topology
advanced fields
advanced fields unsolved

let $ E_1,E_2,\dots$ - open sets in $ R$.
The boundary of $ E_1\cup E_2\cup\dots$ is infinite.
Prove that this boundary is countable.

Hi.

This is wrong: Take $ E_i: \equal{}(i,i\plus{}1)$. Then $ \bigcup_i E_i \equal{} \mathbb{R}\backslash\mathbb{Z}$, which has $ \mathbb{Z}$ as boundary.

The complement of the Cantor set is open and has the entire Cantor set as its boundary, which is uncountable.    Perhaps you want each of the $ E_i$ to be an open [i]interval[/i]?  In this case, the boundary of a union is contained in the union of the boundaries (I hope) and so the result should be immediate.

JBL, any open set in $ \mathbb{R}$ can be written as the countable union of open intervals (i.e. as the union of its connected components).

Right you are ... in particular, it's [i]not[/i] true that the boundary of the union is contained in the union of the boundaries for nonfinite unions.

[quote="primenumber"]Hi :)

the set of integers is countable :P[/quote]

Sorry, I read [i]un[/i]countable...


inequalities
trigonometry
inequalities unsolved

Prove that for any $ x\in \mathbb{R}$ the following inequality holds:
$ cos(cos(cos(cosx)))>sin(sin(sin(sinx)))$  :wink:

This is implicitely proven at
http://www.mathlinks.ro/Forum/viewtopic.php?t=7344
http://www.mathlinks.ro/Forum/viewtopic.php?t=91594

  darij

this is the best solution i've seen to this problem...

it suffices to prove that for any $ x$,

\[ \cos(\cos x) > \sin(\sin x) \mbox{ (*)},
\]

since that then implies

\[ \cos(\cos(\cos(\cos x))) > \sin(\sin(\cos(\cos x))) > \sin(\sin(\sin(\sin x)))
\]

where the first inequality follows by replacing $ x$ with $ \cos(\cos x)$ in (*) and the second follows from the fact that $ \cos(\cos x)$ and $ \sin(\sin x)$ are both in $ [\minus{}1,1]$, and $ \sin(\sin x)$ is increasing on that interval.

note that we only need to consider $ x\in (0,2\pi)$. since $ \sin(\sin x) \leq 0$ and $ \cos(\cos x)) > 0$ on $ [\pi,2\pi)$, it suffices to consider $ x\in (0,\pi)$. and since both sides of (*) remain unchanged when $ x$ is replaced with $ \pi\minus{}x$, it suffices to consider only $ x\in (0,\pi/2]$. now the inequality to be proven is equivalent to $ \sin(\pi/2\minus{}\cos x) > \sin(\sin x)$. by the monotonicity of $ \sin x$ on $ (0,\pi/2)$, it suffices to prove $ \pi/2\minus{}\cos x > \sin x$. and this is evident from

\[ \sin x\plus{}\cos x \equal{} \sqrt{2}\cos(x\minus{}\pi/4) \leq \sqrt{2} < \pi/2
\]


Let $ a_1,a_2,...,a_n$ is permutaion of $ 1,2,...,n$. For this permutaion call the pair $ (a_i,a_j)$ [i]wrong pair [/i]if $ i<j$ and $ a_i >a_j$.Let [i]number of inversion [/i] is number of [i]wrong pair [/i] of permutation $ a_1,a_2,a_3,..,a_n$. Let $ n \ge 2$ is positive integer. Find the number of permutation of $ 1,2,..,n$ such that its  [i]number of inversion [/i]is divisible by $ n$.

Group the $ n!$ permutations into $ (n \minus{} 1)!$ classes of $ n$ elements each where two permutations $ p, q$ are in the same class if and only if $ 1 \leq i < j \leq n \minus{} 1$ implies that $ i$ occurs before $ j$ in $ p$ if and only if $ i$ occurs before $ j$ in $ q$.  In other words, group permutations by what they look like if you erase $ n$.

For a permutation $ p$ in a given class, all inversions fall into two categories: there are $ k$ inversions ($ k$ dependent only on the choice of class) among the entries $ 1, 2, \ldots, n \minus{} 1$ of the permutation and there are some other number of inversions between $ n$ and the entries of $ p$ that follow it.  As a result, there is one permutation in the class with $ k$ inversions (when $ n$ appears last), one with $ k \plus{} 1$ inversions (when $ n$ appears next-to-last), etc., and one permutation with $ k \plus{} (n \minus{} 1)$ inversions (when $ n$ appears first).  Exactly one of these numbers is divisible by $ n$, so exactly $ (n \minus{} 1)!$ permutations have inversion number divisible by $ n$.  (In fact, we proved the stronger result that for any $ i$, there are exactly $ (n \minus{} 1)!$ permutations in $ S_n$ whose inversion number is congruent to $ i$ modulo $ n$.)


[img]http://www.henrydees.com/andys_girl.jpg[/img]

Hi, I'm going to eat you.

What an internet argument probably looks like.

[img]http://www.cokeworld.it/Coke%20Background2.jpg[/img]

The coke bear pulverized you.

Well, well, well! If it isn't the old boomer!
Welcome back, Chocky darling! What a pleasant surprise!
And a nice comeback too!

[img]http://www.pthsd.k12.nj.us/SCH/BMS/graphics2/welcome%20back!.jpg[/img]

If all of your images are coming from one source, it's not really [b]random[/b] is it?

[img]http://img90.imageshack.us/img90/4734/dtep0208vn2.jpg[/img]

If the college admissions next year asks about a character I would like to emulate, I'll say this guy.

momentai
[img]http://membres.lycos.fr/hyna/tommy-chakmon-neemon=eat.jpg[/img]
Edit: grr, this put pictures in post thing is too complicated for me

Centaurs have superior movement but are hindered by deserts and forests. This one is the only one that uses bows instead of lances.

[img]http://www.shiningforcecentral.com/artwork/sf2/93/sf2_book_may.jpg[/img]]

These pics are from completely random sources. Now....

[quote="randomdragoon"][img]http://ps2media.gamespy.com/ps2/image/article/665/665200/guitar-hero-20051107043412575.jpg[/img][/quote]
omggggg I want that game
wahh
wait that is 2 right

ZUO PAI PAI, YOU PAI PAI

WO ZUO KAN YOU KAN SHANG KAI XIA KAN

BU YAO BEI WO DE YANG ZI XIA HUAI
QI SHI WO HEN KE AI

NU HAI MEN DE XING SHI HAI ZHEN QI GUAI

I saw something like this somewhere, but I don't remember where.

[quote="Chocoboom"]o rly? where abouts might you have seen it?  :D  :D[/quote]

lol..i like the "o rry one"

[img]http://www.lunchlady.org/uploaded/images/arbokmuk.jpg[/img]

HAY GUYS DO U LISTEN TO METALICA?

[quote="Treething"]
HAY GUYS DO U LISTEN TO METALICA?[/quote]
YA

[img]http://www.sigg3.net/blogger/spam.jpg[/img]

Um... randomdragoon?  The link is broken.


[quote="linker"]Prove that $ \sum{\cos{(\pm x_1 \pm x_2 \pm \dots \pm x_n)}}\leq 2^n \left( cos{(\sqrt [n]{x_1x_2 \dots x_n})} \right)^n$ , where $ n \in \mathbb{N} , n \geq 2$ , and $ x_1,x_2, \dots . x_n \in (0 ; \frac {\pi}{2})$ , where the sum contains all the $ 2^n$ possible choose for + and - .[/quote]
$ \sum{\cos{(\pm x_1 \pm x_2 \pm \dots \pm x_n)}}\equal{}2^n cos(x_1)\dots cos(x_n) \leq$$ 2^n(\frac{cos(x_1)\plus{} \dots \plus{}cos(x_n)}{n})^n \leq$
$ 2^n(cos(\frac{x_1\plus{}\dots\plus{}x_n}{n}))^n\leq 2^n (cos{(\sqrt [n]{x_1x_2 \dots x_n})})^n$; :)

[quote="linker"][quote="Sergic Primazon"]
$ \sum{\cos{(\pm x_1 \pm x_2 \pm \dots \pm x_n)}} \equal{} 2^n cos(x_1)\dots cos(x_n)$
[/quote]

Can you please explain a little this identity please ?

Thank you :)[/quote]
n=1 : $ cos(x_1) \plus{} cos( \minus{} x_1) \equal{} 2^1cos(x_1)$ ( true );
let it is true for n : $ \sum{\cos{(\pm x_1 \pm x_2 \pm \dots \pm x_n)}} \equal{} 2^n cos(x_1)\dots cos(x_n)$;
n->(n+1) :$ \sum{\cos{(\pm x_1 \pm x_2 \pm \dots \pm x_{n \plus{} 1})}} \equal{} \sum{\cos{(\pm x_1 \pm x_2 \pm \dots \pm (x_n \plus{} x_{n \plus{} 1}))}} \plus{}$ $ \sum{\cos{(\pm x_1 \pm x_2 \pm \dots \pm (x_n \minus{} x_{n \plus{} 1}))}} \equal{}$$ 2^n cos(x_1)\dots cos(x_{n \minus{} 1})cos(x_n \plus{} x_{n \plus{} 1})$ $ \plus{} 2^n cos(x_1)\dots cos(x_{n \minus{} 1})cos(x_n \minus{} x_{n \plus{} 1}) \equal{} 2^{n \plus{} 1} cos(x_1)\dots cos(x_{n \plus{} 1})$; :)


hi.  we want to solve the following system:

hello, i have computed
$ x\equal{}3,y\equal{}\minus{}1,z\equal{}\minus{}2$
$ x\equal{}3,y\equal{}\minus{}2,z\equal{}\minus{}1$
$ x\equal{}\minus{}1,y\equal{}3,z\equal{}\minus{}2$
$ x\equal{}\minus{}1,y\equal{}\minus{}2,z\equal{}3$
$ x\equal{}\minus{}2,y\equal{}3,z\equal{}\minus{}1$
$ x\equal{}\minus{}2,y\equal{}\minus{}1,z\equal{}3$
Sonnhard.

@ Dr. Graubner:

how ?  by guessing ?   :)

There are actually two solutions.

One is where $x,y,z$ are the roots of $a^3-7a-6=0$ which gives $(x,y,z)=(-2,-1,3)$ and its permutations.

The other one is where $x,y,z$ are the roots of $a^3+7a-6=0$, which is quite ugly.

[hide=Solution]We know that if $x+y+z=0$, then $x^3+y^3+z^3=3xyz$. This implies that $xyz=6$.

Let $xy+yz+zx=r$. Then, $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=-2r$.

This means, $x,y,z$ are the roots to the following equation:
$$a^3+ra-6=0$$
Now,
\begin{align*}
x^3+rx-6=0\implies x^4+rx^2-6x&=0\\
y^3+ry-6=0\implies y^4+ry^2-6y&=0\\
z^3+rz-6=0\implies z^4+rz^2-6z&=0
\end{align*}
Adding these three:
$$(x^4+y^4+z^4)+r(x^2+y^2+z^2)-6(x+y+z)=0\implies x^4+y^4+z^4=2r^2$$
And,
\begin{align*}
x^3+rx-6=0\implies x^5+rx^3-6x^2&=0\\
y^3+ry-6=0\implies y^5+ry^3-6y^2&=0\\
z^3+rz-6=0\implies z^5+rz^3-6z^2&=0
\end{align*}
Adding these three:
$$(x^5+y^5+z^5)+r(x^3+y^3+z^3)-6(x^2+y^2+z^2)=0\implies x^5+y^5+z^5=-30r$$
Finally,
\begin{align*}
x^3+rx-6=0\implies x^7+rx^5-6x^4&=0\\
y^3+ry-6=0\implies y^7+ry^5-6y^4&=0\\
z^3+rz-6=0\implies z^7+rz^5-6z^4&=0
\end{align*}
Adding these three:
$$(x^7+y^7+z^7)+r(x^5+y^5+z^5)-6(x^4+y^4+z^4)=0\implies2058-42r^2=0\implies r=\pm7$$
Our final cubic equation is now:
$$a^3-7a-6=0\text{ or }a^3+7a-6=0$$
The first one gives the nice-looking solutions:
$$(x,y,z)=(-2,-1,3)\text{ and its permutations}$$
while the second one gives rather ugly-looking solutions, but it also satisfies the equations.

it's great ！I find （x,y,z）in R  not  C.but  I find  x^2+y^2+z^2=-2r=±14,and why?


I need to calculate the average velocity, final velocity, and acceleration but do not understand the equations. Could someone please help me? Thank you very much.

time in seconds=0.250
distance in meters=0.306

Any help is greatly appreciated. Thanks alot!!

Here are to helpful equations

$ v \equal{} \frac{m}{s}$
$ a \equal{} \frac{m}{s^{2}}$

v is velocity, a acceleration, m metre, and s second

This belongs in the physics subforum.


why is it necessary to remove $ CO$ formed in haber's process

How is carbon monoxide formed in the Haber process?

In Haber Process the catalytic reaction:

$ CO_{(g)}+H_{2}O_{(g)}\rightleftharpoons CO_{2}_{(g)}+H_{2}_{(g)}$

produces the hydrogen gas needed for the reaction.

We know that during the Haber Process, a heterogeneous catalyst is used which provides a reaction surface (with active sites) for the reactants to adsorpt on to.

Substances such as $ S$ and $ CO$ bind strongly to these active sites and hence block the adsoprtion of nitrogen and hydrogen (the catalyst is said to be poisoned).

That's my best guess.

[u]Note:[/u] it was never said that the Haber Process produces $ CO$.

I also never said that it produces CO. Anyway, your explanation makes perfect sense.

[quote="BanishedTraitor"]In Haber Process the catalytic reaction: (...) produces the hydrogen gas needed for the reaction.[/quote]

And from where does the CO come from? That's not the only reaction that produces hydrogen. In fact, the very first reaction in the all process is the reaction between methane and steam to produce hydrogen and CO. Then this CO produced enters the reaction you gave.

even i have no idea at this question it was in my book....
and the chapter had nothing to do with haber's process..not even an intro.it was in surface chemistry :w00t:

 thanks

[quote="Carcul"]
And from where does the CO come from? [/quote]

I think your one would explain it:

$ CH_{4}_{(g)}+H_{2}O_{(g)}\rightleftharpoons CO_{(g)}+3H_{2}_{(g)}$

don't really need the catalytic reaction I wrote.

Both reactions are used in the Haber Process for hydrogen production.


In this age and especiality to futures science studys this question is very importan to decide your subjets. but one think about this problem   the physics work in the energy matter to mechanic and the chemistry work inthe energy and matter to composition and properties changes   :?:

It's tough for me. I was a big physics guy at first, but I didn't know anything about physics until high school....so I became more interested in biology in chemistry. That said, I have a new interest in it, and I'll take the SAT II for it in October without any school course.


Can someone show me a proof of whether or not it is possible to position a rubiks cube so that no two adjacent faces have the same color?

Yes, it's possible.

I've even done it accidentally once. :o

Here are two "opposite" views from the http://www.rubiks.com/cube_online.html virtual cube:

[img]http://img169.imageshack.us/img169/9068/rubikscube1co5.png[/img] [img]http://img156.imageshack.us/img156/3181/rubikscube2en7.png[/img]

Sequence: LR FB RL BF BF

Sounds like a [url=http://en.wikipedia.org/wiki/Four-color_theorem]4-color proof[/url]. Not sure if that helps. but the empirical evidence provided by i_like_pie suffices in this case.

It's very easy:
Top face 180 degrees
Bottom face 180 degrees
Left face 180 degrees
Right face 180 degrees
Front face 180 degrees
Back face 180 degrees

Your cube should look like a checkerboard. Repeat this sequence (U2D2L2R2F2B2) to restore your cube.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let G be a group
what means the Fitting subgroup is hypercentral?
thanks
sastra

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

http://www.google.com.vn/search?hl=vi&q=group++hypercentral&meta=

persevering  :wink:


You could include in the prefix of your document

\newcommand{\tg}{\operatorname{tg}}
\newcommand{\ctg}{\operatorname{ctg}}

and so on. You don't even have to use \tg and \ctg, but how else are you going to remember them?

(Just don't use \operatorname on AoPS/ML - it's not supported.)

[quote="Farenhajt"]Is there ANY LaTeX package that supports European trigonometric notation, namely $ \mathrm{tg}\,x,\mathrm{ctg}\,x,\mathrm{cosec}\,x$?

(Those CAN be simulated via \mathrm, as above, but it would be really nice to have them directly available as commands.)[/quote]

Try to find something similar to hrlatex package, or use that one. ;) Pozdrav!

So basically both of you are telling me it's not going to be seen around here any time soon... Sigh :( Thanks though. Hvala, kokane :)

Hi,

I recommend to use \DeclareMathOperator of amsmath instead of \newcommand:
[code]\DeclareMathOperator{\tg}{tg}[/code]
Stefan

--
[url=http://texblog.net]TeXblog.net[/url]


function
calculus
derivative
calculus computations

Is there a function f: R → R with continuous derivative such that f(x) > 0 and f '(x) = f(f(x)) for all x?

It can be readily shown that $ f'(x),f''(x)>0$ and so $ f(x)-x\to \infty$ when $ x\to \infty$.
Using 'Mean Value',we have
$ f(f(x))=f(x)+f'(\xi )(f(x)-x)>f(x)+f'(x)(f(x)-x)=f(x)+f(f(x))(f(x)-x)$
$ x<\xi<f(x)$.
Thus a contradiction.


geometry
geometric transformation
reflection
function
inequalities
symmetry
LaTeX

[quote="VIMF"]let a,b,c>0. Prove that
$ 3( a^{2} \plus{} b^{2} \plus{} c^{2})( a^{2}b^{2} \plus{} b^{2}c^{2} \plus{} c^{2}a^{2}) \geq ( a^{2} \plus{} ab \plus{} b^{2})(b^{2} \plus{} bc \plus{} c^{2}) ( a^{2} \plus{} ac \plus{} c^{2})$
$ \geq 3(ab \plus{} bc \plus{} ca)( a^{2}b^{2} \plus{} b^{2}c^{2} \plus{} c^{2}a^{2})$[/quote]
Let $ a,$ $ b$ and $ c$ are non-negative numbers, $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2$ and $ abc\equal{}w^3.$
Hence, $ \prod_{cyc}(a^2\plus{}ab\plus{}b^2)$ is linear function of $ w^3.$
Id est, for the proof enough to check two cases:
1) $ b\equal{}c\equal{}1$ ;
2) $ c\equal{}0$ and $ b\equal{}1,$ which gives that an original inequality is true.

I have this one in my note..
\[ (a^2\plus{}ab\plus{}b^2)(b^2\plus{}bc\plus{}c^2)(c^2\plus{}ca\plus{}a^2)\geq (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)(ab^2\plus{}bc^2\plus{}ca^2)\]

[quote="Tourish"]I have this one in my note..
\[ (a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)\geq (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)(ab^2 \plus{} bc^2 \plus{} ca^2)\]
[/quote]

$ (a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)\geq (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)(ab^2 \plus{} bc^2 \plus{} ca^2)$

$ \Longleftrightarrow a^4b^2\plus{}a^2c^4\plus{}b^4c^2\minus{}a^3bc^2\minus{}a^2b^3c\minus{}ab^2c^3\geq 0$

from AM-GM the inequality holds

My solution to the inequality
$ 3( a^{2} \plus{}b^{2} \plus{}c^{2})( a^{2}b^{2} \plus{}b^{2}c^{2}\plus{}c^{2}a^{2})\geq ( a^{2} \plus{} ab\plus{}b^{2})(b^{2}\plus{}bc \plus{}c^{2}) ( a^{2} \plus{} ac\plus{}c^{2})$

Due to symmetry, we may assume that $ a \geq b \geq c$

We have:

$ a^2 \geq ab \geq b^2$

$ b^2 \geq bc \geq c^2$

By using Chebyshev Inequality:

$ \Rightarrow 3(a^2b^2\plus{}ab^2c\plus{}b^2c^2) \geq (a^2\plus{}ab\plus{}b^2)(b^2\plus{}bc\plus{}c^2)$

And thus, it is enough to show that:
 $ (a^2\plus{}b^2\plus{}c^2)(a^2b^2\plus{}b^2c^2\plus{}c^2a^2) \geq (a^2\plus{}ac\plus{}c^2)(a^2b^2\plus{}ab^2c\plus{}b^2c^2)$

$ \Leftrightarrow \dfrac{a^2b^2\plus{}b^2c^2\plus{}c^2a^2}{a^2b^2\plus{}ab^2c\plus{}b^2c^2} \geq \dfrac{a^2\plus{}ac\plus{}c^2}{a^2\plus{}b^2\plus{}c^2}$

$ \Leftrightarrow \dfrac{a^2b^2\plus{}b^2c^2\plus{}c^2a^2}{a^2b^2\plus{}ab^2c\plus{}b^2c^2} \minus{}1\geq \dfrac{a^2\plus{}ac\plus{}c^2}{a^2\plus{}b^2\plus{}c^2}\minus{}1$

$ \Leftrightarrow \dfrac{c^2a^2\minus{}ab^2c}{a^2b^2\plus{}ab^2c\plus{}b^2c^2} \geq \dfrac{ca\minus{}b^2}{a^2\plus{}b^2\plus{}c^2}$

$ \Leftrightarrow (ca\minus{}b^2)(\dfrac{ca(a^2\plus{}b^2\plus{}c^2)\minus{}a^2b^2\minus{}ab^2c\minus{}b^2c^2}{(a^2b^2\plus{}ab^2c\plus{}b^2c^2)(a^2\plus{}b^2\plus{}c^2)}) \geq 0$

$ \Leftrightarrow (ca\minus{}b^2)^2(\dfrac{a^2\plus{}c^2}{(a^2b^2\plus{}ab^2c\plus{}b^2c^2)(a^2\plus{}b^2\plus{}c^2)}) \geq 0$ (it is obviously true)
 The proof is completed. Equality holds if and only if $ latex \Leftrightarrow a\equal{}b\equal{}c$


function
logarithms
calculus
integration
calculus computations

Show that the function $ f(x) \equal{}\ln(x\plus{}\sqrt{x^{2}\plus{}1})$ is odd and find its inverse.

Edit: Sorry, I meant to post this in the algebra forum.

Well it looks like we have some sort of inverse hyperbolic function, but I can't remeber which, I am rusty  with logarithmic equivalents.

Solution:

$ f(x) =\ln{(x+\sqrt{x^{2}+1})}$

$ \Rightarrow f(-x) =\ln{(-x+\sqrt{x^{2}+1})}$

$ =\ln{\left(\frac{(-x+\sqrt{x^{2}+1})(-x-\sqrt{x^{2}+1})}{-x-\sqrt{x^{2}+1}}\right)}$

${ =\ln{\left(\frac{x^{2}-(x^{2}+1)}{-x-\sqrt{x^{2}+1}}\right)}}$

$ =\ln{\frac{1}{x+\sqrt{x^{2}+1}}}$

$ =-\ln{(x+\sqrt{x^{2}+1})}$

$ =-f(x)$

$ f(-x) =-f(x)\Rightarrow f$ is odd.

Let $ y = f(x)$

$ \Rightarrow x+\sqrt{x^{2}+1}= e^{y}$

$ \Rightarrow\sqrt{x^{2}+1}= e^{y}-x$

$ \Rightarrow x^{2}+1 = e^{2y}-2xe^{y}+x^{2}$

$ \Rightarrow 2xe^{y}= e^{2y}-1$

$ \Rightarrow x =\frac{e^{y}-e^{-y}}{2}=\sinh{y}$

$ \Rightarrow f^{-1}(x) =\sinh{x}$

Well, since you did post this in the calculus forum, how about using some calculus?

Note that $ f'(x) \equal{}\frac{1}{\sqrt{x^{2}\plus{}1}}$ and that $ f'$ is even.

Note also that $ f(0) \equal{}\ln(0\plus{}1) \equal{} 0.$

Claim: A function $ f$ that is differentiable on a symmetric interval (an interval of the form $ (\minus{}a,a)$) is odd if and only if $ f(0) \equal{} 0$ and $ f'$ is an even function.

Checking the claim: If $ f$ is odd, the other items do clearly follow.

Suppose $ x > 0.$ Then

$ f(\minus{}x) \equal{} f(0)\plus{}\int_{0}^{\minus{}x}f(t)\,dt.$ Make the substitution $ u \equal{}\minus{}t.$

$ f(\minus{}x) \equal{} f(0)\minus{}\int_{0}^{x}f(\minus{}u)\,du$ Now use that $ f(0) \equal{} 0$ and $ f(\minus{}u) \equal{} f(u).$

$ f(\minus{}x) \equal{}\minus{}\int_{0}^{x}f(u)\,du \equal{}\minus{}f(x).$ Hence $ f$ is odd.


Αν $ xyz \geq (x\plus{}y\plus{}z)^2$
να αποδειξετε οτι:

A)

 $ \frac{z\minus{}4}{8z} \geq \frac{1}{x\plus{}y}$

B)

 $ xyz \geq 27(x\plus{}y\plus{}z)$

[quote="Demetres"][quote="NickNafplio"]Ειναι θετικοι οι αριθμοι? Γιατι αν ειναι θετικοι ειναι αρκετα απλη η ανισοτητα[/quote]

Μάλλον θετικοί είναι. Για (1,-1,-1) ισχύει η συνθήκη αλλά όχι το (Β)[/quote]

Ωραια

Απο τη συνθηκη λαμβανουμε:
$ D \equal{} (2x \plus{} 2y \plus{} 2z)^2 \minus{} 4xyz \leq 0$

Αρα το τριωνυμο:
$ m^2 \minus{} 2m(x \plus{} y \plus{} z) \plus{} xyz$, εχει θετικο μεγιστοβαθμιο συντελεστη και αρνητικη διακρινουσα, αρα ειναι θετικο.

Θετουμε στο τριωνυμο: $ m \equal{} \frac {zx}{2}$ και $ m \equal{} \frac {zy}{2}$ και παιρνουμε:
$ \frac {z^2x^2}{4} \minus{} x^2z \minus{} z^2x \geq 0 \Leftrightarrow xz \minus{} 4z \minus{} 4x \geq 0$ και
$ yz \minus{} 4y \minus{} 4z \geq 0$
Προσθετουμε κατα μελη και εχουμε:
$ zy \plus{} zx \minus{} 4x \minus{} 4y \minus{} 8z \geq 0 \Leftrightarrow (z \minus{} 4)(x \plus{} y) \geq 8z \Leftrightarrow \frac {z \minus{} 4}{8z} \geq \frac {1}{x \plus{} y}$

Για τη 2η τωρα ειναι:

$ x^2y^2z^2 \geq (x \plus{} y \plus{} z)^4 \equal{} ((x \plus{} y \plus{} z)^2)^2 \geq (3(xy \plus{} xz \plus{} yz))^2 \equal{} 9(xy \plus{} xz \plus{} yz)^2 \geq 27(xyxz \plus{} xyyz \plus{} xzyz) \equal{} 27xyz(x \plus{} y \plus{} z) \Leftrightarrow xyz \geq 27(x \plus{} y \plus{} z)$
και τελειωσαμε.

Πολυ ωραια λυση. Εχετε δικιο οι αριθμοι ειναι θετικοι. Sorry που δεν το ανεφερα. :blush:
Νομιζα οτι προκυπτει απο τη συνθηκη αλλα αυτο που εναι σιγουρo ειναι οτι $ xyz \geq 0$


A triangular park is formed by the intersection of three streets, Bridge
Street, Harbor Place, and College Avenue. A walkway parallel to Harbor Place goes through the park. A time
capsule has been buried in the park in a location that is equidistant from
Bridge Street and College Avenue and 5 yards from the walkway. Where is the time capsule buried?

No help here.

What diagram are you talking about?

Thanks!

:gasp:
YOU JUST COMMITTED ONE OF THE CARDINAL SINS OF AOPSING:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=132833


An integer is randomly selected from the integers in the interval $ \minus{}10.5$ to $ 10.5$. What is the probability that it is even? Express your answer as a common fraction.

This is -10 to 10, inclusive. These are 21 numbers. Because of 0, there is one more even than odd. The probability of selecting an even number is $ \boxed{11/21}$


Let a,b>0 . Prove that
$ \displaystyle\frac {a\minus{}b}{lna\minus{}lnb} < \frac{1}{3}(2\sqrt{ab} \plus{} \frac{a\plus{}b}{2})$

[quote="aktyw19"]Let a,b>0 . Prove that
$ \displaystyle\frac {a \minus{} b}{lna \minus{} lnb} < \frac {1}{3}(2\sqrt {ab} \plus{} \frac {a \plus{} b}{2})$[/quote]

That's wrong.

Try $ a\equal{}e,\ b\equal{}\frac{1}{e}$, you will get $ 0<e<1\plus{}\sqrt{2}$. :wink:


Can anyone help with this – I’ve been struggling with it for days.  It relates to a sum in the sigma notation and if someone knew how to convert it into an explicit sum, it would help me hugely.  I have a sequence of 4 terms (‘n’) viz (in ascending order): 5.835, 6.221, 7.081 and 8.733.  I am trying to calculate three values based upon:

             n
a = n-1 Σ  (s -1) x(s)
               j=2     (n -1)

             n
b = n-1 Σ (s -1) (s - 2)  x(s)
               j=3   (n -1) (n - 2)
	  n
c = n-1 Σ  (s -1) (s - 2) (s - 3) x(s)
               j=3   (n -1) (n - 2) (n - 3)

where x(s) denotes the sth element in the sample as arranged in ascending order.

Any help would be very gladly received.

Thanks.

It's quite rude to post the same content in different parts of the forum, as you've done.  Please don't do this in the future.  (Your choice of fora was also questionable.)

Your post is largely incomprehensible -- using a small amount of [[LaTeX]] (see, e.g. that link and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 ) would be extremly helpful.  The codes you need are \sum_{stuff below the sigma}^{stuff above the sigma}, then enclose your code between two dollar signs to get $ \sum_{stuff below the sigma}^{stuff above the sigma}$.

Even though I can't figure out the details of what you're trying to write, it seems pretty clearly to be a question of the form, "Do you understand what sigma notation means?"  You appear to have sums with four or fewer terms, so just do the arithmetic.

[color=red]
Kevin1971, please do not insult others

-Moderator[/color]

Have a little more respect.  Do you expect to get answers for your questions by insulting people who are trying to give you advice?


If a 2 digit number can be divided by the sum of its digits, what is the largest possible remainder?

If I'm not wrong it's 15 but that's through trial and error. I suppose there is another way to do it right?
Not sure if it belongs here, since it's a problem from the national olympiad in my country for the junior section

Do you mean $ 10a \plus{} b$ is dividiable by $ a \plus{} b$

Remainder of what?
If it's divisible by the sum of its digits, the remainder's... kind of zero.


Prove that exist infinite integers such as

5^n-2 -1/n  (the n-2 is the power)

is integer.

take $ n \equal{} 2p$ and use Fermat little theorem   where $ p$ is a prime  :)

it has been posted before :
 http://www.mathlinks.ro/viewtopic.php?t=299455


Can you help me with this one.I want to solve it using Z[i].I know the original solution:

if ab+1|a^2 + b^2 then (a^2 + b^2/ab+1)=k^2 :?

Does any one have any idea ...... :?  :?

This problem is solved by the method of descent...


[quote="AndrewTom"]A postman has seven letters to deliver, one to each of seven neighbours. Unfortunately, he is rather inaccurate in making his deliveries.

(i) In how may ways could he deliver the letters so that exactly two of the neighbours get the wrong letters?

(ii) In how many ways could he deliver the letters so that exactly three of the neighbours get the wrong letters?[/quote]

(i) Number of ways $ \equal{}_7 C_2\equal{}21$

(ii) Number of ways $ \equal{}_7 C_3 \times 2\equal{}70$


I'm lousy at asymptote and a noob at LaTeX :blush: .  If I decide to handwrite my proof, can I print out a picture of the diagram(s) and place it on the solution paper?

[quote="buzzer11"]Since you're handwriting it, why dont you just draw it?[/quote]

Who wants to hand-draw the diagram to problem 2?

I think I did tha last year on one of the problems, but I drew a really really awesome diagram on scrap paper that I was working from, and when I solved the problem I didn't want to draw it again so I just cut it out and pasted it.

I think they scan the papers anyway, so it won't make a difference.

yeah, problem two will be hard to draw exactly... why dont you cut out the diagram from the actual problem then?


There are 1000 lockers numbered 1 to 1000. Suppose you open all of the lockers, then close every other locker. Then, for every third locker, you close each opened locker and open each closed locker. You follow the same pattern for every fourth locker, every fifth locker, and so on up to every thousandth locker. Which locker doors will be open when the process is complete?

this happened to be a thing i did in my 6th grade math class.  So it asks for a number of factors of that number.  Squares have odd factors so they will be open.

1,4,9,...961 so 32 will be open.

hey! i posted this in the basics forum marathon

;) nice problemo

Hint
[hide]

Define a door 'flip' as when it is changed from either open to closed or closed to open.  A door that starts closed (as all the doors are) must be flipped an odd number of times to remain open at the end.  According to the description of the process we do, we flip a door only if it is a multiple of the number that we are currently dealing with.  Conversely looking from the perspective of a single door, it is flipped only if the number we are currently dealing with is one of its factors.

Are there any numbers with an odd number of factors?   :D

Very very interesting. Where did you get this again?

how interesting can it be?
squares have odd factors


The title and the fact that it's a poll pretty much explains the content of this post enough that I don't have to continue typing.

Who voted girl?

On this other forum I'm part of, they have 1 or 2 active girls.

(Pen Spinning Forum)

Read this: http://z15.invisionfree.com/UCPSB/index.php?showtopic=2677&st=60

[b]To the above poster:[/b] Good for you
[b]To KRS:[/b] I don't think [i]miyomiyo[/i] is that desparate, but that's just my perception  :P

Why? What's the point of this? Can't you get a sense of how many girls there are by reading through the threads? I mean, that's not going to be any less accurate than making a poll...

[quote="K81o7"]Maybe if we added back in the option on this forum where we could specify gender with our information?  Seriously, I don't think the reason of this poll was to find out the male-female ratio on AoPS (as many such threads have been made in the past), but rather to find girls on AoPS.  Maybe that's just my suspicion, though...[/quote]

I think you should contact Dateline.....

and dude why isnt there an "i dont know" category because im pretty sure some people have that problem..especially if your initials are vv.............they cant really distinguish themselves.

I just wanted to see the ratio.  Seriously.

yeah, but they said dont revive old threads

[quote="CheeseIsGood"]Who voted girl?[/quote]

Obviously it's Ignite

[quote="miyomiyo"]I was thinking about writing an "other" but I thought people might find that offensive.[/quote]

Nah, I bet a dozen or so people here are just as confused as me.


geometry
3D geometry
trigonometry
geometric transformation
reflection
tetrahedron
vector

The angle between a plane and each of the twelve edges of a cube is $ \alpha$. What is the value of $ \sin \alpha$?

[hide="Big Hint"]Take two diagonally opposite points of the cube, and consider the six edges of the cube that do not contain either point. The midpoints of these edges form the vertexes of a perfect hexagon. Use the plane that contains this hexagon.[/hide]
[hide="Solution 1"][i]Continued from hint[/i]

Let $ A,B,C$ be three consecutive vertexes of the hexagonal cross section of the cube by the plane, and let $ O$ be the vertex common to the edges containing $ A$ and $ B$.

Let $ C'$ be the reflection of $ C$ across $ B$. Note that $ A,B,C'$ define the same plane as $ A,B,C$, so it suffices to find the sine of the angle between $ OA$ and $ \triangle ABC'$. Note that $ OABC'$ is a tetrahedron such that $ OA \equal{} OB \equal{} OC' \equal{} 1$, and such that these three segments are perpendicular.

We know that the volume $ [OABC'] \equal{} \frac {1}{6}$ by considering $ \triangle OAB$ as the base. If we instead consider $ \triangle ABC'$ as the base, then we get $ \frac {1}{6} \equal{} \frac {1}{3} [ABC'] (1 \cdot \sin \alpha) \equal{} \frac {1}{3} \frac {\sqrt {3}}{2} \sin \alpha \implies \boxed{\sin \alpha \equal{} \frac {1}{\sqrt {3}}}$[/hide]
[hide="Solution 2"][i]Continued from hint[/i]

Let the vertexes be $ (\pm 1, \pm 1, \pm 1)$ and let the plane be $ x \plus{} y \minus{} z \equal{} 0$. The distance between the projections of $ (1,0,0)$ and $ (1,1,0)$ is the cosine of the angle.**

$ \left \{ (0,1,1), \; \left( \minus{} \frac {2}{\sqrt {6}},\frac {1}{\sqrt {6}}, \minus{} \frac {1}{\sqrt {6}} \right) \right \}$ is an orthonormal basis of the plane $ x \plus{} y \minus{} z \equal{} 0$. For each $ (1,0,0)$ and $ (1,1,0)$, we project onto both basis vectors individually and add to get the total projections.

For $ (1,0,0)$, we have $ (0,0,0)$ and $ \left( \frac {2}{3}, \minus{} \frac {1}{3}, \frac {1}{3} \right)$ as the two components, and thus its projection onto the plane is $ \left( \frac {2}{3}, \minus{} \frac {1}{3}, \frac {1}{3} \right)$.

For $ (1,1,0)$, we have $ \left( 0, \frac {1}{2}, \frac {1}{2} \right)$ and $ \left( \frac {1}{3}, \minus{} \frac {1}{6}, \frac {1}{6} \right)$, respectively, which sum to $ \left( \frac {1}{3}, \frac {1}{3}, \frac {2}{3} \right)$. The distance between these is $ \frac {\sqrt {6}}{9} \equal{} \sqrt {\frac {2}{3}}$, so this is $ \cos \alpha$. Thus, $ \sin \alpha \equal{} \frac {1}{\sqrt {3}}$

** [i]These two points form a segment of length 1 along one of the edges of the cube.[/i][/hide]

Yup, what I did was consider the plane that cut through the three vertices adjacent to, say, a vertex $ A$. If the vertex diagonally opposite from $ A$ is $ B$, then shifting the plane in the direction of $ AD$ doesn't change the angle at which any of the lines cuts the plane, so this must be the plane we are looking for.

Considering the (irregular) tetrahedron with vertex $ A$ and three adjacent vertices as the base leads to the answer $ \frac {1}{\sqrt {3}}$ as well.


correct but can you do a solution (even though it is obvious)

nice time for the cube
what method do you use?
like steps.

[hide="solution"]The volume of the second hole is 8 times as big as the first hole, so it would take them 16 hours.[/hide]

[quote="moogra"]If it takes two men two hours to dig a hole three meters long, three meters wide, and three meters deep, how long would it take the same two men to dig a hole six meters long, six meters wide, and six meters deep.[/quote]
[hide]$\frac{4}{27}=\frac{2x}{216}$

$x=16$[/hide]


AMC
USA(J)MO
USAMO
number theory
Diophantine equation

Okay, so you're in a text-based RPG. Post the game's response to the previous action, and then input an action of your own.

For example:

Input: GET YE FLASK

Output: You can't get ye flask.

Input: DO MATHS

Output: You do maths in your head, but while you were standing there thinking about maths an apple materialized above your head and bonked you on it. Ouch! You lose one hit point (HP: 9/10)

> LOOK

You eat the math paper. Its equations fill your head and you understand the concept of fluxions.
Knowledge level up 1

> Cast fireball at self

Ouch! 8 hp left! Makes a potion and drinks it. Hit points, 10/10. Alchemy level up. Elemental resistance (fire), level up.

>Turn in to ninja.

You surrender to the nearest ninja.
The ninjas execute you for your crime to ninjas.

         *** You have died ***

Would you like to start over, restore a saved position, or end this session of <insert the name of the game>? (Type RESTART, RESTORE, or QUIT):

    * restart
    * restore
    * quit

>quit.

Start new game. You use a hacking system.

...+

Alchemy + $ \infty$
Str + $ \infty$
Knowledge + $ \infty$
HP + $ \infty$
MP + $ \infty$
Dex + $ \infty$
Luck + $ \infty$
Wisdom + $ \infty$
Stamina + $ \infty$
Lvl + $ \infty$
Mastery + $ \infty$
Gain all equips
Money + $ \infty$
God mode on
Speed + $ \infty$
Charisma + $ \infty$
Archery + $ \infty$
Fishing + $ \infty$
Mining + $ \infty$
Farming + $ \infty$
Herb digging + $ \infty$
Merchandising skills + $ \infty$
Bargaining + $ \infty$
Food + $ \infty$
Ninja Status earned
Dragon Slayer status earned

5 hours later.....

Wood cutting + $ \infty$
Metal working + $ \infty$
Cooking + $ \infty$
Intelligence + $ \infty$
Math + $ \infty$

>You find an ASDF Dragon, the strongest monster in the game.

The dragon turns into a bunny.

>poke bunny

Finds penguin. Penguin executes awesomeness attack.

HP $ \minus{}\omega$

HP: 10

>Attack penguin

Penguin eats you.
GAME OVER

Restart
Restore
Quit

>Restart

Respawn next to penguin

>run away
>attack

>>>ATTACK<<<

Penguin holds a truce with you.

GAINED PENGUIN IN PARTY!

>Kick a broken door

The broken door shatters, revealing a dark, monstrous cave. You enter.

> look at wall

penguin explodes in stomach.
you die.

>restart
>restore
>quit

No I was kidding, you didn't really die. You wake up in a hospital

>COMPLAIN LOUDLY TO HOSPITAL WORKERS

The hospital workers complain about your presence and kick you out. Once outside, you feel the freedom of fresh air rush to greet you. It feels good.

To the east you see the hospital. To the north, south, and west, you see only plains of never-ending grass
[color=#00FF00]The obvious exists are north, south, west, and enter[/color]

> north

you find yourself in lala land
you see ice cream

<eat ice cream

You eat the ice cream and immediately get brain freeze.

HP: 9/10

> look around

You see an AoPS book!
Money - 60
Problem Solving + 100
Math + 100
Number Theory Element Discovered!
New attack learned: Destroy Diophantine Equation
Modular arithmatic skill learned: + 20 attack when facing monsters with number theory attribut
Special Attacks learned: Integer Crusher (insane damage to all monters on the field, especialy against number theory monsters, Meteor Modulo (one hit KO's any non-boss monster number theory monster. if it is a boss monster, it does insane damage)
Nerd level: + 50
Number Theory Apprentice Status earned
New elemental magic: Harness Number Theory: Allows you to use the power of number theory to pwn monsters.
New defensive magic: Ring of Modulos: Surrounds you in a force field of congruencies, making you very hard to damage. However, number theory slayers can do more damage to you.

>Oh no!!! A vicous lvl 173 USAMO jumps out at you!

[quote]>Oh no!!! A vicous lvl 173 USAMO jumps out at you![/quote]
Invalid action. You do not have the ability to summon a vicous lvl 173 USAMO. At current time, your summon level is 10 and the most powerful monster you can summon is the Dire AMC 10.

//For the record, actions are ACTIONS not another description of what happens. Also, you can't force the character to do anything ("You see a cave. You go in"). You have to tell the character to do it, and the next person will say whether your action was successful or not.
//Also, for the record, like a real text-based game you are allowed to be an extreme butt about correct grammar of actions, spelling, etc.

> FLIP TIME TURNER

Okay, so you're in a text-based RPG. Post the game's response to the previous action, and then input an action of your own.

For example:

Input: GET YE FLASK

Output: You can't get ye flask.

Input: DO MATHS

Output: You do maths in your head, but while you were standing there thinking about maths an apple materialized above your head and bonked you on it. Ouch! You lose one hit point (HP: 9/10)

> Wake up

You wake up in dungeon chained to a wall.

>>Escape chains

The chains escape from you. They succeed. You are drawn and quartered.

*** You have died ***

Would you like to start over, restore a saved position, or end this session of <insert the name of the game>? (Type RESTART, RESTORE, or QUIT):

* restart
* restore
* quit

>restore.

Wait, I see nothing wrong with saying "you see a cave. You go in." Most real text-based RPGs do this at some point.

You wake up in lala land and wonder how you got there.

> sleep

A chimpanzee LVL 20000 attacks you

>Run away
>Attack
>Go back to sleep

>>>Go back to sleep<<<

you wake up in a dungeon. you are not chained down. you find food at your feet on a plate. what do you do?

>explore
>stay put
>eat food
>commit suicide

*>eat food.

The food is [size=9] [color=white]not[/color][/size] poisonous. Woohoo.

A prisoner guard is nearby.

>steal keys
>punch guard
>scream
>insult guard
>do nothing

*>punch guard

//Okay, maybe not to the extreme of "you can't force the player to go into a cave" but you know what I mean.

You punch the prison guard.  He dies, and his corpse disappears helpfully leaving a pile of treasure on the floor. In addition, it appears he was guarding a flask.

>GET YE FLASK

The flask contains potion.

GAINED POTION.

>Escape Dungeon.

After you have got out of the dungeon, you find a computer, showing:

"Find out who is doing such evil things to you!"

>type http://www.mathlinks.ro/Forum/viewtopic.php?t=177263 [enter]

Finds this forum.

>post message
>smash computer
>ignore what the forum says and continue

*ignore what the forum says and continue.*

You encounter not one, not two, not three, but FIVE caves.

>go into cave 1
>look into cave 1
>go into cave 2
>look into cave 2
>go into cave 3
>look into cave 3
>go into cave 4
>look into cave 4
>go into cave 5
>look into cave 5
>just walk away

*>look into cave 2


advanced fields
advanced fields unsolved

can someone give me the link with normal math again: "a group is a category with one object in which every every arrow has a two-sided arrow" ?

so the object is $G$  and the arrows are for example $*_{a}$
then $G--*_{a}-\to G$ is an arrow from $G$ to $G$
and so you can form tuples $G \times G$ and associate them with $*_{a}$
and that should somehow represent the group-multiplication GxG -> G

Not sure what you mean by every arrow has a two-sided arrow, but it is easy to see that a group is a category with one object in which all arrows are isomorphisms. The group operation will be composition of arrows, which is always defined as there is only one object. The group axioms are trivially satisfied by the category axioms and the fact that all morphisms are isomorphisms.

On the other hand suppose you have a group $G$ then we can make $G$ in to a category $\mathcal G$ by taking the object of $\mathcal G$ to be $\emptyset$ the empty set, and the morphism are elements of $G$ with composition defined by the group operation.


The point (1,2) has the distance______  from the line defined by
  y=-(x/3)+1

[hide]Shortest DIstance is measured by perpendiculars[/hide]

Given the point $(x_{1},y_{1})$, and the line in the form $Ax_{0}+By_{0}+C = 0$, the shortest distance is:

$\frac{|Ax_{1}+By_{1}+C|}{\sqrt{A^{2}+B^{2}}}$


Let $D$, $E$, $F$ be three points on sides $BC$, $AC$, $AB$ of $\triangle{ABC}$ such that they bisect the perimeter of the triangle. Then $AD$, $BE$, $CF$ intersect at one point $J$.

How do I call this point in English?

Points bisect the perimeter? I guess you mean that the lines AD, BE, CF bisect the perimeter of triangle ABC. In this case, the point J is called the [i]Nagel point[/i] of triangle ABC. See http://www.mathlinks.ro/Forum/viewtopic.php?t=58867 , http://mathworld.wolfram.com/NagelPoint.html and http://faculty.evansville.edu/ck6/tcenters/class/nagel.html .

  Darij


geometry
incenter
geometric transformation
homothety
circumcircle
parallelogram
trigonometry

Let triangle $ABC$ with the incenter$I$. The incircle o triangle touches $BC$ at $X$.From $X$ , $I$ draw $XG//AI$ and it cuts incircle at G , AG cuts circum of triangle $ABC$ at$P$.Prove that $GIXP$ be circle.

[img]6952[/img]
Interesting problem... I tried to locate the center of homothecy of the incircle to the circle through $pigx$ but it wasn't a point I could recognize... it wasn't the excenter let me tell you

It is my problem and my solution use tri.
Please , you post your idea! :lol:

First, define orange circle as circumcircle GIX. Thus we only need to show that P is on the circumcircle.

Using algebra or method of choice, we must prove $AI=GX$
Then AIGX is parallelogram, so AGI is isosceles.
We must show that OI extended is perp. bisector of AP which only requires showing that OI perpendicular to AP (for $OA=OP$)
but this follows clearly since OI perpendicular to IX parallel to AG = AP
Thus AIP is isosceles
Thus $\triangle AGI \sim \triangle AIP$
Thus $AI^{2}=AG AP$
By converse of power of a point, P is on the circle and it is done...

Now the only tough part is the first step, showing AI=XG
but there is most likely a nice and simple proof for this

[quote="me@home"]Using algebra or method of choice, we must prove $AI=GX$[/quote]
[color=darkblue]Sorry, Me@home, but [b][u]Darij G. is right[/u][/b] because $AI=GX\Longrightarrow AG=r$ and $AG\perp BC$ $\Longrightarrow$ the middlepoint $M$ of the side $[BC]$ belongs to the line $GI$ $\Longrightarrow$ the quadrlateral $AGMO$ is parallelogram $\Longrightarrow$ $OM=r$ $\Longleftrightarrow$ $R|\cos A|=r$ $\Longrightarrow$ $1+|\cos A|=\cos A+\cos B+\cos C$, what is falsely in generraly.[/color][hide="Hint."][color=darkred]The rays $[AG$ and $[AX'$are isogonals in the angle $\widehat{BAC}$, where $X'\in (BC)$ and $CX'=BX$, i.e. the ray $[AG$ is the isogonal line of the Nagel's $A$- line in the angle $\widehat{BAC}$. Construct the symmetrical figure of the given figure w.r.t. the bisector of the side $[BC]$ (the perpendicular on the middle of the segment $[BC]$) a.s.o. It is sufficiently ? I think it is O.K.[/color][/hide]

Thanks for pointing out my mistake  :blush: it is much harder than I thought at first

[quote="treegoner"]Inversion and nine point circle of the triangle $XYZ$ where $Y, Z$ are the points of tangency of the incircle with $CA$ and $AB$.  Notice the orthocenter of $XYZ$.[/quote]
you have soln
can you post it

IX cuts the incircle again at D
AD cut BC at E
Angle AIG=Angle IGX (AI//GX) but Angle GID=2Angle IGX so Angle GIA=Angle DIA
Thus AI is the perpendicular bisector of GD. (1)
So Angle AGI=Angle ADI=Angle AEB+90 (2)
Because of (1) Angle BAP=Angle EAC
but Angle BPA=Angle ECA so Triangle BAP and Triangle EAC are similar
Thus BP/EC = AB/AE (3)
Also BX=EC (a beautiful result but not difficult to prove ) thus BP/BX=AB/AE
otherwise Angle PBX=Angle PAC=Angle BAE(because of (1) )
So that Triangle PBX and Triangle BAE are similar
Thus Angle BXP=Angle AEB
so Angle IXP=Angle AEB+90 (4)
From (2) and (4) Angle AGI=Angle PXI
So Angle PXI+Angle PGI=Angle AGI+Angle PGI=180
The result is P,X,I,G are cyclic

Also BX=EC (a beautiful result but not difficult to prove ) thus BP/BX=AB/AE

Does it have to do with the fact:

$M$= midpt $BC$m $AE$ intersect $OM$ is $Y$, then $IX=YM$


Admins, on the google chrome classroom thingy, it says:
When the [b]classroomw[/b] window is blocked, you should see the following at the bottom of your Chrome browser window.


Let n be a positive integer.
Prove that there does not exist any positive integer m such that :

             [size=150]    3n^2+3n+7=m^3[/size]

Well. Here are all cubic residues mod 9:
$0^3 \equiv 0$
$2^3 \equiv 8$
$3^3 \equiv 0$
$4^3 \equiv 1$
$5^3 \equiv 8$
$6^3 \equiv 0$
$7^3 \equiv 1$
$8^3 \equiv 8$
$3n^2 + 3n + 7 = 3n(n+1) + 7 = m^3$

Now, suppose that $n=3k$ or $n=3k+2$ for some integer $k$.
Then: $3n(n+1) + 7 \equiv 7$ mod 9 but 7 isn't a cubic residue modulo 9 - contradiction.

Second case: $n=3k+1$ for some integer $k$.
Then: $3n(n+1) + 7 \equiv 4$ mod 9 (because $3n(n+1) + 7 = 27k^2 + 27k + 13 = 9(3k^2 + 3k + 1) + 4$) but 4 isn't a cubic residue modulo 9 - contradiction and the solution is complete.


The key result is that $ x^5 \equiv 1 \bmod h(x)$, so you can reduce all the exponents $ \bmod 5$.

t0rajir0u means that $ h(x)|x^5 \minus{} 1$ since $ x^5 \minus{} 1 \equal{} (x\minus{}1)h(x)$

$ f(x) \equiv (x^{2} \plus{} x^{3} \plus{} x \plus{} x^{4}) \plus{} 9 \plus{} 2000  \equiv ( \minus{} 1) \plus{} 2009 \equiv 2008 \bmod (h(x))$


Thanks for LynnelleYe.

We know an ordinary one has 12 equivalence classes, how many do we have if each side is painted a different picture instead of color?

The only thing that is new then is the orientation of the middle one pieces.
But you can get that all of them except one oriented the right way, so there are four possibilities how that last one can be, so there are $4\cdot 12 = 48$ equivalence classes, I think.

Do you mean the Rubik's Cube? If so there's a thread about it in the Games and Fun Factory, I think.

[quote="Buffalo"]each side is painted a different picture instead of color[/quote]
I've seen a Rubik's cube with flags of six countries on it. :)

I try to make flags out of the colors on my Rubik's cube sometimes.. the flags end up all pixely. Kinda hard to do some flags with only 3x3..

I saw a Rubik's cube at Chuck-E-Cheese's a long time ago.. it had Chuck-E-Cheese people on it. Their faces or something.
I also saw one at my friend's house with some weird things on it.. I don't remember exactly, but one side had swords, and another had ghosts or something.. simple drawings. And like 2 middles were always facing the wrong way.

[quote="ZetaX"]But you can get that all of them except one oriented the right way, so there are four possibilities how that last one can be, so there are $4\cdot 12 = 48$ equivalence classes, I think.[/quote]
I believe there are more, in order to change one, I have to at least change another two of the six center pieces.

I didn't say 'change', I said 'orientate' (rotate)!
You can rotate two (not more) centers independently from all other pieces, so there are $48$ or less possibilities.
But exactly, there are only $24$ congruence classes, because you can also rotate a single center by $180$ degrees.


group theory
abstract algebra
search
superior algebra
superior algebra unsolved

Let $ G$ be a group, $ H$ be the set of elements of $ G$ which are of finite order.
If $ H$ is finite, show that it is a subgroup of $ G$.
What if $ H$ is not finite?

It's most likely classical but I couldn't find it on this forum

[quote="JC_math"]Let $ G$ be a group, $ H$ be the set of elements of $ G$ which are of finite order.
If $ H$ is finite, show that it is a subgroup of $ G$.[/quote]

[s]This is not true. owk[/s]

EDIT: I do not understand what I was thinking, the counter-example I had in mind does not work for extremely obvious reasons. owk

Sure it is.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=189693973&t=1359[/url]

Thanks grobber,
It surely answers my first question
What about the second one?

Work with $ 2\times 2$ matrices in $ GL(2,\mathbb{Z})$. You will have no problem in finding counterexamples there.

Any free product of two non-trivial torsion groups will do for a counterexample.

$ O(n)$ is another example, it is generated by reflections. owk


Hi my dears...

i want to know how many houres of the day, u study math.

or if not , what do u do?

about 2 hours studing and 1.5 to 3 hours private lessons with mum or dad this year

Not enough!  Actually, I am sometimes surprised I even make it.

When I come back from classes (or do you add that up with too  ;) - in practice like 2.5 hours?

[hide]I think you mean 'hours' instead of 'houres', I hope you don't mind but I find myself improving by being corrected.[/hide]

oh,,,

i remebred a question.

do u know Astrid lindgern ?


After the failure of ctc I've decided to start it up again. Now though we have no almighty president instead we will have 3 groups:

moderators
regulars
watchers

the ratio of mods-watchers-regulars will be 1-7-1 though this is subject to change. It works like this:

Regulars are just regular people eligible in a tournament. Mods are those who create them and maintain them.
(something like how a mafia game starts) Watchers though are a special group not known to anyone but to themselves and a moderator of Aops (probably izzy[because I trust her]). Their job is to watch the mods and contestants to see their behavior. If a tourney mod is abusing his/her privileges he/she can be revoked by izzy (or whoever is in charge then), but only if say 3/4 of them agree that he/she is abusing it. Then a messenger (one of the watchers) is sent to tell izzy(or person in charge) and report whoever is abusing privileges. Then the abuser is revoked and replaced by someone who has been helpful and regularly comes. Watcher jobs are assigned by the person(s) in charge. A watcher is revoked if he she is not participating in the club for over 5 days without notice. If so he/she is revoked and replace by someone who the aops mod(s) choose.

Roles:

Regular: a person eligible for a tourney
Tourney Mod: Someone who creates tournies and maintains them. They may lose their job if their privileges are abuse. Their privileges: boot someone out of a tourney for spamming or not being active. Creates 1-2 tournies at a time and maintains them.
Watchers: secret group of people "hired" to watch mods and regulars. Revoked if inactive for more than 5 days without notice to Aops mod(s).
Aops mod(s): 1-2 real mods whose job is to assign the roles of watcher and tourney mod. Aops mod can only be taken from office by a petition by most of the CTC community or if they resign and give the job to  a tourney mod or a fellow real aops mod.

These rules are subject to change as needed. I will be sending out pms to people I think can take this and then i will post the Aops mod people here once i have chosen.

Actual participation is not needed during school weeks (for the obvious reasons) and the actual tournies will probably not be held until next summer. Sign ups and/or complaints to make or destroy rules start now.

Edit: i have pmed Klebian, Isabella2296, and Vallon22 for chances as aops mods. any others you guys suggest?

OK. Vallon and Izzy you guys are the aops mods for now. Ill pm you each when we have enough people to start 1-2 tournaments.

Stop thinking about it now.  When will this start, like 7 months from now?  Mods, lock this.

fine ernie but i will bring this up in may

edit: whoops forgot to mention that it could start in december during winter break too..

Wait so why cant we just restart CTC when summer comes? The infrastructure is all in place.

look i was just planning ahead because i knew people would have complaints. (or at least i thought so....)

However, CTC's main issue has been solved.  I don't think it's necessary to start it up again.

the old CTC was falling apart
people were quitting
this new one might work out, but time will tell

i suggest this be kept open over thanksgiving break and into winter break
then we have a long period of school, in which if there is a lot of spam in this thread, it will be locked and reopened during the summer
if not, it will remain open for some discussion as long as it does not become spammy

other mods plz review this and post whether and why you agree/disagree to it

We can give it a try.

If there's some spam, we can just delete it but if we're deleting posts every day something's wrong.

Why bother? CTC works, this may or may not. If it ain't broke dont fix it.

Yes but CTC is independent of AOPS mods altogether. Besides in this case you are giving all the power to the mods, which implies you had the power in the first place.

[quote="mathblitz"]MEWTO this IS CTC. I'm just building the foundation for winter break with the time left for Thanksgiving break. Your CTC pretty much gave you all the power. Now I'm transferring that to the Aops mods who shouldn't be unfair because then why would they be mods?[/quote]

I have to agree with this.  However, I think that ALL REGULARS should get a chance to have a say.  In the old one, a few select board members + mewto55555 (with all the power) got to do w/e they wanted.

Yes I did have the power for about 15 min while I decided whether or not to hold elections or give it to someone who was responsible. I decided that an election would be too messy and take too long and maybe some threats would be exchanged. For example in the last CTC election when I wanted to run for both President and Vice and then stop when I was elected one you threatened that if you won the president election you may not stand behind me for vice. What do we need them for anyway? Izzy and Vallon each are mods showing that levans like the posts they posted and wanted to encourage that. They can edit spam or delete it. They have already proven themselves as responsible so they fit the description needed very well. But what are you suggesting mewto? do you want to host elections so you might have a chance as president? This goes onto another thing: I hope that the power is split among multiple people. 2-3 though my first post says 1-2. You say that CTC was independent of mods altogether, but the goverment power had rankings and that most powerful one was controlled by 1 person. Not a good idea.

@Aime: we could actually have representatives. we could have 4-6: 2-3 Aops mods (reason is above) 2-3 elected regulars?

stop inactive tourneys from starting. Active tourneys though will hopefully be started again during winter break.

Mafia-like organization requires http://www.mathlinks.ro/groupcp.php?g=47077 but I wonder how many people would really do their job (I guess less than 5).

ok..
umm...
why don't we just make 4 tournies at any given time, and a person can join at most 2 of them

all agreed?

:o What if a person has never been "modkilled" in any tourney and is omniactive :?:

If we have $ 4$ tournies at a time, and a person is in at most 2 of them, KEEP THE TOURNAMENTS SMALL.  Under 16 players, hopefully.

I sign up for first spot to host a tourney.

Tourney 1: AIME15

What I'm saying I guess is, this isn't like Mafia in the respect that you have to play your game and know youer role and watch everyone. If the mod sends out prod pms, and the player is active, what reason is there to not be able to be in all games?

good point.....
i guess it's just a precaution we put because of things that have happened previously
if we get good participation in the tournies, we can change the rule later

oh blimey...this again?

All this winter ctc madness, I think I'll create Anti-CTC, the goal is to stop CTC, anyone can join with the exceptions of those who can't, and only Anti-CTC members can stop CTC, and ignore all anti-ctc activities done by non-Anti-CTC member, We will have 8 elected board members and a vice president, I'll be the president, and the remaining 2 of you can post spams.

So who's in?

If this is orginized, and doesn't fall apart, and someone responsable leads this, I'll join it.

Ernie suggested that I make a ladder format.  That is what I'll do, after a mini round-robin.  When should I start?

when summer starts and when there are signups :D . DOn't think many people will be active during the school year.

[quote="jjx1"]I think we should have neither. I don't get all of this. just put a limit on number of tournies at a time. Also jsut don't join a tourney unless you know you'll be on a lot on ftw.[/quote]

Yeah, and keep track of inactives, and if someone signs up and doesn't play a couple of tournies, ban him from every other. Simple.

this talk is really getting on my nerves...

i suggest limit on tournies is x (where x is a random number between 5 and 7)
bugi's idea on keeping track of the inactives
and maybe something else i'll think of later

i have locked this thread once before, due to arguing and everything
i'm getting the feeling it's happening again
so unfortunately, this thread is LOCKED

if you have any suggestions on tournies or complaints about this thread being locked or something else, feel free to pm me


geometry rectangle search trapezoid circumcircle parallelogram ratio

I think that this problem is also true in its more general configuration of similar rectangles, instead of squares.

I will search my notes and if I am not mistaken, I will post here next time the synthetic ( but difficult ) solution I have in mind.

Kostas vittas.

We can solve this problem, in its more general configuration of similar rectangles instead of squares, as a direct application of the below two Lemmas:

[b][size=100][color=DarkBlue]LEMMA 1. – A triangle $ \bigtriangleup ABC$ is given and le $ BCAbAc,$ $ ACBaBc,$ $ ABCaCb$ be, three similar rectangles erected outwardly to it, on its sidesegments $ BC,$ $ AC,$ $ AB$ respectively. Through $ Ba,$ $ Bc,$ we draw two lines perpendicular to $ AbBa,$ $ BcCb$ respectively and we denote as $ Q,$ their intersection point. Prove that the line segment $ BQ,$ passes through the point $ P,$ as the center of the rectangle $ BcBaB'aB'c,$ congruent to $ ACBaBc.$[/color][/size][/b]

[b][size=100]PROOF OF THE LEMMA 1.[/size][/b]

Through vertices $ A,$ $ C$ of $ \bigtriangleup ABC,$ we draw two lines perpendicular to $ BcCb,$ $ AbBa,$ respectively.

It is easy to show that these lines are coincided with the medians of $ \bigtriangleup ABC,$ through $ A,$ $ C,$ respectively and so, intersect each other at the centroid $ G$ of $ \bigtriangleup ABC.$

$ ($ We denote as $ A',$ $ B',$ $ C',$ the orthogonal projections of $ A,$ $ B,$ $ C$ respectively, on the line segment $ BcCb$ and from cyclic quadrilaterals $ ABCbB',$ $ ACBcC',$ we have that $ \angle AB'C' \equal{}\angle ABCb \equal{}\angle ACBc \equal{}\angle AC'B'$ and so, $ AB' \equal{} AC'.$

Hence, the line segment $ AA',$ as the midparallel of the trapezium $ BB'CC',$ is coincided with the median of $ \bigtriangleup ABC,$ through vertex $ A$ $ ).$

We denote as $ M,$ the midpoint of side segment $ AC$ and as $ N,$ the center of the rectangle $ ACBaBc.$

It is easy to show that the points $ M,$ $ N,$ $ P,$ are collinear.

Because of $ GA\parallel QBc,$ $ GC\parallel QBa$ and $ AC\parallel \equal{} BcBa,$ we conclude that the triangles $ \bigtriangleup GAC,$ $ \bigtriangleup QBcBa$ are congruent and so, we have that $ GQ\parallel \equal{} ABc$ $ ,(1)$

It is easy to show that $ MP \equal{}\frac{3(ABc)}{2}$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ MP \equal{}\frac{3(GQ)}{2}$ $ ,(3)$

From $ (3),$ because of the collinearity of $ B,$ $ G,$ $ M$ and $ MP\parallel GQ,$ we conclude that the points $ B,$ $ Q,$ $ P,$ are collinear

$ ($ because of $ BM \equal{}\frac{3(BG)}{2}$ $ )$ and the proof of the [b][size=100]Lemma 1[/size][/b], is completed.

[b][size=100][color=DarkBlue]LEMMA 2. – In the configuration of the Lemma 1, we denote as $ R,$ the intersection point of the line segments $ AcBc,$ $ BaCa.$ Prove that the line segment $ BQ,$ passes through the point $ R.$[/color][/size][/b]

$ \bullet$ About the [b][size=100]Lemma 2[/size][/b], we can say that it is also true, in its even more general configuration of arbitrary rectangles, erected on the side segments of $ \bigtriangleup ABC,$ instead of similar ones or squares.

We remind that it has already been posted as a separate problem, in the topic

http://www.mathlinks.ro/Forum/viewtopic.php?t=108818, which unfortunately has not been discussed yet.

By the way of this [b][size=100]Omid Hatami’s problem[/size][/b], I think it is a good time to post there, at least, the proof by a friend of mine [b][size=100]Nikos Kyriazis[/size][/b], the one of two synthetic solutions I have in mind.

$ \bullet$ Because now, of the collinearities of $ B,$ $ Q,$ $ P$ and $ B,$ $ R,$ $ Q,$ we conclude that the line segments $ BP,$ $ CaBa,$ $ AcBc$ are concurrent at one point and the proof of the proposed problem may be completed.

Kostas Vittas.

PS. Although it is a difficult way to solve this problem, however, it is based on two Lemmas, useful ( I think ) to all of us in the future.

I am looking forward in another sorter and simpler proof and I am interesting if has already been arisen such that a solution, during the contest.

Also I am interesting, if the official solution of this problem is available.

[quote="Virgil Nicula"]Is very easy if we"ll use the complex numbers ![/quote]

Although I am not happy because of the "complex numbers" is so much to me, however, I like very much to try for the solutions of the problems, based on classical geometry tools comprehensible to the mid mind and specially, when ( I think ) the only known solutions about some problems, are based on extensive ( to me ) mathematical knowledge.

Best regards, Kostas vittas.

[b]Vittasko[/b], the your extension and the its synthetical proof are very nicely and interestingly.
You are right ! But usually, the first attempt is with the strong instruments and afterwards find a fine jewel - a synthetical proof. Thank you for this present !
And I like the "butterfly" (synth.) and I don't wish to kill it with the "atomic bomb"  (strong instr.) !

[quote="khashi70"]I know a very short and nice solution by Ceva's theorem :)[/quote]
I am eager for such a solution ( when I find a complicated proof of a problem, I am almost sure that there is another clever one, by a simpler way, which someone else younger man, has already found or will find later ).

Dear [b][size=100]Khashi70[/size][/b], would you like to share with us your solution?

Thank you, Kostas vittas.

Dear all my friends.

I would like to post here the solution I have in mind, of the problem "An interesting collinearity",

http://www.mathlinks.ro/Forum/viewtopic.php?t=108818

which It is used as [b][size=100]Lemma 2[/size][/b], for the proof of the extension of the problem presented by [b][size=100]Omid Hatami.[/size][/b]

$ ($ I thought that I had find, one year ago, in [b][size=100]Paul Viu’s[/size][/b] web page, the problem with the erected squares, but now I can’t find it. Probably I am mistaken, but If someone has a reference about it, somewhere else, kindly please inform me $ ).$

[b][size=100][color=DarkBlue]PROBLEM (An interesting collinearity).

Three rectangles $ ABDE,$ $ ACFH,$ $ BCKL,$ are erected on the side segments $ AB,$ $ AC,$ $ BC$ respectively, of a given triangle $ \bigtriangleup ABC,$ outwardly to it. If $ M$ is the intersection point of the line segments $ EK,$ $ HL$ and $ N,$ the one of the lines through $ K,$ $ L$ and perpendicular to $ KF,$ $ LD$ respectively, prove that the points $ A,$ $ M,$ $ N,$ are collinear.[/color][/size][/b]

[b][size=100]PROOF.[/size][/b]

[b][size=100]1).[/size][/b] – At first, we can say that the problem is true, in the particular case when the line segment $ EH,$ is parallel to $ BC.$

[b][size=100]1a).[/size][/b] - $ ($ see the figure [b][size=100]t=164548(b)[/size][/b] $ ),$ We will prove that the point $ N\equiv EK\cap HL,$ lies on the altitude of $ \bigtriangleup ABC,$
 through vertex $ A.$

We denote the points $ B'\equiv AH\cap BL$ and $ C'\equiv AE\cap CK.$

It is easy to show that $ BB' \equal{} CC'$ $ ($ From the cyclic quadrilaterals $ ABCC',$ $ ACBB',$ we have that the points $ B',$ $ C',$ are the antidiametric (=diametric opposite) points of $ B,$ $ C$ respectively, with respect to the circumcircle $ (O),$ of $ \bigtriangleup ABC$ $ ).$

From this fact we have $ B'C'\parallel \equal{} BC\parallel \equal{} KL.$

We see now, that the triangles $ \bigtriangleup AB'C',$ $ NLK,$ are perspective because of the collinearity of the points $ E\equiv AC'\cap NK$ and $ H\equiv AB'\cap NL$ and $ S\equiv B'C'\cap LK,$ where $ S,$ is the point at infinity.

 $ ($ we say that two parallel lines, are intersected at infinity $ ).$

So, by [b][size=100]Desarques’s theorem[/size][/b], we conclude that the line segments $ AN,$ $ B'L,$ $ C'K,$ are concurrent at one point and then we have

$ AN\parallel B'L\parallel C'K.$

Hence, $ AN\perp BC$ and the proof of part [b][size=100](1a)[/size][/b], is completed.

Kostas vittas.

[b][size=100]1b).[/size][/b] - $ ($ see the figure [b][size=100]t=164548(c)[/size][/b] $ ),$ We will prove now, that the point $ N',$ as the intersection point of the lines through $ K,$ $ L$

and perpendicular to $ KF,$ $ LD$ respectively, lie also on the altitude of $ \bigtriangleup ABC,$ through vertex $ A.$

We denote the point $ S\equiv DE\cap FH$ and we will prove that $ EH\parallel DF.$

From $ SE\parallel AB$ and $ SH\parallel AC$ and $ EH\parallel BC,$ we have that the triangles $ \bigtriangleup SEH,$ $ \bigtriangleup ABC,$

are similar and then $ \frac{SE}{AB}\equal{}\frac{SH}{AC}$ $ ,(1)$

But $ AB \equal{} ED$ $ ,(2)$ and $ AC \equal{} HF$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac{SE}{ED}\equal{}\frac{SH}{HF}$ $ \Longrightarrow$ $ EH\parallel DF.$

That is, in the particular case of $ EH\parallel BC,$ we have also $ DF\parallel BC.$

$ \bullet$ We denote as $ Q,$ the orthogonal projection of $ N',$ on $ KL$ and we will prove that the line segment $ NQ,$
passes through vertex $ A$ of $ \bigtriangleup ABC.$

Through $ B,$ $ C,$ we draw two lines and parallel to $ LD,$ $ KF$ respectively and we denote as $ D',$ $ F',$
their intersection points, from the line segment $ DF.$

Also, through $ B,$ $ C,$ we draw two lines and perpendicular to $ BD',$ $ CF'$ respectively and we denote as $ N'',$ their intersection point.

It is easy to show that $ N'N''\perp BC$ $ ($ From $ N''B\parallel N'L$ and $ N''C\parallel N'K$ and $ BC\parallel \equal{} LK,$

we have the congruence of the triangles $ \bigtriangleup N''BC,$ $ \bigtriangleup N'LK$ and so, from the parallelograms $ BN''N'L$ and $ CN''N'K,$

 we conclude tha $ N'N''\equiv QN'N''\perp BC$ $ ).$

$ \bullet$ In order to prove that the line segment $ N'N''$ passes through $ A,$ it is enough to prove that $ A'\equiv A'',$ where $ A'\equiv AB\cap N'N''$ and $ A''\equiv AC\cap N'N''.$

Let $ B',$ $ C'$ be, the intersection points of the line segment $ DF,$ from the line segments $ BL,$ $ CK,$ respectively.

From $ D'B\parallel DL$ $ \Longrightarrow$ $ \frac{B'B}{BL}\equal{}\frac{B'D'}{D'D}$ $ ,(4)$

From $ F'C\parallel FK$ $ \Longrightarrow$ $ \frac{C'C}{CK}\equal{}\frac{C'F'}{F'F}$ $ ,(5)$

From $ B'C'\parallel \equal{} BC\parallel \equal{} LK$ $ \Longrightarrow$ $ \frac{B'B}{BL}\equal{}\frac{C'C}{CK}$ $ ,(6)$

From $ (4),$ $ (5),$ $ (6)$ $ \Longrightarrow$ $ \frac{B'D'}{D'D}\equal{}\frac{C'F'}{F'F}$ $ ,(7)$

From the similar right triangles $ \bigtriangleup B'BD,$ $ \bigtriangleup PBA'$ $ ($ $ A'\equiv AB\cap N'N''$ $ )$ $ \Longrightarrow$ $ \frac{B'D'}{D'D}\equal{}\frac{PN''}{N''A'}$ $ ,(8)$

$ ($ because of the perpendicularities $ AB\perp BD$ and $ N''B\perp BD'$ and $ PB\perp BB',$ where $ P\equiv BC\cap N'N'',$

we have that $ \angle PBN'' \equal{}\angle B'BD'$ and $ \angle N''BA' \equal{}\angle D'BD$ $ ).$

By the same way, from the similar right triangles $ \bigtriangleup C'CF,$ $ \bigtriangleup PCA''$ $ ($ $ A''\equiv AC\cap N'N''$ $ )$ $ \Longrightarrow$ $ \frac{C'F'}{F'F}\equal{}\frac{PN''}{N''A''}$ $ ,(9)$

From $ (7),$ $ (8),$ $ (9)$ $ \Longrightarrow$ $ \frac{PN''}{N''A'}\equal{}\frac{PN''}{N''A''}$ $ \Longrightarrow$ $ A''\equiv A'\equiv A$

Hence, the point $ N,$ lies on the altitude of $ \bigtriangleup ABC,$ through vertex $ A$ and the proof of the part [b][size=100](1b)[/size][/b], is completed.

Kostas Vittas.

[b][size=100]2).[/size][/b] We return now, in to our problem in its general configuration $ ($ see the figure [b][size=100]t=164548(d)[/size][/b] $ ).$

Through the point $ H,$ we draw a line parallel to $ BC,$ which intersects the line segment $ AE,$ at one point so be it $ E'.$

The line through $ E',$ and parallel to $ AB,$ intersects the line segment $ BD,$ at one point so be it $ D'.$

We denote as $ N,$ the intersection point of $ HL,$ $ E'K$ and as $ N',$ the one of the lines through $ K,$ $ L$

and perpendicular to $ KF,$ $ LD',$ respectively.

It has already been proved that the points $ N,$ $ N',$ lie on the altitude of $ \bigtriangleup ABC,$ through vertex $ A.$

We denote as $ M',$ the intersection point of the line segment $ AM,$ $ ($ $ M\equiv EK\cap HL$ $ ),$ from the line through $ L$
and perpendicular to $ LD.$

It is enough to prove that the point $ M',$ lies on the line segment $ KN'\perp KF.$

$ \bullet$ Through $ L,$ we draw the line $ AX,$ perpendicular to $ AL$ and it is easy to see that the homologous rays of the pencils

$ L.AN'M'K$ and $ L.XD'DB,$ form equal angles, because of their perpendicularities.

$ ($ $ AL\perp LX$ and $ N'L\perp LD'$ and $ M'L\perp LD$ and $ KL\perp LB$ $ ).$

Then, we conclude that these pencils have equal cross ratios, that is we have that $ (L.AN'M'K) \equal{} (L.XD'DB)$ $ ,(10)$

The line segment $ BD,$ intersects the pencil $ L.XD'DB$ and so, we have that $ (L.XD'DB) \equal{} (P,D',D,B)$ $ ,(11)$
where $ P\equiv BD\cap LX.$

Let $ S$ be, the intersection point of the line segments $ AK,$ $ HL.$

As the pencil $ A.LNMS,$ is intersected from the line segment $ HL,$ we have $ (A.LNMS) \equal{} (L,N,M,S) \equal{} (K.LNMS)$ $ ,(12)$

But the line segment $ AE,$ intersects the pencil $ K.LNMS$ and so, we have that $ (K.LNMS) \equal{} (Q,E',E,A)$ $ ,(13)$
where $ Q\equiv AE\cap KL.$

$ \bullet$ It is easy to show that the points $ P,$ $ Q,$ lie on the circumcircle $ (W),$ of the triangle $ \bigtriangleup ABL,$ because of the cyclic quadrilaterals $ ABLP$ and $ ABLQ.$

So, we conclude that the points $ P,$ $ Q,$ are the antidiametric (=diametric opposite) points of $ A,$ $ B$ respectively and then the quadrilateral $ ABPQ,$ is a rectangle.

After this result, it is easy to show that $ PQ\parallel \equal{} BA\parallel \equal{} DE\parallel \equal{} D'E'$ and hence, $ (P,D',D,B) \equal{} (Q,E',E,A)$ $ ,(14)$

From $ (10),$ $ (11),$ $ (12),$ $ (13),$ $ (14)$ $ \Longrightarrow$ $ (L.AN'M'K) \equal{} (A.LNMS)$ $ ,(15)$

From $ (15)$ and because of the pencils $ L.AN'M'K,$ $ A.LNMS,$ have the line segment $ AL,$ as their common ray,

we conclude that the points $ N'\equiv LN'\cap AN$ and $ M'\equiv LM'\cap AM$ and $ K\equiv LK\cap AS,$ are collinear.

In other words we conclude that the points $ A,$ $ M,$ $ N,$ are collinear and the proof is completed.

$ \bullet$ This proof is dedicated to [b][size=100]Nikos Dergiades.[/size][/b]

Kostas vittas.

Thank you dear [b]Kostas[/b], for your very nice and hard solution for [b]Lemma 2[/b]  :D . It's really interesting and new to me.

And dear [b]Khashi70[/b], would you like to post your solution here?

It's my solution , I prove this problem by Trigonometric form of Ceva's theorem :
but first notice that : $ \triangle AB_{a}C_{a}$ is similiar to $ \triangle B_{c}AB$ and if $ \angle C_{a}B_{a}A \equal{} x$ then $ \angle BB_{a}A \equal{} x$
and similiary if $ \angle BB_{a}C \equal{} y$ then $ \angle CB_{c}A_{c}\equal{} y$ .
we have $ \frac{\sin{PBB_{c}}}{\sin{PB B_{a}}}\equal{}\frac{\sin{BB_{c}P}}{\sin{BB_{a}P}}$
and we have $ \angle BB_{c}P \equal{} 135\minus{}x$ , $ \angle BB_{a}P \equal{} 135\minus{}y$ thus $ \frac{\sin{PBB_{c}}}{\sin{PB B_{a}}}\equal{}\frac{\sin{(135\minus{}x)}}{\sin{(135\minus{}y)}}$.
$ \angle BB_{c}A_{c}\equal{} 45\minus{}x\minus{}y$ , $ \angle A_{c}B_{c}B_{a}\equal{} 45\plus{}y$ , $ \angle C_{a}B_{a}B_{c}\equal{} 45\plus{}x$ and $ \angle BB_{a}C_{a}\equal{} 45\minus{}x\minus{}y$
now u can check the Trigonometric form of Ceva's theorem :)

[quote="Omid Hatami"]Let $ ABC$ be a triangle. Squares $ AB_{c}B_{a}C$, $ CA_{b}A_{c}B$ and $ BC_{a}C_{b}A$ are outside the triangle. Square $ B_{c}B_{c}'B_{a}'B_{a}$ with center $ P$ is outside square $ AB_{c}B_{a}C$. Prove that $ BP,C_{a}B_{a}$ and $ A_{c}B_{c}$ are concurrent.[/quote]
[b]Solution.[/b] Denote $ M\equal{}BB_c\cap B_aC_a$, $ N\equal{}BB_a\cap B_cA_c$ and $ S\equal{}C_aB_a\cap A_cB_c$. Since two triangles $ ABB_c$ and $ AC_aB_a$ are similar, we conclude that $ \angle AB_cB\equal{}\angle AB_aC_a$, which implies that four points $ A$, $ B_c$, $ B_a$ and $ M$ are cyclic. Similarly, $ N$ also lies on this circle, let it is $ \Gamma$.

Let $ K\equal{}MN\cap B_cB_a$. We have $ PB_c$, $ PB_a$ are two tangents of $ \Gamma$, which follows that $ B_cB_a$ is the polar of $ P$ relative to the circle $ \Gamma$. On the other hand, we have $ K$ lies on $ B_cB_a$, it equivalent to $ P$ lies on the polar of $ K$ relative to $ \Gamma$ $ (1)$.

Let's consider the cyclic quadrilateral $ MNB_aB_c$, we have $ K$ and $ B$ are the intersections of its opposite sides, and $ S$ is the intersection of its diagonals. Therefore the polar of $ K$ relative to $ \Gamma$ is the line $ BS$ $ (2)$.

From $ (1)$ and $ (2)$ we conclude that $ B$, $ S$ and $ P$ are collinear, it equivalent to $ BP$, $ C_aB_a$ and $ A_cB_c$ are concurrent, as desired.

[quote="April"][quote="Omid Hatami"]Let $ ABC$ be a triangle. Squares $ AB_{c}B_{a}C$, $ CA_{b}A_{c}B$ and $ BC_{a}C_{b}A$ are outside the triangle. Square $ B_{c}B_{c}'B_{a}'B_{a}$ with center $ P$ is outside square $ AB_{c}B_{a}C$. Prove that $ BP,C_{a}B_{a}$ and $ A_{c}B_{c}$ are concurrent.[/quote]
[b]Solution.[/b] Denote $ M \equal{} BB_c\cap B_aC_a$, $ N \equal{} BB_a\cap B_cA_c$ and $ S \equal{} C_aB_a\cap A_cB_c$. Since two triangles $ ABB_c$ and $ AC_aB_a$ are similar, we conclude that $ \angle AB_cB \equal{} \angle AB_aC_a$, which implies that four points $ A$, $ B_c$, $ B_a$ and $ M$ are cyclic. Similarly, $ N$ also lies on this circle, let it is $ \Gamma$.

Let $ K \equal{} MN\cap B_cB_a$. We have $ PB_c$, $ PB_a$ are two tangents of $ \Gamma$, which follows that $ B_cB_a$ is the polar of $ P$ relative to the circle $ \Gamma$. On the other hand, we have $ K$ lies on $ B_cB_a$, it equivalent to $ P$ lies on the polar of $ K$ relative to $ \Gamma$ $ (1)$.

Let's consider the cyclic quadrilateral $ MNB_aB_c$, we have $ K$ and $ B$ are the intersections of its opposite sides, and $ S$ is the intersection of its diagonals. Therefore the polar of $ K$ relative to $ \Gamma$ is the line $ BS$ $ (2)$.

From $ (1)$ and $ (2)$ we conclude that $ B$, $ S$ and $ P$ are collinear, it equivalent to $ BP$, $ C_aB_a$ and $ A_cB_c$ are concurrent, as desired.[/quote]
Let us to present a schema, for this beautiful solution.

Best regards, Kostas Vittas.

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 114
Sincerely
Jean-Louis

An generalization can be found in [url]https://artofproblemsolving.com/community/c6h1505582p8907331[/url]


inequalities
trigonometry
inequalities proposed

[quote="SUPERMAN2"]Prove that in any triangle $ ABC$ the following inequality holds:
$ \frac {a^{2}}{b \plus{} c \minus{} a} \plus{} \frac {b^{2}}{a \plus{} c \minus{} b} \plus{} \frac {c^{2}}{a \plus{} b \minus{} c}\geq 3\sqrt {3}R$[/quote]

$ \Leftrightarrow \frac{4R^2}{r}(\sin^2(\frac{\alpha}{2})\sin(\alpha)\plus{}\sin^2(\frac{\beta}{2})\sin(\beta)\plus{}\sin^2(\frac{\gamma}{2})\sin(\gamma))\equal{}2p(\frac{R}{r}\minus{}1)$

$ p^2 \ge \frac{27}{2}Rr$

$ <\equal{}> 3\sqrt{6}\sqrt{Rr}(\frac{R}{r}\minus{}1)\ge 3\sqrt {3}R\Leftrightarrow  2R^2\minus{}5Rr\plus{}2r^2 \ge 0$ , ($ R \ge 2r$ )

hmm ..is the following true  ?

$ \frac{a^2 \plus{}b^2 \plus{}c^2}{a\plus{}b\plus{}c} \geq \sqrt {3}R$

[quote="enndb0x"]hmm ..is the following true  ?

$ \frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c} \geq \sqrt {3}R$[/quote]

It is not true . Try $ a\equal{}b\equal{}1$ and $ c \rightarrow 2_{\minus{}}$


calculus
integration
limit
trigonometry
function
absolute value
calculus computations

Split the integral into two to eliminate the absolute value, then integrating gives:

$\lim_{n\to\infty} n^2\int_{-\frac{1}{n}}^{\frac{1}{n}} \left(2005\sin x + 2006\cos x\right)|x| \ dx =\lim_{n\to\infty}\left[ 4012n\left(\left(\cos\frac{1}{n}-1\right)n+\sin\frac{1}{n}\right)\right]=$$2006$

$2006$ is the right answer, but Jimmy has a very odd looking penultimate step.

$\lim_{n\to\infty}n\left(\cos\frac1n-1\right)=0$ and $\lim_{n\to\infty}n\sin\frac1n=1.$

Should perhaps that $4012$ in there be $2006$?

Simplifying: $\sin(x)|x|$ is an odd function; it integrates to zero over any symmetric interval and can hence be ignored. $\cos(x)|x|$ is an even function, so we write:

$\int_{-\frac1n}^{\frac1n}\cos(x)|x|\,dx=2\int_0^{\frac1n}x\cos x\,dx.$

But for $x$ near zero, $\cos x$ might as well be $1$ and we might as well be integrating $x.$ In more detail, $x\cos x=x+O(x^3)$ and thus

$2\int_0^{\frac1n}x\cos x\,dx=2\int_0^{\frac1n}x+O(x^3)\,dx=\frac1{n^2}+O(n^{-4}).$

Now multiply that by $2006n^2,$ take the limit, and we are done.

[quote="Kent Merryfield"]$2006$ is the right answer, but Jimmy has a very odd looking penultimate step.

$\lim_{n\to\infty}n\left(\cos\frac1n-1\right)=0$ and $\lim_{n\to\infty}n\sin\frac1n=1.$

Should perhaps that $4012$ in there be $2006$?

[/quote]

No, it's right, $\lim_{n\to\infty}n\left(\cos\frac1n-1\right)$ does equal zero, but we want $\lim_{n\to\infty}n^2\left(\cos\frac1n-1\right)$ which equals $-\frac{1}{2}$.

Nice solution though, I wish I saw that the $\sin x$ could be ignored, that would have saved some time. :)

Oh - I missed that you had an extra $n$ multiplying the cosine term. The fact that you wrote it on the far side made it easy to overlook.

i have error just on the last step...
It is obviuous that
\[ \lim_{n\to\infty} n^2 (2005\sin \xi_n+2006\cos \xi_n)\int_{-\frac{1}{n}}^{\frac{1}{n}} |x|dx = \lim_{n\to\infty} (2005\sin \xi_n+2006\cos \xi_n) =\lim_{n\to\infty} 2005\sin \xi_n + \lim_{n \to \infty} 2006\cos \xi_n = 2006 \]
It's just slip of the pen  :)


[quote="saszs"]I doubt any of you can get the poll question right. :rotfl:  :D[/quote]

yea. could it be because

[i]all of the answer choices are wrong????[/i]

[hide=" "]i think so[/hide]

or it could be 1. idk though...

The square root of -1 is equal to [i]i[/i]. To find the absolute value of [i]i[/i], we graph the point on a complex coordinate plane and measure its distance from the origin. This distance is 1 so the absolute value of [i]i[/i] is 1.
(This method will work for any complex number [i]a+bi[/i].)

Remember, absolute value is the distance from 0. Here is a link with a little more information: http://en.wikipedia.org/wiki/Absolute_value#Complex_numbers.

If you mean the modulus/magnitude, the answer is 1.

(The question itself was sort of misleading).

Absolute value is [b]defined[/b] as distance from the origin, I believe.

When you have real numbers, you use a number line. Using complex numbers, like in this case, you use the complex plane.

Also this does not belong in MOEMS.

i = 0 + 1(i).

It's in the form of a complex number a + bi. To find the norm (more common name than absolute value when dealing with complex numbers) of a complex number a + bi, you simply compute the square root of (a^2 + b^2).

Thus the norm of i is the square root of (0^2 + 1^2) = 1.

Well the square root of a negative number (I'm saying -1) is $ i$. The absolute value of i, would be i in which the answer is i...

erm no the absolute value of a complex number $ a\plus{}bi$ is defined as $ \sqrt{a^2\plus{}b^2}$. This gives you the distance from the origin to the point $ a\plus{}bi$  in the complex plane. $ |0\plus{}1i|\equal{}\sqrt{0^2\plus{}1^2}\equal{}\sqrt{1}\equal{}\boxed{1}$

Nice, RoFIL. Didn't think of that. Stupid on my part.

-rts2007

In general a number's norm is the square root of the sum of squares of all independent units. Here, 1 and i are independent units.

Note that the norm we are currently talking about is the [b]Euclidean norm[/b]. The [b]Minkowski norm[/b] is sqrt(x^2-y^2), used in split-complex numbers.

This is true because as said the absolute value is the distance from the origin. So in the absolute value of 5 the point is (0,5) so it is 5 away. So if you know the imaginary number line and vectors your answer is 1.

Oh, my bad, I was thinking of $ i$ XD I must be looking at those problems too much... yeah the answer should be 1 instead.

It just requires a more geometrical perspective

With the Cartesian plane view of all numbers (imaginary,real and complex). I like that sort of approach! Wasn't it shown in algebra videos chapter 11?

[quote=saszs]Now I know I underestimated most of you.[/quote]

Never underestimate AoPS

[quote=jasonhu4][quote=thinkinavi][quote=MathPal]You said only rrusczyk and dpatrick could get at this ,saszs! :dry:[/quote]

Why the revive on a 6.5 year old thread?[/quote]

[quote=djmathman]Even after being on AOPS for this long, the revival power of some users still surprises me.[/quote][/quote]

same

I see that the user who revived this is a relatively new user to AOPS. It is usually appreciated if we stick to current threads and posts, not a post/thread that has been more than a year old. It is okay (still discouraged) to correct a mistake that has not been attentive. However, situations to point out complete random things is highly unsuggested.


geometry
rectangle
combinatorics unsolved
combinatorics

A $3000\times 3000$ square is tiled by dominoes (i. e. $1\times 2$ rectangles) in an arbitrary way. Show that one can color the dominoes in three colors such that the number of the dominoes of each color is the same, and each dominoe $d$ has at most two neighbours of the same color as $d$. (Two dominoes are said to be [i]neighbours[/i] if a cell of one domino has a common edge with a cell of the other one.)

First, color the square black and white as in a checkerboard. Then for each row, label the white squares XYZXYZXYZ etc. A domino has color k iff the white square it covers has label k (and each domino covers exactly one white square). It's easy to verify that this coloring satisfies the conditions.


Is there a nice expression about what (a^3)/(a^2+b^2) is greater than?
i mean, (a^3)/(a^2+b^2) >=???

[quote="Makaveli"]Is there a nice expression about what $\frac{a^{3}}{a^{2}+b^{2}}$ is greater than?[/quote]
\[\frac{a^{3}}{a^{2}+b^{2}}\geq a-\frac{b}{2}. \]
The difference between the left hand and the right hand side is
\[\frac{(a-b)^{2}b}{2(a^{2}+b^{2})}.\]


integration
calculus
calculus computations

For $0\leq{y}\leq{1}$ find the maximum value of $\int_{0}^{y}\sqrt{x^{4}+(y-y^{2})^{2}}dx$

[hide="here is my problem"]
the solution says that we let $I(y)$ be the value of the integral.
however it says that

$I'(y)=\sqrt{y^{4}+y^{2}(1-y)^{2}}+y(1-y)(1-2y)\int_{0}^{y}\frac{dx}{\sqrt{x^{4}+y^{2}(1-y)^{2}}}$

but when i do this i find

$I'(y)=\sqrt{y^{4}+y^{2}(1-y)^{2}}$
[/hide]

what am i doing wrong?

Try to simplify the integrand first by using Cauchy-Schwartz inequality.

i dont know how that helps, or how you would apply it.


[quote="pisqrt314"]Prove that 3^8-1 csn be divided by 8.
What is this question asking for???
 :help:[/quote]

You have to show that $ 8$ divides the number $ 3^8 \minus{}1$ exactly with a remainder of zero.

[hide]If $ 3^8 \minus{} 1$ is divisible by eight, then that means it has a remainder of zero when divided by 8. $ 3^8$ is one larger than $ 3^8 \minus{} 1$, so $ 3^8$ has a remainder of $ 1$. So you now have to prove that $ 3^8$ leaves a remainder of $ 1$ when divided by $ 8$. Using modular arithmetic;

$ 3^8$ (mod 8)
$ 3$^(8 - 8) (mod 8)
$ 3^0$ (mod 8)
$ 1$ (mod 8)
When one is divided by eight you get a remainder of 1. [/hide]

The proof is above. Please do not click on it if you are planning on solving.

[hide="The hard way"]

This is the most obvious wy, but just multiply out $ 3^8$, or $ 81^2$ and subtract $ 1$.  We get $ 6560$, and $ 560$ is divisible y $ 8$ so it is divisible.[/hide]

or $ 3^8\minus{}1\equal{}(3^4\minus{}1)(3^4\plus{}1)\equal{}80\cdot 82$, 80 is clearly divisible by 8

[quote="5849206328x"]or $ 3^8 \minus{} 1 \equal{} (3^4 \minus{} 1)(3^4 \plus{} 1) \equal{} 80\cdot 82$, 80 is clearly divisible by 8[/quote]

Oh, yes that would work and you could also factor out the $ (3^4 \minus{} 1)$ into $ (3^2 \minus{}1)(3^2 \plus{}1)$, and $ (3^2 \minus{}1)$ can be factored into $ (3 \minus{} 1)(3 \plus{} 1)$.


[quote="epitomy01"]There are 1280 pine trees each of diameter 1 metre in the forest where the smurfs live. THe dimensions of the forest are 1001*945 metres. The smurfs would like to build 7 tennis courts of dimensions 20 * 34 metres. Prove this i s always possible without cutting down any pine trees.[/quote]

So there exist 982*912 + 968*926 possible places to build tennis courts.

It also so happens that a pine tree interrupts the building of at most 2 * 34 * 20 tennis courts.

But there are only 1280 pine trees.

So there must be at least 982*912 + 968*926 - 2*34*20*1280 tennis courts to build.

If you multiply that ugliness out you end up with 51152 tennis courts. However, the courts themselves might overlap. But a court can't overlap with more than 2*69*41 other courts, so it follows that there must be at least \frac{51152}{2*69*41} buildable tennis courts. This number is very slightly more than 9, which is of course greater than 7, so we are done.

Oh snap, for some reason I was thinking everything was in a lattice grid. Well, I'm pretty sure you can WLOG to lattice grid. Hopefully.

[quote="epitomy01"]mm seems like a reasonable solution .. but where do you get the ugly numbers? 982*912 + 968*926.2*69*41?[/quote]

I looked at all the points the upper-left corner of a court could start at. Can't start less than 34/20 from the edge or you'd go off the map.

Then 69 comes from the fact that you no longer need upper-left corner, it can be anythinhg, so 2*34 + 1 / 2*20 + 1 ways to overlap.


Find the smallest integral $ x$ such that $ (x-2)(x+2)^{2}+2$ is positive.

(I like this question better than most of the other ones I've written.)

[hide]well i just tried possible values for (x+2) and i got 1 as the asnwer.[/hide]

[hide]Would it be 2? Since x is integral, x-2 has to be bigger than -2. Trying x=1, the value of the expression is -9. Trying x=2, the expression = 2[/hide]

[hide]Oops, I forgot that if part =0 it would also work out, so is the answer -2?[/hide]


Fie P(X)=X^6-X^3+1,un polinom cu coeficienti reali si a o radacina complexa a lui P(X).Sa se demonstreze
a)P(a^5)=0
b)P(X)=(X-a)(X-a^5)(X-a^7)(X-a^11)(X-a^13)(x-a^17)
c)pt orice nr nat m pol Q(X)=X^9m+x^9+1 nu se divide cu P(X)

Avem $p(X)=X^{6}-X^{3}+1=(X^{3})^{2}-X^{3}+1$. Ecuatia $t^{2}-t+1=0$ are ca solutii numerele $\cos (\frac{\pi}{3})+i\cdot \sin (\frac{\pi}{3})$ si $\cos (\frac{\pi}{3})-i\cdot \sin (\frac{\pi}{3})$ de unde putem sa aflam si radacinile polinomului $p$. (altfel  $p(X)=\frac{X^{9}+1}{X^{3}+1}$ de unde rezulta ca radacinile lui $p$ sunt radacinile de ordin $9$ ale lui $-1$ care nu sunt radacini de ordin $3$ ale lui $-1$). Din observatiile astea subpunctele a) si b) rezulta imediat,iar pt subpunctul c) este suficient de observat ca $q(a)=1(mod 2)$ pentru orice $m$ deci diferit de zero asadar $a$ nu poate fi radacina pt $q$.


Two team A and B participated in a tennis competition and each team have 1000 players .Arbirary player in team A played against with arbitrary player in team B exactly once and each match between them wasn't a draw.
 Prove that there exist 10 player in team A such that each player in team B was defeated by one in 10 above player.

1) Please give your topics better names, so that those reading them will be able to discern something of the contents of the question.

2) The question is obviously false without additional conditions.