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advanced fields theorems

Does anyone have this proof? where can I find it?

Which is the relationship between this proof and Fermat's Last Theorem?

EDIT: I foun this, but I don't know what's exactly: http://www.math.uchicago.edu/~chris/topicproposal.pdf#search='Faltings%27%20Theorem'

Have a look Hindry, Silverman, Diophantine Geometry: An Introduction.

[quote="puuhikki"]Hindry, Silverman, Diophantine Geometry: An Introduction.[/quote]Awsome book might I add..  :)

Now there is an even better book available: Heights in Diophantine Geometry, by Enrico Bombieri and Walter Gubler. It has complete and self-contained proofs of Roth's theorem [from Diophantine approximations]; the Subspace theorem (Schmidt's multidimensional generalization of Roth's theorem); the Mordell-Weil theorem; Faltings' theorem [ala Vojta; with glimpses to more advanced results such as the "Big Faltings theorem"]; the main theorems of Nevanlinna theory [on value distributions of meromorphic functions] and their analogy in Diophantine approximations; and outstanding applications.

Unfortunately, it is (still) too expensive...

One more comment: I would say that Roth's theorem is done very sloppily in Hindry-Silverman. Things are not as complicated as they are made look there. And even the formulation there is not the right one: they use the relative height $H_{K}$, while the theorem clearly holds with the absolute height (I find this very strange). And, as I said, I generally find that life is simpler than it is presented in Hindry-Silverman's "Diophantine Geometry." Still, it's a very good expository book.


inequalities function integration logarithms calculus parameterization algebra

I don't know who named me the big master of special cases. Here is another one:
   Let $a_1, a_2,, a_n$ be positive real numbers and S be their sum. Prove that
$\sum_{i=1,n}{\frac{1}{na_i}+\frac{n(n-2)}{S}\geq\sum_{i<>j}{\frac{1}{S+a_i-a_j}}}$. Here $ i<>j$ means that i is different from j. What is nice is that this inequality is stronger than Popovicius inequality in the general case! Prove also that!

I proved the following statement:
If  $f$ is a convex function on $I$ , and $x_1,x_2, ...,x_n \in I$, then

$f(x_1)+f(x_2)+...+f(x_n)+ \frac{n(n-2)}{2}f(a)\geq \frac{n}{2(n-1)}\sum_{i<>j}f(a+\frac{x_i-x_j}{n})$,

where $a=\frac{x_1+x_2+...+x_n}{n}$.

But I think the following problem is more interesting and solves Harazi's Inequality:
Find one additional condition such that the inequality occurs:
$f(x_1)+f(x_2)+...+f(x_n)+ n(n-2)f(a)\geq \sum_{i<>j}f(a+\frac{x_i-x_j}{n})$.

Please, Harazi, can you post your solution?

Thank you very much in advance. ;)

Vasc and Harazi, can you find a compromise to post one of your solution  ;)

This is like the SURPRISING ATTACK DECISION. :lol:  :lol:  :lol:

[quote]Two generals with their troops are drawn up in battle order on the opposite sides of a mountain so that they cannot see each others. They want to coordinate their attack as if only one of the two troops would attack alone it could be destroyed.
Unfortunately they can communicate only using homing-pigeons which can fly from one camp to the other getting over the peak of the mountain.
The first general send this message to the second general:" Attack at 8.00. Confirm the msg received otherwise I will not attack"
The second general receive the msg and he is agree for the attack at 8.00 , but he must send the confirmation as he realizes that the first general will not attack at 8.00 if he will not receive the confirmation from the second.
But the second general cannot know if his pigeon will reach the first, so he will send this msg : " OK for the attack at 8.00. Confirm the msg received ,otherwise I will not attack"
By a little reasoning ,you can deduce that this infinite sequence of msgs will never stop and the two general will never attack the enemy.

Vasc and Harazi are like the two generals, so that.......... :lol:     [/quote]

I think to deserve the proof to have told you this beautiful story.

I am joking :)

The truth is that, as usual,  I am very curious to see the proof as I have spent a lot of days without success time on it. :(  :(  :(  :(  :(  :(

Thank you ;)

Manlio, I have already given you the solution. Try to guess when and in which form. ;)

Manlio, I haven't a solution to Harazi's inequality, yet. ;)

Dear Harazi, for n=3 it can be derived by integrating Shur, but for generic n which is the underlying inequality you used?

Try to derive it! For n=3 you have integrated, for general n just derive to see what inequality I have used.  ;)

Now I see, Harazi ;)

One must integrate this inequality

$ \displaystyle %Error. "displayfile" is a bad command.
x_1^n + ... + x_n^n + n(n - 1)x_1...x_n \geq x_1...x_n(x_1 + ... + x_n)(\frac {1}{x_1} + ... + \frac {1}{x_n})$

with $ x_i = t^{a_i - \frac {1}{n}}$ between $ 0$ and $ 1$

Indeed, Harazi's integrating method is magnificent.
Using this way, from [url=http://www.mathlinks.ro/viewtopic.php?t=14906]my Schur generalization[/url] (for which Harazi posted a beautiful solution on this forum)

$ \displaystyle %Error. "displayfile" is a bad command.
x_1^n + ... + x_n^n + n(n - 1)x_1...x_n \geq x_1...x_n(x_1 + ... + x_n)(\frac {1}{x_1} + ... + \frac {1}{x_n})$,

the required inequality follows. Similarly, from Suranyi's inequality we get

$ \frac {n - 2}{n}\sum\frac {1}{a_i} + \frac {n}{a_1 + ... + a_n} \ge \sum_{i < > j}\frac {1}{a_i + (n - 1)a_j}$.

Also, from [url=http://www.mathlinks.ro/viewtopic.php?t=6026]my inequality[/url]

$ (a^2 + b^2 + c^2)^2 \ge 3(a^3b + b^3c + c^3a)$,

we get

$ \sum \frac {1}{4a} + \sum\frac {1}{b + c} \ge \sum\frac {3}{3a + b}$.

This last inequality was posted by Harazi on this formum. Now it is clear for me why nobody found a solution.
Very, very nice ! :)

There's something I dont understand and I'll be very glad if someone explain it to me clearly. For example here:
[quote="Vasc"]
Also, from [url=http://www.mathlinks.ro/viewtopic.php?t=6026]my inequality[/url]
$ (a^2 \plus{} b^2 \plus{} c^2)^2 \ge 3(a^3b \plus{} b^3c \plus{} c^3a)$,
we get
$ \sum \frac {1}{4a} \plus{} \sum\frac {1}{b \plus{} c} \ge \sum\frac {3}{3a \plus{} b}$.
[/quote]

when I integrate $ (a^2 \plus{} b^2 \plus{} c^2)^2 \ge 3(a^3b \plus{} b^3c \plus{} c^3a)$ I get [I used Derrive to simplify it as much as I can]:

$ 9a^4 \plus{} 10a^2c^2 \plus{} 10a^2b^2 \plus{} 9c^4 \plus{} 10b^2c^2 \plus{} 9b^4 \geq \frac {135a^3b}{8} \plus{} \frac {135ac^3}{8} \plus{} \frac {135b^3c}{8}$
which differs from $ \sum \frac {1}{4a} \plus{} \sum\frac {1}{b \plus{} c} \ge \sum\frac {3}{3a \plus{} b}$. doesn't it ?  :?

...when you think about integrating you mean:

$ \int (a^2 \plus{} b^2 \plus{} c^2)^2 da \ db \ dc \ge \int 3(a^3b \plus{} b^3c \plus{} c^3a) da \ db \ dc$ ?? or am I wrong completely ?  :(

[quote="manlio"]Megus, have you read my previous msg?[/quote]

Yes Manlio, I have but the result is quite strange if I understand it correctly. Maybe I write my steps - it'll be easier:

I take inequality for n=3 and:

$a^3 + b^3 + c^3 + 6abc \geq abc(a + b + c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
I use substitution $a=t^{x-\frac{1}{3}}$, $b=t^{y-\frac{1}{3}}$, $c=t^{z-\frac{1}{3}}$ and I get:

$t^{3x - 1} + 6t^{x + y + z - 1} + t^{3y - 1} + t^{3z - 1} \geq t^{2x - 1}(t^y + t^z) + t^{x - 1}(t^{2y} + 3t^{y + z} + t^{2z}) + t^{2y + z - 1} + t^{y + 2z - 1}$

then i integrate:

$\int _0 ^1 t^{3x - 1} + 6t^{x + y + z - 1} + t^{3y - 1} + t^{3z - 1} dx \ dy \ dz \geq \int _0 ^1 t^{2x - 1}(t^y + t^z) + t^{x - 1}(t^{2y} + 3t^{y + z} + t^{2z}) + t^{2y + z - 1} + t^{y + 2z - 1} dx \ dy \ dz$

to get:

$\frac{t^2 - \frac{1}{t}}{\ln {t}} + \frac{6(t - 1)^3}{t (\ln {t})^3} \geq \frac{3(t^3 - t^2 - t + 1)}{t(\ln {t})^2} + \frac{3(t - 1)^3}{t (\ln {t})^3}$

Now this really look strange, not only it is ugly but also it is just "one variable" inequality. I could substitute some t's as far as $t=e^{\frac{\ln a}{x - \frac{1}{3}}}$ and so on, but does it change anything ?  :?

I'll be very happy if someone explain it to me

sorry for my stupidity   :blush:

t is the variable of integration , while a_i must be considered like costant

Well, now everyone has found my secret.  :( . I just adore this method, it is so powerful. But the show never ends!

Sorry, Harazi. I didn't know it was your secret :blush:

I thought only Vasc had secrets  ;)

Really I don't like secrets ;)

Well, Manlio, life with no secrets is too easy. Come on! Vasc does not have secrets either. They are just temporary. Imagine that everyone offers you the solution even if you did not try sufficiently. Would you be happy? I wouldn't.

However I must say that your integral method, Harazi is really one of the best method on inequalities I have ever seen and only you are the creator of this MAGNIFICENCE

Now I see the mightness of this method  :) Thank you for it,  Harazi and Manlio [for explaining]

[quote="Vasc"]Indeed, Harazi's integrating method is magnificent.
Using this way, from [url=http://www.mathlinks.ro/viewtopic.php?t=14906]my Schur generalization[/url] (for which Harazi posted a beautiful solution on this forum)

$ \displaystyle %Error. "displayfile" is a bad command.
x_1^n + ... + x_n^n + n(n - 1)x_1...x_n \geq x_1...x_n(x_1 + ... + x_n)(\frac {1}{x_1} + ... + \frac {1}{x_n})$,

the required inequality follows. Similarly, from Suranyi's inequality we get

$ \frac {n - 2}{n}\sum\frac {1}{a_i} + \frac {n}{a_1 + ... + a_n} \ge \sum_{i < > j}\frac {1}{a_i + (n - 1)a_j}$.(1)

Also, from [url=http://www.mathlinks.ro/viewtopic.php?t=6026]my inequality[/url]

$ (a^2 + b^2 + c^2)^2 \ge 3(a^3b + b^3c + c^3a)$,

we get

$ \sum \frac {1}{4a} + \sum\frac {1}{b + c} \ge \sum\frac {3}{3a + b}$.(2)

This last inequality was posted by Harazi on this formum. Now it is clear for me why nobody found a solution.
Very, very nice ! :)[/quote]
Sorry,i really don't understand how to obtaint (1) and (2)?I read carefully all post in this topic but i still don't understand what harazi's magnificent method is?

Introduce a parameter t and places your variables in the index. Then perform the integration with respect to t then put t=1. Then you get a new inequality.

first of all, why is the inequality true after integration? (sorry if this question is really stupid)

second of all, can anyone show me a simple example to get the hang of this method?

Perfect-radio, do you accept that if two functions are one greater than the other in each point, then the integral of the greater function is greater than the integral of the second function? If you accept that, here is a simple example:
   Suppose we want to prove that $\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\geq \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$. (of course, this one is trivial, but I present it just to explain the method; you will see some very non-trivial inequalities obtained in the same way on the forum).  We know the inequality $ x^2+y^2+z^2\geq xy+yz+zx$. Thus, we can put $ x=t^{a-\frac{1}{2}}$, $y= t^{b-\frac{1}{2}}$ and $z=t^{c-\frac{1}{2}}$. We find for all $t$ between $0$ and $1$ that $ t^{2a-1}+t^{2b-1}+t^{2c-1}\geq t^{a+b-1}+t^{b+c-1}+t^{c+a-1}$. Now, integrate this inequality with $t$ variable between $0$ and $1$. You will find exactly $\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\geq \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$. Is it clear now?  The motto of the method is: instead of solving fraction inequalities with idiot bunching, think initially whether you can reduce the inequality to some polynomial one without clearing denominators!!!


(U vezi izraza Off-Off-Broadway, konzultirajte svemoćnu Wiki)

Uglavnom, mislim da ovom podforumu treba non-topic topic. Naime, vidim da smo poprilično zaspamali BiH JBMO topic.

Tako da...

 :welcome:

Ne znam baš kako bih počeo, ali ajde, nek piše nešto.

Ja ću sad ponovno probati kontaktirati Valentina u vezi moda, tako da možemo pobrisati sve nebitne poruke iz JBMO (tko god da je smislio ovo ime, nije bio Srbin/Hrvat/Bošnjak/Crnac/Gorac...) topica i ubuduće se brinuti za spam... nemojte se previše nadati.

E, da, Hana, Goblin (i ostali), koliko je poznat taj teorem/lema što ga je Goblin iznio Djuri? Ja si povremeno volim zamišljati da čak i znam nešto iz teorije brojeva, ali nisam za to čuo... Je li dovoljno poznat da se npr. može koristiti na IMO bez dokaza?

Bravo mornik, bas sam jutros htio otvoriti ovakav topic (ali eto, usmeni iz Matematike me pomeo...) Onaj NT teorem meni zvuci poznato, kad se vratim kuci zaviricu u biblioteku... Ima li kakvih vijesti od V.Vornicua?

P. S. Lijep izbor pjesme za potpis... Uvijek su matematicari voljeli Djoleta  :lol:

Ma tek sam sad postao ovo - http://www.mathlinks.ro/viewtopic.php?p=1164335, ak ne bude odgovora kroz par dana još ću malo daviti Valentina.

Još ni riječi od Valentina. Strašno. Mora da se u djetinjstvu bojao nekog Jugoslavena... to je to, razvojna psihologija... niš, sad će biti ovakav do kraja života.  :lol: :lol: Just kidding.

Dokazati ovo nisam danas ni pokušavao, makar sam mislio - rješavao sam nešto drugo, probat ću sutra.

E, da, Harune, od kuda nadimak Kočić, koji vidim da se povremeno veže uz tvoje ime... Ispravi me ako sam u krivu.

Ja i ne ocekujem odgovor od Valentina... posto vidim da moju drugu poruku (poslanu prije sedam dana) nije ni procitao.

Sto se nadimka tice... Vrijeme i mjesto nastanka IMO 2006. Varijacija na temu mog prezimena (Mademoiselle pink floyd je svemu kumovala, ako se dobro sjecam  :lol: )

Tako da ostade Kočić...


geometry
rhombus
perimeter
Pythagorean Theorem

1) How do you find a rhombus' area? Is it just length*width?

2) How would you find a rhombus' area knowing its diagonals?

3) How would you find a rhombus' perimeter knowing its diagonals?

4) Is it "rhombuses" or "rhombi"?

Thanks for the great help!  :)

1) It's not length * width, it's length * height, where height is the perpendicular distance from the bottom base to the top base.

2) It's $ \frac 12 d_1 \cdot d_2$ where $ d_1$ and $ d_2$ are the two diagonals. Here's why:
If you consider the four triangles that result from the diagonals, you can see that each has a base of $ \frac 12 d_1$ and a height of $ \frac 12 d_2$, so its area is $ \frac 18 \cdot d_1 \cdot d_2$. But there are four of those, so the final answer is $ \frac 12 \cdot d_1 \cdot d_2$.

3) Using they Pythagorean Theorem, you can see that the perimeter is $ 2\sqrt {d_1^2 \plus{} d_2^2}$.

4) Both. I prefer rhombi. :D

If it is a rhombus, the intersection of the diagonals will ALWAYS form a right angle.

Okay, then I am pretty sure about my answer.

[quote="Wickedestjr"]the area of a rhombus is the product of the diagonals divided by two. If the lengths of the diagonals of a rhombus are 'y' and 'x', then the perimeter in terms 'x' and 'y' is the following;

2 x [the squareroot of; (X^2 + Y^2)], at least I think that's what it is. This is only a true statement if the intersection of the diagonals intersect to form a right angle.

  I think that it is rhombi.....[/quote]

I'll just latexify this  :lol:

$ 2\sqrt{x^{2}\plus{}y^{2}}$

Please do not revive old threads. This is from over two months ago and responding to a question that is already answered is considered spam.


For $x,y,z$ positive reals prove that

$ ((xyz)^2 + xyx +1)^2 \leq \frac{9}{8}(x^4+1)(y^4+1)(z^4+1) $

This inequality was posted before and not solved because it is in a very recent issue of Gazeta Matematica.

cheers!:D

I enjoy it!

Actually we have to prove:
16(xyz)3+24(xyz)2+16xyz <= (xyz)4+9(x4y4+y4z4+z4x4)+9(x4+y4+z4)+1.
Let t=(xyz)1/3. Using x4y4+y4z4+z4x4 >= 3t8 and x4+y4+z4 >= 3t4 we obtain following inequality:
t12-16t9+27t8-24t6+27t4-16t3+1 >=0
which is true due to following factoring:
LHS=(t-1)4(t8+4t7+10t6+4t5-2t4+4t3+10t2+4t+1).


geometry
function
calculus
geometry proposed

Here's a problem.  Prove that there are two curves that the area bounded between them is $\frac 13$ of a $1\times 1$ square on the same cartesian plane.

Masoud Zargar

Take $x^{\frac 1t}$ and $x^t$, and the square with vertices $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$. As $t\to\infty$, the area bounded between them is between $1$ and $1-\epsilon$ for any sufficiently small epsilon.  As $t\to 1$, the area bounded between them goes to zero.  But the area bounded between them is a continuous function of $t$, so there exists $s\in(1,\infty)$ such that the area bounded between $x^{\frac 1s}$ and $x^s$ is exactly $1/3$.

Of course, you hardly need calculus here; piecewise linear curves would do just as well.  This belongs in Pre-olympiad or lower.


Be k€IN
Be S={a²€ IN/   2k²+k≤a≤2k²+2k}
  Does exist 2 sets A & B Such that:
  A∩B={}  &  A U B=S,
    ∑ a²=∑ b²
   a€A    b€B

You don't have to post it twice or more... :mad:

http://www.mathlinks.ro/Forum/viewtopic.php?t=61722

Pierre.

Is this what you want:
Let $k$ be a positive integer and let $S=\{a^2|2k^2+k\le a\le 2k^2+2k\}$.
Determine if there exists a partition of the set $S$, $S=A\cup B$ such that $\sum_{x\in A}x^2=\sum_{x\in B}x^2$.


what is a lemma? i know i saw it in proofs when i was trying to do some, but i do not understand the concept of it...

A lemma is a little theorem you use in order to prove bigger theorems.  For example, if I know I'm going to have to use some result over and over again in a proof, I might just call it "lemma 1" so I can refer to it easily.  It's just terminology that isn't very important (like the word "corollary").

When dealing with lemmas in proofs, must you prove your lemma in order to use it or can you state it as a result that has been proven (for example say in the USAMO)?

i was told that you don't really have to prove it, as long as it has a legitimate name..

(for example, AM-GM is technically a lemma, but no one bothers to prove it..)

There is really, really no reason whatsoever to use the word "lemma."  It's just something people like to do.  For purposes of writing a competition, you will do just as well by calling things "claims" or by numbering your statements.  Anything that has a name (the AM-GM inequality, for example), should not be a lemma in your solutions -- a lemma is something that you prove and then will use to reach more important results.

Also there are some  very famous 'lemmas' called such by their modest authors that are actually  regarded   higher than some of the so called "major theorems". .. Some times a   theorem is called a lemma just becuase the author chose to do so (more out of modesty than any other technical consideration)  :)


quadratics
analytic geometry
graphing lines
slope
algebra

The tangent line to a circle may be defined as the line that intersects the circle in a single point, called the POINT OF TANGENCY.

If the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b, show that:

(a) r^2(1 + m^2) = b^2

HINT GIVEN FOR PART A:

The quadratic equation x^2 + (mx + b)^2 = r^2 has exactly ONE SOLUTION.

(b) The point of tangency is the point (-r^2m/b, r^2/b).

(c) The tangent line is perpendicular to the line containing the center of the circle and the point of tangency.

(d) Draw a tangent line to a circle in the xy-plane.
Use r to represent the radius.

[hide]Let the point of tangency be $(c,d)$. The center of the circle is $(0,0)$. The slope of tangency is perpendicular to the slope of the line going through the center and the point of tangency, so the slope of the line would be $-\frac{c}{d}$. Since the line goes through $(a,b)$,

$y-d=-\frac{c}{d}(x-c)$

$y=-\frac{c}{d}x+\frac{c^{2}}{d}+d$

So $m=-\frac{c}{d}$, $b = \frac{c^{2}}{d}+d$

Simplifying, $b = \frac{c^{2}+d^{2}}{d}= \frac{r^{2}}{d}$

$b = \frac{r^{2}}{d}$

$m^{2}= \frac{c^{2}}{d^{2}}$

$1+m^{2}= \frac{c^{2}}{d^{2}}+1 = \frac{r^{2}}{d^{2}}$

$r^{2}(1+m^{2}) = \frac{r^{4}}{d^{2}}= b^{2}$

That was the equation. [/hide]

What about the other parts of the question?

(b) The point of tangency is the point (-r^2m/b, r^2/b).

(c) The tangent line is perpendicular to the line containing the center of the circle and the point of tangency.

(d) Draw a tangent line to a circle in the xy-plane.
Use r to represent the radius.


Is there an english version of Gils website? Also do they sell books in english? I noticed many of the links to the english version of GIL you guys posted don't work like when I click on it it says like:

Index of /en

    * Parent Directory
    * .ftpquota

Apache/2.2.9 (Unix) mod_ssl/2.2.9 OpenSSL/0.9.8b mod_auth_passthrough/2.1 mod_bwlimited/1.4 FrontPage/5.0.2.2635 PHP/5.2.6 Server at http://www.gil.ro Port 80

And I can only get to the French site from there.

Try Google Translater, but it won't help very much.


Hi everyone
I'll properly sign up for one of the AoPS math class in the summer. But I will go to Vietnam this summer, in which the time zone will change, and i order to attend the class, I have to wake up very early in the morning. So my question is: can I get all the stuffs taught in class after class, ask the questions afetr class and what is the differnce between "studying after class" like that and "studying in class"? If we can not keep up with the level in class, do the classes have some way in order for me to keep up with?
Thanks in advance.
 :)

As far as I know the transcripts from each session are saved and you can study from there... :|


Here is a fun problem to do. I learned this in MathCircle at the U.

Q: How many non-negative integers satisfy the following?
a+b+c+d+e </ 500 (less than or equal to)

This is very interesting...

Oh! If you get this problem, try this one

Q: How many terms does (a+b+c+d)^24 have when expanded and simplified?

[hide="the second question"]The answer is (27C3), or 27*26*25/6=25*13*9=2925.  You can basically think of it this way:
You have a wall or some object with 27 little slots in it in a line.  You have three pegs and can place them in in whichever three slots you want to.  The section of empty slots to the left of the first (leftmost) peg will be the exponent of a.  The section between the first and second is that of b.  So on.  The total number of empty slots will be 24.  By taking the number of possible ways to insert the pegs, you get every possibility.  That nmber is (27C3)[/hide]
For the first one, is it the number of solutions?  Is it the number of ordered sets (a, b, c, d, e) which fulfill this condition?  Or does order not matter?
For the first problem, you could probably also use a peg/slot approach.  So it would come to something like
[hide](505C5)/24[/hide]
assuming that order does not matter


inequalities
search
inequalities unsolved

[quote="thegod277"]Let a,b,c are non-negative numbers such that ab+bc+ca=3. Prove that
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 7abc\geq 10$ :D[/quote]
After homogenization you'll obtain very easy inequality:
$ \sum_{cyc}(27a^6\minus{}46a^3b^3\plus{}378a^4bc\minus{}300a^3b^2c\minus{}300a^3c^2b\plus{}241a^2b^2c^2)\geq0,$
 which true because $ \sum_{cyc}(a^6\minus{}2a^3b^3\plus{}a^2b^2c^2)\geq0.$ :wink:

posted here in persian:
[url]http://www.mathlinks.ro/viewtopic.php?t=162013[/url]
but you can understand the whole idea...

I think it's not easy problem, thegod277 can you post your solution.I'm waiting for a nice proof from you. :D

:blush: This is my solution:
Apply Schur inequality we have
$ a^3\plus{}b^3\plus{}c^3\plus{}7abc\geq ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)\plus{}4abc$
$ \equal{}(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)\plus{}abc\equal{}3(a\plus{}b\plus{}c)\plus{}abc$
Hence, from ab+bc+ca=3 we have $ a\equal{}\frac{3\minus{}bc}{b\plus{}c}$. We need to prove
$ 3(a\plus{}b\plus{}c)\plus{}abc\geq 10$
or $ 3(\frac{3\minus{}bc}{b\plus{}c}\plus{}b\plus{}c)\plus{}\frac{(3\minus{}bc)bc}{b\plus{}c}\geq 10$
or $ 3(b\plus{}c)^2\minus{}10(b\plus{}c)\plus{}9\minus{}b^2c^2\geq 0$
or $ 3(b\plus{}c\minus{}2)^2\plus{}2(b\plus{}c\minus{}2)\plus{}(1\minus{}b^2c^2)\geq 0$
We can assume that a=min{a,b,c}. It is easy to check that $ a\plus{}b\plus{}c\geq 3$, so $ b\plus{}c\geq 2$.
*If $ bc<1$, it is easy to prove it.
*If $ bc\geq 1$. Let $ \sqrt {bc}\equal{}x$, then
$ LHS\geq 12(x\minus{}1)^2\plus{}4(x\minus{}1)^2\plus{}(1\minus{}x^4)$
$ \equal{}(x\minus{}1)^2(10\minus{}(x\plus{}1)^2)\geq 0$
is true because we have $ 1\leq x\leq \sqrt 3$.
:D

[quote="thegod277"] We need to prove
$ 3(a \plus{} b \plus{} c) \plus{} abc\geq 10$
[/quote]
Let $ f(a,b,c) \equal{} 3(a \plus{} b \plus{} c) \plus{} abc$ and $ a \equal{} \min\{a,b,c\}.$  Then
$ f(a,b,c) \minus{} f\left(a,\sqrt {(a \plus{} b)(a \plus{} c)} \minus{} a,\sqrt {(a \plus{} b)(a \plus{} c)} \minus{} a\right) \equal{}$
$ \equal{} 3\left(b \plus{} c \minus{} 2\left(\sqrt {(a \plus{} b)(a \plus{} c)} \minus{} a\right)\right) \minus{} a\left(\left(\sqrt {(a \plus{} b)(a \plus{} c)} \minus{} a\right)^2 \minus{} bc\right) \equal{}$
$ \equal{} (3 \minus{} a^2)\left(\sqrt {a \plus{} b} \minus{} \sqrt {a \plus{} c}\right)^2\geq0$ because $ 3 \equal{} ab \plus{} ac \plus{} bc\geq3a^2.$
Thus, it remains to prove that $ f(a,b,b)\geq0,$ where $ 2ab \plus{} b^2 \equal{} 3.$
Id est, $ a \equal{} \frac {3 \minus{} b^2}{2b},$ where $ 1\leq b\leq\sqrt3.$
Hence, it remains to prove that $ \frac {3(3 \minus{} b^2)}{2b} \plus{} 6b \plus{} \frac {(3 \minus{} b^2)b}{2}\geq10,$
which equivalent to $ (b \minus{} 1)^2(b^2 \plus{} 2b \minus{} 9)\leq0,$ which true for $ 1\leq b\leq\sqrt3.$

By the way the following stronger inequality is true too:
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\equal{}3.$ Prove that:
\[ 13(a\plus{}b\plus{}c)\plus{}6abc\geq45\]

:D thank you! But I like my solution more. Oh, why do you know the inequality $ 13(a\plus{}b\plus{}c)\plus{}6abc\geq 45$ is stronger?
 :maybe:

[quote="thegod277"] $ 13(a \plus{} b \plus{} c) \plus{} 6abc\geq 45$ is stronger?
 :maybe:[/quote]
Because if $ 13(a \plus{} b \plus{} c) \plus{} 6abc\geq 45$ is true then
$ 3(a \plus{} b \plus{} c) \plus{} abc \minus{} 10 \equal{} \frac {1}{6}(13(a \plus{} b \plus{} c) \plus{} 6abc \minus{} 45) \plus{} \frac {5}{6}(a \plus{} b \plus{} c \minus{} 3)\geq0.$ :wink:
Thus, my inequality is stronger than your. :P

I think it is easy :D See my proof at :
http://www.mathlinks.ro/viewtopic.php?search_id=1852732440&t=156163

[quote=thegod277]Let a,b,c are non-negative numbers such that ab+bc+ca=3. Prove that
$ a^3\plus{}b^3\plus{}c^3\plus{}7abc\geq 10$ :D[/quote]
Let $a,b,c\geq0$  such that $ab+bc+ca=3.$ Prove
$$a^3+b^3+c^3+3\left(2\sqrt3-1\right)abc\geq6\sqrt3$$
When do we have equality?

[quote=WolfusA]By uvw it's sufficient to prove for case $a=b$ and for case $c=0$.[/quote]

Sir, can you please explain this a bit? I see that statement everywhere but I can not understand why.

[quote=WolfusA]By uvw it's sufficient to prove for case $a=b$ and for case $c=0$. The latter is easy: $$a^3+b^3\ge 2\sqrt{a^3b^3}=6\sqrt3.$$The former becomes$$16x^6+(3-x^2)^3+12x^4(2\sqrt3-1)(3-x^2)\ge 48\sqrt3 x^3$$ where $a=c=x\in[0;\sqrt3]$ since $b=\frac{3-a^2}{2a}$.
$$(x-1)^2(\sqrt3-x)\left(3 (8 \sqrt3 - 9) x^3 +3 (6 + 7 \sqrt3) x^2 + 9 (1 + 2\sqrt3) x + 9\sqrt3\right)\ge0$$
The dirty expression has real root between $-3$ and $-2$ then two critical points for some $x<0$ with value greater than $0$ and then it goes to +infinity. Either way it's increasing on $[0;\sqrt3]$ thus
$$x\in[0;\sqrt3]\implies 3 (8 \sqrt3 - 9) x^3 +3 (6 + 7 \sqrt3) x^2 + 9 (1 + 2\sqrt3) x + 9\sqrt3\ge 9\sqrt3>0$$
If you're bored with derivatives you can observe that function below is increasing since some negative number, thus
$$x\in[0;\sqrt3]\implies 3 (8 \sqrt3 - 9) x^2+3 (6 + 7 \sqrt3) x+ 9 (1 + 2\sqrt3) >0$$
QED
Equality iff $a=b,c=0$ (and cyclic permutations) or $a=b=c=1$[/quote]

Bravo!
By Schur, $a^3+b^3+c^3+3\left(2\sqrt3-1\right)abc\geq3\left(a+b+c\right)+3\sqrt3\left(2-\sqrt3\right)abc.$
So we just need to prove
$$\frac{a+b+c}{\sqrt3}+\left(2-\sqrt3\right)abc\geq2.$$
See here http://matinf.upit.ro/MATINF4/index.html#p=9

Note that $a^3 + b^3 + c^3 +7abc = (a+b+c)^3 - 3(a+b+c)(ab+bc+ca)+10abc$.
Consider cubic polynomial $f(t) = (t-a)(t-b)(t-c)$, translate the graph of $f$ upward (consider $f(t)+k$ with $k\geq 0$) unitl
"One root becomes zero" or "Multiple root emerges"
$a+b+c, ab+bc+ca$ is preserved and $abc$ decreases. So it suffices to consider two cases: $a=0$, $b=c$

[b] Case 1. [/b] $a=0$
$bc=3 \Longrightarrow b^3 +c^3 \geq 2(bc)^{3/2} = 6\sqrt{3} > 10$

[b] Case 2. [/b] $b=c$
Since $(2a+b)b=3$, given ineq $\iff a^3 + 2b^3 + 7ab^2 \geq 10 \left (\frac{(2a+b)b}{3} \right)^{\frac{3}{2}} $
Let $t = \frac{a}{b} > 0$. Then, the given ineq is equivalent to
\begin{align*}
&\iff 27(t^3 + 7t+2)^2 \geq 100 (2t+1)^3 \iff 27(t^6 + 14t^4 + 4t^3 + 49t^2 + 28t + 4) \geq 100( 8t^3 + 12t^2 + 6t + 1) \\
&\iff 27t^6 + 378 t^4 - 692 t^3 + 123t^2 + 156t + 8 \geq 0 \iff (t-1)^2 (27t^4 + 54t^3 + 459t^2 + 172 t+8)\geq 0
\end{align*}


LaTeX logarithms calculus function derivative algebra polynomial

How many solutions x > 2/5 does the equation log_x[(5x-2)] = 3 have?

This is your post in $ \text{\LaTeX}$:
[code]$\log_{x}(5x-2)=3$[/code]
which looks like: $ \log_{x}(5x-2)=3$.

[hide="Solution"]$ 5x-2=x^3$

$ x^3-5x+2=0$

We can find minimum and maximum points using differential calculus:
$ 3x^2=5$
$ x=\pm\sqrt{\frac{5}{3}}$. Checking, the negative root gives a maximum and the positive gives a minimum.

At $ x=2/5$ we find that the expression is a positive number. We have a minimum below the $ x$ axis approaching at $ \sqrt{\frac{5}{3}}$, so the function must dip below the $ x$-axis once.

However, the first derivative $ 3x^2-5$ is positive after $ \sqrt{\frac{5}{3}}$, so the function begins increasing (and then continues to do so), meaning that the function will go through the $ x$-axis on its way up, once again.

Therefore the answer is $ \boxed{2}$.[/hide]

Clearly not the best solution.

Because this forum says not to post calculus problems,
I don't want to add to it either by trying to post a
calculus-based solution.

$ x^3 \minus{} 5x \plus{} 2$ has two positive roots because of the
two sign changes as per Descartes' rules of signs.

At x = $ \frac25$, the polynomial has a positive value (sign).

Check x = 1.  It has a negative sign.  There is a sign
change between x = $ \frac25$ and x = 1, so there

is a root between x = $ \frac25$ and 1, and there is a
second root somewhere to the right of this first root.

HAD the sign, instead, at x = $ \frac25$ been negative,
then that would have meant that there would have been
only one root total (speaking on the positive x side, in context)
to the right of that desired x value asked about in the problem.

hello, we have only the two solutions $ x\equal{}2$ or $ x\equal{}\minus{}1\plus{}\sqrt{2}$.
Sonnhard.

You'd get those by considering the rational root theorem to find that $ \plus{}2$ is a root, then using the quadratic formula; but this is unnecessary if you plug in random numbers (which was my approach, before I tried to improve things with calculus).

For me, I did not use a random value of x.

I used the value of contention, x = $ \frac25$,
and the next integer, x = 1, to test for a
sign change.


I got my final pick of books as extra readings/studying from my Physics C class.  It's the Feynman lectures on Physics and Irodov problem book.

How do you think should I approach both books?  What is a good way to study them?

Do you think it's better to do all or most of the Irodov problems from 1 section, and then move on, or make a pick from each sections to get a more comprehensive view of physics?

Thanks.  :D

For Problems in General Physics By I.E.Irodov :->

What i Could Recommend is that Divide The Problems into Categories, DO SOme Of the Probelms and Do Mark The Important Ones. Besides, Rather than Focussing on "Hardcore(Intensive) Problem Solving" One Must Try to Grab the Ideas  Used in SOlving the Problems and Use thiese Ideas as tools in the Future

Also Try to Solve More of Generalized Probelms , as it help you to Understand The Basics Clearly

All The Best!!     :wink:


George has six ropes. He chooses two of the twelve loose ends at random (possibly
from the same rope), and ties them together, leaving ten loose ends. He again chooses two loose ends at random and joins them, and so on, until there are no loose ends. Find, with proof, the expected value of the number of loops George ends up with.

Alright, first I thought this was an accident. But then I saw that you posted all 5 USAMTS Round 1 problems! If you want the solutions, go to http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=125
BTW, it's definitely NOT getting started level either. :roll:

THIS TOPIC AND ALL OTHERS YOU JUST POSTED ARE LOCKED.

[quote="joml88"]Actually, I would appreciate it if someone posted a solution to the USAMTS problems.  For the AoPS-Mathlinks Contest Resources, we need the problems and solutions posted in threads.

Thanks.[/quote]

Ok, [hide="solution"]
For each possible time that a rope is connected with another rope at most 1 loop can be made, and only 1 choice of the end will give this loop.
After you pick any end, you have 11 choices. After you tie that, you have 9 choices, and then 7, and so on until there's only 1 left.
Thus expected value:
$1/11+1/9+1/7+...1/1=6508/3645$[/hide]

[hide="Answer"]George picks any end, then attaches it to another. He has a $\frac{1}{11}$ chance of making a loop on the first rope, but no matter what he does, he then has 5 ropes left. From there, he has a $\frac{1}{9}$ chance of making a loop, and so on, giving him $\frac{1}{11}+\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}+\frac{1}{1}\approx 1.878$ loops.[/hide]

[quote="chess64"]:huh: Well sorry about that. :blush:[/quote]

Don't feel bad. I accidentally criticized Silverfalcon for posting large amounts of AMC problems in the Getted Started Forum. That's even worse. :blush:


can someone please write the code that will make the pdfs have a 1 inch margin on all sides? o, and what is the corresponding 12 font size (in MS Word) in latex?


Let $n\in\mathbb{N}^*$. $2n+3$ points on the plane are given so that no 3 lie on a line and no 4 lie on a circle. Is it possible to find 3 points so that the interior of the circle passing through them would contain exactly $n$ of the remaining points.

let $A, B$ be two points of the set such that all other points of the set lie on one side of the line $AB$. (e.g, take two consecutive vertices of the convex hull.) consider all the $2n+1$ circles through $A, B$ and the other points. the intersections of interiors of these circles with the halfplane on which the points lie can be ordered: $I_1 \subset I_2 \subset ... \subset I_{2n+1}$. the circle whose interior is $I_{n+1}$ contains exactly $n$ points in its interior.

It is a well known problem. It has even appeared in Mathematical Olympiad Challenge, by Titu Andreescu and Razvan Gelca. :)

it was also proposed for the 1995 IMO i think


function
vector
integration
calculus
real analysis
real analysis unsolved

hi all,

here is a cool pb:
find the minimum value of
  int(f^2+f'^2,x=0..1) with f'' continuous, f(0)=a, f(1)=b.

bye

Minimum is obtained for a unique function solution of $y''-y=0$ with
$y(0)=a, y(1)=b$

The answer Moubinool has provided is correct. Here's a proof that it is correct, which will also show how it could have been found.

Consider the vector space of all smooth enough functions on [0, 1] with the following inner product: $<u,v>=\int_0^1 uv+u'v'$.
The norm in this inner product product space is then $||u||^2 = \int_0^1 u^2+(u')^2$ This is a reasonably well known norm: a Sobolev norm.
Let $Z$ be the subspace of this inner product space defined by $y\in Z \iff y(0)=0$ and $y(1)=0$. Let $h$ be any convenient function (the straight line will do) such that $h(0)=a$ and $h(1)=b$.

Our problem can now be phrased as follows: find $y\in h+Z$ that minimizes $||y||^2$.

Suppose $u\in Z$. Then $<u,v>=\int_0^1 uv+u'v'$. Do the following integration by parts: $\int_0^1 u'(x)v'(x)dx = u(1)v'(1)-u(0)v'(0)-\int_0^1 u(x)v''(x)dx$.
Since $u(0)=u(1)=0, \int_0^1 u'v' = -\int_0^1 uv''$.
From this, we get $<u,v>=\int_0^1 u(v-v'')$.

Now, suppose we can find a $y$ such that $y\in h+Z$ and $y-y''=0$ everywhere. Of course, that is exactly what Moubinool has asked us to find. As the unique linear combination of $e^x$ and $e^{-x}$ that satisfies the boundary conditions, it can be written out explicitly without too much trouble. From the previous paragraph, we know that $<u,y>=0 \forall u\in Z$.

If $w\in h+Z = y+Z$, then it can be written as $w=y+u, u\in Z$.
$||w||^2 = ||y+u||^2 = ||y||^2+2<u,y>+||u||^2=||y||^2+||u||^2$. This is minimized when $u=0$, which is the same as $w=y$.


calculus
integration
analytic geometry
trigonometry
logarithms
real analysis
real analysis unsolved

$ I \equal{}\int_{0}^{1}\int_{0}^{1}\frac{dx\, dy}{4 (x\plus{}y)\sqrt{(1\minus{}x)(1\minus{}y)}}$ is after substitution $ x \equal{} 1\minus{}u^{2}\; ,\; y \equal{} 1\minus{}v^{2}$

$ \int_{0}^{1}\int_{0}^{1}\frac{2u\, 2v\, du\, dv}{4\, (2\minus{}u^{2}\minus{}v^{2})\,\sqrt{u^{2}v^{2}}}\equal{}\int_{0}^{1}\int_{0}^{1}\frac{du\, dv}{2\minus{}u^{2}\minus{}v^{2}}$

Because of $ f(u,v) \equal{}\frac{1}{2\minus{}u^{2}\minus{}v^{2}}\equal{} f(v,u)$ the integral can be written as $ 2\int_{0}^{1}\int_{0}^{v}\frac{1}{2\minus{}u^{2}\minus{}v^{2}}du\, dv$

Now use polar coordinates $ u \equal{} r\cos\varphi\; ,\; v \equal{} r\sin\varphi$ .

$ I \equal{}\int_{0}^{\frac{\pi}{4}}\int_{0}^{\sec\varphi}\frac{2r}{2\minus{}r^{2}}dr\, d\varphi \equal{}\int_{0}^{\frac{\pi}{4}}\left[\minus{}\ln(2\minus{}r^{2})\right]_{0}^{\sec\varphi}d\varphi \equal{}\int_{0}^{\frac{\pi}{4}}\left[\ln 2\minus{}\ln(2\minus{}\sec^{2}\varphi)\right] d\varphi$

$ \equal{}\frac{\pi}{4}\ln 2\minus{}\int_{0}^{\frac{\pi}{4}}\left[\ln(2\cos^{2}\varphi\minus{}1)\minus{}\ln(\cos^{2}\varphi)\right] d\varphi$ , whereas $ 2\cos^{2}\varphi\minus{}1 \equal{}\cos(2\varphi)$

Hence $ I \equal{}\frac{\pi}{4}\ln 2\minus{}\underbrace{\int_{0}^{\frac{\pi}{2}}\ln(\cos\varphi)\frac{d\varphi}{2}}_{\minus{}\frac{\pi}{4}\ln 2}\plus{}\int_{0}^{\frac{\pi}{2}}\ln\left(\cos^{2}\frac{\varphi}{2}\right)\frac{d\varphi}{2}\equal{}\int_{0}^{\frac{\pi}{2}}\ln\left(2\cos\frac{\varphi}{2}\right) d\varphi$

$ \equal{}\sum_{k \equal{} 1}^{\infty}\frac{(\minus{}1)^{k\minus{}1}}{k}\int_{0}^{\frac{\pi}{2}}\cos k\varphi\, d\varphi \equal{}\sum_{n \equal{} 0}^{\infty}\frac{(\minus{}1)^{n}}{(2n\plus{}1)^{2}}\equal{} G$


inequalities
invariant
LaTeX
inequalities unsolved

Please, Help me. I can not find such terms. Please see this. http://olympiads.win.tue.nl/imo/imo2003/solutions/sol5.jpg

At the end of the proof we find that $ \Sigma{x_i}\equal{}0$ becasue $ x_i\equal{}\frac{d}{2}(2i\minus{}n\minus{}1)$. This can get by translating each $ x_i$ by the quantity $ \minus{}\frac{1}{2}(x_1\plus{}x_n)$ The question is how did we get this term $ \frac{1}{2}(x_1\plus{}x_n)$ at the end of the page. How about take $ x_i\equal{}\frac{d}{2}(2i\minus{}n\minus{}1)\plus{}\frac{1}{2}(x_i\plus{}x_{i\minus{}n\minus{}1}) , 1 \leqslant i \leqslant n$ or $ x_i\equal{}\frac{d}{2}(2i\minus{}n\minus{}1)\plus{}\frac{9}{10}(x_3\plus{}x_{n\minus{}5})$ And translate it later, we can get $ \Sigma{x_i}\equal{}0$ too????

And about translating. We find that the inequality is invariant (to put $ x\plus{}a$ to each side), so we can choose a constant $ a$ such that $ \Sigma{x_i}\equal{}0$ How can I find such constant that belongs to this IMO problem. And this's example one...

Let $ x_1,x_2,x_3$ be reals, Prove that

\[ (x_1\minus{}x_2)^2\plus{}(x_2\minus{}x_3)^2\plus{}(x_3\minus{}x_1)^2 \geqslant (x_1\minus{}x_2)(x_2\minus{}x_3)\plus{}...\plus{}(x_2\minus{}x_3)(x_3\minus{}x_1)\]

Above problem we know it is true, Can we assume that $ x_1\plus{}x_2\plus{}x_3\equal{}0$ ??? (By choosing the quantity $ k$ to get $ \Sigma{x_i}\equal{}0$ And translate it)

This IMO problem took me so long and I've alrealdy seen this one http://www.mathlinks.ro/Forum/viewtopic.php?p=265#265. (My computer can't show the Latex of this link.) The last question is "Do This IMO problem has a general form"

Sorry for my bad ENG :(  :(


For every real number $a$ find all real numbers $x$ which satisfy the following equation:
\[ (2x+1)^4 + ax(x+1) - \frac{a}{2} = 0  \]

It seems to be a difficult problem.

Is there any hint?  :)

[hide="Am I missing something?"]
Since the equation holds for all real values of $a$, we must have $(2x+1)^4+\frac{x}{2}=0$ and $x(x+1)=0$.

$x(x+1)=0$ gives $x=0,-1$, but neither satisfies $(2x+1)^4+\frac{x}{2}=0$, so we have no solution? :?
[/hide]

Edit: I guess I misread the problem...

I'm terribly sorry, I wrote the wrong equation  :blush:  :blush:  :blush:

Must have been a bad photocopy or something (I now found the correct equation on the net)

But now it's easy

No wonder I got very ugly solutions on the problem....
So, with the corrected problem, we may let $u=2x+1$. Then
$u^4+a\cdot (\frac{u-1}2)(\frac{u+1}2) -\frac a 2 = 0$
$u^4+\frac a 4 \cdot {u^2}- {\frac 3 4} a = 0$, which can be easily solved by quadratic formula.


topology
MIT
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Harvard
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Support

Hello,

I am a rising senior and (of course :P) I love mathematics.  At the moment I have very little knowledge about the top math colleges, their differences, and what chances I really have of getting into them.  I figured now would be a good time to start figuring such things out.  I often use the AoPS forum as a source of contest problems, and I see how much experience people here have, so I thought it would be a good place to ask for some initial advice.  Could someone please give me suggestions about what departments I should look into, and which might make the best fits for my interests?

In my ideal college, I would be able to:

(a) Take informative classes that gave interesting and challenging problem sets

(b) Focus on mathematics with minimal emphasis on general education requirements.  I'd love to learn as much pure math as possible, and also possibly some physics or chemistry.  At least in high school, I find other classes interesting, but extraordinarily time-consuming and not at all as interesting as mathematics.  In a large sense, I think I'd rather skip undergrad and go straight to graduate school, but I doubt that is possible  :( .

(c) Find a professor willing to work with me on an undergraduate research project of some type

(d) Receive credit for or be able to skip mathematics I already know (see below)

I have had the good fortune to be able to take classes at a reasonably rigorous local institution (someone in the department told me it was ranked as one of the top 30 grad programs in pure math but not much higher).  By the end of my senior year, I will have taken official classes on a variety of pure math topics, including Measure Theory, Harmonic Analysis, Functional Analysis, Complex Analysis, Commutative Algebra, Lie Algebras, Lie Groups, Smooth Manifolds, and Algebraic Topology.

Any comments or suggestions about good colleges would be very much appreciated.  Thanks in advance!

I'll give you some short details about MIT, as I think that should be among the list of schools you apply to.

Since flexibility in choosing courses is important, here are the required course MIT students have to take:

Calc I, Calc II (it looks like you've got these covered)
Chemistry
Biology
Physics I, Physics II (you'll get credit for physics I: mechanics but you'll have to test out of physics II if you don't want to take it)

Then there's the HASS requirement (Humanities, Arts, Social Sciences).  You have to take 8 HASS courses.  There are five Hass categories:

Literary and Textual Studies
Language, Thought, and Value
Visual and Performing Arts
Cultural and Social Studies
Historical Studies

It's actually not as bad as it might sound.  There are tons of interesting classes to take.

MIT has a really good program for undergraduate research called UROP (undergraduate research opportunity program).  So a good deal of the professors are used to working with undergrads on research.

I can't really say that MIT is the best fit for you.  There are a lot of other factors that could play a role.  However, I can tell you that MIT would be, at the least, a very good fit for you.

I'm not extraordinarily well versed on all the top math universities in the nation, but I can say that at a tech school you'll be able to focus on math much more than at a liberal arts school.  I'd say to also look into Caltech, UChicago, UMichigan, Harvard, Princeton, Berkeley and maybe Duke.  Harvard and Princeton might require more liberal arts than you're looking for, but I don't know how much they require you to take.  That'd be something I'd be interesting in learning about (either by someone posting here ;) or I'll just go look it up sometime on my own).

Sounds like you should be able to get into almost any college you'd want to. ie. If any college rejected you, it wouldn't be because your accomplishments are deficient, but rather the top colleges' admissions are quite "random"

At Berkeley, the math department belongs to the "Arts and Sciences" college. The "breadth" requirements are seven classes, each from a fairly long list. Some are science, some are humanities- and they won't take transfer/AP for any of that. If you have the equivalent of a course in a breadth list, just take another class. There are also a few university-wide requirements, like a writing class (can be covered by a good enough AP English score) and "American Cultures", which is probably best handled by matching with a breadth class.

As for the math you have- frankly, that's most of the introductory graduate courses already. You need a deep graduate program, and I have no idea what an advisor would do with you.
You will stand out in any applicant pool- you've got the superlatives.

Um. Why don't you go to Princeton grad school?

[i]Nine[/i] graduate-level math courses? [i]Not[/i] self-studied?

[i]How...?[/i]

And I'm no expert in college admissions, but, if you fail to get into a good college with those credentials, I'm pretty much hosed.

It's not impossible to skip undergrad altogether.  I've heard of at least one example (Aleksandr Khazanov, who went straight to grad school at Penn State).  If you're better than most of the grad students at your "top 30" department, then certainly some math grad schools would be happy to have you.  This is especially true if there are some profs who are impressed with your work and support your desire to skip undergrad.  Getting into a "top 5" department might be a bit trickier, but I really don't know.  All that said, I'm a little surprised that anyone would *want* to skip undergrad.  (I'm not judging you.  I just find it surprising.)

Assuming that you do a bachelor's degree, with a background as strong as yours, you basically have two options: (1) try to "hone" your skills with yet more classwork, both wide and deep, or (2) try to jump into research sooner rather than later.  These are not incompatible, but if you want to emphasize (1), then you MUST go to a "top 5" department.  Frankly, other departments will have nothing to teach you, as far as coursework goes.  I do know of people who had super-strong backgrounds and still managed to learn a lot as undergrads at places like Harvard-MIT-Princeton.  If you want to emphasize (2), then you should choose a research area and choose a school with good prospective advisors.  You should even talk to these people before making a decision.

The big thing to note here is that in either case (1 or 2), the gen ed requirements are really of secondary importance.  Even if all you care about is math, you can always just blow off your other classes.  It takes minimal effort to get C's in anything, and grad schools don't care about those grades.

If you don't mind going international(Canada):

 [url=http://www.mun.ca/]Memorial University of Newfoundland[/url] covers your requirements quite well.

[quote](a) Take informative classes that gave interesting and challenging problem sets

(b) Focus on mathematics with minimal emphasis on general education requirements. I'd love to learn as much pure math as possible, and also possibly some physics or chemistry. At least in high school, I find other classes interesting, but extraordinarily time-consuming and not at all as interesting as mathematics. In a large sense, I think I'd rather skip undergrad and go straight to graduate school, but I doubt that is possible Sad .

(c) Find a professor willing to work with me on an undergraduate research project of some type

(d) Receive credit for or be able to skip mathematics I already know (see below) [/quote]

 (a) http://www.mun.ca/regoff/calendar/sectionNo=SCI-1238#SCI-1332

 (b) Its your lucky day, at Memorial University of Newfoundland, you only have to take 2 Arts courses, English 1000 and 1001. The only other requirements for the first year are 2 Mathematics courses(Math 1000 and 1001) and 2 courses in 2 different sciences other than Mathematics(Physics and Chemistry?). All the rest is your choice, however you do have certain Mathematics and Science courses you must complete(All relevant).

 (c) If you are good, you could be drafted for assisting in research. You could also end up working in the Mathematics field in the summer while you are off. Also, students registered under the Honours program in Mathematics may conduct a research project under the supervision of a faculty member. This is a requirement of all Sciences, besides Mathematics. Although you may be able to do it if you have something that you can get faculty support on.

 (d) I have no idea.

http://www.math.mun.ca/index.php3

http://www.mun.ca/regoff/calendar/sectionNo=SCI-1238

[quote="jigger"]If you don't mind going international(Canada):

 [url=http://www.mun.ca/]Memorial University of Newfoundland[/url] covers your requirements quite well.
[/quote]

The OP has already taken more challenging graduate course than are offered at said university, at least from a brief glance.  More appropriate Canadian universities would probably be the University of Toronto or UBC.

[quote]4199

Applied Mathematics Honours Project

requires the student, with supervision by a member of the department, to prepare a dissertation in an area of Applied Mathematics. Although original research work by the student will not normally be expected, the student must show an ability and interest to learn and organize material independently. A one hour presentation at the end of the semester will be given by the student.

Prerequisite: Registration in an Honours or Joint Honours program in Applied Mathematics.
[/quote]

[quote]4399

Pure Mathematics Dissertation

requires the student, with supervision by a member of the department, to prepare a dissertation in an area of Pure Mathematics. Although original research by the student will not normally be expected, the student must show an ability and interest to learn and organize material independently. A one hour presentation at the end of the semester will be given by the student.

Prerequisite: Registration in an Honours or Joint Honours program in Pure Mathematics.
[/quote]

 You can combine the three types of courses into a Mathematics degree. Plus, research is an open opportunity for all science students if they are good enough. The faculty are quite good with projects and such. However, its not a mathematics intensive university, but if you read the course descriptions, there is a bit of subject matter covered in each course(most last about 36 hours over 3 months). Plus, courses are switched up, and only those courses that have been taught last year and this year are displayed. So, they can change up a bit.

 Of course, I personally won't get a mathematics degree, but I thought I'd just put up a university that does meet his criteria.

  Mathematics in Canada is basically dictated at UWaterloo. So, it s a good choice if you want a superior mathematics program.

[quote="jigger"]Mathematics in Canada is basically dictated at UWaterloo. So, it s a good choice if you want a superior mathematics program.[/quote]

This isn't true; Waterloo is great for computer algebra and combinatorics and optimization and the like, but there are definitely better choices for analysis or PDE or probability, for example.

[quote="yenlee"]It's not impossible to skip undergrad altogether.  I've heard of at least one example (Aleksandr Khazanov, who went straight to grad school at Penn State).[/quote]

I hope the original poster is still "listening", but I just thought I'd throw out another guy who went directly from high school to grad school -- IMO Silver Medalist and Putnam Fellow Brian Hunt.  You could probably get in touch with him easily enough if you want to know more of the details of how and why he did it.  He's currently a professor at U. of Maryland.

Brian and I were on the same Putnam team at U. Maryland, and he was indeed brilliant.  I think our team finished 3rd and 5th in the country in those two years.  My last year, the team was me, Brian, and another high-school kid; at age 18, I was the senior member of the team.


Prove that there exists a 101 digit string which contains exactly one copy of every possible 2-digit string over the alphabet {0,1,2,3,4,5,6,7,8,9}.

Are you sure this problem is correct? There are $\frac{10!}{8!}= 90$ possible 2-digit strings, but the string in question has 100 2-digit strings.

mathisfun1, you forgot to count 2-digit strings with two equals digits.

Solution to the original problem.
Let's construct a de Bruijn graph on ten vertices corresponding to the digits from the alphabet that have all possible directed arcs (including self-loops), representing 2-digit strings formed by endpoints of each arc. It is easy to see that in-degree and out-degree of each vertex is 10. Therefore, there exist an Eulerian path in this graph. A required 101-digit string is read along this path.

Could you direct me to a resource where I can get the backup reading to understand your proof?


geometry
geometric transformation
rotation
group theory
superior algebra
superior algebra solved

state true or false and justify.
if a and b are elements of a group G such that o(a)=2 and o(b)=3 then o(ab)=6

False.

[hide]Consider the dihedral group with 6 elements $D_3=<a,b|a^3=e,\ b^2=e, ab=ba^2>$. Then $(ab)^2=ba^2 \cdot ab=beb=b^2=e$. Thus,$|ba^2|=2\not=6$.[/hide]

That's false. There's no good way to predict the order of $ab$ from the orders of $a$ and $b$ in a nonabelian group. For an example, take the smallest nonabelian group, S_3. There are elements of order 2 and elements of order 3, and you can multiply them, but there are no elements of order 6.

Another example to drive the point home. Let $G$ be the group of motions of a Rubik cube. Let $F$ be the action of rotating the front face 90° clockwise, and let $R$ be the action of rotation the right hand face 90° clockwise. Clearly, $o(F)=o(R)=4,$ but $o(FR)=105.$

You can even take the free group generated by $a,b$ which satisfy $a^2=b^3=1$. In that case, $ab$ has infinite order.


If $\tan{(\alpha-\beta)}=\sin{2\beta}$. Prove:
\[\tan{\alpha}+\tan{\beta}=2\tan{2\beta}\]

$\tan \left( {\alpha  - \beta } \right) = \sin 2\beta  \Leftrightarrow \frac{{\tan \alpha  - \tan \beta }}{{1 + \tan \alpha \tan \beta }} = 2\sin \beta \cos \beta  \Leftrightarrow $

$ \Leftrightarrow \tan \alpha  - \tan \beta  = 2\sin \beta \cos \beta  + 2\sin ^2 \beta \tan \alpha  \Leftrightarrow $

$ \Leftrightarrow \tan \alpha  \cdot \left( {1 - 2\sin ^2 \beta } \right) = \tan \beta  + 2\sin \beta \cos \beta  \Leftrightarrow $

$ \Leftrightarrow \tan \alpha  \cdot \left( {1 - 2\sin ^2 \beta } \right) = \tan \beta  \cdot \left( {1 + 2\cos ^2 \beta } \right) \Leftrightarrow $

$ \Leftrightarrow \tan \alpha  = \tan \beta  \cdot \frac{{1 + 2\cos ^2 \beta }}{{1 - 2\sin ^2 \beta }} \Leftrightarrow \tan \alpha  = \tan \beta  \cdot \frac{{3 + \tan ^2 \beta }}{{1 - \tan ^2 \beta }} \Leftrightarrow $

$\tan \alpha  = 4 \cdot \frac{{\tan \beta }}{{1 - \tan ^2 \beta }} - \tan \beta  \Leftrightarrow \tan \alpha  + \tan \beta  = 2\tan 2\beta $


How many of the AoPS books do you own?

I own 2 of them.

[hide="Guess How Many I Own???"]None.  :P [/hide]

I currently own 2; AoPS volumes 1 and 2. I might get Intro to NT for my birthday but it's only intro...

Intro Algebra
intro geo
intro c and p
intro NT

Let me count.
1
1
2
3
5
8
13
21
I own 21.

I own Volume 1, and Intro to Algebra, and I really want Intro to Geometry.


[quote="bpms"]I can't send attachments by pm? Is this my problem?[/quote]
Actually, I'm pretty sure that it's just not allowed to send attachments in a PM, so it's not just you.  :wink:

EDIT: Never mind, DPatrick proved me false.  :roll:

You can send attachments in PMs, but they have to be really small  (less than 0.5 MB).


inequalities
function
algebra
polynomial
inequalities proposed

Try the puzzle method? Maybe you've already tried it but I think it looks promising, you say it is very strong so probably weighted sum of other ineq's won't work...

[quote="Albanian Eagle"]have you tried cauchy ? you will still have some expanding work to do but at least you will be dealing with 4-th degree rather than 6-th degree.[/quote]
Cauchy how? If I do
$\frac{a^{2}+ab+b^{2}}{a^{2}+(b+c)^{2}}= \frac{(a+b)^{2}}{a^{2}+(b+c)^{2}}-\frac{ab}{a^{2}+(b+c)^{2}}$ and do Cauchy on teh first part, I think I lose.

I know that
$\sum_{cyc}\frac{a^{2}+b^{2}}{a^{2}+(b+c)^{2}}\ge \frac{6}{5}$
by methods from before, but that leaves
$\sum_{cyc}\frac{ab}{a^{2}+(b+c)^{2}}\ge \frac{3}{5}$
and I don't think this is true (take one variable equal to 0).

If we can prove (I haven't tried but it's probably true) that
$\sum_{cyc}\frac{ab}{a^{2}+(b+c)^{2}}\le \frac{3}{5}$
then
$\sum_{cyc}\frac{(a+b)^{2}}{a^{2}+(b+c)^{2}}\ge \frac{12}{5}$
may be a more approachable inequality even though it is weaker than the original one, because the numerator is easier to work with. I guess I will work on it...

I have not much time for it but the first should be easy. Without loss of generality, we let $a^{2}+b^{2}+c^{2}=1$, it is suff to prove that \[\frac{ab}{1+2ab}+\frac{ab}{1+2ab}+\frac{ab}{1+2ab}\leq \frac35.\] Note that \[{\frac{ab}{1+2ab}-\frac12}=\frac1{1+2ab}.\] In the end, we need to prove \[\frac1{1+2ab}+\frac1{1+2bc}+\frac1{1+2ca}\geq\frac95.\] which follows from AM-HM and the fact that \[a^{2}+b^{2}+c^{2}\geq ab+bc+ca.\] As for the second one, Cauchy works.

[quote="Xevarion"]Given that $a, b, c \ge 0$, prove that

$\frac{a^{2}+ab+b^{2}}{a^{2}+(b+c)^{2}}+\frac{b^{2}+bc+c^{2}}{b^{2}+(c+a)^{2}}+\frac{c^{2}+ca+a^{2}}{c^{2}+(a+b)^{2}}\ge \frac{9}{5}$

I am pretty sure that this is true, but I think it is pretty strong and I don't know what to do with it. Any ideas that don't involve expanding? :)[/quote]
Xevarion maybe this way can help :blush: .
$\frac{a^{2}+ab+b^{2}}{a^{2}+(b+c)^{2}}+\frac{b^{2}+bc+c^{2}}{b^{2}+(c+a)^{2}}+\frac{c^{2}+ca+a^{2}}{c^{2}+(a+b)^{2}}=\frac{(a^{2}+ab+b^{2})^{2}}{(a^{2}+(b+c)^{2})(a^{2}+ab+b^{2})}+\frac{(b^{2}+bc+c^{2})^{2}}{(b^{2}+(c+a)^{2})(b^{2}+bc+c^{2})}+\frac{(c^{2}+ca+a^{2})^{2}}{(c^{2}+ca+a^{2})(c^{2}+(a+b)^{2})}\geq ...$
Cauchy :D.
Afterward I supposed a^2+b^2+c^2=1,to make easier my work with expretions.Finally I showed that we need to prove that
$2(a^{4}+b^{4}+c^{4})+9(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})+11(a^{3}c+b^{3}a+c^{3}b)\geq 15abc(a+b+c)+7(a^{3}b+b^{3}c+c^{3}a)$
But I'm not quit sure if this is true :maybe:,I even tried to prove it with using SOS,and I meneged to represant it in the basic form of SOS but it didn't make sense :( .

[quote="me@home"] \[2\sum a^{4}+9 \sum a^{2}b^{2}+11\sum a^{3}c \geq 15\sum a^{2}bc+7\sum a^{3}b\] for $\sum$ being cyclic
We are looking for some clever am-gm's to add together \[k\sum a^{4}+l \sum a^{2}b^{2}+m\sum a^{3}c\geq 15\sum a^{2}bc \\ n\sum a^{4}+o \sum a^{2}b^{2}+p\sum a^{3}c\geq 7\sum a^{3}b\] would imply the inequality \[k+l+m=15, n+o+p=7, k+n=2, l+o=9, m+p=11\] I am having issues solving this[/quote]
Are you sure you can find proper solution for this sistem?
Also a note,I'm suer if there is such a solution then in that  solution $k=0,n=2$.

How did you come upon that? (I am curios)

It's not hard to do.First we can easely see that $k,n\geq 0$,because if for

example we have that k<0,leting that $a,b\to 0$,we will see that from the

firs ineq follow that $c^{4}\leq 0$.Now we also have that $k+n=2$,and as

you can see with the first ineq we havn't got any troubles because

$\sum{a^{2}b^{2}}\geq abc(a+b+c)$ and $\sum{a^{3}c}\geq abc(a+b+c)$,the

difficult part is the second ineq,so I'm leting there the coificent of the

bigest sum $\sum{a^{4}}$ to take maximum possible valu and it is 2.


Simplify, where M, N, and R are real numbers such that the following expression is not undefined.

$\displaystyle \left(\frac{M^2(1-R^2)+2MN+N^2}{M^4(1-R)+M^3N}\right)\displaystyle \left(\frac{M(1+R)-N}{M^3(1+2R)+M(M^2R^2-N^2)}\right)$


Is there a "window" in which one can take the AMC 10 test, like with the AMC 8?
I haven't taken the test before, and I would like to take it in 2010, but my high school doesn't offer the test, sadly. =/

My middle school doesn't offer the AMC 8, 10, or 12.  :(

I had to take the AMC 10 at a local university with a homeschooled group.  Is this an option for you?

As for the "window" I don't think so; I think the test dates are more strictly scheduled.

So if they don't offer it, then are we just supposed to take it somewhere else? Where/when/how would you do that? I have not taken the test before, and am thinking about taking it this school year....

I think you have to take it on that exact date. Also, who takes the test where is up to the people proctoring the test. There is a public list of open sites where the B test (for both the 10 and 12) are taken [url=http://www.unl.edu/amc/b-registration/b1-archive/2009-2010/2010-HS-CU-list.shtml]here[/url]


An object with mass $m$ is attached to a spring with spring constant $k$ and made to move in a simple harmonic. The angular speed is $\omega_{1}$. The spring is then cut into three equal parts and connected as in the picture. The mass $m$ is then attached and again made to move in a simple harmonic. The angular speed is $\omega_{2}$. Find $\omega_{1}: \omega_{2}$

$\frac{d^{2}x}{dt^{2}}+\frac{k}{m}x=0$

$w_{1}=\sqrt{\frac{k}{m}}$

$k=\frac{EA}{l_{0}}$

$l_{0}=\frac{EA}{k}$

$\frac{l_{0}}{3}=\frac{EA}{k_{x}}$
E-is Young Modulus
We take the ratio of these to equations and get the following:

$k_{x}=3k$

It means that the new spring constants are $3k$.

Now we add the parallells,

$3k+3k=6k$

$\frac{1}{6k}+\frac{1}{3k}=\frac{1}{2k}$

$\frac{d^{2}x}{dt^{2}}+\frac{2k}{m}x=0$

$w_{2}=\sqrt{\frac{2k}{m}}$


YES ME!!!!!!

my friend literally memmed a whole FFR stage. he drank soda and played at the same time and played cards w/ one hand too--he got a perfect score

lawl, for FFR practice does make perfect.  I'm just about at the stage where I can survive the hardest standards and the easiest tricky's...

i LOVE FFR..
I nearly memorized this one song HTIJ or something like that.. it's page 6 in dance..

if anyone wants to challenge me, i'm kimmystar94...

i can get up to difficult.. :)

Oh my god, I suck so much at these games...I played DDR once, and got 250 misses, 40 perfect, 50 good, 100 bad, or something like that.

i wish i had real DDR.. I'd be so good by now..
and it's REALLY good exercise.. at least if you play standard or higher...

DDR is so much fun.. and it tests your music skills, too.. :)

i'm ok on DDR and FFR

i can do most standards (pass, not get anywhere near 90%  :D )

some trickys

DDR--okay, last time i had as many misses as perfects, so w/e

http://www.stepmania.com/

http://zenius-i-vanisher.com/v4/simfiles.php

+ a USB dance pad

====================

or if you prefer to play on the arrow keys you can ditch the dance pad
(or you can make it to a pseudo-beatmania type thing by assigning the arrow keys to ASKL or something)

====================

FFR is too easy to clear in my opinion - you only need to hit 50% of the notes in order to clear. (real DDR and stepmania are more on the order of hit 90% of the notes at a minimum, assuming no freezes)

A little off topic, but if you have one of those games on computer, you could get a 4 dollar keyboard, do away with the keys, and hook up homemade touch sensors to the 4 correct keys, then play :D Need a hard floor though.

wait.. a USB Dance Pad??

Where can you get one of these??

I'm mediocre at both Stepmania (gr, my keyboard sucks, so I can't play 3 arrows at once) and real DDR. But my real DDR is Mario Mix, which is jacked up and ridiculously easy, so I don't get much good practice.

What's FFR?

I'm okay at DDR, because I have the Wii version at home.

FFR is the online version of DDR with the arrow keys.  It's at flashflashrevolution.com.

lol I'm ranked 158,228th place out of all of FFR  :D

meh, not too good considering I'm asian  :P

EDIT: 153,017 OOOH beat that  :rotfl:

I don't play FFR, but I do have DDR.

I like ITG (in the groove) better though.

Heh, here's a video of me showing off at ITG (I'm not that good).

[youtube]gL7eOH1JTak[/youtube]

meh i'm pretty bad at ddr

like i can AA sightread every song out there probs but i'm feeling out of shape and all

i'm decent at stepmania/ffr though

i'd be impressed if any of you can beat me :)

I pretty much suck at DDR / anything on a pad. I can [i]pass[/i] most songs on standard (unless they're of the evil keyboard sort), but nowhere near a good score.

On a keyboard, it's somewhat different. I can get AA on Tsuhsuixamush on heavy (this is one of my favorite songs), but still only a C on challenge :(. Also, I consistently get Bs on quasar, and I've managed one A.

[quote="infinity4ever"]lol yep i can get like a 150 combo on the normal parts, and then when there's a blur of arrows i just randomly press keys  :rotfl:  :rotfl:  :rotfl:[/quote]

well.. in the blur of arrows part (not the end), i actually got it that time.. it was really lucky...

my finger was dying, and my parents stared at me like i was a freak.. :)

my god, anderson!

you are good...

ITG is like the exact same thing as DDR.. isn't it?? (just different company)

I can do up to standard on DDR. I'm good compared to my friends. (I have physically inept friends)..

HOLY CRAP ANDERSON

that's one of my favorite songs  :D   good job!!!!!!!

I remember randomdragoon talking about a USB dance pad.

is it like normal size?? (like normal DDR size), and if so, how can I get it?
(and please don't say eBay b/c I don't trust ebay)

http://www.flashflashrevolution.com/profile/Potatoes/

yeah, I'm pretty bad :\

oh. my ranking is around

110,000..

teehee!!

http://www.flashflashrevolution.com/profile/kimmystar94/

Ha, my ranking is exactly 86,540 as of now.

[url]http://www.flashflashrevolution.com/profile/ksh10/[/url]

This game is so addicting.  I can do trickies now!!!!  :D

@infinity:
it's b/c you've been playing for a longer time than i have.

i can do tricky + difficult.

did u pass the bonus yoshi round for the zelda song yet?

Ehh whats that?

Also btw how is ranking calculated?  Is it for how many points you have or credits or skillz or what?

Everyone else is doing it...[url]http://www.flashflashrevolution.com/profile/Hamster1800/[/url]

wee! just FCd Hungarian Dance #5 (in classical) 563 steps :)

nice

i'm pretty sure i can AAA songs that (at hardest) difficult and FC like 90% of the songs :P

the 2.something years isn't so accurate though; it's on-and-off playing

by the way, this song called crystalis - house leaves (long) or something like that has a MAD catchy beat

it's second page, lower-right in the arcade section

i'm trying to AAA it but i can't, lol

meh...
I tried it without logging on (my parents don't let me get usernames/passwords except for AoPS).
I AAAed the first song, but that was about it.
I got as high as easy =(


There are several inequalities of Suranyi type around. I've tried to collect them all together and write them in a uniform and homogenous way. May this help for further conjectures  :)  :)

If $p_{k}$ denotes the $k$th power mean of the positive reals $x_{1},...,x_{n}$, we can write them as follows:

(1) [b]Suranyi[/b]: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14008 \[(n-1)p_{n}^{n}+p_{0}^{n}\ge np_{1}p_{n-1}^{n-1}\] (1a) generalized by [b]hungkhtn[/b] http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=turkevici&t=87914
  for $\frac{n}{n-1}\le k\le n$, \[(n-1)p_{k}^{k}+p_{0}^{k}\ge np_{1}p_{k-1}^{k-1}\] (1b) generalized by [b]zaimi-marku[/b] http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=89757 post#5
  for all $k\in\mathbb R$, \[(n-1)p_{k}^{k}+p_{0}^{n}p_{k-n}^{k-n}\ge np_{1}p_{k-1}^{k-1}\] (1c) generalized by [b]Vasc[/b] http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90872
[hide="[color=green"]writing $s_{k}$ for the $k$th symmetric mean,[/color]]defined by $s_{k}^{k}: ={n\choose k}^{-1}\sum_{i_{1}<\cdots<i_{k}}x_{i_{1}}\cdots x_{i_{k}}$[/hide] \[(k-1)p_{k}^{k}+{n-1\choose k-1}s_{k}^{k}\ge np_{1}p_{k-1}^{k-1}\] (2) [b]sss[/b]: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39922 \[\frac{p_{1}p_{n-1}^{n-1}}{p_{0}^{n}}+(n-2)\ge(n-1)\frac{p_{1}}{p_{-1}}\] (3) [b]Vasc[/b]: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14906 \[\frac{p_{n}^{n}}{p_{0}^{n}}+(n-1)\ge n\frac{p_{1}}{p_{-1}}\] (3a) [b]dyd [/b](corollary of that): http://www.artofproblemsolving.com/Forum/viewtopic.php?p=633674#633674 \[\frac{p_{n}^{n}}{p_{0}^{n}}+n\ge (n+1)\frac{p_{0}}{p_{-1}}\] (4) [b]Harazi / Popa Calin[/b]: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=21652#p21652
for $1\le k\le n$, \[p_{n}^{k}-p_{0}^{k}\ge \left(\frac n{n-1}\right)^{k-1}(p_{1}^{k}-p_{0}^{k})\ \Longleftrightarrow \ \left(\frac{p_{n}^{k}}{p_{0}^{k}}-1\right) \ge \left(\frac n{n-1}\right)^{k-1}\left(\frac{p_{1}^{k}}{p_{0}^{k}}-1\right)\] (4a) [b]Vasc [/b] [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123604[/url]
for $2 \leq k \leq n+2$,
\[p_{k}^{k}p_{1}^{n-k}-p_{0}^{n}\ge \left(\frac n{n-1}\right)^{k-1}(p_{1}^{n}-p_{0}^{n}) \ \Longleftrightarrow \ \frac{p_{k}^{k}}{p_{1}^{k}}-\frac{p_{0}^{n}}{p_{1}^{n}}\ge \left(\frac n{n-1}\right)^{k-1}\left(1-\frac{p_{0}^{n}}{p_{1}^{n}}\right)\] (5) [b]Harazi[/b]: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14906 \[\left(\frac n{n-1}\right)^{n-1}\left(\frac{p_{1}^{n}}{p_{0}^{n}}-1\right)\ge n\left(\frac{p_{1}}{p_{-1}}-1\right)\] (6) [b]Vasc:[/b] http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43658 post#3 (still unproved?) \[(n-1)p_{1}^{n-2}p_{2}^{2}+p_{0}^{n}\ge np_{1}^{n}\ \Longleftrightarrow \ (n-1)\frac{p_{2}^{2}}{p_{1}^{2}}+\frac{p_{0}^{n}}{p_{1}^{n}}\ge n\] (7) [b]M.Lascu:[/b] http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3705 \[\left(\frac{np_{1}^{2}-p_{2}^{2}}{n-1}\right)^{n-1}\ge p_{1}^{n-2}p_{0}^{n}\ \Longleftrightarrow \ \frac{p_{2}^{2}}{p_{1}^{2}}+(n-1) \left(\frac{p_{0}}{p_{1}}\right)^{\frac{\scriptstyle n}{\scriptstyle{n-1}}}\le n\] (2) & (3) look like they could be special cases of a more general one, and (5), (6), (7) may probably be generalized as well.  :roll:
Note that (3), if written in the form \[ \frac{p_{n}^{n}}{p_{0}^{n}}-1\ge n\left(\frac{p_{1}}{p_{-1}}-1\right), \] follows directly from (4) & (5).
Also note the interesting asymmetry comparing (6) and (7)  :)

all these inequalities are very nice!
Btw, for #1 the generalisation of hungkhtn is for $k\in [\frac{n}{n-1},n]$
and also (1) and (3) are special cases of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=103737]Zaimi-Marku[/url]
which can be rewritten as $(n-1)p_{n+k}^{n+k}+p_{0}^{n}p_{k}^{k}\geq n p_{1}p_{n+k-1}^{n+k-1}$

Thanks, I’ve edited.
As Vasc has pointed out, yours (=zaimi-marku) holds for all real $k$. I have replaced $k$ by $k-n$ to show the ressemblance with (1a).
Note that (1a) is stronger than it in its whole range, as for $k\le n$ we have $p_{k-n}\le p_{0}$, thus $p_{k-n}^{k-n}\ge p_{0}^{k-n}$.
You are right that (3) is a special case of yours, as it corresponds to $k=0$ with replacing $x_{i}$ by $\frac1{x_{i}}$. Nice!  :winner_first:

As I have posted in the topic of 1(a) it seems like it's a stronger generalisation but in fact is a simple consequence of suranyi itself (+AM-GM)
I hope people will continue to post this kind of inequalities here, they are very interesting and challenging!
[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=variable&t=90872]here[/url]
is another one but with a slightly different style.

[quote="Albanian Eagle"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=variable&t=90872]here[/url]
is another one but with a slightly different style.[/quote]
OK, let's write $s_{k}$ for the "$k$th symmetric mean" defined by $s_{k}^{k}: ={n\choose k}^{-1}\sum_{i_{1}<\cdots<i_{k}}x_{i_{1}}\cdots x_{i_{k}}$.

(MacLaurin tells us $p_{1}=s_{1}\ge s_{2}\ge\cdots\ge s_{n}=p_{0}$.)

Then the one by Vasc you've quoted is equivalent to \[(k-1)p_{k}^{k}+{n-1\choose k-1}s_{k}^{k}\ge np_{1}p_{k-1}^{k-1}.\] Of course it has different equality cases than all the above.

My solution for Suranji and almost kind of these inequalities are using "absolute-sign", do you believe?  :lol:  It's using the "Sysmetrical majorization Criterion" posted by Darij. Anyway, I'm really delighted in these solutions (although it's of mine  :oops: ).


Let a,b,c be positive real number such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$
Prove that: $ 3[\frac{4a\plus{}1}{(a\plus{}1)^2\plus{}1}\plus{}\frac{4b\plus{}1}{(b\plus{}1)^2\plus{}1}\plus{}\frac{4c\plus{}1}{(c\plus{}1)^2\plus{}1}]\geq a(2\sqrt{b}\plus{}1)\plus{}b(2\sqrt{c}\plus{}1)\plus{}c(\sqrt{a}\plus{}1)$

As a start subtract 9 from both sides. If you apply the subtraction correctly it should be pretty doable


What's a good way to prepare for this if your school is too...average to help in any sort of way?

Well, there are several options here.  The first, and my primary suggestion, is not to worry about it.  I have never asked anyone official, but I have a nagging suspcion that the lab barely counts for anything.  Most of the kids from the camp have said they screwed up the lab pretty badly, myself included.  It is much better to focus on theoretical; that is what will get you in.  At the camp, lab becomes more important.

The second option is use the equipment at your school.  Most of the reagents and equipment can be found at an average school.  Talk to your chem teacher, explain what you are doing and that you would really appreciate help.

A third option would be to talk to a chemistry professor at a local university or college (possibly the local USNCO coordinator?).  The university would certainly have what you need.

In any case, you can always look on the site for examples of lab writeups.  You will notice that showing neat, organized data and calcuations seems be worth quite a number of points.

It's ok. I've seen some of the lab questions. They're rather managable :)


function trigonometry logarithms algebra polynomial Euler vector

Is there a proof that if $n$ is odd, then $\zeta (n)$ cannot be evaluted exactly?
I know that for an even number, all you need to use is a fourier series, but it's obvious that this won't work for an odd number.

I'm not aware of any such proof. If you think about it, "evaluated exactly" is a pretty vague criterion. If I were to write a number as $e^{\sin^{-1}(\sqrt{\ln 2})}$ would you accept that as "evaluated exactly"? It has been proved that $\zeta(3)$ is irrational. (That can't come as a big surprise, since almost every real number is irrational.) I don't know one way or another whether $\frac{\zeta(3)}{\pi^3}$ is known to be rational or irrational.

I know of three families of methods (starting with Euler's infinite product method) for evaluating $\zeta(n)$ for even positive $n.$ All three methods share the property that the number of computations needed to compute $\zeta(2k)$ is an increasing function of $k$ - that is, they get harder as they get bigger. All three of these methods work only for the even integers.

Let's put this another way: there are more real numbers than there are real numbers you have names for. An exact, closed-form computation is the exception, not the rule. Each such computation is its own little miracle.

Thanks, you make a good point about how rare an exact solution can be, and yes I would accept $e^{\sin^{-1}({\sqrt{\ln{2}})}}$ as exact, and $2.6743423890041$ as an approximation.

[quote]... and $2.6743423890041$ as an approximation.[/quote]
Is that any different in practice than $1.20205690315959$ for $\zeta(3)$?

(Incidentally, I don't even have anything like Maple or Mathematica loaded on this computer - just whipped that up in a few minutes on Excel, with some partial sums, an estimate of the remainder, and a Richardson extrapolation table.)

And you wrote $2.6743\dots$ for that first one because you used a device that has built-in algorthims for constructing polynomial (or, more likely, rational) function approximations for various functions like that exponential and arcsine. I don't see it as all that different.

[quote="Jimmy"]Is there a proof that if $n$ is odd, then $\zeta (n)$ cannot be evaluted exactly?I know that for an even number, all you need to use is a fourier series, but it's obvious that this won't work for an odd number.[/quote]

Euler discovered, centuries ago, that $\zeta(2k)$ is a (known) rational multiple of $\pi^{2k}$.
Roger Apery proved 30 years ago that $\zeta(3)$ is irrational.
Recently, Tanguy Rivoal showed that at least $C.log(n)$ of the values $\zeta(3), \zeta(5), .... , \zeta(2n+1)$ are rationally linearly independent. I think that he was a French high school teacher at the time he did this, not a regularly-funded academic.
Building on this, Rivoal, Zudilin and possibly other people have proved something like:
at least one of $\zeta(3), \zeta(5), ..., \zeta(11)$ is irrational.
Conjecturally there are no nontrivial linear or polynomial relations between values of $\zeta(n)$ other than the ones known to Euler.

Do you mean at least one of zeta(5), zeta(7),...,zeta(11)?  Because if zeta(3) was proven irrational, that would certainly trivialize that other result.... :)

EDIT:  Sorry, I had meant to reply to this post, but I guess I ended up editing it instead.  I've tried to undo my error - blahblahblah

[quote="Magnara"][quote]Do you mean at least one of zeta(5), zeta(7),...,zeta(11)?  Because if zeta(3) was proven irrational, that would certainly trivialize that other result.... :)[/quote]

Not really.  I guess you can prove that one of $\zeta(3), \zeta(5), \zeta(7),\ldots\,\zeta(11)$ is irrational independently of the proof that $\zeta(3)$ is irrational.  It leaves the possibility that $\zeta(5), \zeta(7),\ldots\,\zeta(11)$ are all irrational, but it's still a step in some direction.[/quote]

Yes, $\zeta(3)$ should have been deleted in the paraphrase of Rivoal and Zudilin's result.
The statements that can currently be proved are lower bounds for the dimension of $\mathbb{Q}$-vector spaces generated by several different odd zeta-values $\zeta(2n_i + 1)$ and $1$.  It is expected that all the generators are linearly, and also algebraically, independent.

[quote="Moubinool"]A link for its article ?[/quote]

See [url=http://arxiv.org/pdf/math.NT/0104249.pdf?front]here.[/url]


$ABCD$ is a quadrilateral and $P,Q$ are the midpoints of $CD, AB, AP, DQ$  meet at $X$ and $BP, CQ$ meet at $Y$. Prove  that $A[ADX]+A[BCY] = A[PXOY]$.

I would very much like to see a soln to this.

[hide]First let find the area of BCY and AXD: doing:$\frac{BY}{BP}=m$ and $\frac{DX}{DQ}=n$ we have:

$\frac{\Delta BCY}{\Delta BCP}=\frac{\Delta BCY}{\frac{\Delta BCD}{2}}=m \implies \Delta BCY=m.\frac{\Delta BCD}{2}$

$\frac{\Delta AXD}{\Delta ADQ}=\frac{\Delta AXD}{\frac{\Delta ABD}{2}}=n \implies \Delta AXD =n.{\frac{\Delta ABD}{2}}$ thus:

$\Delta AXD+\Delta BCY=n.{\frac{\Delta ABD}{2}+m.\frac{\Delta BCD}{2}}$ and:

$\square QYPX=\square ABCD-\Delta CBQ-\Delta ADP-\Delta CYP-\Delta AQX \Rightarrow$

$\square QYPX=\square ABCD-\frac{\Delta ABC}{2}-\frac{\Delta ACD}{2}-(1-m).\frac{\Delta BCD}{2}-(1-n).\frac{\Delta ABD}{2}$

$\Rightarrow\square QYPX=\square ABCD-\frac{\square ABCD}{2}-\frac{\square ABCD}{2}+n.{\frac{\Delta ABD}{2}+m.\frac{\Delta BCD}{2}}$

$\Rightarrow\square QYPX=n.{\frac{\Delta ABD}{2}+m.\frac{\Delta BCD}{2}}$ in consequence:

$\square QYPX=\Delta AXD+\Delta BCY.$[/hide]

Join P and Q. Join A and C. Now divide the triangle into 6 regions: Triangle DXP-1, DXA-2, CYB-3, BQY-4, XAQ-5, PYC-6. Now A(PQC)=A(PQD) (Median Property). This can be written as A(PQX)+1=A(PQY)+6. Same way, A(PAQ)=A(PQB) (Median Property). This can again be written as A(PQX)+5=A(PQY)+4. Subtracting these equation we get 1-5=4-6 which can be written as I+4=5+6. By median property for Triangle ADP, 1+2= A(APC). Same manner, in ABC 3+4=A(AQC). Adding these two, we get 1+2+2+4=A(PXQ)+A(PYQ) which can be written as 1+2+3+4=5+6. We know that 1+4=5+6. Therefore, 3+1+4+2=A(PXQ)+A(PYQ)+1+4. We cancel out 1+4. We get 3+2=A(PXQ)+A(PYQ), exactly what we have to proved. Therefore, we have proved that  the area of Quadrilateral PXQY is sum of the areas of triangles DXA and BYC.


quadratics
number theory proposed
number theory

Let p be the prime number such that 4|(p-1).Prove there are no positive integers x,y,z satisfying $ x^2\plus{}y^2\equal{}pz^2$

SUPERMAN2 wrote
[quote]
Let p be the prime number such that 4|(p-1).Prove there are no positive integers x,y,z satisfying [/quote]

$ 2^2 \plus{} 4^2 \equal{} 5*2^2$, so I think you may have made some typing mistakes.
I guess it should be $ p \equal{} 3(mod \;\; 4)$, which is easy as $ a$ and $ \minus{} a$(where $ p \nmid a$) cannot be both quadratic residue $ mod \;\; p$ by Euler's criterion.

We could use that if p is prime and $ p\equiv3 \left(\mod 4\right)$ than if $ p|a^{2}\plus{}b^{2}$ than $ p|a$ and $ p|b$. Now using the method of infinite descend we could solve the problem.

Sorry.The right problem is:
Let p be the prime number such that 4|(p-3).Prove there are no positive integers x,y,z satisfying $ x^2\plus{}y^2\equal{}pz^2$