modular arithmetic
number theory proposed
number theory

Find all intengers n such that $2^n - 1$ is a multiple of 3 and $(2^n - 1)/3$ is a divisor of $4m^2 + 1$ for some intenger m.

What is problem? n - is even natural, and $m=2^{n/2-1}$.

[quote="jin"]I don't think so .Since that's $4m^2+1$ not $4m^2-1$
The answer is $2^n$
Clearly,$n$ is even for $3|2^n-1$
And $\frac{2^n-1}{3}$ doesn't have a prime factor which $\equiv3\pmod{4}$.
The following proof is easy.[/quote]
Yes. But if n=2k, then $2^{2k-1}-1=(2^k+1)(2^k-1)$ had prime divisor $\equiv 3\pmod{4}$ for all k>2.

That's a IMO shortlist question in 1998.I think.

When $n=2^k$,the only prime factor of $2^n-1$ such that $\equiv3\pmod{4}$ is 3.
$2^{2^k}-1=(2+1)(2^2+1)\cdots(2^{2^{k-1}}+1)$
All of the factors in the right side are Fermat number.

17 year revival :) The condition implies $n=2k$ for some integer $k$. We consider two cases:

[color=#00f][b]Case 1:[/b][/color] $n$ is not a power of 2.

Suppose $k=2^a \cdot b$ for an odd integer $b > 1$. Then
\begin{align*}
\frac{2^n-1}{3} &= \frac 13 \left((4^b)^{2^a}-1\right) \\
&= \frac 13 \left((4^b)^{2^{a-1}}+1\right)\left((4^b)^{2^{a-2}}+1\right) \ldots \left(4^b+1\right)\left(4^b-1\right).
\end{align*}

Thus we must have $2^b-1 \equiv 3 \pmod{4}$ is a divisor of $4m^2+1$, which violates Fermat's Christmas Theorem. ${\color{blue} \Box}$

[color=#00f][b]Case 1:[/b][/color] $n$ is a power of 2.

We aim to construct a sufficient value of $n$. Notice we have $b=1$ in our expression for $\frac{2^n-1}{3}$ from earlier, so the factors $\frac 13$ and  $4^b-1$ just cancel. This makes all factors Fermat numbers, which are known to be pairwise relatively prime. Hence we can construct our value of $m$ using CRT such that
\[m \equiv 2^{2^{c}-1} \pmod{4^{2^c}+1}\]

for all $c \in [0, a-1]$. ${\color{blue} \Box}$

Thus our solutions for $n$ are $\boxed{2^j \text{ for } j \in \mathbb{N}}$. $\blacksquare$


For A,B,C angles of a triangle prove that

(tgA/2)(tgB/2)(tgC/2)>=(cotgA)(cotgB)(cotgC)

I would like a pure trigonometric solution.
I feel it is possible but hard to find.

Well, what do you mean by "hard"? An inequality I am able to prove cannot be hard anyway.

For your one, at first, we will consider only the case of an acute-angled triangle, since for an obtuse-angled triangle the inequality is trivial (cot A cot B cot C is negative). Now, for an acute triangle, your inequality

tan A/2 tan B/2 tan C/2 >= cot A cot B cot C

is equivalent to

cot A/2 cot B/2 cot C/2 <= tan A tan B tan C

(in fact, x >= y is equivalent to 1/x <= 1/y). But according to well-known formulas, we have cot A/2 cot B/2 cot C/2 = cot A/2 + cot B/2 + cot C/2 and tan A tan B tan C = tan A + tan B + tan C (you can easily prove these for yourself by using the tan(x+y) formula). Hence, we have to prove

cot A/2 + cot B/2 + cot C/2 <= tan A + tan B + tan C.

This is obviously equivalent to

2 cot A/2 + 2 cot B/2 + 2 cot C/2 <= 2 tan A + 2 tan B + 2 tan C,

or to

2 cot A/2 + 2 cot B/2 + 2 cot C/2 <= (tan B + tan C) + (tan C + tan A) + (tan A + tan B).

And this follows from

2 cot A/2 = 2 tan ((B+C)/2) <= tan B + tan C,
2 cot B/2 = 2 tan ((C+A)/2) <= tan C + tan A,
2 cot C/2 = 2 tan ((A+B)/2) <= tan A + tan B

(remember that our triangle is acute-angled). Qed.. (I borrowed the proof trick from Treegoner.)

  Darij


I've just realized that the application papers for download have been taken off of MIT's website ( see

[url=http://admissions.mit.edu]How to Apply[/url]

Can someonde send them to my e-mail (richardmath2003@yahoo.co.uk)?

Thank you.

[quote="daveed"]well... isn't the deadline over?

you'd have to kind of wait 'til august until you can apply again... right?[/quote]
I know it. But I suppose the papers for applying for 2007 will be the same, or almost the same, as the ones for applying for 2006. I just want to know what informations I will have to send to them.

Usually, the only things that change from year to year are the essays. The applications for 2007 will be out in the summer/fall of 2006.

[quote="zoogies"]If I recall correctly, the main essay was the same this year and last year.[/quote]

Yeah, that's true for MIT. I'm not sure if this always happens, so I didn't want to say anything.

But generally, if you write a story about cultural diversity and a story about overcoming difficulties, chances are, you'll be able to use them someplace.

[quote="mel"]But generally, if you write a story about cultural diversity and a story about overcoming difficulties, chances are, you'll be able to use them someplace.[/quote]

One of the best things to do to stand out in a college essay is to write about SPECIFICS about your personal interests. Almost everyone writes about what they think college admissions officers are looking for, so most college essays are disappointingly similar to one another. Instead, write about what is unique about your own interests and passions, as you would in a letter to your best friend far away, and you will be much more likely to say something interesting and memorable.


vector
invariant
algebra
polynomial
linear algebra
linear algebra unsolved

Let $T$ be a linear operator in $m$-dimensional vector space $V$ such that for any vector $\psi \in V$ the vectors $\psi,T\psi,...,T^n\psi$ are linearly dependent. Prove that hte operators $I,T,...,T^n$ are also linearly dependent.

I posted it some time ago and I think Alekk and Kent gave some solutions. The result holds even in infinite dimension.

You could get from the finite-dimensional case to the general one like this:

Let $k$ be the smallest positive integer such that for any finite-dimensional subspace $W\le V$ which is $T$-invariant, the restriction $T|_{W}$ of $T$ to $W$ satisfies a monic polynomial equation of degree $k$. The set of such positive integers is non-empty, since it contains $m$ (because of the finite-dimensional case), so the definition of $k$ makes sense.

Suppose now that $T$ satisfies no monic polynomial equation of degree $k$. Let $W$ be a finite-dimensional $T$-invariant subspace of $V$, and let $p$ be a monic polynomial of degree $k$ such that $p(T|_{W})=0$. By asumption, we can find $v\in V$ such that $p(T)v\ne 0$. This means that we can find a monic polynomial $\widetilde p\ne p$ of degree $k$ such that $\widetilde p(T|_{\widetilde W})=0$, where $\widetilde W=W+\langle v, Tv, \ldots, T^{n-1}v\rangle$. This means that $\widetilde p(T|_{W})=0$, so the restriction of $T$ to $W$ satisfies the polynomial equation $p-\widetilde p$, which has degree $\le k-1$. $W$ was arbitrary, so we can do this for every finite-dimensional $T$-invariant subspace $W$. This contradicts the minimality of $k$, and we're done.


abstract algebra
superior algebra
superior algebra unsolved

I'm wondering why this is true:

$ A$ any commutative ring, $ M$ and $ N$ $ A$-modules, and if $ M \otimes_{A} N \cong A$ as $ A$-module, then $ M$ and $ N$ are both isomorphic to $ A$ as an $ A$-module.

Can anyone please tell me?

This is wrong, take a nontrivial element of the picard group.

@sissa: this just reformulates the problem.

what is an explicit example?

this takes place in the category of sheaves of modules over some proj-spectrum. but what about usual modules?


inequalities
function
Cauchy Inequality
inequalities proposed

Zhaobin,To prove this problem ,I only need to prove:
               ${\sum_{i=1}^n \frac{1}{\sqrt{n}+\sqrt{x_{i}}} \leq \frac{n^2}{n+1}\frac{1}{\sum_{i=1}^n \sqrt{x_i}}}$   when $x_{1}+x_{2}+....+x_{n}=1$ (1)
(because $(\sum_{i=1}^n \frac{1}{\sqrt{1+x_{i}}} \leq \sqrt{n+1}(\sum_{i=1}^n \frac{1}{\sqrt{n}+\sqrt{x_{i}}})$)
with (1),Now I have a solution but it's too long,So i'm confused to post it.I promise I'll post the solution to (1) if everyone can't find the solution to it :)  :)

if $\sum tan^2 a_i=1$ then maybe we can show that $\sum sin a_i\le \frac{n}{\sqrt{n+1}}$

General
Let \[ x_i > 0(i = 1,2, \cdots ,n),2 \le n \in N,\sum\limits_{i = 1}^n {x_i } = 1,2 \le m \in N,0 < \lambda \le \frac{n}{{n - 1}}  \]
that\[ (\sum\limits_{i = 1}^n {\sqrt[m]{{x_i }}} )\sum\limits_{i = 1}^n {\frac{1}{{\sqrt[m]{{1 + \lambda x_i }}}}} \le \frac{{n^2 }}{{\sqrt[m]{{n + \lambda }}}} ..........(3)  \]

with (1),Now I have a solution but it's too long。

thank you to you all.
and thanks to scpajmb's nice solution.

to nhat:can you post solution for (1),because it is also a nice inequality.

to Albanian Eagle:this may be the offical solution:but I think it is not only $\sum sin a_i$

I think we can use AM-GM for the original inequality. It suffices to prove $n \sum \sqrt{x_i}+ \sqrt{n+1} \sum \frac 1{\sqrt{1+x_i}} \leq 2n\sqrt n$ and then apply AM-GM.

This inequality follows from the concavity of the function $n \sqrt x+\sqrt{n+1} \frac 1{\sqrt{1+x}}$ on $(0;1)$

@scpajmb:

I think, your solution isn't true, cause x is not x+1.

felix

[quote="scpajmb"]
so that\[ \[ \begin{array}{l} {\rm{ }}\left( {\sum\limits_{i = 1}^n {\sqrt{y_i - 1} } } \right)\sum\limits_{i = 1}^n {\frac{1}{{\sqrt{y_i } }}} \le \sqrt n \sum\limits_{i = 1}^n {\sqrt{ - y_i + \frac{2}{n}(n + 1) - \frac{{2n + 1}}{{n^2 }} \cdot \frac{1}{{1 + y_i }}} } \\ \overbrace \le ^{{\rm{Cauchy}}}\sqrt n \cdot \sqrt n \sqrt{\sum\limits_{i = 1}^n {[ - y_i + \frac{2}{n}(n + 1) - \frac{{2n + 1}}{{n^2 }} \cdot \frac{1}{{1 + y_i }}} } ] \\ = n\sqrt{ - (n + 1) + 2(n + 1) - \frac{{2n + 1}}{{n^2 }}\sum\limits_{i = 1}^n {\frac{1}{{1 + y_i }}} } \\ = n\sqrt{n + 1 - \frac{{(2n + 1)}}{{n^2 }}\sum\limits_{i = 1}^n {\frac{1}{{y_i }}} } \\ \end{array}  \]
[/quote]

in this part, all the $\frac{1}{1+y_i}$ should be $\frac{1}{y_i}$

then the others is no problem.

[quote="zhaobin"]$x_{1}, x_{2}, \cdots, x_{n}$ are positive numbers such that $\sum_{i=1}^{n}x_{i}= 1$. Prove that \[\left( \sum_{i=1}^{n}\sqrt{x_{i}}\right) \left( \sum_{i=1}^{n}\frac{1}{\sqrt{1+x_{i}}}\right) \leq \frac{n^{2}}{\sqrt{n+1}}\][/quote]
I think it's very easy by using Cauchy-Schwartz  Inequality :
First, we have if $x_i \in [0,1]$ then :
\[\sum_{i=1}^n \dfrac{1}{\sqrt{1+x_i}} \leq \dfrac{n}{\sqrt{1+\sqrt[n]{x_1x_2...x_n}}}\]
So we only prove that:
\[\dfrac{\left(\sqrt{x_1}+...+\sqrt{x_n}\right)^2}{1+\sqrt[n]{x_1x_2...x_n}}\leq \dfrac{n^2}{n+1}\]
\[\Longleftrightarrow \[\dfrac{\left(\sqrt{x_1}+...+\sqrt{x_n}\right)^2}{x_1+x_2+...+x_n+\sqrt[n]{x_1x_2...x_n}}\leq \dfrac{n^2}{n+1}\]
Assume that $x_1x_2...x_n=1$ we need to prove:
\[\dfrac{\left(\sqrt{x_1}+...+\sqrt{x_n}\right)^2}{x_1+x_2+...+x_n+1}\leq \dfrac{n^2}{n+1}\]
By Cauchy-Schwartz Inequality, we have:
\[\dfrac{\left(\sqrt{x_1}+...+\sqrt{x_n}\right)^2}{x_1+x_2+...+x_n+1}\leq \sum_{i=1}^n \dfrac{x_i}{x_i+\dfrac{1}{n}}\]
So we need to prove:
\[ \sum_{i=1}^n \dfrac{nx_i}{nx_i+1}\leq \dfrac{n^2}{n+1}\]
\[\Longleftrightarrow \sum_{i=1}^n \dfrac {1}{nx_i+1}\geq \dfrac{n}{n+1}\]
Let $a_i=\dfrac{1}{x_i}$ we have $a_1a_2...a_n=1$ and we need to prove:
\[\sum_{i=1}^n\dfrac{a_i}{a_i+n}\geq \dfrac{n}{n+1}\]
Using Cauchy-Schwartz inequality, we have:
\[\sum_{i=1}^n\dfrac{a_i}{a_i+n}\geq \dfrac{\left(a_1+...+a_n\right)^2}{a_1^2+...+a_n^2+n^2}\]
We prove that : \[ \dfrac{\left(a_1+...+a_n\right)^2}{a_1^2+...+a_n^2+n^2} \geq \dfrac{n}{n+1}\]
\[\Longleftrightarrow \left(a_1+...+a_n\right)^2 +2n \sum_{i<j}a_ia_j \geq n^3\]
It is true because:
\[ \left\{
    \begin{array}{ll}
       \left(a_1+...+a_n\right)^2\geq n^2\\
       \sum_{i<j}a_ia_j \geq \binom{n}{2}=\dfrac{n(n-1)}{2}
    \end{array}
 \right.\]
\[Q.E.D\]

[quote="ndk09"][quote="zhaobin"]$x_{1}, x_{2}, \cdots, x_{n}$ are positive numbers such that $\sum_{i=1}^{n}x_{i}= 1$. Prove that \[\left( \sum_{i=1}^{n}\sqrt{x_{i}}\right) \left( \sum_{i=1}^{n}\frac{1}{\sqrt{1+x_{i}}}\right) \leq \frac{n^{2}}{\sqrt{n+1}}\][/quote]
I think it's very easy by using Cauchy-Schwartz  Inequality :
First, we have if $x_i \in [0,1]$ then :
\[\sum_{i=1}^n \dfrac{1}{\sqrt{1+x_i}} \leq \dfrac{n}{\sqrt{1+\sqrt[n]{x_1x_2...x_n}}}\]
So we only prove that:
...
\[ \left\{
    \begin{array}{ll}
       \left(a_1+...+a_n\right)^2\geq n^2\\
       \sum_{i<j}a_ia_j \geq \binom{n}{2}=\dfrac{n(n-1)}{2}
    \end{array}
 \right.\]
\[Q.E.D\][/quote]
Nice proof, it is quite similar to mine. Thanks; ndk. :)
By the way, I have posted my proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=353429&p=1923790&hilit=Shanku#p1923790]here.[/url]
:)

Here is a very nice solution: By the Cauchy-Schwarz Inequality, we have
\begin{align*}
\sqrt{x_1} + \sqrt{x_2} + \cdots + \sqrt{x_n} &\le \sqrt{\bigg((1 + x_1) + (1 + x_2) + \cdots + (1 + x_n)\bigg)\left(\frac{x_1}{1 + x_1} + \frac{x_2}{1 + x_2} + \cdots + \frac{x_n}{1 + x_n}\right)} \\
&= \sqrt{(n + 1)\left(\frac{(1 + x_1) - 1}{1 + x_1} + \frac{(1 + x_2) - 1}{1 + x_2} + \cdots + \frac{(1 + x_n) - 1}{1 + x_n}\right)} \\
&= \sqrt{(n + 1)\left(n - \frac{1}{1 + x_1} - \frac{1}{1 + x_2} + \cdots - \frac{1}{1 + x_n}\right)}.
\end{align*}
Also by the Cauchy-Schwarz Inequality, we have
\begin{align*}
\frac{1}{\sqrt{1 + x_1}} + \frac{1}{\sqrt{1 + x_2}} + \cdots + \frac{1}{\sqrt{1 + x_n}} &\le \sqrt{\left(1 + 1 + \cdots 1\right)\left(\frac{1}{1 + x_2} + \frac{1}{1 + x_2} + \cdots + \frac{1}{1 + x_n}\right)} \\
&= \sqrt{n\left(\frac{1}{1 + x_2} + \frac{1}{1 + x_2} + \cdots + \frac{1}{1 + x_n}\right)}.
\end{align*}
Therefore, if we set $t = \sum\limits_{i = 1}^{n} \frac{1}{1 + x_i}$, it suffices to prove that $\sqrt{(n + 1)(n - t)} \cdot \sqrt{nt} \le \frac{n^2}{\sqrt{n + 1}}.$ Upon squaring and rearranging, this is just \[(n - t)t \le \frac{n^3}{\left(n + 1\right)^2} \iff \left(t - \frac{n^2}{n + 1}\right)^2 + \frac{n(n - 1)}{n + 1}\left(t - \frac{n^2}{n + 1}\right) \ge 0,\] where we have completed the square above. Hence, it suffices to prove that $t - \frac{n^2}{n + 1} \ge 0$, but this follows immediately from the convexity of the continuous and differentiable function $f : (0, 1) \to \mathbb{R}^+$ defined by $f(t) = \frac{1}{1 + t}.$ We easily compute $f''(t) = \frac{2}{\left(1 + t\right)^3} \ge 0,$ so $f$ is convex. Therefore, by Jensen's Inequality, we have \[\frac{1}{1 + x_1} + \frac{1}{1 + x_2} + \cdots + \frac{1}{1 + x_n} \ge n \cdot \frac{1}{1 + \frac{x_1 + x_2 + \cdots + x_n}{n}} = \frac{n^2}{n + 1},\] as desired. $\square$

[quote=Jia-Jia-Jia-Jia][hide = Chebyshev + Jensen]
WLOG let $x_1, x_2, ..., x_n$ be in non-decreasing order. By Chebyshev,
\[\frac{1}{n^2} \left(\sum_{i = 1}^n\sqrt{x_i}\right) \left(\sum_{i = 1}^n \frac{1}{\sqrt{1 + x_i}}\right) \le \frac{1}{n} \left(\sum_{i = 1}^n \sqrt{\frac{x_i}{1 + x_i}}\right)\]
Note that $f(x) = \sqrt{\frac{x}{1 + x}}$ is concave, for $x \ge 0$, by Jensen's,
\[\frac{1}{n} \cdot \sum_{i = 1}^n \sqrt{\frac{x_i}{1 + x_i}} = \frac{1}{n} \cdot \sum_{i = 1}^n f(x_i) \le \large f\left(\frac{1}{n} \cdot \sum_{i = 1}^n x_i\right) = f\left(\frac{1}{n}\right) = \sqrt{\frac{1/n}{1 + 1/n}} = \sqrt{\frac{1}{n + 1}}\]
Hence we have
\[\left(\sum_{i = 1}^n\sqrt{x_i}\right) \left(\sum_{i = 1}^n \frac{1}{\sqrt{1 + x_i}}\right) \le \frac{n^2}{\sqrt{n+1}}\]
as desired.
[/hide][/quote]

There is something wrong in your using of Chebyshev.
Maybe Chebyshev Inequality can't be used in this problem.


complex analysis
integration
logarithms
calculus
limit
geometry
geometric transformation

If $f$ is continuously differentiable, and $f \circ g$ satisfies the C-R Riemann equations at the point $z = 0$, and $g$ is holomorphic, does $f$ satisfy the C-R Riemann equations at the point $g(0)$?

If this is not true, consider $\Phi = \psi_{\varphi(w)}\circ \varphi \circ \psi_{w}: \mathbb{D}\to \mathbb{D}$, where $\psi_{w}: \mathbb{D}\to \mathbb{D}$ is defined by $\psi_{w}(z) = \frac{w-z}{1-\overline{w}z}$, and $\varphi : \mathbb{D}\to \mathbb{D}$ is continuously differentiable. If $\Phi$ satisfies the C-R equations at 0, does $\varphi$ satisfy the C-R equations at $w$?

$\mathbb{D}$ is the unit disk.

For the first question, consider $g(z)=z^{2}$: then $f\circ g$ satisfies CR at $0$ no matter what $f$ is, as long as $f\in C^{1}$.

However, the situation is different if $g$ is injective -- and in the second part it is.

Thanks for the response, mlok, your help is greatly appreciated.

So, in the second case, is it true or not?

The original question, from which this inquiry stemmed, as well as my last question in the Real Analysis forum which you responded to two days ago (the one with $|f(z)| = |z|$ and $f \in C^{1}$) was this:

Define $\|w\|_{z}= \frac{|w|}{1-|z|^{2}}$ for $w \in \mathbb{C}$ and $z \in \mathbb{D}$. Let $d : \mathbb{D}^{2}\to \mathbb{R}$ be defined by $d(z_{1}, z_{2}) = \inf_{\gamma}\int_{0}^{1}\|\gamma'(t)\|_{\gamma(t)}\, dt$, where the infimum is taken over all smooth curves $\gamma : [0, 1] \to \mathbb{D}$ joining $z_{1}$ and $z_{2}$.

Show that if $\varphi : \mathbb{D}\to \mathbb{D}\in C^{1}$, and $d(\varphi(z_{1}), \varphi(z_{2})) = d(z_{1}, z_{2})$ for all $z_{1}, z_{2}\in \mathbb{D}$, then either $\varphi$ or $\overline{\varphi}$ is an automorphism of $\mathbb{D}$.

Concerning the second question: since $\varphi = \psi_{\varphi(w)}^{-1}\circ \Phi \circ \psi_{w}^{-1}$, you can compute $\frac{\partial \varphi}{\partial \bar z}(0)$ by the chain rule and find that it is equal to zero. That should help with the original question.

Another approach to the latter is to compose $\varphi$ with an automorphism of $\mathbb D$ chosen so that $\varphi(1/2)=\varphi(1/2)$ and $\varphi(-1/2)=\varphi(-1/2)$. Then use the fact that $d(z,1/2)$ and $d(z,-1/2)$ determine $z$ up to conjugation.

This incomplete solution delineates the impasse which I have reached. We already know that automorphisms of $\mathbb{D}$ preserve the hyperbolic distance $d$ by the Schwarz-Pick lemma, which states that if $f : \mathbb{D}\to \mathbb{D}$ is holomorphic, then $\frac{|f'(z)|}{1-|f(z)|^{2}}\le \frac{1}{1-|z|^{2}}$ for all $z \in \mathbb{D}$. Moreover, for all $z_{1}, z_{2}\in \mathbb{D}$, the hyperbolic distance can be shown to be $d(z_{1}, z_{2}) = \frac{1}{2}\log \frac{1+\rho(z_{1}, z_{2})}{1-\rho(z_{1}, z_{2})}$, where ${\rho(z_{1}, z_{2}) = |\frac{z_{1}-z_{2}}{1-\overline{z_{1}}z_{2}}}|$.

Incomplete solution:

Suppose that $\varphi : \mathbb{D}\to \mathbb{D}$ is continuously differentiable and preserves the hyperbolic distance. Given $w \in \mathbb{D}$, consider $\varphi^{*}= \psi_{\varphi(w)}\circ \varphi \circ \psi_{w}$. We may assume WLOG that $\varphi(0) = 0$, since $\varphi^{*}(0) = 0$, and if $\varphi^{*}\in \mathrm{Aut}(\mathbb{D})$, then so is $\varphi$. [b]Moreover, we only need to check that $\varphi$ or $\overline{\varphi}$ satisfies the Cauchy-Riemann equations at the origin, since if $\varphi^{*}$ or $\overline{\varphi^{*}}$ does, then $\varphi$ or $\overline{\varphi}$ satisfies the Cauchy-Riemann equations at $w$.[/b] (The previous statement is the statement that I cannot verify and which I’m not even sure is true.) Fix $z = (x,y) \in \mathbb{C}\setminus \{0\}$, and let $0 \le \tau < \frac{1}{|z|}$. Define the curve $\gamma \in \mathbb{D}$ by $\gamma(t) = zt$, where $0 \le t \le \tau$, and write $\varphi = u+iv$, where $u = \mathrm{Re}( \varphi)$ and $v = \mathrm{Im}(\varphi)$. Then since $\varphi$ preserves the hyperbolic distance, we have that $d(0, \tau z) = \frac{1}{2}\log\frac{1+\tau|z|}{1-\tau|z|}$
\[= \int_{0}^\tau \frac{|z|}{1-|z|^{2}t^{2}}\, dt = \int_{0}^\tau \|\gamma'(t)\|_{\gamma(t)}\, dt = d(0, \varphi(\tau z)) = \int_{0}^\tau \|(\varphi \circ \gamma)'(t)\|_{(\varphi \circ \gamma)(t)}\, dt, \]
where $(\varphi \circ \gamma)'(t) = x u_{x}(tz)+y u_{y}(tz)+i(x v_{x}(tz)+y v_{y}(tz))$ for $0 \le t \le \tau$. By the fundamental theorem of calculus $\|\gamma'(0)\|_{\gamma(0)}$
\[= \lim_{\tau \to 0^{+}}\frac{1}{\tau}\int_{0}^\tau \|\gamma'(t)\|_{\gamma(t)}\, dt = \lim_{\tau \to 0^{+}}\frac{1}{\tau}\int_{0}^\tau \|(\varphi \circ \gamma)'(t)\|_{(\varphi \circ \gamma)(t)}\, dt \]
$= \|(\varphi \circ \gamma)'(0)\|_{(\varphi \circ \gamma)(0)}$. Since we have $\gamma(0) = 0$ and $\varphi(0) = 0$, it follows that $|z|^{2}= (x u_{x}(0)+y u_{y}(0))^{2}+(x v_{x}(0)+y v_{y}(0))^{2}$. Letting $(x,y) = (1,0)$ yields $u_{x}(0)^{2}+v_{x}(0)^{2}$; letting $(x,y) = (0,1)$ yields $u_{y}(0)^{2}+v_{y}(0)^{2}= 1$; and finally, letting $(x,y) = (1,1)$ yields
\[2 = u_{x}(0)^{2}+2u_{x}(0)u_{y}(0)+u_{y}(0)^{2}+v_{x}(0)^{2}+2v_{x}(0)v_{y}(0)+v_{y}(0)^{2}, \]
which implies that $u_{x}(0)u_{y}(0)+v_{x}(0)v_{y}(0) = 0$. Thus, either $u_{x}(0) = v_{y}(0)$ and $u_{y}(0) =-v_{x}(0)$, or $u_{x}(0) =-v_{y}(0)$ and $u_{y}(0) = v_{x}(0)$, so that either $\varphi$ or $\overline{\varphi}$ satisfies the Cauchy-Riemann equations at the origin. [b]It follows that either $\varphi$ or $\overline{\varphi}$ is holomorphic on $\mathbb{D}$ (again, this statement requires the truth of the problematic statement I highlighted above).[/b] Finally, assume WLOG that $\varphi$ is holomorphic; then for all $z \in \mathbb{D}$, since $d(0, \varphi(z)) = d(0, z)$, we have $|\varphi(z)| = |z|$. By the Schwarz lemma $\varphi$ is a rotation, and we are done.

Let me try the chain rule calculation again with $\varphi = \psi_{\varphi(w)}\circ \Phi \circ \psi_{w}$. I could not get this to work out last night no matter how hard I tried. We need to check that $\frac{\partial\varphi}{\partial\overline{z}}(w) = 0$.

Suggestions: 1) don't use the specific form of $\psi$.
2) split the computation into two steps (one composition at a time).

The other problem with this solution is that even if you could show that $\frac{\partial\Phi}{\partial\overline{z}}(0) = 0$ implies that $\frac{\partial\varphi}{\partial\overline{z}}(w) = 0$, there remains the question about what to do about $\overline{\Phi}= \overline{\psi_{\varphi(w)}\circ \varphi \circ \psi_{w}}$.

For the latter, you can use the identity $\frac{\partial\bar \Phi}{\partial\bar z}=\overline{\frac{\partial \Phi}{\partial z}}$.

Chain Rule computations can be simplified by using differentials like $f=f_{z}dz+f_{\bar z}d\bar z$.


Prove that $ 18 ^{\circ} < \arctan \frac{1}{3}^{\circ} <19 ^{\circ}$

The Taylor series for arctan (this is calculus) says arctan x=x-x^3/3+x^5/5-x^7/7....., where the angle is in radians. The first two terms (this is accurate enough) gives the angle is 26/81 radians, which is a little over 18 degrees.


bar= centroid of

help me please!! excuse me for my bad english

[quote="fjolla"]didn't find a question 1) of this exercise ... Fau it as I finish tonight ... Please help me

ABCD is a parallelogram.
I refers to the middle of [AD]
(I concluded that I bar (A; 1) and (D; 1))
E and the center of gravity of the triangle ACD
(I find that (I doubt if I mi vectors in the right direction ... told me if there is a fault thanks)
CE = 2/3CI
3 EC = 2CI
(I no longer jib, but I always speaks of vectors)
3 EC = 2 EI 2CE
EC 2 IE = 0
THEREFORE E bar (c; 1) and (I; 2))
F is defined by (these are vectors) BF = 1/4BC
(F therefore bar (B; 3) and (C; 1))
K designe the middle of [EB]
(So K bar (E; 1) and (B; 1))

DSL but my scanner does not work I do not send you little figure

1) Prove that K is the centroid of (A; 1), (B; 3), (C, 1) and (D; 1)[/quote]
I don't get what (Capital Letter;(semicolon) Number) means. Could you explain what $ (A;1)$ means?


There are $n$ football teams in [i]Tumbolia[/i]. A championship is to be organised in which each team plays against every other team exactly once. Ever match takes place on a sunday and each team plays at most one match each sunday. Find the least possible positive integer $m_n$ for which it is possible to set up a championship lasting $m_n$ sundays.

It is trivial that m_n >= n-1. If n is even then m_n = n-1 and if n is odd, m_n = n. The main thing we have to do is contruct the adquate schedule. I know it (because I often have to onganize such tournament!) but I leave other members to think about.

Namdung

This is the same problem as :
http://www.mathlinks.ro/viewtopic.php?p=7873#7873

I do not agree with you Namdung : Let's kill all these problems, just to clean up the forum :D

So, let's give full solutions to them, or they will be here for a long time (it's sometimes boring to fill to little gaps, and the problem will stay as 'unsolved' for years....)

Pierre.


Sequences like this can be impossible to decode because there are too many possibilities.
I was on to something with yours, nat mc, but then it stumped me.

How about this one? 1 9 9 8 7 3 7 5 2 7 1 5 ...
Find the next two numbers.

Wouldn't A be
one three one two two ones one two:
13122112
Then, one one one three ones one two twos two ones one two:
111311222112

For nat mc's think [hide]Fibonacci![/hide]
236Factorial's confuses me...I kinda see something but the last few terms tripped me up.

That's what happens with all the sequences I try to figure out.

Sequence problems like this are fascinating and challenging, but you won't see sequence problems in competition because there is no single correct answer.  The poser is really looking for the simplest explanation of the pattern.

[quote="bob123"][quote="uglyman621"]Can you find the next two terms in each of the following sequences?

THESE ARE PRETTY CHALLENGING

A)
2, 12, 1112, 3112, 132112, ..., ...

B) 1, 1, 1, 2, 3, 7, 23, ..., ...

C) 1, 1, 1, 2, 3, 5, 11, ..., ...

$ \text{HINT:}$ patterns B and C are very similar[/quote]

[hide]A)
two, one two, one one and one two, three ones and one two, one three two ones and one two.... it's just describing the numbers before them... so the next two are:

1113122112, 311311222112

B) 2=1*1+1
3=2*1+1
7=3*2+1
23=7*3+2
So it would be the product of the two numbers directly behind add added to the number following those two (actually, the one behind that one) so if you have $x_1,x_2,x_3$, to find $x_4$, you do $x_3\cdot{x_2}+x_1$
That means the next two numbers are:
164, 3779 (WOW I DID IT WITHOUT A CALCULATOR!!! :lol:)

C) To find $x_4$ knowing $x_1,x_2,x_3$, you do $x_1\cdot{x_2}+x_3=x_4$...

So the next two numbers are: 26, 81[/hide][/quote]

Wow, you're wicked good at this...

[quote="zatore"]This can't be right:[quote="polymorphic"]

For nat mc's think
[hide]
Fibonacci![/hide][/quote]
because
[hide]this one is Fibonacci:
[quote="superman"]try this one lol

1 1 2 3 5 8 13 21 __ __[/quote] (BTW, even though you should all know this, it's 34 and then 55.)[/hide]

As for 236factorial,
[quote="236factorial"]
How about this one? 1 9 9 8 7 3 7 5 2 7 1 5 ...
Find the next two numbers.[/quote]
[hide] It is 5, then 8.  You take the units digit of the sum of the four preceding numbers.  I was onto something about taking the units digit, but I was doing some weird complicated multiplication dealy.  Then I asked my friend what she thought, and she showed me that right away.[/hide][/quote]
I think he means it has something to do with Fibonacci.

Nice job on that other pattern

[hide] BAH! this took me over 20 mins.   105!!!![/hide]

[quote="nat mc"]How about this one:
Find next 2:
1 1 3 5 10 18 33 59[/quote]

[hide="The solution"]  Take the sum of 1 and 1.  1 and 3.  3 and 5...  And so on as you would in the fibbanoci sequence.  Then take each sum, and subtract it from the next number in the sequence.  You should get the following results:

1
1
2
3
5
8

  This is obviously the fibbanoci sequence.  The next number in this sequence is 13.  So the last sum  33 + 59= 92, must be 13 smaller than the next number in the sequence.  Thus the next number in the sequence is 92 + 13 or $ 105$ [/hide]


MATHCOUNTS geometry AMC USA(J)MO USAMO AIME

Man I'm not taking it because Redwood is taking the middle school test, and my mom doesn't want me to because I have too many competitions... I mean what the heck, it's free and I'm only wasting an hour of my life... compare that to the nine hours a week of MathCounts practice... geez.

Meh I wish I could go to Gunn. Hmm would either Gunn or Paly accept me on their math team?

Eh, I think all our teams are full. We basically have everything planned out already. That's kind of sad that Redwood won't have a high-school team, since you guys are perfectly capable of beating most high schools, but at least it means that you have another 90 minutes to cram for the USAMO.

No I meant can I permanently join your math team for this year or next year or something, or am I too stupid for it and have to be forced to stay on the stupid Saratoga High team?

Hmm do you really think Redwood can beat Lynbrook, Saratoga High, Paly, or Gunn?

Now what I'd *really* like to see is Palo Alto vs. San Jose area a.k.a. Paly + Gunn vs. Saratoga + Lynbrook.

Saratoga High's going to have something like 5 USAMO qualifiers next year, so what exactly are you worried about? And last time I checked, Redwood *did* beat all of those schools except for Gunn at the SMT.

Yeah well we cheated because we all took the General Test, which in general gives a *much* higher score, unless you're someone who scored three points higher than everyone else on a Geometry test *cough, cough*.

And well one person from Redwood in 8th grade (and the person in 7th grade who made it) only made it because of the AIME that favored people like them.

[quote="bookaholic"]Looks like the Paly-Gunn combo team will be taking it tomorrow. First time collaborating instead of competing with junggi since the iTest last year :o So who's the third representative from Gunn?[/quote]
whoa dude you're in the team?
never knew that
well see you tomorrow

palo alto team got 20/25....much worse than we expected? poop...(we even got #18 right : |)

Wow, what happened? Wait, how did we get #18 right? It was a totally random guess based on an obviously incorrect answer, if I remember right. Hmm. Okay, I probably should have been more helpful with checking answers. It's probably all my fault. Bleh. Team contests always make me feel guilty afterwards.

Yay! We got 13/15 (yes, I'm copying i_like_pie, but i'm on his team anyway...)

I can't believe we tied for first in our division...

@ubemaya: You didn't take it? Are you those other guys that got 13? If so, that would explain how our team even tied redwood... Maybe its a different redwood... :(


Hello guys,

I am thinking of opening a small trading firm, and here is the problem:

Say these are the costs of the products I bought from overseas:
- 10 blouses @ $5.00 each
- 12 pants @ $7.00 each
- 15 jackets @ $10.00 each
What is the percentage I should incorporate into the cost of each group above if the expenses I paid for the freight, taxes and whatnot total up to $142.00, so I know the final cost for each item ?

Thanks in advance
Romp.

I think he actually wasn't trying to $%Error. "latex" is a bad command.
$, but in using the dollar sign in his problem, it interpreted it as $%Error. "latex" is a bad command.
$.  Tip for next time: use "USD" or some other currency symbol instead of the S with a line thru it.


Suppose that f(x) is defined as x^2-2. What is the product of the members of the sequence from x=-5 to x=20?

EDIT: It said product not sum I am an idiot :blush:

You missed 4 letters: I said "product" not "sum".

Can x be any real number? If so, when $x=\pm\sqrt2$, the function comes out to be zero, thus the product is 0.

if you intended to say " numbers " it would be 0

altough

if you meant integers,

-1.88548363E40

The product of 23, 14, 7, 2, -1, -2, -1, 2, 7, 14, 23.... , 398 is something big.....

I hope it didn't just mean integers.

:blush:Originally I did mean integers but now I've decided that x is any real number.


$\Delta$ABC.$(O1);(O2)$ are two circle. $(O1)$ in in ABC and it touches AB;AC . $B;C \in (O2)$ such that two circle touch outside at $T$ . Prove that The incircle of $ABC$ $\in$ the internol bisector of $\widehat e{BTC}$

Let I be the incenter and let $(O_{1})$ touches AB, AC at U, V. By the lemma [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41667]Concyclic points with triangle incenter[/url], quadrilaterals BTIU, CTIV are both cyclic and $\angle BTI = \angle IUA = \angle IVA = \angle CTI,$ hence TI bisects the angle $\angle BTC$ formed by the lines BT, CT.


Prove that if two random variables X and Y are identically distributed (not necessarily independent),
then Cov (X+Y,X-Y)=0

Just write out the definition and follow your nose. No tricks needed.

$\text{Cov}(X+Y,X-Y)=E((X+Y)(X-Y))-E(X+Y)E(X-Y)$

$=E(X^{2}-Y^{2})-(E(X)+E(Y))(E(X)-E(Y))$

$=E(X^{2})-E(Y^{2})-(E(X)^{2}-E(Y)^{2})$

$=E(X^{2})-E(X)^{2}-(E(Y^{2})-E(Y)^{2})$

$=\text{Var}(X)-\text{Var}(Y)$

At this point, we realize that we didn't even need for $X$ and $Y$ to be identically distributed; all we need is for then to have the same variance.


limit
trigonometry
function
real analysis
real analysis unsolved

If $ f_{n}(x) \rightarrow f(x)$ as $ n \rightarrow \infty$ for each $ x \in [0,1]$ and $ f_{n}'(x) \rightarrow g(x)$ as $ n \rightarrow \infty$ for each $ x \in (0,1)$, does it follow that $ g(x)=f'(x)$ for each $ x$ in $ (0,1)$?  Either show why or provide a counterexample.

I have a counterexample for an endpoint, but not for an interior point...

Is there a straightforward counterexample or could this actually be true?

I have a counterexample for,
$ \lim f'_{n}(x) = f'(x)$

Consider, $ f_{n}(x) = \frac{\sin nx}{n}$ on $ \mathbb{R}$.

But, your problem makes the hypothesis that $ f'_{n}(x)$ actually exists.

:lol:

It's actually a two-part question.  The first part asked for a counterexample to $ f_{n}\rightarrow f$ pointwise on $ [0,1]$ implies $ f_{n}' \rightarrow f'$ pointwise on $ (0,1)$.

For this part the counterexample $ \frac{\sin (nx)}{\sqrt n}$ will work, but not for the second part.

The sequence $ f_{n}(x)=\sqrt{x^{2}+1/n}$ converges to $ f(x)=|x|$ on $ [-1,1]$, and the pointwise limit of $ f_{n}'$ exists everywhere on $ [-1,1]$. However, $ f'(0)$ does not exist.

Okay, but since the interval is actually $ [0,1]$ we could shift the function to look like:  $ f_{n}(x)=\sqrt{\left(x-\frac{1}{2}\right)^{2}+\frac{1}{n}}$.

Gee, that took several minutes to come up with... I love this forum.


On the PEN archive at http://ultrametric.googlepages.com/pen, I've just uploaded the draft of PEN Solutions Book, Volume A. The volume A includes the following PEN problems: A7  A22  A57  A60  A64  A66  A67  C2  E14  E17  E24  F12  H15 I10 I11  I12   J1     J3    L1     L13  M7  O45  P21  Q7 R3   and no more probles will be added in the volume A.  [b]More comments, alternative solutions, generalizations and specializations are welcome!!  [/b]

Can you please attach the PDF to this topic Hojoo? :)

Does there exist a book where are all this pen problems with solutions?

All the problems are [url=http://www.problem-solving.be/pen/index.php?f=30]here[/url] and most solutions can be found on this site. :)

I know that, but i thought there exist such book, because i dont realy like sitting in front of the computer, i prefer books, that is also the reason why i print the most problems from the internet.

We are working on a book form, but it requires quite some coding to get on-site solutions directly in a PDF to print, so I won't count on this for the coming months.

Hello! In our city there will be an olympiad, but i couldn't take part in that olympiad last year, so since that time i have been preparing. And also there is an exam every month, according to the results of this exams, students will participate in math camp. (First exam's results are very bad :blush: )
Just I want to prepare this exams very well and i need a help: If you have olympiad books with solutions (PEN,comb,geom,ineq., especially PEN with sol-s)
can you post me, please because i don't have any books that there are problems of national MO, PEN,shortlists and etc.
Please help me
Thank you!

Is there anyone who has books? pls send me. my e-mail: ulanbek11@bk.ru

Is there anyone who has books? pls send me. my G-mail: anh4321@gmail.com

And if someone has such a book, please send to me, too: i_minhhoang@yahoo.com.vn
thanks and nice!

Is there anyone who has books? pls send me. my e-mail: ijso9th@gmail.com


...?  [hide="If I understand the problem correctly"] Obviously not.  Let the positive integer we add be $p$.  Then none of the infinite subsets can have two members of the same equivalence class $\bmod p$ in them or the partition is not disjoint; hence each set can have at most $p$ members. [/hide]

It's really, really annoying that you post all your problems as images.  In addition to the (minor) additional hassle it causes me to have to click on an image to see the problem, as well as the inability to copy or quote the text, it makes everything you post invisible to the search function, which behaves poorly enough as it is.  Please stop doing this.

sorry. I think there is a solution, it's from engel  :)


The digits 1, 1, 2, 2, 3, 3 are arranged to form an odd six-digit integer. The 1's are separated by 1 digit, the 2's by 2 digits, and the 3's by 3 digits. What are the last three digits of this integer?

[hide]The number is 231213, so the last three digits are 213.[/hide]

ok, i won't make mine sound like davidli's :P [hide]
so, because the number for the answer of this question happens to be 231213, the answer for the previous question is 213 as it asks for the last 3 digits to be the final answer[/hide]

-jorian

So, you read and paraphrased his answer?

[quote="jhredsox"]no! how else do u explain the answer shortly (i made mine long)

btw, i did the problem, looked at HIS answer and put mine in different wording...

i did this with a traffic problem but apparently mine was the same, just a summary of is :P

-jorian[/quote]

Ummm...[hide="like this (I actually did it this way)"]Since the integer is odd, it has to end in either a one or a three. Now we break this up into two cases, one where the integer ends in a one and the other where the integer ends in a three.

Let's do the case with the one at the end first. Knowing where the first one is, we can easily place the second one. Now lets take a look at the digit between the two ones. This can either be a 2 or a 3. In both cases it obviously doesn't work, as you can check by yourself.

Now we will proceed to the case with the 3 at the end. We can again easily place the other 3. We know that the first digit must be a 2 because both the twos will not fit otherwise. So we can place the other two and then put in the ones and Voiala! The number is 231213, so the last three digits are 213 and we're done. :) [/hide]

[quote="jhredsox"]no! how else do u explain the answer shortly (i made mine long)

btw, i did the problem, looked at HIS answer and put mine in different wording...

i did this with a traffic problem but apparently mine was the same, just a summary of is :P

-jorian[/quote]

How about you just don't post at all?

but my goal in life is to be smarter than davidli :P

maybe i should look at splashd's sig some day

-jorian


AMC
USA(J)MO
USAMO
Support
geometry
geometric transformation

Who do you think will win the IMO this year? Of course, my guess would be [color=red]China[/color]

China. probably. but keep an eye on New Zealand - not that we'll win or even near it, but I think we got the best record over population or something. I'm pretty sure I'll be NZL005 if anyone was wondering :)

I'll be looking for a whole bunch of countries on the results (including N.Z.), and I made the poll with no preparation, so I apologize for the lack of India and European Countries on the poll. GTG, bye

I really have no clue because I don't really follow competitions, but I took a guess anyway... go losers, all of you...

-Isaac

That made me laugh, although I already expected something like that (well, expected it more from Prashant, but oh well.). well, the thing is, you can't you can't really be sure about these things...

I think the US team will be quite strong this year. Five members from last year's team haven't graduated yet, but one didn't make top 12, so he's out. But Tiankai Liu, Po-Ru Loh, Dani Kani, and Anders Kaseorg should be back. I'm not really sure who the other two will be.

[quote]Tiankai Liu, Po-Ru Loh, Dani Kani, and Anders Kaseorg [/quote]

That's the American team? That's pretty sad... i.e. the names...

-Isaac

Tiankai is NOT sad. he's won like every math contest ever... if u call that sad, then something's wrong

I think he was referring to the names not sounding American for the US IMO team.

Yeah I think all the american team are usually chinese. or something like that. Spose thats how everything works in this world..

yeah, sorry. I didn't mean that the team sounded like a bad team, in fact I think it's very string. I was referring to the fact that thee are no English stereotypes, like "John" or "Susie." Sorry for any hard feelings. Go Americans!

-Isaac

Well, Dani Kane is an American name. I have no clue where Anders is from but it definitely doesn't sound American. I imagine Eve knows and possibly MysticTerminator too.

um...  Well they make sure that everyone on the IMO team is at least a US citizen.  Tianki isn't going to MOP fyi...  He's going to RSI.

how bout po-ru-loh's brother po-ling-loh?

Actually Po-Ling is a girl. It's a shame that Tiankai isn't going to MOP. I noticed that myself, but is there still a possibility of his going to IMO?

that would be sister...  Po-Ling is a girl.

[quote="ComplexZeta"]Tiankai Liu, Po-Ru Loh, Dani Kani, and Anders Kaseorg [/quote]

Dani Kani?   :?:  :shock:  :?:  I think it's Daniel Kane, right?  That probably sounds more American.

But as to whether or not their names SOUND American, WHO CARES?  They LIVE in America, they competed in the USAMO, and they qualified...

By the way, I'm casting my vote for the USA.  Not because I'm certain that it'll win, but I like to be hopeful.  Maybe a win would put more of a spotlight on great math achievers.

sorry about the po-ling-loh as a brother thing. i didnt know. i just thought since their name started with po.

I just hope [b]some[/b] team wins. :mrgreen:

?

Anyway, I vote China even though I will root for the US team.

Actually, I think Po-Ling is his sister.  Even without Tiankai , I still think that the US team will be quite strong.  I bet it is going to be between China and the US this year. China is still going to win.  :lol:

I probably don't even have a chance to get to IMO, maybe 1% in 12th grade if I'm lucky enough and the feds are smart enough. Thing is, I think I can only apply for citizenship about 5 years from now, and I'm only going on to 8th grade, so that means, I'll get citizenship in like college. Why do they even have this rule. Someone isn't going to mess their chances up just because they want some other team to win. So, is it fair???

The voting here really surprised me.  In past years the top two have been Russia and China with the US coming in third, yet somehow the States got a lot of votes.  Man we weren't even close to 2nd last year.

I would assume that's because most of the people here are from the US, and they just [b]want[/b] the US to win, so they vote for it. (which on second thought, is what I should have done) (b/c the US rocks) (:D)

I (meant to) vote for China, but I'll be rooting for Sweden.

When did Iran win?

Iran won in 1998. Bulgaria got 2nd that year, and U.S. got third. Go to http://imo.math.ca/results/TCBY.html if you want to know who won each year.

[quote="Syntax Error"]cmon chinaboy............its all about india............well. i would pick china......but representin the indians............welcome to india where the cows eat hay and we drive auto rickshas everyday.........goat meat yummy sweets while the monkeys roaming............the roostas dont crow till 5 in the mornin.......word........woooooooooooooot.

well, personally i think its gonna be between china and russia....just cause they are the ones who consistently win............go CHINA AND INDIA![/quote]

I LOVE THAT SONG!! Anyways, probably China.

I dont really know, but rumor has it that there is some 8th grader portugese person at MOP who presumably went to IMO last year? OOOhh man...now I have seen everything! I just don't know anymore. If that is true....Although I think USA will come 1st this year just because I am baised (actually its pretty sad, also representin Indian, i should say that India should come 1st)...so on that note, I change my vote...India will be # 1. You just watch, 4 Perfect Papers, THe 2nd Dream Team haha...or the 4 horsemen of Apocalypse

w00t im with realzeta (jai hind!). ..but is that true about the 8th grade kid?

where did u get that from anywho. i agree wit the india thing. welcome to india where the cows eat hay and we drive autorickshas everyday. goat meat yummy sweets while the monkeys roamin. the roostas dont crow till 5 in the mornin. go luda go luda its my birfday. really. ya man

:lol: Personally, I think probably China, because of the pure statistics. China's got 1/6 of the word's population (.166666...), so they'd have more people than other countries, and more smart people, and more geniuses. Besides, the REALLY smart asian people in America aren't even citizens, so based on that one post some time ago, they won't get to participate or something.

When your name is Abhishek Asthana it doesn't get anymore indian than that...

Of course, my name is Ajai Sharma and I guess that's a hardcore indian name too... To be honest, there are also quite a few indians who represent in Biology... my coach at USABO was indian...

Indian HoNoR my friend...

Ok, by being an indian, I have to say that india will take 1st, however I am pretty damn sure that because china's china they have to take 1st otherwise it destroys everything that is good and pure in this world.

Also, one note of caution: The team that takes first isn't usually the team that deserves first. I shall support this thesis with a few concrete arguments.

Take the International Chemistry Olympiad, after hearing countless tales of my coach's  experience in the olympiad [i.e. being an alternate for the US team twice in a row] he has said that his friends who did go to IChO told him that there was once a time when the chinese team, during the translation of the test would essentially take a question like "Where does glycolysis occur?" and retranslate it into chinese into a question such as "Glycolyis is a process which occurs in the cytosol, where does Glycolysis occur?"

Another old tale, a singaporian student paid his teacher 40k yen [or whatever singaporian currency is] to use his biology textbook on his preliminary exams [analogous to AMC for mathematics] When he made the camp, he supposedly paid a large sum of money to his coaches while taking them to an upscale Whorehouse [pardon my french]. It was no doubt that he made the team...

I am not telling this because I am bitter of my experience at USABO, it was one of the funnest experiences of my existence. Rather, this is to let you know that regardless of what the outcome may be, it sometimes doesn't always prove to be the real outcome. And plus... they are kinda funny...

Peace,
The Idiot without a village.

It's a shame that these prestigious olympiads can be so corrupt. But I have never heard of the IMO being at all corrupt (not that I have any experience at this; perhaps TripleM can give us his perspective). It seems to me that the USAMO makes all possible attempts to avoid cheating (although I guess it is possible for a proctor to help a student with a problem, but I doubt that happens often).

We Singaporeans use the dollar. I didn't hear of that case though - I don't know...it's a little difficult to believe...where did you hear of this?

My friend who is going to IBO [international biology olympiad] told me that....

It's a shame I didn't think of that earlier... j/k

I haven't noticed any hint of cheating or anything, but most likely if any occured (which I doubt) it would only be amongst the top teams. Oh wait, that probably includes America..
But the organisers are real strict about their rules and everything. Last year they took all cellphones and locked them in a safe a long time before the exams even started. As for translation and such I wouldn't know but surely they would be checked! Although I suppose that could provide some problems. From what I have read/heard about past IMOs the choosing of the problems is VERY very strict, apparently they have votes on whether they should have a vote about which type of problem they should work on first. Or something ridiculous like that.

I suppose it still could be possible though. In my opinion I kinda doubt it..

I still think it's highly implausible. To have this happen on an international level...wouldn't people be raising a big hue and cry then? I don't have any experience with such things (take pity on the poor freshman) but still I don't think it could occur.

Isn't that cute...

Indeed, the question has arisen in my mind how such a thing could occur. Yet, does it surprise you when you see olympic atheletes having their medals revoked because of steroids, failing a drug test, or shooting up RBCs.

Yes my friend, after passing through the American education system such things do not seem so far fetched...

[b]JAI HIND![/b]

ThE InFaMoUs IW/OAV

I'm in the American education system as of now, and it still seems far-fetched. But forget it. Ca ne fait rien. It doesn't matter.

[quote="RealZeta"]I dont really know, but rumor has it that there is some 8th grader portugese person at MOP who presumably went to IMO last year? [/quote]

I was just reading through this thread because I was curious as to what people were saying before the fact, and I thought I should clarify this.

There was an eighth grade girl from [i]Puerto Rico[/i] at MOP this year who has been to the IMO for Puerto Rico twice now. She actually got a bronze medal this year, which is pretty impressive for an eighth grader. I think there's a good chance that we'll be hearing a lot more about her in the next four years.

In any case, there was no one from [i]Portugal[/i].

when i read the mop qualifier list, i saw sherry gong's state listed as "pr". Thank you for clarifying what pr is, cuz i would have never though it stands for puerto rico.


Let XOY be a right triangle with angle XOY = 90 degrees. Let M and N be the midpoints of legs OX and OY respectively. Given that XN = 19 and YM = 22, find XY.

[hide]

Let $ OM \equal{} x$ and $ ON \equal{} y$. We have, by the Pythagorean Theorem:
\[ x^2 \plus{} 4y^2 \equal{} 484
\]
\[ 4x^2 \plus{} y^2 \equal{} 361
\]
Hence, $ x^2 \equal{} 105$ and $ y^2 \equal{} 64$. We need $ \sqrt {4x^2 \plus{} 4y^2} \equal{} 26$

[/hide]

An even faster way to do it
would be to get the two equations
x^2+4y^2=484 and 4x^2+y^2=361 and add them, getting:
5x^2+5y^2=845
Multiplying by 4/5 then gives
4x^2+4y^2=(169)(4)
And finally, taking the square root of that gives
XY=[u]26[/u]

Of course, another way to do this would be
to assign XO and OY as a and b and then
to solve for a^2+b^2, although it takes a
considerably longer time because of algebra.
Nevertheless, it yields the same answer eventually.


You won, Jeremy won, Jane won, Daniel won. Berman and I lost. Darn. Actually I am kind of happy because now I don't need to spend time working on ToC problems and can do olympiads :wink:

sucks how you guys had to go against each other since you were so closely ranked

Yes, and had one less person entered their AMC score two pairs of NCers would have had to face off.  But I won, so I'm happy with that.  Round 2 is considerably more fun than Round 1.


floor function
number theory
number theory unsolved

This is hardly a Number Theory Unsolved Problem. If you consider it as a challenge to the others, it's too easy; if you consider it your own, it's the wrong forum.

Denote $ a \equal{} 2^kx$, then you ask $ \lfloor a \plus{} \frac {1} {2} \rfloor \equal{} \lfloor 2a \rfloor \minus{} \lfloor a \rfloor$, a particular case of the folklore Hermite identity, also easily provable by casework, for all real $ a$.


limit
induction
algebra unsolved
algebra

First, we have:

$ (U_n ): \left\{\begin{array}{l} U_1 \in R \\
U_n \equal{} U_{n \minus{} 1}^2 \plus{} aU_{n \minus{} 1} \plus{} 1. \\
\end{array} \right.(n \equal{} 1,2,3,4,.......)$

Question: Find $ \mathop {\lim }\limits_{n \to \infty } \left( {\frac {{U_1 U_2 U_3 .....U_{n \minus{} 1} }}{{U_n }}} \right)$ with:

1. $ a \equal{} 1$

2. $ a \equal{} \minus{} 1$

[quote="tchebytchev"]for $ a = - 1$:
if $ u_1 = 1$ then for every $ n$ we have $ u_n = 1$ so $ {\lim }\limits_{n \to \infty } \left( {\frac {{u_1 u_2 u_3 .....u_{n - 1} }}{{u_n }}} \right) = 1$.
now by induction we can prove easily that $ u_1 u_2...u_{n - 1} = \frac {u_n - 1}{u_1 - 1}$ therefore $ \left( {\frac {{u_1 u_2 u_3 .....u_{n - 1} }}{{u_n }}} \right) = \frac {u_n - 1}{u_n(u_1 - 1)} = \frac {1}{u_1 - 1}(1 - \frac {1}{u_n})$.
we have $ u_n = u_{n - 1}^2 - u_{n - 1} + 1 \ge u_{n - 1}$, so $ (u_n)$ is increasing , if it has a limit $ l$ then $ l = l^2 - l + 1$ so $ l = 1$.
if $ {\lim }\limits_{n \to \infty } u_n = 1$ then $ {\lim }\limits_{n \to \infty } \left( {\frac {{u_1 u_2 u_3 .....u_{n - 1} }}{{u_n }}} \right) = 0$.
and if $ {\lim }\limits_{n \to \infty } u_n = + \infty$  then $ {\lim }\limits_{n \to \infty } \left( {\frac {{u_1 u_2 u_3 .....u_{n - 1} }}{{u_n }}} \right) = \frac {1}{u_1 - 1}$.[/quote]

That's okie, that is a great solution! But $ a=1$ :(

With $ a=1$, I cannot find the solution :(. Can you help me? Thank you very much! :)


$ x_1,\dots,x_n>0$ prove
\[ {\root{n}\of{\prod _{k\equal{}1}^{n}\
\sum_{t\equal{}1}^{k}\
\frac{1}{t^2\cdot\sqrt[t]{x_1\cdot\ldots\cdot x_t}}
}}
\cdot
\sum _{t\equal{}1}^{n}\frac{\sum
_{a\equal{}1}^{t}\sum _{b\equal{}1}^{a}{x_b}}{t}
>
\left(\frac{n}{e}\right)^2\]


geometry
calculus
AMC
AIME
AMC 12
AoPSwiki
USAMTS

I'm in 12th grade and have basically done the contests offered by my high school - AMC and a few local things. (I don't have any scores.) I've taken algebra, geometry, precalc, and calculus BC but never took anything like number theory beyond the normal high school courses. I'd like to become more seriously involved with competitive math this year; any tips for where I should start? I'm sorry I can't be more specific about what I'm looking for, but I really have no clue what people do for this sort of thing.

And yeah, I know, it's my last year of high school, but I took BC Calculus in two and a half months over the summer while doing other stuff, so I figured I'm a quick enough study to at least give it a shot. I don't expect to win anything incredible.

The AMC is really the main place to get started. Have you made it to the AIME? (You need a 100 or greater on the AMC12) If you click the resources tab at the top of the page in mathlinks, or the aopswiki, they give past AMC exams to practice. You can also take classes on artofproblemsolving or buy the books. You can also practice the problems on the website forum. The USAMTS is another competition open to all high school students, which introduces you to olympiad type problems. Other than that it really matters what your school offers. Do they have have Mu Alpha Theta, or any other competition they are involved in? Are there any other kids at your school into math competitions? (There may not be, I'm the only one who takes it seriously at my school). Unfortunately, now that you're in 12th grade there won't be much time to prepare...

doesn't really matter how old you are!
yes it would be better for competitions if you were younger BUT BUT BUT BUT, you still have a chance and putnam and IMC.
Even then, you could still have a math based career.

Thanks for the tips!  :D   No, I didn't make it to the AIME, and I think I scored in the high 80's last year, so I don't really expect to make it. We don't have Mu Alpha Theta, but there are state and county math leagues and I might try the USAMTS.

Question about the Art of Problem Solving books - which one? I'm not sure, since I'm done with the regular high school math courses but a total beginner at anything outside high school math.

All the introductory books are fine (I personally own 2/4, but I'm going to get the remaining two soon). But you might want to get Art of Problem Solving: Vol. 1 because that gives you an overall view of what the problems are like. The introductory books just go more in-depth and they are certainly enjoyable!

EDIT: You might want to get Vol. 2/Intermediate Algebra also seeing as you are a senior and can only take the AMC 12.


Harvard college Yale Columbia MIT Princeton

I am planning on running for President of the United States in 2032.  What colleges are godd for politics?

Vote for Me      ;)

There are lots of schools in D.C. that have strong programs along those lines.  Any good college information book should point them out.

If winning the election is your goal, you might want to get a reality check by looking at the (surprising) list of colleges attended by previous presidents.

Hmm...it seems that quite a few presidents have nice connections with Yale. And almost half of the folks on Capital Hill have Harvard on their resume. Political schools may not always be your best choice.

[size=150][b]YALE!![/b][/size]

This is a ture stroy and I am not making this up. When we vistied Yale  eary this year our  guide, a young man from British  Columbia  who  was a Junior  started talking about US elections.  He predicted, the Democratic candidate would be Kerry (This is  when Dean was Front-runner)  as he is from Yale.   And  although contest would be very close, and Kerry would come very close to winning , Kerry  would concede and   Bush would win.  ("No  if, and, buts .. it is all  part of a grand plan"

Kerry's chance would come in 2008... etc.. it's not, Democratic/Republican it would be 16 years of "skull and bones". This is Yale's grand plan.

Everyone laughed, but I wonder ... :) .  His prediction is so true.. and  he was really funny in the way he said all this so confidently.

Don't forget Bush W,  Clinton (He was Yale Law student)  and Bush Sr.  (Also skull and bones) went to Yale... Not to mention Hillary also went to Yale Law school.

My son was impressed, He almost chose  Yale, but   he is interested  in Architecture/Engineering and another Guide (Who said she had the same dorm room as Jodie Foster) told about how Yale Engineers "poured"  sulphuric acid over one building to make it look old. (and in that process,   weakend  the whole  structure made out of bricks- and they have to do major reconstruction). My son asked " And how good is your Engineering school - or something to that effect..." ... But that's another story.. He is  at   MIT where they also have strange buildings, but  at least they do not pour sulphuric acid over it...

 :lol:  :lol:

[quote="Gyan"]and another Guide (Who said she had the same dorm room as Jodie Foster) told about how Yale Engineers "poured"  sulphuric acid over one building to make it look old. (and in that process,   weakend  the whole  structure made out of bricks- and they have to do major reconstruction).[/quote]

I was told this too when I visited Yale.  I know exactly which building you're talking about! :D

One of my counselors at Math path, Willy Myerson, told me that Yale was Harvard for dummies and that Harvard hates Yale.  He graduated from Harvard and is currently studying in Cambridge.

Well, he's quite a reliable and unbiased source in that case.

well, since many presidents have been lawyers before election, i would recommend political science or economics for undergrad. that would mean schools such as yale (the EP&E program is amazing), the wharton school at UPenn (naturally), harvard, princeton, and georgetown (which is 20 minutes away from me.  :) )

Just a piece of trivia -  Some place,  the president of the country happens to be - literally-   a rocket scientist !  (And before the current  job he  was   (apart from holding other posts) , chief scientific adviser to the  Prime Minister,  head of the space agency etc etc..)

And  the day to day affairs of this  country is  taken care  by the  Prime Minister, who  has degrees from Cambridge, and Oxford (PhD) and   before  the current job   had professorship in a  few of the most prestigious  academic institutions ..,, as an Economist,

And, of course, both ,  when they were in school, were excellent in Math

Yes I am talking about the current President and the Prime Minister of India,

So may be you don't    always have to go to a law school .
(Okay I am stretching the point ... You  have to take  a look at Yale, seriously  if  Washington is where you want to go!
 :)  :)  :)


Let $a,b,c$ be the sidelengths of a triangle such that $a^2+b^2 > 5c^2$ holds.
Prove that $c$ is the shortest side of the triangle.

Let $b\le c$, then $a>2c>c+b$ contadition.

[quote="Arne"][quote="Rust"]Let $b\le c$, then $a > 2c$[/quote]

Why? :huh:[/quote]
Because if $b\le c$ and $a\le 2c$ then $a^2+b^2\le 5c^2$. Right? :roll:

Wlog, $ b\leq c$. Then, $ c^{2}\plus{}a^{2}\geq a^{2}\plus{}b^{2}> 5c^{2}$, or $ a > 2c$. $ c\geq b$ follows, that  $ a > 2c\geq 2b$. So, $ 2a \equal{} a\plus{}a > 2b\plus{}2c \equal{} 2(b\plus{}c)$, or $ a>b\plus{}c$-condradiction. So, $ c< b$ and $ c < a$, Q.E.D.

[quote="ZetaX"]Let $ a,b,c$ be the sidelengths of a triangle such that $ a^{2}\plus{}b^{2}> 5c^{2}$ holds.
Prove that $ c$ is the shortest side of the triangle.[/quote]

It is old problem  :)


Πιστεύω πως είναι καιρός να ανοιχτεί το τόπικ. Επί την ευκαιρία καλό μήνα και καλή επιτυχία σε όσους θα δώσουν!

Ξεκινάω με ένα προτεινόμενο θέμα για την 3η λυκείου. Το βλέπω για 1ο θέμα. Όχι για παραπάνω  :P

Έστω η συνάρτηση: $ f(x) \equal{} (\sqrt{5} \plus{} 2)^x \minus{} (\sqrt{5} \minus{} 2)^x$

Να βρεθεί η αντίστροφή της.

Eukola vlepoume oti h f einai 1-1 (giati einai gnhsiws au3ousa me to krithrio prwths paragwgou)kai ara exei anistrofh se ka8e shmeio tou pediou orismou. Twra,
$ y\equal{}(\sqrt5\plus{}2)^x\minus{}(\sqrt5\minus{}2)^x\equal{}>y(\sqrt5\plus{}2)^x\equal{}(\sqrt5\plus{}2)^(2x)\minus{}1$ opote 8etontas $ (\sqrt5\plus{}2)^x\equal{}z$ pairnoume $ yz\equal{}z^2\minus{}1\equal{}>z\equal{}(y\pm \sqrt(4\plus{}y^2))/2$
Gia $ z\equal{}(y\minus{}\sqrt(4\plus{}y^2))/2$ exoume $ (\sqrt5\plus{}2)^x\equal{}(y\minus{}\sqrt(4\plus{}y^2))/2$ alla to prwto meros einai panta 8etiko enw to deutero arnhtiko atopo ara
$ z\equal{}(y\plus{}\sqrt(4\plus{}y^2))/2\equal{}>(\sqrt5\plus{}2)^x\equal{}(y\plus{}\sqrt(4\plus{}y^2))/2\equal{}>$
$ x\equal{}Log[\sqrt5\plus{}2][(y\plus{}\sqrt(4\plus{}y^2))/2]$
ara $ y\equal{}Log[\sqrt5\plus{}2][(x\plus{}\sqrt(4\plus{}x^2))/2]$ pou eukola vlepoume oti epalh8euei to zhtoumeno
Shmeiwsh epeidh den 3erw latex kala me to $ Log[a][x]$ sumvolizw ton logari8mo tou x me vash to a

Για θαλή β' λυκείου πάντως για πρώτο θέμα ειδικά βλέπω να πέφτει καμιά εύκολη ανισότητα του στυλ:

Για κάθε πραγματικούς $ a,b$ αποδείξτε ότι

$ a^2\plus{}ab\plus{}b^2\geq 3(a\plus{}b\minus{}1)$  :P

fernoume ta panta sto prwto meros opote prepei na dei3oume oti $ a^2\plus{}a(b\minus{}3)\plus{}(b^2\minus{}3b\plus{}3)>\equal{}0$
h diakrinousa tou triwnumou einai
$ (b\minus{}3)^2\minus{}4(b^2\minus{}3b\plus{}3)\equal{}b^2\minus{}6b\plus{}9\minus{}4b^2\plus{}12b\minus{}12\equal{}\minus{}3b^2\plus{}6b\minus{}3\equal{}\minus{}3(b^2\minus{}2b\plus{}1)\equal{}\minus{}3(b\minus{}1)^2<\equal{}0$
me thn isothta na isxuei an $ b\equal{}1$.
Etsi loipon an b diaforo tou 1 exoume arnhtikh diakrinousa kai 8etiko suntelesth tou triwnumou, ara to triwnumo einai megalutero tou 0 kai h dosmenh isxuei
An $ b\equal{}1$ tote h anisothta ginetai $ a^2\minus{}2a\plus{}1>\equal{}0 \equal{}>(a\minus{}1)^2>\equal{}0$ pou profanws ixsuei kai ara h askhsh lu8hke!
Exeis kamia allh lush up opsin?
Hlias

Βασικά και εγώ αυτή είχα υπόψην μου (και σίγουρα αυτή θα είχαν και όποιοι βάζανε μια τέτοια άσκηση στον Θαλή) αλλά πριν λίγο βρήκα και μια άλλη λύση χωρίς να χρησιμοποιώ τριώνυμο και διακρίνουσα :wink:

Πολλαπλασιάζουμε επί 2 και τα πάμε όλα στο πρώτο μέλος και έχουμε:

$ 2a^2 \plus{} 2ab \plus{} 2b^2 \minus{} 6a \minus{} 6b \plus{} 6\geq 0\Longrightarrow (a \plus{} b)^2 \plus{} (3 \minus{} a)^2 \plus{} (3 \minus{} b)^2\geq 12$ που ισχύει γιατί από την ανισότητα
$ x^2 \plus{} y^2 \plus{} z^2\geq \frac {(x \plus{} y \plus{} z)^2}{3}$ έχουμε ότι

$ LHS\geq \frac{(a \plus{} b \plus{} 3 \minus{} a \plus{} 3 \minus{} b)^2}{3} \equal{} \frac {36}{3} \equal{} 12$  QED

Ta eixa sumplhrwsei ta tetragwna k egw k meta lew asto 8a to lusw me triwnumo!Mpravo, wraia lush pantws!
K parepiptontws sugxarhthria gia tis diakriseis s!

Σε ευχαριστώ πολύ για τα καλά σου λόγια εύχομαι φέτος να είμαστε μαζί στην Ισπανία :wink:

Στέλνω κι άλλη μία λύση της ανισότητας στο συνημμένο word (δεν γνωρίζω Latex) , όχι πιο σύντομη , αλλά βασισμένη  στη σχολική ύλη της Α΄ Λυκείου.

Diairw to deutero klasma (ari8mhth k paronomasth) me m opote pairnw oti to prwto melos isoutai me
$ a^2/(a\minus{}m)\plus{}b^2/(b\minus{}1/m)$ to opoio logw cauchy einai $ >\equal{}(a\plus{}b)^2/(a\plus{}b\minus{}m\minus{}1/m)$  ara prepei na dei3oume oti $ (a\plus{}b)^2/(a\plus{}b\minus{}m\minus{}1/m)>\equal{}8 \equal{}>(a\plus{}b)^2\minus{}8(a\plus{}b)\plus{}8(m\plus{}1/m)>\equal{}0$ to opoio einai ena triwnumo ws pros a+b opote arkei na dei3oume oti exei diakrinousa $ <\equal{}0$ omws $ D\equal{}64\minus{}32(m\plus{}1/m)$ kai $ m\plus{}1/m>\equal{}2$ ara pairnoume to zhtoumeno...
Kai nai xrhsimopoihsa ola ta dedomena ths askshshs (estw k siwphra!  :wink: )

Hlias

Μπράβο Ηλία. Αυτόι είχα και εγώ στο υπ΄όψη. Όσο για τα δεδομένα που λές ( σίγουρα τα χρησιμοποίησες, δεν το αμφισβητώ ), παίζουν σημαντικό ρόλο για την άσκηση.Γενικά η παράσταση έχει κάτω φράγμα το οποίο εξαρτάται από το m. Αλήθεια βρήκες πότε ισχύει η ισότητα? Μπορεί να φαίνεται γελοία η ερώτηση ( m=1 ) αλλά για ψάξτο λίγο. Έχω καταλήξει σε κάποια συμπεράσματα και για τα α,β.

Αν $ f: (0,\plus{}\infty)\rightarrow(0,\plus{}\infty)$ και ισχύει $ \lim_{x\rightarrow \plus{}\infty}\frac{2x}{x}\equal{}1$, να βρεθεί το:

$ \lim_{x\rightarrow \plus{}\infty}\frac{2008x}{x}$

[quote="bilot4"]Για θαλή β' λυκείου πάντως για πρώτο θέμα ειδικά βλέπω να πέφτει καμιά εύκολη ανισότητα του στυλ:

Για κάθε πραγματικούς $ a,b$ αποδείξτε ότι

$ a^2 + ab + b^2\geq 3(a + b - 1)$  :P[/quote]

$ a^2 + ab + b^2 - 3(a + b - 1) = a^2 + (b - 3)a + b^2 - 3b + 3$

$ = (a + \frac {b - 3}{2})^2 + \frac {3}{4}(b - 1)^2\geq 0.$

[quote="vangelismak"]Εγω προτείνω μια άλλη ανισότητα!

Αν m>0 και α>m , β> 1/m  να αποδείξετε ότι  (a^2)/(a-m)+(m*b^2)/(mb-1) ≥ 8[/quote]

Since $ a > m,\ b > \frac {1}{m}$, we have ${ \frac {a^2}{a - m} + \frac {b^2m}{bm - 1}} = \frac {a^2}{a - m} + \frac {b^2}{b - \frac {1}{m}}\geq \frac {(a + b)^2}{a + b - (m + \frac {1}{m}})$

$ = a - m + b - \frac {1}{m} + \frac {(m + \frac {1}{m})^2}{a - m + b - \frac {1}{m})} + 2(m + \frac {1}{m})$

${ \geq 2\sqrt {(a - m + b - \frac {1}{m}})\cdot \frac {(m + \frac {1}{m})^2}{a - m + b - \frac {1}{m}}} + 2(m + \frac {1}{m})$

$ \geq 4(m + \frac {1}{m})\ (\because m > 0).$

$ \geq 4\cdot 2\sqrt {m\cdot \frac {1}{m}} = 8.$

[quote="mostel"]Αν $ f: (0, \plus{} \infty)\rightarrow(0, \plus{} \infty)$ και ισχύει $ \lim_{x\rightarrow \plus{} \infty}\frac {2x}{x} \equal{} 1$, να βρεθεί το:

$ \lim_{x\rightarrow \plus{} \infty}\frac {2008x}{x}$[/quote]

Ooops! The complete exercise is wrong! Sorry :( The correct one is:

If $ f: (0, \plus{} \infty)\rightarrow(0, \plus{} \infty)$ and $ \lim_{x\rightarrow \plus{} \infty}\frac {f(2x)}{f(x)} \equal{} 1$, compute the following:

$ \lim_{x\rightarrow \plus{} \infty}\frac {f(2007x)}{f(x)}$

Sorry again!

Cheers,
Stelios

Δημήτρη τι θες να πεις για το τρίτο πρόβλημα Β' λυκείου? Εμένα πάντως για τα θέματα της δευτέρας λυκείου η γνώμη μου είναι:
Πρώτο ΠΟΛΥ εύκολο ακόμα και για Θαλή (αν και καλό είναι να ενθαρύνονται όλοι με ένα εύκολο πρόβλημα)
Το δεύτερο κι αυτό εύκολο πάιρνοντας περιπτώσεις...(τα πρώτα 2 δεν μου άρεσαν πολύ...)
Το τρίτο πρόβλημα κατά τη γνώμη μου ήταν καλό στο πρώτο ερώτημά του, το οποίο με δυσκόλεψε πολύ :blush: Απ' την άλλη στο συγκεκριμένο πρόβλημα έλειπε η διευκρίνιση για το ποια γωνία του ΑΒΓ ήταν ορθή... :huh:
Το τέταρτο πρόβλημα μου άρεσε αρκετά :) γιατί δεν απαιτούσε τίποτα ιδιαίτερες γνώσεις και ήταν και σχετικά δύσκολο αν δεν σκεφτόταν κανείς να διαιρέσει με τις μεταβλητές του δεξιού μέρους σε κάθε ισότητα.

Παρακαλώ όποιον μπορεί να βάλει τα θέματα της τρίτης λυκείου,καθώς δεν είχα την ευκαιρεία να τα δω..

Ωραία θεματάκια στην Β Λυκείου ... Το τρίτο ήταν αρκετά καλό .
Στην Γ λυκείου επίσης ωραία θέματα ... (εκτός από το 2ο που είχε κάποιες πραξούλες)

Λοιπόν θα βάλλω τα θέματα 3, 4 της Γ λυκείου

3) Έστω  f : R->R  ώστε $ f(f(x) \minus{} f(y)) \equal{} f(f(x)) \minus{} y$  για κάθε $ x,y\in R$

Να αποδείξετε ότι η f είναι περιττή . Εγώ θα προσθέσω και κάτι άλλο . Να βρείτε τon typo tis  f στο Q και τon typo tis f sto R

4) Για καθε $ a,b,c\in R$ ανα δύο διαφορετικούς μεταξύ τους νδο

$ (\frac {a \plus{} b}{a \minus{} b})^2 \plus{} (\frac {b \plus{} c}{b \minus{} c})^2 \plus{} (\frac {c \plus{} a}{c \minus{} a})^2\geq 2$

Η λύση μου στο 4 .
Προφανώς $ (\frac {a \plus{} b}{a \minus{} b} \plus{} \frac {b \plus{} c}{b \minus{} c} \plus{} \frac {c \plus{} a}{c \minus{} a})^2\geq 0$

Θα δείξω επιπλέον ότι $ \sum\frac {(a \plus{} b)(b \plus{} c)}{(a \minus{} b)(b \minus{} c)} \equal{} \minus{} 1$

Απόδειξη
Θεωρὠ το πολὐώνυμο $ f(x) \equal{} \sum\frac {(x \minus{} a)(x \minus{} b)}{(c \minus{} a)(c \minus{} b)} \minus{} 1$

Το πολυώνυμο αυτό είναι δευτέρου βαθμού . Όμως $ f(a) \equal{} f(b) \equal{} f(c) \equal{} 0$ άρα $ f(x) \equal{} 0$ για κάθε χ
άρα $ f(a \plus{} b \plus{} c) \equal{} 0$  από όπου $ \sum\frac {(a \plus{} b)(b \plus{} c)}{(a \minus{} b)(b \minus{} c)} \equal{} \minus{} 1$

QED   :D
 :D

[quote="silouan"]Ωραία θεματάκια στην Β Λυκείου ... Το τρίτο ήταν αρκετά καλό .
Στην Γ λυκείου επίσης ωραία θέματα ... (εκτός από το 2ο που είχε κάποιες πραξούλες)

Λοιπόν θα βάλλω τα θέματα 3, 4 της Γ λυκείου

3) Έστω  f : R->R  ώστε $ f(f(x) \minus{} f(y)) \equal{} f(f(x)) \minus{} y$  για κάθε $ x,y\in R$

Να αποδείξετε ότι η f είναι περιττή . Εγώ θα προσθέσω και κάτι άλλο . Να βρείτε τon typo tis  f στο Q και τon typo tis f sto R

4) Για καθε $ a,b,c\in R$ ανα δύο διαφορετικούς μεταξύ τους νδο

$ (\frac {a \plus{} b}{a \minus{} b})^2 \plus{} (\frac {b \plus{} c}{b \minus{} c})^2 \plus{} (\frac {c \plus{} a}{c \minus{} a})^2\geq 2$

Η λύση μου στο 4 .
Προφανώς $ (\frac {a \plus{} b}{a \minus{} b} \plus{} \frac {b \plus{} c}{b \minus{} c} \plus{} \frac {c \plus{} a}{c \minus{} a})^2\geq 0$

Θα δείξω επιπλέον ότι $ \prod\frac {(a \plus{} b)(b \plus{} c)}{(a \minus{} b)(b \minus{} c)} \equal{} \minus{} 1$

Απόδειξη
Θεωρὠ το πολὐώνυμο $ f(x) \equal{} \prod\frac {(x \minus{} a)(x \minus{} b)}{(c \minus{} a)(c \minus{} b)} \minus{} 1$

Το πολυώνυμο αυτό είναι δευτέρου βαθμού . Όμως $ f(a) \equal{} f(b) \equal{} f(c) \equal{} 0$ άρα $ f(x) \equal{} 0$ για κάθε χ
άρα $ f(a \plus{} b \plus{} c) \equal{} 0$  από όπου $ \prod\frac {(a \plus{} b)(b \plus{} c)}{(a \minus{} b)(b \minus{} c)} \equal{} \minus{} 1$

QED   :D
 :D[/quote]

[size=150][b][color=red] Εξαιρετική λύση !!!!Εύγε ![/color][/b][/size]

Στο τεταρτο της Β λυκειου αντι να διαιρεσω πολλαπλασιασα καταλληλα ωστε στα δευτερα μελη να εμφανιστει το xyz.
 δεν νομιζω να εχει καποια διαφορα??

[color=red][b]Καλά αποτελέσματα και καλή συνέχεια  στην επόμενη φάση ![/b][/color]

   Μπάμπης

Δημήτρη για το ποια γωνία ήταν ορθή, οι άλλες δεν έβγαιναν στο σχήμα. :wink:  Ρώτησα ένα άτομο της Μαθηματικής Εταιρείας που ήταν εκεί στο κέντρο που έδινα και είπε ότι δε χρεάζεται να το διευκρινίσει ο μαθητής... :)

Για Β Λυκείου τα δύο πρώτα θέματα μου φάνηκαν ότι θα είναι η βάση. Να φανταστείτε άκουγα άτομα που έλεγαν ότι το 2ο ήταν δύσκολο... :maybe:  Το τρίτο σχετικά απλό αν και κι εμένα με δυσκόλεψε λιγάκι το πρώτο ερώτημα στην αρχή. :D  Στο τέταρτο διαίρεσα ανά δύο κατά μέλη κι έθεσα $ (xy)^2\equal{}a$ κτλ. και βγήκε ένα 3 επί 3 εύκολο αλλά μου τελείωσε ο χρόνος πριν κάνω τις πράξεις... :blush:  Είχα χάσει χρόνο να προσπαθώ να βρω έξυπνη λύση... :(

Γενικά δε μου φάνηκαν έξυπνα τα θέματα πάντως...

Πρωτονιε (παρεπιπτόντως το μικρό σου :?: ) συμφωνώ οτι τα δύο πρώτα θα είναι η βάση . Το 3ο θέμα ήταν πιστεύω αρκετά ωραίο δε χρειαζόταν ιδιαίτερες γνώσεις
Γεωμετρίας .Βέβαια το πρώτο ερώτημα ήταν δυσκολούτσικο . Στο 4ο , θα συμφωνήσω πως αν δεν έβλεπες τη διαίρεση τοτε τα πράγματα ήταν αρκετα δύσκολα αλλά και πάλι όχι απιαστα  . Γενικώς φέτος τα θέματα ήταν δυσκολότερα ,ίσως, από πέρυσι .

Ρε παιδιά, που πιστεύετε οτι θα παίξει η βάση για τη Γ Λυκείου???

Μιχάλη εγώ πιστεύω γύρω στα 2 θέματα

Παιδιά θέλω να πω κάτι άσχετο . Έχετε συνειδιτοποιήσει την εξέλιξη που έχει το φορουμ από την ημέρα που ξεκίνησε δηλαδή στις 24 Σεπτεμβρίου 2005 . Έχει πετύχει  το στόχο του αρκεί να συνεχίσουμε ετσι .  :)

το πιστεύεις ρε συ ότι εγραψα φουλ τα δυο ,κ ξεχασα  να απορρίψω στο τέλος του δεύτερου τη ρίζα που μηδένιζε τον παρονομαστη?????? :wallbash_red:  :wallbash_red:

Οχι ρε Μιχάλη όχι όχι όχι  :ninja:  :ninja:  :ninja:
Δεν σε πιστευω

εχεις δικιο ειμαι τραγικός :( !!!! Αλλά never mind ότι γίνει...................

Αν είναι δύο θέματα η βάση και αν λάβουν υπόψη ότι στο 1ο δεν έκατσα να κάνω πράξεις ούτε στο 4ο, την έκατσα  :lol:


trigonometry
number theory
relatively prime
number theory unsolved

Not the same as http://www.artofproblemsolving.com/Forum/viewtopic.php?t=292753, but with similarities.
Find the relatively prime natural integers $ a,b,c$ such that there exists $ n\in \mathbb{N}$ with $ (a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c) \equal{} (2n)^4$.
$ 3, 25, 26$ or $ 9, 10, 17$ are such solutions.

i know that i'm saying the obvious but we can assume that a, b and c are triangle sides and for this triangle we have:
b*c*(sinA)=2(2n)^2 so it follows that sinA, sinB and sinC are rational.

You also have $ c^2\equal{}a^2\plus{}b^2\minus{}2ab\cos(\gamma)$ so $ \cos(\alpha), \cos(\beta), \cos(\gamma)$ are also rational
and from $ \sin^2(\alpha)\plus{}\cos^2(\alpha)\equal{}1$ it follows that $ \sin(\alpha)\equal{}\frac{x^2\minus{}y^2}{x^2\plus{}y^2}$ and $ \cos(\alpha)\equal{}\frac{2xy}{x^2\plus{}y^2} \ x,y \in Z$ or vice versa


Putnam
real analysis
real analysis unsolved

Suppose $ u$ is homogeneous of degree $ k$ and let $ f(\vec{x}) \equal{} u(\vec{x}/r)$.
Show that $ \nabla^2u(\vec{x}) \equal{} k(k \plus{} n \minus{} 2)r^{k \minus{} 2}f(\vec{x}) \plus{} r^k\nabla^2f(\vec{x})$

no one can solve this?

i think this is related to Putnam 2005, B5.

by the way, $ r\equal{}|x|\equal{}\sqrt{x_1^2\plus{}...\plus{}x_n^2}$


If $ a,b,c$ are the  sides of a triangle and $ T$ its area

Prove that: $ a^2\plus{}b^2\plus{}c^2\ge 4\sqrt{3}T$

already posted,
put that:
$ a\equal{}x\plus{}y$ , $ b\equal{}y\plus{}z$  , $ c\equal{}z\plus{}x$
using Heron formula, we get,
$ T\equal{}\sqrt{xyz(x\plus{}y\plus{}z)}$
and the inequality becomes:
$ 2(x^2\plus{}y^2\plus{}z^2)\plus{}2xy\plus{}2yz\plus{}2zx \geq 4\sqrt{3xyz(x\plus{}y\plus{}z)}$
we have $ 2(x^2\plus{}y^2\plus{}z^2)\plus{}2xy\plus{}2yz\plus{}2zx\geq 4(xy\plus{}yz\plus{}zx) \geq 4\sqrt{3xyz(x\plus{}y\plus{}z)}$
from which we deduct our inequality,

IMO 1961 , the famous Weitzenbock's inequality. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14388]here[/url]


Find the value of $ n$ if $ a^5\cdot a^{3n}\equal{}\frac1{a^3}$, where $ a>1$. Express your answer as a common fraction.


function
integration
algebra
rational function
complex analysis
complex analysis unsolved

I really don't know what to do with this question so please help...

[quote]Investigate under which conditions on $ f$ the following equality holds:
\[ \sum_{n \equal{} \minus{} \infty}^{ \plus{} \infty} f(n)\exp(int) \equal{} \minus{} \sum \textrm{Residue}\left(f(z) \frac {2 \pi i \exp(i z t)}{\exp(2 \pi i z) \minus{} 1}\right)
\]
where the sum is taken over all poles of $ f$. [/quote]

If you transfer the terms from the right hand side to the left and assume that $ f(n)$ is well-defined for all $ n$, you'll get the sum of the residues at [b]all[/b] poles of the function $ F(z)\equal{}f(z)\frac{2\pi i e^{i zt}}{e^{2\pi i z}\minus{}1}$. Now we would like to employ the Residue Theorem to conclude that this full sum is $ 0$. To this end, we need to be sure that poles are the only possible singularities, i.e., that $ f$ is meromorphic, that the series in question converges in some sense, and that there is a series of closed contours $ \Gamma_n$ eventually surrounding every compact set for which we can make an easy estimate for $ \int_{\Gamma_n}F(z)\,dz$. If you really want to use this formula for computing the sums of Fourier series, is natural to restrict $ t$ to the range $ [0, 2\pi]$ (because the left hand side is $ 2\pi$-periodic), which makes the second factor in $ F(z)$ bounded outside small disks containing the integers, after which it is not hard to argue that the formula can be used whenever $ f$ is a rational function decaying faster than $ 1/z$ at infinity and having no singularities at the integer points. This is not the most general case, of course, but I think you've got the idea by now.  :)


[quote="jin"]$f(x)=(1+e^x)^{-2},f'(x)=-2e^x(1+e^x)^{-3},f''(x)=-2e^x(1+e^x)^{-4}(1-2e^x)$
So when $x<-\ln2$,$f''(x)<0$
Then we can see the inequality doesn't always holds.[/quote]

did you mean the ineq is wrong, by Jensen's ? ( :D $\ \equiv$ you're wrong)

See
http://www.mathlinks.ro/Forum/viewtopic.php?p=202141#p202141 (post #14) and http://www.mathlinks.ro/Forum/viewtopic.php?t=64933

There was a problem in the last year's TST said that when $xyzt=1$,it holds.

[quote="Lovasz"]Let $ x,y,z,t > 0$ and $ xyzt \ge 1$, then

$ \frac {1}{(1 \plus{} x)^2} \plus{} \frac {1}{(1 \plus{} y)^2} \plus{} \frac {1}{(1 \plus{} z)^2} \plus{} \frac {1}{(1 \plus{} t)^2} \ge \frac {4}{(1 \plus{} \sqrt [4]{xyzt})^2}$.[/quote]
See http://www.mathlinks.ro/Forum/viewtopic.php?t=30724

[quote=Lovasz]Let $x,y,z,t>0$ and $\sqrt[4]{xyzt} \ge 1$, then show that:
\[ \frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{(1+z)^2}+\frac{1}{(1+t)^2} \ge \frac{4}{(1+\sqrt[4]{xyzt})^2}  \][/quote]
http://www.artofproblemsolving.com/community/c6h1186887p5771898
Let $a,\ b$ and $c$ be real numbers such that  $a,\ b,\ c\geq 1.$Prove the inequality$$\frac{1}{(1+a)^3}+\frac{1}{(1+b)^3}+\frac{1}{(1+c)^3}\geq \frac{3}{(1+\sqrt[3]{abc})^3}.$$
http://www.artofproblemsolving.com/community/c6h541865p3123478
If $x,y,z>0$ prove that:$$\frac{1}{(1+x)^3}+\frac{1}{(1+y)^3}+\frac{1}{(1+z)^3}\geq\frac{3}{4(1+xyz)}.$$


hallo,
let $a,b,c$ be sides of a triangle. prove:
$a^{2}(b+c-a)+b^{2}(c+a-b)+c^{2}(a+b-c) \le 3abc.$

If we make the change $a=y+z,\;\;b=x+z\;\;c=x+y$ we only need to prove the resulting equality for $x,y,z\in \mathbb{R}^{+}$.
We get:
$\sum_{sym.}(y+z)^{2}(2x)\leq 3\prod_{sym.}(y+z)$
$2\sum_{sym}x^{2}y+3\times4xyz\leq 3(2xyz+\sum_{sym.}x^{2}y)$
$6xyz\leq \sum_{sym.}x^{2}y$
$\frac{\sum_{sym.}x^{2}y}{6}\geq \sqrt [6]{ \prod_{sym.}x^{2}y}=xyz$

The last by $AM\geq GM$. Seems easy but, as usual, I have hidden various mistakes in the proof, so that the reader (yes, I know I'm too optimistic) can have extra-fun finding them.   :P

Are you using massive algebra to find that they are equal? Or just nice tricks :)  Also is the last one just the trivial case of Schur's, so does this mean that a,b,c need not be triangle side lengths? Or have you already taken account for that in some kind of substitution?

no the variables are the same! if they are sides of a triangle the probem is more simple cuz we can directly verify 2) but this inequality is in fact schur's inequality under disguise :D so true for all positive a,b,c.

PS. it's not massive algebra just tiny manipulations :wink:
BTW all of them are well known ways of expressing schur's ineq.

this question has been discused before (as all other IMO problems) so better than opening this thread doicu could just have gone to the resources section and found [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=346917#p346917]this[/url] :wink:


[quote="D.P.L"]Prove that if $A+B+C=180$

cos Acos Bcos C $\leq \frac{1}{8}$[/quote]
Maybe $A+B+C=180^{\circ}$ $?$ :wink:

[hide]
$\frac12=\cos\frac{(A+B+C)}{3}$ edit: Given that $A+B+C=\pi$

$\cos\frac{A+B+C}{3}\geq \frac{\cos A+\cos B+\cos C}{3}$ by Jenson

$\frac{\cos A+\cos B+\cos C}{3}\geq \sqrt[3]{\cos A \cos B \cos C}$ by AM-GM

$\frac12 \geq \sqrt[3]{\cos A \cos B \cos C}$
$\frac18 \geq \cos A \cos B \cos C$

Alternatively, use $OH^{2}=R^{2}(1-8\cos A \cos B \cos C)$[/hide]

tim1234133!
$\cos A\cos B\cos C\leq\frac{1}{8}$ for all real $A,$ $B$ and $C$ such that $A+B+C=\pi.$ :wink:

There is also an algebraic proof about 5/6 the way down [url=https://nrich.maths.org/discus/messages/27/92060.html?1159731751]this page[/url], although it does contain the words "after a massive amount of rearrangement".  :roll:

And, er yes.. I did steal your proof from that page, but it was such a nice one that I remembered it and couldn't resist reusing it here....  :P

Lol, I thought as much.  :mad:

Also, cos is convex on $(\frac{\pi}{2}, \pi)$. :P - it's a trivial issue, but meh.


Let be $ f: \mathbb{R}\rightarrow \mathbb{R}$ ,

$ f(x)\equal{}\frac{1}{\sqrt[3]{x^2\plus{}2x\plus{}1}\plus{}\sqrt[3]{x^2\minus{}1}\plus{}\sqrt[3]{x^2\minus{}2x\plus{}1}}$

Calculate $ S\equal{}f(1)\plus{}f(2)\plus{}...\plus{}f(2006)$

[quote="alex2008"]Let be $ f: \mathbb{R}\rightarrow \mathbb{R}$ ,

$ f(x) \equal{} \frac {1}{\sqrt [3]{x^2 \plus{} 2x \plus{} 1} \plus{} \sqrt [3]{x^2 \minus{} 1} \plus{} \sqrt [3]{x^2 \minus{} 2x \plus{} 1}}$

Calculate $ S \equal{} f(1) \plus{} f(2) \plus{} ... \plus{} f(2006)$[/quote]

After multiplying the upper and downer part with $ \sqrt [3]{x\plus{}1} \minus{} \sqrt [3]{x \minus{} 1}$ we get:$ f(x) \equal{} \dfrac{\sqrt [3]{x \plus{} 1} \minus{} \sqrt [3]{x \minus{} 1}}{2}$


Ok, I know it sounds like a stupid question, but I know some of my friends support 3rd parties but wouldnt vote for them if they could, because then there's one less vote for Kerry/Bush (who are the main contenders) that could have gone to them.

The big reason I said no is becuase I'm under 18. No, that's actualy the only reason.

I think I will vote for Kerry though I am probably more liberal than that especially in terms of saving the environment.

[quote="Peter VDD"]I guess the most important is that bush doesn't get re-elected... or should I just say elected?  :maybe:[/quote]

Yah...Re-defeat Bush in '04!

I will very likely vote for Bush, even though he is far from my ideal candidate.

(I will spend the time between now and then wondering why a man from Belgium has much of anything to say about the US political process)

[quote="confuted"](I will spend the time between now and then wondering why a man from Belgium has much of anything to say about the US political process)[/quote]

You should then also wonder why some Americans care about Iraq's to-be democracy.

[quote="takuanitromars36"]I support Bush. Kerry changes too much. He changes political opinions just to suit his audience. In other words, he's rather dishonest. For example, Kerry said to go to war with Iraq in like 2001, but now he's against it.[/quote]

Kerry voted against the 80 illion thing because bush has no real plan for peace, and kerry wasn't going to give them a blank check.
And that's the truth. Do we have a plan in Iraq now? No! Is there peace in Iraq? NO!

I still don't understand why we went to war in Iraq.  I mean, capturing Sadaam Hussain and his sons was a good thing, but besides that, the only things that were acomplished was attacks from that on guy, a bigger budget deficit, and less spending in Afgahnistan and stopping Al QUida and other terrorist groups.


Given a circle of radius 2, there are many line segments of length 2 that are tangent to the circle at their midpoints. Find the area of the region consisting of all such line segments.

$\text{(A)}\ \frac \pi 4 \qquad \text{(B)}\ 4 - \pi \qquad \text{(C)}\ \frac \pi 2 \qquad \text{(D)}\ \pi \qquad \text{(E)}\ 2\pi$

[hide]
Since the line segments are of length two and are tangent to the circle, the distance from their end points to the center of the circle will be constant all the way around forming another circle.  Thus, the region should be the area of the outer circle with radius $\sqrt{2^2 + 1^2}$ minus the area of the inner circle of radius two.
$(\sqrt{5}^2)\pi - 2^2\pi$
$\pi$
(D)[/hide]

[quote="Silverfalcon"]Given a circle of raidus 2, there are many line segments of length 2 that are tangent to the circle at their midpoints. Find the area of the region consisting of all such line segments.

$ \text{(A)}\ \frac \pi 4 \qquad \text{(B)}\ 4 \minus{} \pi \qquad \text{(C)}\ \frac \pi 2 \qquad \text{(D)}\ \pi \qquad \text{(E)}\ 2\pi$[/quote]

This problem is ill-defined, as it doesn't specify what the desired area of the region is supposed to be.

[hide]
place a line AB=2 tangent to the circle, and then place another line BC=2 tangent to the circle as well.

consider the center of the circle point O, thus line OC is the radius of the larger circle formed when all the points are placed along the graph.

consider the point M where AB is tangent to the circle. it follows that OMC is a right angled triangle, thus $ r^2 \plus{} 1^2 \equal{} OC^2 \rightarrow OC \equal{} \sqrt {5}$

the area of the larger circle is thus $ 5\pi$

subtracting the smaller form the larger gives us the area of the region of all the line segments, so the area of the region is

$ \boxed{\pi}$
[/hide]


function
integration
logarithms
inequalities
real analysis
real analysis solved

Let $p,q$ positive integers >0

Find all $f:[0,1]\rightarrow ]0;+\infty[$ continous such that

$\int_{0}^{1}(\ln f(x))^qdx = (\int_{0}^{1} \ln f(x)dx)^p$

Are you kidding, Moubinool?
It is just Cauchy $ \int_{0}^{1}(\ln f(x))^2dx \geq (\int_{0}^{1} \ln f(x)dx)^2$ and equality iff $\ln f(x)=C$.

Only the constants. You might as well let $g=\ln f,$ so you are asking for

$\int_0^1 g^2=\left(\int_0^1 g\right)^2.$

Let $m=\int_0^1g$. Then

$\int_0^1(g-m)^2=\int_0^1g^2-2m\int_0^1g+m^2 =\int_0^1g^2-m^2.$

Clearly, this can only be zero if $g=m$ almost everywhere.

You should recognize this calculation as the variance of a random variable.

Ok when $p=q=2$ is it obvious

What about the others cases ? (I edited )

I think, for $p\neq q$ we will be unable to find definite answer, since $f$ can be rather arbitrary. Am I right? I don't know at the moment  :?

Since $g=\ln f$ isn't necessarily positive, the lack of absolute values in Moubinool's new statement mess things up somewhat.

Let me just assume $g\ge0$ for now (or assume appropriate absolute values). If $p=q>1,$, Hölder's inequality limits us to just the constants, since $(\int_0^1g)^q\le\int_0^1g^q.$ We used $1$ as the other function in Hölder's inequality. We get inequality in that iff the two functions, properly scaled by being raised to the appropriate power, are constant multiples of each other - that means equality only if $g$ is constant.

To make Myth's remark more specific, take a $g$ that is postive but otherwise arbitrary. Let $x=\int_0^1g$ and $y=\int_0^1g^q.$ Now multiply $g$ by the positive constant $a$ and ask if we can have Moubinool's equality.

We need $(ax)^p=a^qy.$ This always has a solution, namely $a=(\frac{x^p}y)^{\frac1{p-q}}.$


algebra
difference of squares
special factorizations

Determine all pairs (x,y) of positive integers satisfying the equation
$ (x+y)^{2}-2(xy)^{2}=1$

[hide="A start:"]
Well, this is a Pell's equation. Observe the initial solution $ (x+y,xy)=(3,2)$ giving $ (x,y)=(1,2)$ and permutations. Then, we can use difference of squares, square it, simplify it back with difference of squares to obtain, using $ a=x+y$ and $ b=xy$: \[ (a^{2}+2b^{2})^{2}-2(2ab)^{2}=1 \]
The next solution it gives is $ (x+y,xy)=(17,12)$, which doesn't yield positive integer solutions for $ x$ and $ y$.

We can do this indefinitely; but I don't know when we will obtain positive integer solutions for $ x$ and $ y$, and I've got to go. Can someone continue from here?  :) [/hide]


Find, with proof, the unit decimal digit of
\[ \sum_{k \equal{} 1}^\infty\frac {10^{2008}}{10^k \minus{} 1}
\]
[hide="Answer"]
8
[/hide]
How about the $ n$th decimal digit?

Hmm, this has something to do with the number of divisors...
EDIT[11:10PM]Please refer to this topic (JoeBlow double posted): [url]http://www.mathlinks.ro/viewtopic.php?p=1126439#1126439[/url]


inequalities
number theory unsolved
number theory

Find all natural numbers $n$ such that has exactly $\sqrt{n}$ natural divisors.

If $n = \prod p_i^{a_i}$ then the number of positive divisors of $n$ is $\tau(n) = \prod (1+a_i)$ as well known.

We want to solve $\prod (1+2 \cdot a_i) = \prod p_i^{a_i}$

For $p > 3, a_i > 0, p^{a_i} > (1+2 \cdot a_i)$ (by Bernoulli inequality for instance) so we can't have $p_i > 3$ for all $i$.

$\prod (1+2 \cdot a_i)$ is odd so $p \neq 2$.

$3^a p^b > (1+2 \cdot a)(1+2 \cdot b)$ for $a > 2$

$n = 1, 3^2$ are the only solutions.

[quote="tµtµ"]$n = 1, 3$ are the only solutions.[/quote]

I'm sure you actually meant $n = 1$ and $n = 9$.

Wasn't this a problem in Canadian MO 1999?

Possible, but I'm sure it was a problem from Polish MO

[quote="Arne"][quote="tµtµ"]$n = 1, 3$ are the only solutions.[/quote]
I'm sure you actually meant $n = 1$ and $n = 9$.[/quote]
yes, of course   :oops:


calculus integration limit trigonometry LaTeX

I sent my paper but meanwhile I'm trying to learn more, in order to improve it.
Usually, I have groups of equations, and the group is right numbered, at his middle. (Only the group, not the equations individually).
Three troubles: my equations appear aligned at left, but I saw that the fashion is to align them at right, when splitting a longer one. Second, I split them manually, using \\. I don't know how the command split works. Does it follow the math conventions? For example, + or - has to be at the beginning of the second line, when splitting an equation, and not at the end of the previous line. And third, when fractions, or integrals appear in such groups, the letters are too small. I included two samples which I don't like:

[b]Ex. 1:[/b] [quote] \begin{equation} \label{ZEqnNum440443} \begin{array}{l} {\frac{\Gamma (\alpha (a-1))}{\Gamma (\alpha )} \bar{y}^{i(\alpha a)} =\Gamma (1+\alpha )\mathop{J_{j}^{i} }\limits^{\alpha } (\bar{y}^{\alpha (a-1)} ,\; x)y^{j(\alpha )} +} \\ {\frac{\Gamma (2\alpha )}{\Gamma (\alpha )} \mathop{J_{j}^{i} }\limits^{\alpha } (y^{(\alpha (a-1))} ,\; y^{\alpha } )y^{j(2\alpha )} +...+\frac{\Gamma (2\alpha )}{\Gamma (\alpha )} \mathop{J_{j}^{i} }\limits^{\alpha } (\bar{y}^{\alpha (a-1)} ,\; y^{\alpha b} )y^{j((b+1)\alpha )} +} \\ {...+\frac{\Gamma (\alpha (a-1))}{\Gamma (\alpha )} y^{i(\alpha a)} ,} \end{array} \end{equation}
[/quote]

[b]and Exp. 2: [/b]

[quote] \begin{equation} \label{84)} \begin{array}{l} {L(t,x,y^{(1/2)} )=U_{1} (t,x)-\frac{1}{2} \Gamma (2)(y^{(1/2)} )^{2} } \\ {L(t,x,y^{(1/2)} ,y^{(1)} )=U_{2} (t,x)-\frac{1}{2} a_{1} \Gamma (2)(y^{(1/2)} )^{2} +\frac{1}{2} \Gamma (3)(y^{(1)} )^{2} } \\ {L(t,x,y^{(1/2)} ,y^{(1)} ,y^{(3/2)} )=U_{3} (t,x)-\frac{a_{1} }{2} \Gamma (2)(y^{(1/2)} )^{2} +} \\ {\frac{a_{2} }{2} \Gamma (2)(y^{(1)} )^{2} -\frac{1}{2} \Gamma (4)(y^{(3/2)} )^{2} .} \end{array} \end{equation}
[/quote]

In Exp. 2, I'd like the tail be aligned at right, for the third equation. (I'll correct the + which should be on the last line).
In Exp.1,  similarly should be the tail at right, with the + on the last line. And the letters in fractions are smaller. The letters size problem is more annoying at:

Exp. 3: [quote] \begin{equation} \label{16)} \begin{array}{l} {(D_{x^{i} }^{\alpha } f)(x)=} \\ {\frac{1}{\Gamma (1-\alpha )} \frac{\partial }{\partial x^{i} } \int _{a^{i} }^{x^{i} }\frac{f(x^{1} ,...,x^{i-1} ,s,x^{i+1} ,...,x^{n} )-f(x^{1} ,...,x^{i-1} ,a^{i} ,x^{i+1} ,...,x^{n} )}{(x^{i} -s)^{\alpha } }  ds,} \end{array} \end{equation}[/quote]

How can I solve all three problems and right number each group? Thanks. Jean.

Well, I see you have problems with the alignment.

Use this environment

\begin{eqnarray*}
\end{eqnarray*}

So, for example

\begin{eqnarray*}\lim_{x\to0}\frac{1-\cos x}{x}&=&\lim_{x\to0}\frac{1-\cos^{2}x}{x(1+\cos x)}\\ &=&\lim_{x\to0}\frac{\sin^{2}x}{x(1+\cos x)}\\ &=&\lim_{x\to0}\frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x}\end{eqnarray*}

produced by

[color=blue]\begin{eqnarray*}
\lim_{x\to0}\frac{1-\cos x}x&=&\lim_{x\to0}\frac{1-\cos^2x}{x(1+\cos x)}\\
&=&\lim_{x\to0}\frac{\sin^2x}{x(1+\cos x)}\\
&=&\lim_{x\to0}\frac{\sin x}x\cdot\frac{\sin x}{1+\cos x}
\end{eqnarray*}[/color]

Does that make sense?

[b]Never[/b] use eqnarray, it doesn't work properly and has been superseded by align and other environments.  See [url=http://tug.org/pracjourn/2006-4/madsen/]Avoid eqnarray![/url] for details.

The long equations look best with multline (note it's not [i]multiline[/i] but [i]multline[/i]).
So for the last equation use
[code]\begin{multline} \label{16)}
 (D_{x^{i} }^{\alpha} f)(x)= \\
 \frac{1}{\Gamma(1-\alpha)} \frac{\partial }{\partial x^{i} } \int _{a^{i} }^{x^{i} }\frac{f(x^{1},\dots,x^{i-1},s,x^{i+1},\dots,x^{n})-f(x^{1},\dots,x^{i-1},a^{i},x^{i+1},\dots,x^{n} )}{(x^{i} -s)^{\alpha }} ds
\end{multline}[/code]whch looks like this:
$ \par\begin{multline}\label{16)}(D_{x^{i}}^{\alpha}f)(x) =\\ \frac{1}{\Gamma(1-\alpha)}\frac{\partial }{\partial x^{i}}\int_{a^{i}}^{x^{i}}\frac{f(x^{1},\dots,x^{i-1},s,x^{i+1},\dots,x^{n})-f(x^{1},\dots,x^{i-1},a^{i},x^{i+1},\dots,x^{n})}{(x^{i}-s)^{\alpha }}ds\end{multline}$
Note that \dots should be used instead of ... and that \\ have to be put in manually.

More on multline and similar environments can be found at [url=http://texcatalogue.sarovar.org/entries/voss-mathmode.html]Math Mode by Herbert Voß[/url].

There's nothing I need to add that's not already covered in [url=http://tug.org/pracjourn/2006-4/madsen/]Avoid eqnarray![/url]. See also the comparisons in The LaTeX Companion, 2nd edition, pp 470-471.

Yes, array and eqnarray were the cause for small letters and crowded formulas. Split, multline and align are instead now.
Again, thanks a lot!

Jean


This is a special case ($k=2$) of Walther Janous's  inequality :
If  $x_1+x_2+...+x_n=1,$  where  $x_i$ are non-negative real numbers
and  $2\leq k< n,k\in\mathbb N,$ then
$x_1x_2...x_k+x_2x_3...x_{k+1}+...+x_nx_1...x_{k-1}\leq max\{\frac{1}{k^k},\frac{1}{n^{k-1}}\}.$ ;)

I am not so sure it is a special case: the given question is not fully cyclic.

[quote="JBL"]I am not so sure it is a special case
[/quote]
I think you are not right. ;)
[quote="JBL"]
the given question is not fully cyclic.[/quote]
What do you mean? What is this "not fully cyclic"?  ;)

I'm pretty sure the maximum is $\frac{1}{4}$, and I believe there is some induction proof... I'm not sure it's such an easy inequality though.

[quote="JBL"]In the given question, transforming the $x_i$ by a cyclic permutation ($x_n \to x_1$ and $x_i \to x_{i + 1}$ for other $i$) could change the resulting sum.  In the inequality you give, a cyclic permutation of the variables will result in the same sum.  Thus, it is not clear at all that the given problem is a special case of your formula.

In particular, it is clear that one of the two upper limits from your theorem is no longer valid in the "usual" maximizing case: there are only $n - 1$ terms in the sum, so it should be $\frac{n-1}{n^k}$ in the case where all variables are equal (although it's no longer at all obvious that having all variables equal will be a possible maximizing situation).[/quote]
I am sorry, I don't undestand you. :?
If $k=2$ then $x_1x_2...x_k+x_2x_3...x_{k+1}+...+x_nx_1...x_{k-1}=x_1x_2+x_2x_3+...+x_nx_1$
and $max\{\frac{1}{k^k},\frac{1}{n^{k-1}}\}= max\{\frac{1}{4},\frac{1}{n}\}.$ ;)

Your last post contains no mathematical error but it still does not answer the question because $x_1x_2+x_2x_3+...+x_nx_1 \neq x_1x_2+x_2x_3+...+x_{n-1}x_n$.  Go look at the summation in question again.  (Also, the answer you gave is clearly wrong for $n = 2$ and is easily shown to be wrong for $n = 3$.)  Is that sufficiently clear?

In any event, the solution is not actually very hard: [hide]Suppose we have $n \geq 3$ and compare $x_2$ and $x_{n-1}$.  WLOG $x_2 \geq x_{n-1}$.  But in this case, we can clearly make the overall sum at least as large by making $x_1 \to x_1 + x_n$ and $x_n \to 0$.  Now, we can repeat this as much as we want.  Thus, given any starting point, I can make the final sum at least as big by using only $x_1$ and $x_2$.  But in this case, the maximum is clearly when $x_1 = x_2 = \frac12$, so $\frac14$ is indeed the maximum for any $n$ and is always achievable.[/hide]

[quote="JBL"] Go look at the summation in question again.  [/quote]
OK. He has corrected it, and I have not noticed. :)


vector
inequalities
linear algebra
linear algebra unsolved

Let $S$ be a finite set of vectors in $\mathbb{R}^n$ with the property that for each $\vec{v} \in \mathbb{R}^n$, there exist a $\vec{u} \in S$ such that $\vec{v} \cdot \vec{u} \geq 0$.

Prove (or disprove) that we can correspond each $\vec{u} \in S$ with a non-negative real $\lambda_u \geq 0$ (not all zero), such that $\sum_{\vec{u} \in S} \lambda_u \vec{u} = 0$.

I believe it's been posted before, but with strong inequalities instead of weak ones. Try searching the linear algebra forum for words like "hyperplane" and "polytope" and "convex".

I'll try to find it myself.

Edit:

Hmm.. The hypothesis of this one seems different, but it's not, actually (well, except for the fact that weak inequalities are replaced by the strong versions). Your sign $\ge$ is replaced with $<$ in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=polytope&t=42215]this thread[/url]. However, this is a version of your problem, since for any $v$ there must be a vector $u$ in our set with $(u,-v)<0\Rightarrow(u,v)>0$. Maybe it helps.

Second edit:

Come to think of it, your problem is not true. Basically, the hypothesis is not strong enough to guarantee the existence of $n$ independent vectors in $S$. Think about this simple counterexample: $S=\{v,-v\}$ for some $v\ne 0$. Or simpler yet, how about $V=\{0\}$ (both examples work for any $n\ge 2$)? :)