K pasa con la comunida española!!! Nadie a propuesto problemas!!
Espero k posteen algunos!!

Aki les pongo uno divertido:

Sea A el conjunto de enteros positivos representados de la forma a^2+2(b^2) para cualesquiera "a"y "b" enteros con "b" NO CERO.
Muestra que si p^2 pertenece a A para p primo, entonces p tambien pertenece a A.

El problema que propones ya fue resuelto en el foro:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=75320[/url]
(Es uno de los últimos post)

Pero aquí propongo otro problema:

Suponga que $\mathcal{S}$ es un subconjunto finito de los reales, con más de un elemento, con la siguiente propiedad: si tomamos dos elementos distintos de $\mathcal{S}$, estos forman una progresión aritmética con otro elemento de $\mathcal{S}$. Halle todos los valores que puede tomar el cardinal de $\mathcal{S}$.

(es una ligera variación de un problema de la Olimpiada Nacional de Irán)

$Tipe$

Gracias por el link, vi la solucion y no me gusto el problema.... krei k era mas  bonito!!!
Alguna sugerencia para tu problema?? Ah y a k te refieres con cardinalidad de S????

[quote="conejita"]Ah y a k te refieres con cardinalidad de S????[/quote]

El cardinal de un conjunto es el número de elementos que tiene.

$Tipe$

Por fin alguien con ganas, aqui hay problemas que propuse y nadie resolvio

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77856[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81765[/url]

Veo que has posteado muchos problemas, que te parece si por cada problema mio que resulevas yo posteo la solucion de uno tuyo(ya tengo dos soluciones a dos problemas) y ademas te enseño a resolver problemas con inversion  :roll: que tal?

Aceptas?

$\bigstar$ Capitán Mandarina

PD. $Tipe$ es extraño, no entiendo como le puede gustar la mate discreta  :|

Me parece muy wena tu idea, apenas aga alguno tuyo lo posteo, y de paso me vas a pasando los problemas de inversion!!
Y ojala y la comunidad española se anime a poner problemas, soluciones...
pa k nos envidien las demas... jaaja

 :D

Yo también ando con ganas de resolver problemas!!! pero me parecen dificiles y no los puedo resolver

bueno yo tengo unos problemas q no me salen  :blush:  quiza a ustedes si ( la experiencia  :lol: ) aca van dos , mañana quiza postee más

Weno, ps los ultimos k pusieron, ya los ice, ai pongo los soluciones:

El primero:
La desigualdad es equivalente a:
a/c^2+b/c^2+a/b^2+c/b^2+c/a^2+^b/a^2 >= 2(1/a+1/b+1/c)
Lo anterior es facil de ver multiplicando todo por (abc)^2 y usando muirhead.
Ahora (a+b+c)(1/a+1/b+1/c) >= 9 de aqui que                     1/a+1/b+1/c >= 9/(a+b+c).
Por ultimo 2(1/a+1/b+1/c) >= 1/a+1/b+1/c + 9/(a+b+c)
lo cual queriamos demostrar.

El segundo problema:
Cada una de los factores es equivalente a
((a+b)+(b+c)/2(a+c)). Llamemos x=a+b, y=a+c, z=b+c, y tenemos que probar que
((x+y)/2z)((y+z)/2x)((x+z)/2y) >= 1
Por MG-MA el lado izquierdo es mayor o igual a
 (((xy)^1/2)/z)(((yz)^1/2)/x)(((xz)^1/2)/y)
y Volviendo ausar MG-MA con todo ese termino obtenemos el resultado deseado.

Ya resolvi algunos, xfavor, k pongan los problemas de inversion!!!!! :D

[quote="M4RI0"]En la primera ayudaría demostrar:

$x^{2}+y^{2}+z^{2}\geq 9+2(x+y+z)$, si: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$[/quote]

$\sum x^{2}\geq 9+2(x+y+z)$, claramente, agrupando terminos tenemos que $\sum (x-1)^{2}\geq 12$, haciendo un cambio de variable $a=x-1$, $b=y-1$, $c=z-1$ se sigue que hay que demostrar que $\sum a^{2}\geq 12$. De la condicion tenemos que $\sum \frac{1}{a+1}=1$ y desarrllando se sigue que $a+b+c+2=abc$, de ahi que, por la desigualdad MA y MG se tiene que $(a+b+c)\geq 3\sqrt[3]{abc}$ con lo que, reemplazando $(a+b+c)^{3}\geq 27abc$, esto es $(a+b+c)^{3}\geq 27(a+b+c+2)$. Sea $k=a+b+c$, la ecuacion anterior se reduce a $k^{3}\geq 27(k+2)$ con lo que $k^{3}-27(k+2)\geq 0$ y de ahi que $k\geq 6$. Por la desigualdad de Cauchy se sigue que $(a^{2}+b^{2}+c^{2})(1^{2}+1^{2}+1^{2}) \geq (a+b+c)^{2}$ con lo que $\sum a^{2}\geq (a+b+c)^{2}/3\geq 12$. $\blacksquare$

[quote="conejita"]K onda Hucht!!

Oye ya resolvi alguno de tus problemas, kreo k como 3, pero los otros son algo dificiles!!! Ya vas a poner la solucion a alguno de los mios??

 :lol:[/quote]

Primero postea las soluciones

no se si les parecio faciles los anteriores problemas , aver q les parece este  :blush:  :

se tiene un tablero de nxn pintado como tablero de ajedrez . esta permitido efectuar la siguiente operacion :
escoger un rectangulo en la cuadricula tal que las longitudes de sus lados sean ambas pares o ambas impares , pero que no sean las dos iguales a 1 al  mismo tiempo , e invertir los colores de los cuadritos de ese rectangulo ( es decir , los cuadraditos del rectangulo que eran negros se convierten en blancos y los que eran blancos , se convierten en negros)
encontrar para que valores de n  es posible lograr que todos los cuadritos queden de un mismo color despues de haber efectuado la operacion un numero finito de veces ( Nota : las dimensiones de los rectanglos que se escogen pueden ir cambiando), espero sus soluciones  :D

q paso con las soluciones al problema  :D

[quote="conejita"]A ver, ya avance algo en tu problema,  ya probe que se puede para todos los valores de n, excepto cuando son una potencia de 2. Me imagino k sera probar algo de invarianza.
La forma de cambiar las fichas en los demas arreglos, es muy facil d encontrar. Si kieres k los ponga, dime. (aunk no te daria trabajo buscarla)[/quote]

Postealo! es mejor, al menos para que los demás tratemos de ersolverlo a partir de tu método

Weno, ps te explico, para todos los impares (excepto 1), lo k acemos es, cambiar todas las colunmas impares, y luego cambiar toas las filas pares. Para los numeros pares (no potencias de dos) lo unico k acemos, es dividimos el cuadro en un subcuadro que sea impar, y cada uno de esos subcuadros realizamos el arreglo anterior!!!
Ahora solo falta para 2^n!!!!

Si alguien tiene alguna idea...

Al fin he acabado tu problema.
Escribire la solucion completa:

Para n=1 la forma de acerlo es inmediata.
Para n=2 es facil ver k no es posible de ningna forma.
Ahora, para todos los n impares, primero movemos todas las columnas impares completas, y luego todas las filas pares completas. Siempre obtendremos la cuadricula del mismo color.
Para todos los numeros pares, exceptuando las potencias de 2, la forma es dividir el cuadro en subcuadros impares, y realizar un procedimiento similar al anterior en cada subcuadro.
Ahora, para los n potencias de 2, la formas es:
Fijemonos en la cuadricula de 4 x 4, a esta cuadricula dividamosla en 4 subcuadros de 2 x 2 cada uno. Al cuadro de 2x2 de arriba a la izquierda le acemos el cambio, igual le acemos el cambio al cuadro de 2x2 de abajo a la derecha. Luego acemos el cambio al cuadro central de 2x4 y para acabar se lo acemos al cuadro central de 4x2. Ya visto que se puede para el de 4x4, todos los demas k son potencias de 2, los dividimos un subcuadros de 4x4, y realizamos el procedimiento anterior las veces necesarias.
De esat manera demostramos que para todos los n (excepto 2) se puede acer lo pedido.
 :P

bien, asi me salio a mi tambien , mas bien ,conejita ,sabes como sale el problema 5 y 6 de la OCCM , si te salieron postealas porfavor   :lol: , mañana quiza postee otro problema

Ola, aki te dejo este link, en el se encuentran las soluciones oficiales de la prueba.

http://viiiomcc.opm.org.pa/Soluciondia2.pdf

ok gracias por el link mas bien ahora dos problemas de geo  :D

1) sean AD , BE, y CF las bisectrices interiores del triangulo ABC , demostrar que si uno de los angulos <ADF, <ADE, <BED, <CFE, <CFD mide 30° entonces al menos uno o mas de estos angulos mide 30°

[quote="mhuarancca"]2) sea ABC un triangulo acutangulo escaleno . sean L I O el ortocentro . incentro y circuncentro de dicho triangulo, respectivemente . probar que si uno de los vertices del triangulo ABC esta sobre la circunferencia circunscrita al triangulo LIO , entonces algun otro vertice del triangulo ABC debe encontrarse tambien sobre dicha circunferencia[/quote]

Supongamos que $A$, $H$, $I$, $O$ son conciclicos.

Por simple manipulacion de angulos tenemos que $m\angle HAI=m\angle A/2+m\angle C-90$
y $m\angle IAO=m\angle A/2+m\angle C-90$ con lo que $HI=IO$.

Por manipulacion de angulos denuevo tenemos que $m\angle HBI=m\angle C+m\angle B/2-90$ y $m\angle IBO=m\angle C+m\angle B/2-90$. Aplicando ley de senos en $\triangle IOB$ y $\triangle IBH$ respectivamente tenemos que $\frac{IB}{IO}=\frac{sen \angle IOB}{sen \angle IBO}$ y $\frac{IB}{HI}=\frac{sen \angle IBH}{sen \angle IHB}$ con lo que $sen\angle IBO=sen\angle IHB$ con lo que $m\angle IHB=m\angle IOB$ o $m\angle IHB+m\angle IOB=180$. Desde que $\triangle ABC$ es acutangulo escaleno $m\angle IHB\neq m\angle IOB$ y de ahi que $m\angle IHB+m\angle IOB=180$ con lo que $H$, $I$, $O$, $B$ son conciclicos.


Let $ x,$ $ y$ and $ z$ are positive numbers such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3.$ Prove that:
\[ \frac {x^2 \plus{} y^2}{x \plus{} y} \plus{} \frac {y^2 \plus{} z^2}{y \plus{} z} \plus{} \frac {z^2 \plus{} x^2}{z \plus{} x} \geq3
\]

[quote="arqady"]Let $ x,$ $ y$ and $ z$ are positive numbers such that $ x^2 + y^2 + z^2 = 3.$ Prove that:
\[ \frac {x^2 + y^2}{x + y} + \frac {y^2 + z^2}{y + z} + \frac {z^2 + x^2}{z + x} \geq3
\]
[/quote]
I think it is a problem of Cezar Lupu.
$ \frac {x^2 + y^2}{x + y} + \frac {y^2 + z^2}{y + z} + \frac {z^2 + x^2}{z + x} \geq3$

$ \Leftrightarrow \sum \left(\frac{y^2+z^2}{y+z}-\frac{y+z}{2}\right) \ge \sqrt{3(x^2+y^2+z^2)} - (x+y+z)$

$ \Leftrightarrow \sum \frac{(y-z)^2}{2(y+z)} \ge \frac{\sum (y-z)^2}{\sqrt{3(x^2+y^2+z^2)}+x+y+z}$

Because ${ 2(y+z) < 2(x+y+z)< \sqrt{3(x^2+y^2+z^2)}+x+y+z}$
so we are done!

[quote="dgreenb801"]Can you explain why the RHS of the two middle steps are equal?[/quote]
OK, because:

$ \sqrt {3(x^2 \plus{} y^2 \plus{} z^2)} \minus{} (x \plus{} y \plus{} z) \equal{} \frac{3(x^2\plus{}y^2\plus{}z^2)\minus{}(x\plus{}y\plus{}z)^2}{\sqrt {3(x^2 \plus{} y^2 \plus{} z^2)} \plus{} (x \plus{} y \plus{} z)}$
$ \equal{}\frac {(x\minus{}y)^2\plus{}(y \minus{} z)^2\plus{}(z\minus{}x)^2}{\sqrt {3(x^2 \plus{} y^2 \plus{} z^2)} \plus{} x \plus{} y \plus{} z}$

O, whoops, I saw multiplying by the conjugate, but for some weird reason I didn't think it would work... whatever.

[quote=arqady]The following inequality is also true.
Let $ x,$ $ y$ and $ z$ be positive numbers such that $x^3+y^3+z^3+xyz=4$ Prove that:
\[ \frac {x^2 \plus{} y^2}{x \plus{} y} \plus{} \frac {y^2 \plus{} z^2}{y \plus{} z} \plus{} \frac {z^2 \plus{} x^2}{z \plus{} x} \geq3\]
The equality occurs also for $x=\frac{3}{2}$ and $y=z=\frac{1}{2}$.[/quote]
Very interesting problem :)
Write inequality as
\[\frac {x^2 + y^2}{x + y} + \frac {y^2 + z^2}{y + z} + \frac {z^2 + x^2}{z + x} \geqslant 3\sqrt[3]{\frac{x^3+y^3+z^3+xyz}{4}}.\]
Suppose $p = x + y + z  = 3,\,q=xy+yz+zx=3-3t^2\,(0 \leqslant t < 1)$ and $r = abc$ inequality become
\[\frac{2p^2q-4pr-2q^2}{pq-r} \geqslant 3\sqrt[3]{\frac{p^3-3pq+4r}{4}},\]
equivalent to
\[\frac{36-12r-18t^2(t^2+1)}{9(1-t^2)-r} \geqslant 3\sqrt[3]{r+\frac{27}{4}t^2},\]
or
\[4-\frac{6(1-t^2)(4-t^2)}{9(1-t^2)-r} \geqslant \sqrt[3]{r+\frac{27}{4}t^2}.\]
Because $r \leqslant (1+2t)(1-t)^2,$ so
\[4-\frac{6(1-t^2)(4-t^2)}{9(1-t^2)-r} \geqslant 4-\frac{6(1-t^2)(4-t^2)}{9(1-t^2)-(1+2t)(1-t)^2} = \frac{3t^2+t+2}{t+2},\]
and
\[r+\frac{27}{4}t^2 \leqslant (1+2t)(1-t)^2+\frac{27}{4}t^2 = 2t^3+\frac{15}{4}t^2+1.\]
Therefore we need to show that
\[\frac{3t^2+t+2}{t+2} \geqslant \sqrt[3]{2t^3+\frac{15}{4}t^2+1}.\]
Which is true because
\[\left (\frac{3t^2+t+2}{t+2}  \right )^3 - \left ( 2t^3+\frac{15}{4}t^2+1 \right ) = \frac{(4t^2+5t+6)(5t-2)^2t^2}{4(t+2)^3} \geqslant 0.\]
The proof is completed.

[quote=arqady][quote=Nguyenhuyen_AG]
Because $r \leqslant (1+2t)(1-t)^2,$ so
[/quote]
If $x=\frac{3}{2}$ and $y=z=\frac{1}{2}$ so we get $r=\frac{3}{8}$ and $xy+xz+yz=\frac{7}{4}$.
Hence, $3-3t^2=\frac{7}{4}$ and $t^2=\frac{5}{12}$.
Thus, the equality in $r \leqslant (1+2t)(1-t)^2$ does not occur. ;)[/quote]
But I was homogenized your inequality become
\[\frac {x^2 + y^2}{x + y} + \frac {y^2 + z^2}{y + z} + \frac {z^2 + x^2}{z + x} \geqslant 3\sqrt[3]{\frac{x^3+y^3+z^3+xyz}{4}}\]
and suppose $x+y+z=3.$

Dear Arqady, in my solution equality occur if and only if $t = 0$ or $t = \frac{2}{5}.$

If $t = 0$ then $a=b=c=1.$

If $t = \frac{2}{5}$ then $a = \frac{3}{5}, b = \frac{3}{5}, c = \frac{9}{5}$ or any cyclic permutation. :)

[quote=arqady]Yes, it's obvious that in $r \leqslant (1+2t)(1-t)^2$ the equality occurs for $y=z$ in all case. I am sorry![/quote]
Yes :) and for general
\[\frac {x^2 + y^2}{x + y} + \frac {y^2 + z^2}{y + z} + \frac {z^2 + x^2}{z + x} \geqslant 3\sqrt[3]{\frac{x^3+y^3+z^3+xyz}{4}}.\]
Equality occur when $x=y=z$ or $x = 3y = 3z$ or any cyclic permutation.

I checked your proof, dear [b]Nguyenhuyen_AG[/b]. I think your proof is right!
In my proof I used $uvw$: our inequality is equivalent to $f(w^3)\geq0$, where $f$ is a decreasing function and we can take $y=z$ immediately.

@arqady Could you show your solution with $uvw$?

[quote=Nguyenhuyen_AG]\[\frac {x^2 + y^2}{x + y} + \frac {y^2 + z^2}{y + z} + \frac {z^2 + x^2}{z + x} \geqslant 3\sqrt[3]{\frac{x^3+y^3+z^3+xyz}{4}}.\]
I get ugly SOS form result of $\text{LHS}^3 - \text{RHS}^3 = $ [hide] [img]https://i.postimg.cc/fLwCvqS4/Capture.png[/img] [/hide][/quote]

How to write this?


trigonometry
calculus
calculus computations

I hvae to solve that until monday
Please help!

Problem 18 is really straightforward.  Think about $y''$:
$y'' =-c_{1}\cos x-c_{2}\sin x$ so
$y''+y = (-c_{1}\cos x-c_{2}\sin x)+(c_{1}\cos x+c_{2}\sin x) = 0$


If it takes 6 men 6 days to dig 6 holes, how long will it take one man to dig half a hole?

please hide answer

[hide]
Let m = the number of holes a man can dig in a day.
36m = 6
m = 1/6

m*d = 1/2
1/6 * d = 1/2
m = 3
[/hide]

[quote="randomdragoon"]Clearly 1/2 a day, because everyone knows that in every group, one man does all the work while the other five sit around and do nothing.[/quote]

ur wrong 2 they r all working.

[quote="klutzycherry14"][quote="Ihatepie"][quote="Ciaccona"][hide]It's not possible to dig half a hole - no matter what you dig, it will be a hole...(unless you define a hole to have a specific volume where the material that had been present has been replaced by some other material, or lack of material.)[/hide][/quote]

ur right[/quote]

though it could work, if the other holes that were made before were the same size/volume, and then it was implying how long it would take to dig half of that size hole. but i shouldn't assume that...[/quote]

I won't swear but don't assume because u'll make an *SS out of U and ME if u add those words 2gether it = assume

[quote="Ihatepie"][/quote][quote="klutzycherry14"][quote="Ihatepie"][quote="Ciaccona"][hide]It's not possible to dig half a hole - no matter what you dig, it will be a hole...(unless you define a hole to have a specific volume where the material that had been present has been replaced by some other material, or lack of material.)[/hide][/quote]

ur right[/quote]

though it could work, if the other holes that were made before were the same size/volume, and then it was implying how long it would take to dig half of that size hole. but i shouldn't assume that...[/quote][quote="Ihatepie"]

I won't swear but don't assume because u'll make an *SS out of U and ME if u add those words 2gether it = assume[/quote]

clever....mind if i use that line on people? ...i'll keep this one in mind! :wink:


trigonometry
geometry
circumcircle
geometry unsolved

Let $ P$ be an internal point of triangle $ ABC$ and let $ \alpha, \beta, \gamma$ be defined by

$ \alpha \equal{} \angle BPC \minus{} \angle BAC$, $ \beta \equal{} \angle CPA \minus{} \angle CBA$, $ \gamma\equal{} \angle APB \minus{} \angle ACB$.

Prove that

$ PA \frac{\sin \angle BAC}{\sin \alpha} \plus{} PB \frac{\sin \angle CBA}{\sin \beta} \equal{} PC \frac{\sin \angle ACB}{\sin \gamma}$.

Let $ A',B',C'$ be the 2nd intersections of  rays $ AP,BP,CP$ with the circumcircle $ (O)$ of $ \triangle ABC.$ Angle chasing gives

$ \angle B'CP \equal{} \angle BPC \minus{} \angle BB'C \equal{} \angle BPC \minus{} \angle BAC \equal{} \alpha.$

Similarly, $ \angle C'AP \equal{} \beta$ and $ \angle A'BP \equal{} \gamma.$

By sine law in the triangles $ \triangle PCB',\triangle PAC'$ and $ \triangle PBA',$ we obtain

$ \frac {\sin\widehat{BAC}}{\sin \alpha} \equal{} \frac {PC}{PA} \ , \ \frac {\sin\widehat{CBA}}{\sin \beta} \equal{} \frac {PA}{PC'} \ , \ \frac {\sin\widehat{ACB}}{\sin \gamma} \equal{} \frac {PB}{PA'}$

Then, from power of $ P$ to $ (O) \ , $ $PA \cdot PA' \equal{} PB \cdot PB' \equal{} PC \cdot PC',$ we get

$ PA \cdot \frac {\sin\widehat{BAC}}{\sin \alpha} \equal{} PB \cdot \frac {\sin\widehat{CBA}}{\sin \beta} \equal{} PC \cdot \frac {\sin\widehat{ACB}}{\sin \gamma}$


Sixty-four unit cubes are placed together to create a large cube. How many cubes with integer dimensions are in the $ 4 \times 4 \times 4$ cube?

[asy]import three; size(101);
 currentprojection = orthographic(1/3,-1/2,1/2);
for(int i = 0; i<4; ++i)
{draw((i,0,0)--(i,0,4));
draw((0,0,i)--(4,0,i));

draw((4,0,i)--(4,4,i));
draw((4,i,0)--(4,i,4));

draw((i,0,4)--(i,4,4));
draw((0,i,4)--(4,i,4));}
draw((4,4,0)--(4,4,4)); draw((0,4,4)--(4,4,4)); draw((4,0,4)--(4,4,4));[/asy]

[quote="Brut3Forc3"]There are 64 4x4x4 cubes, 27 3x3x3 cubes, 8 2x2x2 cubes, and 1 4x4x4 cube, making 100 total.[/quote]

You mean, 64 1x1 cubes, right?

wait i dont get the question. so we have to find the number of 1x1x1 cubes, etc?

Well, you want to find the number of cubes that can be made out of that big 4x4x4 cube. Like, there's 64 little unit cubes inside it, and 8 2x2x2 cubes inside it.

i think for problems like this the formula for a $ n\times n\times n$ cube, its $ n^3 \plus{} (n \minus{} 1)^3 \plus{} ... \plus{} 2^3 \plus{} 1^3$

[quote="myyellowducky82"]There are:
$ 64 1\cdot1\cdot1$ cubes (each single cube)
$ 8 2\cdot2\cdot2$ cubes ($ 2^3$ is $ 8$, so you divide 64 by 8 to get 8)
$ 1 4x4x4$ cube (or just the original cube)
I got 73 cubes...did i do something wrong?[/quote]

Try making your LaTeX spaced out...
and there are $ 27$ 2 by 2 by 2 cubes, not $ 8$. And $ 8$  3 by 3 by 3 cubes.
If you bothered to look above    :P .

@mz94: Yes that is correct.

i fixed it.
There are:
$ 64$ $ 1\cdot1\cdot1$ cubes (each single cube)
$ 27$ $ 2\cdot2\cdot2$ cubes
$ 8$ $ 3\cdot3\cdot3$ cubes
$ 1$ $ 4\cdot4\cdot4$ cube (or just the original cube)

thanks e^i_penguin!


real analysis
complex analysis
complex analysis theorems

In "Complex Variables" (Schaum's outline series) I read:

Theorem 4: If f(z) is continuous in a closed region, it is bounded in the region.

Theorem: If f(z) is continuous in a closed region, it is uniformly continuous there.

A closed region is defined as the closure of an open connected set.

IMO the complex plane is a closed region, but f(z)=z is not bounded. I think f(z)=z^2 isn't uniformly continuous on the complex plane.

Isn't an extra boundedness condition neccesary?
And if we replaced 'closed bounded region' with 'closed bounded set', will the same still be true, or will we be running into trouble when there are more than a finite number of "connectivity components"? (I think the latter is true)

These statements hold for any compact region in $\mathbb{C}$, so another boundedness property is needed..


vector
algorithm
algebra
polynomial
function
Support
linear algebra

Not sure if this is the right forum or not...but here's my question:

Given:
n vectors $ \vec{x_1}, \vec{x_2}, ..., \vec{x_n} \in R^n$
with labels $ y_1, y_2, ..., y_n \in \{ \plus{} 1, \minus{} 1\}$
and weights $ w_1, w_2, ..., w_n$ where $ w_i \ge 0$ $ \forall i$,

is there an algorithm to find the best separating hyperplane? (Best in the sense that the sum of the weights of the mislabeled vectors is minimized.) Of course there are likely infinitely many equally good separating hyperplanes, but is there an easy way to find any one of them? If there isn't an easy way, how about if I add the constraint that the hyperplane passes through the origin?

Equivalently (I think), if I constrain the hyperplane to pass through the origin, I'm asking if there is a way to solve:

$ \underset{\vec{a}}{\text{argmin}} \sum_i w_i |sgn(\vec{x_i} \cdot \vec{a}) \minus{} y_i|$

separating hyperplane ?
mislabeled vector  (w_i=0  ? )
argmin ?
Do you really want an answer ?

There is no efficient algorithm - the problem is NP-hard (unless $ P\equal{}NP$). Moreover, there is no good polynomial-time approximation algorithm for it.

Hi Yury,
what is the problem ?
Please give to me, miserable ground worm, at least a reference.

The problem is as follows: given a set of vectors $ x_1, \dots x_n$, numbers $ y_1,\dots, y_n \in \{\minus{}1,\plus{}1\}$ ("labels"), and $ w_1, \dots, w_n\in \mathbb R$ ("weights") find a vector $ a$ that minimizes the objective function ("number of mislabeled samples"):
\[ \sum_{i\equal{}1}^n w_i | \mathrm{sgn} (\langle x_i, a \rangle) \minus{} y_i|.\]

This problem arises in Machine Learning. See [url]http://en.wikipedia.org/wiki/Linear_classifier[/url] and [url]http://en.wikipedia.org/wiki/Supervised_learning[/url] for some background. In practice, Support Vector Machines are often used to solve the problem, see [url]http://en.wikipedia.org/wiki/Support_vector_machine[/url].

A very strong hardness of approximation result is proved in this paper:
V. Guruswami and P. Raghavendra, [i]Hardness of Learning Halfspaces with Noise[/i], In Proc. of the 47th Annual IEEE Symposium on Foundations of Computer Science, 2006, pp. 543-552.
http://www.cs.washington.edu/homes/venkat/pubs/papers/halfspace.pdf


A problem I get from dedicating from another problem:
Prove or disprove:if positive real numbers $ a,b,c$ satisfy that $ abc\equal{}1$,then $ \sum (ab\minus{}1)^{2}\plus{}\sum(1\minus{}c)^{2}\geq \sum ab(a\minus{}b)$
I think it is true but I cannot prove it.
Can anyone help me? :?:

Also there is a direct link with the following problem:
"Prove or disprove: positive real numbers $ a,b,c$ satisfy:
$ \prod (\frac{a^{2}}{bc}\plus{}\frac{a}{c}\plus{}\frac{a}{b}\minus{}1)\leq 8$"
I think both problem is nice :lol:
Anyone can help to post a solution or example that disproved :?:

[quote="tang zy"]Also there is a direct link with the following problem:
"Prove or disprove: positive real numbers $ a,b,c$ satisfy:
$ \prod (\frac {a^{2}}{bc} \plus{} \frac {a}{c} \plus{} \frac {a}{b} \minus{} 1)\leq 8$"
I think both problem is nice :lol:
Anyone can help to post a solution or example that disproved :?:[/quote]
this inequality does not hold,you can check it by Maple

[quote="tang zy"]A problem I get from dedicating from another problem:
Prove or disprove:if positive real numbers $ a,b,c$ satisfy that $ abc \equal{} 1$,then $ \sum (ab \minus{} 1)^{2} \plus{} \sum(1 \minus{} c)^{2}\geq \sum ab(a \minus{} b)$
I think it is true but I cannot prove it.
Can anyone help me? :?:[/quote]

Let $ a\equal{}\frac{y}{z},b\equal{}\frac{z}{x},c\equal{}\frac{x}{y},$then it can be transformed into

$ \sum{x^2yz(x\minus{}y)(x\minus{}z)}\plus{}3\sum{y^2z^2(x\minus{}y)(x\minus{}z)}\plus{}(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2\ge{0}.$

[quote="tang zy"]Also there is a direct link with the following problem:
"Prove or disprove: positive real numbers $ a,b,c$ satisfy:
$ \prod (\frac {a^{2}}{bc} \plus{} \frac {a}{c} \plus{} \frac {a}{b} \minus{} 1)\leq 8$"
I think both problem is nice :lol:
Anyone can help to post a solution or example that disproved :?:[/quote]

It can be transformed into

$ 4\sum{b^2c^2(a\minus{}b)(a\minus{}c)}\plus{}(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\ge{0}.$

[quote="tang zy"]Also there is a direct link with the following problem:
"Prove or disprove: positive real numbers $ a,b,c$ satisfy:
$ \prod (\frac {a^{2}}{bc} \plus{} \frac {a}{c} \plus{} \frac {a}{b} \minus{} 1)\leq 8$"
I think both problem is nice :lol:
Anyone can help to post a solution or example that disproved :?:[/quote]
I think it is true and this is my proof:
$ \Leftrightarrow \prod (a(a\plus{}b\plus{}c)\minus{}bc)\leq 8a^{2}b^{2}c^{2}$
$ \Leftrightarrow abc((a\plus{}b\plus{}c)^{3}\plus{}(a\plus{}b\plus{}c)\sum a^{2})\leq 9a^{2}b^{2}c^{2}\plus{}(a\plus{}b\plus{}c)^{2}\sum a^{2}b^{2}$
$ \Leftrightarrow 2abc(\sum a^{3}\plus{}\sum a^{2}(b\plus{}c))\leq \sum a^{4}(b^{2}\plus{}c^{2})\plus{}2(\sum a^{3}b^{3}\plus{}6a^{2}b^{2}c^{2})$
and this can be prove by AM-GM and Schur!

[quote="kuing"][quote="tang zy"]A problem I get from dedicating from another problem:
Prove or disprove:if positive real numbers $ a,b,c$ satisfy that $ abc \equal{} 1$,then $ \sum (ab \minus{} 1)^{2} \plus{} \sum(1 \minus{} c)^{2}\geq \sum ab(a \minus{} b)$
I think it is true but I cannot prove it.
Can anyone help me? :?:[/quote]

I think this ineq holds for any [b]reals[/b] $ a,b,c$ (don't need abc=1)...[/quote]

It's amazing!
Can you give the prove to me?

[quote="tang zy"][quote="kuing"][quote="tang zy"]A problem I get from dedicating from another problem:
Prove or disprove:if positive real numbers $ a,b,c$ satisfy that $ abc = 1$,then $ \sum (ab - 1)^{2} + \sum(1 - c)^{2}\geq \sum ab(a - b)$
I think it is true but I cannot prove it.
Can anyone help me? :?:[/quote]

I think this ineq holds for any [b]reals[/b] $ a,b,c$ (don't need abc=1)...[/quote]

It's amazing!
Can you give the prove to me?[/quote]

this proof is by Tourish:

$ LHS = \frac {1}{2}\left(\sum{[(ab - 1)^2 + (a - 1)^2]} + \sum{[(ab - 1)^2 + (b - 1)^2]}\right)$

$ \geq |(ab - 1)(a - 1)| + |(ab - 1)(b - 1)| + |(bc - 1)(b - 1)| + |(bc - 1)(c - 1)| + |(ca - 1)(c - 1)| + |(ca - 1)(a - 1)|$

$ \geq (ab - 1)(a - 1) - (ab - 1)(b - 1) + (bc - 1)(b - 1) - (bc - 1)(c - 1) + (ca - 1)(c - 1) - (ca - 1)(a - 1)$

$ = \sum{(ab - 1)[(a - 1) - (b - 1)] = \sum{(ab - 1)(a - b)} = \sum{ab(a - b)} - \sum{(a - b)} = \sum{ab(a - b)} = RHS}$

when $ a = b = c = 1$ get "=".

[quote="tang zy"]A problem I get from dedicating from another problem:
Prove or disprove:if positive real numbers $ a,b,c$ satisfy that $ abc \equal{} 1$,then $ \sum (ab \minus{} 1)^{2} \plus{} \sum(1 \minus{} c)^{2}\geq \sum ab(a \minus{} b)$
I think it is true but I cannot prove it.
Can anyone help me? :?:[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-46-unknown-author.html

[quote="kuing"][/quote][quote="tang zy"][quote="kuing"][quote="tang zy"]A problem I get from dedicating from another problem:
Prove or disprove:if positive real numbers $ a,b,c$ satisfy that $ abc = 1$,then $ \sum (ab - 1)^{2} + \sum(1 - c)^{2}\geq \sum ab(a - b)$
I think it is true but I cannot prove it.
Can anyone help me? :?:[/quote]

I think this ineq holds for any [b]reals[/b] $ a,b,c$ (don't need abc=1)...[/quote]

It's amazing!
Can you give the prove to me?[/quote][quote="kuing"]

this proof is by Tourish:

$ LHS = \frac {1}{2}\left(\sum{[(ab - 1)^2 + (a - 1)^2]} + \sum{[(ab - 1)^2 + (b - 1)^2]}\right)$

$ \geq |(ab - 1)(a - 1)| + |(ab - 1)(b - 1)| + |(bc - 1)(b - 1)| + |(bc - 1)(c - 1)| + |(ca - 1)(c - 1)| + |(ca - 1)(a - 1)|$

$ \geq (ab - 1)(a - 1) - (ab - 1)(b - 1) + (bc - 1)(b - 1) - (bc - 1)(c - 1) + (ca - 1)(c - 1) - (ca - 1)(a - 1)$

$ = \sum{(ab - 1)[(a - 1) - (b - 1)] = \sum{(ab - 1)(a - b)} = \sum{ab(a - b)} - \sum{(a - b)} = \sum{ab(a - b)} = RHS}$

when $ a = b = c = 1$ get "=".[/quote]

[b]VERY NICE[/b] I think this question can change into some very hard problem!


Let's work in $\mathcal M_{n}\left( \mathbb F \right)$, where $\mathbb F$ is some field.
a) Characterize all matrices $A$ that commute with every invertible matrix.
b) Let $d \in \mathbb F$. Characterize all matrices $A$ that commute with every matrix of determinant $d$.

Look at the elementary matrices $E_{ij}$ for $i\neq j$ with entries of 1 in the diagonal and the $ij$ slot, and nowhere else. $A$ commutes with $E_{ij}$ if and only if the $ki$ and $jl$ entries of $A$ are zero whenever $k\neq i$ and $l\neq j$, and also the $ii$ and $jj$ entries of $A$ are equal. If $A$ commutes with all of these, it's a multiple of the identity.

All of the $E_{ij}$ are invertible, so that takes care of your first case. For the second, let $M$ be a matrix of determinant $d$; $AM=MA$ and $MAE_{ij}=AME_{ij}=ME_{ij}A$, so $A$ commutes with $E_{ij}$ as before.

These restrictions are far too weak to actually do anything.


Prove that there are infinitely many twin primes if and only if there are infinitely many integers that cannot be written in any of the following forms: \[6uv+u+v, \;\; 6uv+u-v, \;\; 6uv-u+v, \;\; 6uv-u-v,\] for some positive integers $u$ and $v$.

Note that twin primes have form $6k\pm 1.$

If $6k+1$ is not prime then there exist integers $u$ and $v$ such that $6k+1=(6u+1)(6v+1)$ or $6k+1=(6u-1)(6v-1),$ implying that $k=6uv+u+v$ or $k=6uv-u-v.$
Similarly, if $6k-1$ is not prime then there exist integers $u$ and $v$ such that $6k+1=(6u+1)(6v-1)$, implying that $k=6uv-u+v.$
Therefore, if $k$ cannot be written in any of these forms then both $6k-1$ and $6k+1$ are primes.


Hey,

A statistic shows that the number of accidents per day varies between 0 and 5.

During a certain period of time the number of accidents per day was as follows

Number of accidents per day: 0    1   2    3    4     5
Number of days:                  98  63  31   8    0     1

(It should be a table with 98 under 0, 63 under 1, 31 under 2 etc, hope you get the picture)

Q1: Let the number of accidents per day be regarded as a random variable (X). Calculate the mean and the variance of X.

Q2: What theoretical distribution could describe the observed results?

Any suggestions?

[quote="mcokkie"]Number of accidents per day: 0 1 2 3 4 5
Number of days: 98 63 31 8 0 1[/quote]
I think you mean:
Number of days: 0 1 2 3 4 5
Number of accidents per day: 98 63 31 8 0 1

No, it is to be understood that there have been 98 days with 0 accidents daily, 63 days with 1 accident daily, 31 days with 3 accidents a day etc.

Okay, so how would you compute the mean?  The same way you would always compute the mean.


Source: RockieBill on AddictSports forum, modified by me:

How far is it from the pitcher's mound to third base (to the nearest 16th of an inch)? Assume that each base is 1 inch high.

[hide="Hint"]Pythagoras is your friend.[/hide]

[hide="Reminder"]The field is 3-dimensional.[/hide]

[hide="For those who are too lazy to look up distances"]90' between bases.
60'6" from home to the mound,
Mound is 10.5" high.

Bonus points for a nice graphic solution.
[/hide]

ln (dx/dy), you're off; something went wrong

[hide="answer"]

no solution, but the answer is 67 + 7/16
[/hide]

Ypu're a little bit off, both of you. Remember, to the nearest 16th of an INCH.


A privateer discovers a merchantman $10$ miles to leeward at 11:45 a.m. and with a good breeze bears down upon her at $11$ mph, while the merchantman can only make $8$ mph in her attempt to escape. After a two hour chase, the top sail of the privateer is carried away; she can now make only $17$ miles while the merchantman makes $15$. The privateer will overtake the merchantman at:

$\textbf{(A)}\ 3\text{:}45\text{ p.m.} \qquad
\textbf{(B)}\ 3\text{:}30\text{ p.m.} \qquad
\textbf{(C)}\ 5\text{:}00\text{ p.m.} \qquad
\textbf{(D)}\ 2\text{:}45\text{ p.m.} \qquad
\textbf{(E)}\ 5\text{:}30\text{ p.m.}$

Wait...after the topsail is blown away, the two travel FASTER?

[hide]After two hours, there are only 4 miles in between the two.  This difference can be obtained with two more hours of chasing, so A.[/hide]

azn6021023, the privateer doesn't go faster. It moves 17 miles for [b]every[/b] 15 miles the merchant ship moves. The merchant ship still travels at the same speed as before. It does not mean it sails 17mph.

After the first two hours they are 4 apart.

In one time-group, they will catch up 2 more miles so it will take 2 time-groups. In that time, the merchant goes 30 miles, which take it 30/8=3 hours 45 min, so 5:45 total.

11:45+5:45=5:30 so E


geometry
trigonometry
trig identities
Law of Sines

I am not good in english ... so maybe I will do some mistakes in translation.  :blush:
Let $ \omega$ is circle with center $ O$.Let $ \overline{AB}, \overline{CD}$ are two diameter such that $ \angle AOC\equal{}90^\circ$. Let $ M \in \overline{AB}$ with $ M\neq A, M\neq B$. The line $ CM$ cut the circle again in $ N$. Then we draw tangent to the circle in $ N$. Let $ P$ is the intersection of the tangent and the perpendicular line from $ M$ to $ AB$. Prove that $ OP\equal{}CM$. (too long problem  :D .. did you understand  :D ) here is the diagram.

[hide]
Because P is on the tangent from N, angle ONP = 90 degrees. Also, PMO = 90, which implies that OMNP is cyclic, which implies that the measure of angle MPO = angle MNO. Since O is the center of the circle, ON = OC, which implies that angle MNO = angle  MCO, so the measure of angle MPO = the measure of angle MCO. Since OD and MP are parallel, the measure of angle CMP = 180 - the measure of angle MCO = 180 - angle MPO, which implies that MC is parallel to OP, and CMPO is a parallelogram. Therefore, OP = CM. [/hide]

[hide="Trig alternative"]
We prove MP=r
Let F be a point on the extension of PN after N., and let $ \angle PMN \equal{} n$. $ \angle PMN \equal{} \angle OCM$, and $ \angle CNF \equal{} \angle CDN \equal{} 90 \minus{} \angle DCN \equal{} \angle CMO$. So $ \angle MNF \equal{} 90 \minus{} n$. Thus, $ \angle MPN \equal{} 90 \minus{} 2n$. By the law of sines on PMN,
$ \frac {sin(90 \minus{} 2n)}{MN} \equal{} \frac {sin(n \plus{} 90)}{PM}$
$ \frac {cos^2n \minus{} sin^2n}{MN} \equal{} \frac {cos n}{PM}$
$ cos n \equal{} \frac {CO}{CM}$, so since we want to prove, $ MP \equal{} r \equal{} CO$, we have to show
$ \frac {1}{CM} \equal{} \frac {cos^2n \minus{} sin^2n}{MN} \equal{} \frac {\frac {OC^2 \minus{} OM^2}{CM^2}}{MN}$, which simplifies to
$ r^2 \minus{} OM^2 \equal{} OC^2 \minus{} OM^2 \equal{} CM \cdot MN \equal{} BM\cdot AM \equal{} (r \plus{} OM)(r \minus{} OM)$, clearly true.[/hide]


calculus
integration
trigonometry
function
derivative
real analysis
real analysis unsolved

[quote="marshell"]Find $I(\alpha)=\int_{0}^{+\infty}\large(\frac{\sin\alpha x}{x})^{4}dx$
[hide]My result: $I(\alpha)=\frac{\pi}{3}|\alpha|^{3}$???[/hide][/quote]
Ok. if you do next substitution $x\alpha=y$ then your integral will be like this $\alpha^{3}\int_{0}^{+oo}\frac{\sin^{4}y}{y^{4}}dy$
if you use $\sin^{2}y=\frac{1}{2}(1-\cos(2y))$
and again do substitution $2y=z$....
you will be take integral like this $\int_{0}^{+oo}\frac{sin{x}}{x}$
. to be more correct visit this poste
there I have post solution like this intergal. it solved analogisly .
http://www.mathlinks.ro/Forum/viewtopic.php?t=148484
[quote="Extremal"]
FOR Example:
$\int_{0}^{+oo}\frac{\sin^{3}x}{x^{3}}dx=\frac{(\frac{\sin^{3}x}{x^{2}}+3\sin^{2}x\cos{x})|_{0}^{+oo}+3\int_{0}^{+oo}\frac{\sin{x}}{x}dx-9\int_{0}^{+oo}\frac{\sin{x}}{x}dx+9(\frac{3}{4}\int_{0}^{+oo}\frac{\sin{x}}{x}dx-\frac{1}{4}\int_{0}^{+oo}\frac{\sin{y}}{y}dy)}{-2}$
[/quote]

I like my Fourier transforms with $2\pi$ in the exponent of the exponential function, so to reach ground that is familiar to me, let me make these definitions:

$\text{sinc}(x)=\frac{\sin(\pi x)}{\pi x}$ (continuous at $0,$ of course)

$\text{box}(x)=\begin{cases}1,&-\frac12<x<\frac12\\0,&\text{ otherwise}\end{cases}.$

Then $\widehat{\text{box}}=\text{sinc}$ and $\widehat{\text{sinc}}=\text{box}.$

So then $\text{sinc}^{4}{\xi}$ is the Fourier transform of $\text{box}*\text{box}*\text{box}*\text{box}(x).$ From that

$\int_{-\infty}^{\infty}\text{sinc}^{4}(\xi)\,d\xi$ will equal the value at zero of that fourfold convolution.

Now, let $f(x)=\text{box}*\text{box}*\text{box}*\text{box}(x).$ Take the fourth derivative, and use weak (distributional) derivatives:

$f''''(x)=\delta(x+2)-4\delta(x+1)+6\delta(x)-4\delta(x-1)+\delta(x+2).$

The four-times indefinite interal from the left of $\delta(x)$ is $\frac16x^{3}u(x)$ where $u$ is the Heaviside unit step function - $1$ for $x>0,$ $0$ for $x<0.$

From this we get that on the interval $[-1,0]$ we have

$f(x)=\frac16[(x+2)^{3}-4(x+1)^{3}]$ and in particular,

$f(0)=\frac16(8-4)=\frac23.$

So I now claim that

$\int_{-\infty}^{\infty}\text{sinc}^{4}{\xi}\,d\xi=\frac23,$ or

$\int_{0}^{\infty}\left(\frac{\sin(\pi x)}{\pi x}\right)^{4}\,dx=\frac13.$

(One should be able to change variables from here to answer the original problem.)

This would have been easier with the Plancherel theorem, which states that $\|f\|_{2}^{2}=\|\widehat{f}\|_{2}^{2}.$ In particular, in this case we get that

$\int_{\mathbb{R}}\text{sinc}^{4}(x)\,dx=\|\text{sinc}^{2}\|_{2}^{2}= \|\text{box}*\text{box}\|_{2}^{2}.$

Now, it's not hard to compute that $\text{box}*\text{box}(x)=\max(1-|x|,0).$ The square of the 2-norm of that is

$\int_{-1}^{1}(1-|x|)^{2}\,dx=2\int_{0}^{1}(1-x)^{2}\,dx=\frac23.$

So we arrive at the same result as in the previous post:

$\int_{-\infty}^{\infty}\text{sinc}^{4}(x)\,dx=\frac23$ or $\int_{0}^{\infty}\left(\frac{\sin \pi x}{\pi x}\right)^{4}\,dx=\frac13.$

---

Now to finish the problem. Assume for now that $a>0.$ Make the change of variables $\pi x=au$ or $x=\frac{au}{\pi}.$ This gives us

$\frac{a}{\pi}\int_{0}^{\infty}\left(\frac{\sin au}{au}\right)^{4}\,du=\frac13$

$\int_{0}^{\infty}\left(\frac{\sin au}{u}\right)^{4}\,du=\frac{\pi a^{3}}3.$

Noting that the integral is an even function of $a,$ we arrive at the final result of $\frac{\pi|a|^{3}}3,$ as marshell claimed.


could someone find a solution not as the obvious one? thanks.

[hide]
The given expression is equal to $\frac{(2n)!}{n!}$, so dividing it by $2^{n}$ gives $\frac{(2n)!}{2^{n}(n!)}$, which equals
\[\frac{1\cdot 2\cdot 3\cdot \cdots \cdot 2n}{2\cdot 4\cdot 6\cdot \cdots 2n}=1\cdot 3\cdot 5\cdot \cdots \cdot (2n-1)\]
which is clearly an integer (and an odd integer as well, so the expression is not divisible by $2^{n+1}$).
[/hide]

[quote="13375P34K43V312"][quote="Farenhajt"]Induction is quite easy here.[/quote]
[hide="k here we go"] Base case: $n=1$, $2$ clearly divides $2$. Yay. Let's now assume that $(n+1)(n+2)...(2n)$ divides $2^{n}$. Increasing $n$ by $1$ eliminates the $(n+1)$ term, and adds the terms $(2n+1)(2n+2)$, simplified to $2(2n+1)(n+1)$ Thus, increasing $n$ by 1 justs multiplies the whole product by $2(2n+1)$. Clearly, that is divisible by $2$, so we have one more factor of $2$. Thus, the new expression is divisible by $2^{n}\cdot2=2^{n+1}$, and the induction is complete.[/hide][/quote]
that is what Farenhajt means. thanks all. my solution:
$\left\lfloor \frac{2n}{2}\right\rfloor+\left\lfloor \frac{2n}{4}\right\rfloor+...-(\left\lfloor \frac{n}{2}\right\rfloor+\left\lfloor \frac{n}{4}\right\rfloor+...)=n$


calculus
derivative
function
logarithms
calculus computations

Solve the equation $2^{x}=\frac{2}{3}x^{2}+\frac{x}{3}+1$

$x=0, \,1,\, 3$.. looking at the derivative, we see that there are no others..

How actually did you solve the question?
I don't understand how the derivatives could tell these are all there is.

Derivatives just show that there are only 3 roots. I found the roots by inspection..

I wonder how does derivative [i]shows[/i] 3 roots?
$\ln(2) 2^{x}= 4x/3+1/3$

rolle's theorem can help us now: if the original equation had more than three roots, its derivative would have more than two roots, and its second derivative would have more than one... but the second derivative of the function is $2^{x}\log^{2}2-\frac43$, which is increasing, so it has at most one...


How many quintuples of positive integers $ (v,w,x,y,z)$ satisfy both equations: $ v\plus{}x\plus{}y\plus{}z\equal{}35$ and $ w\plus{}x\plus{}y\plus{}z\equal{}30$?

$ x\plus{}y\plus{}z \equal{} 35\minus{}v \equal{} 30\minus{}w \le 29$.  Since the value of $ v$  determines the value of the quantity $ x\plus{}y\plus{}z$, which in turn determines the value of $ w$, we only need to know how many solutions there are in the positive integers to $ x\plus{}y\plus{}z\plus{}v \equal{}30$.

Now lets use a ball and divider argument.  imagine 30 balls in a row.  We are going to place dividers between them to section the balls into groups.  The 1st group will have $ x$ balls, the second, $ y$, etc.  We will need $ 4$ groups, 1 for each variable, so we must place $ 3$ dividers. By specifying the order that the groups are assigned variables, we insure that the variable are treated as distinguishable from eachother.  If we place at most 1 divider between two adjacent balls, we are guaranteeing that each group has at least $ 1$ balls, i.e. every variable is an integer greater than $ 1$.  There are $ 29$ places in which we can place either $ 0$ or $ 1$ dividers, so there are $ \binom{29}{3} \equal{} \boxed{3654}$  to place the dividers, and thus the same number of solutions.

Side note:
Note that if we wanted to treat the variables as indistinguishable (i.e. count the number of partitions of an integer into a specified number of parts), we could not just divide our result by the number of groups factorialed (the standard way of changing from distinguishable to indistinguishable permutations), since this does not consider that there are fewer actual permutations to undistinguish when two or more of the variables are equal.  for example, there are $ 7$ solutions to $ x\plus{}y\equal{}8$ with x,y positive integers: $ (1,7)$ $ (2,6)$, etc.  But if we make x and y indistinguishable, we have the solutions $ 1,7$ $ 2,6$, $ 3,5$, $ 4,4$ - $ 4$ solutions.  Our old method would ignore the fact that $ 4,4$ only appears once in the solutions when $ x$ and $ y$ are considered distinguishable.  The stirling numbers of the second kind should be used if we want to count solutions with variables indistinguishable from one another.

but shouldn't it be 30C3 instead of 29C3?

no definitely 29.  We have 30 balls (imagined in a line, not a circle), so there are $ 29$ spaces between the balls, and choose $ 3$ of these spaces to place dividers.

Btw, I meant to use $ x \plus{} y \plus{} z \plus{} w \equal{} 30$ in my solution not $ v$.  If we wanted to instead solve the problem in terms of $ x\plus{}y\plus{}z\plus{}v \equal{} 35$, we are restricted to $ v \ge 6$ and so have to adjust the method to compensate.  Define $ v'\equal{}v\minus{}5$ then $ x\plus{}y\plus{}z\plus{}v' \equal{} 30$ and we end up finding the number of solutions to this in positive integers, the exact same thing we did before except with $ w$ renamed as $ v'$

[quote="facis"]no definitely 29.  We have 30 balls (imagined in a line, not a circle), so there are $ 29$ spaces between the balls, and choose $ 3$ of these spaces to place dividers.

Btw, I meant to use $ x \plus{} y \plus{} z \plus{} w \equal{} 30$ in my solution not $ v$.  If we wanted to instead solve the problem in terms of $ x \plus{} y \plus{} z \plus{} v \equal{} 35$, we are restricted to $ v \ge 6$ and so have to adjust the method to compensate.  Define $ v' \equal{} v \minus{} 5$ then $ x \plus{} y \plus{} z \plus{} v' \equal{} 30$ and we end up finding the number of solutions to this in positive integers, the exact same thing we did before except with $ w$ renamed as $ v'$[/quote]

i see, if w,x,y,z can be non-negative number, is the answer going to be 33 choose 3?

cdymdcool, yes.

Here's a related question: if $ a$, $ b$, $ c$ and $ d$ are positive integers, how many ordered quadruples $ (a,b,c,d)$ are solutions to $ a\plus{}b\plus{}c\plus{}d \le 12$?

[quote="facis"]cdymdcool, yes.

Here's a related question: if $ a$, $ b$, $ c$ and $ d$ are positive integers, how many ordered quadruples $ (a,b,c,d)$ are solutions to $ a \plus{} b \plus{} c \plus{} d \le 12$?[/quote]

11C3+10C3+9C3......+3C3
What's the fast way to do it?

right idea.  One approach you could take is

$ \binom{11}{3} \plus{} \binom{10}{3} \plus{} \ldots \plus{} \binom{3}{3}$

And if we don't want to compute each of these and add them, we need to use the hockeystick identity: $ \sum_{x\equal{}r}^n \binom{x}{r} \equal{} \binom{n\plus{}1}{r\plus{}1}$

So
$ \binom{11}{3} \plus{} \binom{10}{3} \plus{} \ldots \plus{} \binom{3}{3} \equal{} \binom{12}{4} \equal{} 495$

Alternatively, let $ e \equal{} 13 \minus{} (a\plus{}b\plus{}c\plus{}d)$ so $ e$ must be a positive integer, and you are looking for the number of solutions to $ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}13$, which is $ \binom{12}{4} \equal{} 495$.


Functional Analysis
topology
advanced fields
advanced fields unsolved

let $E$, $F$ be two Banach spaces. Suppose that $T_n$ is a sequence of operators from $E$ to $F$ such that $dim(T_n(E))< \infty$ for any $n \geq 1$ and that $T_n$ converges towards $T$. Then show that the closure of $T(B)$ is compact where $B$ is the open unit ball of $E$.

We only need the fact that $T_n$ are compact (i.e. they take the unit ball onto a relatively compact set). In other words, we can prove that the set of compact operators from a Banach space $E$ to a Banach space $F$ is closed in the norm topology.

If $U$ is the closed unit ball of $E$, then the closure of $T(U)$ is a complete subset of $F$, being a closed subset of a complete metric space, so all we need in order to prove that it's compact is to show that it's totally bounded, i.e. for every $\varepsilon>0$ it can be covered by finitely many balls of radius $\varepsilon$ in $F$.

Fix some $\varepsilon>0$, choose $n$ such that $\|T_n-T\|<\frac\varepsilon 3$, and let $(x_k)_{k=1}^t$ be points in $E$ (we may assume they're in $U$, to keep things simple) such that the $\frac\varepsilon 3$-balls centered at $T_nx_k$ cover $T_n(U)$. Now take some $x\in U$. There is a $k\in\overline{1,t}$ such that $\|T_nx-T_nx_k\|<\frac\varepsilon 3$. We then have $\|Tx-Tx_k\|\le\|Tx-T_nx\|+\|T_nx-T_nx_k\|+\|T_nx_k-Tx_k\|<\frac\varepsilon 3+\frac\varepsilon 3+\frac\varepsilon 3=\varepsilon$, and we're done: for arbitrary $\varepsilon>0$, we were able to cover $T(U)$ with finitely many $\varepsilon$-balls in $Y$.


Let $ABCD$ be a trapezoid with the measure of $AB$ twice that of base $DC$, and let $E$ be the point of intersection of the diagonals. If the measure of diagonal $AC$ is $11$, then find that of segment $EC$.

[hide="hint"]If AB is on the $x$ axis, what is the ratio of the height of $E$ to the height of $CD$?[/hide]
[hide="answer"]The ratio is 1:3 because however long it takes the two diagonals to get together it takes twice as long to get apart. That means $EC=AC/3=11/3$.[/hide]


calculus
integration
algebra
polynomial
derivative
function
calculus computations

#2: You contradict yourself. The only polynomial $ f$ satisfying $ f''\equal{}f$ is identically zero.

[b] for 2 question
 sorry It is not a Polynomial function.[/b]

(2) If f(x) is function  satisfy the following properties
(a) f(0) =2, f '(0) = 3
(b) f '' (x) = f(x).
Then find f(x) =
here f '(x) means first Derivative
and f ''(x) means second Derivative.

hello, for 2)  we get $ f(x)\equal{}\minus{}1/2e^{\minus{}x}\plus{}5/2e^{x}$ as the solution.
Sonnhard.

[quote="jagdish"][b](1) Integration of log(sinx) from 0 to pi/2.
here base = e[/b]

[b](2) If f(x) is polynomial  and satisfy the following properties
    (a) f(0) =2, f '(0) = 3
    (b) f '' (x) = f(x).
Then find f(x) =
here f '(x) means first Derivative
and f ''(x) means second Derivative.[/b][/quote]
for (1)
$ \int _{0}^{\frac {\pi}{2}} ln(sin x) dx \equal{} \minus{}\frac {\pi}{2} ln 2$

2) If f(x) is polynomial and satisfy the following properties
(a) f(0) =2, f '(0) = 3
(b) f '' (x) = f(x).
Then find f(x) =
here f '(x) means first Derivative
and f ''(x) means second Derivative.
[color=blue]
For (2) My solution is as follows:
Let $ f(x)=y$
Now $ y''=y\implies 2y'y''=2y'y\implies ((y')^{2})'=(y^{2})'\implies (y')^{2}=y^{2}+c$
Now by using $ f(0) =2, f '(0) = 3$, we get $ c = 5$
$ \implies (y')^{2}=y^{2}+5\implies y'=\pm\sqrt {y^{2}+5}\implies y'=\sqrt {y^{2}+5}$  as  $ y'(0)$ is positive.
$ \implies\int {\frac {1}{\sqrt {y^{2}+5}}dy =dx}$
$ \implies ln|y+\sqrt {y^{2}+5}|=x+c_{1}$
By using $ f(0) =2, c_{1}=ln5$
$ \implies ln\left|\frac {y+\sqrt {y^{2}+5}}{5}\right|=x$
$ \implies \frac {y+\sqrt {y^{2}+5}}{5}= e^{x}$
$ \implies \sqrt {y^{2}+5}+y =   5e^{x}$
$ \implies \sqrt {y^{2}+5}-y = e^{-x}$
$ \implies \boxed {f(x) = \left(\frac  {5}{2}\right)e^{x}-\left(\frac  {1}{2}\right)e^{-x}}$[/color]


This is a very tabu discussion, and it is impossible to give a perfect answer. If you completely change the wording and the numbers of the problems then you should be ok, but ...  :?  :?

[quote="Silverfalcon"]Greeting,

I'm thinking of making PDF Document with a lot of problems in many areas. I haven't decided yet but I might want to show it on the public? But my question is:

a) If I change the wording of the problem, like Kalva does, and post on the web or do something like that, is it allowed?

b) If a) doesn't work, can I still use them by changing the values?

BTW, how did you get the problems on the aops book? I always wondered how you got those problems.  :?[/quote]Have you tried asking the owners of the problems?  Tell them what you're going to do with the problems, and ask them if it is fine with them to do it.

[quote="Myself"]Have you tried asking the owners of the problems?  Tell them what you're going to do with the problems, and ask them if it is fine with them to do it.[/quote]

I tried that when I was trying to post my Recopilation of Math Conquest Around the World, but people from the USAMO, CMO and APMO told me explicitely that I should prepare the book that I was planning to, and afterwards let the publisher to negociate the copyright. When I pointed out that it was a material that I wanted to put freely available on the web, no one else than CMO answer, and the guy from CMO understood that it was a webpage and asked me to just put a link to their page. When I explained that I wanted to do a PDF, again, they didn't ask (that was a couple of years ago).

Now, I tried to investigate about the international laws of copyright, but it seems like it is very vague, but at least two years ago, even asking to a lawyer that happened to be on campus, it seems that if you find something that is posted online by the authors, and it is freely available in the site, you can make a recopilation without asking anyone, and you are able to place it in a P2P, because that was in some sence your own creation (like if you sing "Billy Jean" by yourself and put in on a P2P), but not in the web, by some weird law, because it is copyrighted material and you don't have the redistribution rights. You can also freely say that you have it, and that if anyone wants it they just need to email you, and send it by email to anyone (that is what I do with my recopilation), because it is not public, is a private comunications.

You can do your own research, and try if you have better luck with the authors of the problems.

Best,

So, how about more SURE one:

I change the wording. I change the value. I change everything. But!!! I get the idea from the other competition's problems.

So, the problem may look similar but not same.

Is this more okay for sure?

Like it's totally different. But I"m getting "ideas" from them. Is "ideas" okay?

I don't see how it can't be okay. It's probably impossible to regulate it anyway since you can't really tell if the author's getting 'inspiration' if everything else is different. Besides, it doesn't seem ethically wrong to get ideas from others. What would this world be like if communication and trading of ideas didn't happen?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hi all,
I have a question that I have diffuculty in solving:

Let G be a finite group and H the subgroup generated by the elements of odd order. Show that H is normal and that the order of G/h is a power of 2.

Thanks

Let $E$ be the set of all elements of $G$ with odd order. Take $g \in G$, and consider $\Delta=gHg^{-1}$. $\Delta$ is the subgroup of $G$ generated by the set $gEg^{-1}$. But $gEg^{-1}\subseteq E$ since $\forall e \in E, geg^{-1}$ has same odd order than $e$. Whence $\Delta \subseteq H$, which proves the normality of $H$ in $G$.

Now if $p|(G: H)$ where $m$ is odd and prime, then $G/H$ contains a subgroup $D$ of order $m$, and the preimage of $D$ by the canonical projection $p: G \rightarrow G/H$ is a subgroup of $G$ with order $m|H|$. The $m$-Sylow of $G$ has odd order but is not included in $H$, which is absurd.


Let n be a positive integer and M be a set containing n^2 + 1 positive integers with the property that in each n+1 numbers from the set M there are two numbers for which one is divisible by the other. Prove that the set M contains different numbers a_1, ..., a_(n+1) for which a_(i+1) divides a_i for i = 1,...,n

For every element $a$ in M, give it a value k which is defined as the maximum integer such that there exists $a_1, a_2, ...,a_k$ such that
$
a_1<a_2<...<a_k<a$, and $a_1|a_2, a_2|a_3,...,a_k|a$
If $a$ is not divided by any other elment in M, its value is 0.

What we have to prove is that at least one element has value greater than or equal to n.
Suppose that all elements have value <n. Then there are only n possible values. Therefore for some non-negative integer c smaller than n we have n+1 elements with value c. Then the n+1 elements form a set that does not satisfy the given condition.

Here's a more general result (and it can be further generalized):

Given a sequence of $n^2+1$ elements from a partially ordered set, there's either a chain or an antichain with length $n+1$ (one small generalization: we can use $mn+1,n+1,m+1$ for our values).

Yes, and it'a been discussed before around here. A few related links have been posted by orl [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Dilworth&t=796]here[/url].

[quote="seshadri"][quote="grobber"]
Given a sequence of $n^2+1$ elements from a partially ordered set, there's either a chain or an antichain with length $n+1$ (one small generalization: we can use $mn+1,n+1,m+1$ for our values).[/quote]
yah,this is Dilworth's theorem-if every antichain in a poset has size $\leq m$, then the poset is a union of $m$ chains.[/quote]
If we consider numbers, it is Erdos's theorem.

Yes, Erdos-Szekeres, I believe. [url=http://mathworld.wolfram.com/DilworthsLemma.html]Here[/url]'s a link.


geometry
parallelogram
angle bisector
geometry unsolved

Let ABCD be a parallelogram with AB>BC.Q and P are points on line segments CD and AD respectively, such that CQ=AP.The intersection point of AQ and CP is S.Prove that S lies on the angle bisector of angle ABC.

I think I have come up with another solution which has not been posted before.  :)
Let the intersection of $BA$ and $CP$ be $X$, that of $BC$ and $AQ$ be $Y$ and that of $BS$ and $CD$ be $R$. It is easy to see that $\angle ABS=\angle CBS\Leftrightarrow RC=AD$.
Consider the transversal $SQY$ in $\triangle RBC$ and $PSC$ in $\triangle ADQ$ respectively. We get, by Menelaus' Theorem, that
$\frac{RS}{SB}\cdot\frac{BY}{YC}\cdot\frac{CQ}{QR}=1$ --- (1)
$\frac{AP}{PD}\cdot\frac{DC}{CQ}\cdot\frac{QS}{SA}=1$ --- (2)
Since $\frac{RS}{SB}=\frac{QS}{SA}$ and $\frac{BY}{YC}=\frac{AY}{YQ}=\frac{DC}{CQ}$, by comparing (1) and (2), we have $\frac{CQ}{QR}=\frac{AP}{PD}$. As $AP=CQ$, we now have $PD=QR$ and so $AD=RC$.


[quote="arbelos08"]Can anyone give an algebraic proof of why F= ma or  can it only be proven using laboratory experiments.[/quote]
You cant really prove laws from physics mathematically. You can sometimes provide explanation to why you expect it to be true mathematically, and in many cases by experiments. But overall there isnt a full rigorous mathematical proof to any of the laws of physics. (That is probably the reason why Hilbert asked, "can we axiomize physics").

There is no proof for the 3 Laws of Newton: they are just principles derived (and confirmed) from experimental observation, which are then used as axioms or first principles in the theory of Physics from which we can then derive logical conclusions. This is how some branches of science work. Another very important example is Thermodynamics, which is founded on four laws derived again from experimental observation (and which have been always confirmed until today): the theory of Thermodynamics on its applications to various systems are then logical conclusions that must follow from those four principles.

i won't tell u and pacify that there are proof s of the newton's law of motion
the Least action principle which agian is a axiom can be used to 'prove' newton's Laws
Again let me stress they are just a def,
have a look at [url=http://en.wikipedia.org/wiki/Least_Action_Principle]here[/url]

[quote="modeler"]Isn't force defined as the change of momentum?[/quote]

Thats impulse, IIRC.

And yes, F=ma is just an idea or concept which 'works' when we look at physics. I mean, we use maths to create physical models, not truths. Out of interest, isn't F=ma for certain situations (like slow speeds and reasonable gravity)

no any force is defined so as to follow this equation except third law which fails under certain situations


This is in a book. I dont know how to organize the information, and also i dont know how to use the info about a loss in the share price, please if you can help me.

An investor has 3 shares, Delta Airlines, Hilton Hotels and Mcdonalds. He says that 2 days ago the shares went down 350 dollars and yesterday they went up 600 dollars. His stock broker remembers that 2 days ago Delta lost 1 dollar per share, Hilton lost 1.5 dollars, and Mc gained 0.5 dollars. He also remembered that yesterday Delta gained 1.5 dollars, Hilton lost again 0.5 dollars and Mc gained 1 dollar. Is it possible to know with this info how many shares of each company has the investor?
If the investor says he has 200 shares of Mc, how many does he have of the other 2?

Thanks to anyone who can give me a hand!

Basically what you have here is a system of linear equations.  For the first day, you have -d - 1.5h + .5m = -350, and for the second day you have 1.5d - .5h + m = 600.  Unfortunately, we have three variables and only two equations, so we cannot solve this system.  But if we know he owns 200 shares of McDonald's, we can substitute this in and get -d - 1.5h + 100 = -350, or -d - 1.5h = -450, and 1.5d - .5h +200 = 600, or 1.5d - .5h = 400.  Solving the first equation for d, we get d=450 - 1.5h.  If we substitute this into the second equation, we have 675 - 2.25h - .5h = 400, which simplifies to -2.75h = -275, or h=100.  If we substitute this back into the first equation, we get -d - 150 = -450, from which we get d = 300. So he owns  [b]100[/b]shares of Hilton and [b]300[/b] shares of Delta.


Let $n\geq 3$ be an integer. Let also $P_1,P_2,...,P_n$ be different two-element-subsets of $M=\{1,2,...,n\}$, such that when for $i,j \in M , i\neq j$ the sets $P_i,P_j$ are not totally disjoint, then there is a $k \in M$ with $P_k = \{ i,j\}$.
Prove that every element of $M$ occurse in exactly $2$ of these subsets.

Just for completness, here goes the 'official' solution:

Let $a_i$ denote the number of sets $i$ is contained in.
Then $a_1+a_2+...+a_n=2n$, since every subset contains two different numbers.
We also have $\binom{a_1}{2} + \binom{a_2}{2} + ... + \binom{a_n}{2} \leq n$, since there are at most $n$ (=all) sets that are 'implied' by the given condition.
Now $(a_1^2-a_1) + (a_2^2-a_2) + ... + (a_n^2-a_n) \leq 2n$ and subtracting $3$ times the equality we get
$(a_1^2-4a_1) + (a_2^2-4a_2) + ... + (a_n^2-4a_n) \leq -4n \iff (a_1-2)^2 + (a_2-2)^2 + ... + (a_n-2)^2 \leq 0$
which gives immediatly $a_1=a_2=...=a_n=2$.


I read darij's post about majorization theory ( http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14975 ).  I think there must be a flaw, because by his definition $(1,1,0) \succ (\frac{2}{3}, \frac{2}{3}, \frac{2}{3})$ and $(\frac{2}{3}, \frac{2}{3}, \frac{2}{3}) \succ (0,1,1)$. So are permutations of an array considered when majorizing?


$E$ is euclidean space such that $E=E_{1}\oplus E_{2}$ just direct sum
$p$ orthogonal projection on $E_{1}$
$q$ orthogonal projection on $E_{2}$

$r=p+q$

prove that $0<det(r) \leq 1$
Study the case $det(r)=1$

Cute problem :).

$p$ and $q$ are symmetric and semipositive-definite, so then so must be their sum. If $(p+q)x=0$, then $px=qx=0$, meaning that $x\in\ker p\cap\ker q\Rightarrow x=0$, so $p+q$ is also invertible. From these two results we find that the eigenvalues of $p+q$ are positive.

We have $\mbox{Tr}(p+q)=\mbox{Tr}\ p+\mbox{Tr}\ q=\dim E_1+\dim E_2=n$, so the arithemtic mean of the eigenvalues of $p+q$ is $1$, meaning tht their product is at most $1$, by AM-GM. Equality occurs iff all the eigenvalues of $p+q$ are $1$. Since $p+q$ is symmetric and thus diagonalizable, this is equivalent to $p+q=I$, or, in other words, $q$ is the projection on $\ker p$, which, in turn, is equivalent to $V_2=V_1^{\perp}$.


Is it possible to become a successful researcher and tenured professor without having a degree from an elite college?

I don't know what will happen in the time between now and college, but I've decided that my friends (and especially my girlfriend of 2 years if we are still together by the time I enter college) mean more to me than "great success," but I would still like to have some semblance of success. I could most likely be accepted to an elite university, but I would prefer at this time to stay in state. I am probably going to attend CU Boulder, but might go elsewhere for graduate study; let's assume, however, that I am going to attend graduate school at CU as well.

Sorry if I made this difficult to understand.

Of course you can -- it's just up to you to get a good research idea. Then, even if your project is funding intensive and your university is not particularly well endowed, there are other options. If you can convince others that your idea is worth pursuing, then you can get grant money from outside sources.

That said, I don't think that where your friends or girlfriend choose to go should compel you to go to a particular college. It's not a good enough reason. In particular, if you and your girlfriend can't work out long distance, you won't work out long term (the distance would just make it end faster), but if you [i]do[/i] work out, then it will strengthen your relationship when you're done.

Additionally, you'll make great friends in college. I'll promise that.


Let $a,b,c,d$ be fixed integers with $d$ not divisible by $5$. Assume that $m$ is an integer for which \[am^{3}+bm^{2}+cm+d\] is divisible by $5$. Prove that there exists an integer $n$ for which \[dn^{3}+cn^{2}+bn+a\] is also divisible by $5$.

[hide="Solution"] Let $P(x) \in \mathbb{Z}_{5}[x] = ax^{3}+bx^{2}+cx+d$ which has some root $m$.

Then $Q(x) \in \mathbb{Z}_{5}[x] = dx^{3}+cx^{2}+bx+a$ has $\frac{1}{m}$ as a root (easy lemma).

[b]Case:[/b]  $m \not \equiv 0$.  Then $\frac{1}{m}$ exists $\bmod 5$.  QED.

[b]Case:[/b]  $m \equiv 0$.  Then $d \equiv 0$, contradiction. [/hide]

maybe a hint to the cool lemma? of course you can use the solutions to 3 degree equation but it's ugly...

[hide="Two lines"] Let $P(x) = a(x-r_{1})(x-r_{2}) ... (x-r_{n})$ be our first polynomial with some roots.

Then $Q(x) = a(r_{1}x-1)(r_{2}x-1)... (r_{n}x-1)$ has roots precisely the inverses of $P(x)$ and coefficients precisely the reverse (up to multiplication by $-1$).

Holds true in an arbitrary field (usually taken to be $\mathbb{R}$ or $\mathbb{C}$; in this case I apply in $\mathbb{Z}_{5}$)  [/hide]


I lend my friend a sum of money of RM 1000. Each month I will be getting a interest of an extra of 5%. Wat will be the total sum of money I will be geting in a 100 years?

Meh, it would be $ 1000(1.05)^{100}$, assuming you can live that long

yea...
he gets interest every month
so it would be...

[hide]100(1.05^1200)
that's 2.67398449*10^27
in other words:
2000000000000000000000000000
about[/hide]
That's a big number
YOU GONNA EARN A LOT OF MONEY!!!

Isn't the compound interest equation $ A \equal{} P(1 \plus{} \frac {r}{m})^{mt}$
where A=Accumulated Money, P=Pricipal, r=rate, m=compundings/year, and t is the number of years.
If so, we have $ A \equal{} 1000(1 \plus{} \frac {.05}{12})^{1200}$. That simplifies to $ \boxed{\$146,879.45}$, way different than your answer.

[quote="thacheezinator"]Mr. Rubix...
Your answer may have worked, but you have to think that over 1200 years of compound interest.  Using the rule of 70, with 5%, his account would double every 14 years.  Over 1200 years, his account would double over 85 times!  If you put this into the calculator (1000*2^85), you would come up with a huge answer.  Even if my answer is off, it can't be off by more than a quintillion...  I did the same thing on another problem, and it worked out perfectly.  I'm pretty sure i'm right, but you can check it if you want to.  Also, this is compound interest, isn't it :huh: ?  If it was simple interest, it would be something like 61000.  what is RM? is it a type of money?

Also, wouldn't the rate be 1.05, and not .05?

 :ninja:[/quote]
Three things:
1) What is the "rule of 70"?
2) I think the problem said 100 years, so it would be $ 1000*2^{\frac{100}{14}}$, which is $ \$141,323.46$ on a calculator. This is very close to my answer of $ \$146,879.45$
3) The reason why I used $ .05$ instead of $ 1.05$ is because we have 5 percent rate of interest, OR because we already added 1:
$ P(1\plus{}\frac{r}{m})^{mt}$

The problem says nothing about annual interest, so you don't divide by 12 :wink:

my fren say is like something (interest)of the powerof x years and he suddenly say he forget d......


[quote="lomos_lupin"]let $a_1,a_2,a_3,...,a_n$ be positive real number. prove

$(\sum_{i=1}^n a_i)(\sum_{i=1}^n a_i^{n-1}) \leq n \prod_{i=1}^n a_i +(n-1)\sum_{i=1}^n a_i^n$[/quote]

This is the Suranyi's inequality. Harazi has given a proof before..

Is this a generalization of Turkevici's inequality?

can some one post the link of mr.harazi solution thanks

[quote="Gil"]This problem can be found in book "Old and New Inequalities" at page 22, problem 116. The book was published by GIL Publishing House in 2004, with authors: T. Andreescu, V. Cirtoaje, G. Dospinescu, M. Lascu.

 :lol:[/quote]
Who can show me the solution of this problem ???
I have'nt this book !!!!!!!

[quote="Soarer"][quote="silouan"][\quote="Gil"]This problem can be found in book "Old and New Inequalities" at page 22, problem 116. The book was published by GIL Publishing House in 2004, with authors: T. Andreescu, V. Cirtoaje, G. Dospinescu, M. Lascu.

 [/quote]
where  could someone find this book? Can I found it at Internet?[/quote]
Yes. Find the website of GIL publishing house and buy it.[/quote]
But I can't buy it !!!!
Can you show me the solution??? :blush:


function
real analysis
real analysis unsolved

Find the real functions f having the intermediate value property ( Darboux) such that $ f(x)=f(x+f(x))$.

Delete this!Q I have already received enough thoughts of the mathlinkers!

It means that before the forum crashed, Jmerry gave a counterexample to the fact that the only such functions are constant. The example was quite nasty, and showed that it's possibly impossible for us to actually find all such functions.


Of the 200 students at Lakeview High School, 170 are taking science and 175 are taking mathematics. What is the fewest number of students at Lakeview who could be taking both?


linear algebra
matrix
combinatorics unsolved
combinatorics

There is a party with $ 12k$ people and every one has exactly $ 3k\plus{}6$ friends. Moreover, every pair of people has the same number of friends in common. Find all natural numbers $ k$ for which the party is possible and for that value give an example of that party,

Daniel

Let $ G(V,E)$ be the graph with the set of vertices $ V$ represent the set of $ 12k$ people and an edge is connected between two vertices if and only if the corresponding two people are friends.  For each $ v\in{V}$, let $ N(v)$ be the set of all vertices adjacent to $ v$.  Suppose that every pair of people has $ d$ friends in common.

Let $ A$ be the adjacency matrix of $ G$.  Thus, if $ \bar{1} : \equal{} (1,1,\dots,1)$, then
\[ A\bar{1} \equal{} (3k \plus{} 6)\bar{1}\,.
\]
Note that the $ (u,v)$-entry of $ A^2$ is the size of $ N(u)\cap{N(v)}$.  If $ u \neq v$, then $ \left|N(u)\cap{N(v)}\right| \equal{} d$.  That is,
\[ A^2 \equal{} dJ \plus{} (3k \plus{} 6 \minus{} d)I\,,
\]
where $ I$ is the identity matrix and $ J$ is the all-one matrix.
Now,
\[ (3k \plus{} 6)^2\bar{1} \equal{} A^2\bar{1} \equal{} dJ\bar{1} \plus{} (3k \plus{} 6 \minus{} d)\bar{1} \equal{} (12kd \plus{} (3k \plus{} 6 \minus{} d))\bar{1}\,.
\]
This shows that
\[ (3k \plus{} 5)(3k \plus{} 6) \equal{} (12k \minus{} 1)d\,.
\]
Since $ \gcd(12k \minus{} 1,(3k \plus{} 5)(3k \plus{} 6))$ is a divisor of $ 175$, therefore, $ 12k \minus{} 1$ must be a divisor of $ 175$ (so that $ d$ would be an integer).  Hence, $ 12k \minus{} 1 \equal{} 1$, $ 5$, $ 7$, $ 25$, $ 35$, or $ 175$.  The only possible case is when $ 12k \minus{} 1 \equal{} 35$ or $ k \equal{} 3$.

Consequently, $ 12k \equal{} 36$, $ 3k \plus{} 6 \equal{} 15$, and $ d \equal{} 6$.  That is, $ G$ is a strongly regular graph $ \mathrm{srg}(\nu,\kappa,\lambda,\mu)$ with $ \nu \equal{} 36$, $ \kappa \equal{} 15$, and $ \lambda \equal{} \mu \equal{} 6$.  There are $ 32548$ examples of such graphs.


I see this problem was posted and no one gave a solution to it.

* Three straws are chosen from a set of nine straws whose lengths are 1 inch, 2 inches, 3 inches, . . . , 9 inches. What is the probability that the three straws, placed end to end, will form a triangle?

I posted the question. But no one answered  :rotfl:

You can choose 3 straws from 9 in 9C3 ways = 84

There are 50 invalid triangles ( 28+15+6+1 )

Probability of making a valid triangle = 34/84 = 17/42

Is it correct?  :blush:

Number of Invalid Traingles
===================
1,2 - 7
1,3 - 6
1,4 - 5
1,5 - 4
1,6 - 3
1,7 - 2
1,8 - 1
Total = 28

2,3 - 5
2,4 - 4
2,5 - 3
2,6 - 2
2,7 - 1
Total = 15

3,4 - 3
3,5 - 2
3,6 - 1
Total = 6

4,5 - 1
Total = 1

From what I can tell, your solution is correct. I would have probably stated some things differently, but your solution is fine.


geometry
circumcircle
geometric transformation
geometry unsolved

A triangle $ ABC$ is given. A point $ R$ belongs to the line $ AC$ and $ C$ is between points $ A,R$. $ R$ belongs to the line intersecting $ AB$ at $ C_1$ and $ BC$ at $ A_1$. Let $ P$ be the midpoint of $ AC$, and $ Q$ be the midpoint of $ A_1C_1$. Prove that three circles, circumscribed around triangles $ ABC,A_1BC_1,PQR$, are concurrent.

I think it's a really nice problem, thank you for it [b]massnet[/b] :lol:

Let $ Z$ be the point other than $ B$ which is common for the circumcenters of triangles $ ABC$ and $ A_{1}C_{1}B$. Then we have $ \angle C_1ZA_1 \equal{} \angle ABC \equal{} \angle AZC$. There is also $ \angle C_1A_1Z \equal{} \pi \minus{} \angle C_1BZ \equal{} \angle ACZ$. So triangles $ C_1ZA_1$ and $ AZC$ are similar. Thus $ \angle A_1C_1Z \equal{} \angle CAZ$ and so we see that quadrilateral $ AC_1ZR$ is circumscrible.

Now lets consider a spiral similarity with center Z taking triangle $ C_1ZA_1$ into triangle $ AZC$. We see now that $ \angle C_1ZA \equal{} \angle QZP$. But since quadrilateral $ ZC_1AR$ is circumscrible, we have $ \angle C_1ZA \equal{} \angle C_1RA$. So there is also $ \angle C_1RA \equal{} \angle QZP$ so the quadrilateral $ ZQPR$ is circumscrible. And with this we're done.


function
trigonometry
limit
geometry
geometric transformation
reflection

something which does not exit how can we say their reciprocal is 0.looks very odd. i need your help on the value of cot pi/2.it is zero but tan pi/2 is not defined.

$ \cot x$ is a function which is defined as $ \frac {\cos x}{\sin x}, \sin x\neq 0$.

$ \tan x\cdot\cot x \equal{} 1$ holds only when $ \sin x\cos x\neq 0$.

And as $ \cos \frac{\pi}{2}\equal{}0$, the formula above doesn't hold (for the product of tan and cot)

This argument may or may not be a "high school intermediate" topic, but since $ \displaystyle\lim_{x\to\frac {\pi}{2}}\tan{x} = - \infty$, you can take the reciprocal and say ${ \cot{\frac{\pi}{2}}}=0$, which I guess is enough for mathematicians.

cotangent and tangent are reciprocals, but only on their domains.
So it is true that $ \tan{\frac{\pi}{2}}$ is undefined. And Stank, $ \cot{\frac{\pi}{2}}\equal{}0$, not just the limit.

There is a totally sensible way to make sense of $ \frac{1}{0}$ if you're willing to drop the field axioms and just think of $ \mathbb{R}$ as a topological space.  Then it has what's called a [url=http://en.wikipedia.org/wiki/One-point_compactification]one-point compactification[/url] which can be identified with the unit circle in $ \mathbb{R}^2$ via projection.  Taking inverses now corresponds to a certain reflection of this circle, and the point opposite $ 0$ is the point $ \infty$ added by the compactification.

$ \infty$ is a meaningful symbol, and you can do some computations with it. $ \frac1{\infty}\equal{}0$ is one of them.


vector
geometry
circumcircle
incenter
analytic geometry
geometry open

$ \triangle ABC$ with circumcircle $ (O)$ , incenter $ I$
$ D$, $ E$, $ F$ = midpoints of $ BC$, $ CA$, $ AB$
$ \overrightarrow {OD}\cap (O)$=$ K$,$ \overrightarrow {OE}\cap (O)$=$ L$,$ \overrightarrow {OF}\cap (O)$=$ M$
Prove that $ \overrightarrow {OI}$=$ \overrightarrow {OK}$+$ \overrightarrow {OL}$+$ \overrightarrow {OM}$

Is the proof easy with no coordinates ? I forget the proof without coordinate   :blush:  thank you

Just now I got idea.
$ I$ is the orthocenter of $ \triangle KLM$, and almost obvious .
Ashamed of myself  :blush:

Oh, that's great!! :cool:
$ A$, $ I$, $ K$ are collinear.
$ B$, $ I$, $ L$ are collinear.
$ C$, $ I$, $ M$ are collinear.
It is simple setting, but is interesting! :lol:

$ J_A$ is the excenter  corresponding to $ A$.
$ J_B$ is the excenter  corresponding to $ B$.
$ J_C$ is the excenter  corresponding to $ C$.

$ J_A$, $ B$, $ C$, $ I$ are collinear.  ($ K$ is circumcenter.)
$ J_B$, $ C$, $ A$, $ I$ are collinear.  ($ L$ is circumcenter.)
$ J_C$, $ A$, $ B$, $ I$ are collinear.  ($ M$ is circumcenter.)

[url=http://mathworld.wolfram.com/Incenter-ExcenterCircle.html]Incenter-Excenter Circle[/url] at MathWorld

Dear Mathlinkers,
the relation proposed by "77ant" and adapted for a triangle, has been discovered by Euler.
Any reference?
Sincerely
Jean-Louis

cf. [url=http://www.mathlinks.ro/viewtopic.php?t=223814]on the line passing through the incenter and circumcenter[/url]


A point N is chosen on the longest side AC of triangle ABC. Perpendicular
bisectors of AN and NC intersect AB and BC at K and M
respectively. Prove that the circumcentre O of ABC lies on the circle
which passes through K,B,M.

[asy]size(200);defaultpen(fontsize(8));
pair A=(0,0), B=(10,8), C=(14,0), N=(8,0), D=midpoint(B--C),E=midpoint(A--C),F=midpoint(A--B), O=circumcenter(A,B,C), K=(4,16/5), M=(11,6);
draw(circumcircle(A,B,C));draw(A--B--C--cycle);draw(O--D..O--E..O--F);draw(O--K--N--M--cycle);
draw(K--foot(K,A,C));draw(M--foot(M,A,C));
dot(A^^B^^C^^D^^E^^F^^O^^N^^K^^M);
label("A",A,(-1,-1));label("B",B,(1,1));label("C",C,(1,-1));
label("D",D,(1,1));label("E",E,(0,-1));label("F",F,(-1,1));
label("K",K,(-1,1));label("M",M,(1,1));label("N",N,(0,-1));label("O",O,(0,2));
label("$\frac{b}{2}-x$",(A+midpoint(A--N))/2,(0,-1));
label("$x$",E/2+midpoint(A--N)/2,(0,-1));
label("$\frac{b}{2}-2x$",E/2+N/2,(0,-3));
label("$x$",N/2+midpoint(C--N)/2,(0,-1));
label("$x$",C/2+midpoint(C--N)/2,(0,-1));[/asy]We want to prove that $ BMOK$ is cyclic, or that $ \angle OMD\equal{}\angle OKF$. We will show that $ \triangle OKF \sim \triangle OMD$.
$ AF\equal{}\frac{c}{2}, AK\equal{}\frac{\frac{b}{2}\minus{}x}{\cos{A}}\implies KF\equal{}\frac{c}{2}\minus{}\frac{\frac{b}{2}\minus{}x}{\cos{A}}$
$ CD\equal{}\frac{a}{2}, CM\equal{}\frac{x}{\cos{C}}\implies MD \equal{} \frac{x}{\cos{C}}\minus{}\frac{a}{2}$
$ OF\equal{}R\cos{C},OD\equal{}R\cos{A}$

$ \frac{KF}{OF}\equal{}\frac{MD}{OD}$
$ \frac{\frac{c}{2}\minus{}\frac{\frac{b}{2}\minus{}x}{\cos{A}}}{R\cos{C}}\equal{}\frac{\frac{x}{\cos{C}}\minus{}\frac{a}{2}}{R\cos{A}}$
$ \frac{c}{2}\cos{A}\minus{}\frac{b}{2}\plus{}x\equal{}x\minus{}\frac{a}{2}\cos{C}$
$ c\cos{A}\plus{}a\cos{C}\equal{}b$


AMC
AIME
USA(J)MO
USAMO
trigonometry
number theory

Depends on your idea of a bad score.  Most people would consider getting to the AIME a big enough accomplishment.  For others 5 is bad and for some they won't take anything but double digits.  Assuming that you are talking about a score in the  0-3 range I would say that a 6 is a reasonably good goal.  Now to qualify for the USAMO it would probably have to be around 7-10.  Anything from 7 or 8 and above I think is unreasonable unless all that you do during the year is study( but then again I wouldn't be suprised if quite a few people on this site do that).

Studying is one thing, but all that jazz comes so much easier if you really enjoy it. Math is fun! You don't have to spend every last moment studying, as long as you understand what you do. I think 6 - 7 is quite reasonable. Last year I kept trying old tests and I kept getting 3's, but over the summer I have brushed up on number theory and trig, and my score went up to 8 on practice tests. I hope for around a 12 on the real test. Too much?

Of course practicing old tests are a good method, but that's not the only part to improving your score. Try to find your weakness and improve on that.


[hide="Brute Force!"]
$ a(1 \minus{} b^2)(1 \minus{} c^2) \plus{} b(1 \minus{} c^2)(1 \minus{} a^2) \plus{} c(1 \minus{} b^2)(1 \minus{} a^2) \minus{} 4abc$

$ \equal{} a \plus{} b \plus{} c \minus{} a^2b \minus{} ab^2 \minus{} a^2c \minus{} ac^2 \minus{} b^2c \minus{} bc^2 \plus{} abc(ab \plus{} bc \plus{} ca) \minus{} 4abc$

$ \equal{} (a \minus{} a^2b \minus{} a^2c \minus{} abc) \plus{} (b \minus{} b^2a \minus{} b^2c \minus{} abc) \plus{} (c \minus{} c^2a \minus{} c^2b \minus{} abc) \plus{} abc(ab \plus{} bc \plus{} ca \minus{} 1)$

$ \equal{} a(1 \minus{} ab \minus{} bc \minus{} ca) \plus{} b(1 \minus{} ab \minus{} bc \minus{} ca) \plus{} c(1 \minus{} ab \minus{} bc \minus{} ca) \plus{} abc(ab \plus{} bc \plus{} ca \minus{} 1)$

$ \equal{} (ab \plus{} bc \plus{} ca \minus{} 1)(abc \minus{} a \minus{} b \minus{} c)$
[/hide]


Show that for each positive integer $ n$, there is a positive integer $ k$ such that decimal representation of each of the numbers $ k, 2k, \dots, nk$ contains all the digits $ 0,1,2,3,4,5,6,7,8,9$.

Just set $ \displaystyle k\equal{}\sum_{1 \le i \le 10^2}{i10^{in^2}}$, see [url=http://bboyjordan.wordpress.com/2009/10/31/every-number-m2m-nm-has-all-digits/]here[/url] for more details  :P


Again, I don't know what the answer is and I don't know whether this is the right level or not.
Given a non-perfect square positive integer, we take its square root.  We then subtract its integer part and take the reciprocal of that.  If we repeat these steps, do we get a cycle for all square rooted integers that satisfy the basic requirements?  Is it true for cubics, quartics,...?

I might be wrong, but yes, i think all roots give an loop. If we have a loop, we can maniputale it so that the next loop replaces the first one, set it equal to the first thingy, and we have a polynomial. I think we can get what ever polynomial we desire, so we get all roots.

No that's not true, for example if you take $2^{1/3}$ the "simple"  continued fraction would go something like:
$\frac 1 {1+} \frac 1 {3+} \frac 1 {1+} \frac 1 {5+} \frac 1 {1+}$ and it would [b] not[/b] go in a cycle. ... Yes square  roots are special that way :)

There is something called "generalized continued fractions" which will show  periodicity  for any root of n degree equation, but those look quite different. (You do not have '1' in the numerator)

Hope this helps.


I'm trying to post a blog entry with two YouTube links, and it only displays the second one and none of the text in between the two links. I checked carefully; I'm pretty sure there is no syntax error. (When I messed up the first tag, it displayed as expected; when I messed up the second tag [changed the first "youtube" to "ayoutube", it still had the same behavior as actually appears in the post with no typo.)

I posted it anyway (with the original text of the post hidden, although the same thing happened when it was not hidden); it's my latest entry.

I will see what's going on very soon (for now I have [b]very[/b] limited access to internet in Vietnam :( ).


farz konid a,b,c,d adaade sahihe mosbati baashand & ad=b^2+c^2+bc
saabet konid a^2+b^2+c^2+d^2 avval nist.

gharar bedid $ S=a^{2}+b^{2}+c^{2}+d^{2}$ darim:

$ S=(a+d)^{2}+(b-c)^{2}-2ad+2bc=(a+d)^{2}+(b-c)^{2}-2(ad-bc)$

amma tebghe sharte masale dashtim:$ ad-bc=b^{2}+c^{2}$ pas age ino jagozari konim darim:

$ S=(a+d)^{2}+(b-c)^{2}-2(b^{2}+c^{2})$

hamchenin darim:$ 2(b^{2}+c^{2})=(b+c)^{2}+(b-c)^{2}$

ke age ino jagozari konim be das miad:

$ S=(a+d)^{2}-(b+c)^{2}=(a+b+c+d)(a+d-b-c)$

amma midoonim $ a+b+c+d>1$ pas age $ S$ bekhad aval bashe paranteze dovom bayad barabare $ 1$ bashe yeni khahim dasht:

$ S=a+b+c+d \Rightarrow a^{2}+b^{2}+c^{2}+d^{2}=a+b+c+d \Rightarrow \boxed{a=b=c=d=1}$

ke inam ba sharte $ ad=b^{2}+c^{2}+bc$ dar tanaghoze...

pas $ S$ nemitoone aval bashe...
[hide="ellate inke natije shod a=b=c=d=1 ?"]
be das avordim $ a^{2}+b^{2}+c^{2}+d^{2}=a+b+c+d$
amma tebghe miangine morabai hesabi darim:
$ \sqrt{\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}}\geq \frac{a+b+c+d}{4}$
pas darim:
$ \sqrt{\frac{S}{4}}\geq \frac{S}{4}$
$ \Rightarrow 4\geq S$
$ \Rightarrow a+b+c+d\leq 4$
amma $ a,b,c,d\geq 1$ pas darim:
$ 4\leq a+b+c+d\leq 4 \Rightarrow a+b+c+d=4 \Rightarrow a=b=c=d=1$[/hide]


For what b and c, $x=sqrt19+sqrt98$ is a root of $x^4+bx^2+c=0$

If $\sqrt{19}+\sqrt{98}$ is a root, then $\pm\sqrt{19}\pm\sqrt{98}$ are roots.
Expanding $(x-\sqrt{19}-\sqrt{98})(x-\sqrt{19}+\sqrt{98})(x+\sqrt{19}-\sqrt{98})(x+\sqrt{19}+\sqrt{98})$, we get the polynomial $x^4-234x^2+6241$.

EDIT: I have many multiplication issues.

Squaring both sides of $x-\sqrt{19}=\sqrt{98}$, we have $x^2-79=2\sqrt{19}x$.Again squaring, we get $x^4-234x^2+6241=0$,yielding $b=-234,c=6241$


inequalities
calculus
integration
geometry
perimeter
triangle inequality

How many scalene triangles have all sides of integral lengths and perimeter less than $13$?

The perimeter must be [b]less than[/b] 13.

[quote="rnwang2"]How do you brute-force this in a systematic way?[/quote]

[hide]
Well first, you realize if $a=1$ the triangle has to be isoceles or equilateral so that doesn't work. When $a=2$ the smallest is 2, 3, 4 which adds up to 9. The difference between 3 and 4 can't exceed 1 or else triangle inequality wouldn't be satisfied. So you add 2 to get 2, 4, 5, which adds up to 11 and can't add anymore. Then you try $a=3$ or 3, 4, 5. That is already at 12 so you can't add anymore and you have 3.[/hide]


i dunno if this is the appropriate forum for this, but...

my math team coach is freaking out because she didnt receive the mandelbrot contests (for the week of 10/17) in the mail yet.  she knows they have a mailing address, but is wondering if they have an email address or some other way they can be contacted.

thanks in advance.

http://www.mandelbrot.org/

Click FAQ'S on lower left

then click Dialogue on upper right


In triangle $ABC$, $\tan \angle CAB = 22/7$, and the altitude from $A$ divides $BC$ into segments of length 3 and 17.  What is the area of triangle $ABC$?

let the lentgh of altitute be x; notice that the altitute divides the angle CAB at two different angle, and tangent of these angle are both 17/x and 3/x. using the tangent sum formula we got: 22/7 = (17/x+3/x)/(1-(17/x)(3/x)) multiplying we got:
 140/x=22-1122/x^2 multiplying x^2 at both side, and then pass everything to left side we get:
22x^2-140x-1122=0
resolve that equation we get the altitute, and as we already got the base which is 17+3=20, we can easily get the area.

(sorry for poor english)

[hide="my solution"]ok i used b instead of x and got 22/7 = (20/b) / [(b^2-51)/b^2]
of course then i got 22/7 = 20b^2 / (b^3 - 51b)
then 11b^3 - 70 b^2 - 561b = 0
i got b = 11
so area is
20 * 11 / 2 = 110[/hide]

[hide="Solution"]Let the altitude from $A$ meet $BC$ at $H$ such that $BH=17,$ $CH=3,$ and $AH=x.$ Since we're given two right triangles and the tangent of an angle, it seems like a good idea to use trig. By the tangent addition formula, \begin{eqnarray*}\tan \angle BAC &=& \tan (\angle BAH+\angle CAH)\\ &=& \cfrac{\frac{17}{x}+\frac{3}{x}}{1-\left(\frac{17}{x}\right)\left(\frac{3}{x}\right)}\\ \\ &=& \frac{20x}{x^{2}-51}= \frac{22}{7}. \end{eqnarray*} Solving, $x=11$ and the area of triangle $ABC$ is $\frac{1}{2}(BC)(AH)=110.$[/hide]

A slightly different way to look at this problem is to note the identity:

$ \tan A\plus{}\tan B\plus{}\tan C\equal{}\tan A\tan B\tan C$. From this, we get

$ \frac{22}{7}\plus{}\frac{h}{3}\plus{}\frac{h}{17}\equal{}\frac{22h^{2}}{7*3*17}\iff 22*3*17\plus{}h*20*7\equal{}22h^{2}$. The answer is $ \frac{1}{2}*h*20$. Note that $ 11|h$, checking this works. So the answer is $ 110$.


inequalities
geometry
3D geometry
inequalities open

Let x = (cube root of 1000) - (cube root of 999). what integer is closest to 1/x?

1/x=(cube root of 1000)^2 + (cube root of 999)(cube root of 1000)+(cube root of 999)^2

100+(cube root of 999)(10+(cube root of 999))

I claim it is closest to 300

so (cube root of 999)(10+(cube root of 999)) is $ \ge 9.99*19.99\ge199.7>199.5$

so $ 300\le 1/x>299.5$

All that has to be shown is that (cube root of 999) is > 9.99 (not too hard)
Q.E.D.


prove there  isn't exist $x,y,z \in Z^+$ and$p\in P;p=3(mod 4)$
         $x^4+py^4=z^4$

Find all$x,y,z \in Z^+$ suchthat
   $x^4+py^4=z^4$     $(p is a prime number ,p=3(mod 4))$


calculus
derivative
calculus computations

If $ z \equal{} f(x,y)$   and   $ x \plus{} y \equal{} (u \plus{} v)^{3}$   and   $ x \minus{} y \equal{} (u \minus{} v)^{3}$ ,
then show that
$ 9(x^{2} \minus{} y^{2}) \cdot (\frac {\partial^{2} z}{\partial x^{2}} \minus{} \frac {\partial^{2} z}{\partial y^{2}}) \equal{} (u^{2} \minus{} v^{2}) \cdot (\frac {\partial^{2} z}{\partial u^{2}} \minus{} \frac {\partial^{2} z}{\partial v^{2}})$

 :coolspeak:

OK
 After keep on trying about an hour, finally I got the answer.

Solution :
solving , $ x + y = (u + v)^{3} , x - y = (u - v)^{3}$

$ x = \frac {(u + v)^{3} + (u - v)^{3}}{2}$ and

$ y = \frac {(u + v)^{3} - (u - v)^{3}}{2}$

$ \frac {\partial z}{\partial u} = \frac {\partial z}{\partial x} \cdot \frac {\partial x}{\partial u} + \frac {\partial z}{\partial y} \cdot \frac {\partial y}{\partial u}$

      $ = 3\cdot \frac {(u + v)^{2} + (u - v)^{2}}{2}\cdot \frac {\partial z}{\partial x} + 3\cdot \frac {(u + v)^{2} - (u - v)^{2}}{2}\cdot \frac {\partial z}{\partial y}$

Let us say that
 $ 3\cdot \frac {(u + v)^{2} + (u - v)^{2}}{2} = k_1$ and
 $ 3\cdot \frac {(u + v)^{2} - (u - v)^{2}}{2} = k_2$

$ \frac {\partial z}{\partial u} = k_1 \cdot \frac {\partial z}{\partial x} + k_2 \cdot \frac {\partial z}{\partial y} ______________________ (1)$

or  $ \frac {\partial }{\partial u} = k_1 \cdot \frac {\partial }{\partial x} + k_2 \cdot \frac {\partial }{\partial y}$

Similarly we get,

$ \frac {\partial z}{\partial v} = k_2 \cdot \frac {\partial z}{\partial x} + k_1 \cdot \frac {\partial z}{\partial y} ______________________ (2)$

or  $ \frac {\partial }{\partial v} = k_2 \cdot \frac {\partial }{\partial x} + k_1 \cdot \frac {\partial }{\partial y}$

solving $ (1)$ and $ (2)$, we get,

$ \frac {\partial z}{\partial x} = \frac { k_1 \cdot \frac {\partial z }{\partial u} - k_2 \cdot \frac {\partial z }{\partial v}}{k^{2}_{1} - k^{2}_{2} }$

$ \frac {\partial z}{\partial y} = \frac { k_1 \cdot \frac {\partial z}{\partial v} - k_2 \cdot \frac {\partial z}{\partial u}}{k^{2}_{1} - k^{2}_{2}}$

$ \frac {\partial }{\partial x}(\frac {\partial z}{\partial x}) = ( \frac {k_{1}\cdot \frac {\partial }{\partial u} - k_{2}\cdot \frac {\partial}{\partial v}}{k^{2}_{1} - k^{2}_{2}}) \cdot ( \frac {k_{1}\cdot \frac {\partial z}{\partial u} - k_{2}\cdot \frac {\partial z}{\partial v}}{k^{2}_{1} - k^{2}_{2}})$

therefore,

$ = \frac {k^{2}_{1}\frac {\partial^{2}}{\partial u^{2}} - 2k_{1}k_{2}\frac {\partial^{2}z}{\partial x \partial y} + k^{2}_{2}\frac {\partial^{2}z}{\partial v^{2}}}{(k^{2}_{1} - k^{2}_{2})^2} ________________________________________(3)$

similarly,

$ \frac {\partial }{\partial y}(\frac {\partial z}{\partial y}) = \frac {k^{2}_{2}\frac {\partial^{2}}{\partial u^{2}} - 2k_{1}k_{2}\frac {\partial^{2}z}{\partial x \partial y} + k^{2}_{1}\frac {\partial^{2}z}{\partial v^{2}}}{(k^{2}_{1} - k^{2}_{2})^2} ________________________________________(4)$

$ (3) - (4)$ gives ,

${ \frac {\partial^{2}z}{\partial x^{2}} - \frac {\partial^{2}z}{\partial y^{2}} = \frac {1}{(k^{2}_{1} - k^{2}_{2})^2} \cdot ((k^{2}_{1} - k^{2}_{2})\cdot \frac {\partial^{2}z}{\partial u^{2}} - (k^{2}_{1} - k^{2}_{2})\cdot \frac {\partial^{2}z}{\partial v^{2} }})$

but, $ k^{2}_{1} - k^{2}_{2} = 9 \cdot (u^{2} - v^{2})^{2}$
and $ x^{2} - y^{2} = (u^{2} - v^{2})^3$.

hence,

$ 9(x^{2} - y^{2})\cdot (\frac {\partial^{2}z}{\partial x^{2}} - \frac {\partial^{2}z}{\partial y^{2}}) = (u^{2} - v^{2})\cdot (\frac {\partial^{2}z}{\partial u^{2}} - \frac {\partial^{2}z}{\partial v^{2}})$

It's lengthy if any body has shorter solution plz post it.

[quote=Anandiscool]OK
 After keep on trying about an hour, finally I got the answer.

Solution :
solving , $ x + y = (u + v)^{3} , x - y = (u - v)^{3}$

$ x = \frac {(u + v)^{3} + (u - v)^{3}}{2}$ and

$ y = \frac {(u + v)^{3} - (u - v)^{3}}{2}$

$ \frac {\partial z}{\partial u} = \frac {\partial z}{\partial x} \cdot \frac {\partial x}{\partial u} + \frac {\partial z}{\partial y} \cdot \frac {\partial y}{\partial u}$

      $ = 3\cdot \frac {(u + v)^{2} + (u - v)^{2}}{2}\cdot \frac {\partial z}{\partial x} + 3\cdot \frac {(u + v)^{2} - (u - v)^{2}}{2}\cdot \frac {\partial z}{\partial y}$

Let us say that
 $ 3\cdot \frac {(u + v)^{2} + (u - v)^{2}}{2} = k_1$ and
 $ 3\cdot \frac {(u + v)^{2} - (u - v)^{2}}{2} = k_2$

$ \frac {\partial z}{\partial u} = k_1 \cdot \frac {\partial z}{\partial x} + k_2 \cdot \frac {\partial z}{\partial y}(1)$

or  $ \frac {\partial }{\partial u} = k_1 \cdot \frac {\partial }{\partial x} + k_2 \cdot \frac {\partial }{\partial y}$

Similarly we get,

$ \frac {\partial z}{\partial v} = k_2 \cdot \frac {\partial z}{\partial x} + k_1 \cdot \frac {\partial z}{\partial y} (2)$

or  $ \frac {\partial }{\partial v} = k_2 \cdot \frac {\partial }{\partial x} + k_1 \cdot \frac {\partial }{\partial y}$

solving $ (1)$ and $ (2)$, we get,

$ \frac {\partial z}{\partial x} = \frac { k_1 \cdot \frac {\partial z }{\partial u} - k_2 \cdot \frac {\partial z }{\partial v}}{k^{2}_{1} - k^{2}_{2} }$

$ \frac {\partial z}{\partial y} = \frac { k_1 \cdot \frac {\partial z}{\partial v} - k_2 \cdot \frac {\partial z}{\partial u}}{k^{2}_{1} - k^{2}_{2}}$

$ \frac {\partial }{\partial x}(\frac {\partial z}{\partial x}) = ( \frac {k_{1}\cdot \frac {\partial }{\partial u} - k_{2}\cdot \frac {\partial}{\partial v}}{k^{2}_{1} - k^{2}_{2}}) \cdot ( \frac {k_{1}\cdot \frac {\partial z}{\partial u} - k_{2}\cdot \frac {\partial z}{\partial v}}{k^{2}_{1} - k^{2}_{2}})$

therefore,

$ = \frac {k^{2}_{1}\frac {\partial^{2}}{\partial u^{2}} - 2k_{1}k_{2}\frac {\partial^{2}z}{\partial x \partial y} + k^{2}_{2}\frac {\partial^{2}z}{\partial v^{2}}}{(k^{2}_{1} - k^{2}_{2})^2}(3)$

similarly,

$ \frac {\partial }{\partial y}(\frac {\partial z}{\partial y}) = \frac {k^{2}_{2}\frac {\partial^{2}}{\partial u^{2}} - 2k_{1}k_{2}\frac {\partial^{2}z}{\partial x \partial y} + k^{2}_{1}\frac {\partial^{2}z}{\partial v^{2}}}{(k^{2}_{1} - k^{2}_{2})^2} (4)$

$ (3) - (4)$ gives ,

${ \frac {\partial^{2}z}{\partial x^{2}} - \frac {\partial^{2}z}{\partial y^{2}} = \frac {1}{(k^{2}_{1} - k^{2}_{2})^2} \cdot ((k^{2}_{1} - k^{2}_{2})\cdot \frac {\partial^{2}z}{\partial u^{2}} - (k^{2}_{1} - k^{2}_{2})\cdot \frac {\partial^{2}z}{\partial v^{2} }})$

but, $ k^{2}_{1} - k^{2}_{2} = 9 \cdot (u^{2} - v^{2})^{2}$
and $ x^{2} - y^{2} = (u^{2} - v^{2})^3$.

hence,

$ 9(x^{2} - y^{2})\cdot (\frac {\partial^{2}z}{\partial x^{2}} - \frac {\partial^{2}z}{\partial y^{2}}) = (u^{2} - v^{2})\cdot (\frac {\partial^{2}z}{\partial u^{2}} - \frac {\partial^{2}z}{\partial v^{2}})$

.[/quote]


My school is taking the AIME II instead of the I due to spring break,
but I still want to take the I test (just for practise purposes :P) unofficially.

Is this possible? or should I just wait until somebody posts the questions after taking the test?

thx.

The questions will definitely be up within a few days afterwards, you'll have plenty of time to take it for practice.

Just avoid looking at any threads right after this forum is unlocked that might spoil any of the questions before you get a chance to take it.

[quote="worthawholebean"]I will be on break, so I'll have the questions up *very* soon after the forum reopens.[/quote]

How come you're the only mod who does this?

I've taken it upon myself to do so. Other people either don't have time or aren't sufficiently obsessive.


The front and back covers of a book are each 3 mm thick, and a
stack of 25 sheets of paper is 4 mm thick. Al's book is 38 mm thick,
and all of the pages are numbered front and back starting with page 1. What is the number of the last page facing the back cover?

$ 38$ mm minus $ 3$ mm for the front and back covers gives $ 35$. $ 35$/$ 4$ gives 8.75. $ 25$ sheets of paper is $ 50$ pages. $ 8.75*50\equal{}\fbox{438}$.

No, $ 3$ mm for each of the front and back.  So there's $ 32$ mm for the pages, dividing by $ 4$ yields $ 8 \times 25\equal{}200$ sheets, for $ \boxed{400}$ pages.


The equation $ (x^{2}\minus{}4)(x^{2}\minus{}1)\equal{}k$ has four non-zero real roots, equally spaced on the number line. If $ x_{4}$ is the largest of the roots, find the value of $ kx_{4}$.

[hide]
$ (x^{2}\minus{}4)(x^{2}\minus{}1)\minus{}k \equal{} 0$ has 4 equidistant roots

Let $ 2a$ be the distance between two of these roots

Because that function is an even function, we can see that the roots must be:
$ a,\minus{}a, 3a,\minus{}3a$
now, we plug in our info to solve for a

$ (a^{2}\minus{}4)(a^{2}\minus{}1)\minus{}k \equal{} (9a^{2}\minus{}4)(9a^{2}\minus{}1)\minus{}k \equal{} 0$

$ a^{4}\minus{}5a^{2}\plus{}4 \equal{} 81a^{4}\minus{}45a^{2}\plus{}4$

$ 77a^{4}\minus{}40a^{2}\equal{} 0$

a isn't 0 (because that wouldn't make sense)

$ 77a^{2}\minus{}40 \equal{} 0$

$ a \equal{}\sqrt{\frac{40}{77}}\equal{} 2\sqrt{\frac{10}{77}}$

The largest root is $ 3a \equal{}\boxed{6\sqrt{\frac{10}{77}}}$

now
$ k \equal{} (x^{2}\minus{}4)(x^{2}\minus{}1) \equal{} (\frac{360}{77}\minus{}4)(\frac{360}{77}\minus{}1)$

multiply those and you got it
[/hide]


superior algebra
superior algebra solved

$G$ a group and $f$ an outer automorphism of $G$.
Prove that $G$ can be embedded in another group $H$ such that $f$ is the restriction of an inner automorphism of $H$.


Hello all,

A little introduction: I'm a junior going into my senior year. I live in the Greater Hartford region and I'm currently in the process of starting a math team at the math and science school I go to. The director of the school is on board with the proposal I had and, if everything works out for next year, we'll be participating in the HMMT. I'm hoping we can extend the list of competitions we participate in, so if any of you have any suggestions, that would be great. After my finals, I'm going to start getting more information on the ARML that I know a lot of you guys participate in.

Also, if you guys are involved in any math teams at your schools that you feel are run and managed particularly well, or if any of you are captains of your math teams and have any tips to share, please do so!

Are you participating in the Hartford area math league?

If that's another name for the Capital Area Math League (CAML), then yes, but not with the math team I mentioned above (I go to two schools). If you're not referring to this, could you give a bit more information about it?

Also, another thing I forgot to mention in my last post: how many students do you guys have in your math teams? And of these, how many are very passionate about the team? Specifically, about how much time do your team members contribute to team meetings as well as self-study? Any information like this would be useful, since I'm conducting a survey about student interest in the math team, and I'd like some data to compare with. Thanks again!

[quote="worthawholebean"]We have probably 6ish people who are really involved in math team activities. But my school isn't really comparable to others, as it's private. There are maybe two or three people who do significant amounts of studying for the competitions we do; the others just do the assigned rounds.

Yeah, I meant the CAML, I just didn't know the name.[/quote]

Who are the 2 or 3?  :maybe:

[quote="Ignite168"]Who are the 2 or 3?  :maybe:[/quote]Dunno, that was just a number I pulled out of nowhere. Do YOU study a lot?

What is the CAML? Like, what kind of problems do you do and how does it work?

Also, is there one near the NW corner?

CAML is the Capitol Area Math League. I'm not sure how far in CT it ranges, but I'm pretty sure it's just schools in the Greater Hartford area. Each participating town sends a team of $ \leq 5$ people to some host school every month (my town this month!), where each member does three of five rounds, with three questions (worth $ 1,2,3$ points respectively) per $ 10$-minute round.

The problems are supposed to increase in difficulty in any given round, but quite frankly they are all quite easy, maybe like AoPS1, some AoPS2, level. For example, the contest writers I think are in love with $ 60^o$ angles and $ 30\minus{}60\minus{}90$ triangles.


[size=150]$ x,y,z\geq 0,  xy\plus{}yz\plus{}zx\equal{}1,$
 proof that:   $ \frac{1}{x\plus{}y}\plus{}\frac{1}{y\plus{}z}\plus{}\frac{1}{z\plus{}x}  \geq\frac{5}{2}$

 :)
when $ x\equal{}y\equal{}o$  ,  $ z\equal{}1$  the inequality becomes equal
[/size]

[quote="admire9898"][size=150]$ x,y,z\geq 0, xy \plus{} yz \plus{} zx \equal{} 1,$
 proof that:   $ \frac {1}{x \plus{} y} \plus{} \frac {1}{y \plus{} z} \plus{} \frac {1}{z \plus{} x} \geq\frac {5}{2}$

 :)
when $ x \equal{} y \equal{} o$  ,  $ z \equal{} 1$  the inequality becomes equal
[/size][/quote]

It's old problem  :) and very easy  :P

[quote="admire9898"][size=150]$ x,y,z\geq 0, xy \plus{} yz \plus{} zx \equal{} 1,$
 proof that:   $ \frac {1}{x \plus{} y} \plus{} \frac {1}{y \plus{} z} \plus{} \frac {1}{z \plus{} x} \geq\frac {5}{2}$
 :)
when $ x \equal{} y \equal{} o$  ,  $ z \equal{} 1$  the inequality becomes equal
[/size][/quote]

You can find it on this uvw paper [url]http://www.mathlinks.ro/viewtopic.php?t=278791[/url], page 8.

[quote="Gibbenergy"]
[size=150]You can find it on this uvw paper [url]http://www.mathlinks.ro/viewtopic.php?t=278791[/url], page 8.[/quote]

using the uvw method,this problem can be transformed to proof
$ 18u^{2} \minus{} 15u \plus{} 5w^{3} \plus{} 2\geq o$
$ 3u \equal{} x \plus{} y \plus{} z     ,      w^{3} \equal{} xyz$
 :P[/size]


$ N$ and $ M$ are consecutive integers such that $ N < \sqrt{942} < M$. Find $ N \plus{} M$.

Educated guestimation.  You [should] know that $ \sqrt{900} \equal{} 30$.  Another nice thing to know is that $ (n\plus{}1)^2 \equal{} n^2 \plus{} 2n \plus{} 1$ (easy-to-prove yet helpful algebra).  This yields $ 31^2 \equal{} 900 \plus{} 2(30) \plus{} 1  \implies 961$.  Since $ \sqrt{900} < \sqrt{942} < \sqrt{961}$, our answer is $ 30 \plus{} 31$ or $ \boxed{61}$.