There is some misunderstanding about the meaning of the documentation licenses of varoius sites.  The GNU (copyleft) documentation license does not mean "you can copy this content to do with as you please."  It requires that those who copy and use the material [b]must accept that license[/b].  AoPS does [b]NOT[/b] accept that documentation licensing and therefore cannot permit reproduction of such material on our site.

All proofs, concepts, and other parts of articles must be written in your own words.  A documentation license will be decided upon later.

I agree with fedja. I belive that the people that wrote GNU were not mathematicians, nor did they have mathematics wiki in mind when they wrote it :)

[quote="fedja"][quote="MCrawford"]
All proofs, concepts, and other parts of articles must be written in your own words. [/quote]
Hmmm... It's true that copying the entire articles doesn't make much sense, but you cannot really rewrite the $\varepsilon-\delta$ definition of a continuous finction, say, in "your own words". You can type it yourself, of course, but it will be identical with that everywhere else, so why not to copy and paste and spare some typing time? The same applies to many standard proofs: they have been shaped for dozens of years and look exactly the same in all textbooks and on the web. Any attempt to improve upon their presentation will, most likely, be futile. And isn't mechanical retyping (from Wikipedia or a Calculus textbook) subject to the same copyleft/right/up/down policies as copying by the cut-and-paste method? Sorry for my naiveness, but it is really hard to understand where the borderline between what is acceptable and what isn't lies here :?[/quote]

You're right that basic definitions are going to look the same or nearly identical.  That part of an article is surely immune to copyright/left documentation -- it is public domain.

However, the rest of any such article should not be copied.

For now, I believe this standard is fine and probably won't change in any major way.  If anyone protests, we'll discuss it again.

in reality, if you copy and paste it to save time, and if no one can tell it was copy and pasted, then it's fine for basic stuff and famous theorems, since some of the notations get annoying,

just warning you, wikipedia copy and paste ain't so straightforward so expect to do some minimal editing and customization anyways,

I agree 100% in terms of copying an entire article though,

Copying from the forums should be OK, unless the original poster explicitly or implicitly states that the text in question cannot go into the Wiki. This is just for Wikipedia and PlanetMath.

As a sidenote, Wikipedia is now under [url=http://creativecommons.org/licenses/by-sa/3.0/]CC-BY-SA 3.0[/url], which is less restrictive than the GFDL; however, the ShareAlike clause prevents us from copying material from it as the copyright on the AoPSWiki is incompatible with SA.


geometry
parallelogram
circumcircle
ratio
geometric transformation
homothety
reflection

Let $ABCD$ be a cyclic quadrilateral and let $H_{A}, H_{B}, H_{C}, H_{D}$ be the orthocenters of the triangles $BCD$, $CDA$, $DAB$ and $ABC$ respectively. Show that the quadrilaterals $ABCD$ and $H_{A}H_{B}H_{C}H_{D}$ are congruent.

To solve the problem, see that $AH_{A}H_{B}B$ is a parallelogram! (and the same for the other similar quadrilaterals)
This is true because $AH_{A}= CD.cotg(DAC) = CDcotg(DBC) = BH_{B}$ and $AH_{A}\parallel BH_{B}$.

One can also use vectors.. and set the orign at O, the circumcenter of $ABCD$

A completely synthetic solution:

let $ G_{A}$, $ G_{B}$, $ G_{C}$ and $ G_{D}$ be the centroids of triangles $ BCD$, $ CDA$, $ DAB$ and $ ABC$. We show that $ G_{A}G_{B}G_{C}G_{D}$ is similar to $ ABCD$ and that the ratio of similarity is $ \frac13$. A homothety with center $ O$ (the centre of the circumcircle) and ratio $ 3$ takes $ G_{A}$ to $ H_{A}$ and so on, so this will imply the desired result.

Let $ M$ be the midpoint of $ AB$. Then $ G_{D}$ divides the segment $ CM$ in the ratio $ 1 : 2$ and $ G_{C}$ divides the segment $ DM$ in the same ratio. Hence $ G_{C}G_{D}\parallel CD$ and $ G_{C}G_{D}= \frac13 CD$. So we're done.

Another solution.

Let $ P$ and $ Q$ be the other interesections of the altitudes from $ B$ and from $ C$, to $ AD$ respectively. We know that $ P$ and $ Q$ are the reflections of $ H_C$ and $ H_B$ respect to $ AD$. So the cuadrilaterals $ BCQP$ and $ H_CH_BQP$ are isosceles trapezoids (this is the synthetic proof for e.lopes' idea). Then $ H_BH_C\equal{}BC$ and $ H_BH_C\parallel{} BC$.

Analogously with the other sides. From here it follows that the cuadrilaterals $ ABCD$ and $ H_AH_BH_CH_D$ have equal sides and equal angles. In conclusion, they are congruent.

:)

Can be done without actually drawing the diagram.

Fix the circumcentre of $ABCD$ at the origin of the complex plane.
Then $H_A=b+c+d$
         $H_B=a+c+d$.
Thus $H_AH_B=\mid H_A-H_B\mid=\mid b+c+d-c-d-a \mid=\mid a-b \mid=AB$ and analogously for the other sides.
So done....

My solution(complex number)
$ H_A=b+c+d $
$ H_B=a+c+d $
$ H_AH_B=|H_A-H_B|=|b+c+d-c-d-a|=|a-b|=AB $


Waiit.

x+2*Sqrt(x-1)+x-2*Sqrt(x-1)+2*Sqrt(x^2-4x+4)
2x+2|x-2|

I don't understand why 2*Sqrt(x^2-4x+4) = 2|x-2|. Isn't it just 2(x-2) = 2x-4, so the squared expression equals 4(x-1)?

no, you need to watch out for that: sqrt(x:^2:)=|x|, not just x.

For example, sqrt((-2)^2)=sqrt 4=2, which is the absolute value of -2.

Oh, I see now. Never thought about that before. Thanks.

[quote="darkquantum"]can that constant be imaginary?[/quote]

I doubt it. It would be interesting, but for a reason why you usually wouldn't do that, consider what would happen if you plugged 0 in. You would get a sqrt(-1) somewhere in the problem. Now, would you replace that by i, or -i? They are both square roots of -1, and there is no reason the choose one over the other. Thus, you probably cannot generalize this to complex variables (darn).

Just an interestings side note: when you have 2 variables, i.e. something of the form a+bi, you lose order. For example, 1 is clearly greater than -1, but is i greater than -i? No, there is no way to tell.
You lose various other properties when you generalize to more dimensions, but Simon should probably be the one explaining that if you really want to know.

Sure, the constant can be imaginary. It just might or might not work if it is. Imaginary numbers pop up in the strangest places sometimes...

[quote="MysticTerminator"][quote="TripleM"]There are numbers a and b (a<b) so that
:sqrt:(x+2:sqrt:(x-1)) + :sqrt:(x-2:sqrt:(x-1))
is constant for a:le:x:le:b!! Find a and b and that constant.[/quote]
I think a is 1, b is 2, and the constant is 2.

If we square our little expression, we get x+2:sqrt:(x-1)+x-2:sqrt:(x-1)+2:sqrt:(x^2-4x+4) = 2x+2|x-2|, so for x:le:2, our expression will equal 2. However, x:ge:1 in order for the expression to make sense in the first place, so we have 1:le:x:le:2, makes the expression a constant value of 2.[/quote]

Yeah, thats right..
Its real interesting how such a strange formula could produce a constant value over some range.
I did this the same way as you, the other way is to add and subtract 1 under each square root, and they turn into perfect squares (eg first one is x-1 + 2:sqrt:(x-1) + 1 = (:sqrt:(x-1) + 1):^2:).

[color=cyan]TripleM, I like your solution with the modification of simply letting x = y^2 + 1, which lets you see that it isn't such a weird function at all.[/color]

Nice problem and solutions!

To also get a graphical perspective on this problem, use a TI-83 or similar calculator to graph these functions in Quadrant I:

Y1 = sqrt(x + 2sqrt(x-1))
Y2 = sqrt(x - 2sqrt(x-1))
Y3 = Y1 + Y2

PS:  Clue me in - where do I 'find' square root and other symbols?

[color=cyan]To the left, where all the smiley faces are, click on the "View More Emoticons" link and a window will appear.  If you click on the math things, they insert automatically.[/color]


Heres another game problem for you.

Two players play this game on a 1x2003 strip of squares. They alternately write either S or O into an empty square (they can write either). The first player (if any) to make three consecutive squares spell "S O S" wins. Does either player have a winning strategy?

Wouldn't a strategy for S be to make O go first?

Each player can write either letter at their turn.

It's easy to play defense and prevent a win.  Prefer to write an O next to another O.  This will render both useless.  After that, prefer to write an S next to another S.  If both players play like this, a win becomes impossible.  I don't know what you could do to develop a winning strategy, because the other player could simply adopt this strategy.

Unfortunately that doesn't quite work.. try it on some smaller size boards and you may see why.

Assuming that either player can place a letter in ANY empty square, and since the given dimensions, I think you should be able to force a draw in most conditions

If you think that, try to prove it. If you want a hint.. [hide]the first player can force a win [/hide]

Arg! I thought that was a hint, not a spoiler... :(
Never mind.

It is not a spoiler. That would be worth no marks at a competition like this, so I wouldn't consider it as such..

The whole thing is proving it, not finding the final answer.

I'm just curious, but would you use Pigeonhole to prove that he can force a win?

Or am I being stupid again?

How would you intend to use pigeonhole?

Everyone please feel free to write any ideas you have had, even if they don't seem to work at all. There is basically one extremely simple question you could ask yourself, and if you start there you should be able to solve it pretty quick.

I don't know, but since the board is finite, then can you prove that no matter wat player 2 does, there will always be at least one segment S0, or 0S, such that player one can win?

But even if there is an OS or SO, the player could still play the other type of letter there and not lose..

OK heres a proper hint. Though it doesn't seem like a hint.. Suppose I just lost a game. Why did I lose?

That;s easy. Because the other player spelled SOS. Or because there was an OS or a SO after your turn.

Think again. Why would I make a move that lets the other person win? Why don't I write the other letter?

Um, because maybe there was 2 OS's or 2 SO's, and i could only stop one?

There can never be two SO or OS 's. You would have won the turn before if there was.

PS: Its not quite as easy as you think. (well it is but you have to think about it carefully). And once you answer that question, you're about 90% of the way to solving the actual problem.

Hmm, good point.
I'm acting like an idiot tonight.

CAn't think of what your looking for..
Perhaps another hint?

Sigh, this is getting pathetic...

Well if you are forced to lose, you can't put an O or S in any square.. what does that mean about those surrounding squares?

I think I got it, I will post in spoiler

[hide]My proof assumes the board has n spaces where n is odd and > 6...

the first player plays an "S" in the middle square.... no matter what the second player does, the first player will place an "S" three places in the opposite direction from his original "S" than the second players move....

this forces "S" _ _ "S".... no matter what is put there by the second player, the first player will win....

3 of the spaces are now used.... and it's the second players turn... he will try to fill up the other spaces but it doesn't matter, there are even amount of squares left and he will be the one to place either an "O" or an "S" in one of those two slots... this assumes the first player does not make a bad move throughout the game... indeed, the first player can force a win.[/hide]

Thats the main idea Spoon. A couple of things still to fix up: with the other squares, how do you know the first player won't be forced to lose first? (if the second player sets up some situation where the first player can't move there either).

your right, the pattern can occur many times, the trick will be to show that no matter how many times it does, the second player will always end up being the one that must place the losing letter.... school starts tomorrow for me, so i need to get some sleep, I will look at it again tomorrow.

That is one thing you have to prove, the other thing you have to think about is: why S--S? are there some other squares that you can never go in? If so, then you will have to take these squares into account.

I think the idea is that the first player should have 2 S's with 2 empty squares between them. When all the other squares are filled the second player will be forced to put an S or an O in one of the 2 empty squares between the 2 S's. Then the first player wins. I swear I didn't look at the spoiler Spoon posted  :) This isn't a complete soln, but I think this is the idea. I'm sorry, but I don't know how to use spoilers (sorry if I spoilt it for anyone).

Here is a full written up solution for those who are interested.

[hide]

Call a 'bad' square a square that you cannot play either an O or an S in. Because we cannot play an O there, there must be an S to one side and a blank on the other. Because we can't play an S there, and there is a blank on one side, we must have another S after that blank, getting a "S - - S" situation. This proves the <only> bad squares are the squares between two such Ss.

So how many bad squares are there at any point in time? An *even* number. How many non-bad squares are left when it is the first players turn? An *odd* number, at any point in time. Therefore the first player can always play when it is his/her turn without losing. So all the first player has to do is make sure there is at least one S--S somewhere - which he/she can do as spoon suggested, an S in the middle to start and then play two away from it. And then the first player must win.

[/hide]

[/hide]


geometry
rectangle
perimeter
Diophantine Equations

If $R$ and $S$ are two rectangles with integer sides such that the perimeter of $R$ equals the area of $S$ and the perimeter of $S$ equals the area of $R$, then we call $R$ and $S$ a friendly pair of rectangles. Find all friendly pairs of rectangles.

$ S[a,b],R[b,c]$
So we has:
$ 2(a\plus{}b)\equal{}cd$
$ 2(c\plus{}d)\equal{}ab$
$ \Rightarrow 4(a\plus{}b)(c\plus{}d)\equal{}abcd$

[quote="Peter"]... and the solutions of the last equality? :)[/quote]
WLOG $ a\ge b$ and $ c\ge d$ and $ b\ge d$
$ abcd \equal{} 4(a \plus{} b)(c \plus{} d)\le 4(a \plus{} a)(c \plus{} c) \equal{} 16ac\Rightarrow bd\le16 \Rightarrow d^2\le bd\le16 \Rightarrow d\le 4$
$ \\
2(c \plus{} d) \equal{} ab\Rightarrow 2c \equal{} ab \minus{} 2d$
$ 2(a \plus{} b) \equal{} cd \Rightarrow 4(a \plus{} b) \equal{} 2cd \Rightarrow 4(a \plus{} b) \equal{} (ab \minus{} 2d)d \Rightarrow abd^2 \minus{} 4ad \minus{} 4bd \minus{} 2d^3 \equal{} 0 \Rightarrow \\
(ad \minus{} 4)(bd \minus{} 4) \equal{} 2d^3 \plus{} 16$
and obviously we can put $ d \equal{} 1,2,3,4$ in the last equality,and find solutions...
$ d \equal{} 1$:
$ (a \minus{} 4)(b \minus{} 4) \equal{} 18$
$ \Rightarrow b \equal{} 5,a \equal{} 22,c \equal{} 54$ or $ b \equal{} 6,a \equal{} 13,c \equal{} 38$ or $ b \equal{} 7,a \equal{} 10,c \equal{} 34$

$ d \equal{} 2$:
$ (2a \minus{} 4)(2b \minus{} 4) \equal{} 32\Rightarrow (a \minus{} 2)(b \minus{} 2) \equal{} 8$
$ \Rightarrow b \equal{} 3,a \equal{}10 ,c \equal{}13$ or $ b \equal{} 4,a \equal{}6,c \equal{} 10$

$ d \equal{} 3$:
$ (3a \minus{} 4)(3b \minus{} 4) \equal{} 70$
$ \Rightarrow b \equal{} 3,a \equal{} 6,c \equal{} 6$

$ d \equal{} 4$:
$ (4a \minus{} 4)(4b \minus{} 4) \equal{} 144\Rightarrow (a \minus{} 1)(b \minus{} 1) \equal{} 9$
$ \Rightarrow b \equal{} 4,a \equal{} 4,c \equal{} 4$
therefore all solutions are:
$ \{\{a,b\},\{c,d\}\} \equal{} \{\{22,5\},\{54,1\}\},\{\{13,6\},\{38,1\}\},\{\{7,10\},\{34,1\}\} \\
,\{\{10,3\},\{13,2\}\},\{\{6,4\},\{10,2\}\},\{\{6,3\},\{6,3\}\},\{\{4,4\},\{4,4\}\}$


function
real analysis
real analysis unsolved

Let $f: [0,1]\to [0,1]$ be an 1-lipschitziene function and $x_0 \in [0,1]$.Define the sequence $x_n$ which verifies ${x_{n+1}}=\frac{x_n+f(x_n)}{2}$.Study its convergence.

That's a very easy one, since the function $g(x)=(x+f(x))/2$ is increasing and therefore you are in classical conditions that will guarantee you the convergence.


This is the "Impossible Test"

I just posted it up because some questions have to do with logic, while others are just pain cool. Don't do this if you have some serious preparing for States/Nationals to do. I couldn't find a General Chat section, so I'm posting it here:

[url=http://uploads.ungrounded.net/content.php?id=365143&name=365143_Impossible_Quiz_Deluxe_NG..swf&title=The%20Impossible%20Quiz&date=1173243600&quality=b&uj=0&w=550&h=400]The Impossible Test[/url][/url]

I'm stuck on 34... I can't find any clickable things.

Nvm its timed.

Just take your mouse off the flash animation and wait...

Oops I meant 35.

I'm also stuck on 40.
EDIT: nvm solved that too.
Now I'm stuck on 42.
EDIT: I also solved 42. :huh:

This post was written on the day of my birthday!

[quote=Pi-dPiper]How are you finding theese?[/quote]

i keep scrolling for a while lol


Hi all, I have a basic algebra problem I am having trouble with and need help desperatly. Here it is:

We are calculating our birthdate by providing the day of the month of our birthday, multiply that by 20, add three, multiply the result by five, add to that the number of the month in which you were born, multiply the result by twenty, add three and multiply that by five. Add the last two digits of the year you were born  to the result. For me, I get 82301. Now when you subtract 1515 from this number, you get my birthday, which is 80786 (July 8, 1986). Now here is where im having trouble. I am now given these instructions:

Write one algebraic expression that represnts the final number used to calculate your birthdate.
Then, Substitute the values of a, b, and c into the expression and simplify, following the order of operations agreements.
A=the day of the month you were born, b=number formed by the month you were born, c=the number formed by the last two digits of the year of your birth.

I am not sure what the expression would be, or how to entirely and accuratly simplify it. some one please HELP!


Find all pair $ (x,y)\in\mathbb{R}^2$ such that
$ \begin{cases}
x\plus{}y\plus{}\sqrt{x^2\minus{}y^2}\equal{}12\\
y\sqrt{x^2\minus{}y^2}\equal{}12\\
\end{cases}$

P/s: I wonder that how many ways you can solve it :lol:

Let $ y\equal{}xcos\phi >0$, then we get $ (1\plus{}cos\phi \plus{}sin\phi )^2\equal{}12cos \phi sin\phi$ or $ sin 2\phi \equal{}\frac{24}{25}$. It give $ x\equal{}\pm 5$. It give $ y^2\equal{}9,16$. Therefore $ x\equal{}5$ and $ y\equal{}3$ or $ y\equal{}4$.

Condition $ y>0$
Let $ y\plus{}\sqrt{x^2\minus{}y^2}\equal{}12\minus{}x$
     $ \Leftrightarrow (y\plus{}\sqrt{y^2\minus{}x^2}\equal{}(12\minus{}x)^2$
      $ \Leftrightarrow x^2\plus{}24\equal{}(12\minus{}x)^2$
      $ \Leftrightarrow x\equal{}5,x\equal{}\minus{}5$
case 1:$ x\equal{}5$
$ y\plus{}\sqrt{25\minus{}y^2}\equal{}7$
     $ 25\minus{}y^2\equal{}(7\minus{}y)^2$
      $ 2y^2\minus{}14y\plus{}24\equal{}0$
       $ y\equal{}4 ,y\equal{}3$
case2:x=-5 $ y\plus{}\sqrt{25\minus{}y^2}\equal{}17$ (has no solution)

"From our TST".

Which country are you from?  :D


limit
modular arithmetic
number theory
real analysis
real analysis theorems

we call a sequence of integers $ (x_n)$  a U.D. mod N sequence iff
\[ \lim_{n\to\infty}\frac{\#\left\{k\in\{1,\dots,n\}: \ x_k\equiv r\pmod{N}\right\}}{n}\equal{}\frac{1}{N}\]

find all $ N$'s for which the Fibonacci sequence is U.D. mod N.

This is number theory, not calculus. It belongs in either Superior Algebra or Olympiad NT.

A [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=16028]related topic[/url] with some useful ideas.
The answer: powers of 5.


$ f: \mathbb{C} \to \mathbb{C}$ differentiable at $ z_{0}$ (in real sense).
Let $ S \equal{} \{z \in \mathbb{C} : z_{n} \to z_{0} \ with \ lim_{n} \frac {f(z_{n}) \minus{} f(z_{0})}{z_{n} \minus{} z_{0}}  \equal{} z\}$.
Show that the set $ S$ is a point or a circumference.

What do you mean "in real sense " ? Analytic functions have a unique defined limit so S is a point .What is left to do is to consider none analytic  functions.Thanks .


I have recently been playing around with Alcumus (and it's really fun!), but one thing annoys me. Sometimes I type things fast and forget a 0 or a digit in my answer and it says I get it wrong. When it takes me back to the problem, can we have it so that it shows what we typed in before so we can compare it with what we got when doing it on paper so that we can correct a typing mistake if we have one?

Alcumus is [b]not[/b] about being fast (ie it is not FTW.) Therefore if you type answers [b]that[/b] fast, that's a problem you have to fix on your own.

what is ftw? for the win? or a short version of something (ie, ttyl, ppl)


algebra
polynomial
number theory proposed
number theory

Does it exist a polynomial $P$ with integer coefficients such that the sequence $P(3^n)$  $mod(n)$ is bounded?

It may be too easy for you, but I did not make any progress on it. And since it is too nice, it should be remembered.

Sorry, Harazi. I made a mistake. It's not too easy. In fact it is an open question for me. :blush:

Suppose we know that $ P \in \mathbb{Z}[x] $ and $ P(3^{n}) mod
(n) $ is bounded by $ C $, independent of $ n $. Take $ n = qm $,
where $ q,m \in \mathbb{N} $ and $ q $ is prime. Then $ P(3^{mq})
mod(mq) $ is bounded by $ C $, so obviously $ P(3^{mq}) mod(q) $
is bounded by $ C $. But now, because of Fermat little theorem, $
P(3^{mq}) \equiv P(3^{m}) (mod q) $, so we get that for every $ q
$ prime and $ m $ natural, we have that $ P(3^{m}) mod(q) $ is
bounded by $ C $, which cannot be true. Indeed, we have that $
\frac{P(3^{m+1})}{P(3^{m})} \rightarrow 3 $, when $ m \rightarrow
\infty $, so for a prime $ q $ large enough, if we take  $ m \in
\mathbb{N} $ first, such that $ |P(3^{m})| > q/5 $, then $
|P(3^{m})| < 4q/5 $, so we get that if for this $ m $, $ k =
P(3^{m}) mod(q) $, then $ k > q/5 $, and we get a contradiction.

Sorry, I made a stupid mistake, saying that $
\frac{P(3^{m+1})}{P(3^{m})} \rightarrow 3 $, when $ m \rightarrow
\infty $. In fact, $ \frac{P(3^{m+1})}{P(3^{m})} \rightarrow
3^{deg P} $, when $ m \rightarrow \infty $, but the rest of the
proof can be corrected.

I don't understand the end of your proof, Leva.

I agree with you that $ (P(3^m) \, \text{mod} \, (q))$ is bounded.

Contradiction using prime number q such that $ q>P(3^m)$. Am I wrong ?  :maybe:

Suppose $ [P(3^n)\mod n]$ is bounded by $ M$ for all $ n\in \mathbb{N}$.
Let $ n \equal{} pq$ where $ p,q$ are primes. Then $ [P(3^{pq})\mod pq]\leq M$, and also $ [P(3^{pq})\mod p] \equal{} [P(3^q)\mod p]\leq M$.
Now $ 3$ is a unit in $ \mathbb{Z}_p$ and so $ \{3^k \mod p | k\in \mathbb{Z}\} \equal{} \mathbb{Z}_p^*$ and so $ [P(x)\mod p]\leq M \,\, \forall x\in \mathbb{Z}^*_p$ which is clearly impossible as $ p\to \infty$.


farz konid $ABC$ yek mosalas bashad va noghteye $E$ rooye $BC$ tagheer
konad.farz konid $I_1$ va $I_2$ marakeze dayere haye mohatye kharejie mosalas haye
$ABE$ va $ACE$ nesbat be rase $A$ bashand.sabet konid dayereye mohiti mosalase
$EI_1I_2$ az noghte ee sabet rooye $BC$ migozarad.

salam
sabet mikonim noghteye sabet mahale tamase dayereye mohatiye kharegiye$abc$
ba zelye $bc$ ast.ke ba $x_a$neshan midahim.
badihi ast ke $<I_1ei_2=90$ pas bayad sabet konim$<i_1x_ai_2=90$
(namgozariha:mahale tamase mohatiye kharegiye $abe$ ba $be,ab,ae$be tartib
$m_1,m_2,m_3$minamimva hamchenin mahale tamase mohatiye kharejiye $ace$ba $ec,ac,ae$rabetartin$n_1,n_2,n_3$minamim.mahale tamate momas moshtarake dakheliye digareha ra ba dayereye$i_1,i_2$va zelye$bc$be tartim ba $s,t,r$neshan midahim.)
barihi ast ke $<i_1ri_2=90$mikhhim sabet konim $r$ haman $x_a$ast.yani
$rb+ba=rc+ca$ pas bayad $rm_1+am_2=rn_1+an_2$ pas bayad
$am_3+rk=an_3+rt$ pas bayad$m_3n_3=kt$ ke rabeteie dorost ast pas masale sabet shod.

dar mosalase $\triangle{ABC}$khate semsone motanazer ba noghte $P$az noghte $P'$migozarad
sabet konid in khat az $G$ham migozarad
[hide]$P'$noghte rooberoo be ghotrie $P$ast
$G$markaze seghle mosalase[/hide]

[quote="jensen"]besiar ziba.kheili lezzat bordam. :D[/quote]
baba joonam man begid ba ki hastid(quote)

[quote="mohamadhosein"]malome dige ba toe rahehale man ke kheili maskhare bood.[/quote]
sharmande motevajeh nashodam
mitoonam bedoonam az che rahi halklesh kardi jensen??


The Fibonacci sequence appears in the family tree of the drone, or male bee.

The drone has a mother, but no father

Female bees are worker bees or queens

They have a mother and a father

The family tree of a male bee bee back to its grandparents is shown below

              :D ---------------------  :!:

                              -
                              -

                              :D

                                :!:

 :!: = drone
 :D = Queen

a) Copy the diagram. Trace the male bee's ancestors back 5 more generations.
b) Explain how you can use this pattern to find the Fibonaccui numbers.

I had trouble typing the chart into the message body. The chart starts off with a queen and a drone bee. The two then create a queen and a drone bee. I need to find the next five generations and then find the pattern. Can anyone help? Thanks

Here it is:

We can quite easily just go up the family tree. Let's make sure we correctly follow the great great great great grandparents part.

1 bee(male)
1 parent(female)
2 grandparents (1M, 1F)
3 great's (2F, 1M)
5 great great's

It's the awesome fibonacci sequence!
8 great great great
13 quadruple greats


analytic geometry
geometry
conics
hyperbola
geometry unsolved

Let B,C be fixed points and A be roving point. Let H, G be orthecentre and centroid of triagle ABC. Known midpoint of HG lies on BC, find locus of A

It is only an easy problem, all is only calculation  :D and I don't think it is for VMO.  Contruct a Cartersian coordinate $Oxy,O$ is midpoint of $BC$, $Ox\equiv BC$, $Oy\perp BC$ assume $B(-a,0),C(a,0)$ and $A(x,y)$ thus $G(\frac{x}{3},\frac{y}{3})$ let midpoint of $GH$ be $M\in BC\Rightarrow M(m,0)\Rightarrow H(2m-\frac{x}{3},-\frac{y}{3})$ because $AH\perp BC\rightarrow x=2m-\frac{x}{3}$ and $BH\perp AC\Rightarrow \vec{BH}\cdot\vec{AC}=0\Rightarrow (2m-\frac{x}{3}+a,-\frac{y}{3})\cdot (x-a,y)=0\Rightarrow (x-a)(2m-\frac{x}{3}+a)-\frac{y^{2}}{3}=0$ now from $x=2m-\frac{x}{3}$ we have $x^{2}-a^{2}-\frac{y^{2}}{3}=0\Rightarrow x^{2}-\frac{y^{2}}{3}=a^{2}$ thus the locus is hyperbolla, that's main ideal but here we need to know $M\in$ segment $BC$ or line $BC$.
 :D

[quote="hien"]Are you sure that have a elemtary solution that don't use Cartersian coordnitate system?[/quote]
Do you know idiom ? Appetite comes with eating  :D surely? It is style of Vietnamese!
And actually I don't understand what do Descartes and coordinate method have guilt ? when all geometer I known they always say synthetic geometry is beautiful while coordinate is only ugly method. :(

If gemath's calculation is flawless, the locus is a hyperbola. How can you prove it in synthetic way? The proof, if it exists, will be complicated. Anyway, I think VMO 's problems for geometry this year (and generally in many past years) are EXTREMELY RIDICULOUS.

[quote]Do you know idiom ? Appetite comes with eating  surely. It is style of Vietnamese[/quote]
It is no joke!
Because i think it have a optimal project to solve this problem.
Fisrt i think author of one discover it by using Descartes Coordinate. And i also realize that if there is a pure solution then  is is very complicated, because of needed focus is hyperbola

[quote="treegoner"]Anyway, I think VMO 's problems for geometry this year (and generally in many past years) are EXTREMELY RIDICULOUS.[/quote]

Are you sure that? You had better consider some of geometry problems [b]in many past years[/b], for example, year 2002, it is very difficult.
This year, we have to solve seven problems in three hours. Is it so easy?

No, doing 7 problems in 3 hours is not easy.

Yet, a good problem is a problem that has an underlying insight and beauty. Well, one thing I need to say about these kinds of VMO geometric problems is that they are so often the locus or "passing through the fixed point" kind of problem. These types of problem are merely hard about predicting and if you have some experience, you can do it. However, the main point I want to address here is not about their level of difficulty. I want to emphasize about the fact that there is nothing fresh in these problems. It is just a game of hide and seek. The more the proposer can hide, the harder the problem becomes. But overall, there is nothing new.


Let $ f(x)\equal{}\frac{1\plus{}x}{1\minus{}x}$. A sequence $ a_{0},a_{1},\dots$ is defined recursively by $ a_{0}\equal{}a\neq 1$ and $ a_{n}\equal{}f(a_{n\minus{}1})$ for $ n\ge 1$. Prove that the sequence is periodic.

[hide="proof"]
observe that no $ a_{i}$ is $ 0$ as
$ f(f(x)) \equal{}\frac{\minus{}1}{x}$
also this gives
$ a_{0}\equal{} a$
$ a_{1}\equal{} f(a_{0}) \equal{} f(a)$
$ a_{2}\equal{} f(a_{1}) \equal{} f(f(a)) \equal{}\frac{\minus{}1}{a}$
$ a_{3}\equal{} f(a_{2}) \equal{} f(\frac{\minus{}1}{a})$
$ a_{4}\equal{} f(a_{3}) \equal{} a$
and so on so the the sequence is periodic with period $ 4$
 [/hide]

[quote="SplashD"]Since $ f(x)\equal{}\tan(\frac{\pi}{4}\plus{}\arctan(x))$, you can clearly see it's pediodic with period 4[/quote]
but $ x$ does not belong to the domain of real no.s
for example if i define $ f(x) \equal{}\tan x$ for $ x \equal{} 0.1,0.4,0.9....0.25$ this is not periodic though $ \tan x$ itself is periodic

[quote="pardesi"][quote="SplashD"]Since $ f(x) \equal{}\tan(\frac{\pi}{4}\plus{}\arctan(x))$, you can clearly see it's pediodic with period 4[/quote]
but $ x$ does not belong to the domain of real no.s
for example if i define $ f(x) \equal{}\tan x$ for $ x \equal{} 0.1,0.4,0.9....0.25$ this is not periodic though $ \tan x$ itself is periodic[/quote]

I don't understand what you mean.

The sequence would go

$ a,\tan(\frac{\pi}{4}\plus{}\arctan(a)),\tan(\frac{\pi}{2}\plus{}\arctan(a)),\tan(\frac{3\pi}{4}\plus{}\arctan(a)),\tan(\pi\plus{}\arctan(a))\equal{}a$

[quote="splashD"]
$ a,\tan(\frac{\pi}{4}\plus{}\arctan(a)),\tan(\frac{\pi}{2}\plus{}\arctan(a)),\tan(\frac{3\pi}{4}\plus{}\arctan(a)),\tan(\pi\plus{}\arctan(a)) \equal{} a$[/quote]
is periodic.
but what u wrote before
[quote="splashD"]Since $ f(x) \equal{}\tan(\frac{\pi}{4}\plus{}\arctan(x))$, you can clearly see it's pediodic with period 4[/quote]
this is [b]wrong[/b]  without showing the values
the proof [b]does not[/b] follow fro m this u have to show that the given sequence is periodic by [b] plunging[/b] the values .
for example see the example i stated which someone marked as a spam   :mad:   :huh: !!!
see define $ f(x) \equal{}\frac{1\minus{}x}{1\plus{}x}$ for $ x \equal{} 0.1,0.11,0.111....0.11111111$
although $ f(x) \equal{}\tan(\frac{\pi}{4}\plus{}\arctan(x))$ the seuqence is [b]not periodic[/b].what is important is the domain

you're arguing a mute point

no one's going to be stupid enough to interpret the problem that way

[quote="pardesi"][quote="SplashD"]Since $ f(x) \equal{}\tan(\frac{\pi}{4}\plus{}\arctan(x))$, you can clearly see it's pediodic with period 4[/quote]
but $ x$ does not belong to the domain of real no.s
for example if i define $ f(x) \equal{}\tan x$ for $ x \equal{} 0.1,0.4,0.9....0.25$ this is not periodic though $ \tan x$ itself is periodic[/quote]

I think SplashD did not want to say that $ f(x)$ was periodic.

He said that $ a_{n}\equal{}f^{n}(x)\equal{}\tan(\frac{\pi}{4}n\plus{}\arctan(x))$ was a periodic sequence (periodic in regard of $ n$), which is obviously true.


If $ a,b,c$ are nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}3$, then

(a) $ \sqrt{a(4a\plus{}5b)(4a\plus{}5c)}\plus{}\sqrt{b(4b\plus{}5c)(4b\plus{}5a)}\plus{}\sqrt{c(4c\plus{}5a)(4c\plus{}5b)} \le 27$;

(b) $ \sqrt{a(2a\plus{}b)(2a\plus{}c)}\plus{}\sqrt{b(2b\plus{}c)(2b\plus{}a)}\plus{}\sqrt{c(2c\plus{}a)(2c\plus{}b)} \ge 9$.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers such that $ a + b + c = 3$, then

(b) $ \sqrt {a(2a + b)(2a + c)} + \sqrt {b(2b + c)(2b + a)} + \sqrt {c(2c + a)(2c + b)} \ge 9$.[/quote]
By Holder we obtain:
$ \left(\sum_{cyc}\sqrt{a(2a+b)(2a+c)\right)^2\sum_{cyc}\frac{a^2}{(2a+b)(2a+c)}\geq(a+b+c)^3.}$
Thus, it remains to prove that $ \sum_{cyc}\frac{a^2}{(2a+b)(2a+c)}\leq\frac{1}{3},$ which a known Tiks's inequality.
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66660

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

(a) $ \sqrt {a(4a \plus{} 5b)(4a \plus{} 5c)} \plus{} \sqrt {b(4b \plus{} 5c)(4b \plus{} 5a)} \plus{} \sqrt {c(4c \plus{} 5a)(4c \plus{} 5b)} \le 27$;

(b) $ \sqrt {a(2a \plus{} b)(2a \plus{} c)} \plus{} \sqrt {b(2b \plus{} c)(2b \plus{} a)} \plus{} \sqrt {c(2c \plus{} a)(2c \plus{} b)} \ge 9$.[/quote]

By Cauchy Schwarz we obtain
\[ \sum {\sqrt {a\left( {4a \plus{} 5b} \right)\left( {4a \plus{} 5c} \right)} }  \le \sqrt {\left( {\sum a } \right)\left( {\sum {\left( {4a \plus{} 5b} \right)\left( {4a \plus{} 5c} \right)} } \right)}
\]
Thus it remains to show that
\[ \sqrt {\left( {\sum a } \right)\left( {\sum {\left( {4a \plus{} 5b} \right)\left( {4a \plus{} 5c} \right)} } \right)}  \le 27
\]

\[ \Leftrightarrow ab \plus{} bc \plus{} ca \le 3
\]

Very nice your solutions, Arqady and Honey_s. :lol:


In Clara county, 25$ \%$ of households earn less than $ \$30,000$ per
year, and 65$ \%$  of households earn less than $ \$80,000$ per year.
What is the largest possible percent of households that could
earn between $ \$30,000$ and $ \$80,000$ per year?

This is an easy problem in disguise.

We're just looking for the overlap, so we subtract 25 from 65 to get 40.


Hi everyone,

Could someone please help me prove this code?

RDSXCVIWTDGNXHUJCLTLXAAATPGCBDGTPQDJIXIAPIIG

The message is encoded using a shift in the alphabet. I tried it when n=4 and 9 and it didn't seem to work.

Any sugestion? I don't think it is all one word because of the three A's in a row. So, I thought that one of them must be a one letter word (I, a) but neither worked.

Thank you very much

I decoded it in less than five minutes and it looks amazingly like homework.

[hide="Hints"]
The AAA in the code is part of one four-letter word and one five-letter word.
Think:
something like "...rate eerie..."
"bell large"
"hiss snake"
Not that those words made any sense.
[/hide]
[hide="Spaced"]
RDSXCV IWTDGN XH UJC! LT LXAA ATPGC BDGT PQDJI XI APITG.
Yes, the second-last letter should be T.
[/hide]


Find "k", to coefficients of \[ x^k
\] and \[ x^{k \plus{} 1}
\]
 in \[ \left( {3x \plus{} 2} \right)^{19}
\]
 be the same.


Let $ a,b,c$ positive reals such that $ \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\equal{}1$. Prove that

$ \frac{b(a\plus{}ab\plus{}bc)}{(a\plus{}c)(b\plus{}c)}\plus{}\frac{c(b\plus{}bc\plus{}ca)}{(b\plus{}a)(c\plus{}a)}\plus{}\frac{a(c\plus{}ca\plus{}ab)}{(c\plus{}b)(b\plus{}a)}\geq\frac{5}{4}$


What is the integer part of

$ \frac{1}{\frac{1}{1990}\plus{}\frac{1}{1991}\plus{}\cdots\plus{}\frac{1}{2008}}$

:play_ball:

[hide="As far as I got"]We can say that the given expression is greater than:

$ \frac{1}{\frac{1}{1990} \plus{} \frac{1}{1990} \plus{} \cdots \plus{} \frac{1}{1990} }$

$ \equal{} \frac{1}{\frac{19}{1990}} \equal{} \frac{1990}{19} \equal{} 104 \frac{14}{19}$

Similarly, it is less than:

$ \frac{1}{\frac{1}{2008} \plus{} \frac{1}{2008} \plus{} \cdots \plus{} \frac{1}{2008} }$

$ \equal{} \frac{1}{\frac{19}{2008}} \equal{} \frac{2008}{19} \equal{} 105 \frac{15}{19}$

So the floor of our sum is either 104 or 105. I can't quite see how to restrict it any more.[/hide]

If $ a$ is fixed and $ x_i \ge y_i$, how would we prove that
$ \sum_i \frac1{a\plus{}x_i} \plus{} \frac1{a\minus{}x_i} \ge \sum_i \frac1{a\plus{}y_i} \plus{} \frac1{a\minus{}y_i}$?

[hide="Intuitively"]$ \frac {1}{19} \cdot \text{HM} [1990 \ldots 2008] \approx \frac {1}{19} \text{AM}[1990 \ldots 2008] \equal{} \frac {1999}{19} \equal{} 105.21$

We have a [i]very[/i] comfortable margin of error here -- for the value to be under 105, we'd need the HM of 1990..2008 to be under 1995, which is already 2.6 times closer to the lower bound than to the upper bound. For a set of large numbers that close together, such an imbalance is out of question.[/hide]

[b]@mathwizarddude:[/b] $ \frac {1}{a \plus{} x} \plus{} \frac {1}{a \minus{} x} \equal{} \frac {a \minus{} x}{(a \plus{} x)(a \minus{} x)} \plus{} \frac {a \plus{} x}{(a \plus{} x)(a \minus{} x)} \equal{} \frac {2a}{a^2 \minus{} x^2}$

Where the denominator is positive, this is minimized when $ x \equal{} 0$, and increases as $ |x|$ increases.

As a clarifier, Palmer meant $ b > c > 0$ up there.

Oh yeah, but how would we prove $ \frac {1}{\frac {9}{1990} \plus{} \frac {1}{1999} \plus{} \frac {9}{2008}} > 105$ (by hand)?


email
geometry
algebra
function
domain
number theory

Kiries kai Kirioi,

Eimai stin euxaristi thesi na sas anakoinwsw oti einai etoimi h istoselida pou legame sta proigoumena post opou tha prosthetoume ta vivlia pou exei se ilektroniki morfi o kathenas apo emas (eite nomima eite paranoma :) ). Tin teleiwsa molis twra kai vrisketai sti dieuthinsi:

             http://www.stfu.gr

Gia na to egainiasoume exw prosthesei ena poli wraio vivlio, apo ekeina pou exei proteinei o Mpampis. Otan thelete na valoume kapoio vivlio 'h simeiwseis apo kapoio kalo vivlio (ellhniko alla kiriws kseno) mporeite na to steilete eite stin dieuthinsi diagonismoi@gmail.com eite se PM se emena kai katopin tha to sikwnw stin sigekrimeni istoselida pros kalo olwn mas. Euxomai na ploutistei oso to dinaton perissotero auti i sillogi wste emeis alla kai epomenoi diagwnizomenoi na vriskoun eukola merika vasika voithimata.
Gia opoiadipote sxolia exw ftiaksei ena vivlio episkeptwn opou mporeite na grapsete tis skepseis sas gia na min kourazoume allo auto to topic :) Elpizw na sas aresei i idea... Provlima xwrou sto parapanw email den iparxei opote steilte o,ti thelete oso megalo ki an einai (Nomizw to gmail epitrepei mexri kai 10 MB email. Ean auto pou thelete na steilete einai megalitero tote mporoume na vrethoume sto msn kapoia stigmi kai na to steilete apeutheias apo ekei...)

filika,
Alexandros

Prosthethikan simera oi "Mathimatikes Olimpiades ana ton kosmo" No1 (1996-1997) kai No2 (1997-1998) me provlimata kai liseis apo polles ethnikes mathimatikes Olimpiades (periexei kai ton diko mas Arximidi twn sigekrimenwn etwn). Opoios exei tin sinexeia autwn twn vivliwn as mou stelnei.

 Kali diaskedasi... :)

Alexandros

Αλέξανδρε,
 σου έστειλα χθες ένα cd με πο΄λλά βιβλία και αυτό που ζητάς(Andreescu ...).Το Νο 1 θα το βρεις στη διεύθυνση που σου στέλνω.Εκείέίδα γράφει για δύο βιβλία , αλλά δες ποια είναι.Φοφάμαι μηπως είναι αυτά που έχεις. Κατέβασέ τα και τα δύο και κρέμασέ τα . Είναι πολύ καλά βιβλία, το ξέρεις άλλωστε !, αν είναι αυτά που θέλουμε. Αλλίως βάλε μόνο αυτό που σου έστειλα.

    http://www.diamante.com.pe/file/carlosbr/ebooks/

          Εγώ δεν έχω γρήγορη γραμμή.Μόλις τα πάρεις στείλτα μου σε cd.

Μπάμπης

Exei prostethei stin parapanw selida ena vivlio tou H-o-j-o-o   l-e-e  to opoio periexei anisotites kai texnikes epilisis provlimatwn pou sxetizontai me anisotites eidika prosarmosmeno gia mathimatikous diagwnismous. Euxaristw ton Stelio pou mou to esteile kai to evala sti selida.

Perimenw kai dika sas sto parapanw email

Alexandros

Πολύ καλη ιστοσελιδα με εκπληκτικο υλικό!Μπραβο cretanman. :)

katarxin sinxarititia gia to site :!:
ego tha ithela,an sas einai eykolo,na valete kapoia vivlia sxetika me ylh likeiou,kirios tritis.

H istoselida exei ananewthei (pros to kalitero nomizw), exei xwristei se merikous tomeis opou tha vazoume ta kainouria vivlia/simeiwseis gia tous diagwnismous kai epipleon exoun prostethei ston tomea tis Gewmetrias to vivlio tou Πρασολοβ pou anefere kai o mpampis (einai pragmatiki vivlos gia ti gewmetria auto to vivlio). Osoi den to exete na to katevasete kai na to ektipwsete me tin prwti eukairia. Tha sas xrisimeusei para poli...

Episis exoun prostethei sto kommati twn mathimatikwn diagwnismwn, ta provlimata kai oi liseis twn Olimpiadwn tou Kanada (CMO) apo ta eti 1969 ews 2002. Ean kapoios diathetei ta provlimata kai tis liseis 1969-1993 se pdf me mikri xwritikotita (2-3MB) as ta aposteilei dioti auto pou exw egw einai peripou 12 MB. Apo 1993-2002 den iparxei kanena provlima.

Perimenw ta sxolia sas oxi edw alla sto Vivlio episkeptwn gia na mi fortwnoume adika me posts tin lista auti.

Episis ean kapoios diathetei ena kalo eisagwgiko vivlio gia thewria arithmwn eite provlimata thewrias arithmwn parakaleitai na to steilei sto parapanw email.

Filika,
Alexandros

Y.G. 1 Euxaristw poli ton Mpampi ton Stergiou pou mou esteile kapoia vivlia (ta opoia tha mpoun siga siga) kathws episis kai ton sinadelfo Kwsta Panagiwtakopoulo o opoios episis mou esteile iliko.

Y.G. 2 Nomizw oti o skopos tis sigekrimenis selidas einai gia mathimatikous diagwnismous kai oxi gia tin kalipsi tis sxolikis ilis kai twn panelliniwn gia tis opoies iparxei arketo ililko se polla vivliopwleia. As kratisoume kalitera auto to xaraktira gia ti selida auti.

Ρε φίλε θα ήθελα να μου πεις γιατί δεν δουλεύει η ιστοσελίδα σου κι να μου την δώσεις σε προσωπικό μηνυμα σε ευχαριστώ θεώρω οτι έχεις κάνει καλή δουλεια αν κρίνω από τις κριτικές

Ambros katarxin elpizw twra na paizei kanonika i istoselida kai na min exeis kanena provlima!

Episis tha ithela na sas enimerwsw oti prosthesa simera 4-5 kainouria poli poli endiaferonta vivlia, ena ek twn opoiwn (To vivlio me rwssikes olimpiades) kathisa kai to scannara gia na mporeite na to apolausete. Ean kapoios exei kanei tin idia douleia se dika tou vivlia as mou ta steilei. Episis ean kapoios endiaferetai na to kanei gia kapoio vivlio kai den kserei me poio tropo tha to kanei wste na MHN piasei poli ligo xwro ston ipologisti tou, as mou steilei ena PM na tou eksigisw. Omoia ean kapoios exei scannarei kapoio vivlio kai den kserei pws tha mikrinei ton xwro wste na ginei ena poli wraio (kai mikro kiriws) .pdf

Einai poli kalo na metatrepsete ta vivlia sas se ilektronika vivlia (ebooks) kathws etsi, ean pathei kati to vivlio sas tha exete tin dinatotita na to ektipwsete me poli mikro kostos. Episis ean xreiasteite kapoia stigmi merikes mono selides apo to sigekrimeno vivlio, anti gia na to pate na to fwtotipisete mporeite na ektipwsete autes tis selides ston ektipwti sas. Thimitheite oti oso pio polles fores anoigete kai vgazete fwtotipies to vivlio toso pio poli ftheirete kai xalaei. Giati na min to kanete mia fora kai meta terma? (afiste pou me auto to tropo tha prosferete kai se allous pou gia ton x y logo den mporoun na agorasoun to sigekrimeno vivlio 'h agnooun akoma kai tin iparksi tou) Skefteite to kai peite mou...

Alexandros

Αγαπητέ cretanman η ιστοσελίδα τώρα που την κοίταξα ειναι παρα πολυ ανανεωμένη κι πλούσια κι με καλά βιβλία ειδικά το συνδυαστικό του andreescu συγχαρητήρια συνέχισε έτσι

Πραγματι Αμβρσιε πολυ καλα βιβλια με πολλες ωραιες ασκησεις.Τα μελεταω προσεκτικα....

φίλε cretanman μήπως είναι δυνατόν να βάλεις στο σιτε την θεωρία αριθμών του πουλάκη που πρότεινε ο φίλτατός Δημήτρης ? :maybe: θα το εκτιμούσα πολύ επίσης αν ειναι δυνατόν θα ήθελα να βρείς κι άλλα βιβλία συνδυαστικής κυρίως αυτά που είναι στα ελληνικά
σε ευχαριστώ που με άκουσες ελπίζω να το λάβεις υπόψη σου

φιλε cretanman πρέπει αμεσα να κοιτάξεις την ιστοσελίδα σου διοτι έχει προβλημα με την κωδικοποίηση

Ola einai ok twra! Euxaristw poli amvrosie gia tin episimansi! Pragmati enw to vivlio me ta poliwnima tou Barbeau einai morfis .djvu, kata lathos fainotan san arxeio .pdf. Gia to vivlio tou Lee, to lathos episis diorthwthike. Enjoy...

Oson afora to vivlio tis Thewrias arithmwn tou D. Poulaki to exw alla prepei na kathisw na to scanarw kai den exw auti ti periodo arketo xrono. Euxomai omws na milame gia to idio vivlio giati an thimamai kala o Dimitris milouse nomizw gia to vivlio thewrias arithmwn tou Poulou. O Poulakis einai kathigitis sto panepistimio tis Thessalonikis kai to vivlio tou einai arketa kalo kai eisagwgiko alla einai grammeno gia tous prwtoeteis foitites tou mathimatikou kai oxi gia kapoio pou proaitoimazetai gia olimpiades. Einai omws arketa euanagnwsto.

Gia tin sindiastiki den exw alla vivlia se morfi .pdf. Kai auto pou evala mou to esteile o kos Stergiou. Gi'auto lew se opoiondipote pou exei tetoiou eidous vivlia na ta aposteilei gia na megalwseis i sillogi...

Filika,
Alexandros

ισως τα ΔΙΑΚΡΙΤΑ ΜΑΘΗΜΑΤΙΚΑ του Παπαιωανου εκδοσεις ΕΜΠ για μερικη συνδιαστικη.εχει τo ιδιο υφος με το advanced section ειναι καλο ...οσο για το   βιβιο του Πουλακη ειναι πολυ καλο πιστευω πως αν και ειναι πανεπιστημιακο πρεπει να το διαβαζουν ακομα και απο α λυκειου. κατευθειαν στα βαθια  :wink:

Ναι, σε λίγες μέρες θα 'ναι κανονικά up, αν όχι σήμερα. Υπόψη ότι οι σύνδεσμοι με τα βιβλία θα 'ναι για λίγες μέρες down, μέχρι να δούμε που θα τα βάλουμε.

Προς το παρόν, μπορείτε να κοιτάξετε τη νέα άσκηση του μήνα!

Πολύ ωραία η άσκηση του μήνα :) . Δεν ξέρω ποια λύση έχετε υπ' όψιν σας Στέλιο και Ράλλη αλλά έχω βρει μια σύντομη και κομψή λύση. Θα τη στείλω με e-mail για να μην πούμε αμέσως τη λύση. Να την προσπαθήσουν κι άλλοι :wink: .

Ένα μικρό hint:

Είναι πολλές οι γραμμές στο σχήμα, πράγματι, αλλά αν αρκεστείτε σε αυτές δε θα βρείτε άκρη. Εγώ στην αρχή όλο κύκλους έκανα. Αρκεί να κάνετε μία κίνηση :idea:  και θα το λύσετε χωρίς να το καταλάβετε.

Εγώ δεν καταλαβαίνω που χρειάζεται το έγγεντρο  :wink:
Η πρόταση ισχύει γενικά  :)

Σιλουανέ όταν λες γενικά...τι ακριβώς εννοείς???
Πώς δηλαδή δεν χρησιμοποιείς το έγκεντρο αφού μία απ' τις ευθείες που συντρέχουν είναι και η ΜΔ, όπου Δ σημείο επαφής του έγκεντρου με ΒΓ... που σημαίνει ότι εξαρτάται απ' το έγκεντρο :oops:
plz εξήγησε λίγο τη σκέψη σου...

Συμφωνώ με τον Σιλουανό.

Η αναφορά στον εγγεγραμμένο κύκλο είναι λίγο παραπλανητική, αφού πρόκειται για ειδική περίπτωση γενικού προβλήματος.

Τα σημεία $E,$ $Z,$ μπορεί να είναι τυχαία πάνω στις πλευρές $A\Gamma,$ $AB$ αντιστοίχως και το σημείο $\Delta,$ ως το σημείο τομής των $B\Gamma,$ $AM,$ όπου $M,$ ο σημείο τομής των $BE,$ $\Gamma Z.$

[hide]Αρκεί να δούμε ότι το σημείο τομής των $B\Gamma,$ $EZ,$ είναι γνωστό σημείο και το πρόβλημα λύθηκε.[/hide]

Κώστας Βήττας.

Το εγγεγραμμένο το βαλα για να διευκολύνω τους λύτες όχι για να τους παραπλανήσω  :o  :(

Πάντως έχετε δίκαιο κύριε Βήττα και Σιλουανέ για τη γενίκευση του προβλήματος.

Πάραυτα, με τον εγγεγραμμένο η λύση απλουστεύεται κατά πολύ! Ο Κώστας ας πούμε έστειλε μία πολύ καλή λύση, κάνοντας χρήση του εγγεγραμμένου κύκλου!

Αυτά ! :)

Ακόμα και η γενίκευση λύνεται σε μία γράμμη . :wink:
Όπως είπε και ο κ. Κώστας πάτρε αυτό το σημείο και μετά
[hide]σκεφτείτε το γράμμα Η   :D [/hide]

[quote="vittasko"]....
Η αναφορά στον εγγεγραμμένο κύκλο είναι λίγο παραπλανητική, αφού πρόκειται για ειδική περίπτωση γενικού προβλήματος.....[/quote]
[quote="mostel"]Το εγγεγραμμένο το βαλα για να διευκολύνω τους λύτες όχι για να τους παραπλανήσω  :o  :( ....[/quote]

Στέλιο, ζητώ συγνώμη γιατί δεν εκφράστηκα καλά. Λέγοντας "παραπλανητική", δεν εννοούσα ( ούτε για αστείο ) ότι κάποιος θέλησε να μας μπερδέψει.

Απλά στη λύση που σκέφτηκα, τα σημεία $\Delta,$ $E,$ $Z,$ δεν έχει σημασία αν είναι τα σημεία επαφής του εγγεγραμμένου κύκλου, στις πλευρές του δοσμένου τριγωνου $\bigtriangleup AB\Gamma.$

Αρκεί το να είναι συντρέχουσες οι ευθείες $A\Delta,$ $BE$ $\Gamma Z$ ( στην περίπτωση του εγγεγραμμένου κύκλου ως γνωστό, συντρέχουν στο σημείο [b][size=100]Gergone[/size][/b] του $\bigtriangleup AB\Gamma$ ).

Μερικές φορές η σκέψη μας επικεντρώνεται στα δεδομένα της εκφώνησης, όπως αυτή εμφανίζεται ( ειδική περίπτωση γενικού προβήματος ), και μας ξεφεύγει η βασική ιδέα που λύνει το πρόβλημα στη γενική του μορφή ( άρα και στην ειδική ).

Όπως ξέρουμε, στην προσπάθειά μας να λύσουμε κάποιο πρόβλημα, επιδιώκουμε να χρησιμοποιήσουμε τις γεωμετρικές ιδιότητες που απορρέουν από αυτήν την ίδια την εκφώνηση, πράγμα που οδηγεί στη λύση τις περισσότερες φορές.

Κώστας Βήττας.

ΥΓ. Καλή επιτυχία στο διαγώνισμα της Γεωμετρίας :)

Στέλιο όντως η λύση που έδωσα χρησιμοποιεί εφαπτόμενα τμήματα όμως έβγαινε και με χρήση του Σεβά (όπως είπε και ο κ. Βήττας gergonne point) γι' αυτό άλλωστε και γενικεύεται η άσκηση :wink: . Δεν κατάλαβα πάντως αυτό που είπε ο Σιλουανός για το όνομα του σημείου :?: . Νομίζω και η γενική περίπτωση λύνεται αποκλειστικά με Σεβά και Μενέλαο όπως και η ειδική άλλωστε.

Ωραίος Σιλουανέ  :lol:

[quote="silouan"]Η λυση σε μία γραμμη
[hide][b]H[/b]armonic -Αρμονικά και διπλός λόγος . Τελος (σε μια γραμμή  :wink: )[/hide]  :wink:[/quote]

Θα έλεγα σε μισή γραμμή....

[hide][b]Harmonic[/b] -Αρμονικά. Τέλος. :wink: [/hide]
Κώστας Βήττας.

[quote="vittasko"] Θα έλεγα σε μισή γραμμή....
[/quote]

Μάλλον ναι  :D  .Να είναι καλά ο βασιλιάς των αρμονικών σημείων Nick Rapanos
που μου τα δίδαξε . Thanks Nick .

Παιδιά που βρίσκονται οι ασκήσεις του Μήνα ?  :(

Kυρίες και κύριοι,

Είμαι στην ευχάριστη θέση να σας ανακοινώσω την λειτουργία της σελίδας μας και πάλι, μετά από απουσία περίπου 2 μηνών. Μπορείτε να δείτε τη διεύθυνση της σελίδας στο πρώτο post του συγκεκριμένου μηνύματος. Να αναφέρω από τώρα ότι η σελίδα αυτή ΔΕΝ είναι η σελίδα που περιέχει τα βιβλία. Θα μπείτε όμως σε αυτή, θα ακολουθήσετε τις οδηγίες για να βρεθείτε στην εν λόγω ιστοσελίδα. Αυτό έγινε για να παραπλανήσει τους ξένους εκείνους που είδαν αυτό το post και θέλουν να αναφέρουν την λειτουργία της ιστοσελίδας μας σε εκδοτικούς οίκους με σκοπό το κλείσιμό της. Γι'αυτό τα πάντα είναι γραμμένα στα Ελληνικά!

***
Η σελίδα είναι

http://homepages.pathfinder.gr/stfu/

Παρακαλώ τον κο Στεργίου να αλλάξει το όνομα της σελίδας στο πρώτο post, να σβήσει και το κείμενο που περιέχεται ανάμεσα στους αστερίσκους στο παρόν post και τέλος να αλλάξει εαν μπορεί τον τίτλο του μηνύματος αυτού σε μικρούς χαρακτήρες χωρίς θαυμαστικά (μην καρφωνόμαστε). Επίσης γράψτε στο πρώτο μήνυμα  μερικά πράγματα από την πιο πάνω παράγραφο, για να μη νομίζουν εκείνοι που μπαίνουν ότι βρίσκονται στην ιστοσελίδα με τα βιβλία.

***

Παρακαλώ πολύ όλους για να μην κλείσει και πάλι η σελίδα (αν και αργά ή γρήγορα όταν καταλάβουν και πάλι οι εκδοτικοί οίκοι ότι βάζουμε τα βιβλία τους για κατέβασμα), να μην αναφέρουν πουθενά σε αυτό το μήνυμα την διεύθυνση της ιστοσελίδας και να μην της κάνουν διαφήμιση σε διάφορα forum κτλ κτλ. Ας την διατηρήσουμε όσο το δυνατόν πιο μυστική. Επίσης όπως θα δείτε (για τους καινούριους) όλα τα ονόματα στην σελίδα είναι στα ελληνικά για να μην καρφωθούμε!! Τις επόμενες ημέρες θα βάζω σιγά σιγά κάποια βιβλία που έχω συγκεντρώσει αυτό τον καιρό. Μπορείτε όμως να στείλετε και τα δικά σας βιβλία/σημειώσεις κτλ κτλ στο email stfu.gr@gmail.com (Να σβηστεί και αυτό εδώ και να αλλάξει στην πρώτη σελίδα αντί του diagonismoi@gmail.com). Δεν μπορεί αυτή η προσπάθεια να συνεχιστεί εαν δεν συμβάλλετε όλοι.

Για να σας ανοίξω την όρεξη βάζω στον τομέα της Γεωμετρίας, την περίφημη Ελληνική Γεωμετρία ΘΡΥΛΟ του Σ. Κανέλλου (549 σελίδες, 1η έκδοση 1970) "Ευκλείδειος Γεωμετρία Επίπεδος" που έκανα scan μία μία σελίδα. Σας συνιστώ να την εκτυπώσετε και να την διαβάσετε προσεχτικά (είναι σε πολύ καλή ποιότητα) καθώς πρόκειται για μία από τις καλύτερες γεωμετρίες που πέρασαν ποτέ από τη χώρα μας. Σίγουρα οι γονείς σας εαν ήταν σε σχολές που χρειαζόταν Μαθηματικά, θα το έχουν ακούσει! Απολαύστε το!!

Μπορούμε να ακούσουμε σχόλια και από τους μεγαλύτερους (τον κο Στεργίου ή τον κο Βήττα).

Ίσως είναι πολύ καλή ιδέα να φέρουμε σε ψηφιακή μορφή τα καλύτερα ελληνικά βιβλία που έχουμε. Το ξέρω ότι είναι κόπος αλλά σκεφτήτε ότι πριν κατεβάσετε κι εσεις αυτά τα βιβλία κάποιος άλλος κόπιασε για να τα φτιάξει. Εαν κάποιος κάνει scan κάποιο βιβλίο και δεν ξέρει πώς να το κάνει σε .pdf και να είναι μικρό σε χωρητικότητα (Φυσικά εαν κάνετε scan, να το κάνετε σε ασπρόμαυρο και προσοχή στις σελίδες να είναι ίσια για να μπορεί κάποιος που θα το κατεβάσει να το εκτυπώσει), μπορεί να μου στείλει τις εικόνες σε ένα CD με το ταχυδρομείο και θα το κάνω εγώ. (Διαφορετικά εαν έχει γρήγορο Internet μπορεί να τις συμπιέσει και να μου τις στείλει από το MSN).

Φυσικά δεν πρέπει να ξεχάσουμε και τον Στέλιο (mostel) του οποίου η βοήθεια ήταν μεγάλη στο σχεδιασμό της σελίδας και στο να επανέλθει και πάλι.

Φιλικά,
Αλέξανδρος


Prove that if p(x) is polynomial with integer coefficients and p(squre root of 2) = 0 then p(− square root of 2) = 0.

Use conjugate: for any polynomial with integer coefficient, if x and y are a pair of conjugate, then P(X) and P(y) are also a pair of conjugate.

Note that the odd terms of $ P(\sqrt{2})$ will always amount to a multiple of $ \sqrt{2}$ and the even terms will always amount to an integer. Thus, the sum of the even terms of $ P(\sqrt{2})$ will be 0, and the sum of the odd terms will too. For $ P(\minus{}\sqrt{2})$, the even terms will remain the same, which is 0, and the sum of odd terms will become the opposite, which is still 0.


Compute the maximum value of each of the following (in radians).
$sin\theta cos\theta$
$sin\theta+cos\theta$
$sin\theta-cos\theta$
$cos\theta-sin\theta$
$sin(\theta sin\theta)$
$cos(\theta cos\theta)$
$sin(\theta cos\theta)$
$cos(\theta sin\theta)$

[quote="miyomiyo"]Compute the maximum value of each of the following (in radians).
$sin\theta cos\theta$[/quote]
$=\frac{1}{2}\sin 2 \theta$
Thus, $\frac{1}{2}$ is maximum.

The other ones are almost similar.

[quote]
$sin(sin\theta)$
[/quote]
$\sin 1$


limit ratio inequalities factorial Euler logarithms integration

Show that $\lim_{n\to\infty}(\frac{n^{n}}{n!})^{\frac{1}{n}}< 3$ in an elementary way

if stirling's approximation is to be considered elementary then yes
the problem can be done very easily

I interpret "in an elementary way" as "not using the letter $e$."

It suffics to prov that th squnc $a_{n}=\frac{n^{n}}{c^{n}n!}$ is boundd for som $c<3$. Notic that
\[\frac{a_{n+1}}{a_{n}}=\frac{1}{c}\left(\frac{n+1}{n}\right)^{n}\le \frac{1}{c}\left(\frac{n+1}{n}\right)^{n+1}\]
Lt $b_{n}=\left(\frac{n+1}{n}\right)^{n+1}$. I claim that th squnc $(b_{n})$ is dcrasing, that is,
\[\left(\frac{n+1}{n}\right)^{n+1}\ge \left(\frac{n+2}{n+1}\right)^{n+2}\quad (1) \]
Indd, (1) is quivalnt to
\[(n^{2}+2n+1)^{n+1}(n+1)\ge (n^{2}+2n)^{n+1}(n+2), \]
which follows from
\[(n^{2}+2n+1)^{n+1}\ge (n^{2}+2n)^{n+1}+(n^{2}+2n)^{n}(n+1) \]
This provs that $(b_{n})$ is a dcrasing squnc. But $b_{5}=(6/5)^{6}<3$, bcaus $6^{6}=46656<46875=3\cdot 5^{6}$. Thrfor, $c=b_{5}$ works.

I almost stepped in to edit your post to compensate for your sticky keyboard - then I realized what you were doing and that it was intentional.

To prove that $a_{n+1}\le a_{n}$, I wrote this inequality in the form $\frac{a_{n+1}}{a_{n}}\le 1$. This is algebra, not the Ratio Test.

[quote="steven128"]Is there also a solution not using something like the ratio test?[/quote]
Here is a completely different solution using just basic combinatorics and the binomial formula. Recall that $n!$ is the number of rearrangements of $n$ objects and $n^{n}$ is the number of ways to put $n$ objects into $n$ boxes labelled $1,\dots,n$. Now let $1\le m\le n$. Suppose we have a way to put $n$ objects (say, the same $n$ first positive integers) into $n$ boxes with exactly $m$ boxes being non-empty. Then, we can create a rearrangement that just lists the content of the first nonempty box  followed by the content of the second non-empty box followed by that of the third and so on. Moreover, if $n_{1},\dots,n_{m}$ are the numbers of the objects in our non-empty boxes, then the number of the corresponding rearrangements is $n_{1}!\cdot\dots\cdot n_{m}!\ge 2^{n-m}$ (because in the product of factorials $m$ factors are $1$ and all other factors are at least $2$). Of course, just knowing the rearrangement does not allow us to recover the way the objects are put into boxes. One needs to know where each box content list ends, which gives us $\binom nm$ possibilities, and what $m$ numbers out of $n$ were used for labelling non-empty boxes (another $\binom nm$ possibilities). Thus, the number of ways to put $n$ objects into $n$ boxes with exactly $m$ boxes nonempty does not exceed ${\binom nm}^{2}2^{-(n-m)}\cdot n!\le \binom nm\cdot\left(\frac{3}{2}\right)^{n}\cdot n!$ (because $\binom nm 2^{-(n-m)}\le\sum_{m}\binom nm 2^{-(n-m)}=\left(\frac{3}{2}\right)^{n}$). Summing over $m$, we get $n^{n}\le 3^{n}n!$. This proof uses a lot of letters $e$, of course, but no Euler constant either explicitly or in disguise  :P.

Let $y: =\lim_{n\to\infty}\left(\frac{n^{n}}{n!}\right)^{\frac{1}{n}}$. Then

$\log y=\lim_{n\to\infty}\frac{1}{n}\log\left(\frac{n^{n}}{n!}\right) = \lim_{n\to\infty}\frac{1}{n}\log\left(\prod_{k=1}^{n}\frac{n}{k}\right)$
$=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left(\frac{n}{k}\right) =-\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left(\frac{k}{n}\right)$
$=-\int_{0}^{1}\log x \,dx =-\lim_{\delta\to 0^{+}}\left[x\log x-x\right]_{x=\delta}^{1}= 1\Rightarrow y: =\lim_{n\to\infty}\left(\frac{n^{n}}{n!}\right)^{\frac{1}{n}}=e$


linear algebra
matrix
geometry proposed
geometry

[color=darkred]Given are the triangle $ABC$ and three lines $d_{k}\ ,\ k\in \overline{1,3}\ .$

Denote the following intersection points and define its positions :

$\{\begin{array}{ccccccc}M\in d_{1}\cap AB & \mathrm{\ ,\ }& \overline{MB}=m\cdot \overline{MA}& \mathrm{\ \ \ ;\ \ \ }& N\in d_{1}\cap AC & \mathrm{\ ,\ }& \overline{NC}=n\cdot \overline{NA}\ .\\\\ P\in d_{2}\cap BC & \mathrm{\ ,\ }& \overline{PC}=p\cdot \overline{PB}& \mathrm{\ \ \ ;\ \ \ }& Q\in d_{2}\cap BA & \mathrm{\ ,\ }& \overline{QA}=q\cdot \overline{QB}\ .\\\\ R\in d_{3}\cap CA & \mathrm{\ ,\ }& \overline{RA}=r\cdot \overline{RC}& \mathrm{\ \ \ ;\ \ \ }& S\in d_{3}\cap CB & \mathrm{\ ,\ }& \overline{SB}=s\cdot \overline{SC}\ . \end{array}$

Prove that the lines $d_{k}\ ,\ k\in\overline{1,3}$ are concurrently $\Longleftrightarrow$ $\boxed{\ mq+ps+rn=1+mpr+nqs\ }\ .$[/color]

[color=darkblue][b]Proof.[/b] Given is the fixed point $O$- the origin of the vectorial system. Then for any $X$ define $X=\overrightarrow{OX}$ [b][u](from this moment l"ll not use the arrows).[/u][/b] Thus, for any $X$ and $Y$ we have $XY=Y-X\ .$ Denote the intersections $V\in d_{3}\cap AB$ , $Y\in d_{1}\cap d_{3}\equiv MN\cap SR$ , $Z\in d_{1}\cap d_{2}\equiv MN\cap PQ\ .$ From assumption we have :

$\{\begin{array}{c}MB=m\cdot MA\ ,\ NC=n\cdot NA\ .\\\\ PC=p\cdot PB\ ,\ QA=q\cdot QB\ .\\\\ RA=r\cdot RC\ ,\ SB=s\cdot SC\ .\end{array}$ $\Longrightarrow$ $\{\begin{array}{ccc}(1-m)M=B-mA & \mathrm{\ \ ;\ \ }& (1-n)N=C-nA\ .\\\\ (1-p)P=C-pB & \mathrm{\ \ ;\ \ }& (1-q)Q=A-qB\ .\\\\ (1-r)R=A-rC & \mathrm{\ \ ;\ \ }& (1-s)S=B-sC\ .\end{array}$

$1\blacktriangleright$ Apply the [u]Menelaus' theorem[/u] to the transversal $\overline{VRS}$ in the triangle $ABC\ :$

$V\in RS$ $\Longrightarrow$ $\frac{VB}{VA}\cdot \frac{RA}{RC}\cdot \frac{SC}{SB}=1$ $\Longrightarrow$ $\frac{VB}{VA}=\frac{RC}{RA}\cdot\frac{SB}{SC}$ $\Longrightarrow \frac{VB}{VA}=\frac{s}{r}\ .$

$\{\begin{array}{ccccccc}R=\frac{A-rC}{1-r}& \mathrm{\ ,\ }& N=\frac{C-nA}{1-n}& \Longrightarrow & \frac{RA}{RN}=\frac{A-R}{N-R}& \Longrightarrow & \boxed{\ \frac{RA}{RN}=\frac{r(1-n)}{1-nr}\ }\ .\\\\ V=\frac{sA-rB}{s-r}& \mathrm{\ ,\ }& M=\frac{B-mA}{1-m}& \Longrightarrow & \frac{VM}{VA}=\frac{M-V}{A-V}& \Longrightarrow & \boxed{\ \frac{VM}{VA}=\frac{s-mr}{r(1-m)}\ }\ . \end{array}\ .$

Apply the[u]Menelaus' theorem[/u] to the transversal $\overline{VYR}$ in the triangle $AMN\ :$

$V\in YR\Longrightarrow$ $\frac{YM}{YN}\cdot\frac{RN}{RA}\cdot\frac{VA}{VM}=1$ $\Longrightarrow$ $\frac{YM}{YN}=\frac{RA}{RN}\cdot \frac{VM}{VA}$ $\Longrightarrow$ $\boxed{\ \frac{YM}{YN}=\frac{(1-n)(s-mr)}{(1-m)(1-nr)}\ }\ \ (1)\ .$

$2\blacktriangleright$ Apply the [u]Menelaus' theorem[/u] to the transversal $\overline{UPQ}$ in the triangle $ABC\ :$

$U\in PQ\Longrightarrow$ $\frac{UA}{UC}\cdot\frac{PC}{PB}\cdot\frac{QB}{QA}=1$ $\Longrightarrow$ $\frac{UA}{UC}=\frac{PB}{PC}\cdot\frac{QA}{QB}$ $\Longrightarrow$ $\frac{UA}{UC}=\frac{q}{p}\ .$

$\{\begin{array}{ccccccc}U=\frac{pA-qC}{p-q}& \mathrm{\ ,\ }& N=\frac{C-nA}{1-n}& \Longrightarrow & \frac{UA}{UN}=\frac{A-U}{N-U}& \Longrightarrow & \boxed{\ \frac{UA}{UN}=\frac{q(1-n)}{p-nq}\ }\ .\\\\ Q=\frac{A-qB}{1-q}& \mathrm{\ ,\ }& M=\frac{B-mA}{1-m}& \Longrightarrow & \frac{QM}{QA}=\frac{M-Q}{A-Q}& \Longrightarrow & \boxed{\ \frac{QM}{QA}=\frac{1-mq}{q(1-m)}\ }\ . \end{array}\ .$

Apply the [u]Menelaus' theorem[/u] to the transversal $\overline{UQZ}$ in the triangle $AMN\ :$

$U\in QZ$ $\Longrightarrow$ $\frac{ZM}{ZN}\cdot\frac{UN}{UA}\cdot\frac{QA}{QM}=1$ $\Longrightarrow$ $\frac{ZM}{ZN}=\frac{UA}{UN}\cdot\frac{QM}{QA}$ $\Longrightarrow$ $\boxed{\ \frac{ZM}{ZN}=\frac{(1-n)(1-mq)}{(1-m)(p=nq)}\ }\ \ (2)\ .$

$\blacktriangleleft\blacktriangleright$ From the relations $(1)$ and $(2)$ obtain :

$d_{1}\cap d_{2}\cap d_{3}\not = \emptyset$ $\Longleftrightarrow$ $Y\equiv Z$ $\Longleftrightarrow$ $\frac{YM}{YN}=\frac{ZM}{ZN}$ $\Longleftrightarrow$ $\frac{s-mr}{1-nr}=\frac{1-mq}{p-nq}$ $\Longleftrightarrow$ $\boxed{\ mq+ps+rn=1+mpr+nqs\ }\ .$[/color]

$\mathrm{(\ continuation\ of\ the\ previous\ message\ )}$

[color=darkblue] [b]Two remarkable particular cases.[/b]

$1\blacktriangleright$ $R: =A\ ,$ $M: =B$ and $P: =C\ ,$ i.e. $d_{1}=BN\ ,$ $d_{2}=CQ$ and $d_{3}=AS$ $\Longrightarrow$

$m=p=r=0$ $\Longrightarrow$ $1+nqs=0$ $\Longrightarrow$ $\frac{NC}{NA}\cdot \frac{QA}{QB}\cdot\frac{SB}{SC}=-1$ [b](the Ceva's theorem.)[/b]

$2\blacktriangleright$ $P: =C$ and $S: =B\ ,$ i.e. $d_{2}=CQ$ and $d_{3}=BR$ $\Longrightarrow$ $p=s=0$ $\Longrightarrow$ $mq+rn=1$ $\Longrightarrow$ $\boxed{\ \frac{MB}{MA}\cdot \frac{QA}{QB}+\frac{RA}{RC}\cdot \frac{NC}{NA}=1\ }\ .$

Apply the [u]Menelaus' theorem[/u] to the transversals $CXQ$ and $BXR$ in the triangles $ABY$ and $ACY$ respectively :

$\{\begin{array}{c}\frac{CY}{CB}\cdot \frac{QB}{QA}\cdot \frac{XA}{XY}=1\Longrightarrow\frac{QA}{QB}=\frac{CY}{CB}\cdot\frac{XA}{XY}\\\\ \frac{BY}{BC}\cdot\frac{RC}{RA}\cdot\frac{XA}{XY}=1\Longrightarrow\frac{RA}{RC}=\frac{BY}{BC}\cdot\frac{XA}{XY}\end{array}\Longrightarrow$ $\frac{MB}{MA}\cdot\frac{CY}{CB}\cdot\frac{XA}{XY}+\frac{NC}{NA}\cdot\frac{BY}{BC}\cdot\frac{XA}{XY}=1\Longrightarrow$ $\boxed{\ \frac{MB}{MA}\cdot YC+\frac{NC}{NA}\cdot BY=\frac{XY}{XA}\cdot BC\ }\ .$[/color]

That's very nice theorem Virgil! . we can have another proof by the concurrent condition of three lines in barycentric equation!
We can use onther notation:
${d_{a}\cap AB=\{C_{a}\},d_{a}\cap AC=\{B_{a}]}$ and $B_{a}(x_{a},0,z_{a}),C_{a}(x_{a},y_{a},0)$ thus equation of $d_{a}$ is
$z_{a}y+y_{a}z=0$ here we don't need discuss $x_{a}$
similar equation of $d_{b}: z_{b}x+x_{b}z=0,d_{c}: x_{c}y+y_{c}x=0$ and we have the concurrent condition of $d_{a},d_{b},d_{c}$ is:
$\begin{vmatrix}0&z_{a}&y_{a}\\z_{b}&0&x_{b}\\x_{c}&y_{c}&0\end{vmatrix}=0$
we have similar your condition!  :)


quadratics
algebra
polynomial
parameterization
function
logarithms
linear algebra

hey Guys, im having a bit of trouble working out some questions i need to how to work out by tomorrow. I feel stupid for not thinking of this earlier but i have been flat out with study.

can you help me work out these questions and show me how its done.

Solve for X in the quations.

1)  2^4x  = 16        2)  ln(2x-5) = 1.9        3)  logbase7(x) = 5

4) the intensity  I of light passing through a window, measured in lumens, is given by the formula

I(d) = 52e^-4.5d

(i) where d is the thickness (in centimeters) of glass in the window.
find the intensity if the thickness is 2.2 cm

(ii) find the thickness required to give an intensity of 12 lumens. give answer in centimeters to at least three decimal places.

5)  Explain why the quadratic equation x^2-6x+140=0 has no real solution for x.

6) on dividing (x+2) into 2x^5-3x^2+x-1 there is both a quotient and remainder.

Thus  2x^5-3x^2+x-1 = q(x)(x+2) + R

where q(x) is a polynomial in x of degree 4, and R is a constant. Find R?

7)

Find all solutions to the system.

x-2y+2z = -6
3x-5y+6z = -15
2x+4y-12z = 44

8)  write the equation 5x+3y = 4 in parametric form. That is express both x and y in terms of the same parameter

9)  solve x(2x-3)^2 = 4x^3+7x-60 for x

Hope this isnt to much to ask. I will be so very thankfull to anybody who can help me.

Thanx in advance

Hayz

Urgent help i ment, sorry for the typo. bit spaced out today, to much study.

Homework, mmm?

1)  $2^{4x} = 16$
You should express $16$ as a power of $2$, and then the exponents must be equal.

2)  $\ln(2x-5) = 1.9$
You should use the function $e^\dots$ at every side of the equation, and then see that $e^\dots$ and $\ln \dots$ cancel each other.
At the end, you must check that your solution $x$ has $2x-5>0$

3)  $\log_7(x) = 5$
Use the definition of logarithm.

5)  Explain why the quadratic equation $x^2-6x+140=0$ has no real solution for $x$.
Look at the value of the discriminant $\Delta=b^2-4ac$ and discuss it.

8)  Write the equation $5x+3y = 4$ in parametric form. That is express both $x$ and $y$ in terms of the same parameter.
Write, for instance $5x=t$, $t+3y=4$, and isolate $x$ and $y$.

This needs to be in a different forum.
1) Apply $\log_2$ to both sides and then solve"
$2^{4x}=16\Rightarrow \log_2{2^{4x}}=\log_2{16}\Rightarrow 4x=4 \Rightarrow x=1$
2) Raise $e$ to the power of each side to eliminate the logarithm; then solve.
3) Same as #2, but use $7$ instead of $e$.
4)(i) very easy
   (ii) Isolate $52e^{-4.5d}$ and apply the natural logarithm to each side.
5) $x^2-6x+140=(x-3)^2+131$  which is clearly always $\ge131$.  (Or use the quadratic formula.)
6) When dividing any polnomial $q(x)$ by any $x-a$, the remainder ($R$) is $q(a)$.
7) There are several ways.  You could make a 3x3 matrix of the coeffcients, find its inverse and then multiply it by a 3x1 matix of the right-hand constants.  You would get a matrix with the the values of $x,y,x$ in (order).
8) There are an infinite number of ways.  You could let $x=t$ and then substitute and solve for $y$ to find it in terms of $t$.
9) Expand the left side and you will see that the $4x^3$ terms cancel on both sides.  What is left is quadratic.

thanx Guys you are life savers. its not homework, got an exam in 12 hours so im clearin up the stuff i was stuck on. sorry draperp im  new to this forum so i didnt know where to post.

At least i get help in this forum, thanx again

Hayz


given a,b,c,d>0 and
$ a \plus{} b \plus{} c \plus{} d \equal{} 4$
prove that:
$ \frac{1}{{5 \minus{} abc}} \plus{} \frac{1}{{5 \minus{} bcd}} \plus{} \frac{1}{{5 \minus{} cda}} \plus{} \frac{1}{{5 \minus{} dab}} \le 1$

This is the inequality of Vasile Cirtoaje :wink:  and I found the solution is long and it's also difficult.I'll try  to solve it in another way.

[quote="Bugi"]Of course, if nothing works:
[hide="Brute Force"]
Homogenize, clear the denominators, use Schur or AM-GM[/hide][/quote]
I think that will me much more long and difficult since when you expand it,it will become huge. :wink: Would you post sth more about your idea? :)

[hide]
Let $ w\equal{}abc, x\equal{}bcd, y\equal{}cda, z\equal{}dab$ w.l.o.g $ w\le x \le y \le z$

$ g(w,x,y,z) \equal{} \sum \frac{1}{5\minus{}x}$

$ g\left(\frac{w\plus{}z}{2},x,y,\frac{w\plus{}z}{2}\right) \minus{} g(w,x,y,z) \equal{} \frac{4}{10\minus{}(x\plus{}y)}\minus{}\frac{1}{5\minus{}x}\minus{}\frac{1}{5\minus{}y}$

And $ 4(5\minus{}x)(5\minus{}y)\minus{}[10\minus{}(x\plus{}y)]^2 \equal{} 4xy \plus{} (x\plus{}y)(16\minus{}(x\plus{}y))\ge 0$ because $ 0\le x,y\le 4$ (proven below)

So from the Stronger Mixing Variables inequality.

$ g(w,x,y,z) \le g(a',a',a',a')$ where $ a' \equal{} \frac{w\plus{}x\plus{}y\plus{}z}{4}$

$ \Rightarrow g(w,x,y,z) \le \frac{4}{5\minus{}\frac{w\plus{}x\plus{}y\plus{}z}{4}}\qquad(1)$

So we wish to maximise $ w\plus{}x\plus{}y\plus{}z\equal{}abc\plus{}bcd\plus{}cda\plus{}dab$

Let $ f(a,b,c,d) \equal{} abc\plus{}bcd\plus{}cda\plus{}dac$ and w.l.o.g $ a\le b\le c\le d$

$ f\left(\frac{a\plus{}d}{2},b,c,\frac{a\plus{}d}{2}\right)\minus{}f(a,b,c,d) \equal{} \left(\frac{a\plus{}d}{2}\right)^2(b\plus{}c) \plus{} \left(\frac{a\plus{}d}{2}\right)(bc\plus{}bc) \minus{} (ad)(b\plus{}c) \minus{} abc\minus{}dbc\equal{}(b\plus{}c)\left(\frac{a\minus{}d}{2}\right)^2\ge 0$

So, again by the SMV inequality

$ f(a,b,c,d)\le f(a',a',a',a')$ where $ a' \equal{} \frac{a\plus{}b\plus{}c\plus{}d}{4} \equal{} 1$

$ \Rightarrow f(a,b,c,d)\le 4\qquad(2)$

Combining (1) and (2) yeilds

$ \sum \frac{1}{5\minus{}abc} \le \frac{4}{5\minus{}\frac{w\plus{}x\plus{}y\plus{}z}{4}} \le 1$

Equality holds when $ (a,b,c,d) \equal{} (1,1,1,1)$ and $ (4,n,\frac{1}{n},0)$ upon permutation for $ n\in \mathbb{R}$
I hope i used SMV correctly :P

[/hide]


geometry
3D geometry
arithmetic sequence
algebra unsolved
algebra

Find an infinite non-constant arithmetic progression of natural numbers such that each term is nithe a sum of two squares,
nor  a sum of two cubes(of natural numbers).

A square (of an integer) is congruent to either 0 or 1 modulo 4. Therefore the sum of two squares can be 0, 1 or 2 but never 3 modulo 4.

A cube (of an integer) is congruent to 0, 1 or -1 modulo 9 (to see this, it's enough to check the cubes of 0, 1, 2, ..., 8). Therefore the sum of two cubes can never be 3 modulo 9.

So, any number which is congruent to 3 modulo 4 and to 3 modulo 9 is neither a sum of two squares nor a sum of two cubes.

So $ 3\plus{}36n$ works, but $ 3\plus{}72n$ works just as well of course.


1) Let $f: R\to R$ sastisfying $|f(a)-f(b)| < |a-b|$ $(a \neq b )$. Prove that if $f(f(f(0)))=0$ then $f(0)=0$

2) Find $f: R\to R$ sastisfying for all $x,y,z \in R$ we have $\frac{1}{2}f(xy)+\frac{1}{2}f(yz)-f(x)f(yz) \geq \frac{1}{4}$

1)  Suppose   $f(0)\neq0$. First notice that $f(0) \neq f(f(0))$, because otherwise we would have $0=f(f(f(0)))=f(f(0))=f(0)$  and also that  $f(f(0))\neq0$, because we would have again $0=f(f(f(0))=f(0)$,  then. This means
$|f(f(0))|=|0-f(f(0)|<|f(f(f(0))-f(f(0))|<$$<|f(f(0))-f(0)|<|f(0)-0|=|f(0)|$,
as well as  $|f(0)|<|0-f(0)|=|f(f(f(0)))-f(0)|<|f(f(0))-0|=|f(f(0))|$,  which is a contradiction.

2) Letting $x=y=z=1$ we get  $f(1)-f(1)^{2}-\frac{1}{4}\geq 0 \leftrightarrow (f(1)-\frac{1}{2})^{2}\leq 0$, so  $f(1)=\frac{1}{2}$. Letting  $x=y=z=0$  yields  $f(0)=\frac{1}{2}$  in the same way. Substituiting (does that exist in English or I'm inventing?) only  $y=z=0$  gives  $\frac{1}{2}-\frac{1}{2}f(x)\geq \frac{1}{4}\leftrightarrow f(x)\leq \frac{1}{2}\;\; \forall x\in R$, while letting  $x=1$  we obtain $\frac{1}{2}f(y)+\frac{1}{2}f(yz)-\frac{1}{2}f(yz)\geq \frac{1}{4}\leftrightarrow f(y)\geq \frac{1}{2}\;\; \forall y \in R$, so $f(x)=\frac{1}{2}\;\; \forall x \in R$ is the only solution. I hope.


function
algebra
polynomial
limit
probability and stats

Let $ X$, $ Y$ be bounded real random variables. Prove that $ E(X^{n}Y^{m}) = E(X^{n})E(Y^{m})$ for all m,n implies that $ X$,$ Y$ are independent.

I don't want anyone to post the solution here - I'd only like someone to tell me if the following idea is correct:

-if $ E(X^{n}Y^{m}) = E(X^{n})E(Y^{m})$, then for any analytic functions $ f, g$ we get $ E(f(X)g(Y)) = E(f(X))E(g(Y))$
- fix $ k, l$; we want to find such random variables $ f(X), g(Y)$ that: a) $ E(f(X)g(Y)) = P(X = k, Y = l)$ and $ E(f(X) = P(X = k), E(g(Y)) = P(Y = l)$, b) f and g can be approximated by analytic functions
- then we use the fact that ${ E(lim f_n}) = limE(f_{n})$ to obtain the independence of X and Y
Does it make sense?

1/ how do you prove the things with analytic functions? it is true, but maybe it would be simpler to use polynomials
2/ I do not really understand what you said with $ k,l$ etc..
3/ you need some conditions for $ \lim E[f_n] \equal{} E[\lim f_n]$ te be true
4/ did ou use the fact that $ X,Y$ are bounded ?

[quote="alekk"]4/ did you use the fact that $ X,Y$ are bounded ?[/quote]
I can think of a number of ways to use that, some of which lead to completely different proofs. For instance, the characteristic function (Fourier transform) $ \chi(t_1,t_2)\equal{}E(e^{i(t_1X\plus{}t_2Y)})$ would then be analytic - in fact, it would be entire as a function of either $ t_1$ or $ t_2.$

1/ how do you prove the things with analytic functions? it is true, but maybe it would be simpler to use polynomials

If $ f(X)\equal{}\sum a_{k}X^{k}$ and $ g(Y)\equal{}\sum b_{l}Y^{l}$, then $ E(f(X)g(Y))\equal{}E(\sum a_{k}X^{k})(\sum b_{l}Y^{l})\equal{}\sum c_{n,m}E(X^{n}Y^{m})\equal{}\sum c_{n,m}EX^{n}EY^{m}\equal{}(\sum a_{k}EX^{k})(\sum b_{l}EY^{l})\equal{}Ef(X)Eg(Y)$

2/ I do not really understand what you said with k,l etc..

To prove that $ X, Y$ are independent we have to show that for any $ k,l$ $ P(X\equal{}k, Y\equal{}l)\equal{}P(X\equal{}k)P(Y\equal{}l)$.

4/ did ou use the fact that X,Y are bounded ?

The assumption ensures that things like $ EX^{k}$ are finite.

[quote="neworder"]1/ how do you prove the things with analytic functions? it is true, but maybe it would be simpler to use polynomials

If $ f(X) \equal{} \sum a_{k}X^{k}$ and $ g(Y) \equal{} \sum b_{l}Y^{l}$, then $ E(f(X)g(Y)) \equal{} E(\sum a_{k}X^{k})(\sum b_{l}Y^{l}) \equal{} \sum c_{n,m}E(X^{n}Y^{m}) \equal{} \sum c_{n,m}EX^{n}EY^{m} \equal{} (\sum a_{k}EX^{k})(\sum b_{l}EY^{l}) \equal{} Ef(X)Eg(Y)$
[/quote]

not that simple I think. For example $ f(x) \equal{} \frac{1}{x^2\plus{}1}$ is analytic over $ \mathbb{R}$ but
$ f(2) \neq \sum_{k} a_k 2^k$ where $ f(x) \equal{} \sum_k a_k x^k$ near $ 0$.

[quote="neworder"]2/ I do not really understand what you said with k,l etc..

To prove that $ X, Y$ are independent we have to show that for any $ k,l$ $ P(X \equal{} k, Y \equal{} l) \equal{} P(X \equal{} k)P(Y \equal{} l)$.
[/quote]
It is true for discrete random variables but absolutely not true for continuous random variables! (in the contiunous case, it is possible that $ P(X\equal{}k, Y\equal{}l) \equal{} P(X\equal{}k)P(Y\equal{}l)\equal{}0$ for every $ k,l \in \mathbb{R}$ but $ X,Y$ not independant)


Determine all integers $ n>1$ for which the inequality \[ x_1^2\plus{}x_2^2\plus{}\ldots\plus{}x_n^2\ge(x_1\plus{}x_2\plus{}\ldots\plus{}x_{n\minus{}1})x_n\] holds for all real $ x_1,x_2,\ldots,x_n$.

[quote="Johan Gunardi"]Determine all integers $ n > 1$ for which the inequality
\[ x_1^2 \plus{} x_2^2 \plus{} \ldots \plus{} x_n^2\ge(x_1 \plus{} x_2 \plus{} \ldots \plus{} x_{n \minus{} 1})x_n\]
holds for all real $ x_1,x_2,\ldots,x_n$.[/quote]
Id est, it must be $ \left(\sum_{i \equal{} 1}^{n \minus{} 1}x_i\right)^2\leq4\sum_{i \equal{} 1}^{n \minus{} 1}x_{i}^2,$
which after assumption $ x_1 \equal{} x_2 \equal{} ... \equal{} x_{n \minus{} 1} \equal{} 1$ gives $ n\in\{2,3,4,5\}.$

[hide]The inequality is equivalent to \[ \left(x_1\minus{}\frac{x_n}2\right)^2\plus{}\left(x_2\minus{}\frac{x_n}2\right)^2\plus{}\ldots\plus{}\left(x_{n\minus{}1}\minus{}\frac{x_n}2\right)^2\ge\frac{n\minus{}5}4x_n^2\] For $ x_1\equal{}x_2\equal{}x_3\equal{}\ldots\equal{}x_{n\minus{}1}\equal{}1,x_n\equal{}2$, we get $ 0\ge n\minus{}5$, so $ n\le5$. For $ n\le5$, the inequality is obviously true because $ LHS\ge0\ge RHS$. So $ n\in\{2,3,4,5\}$.[/hide]


I don't know any theory that would be particularly helpful here, but the examples are small enough to do a quick reduction: choose a row with not many allowed entries and consider the possible positions for the rook in that row.  For each position, if you erase the board and column the rook sits in you get a $ 5 \times 5$ board on which you have to place five rooks.  If you do this once or twice, you get down to small enough boards that it's easy to see the answer (e.g. any $ 3 \times 3$ board, or certainly any $ 2 \times 2$ board).

Your example number 2 is equivalent to the number of permutations in which neither $ 2i \plus{} 1$ nor $ 2i \plus{} 2$ get sent to either $ 2i \plus{} 1$ or $ 2i \plus{} 2$, for each $ i$.  This sequence doesn't exist in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url], which surprises me, but there are some signed sequences that might be the same thing.


I have 56, 923 ants that need to get to a picnic,I need to put them in groups to cross the road, the groups can be no longer than 3 feet and explain why I grouped them.  HELP ME!

Can you elaborate? I don't really understand what the problem is asking for.

How long are the ants and please explain futher.

Assuming the ants are about 1 inch (actually the biggest size of an ant but for convenience purposes), we know that 3 feet is 36 inches, and thus 36 ants.

So you will have $ 36$ ants in groups of $ 1581$ and one group of $ 7$ ants.


modular arithmetic
number theory proposed
number theory

positive integer $n$ is triperfect if the sum of its divisors is $3n$.an odd triperfect number must be a square.prove that the square root of an odd triperfect number cannot be square-free.

Let $a = \prod_{i}p_{i}^{l_{i}}$, the $p_{i}$ being prime numbers.
Then the sum of divisors of $a$ is $\sigma(a) = \prod_{i}\sum_{k_{i}=0}^{l_{i}}p_{i}^{k_{i}}.$

[quote="sam-n"]An odd triperfect number must be a square.[/quote]
A triperfect number is such that $\sigma(a) = 3a.$ If it is odd, all the $p_{i}$ are odd and $\sigma(a)$ is odd as well. Hence all the $l_{i}$ are even and $a$ is a perfect square.

[b]Claim 1 :[/b] For $n \in \mathbb{N}, \ \ 1+n+n^{2}$ does not have any odd prime factor congruent to $2 \mod 3.$
[b]Claim 2 :[/b] If an odd triperfect number which is the square of a squarefree number exists, then it does not contain any prime factor congruent to $2 \mod 3.$
[b]Claim 3 :[/b] If an odd triperfect number which is the square of a squarefree number exists, then it is not divisible by $3.$

[quote="sam-n"]The square root of an odd triperfect number cannot be square-free.[/quote]
Let us assume that $a,$ odd squarefree triperfect number exists.
Let $p_{1}$ be a prime factor of $a.$ Thus, from Claim 2 and Claim 3, we know that $p_{1}\equiv 1 \pmod 3.$
$1+p_{1}+p_{1}^{2}\equiv 3 \pmod 9$ and hence $1+p_{1}+p_{1}^{2}$ is divisible by $3,$ but not by $9.$ Thus there exists a divisor $p_{2}\neq 3$ of $1+p_{1}+p_{1}^{2}.$ So $p_{2}|a,$ and again, $p_{2}\equiv 1 \pmod 3.$
Hence $3|1+p_{2}+p_{2}^{2}.$
Therefore, $9|\sigma(a)$ so that $a$ has to be divisible by $3,$ which contradicts Claim 3 and finishes the proof.


[b]Γεια σε ολους!!!!!Βλεπω μετα τις πανελληνιες το forum ξαναζωντανεψε!!!!![/b]
Εστω α,β θετικοι ακεραιοι τετοιοι ωστε
15α+16β και
16α-15β να ειναι τελεια τετραγωνα.
Να βρεθει το ελαχιστο απο τα δυο τελεια τετραγωνα.

ΡΑΛΛΗΣ :D

Πραγματι η ασκηση σου Ραλλη ειναι καλη.Πρεπει να ειναι παλιο θεμα σε ΔΜΟ.Οσο για τα θρησκευτικα παιδεια ελπιζω να  πιασατε την βαση! :rotfl:

Φίλε  Ράλλη  μου  φαίνεται  οτι  την  άσκηση  την  έχεις  διατυπώσει  λίγο  λάθος  ......... ;)  :(  Αν  θυμάμαι  καλά  ήταν  θέμα  του  Imo το  1996 )..... Δεν  έλεγε  λοιπόν  η  άσκηση  να  βρεθεί  ο  μικρότερος  από  τους  δύο  αλλά   να  βρεθεί  η  ελάχιστη  δυνατή  τιμή  που  μπορεί  να  πάρει  το  μικρότερο  από  τα  δύο τετράγωνα   )

Sorry παιδια,εχει δικιο ο Δεσκος.Η σωστη εκφωνηση ειναι οπως την ειπες,και το θεμα ειναι οντως απο την ΙΜΟ 1996 απλως το εβαλα γιατι μου αρεσε σαν θεμα, γιατι ειναι οπως θα επρεπε να ειναι ολα τα ολυμπιακα θεματα.Απαιτει ελαχιστες θεωρητικες γνωσεις(ενα θ.Fermat το πολυ-πολυ, αν και λυνεται και χωρις αυτο)και ομως δεν ειναι θεμα ΘΑΛΗ!!!,αλλα 4ο θεμα της ΔΜΟ 96!Δεν θελω να το θεοποιησω, αλλα προσωπικα πιστευω οτι μονο τετοιου ειδους θεματα πρεπει να ειναι χαρακτηριστικα των ΔΜΟ!(ΚΑΙ ΣΙΓΟΥΡΑ ΟΧΙ ΕΠΑΝΑΛΗΨΕΙΣ ΑΥΤΩΝ ΟΠΩΣ Π.Χ. ΣΤΗΝ ΒΑΛΚΑΝΙΑΔΑ!)

ΡΑΛΛΗΣ :D

Έχεις δίκιο .Και νομίζω ότι θα συμφωνήσετε και στο γεγονός ότι στην ΙΜΟ υπάρχουν πολύ ωραία προβλήματα τα οποία θα είναι οποιασδήποτε δυσκολίας.....Ας πούμε εχουν μπει  εύκολα (σχετικά πάντα) θέματα αλλά και εξαιρετικά δυσκολα .Όμως και στις ΙΜΟ σεν υπάρχει πάντα πρωτοτυπία.....Δυστυχώς το πιο πρόσφατο παράδειγμα γνωστού προβλήματος σε ΙΜΟ είναι στην ΙΜΟ που έγινε στη χώρα μας το 2004 .Το τελευταίο θέμα ήταν γνωστό (ήταν στη Shortlist 1994 νομίζω) και ήταν ένα από τα λίγα έκτα προβλήματα ολυμπιάδας που λύθηκαν από τόσο μεγάλη πλειοψηφία μαθητών.  ;)
Πέρσυ όλα ήταν ωραία πάντως .................
 :lol:


function
topology
abstract algebra
group theory
advanced fields
advanced fields unsolved

can the group $S_{\infty}$ of all permutation of integers be so metrized that the the group of operation (composition of permutations ) is continuous function and the set $S_{\infty}$ becomes under this metric a compact space?

I believe the answer is negative, and, moreover, no metrizability assumptions need to be made. In other words:

$S_\infty$ cannot be topologized so as to become a compact Hausdorff space with continuous multiplication.

First let’s reduce the problem to the case when $S_\infty$ is a topological group, i.e. the map $x\mapsto x^{-1}$ is also continuous.

Suppose we can find a convergent [url=http://en.wikipedia.org/wiki/Net_%28mathematics%29]net[/url] $(x_\alpha)_\alpha\to x$ such that $x^{-1}_\alpha$ lies outside some fixed open neighborhood $U$ of $x^{-1}$ for all $\alpha$ (the existence of such a net is equivalent to the assumption that $x\mapsto x^{-1}$ is not continuous). By the compactness of $S_\infty\setminus U$, we can find a subnet $\left(x^{-1}_\beta\right)_\beta$ of $\left(x^{-1}_\alpha\right)_\alpha$ converging to some $y\in S_\infty\setminus U$. By the continuity of multiplication, we have $1=x_\beta x^{-1}_\beta\to xy$, so $y=x^{-1}\in U$. We have a contradiction.

Now assume $S_\infty$ is a compact Hausdorff group.

From the representation theory of compact groups, it follows that for every neighborhood of the identity in such a group, there is a finite-dimensional representation whose kernel is contained in that neighborhood. $S_\infty$ only has two proper non-trivial normal subgroups: the finite permutations and the even finite permutations (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=group+symmetric&t=113751]this[/url] topic; there's no proof there, but mlok guessed pretty well :)). From these two results it follows that $S_\infty$ has a faithful finite-dimensional representation, and is hence a subgroup of $\mbox{Gl}(n,\mathbb C)$ for some $n$. Any subgroup of $\mbox{Gl}(n,\mathbb C)$ of exponent $2$ is finite (of size $\le 2^{n}$, actually); $S_\infty$ is easily seen to have an infinite subgroup of exponent $2$, so we have our contradiction.
/url]).


When the three integers $ 618,343,277$ are divided by a positive integer, $ d$, where $ d>1$, the remainders are the same. What is the smallest possible value of $ d$?

[hide="Answer"]$ 11$[/hide]

[quote="xpmath"][hide] We can write
618=ad+m
343=bd+m
277=cd+m.
Subtracting the second from the first, we have 275=d(a-b) and subtracting the third from the second, 66=d(c-b). Clealry, d must divide each of 275 and 66, and the largest is 11.[/hide][/quote]
Ah... I get it. Wow.. I couldn't see that..


[quote="can_hang2007"]Let $ a,b,c$ be [b]real [/b]numbers. Prove that
\[ \sum_{cyc} a^4 \minus{} 2\sum_{cyc}a^3b \minus{} 4\sum_{cyc}a^3b \plus{} 4abc\sum_{cyc}a \plus{} 5\sum_{cyc} a^2b^2 \ge 0
\]
:)
P/s: Notice that the equality does not hold for $ a \equal{} b \equal{} c$. ;)[/quote]
Maybe we need to prove that $ \sum_{cyc}(a^4\minus{}2a^3b\minus{}4a^3c\plus{}2a^2bc\plus{}5a^2b^2)\geq0$ $ ?$

[quote="arqady"][quote="can_hang2007"]Let $ a,b,c$ be [b]real [/b]numbers. Prove that
\[ \sum_{cyc} a^4 \minus{} 2\sum_{cyc}a^3b \minus{} 4\sum_{cyc}a^3b \plus{} 4abc\sum_{cyc}a \plus{} 5\sum_{cyc} a^2b^2 \ge 0
\]
:)
P/s: Notice that the equality does not hold for $ a \equal{} b \equal{} c$. ;)[/quote]
Maybe we need to prove that $ \sum_{cyc}(a^4 \minus{} 2a^3b \minus{} 4a^3c \plus{} 2a^2bc \plus{} 5a^2b^2)\geq0$ $ ?$[/quote]
No, arqady. It must be
\[ \sum_{cyc}(a^4 \minus{} 2a^3b \minus{} 4a^3c \plus{} 4a^2bc \plus{} 5a^2b^2)\geq0\]
:)


I need to use $ \text{\LaTeX}$ in a thing I'm writing, and I use OpenOffice Writer as a word-processor. Does anyone know how get $ \text{\LaTeX}$ on to my document?

Have a look at [url=http://ooolatex.sourceforge.net/]OOoLatex[/url].

However when I first used $ \text{\LaTeX}$ I just went to texer and copy pasted images i needed in.

I'm using a LOT of $ \text{\LaTeX}$ and that would be too cumbersome.

Actually in my opinion, TeXnic center is almost as fast and way better looking than microsoft.


Ascertain the general term of the sequence
$\boxed {\ x_1\ge 0,\ x_{n+1}=|x_n-n|,\ n\in N^*\ }$
and prove that $\lim_{n\to \infty} \frac{x_n}{n}=\frac 12.$

Hmm.... I think that is

x(i) + x(i+1) = n

x(i) = x1 (mod n)
x(i) >=0
x(i) < n

(if n divides x1, then x(i)=0=x(i+1))

Angel "Java" Lopez

[i]Ajlopez and JohnConstantine,[/i] $x_i$ are real non-negative numbers for any $i\in N^*$ !


Here is an interesting fact :

Möbius strip :
http://en.wikipedia.org/wiki/M%C3%B6bius_strip

The Canadian Mathematics Society's (CMS) Symbol is a Mobius Strip with a half maple leaf next to it.

i didnt read all of it..but ya that is quite interesting..nice find lomos_lupin..


Let equilateral triangle $ABC$ inscribed circle (C) with any $M \in (C)$ denote $d_{a}$ is distance from $M$ to line $BC$ similar denote $d_{b},d_{c}$ prove that
$d_{a}^{2}+d_{b}^{2}+d_{c}^{2}=const$ when $M$ move on $(C)$. :)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58199]Topic 58199 - Another incircle[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=59086]Topic 59086 - Locus[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=849]Topic 849 - equilateral triangle[/url]

and I'm sure that there are more  :D

Hi pontios I think you read wrong  :D because $d_{a},d_{b},d_{c}$ is distance from $M$ to [b]lines[/b] $BC,CA,AB$ not to $A,B,C$, in your links I don't see my problem  :D  :D

We can set the given  equilateral triangle such that three points $A(0,\ r),\ B\left(-\frac{\sqrt{3}}{2}r,\-\frac{1}{2}r\right),\ C\left(\frac{\sqrt{3}}{2}r,\-\frac{1}{2}r\right)\ (r>0),$ on $x-y$ plane,

we have the equations of three lines $BC,\ CA,$ and $AB$ as follows.

$BC: \ y=-\frac{1}{2}r,$

$CA: \ y=-\sqrt{3}x+r\Longleftrightarrow \sqrt{3}x+y-r=0,$

$AB: \ y=\sqrt{3}x+r\Longleftrightarrow \sqrt{3}x-y+r=0.$

Use the formula of the distance between point and line,

$d_{a}=r\left|\cos \theta+\frac{1}{2}\right|,$

$d_{b}=\frac{1}{2}r\ |\sqrt{3}\cos \theta+\sin \theta-1|,$

$d_{c}=\frac{1}{2}r\ |\sqrt{3}\cos \theta-\sin \theta+1|.$

Thus $\frac{d_{a}^{2}+d_{b}^{2}+d_{c}^{2}}{r^{2}}=\left(\cos \theta+\frac{1}{2}\right)^{2}+(\sqrt{3}\cos \theta+\sin \theta-1)^{2}+(\sqrt{3}\cos \theta-\sin \theta+1)^{2}$

$=\frac{3}{2}(\sin^{2}\theta+\cos^{2}\theta)+\frac{3}{4}=\frac{9}{4},$ yielding $d_{a}^{2}+d_{b}^{2}+d_{c}^{2}=\frac{9}{4}r^{2}.\ \ Q.E.D.$

Thank Virgil Nicula and kunny for two very good solutions ! Really I found this problem base on general with any $M$ on plane (don't need $\in (C)$) then $d_{a}^{2}+d_{b}^{2}+d_{c}^{2}=\frac{3}{2}MO^{2}+\frac{3}{4}R^{2}$ I think it quite nice I found it but I don't know am I a first man found it ?, I will post proof of this problem later !!Thank you again! :)  :)


I hope it's not too easy problem to post on here. :? But I just thought I'll get better help from here.

Problem #3

Determine, with a mathematical proof, the value of $\lfloor x \rfloor$; i.e., the greatest integer less than or equal to $x$ where:

$x = 1 + \frac {1}{\sqrt 2} + \frac {1}{\sqrt 3} + \frac {1}{\sqrt 4} + ... + \frac {1}{\sqrt {1000000}}$

[hide="Um"]

I know how to write this in sigma notation.

$\sum^{1000000}_{i=1} \frac {1}{\sqrt i}$

But this doesn't help me how to get the answer...[/hide]

Help please! Do not give the answer away! :lol: This problem looks cool. I want to try it and get it right by myself!

Hint: this is an inequality (well, xxreddevilzxx answered it completely)

Sorry, silverfalcon. I didnt read what you said carefully. now i feel like i ruined your studying.... :(

BTW, ask any question about that problem if you have any, through aim.


inequalities proposed
inequalities
algebra

[quote="SUPERMAN2"]Let a,b,c be the positive real number.Prove that:
$ \frac {a}{(a \plus{} 1)(b \plus{} 1)} \plus{} \frac {b}{(b \plus{} 1)(c \plus{} 1)} \plus{} \frac {c}{(c \plus{} 1)(a \plus{} 1)}$≥$ \frac {3}{4}$[/quote]
Try $ a\equal{}b\equal{}c\equal{}2.$ :wink:

Sorry,there are some mistakes.I have corrected.

If so it's obvious after expanding: $ a\plus{}b\plus{}c\geq3$ and $ ab\plus{}ac\plus{}bc\geq3.$

[quote=SUPERMAN2]Let $a,b,c$ be the positive real number satisfying $abc=1.$  Prove that:
$$\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\geq \frac {3}{4}$$[/quote]

Let $a,b,c$  be the positive real number satisfying $a+b+c=1.$ [url=https://artofproblemsolving.com/community/c6h2520953p21391892]Prove that[/url]$$\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\le\frac{9}{16}$$
[url=https://artofproblemsolving.com/community/c3h1092498p4868541]h[/url]
If $ a,b,c>0$ and $ abc=1$. [url=https://artofproblemsolving.com/community/c6h268304p1454338]Prove that\[ \frac {a^3}{(b+1)(c+1)} +\frac {b^3}{(c+1)(a+1)}+\frac {c^3}{(a+1)(b+1)} \ge \frac 34 \]
For $ a_1,\ldots,a_n>0$ with $ a_1\ldots a_n=1$ and $ p,q>0$ with $ (p+1)(q+1)\ge (p+q+1)n$ [url=https://artofproblemsolving.com/community/c6h223573p1242805]Prove that\[ \sum_{i=1}^n\frac1{(a_i+p)(a_i+q)}\le\frac{n}{(p+1)(q+1)}.\]

[quote=SUPERMAN2]Let a,b,c be the positive real number satisfying abc=1.Prove that:
$ \frac {a}{(a \plus{} 1)(b \plus{} 1)} \plus{} \frac {b}{(b \plus{} 1)(c \plus{} 1)} \plus{} \frac {c}{(c \plus{} 1)(a \plus{} 1)}$≥$ \frac {3}{4}$[/quote]

we use $abc=1$ in $(a,b,c)=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})$
Then its sufficient to prove $\frac{zx(z+x)+yz(y+z)+xy(x+y)}{(x+y)(y+z)(z+x)}\ge\frac{3}{4}$
let $k=\prod_{cyc}(x+y)$ we get:
$\frac{k-2xyz}{k}\ge\frac{3}{4}$ from this we get: $k\ge8xyz$ which is true


Πρόβλημα:
The set of all positive integers is partitioned into several arithmetic progressions. Show that there is at least one among these arithmetic progressions whose initial term is divisible by its difference.

Και μια δική μου απορία :blush: :
Μπορείτε να βρείτε παράδειγμα τέτοιων προόδων που να μην προέρχονται από χωρισμό του Ζ σε κλάσεις ισουπολοίπων όπως στις ακολουθίες :$ a_{n}\equal{}0\plus{}3(n\minus{}1), a_{n}\equal{}1\plus{}3(n\minus{}1), a_{n}\equal{}2\plus{}3(n\minus{}1)$;

Πρώτ' απ' όλα η ασκηση δεν ισχύει στην περίπτωση που επιτρέπουμε στο  $ \mathbb{N}$ να διαμεριστεί σε άπειρες προόδους. Ένα αντιπαράδειγμα δίνεται από τις εξής προόδους:

\[ A_k\equal{}\{2^k(1\plus{}2n|n\geq 0\}\quad k\equal{}0,1,2,...\]

Η διαφορά της $ A_k$ είναι $ 2^{k\plus{}1}$ και ο πρώτος της όρος είναι $ 2^k$.

Ας περάσουμε τώρα στην περίπτωση που η διαμέριση είναι σε πεπερασμένο πλήθος προόδων. Έστω $ a$ το ελάχιστο κοινό πολλαπλάσιο των διαφορών των προόδων αυτών. Τότε το $ a$ ανήκει σε μια από τις προόδους αυτές. Αν είναι ο πρώτος όρος της προόδου αυτής , τότε τελειώσαμε. Αν όχι τότε ας συμβολίσουμε με $ d$ τη διαφορά της προόδου αυτής. Τότε ο πρώτος όρος της προόδου θα έχει τη μορφή $ a\minus{}nd$ και είναι συνεπώς διαιρετός από το $ d$.

Για τη δεύτερη ερώτησή σου, πρέπει αρχικά να παρατηρήσεις ότι κάθε αριθμητική πρόοδος είναι το τελικό κομμάτι μιας κλάσης ισουπολοίπων φυσικών αριθμών. Παρ' όλα αυτά μπορείς εύκολα να διαμερίσεις το  $ \mathbb{N}$ σε αριθμητικές προόδους με διαφορετικές διαφορές:

\[ 3n,1\plus{}3n,2\plus{}6n,5\plus{}6n\]

Ευχαριστώ Jondrette! Η άσκηση ήταν για διαμέριση σε πεπερασμένο πλήθος προόδων-συγγνώμη που δεν το τόνισα.

[quote="Demetres"]Ισχύει όμως κάτι άλλο.

Όπως και να διαμεριστεί το σύνολο των ακεραίων σε πεπερασμένο πλήθος αριθμητικών προόδων, θα υπάρχουν τουλάχιστο δύο προόδοι με την ίδια διαφορά.

(Όμορφο αλλά αρκετά δύσκολο πρόβλημα.)[/quote]

Αυτό είναι ισοδύναμο με το ότι μια κυκλική ομάδα δε μπορεί να διαμεριστεί σε  σύμπλοκα τα οποία ανά δύο να έχουν διαφορετικές πληθικότητες...

[quote="george2"]Α, ωραία. Κάτι μου φάνηκε ότι γίνετε με τις διαφορές σε κάποιες δοκιμές που έκανα αλλά δεν μπορούσα να καταλήξω σε κάτι συγκεκριμένο. Demetres έχεις κάποιο link για αυτό το θεώρημα ή γενικά για κάτι σχετικό με τη διαμέριση σε προόδους?[/quote]

Γιώργο, έχω κάποιο link αλλά θα προτιμήσω να την αφήσω για λίγο πρωτού το δώσω. Πιστεύω ότι όσο δύσκολο και να είναι ένα πρόβλημα πάντα κερδίζουμε κάτι όταν αφιερώσουμε χρόνο να το σκεφτούμε άσχετα με το αν τα καταφέρουμε να το λύσουμε ή όχι.

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

Να προσθέσω ότι όπως έγραψα την άσκηση είναι λανθασμένη. Πρέπει να προστεθεί ότι οι ακέραιοι διαμερίζονται σε τουλάχιστον δύο αριθμητικές προόδους.

Διάβασα για το πρόβλημα αυτό και την ιστορία του στο πρώτο κεφάλαιο του "The Mathematical Coloring Book". Νομίζω ότι οι περισσότερες σελίδες αυτού του κεφαλαίου μπορούν να διαβαστούν μέσω του googlebooks.

Ευχαριστώ Demetres!  :lol:  Όντως όλες οι σελίδες του πρώτου κεφαλαίου μπορούν να διαβαστούν μέσω του googlebooks...


First round is this week.  Who is participating?

Good luck to everyone who hasn't done it yet...

And I guess that after Friday this can be a discussion thread, but not until....  So I won't say anything else, yet...

Same here, Mandelbrot needs way more publicity...some people even go as far to think it's a llama! :lol:

Don't discuss the problems as soon as the week is over. I know of some schools that don't take the contests until several weeks later. At my school, we sometimes took the team contest late since it was hard to fit the schedules of all the people on our team.

i think im doing team next week.  Anyone doing regional div?

Several weeks later?  Wow, that is asking for trouble.  I can see doing the team contest the following week because of time problems, but when the rules say that you have to do the contest within the official week, doing it that much later is really pushing it.  Not that I am saying that because such action is not very reasonable we should still discuss things immediately, but I do think that something is very wrong with delaying that long.  Admittedly, it's not the AMC, but taking it so late does bring the validity of some results into question.

Well said, setofallsets.

But I would advise against discussing the problems for a few weeks anyway in case there are people who are taking the Mandelbrot late.

I suppose we could have a score poll, i.e. how many points did you get on individual...

Given how well we didn't do on Mandelbrot considering that our team had a 3-time USAMO qualifier, a 2-time USAMO qualifier, and two more who could make USAMO on a good day but didn't, I don't think there's much of a reason to question the validity of our results.

I got an 11. We had 2 perfs in my school, and I am kinda disappointed cuz i didnt notice something so simple. I knew the method too. Ill talk more about this when all of you guys have taken it. But if u get a hold of a test, its the last one. Btw, anyone know wat the prizes they give out (to the top people) are?

They give Tshirts and frisbees as prizes.

The frisbees are really cool.  That T-shirt is definitely the nerdiest piece of clothing I own, though :D.

jelyman, are you talking about the regional or national division?

Simon, I don't think you can be correct about the Mandelbrot.  The official grading day for the team round is November 1.  If anyone takes it after that, there score can't really matter, anyhow.

If someone takes it after that, it will get graded with the next round, unless we're talking about the last round. I don't know what happens in that case. Anyway, the rules aren't strictly enforced.

Got an 11, made a moronic mistake. Bah. Somehow this round seems easier than any from last year...

If you say what your score is, please also say which contest you took. (They were called A and B last year; I think the names changed to something else this year though.)

I took the regional one, which was B last year. How do u guys think u are going to do by the end? Im predicting, 50-53 for myself.

I got 2 1-point questions wrong on the regional, so whatever that comes out to be.  (edit: just found out that that's a 12).

Do they get harder throughout the year, or are they about the same?

They generally don't change very much from contest to contest.  What is this regional-national bit y'all are talking about?

Each round is worth 14 points. I think they do get a bit harder as the year goes on, but my scores last year decreased through the first three rounds and then increased for the fourth round, so maybe it's not quite that simple.

Oh.  Does the fact that they call it regional mean anything?  Like, are you only competing against teams from your own region?

no, it's just a name change, i believe.  Regional is the same as the previous B division, etc.

Well, regional splits into two sections: East and West.  National has only one (as the name suggests) section.

Got 14 on B/regional.

[quote="MysticTerminator"]blah. my school doesn't do it but i really want to do the questions :(. would it be illegal or something to post them here after the first round?[/quote]0

um . . . so . . . could you post the problems?

I just realized:

[quote="Mandelbrot Rules"]Individual scores should be entered at any time before the corresponding grading weekend. (Late scores will be recorded but may not appear with the results for that round.) For example, scores for round one should be reported no later than Friday, November 14, 2003.[/quote]
(This is taken directly from [url]http://www.mandelbrot.org/[/url].)

As far as I understood, people can still submit results for round 1 up until November 14th.  Therefore we shouldn't discuss the problems for a while still.[/quote]

That's submitting results, not taking the exam.  Different events.  Everyone should have taken the exam a long time ago.

[quote="cats..."]so can we post questions, even after the deadline?

I remember most questions from the regional exam and would post them if it's alright.[/quote]

What do you mean by [i]even[/i]?  I think it's more like "let's not post questions [i]until[/i] after the deadline just in case".

I have the questions right here, but I'm not about to do anything potentially illegal [i]just yet[/i]. ;-9

I say 'even' cause isnt this what you were trying to prevent in the other threads about the MAML Olympiad?  They actually sell these contests in books on their website; they're not distributed freely like the maml olympiad.  I dont understand your logic, but whatever.  I personally dont have a problem with people posting questions, but if we're going to make up rules they should be enforced pretty consistently.

Mmm, I guess this is a special case since we are in a site made by the co-creators of Mandelbrot...I guess it's all up to them.

Notice the connection in between here and Mandelbrot through the Virtual Classroom? :wink: :wink: