Holap todos, por se acaso alguien saben algo de la ibero de este año? supuestamente ya comenzo la prueba... alguien puede publicar los problemas?

faltan varias horas para que la prueba comienze.

[quote="pacman2812"]hmmm... esto no debería estar en el foro de "duscusiones generales"?[/quote]
E susede q no nos kieren  :rotfl:
dale, aca tengo el examen .

Problema 1

Se distribuyen los numeros 1, 2, 3, ... , 2008² en un tablero de 2008 x 2008, de modo que en cada casilla haya un número distinto. Para cada fila y cada columna del tablero se calcula la difrencia entre el mayor y el menor de sus elementos. Sea S la suma de los 4016 números obtenidos. Determine el mayor valor posible de S

Problema 2

Sean ABC un triágulo escaleno y r la bisectriz externa al ángulo ABC. Se consideran P y Q los pies de las perpendiculares a la recta r que pasan por A y C, respectivamente. Las rectas Cp y AB se intersectan en M y las rectas AQ y BC se intersectan en N. Demuestre que las rectas AC, MN y r tienen un punto en común.

Problema 3

Sean m y n enteros tales que el polinomio P(x) = x³ + mx + n tiene la siguiente propiedad: si 'x' y 'y' son enteros y 107 divide a P(x) - P(y), entonces 107 divide a x - y. Demuestre que 107 divide a m.

PD: Qi,  Xavier te manda saludos, obviamente yo tb loko, cuidate.
 Qi!! era pa q vengas!! caraxo!

[quote="cvaneg"][quote="chichi"]gracias loco!!
esta medio facil eso jajaja(casi como el ibero de ecuador jaja) a peru como les fue?[/quote]

[color=green]todos los del equipo de peru hicieron los 3
los de brasil tambien[/color][/quote]

nadie sabrá cómo le habrá ido a los de colombia??? espero que les haya ido bien...

:)

El 1, creo que la respuesta seya 2*2008²*2007 y el 2, es una buena idea usar Analitica.

Que bueno que los brasileños hicieron los 3!
Ojala saquemos 4 oros !   :D

Saludos

jeje, si, coincido con algunos q el examen esta un poco facil... jeje, realmente no soy bueno (por no decir q muy malo  :P ) para combinatoria, pero pienso q el 1 esta muy por debajo de lo q siempre han sido problemas 1 de la ibero... es q este problema ni siquiera requiere el trabajo "tedioso" q hay q hacer para resolver algunos 1's y 4's... del 2 y el 3 no me quejo mucho, pero me parece q ha estado mas sencilla q otras olimpiadas anteriores...

Es notable que hay que probar que APM y CQN estan en perspectiva, y como AP//CQ hay que probar que si X es la interseccion de PC y AQ entonces BX es bisectrz de <ABC o BX//CQ//AP.
Sea Y la interseccion de CQ con AB.
APB y CQB son semejantes, entonces: AB/BC=AP/CQ
APX y CQX son tambien semejantes:  AB/BC=AX/QX
Pero CBY es isosceles y asi BC=BY, AB/BC=AB/BY=AX/QX y BX//CQ

Wow... los cortes van a estar altos :|
el uno parece un uno del cono jajajja
mi solucion del 2 es igualito que Concyclicboy imagino todo el mundo lo hizo asi... pregunto si existe solucion sin usar desargues...

y por favor publique los de segundo dia cuando tengas... gracias.

jeje, por supuesto q si existe una solucion sin desargues...  :D  de hecho, jutaro ya subio los probs (del 1er dia)... busqn los ultimos posts de jutaro y ahi van a encontrarlos... otra gente ha dado otras soluciones...

Hay una buena soluccion para el 2 con Analítica.

en el prob 5, X,Y,Z estan sobre BC,CA,AB o no necesariamente?

[quote="msecco"]EL segundo dia:

4) prove que a equação x^2008 + 2008! = 21^y não tem soluções inteiras

5) pega um triângulo ABC e escolhe pontos X, Y, Z no interior de cada um de seus lados.
Seja A' o circuncentro de AYZ e analogamente para B',C'.Prova que [A'B'C'] >= [ABC]/4 e que a igualdade ocorre se e só se AA', BB' e CC' forem concorentes.

6) Beribol é um jogo jogado por 2 equipes de 4 jogadores cada (que não são fixas).Para quais valores de n é possível formar um campeonato com n pessoas de tal forma que cada par de pessoas jogue em equipes adversárias exatamente uma vez?[/quote]

Disculpa, msecco, ¿podrias reenunciar el problema 5? Creo que asi como esta planteado es ambiguo, no entiendo como se definen análogamente B' y C', y tampoco se indican los lados a los que pertenecen X, Y, Z

De antemano gracias

Problema 4
Demostrar que no existen enteros positivos x e y tales que

            x^2008 + 2008! = 21^y

Problema 5
Sean ABC un triángulo y X, Y, Z puntos interiores de los lados BC, AC, AB respectivamente. Sean A', B', C' los circuncentros correspondientes a los triángulos AZY, BXZ, CYX. Demostra que:

         (A'B'C') >= (ABC)/4

 Observación: Para un triángulo cualquiera RST, denotamos su área por (RST).

Problema 6
En un partido de biribol se enfrentan dos equipos de cuatro jugadores cada uno. Se organiza un torneo de biribol en el que participan n personas, que forman equipos para cada partido (los equipos no son fijos). Al final del torneo se observó que cada dos personas disputaron exactamente un partido en equipos rivales. Determinar para qué valores de n es posible organizar un torneo con tales caracteristicas.

La prueba de secundo dia estás más dificil.

Por Mikel o deduciendolo es facil ver ABC es semejante A'B'C',entonces hay que probar que alguno de los lados de por ejemplo A'B' es mayor o igual que que la mitad de AB.
Sean M1 y M2 los puntos medios de BX y CX resp.,sea M el pie de la perpendicular desde C' a M1B'o desde B' a M2C',en cualquiera delos casos MC=M1M2 o MB=M1M2,Y  por pitagoras seria obvio ver que B'C'>= M1X+M2X=BC/2
Y aSI LA DESIGUALDAD YA ESTA PROBADA.
Pero la igualdad se cumple si y solo si B'C'=M1M2, esto si y solo si B'C'//BC, y si esto sucede ABC y A'B'C' son homoteticos,entonces AA',BB', y CC' concurren.

[quote="Concyclicboy"]
Hey que onas Gabriel, Chele, Aaron y demas gente, ya casi puedo usar las computadoras y esta pagina. Salu[/quote]

hola arnoldo me da gusto saludarte. :D tus soluciones son muy buenas como siempre. :wink:

[quote="msecco"]Buenas noticias de Salvador:

Brasil quédo en primer lugar.El ranking:

BRA - 155
PER - 150
CUB - 130
ARG - 124
ESP - 121

Resultos de Brasil:

BRA1 (Pondé) 7 7 7 7 7 7 = 42 (oro perfecto)
BRA2 (Régis) 7 7 7 7 7 2 = 37  (plata)
BRA3 (Finder) 7 7 7 7 7 2 = 37  (plata)
BRA4 (Ramon) 7 7 7 7 7 4 = 39 (oro)

La informácion que tengo es que el oro de Peru fue PER3(creo que Cesar Cuenca).Además, los otros oros fueran 1 de Mexico y 1 de Cuba.Argentina saco 3 platas e ARG 3 no saco nadita.[/quote]

[color=green]como que ARG3 (Maite) no saco nada??
es para matarlo

cuales eran los cortes?[/color]

Fue el que mi profesor(tutor de Brasil) dije-me.
Los cuertes fueran: Oro-39 Plata - 30 y Bronce - 21.

ARG 3 es Cogorno?
Voy a matá-lo.
Coñoci él en la Cono!

Correcion:

ARG 3 saco mencion.
Sus pontajes: 7 0 0 7 0 4

Sei que la mayor nota de Paraguay fue 14 puntos.

yo recién acabo de leer los resultados de El Salvador :D

                P1     P2      P3    P4    P5    P6    TOTAL
SAL 1        7       7        1      7      2      0        24       Bronce
SAL 2        7       7        0      7      7      0        28       Bronce
SAL 3        5       6        0      7      2      0        20       Mención
SAL 4        7       7        1      6      2      0        23       Bronce

como delegación tuvimos el puntaje histórico de 95 puntos y quedamos en séptimo lugar... realmente estamos muy contentos :) bueno en todo caso, felicidades también a nuestros colegas de otros países :coolspeak:

En algún lado estan los resultados completos?

[quote="cvaneg"][quote="R.G.A.M."][color=blue]No necesitan matarla.

Solita se mata, por lo cerca que estuvo de Bronce.

De paso, ¿qué saben de Paraguay?[/color][/quote]

[color=green]que yo sepa maite es hombre... :maybe:

pero bueno, nunca deja de sorprendernos  [/color]:P[/quote]
[color=blue]¡Ja, ja, ja…!  :blush:

Sucede que he conocido mujeres con Maite como nombre de pila.

¿Dicen que hubo Menciones de Honor para Paraguay?[/color]

Alguien sabrá como le fue a Colombia? :)

Agradecería que manden los resultados completos   ... si los tienen

Gracias

Daniel

Alguien sabe los resultados de Chile? Supe que CHI1 (Anibal Velozo) tuvo plata....

Gracias.

[quote="tipe"]Per 1 :  7 7 7 7 7 2
Per 2 :  7 7 7 7 7 2
Per 3 :  7 7 7 7 7 4
Per 4 :  7 7 7 7 7 2

Peru sumó 150 puntos, pero quedo segundo por paises luego de Brasil.

ya que tengo los resultados de chile a la mano:

Chi 1 : 7 7 7 7 7 2
Chi 2 : 7 4 3 7 0 0
Chi 3 : 7 1 0 0 0 0
Chi 4 : 7 0 0 5 0 0

Otra información: los problemas fueron propuestos por Brasil, Perú, Brasil, México, Colombia, Argentina (en ese orden)[/quote]

gracias por la informacion y felicitaciones al equipo peruano!! :)

Saludos  :)

de costa rica saben algo? la pag "oficial" parece q no tiene informacion al respecto...

Haber ya he resuelto algunos de los problemas del 1 al 5 hehe  : ni tan difciles diria yo pero estaban en algo . Solo comentare lo que hice pero no dare extension de ellas .

Problema 1 : nada que decir solo se trata de colocar los mayores numeros en una diagonal y los menores en la otra diagonal . la acotacion es mas que evidente .

Problema 2 : Como cualquier otro problema de colinealidad la tecnica de Usar el teorema de Menelao es muy buena y sale rapido . Tambien he probado haciendo conjugacion armonica y sale rapido tambien . No la veo por geometria analitica pero creo que saldria mucho mas largo que lo habitual

Problema 3 : El 107 es primo = 4k+3 asi que los restos cuadraticos en el modulo 107 son : $ 0^2$,$ 1^2$,$ 2^2$,$ 3^2$,$ 4^2$,....,$ 53^2$ y los que no son restos cuadraticos son $ \minus{}1^2$,$ \minus{}2^2$,$ \minus{}3^2$,$ \minus{}4^2$,....,$ \minus{}53^2$ luego la idea es hacer : $ m \equal{} x^2 \plus{} xy\plus{}y^2$ equivalente a $ m \equal{} 3a^2 \plus{} b^2$ con b distinto de 0 . luego si m es resto cuadratico facil con a=0 y si no lo es facil usando que : $ (3)3^2 \plus{}(1)20^2\plus{}1\equal{}4(107)$ se encuentra el a y el b .

Problema 4 : se prueba que x es multiplo de 21 luego y>=2008  entonces con ello 2008! deberia ser multiplo de 3^(2008) lo cual es falso .

Problema 5 : prolongas AA' hasta un punto M tal que AA'=A'M de igual manera con los demas BB' hasta un punto N y CC' hasta un punto P . Luego en mi caso que soy muy asqueroso cuando uso mis tecnicas geometricas usando geometria de los complejos probe que :
(A'B'C')=1/4(ABC)+1/4(MNP) luego la desigualdad se sigue de esta relacion .

Aun no he leido el problema 6 .

[color=blue]
Contestando la duda de Campos los resultados de Costa Rica (extrañamente con codigo COS en esta olimpiada) son los siguientes

COS1   7  7  5  7  3  0    29 Bronce
COS2   7  1  0  7  2  0    17 Mención
COS3   7  0  0  7  0  0    14 Mención
COS4   7  3  0  7  0  0    17 Mención

  :ninja:
[/color]

gracias... felicidades al equipo y a seguir adelante!  :)

Ese problema 6 esta muy bueno digno de ser un problema 6 .
Note que los puntajes comunes eran 2,4,7 lo justo para separar platas de oros y sobre todo esta como para decir " No importa cuanta experiencia tengas en olimpiadas matematicas Ese problema 6 pone de cabeza inclusive a los mas experimentados " .

Mis felicitaciones a los equipo de Brasil y Peru Por haber mostrado gran solidez de hecho ambos hicieron lo que debieron haber hecho .

Aun asi lo mas importante es la experiencia ganada en esta olimpiada

Y a seguir entrenando en matematicas .


Consider the lesson to be learned from a recent thread of mine, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129194 , which is this:

[b]Write numbers as variables.[/b]

[hide="Specifically"] Let $x = 12342006$.  Then the denominator is

$x^{2}-(x-1)(x+1)+1 = 2$ [/hide]


inequalities
real analysis
real analysis solved

Let (an) and (bn) be two sequences both in (0,oo) with the property that (an) has no subsequence that converges to 0 and lim(bn)=oo and lim(b_(n+1)-bn))=0. Prove that lim [b1*b2*..*bn]/[(a1+b1)(a2+b2)*..*(an+bn)]=0.

cheers! :D :D

We can rewrite what we must prove as

P_n=Product {1<=i<=n} (1+a_i/b_i) -> infinity as n->infinity. Since (a_n) has no subsequence which tends to 0, it means that (a_n) has a lower bound, so there exists m<=a_n for all n>=1. This means that P_n >= Q_n=Product {1<=i<=n} (1+m/b_i). It's enough to prove Q_n->00 iff lnQ_n->00 but lnQ_n= \sum {1<=i<=n} ln(1+m/b_i). We can use the inequality ln(1+x) >= x/(1+x) (it can be proven with derivatives). We get lnQ_n >=  \sum {1<=i<=n} (m/(m+b_i)). Let m+b_i=c_i. We can see that (c_n) satisfies the same conditions as (b_n), namely c_n->00 and (c_(n+1)-c_n)->0. So all we need now is m \sum 1/c_i ->00 iff  S_n=\sum 1/c_i ->00.

We can use Cesaro to prove that c_n/n->0, so there exists n0 s.t. c_n<=n foa any n>=n0. For n>=n0 we get S_n > 1/n0+1/(n0+1)+..+1/n->00, so S_n->00, which is what we needed.

I'm not sure right now, but I think the condition b_(n+1)-b_n->0 is too strong. I think all we need is b_(n+1)-b_n has a finite limit. Besides that, I don't think we need b_n->00. What do you think? Is this correct?


analytic geometry
trigonometry
geometry
geometry proposed

[quote][color=darkred]Construct the triangles $ XBC$ , $ AYC$ , $ ABZ$ outside of the triangle $ ABC$ so that

$ XBC\sim AYC\sim ABZ$ . Prove that the lines $ AX$ , $ BY$ , $ CZ$ are concurrently.[/color] [/quote]
[b]Remark.[/b] We can obtain new and interesting problems by studying some particular cases of this problem.

In fact, this is Jacobi's theorem. A proof with barycentric coordinates goes this way:

$ \angle BXC \equal{} \angle YAC \equal{} \angle ZAC \equal{} \omega$

$ \angle XBC \equal{} \angle AYC \equal{} \angle ABZ \equal{} \theta$

$\angle BCX \equal{} \angle ACY \equal{} \angle AZB \equal{} \varphi$

Using Conway's formula the barycentric coordinates of $ X,Y,Z$ are given by

$ X \ ( \minus{} a^2: S_C \plus{} S_{\varphi}: S_B \plus{} S_{\theta}) $

$ Y \ (S_C \plus{} S_{\varphi}: \minus{} b^2: S_A \plus{} S_{\omega}) $

$Z \ (S_B \plus{} S_{\theta}: S_A \plus{} S_{\omega}: \minus{} c^2)$

Hence, $ AX,BY,CZ$ concur at a point $ J$  (Jacobi's perspector) with barycentric coordinates

$ J \left ( \frac {1}{S_A \plus{} S_{\omega}}: \frac {1}{S_B \plus{} S_{\theta}}: \frac {1}{S_C \plus{} S_{\varphi}} \right )$

Dear Mathlinkers,
A simple proof of this problem is to use Ceva-sine theorem and some area ratios. :)

Indeed, my proposed problem is [url=http://www.gogeometry.com/geometry/isogonic_jacobi_theorem.htm][color=red][b]Jakobi's theorem[/b][/color][/url]. Sorry, I didn't know it. Thank you, [b]Luis[/b] !
[quote][color=darkred]Construct the triangles $ XBC$ , $ AYC$ , $ ABZ$ outside of the triangle $ ABC$ so that

$ XBC\sim AYC\sim ABZ$ . Prove that the lines $ AX$ , $ BY$ , $ CZ$ are concurrently.[/color] [/quote]
[u][b]Method I[/b][/u] ([b]V.N.[/b] & [b]livetolove212[/b]). Denote $ \left\|\begin{array}{c} M\in AX\cap BC\ ,\ N\in BY\cap CA\ ,\ P\in CZ\cap AB \\
 \\
m(\angle BAZ) = m(\angle CAY) = x \\
 \\
m(\angle CBX) = m(\angle ABZ) = y \\
 \\
m(\angle ACY) = m(\angle BCX) = z \\
 \\
x + y + z = 180^{\circ}\end{array}\right\|$ . Observe that

$ \frac {BX}{CX} = \frac {\sin \widehat {BCX}}{\sin\widehat {CBX}} = \frac {\sin z}{\sin y}$ . Therefore,  $ \frac {MB}{MC} = \frac {[ABX]}{[ACX]} = \frac {BA\cdot BX\cdot\sin\widehat {ABX}}{CA\cdot CX\cdot\sin\widehat {ACX}}$ $ \implies$ $ \boxed {\frac {MB}{MC} = \frac cb\cdot\frac {\sin z}{\sin y}\cdot\frac {\sin (B + y)}{\sin (C + z)}}$ .

Prove analogously that ${ \frac {NC}{NA} = \frac ac\cdot\frac {\sin x}{\sin z}\cdot\frac {\sin (C + z)}{\sin (A + x)}}$ and ${ \frac {PA}{PB} = \frac ba\cdot\frac {\sin y}{\sin x}\cdot\frac {\sin (A + x)}{\sin (B + y)}}$ . In conclusion,

$ \frac {MB}{MC}\cdot\frac {NC}{NA}\cdot\frac {PA}{PB} = 1$ and using the [u][b]Ceva[/b]'s theorem[/u] obtain that the lines $ AX$ , $ BY$ , $ CZ$ are concurrently.

[u][b]Method II[/b] ([b]V.N.[/b])[/u]. $ J\in BY\cap CZ\implies AYC\sim ABZ$ $ \implies$ $ \left\|\begin{array}{c} \frac {AY}{AC} = \frac {AB}{AZ} \\
 \\
\widehat {CAZ}\equiv\widehat {YAB}\end{array}\right\|\implies$ $ AYB\sim ACZ\implies$ $ \left\|\begin{array}{c} \widehat {ABY}\equiv\widehat {AZC} \\
 \\
\widehat {AYB}\equiv\widehat {ACZ}\end{array}\right\|\implies$

$ \left\|\begin{array}{c} \widehat {ABJ}\equiv\widehat {AZJ} \\
 \\
\widehat {AYJ}\equiv\widehat {ACJ}\end{array}\right\|\implies$ $ AJBZ$ , $ AJCY$ are cyclically $ \implies$ $ \left\|\begin{array}{c} m(\angle ZJA) = m(\angle ZBA) = y \\
 \\
m(\angle YJA) = m(\angle YCA) = z \\
 \\
\left(x + y + z = 180^{\circ}\right)\end{array}\right\|$ $ \implies$ $ m(\angle CJY) = x\implies$ $ BXCJ$

is cyclically $ \implies$ $ m(\angle CJX) = m(\angle CBX) = y\implies m(\angle AJX) = x + y + z =$ $ 180^{\circ}\implies$ $ J\in AX$ $ \implies$  $ AX\cap BY\cap CZ\ne\emptyset$ .


think of any 2 numbers. add them to get a 3rd number. now add second and third to get a fourth. continue this procedure untill you get to tenth number. add all ten numbers if you tell me your 7th number i will by multiplying by 11 tell u wat ure sum is. why does this always work?

i thought of the fibonacci sequence but i couldnt explain it  ;)

Work it out algebraically... if $x$ and $y$ are the original two numbers, we can write each of the other terms in terms of $x$ and $y$. You get something like $x, y, x+y, x+2y, 2x+3y, \ldots$ and the result follows by adding up all the terms and comparing that with $11$ times the $7$th term.


A regular pentagon is circumscribed about a circle with radius 25. What is the area of the pentagon?

[hide="hint 1"]Use the formula $A=\frac{aP} 2$... where A is area, a is apothem and P is perimeter.[/hide]
If you dont understand the formula...
[hide="hint 2"] Fin the area of one of the inner regular triangles from the pentagon and multiply by how many of those triangles are there in the pentagon.[/hide]
[hide="hint 2.1"] The area of a regular triangle is $\frac{s^2\sqrt 3}{4}$[/hide]

[quote="Jhonny Ruiz"][hide="hint 2"] Fin the area of one of the inner regular triangles from the pentagon and multiply by how many of those triangles are there in the pentagon.[/hide]
[hide="hint 2.1"] The area of a regular triangle is $\frac{s^2\sqrt 3}{4}$[/hide][/quote]
The innter triangles are isosceles but not equilateral.

[hide]The exterior angles of a regular pentagon are $\frac{360}{5}=72$, so the interior angles are $180-72=108$.  Draw a segment connecting the center and a vertex of the pentagon.  This segment bisects the vertex, so each of the smaller angles formed have a measure of $52$ degrees.  Then $\tan 52=\frac{25}{\frac{1}{2}x}$, where $x$ is the side length of the pentagon.  Simplifying, $x=\frac{50}{\tan 52}=50\cot 52$.  The semiperimeter is $s=\frac{5x}{2}=125\cot 52$.  By the area formula $A=rs$, we have $A=25\cdot 125\cot 52=\boxed{3125\cot 52}$.[/hide]

That is true guys for a second I though of the Pentagon to have six sides.... Studupid isn't it!!!!!

Anyways, we can find the measure of the interior angles by the formula $\frac{180(n-2)}{n}$... .... we get 108
Now we want to make an isoceles triangle inside the pentagon... the triangle will have measurements of 54, 54 and 72.

In order to use my formula, we calculate the apothem to be $25\sin 54$, also we calculate one side of the triangle to be $50\sin 36$... since there is 5 sides... the perimeter would then be $250\sin 36$... now, lets apply the formula...

$A=\frac{(25\sin 54)(250\sin 36)} 2$
$A=3125(\sin 54)(\sin 36)$

I think there should be a way to simplify this though


linear algebra
matrix
induction
inequalities
calculus
integration
vector

Arne wrote:Consider a nxn-matrix A whose entries are given by

aij = (i + j - 1)-1.

One chooses n entries from A, no two of which are from the same column or row. Prove that the sum of the chosen numbers is at least 1.

[hide]As no two are from the same row, there must be one from every row. So, the sum will be of the form :Sigma:1/(i+ji-1) for i from 1 to n, where all the ji are distinct. Ok, now one possible arrangement for the ji is that ji=n+1-i. This will give 1 for the sum. Now, suppose there exists a ja such that ja does not equal n+1-a but instead equals n+1-b. So then either jb equals n+1-a or it equals something else. Well, in the former case, 1/(a+n-b)+1/(n-a+b)>1/(n)+1/(n), so the sum will increase. This was an example of a cycle of 2. I know that this argument will still work if, say jb=n-c, jc=n-a, or even bigger cycles, but I don't know how to rigorously prove it. Induction?[/hide]

Well ...

My original solution is very simple. It is just Cauchy's inequality.

Let (1/ai), i = 1 .. n the n chosen entries.

The key observation is that :Sigma:ai is constant and is n

Then, of course, Cauchy rules :

:Sigma:ai :Sigma:1/ai >= n

I thought this inequality, which you call Cauchy's inequality, was called the AM-HM (arithmetic-harmonic mean) inequality.  Am I mistaken?

We are both right I think   :D :

:Sigma:ai :Sigma:1/ai >= n is a special case of :
http://mathworld.wolfram.com/CauchysInequality.html
with bi=1/ai

AM = :Sigma:ai/n
HM = n /:Sigma:1/ai

Since we have AM >= HM, :Sigma:ai :Sigma:1/ai >= n

I prefer to remember it as a special case of Cauchy, but all inequalities are closely linked to each other (it's possible to prove one using another one).

Can somebody tell me about Cauchy's Inequality, please?
Or in what Grade's book can I find it? :|
Thank you
[/quote]

Cauchy's Inequality can be expressed in many many ways.  For example, there is a Cauchy's Inequality for integrals of continuous functions.  But the one people here are talking about is this:
given any two sequences of real numbers, x[size=75]1[/size], x[size=75]2[/size], . . . x[size=75]n[/size] and y[size=75]1[/size], y[size=75]2[/size], . . . y[size=75]n[/size], then

(x[size=75]1[/size]:^2: + x[size=75]2[/size]:^2: + . . . + x[size=75]n[/size]:^2:)(y[size=75]1[/size]:^2: + y[size=75]2[/size]:^2: + . . . + y[size=75]n[/size]:^2:) :ge: (x[size=75]1[/size]y[size=75]1[/size] + x[size=75]2[/size]y[size=75]2[/size] + . . . + x[size=75]n[/size]y[size=75]n[/size]):^2:
with equality holding if and only if x[size=75]1[/size]/y[size=75]1[/size] = x[size=75]2[/size]/y[size=75]2[/size] = . . . = x[size=75]n[/size]/y[size=75]n[/size]

I prefer to think about the Cauchy-Schwarz Inequality in terms of vectors: the dot product of two vectors in Rn is less than or equal to the product of the magnitudes of the vectors since the cosine of the angle is between -1 and 1.


[quote="george2"]Χρωματίζουμε τα σημεία με ακέραιες συντεταγμένες του επιπέδου με ένα από 6 διαφορετικά χρώματα. Να δείξετε ότι υπάρχει παραλληλόγραμμο του οποίου οι κορυφές έχουν το ίδιο χρώμα.

Με αφορμή αυτή την άσκηση μου δημιουργήθηκαν ορισμένα ερωτήματα.Αν χρωματίζαμε όλα τα σημεία του επιπέδου με πεπερασμένο αριθμό χρωμάτων θα ήταν π.χ. υποχρεωτικό να υπάρχει κανονικό πεντάγωνο ώστε όλες του οι κορυφές να έχουν το ίδιο χρώμα;Αν ναι πως αποδεικνύεται; Υπάρχει περιορισμός στα πολύγωνα και γενικότερα στα μονοχρωματικά σχήματα που μπορούν σχηματιστούν;[/quote]

Η απάντηση είναι ναι. Γενικά, το θεώρημα των Gallai-Witt λέει πως για κάθε πεπερασμένο σύνολο Α του $ \mathbb{R}^n$, αν το $ \mathbb{R}^n$ χρωματιστεί με κ χρώματα, τότε υπάρχει ένα σύνολο Β ομοιοθετικό (=επιτρέπονται οι μεταφορές και οι διαστολές αλλά όχι οι περιστροφές) με το Α το οποίο είναι μονοχρωματικό (= όλα τα σημεία το ίδιο χρώμα)

Αυτά είναι κλάδος της θεωρίας Ramsey. Βάζω σε συνημμένο σημειώσεις από ένα μάθημα τέταρτου (!) έτους στο Cambridge. Είναι το θεώρημα 11. (Αν και το θεώρημα 11 είναι πιο αδύνατο από αυτό που έγραψα εδώ, η απόδειξη είναι η ίδια.) Για το πρώτο κεφάλαιο δεν χρειάζονται ιδιαίτερες γνώσεις. (Αυτό όμως δεν σημαίνει πως είναι ευκολοδιάβαστο.)

EDIT: Αντί για συνημμένο, μπορείτε να βρείτε τις σημειώσεις [url=http://www.dpmms.cam.ac.uk/~par31/notes/]εδώ[/url]

Για το ειδικο προβλημα το μυαλο μου πηγε σε αυτο
http://www.geocities.com/liviozuc/hyperimages/acolori.gif

Ειναι η δισδιαστατη απεικονιση του υπερκυβου.

Demetres ευχαριστώ για την απάντηση και τις σημειώσεις :)

AlexandrosG δεν κατάλαβα τι ακριβώς εννοείς με τον υπερκύβο.Το πρόβλημα πάντως λύνεται σχετικά εύκολα με αρχη της περιστεροφωλιάς...

[quote="george2"]AlexandrosG δεν κατάλαβα τι ακριβώς εννοείς με τον υπερκύβο.[/quote]

Ουτε εγω :rotfl:

Απλα ειπα οτι εκει πηγε το μυαλο μου μιας και τα χρωματα ειναι 6. Αυτο γιατι ξερω την πολυ απλη εκδοχη του προβληματος
που λεει να δειξουμε οτι αν το $ R^2$ βαφτει με 2 (η λιγο πιο δυσκολα με 3) χρωματα τοτε υπαρχουν δυο σημεια ομοχρωματικα που απεχουν αποσταση 1.
Σε τετοια λοιπον προβληματα πρεπει να βρεις το καταλληλο σχημα και μετα να εφαρμοσεις την αρχη της περιστεροφωλιας οπως λες.
Ετσι ειπα στην τυχη μηπως το σχημα που εδωσα δουλευει.  :)

Επισης (ηταν και θεμα Putnam) να εξετασετε αν το επιπεδο βαφτει με 9 χρωματα τοτε ειναι σιγουρο οτι θα υπαρχουν σιγουρα δυο σημεια αποστασης 1
ομοχρωματικα? Χωρις να ειμαι καθολου σιγουρος νομιζω οτι για 4,5,6,7,8 χρωματα το προβλημα ειναι ανοιχτο η τουλαχιστον πολυ δυσκολο.

[quote="AlexandrosG"][quote="george2"]AlexandrosG δεν κατάλαβα τι ακριβώς εννοείς με τον υπερκύβο.[/quote]

Ουτε εγω :rotfl:
[/quote]

Ένας τρόπος να δεις ότι αυτό δεν δουλεύει είναι ότι έχεις μόνο 16 κορυφές. Άρα μπορώ να τις χρωματίσω με 6 χρώματα ώστε κανένα χρώμα να μην χρησιμοποιηθεί σε 4 κορυφές. Άρα σίγουρα δεν μπορεί να υπάρξει μονοχρωματικό παραλληλόγραμμο.

[quote="AlexandrosG"]
Επισης (ηταν και θεμα Putnam) να εξετασετε αν το επιπεδο βαφτει με 9 χρωματα τοτε ειναι σιγουρο οτι θα υπαρχουν σιγουρα δυο σημεια αποστασης 1
ομοχρωματικα? Χωρις να ειμαι καθολου σιγουρος νομιζω οτι για 4,5,6,7,8 χρωματα το προβλημα ειναι ανοιχτο η τουλαχιστον πολυ δυσκολο.[/quote]

Με 9 γίνεται διαμερίζοντας το επίπεδο σε τετραγωνάκια (με το κατάλληλο μήκος). Μπορεί να γίνει και με 7 διαμερίζοντας το επίπεδο σε εξάγωνα (με το κατάλληλο μήκος). Ποιο από τα 4,5,6,7 είναι το σωστό, είναι όντως [url=http://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem]ανοικτό πρόβλημα[/url].


Let ABCD be a quadrilateral such that the internal bisectors of B and C intersects on the segment AD at E. If AE = AB = 13, DE = CD = 15, and [BCE] = 84  (area of BCE), then find the length of the segment CE.


Just wondering, is anyone teaching their younger brother/sister math/french/whatever over the summer?  Or am I the only cursed one?

Here's another one... are you teaching yourself anything over the summer?  Answer the poll also, please...

Me - learning Chem, Calc B, doing a little Phys.
I'm coaching my little brother in Bio...if he likes. Which is unlikely - he prefers electronics. Else we're doing Lit/Writ.

Cool.  Chem is fun.  You thought about taking the SAT IIs?  I will probably take both Sat2 Bio and Chem next year...

Bio?  How old is your 'little brother'?  Mine is gonna be 5th grade, and I'm teaching him Algebra.  Lit./Writ. sounds like a pain to learn and teach... especially because we math types aren't the greatest at English.

- Mr. Psycho
a.k.a. mathFANATIC

About time I learned what a proof is -- I'm going to teach myself some real analysis.

Mine's two years older - entering 7th grade...

Wow.  I remember learning bio in 7th grade over at the high school and it was HARD!  But fun, naturally.  Bio was a bunch of facts, but it was a bunch of [i]cool[/i] facts.

Of course, I would not relish the thought of teaching my own little monster about the reproductive system...that's bio, no?

*cough* I will skip over that. I don't imagine he wants to hear it either...

<laughs> I guess that's good that he wouldn't want to know... but then again, kids at that age are [i]quite[/i] curious about the facts of life, even if they won't admit it... But as long as he [i]says[/i] he doesn't want to learn about that sort of thing, then you're in the clear.

Which brings this up...That's one job I wouldn't relish... a SHARE teacher... teaches little kids about delicate subjects. :oops:

Good luck.  May the force be with you, Hann (my apologies! :wink: )

Same to you.  :D
*blink* There's no need for apologies...

Just wondering, but is your brother doing mathcounts, or AMC, or math league, or any other math competition?  My little bro is probably gonna do mathcounts next year (6th grade - the first time he can really participate).

Probably not. My brother's not a math person...which is highly unfortunate, but ohwell.

First of all everyone is a math person, or at least they have the potential to be. Although it sounds cliche, my parents did think I was mentally challenged until I entered HS. As long as you have the determination and a patient mentor you can succeed in mathematics or whatever *slaps hand for being cliche again*...

As per the topic of the post, I'm tutoring my little bro in Biology (he wants to do the olympiad program) and mathematics. I think what I'm doing with bio is having him read the Campbell text and ask me questions on what he doesn't know, and then I will test him on what was in the chapter. For mathematics I have him read AoPS I a little at a time, and to him it's like a conversation... a technique that more textbooks ought to employ. I think he has picked up on a lot of mathematics by using the books, and I will continue to see what happens.

Stay Patient!
IW.OAV

True, we all have the potential, but whether we have the interest is another matter. Good luck to your little bro - Bio is great.

idiot - just wondering, are you using the campbell 4th ed bio book or the campbell/mitchell/reece bio book.  I used the latter, and thought it was good, but I've heard the former is good too.

and I agree with you.  Everyone's got the potential, it's whether you have the will and the attention span.  Take me and English Class, for example.  If I put my mind to it I could be an English wiz, the expert on some obscure literary work at age 20 - but I like math more.  And running.

Still, Hann, I'll bet your brother feels a lot of pressure to do well at math, since you're good at it.  Mine does - maybe for that reason he sometimes tries not to like math??  To be different?

*blinkityblink* -_-" I'm not good at math. Sorry if I've been giving that impression. I'm okay at it, but I'm not even close to ComplexZeta's level. I like math and I enjoy working at it, but bio's my first love.  Even though for some reason I'm better at Lit than anything else...weird.
Actually I'm using a few S'porean texts now - but I've heard of the campbell 4th ed book - it's good. I should try the other one sometime...Perhaps you're right. My brother enjoys all the areas I'm hopeless in. Such as electronics and programming - although I'm working on the latter.

hm...I have a little 4-year-old brother to whom I'm trying to teach some things.  He's watching way too much TV everyday (like 10 hours), so I have to limit that by teaching him something everyday.  However, I forget on many days...Anyways, I just teach him out of a beginning math book, and another preschool book.

-interesting_move

[quote="rippledance"]I'm not even close to ComplexZeta's level. [/quote]

But neither am I.  Nor are a lot of people.  Fact is, ComplexZeta, TripleM, and some of the other vets here are really good at math.  Y'don't have to be as good as them to be here.  In fact, you could hate math and still be here.

But anyway, my point is this: you're good at something - not good-for-nothing ruffians like me - and that's what matters.  Find what you like.  And do it.  Same for your brother.  He should find what he likes and do that - whether he likes math or not isn't very important.  It's like somebody said before: everyone's got the potential, it's what they do with it.

Aaaand... that concludes my pep talk.  But it doesn't matter to us whether you like math or are good at it, because you're a great person to [cybernetically] be around.  Just keep up your interest in bio - it's a great subject and there are lots of opportunities out there for aspiring biologists (think of the scholarships! the research opportunities!  the jobs!) - and you'll do great.  And keep hanging at artofproblemsolving.com!

It's amusing for me to hear people complaining (or maybe just stating) that they aren't at my level at math after years of getting annoyed with myself for not being at the level of the USAMO winners and people like that.

well  you worked hard and did it. and hopefully other people (including me) will work hard to and be as good as you. or shall i say [i]better[/i]

muachachachachachachachacha

I own like 3 editions of that book, 4th, 5th, and 6th... It makes me feel special. Personally I just think they have gotten better and better. I feel that the 6th employs a lot more technology with regards to its uses in Biology and was kinda interesting. Plus it has some new diagrams. And anyways its green colour suits much more than that red or pink.

I bow to you Neil Campbell.


function trigonometry algebra polynomial

Solve the following system of equation in $ x,\ y,\ z\  \left(0\leq x\leq y\leq z\leq \frac{\pi}{2}\right)$.

$ \left\{
\begin{array}{ll}
\sin x\plus{}\sin y\plus{}\sin z\equal{}\frac{1}{2}(1\plus{}\sqrt{2}\plus{}\sqrt{3}) &\quad  \\
\sin ^ 2 x\plus{}\sin ^ 2 y\plus{}\sin ^ 2 z\equal{}\frac{3}{2} &\quad  \\
\sin ^ 3 x\plus{}\sin ^ 3 y\plus{}\sin ^ 3 z\equal{}\frac{1}{8}(1\plus{}2\sqrt{2}\plus{}3\sqrt{3})&\quad
\end{array}
\right.$

hello kunny, i have found $ x\equal{}\frac{\pi}{6},y\equal{}\frac{\pi}{4},z\equal{}\frac{\pi}{3}$.
Sonnhard.

hello kunny, from where do you get such nice problems?
Sonnhard.

oh, I see. its easy guessing. a more rigorous solution would involve solving a cubic equation.

We use [url=http://www.mathlinks.ro/Wiki/index.php/Newton%27s_sums]Newton's sums[/url].

Consider a polynomial $ P(x)$ of degree $ 3$,
\[ P(x) = a_3x^3 + a_2x^2 + a_1x + a_0\]
Let $ P(x) = 0$ have roots $ \sin x$, $ \sin y$, $ \sin z$.
Define the sums: $ S_n = \sin^n x + \sin^n y + \sin^n z$
Newton sums tell us that,
$ \left\{\begin{array}{lllll}
a_3S_1 + a_2 & = & \frac{1+\sqrt{2}+\sqrt{3}}{2}a_3 + a_2 & = & 0 \\
a_3S_2 + a_2S_1 + 2a_1 & = & \frac{3}{2}a_3 + \frac{1+\sqrt{2}+\sqrt{3}}{2}a_2 + 2a_1 & = & 0 \\
a_3S_3 + a_2S_2 + a_1S_1 + 3a_0 & = & \frac{1+2\sqrt{2}+3\sqrt{3}}{8}a_3 + \frac{3}{2}a_2 + \frac{1+\sqrt{2}+\sqrt{3}}{2}a_1 + 3a_0 & = & 0
\end{array}\right.$
$ \iff\left\{\begin{array}{l}
a_2 = - \frac{1+\sqrt{2}+\sqrt{3}}{2}a_3 \\
a_1 = \frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{4}a_3 \\
a_0 = -\frac{\sqrt{6}}{8} a_3
\end{array}\right.$

So, $ P(x)=a_3\left(x^3-\frac{1+\sqrt{2}+\sqrt{3}}{2}x^2+\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{4}x-\frac{\sqrt{6}}{8}\right)$

\begin{eqnarray*}
P\left(\frac{1}{2}\right)
& = & a_3\left(\left(\frac{1}{2}\right)^3-\frac{1+\sqrt{2}+\sqrt{3}}{2}\left(\frac{1}{2}\right)^2+\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{4}\left(\frac{1}{2}\right)-\frac{\sqrt{6}}{8}\right) \\
& = & 0
\end{eqnarray*}
So, $ \left.\left(x-\frac{1}{2}\right)\right|P(x)$.

Hence,
\begin{eqnarray*}
P(x)
& = & a_3\left(x-\frac{1}{2}\right)\left(x^2-\frac{\sqrt{2}+\sqrt{3}}{2}x+\frac{\sqrt{6}}{4}\right) \\
& = & a_3\left(x-\frac{1}{2}\right)\left(x-\frac{\sqrt{2}}{2}\right)\left(x-\frac{\sqrt{3}}{2}\right)
\end{eqnarray*}

Therefore, $ P(x)=0 \iff x=\frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}$.
Thus,
$ \sin x = \frac{1}{2} \iff x = \frac{\pi}{6}$,
$ \sin y = \frac{\sqrt{2}}{2} \iff y = \frac{\pi}{4}$,
$ \sin z = \frac{\sqrt{3}}{2} \iff z = \frac{\pi}{3}$.

Here is my solution:

$ a = \sin x,\ b = \sin y,\ c = \sin z$

$ \Longrightarrow a + b + c = \frac {1 + \sqrt {2} + \sqrt {3}}{2},\ a^2 + b^2 + c^2 = \frac {3}{2},\ a^3 + b^3 + c^3 = \frac {1 + 2\sqrt {2} + 3\sqrt {3}}{8}$.

$ \therefore ab + bc + ca = \frac {1}{2}\{(a + b + c)^2 - (a^2 + b^2 + c^2)\} = \frac {\sqrt {2} + \sqrt {3} + \sqrt {6}}{4}$, we have that

$ a,\ b,\ c\in{t^3 - \frac {1 + \sqrt {2} + \sqrt {3}}{2}t^2 + \frac {\sqrt {2} + \sqrt {3} + \sqrt {6}}{4} - abc = 0}$.

Since $ 3abc = a^3 + b^3 + c^3 - \frac {1 + 2\sqrt {2} + \sqrt {3}}{2}(a^2 + b^2 + c^2) + \frac {\sqrt {2} + \sqrt {3} + \sqrt {6}}{4}(a + b + c)$

$ = \frac {3\sqrt {6}}{8}\Longleftrightarrow abc = \frac {\sqrt {6}}{8}$.

Solving the cubic equation $ t^3 - \frac {1 + \sqrt {2} + \sqrt {3}}{2}t^2 + \frac {\sqrt {2} + \sqrt {3} + \sqrt {6}}{4} - \frac {\sqrt {6}}{8} = 0$

$ \Longleftrightarrow (t - \frac {1}{2})(t - \frac {\sqrt {2}}{2})(t - \frac {\sqrt {3}}{2}) = 0$ gives $ t = \frac {1}{2},\ \frac {\sqrt {2}}{2},\ \frac {\sqrt {3}}{2}$.

Since $ 0\leq x\leq y\leq z\leq \frac {\pi}{2}\Longrightarrow 0\leq \sin x\leq \sin y\leq \sin z$, we have ${ (\sin x,\ \sin y,\ \sin z) = (\frac {1}{2},\ \frac {\sqrt {2}}{2},\ \frac {\sqrt {3}}{2}})$, yielding $ \boxed{x = \frac {\pi}{6},\ y = \frac {\pi}{4},\ z = \frac {\pi}{3}}$

Comment: You can find the value of $ abc$ by $ \boxed{a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}$


linear algebra
matrix
linear algebra unsolved

Prove if S and T have the same eigenvectors
then ST =TS

are you working over an algebraically closed field?
because over the reals it is incorrect?

A counterexample over any field (with at least three elements)
Let $a$ and $b$ be two distinct nonzero elements of the field.
$S=\begin{bmatrix}0&a&0\\0&0&1\\0&0&0\end{bmatrix}$, $T=\begin{bmatrix}0&b&0\\0&0&1\\0&0&0\end{bmatrix}$.
Then $ST=\begin{bmatrix}0&0&a\\0&0&0\\0&0&0\end{bmatrix}$ and $TS=\begin{bmatrix}0&0&b\\0&0&0\\0&0&0\end{bmatrix}$.
On the other hand, both $S$ and $T$ have multiples of $\begin{bmatrix}1\\0\\0\end{bmatrix}$ as their only eigenvectors.

This works if there exist $n$ linearly independent eigenvectors for each matrix, but that's certainly not always true.

(Thanks for the correction)

$\begin{bmatrix}1\\0\\0\end{bmatrix}$

you probably meant this?


How many ordered triples of integers $ (a,b,c)$, with $ a \ge 2$, $ b\ge 1$, and $ c \ge 0$, satisfy both $ \log_a b \equal{} c^{2005}$ and $ a \plus{} b \plus{} c \equal{} 2005$?

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

[hide="Answer"]$\log_ab=c^{2005}\Rightarrow a^{c^{2005}}=b$
Obviously, if $c>1$, then $b$ will be too large to use, so our options are $c=0$, $c=1$.
For $c=0$, we have $a^0=b$, so $b=1$ and $a=2004$. For $c=1$, we have $a=b$, so $a=1002$, $b=1002$, and there are a total of $2$ ordered triples. $\boxed{C}$[/hide]

[quote="da-kid"]I think c=2003, a=1, b=1 works as well along with the others you mentioned.[/quote]

The value of $ a$ can't be $ 1$, since you're given both $ \log_ab$ and $ a\geqslant 2$

The logarithm of 1 base 1 is undefined anyways. The reason: Using the conventional definition $ \log_b{a} \equal{} c \Leftrightarrow b^c \equal{} a$ with $ a \equal{} b \equal{} 1$ allows $ c$ to be any value. If we allowed $ \log_1{1}$, we could use the transitive property of equality to show all real numbers are equal to each other, and that's bad.

This problem was kinda easy. How come the transition comes so fast (from relatively easy AMCs to the hard one last year)?


Greetings,  :bye:

Before anything else I would like to thank the author of this site for the excellent LaTeX tutorial, which enabled me to acquire basic skills in the language in a short time.

I am writing to ask if anyone could point me to freeware which can plot elegant mathematical graphs and save them as images.

I use Windows XP.

Thank you very much,  :blush:

Grumpy Troll

A few programs to get you started (in no particular order):
1. [url=http://www.gnuplot.info/]Gnuplot[/url] is popular with LaTeX users
2. [url=http://www.graphcalc.com/]GraphCalc[/url]
3. [url=http://math.exeter.edu/rparris/winplot.html]Winplot[/url]
4. [url=http://tug.org/PSTricks/]PSTricks[/url] and in particular [url=http://tug.org/PSTricks/main.cgi?file=pst-plot/FunctionExamples]pst-plot[/url] will allow you to write code inside a TeX file to create plots. For example,
[img]http://sixthform.info/steve/wordpress/latexrender/pictures/2b39fedf33217c52d95726162f3cf0fe.gif[/img]
is created using
[code]\psset{unit=0.5cm}
\begin{pspicture}(-4,-0.5)(4,8)
\psgrid[subgriddiv=0,griddots=5,gridlabels=7pt](-4,-0.5)(4,8)
\psline[linewidth=1pt]{->}(-4,0)(+4,0)
\psline[linewidth=1pt]{->}(0,-0.5)(0,8)
\psplot[plotstyle=curve,linewidth=0.5pt]{-4}{0.9}{10 x exp}% postscript function
\rput[l](1,7.5){$10^x$}
\psplot[plotstyle=curve,linecolor=red,linewidth=0.5pt]{-4}{3}{2 x exp}% postscript function
\rput[l](2.2,7.5){\color{blue}$e^x$}
\psplot[plotstyle=curve,linecolor=blue,linewidth=0.5pt]{-4}{2.05}{2.7183 x exp}% postscript function
\rput[l](3.2,7.5){\color{red}$2^x$}
\rput(4,8.5){\color{white}change\normalcolor}
\rput(-4,-1){\color{white}bounding box\normalcolor}
\end{pspicture}[/code]

I hope others will add programs they have used and that work well for them.

GS demo works for me, but that's only 30 days.

Thank you very much for your answers: PSTricks and pst-plot are exactly the packages that I needed. :lol:

Please may you tell me how to use those packages to produce a graph such as the one above in a document? :blush:

All you need do is add the line \usepackage{pst-plot} plus the above code to a normal document.
[code]\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{pst-plot}
\begin{document}
\psset{unit=0.5cm}
\begin{pspicture}(-4,-0.5)(4,8)
\psgrid[subgriddiv=0,griddots=5,gridlabels=7pt](-4,-0.5)(4,8)
\psline[linewidth=1pt]{->}(-4,0)(+4,0)
\psline[linewidth=1pt]{->}(0,-0.5)(0,8)
\psplot[plotstyle=curve,linewidth=0.5pt]{-4}{0.9}{10 x exp}% postscript function
\rput[l](1,7.5){$10^x$}
\psplot[plotstyle=curve,linecolor=red,linewidth=0.5pt]{-4}{3}{2 x exp}% postscript function
\rput[l](2.2,7.5){\color{blue}$e^x$}
\psplot[plotstyle=curve,linecolor=blue,linewidth=0.5pt]{-4}{2.05}{2.7183 x exp}% postscript function
\rput[l](3.2,7.5){\color{red}$2^x$}
\rput(4,8.5){\color{white}change\normalcolor}
\rput(-4,-1){\color{white}bounding box\normalcolor}
\end{pspicture}
\end{document}[/code]There is a big BUT. PSTricks relies on using postscript (the PS of PSTricks or pst-plot) so you cannot compile directly to PDF but must use postscript. In TeXnicCenter this is easy - choose LaTeX=>PS=>PDF in the dropdown box. This compiles the file to DVI then to PS then to PDF - sounds complicated but it all goes on in the background so you don't see it.

Thank you very much for your answer.

[quote="stevem"]In TeXnicCenter this is easy - choose LaTeX=>PS=>PDF in the dropdown box.[/quote]

How may I obtain this option? I can only convert from LaTeX to PS and from LaTeX to PDF, not from LaTeX to PS to PDF.

It may be because you don't have Ghostscript installed. Have a search on your computer for a ps2pdf file. If it doesn't exist then install Ghostscript from [url=http://www.cs.wisc.edu/~ghost/]here[/url]. Then re-run the Wizard in TexnicCenter to find it (Build -> Define Output Profiles -> Wizard). I hope that will then give you the extra option.

I found many ps2pdf files, so I guess that I have Ghostscript installed, yet the Wizard does not yield the expected result and still produces only three profiles: from LaTeX to DVI, to PDF or to PS.

Thank you for your patience. :blush:

Try importing my output profile for Latex=>PS=>PDF which I am attaching (unzip then Build, Define Output Profiles, Import). You will probably have to adapt the paths to your computer, for example for gswin32c.exe (which is the main Ghostscript program). If you can't find gswin32c.exe then try installing Ghostscript - it won't do any harm or interfere with anything else.

Your suggestion worked for me — thank you so very much. :lol:

One last question, somewhat loosely related: does Adobe Acrobat Reader 8.0 produce an error when closed on anyone else's computer?


geometry
3D geometry
probability
function
greatest common divisor

hi everyone i got this problem at a problem solving class just wondering if anyone knows how to do it?

'At the vertices of a cube are written eight pairwise distinct natural numbers, and on each of its edges is written the greatest common divisor of the numbers at the endpoints of the edge. Can the sum of the numbers written at the vertices be the same as the sum of the numbers written at the edges?'[/quote]

Locked and moved to Pre-Olympiad from F&G.

Generally, it is an unstated policy not to discuss the contents of online classes outside of the class boards. If you need a solution to a problem, your classmates and the instructor will be able to help you out.

If this problem is from a class outside of AoPS, contact an administrator and this thread can be unlocked upon verification of that claim.

Well, I can tell you that the answer is, with 95% probability, no.  The key reason here is that the problem would be a computing excercize if the answer were yes -- there wouldn't be any problem solving aspect of it.  My guess is that there is a bounding argument -- try to prove that the sum of the edges is always less than the sum of the vertices.  For example, if we call V the value function for edges and vertices, it is possible to show that $V(AB) \leq \min\left(V(A), V(B), \max\left(\frac{V(A)}{2}, \frac{V(B)}{2}\right)\right)$.  It might be possible to use this to show that the sum of the edges is less than that of the vertices.  Or it might not -- I don't know for sure, and it's an ugly tool to use.  There are lots of other ways to try to bound the sum of the edges, of course, and I don't know which (if any) of them will work.

The answer is yes if I'm understanding the problem correctly. I'm not really sure what "pairwise distinct" natural numbers means -- if that means adjacent vertices are distinct, then explore with some simple small numbers.

I assumed it meant that among the 8 vertices, there were no two with the same number on them, which is a stronger condition than what you suggest.

Yes, that would be a harder question. I think that the answer then would be no. Perhaps by considering a "best/worst case scenario" and showing that it can't be improved upon.

Perhaps the closest you can get the two sums is by using the divisors of 24 as the vertices.

My version is definitely right, because I just found a nice proof for it :)[hide]

My general suggested approach was right: it's a bounding argument.  The way it goes is as follows:
If a and b are any two different positive integers, we have $\gcd(a,b) \leq \frac{a+b}{3}$.  This can be shown by an examination of two cases, in which the smaller of a and b divides the larger and in which neither divides the other, in combination with the fact that the largest proper factor of any positive integer is at most half its value.  So, then you just show that equality cannot hold on all edges simultaneously, which is easy, and then sum that expression over all edges and you're done![/hide]


For naturals $ a,b,c,d: \; 2^a\plus{}2^b\equal{}2^c\plus{}2^d.$ Prove that $ a\plus{}b\equal{}c\plus{}d.$

Suppose that $ a$ is the minimum of $ a,b,c,d$. So, we have:
$ 1 \plus{} 2^{b \minus{} a} \equal{} 2^{c \minus{} a} \plus{} 2^{d \minus{} a}$.

So, we have some cases:

1. $ d \minus{} a \equal{} 0$ or $ c \minus{} a \equal{} 0$ (without loss of generality, $ c \minus{} a \equal{} 0$). So:
$ 2^{b \minus{} a} \equal{} 2^{d \minus{} a}$.
So, $ b \equal{} d$ and $ a \equal{} c$.

2. $ c \minus{} a > 0$ and $ d \minus{} a > 0$.
The right side is the number that is divisible by 2, so $ 1 \plus{} 2^{b \minus{} a}$ is even, so $ b \minus{} a \equal{} 0$.
Now, $ b \equal{} a$ and $ 2 \equal{} 2^{c \minus{} a} \plus{} 2^{d \minus{} a}$. It means that $ c \equal{} a$ and $ d \equal{} a$. So, $ a \plus{} b \equal{} c \plus{} d$.

In my humble opinion, binary representation of a number is always unique, and this case hardly deserves special consideration...


Find all solutions of the equation

$ (x^2\plus{}y)(x\plus{}y^2)\equal{}2^n$ where x,y and n [b]positive integers.[/b]

[hide="hint"]$ x^2 \plus{} y \equal{} 2^p$

$ x \plus{} y^2 \equal{} 2^q$[/hide]


geometry rectangle MATHCOUNTS parallelogram

[hide]
"unwrap" the cylinder to get a rectangle with a length of 80 and a width of 30pi
there would be two daignal strips going down the rectangle
each strip has an area of 1200-(1200-120pi) or 120
so 240 total, or C
[/hide]

[quote="#H34N1"][img]http://i96.photobucket.com/albums/l167/adasarathy/P19.gif[/img][/quote]

[hide]If you roll the side of the cylinder out, the strip becomes a parallelagram with base 3 and height 80.  Therefore $3\cdot 80=\boxed{240}$.[/hide]

Does the Silo--- the strip go around too :?:

[hide]Really the diameter of the silo doesnt matter, as the width shown is the [i]horizontal[/i] width, not the width perpendicular to the sides. Thus, you may opt to simply multiply3 by 80, getting C.[/hide]

Easiest way[/i]

Wasn't this also a State Mathcounts problem?

There was a state MC (2008) problem like this except the lengths were 3-4-5 instead of 30-80

Sorry for reviving, but I have a [hide="question."]
How do we know so quickly that the area of the stripe equals the area of the parallelogram that we get by "netting" the cylinder?
[/hide]

Thanks in advance!

Because that parallelogram is the stripe... Or do you mean to ask how we know that "netting" a figure is area-preserving?

I thought the parallelogram represents the net of the cylinder without the two circles.

The net of the cylinder without the two circles on the ends is a rectangle. In this case, since the stripe "slants" down the cylinder in a curving fashion, it has the properties of a parallelogram.

Did that help? Or did I answer the wrong question? :)

Thanks for the response, greatwhiteshark. That, definitely, helps. What I really meant to ask, though, was this. Isn't the parallelogram the silo and other part without the silo? It's not the case that this whole parallelogram is covered by the silo.

You need to clarify which parallelogram you're talking about. The whole lateral face may be netted as a parallelogram (most commonly, a rectangle) depending on how you cut it. In that net, both the stripe and what's left of the lateral face without the stripe are also parallelograms. And what is this "other part"?

What I meant by the "other part" is the part without the silo. However, it looks like the solutions above use the fact that the whole rectangle has the silo on it. Why?

...Where do you see something that isn't a silo? Do you mean "without the lateral face", i.e. just the bases? If so, the bases aren't relevant.

Again, I don't know what whole rectangle you're talking about, but let me see if I can summarize what the solutions above are using. When the cylinder is netted, the lateral face forms a rectangle, with the stripe becoming a parallelogram with width $3$ and height $80$. Thus, we can simply use the formula $A=bh=3\cdot80=240$.

just imagine the strip to be a rectangle  then length=80,breadth=3 .So Area=l*b=80 *3=240

I am unable to visualize what will happen once I unroll the strip. I tried to do it with some paper and a pen and I got a weird figure of which I am unable to compute the area. Can someone please put a diagram of what will happen once the strip is unraveled?

Ok, so when you wrap up a rectangle by taping one of its sides to the opposite, side, it becomes a cylinder. Try drawing a parallelogram like in the problem on a piece of rectangular paper, then tape the opposite sides together. You will get the cylinder shown in the diagram.

Or, you can see that the two stripes are a constant $3$ units apart at that particular angle, so that they're parallel.  The top and bottom of the tube show two more parallel "sides" to the figure.  Since everything is a line when it's unwrapped, you have a figure with two sets of parallel sides - and that's your parallelogram.

I'm sorry for reviving this again, but the problem states that the red strip wraps completely around the silo two times. I had two questions that popped up from this. 1. How do we know that the parallelogram of base 3 and height 80 will wrap 2x around the silo, and 2. How would we find the area of the strip if the problem were altered, and instead of wrapping 2x around the silo, it only wrapped once?

No matter how many times it is wrapped around, the base and height remain constant, so the area would still be the same.


LaTeX
modular arithmetic
combinatorics unsolved
combinatorics

Let $ G \equal{} (V,E)$ be a simple planar graph, properly embedded in the plane. Let C be any cycle in G, and let the faces of G in the interior of C (as drawn in the plane) be
$ F_1, F_2,. . . , F_m$. Show that $ |E(C)| \equiv deg(F_1) \plus{} \hdots \plus{} deg(F_m) (\mod 2)$.

having a little difficulty with this :(

Assuming $ |E(C)|$ means the length of the cycle and $ \deg (F_i)$ means the number of edges in the face $ i$, think inductively. (The claim is obvious when $ C$ bounds a face.  When you add a point on the interior of the face with some new edges, ....)

Note that \deg is a LaTeX command.  Also, \pmod gives you the mod symbol with parentheses properly spaced.

I am not sure how the congruency comes into play though >_<

Why don't you start with my hint, write down where it leads you, and then we can discuss where to go from there.  You might also think a bit about what it means for a number to not change modulo 2.


geometry
circumcircle
geometric transformation
reflection
trigonometry
geometry unsolved

[quote="outback"]Let $ ABC$ be an acute-angled triangle with altitudes $ AD, BE, CF$ and $ O$ is the circumcenter. Let $ M, N, P$ be the midpoints of $ BC, CA, AB$. Let $ D_0$ be the reflection of $ D$ across $ M$, $ E_0$ be the reflection of $ E$ across $ N$, $ F_0$ be the reflection of $ F$ across $ P$. Prove that $ O$ is interior to triangle $ D_0E_0F_0$.[/quote]
$ \frac{CD_0}{CE_0}\equal{}\frac{BD}{AE}\equal{}\frac{ON}{OM}\Rightarrow [COD_0]\equal{}[COE_0]$. thus CO passes through the midpoint I of $ D_0E_0$, we get $ 4CI^2\equal{}AB^2\sin^2C\Rightarrow 2CI\equal{}AB\sin C\equal{}2OC\sin^2 C<2OC\Rightarrow CI<CO$
Therefore, O is exterior to triangle $ CD_0E_0$. Similarly, O is exterior to triangles $ BD_0F_0, AE_0F_0$. because of acute triangle, O is interior to triangle $ D_0E_0F_0$


Hi.  I am running for vice-prez of CTC.  Do you think it's a good idea and will it stop the tourney mania?  This is just a discussion thread.  No spamming.

i dont think ctc is a good idea

wuts the point? if people want to make a BUNCh of tourneys who cares?

stupid double thread :P


Let $ABCD$ be a quadrilateral of sides $a,b,c,d$ inscribed in a circle of radius $R$ and circumscribed in a circle of radius $r$.  We have

$ 64(R^2 +2r^2) \leq \frac{(a+c)^6}{abcd} $

[quote="manlio"]Let $ABCD$ be a quadrilateral of sides $a,b,c,d$ inscribed in a circle of radius $R$ and circumscribed in a circle of radius $r$.  We have

$64(R^2 +2r^2) \leq \frac{(a+c)^6}{abcd}$[/quote]

 :D  :D


Hello, I have a question.

How do I derive the double doppler shift. You know, the one that says f'=f((v+-v_0)/(v-+v_s))

Solution

Before embarking on a proof, I will discuss the importance of an "ether" in the validity of this expression. An ether is a medium through which waves are transmitted, like air for sound waves or water for water waves. The ether can move at a velocity, called the "ether drift". In air, it would be wind, and in water it would be the ebb and flow of the tide. If the ether is moving with respect to a source of waves then the ether velocity combines linearly (as in Newtonian kinematics) with the velocity of the source.  We will try and understand the full implications of this with an example. Consider a bouy fixed to a pier wall by a rope, but allowed to bob in the water. Suppose that the net tidal velocity at the buoy is 1 metre per second away from the wall. The bouy has got some sort of device attached which creates waves moving in all directions at 5 metres per second - when the bouy is in [b]stationary[/b], non tidal waters. Then the speed of the waves in the tidal waters will be different. The ether drift adds to the velocity, so waves move at 6 metres per second away from the wall, and at 4 metres per second towards the wall, as measured by someone standing on the pier (i.e. not moving relative to the bouy). This seems pretty trivial. Now consider a bouy moving towards the pier wall at a velocity of 1 metre per second in stationary waters. If the bouy created just an up and down disturbance when stationary in stationary waters, then when it is moving, this disturbance is a wave which moves at the same velocity of the bouy. Any moving wave produced by the bouy will then be defected by this velocity. So the waves produced have a velocity of 6 metres per second towards the wall and 4 metres per second away from the wall. For convenience say the direction away from the wall is forward. Now remove the pier, such that the bouy is on open sea - the same in all directions. If we move along with the bouy, such that it is always the same distance away, not moving relative to us, then we see everything speeded up by 1 metre per second in the forwards direction. Thus the bouy, which appears to be stationary, must produce waves of 5 metres per second in all directions, and the stationary water appears tidal and moving forward with velocity 1 metre per second. This seems to be the same set up as the case when we talked about a stationary bouy, but yields different observed wave speeds. It shows knowing that the relative velocity of a source and an ether, is not enough! It has something to do with the pier in this example, which acts as a universal rest frame. Einstein completely contradicts this, in the case of electromagnetic radiation, in his two axioms of special relativity. In loose terms they jointly state that "light has the same speed in all directions, and it doesn't matter how fast you are travelling relative to the source". Everything is relative. So any classical formula must be restricted to cases like water and air. The universal frame of rest is taken to be that in which fixed distant stars are at rest.

I will complete a proof later.[/b]


Euler
function
group theory
superior algebra
superior algebra solved

let G a cyclic group with |G|=n. Show that for all integer m that m | n (including m=1 and m= n), exists \[ \varphi(m)\] elements with order  m

Obvious, $ G$ is isomorphic with $ \mathbb{Z}_{n}$. So we can prove it on $ \mathbb{Z}_{n}$.
Let $ \hat{x}$ be an element with order $ m$. This means, there is $ k\in\mathbb{Z}$ such that $ mx\equal{}nk$. Also, $ m$ is the smallest number with this property.
First, we can consider $ 0\leq x < n\Rightarrow 0\leq k < m$.
I will prove that $ gcd(m,k)\equal{}1$. Suppose $ d\equal{}gcd(m,k)$ with $ d>1$. Then, $ \frac{m}{d}x\equal{}n\frac{k}{d}$, which contradicts the minimality of $ m$ with that propery. So $ d\equal{}1$.
Also, for any $ k$ with $ gcd(k,m)\equal{}1$, there is $ x\equal{}\frac{nk}{m}\in\mathbb{Z}$ such that order of $ \hat{x}$ is $ m$ (and that $ x$ in $ [0,n)$ is unique).
From these, we conclude that the number of elements with order $ m$ is $ \varphi(m)$.

Try to prove the following theorem first.

[b]Theorem[/b]: Let $ G\equal{}\mathbb{Z}_{n}$ be a cyclic group. If $ a\in G$ then the order of $ a$ is $ n/\gcd (a,n)$. So if we want the order to be $ n$ then $ \gcd(a,n)\equal{}1$ then the number of such integers is $ \phi(n)$ by definition.


I got a question on how to solve this.

If $\arctan x + \arctan k = 45^\circ$, then $x$ is what?

Can anyone help me find the solution to this problem? Thanks.

what is k?? can we solve it w/o knownig what k is??

[hide]Take the tangent of both sides.
$\tan (\arctan x + \arctan k)=1$
$\frac{x+k}{1-xk}=1$
$x+k=1-xk$
$x=\frac{1-k}{1+k}$[/hide]

Heh. That solution above is one of the great things about Art of PRoblem Solving Games.  :)

I'm not too familiar with arctan.

Can you explain how you get the second step and so forth?

Thanks. :)

Just apply the tangent sum formula, $\tan{(x+y)} = \frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}$ and know that $\tan{(\arctan{x})} = x$.


Prove, for any positive reals a, b and c:
(a/(b+c) + 1/2)*(b/(c+a) + 1/2)*(c/(a+b) + 1/2) \geq 1.

Very easy:
  Put a+b=z, b+c=x and c+a=y. It becomes (x+y)(y+z)(z+x)>=8xyz.


[quote="frt"]In $\triangle ABC$, $\angle A=45^\circ$ and $\angle B=75^\circ$. The altitude from $A$ meet $BC$ at $D$, and the altitude from $C$ meet $AB$ at $E$. If $AC=2$, find $DE$.[/quote]
[hide]
Answer: $.5176380902$?

I just constructed angles from the intersection of AD and CE and used $\sin$ and $\cos$ formulas. [/hide]

you can easily find the sides using standard trig values, then there is a nice pair of similar triangles to see that finishes the problem

[quote="agolsme"][hide]
$\triangle ACD$ is a 30-60-90 triangle, so $CD = 1$. $\triangle ACE$ is a 45-45-90 triangle so $CE = \sqrt{2}/2$. Using the law of sins: $DE/\sin{15^{circ}} = CE/\sin{135^{\circ}}\implies DE = 2\sin{15^{\circ}}$. We can find $\sin{15}$ using the half angle formula: \[ \sqrt{2 - \sqrt{3}}/2  \] so $DE = \sqrt{2 - \sqrt{3}}$[/hide][/quote]

careful! that simplifies still

(think about when you find $\sin 15^{\circ}$ using the difference formulas)

alternate solution:

$AE=\sqrt{2}=EC$, $DC=1$, and $AD=\sqrt{3}$ by trig, $\angle AEC=\angle ADC=90$, so $AEDC$ is cyclic, by ptolemy, $2ED+\sqrt{2}=\sqrt{6}\implies ED=\frac{\sqrt{6}-\sqrt{2}}{2}$

[quote="agolsme"][hide]
$\triangle ACD$ is a 30-60-90 triangle, so $CD = 1$. $\triangle ACE$ is a 45-45-90 triangle so $CE = \sqrt{2}/2$. Using the law of sins: $DE/\sin{15^{circ}} = CE/\sin{135^{\circ}}\implies DE = 2\sin{15^{\circ}}$. We can find $\sin{15}$ using the half angle formula: \[ \sqrt{2 - \sqrt{3}}/2  \] so $DE = \sqrt{2 - \sqrt{3}}$[/hide][/quote]

That does indeed simplify to $\frac{\sqrt{6}-\sqrt{2}}{2}$


Let A,B,C be the three angles of a triangle Prove that\[ \displaystyle\sum\sqrt{tan^2A+tan^2B}\geq \sum\sqrt{cot^2{A/2}+cot^2{B/2}}  \]

This is not true, for example, a triangle with side-lengths a=358, b=373, c=572. :roll:

The inequality, however, holds for all acute triangles. :roll:


inequalities
rearrangement inequality
inequalities proposed

the proof i gave above generalizes to show:

if $ a_{i,j}$, $ 1\leq i,j\leq n$ are positive real numbers such that for every $ 1\leq i\leq n$, $ a_{i,1} > a_{i,2} > \cdots > a_{i,n}$, then

$ (n \minus{} 1)!\sum_{i \equal{} 1}^{n}a_{1,i}a_{2,i}\cdots a_{n,i} > \sum_{\sigma\in S_n} a_{1,\sigma(1)}\cdots a_{n,\sigma(n)}$


inequalities
logarithms
calculus
derivative
integration
calculus computations

[quote="AndrewTom"]If $ x > 0$, prove that $ x \minus{} \frac {x^{2}}{2} < \ln (1 \plus{} x) < x \minus{} \frac {x^{2}}{2} \plus{} \frac {x^{3}}{3}$.[/quote]
Prove, by using derivative, that $ ln (1 \plus{} x) \minus{}(x \minus{} \frac {x^{2}}{2})$ is increasing and $ ln (1 \plus{} x) \minus{}(x \minus{}\frac {x^{2}}{2}\plus{}\frac {x^{3}}{3})$ is decreasing .
More general : $ \sum_{k\equal{}1}^{2n} (\minus{}1)^{k\minus{}1}\frac {x^k}{k} < \ln (1 \plus{} x) < \sum_{k\equal{}1}^{2n\plus{}1} (\minus{}1)^{k\minus{}1}\frac {x^k}{k}$
 :cool:

$ x>0\Longrightarrow 1\minus{}x^2<1<1\minus{}x^3\Longleftrightarrow 1\minus{}x<\frac{1}{1\plus{}x}<1\minus{}x\plus{}x^2$

$ \therefore \int_0^x (1\minus{}x)\ dx<\int_0^x \frac{1}{1\plus{}x}\ dx<\int_0^x (1\minus{}x\plus{}x^2)\ dx$, yielding

$ x\minus{}\frac{x^2}{2}<\ln (1\plus{}x)<x\minus{}\frac{x^2}{2}\plus{}\frac{x^3}{3}$


Bueno, este es un conjunto de problemas que colgaremos semanalmente.
Grupo: Pascual, Leonardo, Manuel, Carlosbr :lol:

Quisiera comenzar definiendo las reglas para este post,
pidiendoles a todos que las respeten..

 [hide="REGLAS:"]
- No colgar las soluciones y/o comentarios en este topic.
- Enviar sus soluciones a Email :  jibarotimbalero@hotmail.com (manuel)
- Fecha: (Viernes 21 a Jueves 27 de Octubre)
- Luego de la semana, se publicaran las soluciones mas sobresalientes.
- Tema: Inecuaciones.
  [/hide] :!:

[color=blue][size=150]WEEK 2: INECUACIONES[/size][/color]
[color=darkblue](Viernes 21 a Jueves 27 de Octubre)[/color]

[color=blue][size=134]NIVEL 1[/size][/color]

1. Probar que, si $a,b,c$ reales positivos, tal que $ab+bc+ca=1$, entonces
$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 3(a+b+c).$

2. Sean $a,b,c >0$ .Probar que:
   (a) $\frac{a}{c} + \frac{b}{a} + \frac{c}{b} \geq a+b+c.$ (Si ademas $abc \leq 1$)

   (b) $\frac{a^3}{b+c} + \frac{b^3}{c+a} +\frac{c^3}{a+b} \geq \frac{a^2 + b^2 + c^2}{2}.$

3. Sean $a,b,c>0$ numeros reales.
Probar que:

 $\frac{a}{b+2c} + \frac{b}{c+2a} +\frac{c}{a+2b} \geq 1.$

[color=blue][size=134]NIVEL 2[/size][/color]

1. Sean $a,b,c$ las longitudes de los lados de un tri\'angulo. Sea $p$ el semiperimetro.
Probar lo siguiente:
        (a) $\frac{1}{p-a} + \frac{1}{p-b} + \frac{1}{p-c} \geq \frac{9}{p}.$

        (b) $\frac{3}{2} \leq \frac{a}{b+c} + \frac{b}{c+a} +\frac{c}{a+b} < 2.$

2. Sea $n>3$ un entero y sean $x_1, \cdots, x_n$ reales positivos, tal que $x_1 \cdots x_n =1$.
Probar que:
\[ \frac{1}{1 + x_1 + x_1x_2} + \frac{1}{1 + x_2 + x_2x_3} + \cdots + \frac{1}{1 + x_n + x_nx_1} >1.  \]

3.  Sea $n>2$ un entero y sean los reales $a_1, a_2, \cdots, a_n$. Probar que para todo subjoncunto no vacio $S \subset \{ 1,2,3,\cdots,n \}$, tenemos:

\[ \left ( \sum_{i \in S}a_i \right ) ^2 \leq \sum_{1 \leq i \leq j \leq n}(a_i+\cdots + a_j)^2.  \]

[color=blue] PD. Tema Week3: Geometria[/color]

Good Luck ;)

Carlosbr / Pascual   / Leonardo    / Manuel            :)
     Peru    / Colombia / Venezuela  / Puerto Rico

Supongo que el nivel uno los escoge ... o de acuerdo a que se determina el nivel?

En la b de 1, nivel 2, no debe ser $\frac{15}{4}$ debe ser $\frac{3}{2}$

En primer lugar, gracias por las observaciones Tipe....

(1) Bueno .... el nivel a elegir depende de cada uno .............
pero siendo eticos ...
los problemas del nivel 1 son para los mas pequeños.

(2) Respecto a la pregunta del Nivel2 - P1.b la cota inferior definida por 15/4 es correcta,
como veran es una cota mas refinada de esta desigualdad que muchos ya deben conocer!!

Bueno a intentarlas y suerte!! :D

Carlos Bravo
Lima - Peru

ahhh en el problema 3 no es: "para algun" es "para todo"

En el problema 3... Sean A1, A3.... An que?

entonces es claro que quedamos en 3/2......

Bueno .. gracias nuevamente por el aporte de todos ....

Respecto a: [b]N2-P1.2[/b], la cota inf es ciertamente $\frac{3}{2}$,
confundi mis apuntes ... :P

Respecto a: [b]N2-P3[/b], $a_1,a_2, \cdots, a_n$ son numeros reales.....

Espero que esto responda sus inquietudes ....

Carlos Bravo :)
Lima -PERU


geometry
3D geometry
rotation
octahedron

What is the smallest cube that can touch all the faces of a larger cube? (one inside the other) :)

if the smaller cube touches the midpoints of the faces of larger cube, to show this is the smallest, consider
if we translate the smaller cube, then we have to make the smaller cube larger to touch the side is is being pulled from
and, if we rotate the cube, then the cube must get larger so it touches the side it is being rotated away from

another way to think of this is that the smaller cube must have some dimention that is the distance between the cubes, and the largest dimention for a cube is the diagonal, so to minimize, we use the longest dimention of the smaller cube to reach from opposite sides of the large cube

a cube has 8 vertices, 6 faces. No six vertices can touch the centres of the faces of the larger cube. You're talking about an octahedron :huh:


I came up with this today; I hope that it's an original (and correct :)).
Let $ a,b,c>0$ and prove that $ \sqrt{\frac{1}{ab}}\plus{}\sqrt{\frac{1}{bc}}\plus{}\sqrt{\frac{1}{ca}}\ge\sqrt{2}\frac{(a\plus{}b\plus{}c)^2\minus{}a^2\minus{}b^2\minus{}c^2}{\sqrt{(a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2)}}$

First we have
$ (a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2) \geq \frac{8}{9}(a^2\plus{}b^2\plus{}c^2)(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)$
which is equivalent to $ \sum a^4b^2 \geq 6a^2b^2c^2$ which is AM-GM.

So:
$ \sqrt{(a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2)} \geq \frac{2\sqrt{2}}{3}\sqrt{(a^2\plus{}b^2\plus{}c^2)(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)} \geq \frac{2\sqrt{2}}{3\sqrt{3}}\sqrt{(a^2\plus{}b^2\plus{}c^2)}(ab\plus{}bc\plus{}ca)$

So$ RHS \leq 3\sqrt{\frac{3}{a^2\plus{}b^2\plus{}c^2}}$

But $ LHS \geq 3\sqrt{\frac{3}{a^2\plus{}b^2\plus{}c^2}}$ is equivalent to$ \sqrt{a^2\plus{}b^2\plus{}c^2} (\sqrt{a} \plus{} \sqrt{b} \plus{} \sqrt{c}) \geq 3\sqrt{3abc}$
which is AM-GM.


Pokaz ze dla $n \in \mathbb{N}$,
\[ \varphi ( n ) = \pi ( n ) \Longleftrightarrow n \in \{ 2, 3, 4, 8, 14, 20, 90 \}  \]

[quote="xitami"]\[ \varphi ( n ) \ge \frac{4n}{15} \]

\[\pi ( n )\le \frac{1.25506\ n}{\log n} \][/quote]

Can you prove the statements ?

$\varphi ( n )\ge\frac{4n}{15}$ is a false statement. Let $n = 2\times3\times5\times7\times11 = 2310$. $\phi(n) = 1 \times 2 \times 4\times 6 \times 10 = 480$. $\frac{4n}{15} = 616$.


Let be a, b and c the sides of a triangle such that

$ab+ac+bc = a^{2}+b^{2}+c^{2}$

Show that this triangle is equilateral

---------------------------

An easy answer is

(a - b)² + (b - c)² + (c- a)² >= 0 <=> a² + b² + c² >= ab + bc + ac with equality iff a = b = c. Done

But this isn't the motivation of my topic, I've a question.

I found another way :

By Al-Kashi, we have

a² = b² + c² − 2bc * cos A
b² = a² + c² − 2ac * cos B
c² = a² + b² − 2ab * cos C

Adding these relations, we have

a² + b² + c² = 2(a² + b² + c²) − 2(ab * cos C + bc * cos A + ac * cos B)

Equivalent to

ab + bc + ac − 2(ab * cos C + bc * cos A + ac * cos B) = 0

by the fact that ab + ac + bc = a² + b² + c²

And finally,

ab(1 − 2cos C) + bc(1 − 2cos A) + ac(1 − 2cos B) = 0

Now, have I to prove that the only solution it's cos A = cos B = cos C = 1/2 with A = B = C = pi/3 or I can just say "we have this solution for this equation, so we can conclude" ?

If not, how to prove that the only solution is when A = B = C = pi/3 ?

Thanks

There is no simple one step proof using that final equation - in fact I believe there are more solutions. The first way is the best way to go (its really trivial, that's what I did)

I think the more interesting version of this problem is that if $a$, $b$, and $c$ are distinct complex numbers satisfying $ab+bc+ca = a^{2}+b^{2}+c^{2}$, then they form an equilateral triangle in the complex plane. A topic discussing this problem can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44909]here[/url]

[quote="zweig"]ab(1 − 2cos C) + bc(1 − 2cos A) + ac(1 − 2cos B) = 0

Now, have I to prove that the only solution it's cos A = cos B = cos C = 1/2 with A = B = C = pi/3 or I can just say "we have this solution for this equation, so we can conclude" ?[/quote]

This approach is actually harder than the original because what you're really asking for is the only solution to the [b]system[/b]

$ab(1-2 \cos C)+bc(1-2 \cos A)+ac(1-2 \cos B)$
$a^{2}+b^{2}+c^{2}= ab+bc+ca$

So really, you're just doing extra work for no reason.

$\frac{a}{\sin A}=2R$

So your system becomes:

$\sum\frac{1-2\cos A}{\sin A}=0$
$A+B+C=\pi$

I think it just gets messy from here.

$\sum \tan\frac{A}{2}=\sum\cot A$...

Since I just got mathematica I am all pumped to use it...anyway, if you do:

[code]Reduce[{(1 - 2*Cos[A])/(Sin[A]) + (1 - 2*Cos[B])/(Sin[B]) + (1 -
       2*Cos[C])/(Sin[C]) == 0, A + B + C == Pi, A > 0, B > 0, C > 0,
  A < 2 Pi, B < 2 Pi, C < 2 Pi}, x][/code]

Then you get:

[code]B == \[Pi]/3 && C == \[Pi]/3 && A == \[Pi]/3[/code]

So we might be able to solve it...I just don't know how mathematica did it...:P


geometry
3D geometry
tetrahedron
pyramid
calculus
analytic geometry
area of a triangle

Triangle $ ABC$, with sides of length $ 5$, $ 6$, and $ 7$, has one vertex on the positive $ x$-axis, one on the positive $ y$-axis, and one on the positive $ z$-axis. Let $ O$ be the origin. What is the volume of tetrahedron $ OABC$?

$ \textbf{(A)}\ \sqrt{85} \qquad
\textbf{(B)}\ \sqrt{90} \qquad
\textbf{(C)}\ \sqrt{95} \qquad
\textbf{(D)}\ 10 \qquad
\textbf{(E)}\ \sqrt{105}$

[hide="Solution"]
[i]wlog[/i] use A(a,0,0), B(0,b,0), C(0,0,c)

(i) [i]wlog[/i] set $ 5 \equal{} \sqrt{a^2\plus{}b^2} \qquad 6 \equal{} \sqrt{b^2\plus{}c^2} \qquad 7 \equal{} \sqrt{c^2\plus{}a^2}$

(ii) Square for $ 25 \equal{} a^2\plus{}b^2 \qquad 36 \equal{} b^2\plus{}c^2 \qquad 49 \equal{} c^2\plus{}a^2$

(iii) Adding and dividing by 2 gives $ a^2\plus{}b^2\plus{}c^2 \equal{} 55$

(iv) Subtracting each element of (ii) from (iii) gives $ a^2 \equal{} 19 \qquad b^2 \equal{} 6 \qquad c^2 \equal{} 30$

(v) Multiplying gives $ a^2b^2c^2 \equal{} 19 \cdot 6 \cdot 30 \equal{} 19 \cdot 5 \cdot 6^2$

(vi) Squareroot and divide by 6 to get $ V \equal{} \frac{1}{6} abc \equal{} \sqrt{95} \implies \boxed{\text{C}}$
[/hide]

I actually stupidly skipped this problem the second I saw "5-6-7 triangle" (I blow at geometry) and never came back to it... pity.

You have a tetrahedron. The base is a right triangle that has legs a and b, and height is c.

The area of the base is (1/2)*ab by the area of a triangle formula.

The volume of any pyramid-like figure is 1/3 the area of the base times the height.

So, the volume is 1/3*((1/2)*ab)c = (1/6)*abc

There's probably a simpler way to prove this than using calculus, but given an $ n$-dimensional [url=http://en.wikipedia.org/wiki/Simplex]simplex[/url] such that one vertex is at the origin and all of its edges are orthogonal to each other with side lengths $ a_1, a_2, \ldots a_n$, the $ n$-dimensional volume of the simplex is $ \frac{ a_1 a_2 \cdots a_n }{n!}$. (The linked to page generalizes this)

[hide="Really Simple Geometry"]
If the centroid of the triangle has coordinates (5/3 6/3 7/3)
then the height of this pyramid is the distance from the origin to this point,
which is sqrt(110/9)
Since we now know the height and we can find the area of the base triangle by Heron's Formula, we can easily find the volume.
[/hide]

[quote="E^(pi*i)=-1"][hide]
We have $ x^2 \plus{} y^2 \equal{} 25$, $ y^2 \plus{} z^2 \equal{} 36$, and $ z^2 \plus{} x^2 \equal{} 49$.  Summing, we get $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 55$, so $ x^2 \equal{} 30$, $ y^2 \equal{} 19$, and $ z^2 \equal{} 6$.  The volume is

$ \frac {1}{3}z*\frac {xy}{2} \equal{} \frac {\sqrt {30*19*6}}{6} \equal{} \sqrt {19*5} \equal{} \sqrt {95}$.

Answer is $ C$.

[/hide][/quote]

Is there a formula that lets us do $ \frac {1}{3}z*\frac {xy}{2} \equal{} \frac {\sqrt {30*19*6}}{6} \equal{} \sqrt {19*5} \equal{} \sqrt {95}$?

Thanks!

[quote="314.math"]Is there a formula that lets us do $ \frac {1}{3}z*\frac {xy}{2} \equal{} \frac {\sqrt {30*19*6}}{6} \equal{} \sqrt {19*5} \equal{} \sqrt {95}$?
Thanks![/quote]

That's just the standard formula for volume of a pyramid, $ V\equal{}\frac{1}{3}Bh$, except in this problem, we have $ h\equal{}z$ and $ B\equal{}\frac{xy}{2}$.

[size=85]The volume of a tetrahedron with integer veritices is $[\vec{c}\cdot(\vec{a}\times\vec{b}]\frac{1}{6}$ (those familiar with analytic should recognize this as $\frac{1}{6}$ the box product).

$\frac{1}{6}\left[\begin{array}{ccc}x_{1}&y_{1}&z_{1}\\x_{2}&y_{2}&z_{2}\\x_{3}&y_{3}&z_{3}\end{array}\right]= \frac{1}{6}\left[\begin{array}{ccc}5&0&0\\0&6&0\\0&0&7\end{array}\right]$

Which should equal $\sqrt{95}$, though i am to lazy to calculate...[/size]

[hide=solution]Let $A=(0, 0, a), B=(0, b, 0),$ and $C=(c, 0, 0)$. Then \begin{align*}
AB=5\qquad & a^{2}+b^{2}=25 \\
BC=6\qquad & b^{2}+c^{2}=36 \\
CA=7\qquad & c^{2}+a^{2}=49 \\
\end{align*}
Adding all three equations and dividing by $2$ gives $a^{2}+b^{2}+c^{2}=55$. Subtracting the first equation gives $c^{2}=30,$ subtracting the second equation gives $a^{2}=19,$ and subtracting the third gives $b^{2}=6$. The volume of tetrahedron $OABC$ is $\cfrac{1}{6}abc=\cfrac{1}{6}\sqrt{30\cdot 19\cdot 6}=\cfrac{1}{6}\sqrt{3420}=\boxed{\textbf{(C)} ~\sqrt{95}}$[/hide]


The mean of three numbers is $10$ more than the least of the numbers and $15$ less than the greatest. The median of the three numbers is $5$. What is their sum?

A. $5$

B. $20$

C. $25$

D. $30$

E. $36$

[hide]Let $x\leqslant y\leqslant z$. Then $y=5$ and $z=x+25$. Setting up equations and solving, we get $x=0$, $y=5$, and $z=25$. $0+5+25=30\Longrightarrow D$[/hide]


If $ m\geq 0$:
$ A(m) \equal{} \{x| \minus{} m\leq x\leq m\}$.
If $ m\leq 0$.
Then $ A(m)\equal{}\emptyset$.
Hence answer to 2nd question is $ A(1)$.

Hmmm.... I think that if $ m<0$ then $ A(m) \equal{} \emptyset$ :wink: ...

Pierre.

[quote="pbornsztein"]Hmmm.... I think that if $ m < 0$ then $ A(m) \equal{} \emptyset$ :wink: ...

Pierre.[/quote]
Oh, :wallbash: you are right.


calculus
derivative
function
Putnam
quadratics
real analysis
real analysis unsolved

Let $f$ be a real function on the real line with continuous third derivative. Prove that there exists a point $a$ s.t

$f(a)\cdot f'(a)\cdot f''(a)\cdot f'''(a) \geq 0$

Is it true that there exists a point b s.t

$f(b)\cdot f'(b)\cdot f''(b)\cdot f'''(b)>0$?

Generally speaking, Putnam problems don't belong in the "Computations and Tutorials" section. Yes, although it's not as difficult as some, this is a Putnam problem.

[Oh, and periods for multiplication just don't look right. I edited all of your periods and turned each of them into \cdot .]

[quote="jmerry"]The zero function is an easy counterexample to the second question.[/quote]
Or for that matter any quadratic or lower degree polynomial.


real analysis
calculus
derivative
topology
real analysis theorems

Only first principles? I guess I won't quote the Lebesgue theorem on the differentiation of the integral.

Suppose $U$ is an open set which contains $E.$ Then $U$ can be written as a countable disjoint union of open intervals $(c_j,d_j).$

$m(E)=\sum_jm\left(E\cap (c_j,d_j)\right)=\sum_jm\left(E\cap [c_j,d_j]\right)\le \frac12\sum_j(d_j-c_j)=\frac12m(U).$

Thus $m(U)\ge2m(E)$ for any open $U$ containing $E.$ Hence,

$m(U\setminus E)=m(U)-m(E)\ge m(E).$

But given any $\delta>0,$ we can find an open $U$ containing $E$ such that $m(U\setminus E)<\delta.$ This implies from above that $m(E)<\delta.$ Since this is true for all $\delta,$ it follows that $m(E)=0.$

[quote="nfrer"]Suppose E belongs in [0,1],E Lebesgue measurable and m(E intersect [a,b]) <= (b-a)/2
for every interval [a,b] Prove using only first principles that m(E)=0[/quote]
What is first principles?


superior algebra
superior algebra solved

(A,+,*) is a ring and p>=2, q>=1 are integers s.t. (q,p-1)=1 and x2q(2p-1)=x for all x\in A. Show that x2=x for all x\in A.

I used the fact that if (m,n)=1 and x2m=x2n=x then x2=x.

I think that you didn't need (q,p-1)= because you found that x2=y2=>(x+y)2=2, but I don't see why that's true, because you would need xy=yx, which is not given. (x+y)2 isn't x2+y2+xy+xy=0, but x2+y2+xy+yx=xy+yx, and I don't think we can show that A is commutative.

You're right with that.I can' believe i didn't see it.Anyway let me try to fix it:
I  got to the point where:
x 2q =x (1) and
x2p-1=x=>x2p=x2
=>(x2p-1-x)2=x2p+x2-2*x2p-1+1=0 =>x2p-1=x (2)
Now I'll proof that lemma:x2m=x2n=x for (m,n)=1 =>x2=x
Proof:
(m,n)=1 => there are a,b  \in N st am=bn+1
x2m=x=>
x22m=x=....x2am
In the same way :
x2bn=x=>x2bn+1=x2=x2am=x =>x=x2.So the lemma is proofed.

Now from 1 and 2 using the lemma(since (q,p-1)=1) we get the result


number theory
relatively prime
prime numbers
number theory proposed

Let $T$ be a finite set of positive integer numbers such that
a)For every $a,b\in T$ we have $gcd(a,b),lcm(a,b)\in T$ AND
b)For every $x\in T$ exists $y\in T$ such that $gcd(x,y)=1$ and $lcm(x,y)$ is great element of $T$.
Find great valuation of $|T|$ when $|T|<1990$.

[quote="t0rajir0u"]By "great element" and "great valuation" do you mean an upper bound / a maximum?[/quote]

I can't understand you speak  :maybe: Problem is clear  :)

[quote="K81o7"]Actually, the problem is pretty unclear.  I'd assume a "great element" is a label we're putting on certain elements of the set, and we want to determine the greatest number of great elements?  If the "great element" were the maxmimum, then the product of any two relatively prime numbers would be the same...

Can someone clarify this problem?[/quote]

English of mine is poor  :) Problem is: Find max |T|, with T is such set.

|T|=number of element of T

Are you understand?


What happens when a $5$ microcoulomb capacitor is connected to a $10$ ohm resistor in series, and both are connected to a $20$ V voltage? What about in parallel?

There's no such thing as a $5\mu C$ conductor; you probably meant farads instead of coulombs. Anyway, in the first scenario the capacitor will charge and the current will stop flowing through the circuit, and in the second scenario the capacitor will charge again, but the current will keep flowing even when the capacitor is full.


[quote="236factorial"]How many 4-digit palindromes are there?[/quote]

99

[hide]IT has to be in the fom $abba$.  There are $9$ possibilities for $a$ (all but $0$), and $10$ for $b$. $9\cdot10=90$[/hide]

[hide]
i wrote out all the four digit palindromes and i got 84.  That's probably wrong because i could have missed some or have not counted them correctly.[/hide]

Correct, nat mc, notice that it's now many two-digit numbers there are, and you showed that too. :10:

math92, that's what happens when you try list all those numbers out,  ;)

anirudh, how did you get 99?

[quote="236factorial"]
math92, that's what happens when you try list all those numbers out
[/quote]
yeah, i don't know any formula and i have been trying to find one but when i recount my numbers i now have
[hide]89, which is still wrong i think[/hide]

That was hard, correct math92  :rotfl:

Did you see the one you missed?

that's strange..... how did you list them? Systematically or randomly?

i listed them in order and i must've left that one out.


algebra
polynomial
number theory
number theory unsolved

Let $ f(x)\equal{}\sum_{1 \leq k \leq p\minus{}1}{\frac{x^k}{k}}$, where $ p$ is a prime number. Prove that:
\[ f(2) \equiv_p f(\minus{}1)\]

So you are working with fractions modulo $ p$. Thanks, this makes my job easier.

In http://www.mathlinks.ro/viewtopic.php?t=38958 , I showed

$ \frac {2^{p} \minus{} 2}{p} \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}\frac {\left(p \minus{} 1\right)!}{k!\left(p \minus{} k\right)!}$

and

[color=green][b](1)[/b][/color] $ \frac {\left(p \minus{} 1\right)!}{k!\left(p \minus{} k\right)!}\equiv \frac {\left( \minus{} 1\right)^{k \minus{} 1}}{k}\mod p$ for every $ k\in\left\{1,2,...,p \minus{} 1\right\}$.

Thus,

$ \frac {2^{p} \minus{} 2}{p} \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}\frac {\left(p \minus{} 1\right)!}{k!\left(p \minus{} k\right)!} \equiv \sum_{k \equal{} 1}^{p \minus{} 1}\frac {\left( \minus{} 1\right)^{k \minus{} 1}}{k}\mod p$       (by [color=green][b](1)[/b][/color]).

Hence,

[color=green][b](2)[/b][/color] $ \minus{} \frac {2^{p} \minus{} 2}{p}\equiv \minus{} \sum_{k \equal{} 1}^{p \minus{} 1}\frac {\left( \minus{} 1\right)^{k \minus{} 1}}{k} \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}\frac { \minus{} \left( \minus{} 1\right)^{k \minus{} 1}}{k} \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}\frac {\left( \minus{} 1\right)^k}{k}\mod p$       (since $ \minus{} \left( \minus{} 1\right)^{k \minus{} 1} \equal{} \left( \minus{} 1\right)^k$).

On the other hand, in http://www.mathlinks.ro/viewtopic.php?t=247417 I showed

[color=green][b](3)[/b][/color] $ \sum_{k \equal{} 1}^{p \minus{} 1}2^{k}k^{p \minus{} 2}\equiv \minus{} \frac {2^{p} \minus{} 2}{p}\mod p$.

Since $ \frac {2^k}{k}\equiv 2^k k^{p \minus{} 2}\mod p$ for every $ k\in\left\{1,2,...,p \minus{} 1\right\}$ (since $ k^{p \minus{} 2}k \equal{} k^{p \minus{} 1}\equiv 1\mod p$ by Fermat's Little Theorem, so that $ k^{p \minus{} 2}\equiv\frac {1}{k}\mod p$ and thus $ 2^k k^{p \minus{} 2}\equiv 2^k\cdot\frac {1}{k} \equal{} \frac {2^k}{k}\mod p$), we have

$ \sum_{k \equal{} 1}^{p \minus{} 1}\frac {2^k}{k}\equiv \sum_{k \equal{} 1}^{p \minus{} 1}2^{k}k^{p \minus{} 2}\equiv \minus{} \frac {2^{p} \minus{} 2}{p}$       (by [color=green][b](3)[/b][/color])
$ \equiv\sum_{k \equal{} 1}^{p \minus{} 1}\frac {\left( \minus{} 1\right)^k}{k}\mod p$       (by [color=green][b](2)[/b][/color]),

qed.

By the way, this generalizes: $ \sum_{k \equal{} 1}^{p \minus{} 1}\frac {\left(a\plus{}1\right)^k}{k}\equiv\sum_{k \equal{} 1}^{p \minus{} 1}\frac {\left( \minus{} a\right)^k}{k}\mod p$ for every $ a\in\mathbb{Z}$.

  darij

Thank you very much for the fast answer :) it's like you knew this problem was coming, you had all these lemmas already proven ^^ Now it's easy to prove the generalization when you've seen all the links you posted XD. The key is to observe it as a polynomial(well since we know the generalization is true every coefficient of the polynomial must be $ 0$ in $ \mathbb{Z}_p$ :P)

$ \sum_{k \equal{} 1}^{p\minus{}1}\frac{\left(x\plus{}1\right)^{k}}{k} \minus{} \sum_{k \equal{} 1}^{p\minus{}1}\frac{\left(\minus{}x\right)^{k}}{k}$
or, equivalently:
$ \sum_{k\equal{}0}^{p\minus{}1}x^i(\sum_{i\equal{}k}^{p\minus{}1}\frac{\binom{i}{k}}{i}\minus{}(\minus{}1)^k)$

so, we have to prove $ \sum_{i\equal{}k}^{p\minus{}1}\frac{\binom{i}{k}}{i}\minus{}(\minus{}1)^k \equiv_p 0$
but we have $ \sum_{i\equal{}k}^{p\minus{}1}\frac{\binom{i}{k}}{i}\equal{}\sum_{i\equal{}k}^{p\minus{}1}\frac{1}{k}\binom{i\minus{}1}{k\minus{}1}\equal{}\frac{1}{k}\binom{p\minus{}1}{k}\equal{}\frac{1}{k}\frac{(p\minus{}1)(p\minus{}2)..(p\minus{}k)}{k!}\equiv_p \frac{(\minus{}1)^k}{k}$, which is exactly what we wanted to prove :)(here I used some binomial identity)

For $ i\equal{}0$ in the upper sum this doesnt work, but its easy to prove that $ \sum_{k\equal{}1}^{p\minus{}1}\frac{1}{k}\equiv_p 0$ in various ways :)
Thus, the proof is finished. Btw, this problem appeared at some Russian competition last year, and its much easier to prove the general version if you ask me.


function
limit
real analysis
real analysis theorems

Give a proof of the following theorem or show that the theorem is wrong:

[b]Theorem[/b]:
Let function $ f$ be defined on $ \mathbb{R}$ and differentiable in $ c (\in \mathbb{R})$.
Suppose $ b_1,b_2,b_3,\cdots$ and $ c_1, c_2, c_3,\cdots$ are real sequences with
$ \lim_{n \to \infty} b_n \equal{} c$ and
$ \lim_{n \to \infty} c_n \equal{} c$ and
$ c_n < b_n$ for all $ n \in \mathbb{N}$.

Then $ f'(c) \equal{} \lim_{n \to \infty} \frac {f(b_n) \minus{} f(c_n)}{b_n \minus{} c_n}$.

I think you have your letters all mixed up but I'm wondering if you can use the sequential characterization here, that is: $ \lim_{x\to a} f(x) \equal{} L$ iff for all sequences $ (a_n)$ s.t. $ a_n \to a$ with $ a_n \neq a$ for all $ n$, $ f(a_n)\to L$.


[quote="srulikbd"]quite simple when you read the solution  :lol:
I choose polinomial with integer coefficients, and you need to discover it. you choose one X and I tell you the Y, then you choose one more, and I tell you the Y adn then you can discover the polinomial![/quote]
I can't discover the polynomial. You can choose a polynomial of an arbitrary big degree and I don't know when to stop asking for values in some $x$ points.
Of course the problem is trivial when I know the degree, the coefficients are solutions of an easy to create set of linear equations.

[quote="Rzeszut"][quote="srulikbd"]quite simple when you read the solution  :lol:
I choose polinomial with integer coefficients, and you need to discover it. you choose one X and I tell you the Y, then you choose one more, and I tell you the Y adn then you can discover the polinomial![/quote]
I can't discover the polynomial. You can choose a polynomial of an arbitrary big degree and I don't know when to stop asking for values in some $x$ points.
Of course the problem is trivial when I know the degree, the coefficinets are solutions of an easy to create set of linear equations.[/quote]

once the finite differences become constant, you can stop

Still, I don't understand the original post or how it's possible.

it's possible and thats what is amazing  :)

i think it would be more believeable if you would show us..

we have....if you take x as consecutive integers, and use finite differences, you can determine the degree of the polynomial by when the differences become constant, then lagrange interpolate to find the polynomial

from what i understand from the problem, only two numbers are given.

yeah. ok, choose a polinomial, the first point I choose is 1. tell me the Y.

Already posted here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=37312&start=400
see lightrhee's post at the top of the page and the post right after that for the solution.

The key is that the polynomial has integer coefficients. Otherwise, it is not unique.

@rem: your solution in the other thread has some faulty latex. Would you mind posting a more legible version?

Another solution if we are restricted to nonnegative integers:
[hide]
Let $P=a_{0}+a_{1}x+a_{2}x^{2}...$

Pick $P(1)$ and $P(P(1)+1)$

Since $P(1) = a_{0}+a_{1}... > a_{i}$, we know that $P(P(1)+1)$ is uniquely expressed in base $P(1)+1$. Take the values of this expression to be the coefficients of the polynomial. [/hide]


inequalities
algebra
polynomial
inequalities proposed

let $x_1,...,x_n$ be positive reals s.t. $\sum \frac 1{x_i} = n$. prove (or disprove, but again, i'm 99% sure that it's true) that for all $0\leq k\leq n-1$ we have
\[k(n-1)\left(\sigma_{n-k} - { n\choose k } \right) \leq n (n-k) \left(\sigma_{n-k+1} - {n\choose {k-1} } \right)\].

here $\sigma_k$ is the elementarsymmetric polynomial of degree $k$ in $x_1,...,x_n$.

this gives the "very hard and very beautiful harazi inequality" $x_1\cdots x_n-1\geq \left(\frac{n-1}n \right)^{n-1} (x_1+...+x_n-n)$.

the special case $n=3$, $k=2$ is the classical schur inequality.

So is it still a research question? Do you have a complete solution?

now, i have a complete solution - it is true if i didn't make some stupid error.

Let $y_i=\frac 1{x_i}$.
Then $\sum_{i=1}^n y_i=n$.
Now multiplying the inequality by $y_1\cdots y_n$ we get a new inequality where we are allowed even to have $y_i=0$.
Transform our inequality to $f(y_1,y_2,\cdots,y_n)\geq 0$ for some symmetric polynomial.
Now let's try to decrease $f$.
Pick up arbitrary non-zero $y_i,y_j$ and fix the others. Thus $y_i+y_j$ is constant.
Then $f$ will have form $ay_iy_j+b$ linear in $y_iy_j$ and takes its minimum at the extremes, hence when $y_i=y_j$ or one of them is $0$.
Thus we may replace $y_i,y_j$ by their mean (type $1$) or make one of them $0$ (type $2$).
We keep doing transformations. We see that we can perform only finitely many operations of type $2$ because they increase the number of zeros in the sequence.
Thsu at some time we shall have $l$ zeros and some other non-zero number every two of them we can replace by their arithmetic mean non-increasing $f$.
Now by continuity it's obvious we can suppose the non-zero numbers are also equal hence they are $\frac n{n-l}$.
For this very particular case we can perform a manual checking.
Am I right?

ok, you got the main ideas(i put in some other way round, but in the end your way is nicer). however, it's still some work from now - i needed maybe two hours to really complete the proof... but maybe i was just too dumb :)

however, this gives some really sharp inequalities: for $n=4$, $k=2$ we get after substitution the inequality
$a^4+b^4+c^4+d^4+a^3b+a^3c+a^3d+b^3a+b^3c+b^3d+c^3a+c^3b+c^3d+d^3a+d^3b+d^3c+20abcd\geq 3\left( a^2bc+ab^2c+abc^2+a^2bd+ab^2d+abd^2+a^2cd+ac^2d+acd^2+b^2cd+bc^2d+bcd^2\right)$

someone has a direct proof for this one?

I also find very beautiful Iura method. :lol:

from here it's quite simple(modulo tons of algebra) to deduce the inequality

\[l(k+1)\tau_{k+m}\tau_{l+m+1}-k(l+1)\tau_{k+m+1}\tau_{l+m}\leq (l-k) \frac{\tau_{l+m+1}^{k+1}\tau_m}{\tau_{l+m}^k},\]

where $0\leq k\leq l\leq n-m-1$ and $n$ is the number of variables. $\tau_k$ denotes $\frac{ \sigma_k }{ {n\choose k} }$.

i admit it looks ugly :D

however, e.g. the case $k=l-1$ e.g. gives an ensharpment of the classical inequality between $\sigma_{l-1},\sigma_l,\sigma_{l+1}$.

You have made wonders from my poor inequality! Congratulations, Peter, for these magnificient discoveries. And no matter they are ugly, they are sharp and this is what matters. Well done!

Iura's method is so simple that let me get astonished :?

Yes, manlio, but simple is not a suitable word.  ;)


quadratics
number theory unsolved
number theory

Notice that $\phi(p^3) = p^3 - p^2$, so $\frac{1}{k^p} \equiv k^{p^3-p^2-p} \mod{p^3}$
Which means that:
$\sum_{k=1}^{p-1} \frac{1}{k^p} \equiv \sum_{k=1}^{p-1} k^{a} \mod{p^3}$
Where $a=p^3-p^2-p$. Now, we can see that:
$\sum_{k=1}^{p-1} k^{a} = \sum_{k=1}^{\frac{p-1}{2}} k^a + (p-k)^a$
Because $a$ is obviously odd: $\sum_{k=1}^{\frac{p-1}{2}} k^a + (p-k)^a \equiv \sum_{k=1}^{\frac{p-1}{2}} p*a*k^{a-1} \mod{p^3}$
$k^a$ cancels with $-k^a$ and the rest terms vanish $\mod{p^3}$ because the binomial coefficient $\binom{p^3-p^2-p}{t}$ is clearly divisible by $p$. So we are left with the:
$\sum_{k=1}^{\frac{p-1}{2}} p^2(p^2-p-1)*k^{a-1} \mod{p^3}$
and it will be enough if we prove that:
$\sum_{k=1}^{\frac{p-1}{2}} k^{a-1} \equiv 0 \mod{p}$
Now, $a+1 = p^3 - p^2 - p + 1 = (p-1)(p^2-1)$ so $k^{a+1} \equiv 1 \mod{p}$ and $k^{a-1} \equiv \frac{1}{k^2} \mod{p}$
It means that:
$\sum_{k=1}^{\frac{p-1}{2}} k^{a-1} \equiv \sum_{k=1}^{\frac{p-1}{2}} \frac{1}{k^2} \mod{p}$

It's clear that there are $\frac{p-1}{2}$ quadratic residues $\mod{p}$. We will prove that the set $A= { \frac{1}{1^2}, \frac{1}{2^2}, ..., \frac{1}{(\frac{p-1}{2})^2 }}$ is the set of all quadratic residues $\mod{p}$. To do this we will prove that in the set A all residues are different from each other and then we will be done, because $A$ will be a set of distinct $\frac{p-1}{2}$ quadratic residues. So suppose that for some distinct $k, t$:
$\frac{1}{k^2} \equiv \frac{1}{t^2} \mod{p}$
which is equivalent to
$(k-t)(k+t) \equiv 0 \mod{p}$
which is just impossible, $k \equiv t$ because they are distinct, and $k \equiv -t$ because $k, t$ are from the interval $[1; \frac{p-1}{2}]$, so the claim follows, which means that:
$\sum_{k=1}^{\frac{p-1}{2}} \frac{1}{k^2} \equiv \sum_{k=1}^{\frac{p-1}{2}} k^2 = \frac{p(p^2-1)}{24} \equiv 0 \mod{p}$


Set A has two more elements than set B, and set A has 96 more subsets than set B. How many elements are in set A?

I finally know how to do this  :blush:

[hide="solution"]Let $x=|B|$. According to the problem, we have
$2^{x+2}-2^x=96\Leftrightarrow2^x(4-1)=96\Leftrightarrow2^x=32$
Therefore, $x=5$
Thus $|A|=7$[/hide]

[quote="Ars"]Actually you solved the number of elemets in b[/quote]
I have corrected my solution.

Actually, I don't quite understand ifai's solution, though he does have the correct answers :huh:

My solution is a lot longer, but makes more sense to me.

[hide]

This is a problem involving permutations and combinations.

The formula for finding the number of combinations $_nC_r$ for a group of $ n $ things taking $ r $ things at a time is:

$_nC_r = \frac {_nP_r}{n!}$ where $_nP_r $ is the number of permutations of $ n $ things taking  $ r $ things at a time.

starting with $_2C_1$ (2 things taken 1 at a time):

$_2C_1 = \frac {_2P_1}{1!} = \frac {2}{1!} = 2$

3 things taken 1 at a time:

$_3C_1 = \frac {_3P_1}{1!} = \frac {3}{1!} = 3$

3 things taken 2 at a time:

$_3C_2 = \frac {_3P_2}{2!} = \frac {(3)(2)}{2!} = 3$

The total number of combinations of a group of three is $3 + 3 = 6$

4 things taken 1 at a time:

$_4C_1 = \frac {_4P_1}{1!} = \frac {4}{1!} = 4$

4 things taken 2 at a time:

$_4C_2 = \frac {_4P_2}{2!} = \frac {(4)(3)}{2!} = 6$

4 things taken 3 at a time:

$_4C_3 = \frac {_4P_3}{3!} = \frac {(4)(3)(2)}{3!} = 4$

The total number of combinations of a group of four is $4 + 6 + 4 = 14$

5 things taken 1 at a time:

$_5C_1 = \frac {_5P_1}{1!} = \frac {5}{1!} = 5$

5 things taken 2 at a time:

$_5C_2 = \frac {_5P_2}{2!} = \frac {(5)(4)}{2!} = 10$

5 things taken 3 at a time:

$_5C_3 = \frac {_5P_3}{3!} = \frac {(5)(4)(3)}{3!} = 10$

5 things taken 4 at a time:

$_5C_4 = \frac {_5P_4}{4!} = \frac {(5)(4)(3)(2)}{4!} = 5$

The total number of combinations of a group of five is $5 + 10 +10 + 5 = 30$

6 things taken 1 at a time:

$_6C_1 = \frac {_6P_1}{1!} = \frac {6}{1!} = 6$

6 things taken 2 at a time:

$_6C_2 = \frac {_6P_2}{2!} = \frac {(6)(5)}{2!} = 15$

6 things taken 3 at a time:

$_6C_3 = \frac {_6P_3}{3!} = \frac {(6)(5)(4)}{3!} = 20$

6 things taken 4 at a time:

$_6C_4 = \frac {_6P_4}{4!} = \frac {(6)(5)(4)(3)}{4!} = 15$

6 things taken 5 at a time:

$_6C_5 = \frac {_6P_5}{5!} = \frac {(6)(5)(4)(3)(2)}{5!} = 6$

The total number of combinations of a group of six is $6 + 15 + 20 + 15 + 6 = 62$

7 things taken 1 at a time:

$_7C_1 = \frac {_7P_1}{1!} = \frac {7}{1!} = 7$

7 things taken 2 at a time:

$_7C_2 = \frac {_7P_2}{2!} = \frac {(7)(6)}{2!} = 21$

7 things taken 3 at a time:

$_7C_3 = \frac {_7P_3}{3!} = \frac {(7)(6)(5)}{3!} = 35$

7 things taken 4 at a time:

$_7C_4 = \frac {_7P_4}{4!} = \frac {(7)(6)(5)(4)}{4!} = 35$

7 things taken 5 at a time:

$_7C_5 = \frac {_7P_5}{5!} = \frac {(7)(6)(5)(4)(3)}{5!} = 21$

7 things taken 6 at a time:

$_7C_6 = \frac {_7P_6}{6!} = \frac {(7)(6)(5)(4)(3)(2)}{6!} = 7$

The total number of combinations of a group of seven is $7 + 21 + 35 + 35 + 21 + 7 = 126$

The number of combinations for a group of seven is greater than $ 96$ so  the group of seven must be the larger set and the group of five is the smaller set since $126 - 30 = 96$

[/hide]

Does anyone know a way of abbreviating that solution?

[quote="JesusFreak197"]The number of subsets in a set is $2^x$, where $x$ is the number of elements in that set. I'm not sure how to prove it rigorously, but if you test out any number, it will work that way.[/quote]

Here's a hint.

Note that number of subsets = (number of ways to choose 0 element in a set to form a subset) + (number of ways to choose 1 element in a set to form a subset) + (number of ways to choose 2 elements in a set to form a subset) + .... (number of ways to choose n *that's the number of elements in original set* elements in a set to form a subset).

Mathematically, it's basically:

$2^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \cdots \binom{n}{n}$

which I believe was a Master Round problem for MC one year I think.

[hide]A=B+2 (First clue)
A^2 = B^2 + 96 (Second Clue) (x^2 becuase that's how you can figure out the # of subsets)
128-32=96
A=7, B=5
>>7<<[/hide]