floor function
number theory
prime factorization

Hello,

Find all positive integers where n!/[(n-k)! * k!] =5005

you can factor $ 5005 \equal{} 5\cdot 7\cdot 11\cdot 13$. at the very least, this tells us that $ n\geq 13$.

[url=http://www.artofproblemsolving.com/Wiki/index.php/Factorial#Prime_Factorization]this[/url] looks like it will probably be very useful

as well as the fact that $ \lfloor x \rfloor \plus{} \lfloor y \rfloor \leq \lfloor x\plus{}y \rfloor \leq \lfloor x \rfloor \plus{} \lfloor y \rfloor \plus{}1$

to restate the problem with this new information, you need to find $ n,k$ such that for every prime $ p\leq n$, not equal to $ 5,7,11,13$ and natural number $ r$ we have

$ \left\lfloor \frac{n}{p^r} \right\rfloor \equal{} \left\lfloor \frac{n\minus{}k}{p^r} \right\rfloor \plus{}  \left\lfloor \frac{k}{p^r} \right\rfloor$

and

$ \left\lfloor \frac{n}{p^r} \right\rfloor < \left\lfloor \frac{n\minus{}k}{p^r} \right\rfloor \plus{}  \left\lfloor \frac{k}{p^r} \right\rfloor$

for $ p \equal{} 5,7,11,13$.

i found a pair of symmetric solutions like this, not sure how to show that they are the only solutions

[hide="solutions"] $ \binom{15}{6},\binom{15}{9}$[/hide]

[quote="dgreenb801"]Of couse $ {5005 \choose 1}$ and $ {5005 \choose 5004}$ are also solutions.[/quote]

Yeah, don't know why I forgot that one. I ran a quick script and those four are the only solutions. Still no idea about showing uniqueness properly.


Question the first:
Do you recommend retaking classes? For example, I just finished the Olympiad Geometry class. If I took it again when you next offer it, would the material/questions/etc be too redundant to get anything out of it?

Question the not-quite-first:
If I want to take a class but I will miss a good portion of it (about 4-5 weeks), is it still worth taking? For example, the AIME/Olympiad counting class falls during the 5 weeks I am gone for mathcamp. Would you recommend reading the transcripts from the missed classes or just forfeiting the entire class?

[quote="Rep123max"]I know you want an answer probably from Mr. R or Mr. C, but I know from taking classes that missing more than one can be kinda bad. I missed two Number Theory Classes and now I am trying to learn those so I can get back on schedule, but it is very hard. Somehow, stuff keeps piling up and I dont have time to go over it. But since its during the summer, you probably will have a lot more free time so I would say that it would be feasible to learn information even if you miss 5 classes.[/quote]

Depends on the class. I missed about 4-5 of those Intermediate Counting Classes (now named AIME/Olympiad Counting I believe) due to conflicts with Youth in Government trials and I still think it's one of the clases I learned the most out of.

[quote]Depends on the class. I missed about 4-5 of those Intermediate Counting Classes (now named AIME/Olympiad Counting I believe) due to conflicts with Youth in Government trials and I still think it's one of the clases I learned the most out of.[/quote]

good to know

But that could be because that was the best class that has been on this site (in my opinion anyways). For people who were at that level, or even a bit above it, it was just amazing. Ive taken 4 classes, with that being my first, and it is still the best experience ive had learning (excluding self teaching).

[quote="jelyman"]But that could be because that was the best class that has been on this site.[/quote]

ya true...that COULD have something to do with it :-D ...lol...it most certainly was VERY good... :-) ...

Thanks for the props on the counting class.

As for missing classes - it highly depends on whether or not you'll be able to follow along with the transcripts or whether you'll try to catch up on everything at once in the end.  If it's a matter of 'I can't make it to the class, but at some point during the week, I'll take a look at the transcript and catch up,' you should be OK - provided you actually do it.  We have many students take whole courses this way.  For those who do the work of reading the transcripts later, it works just fine.

As for re-taking courses, it largely depends on how challenging the class was for you the first time.  It would certainly help reinforce the concepts of the class from the first time around, and we've had a few students do just what you suggest.  You could always try the first couple classes again, and if you find you've mastered it enough the first time around that you don't need the reinforcing, then drop the class for a full refund.

If you know that you will be missing a week due to travel or something else can you get the lesson plan or problems before you go out for that week?

Generally, you can find out what subjects we will discuss, but usually you'll have to wait until afterwards to get specific questions that are addressed in the class.


Is there a power of $2$ such that it is possible to rearrange the digits giving another power of $2$?

[quote="Robert Gerbicz"]It is impossible. Indirectly suppose that it is possible to rearrange the digits of $ {2}^{n}$ to give another larger power of two, we can assume that this is $ {2}^{n+c}$ where $ 0<c<4$, because $ {2}^{n+4}>10*{2}^{n}$ so this has got more digits than $ {2}^{n}$. But if we rearrange the digits of a number then it is giving the same remainder modulo 9, because this is the sum of the digits of the number modulo 9. So here $ {2}^{n}\equiv{2}^{n+c}\; modulo \; 9$, from this $ {2}^{n}*({2}^{c}-1)\equiv 0 \; modulo \; 9$ and this is impossible, because $ {2}^{n}$ isn't divisible by 3 and $ {2}^{c}-1$ isn't divisible by 9.[/quote]


[quote="zhaoli"]中国被大家公认为数学竞赛的一个强国，中国数学奥林匹克竞赛的试题也被认为是世界上最难的题目，但是在IMO试题中，中国被选中的题目?是很少，?有??几年?有常庚哲，??章教授的共3?题目被选中。近年?，甚至都没有??供备选题(从Mathlink的download部分中近年shortlist手册的扉页?以看?这一现象)，这是为什么呢？而且，没有年轻人(?韩国的Hojoo Lee仅为20几?，?已?为IMO,APMO??供多?试题)??供题目，有?知?这其中的缘由？[/quote]
I think this is because of the beaty of the problems. Maybe I take problem 4&6 from the just-past cmo to explain this.
For problem 4, indeed, it's very hard. However, the problem itself is extremely ugly-see how complicated the ineq is! The official solution is not any better-it uses Abel's Ineq and takes the first 3 terms in the binomial expansion. This type of solutions are considered as the most ugly one by the most poeple, which is unlike from the nice and simple solutions of the ineq in imo.
For problem 6, nothing special but just to take mod repeatedly. It sounds ridiculous to me....

中国的题目很?美好...你看CMO的题目 就是硬算..想了???样的办法折腾?折腾去..最?把答案?出?...没有数学的味?.

[quote="zhaoli"]中国被大家公认为数学竞赛的一个强国，中国数学奥林匹克竞赛的试题也被认为是世界上最难的题目，但是在IMO试题中，中国被选中的题目忴是很少，忪有忿几年剿有常庚哲，濎房章教授的共3鿓题目被选中。近年濥，甚至都没有濿供备选题(从Mathlink的download部分中近年shortlist手册的扉页忯以看觿这一现象)，这是为什么呢？而且，没有年轻人(僿韩国的Hojoo Lee仅为20几岿，忴已绿为IMO,APMO濿供多鿓试题)濿供题目，有谿知鿓这其中的缘由？[/quote]

这里有Hojoo Lee向IMO,APMO提供的试题及备选题

http://my.netian.com/~ideahitme/orange.html


Find $ f(2)$ if $ f(x)\equal{}\frac{(5x)}{(x^2\minus{}1)}$. Express the answer as a common fraction.


Let f be defined on the natural numbers as follows: $ f (1) = 1$ and for $ n > 1$,
$ f (n) = f ( f (n-1))+f (n-f (n-1))$. Find, with proof, a simple explicit
expression for $ f (n)$ which is valid for all $ n = 1,2, . . .$

[hide]This is relatively simple by induction.
f(1)=1 is the base case.
Assume f(n)=n.
f(n+1)=f(f(n))+f(n+1-f(n)), but f(n) is n and f(1) is 1, so f(n+1)=f(n)+f(1)=n+1.
Thus, the inductive step is complete.[/hide]


Can $ 1$ be written as a sum of $ 2005$ different terms of the form $ \frac{1}{3n\minus{}1},$ where $ n$ is a positive integer?

not, because $ \frac{1}{3n\minus{}1}\equal{}\minus{}1\mod 3$ and $ 2005\equal{}1\mod 3$, $ (\minus{}1)*1\not \equal{}1\mod 3.$


What is the smallest positive integer which uses only the digits 0 and 1 and is divisible by 225 (Note: your answer must be exact)?

the answer is [hide="this"] 11,111,111,100 [/hide]but i have no idea how to get it

225=9*25. Now use your divisibility rules  :)

Ooooohh. I get it, it has to be divisible by 9, so thats why there are 9 1s. and it has to be divisible by 25, so there are two 0s at the end.

Thanks!


[b]Grigory Perelman (THE GUARDIAN | EL MUNDO)[/b]

[b]LONDRES.[/b]- Es tal vez el hombre más inteligente del planeta: un genio enigmático y dado a recluirse que asombró al mundo académico al afirmar haber resuelto uno de los problemas más difíciles de las matemáticas. Es candidato al equivalente al Nobel de las matemáticas por su trabajo sobre las posibles formas del universo. Pero los rumores aseguran que el brillante matemático ruso desdeñará el grandioso galardón que tanto codician otros científicos.

Desde que [b]Grigory Perelman[/b] revelara la solución en 2002 a un problema matemático de un siglo de antigüedad, se ha visto sometido a un escrutinio sin precedentes por parte de las más despiertas mentes académicas. Ninguna ha logrado encontrar un solo error.

El próximo martes se espera que sea premiado con una [b]Medalla Fields[/b], el máximo galardón matemático, en el [b]Congreso Mundial de las Matemáticas[/b] que se celebra en Madrid. Pero incluso virtuosos de las matemáticas como John Nash, le describen como "poco convencional".

Este genio ha dicho que rechazará el premio de 782.000 euros ofrecido por un instituto privado de EEUU, que sería suyo de confirmarse la resolución del problema. Y los niveles más elevados del colectivo aseguran que no aceptará la medalla, galardón que es sólo para matemáticos menores de 40 años. Perelman ya los ha cumplido, por lo que es el último año que opta a tal premio.

También ha rechazado otras destacadas distinciones, alegando que no creía que el comité seleccionador del ganador estuviese lo bastante calificado para juzgar su trabajo. "Creo que es una persona muy poco convencional. Está en contra de todo lo que implique fausto e idolatría", ha
declarado Arthur Jaffe, de la Universidad de Harvard.
Se sabe poco de Perelman, que se niega a hablar con los medios de comunicación.

Nació el 13 de junio de 1966 y su prodigioso talento le llevó a matricularse en la escuela de San Petersburgo, especializada en física y matemáticas. [b]A los 16 años, ganó una medalla de oro en las Olimpiadas Internacionales de Matemáticas - IMO[/b], un concurso para escolares. Tras acabar su doctorado en San Petersburgo, trabajó en el Instituto Steklov para las matemáticas, antes de irse a EEUU a finales de los 80.

Regresó al Steklov hace 10 años para trabajar en su demostración de la forma del universo.

El mundo de las matemáticas se revolucionó en 2002 al saber de su trabajo sobre el problema enunciado por el matemático [b]Jules Henri Poincaré en 1904[/b]. La conjetura, difícil de entender para los neófitos, ejercitó a muchas mentes deslumbrantes del siglo XX.

El problema trata de la geometría de los espacios multidimensionales y resulta clave para la topología. Perelman asegura haber resuelto una versión general del problema, denominado la [b]conjetura de la geometrización de Thurston[/b], del que la conjetura de Poincaré es un caso especial.

"Se trata de un problema fundamental tanto en matemáticas como en física, dado que busca la comprensión de la forma que pueda tener el universo", explica Marcus Du Sautoy, de la Universidad de Oxford. "Su definición es muy complicada. Mucha gente ha anunciado con anterioridad pruebas de su resolución, que han resultado falsas".

La obsesión con el problema, compartida por varios cerebros de las matemáticas, ha sido jocosamente denominada 'Poincaritis'. Pero Perelman parece haber tenido éxito. "Durante muchos meses, incluso años, la gente está convencida de su argumento", comenta Nigel Hitchin, profesor de matemáticas en Oxford. "Creo que la solución es correcta".

[b]473 páginas [/b]

Incluso la forma de anunciar su demostración, que le costó ocho años de trabajo, resultó inusual.

En lugar de publicarla en una revista especializada, envió tres manuscritos a un archivo 'on line' de textos matemáticos. "Fue necesario rellenar muchos detalles, lo que produjo peleas por decidir quién fue el primero en completarlo", afirma Hichin. El documento más reciente en el que resuelve su demostración tiene ¡473 páginas!

En juego está mucho más que una aclamación profesional. En el año 2000, [b]el Instituto Clay de Boston[/b], una organización de investigación matemática privada, estableció los siete '[b]problemas del Milenio[/b]', cada uno de ellos con una recompensa de un millón de dólares para quien los solucione.

La conjetura de Poincaré es una cosa, y el dinero, en el que Perelman no tiene interés, otra.

"Corren muchas bromas que sugieren que tener un millón de dólares en San Petersburgo es peligroso", comenta Hitchin.

Nadie sabe qué ocurrirá si se hace con la medalla. "Si la rechazara, sería un insulto", afirma Du Sautoy. Pero es poco probable que a Perelman, que hace poco renunció a su puesto académico, le preocupe ofender a sus coetáneos. "Se ha marginado de la comunidad matemática", continua Du
Sautoy. "Se ha desilusionado con las matemáticas. Su mayor recompensa es la solución del teorema".

Saludos,

[quote="José"]Es decir que, oficialmente, su solución es aceptada en la comunidad matemática? Asi de oficial como la prueba de Wiles?[/quote]

Me autorespondo al final del segundo párrafo de este [url=http://www.icm2006.org/dailynews/fields_perelman_info_en.pdf]link[/url]. Todavía no hay nada "100% oficial" ni de la Conjetura de Poincaré ni de su generalización.

[quote="hucht"]Pues aqui algo de umor para darle otro color a esta celebracion

Ustedes saben porque a los mejores matematicos se les da elo premio Field (maxima distincion) y porque no existe un premio nobel en Matematica?????

jejeje  :D[/quote]

Resulta que Nobel fue el que inventó la dinamita, la cual le dio mucho dinero... con un poco de cargo de conciencia por su invento, creó la fundación nobel, que premiaría a los más destacados cientificos de todas las áreas, MENOS la de matemáticas, puesto que su mujer le había engañado con un matemático, JAJAJA  :P  :rotfl:

Saludos  :lol:

[quote="hucht"][quote="José"]en serio???? o es solo un rumor?[/quote]

Es verdad[/quote]

Que gran anécdota! Jajaja increíble! siempre me pregunté porqué no había Nobel de matemática... gracias!

[quote="Dave Vather"]De hecho.... creo que no se puede mostrar que sea verdad dicha anécdota... Pero suena a algo muy divertido para contar :wink:[/quote]

Si, obvio, pero como anécdota o "rumor", está muy bueno como para contarlo

Me refería más bien a que creo que es falsa, pero invariantemente, tienes razón: para contarla está muy buena

[color=blue]Dicho sea de paso…

Ya no corren noticias del señor Perelman. ¿Qué fue al final de él? ¿Recibió finalmente su medalla?[/color]

jeje, no, como era de esperarse... no se si la rechazo o qué, pero supe que el no la recibio... creo que se la dieron a terrence tao (el chiquito de australia que en los 80's gano bronce, plata y oro en las imo's) y a otros dos matematicos...


function
logarithms
calculus
calculus computations

Show that the succession
$ f_n(x)\equal{}\frac{1}{1\plus{}2^x\plus{}3^x\plus{}...\plus{}n^x}$ converges uniformly on $ \mathbb{R}$.
Discuss the convergence of the series
$ \sum_{n\equal{}1}^{\plus{}\infty}f_n(x)$.

[quote="Spasty"]Show that the succession
$ f_n(x) \equal{} \frac {1}{1 \plus{} 2^x \plus{} 3^x \plus{} ... \plus{} n^x}$ converges uniformly on $ \mathbb{R}$.
Discuss the convergence of the series
$ \sum_{n \equal{} 1}^{ \plus{} \infty}f_n(x)$.[/quote]
for $ \minus{} 1\geq x < \plus{} oo$ it's obviusly because  because your function wiil coverge to $ 0$ because
we have
$ f_{n}(x) < \frac {1}{\ln n \plus{} \gamma \plus{} o(1)} \minus{} > 0$

and for $ x < \minus{} 1$ we have $ f_{n}(x)_{n \minus{} > \plus{} oo} \minus{} > \frac {1}{\zeta (x)}$ and after is obviusly

seccond question:
for $ x>1$ serie converge and for $ x\leq 1$ serie diverge


inequalities
calculus
inequalities proposed
algebra
High school olympiad

Hmm, I guess you meant

$\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq a+b+c$.

Oh yes, nice joke.  ;)

  Darij

By Cauchy Schwartz
$(\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b})(a+b+c)=(\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b})[(b+c)+(c+a)+(a+b)] \geq  (a+b+c)^2$
and the result follows

By Chebyshev and then Nesbitt,

$LHS\geq \frac {1}{3}\left(\sum a\right)\left(\sum \frac {2a}{b+c}\right)\geq \sum a$

Indeed very easy! I killed this in 10 seconds.

$2\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right) \geqq 2\cdot \frac{(a+b+c)^2}{(b+c)+(c+a)+(a+b)}=a+b+c$ Q.E.D. :lol:

How can you solve it in such a very short time, spider_boy? ;)

I have for this inequality at least five solutions.For example you can use Chebysev,C.B.S,AM-GM,Jensen or as it follows:

     a^2/b+c=(a(a+b+c)-(ab+ac))/b+c and by calculus and well known inequality of Nesbitt the conclusion follows. :P

Here comes two other solutions:

(1) Without loss of generality, we can assume [tex] a \geq b \geq c [/tex], since the inequality is symmetric. Then [tex]a^2 \geq b^2 \geq c^2 [/tex] and [tex] \frac {1}{b + c} \geq \frac {1}{a + c} \geq  \frac {1}{b + c} [/tex]. Therefore, by the rearrangement principle, [tex] \frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq \frac{b^2 + c^2}{b+c} + \frac{c^2 + a^2}{c+a} + \frac{a^2 + b^2}{a+b} = [/tex]
[tex] = (b + c) + (c + a) + (a + b) - \frac{2bc}{b+c} + \frac{2ca}{c+a} + \frac{2ab}{a+b} \geq a + b + c [/tex], since [tex] \frac{b + c}{2} \geq \frac{2bc}{b + c} [/tex] etc.

(2) If we multiply [tex] a, b, c[/tex] with a positive number [tex]t[/tex], the inequality does not change validity. Therefore, we can assume [tex]a + b + c = 1 [/tex]. Now, let [tex] f(x) = \frac{2x^2}{1 - x} [/tex]. Via calculus we then see Jensen coming, and [tex]f(a) + f(b) + f(c) \geq 3f(\frac{a + b + c}{3}) = 1[/tex].

[quote="ehku"]Do :
[tex]\frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b} \ge \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}[/tex][/quote]

Posted many times... once by me.

Chebychev and Nesbitt (again)

$\sum{\frac{a^n}{b+c}} \ge \frac{1}{3}\sum{a^{n-1}}\sum{\frac{a}{b+c}} \ge \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}$

Okie! But there is not only one solution !

Prove that
  [tex]\sum \frac{a^n}{pb+qc} \ge \frac{a^{n-1}+b^{n-1}+c^{n-1}}{p+q}[/tex]
a, b, c, p, q >0.

[quote="ehku"][hide]We have :
[tex]\frac{a^2}{b+c} + \frac{b+c}{4} \ge 2\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}} =a[/tex]
or [tex]\frac{a^2}{b+c} \ge a -  \frac{b+c}{4} [/tex].Do :[tex]\frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b} \ge \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}[/tex][/hide][/quote]
${\frac{a^2}{b+c} + \frac{b+c}{4} \ge 2\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}} =a\Rightarrow\sum\frac{a^2}{b+c} \ge \sum(a -  \frac{b+c}{4})}=\frac{1}{2}\sum a.$

or  $\frac{a^2}{b+c} \geq \frac{4a-b-c}{4}\Rightarrow\sum\frac{a^2}{b+c} \geq \sum\frac{4a-b-c}{4}=\frac{1}{2}\sum a.$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=3028019

[b]Tangent Line method[/b]

Let $a+b+c=1, 0< a,b,c< 1$.The inequality is equivalent to

$\sum_{cyc}\frac{2a^2}{1-a}\geq 1$.

Let $f(x)=\frac{2x^2}{1-x}$.Then

$f(x)\geq \frac{5}{2}x-\frac{1}{2}$
$\Leftrightarrow$
$(3x-1)^2\geq 0$

This holds when $0< x< 1$.So

$f(a)+f(b)+f(c)\geq \frac{5}{2}-\frac{3}{2}=1$ :D

[quote=arccosinus]Let $a,b,c$ be positive real numbers. Prove that \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq a+b+c\](this is, of course, a joke!)

[b]EDITED with exponent 2 over c[/b][/quote]
See also here
B-II. Inequalities  , Po-Shen Loh , 15 June 2004:
8. ([b]88 Friendship Competition[/b]) For $a, b, c > 0:$
$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{a+b+c}{2}.$$

Let $a, b$, and $c$ be positive real numbers. [url=https://artofproblemsolving.com/community/c6h2088622p15081740]Prove that[/url]$$\frac{a^2}{b+c } + \frac{b^2}{c+a} + \frac{c^2}{a+b+ \frac{\left(a-b \right)^2}{4(a+b)}} \ge \frac{a+b+c}{2} $$

titu's lemma: $\sum_{cyc} \frac{a^2}{b+c} \geq \frac{(a+b+c)^2}{2(a+b+c)} = \frac{a+b+c}{2}$, multiply by 2 and done

I used Cauchy Schwarz, one term being the LHS, and the other being the cyclic sum of 2(a+b).

[quote=BestChoice123][quote=arccosinus]Let $a,b,c$ be positive real numbers. Prove that \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq a+b+c\](this is, of course, a joke!)

[b]EDITED with exponent 2 over c[/b][/quote]

Does anyone know the source of this stronger one?

For $a,b,c>0$ Prove that $$\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq \sqrt{3(a^2+b^2+c^2)}$$[/quote]

This one is stronger ( I found it independently in 2018 but it seems that it was posted by [b]Cezar Lupu[/b] earlier than that...)
$$\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a} \geq \sqrt{3(a^2+b^2+c^2)}$$
[url]https://artofproblemsolving.com/community/c6h1761159p11513999[/url]

[quote=bel.jad5][hide=*][quote=BestChoice123][quote=arccosinus]Let $a,b,c$ be positive real numbers. Prove that \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq a+b+c\](this is, of course, a joke!)

[b]EDITED with exponent 2 over c[/b][/quote]

Does anyone know the source of this stronger one?

For $a,b,c>0$ Prove that $$\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq \sqrt{3(a^2+b^2+c^2)}$$[/quote][/hide]

This one is stronger ( I found it independently in 2018 but it seems that it was posted by [b]Cezar Lupu[/b] earlier than that...)
$$\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a} \geq \sqrt{3(a^2+b^2+c^2)}$$
[url]https://artofproblemsolving.com/community/c6h1761159p11513999[/url][/quote]
[url]https://artofproblemsolving.com/community/c6h43001p271771[/url]
[url]https://artofproblemsolving.com/community/c6h1438484p8166663[/url]

[quote=BestChoice123][quote=arccosinus]Let $a,b,c$ be positive real numbers. Prove that \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq a+b+c\](this is, of course, a joke!)

[b]EDITED with exponent 2 over c[/b][/quote]

Does anyone know the source of this stronger one?

For $a,b,c>0$ Prove that $$\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq \sqrt{3(a^2+b^2+c^2)}$$[/quote]

$$ LHS-RHS=\sum \frac{(a-b)^2(2a^2+2b^2+3(ab+bc+ca))}{2(a+b+c)(b+c)(c+a)} + [(a-b)^2+(b-c)^2+(c-a)^2]\frac{\sqrt{3(a^2+b^2+c^3)}-(a+b+c)}{2(a+b+c) (a+b+c+\sqrt{3(a^2+b^2+c^2)})}\geqq 0$$

[quote=BestChoice123]

For $a,b,c>0$ Prove that $$\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq \sqrt{3(a^2+b^2+c^2)}$$[/quote]

$$\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq\sqrt[5]{81(a^5+b^5+c^5)}$$ for positives $a$, $b$ and $c$ a bit of stronger.
There is a nice proof.

Let $a,b,c$ be positive real numbers. Prove thatLet $a,b,c$ be positive real numbers. Prove that$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{\sqrt{3(a^2+b^2+c^2)}}{2}\geq \frac{a+\sqrt{2(b^2+c^2)}}{2}$$
$$a^2+b^2+c^2\geq\frac{1}{3} \left(a+\sqrt{2(b^2+c^2)}\right)^2\iff 2a^2+b^2+c^2\geq 2a\sqrt{2(b^2+c^2)}$$

[hide=Solution]By Titu, we have $\frac{a^2}{\frac{b+c}{2}}+\frac{b^2}{\frac{c+a}{2}}+\frac{c^2}{\frac{a+b}{2}} \geq \frac{(a+b+c)^2}{a+b+c}=a+b+c$, as desired.[/hide]

Let $a,b,c$ be positive real numbers. Prove that$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{a+\sqrt[n]{2^{n-1}(b^n+c^n)}}{2}$$Where $5\geq n\in N^+.$

[b]#27[/b]
Let $a,b,c$ be positive real numbers. Prove that$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{a}{2}+2\sqrt[3]{\frac{b^3+c^3}{2}}$$

[quote=sqing]Let $a,b,c$ be positive real numbers. Prove that$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{a+\sqrt{2(b^2+c^2)}}{2}$$[/quote]

[quote=sqing]Let $a,b,c$ be positive real numbers. Prove that$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{a+\sqrt[n]{2^{n-1}(b^n+c^n)}}{2}$$
Where $5\geq n\in N^+.$

[b]#27[/b][/quote]
For $n=4$ there is a nice solution.

Let $a, b,c>0.$ Prove that
$$\frac{a+k}{b+c+2k} + \frac{b+k}{c+a+2k} +\frac{c+k}{a+b+2k} \geq \frac{3}{2} $$
$$\frac{a(a+k)}{b+c+k} + \frac{b(b+k)}{c+a+k} +\frac{c(c+k)}{a+b+k} \geq \frac{a+b+c}{2} $$
Where $k\geq 0.$

[hide=#20][quote=sqing]For $a, b, c > 0:$
$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq \frac{a+b+c}{2} $$
$$\iff$$
$$\frac{a }{b+c} + \frac{b }{c+a} + \frac{c }{a+b} \geq \frac{3}{2} $$[/quote][/hide]
Let $a, b, c > 0$. [url=https://artofproblemsolving.com/community/c6h619569p3699275]Prove that[/url]
 \[\frac{a^2+2bc}{b+c}+\frac{b^2+2ca}{c+a}+\frac{c^2+2ab}{a+b}\geq \frac{3}{2}(a+b+c).\] $$\frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c$$
[url]https://artofproblemsolving.com/community/c6h1070208p4651461[/url]
[url]https://artofproblemsolving.com/community/c6h1438484p15117577[/url]

[hide='Solution']By T2's-Lemma: $$\sum_{cyc} \frac{2a^2}{b+c} \geq \frac{2(a+b+c)^2}{2(a+b+c)} \geq a+b+c$$[/hide]

The inequality is homogenous; WLOG $a+b+c=1$
It sufficies to prove for $a,b,c \in (0,1)$ that,
\[\frac{2a^2}{1-a} + \frac{2b^2}{1-b} + \frac{2c^2}{1-c} \geq 1\]
provided $a+b+c=1$.
Consider $f(x)=\frac{2x^2}{1-x}$. Note that $f''(x)=\frac{4}{(1-x)^3}$ so $f(x)$ is convex on this interval ($(0,1)$).  Jensen tells us that,
\[\frac{2a^2}{1-a} + \frac{2b^2}{1-b} + \frac{2c^2}{1-c} \geq 3f(\frac{1}{3})=1\]
and we are done

Trivial by holder
$$(\sum_{cyc}\frac{a^2}{b+c})(\sum_{cyc}b+c) \geq  \left(\sum _{cyc}a\right)^2$$
$$\implies 2\sum_{cyc}\frac{a^2}{b+c} \geq \sum_{cyc}a$$

[quote=nathantareep][quote=oolite][quote=nathantareep]\[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b}>\frac{4(a+b+c)^2}{2(a+b+c)}=2(a+b+c)>a+b+c\][/quote]Sorry, no.  See [url=https://artofproblemsolving.com/community/c6h33198p15095128]#18[/url] for Titu done properly.[/quote]

Ohh I see it was a little error,I have corrected, it see now[/quote]

Well actually; If we were to use C-S in Engel form on $\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} $ :

[hide=step-by-step]
$\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} = \frac{\sqrt{2a^2}^2}{b+c} + \frac{\sqrt{2b^2}^2}{c+a} + \frac{\sqrt{2c^2}^2}{a+b} \geq  \frac{(a\sqrt{2} + b\sqrt{2} + c\sqrt{2})^2}{2(a+b+c)} = \frac{(\sqrt{2}(a+b+c))^2}{2(a+b+c)} \geq a+b+c$
[/hide]

Let $a,b,c$ be positive real numbers. Prove that\[\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} \geq  \frac{\sqrt{3( a^2+b^2+c^2)}}{2}\]
Let $a,b,c,d$ be positive real numbers. Prove or disprove $$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{d+a}+ \frac{d^2}{a+b}  \geq \frac{a+b+\sqrt{2(c^2+d^2)}}{2}$$

Isn't simple Titu enough?

\[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} =\frac{a^2}{\frac{b+c}{2}} + \frac{b^2}{\frac{c+a}{2}} + \frac{c^2}{\frac{a+b}{2}}\]
It is easy to see that the denominators add up to $a+b+c$. By [b]Titu's Lemma [/b], we have:

\[\frac{a^2}{\frac{b+c}{2}} + \frac{b^2}{\frac{c+a}{2}} + \frac{c^2}{\frac{a+b}{2}} \ge \frac{(a+b+c)^2}{a+b+c} = a+b+c\]

So, we are done? $\blacksquare$

[quote=wardtnt1234]this problem must be a joke. It's direct by Titu's lemma[/quote]

[hide=not important] There are even funnier jokes: [url]https://artofproblemsolving.com/community/q1h56230p347811[/url]

[quote=arccosinus]Let $a,b,c$ be positive real numbers. Prove that \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq a+b+c\](this is, of course, a joke!)[/quote]
A joke for a joke ...

WLOG $a\le b\le c$.  Write $b=a+x,c=a+x+y$ for $x,y\ge 0$.  Substituting and clearing denominators (don't try this at home, kids), we need to prove
\begin{align*}
12a^2x^2 + 12a^2xy + 12a^2y^2 + 14ax^3+ 21ax^2y + 27axy^2 + 10ay^3&\\
+ 4x^4 + 8x^3y + 13x^2y^2 + 9xy^3 + 2y^4 &\ge 0,
\end{align*}which is obviously true   ;)


Everybody in my school who took the AMC took the B, and the teacher in charge sent them off. Several of us in the school know for sure that we made AIME, but the teacher still hasn't received any information, and hasn't gotten emailed back by those AMC director types he contacted. Has everybody else already been notified about AIME?  :maybe:

Yes, emails about the B-date scores were sent earlier this week.  All of the reports have gone out by either 2-day (early in the week) or 1-day Priority UPS.  An email with a UPS tracking number is generated and sent to the receiving teacher.   As of the end of business on Friday, all potential email addresses where this message could have been directed have been cleaned of these types of problems from teachers.  If your teacher has not received any information through any of these channels, then have your teacher call our 800 number (printed on everything we send and on our website) during business hours (Central time) on Monday.

If your teacher communicated with us, the problems should already be resolved.

Steve Dunbar
MAA Director, American Mathematics Competitions


function
calculus
integration
complex numbers
algebra
functional equation

The Riemann Zeta function is defined as

[tex]

\zeta (t) = \sum_{n=1}^{\infty} \frac{1}{n^t}

[/tex]

The "trivial zeroes" are supposed to be the negative even numbers.  But

[tex]

\sum_{n=1}^{\infty} \frac{1}{n^{-2}} = \sum_{n=1}^{\infty} n^2 \neq 0

[/tex]

Could someone explain this?

The Riemann Zeta function, as you wrote it, is only defined for complex numbers with real parts greater than 1.  You need to know something about "analytic extensions" to understand how it gets expanded to cover the rest of the complex plane.

I can actually tell you a little bit more about it, so it doesn't seem completely meaningless to you.  The function, as you defined it, is pretty clearly only going to be valid on the reals for number greater than 1, since the sum that you described when t = 1 diverges.   In the real numbers, there's nothing we can really do about this.  In the complex numbers, we see that the sum is still defined as long as t (which is now a complex number) has real part greater than 1.  But it's undefined (due to the same divergence problems) for the rest of the plane.  In that half-plane, however, it is what is known as an "analytic function."  Roughly, this just means that it is defined and integrable (and, as a consequence, differentiable) everywhere.  However, for complex functions, the property of differentiability is far more powerful than it is for real functions -- when you know something is analytic, you know a lot about it, including the fact that it is incredibly regular in many respects.  (It's much the same way that differentiability of a real function implies continuity, but the results of analyticity are far more powerful than just continuity.)

As a result of this nice property that the Zeta function has over half of the plane, it turns out that we can make a function that covers the entire plane and which exactly equals the values for that sum over the region we already know it on.  (This is true of any analytic function -- if you know its values in a nice region, you can find a new function that is analytic on the entire plane and which equals your given function on the region on which it was defined.)  Furthermore, this extension is unique.  Thus, we can find a new definition for the Zeta function that will allow it to cover the entire plane.  Under this new definition, we end up having "obvious" roots at all the negative even numbers.  Meanwhile, we end up with lots of other roots which seem to lie along the line [tex]\Re(z)=\frac{1}{2}[/tex].  The Riemann hypothesis claims that this is in fact the case, and that the Zeta function has no other zeroes.

So what is this other definition of the Riemann Zeta function that covers the whole plane?

Go to Mathworld and look up the entry on the Riemann zeta function.  Starting just above equation 13, it shows you how to extend it to the whole right half-plane.  Then it gives you links to see how to extend it to the whole plane after that.  I dunno if they'll make any sense to you, and I have to go so I can't look at them myself.

Ah.  I understand most of it untill we get to the little integrals that look like this:

\[
\oint
\]

Off the top of my head, I can't give a proof of the analytic continuation for $\zeta(s)$, but here's the way it works:

First, we extend $\zeta(s)$ to the half-plane $\Re(s)>0$ by turning it into the alternating $\zeta$ function:

\[\zeta_a(s)=\sum_{n=1}^\infty\frac{(-1)^n}{n^s}.\]

We do this by subtracting off the powers of 2 and then factoring. I'll leave the details to you; it's not hard to work it out.

After that, we start using magic. We define the function

\[\xi(s)=s(s-1)\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s).\]

$\xi(s)$ satisfies the functional equation $\xi(s)=\xi(1-s)$. Using the $\xi$ function, we can analytically extend the $\zeta$ function to the punctured plane $\mathbb{C}\setminus\{1\}$.


articles Princeton college Putnam MIT Harvard

Has anybody read this week's [url=http://www.time.com/time/magazine/article/0,9171,1226150,00.html][i]Time[/i] magazine cover story[/url]?

If so, what did you think of it?

I read the amount of it that was posted on CNN, which really just told me what the article was about.

I don't think going Ivy League is the only way you can get a good education. In fact, when I was applying for college, I chose between the University of Texas science honors program (the Dean's Scholars program, which admits around 30 people per year, I think) and Princeton. I chose Princeton because I got good financial aid and because of the name. I don't regret it.

My feeling is that a motivated student can get as good an education at most large universities as at the top schools in the nation. But the difference is that at those schools, the initiative has to come from the student. At Princeton, students can still get a mediocre education. But at Princeton, it's a matter of avoiding a good education, whereas at other schools, it's a matter of seeking a good education, if you know what I mean.

Also, there's a difference in the caliber of students. Some honors programs at large universities sort of isolate the honors students from all the others, which I would consider to be good for teh strong students (it might also be good for the weak students -- it makes them less inclined to cheat, ask for help, or get discouraged). One of the great things about Princeton is that [i]everyone around me[/i] is brilliant. This is to the extent that when I emailed people in my REU to introduce myself at teh beginning of the summer, what I considered a normal description of my activities of the past year was considered extremely arrogant by my audience. (This was slightly embarrassing.) When my Fourier Analysis class was half freshman and I got the lowest Putnam score out of all of my friends at Princeton, it doesn't feel like arrogance to mention those things. Another anecdote: a friend of mine was surprised one time when she came across the varsity basketball team returning from practice, and the crowd of "dumb jocks" was arguing about Nietzsche. Basically, what I mean is that this environment simply inspires excellence, effort, deep thinking. It makes a difference.

Finally, some people seem to think that top universities are overpriced. This is true at some places, I'm sure. But there are two things to keep in mind: (1) "You get what you pay for" -- there's a reason those places cost a lot; and (2) I am paying less for Princeton than I would have for University of Texas, even though I'd have had the UT National Merit scholarship (and Princeton doesn't have merit-based scholarships at all).

Newsweek also just posted a college related article along the same lines: [url=http://www.msnbc.msn.com/id/14325172/]25 New Ivies[/url].

[quote]Princeton turned down 4 of every 5 of the valedictorians who applied last year...[/quote]

So that's why I got rejected :P (jk)  Actually, reflecting on it, I believe Princeton made the right decision turning me down.  It's wouldn't have been all that great of a fit for me.  I'm just glad MIT accepted me :)

Ok, now what's the deal with this:

[quote]So how do the private consultants fit into all this? As many as 1 in 5 applicants to private four-year colleges get some kind of independent coaching, which can range in price from $\$$469 for Kaplan's three-hour consultation by webcam to $\$$36,000 for four years of hand holding offered by superconsultant Michele Hernandez.[/quote]

$\$$36,000???  That's insane.

Basically, my philosophy is that if you aren't having fun with what you're doing, [i]stop doing it and do something you like[/i]!  If you happen to be a match for an elite school, awesome.  If not, then realize that you wouldn't have liked it there anyway and there are other places that fit your personality much better.  Thus I definitely agree with the idea of looking for a college that fits you (instead of one that everyone raves about).

My own thoughts (self-serving, perhaps, but sincere) are [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=603176#603176]here[/url].

From what I've been told, the quality of education at the best canadian universities ( UofT, McGill, Waterloo, etc ), is the same quality as MIT or Harvard or any of the other ivy league schools. If this is true, then we really do have you beat up here  :P , since the price of going to waterloo for four years is less than two years of tuition at MIT.

No wonder there are so many of you flocking up to McGill. Must be more than just the wild parties?  :D

[url=http://www.slate.com/id/2066353/]Here[/url]'s an interesting article about the high sticker price of elite universities.

Very nice! I like the system, actually. I mean, I'm sort of biased because my family is very low income so I benefit quite a bit from it. But I like the idea that merit is for admission and then they just try to make sure you can afford to go there.

Dave, thanks for the link to the article. In my opinion so far the discussion has rather been limited to the view whether I should go abroad or to an elite college based on tuition fees. Thus I want to offer some random thoughts from my side.

From my point of view you roughly have to distinguish two cases. You are rather biased towards research [b](1)[/b] or you want to get straight into industry [b](2)[/b] and possibly make loads of money.

[b]Case 1:[/b] [i]<[/i]Don't just look at ranking without distinction whether the actual department in question is really good and whether there is a professor whose research interest I match. Younger professor are mostly more motivated and active. You probably get better direction on which approaches may be worth following. But with more experienced professors you often have a "brand mark" and get easier accepted at top-notch journals and conferences though your work might not be as good as in the first case. ;-) Some excellent professors deny professorships to elite colleges as they prefer to stay with their family where they are.

Of course, lots of the best professors will join. But they will more or less pay little attention to you until you are a graduate student. I mean low quality work and making fun of other colleges ends at M.Sc. level, researchers are smart people and know who does a good job. No matter which university they are at, whether they have a good web site, they or their professors are responsive etc.

Elite colleges have much more money to generate a hype attracting rich undergraduate students in order to offer many scholarship to really talented students. You see, some people (parents, alumni, funds) pay the money, the others secure the reputation by top class research.
At lower ranked universities you might have access to more interesting projects, e.g. collaborating with the NASA as mentioned in the article. Of course, most elite colleges offer standardized ways to take part in undergraduate research opportunities programmes. And the professors at smaller and possibly lower ranked universities might be more accessible so you can publish your first research results. But I think the good relationship is more important at graduate level. At the undergraduate level you probably benefit more from your possibly outstanding peer experience which often cannot be accomplished at lower ranked colleges.

Undergraduate curriculum is probably more less the same everwhere, just distinguished by the number of "special" courses you can take. Maybe the factor is whether you prefer great peer students or would enjoy the research experience with a professor at undergraduate level. As said before it comes down how much you make out of your options but as said before it is easier to take advantage of that at good colleges and it takes less initial effort there to get a project started. But in the end you first need to be admitted to the college of your choice. ;-) [i]>[/i]

[b]Case 2:[/b] [i]<[/i]I think the brand mark is a really a huge factor. I guess the last college on your cv, let it be from your B.Sc., M.Sc., MBA or Ph.D. degree, should be an elite college in order to get good jobs. Most people from the human resources don't bother which professor you studied with or how many publications you have.

They check the rankings and tell you whether you are any good. That is go for the universites which are recognised by industry. For high positions they probably receive huge piles of applications. In order to save money and work, they probably check the applications for positive/negative word lists and then everybody who did not study in the "right" college gets dismissed. After that the actual application process with interviews, tests etc. will start.

They mention that relationships became less important as new technologies and branches evolve. Well, I do not really agree. Look, recruiters get so many applications and don't know what to do about it. Then a friend whom you trust comes along and recommends you an applicant. Guess what, his chances of getting a job get a huge boost. It is the same with product advertisements in markets which are spammed with advertisements. What to do that people still listen to you though they get flooded with stimuli? Of course, it is buzz marketing. People who have many friends and are often bloggers talk about products and organize parties involving those products without getting paid. Well, it does not matter, they want to have something to talk about and often are interested in being a trend setter. Same goes for most tradition working fields as law. I don't think it is changing so much that relationsship do not count anymore. Same goes for triggering a start-up. You are in need of venture capital to get things started. Knowing people who are persuaded that you probably will offer a good return later cannot be a disadvantage. Having contacts, it is easier to learn about new opportunities. It is not by chance that online networks as the "open business club" generated such a surge of interest. It might not be always true. But the better the college the more influencing the contacts I assume. [i]>[/i]

I liked the article because it takes pressure from students to get into the most selective college and rather emphasizes issues that makes other colleges worth considering as well. Education is the best insurance to have a job later and to deepen your knowledge and problem-solving skills. In the end it is up to you how you will use your options. I wish you good joy and luck in achieving your goal.

[quote="Xevarion"][quote="rzsolt"] the price of going to waterloo for four years is less than two years of tuition at MIT. [/quote]
I'll probably pay about $\$$8,000-12,000 (US) for four years at Princeton. That's including room and board. How far would that get an international student Waterloo? $\frac{1}{2}$ year? Make sure your axioms are legitimate before you start proving theorems with them.[/quote]

My mistake. To be fair, the cost for myself to go to an american university IS substancialy higher than for me to go to a canadian uniersity. I didn't take into account the situation would reverse for american students  :blush: .

Unfortunately I forgot to purchase the journal. Does anybody have a subscription to the journal or can copy paste it here or pm it to me, please? The freely visible part does not cover the whole article for obvious reasons. :)

BTW I just bought the latest edition of the Newsweek offering a ranking from the global top 50 universities. Maybe I am wrong but ÍMO it seems pretty lazy just to average the [url=http://www.its.caltech.edu/~xuanluo/random/documents/worldrankings2005.pdf]THES[/url] and the [url=http://ed.sjtu.edu.cn/ranking.htm]Shanghai Jiao Tong University[/url] ranking which are pretty U.S. biased in the way they establish their criteria, using English citation indices, consider nobel prize winners which are just a tiny fraction of the researchers etc. Of course, I agree that the top U.S. institutions are absolutely worth being considering as top class but there are quite a few U.S. institutes among the places 10 - 50 that I would not necessarily set as top-notch. There are some UK and German universities which I would place beyond those institutes but well I am not making up the ranking...  :wink:


Not have seemed to be posted. On posting the problem, Valentine would inform us.
Shall we wait for a while. :)

People are supposed to get 2 weeks to solve 3 problems. It doesn't matter if it's a bit late, does it?

sure dosent  :blush:
sorry for the impatience

I'm sorry it's going to take a few hours more to prepare it. Have some patience please (probably the time will be 23.00 GMT).

No problem Valentin. :)  I am grateful that the 7th mathlink contest is actually being held. :)  I can sure wait. :)

[quote="Valentin Vornicu"]Have some patience please ([b]probably the time will be 23.00 GMT[/b]).[/quote]

sorry i didnot know it was for the same reason i intended to see the 6th edison


linear algebra
matrix
vector
superior algebra
superior algebra solved

Let $A_1,A_2,...,A_{n+1}$- matrices of the form $n\times n$. Prove that there is $n+1$ numbers $x_1,x_2,...,x_{n+1}$ don't equal to zero simultaneously such that the matrix $x_1A_1+...+x_{n+1}A_{n+1}$ is singular

This is really weird... Maybe you meant something else? This seems way too easy :?

The first column vectors of the matrices are linearly dependent, so there is a non-trivial linear combination of these vectors which equals the null vector. Just pick the coefficients in this linear combination. The resulting matrix will have a null first column, so it will be singular.

What "numbers" are we working with, by the way? Real? Complex? It's always necessary to specify that.

You solution is ok. I didn't mention the set we are working over, because in the problem condition it wasn't specified, i think it is real $R$
so - such an easy problem


There is a race consisting of 9 horses.In how many different ways can the horses finish 1st, 2nd, and 3rd?

What do you mean, which horse.  Is it just [hide] 9 P 3 [/hide]?

could it be something like [color=white]9C3[/color]?

[hide]9p3[/hide] is right.

Can you see why [hide]9c3[/hide] is not right?

9p3 is not really right, it is 9!, considering that it is possible (although not likely) that the horses tie. Are ties allowed?

generally in such questions it is assumed ties cannot happen.

Hmmm, should math problems really have to assume anything? They should clearly state exactly what the problem is.

If you would like go into the details of problem writing, then yes, problems ideally would state "and no ties are allowed."  However, it's standard practice in questions like this for there to be no ties allowed -- in fact, I can only remember 1 question I've ever seen that allowed ties in a competition.  Thus, as a problem solver, it is reasonable to assume that the problem writer meant to include the fact that there are no ties allowed.

I have seen more than one problem with ties allowed I think. One that comes to mind is from BAMM Test of Ingenuity 2002 (#9 or so if I recall correctly). I think that the writer should state whether or not ties are allowed because it is quite ambiguous otherwise.

Ties are not allowed.I seem to think it is 63.

[quote="scm007"]Are ties allowed?[/quote]

I don't think that there is any ambiguity here, as a matter of English usage. The phrasing "In how many different ways can the horses finish 1st, 2nd, and 3rd?" makes clear that there aren't ties. If there were a tie, one of the places would be missing. You can only have a finish with horses finishing first, second, and third if the three horses finish in an order without ties. (Think about it.)

I'm the verbal guy here. I frequently get flummoxed by math that a lot of the fourteen-year-olds here find routine, but I don't usually make unwarranted interpretations of written English.

Hope this helps!


quadratics
irrational number
number theory
relatively prime

let $ k $ and $ m $ are natural numbers, and $ \frac{1}{2} (\sqrt{k+4 \sqrt{m}} - \sqrt{k}) $ is integer.
prove that $ \sqrt{k} $ is rational and natural number.

[b]As k and m are natural numbers so √(k+4√m) > √k
So ½(√(k+4√m)- √k) is a positive integer number.
Suppose,
              [b]½(√(k+4√m)- √k) =x     such that x belongs to N

      Or, (√(k+4√m)- √k)=2x

      Or, √k=√(k+4√m)-2x

 √(k+4√m)>2x. Now if √(k+4√m) belongs to N then √k is a natural number and as all natural numbers are rational numbers so the proof is complete. So suppose √(k+4√m) does not belong to N. So it is either an irrational number or a fraction.
Suppose it is an irrational number. So √(k+4√m) ≠p/q such that p,q belong to N, q≠0, and p,q are relatively prime.
      (k+4√m)≠ (p/q)²

Or, 4√m≠ (p/q)²-k

Or, 16m≠ [(p/q)²-k]²

But 16m is a natural number. And therefore 16m can be expressed as the subtraction of two natural numbers. So if p/q is a natural number then it is possible to express 16m as [(p/q)²-k]²
So  √(k+4√m) is not an irrational number
Now suppose
So  √(k+4√m) is a fraction
So    √(k+4√m)=a/b such that a,b belong to N, b≠0 b>1 an a,b are relatively prime. So we can write

      (k+4√m)= (a/b)²
Or, 4√m=(a/b)²-k
Or, 16m =[(a/b)²-k]²
As (a/b)² is a fraction and k is natural number so  [(a/b)²-k]² is a fraction. But 16m is a natural number.
16m ≠[(a/b)²-k]²
So  √(k+4√m) is not a fraction.
So √(k+4√m) is a natural number.
So √(k+4√m)-2x is a natural number.
Therefore √k is a natural and a rational number.[Proved]

Please write as soon as possible confirming my proof correct or not.

 [/b]

[quote]So if p/q is a natural number then it is possible to  express 16m as [(p/q)²-k]²
[/quote]
p and q are relatively prime..right so p/q is in its lower term. then it is a fraction.
$(\dfrac{p^2}{q^2} - k)^2$ is not a subtraction of two natural numbers. but it is the quadratic of the substraction of fraction $\dfrac{p^2}{q^2}$ and natural number $k$.

But isn't it possible to write a natural number as p/q when p,q are relatively prime and  q=1. I haven't write that q≠1

Because according to the rule

p and q are relatively prime if and only if they have no common factor other than 1.

Besides this, in the second part of the proof I have shown that p/q cannot be a fraction.

 :o  :?  :roll:  :huh:

Please answer me as soon as possible.

of course it is posible for $\dfrac{p}{q}$ to be the natural number. but, it is not always the case, it is posible for $q$ has the value other than 1, you must prove that even $\dfrac{p}{q}$ is not natural number, $\sqrt{4 \sqrt{m}+k}$ still hold not irrational number. althought you think it is obvious.

BUT, according to your second part proof. You're right. The problem is you should write this part first. :)  so there is no danger of confusion.


trigonometry
parameterization
real analysis
real analysis unsolved

Determining the sum of

\begin{eqnarray}\sum_{k=1}^{\infty}3^{k-1}\sin^{3}\left(\frac{x}{3^{k}}\right)\end{eqnarray}

is a rather well-known problem in most Math training circles. Here you won't be asked to compute that sum.

You'll be asked, instead, to show whether the series in $ (1)$ [b]coverges uniformly[/b] or not.

What is the set on which we are to study uniform convergence? Probably not $ \mathbb R$, since the supremum of the $ k$th term on $ \mathbb R$ is $ 3^{k-1}$.

Hi there, [b]mlok[/b]!

You're to specify that set where [color=red][b]UC[/b][/color] holds or prove its inexistence... That's other way to state the question, my friend!

"the set" of uniform convergence is not a well-defined notion. The series converges uniformly on any compact set; if you specify any set on which the series converges uniformly, I can make a bigger one by adding in some $ [-M,M]$.

[quote="jmerry"]"the set" of uniform convergence is not a well-defined notion. [/quote]
Well, I used to give similar problems to my students in Leningrad. The rules of the game were that they should exhibit a family of sets $ \Omega_{t}$ such that the series/integral converges uniformly on every $ \Omega_{t}$ and every set $ \Omega$ of parameters for which one has uniform convergence is a subset of one of $ \Omega_{t}$. The nicer the family, the more points were given for the answer. "All compact sets" would correspond to $ \Omega_{t}=[-t,t]$ ($ t>0$) and so on. I think coquitao had a similar idea in mind ;)

[quote="fedja"]... I think coquitao had a similar idea in mind ;)[/quote]

Exactly so, [b]fedja[/b]. Heartfelt thanks for all the labour on my behalf, man.


calculus
integration
conics
ellipse
trigonometry
analytic geometry
graphing lines

Draw the perpendicular to the tangent line of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\ (a>0,\ b>0)$ from the origin $O(0,\ 0).$

Let $\theta$ be the angle between the perpendicular and the positive direction of $x$ axis. Denote the length of the perpendicular by $r(\theta).$

Calculate $\int_{0}^{2\pi}r(\theta )^{2}\ d\theta.$

I get that $r(\theta) = b$ and that the integral is $2\pi b^{2},$ but it can't right...

What did you mean by "from the origin"?

-Ben

[quote="benorin"][quote="kunny"]That means $O(0,\ 0).$[/quote]

but the perpendicualars to the tangent lines to the ellipse only go through the origin if it is a circle or if the tangent is drawn at x=a or x=-a[/quote]

Sorry,　I don't think so. :maybe:

[quote="FieryHydra"][quote="benorin"][quote="kunny"]That means $O(0,\ 0).$[/quote]

but the perpendicualars to the tangent lines to the ellipse only go through the origin if it is a circle or if the tangent is drawn at x=a or x=-a[/quote]
I think from the problem it is implied that the perpendiclar from the origin and the tangent line on the ellipse do not necessarily meet at some point on the ellipse. The point of intersection of those two lines could lie outside the ellipse. At least, that is how I understand the problem.[/quote]

That is what I want to say. Thank you for the explanation in replace of me, FieryHydra.

kunny

I got $\pi(a^{2}+b^{2})$. Is it correct, [b]kunny[/b]? I will post the solution later.

[quote="kunny"]Your answer is correct! :)

I'm looking forward to your solution, FieryHydra.[/quote]
Cool! :) Here is my solution.

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \Rightarrow \frac{dy}{dx}=-\frac{x}{y}\cdot\frac{b^{2}}{a^{2}}\Rightarrow \frac{dy}{dx}\Big|_{(a\cos t, b\sin t)}=-\frac{b\cos t}{a\sin t}.$

$\therefore$ Slope of the perpendicular = $\frac{a\sin t}{b\cos t}= \tan \theta$ (from the given condition.)
Or, $\tan t = \frac{b\tan \theta}{a}$ ... $(*)$

Now, equation of the perpendicular is $y = \left( \frac{a\sin t}{b\cos t}\right) x$ ... (I), and

the equation of the tangent line is $y-b\sin t =-\frac{b}{a\tan t}(x-a\cos t)$ ... (II)

If $(h,k)$ is the point of intersection of the above two lines, then solving those equations yields

$h = \frac{ab^{2}\sec t}{a^{2}\tan^{2}t+b^{2}}$ and $k = \frac{a^{2}b\tan t \sec t}{a^{2}\tan^{2}t+b^{2}}$ ... (this is fairly mechanical, so I won't write all the details)

Now, if $r(t)$ is the length of the perpendicular, we have $r(t)^{2}= h^{2}+k^{2}= a^{2}b^{2}\cdot \frac{\tan^{2}t+1}{a^{2}\tan^{2}t+b^{2}}$ ... (again the computation is mechanical)

Using $(*),$ we get $r(\theta)^{2}= \frac{b^{2}\tan^{2}\theta+a^{2}}{\sec^{2}\theta}.$

Finally, using symmetry, we have $\int_{0}^{2\pi}r(\theta )^{2}\ d\theta = 4 \int_{0}^{\frac{\pi}{2}}r(\theta )^{2}\ d\theta.$

$\therefore \int_{0}^{2\pi}r(\theta )^{2}\ d\theta = 4\int_{0}^{\frac{\pi}{2}}\frac{b^{2}(\sec^{2}\theta-1)+a^{2}}{\tan^{2}\theta+1}\, d\theta$

$= 4b^{2}\int_{0}^{\frac{\pi}{2}}\frac{\sec^{2}\theta}{1+\tan^{2}\theta}\, d\theta+2(a^{2}-b^{2})\int_{0}^{\frac{\pi}{2}}(1+\cos 2\theta)\, d\theta$

$= 4b^{2}\left( \tan^{-1}(\tan\theta) \right)\Big|_{0}^{\frac{\pi}{2}}+2(a^{2}-b^{2})\left( \frac{\pi}{2}+\frac12 \sin 2\theta \Big|_{0}^{\frac{\pi}{2}}\right)$

$= 2b^{2}\pi+(a^{2}-b^{2})\pi = \boxed{\pi(a^{2}+b^{2})}.$

Perfect! :)

You can rewrite $r(\theta)^{2}$ as $r(\theta)^{2}=a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta.$

We have more solutions.

By the way, $\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)(a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta )^{2}\geq (x\cos \theta+y\sin \theta )^{2},$ which is obtained by Cauchy-Schwarz's inequality.

I wonder this relation has something with the result.


How many digits are there in the expanded form of 2 to the 40th power?

[hide]2 to the 10th is 1024, square that to get $2^20$ as 1049600.
Round that to 1,000,000. 6 0's, squaring it gets you 12 0's. so 13 digits.[/hide]

[hide="My answer"]I would do it the way Klebian does if you don't memorize logs of common numbers.  So you just keep on raising it to powers until you arrive at an accurate number of digits for your answers.[/hide]


function
real analysis
topology
integration
combinatorial geometry
Functional Analysis
real analysis unsolved

I liked this one very much when at first i saw it in Rudin:
Let $K$ be compact hausdorff space and $f$,$f_n(n=1,2,3,...)$-continious complex functions on $K$ such that $f_n(x) \to f(x)$ when $n \to \infty$ for every $x \in K$, and $|f_n(x)|\leq 1$ $\forall n,x \in K$. Prove that there exist a sequence of functions, which are convex combinations of $f_n$, which converges to $f$ uniformly.

First of all, notice that the hypothesis implies the fact that $f_n\to f$ weakly. That is, for any functional $\Lambda\in C(K)^*$ (the dual of $C(K)$) we have $\Lambda(f_n)\to \Lambda(f)$. This follows from the characterization of $C(K)^*$ (for any such $\Lambda$ there is a complex regular Borel measure $\mu=\mu_\Lambda$ such that $\Lambda(g)=\int_Kg\ \mbox d\mu,\ \forall g\in C(K)$) and Lebesgue's Dominated Convergence Theorem.

Now the result we're after follows from a general fact that holds in real normed spaces: given a weakly convergent sequence $x_n\to x,\ x$ lies in the closed convex hull of $\{x_n\}_n$ (i.e. there is a sequence of convex combinations of $x_n$ convergint to $x$ in the strong or norm topology). Another way to state this would be to say that a strongly closed convex subset of a normed space is weakly closed, and in this formulation it's clearly equivalent to the separation version of the Hahn-Banach Theorem: given a closed convex subset $C$ of a normed space $X$ and a point $x\in X\setminus C$, there is a hyperplane separating $x$ from $C$, i.e. there is a continuous linear functional $x^*\in X^*$ such that $x^*(x)>\sup_{y\in C}x^*(y)$.

[quote="alekk"]very nice solution ![/quote]
Yes, indeed. This is the same solution i knew. Well done,Grobber.


modular arithmetic
number theory
prime factorization
number theory unsolved

Hi all

How do I prove the following

1) Prove that there is no prime p such that p^n=(2^m)-1 where n and m
are integers greater than 1.

2) Prove that there is no prime p such that p^n=(2^m)+1 where n and m
are integers greater than 2.

I was told to let p be either 2k+1 or 2k-1 whichever is more convenient and prove by contradiction using prime factorization, but I don't know how.

Help greatly appreciated!

[quote="enoughsaid05"]
How do I prove the following

1) Prove that there is no prime p such that p^n=(2^m)-1 where n and m
are integers greater than 1.

2) Prove that there is no prime p such that p^n=(2^m)+1 where n and m
are integers greater than 2.
[/quote]

[b]1) [/b]Clearly For $ p\equal{}2$ we can't find a counterexample.
For $ p$ odd, $ n$ even, we have $ p^n\plus{}1\equal{}1\plus{}1\equal{}2 \pmod{8} \Rightarrow m\equal{}1$, contradiction.
For $ p$ odd, $ n$ odd, $ p^n\plus{}1\equal{}(p\plus{}1)(p^{n\minus{}1}\minus{}p^{n\minus{}2}\plus{}....\plus{}p^2\minus{}p\plus{}1)$,
but $ p^{n\minus{}1}\minus{}p^{n\minus{}2}\plus{}....\plus{}p^2\minus{}p\plus{}1$ is odd, contradiction, done.

[b]2)[/b]Similarly for $ p\equal{}2$ we can't find a counterexample, so in the following cases we only consider odd $ p$.
[b]Case 1:[/b] $ m$ odd, then $ 3|2^m\plus{}1 \Rightarrow p\equal{}3$, so $ n$ is even or $ 3^n\minus{}1\equal{}2 \pmod{4}$, impossible.
Now let $ n\equal{}2k$ we get $ 3^{2k}\minus{}1\equal{}2^m \Rightarrow 2^x\equal{}3^k\minus{}1|2^m$ for some $ x$, contradiction (due to result of $ 1)$).

[b]Case 2:[/b] $ m\equal{}2x$ even, then $ p^n\minus{}1\equal{}(p\minus{}1)(p^{n\minus{}1}\plus{}p^{n\minus{}2}\plus{}....\plus{}p^2\plus{}p\plus{}1)$.
If $ n$ odd then $ p^{n\minus{}1}\plus{}p^{n\minus{}2}\plus{}....\plus{}p^2\plus{}p\plus{}1$ is odd , impossible.
If $ n\equal{}2k$ is even then the equation becomes $ (2^x)^2<(p^k)^2\equal{}(2^x)^2\plus{}1<(2^x\plus{}1)^2$, but there can't be a square between two consecutive squares , contradiction, done.


In a 14  baseball team league, each team plays each of the the others 10 times. At the end of the season, the baseballs teams wins formed an arithmetic sequence. Determine the greatest number of wins one team could have, assuming no ties.

well, isn't the largest possible number of wins 130?

Sorry, greatest number of wins the last team could have won.

wouldnt it be 65 then?? if all the teams win half of the games.


There are two doors and two soldiers. One soldier always lies and another always tells the truth. One door leads to bliss, another to torture. You have one question to ask one soldier. You do not know which soldier is which or which door is which. You do not know which soldier is standing in front of which door.

What question do you ask?

"If asked, would the other soldier say that this was the door to bliss?"

[quote="Brut3Forc3"][quote="1=2"]Ask "Where do you deserve to go?". They will both point towards the door with Bliss behind it.[/quote]
This, of course, assumes certain things. :P[/quote]

Such as the truth= good, the lies = bad.

I know some good liars :)

alternatively, ask "if i asked you which door was bliss, which would you reply?"
if they lie, then they would say the torture door if you asked, so they reply bliss door. if they tell the truth, then they will say the truth.

"If I asked your partner for the bliss door, what would he answer?" They would signal a door. Choose the other door. WHEEPEE!!!

How about this? Randomly choose a door. If you don't like it, step out and go to the other one.  :D

[quote="isabella2296"]If you don't like it, step out and go to the other one.  :D[/quote]

No one said anything about not being able to do that  :P .


You can kinda reassemble the faces, and drag them out. So the inside out cube (the place where it is cut out) can be dragged and form the original cube. Thus, the surface area is $ 6\times (4.7)^2\equal{}6\times 22.09\equal{}132.54$


function
limit
integration
real analysis
real analysis unsolved

$ f$ is a real valued integrable function on measure space $ (X,M,\mu)$ and $ \{E_i\}_{i\equal{}1}^\infty$ is measurable subset of $ X$ such that $ \mu(\bigcap_{i\equal{}1}^\infty E_i)\equal{}0$

Show that $ \lim_{n\to\infty} \int_{\bigcap_{i\equal{}1}^n} f d\mu\equal{}0$

are you trying to fill in the gaps [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=219429]here[/url]?

the idea is simple.

1)  outside of $ E\triangleq\cap^\infty_{j\equal{}1} E_j$, $ 1_{E_n}$ converges to $ 0$ pointwise.  So $ 1_{E_n}\to 0$ pointwise a.e., and $ |f|1_{E_n}\to 0$ pointwise a.e.

2)  $ |f|1_{E_n}\leq |f|$ for all $ n$ (alternately, $ |f|1_{E_n}$ is a monotone decreasing sequence which belongs to $ L^1$ for all $ n$)

3)
\[ \int_X |f|1_{E_n}d\mu \equal{} \int_{E_n} |f|d\mu
\]

4) by 2) the hypotheses of the dominated convergence theorem are satisfied.  what does it yield? (alternately, the hypotheses of the monotone convergence theorem are satisfied)

Can you justify  (1) ?
Because we don't know whether $ E_n$'s are incereasing or decreasing.

Replace $ E_n$ with $ E_n' \equal{} \cap_1^n E_j$


Ive seen a variety of differences between the lowest rating and the highest rating on users.

I think it will be interesting to know! :)

[quote="Berzerk"][quote="Mewto55555"]2027=overall record.[/quote]
How long did it took for you to become rating of "1"?[/quote]
That was a glitch.

yeah... I want that glitch  :|

$ 1603 \minus{} 1184 \equal{} \boxed{419}$

Nice ernie.

I was wondering...does anyone know what the all-time lowest rating is on FTW?

Does anyone have rating goals for FTW?

If so, you are welcome to post them here!

Highest - $ 1616$
Lowest - $ 1180$

Not much difference, just 436.

[quote="KingSmasher3"]that was a glitch.
what is the REAL all-time lowest rating?
 :D[/quote]

I think Pink Panther got to about 550~.....

Whoa.
Has anyone ever gotten <500?
I seem to rememer romandragon getting in the 400s.  :lol:

Goal: 14 - 1500
Worst seen... PinkPanther 676

Pink Panther has gone down to 510, I believe.

Just wondering...
Is one able to veiw anyone's FTW profile?

You click on their name in the lobby....

My lowest was 916 :(
I now have 1563..
But ya..I want to get 1600..so close :(

[quote="KingSmasher3"]that was a glitch.
what is the REAL all-time lowest rating?
 :D[/quote]

I think Pink Panther had like 400-500 something rating

[quote="rizikin"]Lol. Nice.
RATING HIGH: 1345
RATING LOW:   978
DIFFERENCE: 367
Yeah. LOL I used to have 1337 rating,twice in my whole "FTW"[/quote]

Hmm...I've been wondering all along,
Why does everyone love 1337 rating? WHat is so great about it?  :)

1337=l33t.......what's so good about THAT?

oh wait.my rating change changes :D
1595-916=679 :o
Need to get +5 :P

:laugh: Yay, my rating rating difference became 800+.

I  was on FTW ont time and realized one guy's rating was "null."  Does that mean 0?

I think null rating is a glitch.
My difference is 1873 :laugh:-1200=673, but it could change.

[quote="k00lperson"]No, I'm pretty sure he had like 516 lol. He was purposely doing that[/quote] The last time I saw Pink Panter, the guy had rating 520-530.

dy317, to you realize the last post in this thread was a year and a half ago?

Also, your post is a little off topic. It's about the [b]difference[/b] between highest and lowest.

Mods, I would like to request a lock since this thread was revived.


articles
geometry
geometric transformation
rotation

This is a pretty cool article I found on Physlink. Its a bit long but it is very good.*warning:Takes all the fun out of Christmas*

Consider the following:

1) No known species of reindeer can fly. But there are 300,000 species of living organisms yet to be classified, and while most of these are insects and germs, this does not COMPLETELY rule out flying reindeer which only Santa has ever seen.

2) There are 2 billion children (persons under 18) in the world. BUT since Santa doesn't (appear) to handle the Muslim, Hindu, Jewish and Buddhist children, that reduces the workload to 15% of the total - 378 million according to Population Reference Bureau. At an average (census) rate of 3.5 children per household, that's 91.8 million homes. One presumes there's at least one good child in each.

3) Santa has 31 hours of Christmas to work with, thanks to the different time zones and the rotation of the earth, assuming he travels east to west (which seems logical).

This works out to 822.6 visits per second. This is to say that for each Christian household with good children, Santa has 1/1000th of a second to park, hop out of the sleigh, jump down the chimney, fill the stockings, distribute the remaining presents under the tree, eat whatever snacks have been left, get back up the chimney, get back into the sleigh and move on to the next house.

Assuming that each of these 91.8 million stops are evenly distributed around the earth (which, of course, we know to be false but for the purposes of our calculations we will accept), we are now talking about .78 miles per household, a total trip of 75-1/2 million miles, not counting stops to do what most of us must do at least once every 31 hours, plus feeding and etc.

This means that Santa's sleigh is moving at 650 miles per second, 3,000 times the speed of sound. For purposes of comparison, the fastest man- made vehicle on earth, the Ulysses space probe, moves at a poky 27.4 miles per second - a conventional reindeer can run, tops, 15 miles per hour.

4) The payload on the sleigh adds another interesting element. Assuming that each child gets nothing more than a medium-sized lego set (2 pounds), the sleigh is carrying 321,300 tons, not counting Santa, who is invariably described as overweight.

On land, conventional reindeer can pull no more than 300 pounds. Even granting that 'flying reindeer' (see point #1) could pull TEN TIMES the normal amount, we cannot do the job with eight, or even nine.

We need 214,200 reindeer. This increases the payload - not even counting the weight of the sleigh - to 353,430 tons. Again, for comparison - this is four times the weight of the Queen Elizabeth.

5) 353,000 tons traveling at 650 miles per second creates enormous air resistance - this will heat the reindeer up in the same fashion as spacecraft re-entering the earth's atmosphere. The lead pair of reindeer will absorb 14.3 QUINTILLION joules of energy. Per second. Each.

In short, they will burst into flame almost instantaneously, exposing the reindeer behind them, and create deafening sonic booms in their wake. The entire reindeer team will be vaporized within 4.26 thousandths of a second.

Santa, meanwhile, will be subjected to centrifugal forces 17,500.06 times greater than gravity. A 250-pound Santa (which seems ludicrously slim) would be pinned to the back of his sleigh by 4,315,015 pounds of force.> In conclusion - If Santa ever DID deliver presents on Christmas Eve, he's dead now.

[quote]This works out to 822.6 visits per second.[/quote]

You could have just stopped there.  :D

what if there are many Santas helping him?

Heh... That's all well, but actually Santa Claus [i]can[/i] be modelled -- and rather effectively to that! -- using quantum mechanics. For example (I'll mention just two rather well known feats of quant. mech.):

- the problem of having Santa visit a thousand locations in one second can be solved by clever use of the Schroedinger's Cat paradox (and the double-slit experiment, which is a bit less famous): Santa is actually at [i]every[/i] home simultaneously since no one sees him. This will allow Santa to deliver a gift to every child instantenously. This leaves Santa an entire Christmas night to deliver but one present (and eat all the snacks in the house)!

- the problem of the weight of all the presents: what presents? Santa has only one present which he leaves beneath every tree -- just like him, these presents exist at several locations in the same time (a more thorrough approach would find this explanation incorrect, yet it can be more subtly modified so that it can't be (that easily) suspected.

Of course, there are several catches, for instance Santa's flying sleigh; but, working within the quantum-mechanical framework, even that could be modelled ;) So who says there's no Santa Claus?

Good point, but no -- there are macroscopic quantum fenomena all around us. Magnets, for instance, can't be explained with classical elecrodynamics. The same goes for the superfluidity of helium.

[quote="Sir.Aaron"]

5) 353,000 tons traveling at 650 miles per second creates enormous air resistance - this will heat the reindeer up in the same fashion as spacecraft re-entering the earth's atmosphere. The lead pair of reindeer will absorb 14.3 QUINTILLION joules of energy. Per second. Each.

In short, they will burst into flame almost instantaneously, exposing the reindeer behind them, and create deafening sonic booms in their wake. The entire reindeer team will be vaporized within 4.26 thousandths of a second.

[/quote]

 :P

now, who can explaine how easter rabbit can hold a basket full of eggs in his "hands"?

it doesnt holds them with its hands but it uses a very strong electromagnetic field


Determine $F(x)$ if, for all real x and y, $F(x)F(y)-F(xy)=x+y$.

Put $y=1$:

$F(x)F(1)-F(x)=x+1\iff F(x)(F(1)-1)=x+1$

We can't have $F(1)=1$ since that would give $(\forall x\in\mathbb{R})x+1=0$. Therefore

$F(x)=\frac{x+1}{F(1)-1}$

Now

$F(1)=\frac{1+1}{F(1)-1}\iff F^{2}(1)-F(1)-2=0\iff (F(1)=2\lor F(1)=-1)$

That gives two solutions: $F(x)=x+1, F(x)=-\frac{x+1}{2}$, but plugging both into the initial equation, we see that the latter doesn't work. Hence the only solution is $F(x)=x+1$

It's clear that $F(0) \neq 0$. Now put $x=0$.
Note that $f$ is surjective, since $F(y)=\frac{y+F(0)}{F(0)}$

So there exists a real number $q$ such that $F(q)=1$.
Now, $F(0)F(q)-F(0)=q \iff F(0)(F(q)-1) = q$, so $q=0$

Plug this in: $F(x)-1 = x$ for all real $x$.


i  desperately need sum help on this one.i am not even able to start on this one.
consider a finite set of segments in a plane whose lengths have sum less that  \sqrt 2 .prove that there exists a infinite square grid whose lines do not intersect any of these segments
bye

c'mon guys,hasnt ne1 cum up with ne thing that wud initiate the thought process and help me think in the rite direction
best regards

I know the soln, but it's quite large.
I'll give you some ideas.
At first, prove, that if there are some segments of a line (!!!) whose lengths sum <  1 then there is a unit grid of that line that doesnt intersect with segments.
(a grid in a line is like 0.2, 1.2, 2.2, ...)
The next step is to find a pair of orthogonal lines such that the sum of the our segments projections onto each line < 1.
Find a grid in each line and obtain a grid in the plane!


quadratics
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

I think this should be easy, but I can't get a hold of it.

Show that if $a \in \mathbb{Q}$, then $\mathbb{Q}(\sqrt{a})$ is contained in some cyclotomic extension of $\mathbb{Q}$.

By gaussian sum, we have $g_p : = \sum_{k=1}^p \zeta_p^{k^2} = \sum_{k=1}^p \left( \frac{k}{p} \right) \zeta_p^{k}= \begin{cases} \sqrt{p} & \text{ iff } p \equiv 1 \mod 4 \\ i \cdot \sqrt{p} & \text{ iff } p \equiv 3 \mod 4 \end{cases}$ for all odd primes $p$ and $\zeta_p = e^{\frac{2 \pi i}{p}}$ being the "first" $p$-th root of unity.
The last equality is not that easy to show, but the equality $T^2 = \left( \frac{-1}{p} \right) p$ is a lot easier and suffices.

Now just take the extension by $\zeta_{4a}$:
Then $\sqrt{\pm p}$ is contained for all primes dividing $a$ (note that we can either assume $a$ to be an integer or just look at the factorisation of a rational), thus $\sqrt{\pm a}$ is also contained. But $i$ is also contained, thus we can change the sign when necessary, giving us finally $\sqrt{a}$.

But this is just a special case of the Kronecker-Weber Theorem stating that any abelian extension is part of some cyclotomic extension.

I should have stated this earlier, but I was wondering if there's a way to prove this without using Gaussian sums.

Since the needed things about them are purely elementary, I see no reason to avoid them...

Well, quoting Kronecker-Weber will then not suffice, too, I think.

The following isn't elementary, but more in the general manner:
We know that every cyclotomic field generated by $\zeta_p$ has (cause of an even degree over the rationals) a quadratic subfield.
And the cyclotomic field has (up to sign) as discriminant a power of $p$, thus every subfield has that property, too. But the only quadratic extension with discriminant being a prime $q$ (well, for any discriminant there is at most one quadratic extension) is that one generated by $\sqrt{q}$ for $q \equiv 1 \mod 4$ and by $\sqrt{-q}$ for $q \equiv 3 \mod 4$. Thus $\sqrt{\pm p}$ is contained in it.
But I doubt this is what you wanted...

That isn't quite what I was looking for, but that is probably my fault.  Thanks for your help, though.

This is a problem in a Field and Galois theory by Morandi.  It follows the chapter on cyclotomic extensions (chapter 7).  The problems actually mentions it as a special case of Kronecker-Weber.

Our professor told us there is a relatively straightforward way to show this using tools in the chapter (or earlier), but the students in my class are still stuck.  We've seen proofs with Gaussian sums, but Gaussian sums are not mentioned anywhere in the book.

In short, I thought I'd mention the problem and see if anyone knew of concise proof using cyclotomic extensions.  Your last reply was along the lines of what I was thinking, but I don't want to get into using discriminants (not covered until chapter 12).  I'm sorry for this rather picky question; it's possible the text pre-supposes knowledge of Gaussian sums and I'm just dense.  I don't know.  Thank you for your replies nonetheless.

I found one more approach:
We know that $x^{p-1}+...+x^2+x+1= \prod_{k=1}^{p-1} (x-\zeta_p^k)$ is the $p$-th cyclotomic polynomial. Setting $x=1$ we get $p=\prod_{k=1}^{p-1} (1-\zeta_p^k) = \prod_{k=1}^\frac{p-1}{2} (1-\zeta_p^k) (1-\zeta_p^{p-k})$ for $p$ odd.
Now using that $(1-\zeta_p^{p-k}) = (1-\zeta_p^{-k}) = -\zeta_p^{-k}(1-\zeta_p^k)$ we get $p = \prod_{k=1}^\frac{p-1}{2} (1-\zeta_p^k)(1-\zeta_p^{p-k}) = \prod_{k=1}^\frac{p-1}{2} -\zeta_p^{-k} (1-\zeta_p^k)^2 = (-1)^\frac{p-1}{2} \cdot \left( \prod_{k=1}^\frac{p-1}{2} (1-\zeta_p^k) \right)^2$.
And this gives what we need :)


If z = r ( cosx + i sinx) how does 1/ z = r ( cosx - i sinx)?

By Demoivres theorem we can take z to the negative first power so

;z^(-1) = ( r (cosx + isinx))^(-1)

so wouldn't the expression be equal to

;1/z = (cosx - isinx)/r ?

Hmm... I'm not sure why the latex didn't work.

Heres the link to the problem by the way: http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2000

Let's see... $z=r(\cos x+i\sin x)=r(e^{ix})$, so $\frac{1}{z}=\frac{1}{r}(e^{-ix})=\frac{1}{r}(\cos x-i\sin x)$.

So... what does $r$ stand for in this equation?

[b]EDIT:[/b] As for the problem, here's my solution:

[hide="Answer"]$z+\frac{1}{z}=e^{i\theta}+e^{-i\theta}=\cos \theta +i\sin \theta +\cos \theta -i\sin \theta =2\cos \theta =2\cos 3^\circ\Rightarrow \theta =3^\circ$
Therefore, $z^{2000}+z^{-2000}=e^{i\cdot 6000^\circ}+e^{-i\cdot 6000^\circ}=2\cos 6000^\circ =2\left(-\frac{1}{2}\right)=-1\Rightarrow \boxed{0}$.[/hide]


Let $a_n$ be an arithmetic progression containing only natural numbers. Prove that for any $p$ in the sequence $\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3},....$ there exist $p$ terms in arithemtic progression.

[quote="Boy Soprano II"][hide="Solution"] Consider $a_{1}(p!)$.  The arithmetic sequence is \begin{eqnarray*} \frac{1}{a_{1}(p!)},\frac{2}{a_{1}(p!)},\cdots, \frac{p}{a_{1}(p!)} &=&\frac{1}{a_{1}\frac{p!}{1}},\frac{1}{a_{1}\frac{p!}{2}},\cdots, \frac{1}{a_{1}\frac{p!}{p}}\\ \end{eqnarray*}

since clearly $a_{1}\frac{p!}{1},a_{1}\frac{p!}{2},\cdots, a_{1}\frac{p!}{p}$ are natural numbers greater than or equal to $a_{1}$.[/hide][/quote]

How do you know that $a_{1}\frac{p!}{1},a_{1}\frac{p!}{2},\cdots, a_{1}\frac{p!}{p}$ are part of the given arithmetic progression?


ELON!!!

...

...

Elon is tomorrow... so goodluck everyone going!

My team:

Me
Calvin
Nick
[hide="do not ever look.. it is a shame..."]
Emily Chen
Lisa Zheng

told you not to
[/hide]

I'll put rankings 2morrow if i have time

Yeah.. sorry for the delay  :D

[hide="individual"]
1. Calvin D. - Ligon
2. Nick T. - Ligon
3. Allen Y. - Ligon
4. Arjun B. - Enloe
5. Andrew - Chapel Hill High
6. Michael (zijing) Gao - Smith
7. - Enloe
8. Michael Bai - Enloe
[/hide]

My brain died on #7 sorry. It was a guy from enloe... I think it started with an e....

[hide="team"]
1. Enloe
2.
3. East Chapel Hill
[/hide]

Brain stump again.

Good job Enloe! 3 people going to states and 1st place team!!

Plaques were for top 3 individual, top 3 team.
Certificates were for best in school, and top 6.

This is comprehensive? Middle schoolers win again. Good job guys.

I did not know I was qualified because I went to perform in Raleigh.

Emil Mayev was 7th place, if indeed the name started with an "E".  He's pretty good, so I'm not surprised.

What does surprise me: Ligon gets top 3 individuals and [i]still[/i] Enloe gets top team.  Pretty odd, guys; pretty odd.

They said that there has to be four people to make a team; we were three. :mad:  :|  :fool:


Are any of you in the situation where you have so much school work and extra-curricular work that you don't have time to do the math you want to do? During weekdays, I'm glad if I can find 1/2 hour a day, and most days it's not even that. Weekends, maybe 1 hour? During the summer I did about 5 hours a day... what a change. Life is pretty busy... :mad:  :(

Yep, that's how it is for me right now. I've got my regular IB schedule, along with classes from college going on, and doing cross country during the week on the scool team. I play tennis on the weekends, and spend about 10 hours a week at church (drama practice/small groups/helping out with cafe thing/actual church). And with other activities I'm involved with in school, jazz band, math team and student council, it's very hard to make time for the math that I like to do.

This is somewhat helped out by the fact that I'm in AOPS classes, but the fact still remains that if I don't have any time to devote to that class/IS then I won't get that much out of it. I get to sleep at about 12 every night (I have to wake up at 5:15 so if I went to sleep any earlier I wouldn't be able to function the next day) and most of the time I still have work to do for school the next day. And note that I have little to no free time during the week (but don't get me wrong, I enjoy all of the extracurriculars that I do, so the work that I have to put into them is not so bad because of this fact.. but it does remain time consuming.

But everything is starting to level out, and when I get my license in November (which will be after cross country season) that will save me a ton of time (i lose countless hours every week due to the buses) so hopefully I will start being able to dedicate some of my time towards problem solving.

I wish everyone good luck with the balancing of their schedules to accomodate their math needs ^^. Currently I'm trying to convince the school to drop my history class and take another math class in its stead (a "loophole" in IB system so I don't technically need the History class) so if that goes through, it will definitely lighten the load!

Oh man, I feel your pain.
I have to stay afterschool every single day for one thing or another.  I'm schooling the Latin 1's for Junior Classical League (the latin equivalent of competition) and practicing for brainbowl/quiz bowl even though I'm not any good.  I have math team practice 2-3 times a week and I really really need to start taking advantage of that.  Plus, I'm taking the most rigorous courses offered for sophomores at Rickards.  AP Physics is really difficult without a competent teacher.  Ah well, I've adjusted to that at least, so things should get easier.  All the homework has been a pain.  I've ordered books and they've just been sitting on my desk for a couple days now, yeesh.  I need to finish doing all the problems in AoPS 1 so I can really start my AIME studying.  Plus I've been doing AoPS classes alot.  That takes alot of my homework time, which I'm okay with, but it often keeps me up till 1.  Hopefully things will settle down soon.  Winter break will hopefully be a good time to do alot of math, particularly for AMC/Mu Alpha Theta.

Wow, someone who can read my mind....

  In times like these, homeschooling looks more promising than ever........

Yep.  I'm in a similar situation.

In fact, right now, I'm supposed to be doing 8 journals on reading books, and I am supposed to have annotated 69 pages of 1984 which I haven't even started.

Mostly for me, its a ton a school work.  Normally the only math  I ever do is that on this site and the problems from NOML...I don't have enough time for anything else.  Weekends are usually my time to do more math, but usually that gets destroyed by a bunch of projects that have been saved for the weekend because of to much stuff going on during the weekdays.

Most days I have something before/after school, and I am the type of person that can barely function with about 7 hours of sleep, 10 is preferred.  So I'm always really tired during the school day even with about 6-7 hours of sleep, sometimes resulting in me not paying very much attention or putting my head down in my blowoff classes.  Especially my AP World History and Pre-AP English II teachers seem to be addicted to projects and long term assignments, I probably have about 6 major projects for those two classes alone.

Winter break normally is a good time to do math (I once went through an entire Algebra II book a few winters ago in 1.5 weeks of several hours of math each day), but this year I have a huge English project that will take up most of my time during Winter Break.  Plus I have to start thinking about next year's schedule, whether or not I want to take AP US History, etc.etc so Winter Break won't be much better.

But on the plus side, it seems being really busy has made me prioritize things and use my time more wisely.  It seems that now I find myself wasting a LOT less time as compared with a few years ago, or even last year.  So I guess that's a plus, learning to work efficiently...but I still don't like being this busy.

[quote="Bictor717"]No. I put math first.

But seriously, sometimes, I do feel like I have too much to do. Have you noticed that you get kind of high after a certain level of sleep deprivation?[/quote]
Yep.

On average, I have between 35 and 40 hours of homework every week. :stink:

Yes, I have the same problem, but I'll rather not even start typing out everything I do. Glad to hear that I'm not the only one. The only time that I get to do the maths I want to, is during holidays.


superior algebra
superior algebra solved

For all n is positive integer number.Prove that there is a ring which its charater is n.

[quote="ZetaX"]$\mathbb{Z} / n \mathbb{Z}$[/quote]
you are true,thank you verymuch!I'm bad :(


What is the smallest positive integer greater than 3/2 + 3/4 + 3/8 + 3/16 + 3/32 + ... + 3/2^n, for a finite value of n?
I don't get it. How are we supposed to calculate that?

akhil0422

Rigorous way:

Well we can factor out a 3:
 We get 3(1/2+1/4....1/2^n). Look! The part in parentheses gets closer and closer to 1 as n grows! In fact, as n approaches infinity the sequence approaches 1 (but is always slightly less). Thus the sequence <3(1)=3 so the answer is 3.

Quick way: It says n is finite. Lets take a nice finite number like 2. We get for our sequence 3/2+3/4=2.25. The smallest integer greater than that is 3.

What if n=0 in the short way? Also, I don't know infinite series. How do you know it is always less than 1 (not equal or greater)?

akhil0422

Hm ok i just realized my short way failed. So dont do that.

Anyway, let S be the sum 1/2+1/4....

It is clear that 2S=1+1/2+1/4....

Now heres the cool part:

Since 1/2+1/4....=S, we get 2S=1+S so clearly S=1

how do u go from S=1/2+1/4.....    to  2S=[b]1[/b]+1/2+1/4.....

I don't get how u multiplied 'S' by 2, and then just added 1 to the other side...

Please help. Thanks

covertcognomen

[b]DON'T PAY ATTENTION TO THIS. READ THE OTHER POST BY ME RIGHT UNDERNEATH THIS POST.[/b][b][/b]

I found out a pretty clear way of doing it, although somewhat rigorous and lengthy.

Let's use mewto55555's variable 'S'.

S= 3/2 + 3/4 + 3/8 + 3/16 + ... 3/2^n. And since n is finite, we can choose any number for it. Let's arbitrarily take 5 for the value of 'n'.

So,

S= 3/2 + 3/4 + 3/8 + 3/16 + 3/32.

And,

2S= 2( 3/2 + 3/4 + 3/8 + 3/16 + 3/32); You can see here that you can multiply everything in parenthesis by 2 and simplify it.

Thus,

2S= 3 + 3/2 + 3/4 + 3/8 + 3/16       (Our equation by multiplying both sides by 2)
 S=       3/2 + 3/4 + 3/8 + 3/16 + 3/32     (The original equation with 'n' being 5)

If u subtract both equations from each other, you get:

2S= 3 + 3/2 + 3/4 + 3/8 + 3/16
-(S=     3/2 + 3/4 + 3/8 + 3/16 + 3/32)

Which equals,

S= 3 - 3/32. With this, you can see that S has to be less than 3 for this to be true.

Mewto, you can't use that for finite geometric series.

Let that S like Mewto.

$ 2S\equal{}S\plus{}1\minus{}\frac{1}{2^n}$, because when multiplied with 2, $ 2^n$ goes back to $ 2^{n\minus{}1}$, and there's none to replace it.

$ S\equal{}1\minus{}\frac{1}{2^n}$ This is basically covert's solution, just short.

Then the sum is $ 3\minus{}\frac{3}{2^n}$

If $ n\ge 2$, the number is 3
If n=1, the number is 2
If n=0, the number is 1
If n is negative, the answer is 1(positive integer)

As in FTW, you can't have 3 answers, it's probably 3 (watch out for conditions!)


limit
logarithms
real analysis
real analysis unsolved

Let $V_{n}$ = $\sum_{i=1}^{n}\frac{x_{i}^{2}}{n}$ - $\frac{\sum_{i=1}^{n}\{x_{i})^{2}}{n^{2}}$.
(1)Compute the value of the value of $V_{n}$ when $x_{i}$=$log i$
(2)Determine the value of ${\lim_{n\to\infty}\{V_{n}}$.

{I have posted this problem in the olympiad forum but since there were no replies I posted it here.}

I assumed that
$V_{n}=\sum_{i=1}^{n}\frac{x_{i}^{2}}{n}-\frac{\left(\sum_{i=1}^{n}x_{i}\right)^{2}}{n^{2}}$.
In this case for $x_{i}=\ln i$ we get, via Stirling`s formula,  that

$\sum_{i=1}^{n}{x_{i}}=\ln n!\approx \frac{1}{2}\ln 2\pi n+n(\ln n-1)$, hence

$\frac{(\sum_{i=1}^{n}x_{i})^{2}}{n^{2}}=\frac{\ln^{2}2\pi n}{4n^{2}}+\frac{\ln 2\pi n \ln n}{n}+\left(\ln n-1\right)^{2}$.
The first two terms of the preceding formula converge to 0, therefore for calculating the limit of $V_{n}$ it is sufficient to calculate the following limit:

$\lim \frac{\sum_{i=1}^{n}\ln^{2}i-n\left(\ln n-1\right)^{2}}{n}$, which can be calculated by an application of Cesaro Stoltz lemma.
I think that, if my calculations were correct, the limit is 1.

As for the second part of the problem, I do not think that $V_{n}$ can be calculated exactly, however I may be wrong.


$ x$ and $ y$ are positive integers, $ x\plus{}y\plus{}xy\equal{}34$. What is the value of $ x\plus{}y$?

hello, $ x\plus{}y$ must be $ 17$.
Sonnhard.

[hide="Solution"] By SFFT, we have $ (x \plus{} 1)(y \plus{} 1) \equal{} 35$. Now, either $ 35 \equal{} (5)(7)$, or $ 35 \equal{} (35)(1)$. However, we may reject the second case, since it forces $ x$ or $ y$ to be $ 0$, and we are restricted to positive integers. Hence, we have either $ (4,6)$, or $ (6,4)$, and in either case $ 6 \plus{} 4 \equal{} \boxed{10}$.[/hide]

[hide]By SFFT, $ (x \plus{} 1)(y \plus{} 1) \equal{} 35$. 35 factors as 1,35 or 5,7. These give 0,34 and 4,6 as the values of x,y. We are given that x,y are positive, so (x,y)=(4,6),(6,4). Thus, the sum is $ \boxed{10}$.

btw: check out my sig  :P [/hide]

$ x\plus{}y\plus{}xy\equal{}34$ so $ x\plus{}y\plus{}1\plus{}xy\equal{}35$ and
$ (x\plus{}1)(y\plus{}1)\rightarrow x\plus{}y\plus{}1\plus{}xy\equal{}35$
$ (x\plus{}1)(y\plus{}1)\equal{}35$ $ \rightarrow$ $ x,y\equal{}(6,4);(34,0);(0,34);(4,6)$

[quote="yildobekir"]$ x \plus{} y \plus{} xy \equal{} 34$ so $ x \plus{} y \plus{} 1 \plus{} xy \equal{} 35$ and
$ (x \plus{} 1)(y \plus{} 1)\rightarrow x \plus{} y \plus{} 1 \plus{} xy \equal{} 35$
$ (x \plus{} 1)(y \plus{} 1) \equal{} 35$ $ \rightarrow$ $ x,y \equal{} (6,4);(34,0);(0,34);(4,6)$[/quote]

Besides prior users already working it the same way as you did,
you went backward in your work when you typed the second line,
because you already established in your first line that $ x \plus{} y \plus{} 1 \equal{} 35$,
so you would not expand it in your third line.
Also you did not truncate your solutions.  The original problem stated that
x and y are positive integers.


Solve cos(cos(cos(cos x))) = sin(sin(sin(sin x))).


geometry
geometric transformation
reflection
rotation
complex analysis
real analysis
linear algebra

Show that the reflection $ z$ in the line $ ax\plus{}by\equal{}c$ ($ a,b,c\in\mathbb{R}$) is given by
$ \frac{2ic\plus{}(b\minus{}ai)\bar{z}}{b\plus{}ai}$

this is high school mathematics, not complex analysis.

and try using a transformation to move your line to the real axis.

It's from a complex analysis book.  I am thinking that I should rotate it and then shift the line up or down but I can't quantify this.

[quote="JRav"]It's from a complex analysis book.[/quote]
For this particular one it may be ok the remain here (as it isn't any better suited in other places). But note that introductory sections of X may not always belong to X ;)
Surely a lot of complex analysis textbooks have some chapters about real analysis and still this chapters are not complex analysis.

I came up with a solution for this problem.  Unfortunately, it requires a change of basis/linear algebra argument and I'm almost positive that is unnecessarily complicated.

I would like to see another argument.

Ok, I didn't choose particular good values to make the calculations easier, but something like this is probably what is intended in a complex analysis book

With $ k \equal{} \frac {ab}{iac \minus{} bc}$ and $ d \equal{} \minus{} \frac {b}{ia \minus{} b}$ the map $ w \equal{} kz \plus{} d$ is a möbius transform that takes $ \frac {c}{a}$ to 0, $ i\frac {c}{b}$ to 1 and $ \infty$ to $ \infty$. Hence it maps the given line to the real line. Conjugating $ w$ and taking it back with the inverse transform gives $ \frac {\overline{kz}}{k} \plus{} \frac {\overline{d} \minus{} d}{k} \equal{} etc$

It's really the exact same thing as the linear algebra argument though... Jeez, why are you other guys so unfriendly about this when you know what was intended with the problem


For each integer $n > 1$, let $p(n)$ denote the largest prime factor of $n$. Determine all triples $x, y, z$ of distinct positive integers satisfying
(i) $x, y, z$ are in arithmetic progression, and
(ii) $p(xyz) \leq 3$.

so each of x<y<z are of the form $2^{a}3^{b}$
$x+z=2y$
Case 1, x, z are both odd
Then $x=3^{X},z=3^{Z}, y=2^{Y_{1}}3^{Y_{2}}$
divide thru by $3^{Y_{2}}$ and relabel:
$3^{X}+3^{Z}=2^{Y}$ which means X or Z =0  because lhs isn't div. by 3
sry, no time now but you see how to continue?

If $x,y,z$ have a common factor, say $h$, then we divide through by $h$. Hence, for know, we assume that $gcd(x,y,z)=1$. Also we assume that $x<y<z$.

We know that $x+z=2y$, since $x,y,z$ are in arithmetic progression.
From this we can infer that $gcd(x,y)=gcd(z,y)=1$ and that $gcd(x,z) \leq 2$.

So $y=2^{b}$ or $y=3^{b}$, since $z$ is not 1.

[b]Case 1: $x=3^{a}, y=2^{b}, z=3^{c}$[/b]

Let the common difference be $d$. We know $x+2d=z$, and if $a>0$, $3|x$ and $3|z$. But then $3|d$ and also $3|y$. But $y$ is a power of 2- contradiction.
So $a=0$ The equation $x+z=2y$ becomes $3^{c}+1=2^{b+1}$. But by Mihalescu's theorem this has no solutions except $c=1, b=1$.
This gives us the triple $(1,2,3)$

[b]Case 2: $x=2^{a}, y=3^{b}, z=2^{c}$[/b]
If $a>1$ then $4|x$,$4|z$ and hence $4|2y$ and from this we get $y$ even: but it is a power of 3, so contradiction.
If $a=0$, then $x$ and $y$ are both odd, from which we have $d$ and hence $z$ odd. But $z$ is a power of 2 greater than 1, so this is a contradiction.
If $a=1$, then $2^{c}+2=2\cdot3^{b}$,
$2^{c-1}+1=3^{b}$. By Mihalescu's theorem, the only solutions are $b=1,2$ with $c=2,4$.
So the only triples we get for case 2 are $(2,3,4)$ and $(2,9,16)$.

Now we must remember to multiply in by the common factor we removed at the start: the general solutions are $(h,2h,3h)$ $(2h,3h,4h)$ and $(2h,9h,16h)$ where $h$ is any positive integer of the form $2^{m}3^{n}$.

You could also use [quote="tim1234133"]From this we can infer (...) that $gcd(x,z) \leq 2$.[/quote]to prove that $a=0$ in Case 1 and $a\leq 1$ in Case 2 :wink:

But to be honest; wow, nice solution. I had a quite long solution, but this is far better. This is the first time I see Mihaelescu's theorem used in problem solving!  :D

[quote="me@home"]Does Mihalescu state in general if $(a,b)=1$ then $a^{x}+1\not=b^{y}$? How is this proven?[/quote]
No, it's really a great theorem, because it says that $a^{x}-b^{y}= 1$ does not have any integer solutions except for the famous $3^{2}-2^{3}= 1$. It was called Catalan's Conjecture (who was Belgian  :D ) for a while, which indicates that the proof is non-trivial  :P  I think [url=http://www.dpmms.cam.ac.uk/Seminars/Kuwait/abstracts/L30.pdf]this[/url] is a sketch of the proof.

[quote="Mathworld"]Interestingly, more than 500 years before Catalan formulated his conjecture, Levi ben Gerson (1288-1344) had already noted that the only powers of $2$ and $3$ that apparently differed by $1$ were $3^{2}$ and $2^{3}$ (Peterson 2002).[/quote]
Ya! Jewish guy!!