If $ a,$ $ b,$ $ c,$ $ d$ are positive real numbers such that
$ a\plus{}b\plus{}c\plus{}d\equal{}abc\plus{}bcd\plus{}cda\plus{}dab,$
then
$ \left(\sqrt{a^2\plus{}1}\plus{}\sqrt{b^2\plus{}1}\right)^2\plus{}\left(\sqrt{c^2\plus{}1}\plus{}\sqrt{d^2\plus{}1}\right)^2 \le (a\plus{}b\plus{}c\plus{}d)^2.$

Observe that it's equivalent to

$ 2 \plus{} \sqrt {a^2 \plus{} 1}\sqrt {b^2 \plus{} 1} \plus{} \sqrt {c^2 \plus{} 1}\sqrt {d^2 \plus{} 1}\leq ab \plus{} bc \plus{} cd \plus{} ac \plus{} bd \plus{} ad$

After the substitution a = $ \frac {x}{y}$  b = $ \frac {y}{x}$   c =$ \frac {z}{w}$   d =$ \frac {w}{z}$,

(and this substitution make a+b+c+d=abc+bcd+cda+dab true)

it becomes

$ \sqrt\frac {x^2 \plus{} y^2}{x^2}\sqrt\frac {z^2 \plus{} w^2}{z^2} \plus{} \sqrt\frac {x^2 \plus{} y^2}{y^2}\sqrt\frac {z^2 \plus{} w^2}{w^2}) \leq (\frac {y}{x} \plus{} \frac {x}{y})(\frac {w}{z} \plus{} \frac {z}{w})$

or

                    $ (\sqrt {(x^2 \plus{} y^2) \plus{} (z^2 \plus{} w^2)}(\frac {1}{xz} \plus{} \frac {1}{yw}) \leq (\frac {x^2 \plus{} y^2}{xy})(\frac {z^2 \plus{} w^2}{zw})$

or

                    $ (\sqrt {(x^2 \plus{} y^2) \plus{} (z^2 \plus{} w^2)}(xz \plus{} yw) \leq (x^2 \plus{} y^2)(z^2 \plus{} w^2)$

or

                    $ xy \plus{} yw \leq \sqrt {(x^2 \plus{} y^2) \plus{} (z^2 \plus{} w^2)}$

By the Cauchy-Schwarz inequality, we have

$ (xz \plus{} wy)^2\leq(x^2 \plus{} y^2)(z^2 \plus{} w^2)$

clear~ :lol:

ㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡ

ㅜ.ㅜ I'm wrong
I'm correcting my sloution. :P

[quote="Sylvester07"]
After the substitution a = $ \frac {x}{y}$  b = $ \frac {y}{x}$   c =$ \frac {z}{w}$   d =$ \frac {w}{z}$,

(and this substitution make a+b+c+d=abc+bcd+cda+dab true)[/quote]
From this substitution, we get $ abcd\equal{}1$ (?)

V. Q. B. Can:
    Learning your question, Does easy to do this ineq?

Let $ x,y,z\geq 0$,such that $ x\plus{}y\plus{}z\equal{}xy\plus{}yz\plus{}zx$,prove that:
$ (x\plus{}y\plus{}z)^2 \geq \frac{3}{2}(\sqrt{(x^2\plus{}1)(y^2\plus{}1)}\plus{}\sqrt{(y^2\plus{}1)(z^2\plus{}1)}\plus{}\sqrt{(z^2\plus{}1)(x^2\plus{}1)}).$

[quote="xzlbq"]V. Q. B. Can:
    Learning your question, Does easy to do this ineq?

Let $ x,y,z\geq 0$,such that $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx$,prove that:
$ (x \plus{} y \plus{} z)^2 \geq \frac {3}{2}(\sqrt {(x^2 \plus{} 1)(y^2 \plus{} 1)} \plus{} \sqrt {(y^2 \plus{} 1)(z^2 \plus{} 1)} \plus{} \sqrt {(z^2 \plus{} 1)(x^2 \plus{} 1)}).$[/quote]
The condition should write as $ x\plus{}y\plus{}z\equal{}xy\plus{}yz\plus{}zx>0.$ Otherwise the inequality will false for $ x\equal{}y\equal{}z\equal{}0.$ I proved this inequality by Cauchy-Schwarz and AM-GM, but before posting the solution, I would like to see a complete solution for my initial inequality first.

[quote="can_hang2007"][quote="xzlbq"]V. Q. B. Can:
    Learning your question, Does easy to do this ineq?

Let $ x,y,z\geq 0$,such that $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx$,prove that:
$ (x \plus{} y \plus{} z)^2 \geq \frac {3}{2}(\sqrt {(x^2 \plus{} 1)(y^2 \plus{} 1)} \plus{} \sqrt {(y^2 \plus{} 1)(z^2 \plus{} 1)} \plus{} \sqrt {(z^2 \plus{} 1)(x^2 \plus{} 1)}).$[/quote]
The condition should write as $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx > 0.$ Otherwise the inequality will false for $ x \equal{} y \equal{} z \equal{} 0.$ I proved this inequality by Cauchy-Schwarz and AM-GM, but before posting the solution, I would like to see a complete solution for my initial inequality first.[/quote]

yes,$ x,y,z>0.$
BQ

[quote="Sylvester07"][quote="can_hang2007"][quote="Sylvester07"]
After the substitution a = $ \frac {x}{y}$  b = $ \frac {y}{x}$   c =$ \frac {z}{w}$   d =$ \frac {w}{z}$,

(and this substitution make a+b+c+d=abc+bcd+cda+dab true)[/quote]
From this substitution, we get $ abcd \equal{} 1$ (?)[/quote]

Oh.....

this substitution is wrong?

plz tell me ㅜ.ㅜ[/quote]
Because when you put a = $ \frac {x}{y}$  b = $ \frac {y}{x}$   c =$ \frac {z}{w}$   d =$ \frac {w}{z}$ we can retain thats
$ ab \equal{} cd \equal{} 1$.
And the question is: Why $ ab \equal{} cd \equal{} 1$?  :D

Let $ S\equal{}a\plus{}b\plus{}c\plus{}d\equal{}abc\plus{}bcd\plus{}cda\plus{}dab$. Then $ a^2\plus{}1\equal{}\frac{1}{S}(a^2(a\plus{}b\plus{}c\plus{}d)\plus{}abc\plus{}bcd\plus{}cda\plus{}dab)\equal{}\frac{1}{S}(a\plus{}b)(a\plus{}c)(a\plus{}d)$, and similar results apply for $ b^2\plus{}1$, $ c^2\plus{}1$, and $ d^2\plus{}1$.

Therefore $ \left(\sqrt{a^2\plus{}1}\plus{}\sqrt{b^2\plus{}1}\right)^2\plus{}\left(\sqrt{c^2\plus{}1}\plus{}\sqrt{d^2\plus{}1}\right)^2$
$ \equal{}\frac{1}{S}\left[(a\plus{}b)\left(\sqrt{(a\plus{}c)(a\plus{}d)}\plus{}\sqrt{(b\plus{}c)(b\plus{}d)}\right)^2\plus{}(c\plus{}d)\left(\sqrt{(c\plus{}a)(c\plus{}b)}\plus{}\sqrt{(d\plus{}a)(d\plus{}b)}\right)^2\right]$
$ \leq\frac{1}{S}\left[(a\plus{}b)\left(\frac{a\plus{}c}{2}\plus{}\frac{a\plus{}d}{2}\plus{}\frac{b\plus{}c}{2}\plus{}\frac{b\plus{}d}{2}\right)^2\plus{}(c\plus{}d)\left(\frac{c\plus{}a}{2}\plus{}\frac{c\plus{}b}{2}\plus{}\frac{d\plus{}a}{2}\plus{}\frac{d\plus{}b}{2}\right)^2\right]$
$ \equal{}(a\plus{}b\plus{}c\plus{}d)^2$, as required. Equality occurs if and only if $ a\equal{}b\equal{}\frac{1}{c}\equal{}\frac{1}{d}$.

Nice, BakerBakura. This is what I am waiting for.

[quote="can_hang2007"]If $ a,$ $ b,$ $ c,$ $ d$ are positive real numbers such that
$ a \plus{} b \plus{} c \plus{} d \equal{} abc \plus{} bcd \plus{} cda \plus{} dab,$
then
$ \left(\sqrt {a^2 \plus{} 1} \plus{} \sqrt {b^2 \plus{} 1}\right)^2 \plus{} \left(\sqrt {c^2 \plus{} 1} \plus{} \sqrt {d^2 \plus{} 1}\right)^2 \le (a \plus{} b \plus{} c \plus{} d)^2.$[/quote]
See my proof here: http://can-hang2007.blogspot.com/2010/01/inequality-58-v-q-bcan.html


The sign is reversed. For the corrected version see here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=26103[/url].

[quote="amgaa36"]Who find this wonderful inequality Is there any easy method?  :oops:  :o[/quote]
V.Adyaasuren found it.

[quote="amgaa36"]Who find this wonderful inequality Is there any easy method?  :oops:  :o[/quote]

Anh actually, we also have an easier solution to this one. The solution in the article is too hard.


Hello!
I've recenetly passed mensa qualification test and received quite a nice result:) I was sent an invitation to become a member, but in my opinion there aren't many benefits from it. I'd have to pay about 20$\$$ for it every year (it's not so much). Now my question is: I'd like to apply to top US university (MIT is my dream), but I'm international, so that makes it very hard. So, do you think that being a Mensa member would help me in admission? Would adcoms consider my mensa membership?

A little about Mensa: it's an international organization that gathers people with high iq (top 2% of population). There are only about 100,000 Mensans around the world.
http://www.mensa.org

Being a Mensa member would not help you at all in MIT admission, and mentioning Mensa membership on your MIT application just might hurt you. That's because admission to an elite college is based more on your I-did and your I-will than on your IQ.

[quote="tokenadult"]That's because admission to an elite college is based more on your I-did and your I-will than on your IQ.[/quote]

This is just excruciatingly cheesy...

Other than that, tokenadult hit the nail on the head.

I hope this isn't offensive, but is there any point whatsoever to joining Mensa?

I didn't realize they charged a fee either...that's a little disturbing.

Years ago, the guy who founded Mensa hoped that it would be a kind of brain trust for the world, a group of people to whom world leaders could go for a resolution to the ongoing disputes in the Middle East, for instance. He was very disappointed to see it turn into a club for publishing puzzle books.

If you like Mensa as a social organization, go ahead and join. And if you like smart people, definitely apply to MIT. Just don't join Mensa to increase your chances of getting into MIT.

Fact of the day:

Shakira (the Columbian pop singer) is a member of Mensa (with an IQ of 140).

That's kinda neat.

Sad though it is, I really do like Mensa's Sudoku book.


Is it possible to make colored $ \text{\LaTeX}$? If so, how?

Do you mean like this $ \color{red}x \color{green}^2$?

Yes, thank you.

$ \color{blue}\textsf{testing...}$

Cool!

$ \color{red}{x}^2$

So does the color stay applied until you change it?

Yes, though you can change this by putting the \color command and text inside {...} as in $ {\color{red}x^2} x^2$.
You can also return to the normal colour of the text with \normalcolor $ \color{red}x^2 \normalcolor x^2$.

$ x\equal{}\frac{{\color{red}\minus{}b\pm}\sqrt{b^2\minus{}4ac}}{2a}$

Cool!


function
real analysis
inequalities proposed
inequalities

Let $x,y$ be positive reals such that $x^{3}+y^{3}=x-y$. Prove that \[x^{2}+y^{2}<1.\]

[quote="N.T.TUAN"]Let $x,y$ be positive reals such that $x^{3}+y^{3}=x-y$. Prove that
\[x^{2}+y^{2}<1. \]
[/quote]
Since, $x$ and $y$ are positive numbers then $x-y>0.$
Let $x^{2}+y^{2}\geq1.$ Hence, $x^{3}+y^{3}=x-y\leq(x-y)(x^{2}+y^{2}).$
Thus, $2y^{3}-xy^{2}+x^{2}y\leq0,$ which contradiction.

Maybe my solution is same as. Put $k=x^{2}+y^{2}$, then $k(x^{3}+y^{3})=(x-y)(x^{2}+y^{2})$, or $k=f(t)=\frac{(t^{2}+1)(t-1)}{t^{3}+1}$, where $t=\frac{x}{y}>1$. Now use real analysis for function $f(t)$.


For positive $ x_{p}$ such that $ \prod^{n}_{p\equal{}1}{x_{p}}\equal{} 1$  prove that:

$ \sum^{n}_{j \equal{} 1}{\frac{1\plus{}\sum^{j\plus{}m}_{k \equal{} j}{\prod^{k\plus{}m}_{k}{x_{k}}}}{1\plus{}\sum^{j\plus{}m\plus{}1}_{k \equal{} j\plus{}1}{\prod^{k\plus{}m}_{k}{x_{k}}}}}\ge n$

I know it seems very ugly, but if you check the smaller cases you will understand it well.

$ m\leq n\minus{}1$ a non-negative integer. And $ x_{i\plus{}n}\equal{} x_{i}$.

[edited jiji]

 :lol:

I see no equality sign.  What identity did you intend?


integration
calculus
calculus computations

Calculate \[ \int_{\sqrt{3}}^{3 \sqrt{3}} \frac{\arctan x}{1-x^2} \, dx  \]

$\frac{\arctan x}{1-x^2} = \frac{\arctan x }{(1-x)(1+x)} \simeq \frac{1}{1-x}$, $x\to 1$
But \[ \int_0^1{\frac{1}{1-x}}\;dx =\infty \]
so improper integral $\int_0^1 \frac{\arctan x}{1-x^2} \;dx$ is not exists.

:blush:  that was typo (I had another problem in my head..). The integral isn't improper. I've edit it.


We can write $\sqrt{2+\sqrt{2+\sqrt{2+...}}}$ as $a_{1}=\sqrt{2}\ , \ a_{k}=\sqrt{2+a_{k-1}}\ \ \ (k\to\infty)$

but how to write the following in that form

$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt{5+\cdots}}}}}$

The closest thing to what you want is $b_{n,1}=\sqrt n$; $b_{n,k}=\sqrt{n+1-k+b_{n,k-1}}$ ($k=2,\dots,n$); $a_{n}=b_{n,n}$ ($n=1,2,3,4,\dots$). There is no way to express $a_{n}$ as a simple explicit function of one or several previous $a_{k}$. If there were such a way, we could use the corresponding formula to find the limit.

How to use Mathematica to compute its value

If your aim is just to calculate its numerical value as accurate as you wish, maybe you can use this method:

$a(n,1) = 2$ and $a(n,k+1) = \sqrt{n-k+a(n,k)}$

and let $b(n) = a(n,n)$. Then for large $n$, we can calculate the numerical value of given iterated sqrts... It is not so good approximation method, but at least it gives result.

Mathematica code:

n = 10;
a = 2;
For[i=0, n>i,
a = Sqrt[n-i+a];
i++;
];
Clear[n,i];
N[a]

Numerical value of 40-digit accuracy is $1.757932756618004532708819638218138527653\cdots$.


superior algebra
superior algebra unsolved

Hi all!

Let [tex]G[/tex] be a non-abelien group with [tex]|G|=p^3[/tex], where [tex]p[/tex] is a prime. Show that [tex]| Z(G)|=p[/tex], where [tex]Z(G):=\{g\in G\vert \forall x\in G: xg=gx\}[/tex] denotes the center of [tex]G[/tex].

Hanno

The center has order $1,p,p^2$ by Lagrange's theorem. Now the class-formula gives $|G| = |Z(G)| + \sum \frac{|G|}{|G_x|}$ where x goes through all representatives of conjugacy classes ( $G_x$ denotes the stabilizer of $x$ ). But $p | \frac{|G|}{|G_x}| , |G|$, so $p| |Z(G)|$. To exclude the case $p^2$, it's now enough to use the following theorem:
When $G/Z(G)$ is cyclic, then already $G$ is commutative (and $G/Z(G)$ by this trivial).
This can be proved by direct calculations I don't want to show now, but immediatly implies the desired result, since $|Z(G)|=p^2$ would imply $|G/Z(G)|=p$ and so $G/Z(G)$ to be cyclic.
By this we get $|Z(G)|=p$.

The $\vert Z\vert = p^2$ case is an nice little argument.  Consider $x \not\in Z$.  The set of elements $C(x)$ that commute with $x$ includes $Z$ and $\langle x \rangle$ which is strictly bigger than $Z$ and hence the whole of $G$.  So $x \in Z$ after all, contradiction.


I am not sure where to post this question...Alright here is my question. I cannot find a formula related to the number of sides in a polygon to the number of total diagonals in that figure. I have the following pattern given to me to make my formula from... 3 sides no diagonals, 4 sides 2 diagonals, 5 sides 5 diagonals, 6 sides 9 diagonals, 7 sides 14 diagonals. My formula would need to be "n" the number of sides equals the total amount of diagonals. Thanks in advance!!

You've already posted this problem here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127964[/url]

Do not double post.


If $ f\left( \frac{x}{x\minus{}1} \right) \equal{} \frac{1}{x}$ for all $ x\neq 0, 1$ and $ 0<\theta<\frac{\pi}{2}$, then find $ f(\sec^2 \theta)$.


function
algebra
domain
calculus
derivative
calculus computations

A function is not a formula. This is one of the cases in which you really should specify the domain.  Please make a habit of specifying the domain.

Guessing at what you probably meant:

If $f(x)$ is the real-valued function defined for $|x|\ge1$ by $f(x)=\sqrt{x^2-1}:$

There is no critical point at $0$; that's not in the domain of the function.

What about $1$ and $-1$? Those are points on the boundary of the domain of the function. Whether you call such points critical points or not, they are answers to the question "what points have any chance to be local minima or local maxima?"

The Mathworld definition says a critical point is one where either the derivative equals 0 or is not defined. So yes, its 1 or -1.

So, can we say that the points on the boundary of the domain of the function are considered as crirtical points?

could you explain more


I'm not sure if this question belongs here...

Do any of you use Reverse Polish Notation (RPN)? I hear it's quite useful...
It's when the operator comes after the operand. For example, instead of 3+6, in RPN it would be 3 6 +.

It doesn't make much sense to use RPN when you are doing math on your own. But it's far easier to program a calculator to use RPN than to use infix. (I have tried; RPN, or postfix, was quite easy, and I never got infix to work.)

[quote="mathfanatic"][quote="Caroline"]Could someone please briefly explain RPN, or point me to a website with an explanation?  Thanks =)[/quote]

[url=http://www-stone.ch.cam.ac.uk/documentation/rrf/rpn.html]This website[/url] keeps it nice and simple.  But if you want something more comprehensive, there's always [url=http://en.wikipedia.org/wiki/Reverse_Polish_Notation]Wikipedia[/url].[/quote]

Awesome, thanks =D

[quote="Nukular"]yeah RPN is remarkably easy to program. It's surprising.[/quote]

I can kind of see how that would be.  Also, it removes the need for multiple parentheses, which is nice.  (Of course, this is from my brief, newfound knowledge based on the first website mathfanatic recommended.)

I love my RPN calculator. It's really a lot easier to use, and no one else can borrow it because they don't know how to use it. I will have to admit, however, that it does take some time to get used to. :)

the computer language FORTH works on this principle
as well, i.e. all data are pushed onto a stack and operations
applied to the items on top.  This makes for a very compactly
implementable computation model, though somewhat weaker
than conventional languages.

[quote="BHorseMath"]I use it on the hps but not on my TI it makes a mess on there.[/quote]

Do the hps also accept normal entry?

[quote="Gyan"]Once you get used to it,  it is faster to use (less   butttens to  be pressed) and   less likely to have  error due to missed  key.. IMO... In the days prior to PC's, when memory was limited,   I was able to  program fairy complicated  programs in my  HP[/quote]But in our days memory is not a problem anymore, not even in hand-held devices :) So the only thing that really matters is how fast you can "write" in the language you are programming in. I think standard logical formatting is faster, especially because it's easier to debug large quantities of code.

But then again I am not an expert in computer science, so I might be wrong :)

sorry i double posted...read the next one

[quote="RC-7th"]
Do the hps also accept normal entry?[/quote]

The newest Hp graphing calculator, if my memory serves me correctly its the 49g and the 49g+, can be changed so that they accept the "normal" entry. I'm not sure about the older ones. However, even though I personally dont use an Hp, i think they are far more useful if you learn to use RPN.


\[ f(x)\equal{}\frac {x\minus{}5}{x^2\minus{}5x\plus{}1992}\]
For which $ x$ does it have its maximum?

$ f(x)\equal{}\frac {x\minus{}5}{x^2\minus{}5x\plus{}1992}\equal{}\frac{1}{x\plus{}\frac{1992}{x\minus{}5}}\equal{}\frac{1}{x\minus{}5\plus{}\frac{1992}{x\minus{}5}\plus{}5}$ $ \le\frac{1}{4\sqrt{498}\plus{}5}$ by AM-GM
Equality occurs when $ x\minus{}5\equal{}\frac{1992}{x\minus{}5}\implies x\equal{}2\sqrt{498}\plus{}5$


Given a positive integer $n$, prove that the following are equivalent:

(a) For any positive integer $a$, $n \mid a^{n}-a$;

(b) For any prime divisor $p$ of $n$, $p^{2}\nmid n$, and $p-1 \mid n-1$.

this is sort of obvious...

put $a=p$ where $p|n$ then we get that if $p^{k}\parallel n$ then $k\le1$ because the greatest power of $p$ dividing $p^{n}-p$ is $p$
on the other side if $p|a^{n}-a$ for all a then $p|a^{n-1}-1$ for all $(a,p)=1$ so $(p-1)|(n-1)$ for all divisors of $n$


[hide="Solution"]Well we have that the last time $ n$ appears in the sequence, it's the $ \dfrac{n(n + 1)}{2}$'th term.

So we need to find and $ n$ such that:

${ \dfrac{n\cdot(n - 1)}{2}} < 2009\le \dfrac{n\cdot (n + 1)}{2}$

It just so happens that when $ n = 63$,

${ \dfrac{63\cdot 62}{2} < 2009\le \dfrac{63\cdot 64}{2}}\implies 1953 < 2009\le 2016 \therefore$ the 2009th term is $ 63$.[/hide]


inequalities
search
inequalities theorems

What is Power Mean Inequality ? I don't know, can you explain for me ?  :blush:

[quote="robinho"]the Power-mean inequality:

$\large x_{1} , .... , x_{n} \geq 0, P_{r}=\sqrt[r]{(\frac{x_{1}^{r}+x_{2}^{r}+...+x_{n}^{r}}{n})}$

If $\large r>s$ -> $\large P_{r} \geq P_{s}$ equation happen $\large x_{1}=x_{2}=....=x_{n}$

 :D[/quote]
Thank you very much !  :blush:

You can also search about the weighted power means inequality :D .


Hola,

Una pregunta, alguien tiene alguna idea de a quien dirigirse para proponer un problema para la IMO o la Ibero (me imagino que los organizadores de la Ibero si podran estar interesados, mientras que los de la IMO, talvez ya tengan suficientes originales para este anio). Es que estaba tratando de resolver un problema abierto en Analisis de Pade, y probe un lema, y me di cuenta que se podia traducir muy bonito en un problema del tipo de Olimpiadas  :D y me gustaria proponerlo!

David los que proponen problemas son los entrenadores  que viajan con su pais a las olimpiadas. si estas interesado en proponerlo deberias hablar con el entrenador de algun pais, darselo y se te dara el credito a ti.

EDIT: Para evitar malentendidos: para proponer problemas se pude hacer directamente con el pais organizador, cada pais organizador tiene seleccioando un jefe de problemas, esta es la persona a la que se le debe enviar directamente, los jefes de delegacion solo eligen la prueba de un banco ya dispuesto. PASCUAL

Me parece que no necesariamente. Por ejemplo, creo que para proponer un problema para la IMO de México necesitas enviárselo a Florian Luca (un investigador bien chingón que te resuelve muchos problemas en una nada de tiempo) y ya el y un comité escoge la lista corta

Gracias,

Estoy intentando por los dos lados. Alli les dire en Agosto si se pudo o no para la IMO, o en Octubre si se pudo o no para la Ibero.

Por cierto que significa chingon ?
es algo parecido a ser excelente

Bueno en eso de crear problemas y proponerlos en la olimpiadas si mal no recuerdo es asi :

IBERO : los problemas propuestos por los paises para que conformen el banco de la IBERO no son enviados por el jefe de delagacion  sino se envian semanas antes via email a traves de la sociedad de matematicas  . Estos problemas se reunen y se arma el file de problemas que seran discutidos y revisados por los jefes de delegacion . Es decir el jurado no tiene conociemiento de este banco .

IMO : Los problemas deben ser enviados al pais organizador via email con meses de anticipacion es decir Existe un plazo maximo de entrega de problemas por parte de los paises participantes . los que envian los problemas es la sociedad de matematica de dicho pais .
Cabe destacar que al pais organizador llegan aproximadamente 200 porblemas los cuales son revisados por un el jurado o sea un equipo chingon de matematica que decidira cuales son aptos para una IMO y despues de ese filtro se obtiene la lista corta de problemas o sea los 30 problemas mas chingones . distribuidos en cada materia .

Chingón:
Significa algo así como ser muy bueno en algo. Mucha gente la considera una mala palabra, pero yo diría más bien que simplemente es muy vulgar.

Qué pasó con ese problema al final?????????


As Karl Frederick was walking up a flight of stairs, he noticed that for every 1 foot that he rose vertically, he covered a horizontal distance of 1 foot. What is the number of degrees in $ m\angle ABC$?
[asy]pair X,Y,B,C,A; C = (2,0); A = (2,2); X = dir(0); Y = dir(90); draw(B--C--A--B--Y--X+Y--X+2*Y -- A); label("B",B,SW); label("A",A,NE);  label("C",C,SE);[/asy]

Since both legs of right triangle $ ABC$ are equal, $ m\angle ABC\equal{}\boxed{45^\circ}$.


calculus
integration
function
limit
trigonometry
calculus computations

\[Let\ f\ :\ (0,\infty)\longrightarrow\ R\ \ be\ a\ function,\ where\  f(\lambda)=\int_{0}^{2\pi}\frac{sinx}{x+\lambda}\ dx.\ Prove\ that:\]
\[(\forall)\ \lambda>0,\ f(\lambda)>0;\ \ \lim_{\lambda\rightarrow\infty}f(\lambda)=0;\ \ \lim_{\lambda\rightarrow\infty}{\lambda}^2f(\lambda)=2\pi.\]

Side note: please don't put that much text inside of $\text{\LaTeX}.$

Integrate by parts, using $1-\cos x$ as the antiderivative of $\sin x.$

$f(\lambda)=\int_0^{2\pi} \frac{\sin x}{x+\lambda}\,dx$

$=\frac{1-\cos x}{x+\lambda}|_0^{2\pi}+ \int_0^{2\pi} \frac{1-\cos x}{(x+\lambda)^2}\,dx=\int_0^{2\pi} \frac{1-\cos x}{(x+\lambda)^2}\,dx$

Since the integrand in this version is positive on $(0,2\pi),$ it is clear that $f(\lambda)>0.$

$\lambda^2f(\lambda)=\int_0^{2\pi} \frac{1-\cos x}{(1+\frac{x}{\lambda})^2}\,dx$

As $\lambda\to\infty,$ the integrand converges uniformly to $1-\cos x.$ Hence,

$\lim_{\lambda\to\infty}\lambda^2f(\lambda)=\int_0^{2\pi}1-\cos x\,dx=2\pi.$

After the fact, the existence of this limit implies that $f(\lambda)\to0$ as $\lambda\to\infty,$ although we could have proved that directly in several other ways.

It is also clear that $f(\lambda)$ is a decreasing function for $\lambda>0$ and that

$\lim_{\lambda\to0^+}f(\lambda)$ exists and equals $\int_0^{2\pi}\frac{\sin x}x\,dx,$ which is a finite positive number.


logarithms
inequalities
inequalities unsolved


15 softball players, each with one ball, are standing at different distances from each other. Each pair (two players) has a different distance between them. Each player throws his ball to the player who is the closest by in distance.

a) Prove that no player receives more than 5 balls.
b) Can this problem be generalized for $n$ players, and what then is the minimal number of balls a player can receive?

Suppose you have a person with $n$ balls coming towards him/her.  If this person receives a ball from two neighboring persons, then the distance between this person and any other $n$ players (throwing balls at him) must be less than any two neighboring throwers.

Let $\gamma$ be the angle formed by two neighboring balls paths, and let $\alpha, \beta$ be the other angles in the triangle formed by the two neighboring ball paths. Then $\gamma>\alpha, \beta\implies 3\gamma>\alpha+\beta+\gamma=\pi\implies \gamma>\frac{\pi}{3}$.  The sum of all the angles meeting at the receiver is $2\pi$.  So the number of balls must be $\leq 5$ for if it were $>5$, then we would have $2\pi>2\pi$.  Contradiction!

Masoud Zargar

How about if this were extended to three-dimensional space. There is still a maximum number of balls. What is it?


Let $ G$ be a finite group of order $ n$,and $ S \equal{} (a_1,a_2,\cdots,a_n)$be a sequence of $ G$ with length $ n$,for $ \forall g\in G$ define $ S(g) \equal{} |\{i|a_i \equal{} g\}|$and
$ \sum_{\le r}(S) \equal{} \{a_{i_1} \plus{} a_{i_2} \plus{} \cdots \plus{} a_{i_l}|1\le i_1 < i_2 < \cdots < i_l\le n,l \le r\}$

Prove,if we let $ h \equal{} \max \{S(g),g \in G\}$,then $ 0 \in \sum_{\le h}(S)$


well you can use that fact that:

(a+b+c)^3=a^3+b^3+c^3+3(a+b)(a+c)(b+c) for any a,b,c from R.

U should calculate the derivate of the left side.
I think x=1 will be a zero of the derivate, so equality holds.
In other points, the function will be either decreasing, or increasing, so smaller or bigger than 2.

DusT from what you are saying it means that this eqs has only one solution: x=1, but it isn't true because it seems that x=0 is also a solution... so you should recheck ....  :D

bye!

If I am correct 9 is also a solution...  :?

tschuess  :D  :D


number theory
greatest common divisor
induction
algorithm
algebra unsolved
algebra

The Fibonacci sequence is defined by \[ a_{n+1} = a_n + a_{n-1}, n \geq 1, a_0 = 0, a_1 = a_2 = 1. \] Find the greatest common divisor of the 1960-th and 1988-th terms of the Fibonacci sequence.

Isn't it well known ?

By induction :

$F_{m+n} = F_{m}F_{n+1} + F_{m-1}F_{n}$

Then using Euclid algorithm and $gcd(F_n,F_{n+1})=1$ (from Cassini relation for instance : $F_{n+1}F_{n-1} - {F_{n}}^2 = (-1)^n$ we have $gcd(F_n,F_m) = F_{gcd(m,n)}$


Does anyone here have any recomendations on what books I should use after I finish AoPS 1 and 2? What book is inbetwen olympiad level and intermediate stuff? I don't want to get a book and be lost in the first chapter.[/quote]

I would put Problem Solving Through Problems (Larson) or Art and Craft of Problem Solving (Zeitz) after AoPS. The former is a bit easier I think, and both are probably below olympiad level, although they include many olympiad problems.

I would keep reviewing AoPS Volume 1 and 2 as well to make sure you understand everything deeply.

But that's not the question he's asking.

after aops 1 and 2, you could start trying to make USAMO and go farther  :lol:

:spam:  :spam:  :spam: Nothing to do with the original question...


:what?: Don't tell me you found it manually?

How about the [url=http://en.wikipedia.org/wiki/Quartic_formula]quartic formula[/url]?

CAS's like Mathematica know the quartic formula, usually, and can solve quartics exactly. The exact solution to a quartic, though, is *usually* (ie, when it looks like that) impractical in its applications.

[b]To the OP:[/b] What are you looking for, and where did that equation come from? All its roots look like the one Sonnhard posted.

You cac see formula of quartic at: http://planetmath.org/encyclopedia/QuarticFormula.html


algebra
polynomial
calculus
integration
algebra proposed

find all the integer $ k$ that the polynomial
$ p(x)\equal{}x^5\plus{}x^4\plus{}x^3\plus{}kx^2\plus{}x\plus{}1$ can be factor in polyinomial with integral coefficients
and with $ deg < p(x)$

If
$ p(x)\equal{}p_2(x)p_3(x)\equal{}(x^2\plus{}ax\pm 1)(x^3\plus{}bx^2\plus{}cx\pm 1)\equal{}x^5\plus{}x^4(a\plus{}b)\plus{}x^3(\pm 1\plus{}ab\plus{}c)\plus{}x^2(\pm1\plus{}ac\pm b)\pm (a\plus{}c)x\plus{}1,$
 then $ b\equal{}1\minus{}a,c\equal{}\mp 1\plus{}a(a\minus{}1),a\plus{}c\equal{}a^2\mp 1$ and we get $ a^2\equal{}\pm 2$ - contradition.
Therefore $ p(x)\equal{}(x\minus{}m)p_4(x)$/ Obviosly $ m\equal{}\pm 1$. It give $ k\equal{}\minus{}5,m\equal{}1$ and $ k\equal{}1,m\equal{}\minus{}1$.


function MATHCOUNTS algebra polynomial FTW calculus integration

What calculator do you use?

i use a Casio fx something something(don't write the model like me :D)  non graphing calc

I use the TI-Inspire I got at MC Nats this year.

OOO my god  i keep hating my self  :censored:  :wallbash_red:
i keep finding out the things about MC nats that make me drool like they're giving out graphing calculators.
And then i keep thinking that i didn't make nats
but then i think that i am still in fifth grade and still have 3 more years to participate but still :wallbash_red:

nspire here.  I had a 4-function before nats :P

I use a TI 84 SE but in my opinion for MATHCOUNTS it doesn't really matter what calculator you use as long as it has the basic features you use regularly and the fact that you are comfortable with it.

but if you want to do anything like AMC 10 or 12 you should have graphing calc, its easier in my opinion

uh, u can cover the full screen by making a program that draws $ y \equal{} 4 (sin x) like a billon times  : D$ or just do something like $ \displaystyle\sum_{x \equal{} 0}^n sin x$ where $ n$ is like a $ BIG$ number (its gonna take like an hour or so to process it though :D)

TI 83

I hate you all who got nSpires at nats

Poincare, I meant just on the Y= menu thingy-ma-bober. I think I have an idea but I'm too lazy to type it or try it...

I don't use a graphing, just a scientific CASIO [i]fx-300ES[/i], which is good enough for me. I really like my calc, and I figured out how to do all these really cool things on it without taking a single look at the mile wide manual thing.

Hmm...  NSpires are very useless in higher-level competitions, which rarely even allow normal calcs, let alone NSpires, especially in high school. :(   Anways, the NSpire we got isn't as useful as the TI-89 Titanium, cuz the latter can manipulate polynomials and solve equations in ways that the former can't.  If you're looking for mathematical functionality, go buy an NSpire CAS, which can do loads of uber-amazing crap.  The NSpire that MC gave is cool and all, but I rarely use it.  Instead, I use my handy-dandy TI-84 SE ftw that I got at Nats last year.  :D

[quote="leoxnlin"]Good ol' TI 83.

In my opinion, the programmable feature is far more important than the graphing feature. (About the only thing I do with the graphing feature is to make cool-looking designs... and find a way to black out the entire screen... haven't figured that out yet)[/quote]

I know how to do that.
Do normal window settings.
$ y\equal{}10$
$ y\equal{}\minus{}10$
NOW FIND SOME INTEGRALS!
calc->(integral feature -- forgot what it's called) go from $ \minus{}15\to15$ for both equations.

I use the TI-84+ SE. I don't like nspire BECAUSE YOU CANNOT PROGRAM IT.

I have one of those simple calculators because its simple to use!

[quote="Poincare"]now, to make this disscussion turn around a bit : how many of you like your calculators?[/quote]

My calculator is my friend.  :wink:

i never use my calculator because im always too lazy to find it. i have an 83+

I use a casio scientific.  I own 3 graphing calcs (from nats) but find them kinda cumbersome, and I haven't really reached the pt where I really need them, I ocassionally use the programs i have downloaded onto it

is it true that  they banned TI-89's from SAT's & other math tests because it can solve equations? :huh:  :huh:  :huh:

Yes. However, 84's can easily be programmed to solve equations and I personally have a CAS system on mines  :P so...not that the SAT really needs a calculator...in fact...you don't even need scratch paper...

is a graphing calc neccessary/ helpful for calculus? (i am taking it in sixth grade)

If you're learning calculus, graphing calculators are helpful for:
i. Visualizing graphs of functions
ii. Visualizing relationships between functions, various derivatives, etc.
iii. Standardized tests, like the AP exam, or school assignments under certain curricula, where the questions are written with the expectation that you have one

(i) is taken care of if you're good with visualizing graphs. (ii) is a one-time deal. Most likely, (iii) is the only real reason to go out and buy one, if it applies.

But if you're taking it in 6th grade, chances are you're either taking it at a college or online, as opposed to at a high school, so even (iii) might not apply.

As for me, my TI 30X IIS has never let me down :) Hooray, scientific calculators!

[quote="herefishyfishy1"]Definitely TI-89 Titanium. The solve feature is very useful, and I use the built in summation function a lot.

Also, you can write FUNCTIONS on it.

It can find out the value of an expression with variables in it without knowing the values of the variables. (for example, x+x typed in on a TI-89 Titanium would return 2x)

It can differentiate and integrate easily, as well.

The Computer Algebra System is the best thing TI has ever invented!

(I don't have a TI-[i]n[/i]spire CAS, and in my opinion, the TI-89 Titanium is more powerful than the TI-[i]n[/i]spire without CAS that I got at National Mathcounts.)

P.S. You CAN program a TI-[i]n[/i]spire[/quote]
Then, do you know how to use ti-89 to graph slope field such as y'=3x+2y?

[quote="Poincare"]yeah
do the differention/derivation first and then put in the thing as y1[/quote]
May you expain in more detail?
The following link is the question I want to ask.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212735[/url]

it looks like Texas Instruments has done very well with making these calculators
im impressed!

[quote="herefishyfishy1"]Definitely TI-89 Titanium. The solve feature is very useful, and I use the built in summation function a lot.

Also, you can write FUNCTIONS on it.

It can find out the value of an expression with variables in it without knowing the values of the variables. (for example, x+x typed in on a TI-89 Titanium would return 2x)

It can differentiate and integrate easily, as well.

The Computer Algebra System is the best thing TI has ever invented!

(I don't have a TI-[i]n[/i]spire CAS, and in my opinion, the TI-89 Titanium is more powerful than the TI-[i]n[/i]spire without CAS that I got at National Mathcounts.)

P.S. You CAN program a TI-[i]n[/i]spire[/quote]

Just found out that my 10 dollar Casio (FX-115-ES) has a solving function!

W00T!

Calculators may be helpful, but they are not good for learning math. Dependency on the calculator is one of the many problems with American math education. Unless you're finding decimal approximations or something, use your hand and your head!


trigonometry
sine double angle
calculus
calculus computations

q1x=sin( 2  tan^-1( 2)),

y=sin(.5tan^-1 (4/3))

then find relation between x and y

Q2   with  attachment

For question number 1) $ x \equal{} 4y^2$.
I used four formulas: half angle formula for sine, double angle formula for sine, $ \cos (\ tan^{\minus{}1}(x)) \equal{} \frac{1}{\sqrt{1 \plus{} x^2}}$, and $ \sin (\ tan^{\minus{}1}(x)) \equal{} \frac{x}{\sqrt{1 \plus{} x^2}}$.


Start with circumscribed right triangle ABC (C=90).
    Select random points X,Y,Z on smaller arcs AC,BC and opposite C
    and draw tangents to circumcircle.
    Find mid-points of segments cut on tangent lines by CA and CB.
    When X,Y,Z  coinside with the mid-points,
    XYZ is regular,  <YCB=1/3(<CAB),  <XCA=1/3(<CBA),
    <ZCA = 1/3(PI - <CBA),  <ZCB = 1/3(PI - <CAB)

     Guess the orientation ( geometrical that is! )  of Morley ?

     Maj. Pestich

[quote="pestich"]Start with circumscribed right triangle ABC (C=90).
    Select random points X,Y,Z on smaller arcs AC,BC and opposite C
    and draw tangents to circumcircle.
    Find mid-points of segments cut on tangent lines by CA and CB.
    When X,Y,Z  coinside with the mid-points,
    XYZ is regular,  <YCB=1/3(<CAB),  <XCA=1/3(<CBA),
    <ZCA = 1/3(PI - <CBA),  <ZCB = 1/3(PI - <CAB)

     Guess the orientation ( geometrical that is! )  of Morley ?[/quote]

Let $X, Y$ be intersections of the trisectors of the angles $\measuredangle A, \measuredangle B$ with the circumcircle $(O)$ of the right angle triangle $\triangle ABC$, such that $\measuredangle BAX = \frac{\measuredangle A}{3}$, $\measuredangle ABY = \frac{\measuredangle B}{3}$. WLOG, assume that $\measuredangle A > \measuredangle B$. Let $X_A, X_B$ be the intersections of the tangent to the circumcircle at the point $X$ with the lines $BC, CA$ and $Y_A, Y_B$ the intersections of the tangent to the circumcircle at the point $Y$ with the lines $BC, CA$. Since the angle $\measuredangle C = \frac \pi 2$ is right, the circumcenter $O$ is the midpoint of the side $AB$.

$\measuredangle AXX_B = \frac \pi 2 - \measuredangle AXO = \frac \pi 2 - \measuredangle XAO = \frac \pi 2 - \frac{\measuredangle A}{3}$
(triangle $\triangle AXO$ is isosceles, with $OA = OX$ being the circumradius)

$\measuredangle AX_BX = \pi - (\measuredangle AXX_B + \measuredangle XAX_B) = \pi - \left(\frac \pi 2 - \frac{\measuredangle A}{3} + \frac{2 \measuredangle A}{3}\right) = \frac \pi 2 - \frac{\measuredangle A}{3}$

This means that the triangle $\triangle AXX_B$ is isosceles, $AX = AX_B$, but we do not need this fact.

$\measuredangle AXC = \measuredangle ABC = \measuredangle B$

$\measuredangle CXX_B = \measuredangle CAX = \frac{2 \measuredangle A}{3}$
(angle between the chord $CX$ and the tangent at its endpoint $X$ equals to the angle spanning the arc $CX$)

$\measuredangle XCX_B = \pi - (\measuredangle AX_BX + \measuredangle CXX_B) = \pi - \left(\frac \pi 2 - \frac{\measuredangle A}{3} + \frac{2 \measuredangle A}{3} \right) =$

$= \frac \pi 2 - \frac{\measuredangle A}{3} = \measuredangle AXX_B$

It follows that the triangle $\triangle CXX_B$ is isosceles, with $XC = XX_B$. Similarly,

$\measuredangle XCX_A = \measuredangle BCX_B - \measuredangle XCX_B = \frac \pi 2 - \left( \frac \pi 2 - \frac{\measuredangle A}{3} \right) = \frac{\measuredangle A}{3}$

$\measuredangle CXX_A = \pi - \measuredangle CXX_B = \pi - \frac{2 \measuredangle A}{3}$

$\measuredangle CX_AX = \pi - (\measuredangle XCX_A + \measuredangle CXX_A) = \pi - \left(\pi - 2 \frac{\measuredangle A}{3} + \frac{ \measuredangle A}{3} \right) =$

$= \frac{\measuredangle A}{3} = \measuredangle XCX_A$

Again, the triangle $\triangle CXX_A$ is isosceles, with $XC = XX_A$. Consequently, $XX_A = XC = XX_B$ and the tangency point $X$ is the midpoint of the segment $X_A X_B$. In exactly the same way, it can be shown that $YY_A = YC = YY_B$, i.e., the tangency point $Y$ is the midpoint of the segment $Y_A Y_B$.

The circumradii $OX, OY$ form the angle

$\measuredangle XOY = \measuredangle XOA - \measuredangle YOA = \pi - \measuredangle XOB - \measuredangle YOA = \pi - 2 \measuredangle XAB - 2 \measuredangle YBA =$

$= \pi - \left(2 \frac{\measuredangle A}{3} + 2 \frac{\measuredangle B}{3} \right) = \pi - \frac 2 3 (\measuredangle A + \measuredangle B) = \pi - \frac \pi 3 = \frac{2 \pi}{3}$

Let the tangents $X_A X_B, Y_A Y_B$ meet at a point $R$. Since these tangents are perpendicular to the radii $OX, OY$, it follows that the angle $\measuredangle XRY$ is equal to

$\measuredangle XRY = 2\pi - 2 \frac \pi 2 - \measuredangle XOY = \pi - \frac{2 \pi}{3} = \frac \pi 3$

Let $Z$ be the intersection of the bisector of the angle $\measuredangle XRY$ with the circumcircle $(O)$, on the opposite side of the diameter $AB$ than the point $R$ and the triangle vertex $C$. Let the tangent to the circumcircle at the point $Z$ intersect the lines $BC, CA$ at points $Z_A, Z_B$ and the tangents $X_A X_B, Y_A Y_B$ at points $Q, P$. The triangle $\triangle PQR$ is obviously equilateral. It remains to show that the tangent $Z_A Z_B$ is parallel to one side of the Morley equilateral triangle and that $ZZ_A = ZZ_B$, i.e., that the tangency point $Z$ is the midpoint of the segment $Z_A Z_B$.

Let $P', Q', R'$ be the intersections of the appropriate trisectors of the angles $\measuredangle B$ and $\measuredangle C$; $\measuredangle C$ and $\measuredangle A$; $\measuredangle A$ and $\measuredangle B$, forming the Morley equilateral triangle $\triangle P'Q'R'$. The angle between the lines $AB$ and $Q'P'$ is equal to

$\measuredangle{(AB, Q'P')} = \pi - (\measuredangle B + \frac 2 3 \measuredangle C) - \measuredangle CQ'P' = \frac{2 \pi}{3} - \measuredangle B - \left( \frac{\measuredangle A}{3} + \frac \pi 3 \right) =$

$= \frac{\measuredangle A - \measuredangle B}{3}$

The triangle $\triangle XYR$ is obviously equilateral, because $RX = RY$ are segment lengths of tangents to the circumcircle from the point $R$ and the angle $\measuredangle XRY = \frac \pi 3$. Hence, the lines $XY \parallel Z_A Z_B$ are parallel. The angle between the lines $AB$ and $YX$ is equal to

$\measuredangle{(AB, YX)} = \measuredangle BYX - \measuredangle YBA = \measuredangle BAX - \measuredangle YBA = \frac{\measuredangle A - \measuredangle B}{3}$

As a result, $P'Q' \parallel XY \parallel Z_A Z_B$. Finally, we will show that the triangles $\triangle CZZ_A, \triangle CZZ_B$ are both isosceles, with $ZC = ZZ_A$ and $ZC = ZZ_B$. The angles $\measuredangle CZ_AZ_B, \measuredangle CZ_BZ_A$ of the rightangle triangle $\triangle CZ_AZ_B$ are

$\measuredangle CZ_AZ_B = \measuredangle B + \frac{\measuredangle A - \measuredangle B}{3} = \frac \pi 6 + \frac{\measuredangle B}{3}$

$\measuredangle CZ_BZ_A = \frac \pi 2 - \measuredangle CZ_AZ_B = \frac \pi 2 - \left( \frac \pi 6 + \frac{\measuredangle B}{3} \right) = \frac \pi 6 + \frac{\measuredangle A}{3}$

The triangle $\triangle CZO$ is isosceles, with $OC = OZ$ being the circumradius. Its angles $\measuredangle COZ$, $\measuredangle OZC = \measuredangle OCZ$ are equal to

$\measuredangle COZ = \measuredangle COA + \measuredangle AOZ = 2 \measuredangle CBA + \frac \pi 2 + \widehat{AB XY} = 2 \measuredangle B + \frac \pi 2 + \frac{\measuredangle A - \measuredangle B}{3} =$

$= \pi - 2 \frac{\measuredangle A - \measuredangle B}{3}$

$\measuredangle OZC = \measuredangle OCZ = \frac{\pi - \measuredangle COZ}{2} = \frac{\measuredangle A - \measuredangle B}{3}$

Consequently,

$\measuredangle CZZ_A = \frac \pi 2 + \measuredangle OZC = \frac \pi 2 + \frac{\measuredangle A - \measuredangle B}{3}$

$\measuredangle CZZ_B = \frac \pi 2 - \measuredangle OZC = \frac \pi 2 - \frac{\measuredangle A - \measuredangle B}{3}$

$\measuredangle ZCZ_A = \pi - (\measuredangle CZZ_A + \measuredangle CZ_AZ_B) = \pi - \left( \frac \pi 2 + \frac{\measuredangle A - \measuredangle B}{3} + \frac \pi 6 + \frac{\measuredangle B}{3} \right) =$

$= \frac \pi 3 - \frac{\measuredangle A}{3} = \frac \pi 6 + \frac{\measuredangle B}{3} = \measuredangle CZ_AZ_B$

$\measuredangle ZCZ_B = \pi - (\measuredangle CZZ_B + \measuredangle CZ_BZ_A) = \pi - \left( \frac \pi 2 - \frac{\measuredangle A - \measuredangle B}{3} + \frac \pi 6 + \frac{\measuredangle A}{3} \right) =$

$= \frac \pi 3 - \frac{\measuredangle B}{3} = \frac \pi 6 + \frac{\measuredangle A}{3} = \measuredangle CZ_BZ_A$

As a result, the triangles $\triangle CZZ_A, \triangle CZZ_B$ are both isosceles, with $ZC = ZZ_A$ and $ZC = ZZ_B$. Hence, $ZZ_A = ZC = ZZ_B$, i.e., the tangency point $Z$ is the midpoint of the segment $Z_A Z_B$.

[quote="sunken rock"]In my books it's called [b]Pollock's Theorem[/b], but I did not succeed to google it!

Best regards,
sunken rock[/quote]

Dear [color=blue][b]SUNKEN[/b][/color], which is your book?
can I buy?

Dear Ligouras,

The book, "Problem Book in Geometry", by G. Titeica, a well-known Romanian geometer. I possess the 6th issue, 1981, and I do not know if other edition has been issued meanwhile. Besides, it is in Romanian.

Best regards,
sunken rock

Dear Mathlinkers,
you can also find the Pollock's theorem in
F.G.M., Exercices de Géométrie, 6th ed., 1920, Rééditions Jacques Gabay (Gabay reprint), Paris, 1991, p. 286.
Sincerely
Jean-Louis


Prove that a penthagon with with sides rational numbers and all angles equal it's regular. ;)  :D

See http://www.mathlinks.ro/Forum/viewtopic.php?t=33920 .

  darij


pigeonhole principle
modular arithmetic
number theory unsolved
number theory

show that for any $ n$ not congruent to $ 0 mod 10$ there exists a multiple of $ n$ not containing  the digit $ 0$ in its decimal representation

[quote="ishfaq420haque"]Show that for any $ n \not \equiv 0 \pmod{ 10}$ there exists a multiple of $ n$ which does not contain the digit $ 0$ in its decimal representation[/quote]
We consider the integers $ 1,11,111,\cdots, \underbrace{111\cdots11}_{n}$, $ \text{modulo} \; n$. If any of these $ \equiv 0 \mod n$ then we are done.
Otherwise, we have $ n$ integers and the possible remainders are $ 1,2,\cdots,n \minus{} 1$.
By Pigeonhole Principle at least two of these integers (say $ a,b$), have the same remainder.
So, we have $ a \minus{} b \equal{} 111\cdots0$.Obviously,  $ n |(a \minus{} b)$ and $ n\not \equiv 0 \pmod{ 10}$.
So, the result follows.

But your proof will work only when $ n$ is coprime to $ 10$. For example, when $ n\equal{}5$, $ 5\mid 110$ but that doesn't necessarily imply $ 5\mid 11$.

[hide="duh... have to think again"][quote="nayel"]But your proof will work only when $ n$ is coprime to $ 10$. For example, when $ n \equal{} 5$, $ 5\mid 110$ but that doesn't necessarily imply $ 5\mid 11$.[/quote]
Ah..thanks. I misread the problem. But still this method works. I mean, if $ (n,10)\not \equal{} 1$

Case 1, $ 2^k\parallel n$: we can consider the integers $ \bar{2^k},\bar{2^k}\bar{2^k},\cdots, \underbrace{\bar{2^k}\bar{2^k}\bar{2^k}\cdots\bar{2^k}}_{n}$

Case 2, $ 5^k\parallel n$:We can consider the integers $ \bar{5^k},\bar{5^k}\bar{5^k},\cdots, \underbrace{\bar{5^k}\bar{5^k}\bar{5^k}\cdots\bar{5^k}}_{n}$

I hope that it works now....

[hide="For Ishfaq and Nayel vi"]Actually I am in a vacation ("sort of")... on a travel to be precise. Last night, I enjoyed the night sitting under the sky...:)[/hide][/hide]

What guarantees that $ \overline{2^k}, \overline{5^k}$ have no zeroes in them?


trigonometry
function
geometry
trig identities
Law of Sines

In $ \triangle ABC$ with $ BC\equal{}a,CA\equal{}b$, and $ AB\equal{}c$, given that $ (b\plus{}c)(b\minus{}c)\equal{}ca$ and $ \angle BAC\equal{}75^{\circ}$, find $ \angle ABC$.

[hide="a very contrived approach"]
The answer is $ \angle ABC \equal{} \boxed{70^{\circ}}$. Then $ x \equal{} \angle ACB \equal{} 35^{\circ}$. We verify that the conditions are satisfied: by the Law of Sines we have $ \frac {\sin x}{c} \equal{} \frac {\sin 2x}{b} \equal{} \frac {\sin (180 \minus{} 3x)}{a}$, that is $ \frac {a}{c} \equal{} \frac {\sin 3x}{\sin x} \equal{} 4\cos^2 x \minus{} 1$, and $ \frac {b}{c} \equal{} 2\cos x$, so $ \frac {b^2}{c^2} \minus{} \frac {a}{c} \equal{} 1$ which re-arranges to the given.

We now show this is the only solution. By Law of Sines we obtain $ \frac {(b \plus{} c)(b \minus{} c)}{c^2} \equal{} \frac {a}{c} \Longrightarrow \frac {\sin^2 (105 \minus{} x)}{\sin^2 x} \minus{} 1 \equal{} \frac {\sin 105^{\circ}}{\sin x}$. That is, \[ \frac {\sin^2 (105 \minus{} x)}{\sin x} \minus{} \sin x \equal{} \sin 105^{\circ}.\] On $ x \in [0, 90^{\circ}]$ we easily verify that the left-hand side is a strictly decreasing function of $ x$, so $ x \equal{} 35^{\circ}$ is the unique root in this interval. On $ x \in [90^{\circ}, 105^{\circ}]$, we re-arrange to get $ \sin^2 15^{\circ} > \sin^2 (105 \minus{} x) \equal{} \sin 105^{\circ} \sin x \plus{} \sin^2 x > 2 \sin^2 105^{\circ}$, so there is no root here.
[/hide]

A manipulative approach would be much nicer, but I can't find any.

[hide]Extend $ AB$ and construct $ D$ on line $ AB$ such that $ BD\equal{}BC$. Then try to confirm the similarity of $ \triangle ABC$ and $ \triangle ACD$.[/hide]

But somehow that's not so satisfying. I am still looking to trigonometrical method.


Suppose that sinA-cosA=1/sqrt[2].
Find the value of (tanA)^3+(cotA)^3

Well...

[hide]I could solve for sin(A) and cos(A) by using their half-angle formulas, and then find tan(A)=sin(A)/cos(A), but...I don't think that is what you were getting at.[/hide]

This is nitpicking a bit, but sin(2A)=1/2 doesn't imply that 2A=30.

yea, it could be 150.

But fiery, just to let u kno, solving for A is perhaps the easiest way.

[quote]yea, it could be 150. [/quote]

....or 390

[quote]But fiery, just to let u kno, solving for A is perhaps the easiest way[/quote]

What is the difficult path? When I come to a fork in the road, I may see two paths. One is a pleasant path, with birds chirping amidst the blueberry shrubs along the trail. The other is through an ominous forest, the trees casting dark shadows upon the muddy pathway, treacherous beasts screaching eerily high tones. But off in the distance, way off in the distance, I can perhaps see a small speck of the paradise at the end of this path, with luscious flowers blooming about, and young maidens dancing in the meadows. I then decide to take the difficult path.

sometmies, the most difficult path isn't the one pointed out to u by a traveler who directs to rocky cliffs and barren deserts, but the one u find urself when u find urself in the middle of nowhere about to be devoured by vultures

[color=cyan]Wait, did that have anythign to do with math at all?

<confused>[/color]

simple, you come to two paths.... take the one less travelled by, and it will make all the difference  :wink:

well, we are talking about math in disguise
this is philosophical math, Joel

So, plainly and simply: what's the hard way?

[quote]Hmm, so u actually solved for A.
Now, can you do it without solving for A?[/quote]

[quote]But fiery, just to let u kno, solving for A is perhaps the easiest way.[/quote]

Huh?

I think not solving for A is the harder way...

P.S. the last part about the paradise should include ripply :), but that's irrelevant. :? *serious face*

Yeah, but Im going to try avoid talking about her on the problem-solving boards from now on. Back to the problem...

I respect your decision and I think it's a wise one.

I have no clue as to how to solve it without A...

yo guys, as moderator, im telling u, amc forum is strictly maht

i don't want to see Fiery and tare hitting on ripply here.

Save it for etc.....

anyway, back the problem

If you haven't noticed, at least half of my posts is about maht, fyi.
[quote="darkquantum"]i don't want to see Fiery and tare hitting on ripply here.[/quote]
Define "hitting"

As I said before, you could leave A as it is and mess around with the equation a bit, and get the solution directly, I think.


function
real analysis
real analysis unsolved

If $F : \mathbb{R}^{p}\rightarrow \mathbb{R}^{q}$ is a smooth function and dF has rank q at a certain point $a \in \mathbb{R}^{p}$, prove that there is a neighborhood of a in which dF has rank q.

More generally, if $dF$ has rank $k$ at $a$, then there is a neighborhood of $a$ in which $dF$ has rank $\ge k$. Indeed, having rank $\ge k$ means having a nonzero $k\times k$ minor. The set of all points where a certain $k\times k$ minor is nonzero is open, by the continuity of the determinant function.


There exists a block of $1000$ consecutive positive integers containing no prime numbers, namely, $1001!+2$, $1001!+3$, $\cdots$, $1001!+1001$. Does there exist a block of $1000$ consecutive positive integers containing exactly five prime numbers?

$f(x)=\pi(n+999)-\pi(n)$ is continuous in the following meaning: if $n$ changes by $1$, $f(n)$ changes by at most $1$.
If it changes from $a$ to $b$, it did have every value $c$ between them at least once (everything integers). With $a=f(2) > 5$, $b=f(1001!+1) =0$, we can find $n$ such that there are exactly $5$ primes in $\{n,n+1,...,n+999\}$.

I apologize to everyone for my solution being very similar to that of [b]ZetaX[/b].

Let $f(n)(n\in \mathbb N)$ denotes the number of prime numbers from $\{ n,\dots ,n+999\}$.It is easy to show that $f(n+1)-f(n)=0,\pm 1$.Obviously $f(1)>5$ and $f(1001!+2)=0$.Then by intermediate value theorem,$1<$∃$m<1001!+2$ s.t. $f(m)=5$.Therefore there exists $1000$ consecutive positive integers containing exactly five prime numbers.$\blacksquare$ :coolspeak:


I will add to this poll later, sorry if this offends anyone playing NeoPets, RuneScape, etc. (It's just that most people stop playing before or when they become teens)

NoTe: Just to tell everyone, I used to go to (dont hilight if u dont want to know)  [hide]Neopets!!![/hide] when I was 11.

http://www.spacepirate.com, to play the demo.

I didn't have the Internet when I was 9.  I used to go to Neopets, but it got boring.  I still play Runescape.

Edit:  Oh yeah, I played around with Beenz, but then they went out of business.  I was about 50 Beenz from getting a Rewardzcard.  I ended up converting the Beenz to Rocketcash, and I eventually spent it on some random CD.  Beenz led to other online rewards programs, all of which are defunct.  I had a huge pile of Blink Rewards points, enough to get about $10.

don't become like my friend... a runescape freak. i hate runescape now.  and steven, u still play neopets and i know that. i'm not that stupid, i'm in like none of ur classes but i see u like everyday

I must say that you are right about Steven playing Neopets. I mean, he actually admits that he plays it, so don't try to deny it, Steven.

[quote="Syntax Error"]I went to ESPN's site like a madman.  My fanaticism for sports, especially basketball, was really high back then.[/quote]

I still do. Especially now that...

THE RED SOX ARE GOING TO THE WORLD SERIES!!!!!!!!!!! TAKE THAT YOU YANKEES FANS!!!!!! 4 IN A ROW!!!!!!!

i'm not a yankee's fan but um... u shouldn't be so mean u could hurt some ppls feelings :(  :(  :(  :(  :(  :(  :(  :(  :(  :(  :(  :D  :D  :D  :D  :D  :lol: [/quote]

i used to play neopets. and the summer b4 7th grade started 4 me my brother started playing runescape so i created an account. it was actually pretty fun until sumone hacked my account. now im straight. but i do remember my level was excetionally high.

Personally, I prefer Cartoon Orbit to Neopets.  My little sister plays Neopets, and it seems to me to be far too time-consuming.  Cartoon Orbit has about the same concept, and it's somewhat safer for children.

ok ok i admit i played neopets when i was in 3rd grade. then i quit in 3rd grade.  it got boring pretty fast.  and i also played runescape from 6-7 grade.  i quit cause 1st it got boring.  2nd my own friend hacked me.  3rd it was a waste of my time

wow neopets has changed alot since i last was addicted. but i admit that SOMETIMES i go on now. when im really really bored.  but in 6th grade i quit and gave rachel (sunnyray) my password and she did a bunch of stuff to it......lol.  but i actually was kinda addicted 2 runescape. i mean REALLY addicted. i bet i would still be playing if the guy didnt hack my account. or maybe not...depends. yes i admit that.

u really don't seem like a computer freak at school.  ur mostly on the quiet side i guess

ermmm.....is that a compliment?  YOU THINK IM ON THE QUIET SIDE??? geez cant u always hear me and linda giggling away in math e? i gess u dont. and we were correcting test papers in the library for a few days and being very retarded and disruptive so the ppl kept telling us 2 shut up. lol. but she started it.

[quote="Syntax Error"]I went to ESPN's site like a madman.  My fanaticism for sports, especially basketball, was really high back then.[/quote]

Like YifMan, i still do, 'cept it's basketball, football, baseball, tennis, anything really to take my mind off things. I don't really don't play much online games, just computer home games, like madden NFL 2005, espn nba 2k5... and more. I also do those sim leagues where you take on a team and do real drafts and trades players and junk. I'm stuck in about 10 of them right now. :?  :?

hey athena i was sitting in the fr..on..t wait a sec. u sat in front of me last year. i gues i don't pay much attention at all

wow. that post was so flattering. really...it is.

ookay...i think i'll back off from this topic for awhile :?

I don't like Neopets because it's hypnotiZing. With a Z. See Duelling Decks for an example.
It's also full of adult ads, so 79% of Neopet players are seeing ads that they're not supposed to.

:rotfl: random revive..  All the people in this thread seem to have known each other.

[quote="You-lost-the-game"]I LOVE NEOPETS
^.^
I'm so childish. But I love neopets. I'm probably younger than most people on this site, though.[/quote]

Probably not younger than me :)

Uh..when I [b]was[/b] 9 or when I [b]am[/b] 9? :)

Anyways, I tried to play Runescape, but I sort of died on controlling.

I prefer tower defense games :D

I'm 13!  j;j;j;j;jjjjjjjjjjjjjjjjjj;!
Has anyone tried conlanging?

what do you mean by dult games? i played club penguin :-D

Runescape is really fun.  I'm 11 and I probably won't get bored.

What do you mean by adult sites?  Like Eharmony or something?

Wats that?

I play all the Artix Entertainment games, and are pretty high levels at taht.

Like aqworlds.com, dragonfable.com, mechquest.com, etc. etc.

I especially like aqworlds.com BECAUSE THEY JUST CHANGED ENGINES!!! THE NEW ENGINE IS BEASTLY!!!

I didn't say adult games, I said adult ads appearing on the screens of their Neopets accounts.

I like neopets

I know, ima guy but the games were pretty fun :P


[quote="shortlist"]let $ P$ be a polynomial of degree $ n$ with real coefficients such that $ \left|f(x) \right| \leq1$ for all $ x \in[0,1]$. Show that
$ \left|f(\frac { \minus{} 1} { n })\right|\geq 2^{ \minus{} n}$[/quote]
What about $ P(x) \equal{} \frac{1}{2^{1000}}x$? Then $ |P(\minus{}1)| \equal{} \frac{1}{2^{1000}} < 2^{\minus{}1}$..

[quote="shortlist"][quote="Mathias_DK"][quote="shortlist"]let $ P$ be a polynomial of degree $ n$ with real coefficients such that $ \left|f(x) \right| \leq1$ for all $ x \in[0,1]$. Show that
$ \left|f(\frac { \minus{} 1} { n })\right|\geq 2^{ \minus{} n}$[/quote]
What about $ P(x) \equal{} \frac {1}{2^{1000}}x$? Then $ |P( \minus{} 1)| \equal{} \frac {1}{2^{1000}} < 2^{ \minus{} 1}$..[/quote]

I am sorry . I repaired again  :blush:[/quote]
The same counterexample works as $ |P( \minus{} 1)| \equal{} \frac {1}{2^{1000}} < 2^2\minus{}1 \equal{} 3$

I am sorry . it must
quote="shortlist"]let $ P$ be a polynomial of degree $ n$ with real coefficients such that $ \left|f(x) \right| \leq1$ for all $ x \in[0,1]$. Show that
$ \left|f(\frac { \minus{} 1} { n })\right|\leq 2^{ n \plus{} 1} \minus{} 1$[/quote]