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(to clear things up, yes I did search DDR using the search function but topic necromancery isn't my speciality, if you get what I mean >_>)

Any of you guys play DDR? Me and my friends play DDR (and some variations, like ITG and StepMania). I've been playing for a while now, and I think I'm getting pretty good at it. At least, I hope so... I also make stepfiles on StepMania... (for those of you who knew me at MathPath, I got DDR in August 2005 :P )

btw, if anybody wants to compare, I can clear every song in DDR on Heavy/Challenge/Oni/whatever you call it except for PSMO. A's on all 9-footers except for Matsuri Japan and Cartoon Heroes, B's and C's on most 10-footers (Sakura I get A's on, PSM I get D's on)

when I get my pad fixed (I use a RedOctane pad), I'll probably post pictures...

I've only ever played DDR Mario Mix, and only for a total of maybe two hours, since my cousins brought it over when they came down from Oregon for Christmas. I only have a GameCube, so the only way I could play any other versions is at an arcade. I'm looking to get DDR: Mario Mix once it's back in stock, at least.

Used to play it. The arcade where I live closed down...

I used to play...but i might be getting back into it...

I remember that at my peak, i could AA most 9-footers.  I had an A on Max 300 (and a B while solving a rubik's cube!  :D ), something like a B or C on MaxX Unlimited...

I rarely went to the arcade, but i remember something like an A on Sakura, don't remember what for Bag, and something or the other for PS.  I might get back into it...

[quote="Treething"][quote="DarkKnight"]Oh. Very nice Treething. I play ITG, SM, DDR, and occasionally 02jam. I'm ok at DDR. Really the only things I can't pass are PSMO and Legend of Max. On Stepmania I'm ok as well compared to the community. If you know these songs, I can AA Air, Ixion, Ageha and a couple more well known songs. I don't step my own files though.[/quote]

Try playing Max Forever (heavy, challenge is insane) on pad...

o_O can't pass LoM but you can pass pandy?

wtf mate.[/quote]

Oh sorry. I meant all the DDR songs except LoM and PSMO. I'm a long way from passing all the ITG songs if I know what "pandy" means.

[quote="DarkKnight"][quote="K81o7"]I also used to play stepmania...of course, the limit there is much higher.[/quote]

Is that where your name comes from? I have a song called K81o7 or something like that.[/quote]

...sorta...ssshhh its a secret

there are other reasons for my name, but that happened to be one coincidence, so i left it that way.

Good song though.  Good stepfile too (the most common one).

[quote="Treething"][quote="DarkKnight"][quote="Treething"]pandy = pandemonium :P .

13-footer, 800+ steps, 330 bpm.

It's actually quite easy if you have enough stamina (I don't. 800+ 8th notes nearly in a row are way too much).[/quote]

Yeah I can't pass pandemonium. I don't really have the stamina either. Btw, I use a cobalt flux. Is your red octane metal or just normal?[/quote]

ignition v3.0, afterburners + me = hate

I believe I can pass most well-known difficult songs I haven't done after a couple tries (HoTN x3, ! x2, Max Forever, Bloodrush x4 or so) on a modded Cobalt Flux (cobalt flux enhanced with foam and duct tape), but I haven't played DDR in three weeks or so due to my faulty RO pad which I'm hopefully getting replaced tomorrow.

Last time I played DDR was on one of those cheap soft pads, AA'd Silent Hill (O) and , cleared MaxX, A'd R&P, CSIFILSM, So Deep, Sakura, and a couple more scores I don't quite remember.[/quote]

Ahh, I see. I just played ITG. I'm a bit shaky.  :( You have very nice scores.

[quote="Treething"]eh

16th streams are fun up to 250, then they become bs[/quote]

no they are best at 250+ like in OML 1.2x ok?

too bad 200 * 1.2 is 240. OML 1.2 is fun though.

alkara get out. out.

Reality.  :D

I actually got to the double jackhammers once in Blur.  But I'm not really good...the best scores I can really boast are maybe A on One more Lovely, FC on Quasar 1.2, B on K81o7, etc...

Well I actually haven't played in a long time so I probably suck now.

FC on p8107 =)

that one part always threw me into a good, i forget where it was but it was near the middle, never had problems getting under 30 greats or so but I'd always have a good or two

B'd vortex challenge (EVEN BATMAN HATES THIS SIMFILE) by zaghurim

fury of the storm is nice... i don't have it on my computer though, anybody willing to email it (meatcar@gmail.com) or provide a link for it to me? I want the full version, not the cut on bemanistyle 7.0 or anything.

If you like ddr, you should try ffr. Its ddr on the computer! It's really cool, fun, and challenging. It has a lot of songs and after you register you can buy some by earning credits for passing songs... You should try it:

[url]http://www.flashflashrevolution.com[/url]

[quote="junggi"]is ffr just ddr on comp?with arrow keys?
then there is a MUCH better game like that online.
It's called O2JAM. I played on the korean server b4 and it was awesome.
I think the US server's here.

O wait you might have to pay to play some harder songs.[/quote]

o2jam is fun but the free songs are incredibly easy,,,

yeah....
do you play the us server?
never played on that before

[quote="K81o7"][quote="laughinghead505"][quote="Elemennop"]FFR is stupid

try searching the stepmaniaonline forums for fury of the storm, I'm sure somebody has a link to them there[/quote]

It's not stupid :mad: Why do you think so?[/quote]

Well last time I checked, you couldn't put in your own songs, the timing system was off, and you couldn't make your own stepfiles.

Maybe I just thought the timing was off becuase it looked weird...[/quote]

Well, how can you put your own songs in on anything? :huh: They would need to have steps with them... The timing thing isn't off... :wink: What's a stepfile?

Well that's the point of stepmania.  You can put your own songs, then make steps to go with them.  And there are lots of options.  You have just about complete freedom.

Next time, actually look at what we're talking about.


Draw a line
draw two diffrent cercles whose centers are on the line
For points A, B, C, D; appear  from the intersection of the circles with the line,from  left to right on the line.
I AND J are the intersection of the two circles.
Line IJ crosses the first line in K.
P is a point on line IJ outside the circles.
Line PB intersects left circle on M
Line PB intersects right cicle on N
Show that lines AM , DN , and IJ converge to a common point ( in french we say that the line are "concourantes"
I proved it using power point and homothetic considerations
But i failed to do it usind Ceva theorem.
How could we do it .

[quote="senouy"]in french we say that the line are "concourantes"[/quote]
In English, "concurrent."  (Also, we would probably say "intersect at a common point" rather than "converge on a common point.")

If I'm not mistaken this is IMO 1995 #1. This is how I solved it:

It's enough to show that $ Q \equal{} AM\cap DN$ lies on $ IJ$, the radical axis of the two circles. For that we need to prove that $ QM\cdot QA \equal{} QN\cdot QD$ i.e. $ AMND$ is a cyclic quadrilateral.

We have, $ BP\cdot PM \equal{} CP\cdot PN\implies \frac {BP}{CP} \equal{} \frac {PN}{PM}\implies BMNC$ is cyclic. Thus

$ \angle CMN \equal{} \angle CBN$

$ \implies \angle CMN \equal{} 90^{\circ} \minus{} \angle CDN$

$ \implies \angle CMN \plus{} \angle AMC \equal{} 90^{\circ} \plus{} 90^{\circ} \minus{} \angle CDN$

$ \implies \angle AMN \equal{} 180^{\circ} \minus{} \angle CDN$

Hence $ AMND$ is cyclic.


[quote="hanschinaman"]a,b,c>0, prove:
$ \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\geq a+b+c+\frac{4(a-b)^{2}}{a+b+c}$[/quote]
It's wrong.  :(
Try $ a=1,$ $ b=4$ and $ c=2.$ :wink:

Actually the inequality is true, $ a=1$, $ b=4$, $ c=2$ implies $ 147 \geq 144$.
This problem is from the BMO 2005. See here: [url]http://www.mathlinks.ro/viewtopic.php?p=225004#225004[/url]

NguyenDungTN, I admit your outstanding talent for inequality :)


Why are there no integer solutions to x^2-5y=27?

This equation is equivalent to $ x^{2}= 5y+27$. Since 2 is not a square residue mod 5, this equation has no integral solutions.


abstract algebra
linear algebra
matrix
superior algebra
superior algebra solved

Hello,

let R be a (not obligatory commutative )ring with unity
let M be a left R module

when M is simple( only has itself and trivial subspace) its endomorphisms form a division ring

now apparently the converse is not true

I was looking for a way to see this but it wouldn't work

All hints are welcome

So any ideas?

Or is this the wrong place?

I did some inquiries and found that every simple module is indecomposable but not the converse.  Then I was hoping to find a nonsimple indecomposable module, but that is no good either, as a module is indecomposable if and only if it only has zero and identity as idempotent endomorphisms.

So despite that effort I am stuck again. :(

take the ring to be the group ring of $Z_2$ in characteristic two. Consider the 2 dimensional module on which the nonidentity element of $Z_2$ acts (with respect to some basis) as $\left(\begin{array}{cc}1 &1 \\0 &1 \\ \end{array}\right)$. This in indecomposable but not irreducible as it  has two factors that are both trivial. the endomorphism ring is a field.

thanks but I am afraid I don't understand much of it.

Isn't $Z_2$ always in characteristic two?

What exactly is a group ring[url]http://en.wikipedia.org/wiki/Group_ring[/url]
didn't make it very clear for me.

Perhaps it is not necessary I fully understand group ring for this problem.  Could you just explicitly state what your ring is, your module, and your scalar multiplication?

sorry ... here's a clearer (hopefully) picture. I will fake my way through gp rings.
 $R=Z_2[X]/I$ there $I=(x^2-1)$. The module is $M=Z_2^2$. Make $x+I$ act via the matrix above and extend lienarly to an action of $R$. Prove $M$ is indecomposable (it only has one submodule) and gives the example you want

Thank you

However , it would be impolite for me not to give one myself if it helps someone
my professor gave me one too

let $A$ be the ring of upper diagonal two dimensional matrices, and consider now $A$ as left module over itself

consider the left module generated by$\begin{bmatrix} 0&0\\0&1\end{bmatrix}$

it is not simple but its endomorphism ring is  isomorphic with the division ring of the reals


inequalities
number theory proposed
number theory

Let $f_n$ be the $n$th Fibonacci number ($f_0 = 0$, $f_1 = f_2 = 1$, $f_3 = 2$, etc).

Find all positive integers $n$ such that \[nf_nf_{n + 1} = \left(f_{n + 2} - 1\right)^2.\]

$ nf_nf_{n+1}=n\sum_{i=1}^{n}f_{i}^2 $ and $ (f_{n+2}-1)^2=(\sum_{i=1}^{n}f_{i})^2 $ and a very well known inequality (and condition when it's a equality) gives the answer  ;)


linear algebra
matrix
vector
geometry
analytic geometry

can u people give me hard matrix problems using nothing but matrix multiplication (any size, but not a brute force multiplication problem), 2x2 and 3x3 determinants, and 2x2 inverses? (pure matrix problems, nothing with vectors or analytic geometry etc.)

like something that requires thinking, not calculations
use only basic stuff

i know thats kinda hard but if you can, thanks!

Problem 1.

Compute $ \left [ \begin{array}{cc} \alpha & 1 \\
0 & \alpha \end{array} \right ]^n$.

[quote="JoeBlow"]Given two squares on a chess board, what is the fewest number of moves that a knight needs to make to get from one square to the other?[hide="hint"]Number the squares from 1 to 64, and let $ A$ be a $ 64\times 64$ matrix with $ A_{ij} \equal{} 1$ if in one move a knight can get from square $ i$ to square $ j$, and $ A_{ij} \equal{} 0$ if it cannot.  What does $ A$ represent?  What does $ A^2$ represent?  What about $ A^n$?

Try starting with a small chessboard; say $ 3\times 3$.[/hide][/quote]
hmm...
i never thought that this could be done with matricies! can someone show a solution?
and btw, too hard HSB...

Matrices are HSB (Alg II, to be precise, at least, that's how I learned them)...

And the idea is that we are giving blackbelt problems, so it kinda ruins it if someone posts a solution.

Uh...Joe, I'm pretty sure that your hint meant "8 X 8" not "64 X 64". Unless I missed something, the matrix should be the same size as the chessboard.

No, the solution I have in mind uses powers of the adjacency matrix corresponding to a knight's possible moves on the chessboard, so its dimension is $ n^2\times n^2$ for an $ n\times n$ board.
[hide="answer"]The fewest number of moves is $ \min \{k\in \mathbb{N}: \left(A^k\right)_{ij}\neq 0\}$ where $ A$ is the board's adjacency matrix, $ i$ is the number of the square currently occupied, $ j$ is the number of the target square.  If $ \left(A^k\right)_{ij}\equal{}0$ for all $ k$, then it is impossible to get from square $ i$ to square $ j$.  Proving these claims requires nothing more than the definition of matrix multiplication.[/hide]

Oh, duh!

Sorry, I completely forgot about that. I was thinking about each square as one element of the matrix.  :blush:


It is a really difficult problem.
Let's consider the equation $1+x+ \cdots + x^{n-1} = y^m$, where $x,y,m,n$ are integers with $n \geq 3, m \geq 2$ and $|x| >1.$

It has been shown that :

1) If $n = 0 \mod [4]$ then the only solution is $n=4,m=2,x=7,y = \pm 20$ (see [1]).

2) If $m = 2$, then the only solutions are the two above and $n=5,x=3, y= \pm 11$ (see [2) and [3]).

3) If $n=0 \mod [3]$ then the only solution is $m=n=3,y=7, x \in \{-19;18 \}$ (see [2]).

4) If $m=3$ and $n  \neq -1 \mod [6]$ then the only solutions are those in 3) (see [2]).

4) For given primes $x,y$, then (see [4]) the above equation has at most one solution $(m,n)$, and if this solution does exist $n$ is a prime and $n$ is the order of $x$ modulo $y$ and $m$ is a multiple of the order of $y$ modulo $x$.

References :
[1] T. Nagell, "Note sur l'équation indeterminée $\frac {x^n - 1 } {x-1} = y^q$", Norsk. Mat. Tidsskr. (1920), 75-78
[2] T. Nagell, "Sur l'équation indeterminée $\frac {x^n - 1 } {x-1} = y^2$", Norsk. Mat. Forenings Skrifter I, n°3 (1921), 17
[3] W. Ljunggren, "Noen setninger om ubestemte likninger av formen $\frac {x^n-1} {x-1} = y^q$ ", Norsk. Mat. Tidsskr., (1943), 17-20.
[4] H.M. Edgar, "The exponential equation $1+a+ \cdots + a^{x-1} = p^y$ ", A.M.M., 1974, 758-759.

Pierre.


geometry
circumcircle
incenter
power of a point
radical axis
geometry proposed

Consider $ \triangle ABC$ and a point $ P$ . Denote the projections $ D$ , $ E$ , $ F$ of $ P$ on $ BC$ , $ CA$ , $ AB$ respectively

and the point $ R\in EF\cap BC$ . The circumcircle of $ \triangle DEF$ cut again $ BC$ in $ M$ . Prove that $ RP\perp AM$ .

[b]Particular cases.[/b]

$ 1\blacktriangleright\ P: \equal{} H$ ([u]orthocenter[/u] of $ \triangle ABC$). In this case the point $ M$ is the midpoint of $ [BC]\ \implies\ RH\perp AM$ .

$ 2\blacktriangleright\ P: \equal{} I$ ([u]incenter[/u] of $ \triangle ABC$). In this case the point $ M$ is $ D\ \implies\ RI\perp AD$ .

Let $ K$ be the projection of $ P$ on $ AM$. Then $ AEPKF$, $ DPKM$ and $ DEFM$ are cyclic. Therefore their radical axis concur, i.e. $ PK$ passes through $ R$, as desired.


[quote="yif man12"]Yankeessuckium[/quote]
lol

iamcoolium :gleam:

[quote="bubala"]:rotfl:  :rotfl:  :rotfl: Wow yifman, all non-NYers seem to have this peculiar hatred of the Yankees.  I wonder why... :)[/quote]

Not just NON-New Yorkers.  ;)

Umm, Metium.

[quote="bleumoose"][quote="GoBraves"]homosexualityiswrongium[/quote]

um. gee. thanks. how about nonpoliticalium? its the periodic table, not a forum to post your political views.[/quote]

The questions is "If you could name a element in the periodic table what would you name it?" , and homosexualityiswrongium is exactly what i would name it.

jellibellium

I don't particularly like jelly bellies, just think it would make a funny name

Muium. It would be the first element whose name is a palindrome.

This would be the second: muiracecarium...whatever that means...

If element 216 is discovered, and there isn't a name ready for it, will it temporarily be called diunihexium? (symbol: Duh)

[quote="yif man12"][quote="math92"]Pistonsshouldhavewonium[/quote]

Hm, this is hard to believe, since it's coming from someone who didn't even know what game of the finals it was yesterday.[/quote]Yeah, but i watched the Finals yesterday


Solve the equation ${{ \frac{\lvert\{x}\rvert \sqrt{x^2+1}-x^2-3+2\sqrt {2}} {\lvert\{x}\rvert\sqrt{x^2+1}-x^2+3-2\sqrt {2}}=x^2$ :?:

hello, can you correct your equation please? i don't no what you mean.
Sonnhard.

I think it is :

Solve the equation $ \frac {\lvert{x}\rvert \sqrt {x^2 \plus{} 1} \minus{} x^2 \minus{} 3 \plus{} 2\sqrt {2}} {\lvert{x}\rvert\sqrt {x^2 \plus{} 1} \minus{} x^2 \plus{} 3 \minus{} 2\sqrt {2}} \equal{} x^2$

Or maybe :

Solve the equation $ \frac {\lvert\{x\}\rvert \sqrt {x^2 \plus{} 1} \minus{} x^2 \minus{} 3 \plus{} 2\sqrt {2}} {\lvert\{x\}\rvert\sqrt {x^2 \plus{} 1} \minus{} x^2 \plus{} 3 \minus{} 2\sqrt {2}} \equal{} x^2$

i have a solution for the first form
setting $ a\equal{}3\minus{}2\sqrt{2}$the equation is equivalent to
$ x\equal{}0$is not the root of the equation so we have
$ \frac{\sqrt{1\plus{}\frac{1}{x^2}}\minus{}1\minus{}\frac{a}{x^2}}{\sqrt{1\plus{}\frac{1}{x^2}}\minus{}1\minus{}\frac{a}{x^2}}\equal{}x^2$
setting $ y\equal{}\sqrt{1\plus{}\frac{1}{x^2}}(y>1)$ our equation is equivalent to
$ \frac{y\minus{}1\minus{}a(y^2\minus{}1)}{y\minus{}1\plus{}a(y^2\minus{}1)}\equal{}\frac{1}{y^2\minus{}1}$
$ \Leftrightarrow(y\minus{}1)(ay^3\plus{}ay^2\minus{}y^2\plus{}2)\equal{}0$
it is not hard to solve this equation

What ever it is, pco could you use Newton's method to show us the root ?


Let $f:R\rightarrow R$ be an injective function.
If $f^-1(x)+f(x)=x \forall  x\in R$ then prove that $f$ is an odd function!
 :)

What is injective function? Is that mean that it has an inverse function $f^{-1}(x)$?

[hide="this is either false or meaningless"]Assume there exists such a function. That function is odd, so $f(-x)=-f(x)$, so clearly $f(0)=0$. Since $f$ is injective and defined for all real $x$, $f$ is either strictly increasing or strictly decreasing. Either case will be analogous so let's suppose it is strictly increasing. $f^{-1}$ will also be strictly increasing (yes, I am making several claims without any proof..). For some arbitrary $x_1>0$ we will have either $0<f(x_1)<x_1$ or $f(x_1)=x_1$ or $f(x_1)>x_1$.

In the first case, $x_1>f(x_1)\Rightarrow{f}^{-1}(x_1)>f^{-1}(f(x_1))=x_1$, so clearly $f(x_1)+f^{-1}(x_1)>x_1$. In the second case, $f(x_1)=x_1=f(f^{-1}(x_1))\Rightarrow{f}^{-1}(x_1)=x_1$, since $f$ is injective. In this case $f(x_1)+f^{-1}(x_1)=2x_1>x_1$. The final case is just like the first but with $f$ and $f^{-1}$ interchanged.

Since we assumed such a function exists and showed that it doesn't satisfy the requirements, we have a contradiction, so no such function exists. The proposition will nevertheless be true as long as no function exists with the given conditions (not including that $f$ is odd; this is what we want to prove). So basically I accomplished nothing :( except to show that if the theorem is true, it is basically meaningless.[/hide]

edit: [hide="meaningless"]Obviously since $f$ is strictly increasing or decreasing there has to be some point at which both $f$ and $f^{-1}$ have the same sign, and then my argument above will suffice, except that nothing I said is proven  :D [/hide]

i think i have a start?

[hide]
rewrite it as:

$f(x) - x = - f^{-1}(x)$
which equals
$f(x) - \displaystyle\frac {1}{2} x = -(f^{-1}(x) - \frac {1}{2} x)$

and now we have some... symmetry?
i tried making a new function, something like

$g(x) = f(x) - \displaystyle\frac{1}{2} x$

so i can try to simplify it to something like $g(-x) = -g(x)$

but then i dunno if that will work

can ne1 help me?

[/hide]

[quote="socrates"]We have to prove that f is odd[/quote]I'm not sure what your point is. I realize that, and I even said it somewhere in my post. I'm saying I don't think such a function $f:\mathbf{R}\to\mathbf{R}$ (satisfying $\forall{x}\in\mathbf{R}:f(x)+f^{-1}(x)=x$) exists. At least, if one does exist, it's not odd. I could very well be wrong though.

Well, it would seem there is no such function.  But.

Suppose we are given some value x.  Let $a_0=x, a_1=f(x), a_2=f(f(x)),$ and so on.  We have $f^{-1}(a_n)+f(a_n)=a_n$, so that $a_{n-1}+a_{n+1}=a_n$, or, $a_n-a_{n-1}=a_n$.

Now, suppose $a_0=m$ and $a_1=n$.  We have:
$a_0=m$
$a_1=n$
$a_2=n-m$
$a_3=n-m-n=-m$
$a_4=-m-(n-m)=-n$
Therefore, $f(a_3)=a_4$ implies $f(-m)=-n$, but that's equivalent to $f(-x)=-f(x)$.

That's a very nice argument but I still want to see an example of such a function  :D

I could similarly say $\forall{x}\in\mathbf{R},f(x)f^{-1}(x)=0\Rightarrow\forall{x}\in\mathbf{R},f(x)=0\text{ or }f^{-1}(x)=0$. This is trivially deduced but there is no such function $f$.

Such a function would be:

$f(x)=\frac{1+\sqrt{3}i}{2}x$

(Okay i know it dosnt map reals to reals...)

[quote="AntonioMainenti"]Since $f$ is injective and defined for all real $x$, $f$ is either strictly increasing or strictly decreasing[/quote]

Did you invent a continuity condition?
 :D

It's pretty trivial, unless I'm sleeping at my keyboard here.

Existence of $f^{-1}$ follows from bijectivity (this is an iff condition, so there's no omitting it).

Now, we get:

$f(x)=x-f^{-1}(x)\rightarrow f^2(x)=f(x)-x\rightarrow f^3(x)=f^2(x)-f(x)=-x$

for all $x$.  Hence $f^3(f(x))=-f(x)$, and also $f(f^3(x))=f(-x)$, so that we get $f(-x)=-f(x)$


linear algebra
matrix
vector
invariant
topology
group theory
abstract algebra

It is welll known that $ m\times n$ matrix can be identified as $ R^{mn}$.
for $ k <{min(m,n)}$, consider the set of all $ m\times n$ matrix with rank $ k$.
Can this set be identified with some $ R^{p}$?
Is it, at least, a submanifold? what is its dimention?

The set of all $ m\times n$ matrices of rank $ k$ is not a vector space.

Example: $ \begin{bmatrix}1&0&0\\0&0&0\end{bmatrix}\plus{}\begin{bmatrix}0&0&0\\0&1&0\end{bmatrix}$

gives two matrices of rank $ 1$ whose sum has rank $ 2.$

Yes, the matrices of rank exactly $ k$ are a submanifold. The matrices of rank $ \le k$ aren't, except in the two extreme cases.

If $ A$ has rank $ k$, there are matrices $ R$ and $ C$ such that $ RAC\equal{}\begin{bmatrix}I_{k}&0\\0&0\end{bmatrix}$ in block form; this is easy to construct using elementary row and column operations.

Since every point looks the same, it's a submanifold. As an exercise, try computing the tangent space at $ \begin{bmatrix}I_{k}&0\\0&0\end{bmatrix}$ and the dimension.

[quote="jmerry"]Yes, the matrices of rank exactly $ k$ are a submanifold. The matrices of rank $ \le k$ aren't, except in the two extreme cases.

If $ A$ has rank $ k$, there are matrices $ R$ and $ C$ such that $ RAC \equal{}\begin{bmatrix}I_{k}& 0\\ 0 & 0\end{bmatrix}$ in block form; this is easy to construct using elementary row and column operations.

Since every point looks the same, it's a submanifold. As an exercise, try computing the tangent space at $ \begin{bmatrix}I_{k}& 0\\ 0 & 0\end{bmatrix}$ and the dimension.[/quote]

Your explanation is vague: what is a chart for this submanifold?

I don't know how to solve your question, can you give some details? Thanks.

[quote="jmerry"]It's a lot easier to find the tangent space than it is to find an explicit chart. What directions can you perturb the matrix in without changing its rank?[/quote]
The way I know is varying each term in the  m x n matrix (total of mn terms) in a small interval.
But rank is lower continiuous, so the rank may go up. The rank will be invariant  if it is maximal rank. But here k is not maximal since k < min(m,n).

Is there a way to perturb the matrix in without changing its rank?

[quote="jmerry"]Yes, of course there is. Changing some entries will increase the rank, while others will not. If I alter an entry in the lower right, the rank goes up. If I alter an entry in the upper left, it won't (This assumes we're basing everything at the specific matrix I mentioned earlier).

For some of the theory on why this works, look up "homogeneous spaces". One theorem: If a Lie group acts smoothly on a manifold, each orbit of this action is an embedded submanifold.[/quote]

OK, I found an chart for this in Hirsch's "differential topology" Them 3.2.6. He used stiefel manifold.
stiefel manifold is homogeneous space.

I was trying to lead you to build a solution yourself- maybe I shouldn't have included the keyword. The dimension is $ (m\plus{}n\minus{}k)k$, and I've got arguments based both on finding the tangent space and finding the subgroup that fixes a point.

Let $ R_{k}=\{A\in\mathcal{M}_{m,n}(\mathbb{R})|rank(A)=k\}$.$ R_{k}$ is a manifold of dimension $ k(m+n-k)$.
Proof: let $ A\in{R_{k}}$; we may assume that the matrix $ B$ formed by the first $ k$ lines and the first $ k$ columns of $ A$ is invertible. A chart of $ Z$, a $ A$-neighbourhood in $ R_{k}$, is as follows:
$ \phi: GL_{k}\times\mathcal{M}_{m-k,k}\times\mathcal{M}_{k,n-k}\rightarrow{Z}$ with $ \phi(s,u,v)=\begin{pmatrix}s&sv\\u&uv\end{pmatrix}.$
Second method: If $ i,j>k$ let $ d_{i,j}$ be the determinant formed with the lines $ \{1,\cdots,k,i\}$ and the columns $ \{1,\cdots,k,j\}$.Let $ Y$ be a $ A$-neighbourhood in $ \mathcal{M}_{m,n}.$ Let
$ \psi: Y\rightarrow\mathbb{R}^{(m-k)(n-k)}$ with $ \psi(U)=d.$
Then $ \psi=0$ (a $ A$-neighbourhood in $ R_{k}$) is a manifold because $ \psi$ is a submersion (the derivative is of maximal rank in the point $ A$). To see this last point, rewrite the components $ U_{i,j}$ of $ U$ s.t. the $ U_{i,j}$ for $ i,j>k$ finish the sequence.
$ Dim(R_{k})=mn-(m-k)(n-k)$.


search
geometry
number theory
advanced fields
advanced fields open

an anyone tell me please about the poincare conjecture?
i m very much interested about it,
help me.  :rotfl:

Well, first of all, this should not appear in a number theory forum.

There are tremendous reference materials on the internet about Poincare Conjecture, so please do some googling (or with any other online search engine) before posting here.

I suggest posting it in the geometry section!
[hide=":D"]Not seriously ;)[/hide]


[quote="randomized"]I`ve found the following inequality in an old book:

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq 3/2$ for every positive $a,b,c$[/quote]

It's Nesbitt's inequality, lots of different proofs have already been posted, use "search"

One also has the generalised Nesbitt's Inequality:
$a_{1}/s-a_{1}+a_{2}/s-a_{2}+...+a_{n}/s-a_{n}\geq n/n-1$

http://www.mathlinks.ro/Forum/viewtopic.php?t=113535 some proofs for Nesbitt inequality!!


[quote="DiscreetFourierTransform"]Each letter is distinguishable, so
[hide="1"]
6! Permutations
[/hide]
[hide="2"]
Fix the P, fix the U. That gives us 4! permutations of the other four letters.
[/hide]
[hide="3"]
The P and the U form a superletter block, so we treat it as one letter:
5! (the super letter, and the other four letters)
[/hide]
[hide="4"]
The vowels are either in places
1, 3, 5 or 2, 4, 6.

Case one:
Vowels in 1, 3, 5
There are 3! ways to arrange the vowels, and 3! ways to arrange the other letters:

Case two: by symmetry, same argument

Total: $ 2(3! * 3!) = 72$
[/hide][/quote]
Your number 4 is wrong, its suppose to be $ 4*3!*3!$, not 2*3!*3! there are 4 ways where they are not together

Its not just 135 it can be 136 246 or 146

[quote="mdk"]How many permutations of the word IMPURE are there? How many permutations start with P
and end in U? How many permutations are there if the P and the U must always be together in the order PU? How many permutations are there in which no two vowels (I,U, E) are adjacent?[/quote]

[hide="my solutions"]1. there are 6 letters that can be put in for the first spot, 5 for the second, and so on. we all know this. its $ 6!$ which equals [b]720[/b].

2. theres 4 spots in the middle for the letters I,M,R, and E. so its 4! because the other 2 dont matter. they always have the same positions. so the answer is [b]24[/b]

3. Asume PU was a different ONE letter of the alphabet, lets say Z. then IMPURE would be IMZRE so it would only be 5! because there would be 5 "letters". so the answer is [b]120[/b]

4. We would have to put the in the order CVCVCV or VCVCVC. c for consonant v for vowel. there would be 3 vowel spaces so 3! ways to set the vowels and the same for the consonants. so it would be $ 3!^{2}$ which is 36. but it can be arranged in 2 ways: CVCVCV or VCVCVC. so wmultiply that by 2 and get [b]72[/b].[/hide]

Number 4 can be arranged in 4 ways...not 2 as some of you think so the answer is 144, not 72.

How do you do that? all we have is VCVCVC and CVCVCV. what else is there?


Let x,y,z positive numbers. Prove that \[ \sum_{sym}{x^4y}\plus{}\sum_{sym}x^2y^2z\ge\sum_{sym}x^3y^2\plus{}\sum_{sym}x^3yz\]

[quote="Marius Mainea"]Let x,y,z positive numbers. Prove that
\[ \sum_{sym}{x^4y} \plus{} \sum_{sym}x^2y^2z\ge\sum_{sym}x^3y^2 \plus{} \sum_{sym}x^3yz
\]
[/quote]
Very easy! SOS kills it:
$ \sum_{sym}{x^4y} \plus{} \sum_{sym}x^2y^2z\ge\sum_{sym}x^3y^2 \plus{} \sum_{sym}x^3yz\Leftrightarrow\sum_{cyc}(x \minus{} y)^2xy(x \plus{} y \minus{} z)\geq0.$ :wink:

[quote="arqady"][quote="Marius Mainea"]Let x,y,z positive numbers. Prove that
\[ \sum_{sym}{x^4y} \plus{} \sum_{sym}x^2y^2z\ge\sum_{sym}x^3y^2 \plus{} \sum_{sym}x^3yz
\]
[/quote]
Very easy! SOS kills it:
$ \sum_{sym}{x^4y} \plus{} \sum_{sym}x^2y^2z\ge\sum_{sym}x^3y^2 \plus{} \sum_{sym}x^3yz\Leftrightarrow\sum_{cyc}(x \minus{} y)^2xy(2x \plus{} 2y \minus{} z)\geq0.$ :wink:[/quote]

    $ \sum_{cyc}(x \minus{} y)^2xy(2x \plus{} 2y \minus{} z)\geq0.$ why?

[quote="Marius Mainea"]

    $ \sum_{cyc}(x \minus{} y)^2xy(2x \plus{} 2y \minus{} z)\geq0.$ why?[/quote]
Let $ x\geq y\geq z.$ Hence, $ S_y\geq0,$ $ S_z\geq0$ and $ x \minus{} z\geq y \minus{} z.$
Thus, $ \sum_{cyc}(x \minus{} y)^2S_z\geq(y \minus{} z)^2\left(S_x \plus{} S_y\right) \equal{}$
$ \equal{} \sum_{cyc}(y \minus{} z)^2(yz(2y \plus{} 2z \minus{} x) \plus{} xz(2x \plus{} 2z \minus{} y)) \equal{}$
$ \equal{} \sum_{cyc}2z(y \minus{} z)^2(x^2 \minus{} xy \plus{} y^2 \plus{} xz \plus{} yz)\geq0.$ :wink:
The maximal $ k$ for which the inequality
\[ \sum_{sym}(x^4y \minus{} x^3y^2 \plus{} kx^2y^2z \minus{} kx^3yz)\geq0
\]
is true for all non-negative $ x,$ $ y$ and $ z$ is $ k \equal{} 1.$

I am sorry, it can be $ \sum_{sym}(x^4y \minus{} x^3y^2 \plus{} x^2y^2z \minus{} x^3yz)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x \minus{} y)^2xy(x \plus{} y \minus{} z)\geq0$ ( I fixed it in my first post ) and the proof for the last inequality is similar.


geometry
circumcircle
combinatorics proposed
combinatorics

Let $S$ be a set of $n\ge 3$ points in the interior of a circle.
$a)$ Show that there are three distinct points $a,b,c\in S$ and three distinct points $A,B,C$ on the circle such that $a$ is (strictly) closer to $A$ than any other point in $S$, $b$ is closer to $B$ than any other point in $S$ and $c$ is closer to $C$ than any other point in $S$.
$b)$ Show that for no value of $n$ can four such points in $S$ (and corresponding points on the circle) be guaranteed.

For the b) part:

Consider some equilateral triangle $ XYZ$ with the side length equal $ 1$, and label the midpoints of its sides as $ A,B,C$. Take the initial circle as the circumcircle of $ XYZ$. And besides, $ A,B,C$, mark any $ n\minus{}3$ points in the interior of $ ABC$.

I do not understand the problem.What if i take all the points in very close proximity near the centre of the circle.
can anyone please explain


algorithm
superior algebra
superior algebra unsolved

Let $ s_m$ be the number of elements of order $ m$ in the  simple group $ Alt_n$. What is the formula of $ s_m$?

Would you settle for an algorithm?

Find all cycle structures that give the right order. Count the number of possibilities for each structure with some annoying combinatorics; we choose which elements go in each cycle, which is a multinomial coefficient, we choose the order within each cycle, which is $ (k\minus{}1)!$ for a cycle of length $ k$, and we permute any cycles of the same length. Finally, add up all the possibilities.

Not all quantities have reasonable formulas.


AMC
AIME
geometric series
algebra
binomial theorem

[quote="matt276eagles"][quote="4everwise"]The polynomial $ 1 \minus{} x \plus{} x^{2} \minus{} x^{3} \plus{} \cdots \plus{} x^{16} \minus{} x^{17}$ may be written in the form $ a_{0} \plus{} a_{1}y \plus{} a_{2}y^{2} \plus{} \cdots \plus{} a_{16}y^{16} \plus{} a_{17}y^{17}$, where $ y \equal{} x \plus{} 1$ and thet $ a_{i}$'s are constants. Find the value of $ a_{2}$.[/quote]

[hide="Solution"]The coefficient of $ y^{2}$ in $ ( \minus{} 1)^{k}(y \minus{} 1)^{k} \equal{} (1 \minus{} y)^{k}$ is $ \binom{k}{2}$ by the Binomial Theorem.  Therefore, $ a_{2} \equal{} \binom{2}{2} \plus{} \binom{3}{2} \plus{} \cdots \plus{} \binom{17}{2} \equal{} \binom{18}{3} \equal{} 816$ by Hockey-Stick.[/hide][/quote]

I don't understand how he got this. Can someone elaborate?

Even easier: Using the geometric series formula, $ 1 - x + x^2 + \cdots - x^{17} = \frac {x^{18} - 1}{x + 1} = \frac {x^{18} - 1}{y}$. Since $ x = y - 1$, this becomes $ \frac {(y - 1)^{18} - 1}{y}$, so we want the coefficient of the $ y^3$ term in $ (y - 1)^{18}$. By the [[binomial theorem]] that is $ {18 \choose 3} = 816$.

What matt did: Again, notice $ x = y - 1$. So
\begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\
& = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}. We want the coefficient of the $ y^2$ term of each power of each binomial, which by the binomial theorem is $ {2\choose 2} + {3\choose 2} + \cdots + {17\choose 2}$. The [[Hockey Stick identity]] gives us that this quantity is equal to $ {18\choose 3} = 816$.


geometry
inequalities
abstract algebra
graph theory
IMO
IMO 2007

Well it appears that only a handfull of persons claim to have solved problem 3. Many contestans solved (at least part (a) of) problem 1. Weird enough, some of the people who solved #2 didn't solve 1(b).

It's hard to judge now but it seems the battle for the medals will be continued tomorrow, as the cut-off limits also seem to be on the fall (the trend being 2005 > 2006 > 2007).

It seems as the internet provides for an abundance of math olympiad problems the jury seems to overestimate the participants' skills and chooses challenging problems to safely discriminate based on skill among the higher scores, i.e. to decrease the amount of luck in the partial credits section. But then as a consequence again it might turn out to be tricky for the lower score regions.

I have solved just 1st problem.Others in my team(Kazakhstan):Bayev Alen #1 and #2,Stassiy Igor #1a and #2,Daliyev Asset,Ismailov Adilzhan and Diyarov Arman #1a.

in your opinion, how many points worth the 1(a)?

Hi.  From our team, Sasha, Mircea Grecu, Iulian Gramatki and I have probably solved problem 1, Freemind solved 1 and 2 but with bugs and we don't know what Andrei Iliasenco did :)

What happened to you Iura

As I see top news is that you have solved just one problem

Salem / Bosnia and Herzegovina

Who knows, how are the Ukrainian or Russian team?

I don't know about the Ukrainian team but in Russian's one four people solved 2 problems, if I am not mistaken, the only girl in the team solved all the problems and one guy solved 1 problem.

In our team I solved 1(a) and 2. The others did almost nothing. :(

[quote="irakli"]I heard that even iuri boreico was not able to solve third problem, is it so?[/quote]

Look above. Reading is a virtue.   :wink:

Interesting...Who is the author of the 3-rd problem...?

It seems that problem this year is really challenge. I think geometry problem is still what the contestants gain more score. The inequality of this IMO is not as everyone expected. It may be understandable since Mathlinkers nowadays are so well-trained in Inequalities.

How many students complete problem 3?

[quote="iura"]Well I'm happy that I finally managed to solve the first problem during the last 15 minutes of the contest. I could easily end up with just 1a). :)[/quote]THat is indeed weird, and I think the fact itself shows the difficulty of this exam.

For 1a) I suspect (judging from my coordinator experience that the marking scheme will be 2-5 points).

[url=http://www.imo2007.edu.vn/index.php?module=ViewRank]You can see it here[/url], but Only the result of Russia have been released.

Then how do people know the total scores of Korea, Vietnam, etc.?

EDIT: Nevermind. Figured out how to use portal properly.

[quote="vulgarfraction"]EDIT: Nevermind. Figured out how to use portal properly.[/quote]Portal? It's all on the official site. ;)

[quote="color"]What is ISL?[/quote]

Congrats to Omid!!!

Check the [url=http://www.mathlinks.ro/viewtopic.php?t=1053]abbreviations for mathematical olympiads.[/url] BTW when I introduced this abbreviation a few years ago some people thought the country code for Israel was meant.  :lol:  If you look at the link you will see that I posted that entry before I actually joined the ML/AoPS forum.  :D

[quote="Omid Hatami"]Hello, everybody
I have a good new.
The problem, I proposed for IMO 2007 was in ISL this year.:)
That's Mathlinks power.[/quote]
Congratulations, too!  :wink:   :P

[quote="ehsan2004"]iran got a fun result!!!!!!!! :huh:[/quote]

I rather think poland to have a funny result (1 2 2)

[quote="Eros88"][quote="delegat"]What happened to you Iura

[/quote]

Vietnamese mosquitos or just a" bad day at  the office".
My guess   :P[/quote]
what I heard is that he messed up with problem 3 and obviously a bad day.But too much bad(day) for a IMO winner.

[quote="Arne"][quote="Valentin Vornicu"]Problem 2 being proposed by Luxembourg is a surprise :)[/quote]
It is not.
Luxembourg proposed already two geometry problems (1988/1 and 1998/1)...
and their team leader, Charles Leytem, is a very good geometer.[/quote]

It indeed is a surprise.

After 1988/1 and 1998/1, everybody was expecting probem 2008/1 to be the
next IMO problem coming from Luxembourg, and now it suddenly is 2007/2.
Strange.

Ah, they just made a mistake, and sent the problem one year early... that's all...  :lol:


Let $ X_{1}, X_{2}, X_{3}, \dots$ be independent random variables with the same distribution. Are variables $ Y_{i} \equal{} X_{i} \minus{} X_{0}$ independent?

Hint: try to calculate $ Cov(Y_{1},Y_{2})$


Hi all,

I have tried hard to solve the [b]part b of the problem 3[/b] of my assigment. A hint is given, but I don't know how to apply it......any idea would be appreciated.

The source of the assignment is in the following file:

http://cicma.mathstat.concordia.ca/faculty/larose/BLD05int_activity.pdf

Thank you.

I'm not positive this is right, but here's my idea.

I think the equation $\frac{X''(x)}{X(x)}=\frac{T''(t)}{a^2T(t)}$ is saying $\frac{X''(\text{any }x)}{X(\text{any }x)}=\frac{T''(\text{any }t)}{a^2T(\text{any }t)}$ because all of the $x$ and $t$ terms are seperated, and therefore both sides must be constant.
The main difference between this and the wave equation is the seperation of the variables, and that is what I think makes it constant.

Let me know if this helps.

I have same thought too..... If we put any values x and t into the equations, the LHS and RHS must be a constant. Also, since LHS is equal to RHS (proven in part a), both sides must equal to a same constant K.

But is this a valid argument? How about the hint that the teacher has given in the problem? How to use the hint?

Well the hint the teacher gave could be applied by saying $\frac{X''(x_1)}{X(x_1)}=\frac{T''(t_1)}{a^2T(t_1)}=\frac{X''(x_2)}{X(x_2)}=k$, where $k$ is some constant and $x_1$ and $x_2$ could be any values of $x$.

Can you provide more explanations of your equation? How did you get x2?
and why the "third side" must equal to LHS?

Thank you

The hint your teacher gave was $f(z)$ is constant iff $f(z_1)=f(z_2)$ for any values of $z_1$ and $z_2$.  I used $x_1$ and $x_2$ to show that $x$ could be any value and equality still holds, and thus the LHS and RHS are constant, which works because of the seperation of variables.  This is a more formal way of writting $\frac{X''(\text{any }x)}{X(\text{any }x)} = \frac{T''(\text{any }t)}{a^2T(\text{any }t)}$.

Good luck on the rest of the assignment.

Take an $(x,t)$ pair $(x_1,t_1)$.  Now consider $(x_2,t_1)$.  The LHS is the same, so we get $X''(x_1)/X(x_1)=X''(x_2)/X(x_2)$.  But $x_1,x_2$ are arbitrary, so the LHS is constant.  Considering the $(x_1,t_2)$, we get that the RHS is constant.  Done.


Let  $x_1,x_2,...,x_n$ is real numbers and $\sum_{sym}x_1x_2\geq0,$ $n\in\mathbb N,$  $n\geq2.$
Prove that $(\frac{\sum_{cyc}x_1}{n})^{2n}\geq\frac{\sum_{cyc}x_1^2}{n}\cdot(\frac{\sum_{sym}x_1x_2}{\frac{n(n-1)}{2}})^{n-1}.$

[quote="Davron"]Hello arqady i think most of the linkers do not like solving problems like that for example i dont love massy problems they are boring for me, sorry for my rudeness  ;)

Davron[/quote]
I understand you.
Game of your soccer team Regar - TadAz  beats off at me any desire to look football. :D

arqady thank you  :)  Sabir acidir meyvesi tatlidir.
                                 Sabir imanin yarisidir.
                                  Sabrin sonu selamettir.
                                 Sabrin sonu tecrubedir, tecrubenin sonuda basaridir.
                                  Al-fakru fahri. Try to understand this ... and my advice never smile like this ok. please  :D
even its too funny...  :)
I am gonna tell a lots of thing by the time. Have a good time.  :)

Davron


[quote="N.N.Trung"]Find all functions $ f: \mathbb R \rightarrow \mathbb R$ satisfying two conditions:
i) $ f(x \plus{} y) \le f(x) \plus{} f(y)$ for every $ x,y\in \mathbb R$
ii) $ lim_{x\rightarrow 0} \frac {f(x)}{x} \equal{} 1$[/quote]

From $ (i)$ it's easy to see that $ f(x) \le pf(\frac{x}{p})\plus{}f(0)$ for any positive integer $ p$.

So $ \dfrac{f(x)\minus{}f(0)}{x} \le\lim_{p \to \infty} \dfrac{f(\dfrac{x}{p})}{\dfrac{x}{p}}\equal{}1$ from $ (ii)$ for $ x >0$
and $ \dfrac{f(x)\minus{}f(0)}{x} \ge \lim_{p \to \infty} \dfrac{f(\dfrac{x}{p})}{\dfrac{x}{p}}\equal{}1$ for $ x<0$

Also from $ (i)$ we have $ f(x) \ge f(x\plus{}y)\minus{}f(y)$
$ \Rightarrow f(x)\minus{}f(0) \ge \minus{}pf(\frac{\minus{}x}{p})$
$ \Rightarrow \frac{f(x)\minus{}f(0)}{x} \ge \lim_{p \to \infty}\frac{f(\frac{\minus{}x}{p})}{\frac{\minus{}x}{p}}\equal{}1$ for $ x>0$
and $ \Rightarrow \frac{f(x)\minus{}f(0)}{x} \le \lim_{p \to \infty}\frac{f(\frac{\minus{}x}{p})}{\frac{\minus{}x}{p}}\equal{}1$ for $ x<0$

So $ f(x)\equal{}x\plus{}f(0)$. Put it into $ (ii)$ we get $ f(0)\equal{}0$. So $ f(x)\equal{}x$ is the only solution.


Lines $ 4x \plus{} 5y \equal{} 40$ and $ Ax \plus{} By \equal{} 10$ intersect on the $ x$-axis. Lines $ 4x \minus{}5y \equal{} 40$ and $ Ax \plus{} By \equal{} 10$ intersect on the $ y$-axis. Find $ A\minus{}B$.

yes but somehow answer is 2, and except me all other 7 people in the game got it as 2? how is it 2?

If that's true, you might wanna report it...


QUERER:
Yo quiero
Tu quieres
Ella quiere
Nosotros queremos
Vosotros/Ustedes   [b]QUEREIS[/b]
Ellos quieren

PODER:
Yo puedo
Tu puedes
Ella puede
Nosotros podemos
Vosotros      [b]PODEIS[/b]
Ellos pueden

Another question:  different sources tell me different things about commands.  For example, one source said:

Tu hable (hablAR)
Nosotros hablen (hablAR)
Tu coma (comER)
Nosotros coman (comER)
Tu viva (vivIR)
Nosotros vivan (vivIR)

However, another source said:

Tu habla (hablAR)
Nosotros hablan (hablAR)
Tu come (comER)
Nosotros comen (comER)
Tu vive (vivIR)
Nosotros viven (vivIR)

They also mentioned something about formality, which I did not understand.  What is the proper way to conjugate a command?

Tu hablas
Nosotros hablamos
Tu comes
Nosotros comemos
Tu vives
Nosotros vivimos

When you speak/write to an unknown or bigger person, yo use formality. For example:

Informal: Cómo estás? Cómo la estás pasando?
Formal: Cómo anda? Cómo la está pasando?

Are those commands?  See what I mean here:

http://www.musicalspanish.com/tutorial/commands.htm

So if I wanted to say "Jose, eat!" I would say "tu comes" or "tu come?"  Which is informal, and which is formal?

Hi, I'm from Spain, so maybe I can help.

FORMAL
At least in Spain, "usted-ustedes" (more formal or polite than the informal second person "tú-vosotros/vosotras"), is always conjugated like the third person "él/ella-ellos/ellas":

IRREGULAR VERBS QUERER / PODER
Yo quiero / puedo
Tú quieres / puedes
Usted/Él/Ella quiere /puede
Nosotros queremos /podemos
Vosotros queréis / podéis
Ustedes/Ellos/Ellas quieren / pueden

The rule that can help you is that you must use the irregular "quie-" or "pue-" (instead of the normal "que-" or "po-") when the syllable is phonetically stressed.

COMMANDS
There are imperative conjugations (used for direct commands) only for the second person "tú-vosotros/vosotras". For the other persons (including "usted/ustedes"), you must use the indirect commands with subjunctive conjugations (like the ones used in subordinate statements: "José, yo quiero que usted hable" $\to$ "Usted/José, ¡hable!")

Tú habla / come / vive (command)
(Usted/Él/Ella hable / coma / viva) (subjunctive)
(Nosotros hablemos / comamos / vivamos)
Vosotros hablad / comed / vivid
(Ustedes/Ellos/Ellas hablen / coman / vivan)

Here you have a link to a useful [url=http://tradu.scig.uniovi.es/conjuga.html]Spanish conjugation website[/url].

[quote="José"]Well the problem is that in Spain they speak Spanish, but we in Argentina use "castellano" (it's 99% the same as english, just some changes, because we don't use "vosotros", we use "ustedes")[/quote]

My teacher said in Mexico they use ustedes, and in Spain vosotros is a less respectful way of saying ustedes.

[quote="mysmartmouth"]My teacher said in Mexico they use ustedes, and in Spain vosotros is a less respectful way of saying ustedes.[/quote]

No is not less respectful...
The same thing happened with american and british english, the language is almost the same, but with slight differences

[quote="mysmartmouth"]Hola,

I am teaching myself Spanish, and I hope to post questions here as I go along.  My first questions are about conjugations of two irregular verbs:

QUERER:
Yo quiero
Tu quieres
Ella quiere
Nosotros queremos
Vosotros/Ustedes [color=red]????[/color]
Ellos quieren

PODER:
Yo puedo
Tu puedes
Ella puede
Nosotros podemos
Vosotros [color=red]????[/color]
Ellos pueden

In both conjugations, I cannot find the present second-person plural form of the verb.  Can someone please help me?  Thank you!!![/quote]

:S What is the purpose

learning maybe, I think that's perfect, so ask us any doubt you've got


function
real analysis
real analysis unsolved

Assume that $ f(x)$ is differentiable,and $ f'(x)$ is strictly increasing for $ x\ge 0$.If $ f(0) \equal{} 0$ prove that $ \dfrac{f(x)}{x}$ is strictly increasing for $ x > 0$

OK, I've been thinking about this while I was walking to college, so it's rather sketchy and quite possibly plain wrong. I obviously don't have much time right now, so, as I said, I'll try to write a proof without going into much detail. Please, let me know if something is wrong or not quite understandable, it really is quite an interesting problem.

Let's take any $ x$ and $ y$ such that $ 0<x<y$. By the mean value theorem, there exist $ c_1$ and $ c_2$ such that $ 0<c_1<x<c_2<y$ and that $ f'(c_1)\equal{}\frac{f(x)\minus{}f(0)}{x\minus{}0}$ and also $ f'(c_2)\equal{}\frac{f(y)\minus{}f(x)}{y\minus{}x}$. Since $ c_1<c_2$, $ f'(c_1)<f'(c_2)$ so $ \frac{f(x)}{x}<\frac{f(y)\minus{}f(x)}{y\minus{}x}$. This is equivalent to $ f(x)y\minus{}f(x)x<f(y)x\minus{}f(x)x$, that is $ f(x)y<f(y)x$. That is obviously equivalent to what we were looking for: $ \frac{f(x)}{x}<\frac{f(y)}{y}$.

There is very nice criteria of convexity function(prove it!): $ f$ is convex on the interval $ (a,b)$ if and only if for every $ x_0\in (a,b)$ the fiunction $ \frac {f(x) \minus{} f(x_0)}{x \minus{} x_0}$ is increasing function. Your statement is immediate consequence of that fact. :)

Let $ x < y$. $ f(x) \equal{} f(\frac {x}{y}y \plus{} (1 \minus{} \frac {x}{y})\cdot0) \leq \frac {x}{y}f(y) \plus{} (1 \minus{} \frac {x}{y})f(0) \equal{} \frac {x}{y}f(y)$, because $ f$ is convex.

Introducing the convexity concept only muddles the waters. Then you'll need to assume (nay, prove), that $ f(x)$ differentiable, with $ f'(x)$ strictly increasing for $ x\ge 0$ and  $ f(0) \equal{} 0$ implies convexity, in order to use that criterium. Now I know that if $ f$ were doubly differentiable, the increasing property of $ f'$ would warrant $ f''(x) \geq 0$, etc., but that is not given. Yes, $ f'$ being monotone, only may have points of discontinuity of first species, but we still are a long way from successful argumentation.
So I would stick with the short and sweet mornik's proof!

The convexity argument generalizes the problem and in my opinion puts it in the right perspective; it doesn't muddle anything! The MVT argument is very close to an argument that $ f$ is convex under the given conditions.


Let $ n$ sets. Each set have $ k$ elements ($ n\geq k\plus{}1$). $ 2$ any sets in $ n$ sets have $ 1$ common uinique element. Prove if $ n$ sets don't have common element then $ k\plus{}1$ don't have common element

assume that the $ n$ sets don't have an element in common but that $ k\plus{}1$ of the sets *do* have an element in common. say that these $ k\plus{}1$ sets are $ A_1, ..., A_{k\plus{}1}$ and that their common element is $ x$. by assumption, there must be a set $ B$ not containing $ x$. then, since $ B$ must have an element in common with every other set, for every $ i$, $ 1\leq i\leq k\plus{}1$, there must exist $ x_i\in A_i\minus{}\{x\}$ such that $ x_i\in B$. but the $ x_i$ are all distinct, so $ B$ contains at least $ k\plus{}1$ elements -- a contradiction.

why [b]the $ x_i$ are all distinct[/b]
please

because of the assumption that any two sets have a unique common element.


What is the $y$-coordinate of the point where the line whose equation is $3x-5y=-15$ intersect the $x$-axis?

[hide]
if it intesects the x-axis then y has to be 0
[/hide]


linear algebra
matrix
vector
linear algebra unsolved

Let $ A\equal{}I\plus{}uv*$. Show that $ A^{\minus{}1}$ has the form $ I\plus{}\alpha u v *$ for some scalar $ \alpha$. (where u and v are m-vectors and A is nonsingular)

Thanks :)

Hint: If $ A \equal{} I \plus{} K$, and $ K$ is a matrix satisfying $ K^2 \equal{} \lambda K$ for a [i]scalar[/i] $ \lambda$, then what is $ A^{ \minus{} 1}$?

  darij


geometry
3D geometry
number theory unsolved
number theory

Prove that every positive rational number can be represented as sum of three cubes of three positive rational number


Prove that for all natural numbers k, k^5 - k is a multiple of 10.

I think, Intermediate is a right place for such problem, instead of Algebra, so I will move it ;)

$k^5-k=k(k^4-1)=k(k^2-1)(k^2+1)=k(k+1)(k-1)(k^2+1)$
One of $\{k,k+1\}$ is divisible by $2$.
Since $2$ and $5$ are relatively prime, if one of $\{k,k+1,k-1,k^2+1\}$ is divisible by $5$, the whole expression will be divisible by $10$ and we'll be happy. ^_^
Let's try some cases then!
$k\equiv 0\bmod 5\longrightarrow 5|k$
$k\equiv 1\bmod 5\longrightarrow 5|k-1$
$k\equiv 2\bmod 5\longrightarrow 5|k^2+1$
$k\equiv 3\bmod 5\longrightarrow 5|k^2+1$
$k\equiv 4\bmod 5\longrightarrow 5|k+1$
Since we have exhausted all the possibilities, and $5$ divides the expression in each of the above cases, $k^5-k$ will be divisible by $10$ for any natural number $k$. Moreover, you could go further, and notice that one of $\{k-1,k,k+1\}$ is divislbe by $3$, so $k^5-k$ is also divisible by $30$ for all natural numbers $k$. And since $k=2$ produces $2^5-2=30$, I don't think you can go any farther than that. :)

Other approach without considering cases is Fermat theorem. Indeed, if $k$ is not divisible by 5, then $k^4-1$ is divisible by 5 by Fermat theorem.

Thanks for solution. Sorry if in wrong forum.


inequalities
logarithms
number theory unsolved
number theory

Let $\omega (n)$ denote the number of prime divisors of the integer $n>1$.Find the least integer $k$ such that the inequality $2^{\omega(n)}\leq k\sqrt[4]{n}$ holds for all $n>1$.

I found [tex]k = 5[/tex] but my "proof" is not rigorous ... it's more of an argument than a proof :(
WLOG take [tex]n = p_1 \cdot \ldots \cdot p_n[/tex] (where [tex]p_j[/tex] are primes) then by taking logarithms we get [tex]n
\leqslant \ln_2 k + \frac{1}{4} \sum_{j = 1}^n \ln_2 p_j[/tex]
Note that if [tex]p_j \geqslant 17[/tex] for every [tex]j[/tex] then we may take [tex]k = 1[/tex]. It is sufficent to find a [tex]k[/tex] such that the relation holds when [tex]p_j < 17[/tex] (and we have 6 primes), more concretely the case with [tex]n = 2 \cdot 3 \cdot \ldots \cdot 13[/tex] is the one with the biggest [tex]k[/tex]. We must have [tex]k = 5[/tex].

Actually, most problems from recent TSTs are usually already discussed on ML. See http://www.mathlinks.ro/Forum/viewtopic.php?t=21497 for this one.

  darij


I can't solve this inequality.It looks nice and hard. Would you please help me?
Let $ a_1,a_2,\cdots a_n$ be positive real numbers.
Prove that: $ \frac{1}{a_1}\plus{}\frac{2}{a_1\plus{}a_2}\plus{}\cdots\plus{}\frac{n}{a_1\plus{}a_2\plus{}\cdots\plus{}a_n} \leq 2\sum_{i\equal{}1}^{n}{\frac{1}{a_i}}$

[quote="joepro"][quote="Allnames"][quote="joepro"]I can't solve this inequality.It looks nice and hard. Would you please help me?
Let $ a_1,a_2,\cdots a_n$ be positive real numbers.
Prove that: $ \frac {1}{a_1} \plus{} \frac {2}{a_1 \plus{} a_2} \plus{} \cdots \plus{} \frac {n}{a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n} \leq 2\sum_{i \equal{} 1}^{n}{\frac {1}{a_i}}$[/quote]
But it is here  http://www.mathlinks.ro/viewtopic.php?t=221923[/quote]
But where is the solution?  :maybe:[/quote]
wait for our ONI 2. :)


Think about  it. Every interval of ten (1,2,3,4,5,6,7,9...) two of them leave 3 mod 5. There are 200 such intervals.

And the other one is 110.

I meant they satisfy both conditions  ;)

$LCM(5,9)=45$, so the first number is $43$, and every $45$ after that. Therefore, the answer is $\boxed{22}$ integers.

[quote="236factorial"]How many of the first 1000 integers have a remainder of 3 when divided by 5, and a remainder of 7 when divided by 9?[/quote]

$x\equiv3(mod5)$
$x\equiv7(mod9)$

$x+2\equiv0(mod5)\equiv0(mod9)$

$x=43$, and since $9(5)=45$, then every $45$ plus $43$ is also with this same condition, so $\frac{1000}{44.something}=\boxed{22}$.

Yeah the first one is easy, how about the second one?

[quote="236factorial"]How many of the first 1000 integers have a remainder of 2 when divided by 5, and a remainder of 7 when divided by 9?

A little changed[/quote]

Here, all you have to do is find numbers that are 7 mod 9 and end in either 2 or 7 (the unit's digit). The first such number is 7. The rest go by increments of 45 (see if you can find out why), so you have the last number being 7+22*45=997, therefore there are 23 such numbers.

that's very smart to use the units digit.

I give you your avatar ( :) )


At the url below
http://library.thinkquest.org/C0110248/trigonometry/sinerule.htm
it wrote
Case 1 (If A is an acute angle)

Consider angle A
Now angle BAC = angle BPC...
Why angle BAC = angle BPC?

well you know one of the angles is 90 degrees since it is trig right?

Thanks, but I can't see that 90 degrees provides any meaningful help in getting angle BAC = angle BPC.
I just knew that both of the angles intercepts the same arc so both of the angles are equal.
Although I have got the solution, but I still can't see 90 degree provides any help on this...

sorry i didn't even read your question i assumed i knew what you were asking :blush:

ic, never mind, thanks for the comment anyway:)

i am bad at any explaining but perhaps
http://www.ucl.ac.uk/Mathematics/geomath/trignb/trig11.html
would help you


geometry
analytic geometry
vector
algorithm

How do you solve problems like:
The vertices of triangle/quadrilateral are  ... What is the area of that triangle/quadrilateral

There is a stupid formula:
S=1/2 abs (x1(y2-y3) + x2(y3-y1) + x3(y1-y2) )

where (xi,yi) are the coordinates of the vertices of a triangle
The formula can be proved in various ways, eg vectors
Quadrilaterals are devided into 2 triangles...

It is not a stupid formula.

It is the [url=http://en.wikipedia.org/wiki/Shoelace_formula]shoelace algorithm[/url].  You will find it very useful.  :)


sin3x+cos3x=2sinx.cosx

Squaring Both sides

$ 1 \plus{} Sin6x \equal{} Sin^22x$
$ 1 \plus{} 3Sin2x \minus{} 4Sin^32x \equal{} Sin^22x$
$ 1 \plus{} 3x \minus{} 4x^3 \equal{} x^2$
$ 4x^3 \plus{} x^2 \minus{} 3x \minus{} 1 \equal{} 0$

which is ofcourse a simple cubic equation you can solve. ;)

Incase you want to know how to solve cubic eqns, here you go. http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method

Nice and easy wasn't it??

Well,A MUCH simpler approach could be to write the left hand side as a function of (3x+ pi/4) and the right as sin 2x, mess around a lil.
cuz, cubics ARE not generally easy to solve other than of course exceptional cases.
And abt the link u had given, first of all, in order to use Cardon's Method, u ll ve to first reduce it to the form (z^3) + 3Hz + G. The procedure is long and bugging :!:

[quote="Euclidean Geometer"]Well,A MUCH simpler approach could be to write the left hand side as a function of (3x+ pi/4) and the right as sin 2x, mess around a lil.[/quote]

Darn thing is that I get an extra BUGGING $ \sqrt2$

[quote]
cuz, cubics ARE not generally easy to solve other than of course exceptional cases.
And abt the link u had given, first of all, in order to use Cardon's Method, u ll ve to first reduce it to the form (z^3) + 3Hz + G. The procedure is long and bugging :!:[/quote]

I was just kidding about Cardan's method.  :rotfl:  :rotfl:


What is the  best algebra, statistics and geometry freeware (totally free software)?

KSEG (http://www.mit.edu/~ibaran/kseg.html)

For drawing geometry


function
calculus
integration
probability
limit
expected value
probability and stats

I'm given the following cumulative distribution function.

F(x)= 0 for x<1
F(x)= (x^2 -2x+2)/2 for   1 less than or equal to x <2
F(x)= 1 for x greater than or equal to 2.

Find the Variance of X.

So, naturally, Var(X)= E(X^2)-E(X)^2.

To get E(X) I need the density function by differentiation.

What I get from the cumulative distribution function is that x cannot be less than 1 and all values of x are less than 2.

So the range I'm concerned with is  1 less than or equal to x <2.

So I differentiate and get x-1 for the density function.

to get E(X) I integrate  x(x-1) from 1 to 2.

My question is, whether that integral includes the probability that x=1. Is it always the case that the border terms are not taken into account?
In other words, if I integrate some other function just like this on some interval a,b, is the probabilty that x=a and x=b excluded from the result?

If $ X$ has cumulative distribution function $ F(x),$ then $ P(X \equal{} a) \equal{} F(a) \minus{} \lim_{x\to a^ \minus{} }F(x).$ That is, it's the height of the jump wherever $ F$ has a jump discontinuity.

Whenever $ F$ is continuous at $ a,$ then $ P(X \equal{} a) \equal{} 0.$ That's what you mean when you say that $ X$ has a continuous distribution. (In practice, you are using the notion that $ X$ has an absolutely continuous distribution, but that's probably more information than you need to know right now.)

(One side note: saying an event has probability zero is [i]not[/i] saying that it is impossible. A nonempty (hence possible) set may still have probability zero.)

But this isn't a continuous distribution, since $ F$ has a jump discontinuity at $ 1.$

$ F(1) \equal{} \frac12$ and $ \lim_{x\to1^ \minus{} }F(x) \equal{} 0,$ so $ P(X \equal{} 1) \equal{} \frac12.$

Since this is neither a continuous distribution nor a discrete distribution, then neither an integral alone nor a sum alone will suffice to compute its expected value; we must use elements of both.

$ E(X) \equal{} 1\cdot P(X \equal{} 1) \plus{} \int_1^2x(x \minus{} 1)\,dx \equal{} \frac12 \plus{} \left.\left(\frac13x^3 \minus{} \frac12x^2\right)\right|_1^2 \equal{} \frac43$

Since that does lie in the interval $ [1,2],$ you don't have to reject it immediately.


1- I need geometry_dev please. Where can i find it.
2- I do not know how to display number on the diagram.

[code]
size(300);import geometry;
pair A,B,C; A=(0,0);B=(2,0);C=dir(60);
dot("$A$",A,S);dot("$B$",B,S);dot("$C$",C,N); draw(A--B--C--cycle,linewidth(bp));
pair D,E; D=midpoint(B--C); E=midpoint(C--A);
dot("$D$",D,NE);dot("$E$",E,NW);
pair K=(2/3,0); pair M,N;M=A+K;N=A+K+K;
dot("$M$",M,SW);dot("$N$",N,SE);
pair P=intersectionpoint(line(D,N),line(E,M)); dot("$P$",P,S);
draw(E--P--D--A,linewidth(bp));
real alpha = angle (line(E,P),line(D,P));

[/code]

I want to know and display the value of alpha on the diagram.

[quote="Ph. Ivaldi"]Hi,

geometry_dev is now the official package of geometry, so it's included in Asymptote and his new name is simply "geometry" ;-)

[code]size(300);import geometry;
pair A,B,C; A=(0,0);B=(2,0);C=dir(60);
dot("$A$",A,S);dot("$B$",B,S);dot("$C$",C,N); draw(A--B--C--cycle,linewidth(bp));
pair D,E; D=midpoint(B--C); E=midpoint(C--A);
dot("$D$",D,NE);dot("$E$",E,NW);
pair K=(2/3,0); pair M,N;M=A+K;N=A+K+K;
dot("$M$",M,SW);dot("$N$",N,SE);
pair P=intersectionpoint(line(D,N),line(E,M)); dot("$P$",P,S);
draw(E--P--D--A,linewidth(bp));
real alpha = degrees(line(P,D),line(P,E));
markangle(string(alpha),line(P,D),line(P,E));[/code][/quote]

Double thanks to you :)

[quote]geometry_dev is now the official package of geometry, so it's included in Asymptote and his new name is simply "geometry" ;-)

[/quote]

I think that's better. I also prefer if all packages are included in Asymptote when you download it, so you don't need to search about packages.

Another problem  :D . If I want to check if the lines A,B are perpendicular or not. How? I kow that we need to use bool perpendicular(A,B) function, but I do not know how to know it's value. Also, what function do I need to check if the point A,B,C,D are cyclic?


inequalities
number theory
prime factorization
number theory proposed

If $ 1 \equal{} d_1 < d_2 < ... < d_k \equal{} n$ are all divisors of natural number $ n$, prove that \[ (d_1 \plus{} d_2 \plus{} ... \plus{} d_k)^2\geq n\cdot k^2\]

We have two cases.

1) $ n$ is a perfect square. Then $ k \equal{} 2t \minus{} 1$, $ t \in \mathbb{N}$. Summing $ \displaystyle d_{t} \equal{} \sqrt {n}$ with following AM-GM inequalities
\[ d_{i} \plus{} d_{2t \minus{} i} \overset{AM\minus{}GM}{\geqslant} 2 \sqrt {d_{i}d_{2t \minus{} i}} \equal{} 2 \sqrt {n} \textrm{, \ \ } i \equal{} 1, 2, \ldots, t \minus{} 1 \textrm{,}
\]
gives
\[ d_{1} \plus{} d_{2} \plus{} \ldots \plus{} d_{k} \geqslant \sqrt {n} \plus{} \left( t \minus{} 1 \right) \cdot 2\sqrt {n} \equal{} k \sqrt {n} \implies \left( d_{1} \plus{} d_{2} \plus{} \ldots \plus{} d_{k} \right)^{2} \geqslant nk^{2} \textrm{.}
\]
2) $ n$ is not a perfect square. Then $ k \equal{} 2t$, $ t \in \mathbb{N}$. Summing following AM-GM inequalities
\[ d_{i} \plus{} d_{2t \plus{} 1 \minus{} i} \overset{AM\minus{}GM}{\geqslant} 2 \sqrt {d_{i}d_{2t \plus{} 1 \minus{} i}} \equal{} 2 \sqrt {n} \textrm{, \ \ } i \equal{} 1, 2, \ldots, t \textrm{,}
\]
gives
\[ d_{1} \plus{} d_{2} \plus{} \ldots \plus{} d_{k} \geqslant t \cdot 2\sqrt {n} \equal{} k \sqrt {n} \implies \left( d_{1} \plus{} d_{2} \plus{} \ldots \plus{} d_{k} \right)^{2} \geqslant nk^{2} \textrm{.}
\]
Equality holds if and only if $ n \equal{} 1$.

Q. E. D.

I will use the prime factorization of n:
$ \prod_{i\equal{}1}^{t} p_{i}^{a_{i}}$
$ \sum_{i\equal{}1}^{k} d_{i} \equal{} \prod_{i\equal{}1}^{t} (1\plus{}p_{i}\plus{}...\plus{}p_{i}^{a_{i}})$
We need to show that: $ (\frac{\sum_{i\equal{}1}^{k} d_{i}}{k})^{2} \ge n$, or:
$ (\prod_{i\equal{}1}^{t} \frac{1\plus{}p_{i}\plus{}...\plus{}p_{i}^{a_{i}}}{a_{i}\plus{}1})^{2} \ge \prod_{i\equal{}1}^{t} p_{i}^{a_{i}}$
it's enough to show that for $ x \ge 2, a \ge 0$:
$ (\frac{1\plus{}x\plus{}...\plus{}x^a}{a\plus{}1})^{2} \ge x^a$
but it follows directly from AM-GM. equality iff $ x\equal{}1$ or $ a\equal{}0$, but $ a_{i}>0$ and $ p_{i} \ge 2$, except for [u]n=1[/u].

Solution that I had in mind when I came up with this (easy) inequality is \[ (d_1\plus{}d_2\plus{}...\plus{}d_k)(\frac1{d_1} \plus{} \frac1{d_2} \plus{} ... \plus{}\frac 1{d_k})\geq k^2\] by $ AM\minus{}HM$, and $ (\frac 1{d_1} \plus{} ... \plus{} \frac1{d_k})\cdot n \equal{} (d_1 \plus{} ... \plus{} d_k)$ :)
but you both solved it :)
bye


geometry perimeter calculus inequalities logarithms trigonometry circumcircle

Let consider a $ \Delta ABC$. We can take $ A_1$ in $ BC$ such that $ AA_1$ is a bisector of the angle $ B\hat{A}C$.

Applying sine's law in $ \Delta ABA_1$, we'll have: $ \dfrac{\sin(A/2)}{BA_1} \equal{} \dfrac{\sin(B)}{AA_1}$. (*)

But i) $ BA_1 \equal{} \dfrac{ac}{b \plus{} c}$, ii) $ AA_1 \equal{} \dfrac{2\sqrt {bcp(p \minus{} a)}}{b \plus{} c}$ and iii) $ \sin(B) \equal{} \dfrac{b}{2R}$.

Substituting i), ii) and iii) in (*): $ sin(A/2) \equal{} \dfrac{abc}{4R\sqrt {bcp(p \minus{} a)}}$.

Remember that $ S \equal{} \dfrac{abc}{4R}$, we finally have: $ sin(A/2) \equal{} \dfrac{S}{\sqrt {bcp(p \minus{} a)}}$.

So we must prove that: $ \dfrac{S^3}{abcp\sqrt {p(p \minus{} a)(p \minus{} b)(p \minus{} c)}} \le \dfrac{1}{8}$ $ \Leftrightarrow$ $ \dfrac{(p \minus{} a)(p \minus{} b)(p \minus{} c)}{abc} \le \dfrac{1}{8}$. (**)

Now we can a change of variables:

I) $ b \plus{} c \minus{} a \equal{} tm$
II) $ c \plus{} a \minus{} b \equal{} tn$
III) $ a \plus{} b \minus{} c \equal{} tp$

In this equations, we have $ t \equal{} a \plus{} b \plus{} c$.

Substituting I), II) and III) in (**), we have to prove that: $ \dfrac{mnp}{(m \plus{} n)(n \plus{} p)(p \plus{} m)} \le \dfrac{1}{8}$ $ \Leftrightarrow$ $ (m \plus{} n)(n \plus{} p)(p \plus{} m) \ge 8mnp$, which is true if we aplly AM-GM three times.

may I know how you get i) ii) and iii) written on your third line????
And what is a, b, c, p ,R???

. a, b and c are the sides of $ \Delta ABC$.

. p is the semi-perimeter of $ \Delta ABC$.

. R is the radius of the circumscribed circle to $ \Delta ABC$.

. the formulae i) can be obtained from a similarity of triangles.

. the formulae ii) can be obtained from Stewart's Theorem and apllying i).

. the formulae iii) is the Sine's Law.

[quote="math beginner"]Show that for every triangle, with interior angles A, B, C, we have
sin(A/2) x sin(B/2) x sin(C/2) is less than or equal to 1/8.

 :?:[/quote]
Remember that:
$ \sum cosA\equal{}1\plus{}4*sin(A/2)*sin(B/2)*sin(C/2)$
but
$ \sum cosA \leq \frac{3}{2}$

$ {sin\frac{A}{2}}.{sin\frac{B}{2}}.{sin\frac{C}{2}}$
 =$ \frac{1}{2}$.$ {sin\frac{A}{2}}.({cos\frac{B-C}_{2}}-{cos\frac{B+C}_{2}})$
=$ \frac{1}{2}$.$ {sin\frac{A}{2}}.(cos\frac{B-C}_{2}}-{sin\frac{A}{2}})$
$ \leq$ $ \frac{1}{2}$.$ {sin\frac{A}{2}}.(1-sin\frac{A}{2})$ , with equality when B=C
$ \leq$ $ \frac{1}{2}$.$ \frac{1}{4}$=       $ \frac{1}{8}$, with equality when $ sin\frac{A}{2}$ =$ \frac{1}{2}$ and B=C that is when A=B=C=60.

t(1-t) $ \leq$ $ \frac{1}{4}$ when 0$ \leq$ t $ \leq$ 1 , the maximum occuring when t=$ \frac{1}{2}$.   Use calculus or the inequality $ (\frac{p+q}{2})^2$ $ \geq$ pq with p=t, q=1-t.

By Jensen's ineq, we have a good solution.

Let $ f(x) \equal{} \ln(\sin(\dfrac{x}{2}))$ $ \Rightarrow$ $ \dfrac{d^2f}{dx^2} \equal{} \minus{} \dfrac{1}{4\sin^2(\dfrac{x}{2})} \le 0$.

So we have: $ \dfrac{1}{3}\ln(\sin(\dfrac{A}{2})) \plus{} \dfrac{1}{3}\ln(\sin(\dfrac{B}{2})) \plus{} \dfrac{1}{3}\ln(\sin(\dfrac{C}{2})) \le$

$ \le \ln(\sin(\dfrac{\dfrac{A \plus{} B \plus{} C}{3}}{2})) \equal{} \ln(\sin(\dfrac{A \plus{} B \plus{} C}{6})) \equal{} \ln(\sin(\dfrac{\pi}{6})) \equal{} \ln(\dfrac{1}{2})$ $ \Leftrightarrow$

$ \ln(\sin(\dfrac{A}{2})) \plus{} \ln(\sin(\dfrac{B}{2})) \plus{} \ln(\sin(\dfrac{C}{2})) \le 3\ln(\dfrac{1}{2}) \equal{} \ln(\dfrac{1}{8})$.

And we have a more one proof.


You just want to evaluate that?

[hide]$\log_{3}81-\ln e^{-7}=4-\left(-7\right)=\boxed{11}$[/hide]

The rules of logs used in this evaluation are:

1) $\log_{x}{x}=1$
2) $\log_{x}{x^y}=y$

Although 1) is really just a specific instance of 2).  $\ln$ means $\log_{e}$, so these rules apply for that as well.


What is the product of the reciprocals of $ \frac{5}{4}$ and $ \frac{10}{3}$, expressed as a percent?

The reciprocal of $ \frac{5}{4}$ is $ \frac{4}{5}$, and the reciprocal of $ \frac{10}{3}$ is $ \frac{3}{10}$. Multiply them together to get $ \frac{12}{50}\equal{}\boxed{24}\: \text{percent}$


For which n is x5 - nx - n - 2  a reducible polynomial in Z[x]?

[i]Greece 2001[/i]

It's nice problem.

The answer is: n=-2, 10, 34, 342 and n = -1, 19.
We have to consider two case:
(1) P(x) = x^5 - nx - n - 2 has factor of the form x-a (easy) then we have first group.
(2) P(x) = (x^2+ax+b)(x^3+cx^2+dx+e)
Then by comparing with P(x) = x^5 - nx - )n+2), we get
(2a+1)b^2 - (3a^2-a^3)b + 2 + a^4 = 0
And we come to following problem
"Find all integer a for which  a^6 - 2a^5 + 5a^4 - 16 - 8 is square of an integer".

I solved it, but I leave it to my dear readers. The hint:  a^6 - 2a^5 + 5a^4 - 16 - 8 = (a^3-a^2+2a+2) - 24a - 12.

Namdung


After using Cauchy in Engel form, it is sufficient to prove that
\[ (a+b+c+d)^2\geq 2\{a(b^2+1)+b(c^2+1)+c(d^2+1)+d(a^2+1)\}.  \]
Since sum of the four numbers equal to $4$, we need to show that
\[ ab^2+bc^2+cd^2+da^2\leq4.  \]
But if we choose $b=3$, $a=1$, and $c=d=0$, then this inequality is reversed.

I failed this way. Has anyone here solved it?

[quote="pavel kozlov"]Sum of any positive numbers $a,b,c,d$ equals $4$. Prove the inequality
$\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{d^2+1}+\frac{d}{a^2+1}\geq 2$.[/quote]

Indeed very nice, pavel.
and besides congrats for participating in IMO 2005  :)

well, this reminds me some old nice problem....

It is $\frac{a}{b^2+1}=\frac{ab^2+a-ab^2}{b^2+1}=a-\frac{ab}{b^2+1} \cdot b \geq a-\frac{ab^2}{2b}=a-\frac{ab}{2}$

so
$\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{d^2+1}+\frac{d}{a^2+1}\geq a+b+c+d-\frac{ab}{2}-\frac{bc}{2}-\frac{cd}{2}-\frac{da}{2}$

$=4-\frac{ab}{2}-\frac{bc}{2}-\frac{cd}{2}-\frac{da}{2}$

It remains to show that

$4\geq ab+bc+cd+da$ $\Longleftrightarrow$ $4^2\geq 4(ab+bc+cd+da)$

$(a+b+c+d)^2\geq 4(ab+bc+cd+da)$

$(a-b+c-d)^2\geq 0$

$\mathbb{QED}$

[quote="socrates'']
well, this reminds me some old nice problem.... [/quote]

Indeed socrates it is very similar to Meditterenean Contest 2003 inequality where we had

$a,b,c>0$  and  $a+b+c=3$.  and to prove that $\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\geq\frac{3}{2}$

It is solved in the same way as you solve this one.

Now i will post the generalisation of this nice problem

Let  $a_1,a_2,....,a_n>0$  and $a_1+....+a_n=n$
Prove that $\frac{a_1}{a_2^2+1}+....+\frac{a_n}{a_1^2+1}\geq\frac{n}{2}$

 ;)

[quote="silouan"]
Now i will post the generalisation of this nice problem
Let  $a_1,a_2,....,a_n>0$  and $a_1+....+a_n=n$
Prove that $\frac{a_1}{a_2^2+1}+....+\frac{a_n}{a_1^2+1}\geq\frac{n}{2}$
[/quote]
If $n\geq6,a_1=a_2=a_3\to\frac{n}{3},a_4=...=a_n\to0,$ then
$\frac{a_1}{a_2^2+1}+....+\frac{a_n}{a_1^2+1}<\frac{n}{2}$ ;)
Remain $n=5$ only! :lol:

We have $\frac{a}{b^2+1}\ge\frac{1}{2}(a-ab+1)$, result follows.

[quote="mecrazywong"]We have $\frac{a}{b^2+1}\ge\frac{1}{2}(a-ab+1)$[/quote]
are you sure? :?:  let $a \to 0$...

Anyway I think it is right that:
$\frac{a}{b^3+1}+\frac{b}{c^3+1}+\frac{c}{a^3+1}\geq\frac{3}{2}$
with the same condition.

:oops_sign:
Silly me, I have made some computation mistakes...

My generalisation was wrong sorry :oops: But I think the followinf is nice

Let  $a_1,a_2,....,a_n>0$  and $a_1+....+a_n=n$
Find the min and the max of

$\frac{a_1}{a_2^2+1}+....+\frac{a_n}{a_1^2+1}$

If there is so many different couterexamples with different answers such as $2.4$, then finding the min or the max of this inequality is rather difficult problem.

[quote="arqady"][quote="zhaobin"]
Anyway I think it is right that:
$\frac{a}{b^3+1}+\frac{b}{c^3+1}+\frac{c}{a^3+1}\geq\frac{3}{2}$
with the same condition.[/quote]
I think the next true:
If $a>0,b>0,c>0,a+b+c=3,0\leq n\leq7,$ then
$\frac{a}{b^n+1}+\frac{b}{c^n+1}+\frac{c}{a^n+1}\geq\frac{3}{2}$[/quote]
nice!!!arqary,can you post your solution? :)

[quote="zhaobin"]
nice!!!arqary,can you post your solution? :)[/quote]
I think and I guess that this is true. Thank you for "nice!!!" :lol:

[quote="zhaobin"]
Anyway I think it is right that:
$\frac{a}{b^3+1}+\frac{b}{c^3+1}+\frac{c}{a^3+1}\geq\frac{3}{2}$
with the same condition.[/quote]
Yes, you are right! I found the nice proof. ;)

[quote=pavel kozlov]Sum of any positive numbers $a,b,c,d$ equals $4$. Prove the inequality
$\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{d^2+1}+\frac{d}{a^2+1}\geq 2$.[/quote]

This reminds me of Shapiro Inequality.

[quote=Vasc]If $a,b,c$ are non-negative nummbers such that $a+b+c=3$, then for any $0<p \leq 3$ the inequality holds
$\frac{a}{b^2+p}+\frac{b}{c^2+p}+\frac{c}{a^2+p} \geq \frac{3}{1+p}$.
Moreover, I conjecture that the inequality is valid for $0<p\leq 3+2\sqrt3$.[/quote]
At least, for $p=3+2\sqrt3$ it's true.
Indeed, it's enough to prove this inequality for $a\leq b\leq c$.
Let $b=a+u$ and $c=a+u+v$.
Thus, we need to prove that:
$$f_1(u,v)a^4+f_2(u,v)a^3+f_3(u,v)a^2+f_4(u,v)a+(2u+v)(4u+(5-\sqrt3)v)(u-(1+\sqrt3)v)^2(83u^2+4(5+6\sqrt3)uv+(5+6\sqrt3)v^2)\geq0,$$
which is true because $f_i(u,v)\geq0.$
For example, $$f_1(u,v)=432\cdot166(2+\sqrt3)(u^2+uv+v^2)\geq0,$$
$$f_2(u,v)=36\cdot166((57+26\sqrt3)u^3+6(12+5\sqrt3)uv^2+3(31+16\sqrt3)uv^2+(39+22\sqrt3)v^3)\geq0,...$$


Does anyone know if Invitation to Number Theory by Oystein Ore or Introduction to Inequalities by E.F. Beckenbach is good or not?

Would they be useful in preparing for AMC10/AMC12?

[quote="13375P34K43V312"]You definitely don't need Inequalities for AMC 10 or 12, maybe a little experience with 2 variable AM-GM, but nothing else is needed.[/quote]
Not just two variable AM-GM, the entire form of it is needed.

I kinda disagree with that...give me an example and we'll talk.


function
algebra
polynomial
search
complex analysis
complex analysis unsolved

An entire function which has pole at infinity is a polynomial

this has been solved a lot of times on the forum...i suggest that you search for this topic.


Do you use the keypad (located on the right) or the numbers bar on the top?

For me, I use the numbers bar. It will be interesting to see how the FTW pros input their answers.

I use the bar too, seeing as that I use my laptop.

I use the numbers bar; I keep typoing for the number pad for some reason.

Number bar - it's easier to type in word answers if needed.  Also, I never learned to use the number pad correctly.

Number row, for the same reasons as the person above.

i use the keypad

for me, it's easier having the numbers close to each other and simply moving my fingers than if i used the number row and had to move farther
plus, i've never really been quick at entering numbers using the number row

I use the number bar at the top simply for the reason that it is easier to type letters that way, and the fact my laptop number doesn't have a number row thingy.

Number bar, only because my laptop doesn't have a keypad.
Otherwise, I would be using the keypad.

i use number row because
1) i have a laptop
2) i hate using keyboard
3) which should be 2 but w/e i don't know how :P

lulz but how does someone use the onsceen one it seems SOOOOO HARD.

I think they picked that choice just to screw up the poll.

[quote="AceOfDiamonds"]i use the number pad with my right hand, left hand on ctrl v in case i need to put a square root or type stuff[/quote]

SMART :)

Never thought of copying sqrt and pasting instead of typing it lol

I use number row because I use a laptop and I'm not used to number pad.

Whats a tablet?

A tablet is a computer with a touch screen.

For anyone with Windows XP(and maybe other Windows things too)
to get an onscreen keyboard:
Click on the green start menu, press "All Programs", then click on "Accessories", than "Accessibility", then "On Screen Keyboard"

Actually I copy and paste square root symbols etc.

[quote="Maybach"]I use keypad.  I can type 1000 times faster with it.  My rating would be like 500 if I used the number row.[/quote]

Me too!

where do you find the sqrt signs i can't

For the square root sign, put "sqrt", not $ \sqrt{}$.

@HappyLand:nice

I think keypad is a lot faster though

Ya, I agree. Keypad is what I use too. Well, I use a laptop, so it's really easy.

Number row for the same reasons.
It's not slow. I rely on it pretty well and it does fine.

I use the number row.  I have no experience with the number keypad.


Find the area of a $ 45\minus{}45\minus{}90$ right triangle with inradius $ 9$.

[hide]The formula for inradius is $ \frac {ab}{a \plus{} b \plus{} c}$, but since this is an isosceles triangle we can state $ a \equal{} b$ and $ c \equal{} a\sqrt{2}$

Which gives us $ \frac {a^{2}}{2a \plus{} a\sqrt {2}} \equal{} 9$

Multiplying and subtracting gives us $ a^{2} \minus{} 18a \minus{} 9a\sqrt {2} \equal{} 0$

We know $ 0$ isn't a solution so get rid of one $ a$ and get $ a \minus{} 18 \minus{} 9\sqrt {2} \equal{} 0$

Therefore $ a \equal{} 18 \plus{} 9\sqrt {2}$

Then area of an $ 45 \minus{} 45 \minus{} 90$ triangle is $ \frac {a^{2}}{2}$

and the area of the triangle is $ 243 \plus{} 162\sqrt {2}$ [/hide]

[quote="CatalystOfNostalgia"]What exactly makes this 'neat?' To me it seems pretty standard. Just find the area two ways.[/quote]

if you dont have something nice to say, dont say it :)

[quote="CatalystOfNostalgia"]What exactly makes this 'neat?' To me it seems pretty standard. Just find the area two ways.[/quote]

Well, usually I make up really bad clunky geometry problems, which you brute force for a while and get the answer, and I thought this one was a bit better.

[quote="Temperal"]Find the area of a $ 45 \minus{} 45 \minus{} 90$ right triangle with inradius $ 9$.[/quote]
[hide]A=rs
So let the side of the triangle be x. So the area is $ \frac {x^{2}}{2}$. Semi-perimeter of the triangle is $ \frac {x \plus{} x \plus{} x\sqrt {2}}{2}$, and inradius is 9. Plug and chug.
$ \frac {x^{2}}{2} \equal{} \frac {18x \plus{} 9x\sqrt {2}}{2}$
$ x^{2} \equal{} 18x \plus{} 9x\sqrt {2}$\
$ x \equal{} 18 \plus{} 9\sqrt {2}$
Since area is that squared over 2, simplify
$ (18 \plus{} 9\sqrt {2})^{2} \equal{} 486 \plus{} 324\sqrt {2}$
Dividing by two....
$ \boxed{243 \plus{} 162\sqrt {2}}$[/hide]

[quote="CatalystOfNostalgia"]What exactly makes this 'neat?' To me it seems pretty standard. Just find the area two ways.[/quote]
Not everyone is as experienced as you are jclarke.
When I first saw a problem like this (it was a 30-60-90 instead I think) I was unsure if there was enough information.


Let $ f(n)\equal{}$ the number of odd coefficients at the n-th row of pascals triangle.

[example: $ f(1)\equal{}1,f(2)\equal{}2, f(3)\equal{}2$ etc.]

(a)find a formula or algorithm to find $ f(n)$
(b)find the values of $ n$ for which $ f(n)\equal{}n$

[quote="Moonmathpi496"]Let $ f(n) \equal{}$ the number of odd coefficients at the n-th row of pascals triangle.

[example: $ f(1) \equal{} 1,f(2) \equal{} 2, f(3) \equal{} 2$ etc.]

(a)find a formula or algorithm to find $ f(n)$
(b)find the values of $ n$ for which $ f(n) \equal{} n$[/quote]
I think $ f(0) \equal{} 1,f(1) \equal{} 2 \equal{} f(2),f(3) \equal{} 4$.
These problem posted many times and $ f(n) \equal{} 2^{k}$, were number of 1 for 2-adic coefficients $ n \equal{} a_{0}\plus{}a_{1}2\plus{}a_{2}2^{2}\plus{}...$.
Therefore $ n\equal{}f(n)\to n\equal{}2^{k}\to f(n)\equal{}2\to n\equal{}2$.

[quote="Rust"]
I think $ f(0) \equal{} 1,f(1) \equal{} 2 \equal{} f(2),f(3) \equal{} 4$.
These problem posted many times and $ f(n) \equal{} 2^{k}$, were number of 1 for 2-adic coefficients $ n \equal{} a_{0}\plus{}a_{1}2\plus{}a_{2}2^{2}\plus{}...$.
Therefore $ n \equal{} f(n)\to n \equal{} 2^{k}\to f(n) \equal{} 2\to n \equal{} 2$.[/quote]
ok..actually i told that $ n\equal{}$row #,not the power of the $ (x\plus{}y)$.but if u assume the 1st row=0 then u r correct.

anyway..if u know the link or complete solution please send.

Use Lucas's theorem //http://en.wikipedia.org/wiki/Lucas'_theorem .


inequalities
function
rearrangement inequality

Prove a⁴ + b⁴ > a³b + ab³, given that a is not equals to b.

Do you want to say $ a^4 \plus{} b^4 > a^3b \plus{} ab^3$?
We have $ a^4 \plus{} b^4 \minus{} a^3b \minus{} ab^3 \equal{} a^3(a \minus{} b) \minus{} b^3(a \minus{} b) \equal{} (a \minus{} b)^2(a^2\plus{}b^2\plus{}ab) \geq 0$
Equality occurs when $ a \equal{} b$,but $ a \neq b$ so $ a^4 \plus{} b^4 > ab^3 \plus{} a^3b$

This is an instance of the Rearrangement Inequality on the two similarly sorted lists $ a^3,b^3$ and $ a,b$. We know the two lists are similarly sorted because the cubing function is monotone increasing.

You can also use AM-GM

$ a^4 \plus{} b^4 \equal{} \frac{a^4\plus{}a^4\plus{}a^4 \plus{} b^4}{4} \plus{} \frac{a^4\plus{}b^4\plus{}b^4\plus{}b^4}{4} \ge \sqrt[4]{a^{12}b^4} \plus{} \sqrt[4]{a^4b^{12}} \equal{} a^3b \plus{} ab^3$


vector
algebra
polynomial
superior algebra
superior algebra unsolved

I don't understand how to do this problem:

Theorem 1. Let $ F$ be a finite field with characteristic $ p$. Then the number of elements is of the form $ p^n$. (I think one can choose $ n \equal{} [F: F_0]$ where $ F_0$ is a prime subfield of $ F$.)

Problem 1: Prove by using theorem 1 that if $ F_{q^n}$ is an extension of $ F_{q}$ with degree $ n$.

In my opinion, it is easier to reason that $ F_{q^n}$ is an $ n$-dimensional vector space over $ F_q$ but how to do the problem 1 using theorem 1?

You need theorem 1 to define $ \mathbb{F}_{q^n}$.
And problem 1 depends on that definition.
Different definition needs different solution for problem 1.

The keyword here is [url=http://en.wikipedia.org/wiki/Frobenius_endomorphism]Frobenius endomorphism[/url].  On a commutative ring of characteristic $ p$ the map $ \phi(a) \equal{} a^p$ respects both addition and multiplication, and on a field also respects inverses.  (Edit:  The argument that used to be here was flawed.)  The Frobenius map is also injective over a field, since $ a^p \equal{} b^p \Leftrightarrow (a \minus{} b)^p \equal{} 0 \Leftrightarrow a \minus{} b \equal{} 0$, and over a finite field is automatically surjective, hence bijective.  Some finite power of the Frobenius map is therefore the identity map (since the Frobenius map is just a permutation), so there exists a minimal $ n$ such that $ a^{p^n} \equal{} a \forall a \in F$.  The $ p^n$ roots of this polynomial are the $ p^n$ elements of $ F$.

[quote="t0rajir0u"]On a commutative ring of characteristic $ p$ the map $ \phi(a) \equal{} a^p$ respects both addition and multiplication, and on a field also respects inverses.[/quote]
That was nice proof.  I'm non-native English speaker so does that sentence mean $ \phi (a\plus{}b)\equal{}\phi (a)\plus{}\phi (b), \phi (ab)\equal{}\phi (a)\phi (b),$ and $ \phi (a^{\minus{}1})\equal{}\phi (a)^{\minus{}1}$? I heard that term first time.

The third condition isn't really necessary. We call that a ring homomorphism.


In the equation $ 2x^2 \minus{} hx \plus{} 2k \equal{} 0$, the sum of the roots is $ 4$ and the product of the roots is $ \minus{}3$. Then $ h$ and $ k$ have the values, respectively:

$ \textbf{(A)}\ 8\text{ and }{\minus{}6} \qquad
\textbf{(B)}\ 4\text{ and }{\minus{}3}\qquad
\textbf{(C)}\ {\minus{}3}\text{ and }4\qquad
\textbf{(D)}\ {\minus{}3}\text{ and }8\qquad
\textbf{(E)}\ 8\text{ and }{\minus{}3}$

[hide]
Just use Vieta. h/2 = 4, h = 8. k/2 = -3, k = -6.
$ \boxed{\text{A}}$[/hide]

Actually...

[hide="My Solution"]
The sum of the roots of this quadratic is $ \frac{h}{2} \equal{} 4\implies h \equal{} 8$ and the product of the roots of this quadratic is $ \frac{2k}{2} \equal{} \minus{}3\implies k \equal{} \minus{}3$. Therefore, the answer is $ \fbox{(E)}$.
[/hide]

Oh whoops, silly mistake from me there. I didn't see the 2 before the k.


Columbia
geometry
calculus
number theory

Hi im from colombia and firts im sorry because my level english is not good.look the problem is that en march of the next year i want to participate in the universitarian maths olympiad of my country but in this moment i dont have the knowledge to solve the problems.i olny know what topics i need to study: number theory,abstrac algebra,analysis(calculus),geometry,linear algebra. moreover i study chemical engineer so i dont have the preparetion that a student of mathematics but i wish study hardest to be a good mathematical problem solve.i want that you help me saying me what books you recomended me to preparate me in each topic?.thanks

A couple general problem solving books that could help with the olympiad are:
Problem Solving Through Problems by Larson or Problem Solving Strategies by Arthur Engel.

I cannot recommend books on every topic, but for calculus my favorite text is Calculus by Spivak.


Este el cronograma de olimpiadas para [b]Perú[/b], espero que sea de ayuda a todos los peruanos que esten en el foro  :D

[color=blue][size=150]SOCIEDAD MATEMATICA PERUANA
COMISION DE OLIMPIADAS
CRONOGRAMA DE ACTIVIDADES 2006[/size][/color]

[color=blue][size=150]OLIMPIADA : APMO[/size][/color]
[color=blue]Prueba de selección : [/color]
----------------
[color=blue]Prueba oficial:[/color]
 - Fecha : lun 13 marzo
 - Hora: 15:00 hrs
 - Lugar: PUCP

[color=blue][size=150]OLIMPIADA : Cono Sur[/size][/color]
[color=blue]Prueba de selección : [/color]
 - Fecha : sáb 25 marzo
 - Hora: 9:00 hrs
 - Lugar: PUCP
[color=blue]Prueba oficial:[/color]
 - Fecha : del 5 al 11 de mayo
 - Hora: --
 - Lugar: Escobar-Argentina

[color=blue][size=150]OLIMPIADA : de Mayo[/size][/color]
[color=blue]Prueba de selección : [/color]
 --------
[color=blue]Prueba oficial:[/color]
 - Fecha : sáb 13 mayo
 - Hora: 9:00 hrs
 - Lugar: PUCP

[color=blue][size=150]OLIMPIADA : IMO[/size][/color]
[color=blue]Prueba de selección : [/color]
 - Fecha : sab 20 mayo
 - Hora: 9:00 hrs
 - Lugar: PUCP
[color=blue]Prueba oficial:[/color]
 - Fecha : del 6 al 18 de julio
 - Hora: --
 - Lugar: Eslovenia

[color=blue][size=150]OLIMPIADA : IBERO[/size][/color]
[color=blue]Prueba de selección : [/color]
 - Fecha : sáb 12 agosto
 - Hora: 9:00 hrs
 - Lugar: PUCP
[color=blue]Prueba oficial:[/color]
 - Fecha : del 22 de sept al 1 oct
 - Hora: --
 - Lugar: Guayaquil- Ecuador

[color=blue][size=150]OLIMPIADA: ONEM 2006[/size][/color]
[color=blue]1ra Etapa	. [/color]
 - Fecha : vier 26 may
 - Hora: 10:00 hrs
 - Lugar: Centro educativo
[color=blue]2da Etapa. [/color]
 - Fecha : vier 14 jul
 - Hora: 10:00 hrs
 - Lugar: UGEL
[color=blue]3ra Etapa. [/color]
 - Fecha : vier 29 sept
 - Hora: 10:00 hrs
 - Lugar: DRE
[color=blue]4ta Etapa. [/color]
 - Fecha : sáb 28 oct
 - Hora: 10:00 hrs
 - Lugar: LIMA

$Tipe$

[quote="mcphisto"]Tipe una pregunta...como sera ahora el sistema de la onem? por equipos? o se mantiene el sistema del año pasado?[/quote]

Se mantiene el del año pasado...
Pero... como tú vas a ir a la IMO ... no podrás ir a la ONEM.

$Tipe$

Alguien sabe por que se cambio la fecha de la segunda etapa de la ONEM?


honestly speakin i'm yet to start 4 bits
so just thot of searchin 4 prev yr papers if available online to've an idea of d q paper's level of difficulty
does anyone know a good source 4 d same (e1 mock tests'l do coz i doubt if prev yr papers'l be available as evry yr nearly 10++ papers come out each day new paper)

god, ppl shud learn to chill. :o  take a break da. you wrote EEE just yesterday.

honestly speaking, my EEE was not that good, so i m now preparing for my last frontier, BITSAT

i hv bought a book for it, by Arihant. dont mind it, coz anyway i didn't have any material for any exam, except for JEE.

can u guys post ur BITSAT dates here. we will know who all r going in first, and going to face the music!

mine is on 10th may, morning slot!

I'm on 15th. From what I hear, BITSAT will also be of NCERT Portions but much more interesting and tougher than EEE.

wth? after flopping in the aieee n jee, i cant do with one more :( im writing on the 16th evening btw :)

Well, I can ask all of you to chill, because irrespective of whether Shreyas agrees with me or not, I am chilling out. Relax for a day or two, start again this weekend or so....

Good luck to all you guys!!

jus spoke to shre yest evening...he sure will agree wid me dat i bearly study these days...
i dunno wats seriously goin wrong wid all u guys...
u ve jus finished one exam.. and u want to study for da next  :!:
Lookamme...
i ve already gone outta stn and even cm back after eee...
tho jus go out n njoy...
wats der afterall... we r jus 12grade passouts.. (i hope).
so njoy... u dont hv to study so much..
now me sayin all dis dusnt mean i did in eee or cet or jee like sm god...
but u c da point is...
[hide]i am not xactly bothered...[/hide]

so
[hide]jus chillax and enjoy.  :) [/hide]

btw i am writin bits on 20 may in da morning.  :)

Look man, when the weight of having flopped in 2 serious exams rests on your shoulders (mine :D), I think that by itself is enough driving force for me to study.

I second Sachin, I too don't have any seats confirmed.

me three, i dont have ne seats confirmed newhere either. but that by itself hasnt proved to be enuf driving force. coz (to use words frm ma fav book)'frankly, i dont give a damn'. :P and ofc i never claimed that i wud do well in the exams.
[quote="sachin a S"][quote]god, ppl shud learn to chill.  take a break da. you wrote EEE just yesterday[/quote]

as if u r chillin a lot  :P  . u don't even spend 3 hours 4 a movie wen i called u 4 d same & den tel me how to chill  :mad:
furthermore i don't 've seats guaranteed at iits like u do  :maybe:  so i've to consider bits seriously too [/code][/quote]
well you cant expect me to watch some stupid bollywood flick, that too at ten in the morning when i wud rather be snoozing in my bed. and i cant help it if u stick to your regimen of getting up at 5 even during the holidays.
and im not worried abt the exams coz i know im gonna flunk neways, unlike u ppl over here.  :( im going out to some place for more than a week, just in case you dont know while you are holed up in your room and studying like mad. :rotfl:
btw, im writing bitsat on the 14th of may.

read some english thats all :D and ofc remember to nv the ncert text book if u want :D

i dint even know we had english. do we have GK or something liek dat too. my friend told me we had logic too? :maybe:

yes 30 ques on eng and logic - 10 eng and 20 logic or sth :)

[quote="Chronoz"][quote="sachin a S"]i'm takin bits on 9th (i'l not be in d city after dat) . so think i'l be d 1st to give u ppl an idea of exam  :D

i do feel d topics gettin swayed a lot infact i stil didn't get any clear idea of whivh source to depend upon 4 bits. i mean wil books meant 4 eee wud be enough or i need to go to d jee level??[/quote]

Nah, portions are more or less same as EEE, though a bit tougher. The toughness is between EEE and JEE, but beware of stupid school portions like semiconductors. Only thing which might be difficult is the English but if you are generally good in English and have a decent vocabulary, its passable.

OR else take a dictionary and start mugging today. :)[/quote]

wth i never knew vocabulary'l take d centrestage in english. if dey check our standards in eng den dey should focus on might be grammar and comprehensions . neways i never considered eng to be a serious obstacle 4 selection but considerin d importance dat u ppl r givin 4 it i'l 've to think again

@varun . though there's no part meant 4 g k  nearly 10 questions r meant to check our logic mainly consistin of analogy & sequence

btw does neone know d diff b/w d last 12 questions & d rest except dat dey'l be a bit tough

oh then english shall be death for me....  :(

me on 22nd.................guys do post ur scores.......... :maybe:
so that I realise how futile my attempt is going to be :rotfl:


$ \text{what are the inequalities you should know for IMO inequalities problem solving}\\
\text{MA,MG,MH}\\
\text{Bernoulli}\\
\text{Besnitt}\\
\text{Cauchy\minus{}Schwarz}\\
\text{what else?}$

well,
AGM inequality (,of course)
holder inequality
rearrangement inequality
Chebyshev
Jensen

no idea...

Sorry to make an offtopic, but what is the Besnitt's inequality? Is there really one like that or is it a typo and should be written as Nesbitt's :?:

I think it's Nesbitt:D. Use this pdf for information


The product of three consecutive integers is 990. What is their sum?

It looks close to $ 10^3$, so we try $ 9 \times 10 \times 11$, which is $ 99 \times 10\equal{}990$.  The sum of these is $ \boxed{30}$.


[quote="basil mohammed"]the  number of positive roots of a polynomial cannot be more than the number of pairs of consecutive terms with opposite signs in the sequence of its coefficents[/quote]

[url]http://www.mathlinks.ro/viewtopic.php?p=1489506&search_id=1832656894#1489506[/url]

i discovered this while solving this problem. it leads to a very nice solution

[url]http://www.mathlinks.ro/viewtopic.php?t=274748[/url]


For any sets C and D, prove that:

$ i) C \equal{} (C\cup D)\cap D$

$ ii) C \equal{} (C\cap D)\cup C$

I think I should be using truth tables but I get sooo confused by them. I don't know which tables I should be doing. Or even if I should be using truth tables at all. Help!

Well, what can I use in the proof? If you can use the axioms of Boolean algebra, then simple absorption (or the distribution) rule should do the trick-they're both applicable here (BTW, check your i) ;) ).
On the other hand, if you want to use truth tables instead of axioms (although the tables are generally proved using the axioms of Boolean algebra), then you can do the same as in logic (just take the cap as conjunction, and cup as disjunction), or you can simply do the following: draw a Venn diagram. Then you have four possibilities where to put an element (just in C, just in D, in the common part of C and D, or outside both of them), and show that if the elemnt is on the LHS of your equality, then and only then is on the RHS. That's basically same as truth table.


M is the midpoint of the side BC of the triangle ABC, and   :ang: CAB=45 :^o:,  :ang: ABC=30 :^o:
1. Find   :ang: AMC
2. Show that AM=AB*BC/2AC

Let $D$ be the foot of a normal from the vertex $A$ to the line $BC$.

${{{\tan{\widehat{MAH}} - \tan(\widehat{CAH}} = \frac 1 2 (\tan(\widehat{BAH}} - \tan(\widehat{CAH}})$

$\tan{\widehat{MAH}} = \frac 1 2 (\tan{\widehat{BAH}} + \tan{\widehat{CAH}}) =  \frac 1 2 (\tan{60^o} + \tan{15^o})$

$\tan{15^o} = \frac{\sin{45^o} \cos{30^o} - \cos{45^o} \sin{30^o}}{\cos{45^o} \cos{30^o} + \sin{45^o} \sin{30^o}} =$

$= \frac{\cos{30^o} - \sin{30^o}}{\cos{30^o} + \sin{30^o}} = \frac{\sqrt 3 - 1}{\sqrt 3 + 1} = 2 - \sqrt 3$

$\tan{\widehat{MAH}} = \frac 1 2 (\sqrt 3 + 2 - \sqrt 3) = 1$, $\measuredangle MAH = 45^o$, $\measuredangle AMC = 45^o$

The triangles $\triangle ABC \sim \triangle MBA$ are similar, because the angle $\measuredangle C = 105^o$ is common and the angles $\measuredangle BAC = \measuredangle AMC = 45^o$ are equal.

$\frac{AB}{AC} = \frac{AM}{MC} = 2 \frac{AM}{BC}$

$\frac{AB \cdot BC}{2 \cdot AC} = AM$


Fifty percent more than what number is 25 percent less than 60 percent more than 10?

In this problem, all we have to do is work backwards.  First, we do $ 60%
$ of $ 10$, which is also $ 16$.  Then, we find $ \frac{3}{4} \cdot 16$, which is $ 12$.  Finally, we find $ \frac{2}{3}$ of $ 12$, which is $ \boxed{8}$.


Prove that for any odd prime number $p$ there exist $p$ positive integers not exceeding $2p^2$ for which the pairwise sums of the integers are all different.

I think it means the sums of the $p$ integers taken by pairs, for example if the integers are 1,2,3 their pairwise sums are 1+2,2+3,1+3