discussion
vector
geometry
linear algebra
number theory
complex analysis

Hi! I didnt find any other place to post this so I try here.
What do you need to study advanced analytic number theory, which courses etc. is necessary? Which analysis-courses? I know complex analysis is necessary, but I wonder if fourier-analysis is also necessary or function-analysis? Do you need vector and tensor analysis?

Fourier analysis is definitely needed for studying analytic number theory since many times you will want to take a Fourier transform of a function you have to end up with a nicer one. Basically analytic number theory is so hard that you'll need to be pretty sharp on everything else (algebra, complex analysis, Fourier analysis, elementary number theory, and probably a whole lot more) to do a really good job. My advisor told me to study algebraic geometry over the summer since it will help me for analytic number theory, so we'll see how that goes.

Ok. Thanks for your answer! Du you know how much combinatorics is necessary.
I am going to take on of the two courses in combinatorics they offer in Bergen this autumn, and I wonder if I should take the second one next spring... (the first one use the book "A WALK THROUGH COMBINATORICS An Introduction to Enumeration and Graph Theory by Miklós Bóna" The second one contain more about ramsey-theory, matrix-theory etc.)

You probably won't need that much actual combinatorics in the sense of actually counting things, but you'll need to be able to use generating functions.

Ok. Then I think the first course is enough.


$Let a,b,c be positive numbers.
Prove that $$ ￥frac{(a+b)＾｛669｝(b+c)＾｛669｝(c+a)＾｛669｝}{(a＾｛669｝+b＾｛669)(b＾｛669｝+c＾｛669｝)(c＾｛669｝+a＾｛669｝)｝￥leq 2＾｛2004｝$

[quote="kunny"]$Let a,b,c be positive numbers.
Prove that $$ ￥\frac{(a+b)＾｛669｝(b+c)＾｛669｝(c+a)＾｛669｝}{(a＾｛669｝+b＾｛669)(b＾｛669｝+c＾｛669｝)(c＾｛669｝+a＾｛669｝)｝￥leq 2＾｛2004｝$[/quote]

Hmm... not so nice. ;)

I think you're:
a) misplacing some dollar signs
b) using strange symbols we can't display
c) not able of placing some english chars: it's  ^  and not  ＾, it's { or }  and not ｛ or ｝, and it's \ instead of ￥
d) you made some parse errors with the parentheses
(yes, you did each of them :) but don't worry, we all had to learn once)

So this would be the corrected wording:
[quote]Let $a,b,c$ be positive reals. Show that we have:

$\frac{(a+b)^{669}(b+c)^{669}(c+a)^{669}}{(a^{669}+b^{669})(b^{669}+c^{669})(c^{669}+a^{669})}\leq 2^{2004}$[/quote]

The solution is easy (rather pre-olympiad level) and goes as follows:
We know that $a^{669}+c^{669} \geq a^nc^{669-n} + a^mc^{669-m}$ with $m,n \in \left[ 0,669 \right]$. Newton's binomium tells us that $(a+c)^{669}$ contains $2^{669}$ such terms. Applying this to each of the 3 separate things, we get that the LHS can be at most $(2^{668})^3$, which is $2^{2004}$.


Esta es una idea que se me ocurrio debido a lo del temario, y me parece que si lo pudiesemos incluir ahi seria excelente, digamos, para cada area o tema que aparezca en el temario colocar algunos libros con los que hayan trabajado donde salga eso y hacer referencia al nivel de complejidad y que tan completo es. Que piensan?

Me parece bien: voy a poner una lista de algunos de mis favoritos por si alguien esta interesado:

[b]Combinatoria:[/b]

Generatingfunctionology. Herbert S. Wilf. [url]http://www.math.upenn.edu/~wilf/gfology.pdf[/url]

Enumerative combinatorics. Richard Stanley

Introductory Combinatorics. No me acuerdo el autor

Graph Theory. Diestel. [url]http://www.math.uni-hamburg.de/home/diestel/books/graph.theory/GraphTheoryII.pdf[/url]

101 combinatorial problems. Titu Andrescu

[b]Geometria:[/b]

Un recorrido por al geometria. German Rincon (colombiano :P )

Geoemtry Revisited. Coxeter

College Geometry. (aunque no es tan recomendable ya que parece un librod e escuela, el mismo ejercicio 10 mil veces!)

Y deberian ver el sitio de Darij Grinberg: [url]http://de.geocities.com/darij_grinberg/[/url]

[b]Teoria de Numeros[/b]

Introduccion a la teoria de numeros. Ivan Niven

A classical Introduction to modern number theory. Ireland y Rosen (libro que empieza desde lo clasico hasta las mas avanzadas herramientas en teoria de numeros, recomiendo para olimpiadas los primeros 5 capitulos)

An introduction to the theory of numbers. Hardy y Wright (buen libro, basico, tal vez demasiado clasico)

[b]Algebra[/b]

Un recorrido por el Algebra. Madroñero (colombiano otra vez :P )

En algebra no tengo buenos libros, depronto en la web se encuentra gran literatura acerca de desigualdades, induccion, generatrices, polinomios, les doy 2 links donde pueden hallarlas:

[url]http://www.people.fas.harvard.edu/~jorza/[/url]

[url]http://www.madras.fife.sch.uk/maths/enrichment/[/url]

[b]Problemas:[/b]

hay buenos libros dedicados a la resolucion de problemas, recomiendo la coleccion:

Problems from arround the world de Titu Andrescu. (problemas y soluciones de las diferentes olimpiadas del mundo)

Mathematical Olympiad challenges. Titu Andrescu (ideas y problemas, muy bueno!)

Problem solving Strategies. Arthur Engel (libro basado en el entrenamiento del equipo aleman y las estrategias entrenadas para resolver problemas, lo malo es que algunos de los problemas son poco significativos, predomina la cantidad sobre la calidad...)

y no se, hay muchos sitios webs con problemas y soluciones, como por ejemplo

[url]http://www.madras.fife.sch.uk/maths/enrichment/[/url]

y el mundialmente conocido, pero ultimamente desactualizado:

[url]http://www.kalva.demon.co.uk/[/url]

Ojala sea de ayuda para los latinos en el site!

Entrenamiento en General (Bastante basico pero bueno)

Solo para Olimpicos, Esteban Meneses, Rodrigo Cruz (Costa Rica) Tengo el libro en formato electronico (bueno, uno de los borradores iniciales)!!

pascual: muchas gracias por tus referencias en libros. ahora de tus dos libros de graph theory y generating functions mil gracias lleva mucho tiempo que busco informacion sobre esta materia(combinatoria) ya que es mi debilidad en los examenes. ahora algunas preguntas: con el fin de obtener la informacion necesaria y suficiente para poder tener las herramientas para poder resolver problemas tipo olimpiada que me recomiendas que lea, es decir todo el libro, o que capitulos no etc???? acerca de tu libro de mathematical olympiad challenges yo lo tengo como es que te gusta??? yo lo encuentro demasiado dificil y mui enfocado en temas de algebra? ya que carece de problemas de combinatoria y los de teoria de numeros son demasiado faciles comparado y pocos comparado con la seccion de algebra y analisis. mi pregunta e: la sección de algebra y análisis es util para mi en la olimpiada (aparte de los problemas de desigualdades) muy pocas veces he visto problemas de ese tipo en las olimpiadas actuales no crees? y en caso de que sea util: donde puedo aprender la teoria necesaria para poder atacar esos problemas ( en el libro claramente no está).
bueno y para finalizar pondre los librosque me han sido util a mi:

bueno varios de los libros que puso pascual yo los tengo asi que no los citaré.

mathematical circles (russian experience)
toda la serie de libros de la olimpiada mexicana de matematicas incluye: (geometria , teoria de numeros, combinatoria, desigualdades)
mathematical problems and proofs ( el autor es polaco solo eso sé jeje)
y bueno ya es todo.

Los de pascual los que conosco son muy buenos.
Particularmente yo uso mucho Engel, el de grafos ese de Diestel es muy bueno. Y pues tambien uso THE USSR OLYMPIAD PROBLEM ese tiene problemas clasicos rusos. Hace poco un pana me regalo uno que se llama An introduction to ineqialities que es bueno para desigualdades. Para geometria hay uno que se llama Problemas de Planimetria de Shariguin, un ruso muy bueno, pero en ese para mi nivel en geometria es bastante avanzado (solo soy capaz de hacer los primeros prblemas). Tambien estudio de los problemas mexicanos, argentinos y los mas bonitos son los problemas rusos.

[quote="manuel"]Marko del libro
 Graph Theory. Diestel. http://www.math.uni-hamburg.de/home/diestel/books/graph.theory/GraphTheoryII.pdf
 yo te diria que para olimpiadas los capitulos mas importantes son del 1-5 y despues que domines bastante esos el 9 y 10 . pero todos son muy interesantes y no pierdes nada en leerlos![/quote]

Manuel... Ese es un libro algo pesado. Yo he leido un par de capitulos (Como el uno y el dos :P ) y esta super bien escrito, pero aqui ese es el libro que Robin Thomas usa para el curso de nivel graduado de Teoria de Grafos, y la gente se queja que despues del capitulo dos, todo es bien complicado!! Honestamente no creo que sea realmente adecuado como para olimpiadas, aunque si alguien de verdad se quiere partir la testa tratando de entender teoria de grafos (que de por si es bastante bonita) pues si, es una obra de arte!

Si es cierto, yo comenze a estudiar teoria de grafos por interes propio y no por olimpiadas y si hay algunas cosas aqui que no son necesarias para olimpiadas y aveces muy pesadas. Pero los capitulos de Ramsey y de ciclos hamiltonianos pienso que no estan demas para las olimpiadas y son leibles. Lo que se ve bn dificil es lo de minors y de las propiedades topologicas de grafos. Pero pienso que si los primeros 5 capitulos y el 9 pueden ser importantes y si se leen con calma se entiende bien.

Alguien sabe donde se puede conseguir libros acerca de la teoría de números irracionales tratado de manera elemental.
Se que hay el libro: Irrational Numbers de Ivan Niven, pero no se de otros libros.
Ayuda por favor.  :(


Let $a,b,c$ are non negative satisfy $a+b+c=3$ Prove that:
$1 \geq \frac{ab}{a+b+1}+\frac{bc}{b+c+1}+\frac{ca}{c+a+1}$

[hide]By AM-HM,
$\frac{a+b+ab}{3}\geq\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}}=\frac{3ab}{a+b+1}$
$\iff \frac{a+b+ab}{9}\geq\frac{ab}{a+b+1}$
Similarly,
$\frac{b+c+bc}{9}\geq\frac{bc}{b+c+1}$
$\frac{a+c+ac}{9}\geq\frac{ac}{a+c+1}$
Adding up the 3 inequalities,
$\frac{2a+2b+2c+ab+bc+ca}{9}\geq\frac{ab}{a+b+1}+\frac{bc}{b+c+1}+\frac{ca}{c+a+1}$
$\iff \frac{6+ab+bc+ca}{9}\geq\frac{ab}{a+b+1}+\frac{bc}{b+c+1}+\frac{ca}{c+a+1}$
But by Cauchy,
$(ab+bc+ca)^{2}\leq(a+b+c)(b+c+a)=9$
$\iff ab+bc+ca\leq3,$
So
$1=\frac{6+3}{9}\geq\frac{6+ab+bc+ca}{9}\geq\frac{ab}{a+b+1}+\frac{bc}{b+c+1}+\frac{ca}{c+a+1}$
$\iff 1 \geq \frac{ab}{a+b+1}+\frac{bc}{b+c+1}+\frac{ca}{c+a+1}$
[/hide]

Your application of Cauchy is incorrect.

Oh right, lol....
But is that part true? Not sure how to prove it....

use AM-GM along with Cauchy for that step.  You'll then get the result.

[hide]
WLOG let $a\geq b\geq c$. Then $a+b+1\geq a+c+1\geq b+c+1$. Also,
\[\frac{ab}{a+b+1}\geq \frac{ac}{a+c+1}\geq \frac{bc}{b+c+1}\]
which is evident from $\frac{ab}{a+b+1}=\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}}$ etc. So we can apply Chebyshev:
\[\sum_\text{cyc}\frac{ab}{a+b+1}=\frac{1}{9}\left( \sum_\text{cyc}a+b+1 \right) \left( \sum_\text{cyc}\frac{ab}{a+b+1}\right) \leq \frac{ab+bc+ca}{3}\]
But from $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$ (this is just $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\geq 0$ ), we get $(a+b+c)^{2}\geq 3(ab+bc+ca)$, so $ab+bc+ca\leq 3$ showing that the right hand side of the above is less than or equal to 1.
[/hide]

The last part could be $ab+bc+ac\leq \sum_{\text{cyc}}\left(\frac{a+b}{2}\right)^{2}\leq\sqrt{\left(\sum_{\text{cyc}}\frac{a+b}{2}\right)^{2}}=3$,
which follows by Cauchy-Schwarz and AM-GM.

We are to prove that $\frac{1}{4-a}+\frac{1}{4-b}+\frac{1}{4-c}\leq 1.$

How do you get that? I get that it is equivalent to:
$1\geq \sum \frac{bc}{4-a}$

and besides it has already been proven in the standard way that was expected :D

He got it using Chebyshev, and $(a+b+c)^{2}\geq 3(ab+bc+ac)$, that's clear.

But there is a little problem with it: $\sum_{cyc}{\frac{1}{4-a}}\geq 1$ by Jensen's inequality  :wink:

I really like your solution, Alexander Remorov!

(You have a small typo in the denominator of the first and second line)


Prove by induction that
sin(x) + sin(2x) + sin(3x) + ... + sin(nx) = sin(nx/2)sin((n+1)x/2) / sin(x/2)

Convert everything to complex numbers, factor out $\text{Im} \ (e^{i(n+1)x/2})$, and collect terms?

If you use \sin for sine and \cos for cosine you get $\sin$ and $\cos$ which look much nicer. Just a friendly $\LaTeX$ tip ;) .

aah, thanx!
Where did you guys learn all this stuff? Or did you do what I did, (Click on interesting formats in posts by other authors)?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5261]Here[/url] or [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]here[/url] are good places to learn.


Let $ A=\begin{pmatrix}\n p & q & 0\\ \n r & s & 0\\ \n 0 & 0 & 1\n\end{pmatrix}$($ p,q,r,s$ are real numbers). Find all $ 3\times 3$ matrices $ B$ that satisfies $ AB=BA$ regardless of $ A$.

I had an ugly solution that setting $ B=\begin{pmatrix}\n b_{11} & b_{12} & b_{13}\\ \n b_{21} & b_{22} & b_{23}\\ \n b_{31} & b_{32} & b_{33}\n\end{pmatrix}$, and solve the simultaneous equations of $ b_{ij}$. I got the answer $ B=\begin{pmatrix}\n a & 0 & 0\\ \n 0 & a & 0\\ \n 0 & 0 & i\n\end{pmatrix}$($ a$ is a real number, and $ i=\sqrt{-1}$) and the trivial solution $ B=\begin{pmatrix}\n a & 0 & 0\\ \n 0 & a & 0\\ \n 0 & 0 & a\n\end{pmatrix}$(also, $ a$ is a real number). But I want to know about pretty approach that gives a new insight.  Thanks in advance for your nice solution.

Your ugly solution seems to be wrong as well. The correct answer is: all block matrices of the form $ \left(\begin{array}{cc}\lambda I_2 & 0 \\
0 & \mu I_1\end{array}\right)$ for $ \lambda\in\mathbb{R}$ and $ \mu\in\mathbb{R}$.

"Block matrices" also is a good hint on how to prove this...

  darij

The answer is $ B=\begin{pmatrix}\n a & 0 & 0\\ \n 0 & a & 0\\ \n 0 & 0 & i\n\end{pmatrix}$ where $ i$ is not $ \sqrt{-1}$, but just a real number,and this contains the trivial solution.


Consider a quarter of a circle of radius 18. Let r be the radius inscribed in this quarter of a circle. Find r. Round your answer to the nearest integer.

[quote="ritchjp"][hide]

$r = \frac{18}{1 + \sqrt{2}} = 7.456$

nearest integer is $7$

 [/hide][/quote]

umm
can you explain
i'm not sure that's the same way I did it, but you got the answer correct  :)

[quote="JesusFreak197"]The small circle is internally tangent to the large circle, and also tangent to two radii of the large circle which form a $90^\circ$ angle. This means that a square is formed with $s=r$, so the distance from the center of the smaller circle to the center of the larger is $r\sqrt{2}$, and the length of the radius of the large circle, in terms of that of the smaller circle, is $r+r\sqrt{2}$. Therefore, $R=18=r(1+\sqrt{2})\Rightarrow r=\frac{18}{1+\sqrt{2}}$. ;)[/quote]

oh.  :P sorry

well i had the radius of the large quarter circle go through the diameter of the smaller circle. This radius was separated into two parts, r, and 18-r. The center of the smaller circle with its perpendiculars form a right triangle. $(18-r)^2=2r^2$ and solving we get $r\approx7$...which is the same thing.  :D


1. In how many ways can 12 pennies be put in 5 purses?

2. A bookbinder must bind 12 identical books using red, green, or blue covers. In how many ways can this be done?

3. How many ways are there to arrange 5 red, 5 green, and 5 blue balls in a row so that no two blue balls lie next to each other?

Balls and urns!

1. 16c12=16c4=1820. More when im not lazy.

can someone explain to me how balls and urns work especially having urns divided and adding those to balls- thats the part i dont get.

Also, can someone do 3?

[hide="Hint for 3"]
Let X represent the red and green books. Lay them out with space in between:

_ X _ X _ X _ X _ X _ X _ X _ X _ X _ X _

Ignoring the blue balls for now, how many ways are there to arrange the red and green ones?

Now you add in the blue ones. There are 11 spaces, and at most one blue ball can go into each of these spaces. How many ways are there to place 5 blue balls into 11 spaces?
[/hide]

what is balls and urns??????

you dont need to tell me the formula, i know the formula.

I just don't understand why you have to place dividers between urns, then add their number to the number of balls and then take (that choose number of balls).

[quote=First]#2 and #1 are the same concept you have 12 objects and you need 2 dividers so it's just $\binom{14}{2}$[/quote]

Thanks, that was what I thought, needed a little verification.

[quote=sophie8]Could someone explain #2 (sorry for the 8-year revive!)?[/quote]

Just a curious question, how do you all find threads this old when they are not linked to a contest, do you scroll through forums that long

(just wondering so I can find these old threads)

Thanks


Prove: $\tan(\alpha+\beta)=[\sin(2\alpha)+sin(2\beta)]/[\cos(2\alpha)+cos(2\beta)]$

I hope I put it in the right forum as this problem is quite easy with the sum/product identity, but is there another way?

:( no solutions?
I got one anyways.
Thanks for trying :rotfl:


calculus
integration
algebra unsolved
algebra

find all postive integral solution to
$x^{2}y-y^{3}=105$


Euler
function
algebra
functional equation

Thanks for coming out everyone. I hope that you guys had as much fun attending as we did putting it on. I just wish I could have heard more of the talks. What I saw seemed really good and I'm particularly sorry that I sorry that I missed Chuck Garner's talk on Euler!

The results of the tournaments were:

JV:
1st - Deep Datta Roy - Parkview - 162
2nd - Nathan Li - Northview - 156
3rd - Pranav Ramesh - Parkview - 146
(Arvind was 5th! Pretty good for a 7th grader! :-))

Varsity:
1st - Vibin Kundukulam - Northview - 108
2nd - Heng Tong - Northview - 106
3rd - Eric Morphis (Twiz!) - Woodward Academy - 100

Super Varsity:
1st - Eddy Ferreira - Roswell - 118
2nd - Harrison Brown - Centennial - 116
3rd - Miles Edwards (Boy Soprano) - 114

The Super Varsity scores are slightly different since I threw out #25. The correct wording of the problem was:
The function $f$ satisfies the functional equation $f\left(x\right)+f\left(y\right)=f\left(x+y\right)-xy-1$ for every pair $x$,$y$ of real numbers. If $f\left(1\right)=1$, then the number of integers $n\neq 1$ for which $f(n)=n$ is
a) 0 b) 1 c) 2 d) 3 e) infinite

So is this something that you would come to again? What should we change. My list of changes starts with:
1) Order more food!
2) Make the varsity test a little easier.
3) Loosen up the schedule some. Don't pack so many things so close! (That's probably #1 on my list.)

Take care,
Fulton

Yeah, I was one of the people who missed out on food at first. I didn't think that the varsity test was too hard; I think that the difficulty will help me learn in reviewing the test (but it did kind of scare me at first that I could only solve about half of the problems; I thought that I had gotten stupid or something). Anyways, the only two things that I didn't particularly like about the test were the kind of advanced terminology (I, as well as others that I heard talking, don't know what a locus is), and the lack of numbers (many problems involved variables instead of numbers, and I prefer numbers), but in general, it was a fun tournament, so thanks!

I lost to Miles by 2 points  :( There goes another chance to beat him... I misunderstood the logic problem involving freshies and people and thinking, so that would've probably gotten me a tie with him. Ack.

But congratulations to the winners!  :)

[quote="Karth"]I lost to Miles by 2 points  :( There goes another chance to beat him... [/quote]

But you DID beat Phillip. And you didn't have to pick up a trophy for him!  :D

It was pretty fun...too bad i missed the talks, other than euler. i did scavenger hunt...was really cool and fun. the fake problems threw off our team like anything, and we still won. did anyone get the spy problem? thats what messed us up. it was pretty easy, but the wording was strange. when the spy ran out of the room, did that count as exiting the room, or as running out of the room? also it said "this happens again". what happens again? haha, ben hu and i had a laugh when both our two answers for that problem led to doors with problems. they should do scavenger hunt for academic bowl, too...

Wait a second here...Deep is a 10th grader! Why'd he take JV??? :huh:

I, like Karthik, missed getting a place by two points (on Varsity). Indeed, the scavenger hunt was crazy. We did end up getting first place, however (even though no one got all the real answers...hahaha).

I liked the Scavenger Hunt too, it was pretty well thought out to post [i]that[/i] many misleading problems. We were running everywhere.. good exercise.


number theory
prime numbers
number theory proposed

Let $n+1 (n \geq 1)$ positive integers be given such that for each integer,the set of all prime numbers dividing this integer is a subset of the set of $n$ given prime numbers.Prove that among these $n+1$ integers one can find numbers (possibly one number) whose product is a perfect square

Posted before (easy combinatorical problem, no number theory).
But I would appreciate it very much if you name all those topics using the content, what you used as title is supposed to get into the source field.


Consider all line segments of length $4$ with one endpoint on the line $y=x$ and the other endpoint on the line $y=2x$. Find the equation of the locus of the midpoints of these line segments.

[hide]$y=\frac{4}{3}x$

Just a guess...[/hide]

[hide="Brute forcins"] Let's explicitly calculate such a locus.  Now, we want the set of all $a, b$ such that the distance between $(a, a)$ and $(b, 2b)$ is $4$.  This gives

$\sqrt{ (2b-a)^{2}+(b-a)^{2}}= 4$

Our midpoint has coordinates $x = \frac{a+b}{2}, y = \frac{a+2b}{2}$.  Then

$b = 2y-2x$
$a = 4x-2y$
$b-a = 4y-6x$
$2b-a = 6y-8x$

Then our locus has the equation

$(6y-8x)^{2}+(4y-6x)^{2}= 16$

Which is an arc of some ellipse. [/hide]

The locus is indeed an ellipse.

Can you explain why the axes of the ellipse are the bisectors of the angles formed by the two given lines?


a^2 + b^2 +c^2 +(abc)^1/2 *(a^1/2  +b^1/2 +c^1/2 ) >=2( ab+ bc +ca )
a,b ,c >0  and be the lenght sides of a triangle[/list][/code]

It's not hard at all. Actually, much more is true, from Schur's inequality $ 2(ab+bc+ca)-(a^2+b^2+c^2)\leq\frac{9abc}{a+b+c}$. I think the one you wrote was in romanian TST, but I'm not sure.


The sequence of digits \[ 123456789101112131415161718192021 \, \ldots \]is obtained by writing the positive integers in order. If the $10^n$th digit in this sequence occurs in the part of the sequence in which the $m$-digit numbers are placed, define $f(n)$ to be $m$. For example, $f(2)=2$ because the $100$th digit enters the sequence in the placement of the two-digit integer $55$. Find, with proof, $f(1987)$.

For all $n \geq 1$, there are $9 * 10^{n-1}$ numbers with $n$ digits. The last digit of the last $n$-digit number is therefore at position $9 \sum_{k=0}^{n}10^{k-1}k$.

We can write this sum as \[ 9 [(10^{n-1} + 10^{n-2} + ... + 10^{0} ) + (10^{n-1} + 10^{n-2} + ... + 10^1) + ... + (10^{n-1} + 10^{n-2}) + (10^{n-1})] \]

or more formally $ 9 \sum_{j=0}^{n-1} \sum_{k=j} ^{n-1} 10^k $. We find that this number is lower-bounded by $9n * 10^{n-1}$ and upper-bounded by $9n(10^{n-1} + 10^{n-2} + ... + 10^0) = n(10^n - 1) $. For convenience we'll say the upper bound is $n 10^n$.

We notice that $1987$ is on the order of $10^3$, so we inspect the last digit of the last $1984$-digit number. We find that it is in the range of $[1984 * 9 * 10^{1983}, 1984 * 10^{1984}]$. Now $1984 * 9 > 10^4$, so the lower bound is higher than $10^{1987}$. Now we check the last digit of the last $1983$-digit number. We find that the upper bound is $1983 * 10^{1983} < 10^4 * 10^{1983} = 10^{1987}$.

Thus, the $10^{1987}$'th digit in the sequence is for a $1984$-digit number. $f(1987) = 1984$.


Suppose  $ \lfloor x \rfloor$ is floor function of $ x$. There is  a sequence  that $ \lfloor \frac{2070}1\rfloor ,\lfloor \frac{2070}2\rfloor, \lfloor \frac {2070}3\rfloor,\cdots ,\lfloor \frac {2070}{2069}\rfloor,\lfloor \frac {2070}{2070}\rfloor$
How many different integers represent? ?

Every number from $ \displaystyle \lfloor \frac{2070}{1} \rfloor$ to $ \lfloor \frac{2070}{45} \rfloor$ will be different, which is 45 values. The numbers from $ \displaystyle \lfloor \frac{2070}{46} \rfloor$ to $ \lfloor \frac{2070}{2070} \rfloor$ will contain every number from 45 to 1. Thus, there are 90 possibilities. Is this right?


Determine all monic polynomials $ p(x)$ of fifth degree having real coefficients and the following property: Whenever $ a$ is a (real or complex) root of $ p(x)$, then so are $ \frac{1}{a}$ and $ 1\minus{}a$.

[quote="moldovan"]Determine all monic polynomials $ p(x)$ of fifth degree having real coefficients and the following property: Whenever $ a$ is a (real or complex) root of $ p(x)$, then so are $ \frac {1}{a}$ and $ 1 \minus{} a$.[/quote]

If $ a$ is a root, then so are $ \{a, \frac 1a,$ $ 1\minus{}a, \frac{a\minus{}1}{a},$ $ \frac{1}{1\minus{}a}, \frac{a}{a\minus{}1}\}$ and so $ a\notin\{0,1\}$ and at least two of these six numbers must be identical. So :

1) $ a\equal{}\frac 1a$. Then $ a\equal{}\minus{}1$ and the set is $ \{\minus{}1, \minus{}1, 2, 2, \frac 12, \frac 12\}$

2) $ a\equal{}1\minus{}a$. Then $ a\equal{}\frac 12$ and we have the same set as 1) above

3) $ a\equal{}\frac{a\minus{}1}{a}$ then one set : $ \{\frac{1\minus{}i\sqrt 3}{2}$, $ \frac{1\plus{}i\sqrt 3}{2}$, $ \frac{1\plus{}i\sqrt 3}{2}$, $ \frac{1\minus{}i\sqrt 3}{2}$, $ \frac{1\minus{}i\sqrt 3}{2}$, $ \frac{1\plus{}i\sqrt 3}{2}\}$

4) $ a\equal{}\frac{1}{1\minus{}a}$ and we have the same set as 3) above.

5) $ a\equal{}\frac{a}{a\minus{}1}$ and we have the same set as 1) above.

Result :

We cant have a polynomial with roots only in the second set since we have real coefficients. So the solutions are :

Roots only in first set : $ P(x)\equal{}(x\plus{}1)^p(x\minus{}2)^q(x\minus{}\frac 12)^r$ with $ p,q,r>0$ and $ p\plus{}q\plus{}r\equal{}5$

Roots in both sets : $ P(x)\equal{}(x\plus{}1)(x\minus{}2)(x\minus{}\frac 12)(x^2\minus{}x\plus{}1)$


[color=purple] Find the best constant $ K$ such that the following inequality is true for arbitrary triangle.
 $ \frac {1}{l_a} \plus{} \frac {1}{l_b} \plus{} \frac {1}{l_c}\ge\frac {1}{R} \plus{} \frac {1}{2r} \plus{} K(\frac {1}{R} \minus{} \frac {1}{2r})$
[/color]

It's quite a descent type of problem. Just one thing,
$ R\geq{2r}\Leftrightarrow{\frac{1}{R}\leq\frac{1}{2r}}$

which shows that $ (\frac {1}{R} - \frac {1}{2r})\leq{0}$. Is $ K\geq{0}$ or $ \leq{0}$ ??


summer program
MathPath
AMC
AIME
USA(J)MO
USAMO
geometry

Hopefully somebody actually posts here

so who is coming?

also post(totaly copied from AMP forum):
Name
grade
state
what you look like so i can identify you

MC results
AMC 8 score
AMC 10/12 score
AIME score
USAMO(if you qualified)

What you are bringing

so i will start

Ray Li
grade 7
Pennsylvannia(baller)
Azn w/ glasses

5th chapter 2nd team
8th state 1st team
AMC 8:21
AMC 10B:123
AIME:5
USAMO: did not qualify

Bringing:
ipod
chess
cards
other board games
a math book
whatever else i can think of(no this is not my entire list)

I'm trying to cram the Qualifying test.
:o
Eh, I'm probably gonna bring some books, and like, some AoPS volumes if I get bored.

I hope I pass the test... waht do you think the qualifying score

will be?

I'm sure that I got *EDITED OUT* right, *EDITED OUT* partially, and working on *EDITED OUT*

If I did *EDITED OUT* perfectly, I get *EDITED OUT*  :lol:

Name: *Will not publically announce...*

I'll look for you there...

Grade: 5

Score: N/A so far, hoping to make it...

State: NJ

Short, Brown hair/eyes, most likely carrying something to do wtih

math

Contest stuff:

MC: too young

AMC 8: 22 (...)

AMC 10: 108 A/ 114 B

AIME: (taken unofficially) 5

USAMO: N/A

IMO: lol...

Bringing:

Math stuff,

Chess set (?)

cards

some other games

...

Did you qualify some other year? If so what was your score?

EDITS: we weren't supposed to discuss the test... sry

Well this year's anyway

What do you THINK you got? About

i qualified last year, they didnt send me the results(or maybe my parents just havent told me), but i didnt go because i had some other conflict. i am going this year for sure. i dont need to qualify again, but they still want me to take the test.

Carl~No worries. Vermont isnt on the east coast. :)

[quote="CatalystOfNostalgia"]hmmm guys if you choose to bring glasses don't go in the ocean wearing them[/quote]

Or else you will have to sit very close to the chalkboard in Dr. T's geometry class.

um

Anirudh Prabhu
Grade 8
Indiana

MC: 29/16(3rd yeah, justin ahmann is in my chapter) 2nd place team
MC state: rather not talk about it.
AMC 8: 23
AMC 10A/B = 121.5/126
AIME: 5 could have gotten 7 though. argh, i missed number 3.  :rotfl:
USAMO(SHOULD HAVE QUALIFIED)

i missed 2, which is even more pathetic. i put 625

I am going, but I don't think you guys are going to care since I'm a counselor. ^_^
Oh and shameless plug: please also visit the new mathpath forums at http://www.mathpath.org/forum20/index.php
It gets kind of boring when all I can do is moderate my own posts

-IntoTheCloudS-

Hmm well I guess might as well post my profile too
Name: Charles Song
Grade: College sophomore
MathPath '02 Alum and '08 Counselor

I shall not complain on how you stole my topic but still.

Name: Ang Yan Sheng (my surname is Ang, so you might not get a response when you call me Yan, Sheng, Ang Yan, etc.)
Grade: Secondary 2 (which would be grade 8 in the US system I think)
State: Singapore (I am so tempted to write "solid" here)
What I look like: Well, black hair, black eyes, the asian type. My glasses are purple and green, so they look dark blue-green from afar.

MC results: none, I'm not from the US
AMC 8 score: 25
AMC 10A score: 126
AIME score: 9 (ha!)
USAMO score: haven't taken it yet

Bringing: Erm... still thinking.

I remember last year when I got in only answering 2 and a third questions...

I'm going again assuming I don't die or anything that would prevent me from going.

If you want to know what I look like, figure that out yourself, I don't feel like putting in the description

Don't feel like doing name.... But I'm pretty recognizable.
7th
Georgia
Asian, glassess, with a penchant for Pokemon T-Shirts

MC- 2nd at chapter, 20th at state
AMC 8- 21
AMC 10/12- N/A
AIME- N/A
USAMO- N/A

What you are bringing: Assorted Books, Rubix Cube. still thinking....

I don't know if I can go, but I was invited so....

dynamo729 (nice try)
7th
Alabama (hey y'all!)
Probably the only non-Asian going. 6 foot 3 (probably 6 foot 4 by then...) Hard to miss.

MC: 4th school, 5th chapter, (thought I saw a pattern...) 3rd state.
Ummmm we dont do AMC anymore cuz its too expensive. Or AIME or USAMO...
This is what qualified me:  [url]http://www.mathleague.com/reports/2007_08/grade678/AL_7.HTM[/url] Had to get above 38 (YES! 39!)

[quote="alanchou"]did you do the qualifying test and send it in with 2 recommendation letters and that other stuff?[/quote]

no. Ok, first: all advanced 7th grade math students got tooken into a big room (I got demoted ARGH!). They make us take a test, I've prepared and gotten perfects on two past ones (though last year I failed with a 37/40). I miss STUPID NUMBER 39 I THOUGHT IT SAID COUNTERCLOCKWISE!!!!!! but I get a 39/40. (The test are easy, see  [url]http://www.mathleague.com/contests.htm[/url] for sample tests. 40 mins.) Like a month later, I get a packet from my Math Seminar teacher and it talks about MathPath blahblahblah. You've been selected because of your <38 score. Whatever. I didnt know anything about this before I got the packet, but I noticed there was a forum section so......

I got the same packet for getting 39 on Georgia Math League and for getting high score on AMC8. You actually have to send a Qualifying Test and recommendation letters. What you got was only an encouragement to come.

Yeah, I had that thing last year. You [i]do[/i] need to submit everything.
Hmm, my parents said I couldn't go.

I might not be able to go too...  :mad:

Is ANYONE at all from Singapore too? Because my parents say that I can NEVER go without someone from Singapore to accompany me. Argh.

Anyone have any recommendations for the classes I should take?
Like who's a good teacher.

Zeitz isn't going back I hear. That's too bad. But here's a suggestion for you personally, not necessarily like, everyone. You, Peijin, have worked very hard on MathCounts this year, and will do the same next year. With how hard you guys work at school, I don't see that much benefit in you taking the MathCounts classes at MathPath. Again, I want to emphasize, this probably isn't the case for everyone going to MathPath. But pick the classes that you'll find interesting and the classes that you think you will learn a lot. Don't worry too much about picking the hardest class (if that's what you want to do). If you think you can handle the difficulty of every class, then pick the one you will be most interested in. Even if you don't think you can handle the difficulty of a class, you may want to try it anyway.

name: My initials are TT, that should be plenty
grade: 6
state: WA
what you look like so i can identify you
Blackish brown hair(always messy), glasses, 5'1 (ish, im growing)

MC results: got to state, lost because of the flu. wouldave made it too. (yes i know darryl wu, i go to his school, we are friends
AMC 8 score 25 (YES!!!!)
AMC 10/12 score 117 (one stupid eraser mark)
AIME score ____________-
USAMO(if you qualified) __________


$ \phi : V \rightarrow V$ is a linear transformation and $ dimV\equal{}n$. Prove that :  $ im(\phi^{m\plus{}1})\subset im(\phi^{m})$ for any $ m$
Second point:

$ n\equal{}11$ and for $ 1\leq k\leq 4$ holds $ im(\phi^{k\plus{}1})\not\equal{} im(\phi^{k})$ then $ ker(\phi)\leq 7$.

Surely you mean $ \dim \ker \phi \leq 7$, since $ \ker \phi$ is some subspace (and so not naturally a part of any ordered set that includes $ 7$ :wink: ).  Also, \dim and \ker are built-in commands in LaTeX (but \im is not :( ) and your text will look much nicer if you use them.

For the first question, a little extra generality actually makes the question easier: let $ \phi, \varphi$ be any functions $ V \to V$.  Then $ \textrm{im}(\varphi \circ \phi) \subseteq \textrm{im}(\varphi)$.  Your question is the special case when $ \varphi \equal{} \phi^m$.  (Note that we don't need any properties of linearity at all for this result to hold.)

The second question actually requires that you know something about linear algebra, not just about sets :)


Mccain is getting a lead in MO.... looks like our legacy is over.......

......Unless Mccain wins OH, NC, PA, FL, and some other random state that has more than 10 EVs....

McCain's going to win some random states like Nevada that are projected Obama, so he still has a chance...

In other news, Missouri is one state in which everybody's vote REALLY counts.

It'd be funny if everyone in the country thought, "BLAH! One vote won't make a difference. I not going to waste my time waiting in line to vote this year." :rotfl:  :rotfl:  :rotfl:

LOL every time i see EV i think of Pokemon...

How do you not get why he thinks Pokemon? Gosh I thought you were a Mathcounter! EV=Eevee

No, I believe IV and EV are some statistical measurement of Pokemon.

Or else...I am right.

http://answers.yahoo.com/question/index?qid=20070601162028AADYdH0
http://answers.yahoo.com/question/index?qid=20070605223819AAEETEy

I guess your right BUT

my name is AYLsero not ALYsero.

Yeah seriously. Earned values. Get it right. Eevee does [b]not[/b] come to mind.

MO's legacy is officially over.... even though, by law, they must recount all the votes because of a <0.5% win, but since it doesn't matter which way MO goes, no one cares...

MO is such a neglected state.

[quote="tinytim"]Wow, Alysero got pwned..... :wink:  :rotfl:[/quote]
WTF? That's mean...that's not a nice thing to say. :o

MO must be such a loserish state this year...Even worse, MO lost it's legacy for voting for the winning presidential candidate.
THBAPSO Missouri. TY.

Wrong forum. I'm sure your state has a worse record.

Illinois sucks. They vote democrat everytime... LOL

Were you at that loser Obama Rally in Grant Park?

What Socialists....

It'd be funny if Obama had an AoPS account and was reading this right now... :rotfl:

He's probably got someone googling his name repeatedly and seeing what comes up. Anyway, MO failed. Still I think weve got the best record.

Shift bellwether to Ohio.

Edit: This is all a result of my moving here. :maybe: You guys had snow the year I came, etc.

MO failed because we jinxed it. If we never mentioned the "legacy", MO would have gone DEM. When we first mentioned it, the polls showed MO was leaning DEM but them it started shifting to GOP... :rotfl:

Have all of us simultaneously been on for 1+ hour? That's new.

I've been muted. Bonus points for seeing where the flag on my profile is from.

I've been muted for testing out "stfu" and "fu", as well.


Given a sphere of radius 1 inscribed in a cube (tangent to all 6 faces). A smaller sphere is nestled in the corner such that it is tangent to the given sphere and also tangent to three intersecting faces of the cube. The radius of the smaller sphere, to two decimals, is?

I'm not great with  geometry, so I hope this is correct.
[hide]Let the length of the radius of the small sphere be $ s$.
Since the side length of the cube is 2, the distance from the center to the corner is $ \sqrt3$. This is also the sum of the radii of the two spheres and the diagonal of a cube with the side length $ s$. Set up an equation:\[ 1\plus{}s\plus{}s\sqrt3\equal{}\sqrt3\]\[ s\equal{}\frac{\sqrt3\minus{}1}{\sqrt3\plus{}1}\]\[ \equal{}2\minus{}\sqrt3\]\[ \approx\boxed{.27}\][/hide]
I don't know how to make 3-D drawings. Sorry.

[quote="adbh94"]I think you got the correct answer, since .27 was one of the choices :). However, I am confused on how you computed the distance from the center to the corner to be sqrt(3). Wouldn't the distance from the center to the corner be the hypotenuse of the 45-45-90 triangle created by drawing a horizontal line from the center to the side of the cube, and a line from the center to the corner of the cube? The lengths of this triangle would 1, 1, and sqrt (2) right?[/quote]
Since the cube's side length is 2, the space diagonal which runs through the center is length $ 2\sqrt{3}$. Thus the distance from the center to the corner is half of this, or $ \sqrt{3}$.

If we were to use your method, drawing a horizontal line and drawing a line to the corner, it would not be a 45-45-90 triangle because when we draw the horizontal line, its half the diagonal of the face, thus the length would be $ \sqrt{2}$. Then the right triangle would be legs $ 1$ and $ \sqrt{2}$ and hypotenuse $ \sqrt{3}$


abstract algebra
superior algebra
superior algebra unsolved

Let $ I$ be a nilpotent ideal in a commutative ring $ R$, let $ M$ and $ N$ be $ R\minus{}$modules and let $ \phi$ : $ M \rightarrow N$ be an $ R\minus{}$module homomorphism. Show that if the induced map $ \bar{\phi} : \frac{M}{IM} \rightarrow \frac{N}{IN}$ is surjective, then $ \phi$ is surjective.

Did you try to show the claim if $ I^2\equal{}0$?


invariant
number theory solved
number theory

At a round table there n forum members. The number of those members who have the same sex as his/her respective right member at the table equals the number of members for which that is not true.
Prove that n is divisible by 4 !

mr: -
hr: expressing properties as equations, invariants, analysis of the situation

We consider the following problem: we have n numbers x1,..,xn from the set {1, -1}. If x1*x2+x2*x3+..+xn*x1=0 prove that 4|n. We can see that this problem is equivalent to what you posted: We denote sexes by 1 or -1. If 2 neighbours have the same sex then their product is 1, if they have opposite sex then their product is -1. The problem states the equality of the number of 1's and -1's, so the sum x1*x2+..+xn*x1=0, where xi is 1 or -1, depending on whether the i'th person is male or female, counting counter-clockwise on the circle.

Anyway, here's how we prove that 4|n: because the number of terms in the sum S=x1*x2+..+xn*x1 is n and xi*x(i+1) is either 1 or -1 and we know S=0, it means that we have an equal number of 1's and -1's, so 2|n. Then we kave n/2 1's and n/2 -1's.  We calculate the product (x1*x2)*..*(xn*x1)=(-1)^n/2. At the same time (x1*x2)*..*(xn*x1)=x1^2*..*xn^2=1, so n/2 must be even, so 4|n. The problem you posted, being equivalent to this one, is proved.


I wonder if its possicle to inscribe a circle in a circle...
 :|

And you can inscribe a little circle in a fraction of a whole circle.

I thought the definition of inscribed (mathematically) is that all vertices must touch the sides and all points must be within the interior? I might be mistaken.

I thought so too; in that case, basically every shape has been inscribed within itself an indefinite amount of times. That really beats the point of the whole "inscribed" concept. So I would have to go against isabella2296 (which scares me; she's way smarter than me and I may be humiliated by this post) and say no, it's not possible. :maybe:

An "Inscribed Circle" is defined as the largest possible circle that can be drawn interior to a plane figure. You can't inscribe a circle in a circle because the largest possible circle is the original circle, which would make the circle coincide with the original circle. Therefore, you can't inscribe a circle in a circle.

[quote="isabella2296"]Sure. Inscribe the circle inside itself.[/quote]

That is a circle which is inscribed inside the original, but it can't be called inscribed circle, as this stands for something that math154 and later explained.

wow, a really random questioon has turned into A mathmatic discussion. Good job peeps!!!


Let $ A$ is $ n\times n$ matrix. Prove that there is $ B$ matrix such that $ ABA \equal{}A$.

if A is invertible, we have ABA=A is equivalent to BA =I, hence, B=A^{-1}.
if A is not invertible, i guess we'll use continuity. :D

Without the continuity and over any field:

Do row elimination on $ A$ to obtain a matrix in row echelon form, then permute the columns so that the pivot columns are all stacked together first. That is, we find invertible $ P$ and $ Q$ such that
\[ PAQ\equal{}\begin{pmatrix}T&W\\0&0\end{pmatrix}\quad\text{or}\quad A\equal{}P^{\minus{}1}\begin{pmatrix}T&W\\0&0\end{pmatrix}Q^{\minus{}1}\]
where $ T$ is upper triangular and invertible and $ W$ is arbitrary.

Now let $ B\equal{}Q\begin{pmatrix}T^{\minus{}1}&0\\0&0\end{pmatrix}P.$

It follows that

$ ABA\equal{}P^{\minus{}1}\begin{pmatrix}T&W\\0&0\end{pmatrix}\begin{pmatrix}T^{\minus{}1}&0\\0&0\end{pmatrix}\begin{pmatrix}T&W\\0&0\end{pmatrix}Q^{\minus{}1}$

$ \equal{}P^{\minus{}1}\begin{pmatrix}T&W\\0&0\end{pmatrix}Q^{\minus{}1}\equal{}A.$

[b]A solution very nice ! [/b] :read: :read:
We have  :
  $ BAB\equal{}Q\begin{pmatrix}T^{\minus{}1}&0\\0&0\end{pmatrix}\begin{pmatrix}T&W\\0&0\end{pmatrix}\begin{pmatrix}T^{\minus{}1}&0\\0&0\end{pmatrix}P\equal{}Q\begin{pmatrix}T^{\minus{}1}&0\\0&0\end{pmatrix}P\equal{}B$
General , Let $ A \in M_{nxm}(\mathbb{C})$ , allway the exist matrix $ B \in M_{mxn}{(\mathbb{C})}$
Such that : $ ABA\equal{}A$ and $ BAB\equal{}B$

[quote="Kent Merryfield"]Without the continuity and over any field:

Do row elimination on $ A$ to obtain a matrix in row echelon form, then permute the columns so that the pivot columns are all stacked together first. That is, we find invertible $ P$ and $ Q$ such that
\[ PAQ \equal{} \begin{pmatrix}T & W \\
0 & 0\end{pmatrix}\quad\text{or}\quad A \equal{} P^{ \minus{} 1}\begin{pmatrix}T & W \\
0 & 0\end{pmatrix}Q^{ \minus{} 1}
\]
where $ T$ is upper triangular and invertible and $ W$ is arbitrary.

Now let $ B \equal{} Q\begin{pmatrix}T^{ \minus{} 1} & 0 \\
0 & 0\end{pmatrix}P.$

It follows that

$ ABA \equal{} P^{ \minus{} 1}\begin{pmatrix}T & W \\
0 & 0\end{pmatrix}\begin{pmatrix}T^{ \minus{} 1} & 0 \\
0 & 0\end{pmatrix}\begin{pmatrix}T & W \\
0 & 0\end{pmatrix}Q^{ \minus{} 1}$

$ \equal{} P^{ \minus{} 1}\begin{pmatrix}T & W \\
0 & 0\end{pmatrix}Q^{ \minus{} 1} \equal{} A.$[/quote]

How we can find invertible P,Q and T?  :maybe:

[quote="phhj"]How we can find invertible P,Q and T?  :maybe:[/quote]
I said that in my original post - by routine Gaussian elimination. Every time you perform an elementary step of row elimination, you are left-multiplying by an invertible (elementary) matrix. Hence, $ P$ can be obtained from a record of your steps. I then said to permute the columns to get all the pivot elements in the left block. That's right-multiplying by a permutation matrix.

In fact, I could have said more. If you do full Gaussian row elimination and then follow it by full Gaussian column elimination, you can obtain
\[ PAQ\equal{}\begin{pmatrix}D&0\\0&0\end{pmatrix}\]
where $ P$ and $ Q$ are invertible and $ D$ is diagonal and invertible. The size of $ D$ (as was the size of $ T$ in the previous version) is the rank of $ A.$

In linear algebra, there's no substitute for knowing the Gaussian elimination algorithm and understanding what it does and what it can and cannot accomplish.


vector
combinatorics proposed
combinatorics

Given a connected graph $G=(X,U)$ with $n$ vertices and $m$ edges, how many partial graphs $G'=(X,V),\ V\subseteq U$ are there in which every vertex has even degree?

is $X$ the set of vertices and $U$ the set of edges of G?

Would you mind completing it though? :? I'm sure nobody else will post here, since you gave the answer, thus spoiling the fun :).

well, one way of doing it(obtaining the answer) would be to start with a fixed spanning tree $T$ of $G$(which accounts for $n-1$ edges) and for each of the remaining edges $e$, consider $T\cup e$.this being non-cyclic contains a cycle $C_e$(necessarily the cycle contains $e$).Now consider the graphs with edge sets being $\sum_{e \in \mathcal{C}} C_e$ where $\mathcal{C}\subseteq \{C_e| e\notin T\}$.(here by $\sum$  i mean a binary sum,i.e.if a certain edge is added an even number of times then it is not to be included). clearly this accounts for that number and all such graphs of course have all vertex degrees even.all that suffices now is to say that these are all of them. but since each vertex has even degree,every connected component is Eulerian and so the edge set of each connected component of $G'$ is a union of cycles.finally one can get one tree from another by exchanging some edge of the tree with any given edge off the tree(this is in some sense like the property of exchanging a vector in a basis in a vector space).
of course i have not written as detailed as one would have liked but i think this should do the job(or atleast one should be able to take over from here).
hopefully this is not too vague....


function
calculus
integration
real analysis
real analysis solved

Solve the following differentilal equation:
$ f\"(x) = f'(x) f(x) $

I reduced it to a first degree equation, just note that
\[\frac{f''}{(f')}=\frac{f}{f'}\]
since the left side is $-(1/f')'$ and the right side is $(ln f)'$ then exists a constant $c$ such that
\[-\frac{1}{f'}=c+ln f\]

I think you have to do more than that, because the functions might have some $0$'s, and then you wouldn't be able to do those divisions. This argument can be employed on certain maximal intervals on which $f'\ne 0$, but there might be more than one such interval.

First I think that $ ( ln f )' $ is $ \frac {f'}{f} $ and not $ \frac {f}{f'} $.
Does not follow that $f$ is $C^2 (D)$ so $f$ is analytical on some compact set $D$ ? I might be wrong.
If $f$ is analytical then $f = \sum\limits_{n\in N}a_n x^n$ . If we replace in the differential equation results that for any $a_0 , a_1$:
$a_2 = \frac{a_0 a_1}{2}$
$a_3 = \frac{a_1 (a_1 + a_0^2)}{6}$
$a_4 = \frac{a_0 a_1 (4 a_1 + a_0^2)}{24}$
.......
and still does not lead to some explicit answer.
Particular solutions: $f:R->R$ , $f(x) = 0$ and $f:R -${$-a$}$ -> R^*$ , $f(x) = \frac {-2}{x+a}$ .

I know almost nothing about diferential equations, but I will try an approach. First, write the equation in the form $ (f^2(x)-2f'(x))'=0 $, thus there exists a constant $c$ such that $ f^2(x)=2f'(x)+c$. This sounds quite classical ( though I do not know what is classical and what is not in this field). The only problem I also have is with the possible denominators of the fractions that appear in the problem. For $ c>0$  it is not difficult, I think, since we cand deduce that for a convenient constant $A$ we have ${ arctg{\frac {f}\{A}} $ is constant from where a first class of solutions. The case $ c<0$ seems to be the most difficult and I do not know what to do with the points in which $ f^2-c$  is zero. But anyway, this new problem seems a little bit more comestible.

I found immediately after my last post how to solve it, but my isp was done for a while.
$f\"(x) = f(x)f'(x) $ , then:
$ (f'(x))' = (\frac{1}{2} f^2 (x))' $
by integration, then by separation of the variables and again by integration :
$ \int \frac{df}{f^2+ 2c} = \frac{1}{2}x + K $
Appear 3 cases, each solvable:
1) $c=0$
$ \int \frac{df}{f^2} = \frac{1}{2}x + K $
then something like before $f(x) = - \frac {2}{x+a}$.
2) $c>0$
$ \int \frac{df}{f^2 + b^2} = \frac{1}{2}x + K $
like $arctan$ and so on...
1) $c<0$
$ \int \frac{df}{f^2 - b^2} = \frac{1}{2}x + K $
like $ln$ and so on...
Indeed it is classical and more like a simple exercise in the field of differential equations.


function
analytic geometry
graphing lines
slope
algebra
linear equation

According to company records, in 1980 the company earned 3.2 million in earnings and in 1985 earned 4.7 million. Assume a linear model for earnings in millions of dollars, E, versus time in years,t. Let 1980 be year t=0

a. Write an equation for E as a function of t.
b. Estimate the company's gross profits in the year 2000.
c. What is the slope of the line?
d. What information does the slope convey to the company?

My answer:
1. E= 3.2e^(.0769t)
2. 14.9 million
3. 1.71

But then I stopped because 1.71 was an odd slope, so I want to know if I am doing this right...

[hide]
The first equation you made is an exponential equation; the question asks for a linear equation. A linear equation is in the form y = mx + b.

The y-intercept, b, is 3.2, so the b is 3.2.

The slope is found by using the slope formula for the two points corresponding to the two years and the two earnings.
1st point: (0, 3.2), 2nd point: (5, 4.7)

(y2 - y1) / (x2 - x1) = (4.7 - 3.2) / (5 - 0) = 1.5 / 5 = [b]0.3[/b]

In this case, y = E and x = t.  a.) So, the equation is E = [b]0.3t + 3.2[/b]

In the year 2000, t = 20, so plug in 20 in the first equation above:
E = 0.3t + 3.2
E = 0.3(20) + 3.2
E = 6 + 3.2
E = 9.2

b.) So, the earnings will be [b]9.2 million.[/b]

c.) The slope m is 0.3, according to the above equation.

d.) Since the slope is positive, this shows that the earnings are going up.

[/hide]


Let $ a$ be nonzero complex number.If $ f$ and $ g$ are polynomials with comlex coefficients and $ (f(x))^3\equal{}(g(x))^2\plus{}a$ holds for all complex $ x$ then show that $ f$ and $ g$ are constant.

$ f(x)^3\equal{}(g(x)\minus{}ia)(g(x)\plus{}ia)$ and we have $ GCD(g(x)\minus{}ia,g(x)\plus{}ia)\equal{}1$
then $ \exists P,Q\in\mathbb{C}[X]: \ GCD(P,Q)\equal{}1,f\equal{}PQ,(P^3,Q^3)\equal{}(g\plus{}ia,g\minus{}ia)$
then $ 2ai\equal{}P^{3}\minus{}Q^{3}\equal{}(P\minus{}Q)(P^2\plus{}PQ\plus{}Q^2)$
so $ \exists c,d\in\mathbb{C}: \ P\minus{}Q\equiv c,P^2\plus{}Q^2\plus{}PQ\equiv d$
so $ 3Q^2\plus{}3cQ\plus{}c^2\equiv 0$ then $ Q,P$ are constant
then $ f\equal{}PQ,g\equal{}Q\plus{}ia$ are constant


I don't know how to use Asymptote so here is the problem source

[hide]
Introduction to Geometry
Problem 3.33 (page 77)[/hide]

Not that we would need it.
Hint for (a) and (b)
[hide]Triangles BOC and ABO are isosceles.[/hide]Hint for (c)[hide]Since we know that CD and OA are parallel, we know something else about the measures of the angles they form with OC (AOC and OCD).[/hide]If we needed a further hint
[hide]Just call the measures of angles that you still don't know x and y. Then write a couple of equations based on what you know of geometry and solve them.[/hide]


Let $ f: \mathbb{R}\to \mathbb{R}$ such that $ |f(f(x))\minus{}f(f(y))|\le \frac{1}{2}|x\minus{}y|$,must $ f$ be continuous?

No. A silly example: $ f(x)\equal{}\begin{cases}1&x\text{ is rational}\\0&\text{ otherwise}\end{cases}$. $ f\circ f$ is identically $ 1$.


[i]"During a court case the jury has to decide whether or not an unlikely event has occurred based upon the testimony of two independent witnesses. Let:
$ \tau$ be the probability of the unlikely event occurring
$ \alpha$ be the probability witness A tells the truth
$ \beta$ be the probability witness B tells the truth
Given that both witnesses claim the event occurred, what is the probability that the event occurred?"[/i]

I've been pishing about for about an hour with various things. I said let:

$ X$ be the event that the unlikely event occurs $ (P(X) \equal{} \tau )$
$ X^C$ be the event that the unlikely event doesn't occur $ (P(X^C) \equal{} 1 \minus{}\tau )$
$ A$ be the event that witness A tells the truth $ (P(A) \equal{} \alpha )$
$ A^C$ be the event that witness A lies $ (P(A^C) \equal{} 1 \minus{}\alpha )$
$ B$ be the event that witness B tells the truth $ (P(B) \equal{} \beta )$
$ B^C$ be the event that witness B lies $ (P(B^C) \equal{} 1 \minus{}\beta )$
$ C$ be the event that witness A claims $ X$
$ C^C$ be the event that witness A claims $ X^C$
$ D$ be the event that witness B claims $ X$
$ D^C$ be the event that witness B claims $ X^C$

I end up messing around with a bunch of stuff but then I just get lost. Can anyone help?

[hide]Since both witnesses said the same thing, either both must be telling the truth or both must be lying.

Case 1:  Both witnesses are telling the truth.
Then the event occurred.  Since witnesses A and B are independent of each other, this occurs with probability $ \alpha\beta$.

Case 2:  Both witnesses are lying.
Then the event did not occur.   This occurs with probability $ (1\minus{}\alpha)(1\minus{}\beta)$.

So, the probability that the event occured is $ \frac{\alpha\beta}{\alpha\beta\plus{}(1\minus{}\alpha)(1\minus{}\beta)}$.[/hide]

Where are you getting $ \frac {\alpha\beta}{\alpha\beta \plus{} (1 \minus{} \alpha)(1 \minus{} \beta)}$ from?

It's the desired outcome divided by all possible outcomes.

Hold on, are you sure that's correct? Because this is part b) of the question:

[i]"Given that $ \tau \equal{} 0.001,\alpha \equal{} 0.95, \beta \equal{} 0.8$, what is the probability that the event occurred?"[/i]

If we insert the $ \alpha$ and $ \beta$ values into the equation you gave, we get 76/77. But where does $ \tau$ come into it? Why have they given a value for that?

That answer is actually not right. See [url=http://en.wikipedia.org/wiki/Bayes_theorem#Example_1:_Drug_testing]this example[/url] of a problem like this having a deceptive answer. Bayes' Theorem, what the rest of the page is about, is a way of fixing the argument.

Ah right, I was afraid it was something to do with Bayes' Theorem. I remember our lecturer going over that, but it was the morning after a flatmates' 19th birthday so I was in no decent state to take it in. I'll read up on that then, if I can't do it still then I'll be back! Thanks

hmmm...

There is a probability of $ \tau$ that the event occured.  If both witnesses tell the truth (that it occured), then we have $ \alpha \beta \tau$; otherwise, if they both lie (then the event did not happen), we have $ (1\minus{}\alpha)(1\minus{}\beta)(1\minus{}\tau)$ probability.  So the chance the event occured is:

$ \frac{\alpha \beta \tau}{\alpha \beta \tau \plus{} (1\minus{}\alpha)(1\minus{}\beta)(1\minus{}\tau)}$

[quote="facis"]hmmm...

There is a probability of $ \tau$ that the event occured.  If both witnesses tell the truth (that it occured), then we have $ \alpha \beta \tau$; otherwise, if they both lie (then the event did not happen), we have $ (1 \minus{} \alpha)(1 \minus{} \beta)(1 \minus{} \tau)$ probability.  So the chance the event occured is:

$ \frac {\alpha \beta \tau}{\alpha \beta \tau \plus{} (1 \minus{} \alpha)(1 \minus{} \beta)(1 \minus{} \tau)}$[/quote]

This is wrong.

what's wrong with it?  I think that it is right, so please explain your claim.

Sorry my mistake, you are right. A later part of the question gave values for $ \tau, \alpha, \beta$ and asked to calculate something else and I ended up with the square root of a negative...which is why I thought your answer was wrong, but it turns out I just made a +/- error in my calculations. Sorry for the accusation and thanks for the answer!

no problem.  I've recently realized that about half of my mistakes are forgetting minus signs, so I completely understand.

hey is there any way to prove that formula is right? i've tried to use bayes theorem to derive it but i keep endung up with the wrong thing on the bottom of the equation?!

It is not possible for the witnesses to have told the truth and the event not happen, or vice versa.  So though the probabilities $ \alpha$, $ \beta$, and $ \tau$ are independent, knowing that both witnesses say the same thing - that the event occured - changes the probability that the event occured, as we must remove cases in which either witness says the event didn't happen.

Formally, we proceed as follows.  We want to find $ P(X | C \cap D)$.

$ P(X | C \cap D) \equal{} P(X \cap C \cap D)/P(C \cap D)$ by bayes theorem.

Since both people say it happened, the event can only have happened if they were telling the truth, so $ P(X \cap C \cap D) \equal{} P(X)P(A)P(B) \equal{} \alpha \beta \tau$ (we must assume independence of the event and the two witnesses' testimony here).

$ P(C \cap D) \equal{} P(X)P(C \cap D|X) \plus{} P(X^C)P(C \cap D|X^C)$

Here, we are merely partitioning the set $ C \cap D$ into two subsets which total to the whole to compute the probability.  If the event occurs ($ X$), both witnesses tell the truth; hence, $ P(C \cap D|X) \equal{} P(A \cap B) \equal{} \alpha \beta$.  If the event does not occur ($ X^C$), both witnesses lie and hence $ P(C \cap D|X^C) \equal{} P(A^C \cap B^C) \equal{} (1\minus{}\alpha)(1\minus{}\beta)$.  Thus we find $ P(C \cap D) \equal{} \alpha \beta \tau \plus{} (1\minus{}\alpha)(1\minus{}\beta)(1\minus{}\tau)$.

Plugging this back in, we have $ P(X | C \cap D) \equal{} \frac{\alpha \beta \tau}{\alpha \beta \tau \plus{} (1\minus{}\alpha)(1\minus{}\beta)(1\minus{}\tau)}$


[quote="cimabue"]For $n\geq 2$, compute
\[ s_n=\sum_{k=0}^n\sum_{j=0}^{n-k}\frac{(-1)^j}{j!}\cdot\frac{k^2}{k!}  \][/quote]

The sum is the coefficient of the term of degree $n$ in $\left(\sum_{j\geq 0}\frac{(-1)^j}{j!}X^j\right)\cdot\left(\sum_{k\geq 0}\frac{k^2}{k!}X^k\right)$. After some computations this product can be rewritten as $e^{-X}\cdot(X^2+X)e^X=(X^2+X)$.

The conclusion:
\[ s_n=\left\{\begin{array}{cc}0,&\mbox{if }n=0\mbox{ or }n\geq2\\1,&\mbox{if }n\in\{1,2\}\end{array}\right. \]


[quote="chien than"]Let $a;b;c$ are non-negative.Prove thatL
$\frac{a^{4}}{b+c}+\frac{b^{4}}{a+c}+\frac{c^{4}}{a+b}\geq \frac{1}{2}(a^{3}+b^{3}+c^{3})$[/quote]

Use Chebishev :wink:

Very nice solution, scorpius 119 !  :lol:

It can be easily proved that if $a_{1},\ldots,a_{n}$ and $x_{1},\ldots,x_{n}$ are positive real numbers and $(a_{i})$ are $(x_{i})$ are equally sorted, then for any positive real number $k$,
\[\sum_{i=1}^{n}\frac{x_{i}a_{i}^{k}}{(S-a_{i})^{k}}\ge\left(\frac{n}{n-1}\right)^{k}\frac{x_{1}a_{1}^{k}+\cdots+x_{n}a_{n}^{k}}{(a_{1}+a_{2}+\cdots+a_{n})^{k}}\]
where $S=a_{1}+\cdots+a_{n}$.

[quote="Vasc"]Nice solution, April.
Is the below inequality known ?

$\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}\geq \frac{3}{2}.\frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}$.[/quote]

This inequality was posted on this forum by Pham Huu Duc (see problem 65 in Algebraic Inequality, p. 381 and 458).


The symmetric point of $ E_a$ wrt $ N$ is just the midpoint $ M_a$ of $ BC$. Construct the nine-point circle $ (N)$ of triangle. Draw the line $ d$ through two points $ X_a$ and $ M_a$. The perpendicular line from $ N$ to $ d$ meets $ (N)$ at $ T$ ($ d$ cuts the segment $ NT$). The line $ TX_a$ meets $ (N)$ at $ F$ again. Easy to check that $ F$ is the Feuerbach point. The perp line to $ d$ at $ X_a$ meets $ NF$ at the incenter $ I$. And we continue work with http://www.mathlinks.ro/viewtopic.php?t=287425   :)


καλησπέρα παιδιά!
Μπορείτε να μου πείτε ποιές είναι οι πιο βασικές ανισωτικές σχέσεις (σαν εργαλεία) που χρησημοποιούμε για
για να αποδείξουμε ασκήσεις με ανισώσεις μέτρων (εκτός βέβαια την τριγωνική ανισότητα) στους μιγαδικούς στην Γ΄Λυκείου?
Καταλαβαίνω πως η ερώτηση μου είναι πολύ γενική,
για να γίνω πιο συγκεκριμένος με ενδιαφέρει το επίπεδο καλής προετοιμασίας για πανελλαδικές εξετάσεις και νομίζω πως αυτές είναι οι ασκήσεις που χρειάζεται μια πιο ώριμη μαθηματική σκέψη.
Αν δεν σας είναι κόπος στείλτε μου ένα γενικό κατάλογο από αυτές τις ανισώσεις, και αν έχεται τέτοιου είδους ασκήσεις που μπορώ να τις χρησιμοποιήσω
Σας ευχαριστώ!

Θα σου παραθέσω τις πιο βασικές ανισότητες που χρησιμοποιούνται σε μαθηματικούς διαγωνισμούς:

ΑΝΙΣΟΤΗΤΕΣ ΣΤΟΥΣ ΠΡΑΓΜΑΤΙΚΟΥΣ

Οι παρακάτω δεν έχουν όνομα αλλά βοηθάνε στην απόδειξη των περισσότερων ανισοτήτων):

1. $ (a \plus{} b)^2 \ge 0$
2. $ (a \plus{} b)^2 \ge 4ab$
3. $ (a \plus{} b)^2 \ge 2ab$
4. $ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca$
5. $ |\frac {a}{b} \plus{} \frac {b}{a}| \ge 2$

[size=150]Ανισότητα Buniakowsky - Caushi - Shwarz (B. C. S.):[/size]

$ (a_1^2 \plus{} a_2^2 \plus{} ... a_n^2)(b_1^2 \plus{} b_2^2 \plus{} ... b_n^2) \ge (a_1b_1 \plus{} a_2b_2 \plus{} ... a_nb_n)^2$

(η ισότητα ισχύει όταν $ \frac {a_1}{b_1} \equal{} \frac {a_2}{b_2} \equal{} ... \frac {a_n}{b_n}$)

[size=150]Ανισότητα αναδιάταξης (Rearangement Inequality):[/size]
αν οι n - άδες a, b έχουν την ίδια διάταξη και w μια αναδιάταξη της b τότε

1. $ a_1b_1 \plus{} a_2b_2 \plus{} ... a_nb_n \ge a_1w_1 \plus{} a_2w_2 \plus{} ... a_nw_n$

2. $ (a_1 \plus{} b_1)(a_2 \plus{} b_2)...(a_n \plus{} b_n) \le (a_1 \plus{} w_1)(a_2 \plus{} w_2)...(a_n \plus{} w_n)$

(και στα δύο η ισότητα ισχύει όταν $ b_1 \equal{} b_2 \equal{} ... b_n$ ή $ a_1 \equal{} a_2 \equal{} ... a_n$, αν είχαν αντίθετη διάταξη τότε η ανισότητα θα ήταν ανάποδα)

ΑΝΙΣΟΤΗΤΕΣ ΣΤΟΥΣ ΘΕΤΙΚΟΥΣ

[size=150]Ανισότητα αριθμητικού - γεωμετρικού - αρμονικού μέσου (AM - GM - HM):[/size]

$ \frac {a_1 \plus{} a_2 \plus{} ... a_n}{n} \ge \sqrt [n]{a_1a_2a_n} \ge \frac {n}{\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} ... \frac {1}{a_n}}$
(η ισότητα ισχύει όταν οι αριθμοί είναι ίσοι)

[size=150]Ανισότητα του Holder:[/size]

αν $ m_1 \plus{} m_2 \plus{} ... m_k \equal{} 1$ με $ m_1, m_2, ... m_k > 0$ τότε

$ (a_1 \plus{} a_2 \plus{} ... a_n)^{m_1}(b_1 \plus{} b_2 \plus{} ... b_n)^{m_2}...(z_1 \plus{} z_2 \plus{} ... z_n)^{m_k} \ge$ $ a_1^{m_1}b_1^{m_2}...z_1^{m_k} \plus{} a_2^{m_1}b_2^{m_2}...z_2^{m_k} \plus{} ... a_n^{m_1}b_n^{m_2}...z_n^{m_k}$

(η ισότητα ισχύει όταν  $ \frac {a_1}{b_1} \equal{} \frac {a_2}{b_2} \equal{} ... \frac {a_n}{b_n}, ... , \frac {a_1}{z_1} \equal{} \frac {a_2}{z_2} \equal{} ... \frac {a_n}{z_n}$)

[size=150]Ανισότητα των βαρών (Weighted Means Inequality):[/size]

αν $ x_1 \plus{} x_2 \plus{} ... x_n \equal{} 1$ με $ x_1, x_2, ... x_n > 0$ τότε

$ x_1a_1 \plus{} x_2a_2 \plus{} ... x_na_n \ge a_1^{x_1} \plus{} a_2^{x_2} \plus{} ... a_n^{x_n}$

(η ισότητα ισχύει όταν $ a_1 \equal{} a_2 \equal{} ... a_n$)

[size=150]Ανισότητα των δυνάμεων (Power Means Inequality):[/size]

αν $ k \ge m$ τότε

$ (\frac {a_1^k \plus{} a_2^k \plus{} ... a_n^k}{n})^{\frac {1}{k}} \ge (\frac {a_1^m \plus{} a_2^m \plus{} ... a_n^m}{n})^{\frac {1}{m}}$

(η ισότητα ισχύει όταν $ a_1 \equal{} a_2 \equal{} ... a_n$ ή $ k \equal{} m$)

ΥΠΑΡΧΟΥΝ ΑΚΟΜΑ:
Η ανισότητα Minkovski που είναι η τριγωνική ανισότητα σε τυχαίο αριθμό διαστάσεων.

Η ανισότητα του Huegens που είναι η εφαρμογή της Holder για $ n \equal{} 2$.

Η ανισότητα του Andreescu (δεν είναι δεκτή σε διαγωνισμούς και πρέπει να αποδεικνύετια κάθε φορά, αλλά είναι πολύ χρήσιμη):
για θετικούς παρανομαστές είναι:
$ \frac {a_1^2}{b_1} \plus{} \frac {a_2^2}{b_2} \plus{} ... \frac {a_n^2}{b_n} \ge \frac {(a_1 \plus{} a_2 \plus{} ... a_n)^2}{b_1 \plus{} b_2 \plus{} ... b_n}$
(η απόδειξη εύκολα για n = 2 και μετά επαγωγή)

Αυτές είναι οι κυριότερες ανισότητες. Υπάρχουν πολλές άλλες που θεωρούνται βασικές, αλλά τα ποιο πολλά θέματα λύνονται μόνο με αυτές.

Βέβαια, χρησιμοποιώντας μαζί με τα παραπάνω και θεωρία ομογενοποίησης ([url]http://www.mathlinks.ro/viewtopic.php?t=176946[/url]) και συμμετρίας (αν η παράσταση είναι συμμετρική μπορούμε να θέσουμε χωρίς βλάβη της γενικότητας μια συγκεκριμένη διάταξη στις μεταβλητές) μπορούμε να λύσουμε τα περισότερα θέματα ανισοτήτων.  :lol:

Σε ευχαριστώ πολύ maniopas! για το πολίτιμο υλικό που μου έδωσες!

Για πανελλαδικες δεν νομίζω ότι χρειάζονται αυτά.Για διαγωνισμούς ή για το πανεπιστήμιο είναι χρησιμότατα.
Να ένα καλό Link για ανισότητες [url]http://www.artofproblemsolving.com/Resources/Papers/KedlayaInequalities.pdf[/url]
που έχει καλες τεχνικες και αποδειξεις αυτων των ανισοτήτων και προβλήματα ανά ενότητα.
Αν θέλεις και άλλο υλικό στείλε μου pm για να στο στείλω.

Ναι, πράγματι. Αλλά γνωρίζοντας κάτι παραπάνω κάποιες φορές κάνει πιο εύκολη την όλη για διαδικασία.

Για παράδειγμα, αν θέλουμε να δείξουμε ότι για $ z \equal{} x \plus{} iy$ ισχύει

$ 4|z| \ge \sqrt {2}(|z \minus{} \bar{z}| \plus{} |z \plus{} \bar{z}|)$ μπορούμε να δούμε ότι αυτή μετασχηματίζεται στην

$ (\frac {|x|^2 \plus{} |y|^2}{2})^{\frac {1}{2}} \ge (\frac {|x|^1 \plus{} |y|^1}{2})^{\frac {1}{1}}$

που ισχύει από την ανισότητα των δυνάμεων

Βέβαια, το να χρησιμοποιηθούν οι παραπάνω ανισότητες στις πανελήνιες πάει πολύ, αλλά δε νομίζω να υπάρχουν άλλες πέρα της τριγωνικής ανισότητας που να μην πηγάζουν άμεσα από αυτές.

Παιδιά ακούστε λίγο κάποια πράγματα ...

1. Στις Πανελλαδικές δεν υπάρχει περίπτωση να μπεί θέμα που να λύνεται [b]μόνο[/b] με τη χρήση κάποιων από τις παραπάνω ανισότητες διότι η φιλοσοφία των θεμάτων των πανελληνίων, είναι διαφορετική από αυτή των θεμάτων σε
διαγωνισμούς όπως ο Ευκλείδης - Αρχιμήδης κ.λ.π.

2.  Θεωρούνται γνωστά οποιαδήποτε Θεωρήματα - Εφαρμογές που υπάρχουν στο βιβλίο της Τρίτης Λυκείου ή στα βιβλία των προηγούμενων τάξεων.
[b]Από τις παραπάνω ανισότητες δεν υπάρχει καμία.[/b]
Φαντάσου να πετάξεις στο Διωρθωτή κάτι σαν << Από την ανισότητα Andreescu προκύπτει το ζητούμενο !!! :o  θα τρελαθεί ο άλλος που θα διορθώνει >>

3.  Αν ένα θέμα λύνεται με τη βοήθεια κάποιας άσκησης που βρίσκεται μέσα στο βιβλίο, τότε η άσκηση πρέπει να αποδειχτεί. Πολύ περισσότερο Θεωρήματα τέτοιου τύπου όπως τα προηγούμενα.Ακόμα και η άσκηση του σχολικού της Γ Λυκείου που αποτελεί κριτήριο για το πότε ένας μιγαδικός z ανήκει στο R και πότε ανήκει στο I πρέπει να δείχνεται, άσχετα άν οι διορθωτές δεν αφαιρούν μονάδες γιατί είναι   one line solution    η απόδειξη και θεωρείται βασική, κάτι σαν τη  $ x^2 \plus{} y^2\ge2xy$.

4.  Αν κάποια άσκηση είναι ανισότητα στους μιγαδικούς, κατά 90% θα θέλει χρήση τριγωνικής ανισότητας ( που υπάρχει μέσα στο βιβλίο ) ή απόδειξη με γεωμετρικό τρόπο και χρήση ανάλυσης.

Κλείνοντας να πώ ότι:  Spinos και όλα τα άλλα παιδιά που δίνετε Πανελλαδικές να προσέχετε να αιτιολογείτε με βάση το βιβλίο. Ειδικά εσύ Σιλουανέ πρόσεχε μή τους πετάξεις καμία "κουφή" λύση και τους στείλεις...  :lol:
Πραγματικά είναι κρίμα να χάσετε μόρια λόγω των πολλών γνώσεων που έχετε. Ακόμη και λάθη του βιβλίου πρέπει να τα αντέξετε  :(

Για παράδειγμα στο βιβλίο της Β Λυκείου Κατεύθυνσης λέει ότι για ένα διάνυσμα που είνα μηδενικό μπορούμε να θεωρήσουμε ότι σχηματίζει οποιαδήποτε γωνία με ένα άλλο μη μηδενικό διάνυσμα  :ninja: (έλεος) ενώ την ίδια στιγμή αν πάρετε και βάλετε δίπλα το βιβλίο της Γ Λυκείου και διαβάσετε για τα ορίσματα μιγαδικών θα δείτε ότι λέει ότι δεν ορίζεται όρισμα όταν ο μιγαδικός είναι ο z = 0,(που είναι και το σωστό ) πράγμα που αναιρεί το προηγούμενο ... ( και αν θυμάμαι καλά μπήκε τέτοιο θέμα κάποια χρονιά )
Περισσότερα σίγουρα μπορεί να σας πεί ο κύριος Μπάμπης, αλλά πιστεύω ότι έχω καλύψει τις γενικές αρχές για το τι πρέπει να χρησιμοποιούμε και τί όχι.

και για να μην αδικήσω τον Μανώλη ... :)

Μανώλη η παρουσίαση που έκανες με τις ανισότητες ήταν πολύ καλή και εγώ προσσωπικά σε ευχαριστώ για τον χρόνο που ξόδεψες για να τα γράψεις. Πιστεύω ότι πρέπει να υπάρχουν τέτοια posts που να δείχνουν τα βασικά θεωρήματα ( και σε άλλα πεδία ) και να πώ ότι όταν πρωτοέγινα μέλος έψαχνα για κάτι τέτοιο. Βέβαι από εκεί και πέρα αν κάποιος θέλει κάτι περισσότερο υπάρχουν Βιβλία που μπορεί να αγοράσει.

maniopas, thnx και απο μένα! Πολύ καλή παρουσίαση...

Να προσθέσω μερικές ακόμα.

[b][size=150]Ανισοτητα Bernoulli:[/size][/b]

Αν $ x > \minus{} 1$ και $ n$ ειναι θετικος ακεραιος, τοτε

$ (1 \plus{} x)^n\ge 1 \plus{} nx$

Επισης ισχυει και για $ \minus{} 2\le x \le \minus{} 1$

Αν $ \minus{} 1 < x\neq 0$ τοτε ισχυουν οι ακολουθες γενικευσεις

$ (1 \plus{} x)^a > 1 \plus{} ax$ αν $ a > 1$ ή $ a < 0$

$ (1 \plus{} x)^a < 1 \plus{} ax$ αν $ 0 < a < 1$

[b][size=150]Ανισοτητα Aczel:[/size][/b]

Αν $ a_1,a_2,...,a_n,b_1,b_2,...,b_n$ ειναι πραγματικοι αριθμοι τετοιοι ωστε

$ a_1^2 \minus{} a_2^2 \minus{} ... \minus{} a_n^2 > 0$ και $ b_1^2 \minus{} b_2^2 \minus{} ... \minus{} b_n^2 > 0$ τοτε

$ (a_1^2 \minus{} a_2^2 \minus{} ... \minus{} a_n^2)(b_1^2 \minus{} b_2^2 \minus{} ... \minus{} b_n^2)\le (a_1 b_1 \minus{} a_2 b_2 \minus{} ... \minus{} a_n b_n)^2$

[b][size=150]Ανισοτητα Schur:[/size][/b]

Αν $ a,b,c$ ειναι θετικοι πραγματικοι αριθμοι και $ r\in R$ τοτε

$ a^r(a \minus{} b)(a \minus{} c) \plus{} b^r(b \minus{} c)(b \minus{} a) \plus{} c^r(c \minus{} a)(c \minus{} b)\ge 0$

Η ισοτητα αληθευει οταν και μονο οταν $ a \equal{} b \equal{} c$

Να προτείνω να γίνει stickied το topic? Είναι πολύ χρήσιμο σαν αναφορά... ίσως σαν sub-topic του stickie για τις ανισότητες?

Maniopas και mailo, μπράβο! Πολύ καλή δουλειά με τα συγκεντρωτικά posts! :coolspeak:  :10:

Cheerio,

Durandal 1707

Να θυμίσω και το παρακάτω πολύ χρήσιμο και εξαιρετικό post που έστειλε ο Σιλουανός πέρισυ τέτοιο καιρό, για όσους δεν ήταν τότε γραμμένοι αλλά και για να θυμηθούν οι διαγωνιζόμενοι τις βασικές ανισότητες μια και σε λίγες ημέρες έχουμε τον Αρχιμήδη!
Καλό διάβασμα

http://www.mathlinks.ro/viewtopic.php?search_id=1619454502&t=131000

Αλέξανδρος

:oops_sign:

sorry  Siluan ... δεν το είχα δεί αυτό ...

είναι απλά ...   [size=150][b][color=darkred]καταπληκτική παρουσίαση[/color][/b][/size]

 :clap:


Not sure what you mean...

If you mean, how would you use $\LaTeX$ on the forum as opposed to how you would usually use $\LaTeX$, then the answer is to enclose your equations in dollar signs (or $\backslash$[ $\backslash$]), just like you usually would in $\LaTeX$.  For example, $\backslash$[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$\backslash$] becomes
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

If you mean you don't know how to use $\LaTeX$ and want to learn, this website has a $\LaTeX$ [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]tutorial[/url] section.

[quote="zengmao"]Sorry for this silly question. But how can I add equations written in Latex when I post a reply in the forum?[/quote]I assume that no one sees this BIG, HUGE announcement on top of these forums, namely [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5261]how to write and use LaTeX on the board[/url]? :)


A because that's the only one without a curve.

A *sigh*

Although if I wasn't sure of the answer on a real test, I'd almost never put A.

no one voted D!

I voted C because A, B and D sucks.

i voted D!

i dunno why i did. it was the first one i thought of.

A because it's the only vowel, but I voted C, I guess because it has more empty space; it's the only one without an enclosed space.

Also, it's "psychological".

None because Z is not one of them... :)  :|  :P


Please note that all ties have been broken, even though listed scores may not indicate this. \begin{tabular}{|c|c|c|c|c|}\hline \multicolumn{5}{|c|}{MIMC Winners}\\ \hline Place & Na{m}e/Userna{m}e & Score & Grade & State\\ \hline 1st & nat mc & 250 & 9th & IN \\ \hline 2nd & Xantos C. Guin & 240 & 9th & GA \\ \hline 3rd & scorpius119 & 224 & 11th & PA \\ \hline 4th & tiredepartment & 222 & 10th & NJ \\ \hline 5th & Avi Levy / me@home & 211 & 9th & OR \\ \hline 6th & John Berman / jb05 & 209 & 9th & NC \\ \hline 7th & arjunlandes & 206 & 11th & DC \\ \hline 8th & Mathlete9366 & 200 & 9th & TX \\ \hline 9th & Adeel Khan / chess64 & 200 & 9th & MD \\ \hline 10th & Simba & 200 & 10th & UK \\ \hline 11th & eeheehee & 199 & 8th & OH \\ \hline 12th & 10000th User & 195 & 10th & NJ \\ \hline 13th & pieterminate & 190 & 9th & OR \\ \hline 14th & Neil Gurram / mathgeniuse^{\ln(x)}& 180 & 8th & MI \\ \hline 15th & bthings\_{2}000 & 163 & 11th & GA \\ \hline 16th & Shogia & 163 & 11th & IA \\ \hline \end{tabular} $\text{Team Awards}$
1st Place - Georgia
2nd Place - New Jersey
3rd Place Team - Oregon
4th Place - Texas

$\text{Highest Scores: 6th Grade \& Under}$
Demetrius - 58
Hypatia 2 - 58

$\text{Highest Scoring 7th Grader}$
ragnarok23 - 78

$\text{Highest Scoring 8th Grader}$
eeheehee - 199

$\text{Highest Scoring 9th Grader}$
nat mc - 250

$\text{Highest Scoring 10th Grader}$
tiredepartment - 222

$\text{Highest Scoring 11th Grader}$
scorpius119 - 224

$\text{Highest Scoring 12th Grader}$
gtpopojp - 124

[color=red][size=150]WORK IN PROGRESS[/size][/color]
Latex is weird due to AoPS slowdown??

The LaTeX problem should be a temporary affect of a mistake by our server host.

[quote="mysmartmouth"][quote="robinhe"]What about top 3 in each grade?[/quote]

Hmm...not really enough entries to do that.  Maybe we will try...[/quote]

OK, but could you please tell me what I got (me=lazy) and/or tell me how I ranked among fellow 8th graders?

Thanks. :)

My goal for this contest was to be the #1 9th grader.

Oh well.  :winner_second: isn't bad either.

its amazing how low my score was comparing to what I thought I got; I had SO many mistakes, and funny enough, my mistakes were little enough to be answer choices. I thought I had gotten a perfect, and I missed 10...

Oh well, I guess I didn't take this contest seriously enough, and congratulations to Xantos and nat mc for pwning me :D

Ha, at least you didn't miss #3.  :oops:  :)
EDIT: Oh, and why is it that most of top 16 are 9th graders or below?

[quote="pieterminate"]Ha, at least you didn't miss #3.
[/quote]

speaking of which...why is my name in that question? ;) I'm very organized :mad:

Hmmm...why are there so many 9th graders?  That's a really good question.  Maybe they're anxious to prove themselves?  Still fresh and into math?  Free time?  Hmm...

Anyway, about Individual results:  We will PM them to you eventually.  We have a LOT of submissions, so it will take a while to get you your scores.  However, once we do, we will tell you both a score, and the position you placed in.  Expect all scores to be distributed by the end of this week.

I'm amazed at the results considering the lack of effort that I have devoted. That is not to say that I didn't take this seriously, but the test I thought was difficult and lost my confidence. I expected my ranking to be around 40+. But I'm happy now :)

[quote="Elemennop"][quote="mysmartmouth"]Still fresh and into math?[/quote]

That and having more free time is probably the bulk of the reason.[/quote]

And being forced to attend Study Hall in school for 2 hours per day... :(  :)

[quote="me@home"]It's funny, in my school district they didn't let me get ahead in math all during elementary and middle school, but when I came to high school they were like "sure take these classes but if you crash and die its your fault." And it helps that we have a senior who just got a 5 on 17 AP tests (from last year).[/quote]
My middle school won't let us go ahead either. Ironically, they will let you slow down in math but won't let you speed up. (ex: We can take Algebra 1, but they won't let you take Geometry or Algebra 2. they do however, let you take a course called Pre-Algebra Advanced which is like halfway between Pre-alg and Alg.  :rotfl:  :rotfl: )

My school let me skip Pre-algebra, but didn't let me skip Alg II.  Good thing too  :D

BTW, there's a Darth Maul avatar directly under Yoda... Yoda wins...  :lol:

My middle school offered Alg I for the "most capable" of 8th graders only.  Otherwise, you would take "everyday math" in 6th, followed by 2 years of pre-alg.

[b](If you're talking about how your school's math education sucks, don't read this)[/b]

[quote="jb05"]How many entries were there?[/quote]

Weren't there 200 or so? Shouldn't that be enough for top 3?

From mysmartmouth's post:

[size=200][color=red][u][b]SINCE I AM NOT ON THE MIMC COMMITEE THESE ARE UNNOFICIAL RESULTS.[/b][/u][/color][/size]

$\text{6th Grade}$
1. Demetrius - 58
1. Hypatia 2 - 58
3.

$\text{7th Grade}$
1. ragnarok23 - 78
2.
3.

$\text{8th Grade}$
1. eeheehee - 199
2. mathgeniuse^ln(x) - 180
3.

$\text{9th Grade}$
1. nat mc - 250
2. Xantos C. Guin - 240
3. me@home - 211

$\text{10th Grade}$
1. tiredepartment - 222
2. Simba - 200
3. 10000th User - 195

$\text{11th Grade}$
1. scorpius119 - 224
2. arjunlandes - 206
3. bthings_2000 - 163

$\text{12th Grade}$
1. gtpopojp - 124
2.
3.

I think that there are enough enteries to fill in the slots not filled.

We had 200 or so sign up, but only about 40 turned in answers.  We'll see after ARML...

[quote="robinhe"]The Michigan legislature passed a bill that's now a law that states that school cannot start until after Labor Day...[/quote]

They did? When was that? I had school the week before Labor Day this year and the year before that and .....every other year basically.

Wait that means we get out later next year?

[quote="not_trig"]Hmm... people....

Joining something and not doing anything? Huh?
What's the point?[/quote]
Many people signed up but were doubtful, then got busy, and thought they missed the deadline anyway (of course the deadline was moved back pretty far...but they wouldn't know). That's almost what happened to me, then I remembered and checked to see who won when I saw there was still time...

Hi any one from India who took the MIMC?

[quote="Captain Sugar"]i thought we were going to get a pm that told us our scores. when will we get that?[/quote]

We already sent these out.  However, we have 3 score sheets that were printed by someone else, and they had no username, just a real name.  We found one of these people, but the other two are still sitting here.  If you didn't receive your results (two people) then please PM me ASAP with your real name.  Thanks!

[quote="CatherineAsaro"][quote="mysmartmouth"]

$\text{Highest Scores: 6th Grade \& Under}$
Demetrius - 58
Hypatia 2 - 58
[/quote]

They will be happy to hear the news.  They were rather daunted by the test!  But then, they have eleven more years before college to develop their problem solving gifts.  :-)

Best,
Catherine[/quote]

yes they will, up until 17th grade.

in defense of those who signed up but didn't turn in answers (me), what happened to the good old excuse of doing the work but forgetting to turn it in?

[quote="joml88"]Actually, it's $\le 17$th grade since the category is $\text{6th grade \& under}$.  It's not impossible that those kids are 6-7 years old (actually I think there were some really young kids taking the test.[/quote]

nice save lol


Hi everyone, I'm new here.

I'm going to be a sophomore next year and I've been trying pretty hard to prepare for college.

I was wondering, to get into a top ivy league college like harvard or princeton, is it best to take as many AP tests and SAT II's as possible? Of course when I say as many as possible I mean the ones within my reach of getting a 5 or a 700-800.

Also, my school isn't the greatest and only offers maybe 15 or so APs. Should I independently study subjects like stats, psychology, euro/world history, etc?

Thanks in advance for the advice.

the more aps and sat IIs you have the better off you are, as long as youre getting mostly 5's on the AP's and 750+ on the SAT II's (at least for highly selective colleges such as Harvard and MIT).  in terms of self-studying some AP's your school does not offer, that's really up to you and what you want to study and learn.  if you're a math person, colleges will prefer to see AP Statistics, but if your school does not offer it, it will not hurt you as much as if your school did offer it and you chose to not take it.  don't stress too much and just plan your sat II's/AP's out for the rest of your high school career.

[quote="h78reg"]isn't 15 AP's good enough? the high school where i'm going to go to has like 10, and it's not a very bad school, though I don't think it offers AMC tests. :( i hope my guess about the AMC's is wrong.........[/quote]
Eh...I'm sure you can go to another high school to take the AMCs.  My high school (huge school with almost 4000 students) messed up and didn't allow us to take the AMC B this year, so some kids went to another high school to take it.

15 is plenty fine. thats really quite a bit, depending on the school size, etc. i know most colleges look for students that take "the most rigorous classes offered." i always thought self-studying for APs was a waste. what do you gain? sure, if you do well, you may get credit at the college of your choice, but a lot of students i've seen do that (on aops) cram for the test the month/week/night in advance (and still end up making a 3, 4, or 5). however, it can be argued that you dont really learn the material that well, so when you get to college and get out of that class, you arent TRULY prepared for the next class in the series (like say for example you self-studied biology like that, you may not be ready to take genetics yet). when you are weighing that kind of decision, dont just think about what the colleges want. think about YOUR education. thats what getting into college is all about, anyways.

15 is a lot. My school offers... *counts on fingers and a toe* 11.

nvm, I recounted carefully we've got only 12

We only have honors physics at our school for some reason so I think I'll self-study those, good preparation anyway for physics SAT II. Also, maybe stats and an additional history class. THanks for the advice.

My school only offers about 6-7 AP's with 1400+ students.

[quote="ffdbzathf"]my school offers 22 and we have about 1000 kids, but the problem is that they put WAY too many kids into AP classes who should not be there, and therefore the classes become more watered down.[/quote]

Same, too much pressure to take APs; but I guess that's where the percentage of 1's and 2's comes from...

to answer the original question, you would [b]never[/b] get rejected because you took only 15 APs. It's nonsense!!

I took 7, three fives and four fours, and I took four SAT IIs, 800, 800, 760, 640.
yet I go to MIT.

[quote="xxreddevilzxx"][quote="cory123"]Should I independently study subjects like stats, psychology, euro/world history, etc?[/quote]

that sounds weird.... your school offers 15 APs, and it doenst include Stats or World History???? those are some of the main AP tests/courses.[/quote]

I think he might mean that he doesn't want to take the class, but wants to take the AP test itself. He probably wants the space in his schedule for other classes or just plain doesn't have room in his schedule.

[quote="cats..."]you have a computer now?[/quote]
nope, but there is computer everywhere at MIT

Nah, our school doesn't have stats or world history. We do have AB, BC, and US history though. That's why I wanted to self-study them. We might be getting euro history in a few years if our budget allows it.

I guess our school has a pretty good amount of AP's now that I think about it, because we only have about 150-200 people per grade.


geometry
3D geometry
sphere
search
combinatorics unsolved
combinatorics

let $ rs \plus{} 1$ points  in the space such that among $ r \plus{} 1$ points we can find  two points with a distance $ 1$.  prouve that  there is  one  sphere with a radius $ 1$ where there is at least $ s \plus{} 1$ points.

Have a look here
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=737660245&t=4886

Pierre.


trigonometry
analytic geometry
geometry
geometric transformation
reflection
rectangle
rhombus

[hide="not really sure"]

Drawing in the line $BD$, we create a $45=45=90$ right triangle, $\bigtriangleup BCD$. If we call the length of $BC$ $m$, then the length of $BD$ is $\sqrt{2}m$. In $\bigtriangleup BAD$, $DA$ is length $m$, and the measure of $\angle BDA$ is $105$. If we call the measure of $\angle DBA$ $x$, with Law of Sines, we can write: $\frac{\sin x}{m}=\frac{\sin(75-x)}{\sqrt{2}m}$
$\sqrt{2}\sin x=\sin{75-x}$.

Did I make any mistakes, the numbers aren't coming out nicely.

[/hide]

[quote="Adunakhor"]Did I make any mistakes, the numbers aren't coming out nicely. [/quote]
You didn't make any mistake, so try to finish your solution.
$x$ turns out to be a nice number.

[hide="Here is a trigonometrical solution."]
[color=red][b]Lemma.[/b] Let $ABCD$ be a convex quadrilateral. Define $I\in AC\cap BD$ and
$m(\widehat {IAB})=x_1\ ;\ m(\widehat {IBC})=x_2\ ;\ m(\widehat {ICD})=x_3\ ;\ m(\widehat {IDA})=x_4\ ;$
$m(\widehat {IAD})=y_1\ ;\ m(\widehat {IBA})=y_2\ ;\ m(\widehat {ICB})=y_3\ ;\ m(\widehat {IDC})=y_4\ .$
Then exists the following trigonometrical relation :
$\sin x_1\sin x_2\sin x_3\sin x_4=\sin y_1\sin y_2\sin y_3\sin y_4\ .$[/color]

[color=blue][b]The solution of the proposed problem.[/b] Denote $m(\widehat {ABD})=x$. Then
$\sin x\sin 75^{\circ}\sin 45^{\circ}\sin 15^{\circ}=\sin 45^{\circ}\sin 15^{\circ}\sin 105^{\circ}\sin (60^{\circ}-x)\Longleftrightarrow$
$\sin x=\sin (60^{\circ}-x)\Longleftrightarrow x=60^{\circ}-x\Longleftrightarrow x=30^{\circ}\Longleftrightarrow B=75^{\circ}\ .$[/color][/hide]

[quote="frt"]In quadrilateral $ABCD$, $\angle C=90^\circ$ and $\angle D=150^\circ$. If $BC=CD=DA$, find $\angle B$.[/quote]

[hide="coordinates"]
so let $C$ be located at $(0,0)$ then we will locate $B$ at $(0,1)$ and $D$ at $1,0$ WLOG because we're just dilating the figure. A is located at $(1+\cos{30},\sin{30})$

Now we can calculate the distance to the origin, distance to B and use law of cosines to find $\angle B$

We find that $BA = \sqrt{2 + \sqrt{3}} = \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2}$
and hen we use the law of cosines we find that $\cos{\theta} = \frac{1}{2BA} = \frac{1}{\sqrt{2} + \sqrt{6}}$

so then $\theta = 75^{\circ}$

[/hide]

[quote]You didn't make any mistake, so try to finish your solution. $x$ turns out to be a nice number.[/quote]

Oh ok, heh, should have kept going,  :blush:

Btw, how do you change the Quote caption to say something else, such as "frt wrote" - I thought it would have been the same as changing it for Hide, but I tried that and it didn't work.

Just write ="something" in the first quote bracket.  Example:
[quote="boxedexe"]Here's hwo it works[/quote].

Or just click on the quote button.

Masoud Zargar

a pure geometric solution exists...
[hide="hint"]what is special about $150^{\circ}$[/hide]

Altheman, I've been thinking about a pure geometric solution for over half an hour...do you have another hint?

Construct a circle with centre $D$ and radius $CD$. Since $CD=DA$, so $A$ is on the circle. Since $\angle{C}=90^o$, so $BC$ is a tangent line to the circle. Because $AD=CD$ and $\angle{ADC}=150^o$, so $\angle{ACD}=\angle{CAD}=15^o$. Thus $\angle{BCE}=75^o$, $\angle{BEC}=180^o-45^o-75^o=60^o$.

Now, construct $AF \perp BD$ at $F$ and reflect $D$ through $AF$ to $D'$. Then $\angle{AD'D}=\angle{ADD'}=\angle{AED}+\angle{EAD}=75^o$. Since $AD'=AD=BC$, so we have $\triangle{BCE} \cong \triangle{AD'E}$, which implies $BE=AE$. Therefore, $\triangle{AEB}$ is an isoceles triangle with an angle being $120^o$. Thus $\angle{ABE}=30^o$, $\angle{ABC}=\angle{ABE}+\angle{CBE}=75^o$.

[hide="another solution"]Draw circle centered at $D$ with radius $AD=CD$. $BC$ is a tangent since $\angle C=90$. Draw line $l$ parallel to $BC$ through $D$, hitting the circle at the same side of $DC$ as $B$ at $Q$.

By parallel lines, $\angle QDC=90$, and equal radii give us $QD=DC=CB$, $angle DBC=\angle DQC=45$ by isosceles right triangles, so $QBCD$ is cyclic by inscribed angles, $\angle DQB=180-90=90$, and $\angle QBC=90$. The right angles make it a rectangle, the equal sides make it a rhombus, thus $QBCD$ is a square.

$\angle ADQ=150-90=60$, and since $AQD$ is isosceles, $\angle DAQ=\angle AQD=60$ so $\triangle AQD$ is equilateral. Then $AQ=QB=BC=DC=AD=QD$. $\angle AQB=60+90=150$. Then $\angle ABQ=15$ since $\triangle QBA$ is isosceles, and $\angle ABC=90-15=75$[/hide]

[hide="comment:"]all this is is just a square and an equilateral triangle sharing a common side, where BA is the line connecting the far vertex of the triangle, to the far vertex of the square[/hide]

[hide="Here is a nice, short and simple solution."]Define the middlepoint $F$ of the side $[BC]$ and the projection $E$ of the point $A$ on the sideline $CD$. The quadrilateral $AECF$ is a rectangle and the triangle $ACB$ is isosceles. Therefore, $\boxed {m(\widehat {ABC})=75^{\circ}}\ .$[/hide]


Determine (ALGEBRAICALLY) whether each function is even, odd or neither.

(1) f(x) = 2x^4 - x^2

(2) f(x) = CUBEROOT{2x^2 + 1}

[quote="Interval"]So, basically is just the same old plug and chug, right?[/quote]

Yes it is.

[b]Lemma:[/b]  Any function $f(x)$ which can be written in the form $g(h(x))$ where $h(x)$ is even is also even.

Consider $h(x) = x^{2}$, which works for both functions.


I have a question about converting radicals. Is there any definitive way to convert radicals of form $k\sqrt{x}$ into those of form $\sqrt{a+\sqrt{b}}$. It's easy the other way around, but I was just wondering if there was a method... :(

Nevermind, I just realized that what i asked is basically a useless question as it isn't really possible...  :blush:

Why not?  Couldn't you just do something like $\sqrt{k^2+\sqrt{x^2}}$?  It's the other way that I can't see how to do it...

That doesnt work because $k\sqrt{x} = \sqrt{xk^2}$

Oh, right.  I added rather than multplied... :blush:

Hmmm, neither one of those is very easy then, it there are ways at all...

There should be a way.  It's in the aops book v1 somewhere.

if you have something like $k\sqrt{x}$, where k and x are integers, then if you square that, it's $k^2x$.  but if you square $\sqrt{a+\sqrt{b}}$, then you  get $a+\sqrt{b}$, which is irrational if b is not a perfect square, and if b was a perfect square you would have simplified it in $\sqrt{a+\sqrt{b}}$ in the first place.

so you can't exactly convert them from one form to another, but there are some numbers of the form $\sqrt{a+\sqrt{b}}$ that you can change into $c+\sqrt{d}$ (you sort of factor it....) but with some radicals you can't do that, and i'm not exactly sure how you can tell if $\sqrt{a+\sqrt{b}}$ is the most simplified it can get...


search
combinatorics unsolved
combinatorics

find the number of subsets B of the set {1,2,.....,2005} such that the sum of the elements of B is congruent to 2006 modulo 2048

Searching for "subsets congruent 2006" yields
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1197697058&t=247071
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1197697058&t=167883
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1197697058&t=39550

Also, please give your posts more informative titles.


[quote="vivemaths"]a,b,c > 0 , prove that :
a^5/(a^4+b^4) + b^5/(b^4+c^4) + c^5/(c^4+a^4) >= (a+b+c)/2
good luck !! :oops:[/quote]
See the third Vasc's post here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23113


Let $n\geq a_1>a_2>...>a_k$ be positive integers such that $lcm(a_i,a_j)\leq n$ for all $i,j$.Prove that $ia_i\leq n$ for $i=1,2,...,k$.


Hi, guys,

today, we only have phones calls for the USAMO winners, TST takers (4 additional students tied with 28), and underclassmen honorable mantions.

By the rule, these are the students who are invited to go to MOSP without computing MOSP index scores, which is not completed. Other phone calls will happen Mon. and Tu.

There are 3 winners could not be able to be on the USA IMO team, this is the reason we invited 4 additional students. (We did the same thing last year.) To avoid rumors on scoring (and time wasting guessing and checking), here is some official info: 41, 35, 34, 34 (top 3). Winner cut off: 29.

The scores are tight, every TST takers have a real chance to make the team. It's time to work hard! Really really hard!!! Many of you guys said that the problems are somewhat easier than expected. Then I would say that there are tons of ground to imporve the performances! So let's enjoy the results for a minute and start to work!

Best,

Zuming

Well, it's nice to know, so we don't think it's the end of the world when we don't get a phone call.  :)  :D

Although it'll be one helluva nerve-racking night for some of us...

There was only one score above 35?

Surprising.

[quote="beta"]Any 9th graders got called today?[/quote]

If there had been, the forums would be surging with activity.  And you would have received one.

I have a curiosity: how many people have solved problem 6? I have seen some remarks saying it was so easy for an USAMO question. Yet, the scores are small, thus it must have been another problem very hard. But I do not see which one...  :?


linear algebra
matrix
vector
linear algebra theorems

.If the triangular matrix A is similar to a diagonal matrix, den A is already diagonal?
2.let V b d space of matrises of order n over F.let A be a fixed matrices of order n pver F.
let T and U be the linear operators on V defined on V defined by
T(B)=AB,U(B)=AB-BA.
a)If A is diagonalizable over F den T is so?
b)if A is diagonalizable den so is U?
3. let T be a lin. operator on a finite dimensional vector space over an algebracally closed

field F.let f be a poly. over F. prove dat c is a characteristic valu of
f(T) iff c=f(t).where t is char.value of T.

1.let V be a finte dimebsnal vector space over F and let W be a subspace of V.if f is a linear functional on W, prove dat dere is a linear functional
g on V suchdat g(a)=f(a) for each a in d space W.
2. T is lin. op. on V if Tsquare=0 what can u say about d relation of range of T to d nullspace of T?
give an xample of alin.operator T on R2 s.t Tsquare=0 but T=/=0

If you could type more clearly, that would be helpful in trying to understand your problems. They seem interesting, though.


induction
number theory solved
number theory

a,b,c are given positive integers such that a is not equal to b.
Prove that if there exists infinite positive integers n such that a^n=b^n(mod c^n), then c|a and c|b

Let p be a prime s.t. p|c. Then pn|an-bn for infinitely many n. Assume p doesn't divide b. Then b is invertible in Zp, and then there is a k s.t. pn|kn-1 for infinitely many n (we take k s.t. k=a*b-1 in Zp). Let t be the order of k in Zp (with respect to the multiplication).

Then all those n's for which the sentence is true must be divisible by t. At the same time, the order of k in Zpn is t*pn-1 for all n (it's not that hard to prove). From all of this we get that n>=t*pn-1 for infinitely many n, which is false, so the assumption that p doesn't divide b is false, so p|b=>p|a, and we can repeat the process for a'=a/p, b'=b/p, c'=c/p.

[quote="grobber"]At the same time, the order of k in Zpn is t*pn-1 for all n (it's not that hard to prove)[/quote]

Could you please prove it?  :)

I think I've made a mistake:

k(p-1)pn-1=1(mod pn), so the order of k divides (p-1)pn-1. Moreover, it's easy to show (by induction; this I won't prove) that pn | ktpn-1-1, so the order of k in Zpn divides tpn-1. If the order has form t'pa with t'<t, then p | kt'-1, which means that t isn't the order of k mod p, contradiction. This shows that the order of k must be of the form tpa.

However, nothing tells us that a=n-1 for al n. For example, we might very well have p2 | kt-1, so the order of k in Zp2 is t, not tp. I think we must continue the solution in some other way.

Ok, so we have pn | kn-1 for infinitely many n. This is where we must get the contradiction, but it seems like my "solution" doesn't work. I'm thinking.. Meanwhile, I think this shouldn't be in the "solved problems" section :):(. I'll be back :D.

The solution to this problem can be found in the solutions of the first round of mathlinks contest.

Ok, I think I have it now:

Assume, as before, that t is the order of k in Zp. Assume, that pa|kt-1 and a is the maximal integer with this property. Then pa+1 | ktp-1=(kt-1)(kt(p-1)+..+kt+1). The last factor is p(mod pa), so if a>1 then we can show by induction that pa+n|ktpn-1-1 and a+n is the maximal exponent of p in ktpn-1-1, and the solution to our problem follows in a similar way to my first.. "solution" (:D).

If a=1 then we can do something very similar. We take n0 to be the first n s.t. there exists a>n s.t. pa|ktpn-1-1 and we continue as we did in the above paragraph (that was a special case, when n0=1).


Hey guys, I am currently learning probability in high school (Grade 12 Principles of Math in BC, Canada) and I was wondering if there are some great math websites that are very detailed in the explanation of probability, such as conditional probability. Thank you very much :)

lol hey, I just started probability in high school principles 12 in BC canada  :P
its easy, just read the textbook. and i guess ask teacher for help or if its something you really dont understand, ask around here.
btw, theres no awesome website that completely teaches you all you need to know about a subject in math, thats what textbooks are for  :)


Find all the integers $ n$ such that the number $ log_ {2n\minus{}1}(n^2\plus{}2)$ is rational.

I think $ n\equal{}5$ is the only solution.

First of all notice that $ n^2 \plus{} 2 > 2n\minus{}1$. Therefore, $ 2n\minus{}1 | n^2 \plus{} 2$ is a necessary condition.

We wish to solve $ n^2 \plus{} 2 \equal{} k(2n\minus{}1)$ for $ n,k \in \mathbb{Z}$. Reordering and solving quadratic for $ n$ we have $ n\equal{}k \pm \sqrt{k^2 \minus{}k \minus{}2} \equal{} k \pm \sqrt{(k\minus{}2)(k\plus{}1)}$

Notice that $ (k\minus{}2,k\plus{}1) \le 3$. Therefore, $ k\minus{}2$ and $ k\plus{}1$ are both perfect squares or 3 times a perfect square.
There are two solutions $ k\equal{}3$ and $ k\equal{}2$.

From this we obtain 3 potential solutions : $ n\equal{}1,2,5$. Only $ n\equal{}5$ works.


Let $ABC$ be an equilateral triangle of side $s$.  Construct a circular triangle as follows.  Consider the circular arc from $A$ to $B$ with center of the arc at $C$.  Another arc is from $B$ to $C$ with center at $A$.  A third arc is from $C$ to $A$ with center at $B$.  Find the formula for the area of the circular triangle, i.e., the area enclosed by the three circular arcs.
\[ \text{(A)}\ \sqrt{3}s^2 \qquad \text{(B)}\ \frac{\pi}{3}s^2 \qquad \text{(C)}\ \left( \frac{\pi}{2} + \frac{\sqrt{3}}{2} \right)s^2 \qquad \text{(D)}\ \left( \frac{\pi}{2} - \frac{\sqrt{3}}{2} \right)s^2 \qquad \text{(E) None of these answers} \]

[hide]The area of each "football" shaped piece is equal to $\frac 16$ of the circle minus the area of the equilateral triangle because each arc is $60^\circ$. This is equal to $\frac 16 s^2 \pi - \frac 14 s^2 \sqrt 3$. Then, we multiply that by 3 and add the equilateral triangle back in to get $s^2(\frac 12 \pi - \frac 12 \sqrt 3)$.[/hide]