There will be a Countdown round for the top 8 who were in my Mock Mathcounts. Qualifiers have been posted in the results page of the Submission/Mock MC website.

The contest will take place in AoPS FTW on Saturday, March 1st, at 9:00 P.M. EST.

Due to the fact that it will be very hard for everyone to make it, the top 12 may be present at the time. Shall a member not be present, then each place will move up one in order to ensure that the bracket is complete.

Ex:

If 2nd place is not present, then 1st will stay at first, 3rd will move to sencond, etc., and 9th place wil move into 8th place.

This is the only time that I can spare for a Countdown.

Sorry in advance for any inconvenience.

I might or might not be there (have three math competitions that day :stretcher: , two of which are the last round, and one is a second round).

I can be there if someone can shout a reminder in my ear at 5:55 :(

Please verify that you are present and will be at the CD.

I'm in 10th place and I'll most likely be there.

[quote="Lazarus"][quote="CircleSquared"][quote="Lazarus"]hey i most likely will not be there because of MUN[/quote]

Noooooo now I have to face stevenmeow in Countdown.  HE'LL MURDER ME[/quote]

And I am less intimidating ;] ?[/quote]

Well...if you stayed maybe I could actually get to the 2nd round.  :o

[quote="Lazarus"][quote="CircleSquared"][quote="Lazarus"]hey i most likely will not be there because of MUN[/quote]

Noooooo now I have to face stevenmeow in Countdown.  HE'LL MURDER ME[/quote]

And I am less intimidating ;] ?[/quote]

That depends, are you going to buzz in, wait 3 seconds and then say "myyyyyyyyy aaaaaannnnnnnsssssswwwwwweeerrrrrr iiiiiiiissssssssssss................."

 :D

[quote="frost13"][/quote][quote="Lazarus"][quote="CircleSquared"][quote="Lazarus"]hey i most likely will not be there because of MUN[/quote]

Noooooo now I have to face stevenmeow in Countdown.  HE'LL MURDER ME[/quote]

And I am less intimidating ;] ?[/quote][quote="frost13"]

That depends, are you going to buzz in, wait 3 seconds and then say "myyyyyyyyy aaaaaannnnnnnsssssswwwwwweeerrrrrr iiiiiiiissssssssssss................."

 :D[/quote]

 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

its on ftw so he can't do that :rotfl:

[quote="#H34N1"][/quote][quote="frost13"][/quote][quote="Lazarus"][quote="CircleSquared"][quote="Lazarus"]hey i most likely will not be there because of MUN[/quote]

Noooooo now I have to face stevenmeow in Countdown.  HE'LL MURDER ME[/quote]

And I am less intimidating ;] ?[/quote][quote="frost13"]

That depends, are you going to buzz in, wait 3 seconds and then say "myyyyyyyyy aaaaaannnnnnnsssssswwwwwweeerrrrrr iiiiiiiissssssssssss................."

 :D[/quote][quote="#H34N1"]

 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

its on ftw so he can't do that :rotfl:

[/quote]

you said it would be on aim?

I'm willing to do it on AIM, BUT everyone then needs to post their aim etc

id highly prefer ftw tho

[quote="xpmath"]But on FTW mem people will win.[/quote]

Does everyone have AIM?

Can I get on AIM at 9:00 EST?

[quote="stevenmeow"][quote="xpmath"][quote="mathcrazed"]AIM MANTHGACRAZED. I THINK.

ya lazarus you are less intimidating no offense... it's just that... facing a 7th grader who's like 4' 11 is really scary since you know he is beast. xD[/quote]
He was talking about stevenmeow.[/quote]
Ahem, [b]less than 4' 8''[/b] and i will most likely be there and... i have no AIM and i'm not willing to create one right at this moment :-([/quote]

Same situation.  Except I'm not less than 4'8".

[quote="unimpossible"][quote="mathcrazed"]LOL THAT'S SAD... I'm somehow like 5'4/5'5 lol. ahaha bobby[/quote]

last time i saw you you were like 3'2"   :D[/quote]

i saw him today and he was like only 4'11" and yet your brain size is disproportional with your height

I'm 8' 16". I know how to convert. And I'm also really a part of this conversation.

But in all seriousness, if you must have "original problems", you could try using the chatroom of FTW in place of AIM(for circlesquared and stevenmeow), although ppls with no lag will have an advantage...

I am sorry, but due to time constraints, this has been cancelled.

Mods - please lock.

Unlocked, so discussion is still possible.


vector
floor function
Olimpiada de matematicas

hola espero que en este topic se planteen preguntas de combinatoria , yo comenzare con una pregunta 1 de la imo del 1987
 :lol:

sea $P_{n}(k)$ el numero de permutaciones del conjunto $[ 1,2,3,4......,n]$ el cual tiene k puntos fijos. probar que :

         $\sum_{k=1}^{n}$$kP_{n}(k) = n!$

Una pregunta, a que se refiere con puntos fijos??
 :lol:

A números que quedaron en su misma posición después de la permutación. Por ejemplo al permutar {1,2,3}, una de las posibles permutaciones es {3,2,1} en donde 2 es un punto fijo

q paso con las soluciones al problema  :blush: ,  tambien seria bueno que pongan mas problemas de combinatoria  :D

La cantidad de permutaciones que dejan fijo el número $i$ es $(n-1)!$, para cada $1\leq i\leq n$.
Si contamos todas estas permutaciones (con repeticiones) obtenemos $n!$
Por otro lado, cada permutación con exactamente  $k$ puntos fijos fue contada exactamente $k$ veces.

[quote="mhuarancca"]bueno un problema de tableros :

se tiene un tablero de 7x7 , pruebe q es posible colocar un cuadrado de 1x1 de tal manera no pueda ser llenado el tablero por una ficha del tipo A y 15 fichas del tipo B

nota:

tipo A (figura en forma de L formadapor 3 cuadraditos de 1x1 )

tipo B (figura en forma rectangular de 1x3)

espero soluciones[/quote]

Probaremos que si cubrimos la esquina con un monominó, es imposible cubrir lo que queda del tablero con una ficha tipo A y 15 fichas tipo B.

Demostración:

Coleramos por diagonales usando 3 colores $c_{1},c_{2},c_{3}$.
Notemos que (si no nos importa el orden de los colores sino de las diagonales) hay dos formas de colorear el tablero. Tomando las diagonales que de izquierda a derecha van hacia arriba o las que van hacia abajo(llamemos a esas coloraciones de tipo W y tipo X). Ahora si en la coloración tipo W una ficha tipo A ocupa una casilla de cada color, en la tipo X ocupa dos de un color y una de otra, pues los dos cuadritos que no son adyacentes estan sobre una diagonal. Ademas en ambas coloraciones las 4 esquinas son monocromaticas.

Ahora volvamos a nuestro tablero y consideremos las dos coloraciones. Al quitar una esquina, quedan entonces, la misma cantidad de casillas de cada color, sin importar cual de las dos coloraciones sea.
Una ficha tipo B ocupa siempre sin importar como se ponga una casilla de cada color. Tenemos que poner 15 fichas de tipo B por lo que en total quedan 15 casillas ocupadas de cada color, es decir que la ficha tipo A tiene que ocupar una casilla de cada color en ambas coloraciones, lo que ya vimos que no es posible.

QED
PS. No pude poner una imagen para mostrar como son las coloraciones.

hola jasamper88, mi solucion es algo similar.
Vamos a colorear el tablero de 7x7, con tres colores.
Sean los colores 1,2,3. los cuales los pintaremos de la siguiente forma.
(1,2,3,1,2,3,1)-(2,3,1,2,3,1,2)-(3,1,2,3,1,2,3)-(1,2,3,1,2,3,1)-(2,3,1,2,3,1,2)-(3,1,2,3,1,2,3)-(1,2,3,1,2,3,1).
donde cada vector es una fila y su respectiva coloracion. podemos notar que tenemos 17 unos, 16 dos y 16 tres, ahora las fichas de 1x3 estan formadas por casillas coloreadas de (1,2,3) , (2,3,1) y (3,1,2), ahora      si necesitamos que se de la imposbilidad para rellenar el tablero tendremos que cubrir almenos 2 casillas coloreadas con el color 2 ó el color 3, ya que no puede ser 3 casillas de 1. ahora notamos que en las esquinas facilmente se observa esto. Ahora si cubrimos con la ficha L la primera esquina cubriremos 3,2,2 y podremos desaparecer la primera casilla de color 1.
Q.E.P.

[quote="jasamper88"]La coloración W es la misma coloración de aev5peru y la coloración X es lo mismo pero empezando poniendo esas filas de abajo hacia arriba.

aev5peru, no entiendo tu solución, podrias explicarla de una manera más clara?[/quote]
que parte no entiendes??
para explicarlo mejor nesecitaria un grafico, alguien sabe como subo un gafico??

ahora del mismo problema
¿ cual es el numero total de cuadraditos que cumplen la condicion ?  :lol:

Usemos nuevamente las dos coloraciones y comparemolas. Si en la coloración W y en la coloración X las dos tienen color $c_{1}$ (que es del que hay 17), entonces funciona el argumento de mi solución y no se puede. Estos son  exactamente 9 cuadros (Las esquinas la mitad de los bordes y la casilla central. Si una casilla coloreada con $c_{2}$ o $c_{3}$ en W es coloreada con color $c_{2}$ o $c_{3}$ en X (no importa si se cambian) funciona el mismo argumeto de mi solución pero pues diciendo que en ambos debe haber una ficha que tenga dos casillas del mismo color. Estas casillas son lo que falta del borde y la columna y la fila central del tablero, que en total son 24 casillas  Nos quedan cuatro cuadrados de $2\times 2$ que no tocan el borde ni la fila ni la columna central. Si quitamos una casilla de esos cuadrados, completamos el cuadrado con la casilla de tipo A, y es fácil cubrir el resto del tablero con casillas de tipo B. (esta tarde trato de colgar un archivo con figuras).

Esto quiere decir que hay 33 casillas que se podrian quitar y no es posible llenar el tablero, y las restantes 16 si se quitan si es posible.

QED

PS. aev5peru creo que tu solución esta mal porque estas considerando solo un caso de poner la ficha de tipo A. Y pues que pasa si la pones en una casilla que si te cumpla. Te lo digo porque ahora que lo lei detenidamente me doy cuenta que tiene el mismo problema que tenia mi solución inicial. Ahi no se esta asegurando sino el caso en particular en que la ficha de tipo A esta donde tu quieres.

jaja bueno, tenias razon, es que no entendi bien el problema. Entonces posteo un problema por mi error jajaj  :rotfl: .
---
Se tiene un tablero de 5x5, donde en cada casilla tiene el número $0$ ó $1$, se llama $paso$ cuando seleccionamos una fila, columna o diagonal(tambien se considera diagonal a las pequeñas y a las esquinas), al cambiar todos sus digitos a otro estado. Por ejemplo : si la casilla antes estaba en $0$ pasa a $1$, ($0\to 1$). y viceversa ($0\to 1$).
Demostrar que con una cantidad par de $pasos$, empezando con un tablero rellenados con solo $0$, es imposible llegar a que el tablero acabe con solo $1$..
--
espero que lo resuelvan. chau

[quote="jasamper88"]
Creo que el problema de "aev5peru" peru da $n+1$ pues es un sistema de ecuanciones con $n+1$ ecuaciones y $n+1$ incognitas. Lo unico que hay que hacer es preguntar los numeros de $1$ hasta $n+1$ y solucionar el sistema de ecuaciones.
[/quote]
"jasamper88", no digo que tu idea esta mal, pero en realidad sale un valor numerico, recuerda que los coeficientes son enteros no negativos, ese dato ayuda bastante.
chau.
PD: no intenten mucho tu problema.

[quote="jasamper88"]Uy eso me suena raro. ¿Existe un valor $k$ independiente de $n$ tal que todo polinomio con coeficientes enteros positivos de grado $n$ se puede determinar conociendo $k$ valores? no se si entendi bien la pregunta.

PS. ¿Piensan mi problema o posteo la solucion?[/quote]
SI :D , recuerda que es un polinomio, cuando evaluas valores enteros :D
chau creo que esa pista es suficiente :D

[quote="jasamper88"]La respuesta al problema de aev5peru es $2$ preguntas.
Preguntamos el valor de $P(1)$ y el valor de $P(P(1)+1)$. COn el resultado de $P(P(1)+1)$ lo escribmos en base $P(1)+1$ y ese es el polinomio que queremos si cambiamos los $P(1)+1$ por $x$.

Esto funciona porque todos los coeficientes del polinomio son enteros y menores o iguales que $P(1)$ pues no son negativos.

QED

PD.
¿Que paso con el problema que puse yo? ¿Nadie a podido? ¿Pongo la solución?

Una pista es que es parecido al de las coloraciones de los enteros del reto 1.[/quote]
Muy buena tu solucion, pero la que yo vi fue la siguiente.
Pregunta 1.  cuanto vale $P(1)=k$-->entonces podemos elegir un numero $10^{m}>k$, entonces si preguntamos por el tendremos los coeficientes en la base $10^{m}$.
Pregunta 2. cuanto vale $P(10^{m})$. y tendremos asi los coefecientes.
.---
Para tu problema jasamper88, le vi la similitud tratando de hacer una biyeccion ya que en una circunferencia hay una cantidad infinita de puntos, aunque lo este diciendo muy a la ligera, pero trate de hacer la biyeccion, pero lo que me complica es eso de los isoceles.
---
Otro problema.
Demostrar que todo numero real puede ser escrito por la suma de $9$ numeros que tienen como digitos solo las cifras $0$ y $a$, donde $a$ puede ser ${1,2,..,9}$.

[hide="una pista para mi problema"] Mira los triangulos isoceles como progresiones aritmeticas de longitud 3 [/hide]

Una pregunta para el nuevo problema.
¿Puede haber números entre los 9 que sean 0?

[quote="jasamper88"]Una pregunta para el nuevo problema.
¿Puede haber números entre los 9 que sean 0?[/quote]
SI.

[quote="aev5peru"][quote="R.G.A.M."][quote="aev5peru"]Demostrar que todo numero real puede ser escrito por la suma de $9$ numeros que tienen como digitos solo las cifras $0$ y $a$, donde $a$ puede ser ${1,2,..,9}$.[/quote]
[color=blue]Se ve sencillo de resolver tu problema, mas me queda una duda. Para los números irracionales, ¿consideramos la parte decimal que se desarrolla?

Noto que podría cumplirse sin problemas para $a=1$, inclusive para cualquier $a$ tal que $(a;10)=1$. Supongo que querrás que se demuestre para todo $a$, ¿no?[/color][/quote]
bueno, la verdad es que solo quiero que demuestres para cuando $a$, puede ser ${1,2,3,...,9}$, si puedes generalisarlo aun mas seria estupendo. Bueno espero soluciones.

Con respecto al problema de Jasamper88 creo que sale considerando, el Teorema de Van Der Warden, que nos dice, que al pintar los numeros naturales con $n$ colores , podemos encontrar pogresiones aritmeticas monocromaticas.
Ahora en el problema, consideramos un punto arbitario y lo etiquetamos con el numero $1$, y en forma horaria vamos etiquetando los demas puntos con los respectivos numeros, y usando la sugerencia que nos das de considerar progresiones aritmeticas de 3 terminos, donde sean los puntos etiquetados $a<b<c$, entonces si los tres estan en progresion aritmetica la distancia del punto $a$ hasta el punto $b$ es la misma que la del punto $b$ al punto $c$. Y usando el teorema esto queda mucho mas facil, bueno se podria decir que queda resuelto.
Perdon, por ser muy ligero en mi solucion, es que no tengo mucho tiempo.[/quote]

Esa es la misma solución que yo tengo. Toca considerar el Van der Waerden $W(n,3)$ y separar el circulo en $W(n,3)$ pedazos escogiendo un punto al azar para hacer la separación y enumerar de alguna forma. Entonces hay 3 numeros en progresión aritmética con el mismo color. Esos 3 numeros son un triangulo isoceles. Como podemos escger el punto inicial de infinitas formas hay infinitos triangulos isoceles monocormáticos.

El unico problema que tengo con el problema de $aev5peru$ son los numeros irracionales. Porque lo unico que hay que hacer realmente es acomodar el digito de mas a la derecha y ya (en caso de racionales). Si es periodico pues se hacen los 9 numeros periodicos y ya. Me da pereza escribirlo bien porque tengo un par de graficas y no se como explicar sin mostrarlas.

[color=blue]En cierto pais hay trayectos de de avión que conectan dos ciudades (puede haber trayectos en ambas direcciones). Algunos de estos trayectos son volados en por $k$ aerelineas de manera que cada par de trayectos volado por una aerolínea tiene al menos una ciudad en común. Pruebe que es posible agrupar las ciudades en $k+2$ grupos de modo que no haya dos ciudades en el mismo grupo que estan conectadas por un vuelo.[/color]

[quote="mhuarancca"]¿ cual es el numeros de permutaciones de ${1,2,3,4,.....,n}$ de tal manera que en las permutaciones no existan dos numeros consecutivos ?[/quote]
A que le llamas consecutivos, a los terminos $a(a+1)$, y/o $(a+1)a$.

si , a eso me refiero , mas bien creoq hay dos formas de hacer esto , una por combinatoria y otra por inclusion y exclusion , seri bueno si puedes hacer las dos formas


induction
calculus
integration
algebra unsolved
algebra

a) Define $H(n)=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}$. For each $n\in N$, let $K_n$ is the least positive integer such that $H(K_n)\ge n$.
Prove that ${lim}\limits_{n\to +\infty}\frac{K_n}{n^2}=\frac{1}{4}$

\[ H(n) \approx \int_1^n x^{- \frac12} dx \approx 2 n^{- \frac12}  \]

Therefore,

\[ K_n \approx \frac14 n^{\frac12}  \]

[quote="tµtµ"]I think this comes directly from :

$2 (\sqrt{n+1}-1) \leq K_n \leq 2 \sqrt{n}$

which may be proved by induction or a little bit of calculus[/quote]

yes, so do I.


Find the minimum of $ x^4\plus{}y^4\plus{}z^4\minus{}4xyz$ over all real numbers $ x$, $ y$, and $ z$.

[hide="outline"]Let xyz=c. The expression will be minimized when, by AM-GM, x=y=z. Let x=k. Then we want to minimize $ 3k^4\minus{}4k^3$.[/hide]

I don't think we can use AM-GM so directly...

[hide]
We consider three cases. The minimum must arise in one of them.

Case I: $ -4xyz$ is positive. Assume temporarily that a minimum occurs with this case. But by negating one of $ x$, $ y$, or $ z$, we can obtain a negative $ -4xyz$ and this a smaller number. A contradiction occurs, so we cannot possibly have a minimum here.

Case II: $ -4xyz$ is zero. Then any or all of $ x$, $ y$, or $ z$ are zero. But then to minimize the given expression, $ x^4$, $ y^4$, and $ z^4$ are as small as possible, and they can only each be as small as zero. This agrees with our zero $ -4xyz$ The minimum, if it exists here, is $ 0$.

Case III: $ -4xyz$ is negative. Since $ x^4$,$ y^4$, and $ z^4$ are all greater than or equal to zero no matter what, we can assume that $ x$, $ y$, and $ z$ are positive in this case. We let $ a=x^4$, $ b=y^4$, and $ c=z^4$. Then our expression becomes $ a+b+c-4\sqrt[4]{abc}$. We add an $ n_0=1$ into the expression. Then we have
$ \&a+b+c+n_0-4\sqrt[4]{abcn_0}-n_0$ and we factor:
$ 4\left(\frac{a+b+c+n_0}{4}-\sqrt[4]{abcn_0}\right)-n_0$
Now we use AM-GM. Clearly our minimum is where equality occurs.
$ 4(0)-n_0$
$ -n_0$.
We defined $ n_0=1$, so our minimum is $ \fbox{-1}$. No other case provides a lower minimum. We see that when $ x$, $ y$, and $ z$ are all $ 1$, this can occur.
[/hide]

You are right that the fact that it is over all real numbers has to be considered. There is a proof that does not use as much casework though.

$ x^4\plus{}y^4\plus{}z^4\plus{}1\ge \sqrt[4]{x^4y^4z^4}\equal{}4|xyz|\ge 4xyz$

where I used the fact that $ |u|\ge u$ (this would not need to be explicitly stated in an actual proof since it is obvious).

Hence $ x^4\plus{}y^4\plus{}z^4\minus{}4xyz\ge \minus{}1$.

By equality conditions of the am-gm and the absolute value thing, equality occurs iff $ x^4\equal{}y^4\equal{}z^4\equal{}1$ and $ |xyz|\equal{}xyz$

So we have $ (x,y,z)\equal{}(1,1,1),(1,\minus{}1,\minus{}1)$ and permutations.

(You don't have to find all equality cases unless it says so, but it is generally a good thing to do.)


calculus integration derivative function

$\int_{0}^{8}T'(x)dx$ when Distance x (cm) = 0,1,5,6,8 when Temperature T(x) (celcius) = 100,93,70,62,55 respectively..

I know the derivitive of an integral, but not the opposite...I was thinking it meant the net change in temperature from 0 to 8 cm?

Any thoughts as to what it means and how to evaluate it?

Thanks!

The indefinite integral is $T(x)$, so

$\int_{0}^{8}T'(x) \, dx = T(8)-T(0)$

Your assessment is correct.

OK Thanks..When I was doing this, I was thinking of the position function and it's derivitive, the velocity function, and the integral of the velocity function, the net distance traveled..Thanks again for the check!


inequalities
function
inequalities proposed

[color=red][This is wrong][/color]

If I am not wrong,
1. $f(a,b,c) = a^3+b^3+c^3 + 15 - 6(ab+bc+ca)$
We prove $f(a,b,c) \ge 0$ for $abc \ge 1$
Let $2t = a+b$, then $f(a,b,c) \ge f(t,t,c)$. So minimum occurs when all variables are equal.

2. $f(a,b,c,d) = a^4+b^4+c^4+d^4 + 20 - 6(abc+bcd+cda+dab)$
We prove $f(a,b,c,d) \ge 0$ for $abcd \ge 1$
Let $3t^2 = ab+bc+ca$, then $f(a,b,c,d) \ge f(t,t,t,d)$. So minimum occurs when all variables are equal.

[quote="arqady"][quote="Soarer"]
Let $2t = a+b$, then $f(a,b,c) \ge f(t,t,c)$[/quote]
Why?
[/quote]
$f(a,b,c)\geq f(t,t,c)$
$\Leftrightarrow a^3+b^3+c^3+15-6(ab+bc+ca)\geq t^3+t^3+c^3+15-6(t^2+tc+ct)$
$\Leftrightarrow a^3+b^3-6ab\geq 2(\frac{a+b}{2})^3-6(\frac{a+b}{2})^2$
$\Leftrightarrow 4a^3+4b^3-24ab\geq (a+b)^3-6(a+b)^2$
$\Leftrightarrow 3(a^3+b^3-a^2b-ab^2)+6(a^2+b^2-2ab)\geq 0$

[quote="arqady"][quote="Soarer"]
Let $2t = a+b$, then $f(a,b,c) \ge f(t,t,c)$[/quote]
Why?
[quote="Soarer"]
2. $f(a,b,c,d) = a^4+b^4+c^4+d^4 + 20 - 6(abc+bcd+cda+dab)$
We prove $f(a,b,c,d) \ge 0$ for $abcd \ge 1$
Let $3t^2 = ab+bc+ca$, then $f(a,b,c,d) \ge f(t,t,t,d)$[/quote]
Why?
 Show something one, please.Thanks. ;)
 Best regards  :)[/quote]

For the second one, by muirhead, $a^4+b^4+c^4 \ge 3t^4$ by muirhead, $6d(ab+bc+ca)$ was artificially made to be fixed, and $6t^3 \ge 6abc$ by amgm. I hope it's correct.

[quote="arqady"][quote="leepakhin"]
$f(a,b,c)\geq f(t,t,c)$
$\Leftrightarrow a^3+b^3+c^3+15-6(ab+bc+ca)\geq t^3+t^3+c^3+15-6(t^2+tc+ct)$
$\Leftrightarrow a^3+b^3-6ab\geq 2(\frac{a+b}{2})^3-6(\frac{a+b}{2})^2$
$\Leftrightarrow 4a^3+4b^3-24ab\geq (a+b)^3-6(a+b)^2$
$\Leftrightarrow 3(a^3+b^3-a^2b-ab^2)+6(a^2+b^2-2ab)\geq 0$[/quote]
But $abc\neq att.$
For instance: $15abc\leq15att.$ ;)[/quote]
We have $t^2=(\frac{a+b}{2})^2\geq ab$, and that's why Soarer proved the inequality with $abc\geq 1$.
P.S. Why did you quote my post? :huh:

OK. Why $f(t,t,c)\geq0?$
But $abc\neq ttc.$
For instance: $15abc\leq15ttc.$ ;)
Or try $a=b=2,c=\frac{1}{3}.$ ;)

arghhhhhhh I see your point.. :blush: Sorry I was wrong

[quote="silouan"]Is the function
$f(x)=x^{3}-\frac{6}{x}+5-10(x-1)+ln\ x$  a way to solve it ?
If my calculations are true then yes .[/quote]
For example, $f'(x)=\frac{(x-1)(3x^3+3x^2-7x-6)}{x^2}$ and what is the rest?

by MV it is enough to check it when $c=d$

[quote="Speed2001"][quote="arqady"]The following inequality is also true.
Let $a$, $b$, $c$ and $d$ are positives such that $abcd=1$. Prove that:
\[a^4+b^4+c^4+d^4+24\geq7\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\][/quote]
With the same conditions , the following inequality also hold : $ a^{4}+b^{4}+c^{4}+d^{4}+44\geq 12\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right) $ :cool:[/quote]
Try $b=c=d=1.5$ and $a=\frac{1}{1.5^3}$. :wink:
[quote="Mirus"][quote="arqady"]1) $a>0,b>0,c>0,abc=1.$ Prove that
 $a^3+b^3+c^3+15\geq6(ab+ac+bc).$
[/quote]
Nice inequality. It isn't solved until now? Arqady, from which inequality of Vasc it was inspired?[/quote]
I don't remember exactly. Maybe the following:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=71532

Thanks Arqady. First inequality can be proved without Mixing Variables  method :)

[quote="arqady"] I don't remember exactly. Maybe the following:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=71532[/quote]
Or, maybe
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=71533

[quote="arqady"]1) $a>0,b>0,c>0,abc=1.$ Prove that
 $a^3+b^3+c^3+15\geq6(ab+ac+bc).$
[/quote]
The following stronger inequality also holds

$a>0,b>0,c>0,abc=1, $ q is real number such that  $ q \le 6.4$ Prove
 $a^3+b^3+c^3+3(q-1) \geq q(ab+ac+bc).$

-------
1/ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2599402#p2599402&hilit=found]See more here[/url]
2/ The maximum value of $q $, for which above inequality holds for all $a>0,b>0,c>0,abc=1$ , is  6.432247012.... (  I have proved it yet).


I think mixing variable method will kill it  :wink:

Don't just say, please post your solution!

No one interested in? May be it is too ugly but the proof is very special
$S_{a}+S_{b}+S_{c}\ge 0$ and $S_{a}S_{b}+S_{b}S_{c}+S_{c}S_{a}\ge 0$


Throughout this post, the word "this" will be used as an antecedent for anything under the innermost quotes.

"[i]This statement is an opinion[/i]" can be expanded to the form:

"The validity of '[i]This statement is an opinion[/i]' is an opinion"

And you would have to agree, because if you disagree, that "'[i]This statement is an opinion[/i]' is an opinion", then you are saying that it is a fact, which is in essense agreement on your part.

So if I change my previous statement to:

"The validity of '[i]every statement in the world besides the one under the outermost set of double quotes is an opinion' is an opinion," then in a weird way I have proved that everything besides what I have just written in "" is an opinion.

Obviously this isn't correct. What's wrong? Open for suggestions.
I thought of this after reading Paul Cooijman's proof of absolute truth, they are similar in a slightly different sense.

[quote="banblink182"]"[i]This statement is an opinion[/i]" can be expanded to the form:

"The validity of '[i]This statement is an opinion[/i]' is an opinion"

And you would have to agree, because if you disagree, that "'[i]This statement is an opinion[/i]' is an opinion", then you are saying that it is a fact, which is in essense agreement on your part.
[/quote]

I disagree that your expansion is valid, but that doesn't have anything to do with its actual content.  Rather, I think "The validity of "X" is an opinion" is a poorly-formed logical statement, since validity (or, truth value) is some real number between 0 and 1 and no real number between 0 and 1 is an opinion.

Logic is essentially meaningless manipulations of symbols -- I think you and Paul are engaging in a lot of meta-logic, but I'm not convinced that you're proving anything.  At the very least, your conclusions seem to me to be just opinions :)

Off-topic, but I remmeber in school when we has to write a pragraph on something with a group and it included saying that fiction books had "false facts", since they don't have have opinions, but aren't nessesarily true.

[quote="filletwho"][quote="confuted"]Not being an opinion does not necessarily imply that it is a fact.[/quote]

Yeah, it could be just a false statement.[/quote]

Thanks for responding to my post, I appreciate that a lot.

On your counter-arguments, I think I understand what you are saying, but I'm not entirely sure. The way I can be sure of what you really mean is to ask you this question, if I said that when I said 'fact' I didn't stress for it to be true or not, all I meant was that it wasn't an opinion, would your argument still stand? If yes, then I'm not understanding your point exactly.

To counter possible future arguments against the paradox along the lines of, "Well, if it is a false fact, then it is wrong, so your argument doesn't stand any longer," I say this:

If you think that the statements I've made in quotes are false facts, then you are saying that "the validity of 'this statement is an opinion' is definitely not an opinion". What this is saying is that "the validity..." is a fact.

Again, you can say, that you are saying that it is a "false fact". This would mean that you think "this statement is an opinion" is false, but this statement is clearly a true fact, because if this statement were a false fact, then it would be a circular thing that would repeat forever. (Of course I'm assuming that there don't exist any truths in the world that do not have a final destination, i.e. have an eventual outcome - true, false, opinion, etc.) Therefore if anything, it is a fact.

Don't be led to think that the fact that this statement is an opinion is also circular and repeats forever, because it doesn't. All this is saying is that one has to agree with the [i]fact[/i] that the statement is an opinion. Or at least that is how I look at it.

Although I sound confident (or maybe I don't), I'm not saying that what I'm saying is indeed a fact. You may contradict me as much as you like. I'm all ears.  :lol:

I concede my point; there is a logical flaw.

Everything doesn't [i]have[/i] to be anything.

Thanks for reading...

 :lol:


Solve the symtem equations:
a)$\{ \begin{array}{l}\sqrt{-1-x}-\sqrt{2y-x}=1 \\ \sqrt{1-2y}+\sqrt{2y-x}=4 \end{array}$
b)$\{ \begin{array}{l}\sqrt[3]{10x-y}+\sqrt{3x-y}=3 \\ 2x+y+\sqrt{3x-y}=5 \end{array}$


geometry
circumcircle
parallelogram
power of a point
radical axis
perpendicular bisector
geometry unsolved

Dear Mathlinkers,
AB intersect the circumcircle of ADC in B'.
B is the midpoint of AB'.
Sincerely
Jean-Louis

[quote="mhdi"]Hello,

Here is a problem i couldn't solve:

$ (C_1)$ et $ (C_2)$ are 2 circles with two common points $ A$ and $ B$.
The tangent at $ A$ of $ (C_1)$ cuts $ (C_2)$ at $ C$.
The tangent at $ A$ of $ (C_2)$ cuts $ (C_1)$ at $ D$.
A line which passes through $ A$ and is inside the angle $ \angle{CAD}$ cuts $ (C_1)$ at $ M$, $ (C_2)$ at $ N$, and the circumscribed circle of the triangle $ \triangle{ADC}$ at $ P$.

Prove that $ AM \equal{} NP$.

http://www.z-upload.fr/fichiers/63/o2.png[/quote]

It is a problem from Romany in some tst . We have discussed it in the greek community, but I can not find the link this moment.If you still have questions, please send me a pm to send you the solution(using similar triangles).

 Babis

@jayme : u mean?

Let $ E,F$ be the centre of $ C_{1}$ and $ C_{2}$ respectively. Now, it is obvious that $ \angle DAC \equal{}90^0$, $ AD$ passes $ E$ and $ AC$ passes $ F$.
Since $ \angle ABD \plus{}\angle ABC \equal{}90^{0} \plus{}90^{0}\equal{}180^{0}$, then $ CD$ passes $ B$.
Since $ AD,AC$ are both diameters, we get $ \angle AMC \equal{}\angle AND \equal{}90^0$
Notice that $ \angle DAN\equal{}90^0 \minus{} \angle AMC \equal{} \angle ACM$. Thus, two triangles $ ACM$ and $ ADN$ are similar. We get, $ \frac{AC}{AD}\equal{}\frac{AM}{DN}$.........(1)
Again, $ \angle ADC\equal{} \angle DPN$, thus two triangles $ DPN$ and $ DCA$ are similar. We get $ \frac{AC}{AD}\equal{}\frac{NP}{DN}$..............(2)
From (1) and (2), we conclude $ AM\equal{}NP$.

@mlequi: Sorry but GeoGebra tells me that ADC isn't equal to 90° :/

Dear Mathlinkers,
noted X, Y the midpoints of AD, AC.
The circle (knowed also as the midpoint circle) through A, X, Y passes through the center of the circumcircle of ADC.
The rest follows...
Sincerely
Jean-Louis

Hello [b]jayme[/b],

Did you mean $ B$ is the circumcenter of triangle $ ADC$ because $ AB$ is the radical axis of the two circles centered at $ C_1$ and $ C_2$.

However weren't we supposed to consider that the line $ AP$ intersects the two circles at distinct points $ M$ and $ N$.

[b]mlequi[/b] are you sure angle $ DAC$ is $ 90$ degrees. It was never mentioned in the question that the circles centered at $ C_1$ and $ C_2$ are orthogonal.

The problem said "please, use inversion! USE! I NEED IT! Pleaaaaaaaaaaaaaase! I BESEECH! I do anything, but invert me!". Then I used! (with center $ A$, of course)... The resulting problem is:

Let $ ACBD$ be a paralelogram (the image of a circle passing through $ A$ is parallel to its tangent through $ A$). Let $ M$, $ N$ and $ P$ be the intersections of a given ray (conteined in the angle $ DAC$) with $ BD$, $ BC$ and $ CD$, respectively. Then $ \frac {1}{AM} \plus{} \frac {1}{AN} \equal{} \frac {1}{AP}$ (note that, in the original problem, $ AM \plus{} AN \equal{} AP \Longleftrightarrow AM \equal{} AP \minus{} AN \Longleftrightarrow AM \equal{} NP$).

To prove this, we can make an affine transformation so that $ ACBD$ becomes a rectangle. Let $ Q$ and $ R$ the projections of $ P$ over $ AC$ and $ BD$, respectively. Now let us compute the distance of $ z \equal{} PQ$ if $ AD \equal{} x$ and $ CN \equal{} y$. Clearly

$ \frac {z}{y} \equal{} \frac {PQ}{CN} \equal{} \frac {AQ}{AC} \equal{} \frac {DR}{DB} \equal{} \frac {PR}{BC} \equal{} \frac {x \minus{} z}{x} \Longrightarrow \frac {1}{y} \equal{} \frac {x \minus{} z}{xz} \equal{} \frac {1}{z} \minus{} \frac {1}{x}$,

q.e.d.

jayme, could you please explain better your solution?

[hide="without inversion"]Let $ X$ be a point on $ C_1$ not on arc $ AMD$. Let $ Y$ be a point on $ C_2$ not on arc $ ANC$.
$ \angle CAD$ is a degenerate inscribed angle (one with a tangent to the circle) of both circles $ C_1$ and $ C_2$, so
$ \angle CAD \equal{} \dfrac{1}{2} AMC \equal{} \dfrac{1}{2} ANC$. $ AMD \equal{} ANC$ so $ AXD \equal{} AYC$
$ \angle AMD \equal{} \dfrac{1}{2}AXD \equal{} \dfrac{1}{2}AYC \equal{} \angle ANC$.
$ \angle ADM \equal{} \angle CAM \equal{} \angle CAN$ because $ \angle CAM$ is a degenerate inscribed angle of circle C.
By AA Similarity $ \triangle ADM \sim \triangle CAN$.
Let $ DM$ meet $ NC$ at $ Z$.
$ \angle CNP \equal{} 180^\circ \minus{} \angle ANC \equal{} 180^\circ \minus{} \angle AMD \equal{} \angle DMP \equal{} \angle AMZ$.
$ \triangle ZMN$ is isosceles, so $ MZ \equal{} ZN$.
The median of $ \triangle ZMN$ from $ Z$ to $ MN$ is perpendicular to $ MN$. Let $ S$ be the point of intersection, so $ MS \equal{} SN$.
Extend ray $ CN$ to meet the circumcircle at $ T$.
$ \angle DTP \equal{} \angle DAP \equal{} \angle DAM \equal{} \angle NCA \equal{} \angle ACT \equal{} \angle TDA$.
$ \angle CTP \equal{} \angle CAP \equal{} \angle ADM$.
$ \angle DTZ \equal{} \angle DTP \plus{} \angle CTP \equal{} \angle TDA \plus{} \angle ADM \equal{} \angle TDZ$.
$ 2\angle DTZ \equal{} \angle DTZ \plus{} \angle TDZ \equal{} 180^\circ \minus{} \angle TZD \equal{} \angle ZNS \plus{} \angle ZMS \equal{} 2\angle ZNS$
so $ \angle DTZ \equal{} \angle ZNS$ and $ DT$ is parallel to $ MN$.
Ray $ ZS$ which is perpendicular to $ MN$, is perpendicular to $ DT$. Since $ \triangle DTZ$ is isosceles with $ \angle DTZ \equal{} \angle TDZ$, ray $ ZS$ bisects $ DT$.
Since line $ ZS$ is the perpendicular bisector of $ DT$, a chord of the circumcircle, line $ ZS$ is along a diameter of the circumcircle. Since line $ ZS$ is perpendicular to $ AP$, it bisects $ AP$. Thus $ AS \equal{} SP$. Since $ MS \equal{} SN$, $ AS \plus{} SM \equal{} SP \plus{} NS$.
$ AM \equal{} NP$, as desired.[/hide]

[quote="feliz"]The problem said "please, use inversion! USE! I NEED IT! Pleaaaaaaaaaaaaaase! I BESEECH! I do anything, but invert me!". [/quote]
 :lol: Nice solution feliz and thanks Mij for a solution without inversion.

Thanks everybody. I don't know what inversion is, so i'll read Mij's solution this eveing - maybe tomorrow.  :)

Ptolemy theorem will do here.

Let the radius of $ C_{1}$ and $ C_{2}$ be $ R_{1}$ and $ R_{2}$ respectively, then
$ AP \equal{} \frac {PD}{CD} \times AC \plus{} \frac {PC}{CD} \times AD\\
     \equal{} \frac {sin\angle PAD}{sin\angle DAC} \times 2R_{2} \times sin\angle DAC \plus{} \frac {sin\angle PAC}{sin\angle DAC} \times 2R_{1} \times sin\angle CAD\\
     \equal{} sin\angle PAD \times 2R_{2} \plus{} sin\angle PAC \times 2R_{1}\\
     \equal{} sin\angle NAD \times 2R_{2} \plus{} sin\angle MAC \times 2R_{1}\\
     \equal{} sin\angle NCA \times 2R_{2} \plus{} sin\angle MDA \times 2R_{1}\\
     \equal{} AN \plus{} AM$

Hence, $ AM \equal{} AP \minus{} AN \equal{} NP$.


Each member of the sequence $112002, 11210, 1121, 117, 46, 34,\ldots$ is obtained by adding five times the rightmost digit to the number formed by omitting that digit.  Determine the billionth ($10^9$th) member of this sequence.

[b]Elementary solution:[/b] the sequence goes 112002,11210,1121, 117, 46, 34, 23, 17, 36, 33, 18, 41, 9, 45, 29, 47, 39, 48, 44,24, 22, 12, 11, 6, 30, 3, 15, 26, 32, 13, 16, 31, 8, 40, 4, 20, 2, 10, 1, 5, 25, 27, 37, 38, 43, 19, 46, and hence repeats per 42 and we only need to investigate the $10^9 \bmod 42$th element.
Since there are 42 integers appearing in the sequence, and the relation is recursive with 1 parent, we can take the index mod 42.

[b]More interesting solution:[/b] Consider the set $A=\{i\in\mathbb{N}|i>0,i<50,7\not|i\}$ and denote $f(x)$ the number after $x$ in the sequence. We can check that $f(x)\in A \Leftrightarrow \forall x\in A$, thus $f$ is bijective. We now know that if $G=\{f^{k},k\in N\}$, then $(G,\circ)$ is a group. As the relation was bijective, and our group is cyclic, $|G|$ divides 42, thus $f^{42}=e$, thus we can take the index mod 42.

*brains start working late today*

This is a very efficient trick to find out if large numbers are a multiple 7. Thanks, didn't know that one yet! :)


$ \triangle ABC$ is isoceles and $ \angle BAC \equal{} 30^o$. $ D$ is a point on the circumcircle of $ \triangle ABC$ such that
the [b]arc[/b] of $ DC \equal{} 30^o$. $ G$ is a point on the [b]arc[/b] of $ BC$ such that $ (arc$ $ of$  $ BG ) >$ ( $ arc$ $ of$ $ AG)$.
The line segment $ DG$ cuts $ AC$ and $ AB$ respectively $ E$ and $ F$.

If $ AC \equal{} AB \equal{} DG$ then $ \frac {area\triangle(AFE)}{area\triangle(ABC)} \equal{} ?$

Let $ R$  the radius of cicumcircle
It is easy to prove that $ \triangle$$ AFG$ equilateral
then $ AF\equal{}2Rsin15^{\circ}$  and   $ AE\equal{}\frac{R\sqrt3}{2cos15^{\circ}}$
so $ \frac{ [AEF] }{ [ABC] }\equal{}\frac{\sqrt3tan15^{\circ} }{4cos^215^{\circ}}\equal{}7\sqrt3\minus{}12$


Given a permutation £=£1 £2 £3...£n of the set {1,2,...,n},

define the distance g(£) of £ by g(£)=max|£(n)-n|. Find the number of

permutations for which

a)g(£)<2

b)g(£)<3

c)g(£)<4

Here is the solution for case (a).

Let $ p$ be a permutation of $ \{1,\dots,n\}$.
$ |p(i)\minus{}i|<2\quad \forall i \quad \Rightarrow |p(i)\minus{}i|\equal{}0\quad or \quad 1\quad  \forall i$. Let $ f(n)$  be the number of permutations with the above mentioned restriction.

If $ p(1)\equal{}1$, we have $ f(n\minus{}1)$ permutations. Else if $ p(1)\equal{}2$ then $ \Rightarrow p(2)\equal{}1$ and we have $ f(n\minus{}2)$ permutations.$ f(1)\equal{}1$ and $ f(2)\equal{}2$. $ f(n)$ is the $ n$th fibonacci number.


If $ x$ is from interval $ (0,\pi)$ and $ f(x) \equal{} sin^{2}(x)$ then determine is $ f$ concave or convex function?

$ f(x)\equal{}\frac{1\minus{}\cos{2x}}{2}$.Hence $ f''(x)\equal{}(\sin{2x})'\equal{}2\cos{2x}$.
If $ x\in (0,\frac{\pi}{4})\cup({\frac{3\pi}{4},\pi})$ then $ f$ is convex,otherwise concave.


How do you type a double digit exponent in latex? For example, when I try to type 10^11, I get $10^{1}1$, and if I insert parentheses, I get $10^{(}11)$.

you can also try 10^1^1 or
$10{}^{1}{}^{1}$. Arne's is much better though. :)

Basically, use the {} whenever you want to put an expression longer than one character in any subscript/superscript/fraction/etc. If you ever want to display a { or }, use \{ or \}.


A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square?

$ \textbf{(A)}\ \frac{\pi}{16}\qquad
\textbf{(B)}\ \frac{\pi}{8}\qquad
\textbf{(C)}\ \frac{3\pi}{16}\qquad
\textbf{(D)}\ \frac{\pi}{4}\qquad
\textbf{(E)}\ \frac{\pi}{2}$

[hide]Let the side of the largest square be x

Then, the diameter inscribed in the circle must be x
Which also means that x is the diagonal length of the square inscribed that circle. and we can find the length of the square by dividing x by √2

Once again x / √2 acts as a diameter for the circle, thus the radius must be x / 2√2
So we do (x / 2√2)^2 \pi : x^2
and we get (B) \pi / 8 [/hide]

uhm sry for the messiness, I haven't figured out how to use latex yet T_T I'll try to learn it now and maybe edit this post.[/hide]

[hide="Here"]
Let the radius of the smaller circle be $ r$.  Then the side length of the smaller square is $ 2r$.  The radius of the larger circle is half the length of the diagonal of the smaller square, so it is $ r\sqrt{2}$.  Hence the larger square has sides of length $ 2\sqrt{2}r$.  The ratio of the area of the smaller circle to the area of the larger square is therefore

$ \frac{\pi r^2}{\left (2\sqrt{2}r \right )^2}\equal{}\frac{\pi}{8}$ or $ \boxed{\textbf{(B)}}$.
[/hide]


function
topology
probability
trigonometry
calculus
derivative
real analysis

The set of real valued functions continuous on $[0,1]$ but nowhere differentiable is dense in $C([0,1] ; \mathbb{R})$ (equipped with the sup-norm).

This was given as the most difficult problem on my topology exam, although with plenty of hints. The solution set out for you in the hints uses Baire's theorem, but maybe there is some more explicit way?

take your favorite example of nowhere differentiable continuous function $g$ on $[0;1]$. You can suppose, changing it a little bit, that $\|g\|_{\infty}=1$. Now, it $f$ is an arbitrary continuous function on $[0;1]$ and $\epsilon > 0$, find a smooth function $\phi$ such that $\|f-\phi\|_{\infty}<\epsilon$ (this is possible by Weierstrass theorem for example). Then $f_{\epsilon}=\phi+\epsilon g$ is nowhere differentiable but $\|f-f_{\epsilon}\| < 2 \epsilon$

Aha, right. So I suppose actually the difficult part of this question is to establish that such a function exists. I didn't think of the fact that the density followed so easily. The hint given in the exam was the following:

Consider the subset $E_{n}$ of $C[0,2]$ consisting of functions $f$ for which there is an $x\in [0,1]$  such that $\left|\frac{f(x+t)-f(x)}{t}\right| \leq n$ for all $t$, $0 < t < 2-x$. Show that $E_{n}$ is closed and nowhere dense and use Baire's theorem.

I guess this is a pretty neat way to show the existence of such functions.

or if you know probability, take the trajectory of a brownian motion  :)

There's also the classic Weierstrass example: consider $f(x)=\sum_{n=0}^{\infty}2^{-n}\cos(4^{n}\pi x)$. This sum converges uniformly to a continuous function by the M-test.
Now, choose $x$ arbitrarily and define $a_{n}$, $b_{n}$ as follows: $a_{n}$ is the largest number of the form $\frac{k}{4^{n}}$ less than or equal to $x$, and $b_{n}$ is the smallest number of the form $\frac{k}{4^{n}}$ greater than equal to $x$. This gives $b_{n}-a_{n}=4^{-n}$. Now, $|f(b_{n})-f(a_{n})|=|\sum_{k=0}^{n-1}2^{-k}\left(\cos(4^{-k}\pi b_{n})-\cos(4^{k}\pi a_{n})\right)+2^{-n}\left(\cos(4^{n}\pi b_{n})-\cos(4^{n}\pi a_{n})\right)$
$+\sum_{k=n+1}^{\infty}2^{-k}\left(\cos(4^{k}\pi b_{n})-\cos(4^{k}\pi a_{n})\right)|$
$|f(b_{n})-f(a_{n})|\ge 2^{-n}\left|\cos(4^{n}\pi b_{n})-\cos(4^{n}\pi a_{n})\right|-\sum_{k=0}^{n-1}2^{-k}\left|\cos(4^{k}\pi b_{n})-\cos(4^{k}\pi a_{n})\right|$
$-\sum_{k=n+1}^{\infty}2^{-k}\left|\cos(4^{k}\pi b_{n})-\cos(4^{k}\pi a_{n})\right|$
The difference of cosines is exactly 2 in the first term. For the second, we use that $\cos$ is Lipschitz with constant 1. For the third, the difference of cosines is always zero.
$|f(b_{n})-f(a_{n})|\ge 2\cdot 2^{-n}-\sum_{k=0}^{n-1}4^{k}\cdot 2^{-k}\cdot 4^{-n}=2\cdot 2^{-n}-4^{-n}\frac{(4/2)^{n}-1}{4/2-1}$
$=2\cdot 2^{-n}-\left(2^{-n}-4^{-n}\right)>2^{-n}$

That makes the difference quotient $\frac{f(b_{n})-f(a_{n})}{b_{n}-a_{n}}$ greater than $2^{n}$ in absolute value. If $f$ was differentiable at $x$, these difference quotients would have to converge to $f'(x)$; they are weighted averages of the difference quotients $\frac{f(b_{n})-f(x)}{b_{n}-x}$ and $\frac{f(x)-f(a_{n})}{x-a_{n}}$, which would converge to $f'(x)$. By our calculation, these difference quotients diverge, and $f$ is not differentiable at $x$.

The choice of $\cos$ is not critical; any periodic Lipschitz function would work, possibly with tweaked coefficients.

Suppose we have set $A$ which is a subset of set $B,$ and the question arises: is $A$ all of $B,$ or is $B\setminus A$ nonempty? In this particular instance, $B$ is the set of continuous functions (perhaps on some interval) and $A$ is the set of continuous functions that are differentiable at at least one point.

There are two main types of proof for this sort of thing:

Type I - construction. Explicitly produce an element of $B\setminus A.$

Type II - size. Argue that by some "size" criterion, $B$ is "bigger" than $A$ and hence must have elements that $A$ does not have. There are three major types of size:

(a) Cardinality. (For instance, this could be used to prove the existence of transcendental numbers).

(b) Baire category.

(c) Measure.

In this instance, the Weierstrass example and its relatives constitute a Type I argument for the existence of nowhere differentiable continuous functions.

II(a) goes nowhere. In particular, $C^{1},$ which is much smaller than the set $A$ we are really interested in, has the same cardinality as $C.$ But II(b) can be used to show there must exist such a function without actually constructing it. That was clearly the intent of the writers of this problem and Kalle has pointed in the right direction.

And then there's II(c), measure. Or probability, which is the same thing, and alekk has mentioned it. Brownian motion is a way to assign a probability measure to a set of continuous functions (with a common base point.) What we need to prove is that the probability of a Brownian path having a derivative at any point is zero.

This follows from Markov's property pretty easily once we show it is almost surely not differentiable in $0$. But one can show that $B_{t}/\sqrt{t}$ has superiour and inferiour limits at $0$ equal to $\infty$ and $-\infty$.


Let $\mathcal{C}_1$, $\mathcal{C}_2$ and $\mathcal{C}_3$ be three circles, of radii 2, 4 and 6 respectively. It is known that each of them are tangent exteriorly with the other two circles. Let $\Omega_1$ and $\Omega_2$ be two more circles, each of them tangent to all of the 3 circles above, of radius $\omega_1$ and $\omega_2$ respectively.
Prove that $\omega_1+\omega_2=2\omega_1\omega_2$

Hi Silouan ;)
In my solution I use this  lemme.
[b]Lemme .[/b]
Let$C_1$, ,$C_2$and$C_3$ be three circles,of radii $a,b,c$.It is known that each of them are tangent exteriorly with the other two circles. Let $\Omega_1$ and$\Omega_2$  be two more circles, each of them tangent to all of the 3 circles($\Omega_1$internaly,$\Omega_2$ extenaly) above, of radius $d$ and $e$ respectively.
Then $(1/a+1/b+1/c+1/d)^2=2(1/a^2+1/b^2+1/c^2+1/d^2)$
and$(1/a+1/b+1/c-1/e)^2=2(1/a^2+1/b^2+1/c^2+1/e^2)$

[quote="Virgil Nicula"]If the circles $C_k=C(O_k,R_k),\ k\in \overline {1,3}$ and $\Omega_e=C(O,R),\ \Omega_i=C(I,r)$
have the mentioned above property, then:
$\blacksquare\ 1^{\circ}.$ This problem makes a sens iff $\frac{1}{\sqrt{R_k}},\ k\in \overline {1,3}$
can be the lengths of the sides of a some triangle.
$\blacksquare\ 2^{\circ}$. If $R_k,\ k\in \overline {1,3}$ are the roots of the equation $x^3-mx^2+nx-p=0$,
i.e. $R_1+R_2+R_3=m$, $R_1R_2+R_2R_3+R_3R_1=n$, $R_1R_2R_3=p$,
then $R=\frac{p}{2\sqrt{mp}-n}$ and $r=\frac{p}{2\sqrt{mp}+n}.$
$\blacksquare\ 3^{\circ}.\ \left(\frac 1R+\frac 1r\right)^2=16\cdot \frac{R_1+R_2+R_3}{R_1R_2R_3},\ \frac 1R+\frac 1r\ge \frac{12\sqrt 3}{R_1+R_2+R_3}.$[/quote]
Have you seen the problem like this one.
.[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64502]www.mathlinks.ro/Forum/viewtopic.php?t=64502[/url]
I can say that this problem is my own problem,but I can't find answer too :rotfl:

[quote="Virgil Nicula"]This problem was published in G.M.\ seria B, 1982 aprox. (author: Virgil Nicula), but I think this problem is very wellknown formerly.[/quote]
You mean the poblem from this topic ?? :?
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64502]www.mathlinks.ro/Forum/viewtopic.php?t=64502[/url]

[quote="Virgil Nicula"]No, but the problem from the my last reply (with the three points).[/quote]
O.K. I see :).

[quote="perfect_radio"]Tiks, how did you prove your lemma?[/quote]
Good question :D.
Let $A,B$ and $C$ the centers of circles$C_1,C_2$ and $C_2$res.
$D$center of $\Omega_1$.Then you can easy see that
$cos^2(\frac{\angle BDC}{2})=\frac{d(b+c+d)}{(b+d)(c+d)}$
$sin^2(\frac{\angle BDC}{2})=\frac{bc}{(b+d)(c+d)}$
similar for angles$\angle ADB$ and $\angle ADC$.
If $x+y+z=180^0$ then $sin^2x=sin^2y+sin^2z-2sinxsinycosz$(laste one equivalent to cosinus's theorem :) ).=>
If  $2x=\angle BDC,2y=\angle ADB,2z=\angle ADC$ then
$sin^2x=sin^2y+sin^2z-2sinxsinycosz$<=>
$\frac{bc}{(b+d)(c+d)} =\frac{ac}{(a+d)(c+d)}+\frac{ab}{(a+d)(b+d)}-2\frac{a\sqrt{bcd(b+c+d)}}{(a+d)(b+d)(c+d)}$<=>$(1/a+1/b+1/c+1/d)^2=2(1/a^2+1/b^2+1/c^2+1/d^2)$ ;)
I hope  you can prove second equation with similar method  ;) .


Maybe this will give an answer to your question. I have to learn it for my school exam.

The registered amount of money is defined as the total amount of money in the form of notes, coins and deposit money which is not in the hands of money creating instances. There are a couple of instances that are able to create or destroy money. The amounts of availible notes and coins are usually managed by the central bank or by the government. In my country for example the ECB (European Central Bank) is responsible for the production of notes and the Chancellor of the Exchequer for the production of coins. The amount of money in account however is managed by credit establishments, which are in most of the cases banks.

There are two special groups of transactions with money creating instances. First you have substitutions, by which the registered amount of money remains constant. This is for example when you deposit money into a bank. Then the amount of notes and coins will decrease and the amount of deposit money will increase, but the sum of both kinds doesn't change. Another kind of transaction is called a transformation, by which the amount of registered money is altered. An bank can buy computers for example and pay by raising the balance of the bank account of the supplier, without lowering the balance of another bank account. In this case the amount of deposited money will raise and the amount of other kinds of money doesn't change.

In general the demand for goods will grow when the amount of money grows. If the supply of goods remains constant, there can occur inflation and raising interests and exchange rate.

Are you sure you got it right? I think depositing money increases the supply and does not make it remain constant, I just don't really understand why.

Depositing money increases the supply of money but not the supply of goods. If the supply of money grows, it will cause that the demand for goods will raise, which will lead to higher prices.

Here is another way to explain the same ideas.

When you deposit money in a bank, it doesn't just sit there.  Banks lend money to other customers with capital needs and charge interest.  Some of the revenues generated from that interest are passed back to you for supplying the capital.  Presumably that interest is the result of greater economic value generated by the availability of cash funds for useful purposes.

The money supply does not need to increase for you to gain more money.  Paper money is not the only form of capital -- all goods and services are capital.  So, while businesses and people who borrowed money generate other forms of capital, the total wealth of the system increases even while the supply of paper money stays the same.  You get more cash in your interest-earning bank account, but that doesn't mean anybody lost in the exchange because of the other forms of capital generated.

Economics gravitates toward an optmization of small win-win scenarios like that.

The [basic] function of paper money is nothing more than liquidity -- greasing the wheels of economic exchange by representing value in a form that can be more easily exchanged, but we shouldn't confuse is with the total capital supply.

I am not sure what you mean by:

When you deposit money in a bank, it doesn't just sit there. Banks lend money to other customers with capital needs and charge interest. Some of the revenues generated from that interest are passed back to you for supplying the capital. Presumably that interest is the result of greater economic value generated by the availability of cash funds for useful purposes.

I am not sure I understand the explanation.

The bank serves as a conduit for cash.

Some people deposit.

Some people borrow.

The interest paid by the borrows goes to the bank who pass most of it off to the depositers for providing the pool of cash.

Ultimately, that's what interest is -- payment for the use of cash.  It is set according to how well cash generates an increase in overall capital from investments.

Does this make more sense?

Yes, I know what you mean.

Well, this is how I learned it. Depositors put money in the bank, which keeps some in reserves and lends the rest to others with interest. This money eventually goes back into the bank increasing the money supply. What I dont understand is why the loans never decrease as they are paid back, where is the more coming from, like logically I don't understand how something is made out of nothing. Sorry for showing my stupidity.

The principle of a loan does not decrease, but the amount owed does, and that's the important part.  The principle is just for initial value calculations.

The formula for Money created over the long term is this:
Money deposited/ percent  required to remain in bank.

i.e. M/a

The reason for this is that it is a geometric series:
1st time deposited: M "created"
Loaned by bank, 2nd time deposited: M(1-a) created
Loaned by bank, 3rd time deposited: M(1-a)^2

Total money "created": M + M(1-a) + M(1-a)^2 + M(1-a)^3......  M(1-a)^infinity

This is a sum of a geometric series.
Sum = G(i)(1-r^n)/(1-r)

In our case, Sum = M(1-(1-a_^infinity)/(1-(1-a))
Since, (1-a) is very small as n goes to infinity (1-a is less than 1), we are left with
= M/(1-(1-a))
=M/a
=Money deposited/ percent  required to remain in bank.

Amazing how math can show this (Thanks to AoPS I proved this!).

Of course, the actual amount created is much less. (Not all is lent, someone stops lending, etc.. loss in transactions)

[quote="hello"]Somehow I never managed to understand how money is created by being deposited into banks? TO me, it seems illogical but can someone give a simple explanation so I can understand?[/quote]

It has probably been explained well enough by others, but banks are like money amplifiers. They are required hold a percentage of their total outstanding loans as cash-on-hand. Given that this percentage is less than 1, the bank's loan to value ratio (LTV) is greater than 1.

If, for example, the cash requirement is 15% then $LTV=\frac{1}{0.15}\approx 7$

That means if you deposit a dollar, the bank can loan 7 dollars to others. This is amplified further by the multiplier effect itself, which is essentially a the sum of an infinite geometric series.

Let us say that you are paid one dollar, in the American economy on average you will spend about 96 cents of this dollar in exchange for goods and services. The recipients of this 96 cents then each spend 96% of that amount, and so on, an "infinite number of times."  The sum of such an infinite series is 1/(1-0.96) = 25.

Putting these two things together, if you deposit a dollar, the bank loans 7 dollars, which in turn gets reinvested in the economy for 25 times growth. Therefore, your 1 dollar deposit = 175 dollars of economic growth.*

*There are 3 kinds of mathematicians: those who can count, and those who cannot.


geometry
perimeter
inequalities
trigonometry
perpendicular bisector
Pythagorean Theorem

Point $B$ is on $\overline{AC}$ with $AB = 9$ and $BC = 21$.  Point $D$ is not on $\overline{AC}$ so that $AD = CD$, and $AD$ and $BD$ are integers.  Let $s$ be the sum of all possible perimeters of $\triangle ACD$.  Find $s$.

[hide]Let $h$ be the height of triangle $\triangle ACD$.  Then:

$h^2+15^2=AD^2$
$h^2+6^2=BD^2$

$189=AD^2-BD^2$
$189=(AD+BD)(AD-BD)$
$3^3*7=(AD+BD)(AD-BD)$

Because $AD$ and $BD$ are positive, $AD+BD>AD-BD$.  So the following two systems can be set up (one can be disregarded because it makes $h=0$):

$AD+BD=189$
$AD-BD=1$
$AD=95$
$BD=94$

$AD+BD=63$
$AD-BD=3$
$AD=33$
$BD=30$

So the perimeters are $9+95+94=198$ and $9+33+30=72$.  Their sum is $\boxed{270}$.[/hide]

[hide]I think you missed a case

$AD+BD=27$

$AD-BD=7$

$\implies AD=17, BD=10$ and rather than finding perimeter of $\triangle ABD$ we need to find perimeter of $\triangle ADC$

so final sum is $95\cdot 2 + 33\cdot 2 + 17\cdot 2 +3\cdot 3 =380$[/hide]

[hide]Let $M$ be the midpoint of $AC$.  Let $x=DM$, $n=AD$, and $m=BD$.  Since $AD=CD$, $D$ lies on the perpendicular bisector of $\overline{AC}$.  Therefore, $\overline{DM}\perp\overline{AC}$.

We have $AM=\frac{9+21}{2}=15$ and $BM=15-9=6$.  By the Pythagorean Theorem on $\bigtriangleup ADM$, we have $n^{2}=x^{2}+225$.  By the Pythagorean Theorem on $\bigtriangleup BDM$, we have $m^{2}=x^{2}+36$.  Subtracting this equation from the previous equation gives $n^{2}-m^{2}=189=3^{3}\cdot 7$.  Factoring, we have $(n-m)(n+m)=3^{3}\cdot 7$.  Since $n$ and $m$ are integers, $n-m$ and $n+m$ are also integers.

Since the hypotenuse of a right triangle is greater than either leg, we have $n>15$ and $m>6$.  Adding these two inequalities gives $n+m>21$.  From the equation above, using only solutions that satisfy this inequality, we have $(n-m,n+m)=(1,189),(3,63),(7,27)$.  From these solutions, we have $(n,m)=(95,94),(33,30),(17,10)$.  The perimeter of the triangle in each case is $2n+30$, so the sum of the perimeters is $2\cdot 95+30+2\cdot 33+30+2\cdot 17+30=\boxed{380}$.[/hide]

[quote="Iversonfan2007"][quote="computer_nuke"][quote="Iversonfan2007"]eh...what about just using plain old stewart's theorem? That'll get to the $x^{2}-y^{2}=189$ in two steps :wink:[/quote]

I did it just as the other's did.
I never remember stewart's theorem, I always have to derive it when I need it again. Is there a easy way to remember it?[/quote]

$man+dad=bmb+cnc$ with $a=m+n$

And this is may sound like a REALLY stupid way to remember it :blush:, but if we have a triangle $ABC$, with $A$ the top vertex, $B$ the bottom left vertex, and $C$ the bottom right vertex, and $AP$ the cevian, then what we have is $c=AB, m=BP, d=AP, b=AC, n=PC$. Notice that the distribution of $c,m,d,b,n$ (in that order) for $\triangle{ABC}$ looks kinda look a $\sin$ curve.  :oops:[/quote]

Man+Dad=Bmb+Cnc, wow nice, no its not stupid, and maybe corny but its good  :)  Thanks, I'll remember that now lol

[quote="Iversonfan2007"][quote="computer_nuke"][quote="Iversonfan2007"]eh...what about just using plain old stewart's theorem? That'll get to the $x^{2}-y^{2}=189$ in two steps :wink:[/quote]

I did it just as the other's did.
I never remember stewart's theorem, I always have to derive it when I need it again. Is there a easy way to remember it?[/quote]

$man+dad=bmb+cnc$ with $a=m+n$

And this is may sound like a REALLY stupid way to remember it :blush:, but if we have a triangle $ABC$, with $A$ the top vertex, $B$ the bottom left vertex, and $C$ the bottom right vertex, and $AP$ the cevian, then what we have is $c=AB, m=BP, d=AP, b=AC, n=PC$. Notice that the distribution of $c,m,d,b,n$ (in that order) for $\triangle{ABC}$ looks kinda look a $\sin$ curve.  :oops:[/quote]

 :D That's how I remember it too.

Couldn't this go off to infinity? Eg, if AD = 1000, let BD = say,  992..

well, realize that the difference between two consecutive perfect squares gets greater and greater as the numbers go up; thus, if you had an outrageously big number like 1000, then the distance between $1000^2$ and $999^2$ would be WAYYY greater than 189; in fact, the difference between those two perfect squares is 1999 by DOTS.

Here's why the difference between consecutive squares $x^2$ and $(x-1)^2$ gets larger as x gets larger:

We are trying to find $x^2 - (x-1)^2$, which, by DOTS, becomes $(2x-1)(1)=2x-1$. Obviously, as x gets infinitely larger, $2x-1$ gets infinitely larger as well.

Just realized. The problem never said that the order of the points on the line had to be A,B,C. It could be in the order B A C for example which yields other results.

Uh, it says $B$ lies on the [i]segment[/i] $\overline{AC}$, which by definition is only defined between points $A$ and $C$.  So the order of the points must be $A$, $B$, $C$.

Can one do this with the triangle inequality?

^^ why does triangle inequality not seem to work here?

This problem is wrong. Why?
Since AD=CD then <DAC=<DCA
Apply the cosine rule in ABD with <DAC to get the value of BD.
Apply the cosine rule in BCD with <DCA to get the value of BD again. See that the two values of BD are different(since <DAC = <DCA). Hence the problem contradicts itself.

This is because $AB$ and $BC$ aren't nessesarily the same.

[b]Solution:[/b] It is easy to notice that $\sqrt{x^2 + 36}$ and $\sqrt{x^2 + 225}$ must both be integers for this to be true. Let $\sqrt{x^2 + 36} = m$ and $\sqrt{x^2 + 225} = n$ which is just begging for a difference of squares exploitation so we get that $n^2 - m^2 = 189 \Rightarrow (n-m)(n+m) = 189$ which gives us that $n = 95, 33, 15$. This gives us three possible values of the perimeter and out answer is the sum of all possible values of $2n + 30$: $2(95) + 2(33) + 2(17) + 3(30) = \boxed{380}$.


$ M$ is the midpoint of side $ BC$ of $ \triangle{ABC}$. $ E$ and $ F$ are on $ AB$ and $ AC$ such that $ ME \perp AB$, $ MF \perp AC$.

Prove that $ AM$ bisects $ EF$ if and only if $ AB\equal{}AC$ or $ AB \perp AC$.

[color=#666666]Hi............[/color]

$ D$ is the point that $ AM$ meet $ EF$.
If $ AB \equal{} AC$ then $ ED\equal{}DF$ easy proved because ... .
If $ AB\perp AC$ then $ ED\equal{}DF$ easy proved because $ AEMF$ is oblong.

IF $ ED\equal{}DF$ (and $ BM\equal{}MC$) then $ EF \perp BC$ so $ \dfrac{ED}{BM}\equal{} \dfrac{AD}{AM}$ and $ \dfrac{DF}{MC}\equal{}\dfrac{AD}{AM}$ $ ,BM\equal{}MC \Rightarrow ED\equal{}DF$

isn't easy(all problem is easy when solved :wink: )

[color=#666666]Bye............[/color]


integration
trigonometry
logarithms
calculus
algebra
partial fractions
geometry

So here are 6 of the 7 problems.  Hoping for solutions:

1) Evaluate $ \int_{0}^{x}\frac{d\theta}{2+\tan{\theta}}$, where $ 0\leq x\leq \pi/2$.  Use your result to show that $ \int_{0}^{\pi/4}\frac{d\theta}{2+\tan{\theta}}=\frac{\pi+\ln(9/8)}{10}$.

2) Given that $ e^{x}=1/0!+x/1!+x^{2}/2!+\cdots+x^{n}/n!+\cdots$ find, in terms of $ e$, the exact values of
a) $ \frac{1}{1!}+\frac{2}{3!}+\cdots+\frac{n}{(2n-1)!}+\cdots$ and
b) $ \frac{1}{3!}+\frac{2}{5!}+\cdots+\frac{n}{(2n+1)!}+\cdots$

3) Solve the initial value problem $ \frac{dy}{dx}=y\ln{y}+ye^{x}$, $ y(0)=1$ (find $ y$ in terms of $ x$).

5) Find the third digit after after the decimal point of $ (2+\sqrt{5})^{100}((1+\sqrt{2})^{100}+(1+\sqrt{2})^{-100})$

6) Let $ n$ be a positive integer, let $ A$, $ B$ be a square symmetric $ n\times n$ matrices with real entries (so if $ a_{ij}$ are the entries of $ A$, the $ a_{ij}$ are real numbers and $ a_{ij}=a_{ji}$).  Suppose there are $ n\times n$ matrices $ X$, $ Y$ (with complex entries) such that $ \det(AX+BY)\neq 0$.  Prove that $ \det(A^{2}+B^{2})\neq 0$.

7) Determine whether the series $ \sum_{n=2}^{\infty}n^{-(1+(\ln(\ln n))^{-2})}$ is convergent or divergent.

The problems can be found [url=http://www.math.vt.edu/people/plinnell/Vtregional/E07/index.html]here[/url]. Here are some of my ideas:
[hide="1"]
Substitute $ u \equal{} \tan{\theta}$. Once you do this, you should get something that can be decomposed into partial fractions and integrated. I got:

$ \int_0^x{\frac{d \theta}{2 \plus{} \tan{\theta}}} \equal{} \frac{1}{5}\ln{\left(\cos{x}\plus{}\frac{1}{2}\sin{x}\right)} \plus{} \frac{2}{5}x$.
[/hide]
[hide="3"]
Substitute $ v \equal{} \ln{y}$. You should get a first-order equation whose solution is $ v \equal{} (x \plus{} c) e^x$. Given the initial value, $ y \equal{} e^{x e^x}$.
[/hide]

Unfortunately, I don't think I'm still in the region of Virginia Tech, but I can still do problems?

1. [hide]Letting $ u \equal{} \tan \theta$, we seek to integrate $ \int_0^{\tan x} \frac {du}{(1 \plus{} u^2)(2 \plus{} u)} \equal{} \int_0^{\tan x} \frac {1}{5} \Big(\frac {1}{2 \plus{} u} \minus{} \frac {u \minus{} 2}{1 \plus{} u^2}\Big)\,du$ $ \equal{}$ $ \frac {1}{5} \ln |u \plus{} 2| \plus{} \frac {2}{5} \arctan u \minus{} \frac {1}{10} \ln |u^2 \plus{} 1| |_0^{\tan x} \equal{} \frac {1}{5} \ln \frac {\tan x \plus{} 2}{2} \plus{} \frac {2}{5} x \plus{} \frac {1}{5} \ln \sec x$. For $ x \equal{} \frac {\pi}{4}$, this yields $ \frac {1}{5} \ln \frac {3}{2\sqrt {2}} \plus{} \frac {\pi}{10} \equal{} \frac {\ln \frac {9}{8} \plus{} \pi}{10}$.[/hide]

2. [hide]The given sum is $ \frac{1}{2} \Big( \sum_{n \equal{} 0}^{\infty} \frac{1}{2n!} \plus{} \frac{1}{(2n\plus{}1)!} \Big) \equal{} \frac{e}{2}$. The other given sum is the same except witha minus sign in the second sum, yielding $ \frac{e^{\minus{}1}}{2} \equal{} \frac{1}{2e}$.[/hide]

3. [hide]Letting $ z \equal{} \ln y$, we have $ \frac {dz}{dx} \equal{} z \plus{} e^x$; we know the homogeneous solution and can implement the initial conditions to find $ z \equal{} xe^x$ and hence $ y \equal{} \exp (xe^x)$.[/hide]

4. [hide] Extending $ QR$ to hit $ BC$ at $ S$, we see $ (B, C; P, S)$ is a harmonic range by a combination of Ceva and Menelaus, so $ (RB, RC; RP, RQ)$ is a harmonic pencil of lines. Intersecting this pencil with a line perpendicular to $ RP$ shows that as the intersection of $ RP$ is the midpoint of the other two intersections, the intersection of $ RQ$ must be infinitely far away and as such, $ RQ \perp RP$. [/hide]

Alternatively, #4 is really just a direct application of Ceva+angle bisector.

[hide]By Ceva, $ 1\equal{}\frac{AR\cdot BP\cdot CQ}{RB\cdot PC\cdot QA}$. Then we notice by the angle bisector theorem that $ \frac{BP}{PC}\equal{}\frac{BR}{RC}$, and plugging into the Ceva, $ 1\equal{}\frac{AR\cdot BR\cdot CQ}{RB\cdot RC\cdot QA}\equal{}\frac{AR\cdot CQ}{RC\cdot QA}$, hence $ \frac{AR}{RC}\equal{}\frac{AQ}{QC}$, which is again angle bisector and means that $ RQ$ bisects $ ARC$. Consequently, $ PRQ$ is half of the (straight) angle $ ARB$, and is thus $ 90^o$.[/hide]
I'm very interested in solutions to 5-7 (particularly 5, 6). Does anybody know of any?

5 looks like a simple estimation. If you add the conjugate in the first factor, you get an integer, and it seems like the terms you're adding all are pretty small in magnitude, so the third (and several more) digits after the decimal point should all be nine, but it's kinda like what why did they ask for the third then. I'll check my work and attempt 6, 7 one of these days.

#7 is kinda cool. I think you can solve it by comparison to the corresponding integral; substitute $ u \equal{} \log(\log(x))$ and you should get

$ \left(e^{e^u}\right)^{\minus{}1 \minus{} \frac{1}{u^2}} \cdot e^{e^u \plus{} u}$

for the integrand, which simplifies to

$ e^{u \minus{} \frac{1}{u^2}e^u}$

We are only concerned with what happens when $ u$ is large, and clearly this integrand decays rapidly when $ u$ is large. So the sum should converge.

#7 is also being discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=172530]here.[/url] jmerry used the condensation test.

for problem 6 see

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=957045#957045

------------------------------------------------------------------------------------------
I am curious to see the official solution
Could someone post it please for problem 6

[quote="MysticTerminator"]Unfortunately, I don't think I'm still in the region of Virginia Tech, but I can still do problems?

1. [hide]Letting $ u \equal{} \tan \theta$, we seek to integrate $ \int_0^{\tan x} \frac {du}{(1 \plus{} u^2)(2 \plus{} u)} \equal{} \int_0^{\tan x} \frac {1}{5} \Big(\frac {1}{2 \plus{} u} \minus{} \frac {u \minus{} 2}{1 \plus{} u^2}\Big)\,du$ $ \equal{}$ $ \frac {1}{5} \ln |u \plus{} 2| \plus{} \frac {2}{5} \arctan u \minus{} \frac {1}{10} \ln |u^2 \plus{} 1| |_0^{\tan x} \equal{} \frac {1}{5} \ln \frac {\tan x \plus{} 2}{2} \plus{} \frac {2}{5} x \plus{} \frac {1}{5} \ln \sec x$. For $ x \equal{} \frac {\pi}{4}$, this yields $ \frac {1}{5} \ln \frac {3}{2\sqrt {2}} \plus{} \frac {\pi}{10} \equal{} \frac {\ln \frac {9}{8} \plus{} \pi}{10}$.[/hide]

[/quote]

how did he/she come up with
$ \int_0^{\tan x} \frac {du}{(1 \plus{} u^2)(2 \plus{} u)} \equal{} \int_0^{\tan x} \frac {1}{5} \Big(\frac {1}{2 \plus{} u} \minus{} \frac {u \minus{} 2}{1 \plus{} u^2}\Big)\,du$
What was the motivation? Explain the details please

It's a substitution.

Let $ u\equal{}\tan{\theta}$.  Then $ du\equal{}\sec^{2}{\theta}\equal{}1\plus{}u^{2}\ d\theta$.

From there, it's just a partial fraction decomposition.


http://www.artofproblemsolving.com/Forum/viewtopic.php?t=317660

Where's this problem from? It's kind of strange that two people are posting it at the same time...

First of all it is well-known that:$ (a\plus{}b)^n\equiv b^n (mod a)$.Hence $ 10\equal{}7\plus{}3$ ,using that theorema we have that:
$ 10^{10}\plus{}{10}^{10^{2}}\plus{}...\plus{}10^{10^{10}}\equiv 3^{10}\plus{}3^{10^{2}}\plus{}...\plus{}3^{10^{10}} (mod 7)$
We observe that $ 3^3\equiv(\minus{}1) (mod 7)$ Hence $ 3^{10}\equal{}(3^3)^3*3\equiv(\minus{}1)^3*3\equiv(\minus{}3)\equiv 4(mod 7)$
In this way we can prove that $ 3^{10}\equiv3^{10^{2}}\equiv...\equiv3^{10^{10}}\equiv 4(mod 7)$,so
$ 10^{10}\plus{}10^{10^{2}}\plus{}...\plus{}10^{10^{10}}\equiv 4*10\equiv5(mod 7)$.This means that the remainder is [b]5[/b].


geometry
geometric transformation
Functional Analysis
advanced fields
advanced fields unsolved

This is essentially the Dixmier approximation theorem. Like everything else in operator algebras, it's in Kadison & Ringrose (Section 8.3). Better still, look up the treatment by Dixmier himself, Chap.III§5 in [i] Les algèbres d'opérateurs dans l'espace hilbertien[/i] (English translation: [i]Von Neumann Algebras[/i]).

[quote="Kuba"]Let $ H$ be an infinite dimensional Hilbert Space and $ x\in B(H)$.
Define $ K(x) \equal{} \overline{co}\{uxu^*: uu^* \equal{} u^*u \equal{} 1, u\in B(H)\}$. (convex (norm)closed envelope).
Prove that $ K(x)\cap \mathbb{C}\cdot 1 \equal{} \overline{\{ < x\xi,\xi > : \xi\in H, \parallel{}\xi\parallel{} \equal{} 1\}}$.[/quote]
[quote="Kuba"]I don't see how you obtain this from Dixmier approximation theorem. Could you give more details please?[/quote]
My mistake — I misread the question. The Dixmier theorem says that $ K(x)\cap \mathbb{C}\cdot 1$ is nonempty. What this question asks for is unclear, since $ K(x)\cap \mathbb{C}\cdot 1$ is a set of operators, while $ \overline{\{ < x\xi,\xi > : \xi\in H, \parallel{}\xi\parallel{} \equal{} 1\}}$ is a set of scalars. The only way I can make sense of the question is to suppose that it should read "Prove that $ \{\lambda: \lambda\cdot 1\in K(x)\} \equal{} \overline{\{ < x\xi,\xi > : \xi\in H, \parallel{}\xi\parallel{} \equal{} 1\}}$." But that result is not true.

Suppose for example that x is a compact operator. Since the compact operators form a norm-closed two-sided ideal, it follows that K(x) will consist of compact operators. But the only compact multiple of the identity operator is 0. Therefore $ K(x) \equal{} \{0\}$. But it is easy to find compact operators whose numerical range contains nonzero numbers.

The weaker result $ \{\lambda: \lambda\cdot 1\in K(x)\} \subseteq \overline{\{ < x\xi,\xi > : \xi\in H, \parallel{}\xi\parallel{} \equal{} 1\}}$ [i]is[/i] true. (It follows easily from the fact that the numerical range is convex.)

Seems you're right Opalg... I found it as an exercise in some notes on operator theory... but maybe I don't remember it well.


[quote="ac1072"]
we have to prove $9x^{2}y^{2}+x^{4}y^{2}+x^{2}y^{4}+1\ge4x^{3}y^{2}+4x^{2}y^{3}+xy$
[/quote]
Can you explain this step?

P/s: I have 2 solutions for this inequality, and both nicer than yours


geometry circumcircle trigonometry ratio Euler incenter

Let I be the center of a circle inscribed in triangle ABC. AI inersect BC at D, intersects the circumcircle of triangle ABC at E. Prove AE/IE=IE/DE.  :?  :(

EDIT: trig is allowed

Try to prove that the ratio is $\frac{b+c}{a}$.

[quote="D.P.L"]what are a, b and c. :blush:[/quote]
Side lengths of triangles.

Hey shobber, can you post your solution. Thnx. ;)

$\frac{AE}{IE}=\frac{AE}{BE}=\frac{AC}{CD}$ and $\frac{AE}{IE}=\frac{AE}{CE}=\frac{AB}{BD}$. Thus
\[ \frac{AE}{IE}=\frac{AB+AC}{BD+CD}=\frac{c+b}{a} \]
Similarly we can also get $\frac{IE}{DE}=\frac{c+b}{a}$.

I'll take the liberty to clarify Shobbers solution. Hope he is not mad with me :D
 [hide]
First and very important fact $IE=BE=CE$.(this is used to prove Euler formula $d^2=R^2-2Rr$)
 To prove this calculate $\angle BIE=\frac{\angle A+\angle B}2$ as an angle with vertex in the circumcircle and $\angle IBE$ as inscribed.
And next note that triangles $AEB$,$ACD$ and $BED$ are simmilar.  Then $\frac{AE}{IE}=\frac{AE}{BE}=\frac{AC}{CD}=\frac{BE}{ED}=\frac{IE}{ED}$.
[/hide]
Hope i got everything right! ;)

Hey Delta, how did you get IE=BE=CE. And can you provide a diagram for the problem, thnx. :D  :P

Can you give me a full proof and be more specific, thnx :D

[quote="John Liao"]I thought he asked to prove AE/IE=IE/DE. Rather than proving the ration is b+c/a. :?  :?:[/quote]
If the two ratios are all equal to a constant ratio, then these two must be also equal. Do you understand it now?

[quote="John Liao"]Hey Delta, what does d stand for in d^2=R^2-2Rr? :?:[/quote]
The distance between Incentre and Circumcentre.


geometry
geometric transformation
algebra theorems
algebra

[quote="edriv"]On Hojo Lee's home page I found an interesting paper about functional equations (I love them  :) ):
http://ultrametric.googlepages.com/

But... unfortunately it's written in Korean!
If Hojo Lee agrees... I would ask someone who speaks korean to translate it in English, to make in available for more people. I would do the translation myself if I knew Korean  :(  , but all I can do is translate from English to Italian.

Is anyone available for this work?[/quote]

Yes please! Could someone translate it? I imagine it is very very good, and would love to see it :lol:

But I can*t find it please help me how I can find them give me a links please

I'm a Korean studying Math Olympiad. ( Of course, Republic of Korea )
    First, I want to say that
  the Korean students are really proud of him Ho-Joo Lee

  Um.... In Korea, When we say a person's name, we say his or her famaily name first , and first name second, such as
  'Lee Ho Joo'  (Lee is his family name, Ho Joo is his first name . )
 For more example,  Ban Ki Moon (Secretary General of UN. He is a Korean. ) and Lee Myung-bak (The President of Korea now )
  But, as the Korean name system is different from many other nations, Koreans usually follows the Western stryle when using English name.

   Anyway..

 I'm only 15 years old ( Actually, whne a Korean is born, he or she is regarded as one year old . It is the Estern style age system. So, I'm 16 years old in Korea. ) ( I was born in 1993 )
 And I do not know Mathematics well. I'm just a middle school student.
 Moreover , I'm not good at using English.  ( You can probably see some strange English in my writing. )
 So I can't Translate it.

 I'll find somone who would tanslate it into English ( Maybe,  the translater will be supposed to get the permission of Lee Ho Joo first )
 Or if Lee Ho Joo translate it by himself, I'll upload it in this site.

  If you want to ask something to me, send an e-mail to "guldam@naver.com"

  Please use easy English  :lol: English doesn't like me  :)

This is the collention. and.. I will translete it a little. ( I'm a Korean. )

Here is the translation for the preface ..

  I made it now.. ( It is 2 am in Korea.. T.T )

   //  This is not a pdf file.
 I don't know how to make a pdf file. ;;  ( The Micro Word file is not avilable here)

   Download this file and copy all the words to Micor word or the other word processor.

  And.. Please pardon my English  :lol:  ( I'm a Korean )

Just searching 'pdf 변환기' in naver -;

I could make translation but it may takes up too long(about 1 week??) because of my middle test  :(


algebra
polynomial
function
algorithm
algebra proposed

Let $ k \equal{} \sqrt[3]{3}$.
a, Find all polynomials $ p(x)$ with rationl coefficients whose degree are as least as possible such that $ p(k \plus{} k^2) \equal{} 3 \plus{} k$.
b, Does there exist a polynomial $ p(x)$ with integer coefficients satisfying $ p(k \plus{} k^2) \equal{} 3 \plus{} k$

a. It's easy to see that P cannot have degree 1, since this would imply that the minimal polynomial of k has degree at most 2. Then check that $ P(X)\equal{}\frac{1}{2}(X^2\minus{}X)$ is the only polynomial of degree 2 fulfilling the condition.

b. Assume that there is some P(X) with integer coefficients. Since the minimal polynomial of $ k\plus{}k^2$ has degree 3 and has integer coefficients, we can assume that the degree of P has degree 2 at most, wlog, using the division algorithm for polynomials in Z[X] (!). Then a. tells us that P cannot have integer coefficients.

[quote="mr.danh"]Let $ k \equal{} \sqrt [3]{3}$.
a, Find all polynomials $ p(x)$ with rationl coefficients whose degree are as least as possible such that $ p(k \plus{} k^2) \equal{} 3 \plus{} k$.
b, Does there exist a polynomial $ p(x)$ with integer coefficients satisfying $ p(k \plus{} k^2) \equal{} 3 \plus{} k$[/quote]
I am deeply convinced that I have solved this problem already and I also remember that there were plenty posts by me and Rust.
I failed to find this thread,probably because the number of my posts seems to get too large.

[quote="Erken"]I am deeply convinced that I have solved this problem already and I also remember that there were plenty posts by me and Rust.
I failed to find this thread,probably because the number of my posts seems to get too large.[/quote]Just found [url]http://www.mathlinks.ro/viewtopic.php?t=138975&search_id=1071128351[/url]


Estimate the quotient $ \frac{301\times 4.9\times 124}{.5107 \times 149\times 24}$ as an integer.

If I understood well:
$ \frac{301}{149}$ is close to 2
$ \frac{124}{24}$ is close to 5
$ \frac{4.9}{0.5107}$ is close to 10
This means that the fraction is near 100.
If someone wanted to check on calculator, the fraction is equal to 100.143068, or 100 as a integer.


Evaluate $ i^{\frac{1}{i}}$

Solve $ ln(log_{i}(e^{\frac{\pi}{2}})\equal{}x$

What are you supposed to solve for in #2?

I believe ckck means simplify...and I believe FantasyLover's approach can also be applied on the second equation. Substitute in $ i^{\frac {1}{i}}$ for $ e^{\frac {\pi}{2}}$ and you see that $ x = - \text{ln}{i}$.

EDIT: Argh! How do you fix that $ \text{\LaTeX}$ there?

This is a [i]great[/i] example of the perils of taking non rational-real roots of numbers.

Fantasy lover basically said, $ i^{1/i} \equal{} e^{(\pi i/2)/i} \equal{} e^{\pi /2}$, a real number.

What if I were to write $ i^{1/i} \equal{} e^{(\pi i/2 \plus{} 2k i \pi )/i} \equal{} e^{\pi /2 \plus{} 2 k \pi}$ for all integers $ k$?

We have a family of infinitely many roots. We're used to seeing multiple roots when we take rational roots of real numbers, where we can define a principle root because the ratio between consecutive roots is a root of unity. Here, though, we have all real values, and consecutive values differ by a factor of $ e^{2 \pi} \approx 535.492$. Which one's the principle root?

[b]Here's another one to ponder. [/b][i]What are the $ \pi$'th roots of unity?[/i]
Clearly, they satisfy $ x^\pi \equal{} 1 \equal{} e^{2 k \pi i}$
Taking the $ \pi$ root of both sides, we get $ x \equal{} e^{2 k i}$
Now, this is basically saying, all unit-length complex numbers whose radian angle measure can be written as an even integer.

But, since 2 and $ \pi$ are algebraically independent, there'll exist such an integer $ k$ such that the angle $ 2k$ falls in any arbitrarily small sub-range of $ 0\ldots 2\pi$. We can make it fall at, say, $ \pi / 8 \ldots \pi/4$.

But then, raising this to the $ \pi$ power produces something with an argument between about $ 1.234$ and $ 2.468$ radians, rather than the $ 0$ radians we expected (we took the $ \pi$ roots of 1; now raising it to the $ \pi$ power should return $ 1$).

What's going on?


In the game of craps, the player makes a bet and rolls a pair of die. If the sum of the numbers is 7 or 11, the player wins. If it is 2,3, or 12, the player loses. If any other number results, say $r$, then $r$ becomes the player's point and he continue to roll until either $r$ or 7 occurs. If $r$ comes up first he wins, and if 7 comes up first, he loses.

Say you bet 9999 dollars in this game. Find the money that you expect to win or lose.

You have to post full solution in this problem and note, this problem is tough.

What is the payoff? If it's 2:1 it'll be very different than if winning on 1 roll is 2:1 and winning on 5 is 5:1


[quote="yunbo"]Um... is it -infinity?[/quote]

The expression, as it is stated now, is equal to $x^{2}\left(\frac{1}{\sqrt{x^{2}+2x}+\sqrt{x^{2}+x}}+1\right)$. Looks quite $+\infty$ to me...

$\sqrt{x^{2}+2x}>\sqrt{x^{2}+x}$

$x(\sqrt{x^{2}+2x}-\sqrt{x^{2}+x}+x)>x(0+x)$
$x(\sqrt{x^{2}+2x}-\sqrt{x^{2}+x}+x)>x^{2}$

Since $x^{2}$ diverges in the positive direction, the given limit, which is greater than $x^{2}$, also diverges in the positive direction.

$\sqrt{x^{2}+2x}>\sqrt{x^{2}+x}$. It's wrong! :blush:


hey!
this is a question that i've been working on a couple of hours,
can anyone help me?

[b]Given E>0 how close to (0,0) should we take (x,y) to make
|f(x,y)-f(0,0)|<E if f(x,y)= x+y/x^2+y^2+1[/b]

You wrote $f(x,y)=x+\frac{y}{x^{2}}+y^{2}+1.$

This is of course, not continuous - and not even defined - at the origin.

Now, I'm sure you actually meant something else - but if you're going to write in-line plain-text algebra, [i]please use appropriate parentheses![/i]

Hint: $\left|\frac{x+y}{x^{2}+y^{2}+1}-0\right|<\epsilon$ whenever $|x+y|<\epsilon$.


function
quadratics
algebra
system of equations

If the function $f$ is defined by
\[ f(x)=\frac{cx}{2x+3} , ~~~x\neq -\frac 32 ,  \] satisfies $x=f(f(x))$ for all real numbers $x$ except $-\frac 32$, then $c$ is
$\text{(A)} \ -3 \qquad \text{(B)} \ - \frac{3}{2} \qquad \text{(C)} \ \frac{3}{2} \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely determined}$

[quote="minsoens"][hide="confused"]
$6x^2+9x=c^2x-2cx^2$
$x=\frac{c^2-9}{2c+6}$
[/hide][/quote]

[hide]You can't always go from the first step to the second one because it requires $2c+6\neq 0$ in order to divide. [size=59][color=white]
b
i
g

h
i
n
t
.
.
.
[/color][/size][/hide]

oh yeah... my answer couldn't be true because
$f(\frac{c-3}{2})=\frac{cx}{2(\frac{c-3}{2})+3}$ which produces an undefined answer.
:blush:

ok..i went to the following website
[url]http://www.math.ksu.edu/main/events/hscomp/samples/amc12/ahsme1980sol.pdf[/url]
and they told me that since $(2c+6)x+9-c^2=0$ is always true,
$2c+6=9-c^2=0$
so $c=-3$

yup... thanks scorpius119

[quote="Kurt Gödel"][hide] Well, as \[ x=f(f(x))=f\left(\frac{cx}{2x+3}\right)=\frac{\frac{c^2x}{2x+3}}{\frac{2cx}{2x+3}+3} = \frac{c^2x}{2cx+6x+9},  \]we know that $2cx^2+6x^2+9x$ must be equal to $c^2x$, or $(2x+6)x^2+(9-c^2)x=0$. This reduces to solving the following system of equations:  \[ \begin{aligned}2x+6 & = 0 \\ 9-c^2 & = 0\end{aligned}  \]which has one solution: $c=-3$. $\square$ [/hide][/quote]
how do you know the numerator must equal the denominator?
and since it was c^2, how do you know its the negative and not the positive answer?

[quote="Kurt Gödel"][hide] Well, as
\[ x \equal{} f(f(x)) \equal{} f\left(\frac {cx}{2x \plus{} 3}\right) \equal{} \frac {\frac {c^2x}{2x \plus{} 3}}{\frac {2cx}{2x \plus{} 3} \plus{} 3} \equal{} \frac {c^2x}{2cx \plus{} 6x \plus{} 9},
\]
we know that $ 2cx^2 \plus{} 6x^2 \plus{} 9x$ must be equal to $ c^2x$, or $ (2x \plus{} 6)x^2 \plus{} (9 \minus{} c^2)x \equal{} 0$. This reduces to solving the following system of equations:
\[ \begin{aligned}2x \plus{} 6 & \equal{} 0 \\
9 \minus{} c^2 & \equal{} 0\end{aligned}
\]
which has one solution: $ c \equal{} \minus{} 3$. $ \square$ [/hide][/quote]

But we are left with $ (2x\plus{}6)x^2\equal{}0$ which implies $ x\equal{}0$ as the only solution... so doesn't $ f(f(x))\equal{}x$ for $ x\equal{}0$ ONLY?

What is the bump though... 11 year bump...

[quote=candyofthesun][quote=Ezbakhe Yassin][quote="Tecnogram888"]how do you know the numerator must equal the denominator?
and since it was c^2, how do you know its the negative and not the positive answer?[/quote]

[b]1.[/b] The numerator is not equal to the denominator. Since we have that $x=f(f(x))=\frac{c^2x}{2cx+6x+9}$, we get that $2cx^2+6x^2+9x=c^2x$.

[b]2.[/b] Resolving the quadratic equation $c^2x-2cx^2-(6x^2+9x)=0$ for $x=-\frac3{2}$, give us $c_1=-3$ and $c_2=0$. Since $c_2=0$ does not satisfy our initial conditions, we conclude that the answer is $c=-3$ => $\boxed{A}$.[/quote]

Why do we plug in x=-3/2 for part 2?[/quote]

well why not :)

We can also simplify the calculations by using the fact when $f(f(x))=x$ then we have $f(x)=f^{-1}(x)$. Since the inverse of the function exists.

well i didn't get how it works (plugging it in)

edit: what's wrong with bumping old threads if i don't understand something?

For any hyperbola of the form $f(x)=\frac{ax+b}{cx+d}$, if we are given with $f(f(x))=x$ then the self inverse of the hyperbola can exist in a certain domain. So, applying the condition of self inverse $a+d=0$, which is infact $\fbox{c=-3}$.
I think this is a pretty easy question if you know this concept.
Thanks @below for explaining it in more detail

[hide="Fun with asymptotes"]The original hyperbola has asymptotes of $x = -\frac{3}{2}$ (vertical) and $y = \frac{c}{2}$ (horizontal).

If $x = f(f(x))$, then taking the inverse of both sides gives $f^{-1}(x) = f^{-1}f(f(x))$.  By using the definition of inverse functions on the RHS, we get $f^{-1}(x) = f(x)$, which means the function is its own inverse.

If the original function had asymptotes at $x = -\frac{3}{2}$ and $y = \frac{c}{2}$, then the inverse has asymptotes at $y = -\frac{3}{2}$ and $x = \frac{c}{2}$.

That pair of lines must be the same because the function and the inverse are the same.  This means $c = \boxed{-3}$.[/hide]


analytic geometry
superior algebra
superior algebra theorems

I want to check whether the given point ( Refer Sample Picture 1 for more clarity)
1. Lies on the Line
2. if so , whether it lies between the start and intersection point on the line or between end and intersection point on the line.

Also When it is a circular path, how to determine whether the given point
1. lies on the circular path
2. if so, how to determine whether it lies between point 1 and point2 or between point 2 and point1.

How to check the above ?

Thank you

How are you representing the lines, segments and circles?

If you can answer that, you're most of the way there; it's pretty much a matter of replacing an equation with an inequality.

I am in need of this concept for software programming.

Line is represented by X and Y coordinate of Start Point and EndPoint of the line.
Circle is represented by its Radius and X and Y Coordindate of Center Point of the circle.

Using the Line/Circle Input I can represent in standard line/circle equation also.

Given Point will be a X and Y coordinate point.
Intersection Point will be also a X and Y coordinate.


Suppose that $ S\equal{}\{1,2,..,n\}$ and that $ A_1,A_2,..,A_k$ are subsets of $ S$ such that for every $ 1 \leq i_1,i_2,i_3,i_4 \leq k$
Wehave : $ |A_{i_1}\bigcup A_{i_2} \bigcup A_{i_3} \bigcup A_{i_4}| \leq n\minus{}2$
 Prove that : $ k \leq n\minus{}2$

it's from Iran 1999. and actually the correct statement is "Prove that $ k\leq 2^{n \minus{} 2}$.

in fact, see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=4400


Solve:

$ 3^x \minus{} 3^{ \minus{} x} \equal{} 10$.

I found this problem when helping someone with Pre-Calc.  I found one solution with the following method:

$ 3^x \minus{} 3^{ \minus{} x} \equal{} 10$
$ (3^x)^2 \minus{} 1 \equal{} 10 \cdot 3^x$
$ y \equal{} 3^x$
$ y^2 \minus{} 10y \minus{} 1 \equal{} 0$
$ y \equal{} 3^x \equal{} \frac {10 \pm \sqrt {100 \plus{} 4}}{2}$
$ x \equal{} \log_3 {(5 \pm \sqrt {26})}$

Just for fun I started using different methods to find solutions (for example, differentiate both sides).  Mathematica (computer program) gave the general solutions:

$ x \equal{} \frac {\log{(\sqrt {26} \plus{} 5)} \plus{} i \pi (2n \plus{} 1)}{\log{3}}$ or $ x \equal{} \frac {\log{(\sqrt {26} \minus{} 5)} \plus{} 2 i \pi n}{\log{3}}$

How would one solve for these general solutions without a computer?

Hint:
[hide]
Your solving it was good, you kept all valid solutions, whereas squaring it you lose solutions (I think, not sure), but just remember that:
$ e^{2pi\imath}\equal{}1\equal{}e^{0}$
So you can expand your solutions.
[/hide]

[quote="monkeythyme"]Hint:
[hide]
Your solving it was good, you kept all valid solutions, whereas squaring it you lose solutions (I think, not sure), but just remember that:
$ e^{2pi\imath} \equal{} 1 \equal{} e^{0}$
So you can expand your solutions.
[/hide][/quote]

This problem was found when someone was helping someone with
precalculus problems, so it should only be the real values of x
that are being sought in the context as exponential and logarithmic
equations are given in that level of math.  In those types of equations,
imaginary solutions of x are not considered.  This would be the same
as stating all the real solutions of x for the intersection of
$ y \equal{} 3^x \minus{} 3^{ \minus{} x}$ and $ y \equal{} 10$.

[quote="Stank"]Mathematica (computer program) gave the general solutions:

$ x \equal{} \frac {\log{\sqrt {26} \plus{} 5} \plus{} i \pi (2n \plus{} 1)}{\log{3}}$ or $ x \equal{} \frac {\log{\sqrt {26} \minus{} 5} \plus{} 2 i \pi n}{\log{3}}$[/quote]
In Mathematica, log means the natural logarithm, so your solutions should be written:
$ x\equal{}\frac{\ln(\sqrt{26}\plus{}5)\plus{}i\pi(2n \plus{} 1)}{\ln{3}}$ or $ x\equal{}\frac{\ln(\sqrt{26}\minus{}5)\plus{}2i\pi n}{\ln{3}}$

However, I used Mathematica to try to find the solutions, and I think you mixed up some signs, and that these are the correct solutions:
$ x\equal{}\frac{\ln(\sqrt{26}\minus{}5)\plus{}i\pi(2n \plus{} 1)}{\ln{3}}$ or $ x\equal{}\frac{\ln(\sqrt{26}\plus{}5)\plus{}2i\pi n}{\ln{3}}$

To find all the solutions, you would need to find the solutions to $ y\equal{}3^x\equal{}5\pm\sqrt{26}$ for complex values. The real solution to $ 3^x\equal{}5\plus{}\sqrt{26}$ is $ x\equal{}\log_3(5\plus{}\sqrt{26})$, and since $ e^{2\pi i}\equal{}1$, it can be shown that the values $ \log_3(5\plus{}\sqrt{26})\plus{}2\pi in\log_3{e}$ also satisfy the equation. Then, you would need to show that these are all the solutions to this equation, which could be done by showing that if $ x$ satisfies the equation, then it must differ from $ \log_3(5\plus{}\sqrt{26})$ by a multiple of $ 2\pi i$. For the remaining solutions, you would need to find the complex solutions to $ 3^x\equal{}5\minus{}\sqrt{26}$ (which I don't know how to do unfortunately). Wikipedia or Mathworld could have more information on complex powers/logarithms.

@monkeythyme: Unless I'm mistaken, you can never [i]lose[/i] solutions by squaring (or taking any function of both sides), but you could gain extraneous solutions.

I know I only needed real solutions, I was thought it would be fun to find more  :D

You're right, MathWise; I think I left out parenthesis from my latex.  Copy and pasted from Mathematica:

In[2]:= Reduce[3^x - 3^(-x) == 10, x]

Out[2]= C[1] \[Element]
  Integers && (x == (2 I \[Pi] C[1])/Log[3] + (
     I (\[Pi] - I Log[-5 + Sqrt[26]]))/Log[3] ||
   x == (2 I \[Pi] C[1])/Log[3] + Log[5 + Sqrt[26]]/Log[3])

I'm probably being ignorant but can you please explain why "it can be shown that the values $ \log_{3}(5\plus{}\sqrt{26})\plus{}2\pi in\log_{3}{e}$ also satisfy the equation."

Thank you for your help.

[quote="kunny"][quote="Stank"]Solve:

$ 3^x \minus{} 3^{ \minus{} x} \equal{} 10$.

I found this problem when helping someone with Pre-Calc.  I found one solution with the following method:

$ 3^x \minus{} 3^{ \minus{} x} \equal{} 10$
$ (3^x)^2 \minus{} 1 \equal{} 10 \cdot 3^x$
$ y \equal{} 3^x$
$ y^2 \minus{} 10y \minus{} 1 \equal{} 0$
$ y \equal{} 3^x \equal{} \frac {10 \pm \sqrt {100 \plus{} 4}}{2}$
$ x \equal{} \log_3 {5 \pm \sqrt {26}}$[/quote]

That's wrong.

$ y \equal{} 3^x > 0$, so $ x \equal{} \log_3 (5 \plus{} \sqrt {26})$

Here is my solution:

$ (3^x \plus{} 3^{ \minus{} x})^2 \equal{} (3^x \minus{} 3^{ \minus{} x})^2 \plus{} 4 \equal{} 104$, since $ 3^x \plus{} 3^{ \minus{} x} > 0$, we have $ 3^x \plus{} 3^{ \minus{} x} \equal{} 2\sqrt {26}$

$ \therefore 3^x \equal{} \frac {3^x \plus{} 3^{ \minus{} x}}{2} \plus{} \frac {3^x \minus{} 3^{ \minus{} x}}{2} \equal{} 5 \plus{} \sqrt {26}$, yielding $ \boxed{x \equal{} \log_3 (5 \plus{} \sqrt {26})}$[/quote]

I don't completely agree with you.  I understand this problem expects the student to find real solutions, but the not-real solutions are still valid.  For instance, $ \log_3 (3\minus{}\sqrt{26})\equal{}\frac{i \pi\plus{}\ln{(\sqrt{26}\minus{}5)}}{\ln{3}}$ is clearly a solution when plugged into the original equation.

Sorry typo.  I meant $ \log_{3}(5\minus{}\sqrt{26})\equal{}\frac{i\pi\plus{}\ln{(\sqrt{26}\minus{}5)}}{\ln{3}}$

There is no room to argue the problem more than this. :wink:

hello, i have got
$ c_1\in \mathbb{Z}\land \left(x\equal{}\frac{2 i \pi  c_1}{\log (3)}\plus{}\frac{i
   \left(\pi \minus{}i \log \left(\sqrt{26}\minus{}5\right)\right)}{\log (3)}\lor
   x\equal{}\frac{\log \left(5\plus{}\sqrt{26}\right)}{\log (3)}\plus{}\frac{2 i \pi
   c_1}{\log (3)}\right)$
Sonnhard.

What's on earth, why do we need to find imarinary root for the easy equation $ 3^x \minus{} 3^{ \minus{} x} \equal{} 10?$


Putnam
probability
induction
expected value
college contests

Let $S=\{1,2\dots,n\}$ for some integer $n>1.$ Say a permutation $\pi$ of $S$ has a local maximum at $k\in S$ if
\[\begin{array}{ccc}\text{(i)}&\pi(k)>\pi(k+1)&\text{for }k=1\\ \text{(ii)}&\pi(k-1)<\pi(k)\text{ and }\pi(k)>\pi(k+1)&\text{for }1<k<n\\ \text{(iii)}&\pi(k-1)M\pi(k)&\text{for }k=n\end{array}\]
(For example, if $n=5$ and $\pi$ takes values at $1,2,3,4,5$ of $2,1,4,5,3,$ then $\pi$ has a local maximum of $2$ as $k=1,$ and a local maximum at $k-4.$)
What is the average number of local maxima of a permutation of $S,$ averaging over all permuatations of $S?$

Count number of times each position is max.
Endpoints: if the value at one endpoint is k, then for it to be max, the neighboring entry has to be $1,2,\ldots,k-1$.
So one endpoint is max for ${\sum_{i=1}^{n}i-1 = \frac{n(n-1)}{2}}$. However, each max occurs for each of the $(n-2)!$ permutation of the other n-2 entries.
=> $\frac{n(n-1)}{2}*(n-2)!$ at one endpoint, and $n(n-1)*(n-2)! = n!$ max at two endpoints.

for the non-endpoint $n-2$ positions
we need the neighboring positions each to be smaller, i.e., if a position has value k, we have $k-1$ choices of value for the position before it $k-2$ choices for that of the position after it.
max occurs at one position, considering the permutation for the other $n-3$ positions, for $(n-3)!*{\sum_{i=1}^{n}(i-1)(i-2)}$.
This occurs at each of the (n-2) center positions.

The average is then $(n-2)*(n-3)!*{\sum_{i=1}^{n}(i-1)(i-2)}+n!$ divided by $n!$ giving us $\boxed{\frac{n+1}{3}}$.

--
i dont think i'll be getting credit for this on the actually exam because i forgot to permute ><
too bad

I did it recursively: every permutation of $[n+1] = \{1, \ldots, n+1\}$ arises via insertion of $(n+1)$ into a permutation of $[n]$.  This either increases the number of local maxima by 1 (if we insert $(n+1)$ between two non-l.m.) or leaves it the same (if we insert it next to a l.m.), and we can count how many ways there are to do each in terms of the number of l.m. of the permutation of $[n]$ we are considering.  Summing over all permutations of $[n+1]$ reduces to an expression involving the sum of the number of l.m.s over all permutations of $[n]$, giving us a relatively simple recursion we can solve to find the result.

[quote="Ravi B"]The probability that 1 is a local max is $\frac{1}{2}$.  Similarly, the probability that $n$ is a local max is $\frac{1}{2}$.  For any other value of $k$, the probability that $k$ is a local max is $\frac{1}{3}$.  Because expectation is additive, the expected number of local maxima is the sum
\[2 \cdot \frac{1}{2}+(n-2) \cdot \frac{1}{3}= \frac{n+1}{3}\, . \]
[/quote]

I can't believe I missed this solution.

Let $a_{n}$ be number of maximums over all $n!$ permutations. First few terms are $a_{0}= 0, a_{1}= 1, a_{2}= 2, a_{3}= 8$. We will establish recurrent formula for $a_{n}$ and number $b_{n}= \frac{a_{n}}{n!}$, that we are looking for.

Consider permutation of $n+1$ elements, and let number one be on position $k+1$. It's easy to see that number of maximums on the left is $a_{k}{n \choose{n-k}}(n-k)!$ and the number of maximums on the right is $a_{n-k}{n \choose k}k!$ over all permutations with fixes $1$. Summing for $k = 0, 1, \dots, n$, we get:
\[a_{n+1}= n! \sum_{k = 0}^{n}\left ( \frac{a_{k}}{k!}+\frac{a_{n-k}}{(n-k)!}\right ).\]

After some manipulations, we have:
\[b_{n+1}= \frac{2}{n+1}\sum_{k = 0}^{n}b_{k}.\]

Now, we can subtract two consecutive equations, to get $b_{n}= \frac{n+1}{n}b_{n-1}.$ Using induction, we easily prove that $b_{n}= \frac{n+1}{3}$.

alexilic

We add the extra number 0 (at the end) and put the $n+1$ points in a circle. Note that the thus obtained circle has the same maxima as the permutation on a line it came from, since maxima remain maxima, and 0 can not be a maximum. The odds for any point on the circle to be local maximal is 1/3 (as in Ravi B's solution), so the total number of expected local maxima is $(n+1)/3$.

Note that $n>1$ is used when we say that the odds of a point on the circle being a local maximum is 1/3.