x$ and $ x\plus{}y$ and $ \theta_2$ be the angle between $ y$ and $ x\plus{}y$, then $ \cos(\theta_1)\equal{}\frac{<x|x\plus{}y>}{\|x\|\|x\plus{}y\|}$ and $ \cos(\theta_2)\equal{}\frac{<y|x\plus{}y>}{\|y\|\|x\plus{}y\|}$. Hence $ \theta_1\equal{}\theta_2$ follows from the assumption that $ x,y$ have the same length.


This isn't perhaps a best way to explain it but is there anyway in $\LaTeX$ to form an upside down Pascal's Triangle?

This is what I mean:

[code]
 1 1 1 1 1 1 1 1 1 1
  1 1 1 1 1 1 1 1 1
   1 1 1 1 1 1 1 1
     1 1 1 1 1 1 1
      1 1 1 1 1 1
       1 1 1 1 1
        1 1 1 1
         1 1 1
           1 1
            1
[/code]

[quote="nsato"]This will get you started:
[code]
\begin{tabular}{ccccc}
1 &   & 1 &   & 1 \\
  & 1 &   & 1 & \\
  &   & 1 &   &
\end{tabular}
[/code]

By the way, if you have any other $\LaTeX$ questions, you should probably post them to the $\LaTeX$ forum.[/quote]

Thanks, Mr.Sato (Is that okay?).

I just posted in this forum because I need to use this for my Latex part in my solutions.

[quote="matt276eagles"]:lol: That's exactly why you're not supposed to post questions about LaTeX here![/quote]

Why? It doesn't reveal anything.. LOL. ;)

Actually, you never know. Someone might be just close enough to a solution that posting something like this could actually help them.


A new math game:
Duck Duck Goose
Here's how you play
I'll pm you if you're duck or the goose.
If you're the goose you pm me who to tap.
If you get tapped i'll pm you.
You and the goose will race to answer the math problem.
I put it on here because we need people to join and the goose and duck answer the problem on here. :D
To sign up for the game, post HERE:

You are always so urgent to get things started.  Wait until 10 people.

does anyone else want to play.  It is going to be stupid but its worth a try.


inequalities
trigonometry
pigeonhole principle

Αν $ a,b,c > 0$
Με $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{} 1$
Νδο:
$ 2(ab \plus{} bc \plus{} ca) \leq 1 \plus{} 4abc$

Xρησιμοποιώντας τη δοσμένη σχέση μπορούμε να αποδείξουμε την ισοδύναμη..:

$ ab+bc+ac+a^2+b^2+c^2 \leq \frac {3} {2}$

Με την αντικατάσταση $ x=a^2, y=b^2, z=c^2, w=2abc$ έχουμε:

$ x+y+z+w=1, w=w(x,y,z)=2\sqrt{xyz}     (\ast)$

Τότε αρκεί να δείξουμε την:

$ x+y+z+\sqrt{xy} +\sqrt{yz} +\sqrt{xz}=x+y+z + \frac {w} {2} \cdot (\frac {1} {\sqrt{x}} +\frac {1} {\sqrt{y}} + \frac {1} {\sqrt{z}}) \leq \frac {3} {2}$

Θεωρώντας το πρώτο μέλος ως συνάρτηση 4 μεταβλητών, παρατηρούμε ότι είναι συνεχής στο συμπαγές [0,1]. Άρα γνωρίζουμε ότι έχει μέγιστο. Θα δείξουμε ότι το μέγιστο αυτό επιτυγχάνεται για x=y=z=w.

Αρχικά ας δεχτούμε $ x \succ y$. Έστω $ 0\prec \epsilon \prec x-y$ (δεν ξέρω το σύμβολο για strict inequality  :oops: )

Τότε αντικαθιστώντας $ x\rightarrow x-\epsilon, y\rightarrow y+\epsilon$ το μόνο που αλλάζει στην συνάρτηση είναι ο όρος με τις ρίζες των x,y στον παρονομαστή.

Όμως η συνάρτηση $ f(x,y)=\frac {1} {x} +\frac {1} {y}$ με τον περιορισμό $ x^2+y^2=c$ έχει μέγιστο στο $ x=y$ (αποδεικνύεται εύκολα με χρήση πολλαπλασιαστών Langrange ) που σημαίνει ότι ${ \frac {1} {\sqrt{x}} +\frac {1} {\sqrt{y}} - \frac {1} {\sqrt{x-\epsilon}} +\frac {1} {\sqrt{y+\epsilon}} \succ 0}$ δηλαδή οποιοδήποτε σημείο για το οποίo $ x\succ y$ δεν είναι μέγιστο. Παρόμοια και για το z, οπότε ουσιαστικά δείξαμε ότι για οποιοδήποτε σημείο με $ x\ne y\ne z$ η συνάρτηση δεν έχει μέγιστο.

Αλλά από την αρχή του Sturm, αν για μια συνάρτηση f σε ένα σύνολο Μ με $ x_0 \in M$ ισχύει:
-H f έχει μέγιστο στο Μ
-Κανένα σημείο εκτός του $ x_0$ δεν είναι μεγιστο της f.
Τότε το $ x_0$ είναι μέγιστο της f στο Μ.

Πράγμα που σημαίνει ότι η συνάρτησή μας παίρνει τη μέγιστη τιμή της στο σημείο x=y=z (και άρα και =w από τις σχεσεις $ \ast$.

Άρα συνολικά:
$ x+y+z + \frac {w} {2} \cdot (\frac {1} {\sqrt{x}} +\frac {1} {\sqrt{y}} + \frac {1} {\sqrt{z}}) \leq \frac {3} {4} + \frac {3}{4} = \frac {3} {2}$

Θα πω τις σκέψεις μου: (πραγματικά δύσκολο πρόβλημα κατ' αρχήν)
Με το που βλέπει κατ' αρχάς κανείς αυτή τη συνθήκη το μυαλό πάει αμέσως στην αντικατάσταση $ a=\cos{A}$ και λοιπά με $ A,B,C$ γωνίες τριγώνου (την οποία γίνεται να κάνουμε!).

Επειτα μπορεί να φτάσει κάποιος στην ισοδύναμη (που λέει κι ο Λεωνίδας)
$ 2{}^a{}^2+b^2+c^2+ab+bc+ca)\leq 3$ (βολεύει γιατί το κάθε όρος του 1ου μέλους είναι του ίδιου βαθμού) κι αυτή γράφεται:
$ (a+b)^2+(b+c)^2+(c+a)^2\leq 3 \iff$
$ \sum{(\cos{A}+\cos{B})^2}\leq 3$

Αυτό τώρα θυμόμουν ότι κάπου το είχα ξαναδει΄κι έτσι έψαξα λίγο στο forum. Πράγματι βρήκα μια φοβερή απόδειξη του ισοδύναμου:
$ (\cos{A}+\cos{B}+\cos{C})^2\leq \sin^2{A}+\sin^2{B}+\sin^2{C}$

Για τη λύση αυτού κοιτάξτε το γενικότερο που αποδεικνύει στο post toy o treegoner:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6523[/url]

[quote="bilot4"]Θα πω τις σκέψεις μου: (πραγματικά δύσκολο πρόβλημα κατ' αρχήν)
Με το που βλέπει κατ' αρχάς κανείς αυτή τη συνθήκη το μυαλό πάει αμέσως στην αντικατάσταση $ a = \cos{A}$ και λοιπά με $ A,B,C$ γωνίες τριγώνου (την οποία γίνεται να κάνουμε!).

Επειτα μπορεί να φτάσει κάποιος στην ισοδύναμη (που λέει κι ο Λεωνίδας)
$ 2{}^a{}^2 + b^2 + c^2 + ab + bc + ca)\leq 3$ (βολεύει γιατί το κάθε όρος του 1ου μέλους είναι του ίδιου βαθμού) κι αυτή γράφεται:
$ (a + b)^2 + (b + c)^2 + (c + a)^2\leq 3 \iff$
$ \sum{(\cos{A} + \cos{B})^2}\leq 3$

Αυτό τώρα θυμόμουν ότι κάπου το είχα ξαναδει΄κι έτσι έψαξα λίγο στο forum. Πράγματι βρήκα μια φοβερή απόδειξη του ισοδύναμου:
$ (\cos{A} + \cos{B} + \cos{C})^2\leq \sin^2{A} + \sin^2{B} + \sin^2{C}$

Για τη λύση αυτού κοιτάξτε το γενικότερο που αποδεικνύει στο post toy o treegoner:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6523[/url][/quote]

Το αστειο ειναι οτι ο καθηγητης που μας κανει Μαθηματικη Αναλυση, την ασκηση αυτη την εχει στο βιβλιο του αναμεσα στις ασκησεις - εφαρμογες στην ΑΜ-ΓΜ/Holder/Weighted-Means :P

Μου φαίνεται δύσκολο αυτό γιατί για να σας δώσω να καταλάβετε πόσο sharp είναι αυτή η ανισότητα, αν εκτελέσουμε υπομονετικά πράξεις (παρακαλώ τα υπόλοιπα μέλη της ομάδας να μη σχολιάσουν) :) , αφού εφαρμόσουμε νόμο συνημιτόνων παίρνουμε:

$ \sum{cyc}(a^6) \plus{} \sum{cyc}(a^4bc) \plus{} \sum{sym}(a^3b^3)/\leq\sum{sym}(a^5b) \plus{} \sum{sym}(a^4b^2)$

Παρατηρήστε ότι οι μεγάλες δυνάμεις είναι αριστερά άρα η ανισότητα δύσκολα λύνεται με τη χρήση βασικών ανισοτήτων. Λύνεται μόνο με delicate arguments... Από εδώ η ανισότητα αγγίζει το ακατόρθωτο...

[quote="Protonios"]παρακαλώ τα υπόλοιπα μέλη της ομάδας να μη σχολιάσουν:
[/quote]

Τώρα είναι πράγματα αυτά που μας κάνεις :P  (χαχααχα! ΜΕΓΑΛΟΣ)

Πάντως είναι αλήθεια ότι η ανισότητα είναι ΠΟΛΥ δύσκολη!!!

[quote="Dimitris X"]Nick, γιατί δεν τη βαζεις στο αγγλικό να δούμε εαν θα έχουμε κάποια elementary solution???[/quote]

Οσο για αυτό δεν σε καλύπτει το link που έστειλα? Χρησιμοποιεί μόνο Cauchy! Ποσο πιο elementary...  :P

Να σημειώσω επίσης ότι έφτασα σε μια ακόμα ισοδύναμη μορφή (τελείως διαφορετική, αλλά ενδιαφέρουσα!)
$ 7R^2\plus{}4Rr\leq 2s^2$ όπου s ημιπερίμετρος ενός τριγώνου και τα άλλα, τα γνωστά!

Α! Νίκος, μήπως έχει και την αντίστοιχη λύση ?

[quote="silouan"][quote="Dimitris X"]Nick, γιατί δεν τη βαζεις στο αγγλικό να δούμε εαν θα έχουμε κάποια elementary solution???[/quote]

Θα προσπαθήσω να καλύψω το αίτημα  :P  (αν και ο πιο φυσιολογικός τρόπος είναι του mim . Κώστα τον δικό σου δεν τον σχολιάζω)

Λοιπόν παρατηρούμε ότι η ανισότητα προς απόδειξη είναι γραμμική . Επομένως τη γράφουμε στην ισοδύναμη μορφή
$ z(x \plus{} y \minus{} 2xy)\leq\frac {1}{2} \minus{} xy$  . (1)

Άρα ψάχνουμε ένα πάνω bound για το z για να την κάνουμε δύο μεταβλητών. Πράγματι  $ 1 \equal{} x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz\geq 2xy \plus{} z^2 \plus{} 2xy$
άρα $ 2xy(1 \plus{} z)\leq 1 \minus{} z^2$ από όπου $ 2xy\leq 1 \minus{} z$ άρα $ z\leq 1 \minus{} 2xy$

Βάζοντας το στην (1) αρκεί νδο $ x \plus{} y \minus{} 2xy\leq\frac {1}{2}$ ή ισοδύναμα $ (2x \minus{} 1)(2y \minus{} 1)\geq 0$ το οποίο ισχύει αφού apο pigeonhole  υπάρχουν δύο που είναι μεγαλύτεροι από 1/2 ή   δύο που είναι μικρότεροι από 1/2 . Λόγω συμμετρίας μπορούμε να υποθέσουμε ότι αυτοί είναι οι x,y . QED[/quote]

Καλα ρε δεν παιζεσε... αυτο το προβλημα το προσπαθησα αρκετες φορες τους τελευταιους μηνες (περισσοτερο με διαφορες αντικαταστασεις οι οποιες συνηθιζονται σε ανισοτητες αυτης της μορφης και σε αρκετες ανισοτητες με τη συγκεκριμενη συνθηκη) αλλα χωρις αποτελεσμα... και εσυ το σκοτωσες σχεδον αμεσως

Sry για το spam (!) αλλά αυτή η λύση σου ρε Σιλ είναι ΟΛΑ ΤΑ ΛΕΦΤΑ!!!!!!  :omighty:

Πραγματικά έψαχνα το ισοδύναμο πρόβλημα στο forum και είδα κάμποσες περίεργες λύσεις (με ευκολότερη αυτή που έγραψα). Ομως τέτοια λύση δεν περίμενα με τίποτα!

Special prize :wink:

:10:

Ρέστα.... Δε θα μάθω ποτέ ανισότητες...

Sil, ρέστα κι από μένα :thumbup: τόσο elementary λύση δεν την είχα καθόλου!

Cheerio,

Durandal 1707

Επειδή έχω γίνει πολύ κακόφημος λύτης ανισοτήτων :P βάζω μια νομίζω καλή λύση.

Έστω πάλι η ίδια τριγωνομετρική αντικατάσταση.

Θα χρησιμοποιήσω δύο trivial ανισότητες ως λήμματα:

$ cosA\plus{}cosB\plus{}cosC\leq\frac{3}{2}$ (αν κάποιος δεν γνωρίζει τη συγκεκριμένη να μου πει να την αποδείξω αλλά είναι γνωστή γενικά)

και $ cosBcosC\leq\frac{1\minus{}cosA}{2}$ διότι ισοδύναμα έχουμε

$ 2cosBcosC\leq1\plus{}cos(B\plus{}C)$
$ cos(B\minus{}C)\leq1$

που ισχύει προφανώς.

Στο πρόβλημα τώρα:

Μα το 1το λήμμα:

$ LHS\equal{}2[(cosB\plus{}cosC)cosA\plus{}cosBcosC]\leq2[(\frac{3}{2}\minus{}cosA)cosA\plus{}cosBcosC$

Άρα αρκεί να δείξουμε την εξής ανισότητα:

$ cosA(\frac{3}{2}\minus{}cosA)\plus{}cosBcosC\leq\frac{1}{2}\plus{}2cosAcosBcosC$

Ισοδύναμα:

$ \frac{3}{2}cosA\minus{}cos^2A\plus{}cosBcosc(1\minus{}2cosA)\leq\frac{1}{2}$

Με το 2ρο λήμμα όμως:

$ LHS\leq\frac{3}{2}cosA\minus{}cos^2A\plus{}(\frac{1\minus{}cosA}{2})(1\minus{}2cosA)\equal{}\frac{1}{2}$

Q.E.D.


calculus
integration
LaTeX
calculus computations

F(x)= integral of (1+t^16)^0.5  (the integral is from 3 to x), find F'(x)?

[url=http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus]Fundamental Theorem of Calculus[/url].

Also, take a look at [url=http://www.mathlinks.ro/index.php?f=123]LaTeX[/url].


geometry
inequalities
inequalities proposed
n-variable inequality

Let $x_1,x_2,\ldots,x_n$ be positive reals. Prove that
\[ \frac 1{1+x_1} + \frac 1{1+x_1+x_2} + \cdots + \frac 1{1+x_1+\cdots + x_n} < \sqrt { \frac 1{x_1} + \frac 1{x_2} + \cdots + \frac 1{x_n}} . \]

[i]Bogdan Enescu[/i]

For $n=1$ we have that $\frac{1}{1+x_1} < \sqrt{\frac{1}{x_1}}$ which is obvious. Suppose that the inequlity it true for $n-1$.
$ \frac 1{1+x_1} + \frac 1{1+x_1+x_2} + \cdots + \frac 1{1+x_1+\cdots + x_n} < \sqrt { \frac 1{x_1} + \frac 1{x_2} + \cdots + \frac 1{x_n}}$ We will multiply $LHS$ and $RHS$ with $n-1$ and we will group the $LHS$ in $n-1$ groups with $n-1$ elements such that $\frac{1}{1+x_1+x_2+\cdots+x_i}$ appears in all groups but one.
${{\frac{1}{1+x_1}+\cdots+\frac{1}{1+x_1+\cdots+x_{i-1}}+\frac{1}{1+x_1+\cdots+x_{i+1}}+\cdots+\frac{1}{1+x_1+\cdots+x_n}}} < \sqrt{\frac{1}{x_1}+\cdots+\frac{1}{x_{i-1}}+\frac{1}{x_i+x_{i+1}}+\cdots+\frac{1}{x_n}} < \sqrt{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}$. And summing all up get what we are looking for ;)  SO I REALLY HOPE I DIDNT MESS EVERYTHING UP!!! :lol:

Let $S_0=1$ and let $S_i=1+x_1+x_2+...x_i$ We have by cauchy that:

$\left(\sum_{i=1}^{n}\frac{1}{S_i}\right)^2\leq \left(\sum_{i=1}^n\frac{1}{x_i}\right)\left(\sum_{i=1}^n\frac{x_i}{S_i^2}\right)$

So it is enough to prove $\left(\sum_{i=1}^n\frac{x_i}{S_i^2}\right)<1$ but clearly, this sum is less than the area under the curve $\frac{1}{x^2}$ between $1$ and infinite, but this area converges to $1$, so thats it....

I know, anyway it is a nice method and one can use it in any olympiad any grade....

[quote="RaduSorici"][quote="Pascual2005"]
So it is enough to prove $\left(\sum_{i=1}^n\frac{x_i}{S_i^2}\right)<1$ but clearly, this sum is less than the area under the curve $\frac{1}{x^2}$ between $1$ and infinite, but this area converges to $1$, so thats it....[/quote]
I think this part is too much for a 9th grader  :?[/quote]

It may be said in other words: $\frac{x_i}{S_i^2}<\frac{1}{S_{i-1}}-\frac{1}{S_{i}}$
and then sum up.

Could anyone post Enescu's solution (official solution)?

I posted it once. :P . Anyway, here it goes again:

Let $s_{k}=\sum_{k=1}^{n} x_{k}.$ Now it is obvious that $s_{1} <s_{2} <\ldots <s_{n}$ and $x_{k}=s_{k}-s_{k-1}.$ Now for any $k\geq 2$ and by Cauchy-Schwarz inequality, we have

\[ (\sum_{k=1}^{n} \frac{1}{1+s_{k}})^{2}\leq (\sum_{k=1}^{n}\frac{1}{x_{k}})\cdot(\sum_{k=1}^{n} \frac{x_{k}}{(1+s_{k})^{2}}. \]

But, from $\sum_{k=1}^{n} \frac{x_{k}}{(1+s_{k})^{2}}<1$, the conclusion becomes obvious. ;)

Looks like [url=https://artofproblemsolving.com/community/c6h17449p14971281]this one[/url] (but not the same).


geometry
circumcircle
parallelogram
geometry proposed

Let a triangle $ ABC$ with $ O,H$ are the circumcenter and orthocenter respectively, $ OH\parallel BC$. Denote $ M,P,I$ be the midpoints of segments $ BC,BA,AH$; $ D,E,F$ be respectively the feet of the altitudes from $ A,B,C$ of triangle $ ABC$; $ H$ be the orthocenter of triangle $ BDF$. Prove that $ PD,MF,IT,OH$ are concurrent.

Call $ OH\cap AB\equal{}K$ and $ N$ is midpoint of $ HK$. We prove that: $ PD,MF,IT,OH$ pass over $ N$.
In fact:
-) $ OH\parallel{}BC\Leftrightarrow GH\parallel{}BC \Leftrightarrow AH\equal{}2HD\Leftrightarrow AI\equal{}IH\equal{}HD$
Since $ T$ is orthocenter of $ BDF$ then $ FTDH$ is a parallelogram, it mean $ FTIA$ is a parallelogram, too.
Now $ IT\parallel{}AB$ and $ I$ is midpoint of $ AH$ then $ IT$ pass over midpoint $ N$ of $ HK$.
-) Consider triangle $ BCF$ with median $ FM$ and $ HK\parallel{}BC$ then $ FM$ pass over $ N$.
-) Finally, apply Menelaus theorem with triangle $ AHK$ with line $ DN'P$ ($ DP\cap HK\equal{}N'$) we have:
$ \frac{DA.N'H.PK}{DH.N'K.PA}\equal{}1\Leftrightarrow \frac{N'H}{N'K}\equal{}\frac{DH.PA}{DA.PK}\equal{}\frac{1}{3}.3\equal{}1\Rightarrow N\equiv N'$

Let a triangle $ ABC$ with $ O,H$ are the circumcenter and orthocenter respectively, $ OH\parallel BC$.
Denote $ a \equal{} BC ,b \equal{} CA,c \equal{} AB$ .Prove that:  $ a^2 \equal{} \frac {(a^2 \minus{} b^2 \plus{} c^2)(a^2 \plus{}b^2 \minus{} c^2)}{b^2 \plus{} c^2 \minus{} a^2}$ :thumbup:


[quote="mr.danh"]Given four positive intergers a,b,c,d such that a>b>c>d>0 and ac+bd is divisible by b+d+a-c. prove that ad+bc is not a prime.[/quote]
http://www.mathlinks.ro/viewtopic.php?t=150525


given $ a,b,c> 0$ and $ ab\plus{}bc\plus{}ca\plus{}2abc\equal{} 1$.Prove that:
$ \frac{a}{1\minus{}bc}\plus{}\frac{b}{1\minus{}ca}\plus{}\frac{c}{1\minus{}ab} \ge\ 2$

[quote="quykhtn-qa1"]given $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \plus{} 2abc \equal{} 1$.Prove that:
$ \frac {a}{1 \minus{} bc} \plus{} \frac {b}{1 \minus{} ca} \plus{} \frac {c}{1 \minus{} ab} \ge\ 2$[/quote]
If we set $ a \equal{} \frac {x}{y \plus{} z}, b \equal{} \frac {y}{z \plus{} x}, c \equal{} \frac {z}{x \plus{} y}$, then it simplifies to:
$ \sum_{cyc} \frac {x^2 \plus{} yz}{y \plus{} z} \ge \sum_{cyc} x \iff$
$ \sum_{cyc} x^4 \ge \sum_{cyc} x^2y^2$ and we are done :)

[quote="Mathias_DK"][quote="quykhtn-qa1"]given $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \plus{} 2abc \equal{} 1$.Prove that:
$ \frac {a}{1 \minus{} bc} \plus{} \frac {b}{1 \minus{} ca} \plus{} \frac {c}{1 \minus{} ab} \ge\ 2$[/quote]
If we set $ a \equal{} \frac {x}{y \plus{} z}, b \equal{} \frac {y}{z \plus{} x}, c \equal{} \frac {z}{x \plus{} y}$, then it simplifies to:
$ \sum_{cyc} \frac {x^2 \plus{} yz}{y \plus{} z} \ge \sum_{cyc} x$[/quote]
yes. :)


geometry
geometric transformation
rotation
dilation
complex numbers
complex analysis

Show that a Mobius transformation $ \phi$ is a composition of translations, rotations and dilations if and only if $ \phi(z)\equal{}az\plus{}b$, for some complex numbers $ a,b$ with $ a\neq 0$.

$ (\Leftarrow)$ is straightforward
$ (\Rightarrow)$
suppose $ \Phi(z) \equal{} \frac {Az \plus{} B}{Cz \plus{} D}$, $ AD\minus{}BC\neq 0$
if $ C \equal{} 0$ , we are done
if $ C\neq 0$, put $ S_{1} \equal{} z \plus{} D/C; S_{2} \equal{} 1/z; S_{3} \equal{} \frac {BC \minus{} AD}{C^2}z ; S_{4} \equal{} z \plus{} A/C$, then from here you have $ \Phi \equal{} S_{4}oS_{3}oS_{2}o S_{1}$.  Thus we have a contradiction


[quote="mdk"]The sequence a_1, a_2,a_3,.. satisfies a_1=19, a_9=99, and, for all n>=3, a_n is the arithmetic mean of the n-1 terms. Find a_2.[/quote]

[hide]Notice that $a_{3}=a_{4}=\ldots = \frac{19+a_{2}}{2}$. Then $\frac{19+a_{2}}{2}=99$ and $\boxed{a_{2}=179}$.[/hide]


[quote="arinthros"]Ok, this is crazy.  :o I can't believe that people actually like this stuff, not to mention doing it for fun. So, if you want to step over to the geek side, and stop succeeding at life, like me, play runescape. http://www.runescape.com and you cant go wrong. My username is rhfaboo, if you need help getting started. I am usually on world 50, but sometimes I am on world 55. Chances are I'll be near Falador. Soo, stop being nerds! Try to be just a little bit geeky for a change. :|[/quote]

You must be a complete loser and idiot to come to this website and diss math...

MOD PLEASE DELETE THIS

What's wrong with being a nerd? :?

Personally, I'd rather be a nerd than waste my life away playing a game that has horrendous graphics and is more addictive than alcohol.

This hardly qualifies as intermediate math.  It was pretty funny, though.

[quote="arinthros"]Alcohol destoys brain cells. Runescape doesn't. Besides, video games help increase the muscle size on your opposable thumbs.[/quote]

 :o Whoa man, you better get rid of that avatar, for real, quit being a dumb@ss

[quote="arinthros"]Alcohol destoys brain cells. Runescape doesn't.[/quote]

That doesn't make it good. Why is math crazy?

[quote="arinthros"]Alcohol destoys brain cells. Runescape doesn't. Besides, video games help increase the muscle size on your opposable thumbs.[/quote]

Runescape sure doesn't do anything [i]good[/i] for your brain cells.

[quote="arinthros"]Math isn't crazy. A forum website devoted to its propogationg is.[/quote]

Some of us want to go to a good college while others like you live in their parent's basements playing runescape till they're 80.

BTW, nr(not)1337, the formal definition of a limit is pretty easy to comprehend. If you don't understand that, then you'll never make it to maximizations and minimizations. lol. :P

Let me know when you get money or scholarships for über Runescape skillz.

[quote="arinthros"]BTW, nr1337, the formal definition of a limit is pretty easy to comprehend. If you don't understand that, then you'll never make it to maximizations and minimizations. lol. :P[/quote]

Yes, it is easy to comprehend.  Just doing it in calc class was boring.

[quote="arinthros"]Math isn't crazy. A forum website devoted to its propogationg is.[/quote]

Runescape isn't crazy.  A website devoted to it is.

...oops :P

[quote="arinthros"]Oh, I will. Just you wait.[/quote]

Administrator,  please ban this dumb@ss... :mad:

hey what's that jerk's runescape acct name again? i need to test out a hack  :D

username: rhfaboo
password: *******

 :lol:  :rotfl:

click here to see what level i am on :) [hide][hide][hide] :rotfl:  :rotfl:  :rotfl: still clicking!  :) hi incase you were wondering i dont play runescape :rotfl: [/hide][/hide][/hide][/hide]

YEAH!!! I got to the end of your hides!!!  [url=http://www.homestarrunner.com/sbemail3.html]Click to reveal hidden content[/url] :rotfl:

[hide]i just looked in the souce after the big block of hide html after clicking ~10 times. jerks.[/hide]

Actually, the definitions of 1, +, =, and 2 imply that 1+1=2 is always true... so I guess you can't even comprehend simple math. :P

[quote="arinthros"]LOLOLOLOL For those of you who think I cant comprehend your simple math. I got a 5 on the BC test.[/quote]

LOLOLOLOL so what?

[quote="arinthros"]LOLOLOLOL For those of you who think I cant comprehend your simple math. I got a 5 on the BC test. Oh yeah, btw, 1+1 does not always equal 2. Comprehend that.[/quote]
Of course it doesn't. For example, in ${\bf Z}_2$, 1+1=0, and in Boolean algebra, 1+1=1. What's your point? Further, the type of math we do here is much more complex than BC calc. We discuss problems which require much critical thinking, the type which are not usually taught in school. Working these types of problems increases your lateral thinking ability, which is one of the reasons this site exists. So I don't see what you have against a forum that promotes good critical thinking and problem solving skills.

For more about this, see [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]What is Problem Solving?[/url]

arinthros:

it says you are jewish and your avatar is nazi?!?!

you know there actually are some jewich peoples on this site

You can be Jewish and support Nazis...I dunno why you would. Isnt it obvious that he is just trying to be funny?

Arthinos should be condemned for putting the Nazi sign because even his rights of free speech cant be used to do that. It doesnt make sense to have a Nazi sign and be Jewish. Nazism is condemned by everyone in this world, and Arthinos should have his posting rights denied, (I think Valentin might not like me saying this ) because thoughts for destroying humanity should be condemned.

GoBraves, even if he's trying to be funny, he's not. I agree with hello.

Yeah, I'm jewish, & the Nazi symbol isnt funny at.. all. To tell u what i mean, i lost all my great uncles and aunts and all that family, and any family that would have come from them, in the Holocaust (WWII Jewish (and a little bit of a few other groups) massacre by Nazis). So yeah, not very amusing.

[quote="arinthros"]Ok, this is crazy.  :o I can't believe that people actually like this stuff, not to mention doing it for fun. So, if you want to step over to the geek side, and stop succeeding at life, like me, play runescape. http://www.runescape.com and you cant go wrong. My username is rhfaboo, if you need help getting started. I am usually on world 50, but sometimes I am on world 55. Chances are I'll be near Falador. Soo, stop being nerds! Try to be just a little bit geeky for a change. :|[/quote]

[size=200]LOL UR COOL[/size]


wow that makes it a thousand times easier... thank you.


How can I measure the height of a building?