calculus
integration
function
geometry
area of a triangle
real analysis
real analysis solved

I didn't want to post it to the Calculus section because I think it's to easy for it.

Let int(a,b)(f(x)dx) be the integral of the function f(x) in the interval [a,b].

g(x,y) is defined as follows: g(x,y)= :sqrt: (1-x^2-y^2) if x^2+y^2<=1 and
g(x,y)=0 else.

Calculate the integral:

int(-1,1)(int(-1,1)(g(x,y)dx)dy)

Today I found this problem in a Calculus problem book and I laughed a lot because I have a two-lines-solution and the author of the book had a two-sides-solution :D  :D  :D .

Misha

Helping Misha123 out with showing the problem:

$g(x,y)=\sqrt{1-x^2-y^2}$ on $\{(x,y) : x^2+y^2\le 1\}$ and $g(x,y)=0$ otherwise.

Compute $\int_{-1}^1\int_{-1}^1 g(x,y) dxdy$.

Archimedes wouldn't have understood the notation, but he did know the answer.

[b] Moderator edit: edited for readability (fixed a problem with ":" in math. mode) [/b]

Isn't it the volume of the hemisphere having the circle inscribed in the square $(1,1),(-1,1),(-1,-1),(1,-1)$ (i.e. the unit circle) as a great circle? And that's $2\pi$.

Uh, grobber... that's volume, not area. Look at your answer again.

Sorry for the bad notation. I had really no time to TEX it.

And I'm sorry for the bad answer. I was thinking "volume", and I computed the area. It should be $\frac{2\pi}3$.

Yeah, it's the right result but I'm much more interested in your solution than in the answer. I'm really curious if you used the same idea as I did.

Misha

You can see my solution in one of the previous posts: it's the volume of the hemisphere having the unit circle as a great circle.

Yes , of course. Your solution is much better than mine. I just integrated the length of a circle with radius r multiplied with  :sqrt: (1-r^2) from 0 to 1. The straigth-forward solution in the book was really terrible, they calculated both integrals using methods of integrating the functions of the form  :sqrt: (a-x^2) :D  :D  .

Thanks for yuor interest.

Misha

One of the pieces of advice I believe in most strongly is this: don't compartmentalize your knowledge of mathematics! The things you know are yours to use, no matter what the branch of mathematics the problem at hand ostensibly belongs to. The best solutions are the ones that connect knowledge together, that make use of a technique or result far from its usual home.

As a minor part of this: the fact that you are learnng calculus does not remove or replace the geometry you already new. As soon as I saw your problem, my reaction was instantaneous: [i]of course[/i] it's the volume of a hemisphere and I know (from the geometry I know long before I learned calculus) what the volume of a hemisphere is. Further examples:

$\int_0^2 (4-2x) dx$: Visualize it! It's the area of a triangle, base 2, altitude 4, hence area 4.

$\int_{-r}^r \sqrt{r^2-x^2} dx$: Are you scrapping your way through the trigonometric substitution and the half-angle formulas? Or are you saying "I know the area of a circle"?


Let (a) be a sequence: a(1)=sqrt2, a(2)=pi, a(n)=a(n-1)-a(n-2) for all succeeding integral n. Computs a(2008).

[hide="Solution"]Let $ x\equal{}\sqrt{2}, y\equal{}\pi$. Compute the sequence for $ a_1$ through $ a_8$. It comes out to be $ x,y,y\minus{}x,\minus{}x,\minus{}y,\minus{}y\plus{}x,x,y$. From this, we can see that the sequence will repeat with a period of 6. Therefore, $ a_{2008}\equal{}a_4\equal{}\minus{}x\equal{}\minus{}\sqrt{2}$.[/hide]


The equation \[ x^{10} + (13x-1)^{10} = 0 \] has 10 complex roots $r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5}$, where the bar denotes complex conjugation. Find the value of \[ \frac{1}{r_1\overline{r_1}} + \frac{1}{r_2\overline{r_2}} + \frac{1}{r_3\overline{r_3}} + \frac{1}{r_4\overline{r_4}} + \frac{1}{r_5\overline{r_5}}. \]

You can completely avoid trigonometry:

We devide both sides of the equation by $x^{10}$ and we get
(1)  $(\frac{1}{x}-13)^{10}=-1$.
Denote $t=\frac{1}{x}-13$. The equation becomes
(2)  $t^{10}=-1$.
Let $S$ be the sum in the original statement, $x_1,x_2,...,x_{10}$ the solutions of equation (1) and $t_1,...,t_{10}$ the solutions of the equivalent equation (2).
Then
$S=\frac{1}{2}\sum_{k=1}^{10}\frac{1}{x_k}\frac{1}{\overline{x_k}}=\frac 12\sum_{k=1}^{10}(13+t_k)\overline{13+t_k}=\frac 12(\sum_{k=1}^{10}13^2+13\sum_{k=1}^{10}t_k+13\sum_{k=1}^{10}\overline{t_k}+\sum_{k=1}^{10}t_k\overline{t_k})$.
From (2) we get $\mid t_k\mid=1$. Since in general $\mid z\mid^2=z\overline{z}$, we get $t_k\overline{t_k}=1$.
The first Viete relation for equation (2) yields  $\sum_{k=1}^{10}t_k=0$, so we have also $\sum_{k=1}^{10}\overline{t_k}=0$.
We obtain $S=\frac 12(1690+0+0+10)=850$


[quote="jonfv"]Sea $n$ un numero natural. Pruebe que: \[\sum_{k=0}^{k=n}\binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}\] Convencion: $\binom{0}{0}=1$[/quote]
HOLA AMIGO, JEJEJ ESE PROBLEMA SALE POR FUNCION GENERATRIZ, ME LO DEJARON DE TAREA, DONDE VISTE EL PROBLEMA??,
CHAUFAS.

[quote="jonfv"]Me lo dio un amigo, lo resolvi usando el siguiente problema:
Sea $n$ un numero natural. Cuantas $n$-uplas $(x_{1},\dots,x_{n}$ existen tales que:
\begin{itemize}
\item $x_{k}\in \{-1,1\}$ para todo $1\leq k \leq n$.
\item $S_{k}=\sum_{r=1}^{r=k}x_{r}\geq 0$ para todo $1\leq k \leq n$.
\end{itemize}
este problema esta relacionado con el problema que puse en el post "Conteo".[/quote]
BUENO ACA PONGO UNA SUGERENCIA PARA LA SOLUCION QUE ES POR FUNCION GENERATRIZ.
BUENO RECORDEMOS ESTO
$\dbinom{-1/2}{k}(-4)^{k}= \dbinom{2k}{k}$.
ahora sea AHORA FORMEN sus funcionges generatrices, y eleven al cuadrado. y desps analisen el coeficiente de grado $x^{n}$, y por ser una identidad se tendra lo pedido, espero k alguien la postee, no termino la solucion por falta de tiempo.

SOLUCION POR FUNCION GENERATRIZ.
sea $F(x)=\sum_{k\geq 0}\dbinom{2k}{k}x^{k}$.
sea $G(x)=\sum_{k\geq 0}{4^{k}x^{k}}$
ahora bastaria con demostraremos que $G(x)=F(x)^{2}$
ya que en analisariamos el coefiente de $x^{n}$, y tendriamos lo pedido.
AHORA RECORDEMOS LO SIGUIENTE.
$\dbinom{2k}{k}=\dbinom{-1/2}{k}(-4)^{k}...(1)$
dem.-
solo transformaremos.
$\dbinom{-1/2}{k}(-4)^{k}=((-4)^{k})[(-1)^{k}(1/2)(3/2)...((2k-1)/2)]/k!$ ahora le multiplicaremos por $k!$ en el numerador y denomidar.
$\Rightarrow \dbinom{-1/2}{k}(-4)^{k}=[(-1)^{2}k(1)(3)...(2k-1)(2)(4)...(2k)]/k!k!=(2k)!/(k!)^{2}=\dbinom{2k}{k}$. $\blacksquare$
ahora a la ecuacion (1), le multiplicaremos por $x^{k}$ y tomaremos sumatoria $\sum_{k\geq 0}$.
$\to  \sum_{k\geq 0}\dbinom{2k}{k}x^{k}=\sum_{k\geq 0}\dbinom{-1/2}{k}{(-4)^{k}x^{k}}$, ahora notemos que el lado derecho es $F(x)$, y $\sum_{k\geq 0}\dbinom{-1/2}{k}{(-4)^{k}x^{k}}=(1-4x)^{-1/2}=G(x)^{1/2}$.
$reemplasando\to  F(x) = G(x)^{1/2}\to F(x)^{2}=G(x)$, que es lo que queriamos demostrar.  $\blacksquare$

Ola!
Alguien tiene alguna solucion puramente combinatorica??
Es que como que no entendi mucho la de generatrices...
jeje
 :oops:

Otra solucion:
Considera las $2n$-uplas $(x_{1},\dots,x_{2n})$ de vectores en $R^{2}$ tales que $x_{k}\in\{(1,0);(0,1)\}$ para todo $0\leq k\leq 2n$. La cantidad de tales $2n$-uplas es $2^{2n}$. Por otro lado, si para cada $1\leq k\leq n$, $a_{k}$ es la cantidad de dichas $2n$-uplas que satisfacen $S_{2k}=\sum_{r=1}^{2k}x_{r}\in l$ y $S_{m}=\sum_{r=1}^{m}x_{r}\notin l$ para todo $m>2k$ (donde $l$ es la recta identidad); entonces $a_{k}=\binom{2k}{k}\binom{2(n-k)}{n-k}$.
La parte dificil en la ultima igualdad, es provar que la cantidad de $2(n-k)$-uplas $(x_{2k+1},\dots,x_{2n})$ que satisfacen $\sum_{r=2k+1}^{m}x_{r}\notin l$ para todo $m>2k$ es $\binom{2(n-k)}{n-k}$.
Esto ultimo se puede probar usando el principio de reflexion (cuya prueba es muy simple): "Sean $P, Q\in Z^{2}$ y $\tilde{P}$ la reflexion de $P$ con respecto a $l$ (o a una paralela a $l$) entonces la cantidad de caminos (sobre la cuadricula generada por $Z^{2}$) que unen $P$ y $Q$, intersectan $l$ y tienen longitud minima es igual a la cantidad de caminos que unen $\tilde{P}$ y $Q$, intersectan $l$ y tienen longitud minima


logarithms
inequalities
inequalities unsolved

a, b, c >1

Prove that:  log_a〖a^2*b〗 +log_b〖b^2*c〗 +log_c〖c^2*a〗<1

SORRY. THIS is the good problem

a, b, c >1
Prove  that  :  log_(a^2*b)a +log_(b^2*c)b +log_(c^2*a)c <=1

This inequality was given in a Romanian contest.
 It's equivalent to
\[\frac{1}{2+x}+\frac{1}{2+y}+\frac{1}{2+z}\leq 1 \]
where $x=log_{a}b$,$y=log_{b}c$ and $z=log_{c}a$.From now i think you will be able to solve it having in mind that $xyz=1$

From Ma>=Mh we have  1/2+x   +  1/2+y  +  1/2+z>=9/6+x+y+z

Because x+y+x>=3   ( because xyz=1) we have 9/6+x+y+z<=1

But from here we cam say that 1/2+x   + 1/2+y  + 1/2+z<=1   ?????

I don't think so.My proof it's a little bit ugly but is't correct.
\[\sum_{cyc}\frac{1}{2+x}\leq 1\] $\leftrightarrow$
\[\sum_{cyc}(2+y)(2+z)\leq(2+x)(2+y)(2+z)\] $\leftrightarrow$
\[12+4(x+y+z)+xy+yz+xz\leq 2(xy+yz+xz)+4(x+y+z)+8+xyz\] $\leftrightarrow$
\[4\leq xy+yz+xz+xyz\] whic is true from AM-GM and using $xyz=1$


induction
combinatorics unsolved
combinatorics

A graph has $ 2002$ points. Given any three distinct points $ A$, $ B$, $ C$, there is a path from $ A$ to $ B$ that does not pass through $ C$. A move is to take any cycle (a set of distinct points $ P_1$, $ P_2$, ... , $ P_n$ such that $ P_1$ is joined to $ P_2$, $ P_2$ is joined to $ P_3$, ... , $ P_{n \minus{} 1}$ is joined to $ P_n$, and $ P_n$ is joined to $ P_1$), remove its edges and add a new point $ X$ and join it to each point of the cycle. After a series of moves the graph has no cycles. Show that at least $ 2002$ points have only one edge.

I'll just sketch it. I may be wrong, but right now, it looks like this should work.

You can prove what you want by induction on the number $|G|$ of vertices. More precisely, what we want to prove is that if we perform several moves such as the ones described on a $2$-connected graph $G$ with $|G|$ vertices until it becomes a tree, this tree will have at least $n$ leaves.

After several moves, we'll reach the first graph $\widetilde G$ which is no longer $2$-connected. We may as well assume that this happens after the first move, in which we've eliminated the edges in the cycle $(x_1,\ldots,x_t)$, and we've added the vertex $x$, connected to all the $x_i,\ i=\overline{1,t}$. It shouldn't be hard to see now that if $H$ is a connected component of $\widetilde G-x$ (the graph you obtain by deleting $x$ and the edges $xx_i$) with more than one vertex (i.e. if it's not of the form $\{x_i\}$), then the graph $\widetilde H$ obtained from $H$ by adding $x$ and the edges $xx_i$ for $x_i\in H$ is $2$-connected. Furthermore, all subsequent moves must be performed in one of the $\widetilde H$ (i.e. when we choose a cycle, it must be contained completely inside one such $\widetilde H$; try making a drawing).

By the induction hypothesis, in the end, each $\widetilde H$ (including those consisting of a single edge $xx_i$) will have at least $|\widetilde H|-1$ vertices of degree $1$ in common with $\widetilde G$ (that $-1$ is there because all $\widetilde H$ contain $x$, but we're only interested in vertices which belong to the initial graph $G$ as well). Since the sum of all $|\widetilde H|-1$ is precisely $|G|$, we're done.

[quote="grobber"]

By the induction hypothesis, in the end, each $ \widetilde H$ (including those consisting of a single edge $ xx_i$) will have at least $ |\widetilde H| \minus{} 1$ vertices of degree $ 1$ in common with $ \widetilde G$[/quote]

I don't get it. What if $ \widetilde H\equal{}\widetilde G$? How can you use induction hypothesis? And what if the $ \widetilde H$ we use overlap each other? Are you sure we are done?


geometry
trigonometry
APMO
vector
triangle inequality

Two equal-sized regular $n$-gons intersect to form a $2n$-gon $C$. Prove that the sum of the sides of $C$ which form part of one $n$-gon equals half the perimeter of $C$.

[i]Alternative formulation:[/i]

 Let two equal regular $n$-gons $S$ and $T$ be located in the plane such that their intersection $S\cap T$ is a $2n$-gon (with $n\ge 3$). The sides of the polygon $S$ are coloured in red and the sides of $T$ in blue.

Prove that the sum of the lengths of the blue sides of the polygon $S\cap T$ is equal to the sum of the lengths of its red sides.

Let the n-gons be A1 A2 ... An and B1 B2 ... Bn, such that clockwise we encounter them in this order : A1, B1, A2, B2, ...

{C1} = B1Bn \cap A1A2
{C2} = B1B2 \cap A1A2
{C3} = B1B2 \cap A2A3
...................................

Let a = length of perimeter of C1 C2 .. C2n which belongs to A1 A2 .. An, and b the similar for the second n-gon. Let p be the common value of the perimeters of the n-gons.

Triangles Bi C(2i-1) C2i and Bj C(2j-1) C2j are similar, for all i,j. Therefore : (C(2i-1)Bi+BiC2i)/C(2i-1)C2i = (C(2j-1)Bj+BjC2j)/C(2j-1)C2j = (sum C(2i-1)Bi+sum BiC2i)/(sum C(2i-1)C2i) = (p-b)/a

Also, notice that in a similar manner for the other triangles, we obtain that each ratio like above is equal to (p-a)/b. But by the sine law the two ratios are equal, and therefore :

(p-b)/a = (p-a)/b ===> pb - b^2 = pa - a^2 ===> (a-b)(a+b) - p(a-b) = 0 ===> (a-b)(a+b-p)=0

But by triangle inequality we have C(2i-1)C2i<C(2i-1)Bi + BiC(2i+1). Summing up yields b<p-a, therefore a+b<p. So a=b, and we are done.

I have seen a rather nicer solution:

Let O1 and O2 be their centers. Shift the first one with the vector O1O2 such that the problem becomes isomorphic. Note that through this transfer the share of each n-gon in the total perimeter doesn’t change and that in the latter situation the problem is trivial.

This problem for n=2 has been used in USSR math Olympiad.

What you mean when you say that the problem becomes isomorphic?

Lich: Chamos literally traduced is a way to refer to young people here im venezuela. So kinda everybody that participates on the IMO are chamos :D:D:D

I meant by isomorphic that:

If you look the figure in every direction its the same thing so the problem becomes trivial. :)

I don't know why it lies in Combinatorics. I will move it Geometry. Though, I think problem was solved.

Here is the solution given in our yearly collection of training problems. It was probably copied from the official solution.

It is easy to see that the intersection $S\cap T$ has $2n$ sides if and only if the sides of $S\cap T$ alternate: blue, red, blue, red, etc.

Denote the vertices of $S\cap T$ as $C_1D_1C_2D_2\ldots,C_nD_n$ so that the sides $C_1D_1$, $C_2D_2$, $\ldots$, $C_nD_n$ are blue. Denote the vertices of $S$ by $A_1, A_2,\ldots, A_n$ and the vertices of $T$ by $B_1,B_2,\ldots,B_n$ so that $C_1D_1\subset B_1B_2,\ldots, C_nD_n\subset B_nB_1$ and $D_nC_1\subset A_1A_2,\ldots, D_{n-1}C_n\subset A_nA_1$. One can check that all the triangles $D_nB_1C_1, D_1B_2C_2,\ldots, D_{n-1}B_nC_n$ and $C_1A_2D_1, C_2A_3D_2,\ldots, C_nA_1D_n$ are similar.

Therefore
$$
\frac{D_nC_1}{D_nB_1+B_1C_1}=\frac{D_1C_2}{D_1B_2+B_2C_2}=\cdots=\frac{D_{n-1}C_n}{D_{n-1}B_n+B_nC_n}$$
$$=\frac{C_1D_1}{C_1A_2+A_2D_1}=\frac{C_2D_2}{C_2A_3+A_3D_2}=\cdots=\frac{C_nD_n}{C_nA_1+A_1D_n}.
$$

Hence
$$\frac{D_nC_1+\cdots+D_{n-1}C_n}{D_nB_1+B_1C_1+\cdots+D_{n-1}B_n+B_nC_n}=\frac{C_1D_1+\cdots+C_nD_n}{C_1A_2+A_2D_1+\cdots+C_nA_1+A_1D_n}.$$
Let $x=D_nC_1+D_1C_2+\cdots+D_{n-1}C_n$ and $y=C_1D_1+C_2D_2+\cdots+C_nD_n$, so that $x$ is the sum of the red sides of $S\cap T$ and $y$ is the sum of the blue sides. If $a$ is the length of a side of $S$ (or $T$), then the equality above can be written in the following form
$$\frac{x}{na-y}=\frac{y}{na-x}$$
or equivalently
$$(na-x-y)(x-y)=0.$$
Since the perimeter of $S\cap T$ is strictly less than the perimeter $na$ of $S$ or $T$, we obtain $x=y$.


geometry
3D geometry
sphere
integration
trigonometry
calculus
analytic geometry

Find the volume of the region that is under the sphere $x^{2}+y^{2}+z^{2}=1$ and above the cone $z=\sqrt{x^{2}+y^{2}}$.

$\cdots = \int_{0}^{\frac1{\sqrt{2}}}\int_{r}^{\sqrt{1-r^{2}}}\int_{0}^{2\pi}r\,d\theta\,dz\,dr$

Take your pick between spherical and cylindrical coordinates. Works either way. Or go back to first or second semester calculus and call it a solid of revolution. (My setup amounts to using the "shell method.")

could you explain how you apply triple integrals to this. my teacher skipped this section in the text and i haven't had the time to get around to reading it yet.
he put this question on a quiz and wanted us to use double integrals and polar coordinates.

Then
$\mathrm{volume}(E)=\iint_{D}(\mathrm{top}-\mathrm{bottom})dA= \iint_{D}(\sqrt{1-x^{2}-y^{2}}-\sqrt{x^{2}+y^{2}})dA= \int_{0}^{2\pi}\int_{0}^{1/\sqrt{2}}(\sqrt{1-r^{2}}-r)r\,dr\,d\theta$
where $D$ is the projection of $E$ onto the $xy$-plane, namely a disk of radius $1/\sqrt{2}$.

how did you get the limit of $r: 0\to\frac{1}{\sqrt 2}$?

We have to find out where the cone intersects the sphere. The cone is $z=r.$ The (upper half of the) sphere is $z=\sqrt{1-r^{2}}.$ They intersect when $r=\sqrt{1-r^{2}}$ which gives us $r=\frac1{\sqrt{2}}.$

[quote="maokid7"][quote="mlok"]$\mathrm{volume}(E)=\iiint_{E}1\,dV=\int_{0}^{\pi/4}\int_{0}^{2\pi}\int_{0}^{1}\rho^{2}\sin\phi\,d\rho\,d\theta\,d\phi=\cdots$[/quote]

so i finally had time to go over triple integrals. what I am wondering is why are the limits on the last integrand are $0\to\frac{\pi}{4}$ and not $\frac{\pi}{4}\to\frac{\pi}{2}$?[/quote]

The angle $\phi$ denotes the angle of declination from the positive z-axis so that the eqauations: $\phi = 0$ is the z-axis, $\phi = \frac{\pi}{2}$ is the xy-plane, $\phi = \frac{\pi}{4}$ in spherical coordinates is the cone given by $z=\sqrt{x^{2}+y^{2}}$ in rectangular coordinates, and $0\leq \phi \leq \frac{\pi}{4}$ is the solid version of this, which is given by $z\geq\sqrt{x^{2}+y^{2}}$ in rectangular coordinates.

o sorry. for some reason i got confused and thought that 0 was the xy-plane.


The AMP Olympiad is an annual competition and had its second edition in 2007. We say a year is a tuanis year if the product of it and the number of the correspondent edition is a perfect square.
For example, 2008 is not a tuanis year because $ 2008\cdot 3 \equal{} 2^3\cdot 3\cdot 251$ is not a perfect square.

a) Prove that there is no tuanis year in this millenium.

b) Determine a previous year (in this millenium) in which the second edition should have been held such that there would have existed at least one tuanis year in this millenium.

[hide] sorry for the writing of b), i didn't find any better[/hide]

The point is to find $ n \in \mathbb{N}$ such that $ 0 < n < 995$ and $ n(2005 \plus{} n) \equal{} 2005n \plus{} n^2 \equal{} m^2$ whit $ m \in \mathbb{N}$.
We see $ n|m \equal{} n\cdot q$, then
$ 2005n \plus{} n^2 \equal{} (nq)^2$
$ \frac {2005}{n} \plus{} 1 \equal{} q^2$, where $ n|2005$. Also $ n \in \left\{1,5,401,2005\right\}$, but no value satisfies the original equation.
Analog whit these years, when the 2nd olympiad take place before 2005 in this milenium. For the years 1999 and 2003 is easy to check that they dont work (they´re primes).
instead of 2005, if we would write:
1998: $ n \equal{} 666$, :diablo: tuanis year = 2664
2000: $ n \equal{} 250$,   tuanis year = 2250
2001: $ n \equal{} 667$,   tuanis year = 2668
2002: $ n \equal{} 14$,     tuanis year = 2016
2004: $ n \equal{} 668$,   tuanis year = 2672

Sorry for my poor english.
Anybody else has another solution?
[i]tuanis[/i]

Sorry for my last solution, it has a terrible error... i took n as a prime number. Anyway, the part b) is good.
Here have I a good solution.
Resolving the equation $ n^2 \plus{} 2005n \minus{} m^2 \equal{} 0$ as a cuadratic in $ n$ whit $ m$ strict positive, we have
$ n \equal{} \frac { \minus{} 2005 \plus{} \sqrt {2005^2 \plus{} 4m^2}}{2}$, only the positive root.
Let $ k \in \mathbb{N}$ such that $ 5\cdot5\cdot401\cdot401 \equal{} 2005^2 \equal{} k^2 \minus{} 4m^2 \equal{} (k \plus{} 2m)\cdot(k \minus{} 2m)$
the posibilities are: $ k \plus{} 2m > k \minus{} 2m$
$ k \plus{} 2m \equal{} 2005^2$, $ k \minus{} 2m \equal{} 1$, $ m \equal{} 1005006$, $ n \equal{} very big$
$ k \plus{} 2m \equal{} \frac {2005^2}{5}$, $ k \minus{} 2m \equal{} 5$, $ m \equal{} 201000$, $ n \equal{} very big$
$ k \plus{} 2m \equal{} \frac {2005^2}{401}$, $ k \minus{} 2m \equal{} 401$, $ m \equal{} 2406$, $ n \equal{} 1604$
$ k \plus{} 2m \equal{} 401^2$, $ k \minus{} 2m \equal{} 25$, $ m \equal{} 40194$, $ n \equal{} very big$
never in this milenium.
In the part b), note that there are always many finite tuanis years.
I hope, this solution is correct.

We try to all ( :lol: ) integer solutions of the equation $ n(n\plus{}pk)\equal{}m^2$, with $ p>k$ both primes:

$ (n,n\plus{}pk)\equal{}(n,pk)\equal{}d|pk, \text{ so }$ $ (N, N\plus{}\frac{pk}{d})\equal{}(N,\frac{pk}{d})\equal{}1, \text{ with } n\equal{}dN$, but $ d^2|m^2$ then:

$ N(N\plus{}\frac{pk}{d})\equal{}(\frac{m}{d})^2\equal{}t^2 \implies N\equal{}q^2$ for some integer $ q$, then $ t^2\minus{}q^2\equal{}\frac{pk}{d}$, so if $ t>q\plus{}1$ we have necessary $ t\plus{}q\equal{}p$ and $ t\minus{}q\equal{}k$ and $ d\equal{}1$ which give $ q\equal{}\frac{p\minus{}k}{2} \implies n\equal{}dN\equal{}dq^2\equal{}\frac{(p\minus{}k)^2}{4}$; otherwise $ t\equal{}q\plus{}1$ which give us $ q\equal{}\frac{pk\minus{}d}{2d}$ for all $ d|pk \implies n\equal{}dN\equal{}dq^2\equal{}\frac{(pk\minus{}d)^2}{4d}$.

In our case all acceptable solutions are not in [2000,3000).


How can you use modulus to solve diophantine problems, and what is a quadratic residue.

P.S: what does this mean [/quote]x^3 + x + 57 is congruent to 0 (mod 53)

$x^{3}+x+57 \equiv 0 \mod 53$ $\Longleftrightarrow$  $\frac{x^{3}+x+57}{53}\mathbb \in{N}$

Let $n$ be given.
$a$ is called a quadratic residue $\mod n$ if $a$ is coprime to $n$ and there is an $x$ with $x^{2}\equiv a \mod n$.

And see http://www.artofproblemsolving.com/Wiki/index.php/Quadratic_residues .

And to see how to use: just look around in the forum, there are various ways.


ratio
geometry
perimeter
circumcircle
floor function

What is the ratio of the circumference of a circle to the perimeter of an inscribed square?

Let the radius of the circle be $ R$, such that the diagonal of the square is $ 2R$, and the length of a side of the square is $ \sqrt2R$. This means that the ratio is $ \frac R{4\sqrt2R}\equal{}\boxed{\frac{\sqrt2}{8}}$.

[quote="helijet"]What is the ratio of the [b]circumference [/b]of a circle to the perimeter of an inscribed square?[/quote]

So the ratio is actually $ \frac {2\pi R}{4\sqrt 2 R} \equal{} \frac {\pi}{2\sqrt 2} \equal{} \frac {\pi\sqrt 2}4$. :)

[quote="math154"]When I read it, I thought it said "...ratio of the circumradius...".
I hope I am not having hallucinations. :o[/quote]

Circumradius of a circle? Haha :rotfl:

Anyway, how did you get so smart all of a sudden? I distinctly remember you from this summer...

[quote="Yongyi781"]Anyway, how did you get so smart all of a sudden? I distinctly remember you from this summer...[/quote]

What are you talking about... :maybe:

Oh dang I think you just got insulted. Pwned like a newb in teh wildy.

[quote="mewto55555"]Oh dang I think you just got insulted. Pwned like a [b]newb in teh wildy[/b].[/quote]

..... eh-hem... http://www.google.com/search?q=noob+wildy
Care to reveal your username? :D :D

http://hiscore.runescape.com/hiscorepersonal.ws?user1=mewto55555

*Gasp* You were a member?? :o

Oh dang Im ranked in everything but range? DId they expand the highscores?

Hm... they expanded it to top 2 million, I believe. (And you're still not ranked...)

Magic: $ \text{Exp} \equal{} \left \lfloor \frac 14 \sum_{i \equal{} 0}^{L\minus{}1} \left \lfloor i \plus{} 300 \cdot 2^{i/7} \right \rfloor \right \rfloor$

I have a feeling we should be discussing "ratio of circumference to perimeter of square"... perhaps we can continue this "discussion" in FTW... or possibly in-game if I can find the time :maybe:


A base-10 three-digit number n is selected at random.  Which of the following is closest to the probablity that the base-9 representation and the base-11 representation of n are both three-digit numbers?

A. 0.3   B. 0.4   C. 0.5   D. 0.6   E. 0.7

Here's how I did it:

[hide]Sorry if this is a bad explanation, it's hard to explain in words.  For base-9, we need to find the highest and lowest values n can be so n is still 3 digits in base-9.  The lowest 4 digit integer in base-9 is 1000, which is 9^3 or 729 in base 10.  So n has to be less than 729, or else it'll be a 4 digit number in base-9.  Highest 2 digit integer in base-9 is 88, which is 80 in base 10.  So we have 80<n<729.  We do the same thing for base 11 and get 121<n<1331 base 10.  So n needs to be above 121 but less than 729 for this to work.  It's exclusive, so 729-121-1=607 3 digit base 10 numbers n will work.  There are 900 base 10 3 digit numbers, so answer is 607/900 or .67444444.  This is closest to E. 0.7.  Your calculations don't have to be exactly right for you to get the right answer, and I don't think mine are exactly right either.  Also, I solved the problem mentally in a way, so what I wrote here is different from how I actually solved it.[/hide]

Moderators, please don't put this in the Intermediate forum, where it will be easily solved by everyone.  I am a beginner, and by thinking about this problem a little, I was able to solve it.  Thanks.

I don't think the moderators were going to move this anyways, because it isn't too difficult.

Answer:

[hide][tex]100_{9} = 81_{10} \\ 1000_{9} = 729_{10} \\ 100_{11} = 121_{10} \\ 1000_{11} = 1331_{10} \\ [/tex]

Those are the basics. You have the 3 digit number below 729, and above 121, so that leaves 607 numbers inbetween, so 607/900 - which is .674444444.... [/hide]

Both of you made one slight mistake.
You say that 729>n>121, however the number 121 itself works since it is 100 in base 11 and some other three digit number in base 9.  So, strictly speaking, 729>n>=121 or 729>n>120, which changes the probability to 608/900=152/225~=1/3 so E is the answer.  So it really doesn't change the answer, however, if it had been AIME or something where they wanted it in the form a/b and for you to find a+b you would have regretted it afterwards.
Or am I missing something?

And where's this problem from, I think I saw it before.

Oops!!! :blush: You are right, I forgot that 121_{10} was 100_{11} and that 100 was a 3 digit number. Thank you for the correction!

The beginners have missed it, and the mods have moved it :P

Anyway you should be happy your problem get solved easily ;)


limit
logarithms
real analysis
real analysis solved

Compute the limit: $\displaystyle \lim_{x\to \infty} (\sum_{k=1}^{n} n^{\frac{k-x}{x}})^{x}$ where $n \in N$ $\geq 2$.

The limit should be $n^{(n+1)/2}$ (up to errors in calculations).

Setting $y=1/x$, with trivial manipulations you get the limit as $y\to 0$ of
\[\left(\frac{1}{n}\sum n^{ky}\right)^{1/y}=
\left(\frac{1}{n}\sum e^{ky\log n}\right)^{1/y}.\]
Now use Taylor expansion up to order 1. You obtain
\[\sum e^{ky\log n}=\sum (1+ky\log n)=n+\frac{n(n+1)}{2}y\log n,\]
up to terms of order higher than 1. So the initial limit is the limit of
\[\left(1+\frac{ny\log n}{2}\right)^{1/y}\]
which is indeed $n^{(n+1)/2}$.

Another interpretation:
$\displaystyle (\sum_{k=1}^{n} n^{\frac{k-x}{x}})^{x} = (\frac{\sum_{k=1}^{n} n^{\frac{k}{x}}}{n})^{x}$. This is the $\frac{1}{x}$th power mean of $n^k$. As $n\to \infty$, this goes to the 0-power mean, which is the geometric mean $\displaystyle (\Pi_{k=1}^n n^k)^{\frac{1}{n}} = n^{\frac{n+1}{2}}$.


conics
ellipse
analytic geometry
inequalities
geometry

my assingnment is to graph a happy face on a graphing calc, but i need help.  the eyes are suppose to be like minature circle, the nose as an ellipse and a half circle as the smile, i know it sounds ridiculous but can anyone help, i'll appreciate it'

You could use the equations for circles, ellipses, then choose whatever center you want.

For the half circle, I think you could just set restrictions on y.

[hide]The equations for a circle is $(x+h)^{2}+(y+l)^{2}=r^{2}$, where h and l are the x and y coordinates of the center of the circle, and r is the radius.
[/hide]
[hide]The equation for an ellipse is $(x-h)^{2}/ a^{2}+(y-k)^{2}/b^{2}= 1$
Where a is the major axis and b is the minor axis, and h and k are the coordinates of the center.[/hide]

[quote="kyyuanmathcount"]For the half circle, I think you could just set restrictions on y.[/quote]
Don't calculators require $y=\text{stuff}$? So, solving for $y$ from the basic circle equation ($(x-h)^{2}+(y-k)^{2}=r$), take only the negative square root to form a smile.

[quote="mgao"]Mustafa:

I think on a graphing calc you have to [i]express everything in terms of y[/i].[/quote]

On my TI-84 plus silver edition, under apps, you can do stuff like conics, or inequalities, and under inequalities, you can do things like x=.... or x<......

[quote="mysmartmouth"][quote="anirudh"]and also there is a section for X= in my 84, and on my 89. But on 83's you have to rearrange and simplify in terms of y[/quote]

I have both calculators.  How do you get the x= on them?  I see the Y= in the top left, but I can't find the x=...[/quote]

Use parametric equations: Set $y=t$, and then you can have $x=f(t)=f(y)$.  :D

[quote="mysmartmouth"][quote="anirudh"]and also there is a section for X= in my 84, and on my 89. But on 83's you have to rearrange and simplify in terms of y[/quote]

I have both calculators.  How do you get the x= on them?  I see the Y= in the top left, but I can't find the x=...[/quote]

There is a free software for the 84, the Inequalities App. I think the X= button is on the bottom corner. On the 89, I downloaded some app from ticalc.org (dont remember what, renamed it)


Could you please help me in solving the following problem? :blush:
Show that given any set "A" of real numbers (distinct) there exist $ x, y\in A$ such that:
$ 0<\frac{x\minus{}y}{1\plus{}xy} < 2 \minus{} \sqrt{3}$

It is a problem from my book and it's in the chapter "Pigeon Hole Principle". :)

[hide="Large hint"] Write $ x \equal{} \tan \alpha, y \equal{} \tan \beta$ for some angles $ \alpha, \beta \in \left( \minus{} \frac {\pi}{2}, \frac {\pi}{2} \right)$.  What happens when you apply Pigeonhole to this interval?  (You will probably want to know what the arctangent of $ 2 \minus{} \sqrt {3}$ is.) [/hide]


probability
AMC
AIME
geometry
AMC 10
\/closed

I was looking at the class schedules and what classes I was thinking about taking and I realized that a few of the classes conflict with each other.  Lets say I was taking Intermediate Counting/Probability and I also wanted to take AMC 10 Problem Series. That'd be fine for the first month or so, but what happens when the AMC 10 Problem Series starts?  Intermediate Trig/Complex is on Thursdays from 8:00-9:30 and AMC 10 Problem Series is on Thursdays at 7:30-9:00.  More likely though, the Intermediate Counting and Probability and AIME Problem Series which are both on Mondays from 7:30-9:00 and 8:00-9:30 respectively rather concerns me.  The schedule would conflict for a 3-week period.  My main concern is these two,  although another time where this would happen would be Olympiad Geometry and the AMC 10 Problem Series though unlikely.  I just think that students should be able to take any and all classes that are offered.

Now the questions

What happens if you want to take both?
Can you just not take both then?
Will an alternate class period be offered or the schedule changed entirely?
Can you take both classes at the same time? (i.e., simultaneously)
Would you just have to read transcripts of classroom time then?

Thank you

You can take both classes.  You can, in fact, take both of them at the same time!   A little while ago they made it so that you could be in multiple classrooms at the same time.  Although, I have to say from experience that it is quite hard to follow along two discussions at the same time.  You'll probably just end up deciding which class to focus on that day and looking at the other's transcript.

You have to understand that due to time constraints they cannot offer all of the class at independent times.  But they have made it so that most people who want to double/triple up can.  For example, I honestly don't think there are many people who want to take Olympiad Geom. and AMC 10 PS at the same time.

Also, I may be wrong, but I believe that a 3 week overlap isn't as bad as it has been in the past.

What joml88 said is basically correct.  Fortunately and unfortunately the online school has grown so quickly that it is simply impossible to run classes without any conflicts at all.  Short overlaps are not so bad.  Decisions are left to the students, though I agree with joml88 that it's best to focus on one ongoing class and read the transcript for the other.

Overall, we try to do our best to schedule classes at the same time that would have few students overlapping, though instructor schedules dictate some of the overlapping, so the only way to make it perfect would be...to do away with your schooldays altogether -- not something we have control over.

What you can do, something I've done repeatedly when there's class and a Math Jam at the same time, is be in both classrooms and resize the windows so you can see both windows at the same time. It's all right if the classes don't go insanely fast.


calculus
integration
trigonometry
LaTeX
derivative
logarithms
calculus computations

Can some one help me explaining the following two integration? Thanks

a. [Integral] SIN^2 (pi/6)
b. [Integral] SIN (pi/2 + pi/6 t)

For a)
Do you mean $\int{\sin}^{2}({\pi \over 6}) dt$?
If yes, then it is just the integral of a constant and the answer is ${\sin}^{2}({\pi \over 6})t$

For b)
Do you mean $\int \sin({\frac{\pi}{2}+\frac{\pi}{6}t) dt}$
If yes, then

$\int \sin({\frac{\pi}{2}+\frac{\pi}{6}t) dt = \int \cos(\frac{\pi}{6}t) dt =-\sin(\frac{\pi}{6}t)\cdot \frac{6}{\pi}}$

By the way, could you please write in LATEX next time. I gurantee you that more people would answer your question if you would just make an effort to write in LATEX. If you don't know how to write in LATEX then you can learn about LATEX in our LATEX forum
http://www.artofproblemsolving.com/Forum/index.php?f=123

Also, you can also learn by emulating the various posts that are posted in any forum. Just move you mose over a specific mathematical expression and you should see the LATEX code. When you post your message, just put that code with a $\$$ sign around each end and it should display that code in LATEX when you hit 'Preview' button(or when you submit your post for that matter).

Thanks that was fast!
Actually I wanted to do integrate SIN ^3 (pi/6) dt
instead of square.  But regardless why do u think, it is just integral of the constant?  Can you explain please?

b. On the second one, I was thinking there is a formula that I need to evaluate first. Some thing like SIn (Ax + B) =1/a (sin bla bla * Cos bla bla).  MAy be I wrong.

[quote="bengalliboy"]Thanks that was fast!
Actually I wanted to do integrate SIN ^3 (pi/6) dt
instead of square.  But regardless why do u think, it is just integral of the constant?  Can you explain please?[/quote] The term 'constant' in the context of integrals/derivatives just means that it has no dependence on the variable you integrate or differentiate over. In this case, your variable $t$, does not appear in ${\sin}^{3}({\pi \over 6})$ and thus you know that it must be a constant.

[quote="bengalliboy"]b. On the second one, I was thinking there is a formula that I need to evaluate first. Some thing like SIn (Ax + B) =1/a (sin bla bla * Cos bla bla).  MAy be I wrong.[/quote]I guess you could use that formula but the easiest one I used was the identity $\sin({\pi \over 2}+x) = \cos(x)$. Here is a link that lists all the trignometric identities that you would ever encounter in any of you math classes (unless you plan to become a mathematician ofcourse). :)
http://www.mathwords.com/t/trig_identities.htm

edit: sorry, I forgot to put the link before :blush:

:mad:  What are the derivatives for $\ln 2$ and $\sqrt 2$ ?  :P


Hello
I am taking AIME problem series A class.
i had my first class last week and this week the class was cancled due to some technical difficulties..?
When i signed up for the class i thought i would see how i like it in the first two sessions and decide whether or not i want to continue because according to the website full refund is given when a student drops a class within 10 days after the first session. (to let the student attend 2 sessions)
But in this case I do not get to attend 2 sessions in the 10 days since the class was cancled this week.
Is it possible that you extend the time period to two weeks so i'll be able to attend one more session before i make my decision?

Absolutely.  You will be able to make your decision during the week following the class next Tuesday.


Is it possible to have an equation for a circle that has an xy term in it??? If so, what is the equation?

Nope.  Every circle in the plane is given by an equation without an $ xy$ term in it, and the equation for a conic section is unique.  The closest thing you'd get is an ellipse such as $ (x \plus{} y)^2 \plus{} y^2 \equal{} r^2$.

you guys are great!!!!im just new here..hehehe..wow...


This error is boring:

[b]Fatal error: [/b]Allowed memory size of 8388608 bytes exhausted (tried to allocate 11520 bytes) [b]in /data/home/mathlinks/www/includes/mods_settings/mod_announces.php on line 92[/b]

I had the same stuff before, but then with lang_main.php

so I hope you'll be fixing this? :D

thx

yes. it means that there isn't sufficient RAM memory allocated. the php default is 8Mb, but I think it needs a little "update" :)

it will be fixed, just hit refresh when this happens, ok ?

sorry about it.

ya i did :D I know how to get rid of these, but I know other users who don't

got it again while replying :(

Fatal error: Allowed memory size of 8388608 bytes exhausted (tried to allocate 184320 bytes) in /data/home/mathlinks/www/language/lang_english/lang_main.php on line 474

and now again for line 136 :?

the occurence will be always random, as it inputs the error message when you filled up the memory :? well I need to get down to the server in person ... to fix some stuff.

just drop them an e-mail

piece of cake for the server gurus :D

yes, unfortunately I've been e-mailing them about a lot of things and I got the reply-s like "sure we'll do it when we have time" -> nothing done yet, so I have to do it personally.

:D why does this give me a deja-vu feeling regarding to my host here :D

btw, you certainly need to add the razz smiley  :P

This is the reason why I developed the habit of temporarily copying each of my posts so that I wouldn't lose it in case of internet connection problems or a problem on the server side...

well, luckily this one doesn't make you lose posts

elseway it'd have been QUITE boring :D


group theory
abstract algebra
superior algebra
superior algebra unsolved

See http://www.mathlinks.ro/viewtopic.php?t=121274 for a general description of Sylows of $ S_n$.


greatest common divisor
number theory unsolved
number theory

Prove:
(1) If $2n-1$ is a prime number, then for any n distinct positive integers $a_{1},a_{2},...,a_{n}$, there exist $i,j\in\{1,2,...,n\}$ such that
\[\frac{a_{i}+a_{j}}{(a_{i},a_{j})}\geq2n-1; \]
(2) If $2n-1$ is a composite number, then there exist n distinct positive integer $a_{1},a_{2},...,a_{n}$, such that for any $i,j\in\{1,2,...,n\}$,
\[\frac{a_{i}+a_{j}}{(a_{i},a_{j})}<2n-1, \]
where $(x,y)$ denotes the greatest common divisor of positive integer $x,y$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130463]Whole Paper[/url]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=130501
Please do not post something that has been posted on the same page.


Asymptote
geometry
ratio
analytic geometry

[geogebra]c344697c7da7d2f127f25b78690c399816fc9269[/geogebra]

The blue dots are connected to form a triangle, and the grey points are the midpoints of the sides of the triangle. Do you see anything interesting in the picture? Try moving the blue dots.

What do you mean "something interesting"? There's the fact that all the lines coming from the midpoints (I forgot what they're called) converge at one point...

Is that what you meant?

Yes, that is what I meant.
Can you prove that they always converge?

Ugh...I forgot the theorem...I think I'll remember if you tell me what those lines are called... :wallbash_red:  :wallbash_red:  :wallbash_red:

can someone put an asymptote picture? I can't view geogebra

Try getting the latest Java. It shouldn't harm your computer.

And they are called medians, and they intersect at the centroid.

Can you prove that they intersect at one point for any triangle?

(without using [[Ceva's Theorem]] )

I clicked on the first link I found in Google. There may be others.

http://math.coe.uga.edu/jwilson/EMAT6680Fa05/Hawkins/Assignment%204/triangleproofs.html

I have had this idea:

To proof: The median lines of every triangle cross in exactly one point.

1) Every triangle has exactly one centroid.

There must be a positive number of centroids. Two or more centroids imply infinite area and thus cannot occur.

2) The centroid is the point where all median lines cross.

Each median line splits a triangle in two triangles with equal size. The intersection point has the property of balancing the triangle to all three sides.

What do you think?

this fact can also be proven with analytic geometry.

or alternatively, take two medians and show that if the third passes through their intersection, then it must also be a median.

[quote="violin321"]couldn't it be proven with the definition of centroid? the point at which the medians of a triangle converge?[/quote]
There exists a elementary and easy proof using euclidean plane geometry.

[hide]
Let $ BE,CF$ be medians of $ ABC$ intersecting at point $ G$. Let $ AG$ produced meet $ BC$ at $ D$.

We need to prove $ BD\equal{}DC$

Produce $ AG$ to $ P$ such that $ AG\equal{}GP$, join $ BP,CP$.

 [geogebra]50f1ad1ad81ad2ddeb3e4bef4c2124ed4cd25c99[/geogebra]

In the triangle $ ABP$, $ F,G$ are respectively the midpoints of $ AB,AP$. Hence $ GF \parallel{} PB\implies CG \parallel{} PB$

Similarly it can be proved that $ BG \parallel{} PC$.

Hence $ BGCP$ is a parallelogram. Thus It's diagonals bisect each other.

We have 2 direct consequence of this: $ AD$ is the third median (QED) + the median is divided into $ 2: 1$ ratio by the Centroid.[/hide]

Very elegant! Thanks! I was looking for a proof like that ...

oh wait-now that i look at that proof and think about it, i think we did that in geometry...rofl

Yeah, that's kinda of like the proof I was trying to remember (or is it the same...)

Anyways, going back to some of the earlier posts, some people were asking why this couldn't be proved using the definition of a centroid, etc.

The problem is that that is what this proof is about: showing that such a point always exists.
If we tried to prove it like that, then we'd be stuck in circular reasoning.

yes the proof was to prove the definition of centroid.

I thought I could establish some reasoning that goes backwards.

In general: If we want to prove something, we could start out with a proof for one result to which the relationship of interest is necessary to hold and then proceed towards the relationship of interest.

I agree that in my statement above, I would need to give a definition of centroid that is different from "the point where all medians cross" otherwise it would be circular reasoning. But why should such a definition not exist? If one starts out very vague by asking whether [i]for every finite 2D object there exists a point A such that, if we poise the object on that point and let a "force" push orthogonal to the area and targeted against the "force" of the poising the object will remain constant[/i]

Not very sophisticated, I agree, but we should be able to prove that such a point exists for every triangle and that it must satisfy the conditions that all medians cross in that point.

If Andrew Wiles had found a different proof of Fermat's last theorem (the one that Fermat himself had found - if he was right), he could have concluded that some of the relationships he proved in his (Wiles') version of the proof of that theorem must hold.

Sorry if I'm a bit confusing here... you are welcome to help me out!

Your second sentence was a tad confusing, but I understood the rest of it. The question arises whether you can show that such a point exists. (I am referring to the 3rd paragraph.) It would be nice to show this without using vectors/physics. Also, how do you know that the centroid is the point where this phenomenon occurs, without knowing the definition of the centroid, and that it exists?

I don't actually have to know that the point I am looking for is what we call "centroid", do I? Given my unsophisticated definition of the centroid, every line that goes through that point must cut the triangle in two tri- or quadrangles of equal area. That would imply as a special case, the point where the three medians cross, because every median splits each triangle in two equally sized (=same area) triangles.

Therefore it should be possible to proceed in the way I indicated with my first reply to this problem.

Is there a proof, other than the one given above, that there is such a point, though? (Following your idea, of course.)
I'm not that familiar with point-masses, so I don't know if you can show that there is a center of mass for every finite 2D figure.

Hmmm...

Couldn't you also use coordinate geometry? (i.e. points $ (0,0)$, $ (a,0)$, and $ (b,c)$)

Yeah, but coordinate geometry isn't always the most elegant. :wink:

I guess at least for every convex figure there should be a point through which all lines that seperate the object in two objects of the same area must go...

This is one of the hardest parts about geometry. Knowing something intuitively and saying that it's obvious is not reason enough for saying it's true. You must prove it first, before I will accept this. (And don't even try to say that it could be accepted as a postulate. :wink: )


group theory
abstract algebra
complex numbers
superior algebra
superior algebra unsolved

Related to http://www.mathlinks.ro/Forum/viewtopic.php?t=112354

What are the $m\in \mathbb{N}^{*}$ s.t. $S_{n}$ (resp. $A_{n}$) could be embedded in $\mbox{GL}_{m}(\mathbb{C})$?

Note: I know the answer for $\mbox{GL}_{m}(\mathbb{Q})$ or $\mbox{GL}_{m}(\mathbb{Z})$ and I think we treated that one on this forum, but not for $\mbox{GL}_{m}(\mathbb{C})$.

Hi.

Let $m_{n}$ be this minimal m s.t. there is an embedding $A_{n}\to GL_{m}(\mathbb{C})$. Than there is of course such an embedding for all $m\geq m_{n}$.
And since $A_{n}$ can be embedded into $A_{n+1}$ we have $m_{n}\leq m_{n+1}$ for all n. We now $m_{1}=m_{2}=m_{3}=1$ and with the help of the charactertables one can find $m_{4}=m_{5}=3$.

The group of permutation matrices yield embeddings of $S_{n}$ into $GL_{n}(\mathbb{C})$, so $m_{n}\leq n$.

But what next?

[quote="Gockel"]...
The group permutation matrices yield embeddings of $S_{n}$ into $GL_{n}(\mathbb{C})$, so $m_{n}\leq n$.

But what next?[/quote]
Gockel, since $S_{n}$ can be embedded in $\mbox{GL}_{n-1}(\mathbb{Z})$, you also have $m_{n}\leqslant n-1$. But unfortunately $S_{n}$ cannot be embedded in $\mbox{GL}_{n-2}(\mathbb{Z})$ .
So then what's next :?:

Let n be greater than 5.
If there is an embedding $A_{n}\to GL_{m}$, there is also an embedding into $U_{m}$ since every finite subgroup of $GL_{m}$ is conjugated to a subgroup of $U_{m}$ and because we know that all embeddings of $A_{n}$ are in $SL_{m}$, we get that $A_{n}$ can be embedded into $SU_{m}$.

Since $m_{n}$ is minmal and $A_{n}$ is simple, any embedding into $GL_{m_{n}}$ is an irreducible representation of $A_{n}$, so knowledge about caractertables would be very useful...

Actually, from the general theory of representations of the symmetric groups (look the topic up in any book dealing with representation theory), it follows that all irreducible representations of $ S_{n}$ (and hence all representations) over the complex numbers are equivalent to rational representations (and the same is true for the alternating groups). Thus the question asked by JC_math has the same answer whether we work in $ \mathrm{Gl}_{n}(\mathbb C)$ or in $ \mathrm{Gl}_{n}(\mathbb Q)$. Furthermore, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=27907]this[/url], the answer is the same if we work in $ \mathrm{Gl}_{n}(\mathbb Z)$ as well.

Wow :!:  :!:  :!:
That's a fantastic result, we are saying that $S_{n}$ can't be embedded in $\mbox{GL}_{n-2}(\mathbb{C})$ (because it can't be embedded into $\mbox{GL}_{n-2}(\mathbb{Z})$), which is far from obvious since there are many more finite groups in $\mbox{GL}_{n-2}(\mathbb{C})$ than in $\mbox{GL}_{n-2}(\mathbb{Z})$.


group theory
abstract algebra
number theory
relatively prime
superior algebra
superior algebra unsolved

Let $ G$ be a finite group and let $ H_1,H_2,...,H_r$ be normal subgroups s.t. the order of $ H_i$ is relatively prime to the order of $ H_j$ for $ i\neq j$. Prove that $ H_1\cdots H_r\cong H_1\times \cdots \times H_r$.

Just http://www.mathlinks.ro/viewtopic.php?t=189690 and induction.


calculus
integration
inequalities
function
real analysis
real analysis unsolved

Let $f,g$ be two continuous function with $g$ of class $C^1$ incresing and positive and $\int_0^x f(t)dt \le\int_0^xg(t)dt,\forall x\in [0,1]$. Prove that $\int_0^x f^\alpha(t)dt\le \int_0^x g^\alpha(t)dt$ where $\alpha\le 1$.


function
trigonometry
inequalities unsolved
inequalities

[u][i] Problem  [/i] [/u]  :  [i]Find the least value of the function :
  $ f(x) \equal{} cos^{4}x \plus{} sin^{2}x \plus{} sinxcosx$[/i]

$ \mathrm{f}(x)\equal{}\frac{(1\plus{}\cos{2x})^2}{4}\plus{}\frac{1\minus{}\cos{2x}}{2}\plus{}\frac{\sin{2x}}{2}$

$ \mathrm{f}'(x)\equal{}\minus{}\sin{2x}(1\plus{}\cos{2x})\plus{}\sin{2x}\plus{}\cos{2x}$

$ \color{white}.\quad\ \,.$ $ \equal{}\cos{2x}(1\minus{}\sin{2x})$

We need only consider $ x\in[0,\pi)$ as f has period $ \pi$. It is a simple matter to check that the minimum value is at $ x\equal{}\frac{3\pi}{4}$, where $ \mathrm{f}(x)\equal{}\frac{1}{4}$.


inequalities
vector
induction
function
ceiling function
inequalities unsolved

Suppose $A\subset \{(a_1,a_2,\dots,a_n)\mid a_i\in \mathbb{R},i=1,2\dots,n\}$. For any $\alpha=(a_1,a_2,\dots,a_n)\in A$ and $\beta=(b_1,b_2,\dots,b_n)\in A$, we define
\[ \gamma(\alpha,\beta)=(|a_1-b_1|,|a_2-b_2|,\dots,|a_n-b_n|), \] \[ D(A)=\{\gamma(\alpha,\beta)\mid\alpha,\beta\in A\}. \] Please show that $|D(A)|\geq |A|$.

Does this work?  :o

Let $|A| = a$.  Playing around with the cases $n = 1, 2$, we are inspired to induct on $a$.

For $a=1$, the result is trivial because taking $\alpha = \beta$, we deduce that the zero vector is in $D(A)$.

Assume that $|D(A)| \geq a$ some integer $a \geq 1$.  For an element $\alpha$ in $A$, let $f_k(\alpha)$ denote the $k$th term in the $n$-tuple $\alpha = (a_1,\dots , a_n)$.  (Of course, $k = 1,\dots , n.$)  Now, consider the element $X = (x_1, \dots , x_n) \in A$ such that $f_1(X) > f_1(Y)$ for any $Y \in A$ such that $X \neq Y$.  If $f_1(X) = f_1(Y)$ for some $X\neq Y$, then out of all the elements $Z \in A$ with $f_1(Z)$ maximal, we choose an $X$ such that $f_2(X) > f_2(Y)$ for any $Y \in A$ such that $X \neq Y$.  If instead $f_2(X) = f_2(Y)$ for some $X \neq Y$, we may continue this procedure and move on to $f_3(X)$.  Note that eventually, we can obtain an $X$ such that for some $Y \in A$ with $X \neq Y$, we have $f_k(\gamma (X,Y)) > 0$ for some $k \in \{1,\dots , n\}$ and $f_j(\gamma (X,Y)) = 0$ for all integers $0 < j < k$.  This will hold, unless $k=1$, but $f_1(\gamma (X,Y)) > 0$ is all we will need in this case.  We can choose such an $X\in A$ satisfying these because otherwise, if there exists $X,Y \in A$ with $X \neq Y$ and $f_k(X) = f_k(Y)$ for all $k = 1, \dots , n$, then $X=Y$, contradiction.

Now, consider the set $A/ \{X\}$.  We apply the induction hypothesis on this set, so we have $|D(A/ \{X\})| \geq a-1$.  Now, pick an element $X' \in A/ \{X\}$ as we did when we chose an $X \in A$, except in the exact opposite way.  (Essentially, we reverse all of the inequality signs.)  Thus, there will exist some $k \in \{1,\dots , n\}$ such that $f_k(\gamma (X, X')) > f_k(\gamma (X, Y))$ for any $Y \in A/ \{X\}$ satisfying $f_j(\gamma (X, Y)) = f_j(\gamma (X, X'))$ for $j = 1,\dots ,k-1$.  Unless, of course $k=1$ (when the result is pretty much clear), this procedure is valid.  Thus, we have $|D(A)| > |D(A/ \{X\})| \geq a-1$.  This completes the induction.

(Yeah, that was kind of hard to write-up so it's kind of vague... I'll probably rewrite it later, but feel free to object -- I'm not sure at all if this is even close to being valid. :P )

Actually, the last part isn't completely true.  Take, for instance (5, 3, 3), (5, 1, 1), (3, 2, 2).  So, X = (5, 3, 3) and X' = (3, 2, 2) but nothing new in D(A) is added for this pair.  (Thanks to darktreb for pointing that out.)

Any ideas?  :?:

OH SWEET I THINK I GOT IT  :D

Let $m = |A|$ and let the $n$-tuples of $A$ be $(a_{i,1}, a_{i,2}, \dots , a_{i,n})$ where $1 \leq i \leq m.$
Without loss of generality, assume $a_{1,1} \leq a_{2,1} \leq \cdots \leq a_{m,1}$.

We proceed by induction on $n$.  For $n=1$, our ordering must be strict, so $a_{1,1} < a_{2,1} < \cdots < a_{m,1}$.  Note that
\[ 0 = a_{m,1} - a_{m,1} < a_{m,1} - a_{m-1,1} < \cdots < a_{m,1} - a_{1,1}  \]
are $m$ elements all in $D(A)$ so it follows that $|D(A)| \geq m = |A|$.  Now assume that the result holds for some positive integer $n$.  Let $X_1$ be the set of all distinct values of $a_{i,1}$ for $1 \leq i \leq m$.  Let $r = |X_1|$ and let $x_1 < x_2 < \cdots < x_r$ be the elements of $X_1$.  Moreover, define $t_j$ to be the number of $a_{i,1}$ equal to $x_j$ for $1 \leq j \leq r$ .  In other words, we have
\[ a_{1,1} = \cdots = a_{t_1,1} < a_{t_1+1,1} = \cdots = a_{t_1+t_2,1} < \cdots < a_{m-t_r+1,1} = \cdots = a_{m,1}.  \]
Clearly, we have $t_i \geq 1$ for $1\leq i \leq r$ and also $t_1+t_2+\cdots + t_r = m$. Now, consider the sets of $(n-1)$-tuples $B_k$ for $1 \leq k \leq r$, where an element $(b_1, b_2, \dots , b_{n-1})$ is in $B_k$ if and only if $(x_k, b_1, b_2, \dots , b_{n-1}) \in A$.  We have at least $r$ distinct values of the first entry in the elements of $D(A)$, namely,
\[ 0 = x_1 - x_1 = x_2 - x_2 = \cdots = x_r - x_r < x_r - x_{r-1} < \cdots < x_r - x_1.  \]
If the first entry of an element $\alpha \in D(A)$ is $0$, then there are at least $|D(B_k)| \geq |B_k|$ (by the induction hypothesis) of these elements for all $1 \leq k \leq r$.  If the first entry of an element $\alpha \in D(A)$ is $x_r - x_\ell$ for some $1\leq \ell < r$, then there are at least $|D(B_r)| \geq |B_r|$ or at least $D(B_\ell) \geq |B_\ell|$ of these elements in $D(A)$, where the inequalities again follow from the induction hypothesis.

Thus, we have
\[ |D(A)| \geq \sum_{i=1}^{r-1} \textrm{max}\{t_i, t_r\} + \textrm{max}\{t_1,t_2,\dots , t_r\} \geq \sum_{i=1}^{r-1} t_i + t_r = m = |A|,  \]
as desired.

[edit] nvm maybe this is wrong?

[quote]If the first entry of an element  is , then there are at least  (by the induction hypothesis) of these elements for all . If the first entry of an element  is  for some , then there are at least  or at least  of these elements in , where the inequalities again follow from the induction hypothesis.

[/quote]

Not true.  Suppose A contains the four triples

(-1,-1,0)
(-1,1,0)
(1,0,-1)
(1,0,1).

Then D(A) only has one element whose first element is 2.

[hide="Solution"]
Induct on $n$. For the base case $n = 1$ let the elements of $A$ be $z_1, z_2, \cdots, z_{|A|}$ and, without loss of generality, $z_1 < z_2 < z_3 < \cdots < z_{|A|}$. Thus, for all $1\le j\le |A|$, $(z_i - z_1)$ is an element of $D(A)$.  Since $z_i - z_1\ge 0$ and $z_i\ne z_j$ when $i\ne j$, $D(A)$ has at least $|A|$ elements. Now suppose the inequality is true when $A$ consists of at $n-1$-tuples and we will prove the inequality if $A$ consists of $n$-tuples.

Let $A$ consist of the elements $e_i = \{a_{i, 1}, a_{i, 2}, a_{i, 3}, \cdots , a_{i, n} \}$ for all $1\le i\le a = |A|$. Consider the $n-1$-tuples $e_i' = \{a_{i, 2}, a_{i, 3}, a_{i, 4}, \cdots , a_{i, n} \}$ for all $1\le i\le a$, and let $f_1, f_2, f_3, \cdots, f_t$ be the set of distinct elements $e_i'$, ordered such that if $n_j$ is the number of $e_i'$ so that $e_i' = f_j$ for each $1\le j\le t$, then $n_1 \le n_2\le n_3\le \cdots \le n_t$. Set $A_j$, for all $1\le j\le t$, to be the set of $e_i$ so that $e_i' = f_j$. By definition, $\sum_{i=1}^t n_i= a$. Define the function $f(f_i, f_j) = \{ |a_{u, 2} - a_{v, 2}|, |a_{u, 3} - a_{v, 3}| , \cdots , |a_{u, n} - a_{b,n}| \}$ if $e_u' = f_i$ and $e_v' = f_j$.

Correspond these sets to vertices of a graph; let vertex $v_i$ correspond to the set $A_i$ in our graph. Now define the following process. Start at $v_t$, draw an edge to itself, and record the $n-1$-tuple $f(f_t, f_t) = \{ 0, 0, 0, \cdots , 0 \}$. Then, draw an edge between $v_t$ and $v_{t-1}$ and record the difference $f(f_{t-1}, f_t)$. Now, if $f(f_t, f_{t-2})$ has not yet been recorded, draw an edge between $v_t$ and $v_{t-2}$ and record $f(f_t, f_{t-2})$. Otherwise, do nothing and proceed. If $f(f_{t-1}, f_{t-2})$ has not been recorded, draw an edge between $v_{t-1}$ and $v_{t-2}$ and record $f(f_{t-1}, f_{t-2})$. Otherwise, do nothing and proceed. Do the same for the pairs $(v_t, v_{t-3})$, $(v_{t-1}, v_{t-3})$, and $(v_{t-2}, v_{t-3})$ in that order. Keep doing this for $t - 4, t - 5, \cdots , 1$. Say that once we have determined whether to draw an edge between $v_{i-1}$ and $v_i$, we have "completed the $i$th set."

By Lemma 1 below, there are at least $\left\lceil \frac{n_x + n_y - 1}{2} \right\rceil$ differences between elements of $A_x$ and $A_y$ (upon taking differences between the first element of $A_x$ and the first of $A_y$). Moreover, by the base case of the original problem, there are at least $n_t$ differences between elements of $A_t$. Let $n(A_x, A_y)$ be the number of distinct $n$-tuples in the set of $f(s, t)$, where $s$ ranges over all elements of $A_x$ and $t$ ranges over all elements of $A_y$. Summing this over all sets $(x, y)$ so that $v_x$ and $v_y$ are connected gives that \[|D(A)| \ge \sum_{v_i, v_j connected} n(A_i, A_j) + n_t \ge \sum_{v_i, v_j connected} \left\lceil \frac{n_i + n_j - 1}{2} \right\rceil + n_t\] because differences between $A_i$ and $A_j$ will form a "new difference" in the last $n - 1$ elements of the $n$-tuple if $v_i$ and $v_j$ are connected.  By the inductive hypothesis, after completing the $j$th set, there are at least $j$ edges for all $1\le j\le n$. Hence (by induction, for instance), \[\sum_{v_i, v_j connected} \left\lceil \frac{n_i + n_j - 1}{2} \right\rceil \ge \sum_{i=2}^t \left\lceil \frac{n_i + n_{i-1} - 1}{2} \right\rceil + n_t\] which is greater than or equal to $ \sum_{i=1}^t n_i = a = |A|$ by Lemma 2 below. Hence, $|D(A)| \ge |A|$, as desired.

[b]Lemma 1:[/b] Given two sets of reals $X = \{x_1, x_2, x_3, \cdots , x_k\}$ and $Y = \{y_1, y_2, y_3, \cdots , y_l\}$, the set containing the distinct values of $|x_i - y_j|$, where $i$ ranges from $1$ to $k$ inclusive and $j$ ranges from $1$ to $l$ inclusive, contains at least $\left\lceil \frac{k+l- 1}{2} \right\rceil$ elements.

[b]Proof:[/b] It suffices to show that there are at most $k + l - 1$ elements in the set consisting of distinct values of $x_i - y_j$. Proceed by induction on $k + l$ to prove this. If $k +l = 2$, this is clear. Suppose it is true for all $k, l$ so that $k + l\le r - 1$, and we will show that it holds for $k, l$ so that $k + l = r$. Assume that $x_1 < x_2 < x_3 < \cdots < x_k$ and $y_1 < y_2 < y_3 < \cdots < y_l$. If $\min\{k , l \} = 1$, suppose $k = 1$, so the elements $(y_i - x_1)$, for all $1\le i\le l$ are pairwise distinct and lie in our set of differences, thereby yielding $l$ differences, as claimed. Now, suppose $\min\{k, l\} > 1$. If $X' = X - x_k$, then there are at least $k + l - 2$ distinct values among differences between elements of $X'$ and $Y$ due to the inductive hypothesis. Now, $x_k - y_1 \ge x_k - y_j \ge x_i - y_j$ for all $1\le i\le k$ and $1\le j\le l$, where equality only occurs when $i = k$ and $j = 1$. Thus, $x_k - y_1$ is a new difference, so our set has at least $k + l - 1$ elements, as claimed.

[b]Lemma 2:[/b] Given integers $t\ge 2$ and $1\le n_1 \le n_2 \le n_3 \le \cdots \le n_t$, \[\sum_{i=2}^t \left\lceil \frac{n_i + n_{i-1} - 1}{2} \right\rceil + n_t \ge \sum_{i=1}^t n_i\]

[b]Proof:[/b] Proceed by induction on $t$. Consider the base case, when $t = 2$. If $n_1 = n_2$, then $\left\lceil \frac{n_1 + n_2 - 1}{2} \right\rceil = n_1$, so \[\left\lceil \frac{n_1 + n_2 - 1}{2} \right\rceil + n_2 = n_1 + n_2\] as claimed. If $n_1 > n_2$, then \[\left\lceil \frac{n_1 + n_2 - 1}{2} \right\rceil + n_2 \ge \frac{n_1 + n_1 + 1 - 1}{2} + n_2 = n_1 + n_2\] The base case is thus proven. Now suppose that result holds for $t = r - 1$, and we shall prove the result for $t = r$. If $n_{r-1} = n_r$, then \[\sum_{i=2}^r \left\lceil \frac{n_i + n_{i-1} - 1}{2} \right\rceil + n_r = \sum_{i=2}^{r-1} \left\lceil \frac{n_i + n_{i-1} - 1}{2} \right\rceil + n_{r-1} + \left\lceil \frac{n_{r-1} + n_r}{2} \right\rceil \] \[\ge \sum_{i=1}^{r-1} n_i + n_r = \sum_{i=1}^r n_i\] If $n_r > n_{r-1}$, then \[\sum_{i=2}^r \left\lceil \frac{n_i + n_{i-1} - 1}{2} \right\rceil + n_r = \sum_{i=2}^{r-1} \left\lceil \frac{n_i + n_{i-1} - 1}{2} \right\rceil + n_r + \left\lceil \frac{n_{r-1} + n_r - 1}{2} \right\rceil \] \[\ge \sum_{i=2}^{r-1} \left\lceil \frac{n_i + n_{i-1} - 1}{2} \right\rceil  + \frac{n_{r-1} + n_{r-1} + 1 -1}{2} + n_r \ge \sum_{i=1}^r n_i \] by the inductive hypothesis. Hence, the lemma is proven.

[/hide]

Sorry to revive this, but I had an incomplete idea, could someone help me complete it?

The condition is very much suitable for using the Combinatorial Nullstellensatz but I probably cant find any nice polynomial.


AMC
AIME
trigonometry
AMC 12
AMC 10
USA(J)MO
USAMO

I'm currently a seventh grader and I've practiced the AMC 8 several times already. I have noticed that I nearly get perfect scores every time. But I get frustrated on the AMC 10. I know I have to do well on it to make AIME, and I know it's easier to qualify through the AMC 10 than through AMC 12. I also know that AMC 10 has a test A and a test B. When I do the real thing, am I allowed to take the AMC 10 twice in one year and use my best score to qualify to AIME?

And what materials other than the 21st Century Contests CD do you prefer using for AMC 10 prep?

um..yeah..you can take both on each date.
that's what FAQ pages are for.

as for what materials to use...AoPS vol 1 is good for AMC 10..really good...a little bit of AopS 2 is good for the last few problems...the intro textbooks as well...
speaking from personal experience, i think vol 2 is especially good for AMC 12/AIME/early USAMO...
i just took my first AMC 12 today...practice...i bombed... :D

You cannot take two A date contests or two B date contests- they're offered at the same time, and have some of the same questions. If your school happens to offer both test dates, you can take one contest on each date, in any combination of 10/12 (as long as you're eligible for both). You qualify for the AIME if either score meets the qualifying threshold.

[quote="gauss1181"]I know it's easier to qualify through the AMC 10 than through AMC 12.[/quote]
This is not at all obvious. The AMC 12 isn't worth it if you don't know the algebra and trigonometry that it uses, but it's probably easier to qualify for the AIME through the 12 if you have had that material. There is a very big difference between needing 100 points and needing 120.

Oh...so there are lower point requirements on the AMC 12, right? But even though the content on the AMC 10 is easier, it requires more points.

20 points is more than 3 problems you know;even if the difficulty is easier in the 10, I don't know if 3 questions are easier to get.
Also, I've found that the 10A is easier than the 10B and the 12B is easier than the 12A (but that's just my opinion), so I think it's best if you take 1 10 and 1 12.

BTW, does anyone know if you make it through both 10 and 12, which score they consider for index? The higher one?

They consider both. So if you are attempting to qualify through the index cutoff, you have to use your AMC12 score. However, with the floor, they will use the higher of the two for tiebreaking.

Personally, I got a 146.5 on the AMC10 in 8th grade and so in 9th grade I didn't see any reason other than a perfect score (which is unlikely due to stupid errors) to take the AMC10, so I took 12/12 and got 138/144.

Granted, I'm not at all a common case. It goes with how much you've learned. If you don't have all the materials for the AMC12, then take the AMC10. Of course, if you can take 2, I recommend taking one of each so that you're at least exposed to the harder problems.

Be sure to read Frequently Asked Questions AMC 10/12 #12, AIME #2 and AIME #8 on our Frequently Asked Questions Page:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions

im scoring 3-5 average on AIME but the thing is, i cant even qualify. i score around 90 AMC 12 and 110-15 AMC 10. WTF!

I guess you just need to practice. For me, it's the other way, I can usually get ~130 on AMC 10, ~110 on AMC 12, but only 4-6 on AIME, which won't qualify me for USAMO safely.

[quote="xpmath"]Oh...that's not good. If I try to qualify by index, my AMC 12 score will probably be below 120, which probably won't get me in to USAMO...so I'll just have to start practicing AIMES!
Thanks.[/quote]

Uh, I'm not sure how you didn't know what the AMC index was, as it's in your signature. Also, whether you take AMC 10 or 12 sometimes can decide which color MOP you end up in, and indirectly qualification for IMO.

Also, what's the difference between AMC A and B? I know AIME I has answers of integers in $ [1,999]$, AIME II has all answers, but what's the difference for AMC?

I didn't really know what they used for index (as in AMC 12 or AMC 10 score), just wanted to make sure.

There's no difference, just different problems.

Do they exist so that people could take both AMC 10 and 12?

[quote="gauss1181"]Errr, i once did a practice test on the AMC 10 and scored a 115. So do you think that means I can also exceed 100 points on AMC 12? cuz aren't the AMC 10 and 12 problems inter-related in some way by difficulty or something?[/quote]

Try not to assume things like that. Best thing to do is to take practice tests for both. Also, practice tests scores are almost invariably higher than the real ones, so keep that in mind.
By the way, in 8th grade, I managed a 138 on the AMC 10, but the next year, I only scored a 112.5 on the AMC 12.

[quote="Temperal"]Also, whether you take AMC 10 or 12 sometimes can decide which color MOP you end up in, and indirectly qualification for IMO.[/quote]
This is false.  MOP color distinctions are based on USAMO performance, with modifications as the MOP directors see fit.  And if you have a chance at qualifying for the IMO, it is probably irrelevent whether you take the AMC 10 or 12.

(And please read the [url=http://www.unl.edu/amc/f-miscellaneous/faq.shtml]FAQs[/url] on the AMC website, as they answer most questions asked here.)

Also, nobody can say for sure whether you will qualify for the AIME or USAMO before you actually take the tests.  So just practice and do the best you can, and try not to worry about it.  If you practice, you'll be better then than you are now, anyhow.  Also, practice scores are [b]not[/b] almost invariably higher than the real ones.

A 115 on AMC 10 is actually quite good but you need to keep in mind that this year they aren't allowing calculators and much like last year when they changed the grading system, the score to qualify for AIME will most probably not be 120 for AMC 10 or 100 for AMC 12.

I guessed that the qualifying scores would dip down last year- and I'd guess that they'll go back up this year, just because it's a little easier to calibrate the difficulty of the questions with a year of data in the current format.
Don't forget that the AMC also switched from 2.5 points per blank question to 1.5 last year- that's an average of seven points for all AMC 12 AIME qualifiers last year, more than enough to account for the change in qualifying levels.
(Data extracted from the AIME qualifier item difficulty document)

[quote="davidyko"]practice tests scores are almost invariably higher than the real ones[/quote][quote="Boy Soprano II"]practice scores are [b]not[/b] almost invariably higher than the real ones.[/quote]
This is an interesting issue- you can simulate almost everything execpt the adrenaline of the real contest in a practice environment. How do you react to that adrenaline? Some people are destroyed by it, and have trouble with all sorts of tests. Some feed on it, and do their best under that pressure. Most people lie somewhere inbetween. There's no way to know where you are except experience.

Where do I fit? I think faster and more clearly with the adrenaline running through me. I consistently outperformed my practice runs in the real competition.

I got a better score on the AIME last year than I had ever gotten in practice, so practice tests are NOT invariably higher/lower than the real thing.

Oh. Ok, i got it now. Thanks for the advice.

[quote="temperal"]

Also, what's the difference between AMC A and B? I know AIME I has answers of integers in $ [1,999]$, AIME II has all answers, but what's the difference for AMC?[/quote]

Actually, AIME I and II are exactly the same except for the problems (and the date) :wink:

Speaking wrong, authoritatively, does not look very good.

[quote="Temperal"][quote="Boy Soprano II"][quote="Temperal"]Also, whether you take AMC 10 or 12 sometimes can decide which color MOP you end up in, and indirectly qualification for IMO.[/quote]
This is false.  MOP color distinctions are based on USAMO performance, with modifications as the MOP directors see fit.  And if you have a chance at qualifying for the IMO, it is probably irrelevent whether you take the AMC 10 or 12.
[/quote]

What is your source for this? This article: http://en.wikipedia.org/wiki/USAMO seems to contradict that.[/quote]
No, it's purely USAMO score. Maybe if you tie with someone for the last spot, but that's way too rare a case to consider.

[quote]Actually, AIME I and II are exactly the same except for the problems (and the date) :wink:

Speaking wrong, authoritatively, does not look very good.[/quote]

Mmm, it doesn't. does it? Another mistaken idea of mine shot down.

No, it's purely USAMO score. Maybe if you tie with someone for the last spot, but that's way too rare a case to consider.

[quote]No, it's purely USAMO score. Maybe if you tie with someone for the last spot, but that's way too rare a case to consider.[/quote]

That's strange, I seem to have read otherwise somewhere, but I'll take your word for it. Wasn't there at least argument over the issue of qualification through 10/12, though?

It USED to be something convoluted like AHSME+10*AIME+4*USAMO (The formula structure is right, the coefficients are probably wrong), but they scrapped it for just the USAMO score.


In a triangle ABC, AB is smaller then BC and BC is smaller than AC. The points $A', B', C'$ are such that $AA'$ is perpendicular to $BC$ and $AA' = BC$, $BB'$ is perpendicular to $AC$ and $BB' = AC, CC'$ perpendicular to $AB$ and $CC' = AB$ .
If $<AC'B = 90$ degrees, prove that $A', B'$ and $C'$ are colinear. (lies on a straight line)

Just posted here:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86611]www.artofproblemsolving.com/Forum/viewtopic.php?t=86611[/url]

What is OMK?

[b]Angle-Chasing Method [/b]

If we draw out , we can see that $C'$ lies in $ABC$ while $A',B'$ lie outside $ABC$ . Let the orthocenter be $H$ and let $BB'$ and $AC'$ intersect at $M$ ; $AA'$ and $BC'$ intersect at $N$ . Also let $\angle BC'A'=x$ ,$\angle AC'B'=y$ ,$\angle C'AH=a$ ,$\angle C'BH=b$ ,then

$\angle AA'C'=180^{\circ}-\angle C'AH-\angle AC'A'\\=180^{\circ}-a-(\angle AC'B+\angle BC'A')\\=180^{\circ}-a-(90^{\circ}+x)\\=90^{\circ}-a-x$ .....(i)

$\angle BB'C'=180^{\circ}-\angle C'BH-\angle B'C'B\\=180-b-(\angle B'C'A+\angle AC'B)\\=180^{\circ}-b-(y+90^{\circ})\\=90^{\circ}-b-y$.....(ii)

$\angle C'MB=90^{\circ}-\angle C'BM=90^{\circ}-b$
$\angle C'NA=90^{\circ}-\angle C'AN=90^{\circ}-a$

$\\\implies \angle B'HA'=360-\angle C'MB-\angle C'NA-\angle AC'B\\=360-(90^{\circ}-b)-(90^{\circ}-a)-90^{\circ}\\=90^{\circ}+a+b$ .....(iii)

Also $\angle AA'C'+\angle BB'C'+\angle B'HA'=180^{\circ}$
so from (i),(ii),(iii) $(90^{\circ}-a-x)+(90^{\circ}-b-y)+(90^{\circ}+a+b)=180$ $\Longleftrightarrow x+y=90^{\circ}$ $\implies \angle BC'A'+\angle B'C'A+\angle AC'B=x+y+90^{\circ}=180^{\circ}$ .

Hence $A',B',C'$ collinear .

Very nice solution Shyong. You kill it :D

Can some moderator please merge the 3rd post of this topic into http://www.mathlinks.ro/Forum/viewtopic.php?t=88609 and delete the rest?

  darij


Which weapon is better- a broadsword or a rapier ?
What is the absolute best non-projectile weapon?
Post your answer and defense for it.

Anyway who has played Dota will give a direct answer: rapier :D

I was speaking of superiority in full-contact live steel combat between two equally skilled opponents. In a couple of days I will post my defense of the better weapon.

rapier in my opinion. although that's because i'm too small to wield a broadsword.

Actual broadswords were expertly weighted so that they would balance at the hilt. This made their average weight of 2 to 3 pounds seem even lighter than it was. Most rapiers, however, were a bit too long to be maneuvered comfortably. Plus, there was the added advantage of having a cutting/chopping blade on ye olde english broadsworde.


In quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $P$. $AC$ bisects $\angle C$ and $BD$ bisects $\angle B$. If $\angle APB=45^\circ$ and the area of $\triangle PBC$ is $12$, find the area of quadrilateral $ABCD$.

Hmmm.... I can imagine a quadrilateral with such conditions, but with D corner undetermined..... English is not my mother tongue, but I understand that quadrilateral is a convex polygon with four sides (any length, no restrictions).

For example, angle B is a right angle, angle C too. PA is parallel to BC, AB = 1/2 BC, D can be in any place, I think.

Sorry, I just found a mistake on the problem :blush:  I forgot to mention that the two angle bisectors are the diagonals of the quadrilateral...

It's fixed now.

[quote="ajlopez"]Ok, then, an informal proof, yields to me an total area of 24 for the quadrilateral.

Angel "Java" Lopez[/quote]
can you post your solution?

Yes, I can post an informal and incomplete solution. I think that or this is the way, or there is another simpler way....

First, an comment: prolong BA, CD, so these lines cut on, say, point R. Then BRC is a right angle (90 degrees). Because, angle BPA = 45, then

BPA + BPC = 180
BPC = 135
PBC = angle B/2
PCB = angle C/2

then B/2 + C/2 + BPC = 180
B/2 + C/2 = 45
B + C = 90
B + C + R = 180
R = 90

Then, all quadrilaterals of the problems, are obtained of the intersections of bisectrix B o CR, and bisectrix C on BR. All point Rs are on a semicircle.

After this intro, some semiproof:

Select point A' on BC, such BA' = BA.

Then, it can be prove:

triangle BPA = BPA'

The same with D' on BC, such triangle CPD = CPD'

My proof is based on the hypothesis triangle APD has the same area of  triangle PA'D'.

If we can proof that, then triangle BPC has the same area of the rest of ABCD.

To prove triangle APB = PA'D', select E on AB such AE = EB (the middle point of AB).

And select E', the middle point of A'D'.

Then AEP is similar triangle to PA'E'.
And DEP is similar to PD'E'.

I need to proof that are equals.

It can be easily proved in case AD is parallel to BC.

Angel "Java" Lopez

Gotcha!

To prove triangle APD has the same area that triangle PA'D', the procedure is:

angle APD = angle PA'D' + angel PD'A'

Yes, because, with angles:

APA' + A'PD' + D'PD + APD = 360

but APA' = 2 APB = 2 * 45 = 90 according to the initial conditions, then. Same for D'PD, then

A'PD' + APD = 180
but A'PD' + PA'D' + PD'A' = 180 (are the angles of a triangle)

APD = PA'D' + PD'A'

By a similar chain, it can be proved for angles

A'PD' = PAD + PDA

Now, select E' on segment A'D' such angle A'PE' = angle DAP. And select E on segment AD such angle APE = angle PA'D'. Then the triangles APE and A'P'E' have two equal angles and one side (AP = A'P). Then are equals.

It can be noted than angle D'PE' = angle ADP. And, following the same process, it yields triangle DPE is equal to triangle PBE'.

Then triangle APD has the same area that triangle PA'D'

Angel "Java" Lopez

Nice job! That's basically the same as one of my solutions.

[quote="ajlopez"]My proof is based on the hypothesis triangle APD has the same area of triangle PA'D'.

If we can proof that, then triangle BPC has the same area of the rest of ABCD. [/quote]

Here's an interesting way to prove $\triangle APD=\triangle PA'D'$.

Rotate $\triangle PA'D'$ $90^\circ$ clockwise about $P$ so that $PA$ overlaps $PA'$. Note that three points $D$, $P$ and $D'$ are collinear. Now we have $\triangle D'AD$ where $P$ is the midpoint of $D'D$. Hence $\triangle APD=\triangle PA'D'$ ;)

Or, if you know trigonometry, you can show it simply by using $\text{Area}=\frac{1}{2}ab\sin{C}$, where $a$ and $b$ are two side lengths and $C$ is the angle formed by the two sides. Since we have two congruent sides and $\sin{45^\circ}=\sin{135^\circ}$, the two areas are equal.


I think that the "limiting sum" is just the "infinite sum" of the sequence.

er yes, you guys are right. I think the term arises from the way an infinite sum is defined. Obviously, an "infinite sum" is not a very rigorous defenition. So, the actual defenition is the limit as n goes to infinity of the partial sums, where a partial sum is defined as the sum of the terms from 1 to n. So that's where the term comes from: the limit of the partial sums.

[quote="MysticTerminator"]So that's where the term comes from: the limit of the partial sums.[/quote]

Exactamundo!

[quote="gauss202"]Exactamundo![/quote]
You mean "exactamente" :)

OK, I don't like where this is going, so let's just end it here by saying "You say potato salad, I say Mr. Potato Head salad :twisted: "

[quote="mathdude"]I was recently working through a past test of a local math contest and came across the unfamiliar term of "limiting sum".  More specifically the question asked:
[color=red]"Find the limiting sum of the series
                       4-8/3+16/9-32/27+....."[/color]

My best guess would be to just find the sum of the series, which I calculated to be 12/5 (I do not have the key).  Is limiting sum just a fancy way to describe what I calculated or am I completely off track?  Any info would be appreciated.[/quote]

Oh, by the way. I don't think that answer of 12/5 is correct.

er. i think it's correct

5 BUCKS SAYS IT'S CORRECT!!! ok, jk

4/(1-(-2/3)=12/5

I am pretty sure he is right... We can break this down into two infinite geometric sequences with the ratio of 4/9...

4/(1-4/9) - 8/(3(1-4/9)) = 12/5

[quote="Fierytycoon"]MysticTerminator: I don't think you have taken into account that the series has alternating + and - signs.[/quote]

yes I did. note that the denominator is 1-(-2/3) not 1-(2/3). So, Tare, now I get to ask the question. Cash or check?

[quote]yes I did. note that the denominator is 1-(-2/3) not 1-(2/3). So, Tare, now I get to ask the question. Cash or check?[/quote]

Whoopsiedoo.

ah okay, I couldn't see the minus signs. It should be correct then. *thumbs up*

[quote="Ragingg"]Do you accept Visa or American Express?[/quote]
You're not paying. You sided with Mystic. gauss202, Fiertycoon, and I would be the ones paying.
Course, we don't owe as much as teratomato. He owes a grand to TripleM. (Look [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=920]here[/url])

[quote="TripleM"]Yep. I'm waiting patiently :)[/quote]
You know him, he'll just bilk you.


induction
Pascal\u0027s Triangle
\/closed

Here are a couple extension questions from [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=30]tonight's Math Jam[/url].

Click the link above to see the problem set-up.

1) Show that

[tex]T(1) + T(2) + T(3) +\cdots +T(10^k-1) = 55^k - 1[/tex]

2) Find a closed expression for

[tex] [T(1)]^2 + [T(2)]^2 + [T(3)]^2 + \cdots +[T(10^k-1)]^2[/tex]

3) Find a closed expression for

[tex] [T(1)]^r + [T(2)]^r + [T(3)]^r + \cdots +[T(10^k-1)]^r[/tex]

for any general integer r.

4) Suppose we write N in base p, where p is prime, and we define T(N) as before.  Prove that there are exactly T(N) terms in the row of Pascal's triangle starting '1 N ...' which are [b]not [/b]divisible by p.

thoughts... [hide]1. I could do another of my huge non-induction proofs, but yes, induction is easier.

Just for simplicity, let us prove instead that [tex]\displaystyle \sum_{i=0}^{10^k-1} T(i) = 55^k[/tex] (notice I added in the T(0) term).

Anyway, the basecase is k=1, in which case we are summing from 0 to 9. However these are all 1-digit, so T(n)=n+1, so we are adding the integers from 1 to 10, which is clearly 55=55^1.

So that takes care of the base case.

Now for the inductive step, assume the equation is true for k. We wish to prove it for k+1.

Well, we split up the LHS of the sum based on the first digit as so: [tex]\displaystyle \sum_{i=0}^{10^{k+1}-1} T(i) = \sum_{i=0}^{10^k-1} T(i) + \sum_{i=10^k}^{2\cdot10^k-1} T(i) +\ldots[/tex][tex]\displaystyle + \sum_{i=9\cdot10^k-1}^{10^{k+1}-1} T(i)[/tex].

Now, the first term, by the inductive hypothesis, is just 55^k.

The second term is almost identical to the first term except all terms have a 1 at the beginning. This will cause each term to be multiplied by (1+1), i.e. 2. So the second term is 2*55^k, etc.

So the sum is 1*55^k+2*55^k + ... + 10*55^k = (1+2+...+10)(55^k) = 55^(k+1), as was to be shown.

3. Define f(r)=[tex]\displaystyle \sum_{i=1}^{10} i^r[/tex]. As a few examples, f(1)=55 and f(2)=385. Then I claim the closed form we seek in number 3 is [f(r)]^k-1. We use the same factoring technique as in the previous solution - everywhere there is a 55, replace it with a f(r), basically. I do confess that, as of right now, I have not found a closed form for f(r).

4. Lucas' theorem states that if p is a prime, and the base-p representation of n is n_m, n_(m-1), n_(m-2), ... , n_0, and the base-p representation of i is i_m, i_(m-1), ..., i_0 (where n_0 and i_0 are the units digits of n and i respectively), then [tex]\displaystyle {n \choose i} \equiv \prod_{k=0}^m {n_k \choose i_k} [/tex] (mod p). We have that p|C(n,i) iff p|C(n_k,i_k) for some k. Since n_k, i_k :le: p, we have that C(n_k, i_k)=0 only if i_k>n_k for some k. We will count the complement here, since it's easier: this will not happen only if for every k, 0<=i_k<=n_k, and as we have (n_k+1) choices for each i_k, we have T(N) choices total for the number of ways in which p does not divide C(n,i), which leads me to believe the problem was stated wrong?[/hide]

Sure. p=3, N=510=123 so that T(Np) = 6. The desired row of Pascal's triangle is 1, 5, 10, 10, 5, 1. Note that 0 terms are divisible by p=3, while 6 terms, or T(Np) terms, are NOT divisible by p=3. This is a specific counterexample to the claim that "T(N) terms will be divisible by p", and confirms our suspcions that "T(N) terms will NOT be divisible by p", if my superb proof wasn't enough to remove all suspicions ;)

Now I see the confusion - I left a word out (you're supposed to read my mind, Mystic).

I've added the missing [b]not[/b]

[quote="rrusczyk"](you're supposed to read my mind, Mystic).
[/quote]

ooh, my bad.

By the way, in problem number 3, are we supposed to be able to find a closed form for f(r) or will they be happy if we just defined in terms of the sum ([tex]\displaystyle \sum_{i=1}^{10} i^r[/tex])

Yeah, for #3, you can give me a semi-closed form like that.


Hi there!

I was wondering whether someone could explain how to answer the following questions:

[quote="Question 1"]A three-legged table never wobbles, but a four-legged one may do so. Why? Why do people make four-legged tables, then?
[/quote]

[quote="Question 2"]A finite geometry is defined as follows. There are eight points, $ {P_{0}}, ... ,{P_{7}}$. A line is a set of two points. A plane is a set of four points $ \{{P_{a}},{P_{b}},{P_{c}},{P_{d}}\}$ for which the Nim-sum $ a\otimes b\otimes c\otimes d$ is zero. (the Nim-sum of two numbers is defined by writing them out in binary notation and then adding them in binary but ignoring all carries.) Show that this finite geometry obeys all the axioms of incidence.[/quote]

Can anyone explain what Nim-sums are and what all the stuff inside the brackets is trying to tell me?

Thanks
 :)

[hide="Question 1"]
A three legged chair never wobbles (though it may be slanted). This is because in euclidean geometry, three points always define one and only one plane. This means that a three leggen chair (or table) will rest with all three legs on any flat floor.

However, if the legs of a four legged chair are not in a plane, the four legged chair will wobble, since its legs will determine two planes instead of one. then the chair can alternate between resting on one plane, then the other. this is the "wobble".

Hm i dont know why people make 4 legged chairs though...[/hide]


trigonometry
inequalities
inequalities proposed

(1)If $x$, $y$ are real numbers, then
$\cos x+\cos y-\cos(x+y) \leq \frac 3{2}$;

(2)If $x$, $y$, $z$ are real numbers, then
$\cos x+\cos y+\cos z-\cos(x+y+z) \leq 2\sqrt 2$;

(3)If $x_{1}$, $x_{2}$, ..., $x_{n}$ are real numbers, then
$\cos x_{1}+\cos x_{2}+...+\cos x_{n}-\cos(x_{1}+x_{2}+...+x_{n}) \leq (n+1)\cos\frac \pi{n+1}$.

I think rather easy than hard. (that is, if my soln is correct)  :)

Put $m: =n+1$ and $x_{m}: =\pi-(x_{1}+x_{2}+...+x_{n})$. Then the ineq becomes $\sum_{k=1}^{m}\cos x_{k}\le m\cos\frac \pi{m}$ with the condition $\sum_{k=1}^{m}x_{k}=\pi$. Wlog we can assume $x_{i}\in[-\pi,\pi]$.

If $n\ge4$ thus $m\ge5$, it is easy to see that $m\cos\frac \pi{m}\ge m-1.$ So if one or more of the $\cos x_{k}$ are negative, then $\sum_{k=1}^{m}\cos x_{k}\le m-1$ and done. So we can suppose all $x_{i}\in[-\frac\pi2,\frac\pi2]$. But then the result follows by Jensen or mixing of variables via the concavity of $\cos$.

The remaining cases $n=2$ and $n=3$ are easy by mixing of variables.

I think your solution is valid if we replace  $x_{i}\in[-\pi,\pi]$ with $x_{i}\in[-3\pi/2,3\pi/2]$. :lol:
Can you prove that
$2\sqrt{2+\cos x+\cos y}\leq 5\cos\frac \pi{5}+\cos(x+y)$,
for any reals $x$ and $y$?

[quote="Vasc"]I think your solution is valid if we replace  $x_{i}\in[-\pi,\pi]$ with $x_{i}\in[-3\pi/2,3\pi/2]$. :lol:
[/quote]Of course, but what's the use please  :lol: :rotfl: ?
[quote="Vasc"]Can you prove that
$2\sqrt{2+\cos x+\cos y}\leq 5\cos\frac \pi{5}+\cos(x+y)$,
for any reals $x$ and $y$?[/quote]OK, thinking about that one now. Sure enough, the (?) equality case is $x=y=\frac{2\pi}5$ :)

[quote="Vasc"][quote="spanferkel"] Wlog we can assume $x_{i}\in[-\pi,\pi]$.[/quote]
Is it possible?
Can you do it in the case $a_{1}=-\pi/3, a_{2}=...=a_{n-2}=0, a_{n-1}=\pi/6, a_{n}=7\pi/6$ such that
 $x_{1}+...+x_{n}=a_{1}+...+a_{n}=\pi$ and $\cos x_{1}+...+\cos x_{n}=\cos a_{1}+...+\cos a_{n}$ with $x_{i}\in[-\pi,\pi]$?[/quote]

Oh, tell me Vasc, why part (b) is open? Should we make a try? Solution of spanferkel has problem?

Well OK, I should have said that we deal [i]first [/i]with the case that some of the $cos x_{i}$ are negative. And only after that we do a "wlog" assumption that can lead us then right away to assume $x_{i}\in[-\frac\pi2,\frac\pi2].$  :blush:  Not a big deal anyway I think.
I wonder if apart from that, we could [i]still [/i]apply Jensen, even if $\sum x_{i}=-\pi$ or $3\pi$ or whatever odd multiple. (supposed of course $x_{i}\in[-\frac\pi2,\frac\pi2].$)  <What do you think?

[quote="spanferkel"] I wonder if apart from that, we could [i]still [/i]apply Jensen, even if $\sum x_{i}=-\pi$ or $3\pi$ or whatever odd multiple. (supposed of course $x_{i}\in[-\frac\pi2,\frac\pi2].$)  <What do you think?[/quote]Actually, I published in 1991, in Gazeta Matematica, the problem:

If $x_{1}$, $x_{2}$, ..., $x_{n}$ are real numbers such that $|x_{1}+x_{2}+...+x_{n}| \leq \pi$, then
$\cos x_{1}+\cos x_{2}+...+\cos x_{n}\leq n\cos\frac{x_{1}+x_{2}+...+x_{n}}{n}$.

I have proved it by induction. But the Spanferkel's solution also works. :lol:

[quote="Vasc"]But the Spanferkel's solution also works. :lol:[/quote]Yeah, with your condition $|x_{1}+x_{2}+...+x_{n}| \leq \pi$  it does work. But I have just realized that in your original problem (3), if we have e.g. $n=8$ and $x_{1}=...=x_{8}=\frac\pi3$, we would obtain $x_{9}=\frac{-5\pi}3$. What then? If we replace by $x_{9}=\frac{\pi}3$ (and even without that), my solution with the "mixing variables" approach doesn't work, nor would Jensen of course.  :(
So I wonder if my method is even "recuperable" at all for problem (3), or do we need a different approach. Note that the ineq becomes weaker as $n$ grows - may that help?

I think your method is recuperable.
If $k$ of $x_{i}$ are in $[-2\pi,-3\pi/2]$ and the other $n-k$ in $[-\pi/2,\pi/2]$, then $n\geq 4k+2$ from condition $x_{1}+...+x_{n}=\pi$.
Replacing each of these $x_{i}$ by $x_{i}+2\pi$ and applying Jensen, we still have to show that $\cos\frac{\pi+2k\pi}{n}\leq \cos \frac \pi{n}$.
This is true, since $\frac \pi{n}\leq \frac{\pi+2k\pi}{n}\leq \frac \pi{2}$. :lol:

Fine. I have repaired it in a slightly different way. For the case that all $\cos x_{i}\ge0$, we can modify all but one $\mod 2\pi$, i.e. assume wlog $x_{1},...,x_{n-1}\in[-\frac\pi2,\frac\pi2].$ By concavity, it is now sufficient to prove the ineq for $x_{1}=...=x_{n-1}=: x$, thus to prove $(n-1)\cos x-cos(n-1)x\le n \cos \frac \pi{n}$, which is straightforward. :)

[quote=Vasc](1)If $x$, $y$ are real numbers, then
$\cos x+\cos y-\cos(x+y) \leq \frac 3{2}$;[/quote]
http://www.artofproblemsolving.com/community/c6h496087p2785606:
Let $\alpha,\beta,\lambda$ be real numbers . Prove that  $$\lambda (cos\alpha+cos\beta)-cos(\alpha+\beta)\le1+\frac{\lambda^2}{2}.$$


Find all function continous and bound f:R to R such that :
  abs (f(x+y+z)) >= abs ( f(x)+f(y)+f(z)) for all x,y,z in R

Actually only the condition " bounded function" is sufficient .
$x=y=z=0 \Longrightarrow |f(0)| \geq 3|f(0)|\Longrightarrow f(0)=0$
Now it's easy to see that $|f(2x)|\geq 2|f(x)|,...., |f(2^{n}x)|\geq 2^{n}|f(x)|...$
So $f(x)=0$ because $|f(2^{n}x)|$ is bounded ..

 :cool:


Prove that for every fixed number $k$ ( $k \in N$ ) there is only finite number of positive integers like $n$, which can be written in the form $n = a_{1}+a_{2}+\cdots+a_{k}$ where $a_{i}$ are positive integers ( not necessarily distinct ) and
$\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{k}}= 1$.
( We call $n$ a fantastic number then :D  )

Consider the set $F=\{ f: \mathbb{N}_{k}\to \mathbb{N}\;/\;\sum_{i=1}^{k}\frac{1}{f(i)}=1 \wedge i\leq j\rightarrow f(i)\leq f(j) \}$. It will be sufficient to show $|F|\in \mathbb N$. To do so is enough to demonstrate that $\forall i\in \mathbb{N}_{k}\;|\{ f(i) \;/ \; f\in F \}|\in \mathbb N$. We will procede by induction. Suppose $\forall i \in \mathbb{N}_{m}\; |\{ f(i) \;/\; f\in F \}|\in \mathbb{N}$, then $| \{ g: \mathbb{N}_{m}\to \mathbb{N}\;/ \; \exists f\in F\;\! g\subset f \}|\in \mathbb{N}\Rightarrow \exists M\in \mathbb{Q},\; \forall f \in F \;\;\sum_{i=1}^{m}\frac{1}{f(i)}<M<1$; but since $\forall f\in F$,  $(1-M)<\sum_{i=m+1}^{k}\frac{1}{f(i)}\leq \frac{k-m}{f(m+1)}$, we have to conclude $\forall f\in F \;\;f(m+1)<\frac{k-m}{1-M}\Rightarrow |\{ f(m+1) \;/\; f\in F \}|\in \mathbb{N}$. Take $M=m=0\;\;\;,\sum_{i=0}^{0}\frac{1}{i}=0$,  for the base case and we are done, I suppose.

[ Sorry for using  $f(i)$  instead of  $a_{i}$; I hope it is not a problemo, baby.]

Hmmm.. it's basically the same idea, but it is easy to see that if, say $a_{k}$ is the least positive integer among the $a_{i}$'s then $a_{k}\leq k$. Thus there is only a finite number of possibilities for $a_{k}$. For each of them, that is for a fixed $a_{k}$, the same reasoning leads to a finite number of $a_{k-1}$ and so on..
This gives only a finite number of $k$-uples such that the sum of their inverses is $1$. This leads to a finite number of $n$.

Pierre.

Ook Guys, thank you very much, Nice solutions, but i think there's another solution too, it's easy to prove that the smallest value of $n$ is $k^{2}$, but it seems it's really difficult to find the biggest..... :lol:    Can anyone find it?


It takes exactly 74 colored beads on a string to make a necklace. The beads are strung in the following order: one red, one orange, two yellow, one green, and one blue. Then the pattern repeats starting again with one red bead. If the first bead of the necklace is red, what is the color of the last bead used to make the necklace?

$ 74\equiv2\pmod{6}$. Therefore the color of the bead is orange.