Folks, I would like to know how many IMO medalists (Gold/Silver/Bronze) are there in this community? How many Putnam winners are there in this community?

I know that this is a  silly query, but I am curious to know.

I am pretty sure you can find that in Aopswiki.

i think we should also include all the participants of aops in intel isef

If there's something wrong with it, fix it. That's the point of having a wiki.

Do you recommend we ask someone before putting their name on the list?

Where [in this forum] can I expect to chat [ regarding problems] with a Putnam winner/IMO Medalist?

I want to know if there are any Putnam winners, IMO medalists in this forum as of now.

Please remember to keep in mind the privacy of our members in this thread.  Listing the winners on the AoPSWiki, or their real names in this forum, is fine -- these names are public record.  Don't link usernames and real names unless you can provide links to places on the site in which the people have done so themselves.  Also, do not PM these users out of the blue; respect their privacy!

[quote="orl"][quote="cf249"]Specifically, look at [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPS_Community_Awards]this page[/url].[/quote]

Actually there are also people as Neal Wu who went to IOI. And for example [url=http://www.imo-official.org/participant_r.aspx?id=7901]Darij Grinberg[/url] went to IMO 2004-2006 and not only in 2006.[/quote]

All of last year's USA IOI team have AoPS accounts (they were all '07 MOPpers :O)

If you want to find IMO medalists, you can... snoop around? Or you can just make lots of good posts and one could eventually respond to you.

Yeah, I am trying that in the hope of invoking a response from any one of these people.  :lol:

Maybe chances get better if you tell us why you are so keen on this...

Well, My only intention is: I would like to communicate with these people [Putnam winners/IMO Medalists] because of the fact that I perceive them to be extremely adept at solving intricate,knotty problems [ and they usually are very good at problem-solving], in a way one can learn many things pertaining to problem solving from these people.

Now that I had conveyed my intention,I would like to invoke a response from any of these winners, and also from people who participated in the Putnam or the IMO,IPho.

As other people already pointed out, most people on this site are pretty good at solving problems, and most of them are not IMO medalists or Putnam fellows.  Why don't you try to get into correspondence with a university professor?  Most of them are pretty quick and talented, and most are not IMO medalists or Putnam fellows.

It's still utterly unclear what exactly you are trying to make happen.  If there is some particular piece of information that you are trying to glean, you're better off starting by asking for that information, rather than asking people to announce themselves so that you can ask them for information.  If all you want to do is see them in action, you're in the wrong section of the forum -- go to the college and olympiad sections and read through the old posts there.  There's an abundance of adept solutions to intricate problems just stored on a server somewhere waiting for you to take a peek ....

That list from the AoPSWiki doesn't contain all of the Romanian users who are IMO medalists. It seems that this site contains more IMO medalists, maybe twice, but surely more than that. I agree with the fact that the IMO medalists are good problem solvers, but there are a lot of other users here who are good enough or better. It would be very interesting to know the exact number of IMO participants from Mathlinks. I see that here are users who have perfect scores at IMO. Congratulations!!!
 :roll:

I'm trying not to word this too strongly, but if I were a Putnam fellow or an IMO contestant, I wouldn't talk to you based on your post history.  You seem to have some silly views on giftedness and intelligence as well as some need to prove a pet theory that 'ordinary people' are just as capable as Putnam fellows or whatever, and I'd suspect that you just want to talk to these people to prove a point (to yourself or others).  It's kind of weird, at least as far as I'm concerned.

blahblahblah

Well, As far as I am concerned, the only reason I have in interacting with these people is to discuss certain aspects of problem-solving with them; I don't have any intention to prove a so-called "pet theory" that you say.

Any person with so-called above-average  problem solving abilities would like to hone his/her skills by interacting with much people who are much better than them.

The so-called pet-theory that you said I seemed to possess was actually discussed in this forum by many respected people (of this forum), I was just trying to elaborate what they had said.

But I would like to say the above post by you is "abrasive, harsh" because you have given people an inkling to view my personal posts and categorize me as a silly person, so I request the administrator/moderator to delete the above post by this person.  :maybe:

It's a valid concern given your posts.  You've gone [i]out of your way[/i] to mention your opinions on intelligence and giftedness in several posts here, so I think that anyone who was considering talking with you about this should know what they may be getting into.  It's pretty weird to ask for ways that you can contact people that you don't know at all so you can ask them how they think (this is at least nontrivially personal), especially given that you haven't really participated in any of the mathematical subforums in the slightest.

No, it isn't a trivial query. I haven't been accessing this community regularly so I wasn't been able to 'post' any replies/solutions to any of the mathematical problems that might have cropped up; I had decided to join this forum based on the intellectual discourse that was going on here, and also the problem solving techniques that were being employed to solve problems.

I surmise before you haughtily dismiss my queries as unimportant, you should check Terry Tao's personal blog; The blog maintains a lot of discourse on the nature of giftedness/intelligence/genius. A major aspect of " Cognitive Psychology" is focused on the nature of intelligence/giftedness etc.

I don't view my query as a trivial one, because there's nothing trivial in seeking the replies of superior problem solvers in solving problems, but hey wait, but didn't people ask Einstein how he has conceived the theory of relativity? Would you dismiss that as non-trivial too? :lol:

Every person has his/her own opinions regarding certain issues, you might be having a different opinion on another topic, but a person has no right to criticize in  a harsh manner.

Finally, If you don't want me to see me asking such queries, I urge you not to participate in this thread.

[quote="Gen8"]I surmise before you haughtily dismiss my queries as unimportant, you should check Terry Tao's personal blog; The blog maintains a lot of discourse on the nature of giftedness/intelligence/genius. A major aspect of " Cognitive Psychology" is focused on the nature of intelligence/giftedness etc.[/quote]

Terence Tao asserted that one doesn't have to be a genius to do mathematics (or to contribute to some area of mathematics), but he certainly didn't assert that intelligence doesn't play a role in doing mathematics.

Moreover, the statement that his blog maintains 'a lot' of discourse on the nature of giftedness/intelligence/genius is patently false, and Tao himself describes the blog as  'Updates on my research and expository papers, discussion of open problems, and other maths-related topics').  In the [url=http://terrytao.wordpress.com/career-advice/does-one-have-to-be-a-genius-to-do-maths/]one post where he does talk about whether one has to be a genius to do mathematics[/url], he says that

'As long as you have education, interest, and a reasonable amount of talent, there will be some part of mathematics where you can make a solid and useful contribution. It might not be the most glamorous part of mathematics...'

and later that

'Talent is important, of course'

You might want to keep this in mind before namedropping Terence Tao like he agrees to the letter with everything you say!  By the way, Terence Tao as a child was unbelievably precocious, as can be seen from [url=http://www.jstor.org/pss/3482178]this[/url] article, he clearly is an example of a 'wildly superior intellect' if anyone is.

[b]You might want to keep this in mind before namedropping Terence Tao like he agrees to the letter with everything you say! By the way, Terence Tao as a child was unbelievably precocious, as can be seen from this article, he clearly is an example of a 'wildly superior intellect' if anyone is.

Terence Tao asserted that one doesn't have to be a genius to
do mathematics (or to contribute to some area of mathematics),
but he certainly didn't assert that intelligence doesn't play a role
in doing mathematics.
[/b]

Huh! I never tried to assert that intelligence doesn't play a role, and I advise you to go through Mr Tao's blog completely before saying that there doesn't exist an intellectual discourse over there, and not just" Does one have to be a genius to do Mathematics", I mean the replies he gave to queries asked by some people on that blog.

I never "names drop" Terry, you should go through his entire blog before arriving at such a naive conclusion.

I can't understand why you want to argue over this with me in this way; I had already asked this forum's administrator/moderators before I created this thread, so before this gets any obnoxious, you should stop this!

As a moderator of this respected/wonderful community, I can't comprehend your rather illogical behavior; I hope you mend it. [I can't understand who had made you as a moderator]

I'm locking this thread.  This discussion is no longer fruitful.


[quote="me@home"]Okay, so I happened upon the really nice identity: The constant term of
\[\sum_{i=0}^{n}\left(\sum_{j=i-2n}^{j=2n-i}x^{j}\right)^{k}\]
is $(n+1)^{k}$.

For instance, this claims that the constant term of
\[(x^{-2}+x^{-1}+1+x+x^{2})^{2}+(x^{-1}+1+x)^{2}+1^{2}\]
is $9$.

So how to prove it in a nice way? There's plenty unnice ways.[/quote]
I've posted this elsewhere a while ago but no response. I'll try here I guess...

[quote="me@home"]Okay, so I happened upon the really nice identity: The constant term of
\[\sum_{i=0}^{n}\left(\sum_{j=i-2n}^{j=2n-i}x^{j}\right)^{k}\]
is $(n+1)^{k}$.

For instance, this claims that the constant term of
\[(x^{-2}+x^{-1}+1+x+x^{2})^{2}+(x^{-1}+1+x)^{2}+1^{2}\]
is $9$.
[/quote]

The formula doesn't match the example.  In the example, I'm assuming $n = 2$ and $k = 2$.  Then
\begin{eqnarray*}\sum_{i=0}^{2}\left(\sum_{j=i-2n}^{2n-i}x^{j}\right)^{2}&=& \left( \sum_{j=-4}^{4}x^{j}\right)^{2}+\left( \sum_{j=-3}^{3}x^{j}\right)^{2}+\left( \sum_{j=-2}^{2}x^{j}\right)^{2}\\ &=& (x^{-4}+\dots+x^{4})^{2}+(x^{-3}+\dots+x^{3})^{2}+(x^{-2}+\dots+x^{2})^{2}. \end{eqnarray*}

[hide="Sorry, here it is corrected:"]
The constant term of
\[\sum_{i=0}^{n}\left(\sum_{j=-i}^{j=i}x^{j}\right)^{k}\]
is $(n+1)^{k}$.

For instance, this claims that the constant term of
\[(x^{-2}+x^{-1}+1+x+x^{2})^{2}+(x^{-1}+1+x)^{2}+1^{2}\]
is $9$.
[/hide]

That isn't right either.  The constant term of 1+$(\frac1x+1+x)^{4}$ is 20, not 16.

hmm let me double check... your calculation is correct but I might have had something else in mind. It is certainly correct for exponents 0,1,2,3 so I'm not sure if I even checked four...!


trigonometry
combinatorics unsolved
combinatorics

The distance between any two of 6 given points in the plane is at least $ 1$.Prove that the distance between some two points is at least $ \sqrt{\frac{5\plus{}\sqrt5}{2}}$

$ \cos 144^{\circ}\equal{}\minus{}\frac{1\plus{}\sqrt{5}}{4}$, so i'm guessing we should be able to find a triangle that has an angle of at least $ 144^{\circ}$?


function
calculus
limit
inequalities
geometric series
real analysis
real analysis unsolved

Let $ f$ be a twice differentiable function such that $ f''(x) \plus{} f'(x)$≥$ f(x)^2 \forall x > 0$ and $ f(x) > 0 \forall x > 0$.
Prove that the limit $ limf(x)$(when $ x$ is infintely large) exists and is finte.Evalute the limit.

Wrong forum for a calculus problem - a moderator should move it. [color=blue][Done. -mathmanman][/color]

Not the whole solution but one observation: if $ \lim_{x\to\infty}f(x)$ exists as a finite number, then it can only be zero.

Suppose otherwise - that $ f$ tends to a finite positive limit. Then there exist some $ x_0$ and some $ A > 0$ such that $ f(x)\ge A$ for all $ x\ge x_0.$ Then

$ f''(x) \plus{} f'(x)\ge A^2$ for $ x\ge x_0.$

$ e^xf''(x) \plus{} e^xf'(x)\ge A^2e^x$

$ \left(e^xf'(x)\right)'\ge A^2e^x$

Integrate that from $ x_0$ to $ x$ to get

$ e^xf'(x)\ge A^2(e^x \minus{} e^{x_0})$

$ f'(x)\ge A^2 \minus{} A^2e^{x_0 \minus{} x}.$

But then $ f'(x)$ is greater than some positive constant for all sufficiently large $ x,$ meaning that $ f(x)$ grows at at least a linear rate, meaning that $ f(x)\to\infty,$ contradicting the assumption of a finite limit.

As Kent noticed, $ (e^x f'(x))'\ge e^xf(x)^2>0$, so if $ f'$ is positive at any point, it is positive everywhere beyond that point. Thus, in this case $ f$ is increasing. All we need to show now is a finite time blow-up. Let's use Kent's inequality again. If $ f>A$ at some point $ x$, then $ f'(x\plus{}t)\ge A^2(1\minus{}e^{\minus{}t})$ for $ t>0$. Integrating, we get, $ f(x\plus{}t)\ge A\plus{}A^2(t\minus{}1\plus{}e^{\minus{}t})\ge A\plus{} cA^2t^2$ when $ t<1$ with some $ c>0$. This means that the doubling time from $ A$ to $ 2A$ is of order $ A^{\minus{}1/2}$ for large $ A$ (and, as Kent has shown, $ f$ must be unbounded if it stays separated from $ 0$, so we can start with arbitrarily large level), so it'll take time $ CA^{\minus{}1/2}$ to get from $ A$ to $ 2A$, then $ C(2A)^{\minus{}1/2}$ to get to $ 4A$, then $ C(4A)^{\minus{}1/2}$ to get to $ 8A$ and so on. The geometric series converges, so we get our blow-up.

Thus $ f'\le 0$, so the function is non-increasing and bounded from below. Thus it has a finite limit.

Could someone please elaborate on the last post? I dont quite get where the contradiction comes from.

See Radulescu problems in real analysis.


The screw of a copying press has 2 threads to the cm and the diameter of the wheel to which the effort is applied is 0.5m. If the efficiency is 40% what effort will be required to produce a force of 31.4 kN on the plate?
PS: I don't get what it means in this sentence:" 2 threads to the cm". I know it must relate to the diameter of the screw but can't get anywhere from this sentence.

I am not sure if this correct. But I think that it means that that idf the screw is rotated twice, it will go in (or out) linearly by one centimetre.


In a plane, two congruent squares share a common vertex but have no other points in common. Connect pairs of the remaining six vertices to get 3 parallel segments. If two of these segments have lengths 8 and 6, what are all possible lengths of the third segment?

not sure this is exactly right... in order to have the 3 lines be parallel to one another, they have to be equally spaced apart, the squares intersect at one vertex and forms 4 right angles.. then by SAS, then two of the triangles formed will be congruent, thus the 3rd line has to be 6... anyone know how to draw diagrams on here :P?

Err... i don't want to actually *construct it* on paper and scan it in, but I think that you're wrong about that. We're only given that the squares share one corners, so you could have them look oddly like diagram 1 (I drew that with a mouse, so don't comment.) So, ehh, you might need to move the squares around.

[geogebra]0faf13da80605d6dfbce729fd826cb29c22d370c[/geogebra]

Given the image above, we fix the lower square (S1) in place in the fourth quadrant.

Then the locus of possible vertex choices for S2 are given by the circles C1 and C2 (C1 for the closer two vertices, and C2 for the further vertex).

So now, we've got distances from known points to a locus. It should be a relatively simple matter to determine the other possibilities from here.

But doesn't it say the 3 segments are parallel.. and so the diagram doesn't work that way?... im not really sure..


Let be $ x,y,z\in \mathbb{R}_\plus{}$ such that $ x^2\plus{}xy\plus{}\frac{y^2}{3}\equal{}25\ ,\ \frac{y^2}{3}\plus{}z^2\equal{}9\ ,\ z^2\plus{}zx\plus{}x^2\equal{}16$ . Calculate $ xy\plus{}2yz\plus{}3zx$ .

We see something that looks like cosine law, and the pythagorean triples.

Consider a triangle with a point in the interior with lengths $ x,\frac{y}{\sqrt{3}},z$ to the vertices.

The first equation tells us (by cosine law) that the angle between $ x$ and $ \frac{y}{\sqrt{3}}$ is 150 degrees.

Similarly, between $ \frac{y}{\sqrt{3}}$ and $ z$ is 90 degrees, and the angle between $ x$ amd $ z$ is 120 degrees.

The lengths of the sides of the triangle are $ 3,4,5$, and the Area of the triangle can be found in 2 ways.

Summing up the areas of the smaller triangles formed by the interior point, $ \frac{1}{2}(x\frac{y}{\sqrt{3}}\sin(150)\plus{}\frac{y}{\sqrt{3}}z\sin(90)\plus{}zx\sin(120))\equal{}\frac{1}{4\sqrt{3}}xy\plus{}\frac{1}{2\sqrt{3}}yz\plus{}\frac{\sqrt{3}}{4}xz\equal{}\frac{1}{4\sqrt{3}}(xy\plus{}2yz\plus{}3xz)$.

But the triangle is just a 3-4-5 triangle, so is a right triangle with area $ \frac{1}{2}3\cdot 4\equal{}6$

Hence $ xy\plus{}2yz\plus{}3xz\equal{}24\sqrt{3}$

Note, this is not my solution, mine involved solving for the variables (which worked, but was tedious).

I don't understand your solution.Can you write algebraic solution ?

He has involved a triangle, and a point inside. Then he applied cosine law to 3 smaller triangles, such that it looks like the equations in the problem (note that he could take the triangle absolutely free), and then found the sides, the area, and finally the value of the expression.

What don't you understand?

Yes,I got it now idea of the problem but I don't know how rofler could find numbers $ x,\frac{y}{\sqrt{3}},z$ and then to use cosine law.I think,He had just solved this problem before,or had solved many such problems before.

But however this is algebraic problem.

$ x^2\equal{}(x)^2$
$ \frac{y^2}3\equal{}(\frac{y}{\sqrt3})^2$
$ z^2\equal{}(z)^2$

It doesn't matter if its algebraic, you can solve it anyway correctly :)

Let be $ a \equal{} x^2 \plus{} xy \plus{} \frac {y^2}{3} \equal{} 25\ ,\ b \equal{} \frac {y^2}{3} \plus{} z^2 \equal{} 9\ ,\ c \equal{} z^2 \plus{} zx \plus{} x^2 \equal{} 16\ ,\ d \equal{} xy \plus{} 2yz \plus{} 3zx$

We can observe that $ a \equal{} b \plus{} c\Leftrightarrow xy \equal{} 2z^2 \plus{} zx\Leftrightarrow xy \equal{} z(2z \plus{} x)(1)$

We also know that $ d \equal{} y(x \plus{} 2z) \plus{} 3zx\Rightarrow d \equal{} y\cdot \frac {xy}{z} \plus{} 3zx\Leftrightarrow d \equal{} \frac {3x}{z}\left(\frac {y^2}{3} \plus{} z^2\right)$ .

Now using $ b$ we get that $ d \equal{} \frac {27x}{z}$

Using $ (1)$ we have that $ d \equal{} xy \plus{} 2yz \plus{} 3zx \equal{} 2z^2 \plus{} zx \plus{} 2yz \plus{} 3zx \equal{} 2z(2x \plus{} y \plus{} z)\Rightarrow 2x \plus{} y \plus{} z \equal{} \frac {27x}{2z^2}$

$ a \minus{} b \plus{} c \equal{} 32\Leftrightarrow x(2x \plus{} y \plus{} z) \equal{} 32 \Rightarrow 32 \equal{} x\cdot \frac {d}{2z} \equal{} \frac {27x^2}{2z^2}\Rightarrow \frac {x}{z} \equal{} \frac {8}{3\sqrt {3}}\Rightarrow d \equal{} \frac {27\cdot 8}{3\sqrt {3}} \equal{} 24\sqrt {3}$

[quote="alex2008"]Let be $ a \equal{} x^2 \plus{} xy \plus{} y^2 \plus{} \frac {y^2}{3} \equal{} 25\ ,\ b \equal{} \frac {y^2}{3} \plus{} z^2 \equal{} 9\ ,\ c \equal{} z^2 \plus{} zx \plus{} x^2 \equal{} 16\ ,\ d \equal{} xy \plus{} 2yz \plus{} 3zx$

We can observe that $ a \equal{} b \plus{} c\Leftrightarrow xy \equal{} 2z^2 \plus{} zx\Leftrightarrow xy \equal{} z(2z \plus{} x)(1)$

We also know that $ d \equal{} y(x \plus{} 2z) \plus{} 3zx\Rightarrow d \equal{} y\cdot \frac {xy}{z} \plus{} 3zx\Leftrightarrow d \equal{} \frac {3x}{z}\left(\frac {y^2}{3} \plus{} z^2\right)$ .

Now using $ b$ we get that $ d \equal{} \frac {27x}{z}$

Using $ (1)$ we have that $ d \equal{} xy \plus{} 2yz \plus{} 3zx \equal{} 2z^2 \plus{} zx \plus{} \plus{} 2yz \plus{} 3zx \equal{} 2z(2x \plus{} y \plus{} z)\Rightarrow 2x \plus{} y \plus{} z \equal{} \frac {27x}{2z^2}$

$ a \minus{} b \plus{} c \equal{} 32\Leftrightarrow x(2x \plus{} y \plus{} z) \equal{} 32 \Rightarrow 32 \equal{} x\cdot \frac {d}{2z} \equal{} \frac {27x^2}{2z^2}\Rightarrow \frac {x}{z} \equal{} \frac {8}{3\sqrt {3}}\Rightarrow d \equal{} \frac {27\cdot 8}{3\sqrt {3}} \equal{} 24\sqrt {3}$[/quote]
It is a REALLY brilliant solution!!!! :lol:  :o  :D
Thanks a lot, alex.
PS: There is a small typo in the first line, it should be
$ a\equal{}x^2\plus{}xy\plus{}\frac{y^2}{3}$.


geometry
3D geometry
algebra
factorization
sum of cubes
square of a binomial

Basically by observation. In #1 I started from $x^4-2x^3$, which looked like the beginning of a square of a binomial, and they I noticed $380=20^2-20$, which proved useful. In #2 it's obvious that $27x^3=(3x)^3$ and $8=2^3$, but $3x$ and $2$ don't lead anywhere. Then I tried $8=7+1$ and that proved good, since we get $(27x^3+1)+7(3x+1)$ Then I wrote the factorization in such a way to immediately get the factor $3x+1$ (though I could write it using the sum of cubes, which would lead to the same result)

[b]Farenhajit[/b],Could you please tell me how do you know what to add and what combinations to take,like you did in the 1st one.[b]Pleeaaasee[/b]

A part of the answer is in my previous post :)


function
geometry
3D geometry
analytic geometry
calculus
calculus computations

Describe the set of points whose distance from the y-axis equals the distance from the $ xz$-plane.  Write and equation for this set in the form $ G(x, y, z) \equal{} 0$.  Can the set be described as the graph of a function of two variables?

I am thinking of a plane through the origin that is 45 degrees between the y-axis and the xz plane... something along the line of $ x\equal{}y$... but I'm stuck.

The distance from a point $ (x,y,z)$ to the $ y$-axis is the distance between the points $ (x,y,z)$ and $ (0,y,0)$ (why?).  The distance from a point $ (x,y,z)$ to the $ xz$-plane is the distance between the points $ (x,y,z)$ and $ (x,0,z)$ (why?).  Equating these distances, we find $ \sqrt {x^2 \plus{} z^2} \equal{} |y|$ or
\[ G(x,y,z) \equal{} \sqrt {x^2 \plus{} z^2} \minus{} |y| \equal{} 0
\]
To get an idea of what this looks like, consider the level sets when $ y$ is held constant; they're just circles with radius $ |y|$.  The level sets when either $ x$ or $ z$ are held constant are hyperbolas.  In other words, the set is two infinite cones, and we conclude that no matter how the coordinate system is defined, the set cannot be described as the graph of any function of two variables.  It can, however, be described as the union of two graphs of functions of two variables.  For example,
\[ \left \{(x,y,z)\in \mathbb{R}^3: y \equal{} \sqrt {x^2 \plus{} z^2}\right \}\cup \left \{(x,y,z)\in \mathbb{R}^3: y \equal{} \minus{} \sqrt {x^2 \plus{} z^2}\right \}
\]


Let $ S$ be the set of integers $ x$ such that $ x \equal{} a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc$, for some integers $ a,b,c$ Prove that if $ x,y \epsilon S$, then $ xy \epsilon S$

[hide]$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a \plus{} b\alpha \plus{} c\alpha^2)(a \plus{} b\alpha^2 \plus{} c\alpha^4)$

$ \alpha \equal{} \frac { \minus{} 1 \plus{} i\sqrt {3}}{2}$

So, if we define polynomial $ P(x) \equal{} a_1 \plus{} b_1x \plus{} c_1x^2$ we may write the equation as

$ x \equal{} a_1^3 \plus{} b_1^3 \plus{} c_1^3 \minus{} 3a_1b_1c_1 \equal{} P(1)P(\alpha)P(\alpha^2)$

$ y \equal{} a_2^3 \plus{} b_2^3 \plus{} c_2^3 \minus{} 3a_2b_2c_2 \equal{} Q(1)Q(\alpha)Q(\alpha^2)$

$ xy \equal{} a_3^3 \plus{} b_3^3 \plus{} c_3^3 \minus{} 3a_3b_3c_3 \equal{} S(1)S(\alpha)S(\alpha^2)$  where $ S(x) \equal{} P(x)Q(x)$

There is a problem about degrees of polynomials but i guess we can handle it[/hide]

Hi carlebee,
I don't see any proof in your post.


algebra
polynomial
category theory
calculus
integration
function
domain

Let $ \varphi: \mathbb{A}^n \rightarrow \mathbb{A}^n$ be a morphism of $ \mathbb{A}^n$ to $ \mathbb{A}^n$ given by $ n$ polynomials $ f_1, \ldots, f_n$ of $ n$ variables $ x_1, \ldots, x_n$.  Let $ J\equal{} \det | \partial f_i/\partial x_j|$ be the Jacobian polynomial of $ \varphi$.

If $ \varphi$ is an automorphism, then show that $ J$ is a nonzero constant polynomial.

I've gotten this far:

The functor $ X \mapsto A(X)$ induces an arrow reversing equivalence of categories between the category of affine varieties over $ k$ and the category of finitely generated integral domains over $ k$, that is, there is a bijection $ \mbox{Hom}(X,Y) \xrightarrow{\sim} \mbox{Hom}_k(A(Y), A(X))$.  More precisely, if $ \varphi \in \mbox{Hom}(\mathbb{A}^n, \mathbb{A}^n)$ is an automorphism, it corresponds to an automorphism $ \tilde{\varphi} \in \mbox{Hom}_k(k[x_1, \ldots, x_n], k[x_1, \ldots, x_n])$.  So suppose we're given an automorphism $ \psi$ of $ A: \equal{}k[x_1, \ldots, x_n]$.  We want to find the automorphism of $ \mathbb{A}^n$ that corresponds to $ \psi$, because that's the only way we can get automorphisms of $ \mathbb{A}^n$: from automorphisms of $ A$, and vice versa.  We define the automorphism as follows.  If $ \psi$ is an automorphism of $ A$, it takes units to units, and generators to generators.  Hence, the image of a constant is a constant, and $ \psi(x_i)\equal{}x_j$ for some $ j$ possibly equal to $ i$.  Consider the elements $ \xi_i\equal{}\psi(x_i) \in A$.  These are global functions on $ \mathbb{A}^n$, so we can use them to define a mapping $ \tilde{\psi}: \mathbb{A}^n \rightarrow \mathbb{A}^n$ by $ \tilde{\psi}(P)\equal{}(\xi_1(P), \ldots, \xi_n(P))$ for $ P \in \mathbb{A}^n$.  Clearly, the image of $ \tilde{\psi}$ is all of $ \mathbb{A}^n$, this is a morphism, and since $ \psi$ was an automorphism, so is $ \tilde{\psi}$.  So every automorphism of $ \mathbb{A}^n$ is given by $ n$ polynomials which are just reorderings of the coordinates.  So each $ \partial f_i/\partial x_j$ equals 1 or 0.

But how does this imply that $ J$ is nonzero?  And moreover, is this train of thought sound?

Thanks so much.

Okay, $ J$ is clearly nonzero because none of the columns or rows are identically zero.  But the way I've done this seems to imply that $ J\equal{}\pm 1$.  I think if $ J$ were $ \pm 1$, then it would've said, "show that $ J\equal{}\pm 1$" not "show that $ J$ is a nonzero constant."  Maybe a more pointed question is what are the automorphisms of $ A$?  That would help me immensely because I think my above proof works.

Thanks in advance.

$ \psi(x_i)\equal{}x_j$ is not true.

use the chain rule to show that the jacobian is a invertible matrix.

Why is that not true?  An automorphism should take generators to generators.

Why?  Any invertible linear transformation is an automorphism.

For example, consider $ k[x]$, then by your argument, because $ k[x]$ is the free k-algebra generated by $ \{x\}$, its automorphisms should all be equal to the identity  :(

:( okay, sorry.  I was trying to use the language of Hartshorne I.3, since that's where this problem is coming from.  I guess I wasn't thinking clearly.  Thanks.

indeed, if $ f$ is an automorphism of some structure, then the images of a generating system consitute a generating system. but there is no reason for this to be the same generating system.

If $ \varphi\equal{}(f_1, \ldots, f_n)$ is a polynomial map that is an autmorphism, how can I be sure that its inverse is also a polynomial map?  According to a theorem of Bialynicki and Birula (1962), an injective polynomial map $ F: k^n \rightarrow k^n$, $ k$ algebraically closed, is bijective and its inverse $ G$ is a polynomial map.  It seems like I need this to show that $ J(G(F)) \cdot J(F)\equal{}I_n$ by the chain rule, but it is not apparent to me.

In the case $ k$ is algebraically closed (this is probably true for the case where $ k$ isn't but I'm too tired to think about that), the only invertible polynomial maps have degree $ 1$, otherwise there will be multiple zeros and so when you find the inverse map of a linear polynomial, you will get a polynomial. Example:

$ f: k\rightarrow k$ given by $ x\rightarrow 2x$ is an automorphism (in characteristic not equal to $ 2$) but if it sends $ x\rightarrow (x\plus{}a)(x\plus{}b)$, it will not be an automorphism because it has two zeros, $ x\equal{}\minus{}a$ and $ x\equal{}\minus{}b$. And by algebraic closure any nonlinear polynomial will have more than one zero =(


Has this equation below at least 5 roots in $ [0;\frac{\pi}{12}]$ ?
$ 18.x^{3}\plus{}sin6x\equal{}0$

...Both $ 18x^3$ and $ \sin6x$ are positive on $ [0,\frac{\pi}{12}]$...


Calculate: $\dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{8} - \dfrac{1}{16}$.  Express your answer as a common fraction.


number theory
greatest common divisor
prime factorization

prove gcd(a,b,c) = gcd(a,gcd(b,c)). I can doing this using prime factorization but I was wondering if there was a way concerning linear combinations?

hallo,
Let $\gcd(a,b,c)=d$ and $\gcd(b,c)=d_{1}$ and $\gcd(a,d_{1})=d_{2}$. so there $\exists \alpha, \beta \in \mathbb{Z}$ s.t. $d_{2}=\alpha \cdot a+\beta \cdot d_{1}$.
Beacouse $d \mid d_{1}$ we have $d \mid d_{2}$. From here we write the equation $\mathbb{Z}\ni r= \frac{d_{2}}{d}=\alpha \cdot \frac{a}{d}+\beta \cdot \frac{d_{1}}{d}$ from here we have $r \mid \frac{a}{d}$ and $r \mid \frac{b}{d}$ and $r \mid \frac{c}{d}$ whitch is a contradiction beacouse $\gcd(\frac{a}{d}, \frac{b}{d},\frac{c}{d})=1$ $\Rightarrow$ $r=1$.


Fun time waster: search "fun things to do at walmart" on Google.

WARNING:
Search at your own risk.  Some things may not be appropriate to little kids.

DISCLAIMER: By going on google after reading this post, you agree consequences are in no way my fault.

http://www.funny.co.uk/stuff/art_173-1448-50-Fun-Things-to-do-at-Walmart.html

[quote]14. While walking through the clothing department, ask yourself
loud enough for all to hear, "Who buys this crap anyway?!"[/quote]
Creative.

[quote]19. As the cashier runs your purchase over the scanner say "BEEP" in a loud voice.  Repeat this for every item, and for other customers items.  If the cashier protests, kill them.[/quote]
Mature.

[quote]26. Climb things.[/quote]
Intelligent.

[quote]46. When someone steps away from his or her cart to look at
something, quickly place random combinations of items in their cart, such as 'A Large Cucumber and a Tub of Vasceline'.[/quote]
Sophisticated.

[quote]49. Pay off layaways 50 cents at a time.[/quote]
Witty.

This list must be Britain's pride and joy.


number theory
least common multiple
inequalities
number theory open

Let $ n$ be a given positive integer ($ n\ge 3$ and $ 1\le k <n$ positive integers $ 1\le a_1;a_2;..a_k \le n$ satisfy $ lcm [a_i;a_j] >n$ for all $ 1\le i;j \le k$. Find the least constant $ c$ to the following inequality be true for all $ n$:
$ \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}....\plus{}\frac{1}{a_k} <c$.
I found that $ c<2$ and that is all.

[quote="Allnames"]Let $ n$ be a given positive integer ($ n\ge 3$ and $ 1\le k < n$ positive integers $ 1\le a_1;a_2;..a_k \le n$ satisfy $ lcm [a_i;a_j] > n$ for all $ 1\le i;j \le k$. Find the least constant $ c$ to the following inequality be true for all $ n$:
$ \frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} .... \plus{} \frac {1}{a_k} < c$.
I found that $ c < 2$ and that is all.[/quote]
What 've I done! (Lin kin park  :) ):
When $ c \equal{} 2$.Consider $ S_i \equal{} \{1\le x \le n| a_i| x\}$.Hence $ S_i \cap S_j \equal{} \emptyset$ (since $ lcm[a_i;a_j] > n$). It implies $ S_1 \plus{} S_2 \plus{} ... \plus{} S_k \le n$.
Notice that $ S_i \equal{} [\frac {n}{a_i}]$ thus $ \sum_{i \equal{} 1}^{k}[\frac {n}{a_k}] \le n$
But $ [\frac {n}{a_i}] \ge \frac {n}{a_i} \minus{} 1$ and $ k\le n$ we can complete the proof in this case.
But it is too weak.
I am waiting...... !


$ \equal{} \left( \frac{n(n\plus{}1)}{2} \plus{} \frac{(n\minus{}1)(n\plus{}2)}{4} \plus{} \frac{(n\minus{}2)(n\plus{}3)}{6} \plus{} \ldots \plus{} \frac{(1)(2n)}{2(n\minus{}1)} \right) \minus{} 1$
where $ n \equal{} 100$ in your problem.

That's the best I can do.

[quote="MellowMelon"]$ \equal{} \left( \frac {n(n \plus{} 1)}{2} \plus{} \frac {(n \minus{} 1)(n \plus{} 2)}{4} \plus{} \frac {(n \minus{} 2)(n \plus{} 3)}{6} \plus{} \ldots \plus{} \frac {(1)(2n)}{2(n \minus{} 1)} \right) \minus{} 1$
where $ n \equal{} 100$ in your problem.

That's the best I can do.[/quote]

would you please offer some hints to simply the above form?

[quote="mathgeniuse^ln(x)"]That is his simplified form.[/quote]

how to derive the simplified form?


inequalities
function
calculus
algebra unsolved
algebra

Solution is $x_{0}<x\le \sqrt 2$, were $-\sqrt 2<x_{0}<-1.$

Chek, that
1. $1\le x\le \sqrt 2$ work.
2. $0\le x\le 1$ work.
3. $-1\le x \le 0$ work.
4. $-\sqrt 2<x<-1$ f(x) monotonen.

Hello Beat
[quote="Beat"]can somebody explain me the solutin, please? This problem is form Bulgarian Math Olympiad for grade 10 and no one from the contestants managed with the problem.[/quote]

I think noone knows the solution for the moment.
Rust studied the function $f(x)=4x+8\sqrt{2-x^{2}}-4-(x^{2}-2)2^{x}-2x.2^{x}\sqrt{2-x^{2}}$ in the interval of definition $[-\sqrt{2},+\sqrt{2}]^$ and stated that :
$f(-\sqrt{2})<0$
$f(-1)>0$
$f(x)>0$ $\forall x\in [-1,+\sqrt{2}]$

$f(x)=0$ has a solution $x0$ somewhere in $(-\sqrt{2},-1)$

But the problem remains to find this $x0$. Direct calculus gives something about $-1.099466495463...$ but this seems to give no interesting information.

Are you completety sure about the text of the problem ?

--
Patrick


From an initial triangle $\Delta A_0B_0C_0$, a sequence of triangles $\Delta A_1B_1C_1$, $A_2B_2C_2$, ... is formed such that, at each stage, $A_{k + 1}$, $B_{k + 1}$ and $C_{k + 1}$ are the points where the incircle of $\Delta A_kB_kC_k$ touches the sides $B_kC_k$, $C_kA_k$ and $A_kB_k$ respectively.

(a) Express $\angle A_{k + 1}B_{k + 1}C_{k + 1}$ in terms of $\angle A_kB_kC_k$.

(b) Deduce that, as $k$ increases, $\angle A_kB_kC_k$ tends to $60^{\circ}$.

(a)
[hide] Let $I_k$ be the incircle center, then $C_{k+1}B_kA_{k+1}I_k$ is cyclic, so $\angle A_{k+1}B_{k+1}C_{k+1} = \frac12 \angle A_{k+1}I_kC_{k+1} = 90^{\circ} - \angle \frac12 A_kB_kC_k$[/hide]
(b)
[hide] Let $\alpha_k=\angle A_kB_kC_k$, then $\alpha_{k+1}=90^{\circ} -\frac12\alpha_k=\sum_{i=0}^k (-1)^i \cdot \frac1{2^i} 90^{\circ}+(-1)^{k+1}\alpha_0 \implies \lim_{k\to\infty} \alpha_k = \sum_{i=0}^k (-1)^i \cdot \frac1{2^i} 90^{\circ} =90^{\circ} \cdot \sum_{i=0}^k (-1)^i \cdot \frac1{2^i}$
notice that $(a^0-a^1+a^2-...+a^n)(1+a)=1+a^{n+1}$ for all odd $n+1$, so lets assume that $\infty$ is odd :) $(1+\frac 12) \cdot \sum_{i=0}^{\infty} (-1)^i \cdot \frac1{2^i} = 1+\left(\frac12\right)^{\infty+1}=1\implies \sum_{i=0}^{\infty} (-1)^i \cdot \frac1{2^i} =\frac 23 \implies \lim_{k\to\infty} \alpha_k =\frac 23 90^{\circ} =60 ^{\circ}$[/hide]


group theory
abstract algebra
algorithm
logarithms
superior algebra
superior algebra unsolved

I want to generate all the subgroups of $ S_n$ for a given n. I'll be writing an algorithm for this.

[b]Possible solution 1:[/b]

What I thought of was to try to generate all the cyclic groups $ < x >$ with each $ x \in S_n$ and then all the ones generated by 2
elements $ < x,y >$ and so forth.
The problem is I don't know where to stop because I don't know an upper bound for this.
I can remember the Lagrange theorem that said if $ H < G$ then $ |H|$ divides $ |G|$. But maybe there are 2 subgroups of a given order.
What stop condition do I have with this, when do I know I generated all the subgroups ? And can I generate them all with this method ?
So I don't know what the running time of this would be.

[b]Possible solution 2:[/b]
Getting each subset $ H \subset \{1,2,3,...,n\}$ (so I'd have to check $ 2^n$ subsets) and checking if they don't respect
closure and the other axioms of the group.
The worst case running time(just based on the closure constraint) for this would be
$ \sum_{i \equal{} 1}^{2^n} \frac {i\cdot(i \minus{} 1)}{2}\sim O(2^{2n})$

Then again I don't know how many groups/monoids/semigroups ...over a set there are(I think I've read somewhere they can be
counted) , and they would serve to give a far better approximation.

I'm pretty sure these are very bad solutions in terms of complexity and the first one
I'm not sure if it's even correct , so how would you generate them all for a particular $ S_n$ ?

[i]P.S. A bigger picture for this would be to take the subgroups generated so far and construct the subgroup lattice and draw it with a tool like  [url=http://en.wikipedia.org/wiki/Graphviz]Graphviz[/url].[/i]

Alter your first solution just a little and you get an algorithm that works for n up to 8ish.  (An improved version of it) is called the method of cyclic extension.

Preparation: Find the "zuppos", that is for every cyclic subgroup of prime power order <x>, choose a generator x.  For Sn you can do this abstractly, for a general group it is done by conjugacy class, checking for rational conjugacy.

Step 0: Construct the trivial subgroup as the subgroup containing no zuppos.  This constitutes layer 0.

After executing step i, you have constructed layer i of the subgroups.  Each subgroup in the layer is represented as (the subgroup generated by) a subset of the set of zuppos.

Step i+1:  For every subgroup H in layer i, and for every zuppo x not in H, add the subgroup <H,x> to layer i+1.  Represent <H,x> as the list of all zuppos y contained in <H,x>.

Termination: If after the completion of step i, layer i is empty, then all further steps will produce empty layers, so terminate.  The union of the layers 0 to i-1 constitute a complete list of subgroups.  You still have to eliminate duplicates (a subgroup could even be listed in two different layers).

There are 1,2,6,30,156,1455,11300,151221,1694723,29594446,404126228,10594925360 subgroups of Sn for n=1,2,3,...,12.  The logarithm of the number of subgroups of the symmetric group is bounded above and below by constant multiples of n^2

Holt's textbook on Computational Group Theory would probably be interesting to you.  There is also a book Fundamental Algorithms for Permutation Groups by Butler that is in some ways easier and definitely more focused on permutation groups.  However it does not contain any work from the last 2 decades.

Butler, G. "Fundamental algorithms for permutation groups."
Lecture Notes in Computer Science, 559.
Springer-Verlag, Berlin, 1991. xii+238 pp.
ISBN: 3-540-54955-2
http://www.ams.org/mathscinet-getitem?mr=1225579
http://dx.doi.org/10.1007/3-540-54955-2

Holt, Derek F.; Eick, Bettina; O'Brien, Eamonn A. "Handbook of computational group theory."
Discrete Mathematics and its Applications.
Chapman & Hall/CRC, Boca Raton, FL, 2005. xvi+514 pp.
ISBN: 1-58488-372-3
http://www.ams.org/mathscinet-getitem?mr=2129747


I need a book that covers the theory well, has examples and unsolved problems. That is a must. I want the book to start from newbie and stuff and going over intermediate and advanced stuff.
What do you propose?

I have found this book and I want your opinion:
Probability and Statistics (Paperback) by Murray R Spiegel

My personal choice is "Weighing the odds" by Williams. But this is a matter of taste. Maybe not so good for a newbie, but it assumes no previous knowledge of probabilities and stats. It also depends on the meaning of "advanced stuff"...


You're given a set of towns and you know for each two towns is there a road connecting them or not.
All roads are of the same length 1.
Mirko and Slavko are two characters who are walking between the towns.
you have a begining position of both Mirko and Slavko and some positive integer k which represents the maximum possible distance between Mirko and Slavko in any momment.
If Mirko and Slavko in certain momment find themselves in the beggining position of the other character(they change their positions) in a certain number of steps does that implies that they have been in the same town in any momment.

Enjoy.

I don't understand the problem. Imagine a square ABCD. Mirko is in A, Slavko is in C. They walk like A->B->C and C->D->A, so they don't meet at any town.


Is there a command for multiplying numbers in LateX? I would like to be able to e.g. define a global constant and then use multiples of it in the document. In C I would write something like<pre>
const int p = 42;
...
function1(4*p);
function2(7*p);
</pre>but I don't know how to achieve the same result in LaTeX.

If you install the [b]calc[/b] package, you can do this.

Example:

[code]\documentclass{article}
\usepackage{calc}

\begin{document}

\newcounter{testa}
\newcounter{testb}
\setcounter{testa}{3}
\setcounter{testb}{\value{testa}*4}
\thetestb

\end{document}
[/code]
This will output $12$

A little bit of explanation to what David did there, if you define a new counter with the \newcounter command (as presented above) you can see at one point the value of that counter with the command \the{COUNTERS_NAME}. So if you say \newcounter{foo} then \thefoo will be the command that will output counter's {foo} value at any given time. The \setcounter command is easy to understand. :)


This was my solution for AIME #8, but it's so much simpler then everyone else's that I'm wondering if I didn't just get lucky in getting the right answer.

Solution: Since K is a rhombus, its maximum area comes when K is a square and the diagonals are equal. The area of K is the product of the diagonals. When K is a square, the horizontal diagonal of K (and thus the vertical as well) must be equal to the length of a side of a rhombus; consider 45-45-90 triangles that can be formed from the rhombi, whose angles would be 135 and 45. Since the diagonal of K is a length of the side of a rhombus, half the vertical diagonal would be the height of a rhombus, so we have

rhombus area base x height = (n^2)((n/2)^2) = sqrt(2006)

From this, the area of K is (n^2)(n^2) = sqrt(2006 x 4), and 89<sqrt(8024)<90 so the answer is 89.

It seems sound, and so much simpler than what everyone else is doing. I'm not bragging, though; I got a 5, and most of you guys got 10+. Which brings me to my next question: I'm a Junior with a 5 and I really love math, but with that kind of score should I bother majoring in math?

I just wanted to be thorough, so I said $\sqrt{2006}=x^2\sin{\theta}$, where $x$ is a side of a rhombus and $\theta$ an angle in one of the outside rhombi.  Then the inside rhombus has area $x^2\sin{2\theta}=2\sqrt{2006}\cos{\theta}$, and so we get an area ranging from 0 to $2\sqrt{2006}$, exclusive.  Same idea...I just wanted to be thorough as to not make careless mistakes.

Your AIME score does NOT determine how good you are at math.  It looks at your ability to solve problems, and that kind, specifically.  There's also the issue of careless mistakes on the AIME.  But in terms of mathematical skills, it doesn't really test much conceptualization or inductive thinking.  So I would say that if you love math, your AIME score shouldn't be able to change that at all.  Go for it.


Ara and Shea were once the same height. Since then Shea has grown 20% while Ara has grown half as many inches as Shea. Shea is now 60 inches tall. How tall, in inches is Ara now?


inequalities
geometry
perimeter
incenter
trigonometry
inequalities proposed

Triangle $ ABC$ has perimeter $ 2s$ .if incircle has radius $ r$ and distance from the incenter to $ A,B,C$ is
$ s_{a},s_{b},s_{c}$ respectively. prove that $ \frac{3}{4}\plus{}\frac{r}{s_{a}}\plus{}\frac{r}{s_{b}}\plus{}\frac{r}{s_{c}}\leq\frac{s^{2}}{12r^{2}}$


ratio
geometry
3D geometry
sphere
trigonometry

In a homogenous atomic reactor of radius $R$ the number of the appeared neutrons per unit time per unit volume is proportional to the neutron density. Let there ratio be $q$. The diffusion constant of the neutrons  inside the sphere of radius $R$ is $D$. What conditions should there be so that the number of neutrons grow, so that the chain reaction can be used in macroscopic terms?

Well here is a
[hide="Tip"]
use the $\frac{\partial n}{\partial t}=D \Delta n+qn$
diffusion equation.[/hide]

well ok here is the solution:
The diffusion equation for the $n$ neutron density:
$\frac{\partial n}{\partial t}=D \Delta n+qn$
lets look for the solution in the
$n=e\lambda_{j}t \frac{1}{r}f(r)$
form.
Then we get for $f$- the following equation:
$\frac{\partial^{2}f}{\partial r^{2}}=-\frac{q-\lambda_{j}}{D}f$
Then neutron concentration on the surface of the reactor is negligible, so:
$f(r)=\sin{\bigg (\frac{j\pi r}{R}\bigg )}$ and  $\frac{q-\lambda_{j}}{D}=\frac{j^{2}\pi^{2}}{R^{2}}$
so the general solution:
$n(r.t)=\sum_{j=1}^{\infty}A_{J}exp \bigg [{\bigg ( q-\cfrac{j^{2}\pi^{2}D}{R^{2}}\bigg ) t}\bigg ] \frac{1}{r}\sin{\bigg (\frac{j\pi r}{R}\bigg )}$
The number of neutrons grow if the exponent is positive for at least $j=1$.
So the needed condition:
$R \gneq \pi \sqrt{\frac{D}{q}}$


inequalities logarithms geometry function

[quote="towersfreak2006"]It doesn't seem to be true for \[p=-3\qquad q=\frac{3}{4}\qquad u=1\qquad v=0.\] [/quote]
Sorry, I have edited the original problem.  It should be $\mathbb{R}^{+}$ rather than $\mathbb{R}$.

[quote="amcavoy"]Let $p,q\in\mathbb{R}^{+}$ such that \[\frac{1}{p}+\frac{1}{q}=1\] and let $u\geq 0$ and $v\geq 0$.  Show that \[uv\leq\frac{u^{p}}{p}+\frac{v^{q}}{q}.\] [/quote]
There is a ("well-known") theorem which states,
If $\alpha, \beta > 0$ with $\alpha+\beta = 1,$ then for all $x,y > 0$, we have $\alpha x+\beta y \ge x^{\alpha}y^{\beta}.$

For the given problem, just set $\alpha = \frac{1}{p}$ and $\beta = \frac{1}{q}.$ The rest follows from there.

[quote="amcavoy"]Let $p,q\in\mathbb{R}^{+}$ such that \[\frac{1}{p}+\frac{1}{q}=1\] and let $u\geq 0$ and $v\geq 0$.  Show that \[uv\leq\frac{u^{p}}{p}+\frac{v^{q}}{q}.\] [/quote]
I am not sure about it :blush:  but I remember it was proved by using  AM-GM and it is one of steps to approach Holder Inequality.

You can associate the area of three figures.

[quote="FieryHydra"]I am not sure if the following is always true: $\log{a}< \log{b}\Rightarrow a < b.$[/quote]
Well, let $\log{a}=m, \log{b}=n, m<n$. We have $a=10^{m}<10^{n}=b$ QED

[quote="FieryHydra"][quote="paladin8"][hide="Be cool and use Jensen"]
Let $f(x) = \log{x}$, which is concave. So by Jensen's

$\frac{1}{p}\log{u^{p}}+\frac{1}{q}\log{v^{q}}\le \log{\left(\frac{u^{p}}{p}+\frac{v^{q}}{q}\right)}$

or

$uv \le \frac{u^{p}}{p}+\frac{v^{q}}{q}$

as desired.[/hide][/quote]
I am not sure if the following is always true: $\log{a}< \log{b}\Rightarrow a < b.$[/quote]

$\log{x}$ is a strictly increasing function, so it is true.

[quote="amcavoy"]Let $p,q\in\mathbb{R}^{+}$ such that \[\frac{1}{p}+\frac{1}{q}=1\] and let $u\geq 0$ and $v\geq 0$.  Show that \[uv\leq\frac{u^{p}}{p}+\frac{v^{q}}{q}.\] [/quote]
http://mathworld.wolfram.com/YoungsInequality.html It is Young inequality.


integration
LaTeX
calculus
calculus computations

Solve the differential equations

y' = 2y - y^3

\frac{dy}{dx}=2y-y^{3}

\int \frac{dy}{2y-y^{3}}= \int dx

now u have to worry about integrating  \frac{1}{2y-y^{3}}     thats it

[quote="b555"]$ \$$\frac{dy}{dx}=2y-y^{3}

\int \frac{dy}{2y-y^{3}}= \int dx$ \$$

now u have to worry about integrating  $ \$$\frac{1}{2y-y^{3}}$ \$$     thats it[/quote]

Its good you are now using $ \text{\LaTeX}$ but you need to wrap your $ \text{\LaTeX}$ code in $ \$$ sign :wink: as shown in the quote.

So it'll appear like this:

$ \frac{dy}{dx}=2y-y^{3} \\ \\
\int \frac{dy}{2y-y^{3}}= \int dx \\ \\
\frac{1}{2y-y^{3}}$

ok  so  thats  where my latex went wrong, i will try to implement watever u sid from the next post itself.!

and ya by the way  watever i said is correct only? :?:

[quote="b555"]and ya by the way watever i said is correct[/quote]

Of course!

The solution to the differential equation yields:

$ \frac{y^2}{1\minus{}y^2}\equal{}ce^x$


I'd noticed this a while back, but forgot to ask about it. There used to be 6 problem sets per class (for subject classes). Are these now replaced with additional board problems?


How many set partitions of [8] satisfy the three following properties:

(1) 1 and 2 are in different parts
(2) 2 and 3 are in different parts
(3) 7 and 8 are in different parts

I think it should be
\[ P_{8} \minus{} 3\cdot P_{7} \plus{} P_{6} \plus{} P_{5} \plus{} P_{5} \minus{} P_{4}\]
where $ P_{n}$ denotes the number of ways that [n] can be partitioned into non-empty subset(s).


Is there some set $ E \subset [0,1]$ such that $ m^*(E) = m^*([0,1]\E) = 1$?

I think there should be! though I'm not sure how to show that.  :P

EDIT: I am told that you make use of Dirichlet's theorem in some way...

What condition are you missing? Right now it's obvious, just remove any measure zero set.

Hei Kalle,

thanks for your very quick answer :-)

Just to make sure: So, I begin with $ m^*(E)$, and then I remove a measure zero set from this, or? I think your suggestion is correct--But I'm not sure how to go about it!

Greetings from Porvoo,

Astrid

You don't know any sets of measure 0? A single point is a set of measure 0, so $ m^*([0,1)) \equal{} 1$. I really think the original question asks about something else though, this is too easy.

Show that there exists $ E\subset [0,1]$ such that $ m^*(E) = m^*([0,1])\E = 1.$ [b]Hint[/b]: Use the construction of the following theorem... The theorem which it suggests making use of is: There exists a subset $ A$ of $ \mathbb{R}$ with the property that $ A \bigcap B$ is unmeasurable for every L-measurable set $ B$ with $ m(B) > 0$. The proof is by contradiction, and it uses Dirichlet's theorem.

I agree with Kalle that there seems to be some part of the question missing.

Do we want E to be non measurable?

Otherwise, just take E=[0,1]. (Or E=[0,1) if you want E to be a [i]proper [/i]subset of [0,1])

Maybe it is that easy. I have posted the question as it stands in the text--maybe the 'hint' to look the theorem is just a red herring?  :blush:

I think we want $ E$ to be arbitrary. The only condition that $ E$ has is that it is L-measurable.

Given this, the solution to take $ E\equal{}[0,1]$ is not entirely satisfying. Apparently the text I scanned gives a way to do this for arbitrary sets, but I don't see exactly how to copy it--because in the scanned theorem they are essentially doing just the opposite, I think.

hey all, i have got it. Thanks for your suggestions. i will be sure to post my solution after term is over...which is next week!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


geometry
3D geometry
prism
pyramid
MATHCOUNTS
probability

Can some one give me some good ideas to post for the math bulliten board in school.

My teacher has the "Go figure" section of Sports Illustrated which is like this:
[b]16[/b]games in a NFL season
[b]15[/b]games the Steelers won in that season
[b]18[/b]is the number of Peyton Manning

My teacher has a cool poster of the Pythagoran thereom.
My teacher also has foldable prisms(like where you fold on the line and it makes a prism)
She has hexogal, rectangular, cube, triangular pyramid, etc.

[/b]

Perhaps some mathcounts problems, a solution or 2.
You could also say something about history of mathematics and the like. Finally, you could write a proof of something or the like.

or you could talk about why 0.999999...... (re-occuring 9 ) = 1.  that always seems to spark a debate, then later on you could present a few proofs of why the two numbers are equal. you could present the following probability problems, many similar problems exist:

A box contains two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. One coin is selected from the box at random and the face of one side is observed. If the face is heads what is the probability that the other side is heads

  The answer is in fact 2/3, but most people answer 1/2.  Finally you could post the classic monty hall problem (that never gets old  :P ) but its possilbe most people have already seen it.


[quote="Cliu0301"]1. There are a total of 678 legs on a farm. The farm has horses (4 legs) and chickens (2 legs). How many heads are there all together?

2. What is the sum of all solutions for X to the equation 3x + 4y + 5z = 120?[/quote]
So For #1 we can have a lot of possible heads.....
And #2 i think you meant positive or else there are infinite solutions..

I make up a lot of these problems in my head sooo.....Sorry :D

For #1...I forgot to put that the number of horses is 4 more than the number of chickens.

(or is that still wrong? Idk)

For #2...What is the sum of all the POSITIVE integer solutions for X)

#1 shouldn't be too hard, just set up the system of equations
h=horse c=chickens
4h+2c=678
4+h=c (Edit Read wrong)
This is quite easy to substitute in

Of course, I'm pretty sure there isn't 4 more the number of horses as chickens

#2. Hint: See if Y and Z even matter.

[quote="Cliu0301"]1. There are a total of 678 legs on a farm. The farm has horses (4 legs) and chickens (2 legs). How many heads are there all together?
[quote]
and
[quote="Cliu0301"] ...I forgot to put that the number of horses is 4 more than the number of chickens. [/hide]

I'll try this one:  Set up the equations(x=# of horses, y=#of chickens):
4x+2y=678
x=y+4.

We then multiply the second one by 4 to get:
4x=4y+16.  Subtract the second one from the first to get:
2y= -4y+662
add 4y to get
4y=662

I'm getting y=165.5.
We can't get a decimal answer so I'm pretty sure the question was a dud.[/b]

[quote="maybach"]I'll try this one:  Set up the equations(x=# of horses, y=#of chickens):
4x+2y=678
x=y+4.

We then multiply the second one by 4 to get:
4x=4y+16.  Subtract the second one from the first to get:
2y= -4y+662
add 4y to get
4y=662

I'm getting y=165.5.
We can't get a decimal answer so I'm pretty sure the question was a dud.[/b][/quote]

I got:

$ 0\equal{}\minus{}2y\plus{}662$
$ 2y\equal{}662$
$ y\equal{}113$

I think you made a careless mistake somewhere, but your logic is right.


OK, Apparently there's interest. Sign up [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=143193]here.[/url] Post questions in this thread, not AIM sns.

Let's set a date at next Monday at 8:00 PM, Eastern. Sound good?

[quote="13375P34K43V312"]OK, Apparently there's interest.

Let's set a date at next Monday at 8:00 PM, Eastern. Sound good?[/quote]

Where are you getting the problems?

ok, somehow this got split like instantaneously.

I'll be writing them.

can i do it, or am i banned from all your things?

jorian

ummm...it is a mathcounts math jam...[razz]and you technically are finished with mathcounts forever[\razz]. I would say probably it wouldn't be worth it for any 8th grader who didn't make nats...but I guess that's up to you...and leet...

I really don't care, any middle schooler can do it just for fun.

cool you're the best

i wonder if my great guessing skillz will come in handy...?

6!!!

:rotfl:

thanks,
jorian

BUMP

REMINDER REMINDER REMINDER in 20 min.

Yeah

Hey guys,

my bad.

I was (regretfully) working on my mock exams for the past hour, and it completely slipped my mind. I apologize for the inconvenience. I only had forty problems anyway, that would probably only last around a half hour.

Let's reschedule to Wednesday at 8 PM EST, I won't forget again, I promise. I'll write it on my hand or something, and I'll also try to write some more problems between now and then.

Again, i apologize.

lol i even sent you a pm saying i couldn't make it :rotfl:

BUMP!

Reminder (at least I think it's in 20 minutes) :huh:

Please note: the game will begin in twenty minutes.

Remember, if you wish to join, send me an IM and I will invite you to the chat room.

[quote="13375P34K43V312"]FINAL RESULTS

9 people showed up... :huh: but it was a good game anyway. Note these are AIM sn's, not AoPS usernames, some are quite obvious to link, please don't discuss them anyway because some people don't want that.

SplashZ80: 10
manthgacrazed: 8
preetgpatel: 6
PiCkMan127: 4
infinity77beyond: 3
AMardini: 1
QWang: 1 (coming through on the last problem!)
Aznwithabrain had 1 but shamelessly lost it due to disturbances...[/quote]

 :huh:
Thought I got (-1).
Well doesn't matter
I'm still the dumbest kid on AOPS :rotfl:

When is the next one? (if you're gonna do another one)

I think the problems could be made a little more challenging... but it was pretty good overall


number theory
relatively prime
number theory unsolved

Let $a_1$, $a_2$, $\ldots$, $a_n$ be a sequence of integers with values between 2 and 1995 such that:
(i) Any two of the $a_i$'s are relatively prime,
(ii) Each $a_i$ is either a prime or a product of primes.
Determine the smallest possible values of $n$ to make sure that the sequence will contain a prime number.

We can exhibit a sequence with 13 terms which does not contain a prime: 2·101 = 202, 3·97 = 291, 5·89 = 445, 7·83 = 581, 11·79 = 869, 13·73 = 949, 17·71 = 1207, 19·67 = 1273, 23·61 = 1403, 29·59 = 1711, 31·53 = 1643, 37·47 = 1739, 41·43 = 1763. So certainly n ≥ 14.

If there is a sequence with n ≥ 14 not containing any primes, then since there are only 13 primes not exceeding 41, at least one member of the sequence must have at least two prime factors exceeding 41. Hence it must be at least 43·47 = 2021 which exceeds 1995. So n =14 is impossible.


Is the answer to (x + 22)2(squared)=0

X2(squared)+484? or is it X2(squared) +44X + 484...

or neither.. Can someone please help me and my math partner figure this out, we both think its the other one...Thanks :)

Thanks for the help. Hope I can return it one day... :roll:

A generalization is that

$(a+b)^{2}=a^{2}+2ab+b^{2}$.

If you are solving for $x,$ you don't need to expand.

$(x+22)^{2}=0$

$x+22=0$

$x=\{-22\}$

[quote="13375P34K43V312"]A generalization is that

$(a+b)=a^{2}+2ab+b^{2}$.[/quote]
$(a+b)^{2}$ :wink:

[quote="jbf13"]Is the answer to (x + 22)2(squared)=0

X2(squared)+484? [/quote]

When you are confronted with a question about whether two algebraic expressions are equal, the easiest thing to do is to plug in a simple value for $x$.

For example, letting $x = 1$, we have

$(x+22)^{2}= 23^{2}= 529$

But

$x^{2}+484 = 1^{2}+484 = 485$

Which are clearly not equal.


A teacher finds that when she offers candy to her class of 30 students, the mean number of pieces taken by each student is 5. If every student takes some candy, what is the greatest number of pieces one student could have taken?

She has $ 30(5)\equal{}150$ pieces of candy. We next assume that all but $ 1$ of the students takes $ 1$ piece of candy. Since that group would have $ 29$ students, we have $ 150\minus{}29\equal{}\boxed{121}$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Two new suestions:

The statement: Let $ N$ be a normal subgroup of a group $ G$, then there ist a subgroup $ K$ of $ G$, with $ K\cong G/N$

is not true in general. But I have only counter-examples for infinite groups $ G$ $ (\mathbb{Z})$. So the first question is: Is that statement also for finite groups wrong?

Secondly: By searching for finite counter-examples of question one, I asked: Contains every finite group of even order the subgroup $ \mathbb{Z}_2$? I couldn't find an counter-example, too. Thx.

For finite group $ G$ and any prime $ p$ dividing $ |G|$ there is a subgroup isomorphic to $ \mathbb Z / p \mathbb Z$. This is called Cauchy's theorem.
By the way, your statement is true for finite abelian groups.
A counterexample for nonabelian finite groups is given by $ Q_8 \equal{} \{ \pm 1 , \pm i , \pm j , \pm k\}$ (quaternion group) and $ N\equal{}\{ \pm 1 \}$.

[quote="Tauchender"] By searching for finite counter-examples of question one, I asked: Contains every finite group of even order the subgroup $ \mathbb{Z}_2$? I couldn't find an counter-example, too. Thx.[/quote]

Hello.
The answer is yes. It is a particular case of Cauchy theorem : if $ p$ is a prime number, if a finite group $ G$ has order divisible by $ p$, then $ G$ contains an element of order $ p$.
If $ p \equal{} 2$, the proof is easy. If $ G$ has even order, consider the following equivalence relation between elements $ x,y$ of $ G$ : $ x$ and $ y$ are equal or mutual inverses. An equivalence class has one or two elements. A class with one element corresponds to an element $ x$ such that $ x\equal{}x^{\minus{}1}$, i.e. such that $ x^{2} \equal{} 1$. Since the order of $ G$ is even, the number of classes with only one element must be even. Thus, the number of elements $ x$ of $ G$ such that $ x^{2} \equal{} 1$ is even. Thus, there is at least one such element other than $ 1$, i.e. there is at least one element of order $ 2$.

Friendly,
P.


geometry
geometric transformation
homothety
trigonometry
similar triangles

Let $ C_1$ and $ C_2$ be two given concentric circles with diameters $ 2R_1>2R_2$ respectively.

Draw a straight line, wich cuts $ C_1$ at $ A$ and $ D$, and $ C_2$ at $ B$ and $ C$; such that

              $ AB\equal{}BC\equal{}CD$

 :)

You've neglected the condition $ 3R_2 \geq R_1$.

I know this must hold, but as in a regular construction problem I expected the person who solve it to give the necesary conditions for the construction, anyway this isn't hard to see.

Thanks.

 :)

Nice and easy problem with at least two different solutions. Try it!
[hide="Hint1"]homothety[/hide]
[hide="Hint2"]algeraic method[/hide]

My solution uses only some trigonometry, wich is algebraic but esencially it is almost the same as similar triangles wich is almost the same as homethety.

 :)

Why not post your solution. I don't have time right now to do the same.


I'm trying to prepare for the USABO, first level for now.
Does anyone know a website with previous competitions?

go to the CEE website, they have links to practice exams from britain and australia if i remember correctly. also, when you being your USABO journey, your teacher will get practice questions to give you.

How do you prep for the practical section of the USABO finals?

yikes.....now thats a part that i unfortunately never got to in my USABO journey.....lol. id imagine that youd need straight up lab skills. but i know for a fact that before they compete at USABO, they do lots of lab stuff so that might help.

Yes, I found out about the CEE site, but there's only the British and the Australian ones. The only American one is International level, which I am not interested in for now. Does anyone know a website for the USABO Open Exam past exams?


inequalities
number theory
algebra
functional equation
national olympiad

The first day of ZIMO 2009 was conducted today. Problems were generally favorable, the first problem was a pretty easy number theory problem, the second one was a really surprising functional equation, though it has a really good-looking solution, it demands some unnatural - in the very best sense - ideas. But the cherry on the cake was clearly the thrid problem - an awesome combination of toughness and beauty. Furthermore, I have no idea what topic this problem belong to, though I am more inclined to inequality variant.
I am eager to hear out your impressions and comments concerning everything: organization, problems, jury...

P.S: Have the problems been posted?

okay :wink:
Edit:
Here we go:

[url=http://www.mathlinks.ro/viewtopic.php?t=251954]problem 1[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=251955]problem 2[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=251957]problem 3[/url]

[url=http://www.mathlinks.ro/viewtopic.php?t=251958]problem 4[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=251960]problem 5[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=251961]problem 6[/url]

BIG THANKS FOR ADDING THE PROBLEMS TO THE RESOURCES SECTION, BUT THE LINK TO THE 4TH PROBLEM IS INCORRECT, PLEASE FIX IT.

[quote="chess"]Erken! You posted problems! Thanks.
Can you post results of ZIMO 2009 or
do you know if the official page of ZIMO exists?   :)

P.S. Who knows names of authors of the problems of ZIMO 2009? :)[/quote]

The results are not available so far, they will be published the day after tomorrow, just before the closing ceremony.

I have no information about the official web-page, neither about authors of the problems, except that Nairy Sedrakyan is the author of the third problem, which is, in my opinion, the most beautiful and tough one among all problems.

BIG THANKS FOR ADDING THE PROBLEMS TO THE RESOURCES SECTION, BUT THE LINK TO THE 4TH PROBLEM IS INCORRECT, PLEASE FIX IT.

Tahnks for posting the subjects. Could you also post the results ?

The closing ceremony will start soon - only 2 hours left. I'll post the results as soon as I will come back from there.

I have just come back from the closing ceremony. I don't have any personal results, but I know at least medal cut-offs -

a gold medal was awarded for $ \geq 22$

a silver medal was awarded for $ \geq 16$

a bronze medal was awarded for $ \geq 10$...

Do you know something about informatics?
Where can I find problems and results?

yep, but nothing concerning Georgia team...I will post the results as soon as I will get them.

here are the results (informatics) you were looking for:

http://olympiads.kz/files/zhautykov-2009/index.html

Thank you.

And there will be mathematics results too soon , yes?

Hi, Erken :)
how r u?
didn't you already graduate?
how is our school doing?
how was results? anything interesting or strange?
do we have any olympiad geniuses in kz now?
who was in jury?
who teaches [i]rezerv[/i] in our school now?

[quote="cefer"]What about Azerbaijan?

[u]@Erken:[/u] What did you get ? I hope it is gold :)[/quote]One of them got gold. I heard that the others do not get more than 10 points. I think Erken did not take part in the olympiad, I didn't see his name.

Please, where can we see all the results? Thank you.


[quote="triplebig"]Show that $ CD \plus{} AD \equal{} AB$[/quote]

This triangle is isosceles triangle.
So it's trivial.

Another property (which probably is the property triplebig wanted to prove): $ BD\plus{}AD\equal{}BC$.

Let $ E\in AB$ such that $ CE$ bisects $ \angle ACB$ and $ F\in BC$ such that $ BF\equal{}BD$. Some angle chasing shows that $ DE\equal{}DC$ and $ \triangle ADE,\triangle FDC$ are similar. So $ \triangle ADE,\triangle FDC$ are congruent, and hence $ AD\equal{}FC$. Therefore $ BC\equal{}BF\plus{}FC\equal{}BD\plus{}AD$, as desired.

hello, i have the following solution
$ \overline{BD}\equal{}\frac{b\sin(100^{\circ})}{\frac{\sqrt{3}}{2}}$, $ \overline{AD}\equal{}\frac{b \sin(20^{\circ})}{\frac{\sqrt{3}}{2}}$ and $ \overline{BC}\equal{}b\sqrt{2(1\minus{}\cos(100^{\circ})}$ and we have to show that
$ \sin(100^{\circ})\plus{}\sin(20^{\circ})\equal{}\frac{\sqrt{3}}{2}\sqrt{2(1\minus{}\cos(100^{\circ})}$
this is equivalent with
$ 2\cos^2(40^{\circ})\equal{}1\minus{}\cos(100^{\circ})$ wich is true, since $ \cos(100^{\circ})\equal{}\minus{}\cos(80^{\circ})$.
Sonnhard.

[quote="Johan Gunardi"]Another property (which probably is the property triplebig wanted to prove): $ BD \plus{} AD \equal{} BC$.

Let $ E\in AB$ such that $ CE$ bisects $ \angle ACB$ and $ F\in BC$ such that $ BF \equal{} BD$. Some angle chasing shows that $ DE \equal{} DC$ and $ \triangle ADE,\triangle FDC$ are similar. So $ \triangle ADE,\triangle FDC$ are congruent, and hence $ AD \equal{} FC$. Therefore $ BC \equal{} BF \plus{} FC \equal{} BD \plus{} AD$, as desired.[/quote]

Yes that is very much what I wanted to prove, I typed it wrong (different in figure)

Thanks for your solutions.


$a,b,c,d>0, abcd=1$. Prove that
$(ab+cd)^2+(ac+bd)^2+(ad+bc)^2\ge \sum_{sym}\frac ab$
(the RHS sum has 12 terms)

If $abcd=1$ then
$(ab+cd)^2+(ac+bd)^2+(ad+bc)^2\ge \sum_{sym}\frac ab\Leftrightarrow$
$\Leftrightarrow(a-b)^2(c-d)^2+(a-c)^2(b-d)^2+(a-d)^2(b-c)^2\geq0.$ :)


function
inequalities
geometry
rearrangement inequality

The function $ f(x)\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d$ has $ 3$ positive real roots. If the sum of the roots of $ f(x)$ is $ 4$, what is the largest possible value of $ \displaystyle \frac{c}{a}$?

[hide="Answer"]By the Rearrangement Inequality, $ pq\plus{}qr\plus{}rp \leq p^2\plus{}q^2\plus{}r^2$

$ \therefore \frac{c}{a} \equal{} pq\plus{}qr\plus{}rp \leq \frac{2(pq\plus{}qr\plus{}rp) \plus{} (p^2\plus{}q^2\plus{}r^2)}{3} \equal{} \frac{(p\plus{}q\plus{}r)^2}{3} \equal{} \frac{16}{3}$

Equality occurs when $ p\equal{}q\equal{}r\equal{} \frac{4}{3}$[/hide]

This is just a reworded version of this problem I posted a while back - http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204452

LOL

[hide="Alternate Method"]This problem can also be modeled with the roots as lengths of edges of a rectangular prism. $ \displaystyle \frac{c}{a}$, or half the surface area will be maximized when the lengths of the edges are equal.[/hide]

$ \frac {c}{a} \equal{} pq \plus{} qr \plus{} rp \leq \frac {2(pq \plus{} qr \plus{} rp) \plus{} (p^2 \plus{} q^2 \plus{} r^2)}{3}$

how did you get this inequality from the rearangement inequality


Solve the problem of the week in:

usuarios.lycos.es/redmath

[url=http://translate.google.com/translate?hl=en&sl=es&u=http://usuarios.lycos.es/redmath/&prev=/search%3Fq%3Dredmath%2Blycos%26hl%3Den%26lr%3D%26client%3Dfirefox-a%26rls%3Dorg.mozilla:en-US:official%26sa%3DG]Here[/url] is the translated page for those who can't speak Spanish.  Google is not exactly the best translator, but I dont really know of any one that's better.

There'es also an Alta Vista translator, I don't know if that's better though, I know it has more languages.


function
vector
Functional Analysis
advanced fields
advanced fields unsolved

Let $X$ be a Banach space (more generally, one can consider locally convex space). Let us say that $\Phi\colon X\to\mathbf R$ allows dual representation if $\Phi=\sup_{f\le\Phi}f$ with $f$ being continuous affine functionals.

The following result from convex analysis holds:

$\Phi\colon X\to\mathbf R\cup\{+\infty\}$ is convex lower semicontinuous functional iff it allows dual repsesentation.

Here're the problems:

1. Let $\Phi\colon\mathbf R\to\mathbf R$ be convex then it is upper bounded iff it is a constant. (A convex $\Phi\colon\mathbf R\to\mathbf R$ is always of $C^{0}(\mathbf R,\mathbf R)$ class).

Can you prove the stement not using the dual representation?

2. More generally, let $\Phi\colon X\to\mathbf R$ be [i]any[/i] convex upper bounded functional. Does it follow that $\Phi=\mathrm{const}$?

I am not sure I have understood the question:

1/ if there exists $x < y$ such that $\Phi(x) \neq \Phi(y)$ then $\Phi$ is clearly unbounded because $\Phi(z) \geq L(z)$ for $z \not \in [x;y]$ and $L(z)$ is the affine linear function passing throught $(x, \Phi(x))$ and $(y;\Phi(y))$

2/ in a normed vectorial subspace $X$, if $(e_{i})_{i \in I}$ is a basis, then the functions
$\Phi_{i}(t)=\Phi(t e_{i})$ are also convex and upper bounded so constant to $\Phi(0)$. Hence $\Phi$ is constante.

You understood the question properly. In fact, you have even proved that bounded convex functionals on vector spaces are exactly the constants.

In 2/ you used Hamel basis, right?

I see no need for any basis in (2) :?. Just work on each line passing through the origin. By (1), the restriction of the function $f$ to each such line is constant and equal to $f(0)$ (each line can be naturally identified with $\mathbb R$), hence it's constant on the entire space.


Prove that $ (n\minus{}1)\sqrt{2} < \sqrt{2} \plus{} \sqrt{3} \plus{} ... \plus{} \sqrt{n} < (n\minus{}1)\sqrt{n}$, for all integers $ n > 2$.

This follows rather trivially from the fact that for positive $ m,n$ with $ m>n$ that

$ \sqrt{m} > \sqrt{n}$.

We have:

$ (n\minus{}1)\sqrt{2} \equal{} \sqrt{2} \plus{} \sqrt{2} \plus{} \sqrt{2} ... < \sqrt{2} \plus{} \sqrt{3}... \plus{}\sqrt{n} < \sqrt{n} \plus{} \sqrt{n} \plus{}\sqrt{n} ... \equal{} (n\minus{}1)\sqrt{n}$.

Yes, thanks. I thought I'd posted it in a High School forum (12-16).


geometry
analytic geometry
graphing lines
slope
symmetry
similar triangles
Pythagorean Theorem

Given two externally tangent circles with radii $R$ and $r$, $2$ tangents are drawn from the center of each to the other.  Compute the area of the quadrilateral formed by the $4$ tangents.

Solution:

[hide]
I did this using coordinate geometry. I'd welcome a non-coordinate approach, but the messy nature of the answer suggests it may not be much easier.

Set the centers of the two circles at (-R,0) and (r,0) respectively. (Clearly either one can have the positive x value as long as the other is negative.)

Using similar triangles and the pythagorean theorem we can see that the slope of the tangent line from (-R,0) to the other circle is $\frac{r}{\sqrt{2Rr+R^{2}}}$. Likewise, the slope of the tangent line from (r,0) to the other circle is $-\frac{R}{\sqrt{2Rr+r^{2}}}$.

The area of the desired quadrilateral can be split into two congruent triangles with the x-axis segment from (-R,0) to (r,0). Thus if we find the area of one of these triangles and double it we'll have what we want.

Using point-slope form, the two line equations are:

$y=\frac{r}{\sqrt{2Rr+R^{2}}}(x+R)$ and $y=-\frac{R}{\sqrt{2Rr+r^{2}}}(x-r)$

Setting them equal and solving for the $x$ intersection gives:

$x_{i}=\frac{Rr(\sqrt{2Rr+R^{2}}-\sqrt{2Rr+r^{2}})}{r\sqrt{2Rr+r^{2}}+R\sqrt{2Rr+R^{2}}}$

Substituting this back into either line equation gives the $y$ intersection as:

$y_{i}=\frac{Rr(R+r)}{r\sqrt{2Rr+r^{2}}+R\sqrt{2Rr+R^{2}}}$

Clearly $y_{i}$ is the height of the aforementioned triangle and $R+r$ is the base. Their product gives us the area A of the quadrilateral:

$\boxed{A=\frac{Rr(R+r)^{2}}{r\sqrt{2Rr+r^{2}}+R\sqrt{2Rr+R^{2}}}}$

Dispite being convoluted, I believe this answer is correct; in addition to the deduction with variables, it holds over some calculator examples I did and over the r=R case. The symmetry is quite beautiful :D.
[/hide]


How many p  prime  numbers  can  be found  that  $p^2+23$  has  14  positive  divisor?

From information that it has 14 divisors we know that it's in form $q^{13}$ or in form $q^6r$ where $q,r$ are primes.
$p=2$ and $p=3$ doesn't suffices. If $p>3$ then $p^2+23$ is divisible by 4 and 3. From this we know, it's not in form $q^{13}$, but we see it's in form $q^6r$ and from the fact that it's divisible by $4=2^2$ and $3$ we know that $q=2,r=3 \rightarrow p^2+23=2^6.3 \rightarrow p=13$. So answer is only one (p=13).

I didn't write details, but if anyone wants them I'll write them down.

OK, there's complete solution:

Firstly we look at number of divisors generally. Let $p_1,p_2,\ldots,p_m$ be all distinct prime divisors of $n$. Then we can write $n$ as $n=p_1^{a_1}p_2^{a_2}\ldots p_m^{a_m}$ where $a_1,a_2,\ldots,a_m$ are positive integers. Now consider a divisor of $n$. We know that it don't have other prime divisors than $n$ has. So divisor of $n$ is in form $p_1^{b_1}p_2^{b_2}\ldots p_m^{b_m}$ where $b_i$ is nonegative integer such that $b_i\leq a_i$ for each $i=1,2,\ldots m$. Now a little from combinatoics. We can choose exponent $b_i$ as $0,1,2,\ldots,a_i$ so there's exactly $a_i+1$ ways how to choose exponent $b_i$. So to choose proper exponents $b_1,b_2,\ldots,b_m$ we have exactly $(a_1+1)(a2+1)\ldots(a_m+1)$ ways and it's also number of divisors of $n$.

We know that $p^2+23$ has $14$ divisors. We can try $p=2,3$ and it fails. So suppose that $p>3$ (we will need it later). We know that $14=(a_1+1)(a2+1)\ldots(a_m+1)$ (it's number of divisors). Because $14=2*7$ we know that there are only two ways:

1) $m=1; a_1+1=14$ => $p^2+23$ has only one prime divisor and it's exponent is $13$, so there exists such prime $q$ such that $q^{13}=p^2+23$.

2) $m=2; a_1+1=7; a_2+1=2$ => $p^2+23$ has two prime divisors with exponents $6,1$, so there exists 2 prime $q,r$ such that $p^2+23=q^6r$.

Now let's look at $p^2+23$. We exclude $p=2,3$ so it's known fact (and easy to prove) that $p$ is in form $6k+1$ or $6k-1$:

a) $p^2+23=(6k+1)^2+23=36k^2+12k+1+24=12(3k^2+k+2)$
b) $p^2+23=(6k-1)^2+23=36k^2-12k+1+23=12(3k^2-k+2)$

In both cases $p^2+23$ is divisible by $12=2^2*3$ so it has at least two prime divisors so case 1) cannot happen. Now we see that it must be case 2). We also see that primes $q,r$ are $2,3$. Exponent of $2$ is at least 2 so we know that $p^2+23=q^6r=2^6*3$. From this equality we get $p=\sqrt{2^6*3-23}=13$.

Answer is: There's only one such prime $p=13$.

I hope everything's clear now  :)

I did too get the answerof 13 but just by plugging in the smallest possible values of q and r; 2 and 3.
and I couldnt prove why others didnt exist.

Sorry but can you re-explain why p is in a form of (6k-1) or (6k+1)?
and also, why q and r HAS TO be 2 and 3.

Sorry again.[/quote]

Hi, this was discussed  in details before too  (another thread which also used p^2+23)  . Basically   except for 2, 3 every prime has to be of the form (6k +- 1 )    (Others would be divisible by 2 or 3).

Notice that, as explained in the post,  only way to factor 14 : (1*14) and (2*7). and since p^2+23 is divisible by 12, you have only choice as 2^6*3..

(Re-reading the original  answer may  make it clearer.. as IMO the explaination  has  enough details) - hope this helps.


ratio
quadratics
combinatorics unsolved
combinatorics

Consider the following two person game. A number of pebbles are situated on the table. Two players make their moves alternately. A move consists of taking off the table x pebbles where x is the square of any positive integer. The player who is unable to make a move loses. Prove that there are infinitely many initial situations in which the second player can win no matter how his opponent plays.

We should be able to answer much more difficult questions about this game, like when does the first player win, and when does the second player win?

As for this problem, think about this: if $B$ (the second player) cannot win, then $A$ can win. This means that after $A$ takes a number of pebbles which is a positive square, the position must be one in which the second player wins (because in this new position, $A$ is the second player, so he must have a winning strategy). This shows that $A$'s winning positions are of the form $\mathcal P+n^2$, where $\mathcal P$ is a position from which $B$ wins, and $n^2$ is a positive square. Assume now that we have finitely many positions $\mathcal P_i,\ i\in\overline{1,k}$ in which $B$ wins. It means that the set $(\mathcal P_1+\mathbb N^2)\cup\ldots\cup(\mathcal P_k+\mathbb N^2)\ (*)$ contains all sufficiently large naturals, where, for a natural $x$, we define $x+\mathbb N^2=\{x+n^2|n\in\mathbb N^*\}$.

It's pretty easy to reach a contradiction from here. One way would be observing that a set of the form $x+\mathbb N^2$ has density zero, i.e. the ratio $\frac{(x+\mathbb N^2)\cap\{1,2,\ldots,n\}}n$ goes to $0$ as $n\to\infty$. The union of finitely many sets of density zero can only have density zero, so $(*)$ can't contain all sufficiently large naturals, as was stated.

P.S. By a position $\mathcal P$ I simply mean $\mathcal P$ pebbles at the beginning of a game :).

A litle of quadratic reciprocity  will do the job


HCSSiM
geometry
geometric transformation
reflection

1. What are his opinions to science?
2. Is he as religious (christian? muslim?) as Bush?
3. Also, what are his standings towards questions as stem cells and the research involving them?
4. How does he look at the Iraq war? What will happen if he gets to become the next president?

[quote="daboa"]2. Is he as religious (christian? muslim?) as Bush? [/quote]

Obama is a member of the United Church of Christ, a mainstream Protestant denomination.  (If I'm not mistaken, it traces its history in this country back to the congregationalist tradition in colonial New England.)  Why religion should be considered a significant factor in choosing a presidential candidate is completely beyond me.  The answers to all of your questions are easiest to answer by visiting his [url=http://obama.senate.gov/about/]web[/url] [url=http://www.barackobama.com/index.php]sites[/url].

I would rather like an atheist as president, to whom God does not tell to attack Iraq. Thats why, in my case.
I don't have any problem with religions but in USA they tend to let religion have greater influence than one could wish (that's the impression i've gotten).

[quote="daboa"]I would rather like an atheist as president,[/quote]

I agree. Even though the Founding Fathers built government during a time where Christianity ran unopposed, we should not have a leader that bases his decisions on religion.

Speaking of Obama, it seems that he has quite a substantial momentum going on in the New Hampshire primaries.

I'd much rather have someone who didn't make religion an issue as president of my country (not american). I think that saying "I'd like an atheist" stereotypes towards religion being a issue.

Surely the job should go to the best person for the job whether or not they believe in any religion. (As long as they can separate it from their work)

For the record I'm atheist, but the Dawkin-esque atheism is almost as bad to be as radical $ x$, where $ x \in \{$ world religions $ \}$

SimonM: In the best of worlds. I have not seen anything speaking for a good religious presidency in USA. Too many over there are into it and want much from their leader, a man of faith.

I agree with what you said about Dawkins.

[quote="daboa"] Too many over there are into it and want much from their leader, a man of faith.[/quote]

Is that not how democracy works?

I'd much rather a 'religious' country voted for a religious president than had enforced and atheistic leader.

Also, all American presidents have been religious to some degree, does that mean they were all useless?

Anyone who publicly identified as atheist would lose badly in a national election- a large fraction of the electorate simply wouldn't trust such a person. What we can reasonably demand are leaders who hold to secular policies, and don't endorse a specific religion in their official duties. This is true of all of the major Democratic Party candidates, so it shouldn't matter what their faiths are specifically (and yet it does in the minds of some).

As to why an atheist wouldn't be trusted- those who believe their morality comes from God often also believe that morality can come [b]only[/b] from God, and thus that an atheist can't be a moral person.

[quote="jmerry"]As to why an atheist wouldn't be trusted- those who believe their morality comes from God often also believe that morality can come [b]only[/b] from God, and thus that an atheist can't be a moral person.[/quote]

Yeah, I get that argument all the time for the school Chaplin, my response usually questions why if I have no morals I still haven't gone on a murder spree and refuse to drink Coke because of its treatment of the 3rd world

[quote="7h3.D3m0n.117"]But would you really want a leader that makes a decision based on what a god (that may or may not exist) said? Yes I don't mind religion that much in a president but the recent attack ads on Romney's religion is really pathetic.[/quote]

You've misunderstood me. I said that "someone who didn't make religion an issue". So they obviously wouldn't do something based on what their god said.

[quote="SimonM"]my response usually questions why if I have no morals I still haven't gone on a murder spree and refuse to drink Coke because of its treatment of the 3rd world[/quote]
Perhaps because you are afraid of the police that enforces the laws set up by people with morals (that's about murder spree) and because you want to stand out in some way (that's about coke). :P Now, seriously, I do not really believe that any politician would base his/her decisions on what he thinks the God deems right. What they'll do is to try to please their electors. They'll all become atheists the very moment the majority of the population becomes atheistic (not that it'll happen soon, but...) in exactly the same way the former communist leaders in Russia suddenly became publicly religious in 90s. So, the problem is not with the God but with majority of people that are under the influence of the Church. As long as we have it, our politicians will all be religious as well because being openly non-religious would be just a political suicide.

Well most of you say you want an athiest to be president. However it would be extremely unlikely they could win the white house. This is because of the media out burst of them being an athiest would create outbursts everywhere. There would be rumors, myths, the whole package. I would just perfer a president that had common sense.

Hmm... I'm not sure, having never studied american in great detail, from the sound of it the religious lobbies have a large about of power?

I'm still not convinced that many people in the US actually care about stem cell research etc, or whetehr they are just more concerned about not upsetting a small minority.

[quote="Gallup"]
Between now and the 2008 political conventions, there will be discussion
about the qualifications of presidential candidates -- their education,
age, religion, race, and so on. If your party nominated a generally
well-qualified person for president who happened to be X, would you vote
for that person?

[code]
                   Yes   No
Catholic           95%    4%
Black              94     5
Jewish             92     7
A woman            88    11
Hispanic           87    12
Mormon             72    24
Married 3rd time   67    30
72 years of age    57    42
Homosexual         55    43
Atheist            45    53
[/code]
[/quote]

[quote="SimonM"]I've seen that quoted before, but I'd like to get hold of a link to the real study?

Anyone help?[/quote]

the link was once [url=http://www.galluppoll.com/content/default.aspx?ci=26611&pg=1]here[/url], but is now dead.  i'm sure you can find it somewhere on their site if you look hard enough.

The problem isn't that the country is filled with idiots. The problem is that the country is filled with idiots who are perfectly complacent to remain that way. We've become the epitome of a bread and circus society. The people we elect are all wrong. The seemingly never-ending series of laws compounding upon more laws compounding upon more laws that we continually pass to make us "feel safer" are killing us... But no one notices, because we're too stupid and have no problem being so.

Idiots are always perfectly complacent to remain idiots (basically it is a definition of idiocy; the very moment an idiot recognizes his drawbacks he goes to a completely different category). I do not think most people in the USA are idiots though. True, I have almost never travelled to rural America but I cannot say that my neighbors, colleagues at work, and other people I see and communicate to on an everyday basis are idiots. It's about something else; I do not know how to express it right, but something is rotten in the state of Denmark. I once drove from coast to coast and tried to find a decent radio station: something educational, or somebody reading a good book, or a serious discussion of some topic, or the translation of some good radio play. I found none from the Atlantic to the Pacific: just plenty of stupid songs, equal number of programs advertising some junk, and some rather boring preaching. The few programs playing classical music were the top of the "radio intellectuality". The impression is that somebody (government? corporations?) is deliberately trying to stupefy the nation. This is just a small example which may seem quite irrelevant but I believe that it is a reflection of some bigger picture, of something that is not so easy to see, some big undercurrent that erodes the ground under our feet and prepares a big pit we'll fall into when the thin crust will give way under the heavy burden. I don't know exactly what it is, but it is not individual idiocy, it is some mass psychology phenomenon that brought the USA to the current state. There is some story by J.B. Priestley where the hero notices that almost everybody around is either dead or fast asleep despite they are walking and talking like normal people and ends up coming to the podium (he is some politician running for some office or something like that, if I remember it right) and instead of the prepared speech shouting "Wake up! wake up!" at the audience. Probably this is the closest allegorical description of what is going on I know of.

Contrary to (un)popular belief, most Christians aren't nutjobs who look to the sky every time they are considering invading some Muslim country.

However, a lot of said nutjobs are extremely loud, giving religion as a whole a bad name and lending credibility to nutjob atheists which can be just as bad.

/is an atheist
/more like completely apathetic

I agree with your stance on nutjob atheists, its one of my pet hates at the moment. But having said that, a member of the Church of England (generally regarded as not being radical in any way shape or form) said that flooding in England was caused by a new policy on homosexual marriage.

[code]
                   Yes   No
Catholic           95%    4%
Black              94     5
Jewish             92     7
A woman            88    11
Hispanic           87    12
Mormon             72    24
Married 3rd time   67    30
72 years of age    57    42
Homosexual         55    43
Atheist            45    53
[/code]
That's gotta such for the female Hispanic 72-year old homosexual athiest married for the third time.

[quote="Twiz"]That's gotta such for the female Hispanic 72-year old homosexual athiest married for the third time.[/quote]I wonder how many people have these traits :)

I doubt those figures are true.  I don't believe so few people are not racist. As you ask you get answers. You can't just call someone on the phone and ask "would you vote for a black person?", since it amounts to "are you a racist?"


I'm sure all of you would have heard of the lowest sum rule, which plays a major role in naming of the compounds with branching. However one my IIT JEE books says Lowest Set rule must be followed.

In 99% of the case both give the same answer but there are exceptions again.

Consider that the length of the parent chain is the same and 2 positions of branchings can be obtained when viewing the compound in 2 different ways.

(2,5,9)  (2,6,7)

According to sum rule (2,6,7) is the answer but Set rule says otherwise. What do you think one must follow in JEE??

i think sum rule shud b used

but i don't think jee wud test things like that :mad:

lowest locant rule is more important

see this liiustration

      That set of locants is preferred, which when compared term by term with other set of locants, each in order of increasing magnitude, has the lowest term at the first point of difference. For example the set of locants (2,7,8) is preferred over the set of locants (3,4,9) since 2 comes before 3 even though the sum of locants in the former case is 17 while in the latter case, it is 16

      The correct name is: 2,7,8 - Trimethyldecane and not 3,4,9 - Trimethyldecane

source-http://tutors4you.com/chemorganicbasics.htm

the lowest set rule is applied to alkenes and heteromonocycles i think. otherwise lowest sum rule is applied.

i m pretty sure iit-jee won't test such stuff---and even if it does ;they can't make controversial choices!!!!

@anand --http://tutors4you.com/chemorganicbasics.htm --ofc can b wrong

[quote="anandghegde"]hi varunarasu....do you live in B'lore? sorry for deviating off the topic. :oops:[/quote]
yup. and seeing one of your posts, i think you too live in bangalore. which coll/school r u from?

I go to Vijaya college...... :)

Go to BASE for coaching....you?

lot of off topic discussions here. perhaps we shud continue our discussion by pm. check your inbox. :D


nice problem, considering it was made during your sleep :D [hide]
  We can write an equation using the information given. Since we know that the ERA is 4.32 and we know the formula for ERA we can write this:

$4.32=\frac{x}{y} \cdot 9$
where $x$ is the number of runs allowed by the pitcher and $y$ is the number of innings pitched.
When we divide $4.32$ by $9$ we get $0.48$.  Now the equation is simplified to:

$.48=\frac{x}{y}$
We know that $.48=\frac{48}{100}$
so x can equal to 48 and y is equal to 100
simplifying $\frac{48}{100}$ we get $\frac{12}{25}$

so the answer is $\boxed{\boxed{12:25}}$[/hide]

:clap2: good, that is correct!  And yes, i did make that up in my sleep :blush:

heh.. im gonna post a problem i dreamed up in the basic math forum... It was an awkward dream O.o

:D solve this one first, its a cool problem :)

[quote="jhollenbeck"]A pitchers's ERA(Earned Run Average) is calculated by dividing the number of earned runs the pitcher has allowed by the amount of innings he has pitched and then multiplying this number by 9.  What is the ratio of earned runs allowed to innings pitched for a pitcher who has a 4.32 ERA.
(I dreamed this problem up in my sleep! :blush: )[/quote]

ok..

[hide="solved?"]
earned runs = e
allowed inings = i

9e/i = ERA

9e/i = 4.32

e/i = 0.48

the ratio is 48/100
so its 12/25
[/hide]


find all $ m \in N$,if $ x^m\plus{}y^m\plus{}z^m\minus{}(x\plus{}y\plus{}z)^m$  is divisible by $ (x\plus{}y)(y\plus{}z)(z\plus{}x)$

[quote="apollo"]find all $ m \in N$,if $ x^m \plus{} y^m \plus{} z^m \minus{} (x \plus{} y \plus{} z)^m$  is divisible by $ (x \plus{} y)(y \plus{} z)(z \plus{} x)$[/quote]
All m odd .
$ f(x,y,z)\equal{}x^m\plus{}y^m\plus{}z^m\minus{}(x\plus{}y\plus{}z)^m$
$ h(x,y,z)\equal{}(x\plus{}y)(y\plus{}z)(z\plus{}x)$
$ h|f$ then
$ f(x,\minus{}x,z)\equal{}0$ so m odd.

[quote="TTsphn"][quote="apollo"]find all $ m \in N$,if $ x^m \plus{} y^m \plus{} z^m \minus{} (x \plus{} y \plus{} z)^m$  is divisible by $ (x \plus{} y)(y \plus{} z)(z \plus{} x)$[/quote]
All m odd .
$ f(x,y,z) \equal{} x^m \plus{} y^m \plus{} z^m \minus{} (x \plus{} y \plus{} z)^m$
$ h(x,y,z) \equal{} (x \plus{} y)(y \plus{} z)(z \plus{} x)$
$ h|f$ then
$ f(x, \minus{} x,z) \equal{} 0$ so m odd.[/quote]
why if $ h|f$ then $ f(x,\minus{}x,z)\equal{}0$ :?:

[quote="apollo"][quote="TTsphn"][quote="apollo"]find all $ m \in N$,if $ x^m \plus{} y^m \plus{} z^m \minus{} (x \plus{} y \plus{} z)^m$  is divisible by $ (x \plus{} y)(y \plus{} z)(z \plus{} x)$[/quote]
All m odd .
$ f(x,y,z) \equal{} x^m \plus{} y^m \plus{} z^m \minus{} (x \plus{} y \plus{} z)^m$
$ h(x,y,z) \equal{} (x \plus{} y)(y \plus{} z)(z \plus{} x)$
$ h|f$ then
$ f(x, \minus{} x,z) \equal{} 0$ so m odd.[/quote]
why if $ h|f$ then $ f(x, \minus{} x,z) \equal{} 0$ :?:[/quote]
It is a fundamental of Polynomial.

[quote="apollo"]
$ f(x,y,z) \equal{} x^m \plus{} y^m \plus{} z^m \minus{} (x \plus{} y \plus{} z)^m$
$ h(x,y,z) \equal{} (x \plus{} y)(y \plus{} z)(z \plus{} x)$

why if $ h|f$ then $ f(x, \minus{} x,z) \equal{} 0$ :?:[/quote]
we fixe $ y,z\in R$
we take $ P(x)\equal{}h(x,y,z),\ Q(x)\equal{}f(x,y,z)\in R[X]$
then $ P|Q$
so if $ P(\minus{}y)\equal{}0$ gives $ Q(\minus{}y)\equal{}0$
then $ f(\minus{}y,y,z)\equal{}Q(\minus{}y)\equal{}0$


Se da un triunghi $\ ABC$ cu $\ m(<BAC)=135,m(<ABC)=30$.Fie $\ D$ piciorul inaltimii din $\ A$ astfel incat
$\ D$ sa se afle pe $\ (BC)$ si $\ E$ piciorul bisectoarei din $\ B$, cu $\ E$ pe $\ (AC)$.Atunci vom avea dreptele
$\ AD,BE,CF$ concurente, cu $\ F$ pe $\ (AB)$.
a) Aflati valoarea raportului $\frac{AF}{BF}$.
b) Daca, in schimb, $\ E$ este piciorul medianei din $\ B$, aflati in acest caz valoarea raportului $\frac{AF}{BF}$.

Ba da, dar nu pot sa nu fac la matematica nici macar in vacanta.Sau te referi la altceva?

Pai raportul il poti calcula si din CEVA.Raportul BD/DC=ABsin60/ACsin75=sincsin60/sin bsin75,iar AE/EC=AB/BC=sinc/sina..etc

Da, cred ca e bine ce zici tu.Eu ma gandisem sa aplic $\ Teorema\ Bisectoarei\ Generalizata$ pentru raportul
$\frac{BD}{CD}$ si apoi e simplu. :)


Find the simplified expression for $\sqrt{111\ldots1-222\ldots2}$, where there are $n$ 2's and $2n$ 1's in the expression.

Also, i'm back to AoPS. I havent been here in about 3 months or so.

[hide]
$333...$ there are $n$ threes..
not sure how to prove it though..
[/hide]

[hide="Hint"] The expression is equivalent to $\sqrt{\frac{10^{2x}-1}{9}-2\frac{10^{x}-1}{9}}$. [/hide]


If $ x \plus{} y \plus{} z \equal{} 12$ and $ x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} 96$ and $ \frac{1}{x} \plus{} \frac{1}{y} \plus{} \frac{1}{z} \equal{} 36$, compute $ x^{3} \plus{} y^{3} \plus{} z^{3}$.

I've been trying to solve this problem for hours. Gah!

Hmm, thank you very much! Man, I wish I knew how to start tackling these problems. =)
I have a math competition (bowl) tomorrow. Do you guys have any tips? What should I memorize?

From $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 36$ we have $ xyz \equal{} \frac {2}{3}$.

$ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx) \equal{}$

$ \equal{} (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} \frac {1}{2}((x \plus{} y \plus{} z)^2 \minus{} x^2 \minus{} y^2 \minus{} z^2)) \equal{} 12(96 \minus{} 24) \equal{} 864$.

$ x^3 \plus{} y^3 \plus{} z^3 \equal{} 866$.

$ xy+yz+zx=\frac{1}{2}\{(x+y+z)^2-(x^2+y^2+z^2)\}=48$. Thus $ x,\ y,\ z$ are the roots of the cubic equation : $ t^3-12t^2+24t+k=0$.

Then $ \frac{1}{x},\ \frac{1}{y},\ \frac{1}{z}$ are the roots of the cubic equation : $ kt^3+24t^2-12t+1=0\ \because k\neq 0$.

Vieta's formula gives $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{24}{k}=36\Longleftrightarrow k=-\frac{2}{3}$.

$ \therefore x^3+y^3+z^3=12(x^2+y^2+z^2)-24(x+y+z)-3k$

$ =12\cdot24\cdots (4-1)+2=\boxed{866}$.


[quote="white_horse_king88"]I get a huge error message while compiling my first latex document. What happened?[/quote]Nobody knows as long as you won't tell us what the error was :)

I can't copy and paste it. But, I know that when I installed TenixCenter, it didn't ask for a DVI source code, which could be a major problem. That's the only thing that didn't go along with the website instructions. I followed everything alse perfectly.

If you don't give an error, there will be no answer (at least not one that will have real chances to help you) besides "reinstall everything" more carefully this time.

Oh - I see. I fixed it. I didn't realize that this wasn't like JAVA, and I expected the output to be under the compiler. Whoops. Anyways, how to I transport the output file so that it is readable? So that I can e-mail it?

The output file is a pdf or dvi or postscript file. It will be in the same folder as your source file.