Let n is a positive integer.Prove that:
$ {(n!)}^{n^2 \plus{} n \plus{} 1}|(n^3)!$

plz edit it as soon as possible it is $ (n!)^{n^2\plus{}n\plus{}1} | (n^3)!$ :)  (just take $ n\equal{}2$ to see that the reverse is false!)
as far as i know we can easily prove that $ ((n\minus{}1)!)^{n^2\plus{}n\plus{}1} | (n^3)!$ (combinatorics)
now we have to prove the problem for $ p |n$ (we must prove $ v_p((n!)^{n^2\plus{}n\plus{}1}) \le v_p((n^3)!)$ and then we'll be done)

See [url]http://www.mathlinks.ro/viewtopic.php?t=303975[/url]. It solves the problem very nicely!
[hide="Solution"]
From the link above, we have $ (n^3)!\equal{}(n^2 \cdot n)!$ is divisible by $ (n^2)!^n \cdot n!\equal{}(n \cdot n)!^n \cdot n!$ is divisible by $ ((n!)^n \cdot n!)^n \cdot n!\equal{}(n!)^{n^2\plus{}n\plus{}1}$[/hide]


How many pounds of salt must be added to 75 pounds of a $ 30\%$ salt mixture to obtain a mixture that is $ 40\%$ salt?


[b]Ok here is the problem....It seems like it's easy but I know there has to be an equation out there somwhere? and if you know what i need to do can you please let me know. Thank ya!

[/b]A deck of cards was dropped on the floor. While Amel was out of the room, Naoko picked up at least one card. She may have picked up one, all fifty-two, or any number in between. How many possible combinations are there for what she picked up?

[b]I know I can start making up the combinations by myself but there has to be an easier way, my mind just can't think of it right now. Help me please![/b]

Either a card is picked or it isn't. That would make the number of combinations $2^{52}$ including no card picked up. Since at least 1 card picked up, that possibility is discarded, leaving $2^{52}-1$.


let P(x) be a polynom of degree n such that P(x)>=0 for all x>=0
prove that P(x)+P'(x)+P''(x)+P'''(x)+.......+P^{n}(x)>=0 for all x>=0

$P(x+t)=\sum_{k\ge 0}\frac{P^{(k)}(x)}{k!}t^{k}$. Now integrate against $e^{-t}$ from $t=0$ to $t=\infty$.


Diophantine equation
number theory unsolved
number theory

I have the following equation:

$ X^{2} \plus{} Y^{2} \minus{} 6XY \equal{} 1$

How can I convert it to pell equation of the form: $ 1 \equal{} X^2 \minus{} DY^2$ Where D is some integer (none square).

After converting I can solve it, but I cant seem to figure out a way (like to complete square or something).

Thanks!!!

[quote="fedja"]$ (X \minus{} 3Y)^2 \minus{} 8Y^2 \equal{} 1$[/quote]

Am I allowed to have a Y in the first square in Pell equation?

If $ a \equal{} X \minus{} 3Y, b \equal{} Y$ then $ X \equal{} a \plus{} 3b, Y \equal{} b$, i.e. one can recover the original desired solution from the modified desired solution.

Ok, thanks.

One more thing, on the (a) part of the problem they asked to turn $ \sqrt{2}$ to a continued fraction.

On (b) (this problem), they gave a hint to use (a) to solve it.

 So Im guessing that $ D$ in pell equation should be $ 2$.

Any thoughts?


function
search
algebra unsolved
algebra

[i]This is from the 2008 IMO Problem #4[/i]

Find all functions $ f : \left(o,\i\infty\right) \rightarrow f: \left(o,\infty\right)$ (so, [i]f [/i]is a function from the positive real
numbers to the positive real numbers) such that:

                                                               $ \frac{(f(w))^2 \plus{} (f(x))^2}{f(y^2) \plus{} f(z^2)} \equal{} \frac{w^2 \plus{} x^2}{y^2 \plus{} z^2}$

for all positive real numbers $ w,x,y,z,$ satisfying $ wx\equal{}yz$

[quote="Papalakis"][i]This is from the 2008 IMO Problem #4[/i]

Find all functions $ f : \left(o,\i\infty\right) \rightarrow f: \left(o,\infty\right)$ (so, [i]f [/i]is a function from the positive real
numbers to the positive real numbers) such that:

                                                               $ \frac {(f(w))^2 \plus{} (f(x))^2}{f(y^2) \plus{} f(z^2)} \equal{} \frac {w^2 \plus{} x^2}{y^2 \plus{} z^2}$

for all positive real numbers $ w,x,y,z,$ satisfying $ wx \equal{} yz$[/quote]
Please use search function .See http://www.mathlinks.ro/viewtopic.php?t=215430


geometry
3D geometry
pyramid
inequalities
rectangle
analytic geometry
perimeter

I will (probably in vain) try to get some action in this forum. Here is a problem from my blog.

A pyramid has a rectangular base. The rectangle's diagonal is $ 2\sqrt {106}$ and its area is $ 180$. Given that the height of the pyramid is $ \frac 23$ of the length of the longer side of the rectangular base, find, with proof, the least possible surface area of the pyramid.

Hm, I have not solved this yet, but now that I think about it, I get the feeling that this involves a pretty classical inequality, although this is not hard to solve intuitively without proof.

Ok nevermind it is now a classical inequality...actually it might be. I fail at inequalities really badly, as evident by the fact that I had like 5 people help me with this problem.

[hide="anyways"]
Ok, I need to work on being more concise, since the following is not nearly as long as it looks

outline:so its pretty easy to guess that the SA is minimized when the vertex is directly above the middle of the base, and its also pretty easy to see find the area and height of the base with symmetric manipulations.

hard part (at least for me): proving that the SA is minimized as aforementioned. Ok so lets make the rectangle ABCD and put it on the xy plane such that the sides AB and BC of the rectangle are parallel to the x and y axes respectively. Now consider the vertex E with z coordinate equal to the height of the pyramid. As we slide E sideways (i.e. only changing its x coordinate), we keep the areas of ABE and CDE constant, since the heights of these triangles depends only on the y and z coordinates. The areas of the other 2 triangles change as point E slides in this manner, and the rate of change is proportional to the rate of change of the perimeter of a triangle EFG, where F and G are on  BC and AD respectively and both have same y coordinates E. Similarly, as we slide E forward and backward (i.e. only changing its y coordinate), we keep the area os BCE and ADE constant and we do the same thing in this dimension.

Basically we've boiled this down to proving that that the perimeter of a triangle XYZ with a given base XY and given height from Z is minimized when Z lies on the perpendicular bisector of XY. We can do this by considering the set of ellipses with foci at X and Y and at least one point on the line  m parallel to XY through Z (the locus of points where Z can be). We note that the the perimeter of XYZ is basically the sum of the distances from X to Z and form Y to Z, so we need to find the smallest ellipse in this set. This ellipse is quite clearly the one that is tangent to m - i.e. Z lies on the perpendicular bisector of XY. QED

Ok sorry for that horrible explanation. Feel free to ask for clarifications....[/hide]

Boo Geometry! I felt like spamming just because I know no one ever comes on here. I stopped posting because of that :P


geometry
trigonometry
circumcircle
ratio
trig identities
Law of Sines

A triangle with sides of lengths 6, 10, and 14 is inscribed in a circle.  Find the area of the largest triangle that can be inscribed in the circle

[hide="Hint"]I got the diameter to be 14; the largest triangle should be equilateral[/hide]

[hide]Assuming the diameter is 14 and the largest triangle is equilateral, let $s$ be the length of a side of the triangle. Since the radius is 7, we can divide the triangle into 3 120-30-30 triangles. Using this, Law of Sines tells us that $s=7\sqrt{3}$. Then we find that the area of the triangle is $\frac{147\sqrt3}{4}$.[/hide]

[quote="Apprentice"]I got the diameter to be 14[/quote]
Are you sure that's right? ;)

[hide="A general strategy to find the circumradius of a triangle"]
For any triangle $\triangle ABC$, the circumradius $R$ is calculated by $\frac{abc}{4K}$, where $K$ is the area of the triangle.

To find the area, Hero's formula usually works pretty nicely.
[/hide]

yes you are right; I got the side to be 15; the circumradius 42*sqrt(675)/135; the diameter, then, is $\frac{84sqrt{675}}{135}$ - right?

By $R=\frac{abc}{4A}$, we get $R=\frac{14\sqrt 3}3$.  The largest area is with an equilateral triangle, which will have height $7\sqrt 3$ (medians cut into lengths in ratio 2:1).  The area is $\frac{h^2\sqrt 3}{3} = 49 \sqrt 3$


induction
number theory proposed
number theory

Let $n$ be positive integer,$r$ be positive rational number.
Prove that the number of the pair of positive integer $(x_1,x_2,\cdots,x_n)$  which satisfies the following equation is finite.

\[\sum_{k=1}^n \frac{1}{x_k}=r\]

Assume WLOG that $x_1\le x_2\le\ldots\le x_n$. $x_1$ cannot grow indefinitely, so it's bounded. Assuming there are infinitely many solutions to the equation, there is a value $x$ of $x_1$ which will be taken infinitely many times. This means that the equation $\sum _{i=2}^n\frac 1{x_i}=r-\frac 1x$ has infinitely many solutions. We can now proceed by induction (we have obtained the same problem, but with $n-1$ variables instead of $n$).


Let $a,b,c$  be positive reals satisfying $a^{2}+b^{2}+c^{2}= 1$ . Prove that $a+b+c+\frac{1}{abc}\ge 4\sqrt{3}$ .

[quote="arqady"][quote="Fitim"]
$(a+b+c)\frac{1}{abc}\geq 3\sqrt[3]{abc}\frac{1}{abc}$
[/quote]
The mistake is here.[/quote]

Why?  :maybe:  :wink:


Let ABCD be a quadrilateral with side AB extended to E so that AB=BE.  Segments AC and CE are drawn to form angle ACE.  If ACE is a right angle which of the following must be true about ABCD?

(a) All of the sides are equal.
(b) All of the angles are equal.
(c) ABCD is a parallelogram.
(d) At least two angles are equal.
(e) At least two sides are equal.

Show work.

[quote="lotrgreengrapes7926"][hide] If $\angle ACE$ is a right angle, then $BC$ is the median to the hypotenuse. So, $BC=AB=BE$. The value of $\angle ACE$ is independent to the position of $D$, so $\boxed{D}$?? [/hide][/quote]

D is not the correct answer.

I think you should try it again.

I also think you should fix your Latex.

First of all, is CE an extension of CD or is CE unaligned from CD?

They said something about E in the problem, so I think you should read it again.

[hide]Choice E is correct!

I don't understand your solution, but I can try to put a solution.

Solution
Since <ACE=90 degrees.

Then we can draw a semicircle around ACE, with diameter AE.

B is the midpoint of AE, so B is the center of semicircle ACE.

Hence AB=BC.

We can't figure anything else, so hence it is E

I think this is right.[/hide]

[quote="jclarkemathL314159"][hide]

This is probably wrong, since I'm terrible at geometry. This is going to be hard to explain without a picture. Anyway, call length AB $n$. Then BE also has length $n$. Since ACE is a right angle, BC is perpindicular to AE, BC = $n$, and AC = CE = $\sqrt{2}n$. AB = BC, and ABCD is a trapezoid?

So choice E.[/hide][/quote]

$BC$ is not perpendicular to $AE$, $AC\ne CE$, and I don't think it's a trapezoid.

I meant $E$. I just accidentally wrote $D$.

Yep, that is what I thought.  Your soluton made it seem like your answer is E, but just in case...

And jclarke, yes, you might want to edit your solution somewhat.


calculus
integration
function
calculus computations

Let $ f(x)\equal{}\begin{cases}\ 1\minus{}x & 0\leq x\leq 1\\0 & 1<x\leq 2\\(2\minus{}x)^2 &2<x\leq 3\end{cases}.$

Define the function $ F(x)\equal{}\int_{0}^xf(t)dt$.

$ F(x)\equal{}\left\{\begin{array}{ll}
x\minus{}\frac{x^2}{2}&0\leq x \leq 1\\
\frac{1}{2}&1\leq x \leq 2\\
\frac{1}{2}\minus{}\frac{(2\minus{}x)^3}{3}& 2\leq x\leq 3.
\end{array}
\right.$

I edited my false reply, didnt saw F(0)=0

How do you get $ F(x)=\frac{1}{2} ,\& 1<x\leq 2$


Please put anything funny you saw recently in chat here.

[**:31] LanguageFilter: AmericanPwn was banned for 15 minute(s) for using vulgar/abusive language!
[**:32] AmericanPi: lol americanpwn gj
[**:32] aleph0: really?
[**:32] aleph0: what did he say?
[**:32] AmericanPi: i have no idea
[**:32] AmericanPi: he is my brother
[**:32] AmericanPi: i will ask him
[**:32] LanguageFilter: AmericanPi was banned for 15 minute(s) for using vulgar/abusive language!
[**:32] aleph0: lol?

lol.
5 minute ago Americanpi and AMericanpwn were still arguing about this topic.

Hi, I have a collection of FTW quotes. It is eonic. Pay me \$0.00 for some free samples.

Notice that \$0.00 is a true value deal, as it is 1337.42% off the normal price.

This like just happened
18:46] CRICKET229: do you use calc
[18:46] AIME15USAMO: lol
[18:46] vahalla: no
[18:47] AIME15USAMO: yeah i use a calc
[18:47] AIME15USAMO: but not rlly
[18:47] vahalla: I just am better than you mental math
[18:47] CRICKET229: Aime your just smart
[18:47] jjfun1: lol
[18:47] AIME15USAMO: [18:47] vahalla: I just am better than you mental math
[18:47] rubberglov3: cd kwe?
[18:47] vahalla: those darn smart people
[18:47] AIME15USAMO: heh
[18:47] AIME15USAMO: im a smart person?
[18:47] AIME15USAMO: FAIL

[6:08] * meewhee009 is WAY MORE immature since she thinks its awesome when she teaches Billy Mays how to moonwalk
[6:09] * meewhee009 pwns
[6:09] * monkeygirl13 pwns meewhee009
[6:09] * meewhee009 pwns monkeygirl13 while she attempts to pwn meewhee009'

[quote="aleph0"][**:53] hyperddude: hi phire
[**:53] dragon96: hi phire
[**:54] aleph0: hi phire
[**:54] vahalla: hi phire[/quote]

This is from quite a while ago.

[19:07] jjx1: OMG
[19:08] r15s11z55y89w21: OMG
[19:08] THECHAMP: OMG
[19:08] JKowl: OMG

lol
I've seen stuff about se*
in the chat b4.

[quote="KingSmasher3"][quote="LLin1"]lol what did he get banned for?[/quote]

LanguageFilter bans anyone who uses swear/vulgar words.

Also,

Although I have played FTW for nearly three years, I only recently found out that if one enters " " without the quotes into the chat box, a space will appear. Although this is common in the source code of many documents, I think it is interesting that it works on FTW.  :)[/quote]

Interesting how you have played FTW for nearly three years, but your account was made 6 months ago, and FTW has been out for just over two years...

[17:36] CRICKET229: dang
[17:36] CRICKET229: not 20
[17:37] CRICKET229: fail
[17:37] bluecarneal: 40.
[17:37] CRICKET229: OH!
[17:37] CRICKET229: fail!
[17:37] westiepaw: fail
[17:37] bluecarneal: lol

I was the only one to get that one. ME the only one. wow.  :o

[quote="Berzerk"][17:33] Special: lol
[17:33] Special: lol
[17:33] abhinand: lol
[17:33] Special: lol
[17:33] Special: !@#@#
[17:33] Special: LOL

Everyone spams when there are no moderators. This is NOT fake.[/quote]

Are you therefore suggesting that everyone spams all the time now? :rotfl:


[img]http://i23.tinypic.com/aokhfl.jpg[/img]

I got \alpha but can't do part (2). What does it mean by around? the circumference of both circles?

[hide="answers"]$ (a)1.2661 radians$ $ (b)49.2cm$[/hide]

Er...problems that are blatantly obviously homework? No.

it may be better if the pics were clearer

and as david mentioned this belongs in another place not intermediate tops

[quote="bos1234"]This is not a homework. I'm studying for a test and don't know how to do this
and i photocopied from a book[/quote]

In that case, seek a teacher.

someone jsut answer the question  :roll:  :roll:

school is finished - im studying for HSC (stat) test

Search for "2 Pulleys" for very similar questions.  The discussions in those threads may help you.


[hide]20 percent of 2005 five is: 401

40 percent of 401 is: 160.4

30 percent of 160.4 is: 48.12

Answer: 48.12[/hide]

$\frac{2}{5}*\frac{3}{10}*\frac{1}{5}=\frac{3}{125}$ multiply that times 2005 and the answer is $\fbox{48.12}$


Let  $a,b,c$ sides of triagle. Prove the following ineq

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

Let p the semiperimeter of ABC. Mahe the next changing of variables:

.a=px
.b=py
.c=pz

Note that x,y e z are positive reals and x+y+z=2.

Our inequalitie is equivalente to prove that:

x^3/y^2+y^3/z^2+z^3/x^2>=2*(x/z+z/y+y/x-2) <->
xa^2-2/a+yb^2-2/b+zc^2-2/c>=-4, where a=x/y, b=y/z e c=z/x.

Suppose x>=y>=z and note that we must have a=b=1 -> c=1. Then:

xa^2-2/a+yb^2-2/b+zc^2-2/c>=x-2+y-2+z-2=-4.


geometry
geometric transformation
dilation
LaTeX

Suppose you have a sphrical ball with a mass of 100g(grams) at rest.If you throw the ball at three-quartes the speed of light. What will its mass become,as mursed from a stationary view point.

[hide="Here is what I got."]

$M=(1-0.75^2)^\frac{1}{2}$
  $=(1-0.563)^\frac{1}{2}$
  $=1/(.437)^\frac{1}{2}$
  $=1/0.6610$
  $=1.53$
  $100(1.53)$
  $M=151g$[/hide]

it's easy, because we know that from a stationary point of view, $m(v)=m_0\gamma$, where $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$. So just substitute and find the value

I'm sorry but i'm a noob at physics. I'm assuming Mo is initial mass and the weird y looking thing is the expression for time dilation? (i would ues latex but i can't get it to work on my computer).


algebra
polynomial
invariant
quadratics
Galois Theory
superior algebra
superior algebra unsolved

What's the degree of the extension Q(sqrt(2),sqrt(3)) and how do you know?

it'd be helpful to know what the minimum polynonmial is for the extension and to know for certain that it's irreducible.

thanks,

.

The degree of the extension is $ 4$ and a minimal polynomial is $ (x \minus{} \sqrt {2} \minus{} \sqrt {3})(x \minus{} \sqrt {2} \plus{} \sqrt {3})(x \plus{} \sqrt {2} \minus{} \sqrt {3})(x \plus{} \sqrt {2} \minus{} \sqrt {3})$.  This polynomial is invariant under every member of the Galois group, but this is not true of any other combination of its factors.

Alternately, $ \mathbb{Q}[ \sqrt {3}, \sqrt {2} ] : \mathbb{Q}[ \sqrt {2} ] \equal{} 2$ (can you see why?  Is this intuitive to you?  And can you prove it?) and $ \mathbb{Q}[\sqrt {2}] : \mathbb{Q} \equal{} 2$.

Well in this case, as the previous poster did, it is easy to come up with the minimal polynomial which happens to be a quartic.  To check that the quartic is indeed irreducible, you need to see if it breaks up into either the product of 2 quadratics with coefficients in $ Q$ or a cubic and a linear factor with coefficients in $ Q$. The latter is not possible because none of the roots are in $ Q$. The former is not possible since when you multiply any pairs of $ (x \minus{} r_1)(x\minus{}r_2)$ with $ r_1, r_2$ distinct roots, you don't get a polynomial with coefficients in $ Q$. So you can conclude that the quartic is irreducible and thus the minimal polynomial.  Since the degree of the extension is the degree of the minimal polynomial, you're done.

The other approach like the previous poster sort of suggested is to use the Tower Theorem and consider the degrees of intermediate extensions.  $ [Q(\sqrt2, \sqrt3): Q(\sqrt2)][Q(\sqrt2): Q] \equal{} [Q(\sqrt2,\sqrt3): Q]$. $ [Q(\sqrt2): Q] \equal{} 2$ since the minimal polynomial of $ \sqrt2$ over $ Q$ is $ x^2 \minus{} 2$. The minimal polynomial of $ \sqrt3$ over $ Q$ is $ x^2 \minus{} 3$ so that  $ [Q(\sqrt2, \sqrt3): Q(\sqrt2)] <\equal{} 2$. The only thing to determine is whether or not $ \sqrt3 \in Q(\sqrt2)$.  I'm sure you've proven that $ Q[\sqrt2]\equal{}Q(\sqrt2)$ in your course and so you can consider representing $ \sqrt3$ in the form $ a \plus{} b\sqrt2$ with $ a,b \in Q$. Just using the facts that $ \sqrt2, \sqrt3, \sqrt6$ are irrational it should be easy to come up with a contradiction.

Hope this helps (figured I'd fill in some more of the details for you).


calculus
integration
trigonometry
limit
real analysis
real analysis unsolved

using integral calculus, or otherwise, prove the identity:

$\frac{\sin{\theta}}{\theta}\;=\;\prod_{n=1}^{\infty}\left[ 1-\frac{4}{3}\cdot\sin^{2}{\left( \frac{\theta}{3^{n}}\right)}\right]$

[hide]Let $\phi_{n}=\frac{\theta}{3^{n}}$

$\prod_{n=1}^{\infty}( 1-\frac{4}{3}\cdot\sin^{2}{\phi_{n}})= \prod_{n=1}^{\infty}\frac{3-4\cdot\sin^{2}{\phi_{n}}}{3}= \prod_{n=1}^{\infty}\frac{\sin{3\phi_{n}}}{3\sin\phi_{n}}= \prod_{n=1}^{\infty}\frac{\sin{\phi_{n-1}}}{3\sin\phi_{n}}$

$P_{k}=\prod_{n=1}^{k}\frac{\sin{\phi_{n-1}}}{3\sin\phi_{n}}= \frac{1}{3^{k}}\prod_{n=1}^{k}\frac{\sin{\phi_{n-1}}}{\sin\phi_{n}}= \frac{1}{3^{k}}\cdot\frac{\sin{\phi_{0}}}{\sin\phi_{1}}\cdot\frac{\sin{\phi_{1}}}{\sin\phi_{2}}\cdots \frac{\sin{\phi_{k-1}}}{\sin\phi_{k}}\Rightarrow$

$P_{k}= \frac{1}{3^{k}}\cdot\frac{\sin{\theta}}{\sin\phi_{k}}$

$\lim_{k\to\infty}P_{k}= \frac{\sin\theta}{\lim (3^{k}\sin\phi_{k})}$

But $\lim{(3^{k}\sin\phi_{k})}= \lim[3^{k}\cdot\sin(\frac{\theta}{3^{k}})]= \lim\frac{\theta\cdot\sin(\frac{\theta}{3^{k}})}{\frac{\theta}{3^{k}}}= \theta\cdot\lim \frac{\sin\phi_{k}}{\phi_{k}}=\theta$

so $\lim_{k\to\infty}P_{k}= \frac{\sin\theta}{\theta}$[/hide]


inequalities
LaTeX
inequalities proposed

Let $a, b, c > 0$. Prove that
\[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{c+a}{c+b} + \frac{a+b}{a+c}+\frac{b+c}{b+a}.\]

[color=red]Moderator edit (un-understandable post).[/color]

This has been posted before. In fact, it is problem 2 of the first round of the first edition of the Mathlinks contest so it's easy to find a solution here. And oh yes, please improve your LaTeX skills and try to give your topic decent names or titles. Topic locked.


Problemele(10 probleme) au aparut intr - un topic de pe FORUMUL in lb. maghiara.
Acest concurs este un concurs prin corespondenta, termenul limita de trimitere al solutiilor fiind $ \text{5 noiembrie 2007}$.
Pentru a nu afecta corectitudinea concursului, $ \mathbf{VA\ ROG\ SA\ NU\ POSTATI\ PE\ FORUM\ NICI\ O\ SOLUTIE,}$
$ \text{inainte de expirarea termenului limita de trimitere a solutiilor!}$

$ \mathbf{P2: \ }$ Toate submultinile cu  $ {n}$ elemente ale unei multimi avand $ {n^2 + n - 1}$  elemente le impartimi in 2 grupe (familii).
Demonstrati ca una dintre cele doua familii contine  $ {n}$ multimi disjuncte, 2 cate 2!

$ \mathbf{P3: \ }$ Notam cu $ {\omega(n)}$ numarul numerelor prime din descompunerea in factori primi ai numarului
$ n\in\mathbb{N}$, iar cu:  $ {F(x) = \max_{n\le x}\omega(n),\ G(x)\max_{n\le x}[\omega(n) + \omega(n^2 + 1)]}$.
        Demonstrati ca: $ {\lim_{x \to \infty} [F(x) - G(x)] = \infty}$.

$ \mathbf{P6: \ }$ Care sunt submultimile $ A\subset \mathbb{R}$, cu proprietatea:
$ (\forall) \ 0\le x_0 < x_1 < \dots < x_n\le 1, n = 1, 2, 3, \dots \ (\exists) \ y_j\in A$, avand proprietatea ca:
$ y_{j + 1} - y_j > x_{j + 1} - x_j$  pentru $ (\forall) \ 0\le j < n$.
$ \mathbf{P8: \ }$ Consideram sirul: $ A = (a_i)_{i\ge 0}$  si notam cu
$ SA = (a_0,a_0 + a_1,a_0 + a_1 + a_2, ... )$ sirul sumelor partiale al seriei:
$ a_0 + a_1 + a_2 + a_3 + ...$
   Exista oare vreun sir nenul A, avand proprietatea ca toate sirurile A, SA, SSA, SSSA, ... sunt convergante?.

$ \mathbf{P9: \ }$Fie $ A$ si $ B$ doua triunghiuri situate in acelasi plan a caror interior contine originea si
avand proprietatea ca arcele fiecarui cerc $ C_r$  cu centrul in origine au proprietatea
$ |C_r\cap A| \equal{} |C_r\cap B|$ (am notat aici cu  $ | . |$ masura arcului cercului $ C_r$).
     Demonstrati ca triunghiurile A si B sunt congruente.
Ramane aceasta afirmatie adevarata si in cazul in care originea se gaseste pe perimetrul triunghiurilor $ A$ si $ B$?

Hello

Could we have all problems Schweitzer 2007 in english and in PDF, please ?

Thank you


Prove that if $ a\leq b\leq c\leq d$ and $ b\plus{}c\equal{}a\plus{}d\Rightarrow bc\geq ad$

so $ a \equal{} b \plus{} x$, $ d \equal{} c \minus{} x$, but $ ad \equal{} (b \plus{} x)(c \minus{} x) \equal{} bc \minus{} (b \minus{} c)x \minus{} x^2 \le bc$ since $ b \minus{} c \ge 0$
EDIT: whoops, looks like i switched around the variables but you get the point

I think, we may not subtract two inequalities

For example

$ 5<7$
$ 6<9$
If we subtract them, it will become $ \minus{}1<\minus{}2$, contradiction
I'm sorry if my statement is wrong

 :)

[quote="IW@IT"]I think, we may not subtract two inequalities

For example

$ 5 < 7$
$ 6 < 9$
If we subtract them, it will become $ \minus{} 1 < \minus{} 2$, contradiction
I'm sorry if my statement is wrong

 :)[/quote]

As $ (1)$ is an equality , the solution in my last post is correct. (The sign $ \ge$ has changed during the subtraction)

stephen I don't think you can do that. You don't know which is bigger $ (a\plus{}d)^2$ or $ (b\plus{}c)^2$

We also don't know if $ (d\minus{}a)^2\stackrel{?}{\geq}(b\minus{}c)^2$

So unless you prove that I don't think that is the solution. Sorry if I am mistaken.

Yes he can, a+d=b+c so their squares have to be equal, and clearly a and d are farther apart than b and c.

An easy way to think of this is to let $ p$ be the arithmetic mean of all the numbers (which by the condition is also the mean of b,c and a,d). Let $ b\equal{}p\minus{}x$ $ c\equal{}p\plus{}x$ $ a\equal{}p\minus{}y$ $ d\equal{}p\plus{}y$ with, by the condition, $ 0\le x \le y$

Then we need to show $ p^2\minus{}x^2 \ge p^2\minus{}y^2$ or $ (y\minus{}x)(y\plus{}x)\ge 0$ which is true.

We can replace $ (a,b,c,d)$ with $ (a\plus{}q,b\plus{}q,c\plus{}q,d\plus{}q)$ and the conditions are the same. Therefore, we can take $ q\equal{}\minus{}a$, and the inequality obvious.


You can actually just leave it as.

$x = \frac{1}{3}^3$

You can see that you then have: $\log_3{\frac{1}{3}}^{3} = 3\log_3{\frac{1}{3}}$

Using log properties is important so try to leave it simplified instead of multiplying $\frac{1}{3}^3$ out. Sometimes the numbers are too huge.

[hide]
$\log_2{8}=3, \log_8{2}=\frac13$
So the answer would be $-3$[/hide]


limit
logarithms
integration
calculus
calculus computations

Given - $a_{n+1}= a_{n}+ln(n)$.
Find $ \lim_{n \to \infty}\frac{a_{n}}{nln(n)}$

[quote="N.T.TUAN"][quote="pou"]Given - $a_{n+1}= a_{n}+ln(n)$.
Find $\lim_{n \to \infty}\frac{a_{n}}{nln(n)}$[/quote]

$-a_{n+1}=a_{n}+\ln{n}$ or $a_{n+1}=a_{n}+\ln{n}$ ? :maybe:[/quote]

$a_{n+1}= ...$
that "-" isnt minus.

You can also do it with Sterling's approximation of the factorial.

$a_{n}= a_{1}+\log (n-1)!$ so

$\frac{a_{n}}{n\log n}= \frac{a_{1}+\log{n!}-\log n}{n \log n}= \frac{a_{1}+\log{n^{n}}-\log{e^{n}}+\log \sqrt{n}+\log c(n)-\log n}{n \log n}$ where $c(n)$ approaches a constant (actually $\sqrt{2 \pi}$) as $n \to \infty$, and we can further rearrange this to $1+\frac{a_{1}+c(n)}{n \log n}-\frac{1}{\log n}+\frac{1}{2n}$.

Or this:

$\int_{1}^{n}\ln x\,dx< a_{n}-a_{1}<\int_{1}^{n+1}\ln x\,dx$

Use the fact that we can explicitly compute the integral.


[quote="Kent Merryfield"]¿Cuál es el dígito de las unidades de $S=1!+2!+3!+\cdots+99!$ ?[/quote]

Bueno, este esta bastante facil

[hide="Mi Solucion"]Notese que para $n>4,\ 10|n!$, entonces esos no contribullen con nada al digito de las unidades, y $1!+2!+3!+4!=33$, bueno, tengo que decir mas?[/hide]

[quote="RaMlaF"]bueno ese era facil.. pero vamos a ver algo un poco mas desafiante:
 cual es la cantidad de cifras de $10000!$ ???[/quote]

Tenes vos alguna idea de como calcular el numero de cifras sin calcular el numero mismo?

Bueno, en fin, tengo dos soluciones sin computadora (solo una tabla de logaritmos neperianos, o una calculadora que los tenga), ambas bajo aproximacion, y con tecnicas que escapan un poco a Matematicas de Secundaria. De todas maneras

[hide="Huyuyuy"]$10000!\approx2.846\cdot10^{35659}$, por lo tanto, el numero tiene 35659 cifras[/hide]

Ahora, un problema que posiblemente si este al nivel del grupo: evidentemente $10000!$ tiene muchisisisimos ceros al final de su expresion decimal, ¿Cuantos son, exactamente, esos ceros? ¿Cual es el ultimo digito antes de todos esos ceros?

Bueno, que lo disfruten. Ciao.

Ese tambien es muy facil, mi solucion aca abajo

[hide="Solucion"] Para ver cuantos 0´s tiene 10000!, basta ver cual es la maxima potencia de 10 que lo divide. Ahora, es evidente que entre 1 y 10000 hay mas numeros pares que multiplos de 5, entonces basta ver cual es la maxima potencia de 5 que divide a 10000!. Esto ultimo se hace con una formula sencilla y bastante conocida. Si $k$ es tal que $5^k\|10000!$, entonces se tiene que
$k=\sum_{i=1}^{\infty}\left\lfloor\frac{10000}{5^i}\right\rfloor$
$=\left\lfloor\frac{2^4.5^4}{5}\right\rfloor+\left\lfloor\frac{2^4.5^4}{5^2}\right\rfloor\left\lfloor\frac{2^4.5^4}{5^3}\right\rfloor+\left\lfloor\frac{2^4.5^4}{5^4}\right\rfloor$
$=2^4\frac{5^4-1}{5-1}=2^2.624=2496$
Y listo, el numero de ceros es 2496.[/hide]

Uy, me falto la segunda pregunta, no la lei  :? ...

Ademas Leonardo (Se te puede decir Leo?) una diminuta impresicion, tu solucion es correcta, pero ese es el numero de ceros AL FINAL del numero, no necesariamente el numero de ceros en el numero en si. Posiblemente hay un numero mucho mas grande de ceros en el numero completo. Digo, al menos he escuchado que en la Ibero y en la IMO suelen quitar puntos por impresiciones de ese tipo, asi que, cuidado tio!!

Ciao,

[quote="Leonardo"][quote="djimenez"]pero ese es el numero de ceros AL FINAL del numero [/quote]

Bueno Diego, la cosa es que yo estoy respondiendo tu pregunta

[quote="djimenez"] evidentemente tiene muchisisisimos ceros al final de su expresion decimal, ¿Cuantos son, exactamente, esos ceros? [/quote]

Y claro que en una IMO me quitarian puntos si cometiera el error de no especificar, pero en esa competencia no me daria el lujo de dejar detalles en el aire  ;)[/quote]

Bueno, eso espero. Recuerdo que un jurado de una IMO de hace como doce anios me conto que a un miembro de la IMO de EEUU le quitaron un punto por que dividio una pruega en triangulos "equilateros", "isosceles" y "escalenos", y en la parte de isosceles, el tio uso el hecho de que el triangulo era isosceles no equilatero, pero no especifico que el triangulo lo fuera (los equilateros en particular son isosceles :D ). Por supuesto, espero que dado que sos miembro del equipo de la IMO por segunda vez, pues espero que no fuese asi. Por cierto, buena suerte.

Ahh!!, por que sera que todo el mundo me llama Diego? En serio, ultimamente al menos cuatro o cinco personas me han llamado Diego!!!, mi nombre es David!


algebra
polynomial
group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

1.  Let $p$ be prime and let $F$ be a field of order $p^2$.  Prove there exists $a\in F$ such that $a^2=5$.

2.  Determine the Galois group of $x^4+ax^2+b\in \mathbb{Q}[x]$.

3.  Let $\mathbb{F}_{q}[x]$ be a finite field of characteristic $p$.  If $f\in \mathbb{F}_{q}[x]$ satisfies $f'(x)=0$ then $f$ is the pth power of some polynomial in $\mathbb{F}_{q}[x]$.

2. This depends on the choice of $a,b$, of course. I think every subgroup of $V_4 = \mathbb{Z}/2 \times \mathbb{Z}/2$  occurs as galois group, but also $D_4$ ...

3. This is true for all fields $F$ of characteristic $p>0$.

Let $f \in F[x]$ s.t. $f' = 0$. Writing $f=a_0 + a_1 x + ... + a_n x^n$, this yields $i a_i = 0$ for all $1 \leq i \leq n$. If $i$ is not divisible by the characteristic $p$, it follows $a_i = 0$. So, $n$ is divisble by $p$ (because of $a_n \neq 0$), $n=mp$ for some $m$. Now define $g = a_0 + a_p x + ... + a_n x^m$. Then $f = g(x^p)=g^p$.

1. call $F'= \mathbb{F}_p$ the fundamental field of $F = \mathbb{F}_{p^2}$. if you call $\alpha$ a root of $x^2-5$, the extension $F'(\alpha)/F'$ has degree at most $2$, hence it's contained in $F$. (just recall that there's just one extension of degree $k$ for a finite field, for every $k$)


A and B play a game. in principle, there are some column of coins (not necessary with the same numbers) on the table.
every player, in turn, plays one of i) or ii):
i)let a column of (2k) coin and sobstitute with two columns of (k) coins.
ii)remove all colums with odd numbers of coins.
(naturally if there aren't column with even number of coins, a player cannot choose i), and if there aren't columns with odd number of coins, a player cannot choose ii) ).
the player who removes the last coins wins the game.
A begins the game.

If in principle there are a unique column with $ 2008^{2008}$ coins, what player has a winner strategy?
Generally, on what initial configuration can A win?

We will label each state of the game with an [b]L[/b] or [b]W[/b]

[b]W[/b]: Winning state: 1st player wins
[b]L[/b]: Losing state: 1st player loses

We want to make the opponent play always from an [b]L[/b]-state

So, the property of each [b]L[/b]-state is that, [color=blue]independently of the opponent's move, we can play in such way, that the game turns again to an [b]L[/b]-state[/color]

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

[color=Indigo][size=125][b]Definitions[/b][/size]
$ \bullet$ Let $ P_i$ be the number of the coins that are contained in the pile-$ i$
$ \bullet$ For each pile $ i$ we define the number $ r = r(P_i)$ such that $ r$ is the gratest natural number with the property $ 2^r \mid P_i$
(so, if $ P_i$ is odd then $ r(P_i) = 0$)
$ \bullet$ Let ${ \mathbb{F}_k}$ be the set of all piles having $ r(P_i) = k$ and let $ F_k = \#(\mathbb{F}_k)$

So there are $ F_0$ such that $ P_i$ is odd, $ F_1$ piles with $ 2\mid P_i$ but $ 4\not| P_i$ and so on
$ \bullet$ Finally, let $ S = \sum_{k = 2}^\infty{F_k}$[/color]

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

[color=brown][size=125][b]Theorem[/b][/size]
Any state with $ F_0 = 0$ and $ S$ even is an [b]L[/b]-state[/color]

We have to prove that, if the initial state has the properties $ \{F_0 = 0$ and $ S$ even$ \}$, then, independently of the 1st player's move we can turn the game in a state with the same properties

Notice that the theorem doesn't say anything about $ F_1$. So we just don't care about this value.

[color=darkblue][size=125][b]Proof[/b][/size]

The 1st player (the opponent) will split either some pile $ i \in \mathbb{F}_1$ or a pile of some set $ \mathbb{F}_k$ with $ k\geq 2$

$ \star$ Suppose that he does the first. So, suppose that there is at least one pile in the set $ \mathbb{F}_1$

In the new generation we have $ F'_1 = F_1 - 1$ and $ F'_0 = 2$ (because two odd piles are created)
But the other piles didn't change, so $ S' = S$
So we have again the same conditions, $ S'$ even and $ F'_0 = 0$

$ \star$ Now, suppose that the opponent does the second thing. That is, he splits some pile of the set $ \mathbb{F}_k$ with $ k\geq 2$ (this means that $ S > 0 \Rightarrow S \geq 2$)
In the new generation we have:
$ F'_k = F_k - 1$ and $ F_{k - 1} = F_{k - 1} + 2$

The rest piles remain unchanged. The only change (modulo 2) in the sum $ S$ comes from $ F_k$
(it is $ S' = S - 1$ or $ S' = S + 1$ depending on $ k = 2$ or not)

So the new sum is odd. This means that there's at least one $ k\geq 2$ s.t. $ F_k\neq 0$

Then we split this pile. The resulting sum is $ S'' = (S' + 1) \mod 2 \Rightarrow S''\equiv S \mod 2$
So $ S''$ is even and $ F'_0 = F_0 = 0$[/color]

So, anything the opponent do, we can lead him again to the same situation

[u][b]Marginal states[/b][/u]:

$ S = 0, \ F_1 = 0,\ F_0 \neq 0 \to$ [b]W[/b]-state (because we can remove all the piles)

$ S = 0, \ F_1\neq 0,\ F_0 = 0 \to$ [b]L[/b]-state (the opponent can only split. Then remove the two splitted piles. Repeat until finished)

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

[color=darkred][b][u]Note[/u][/b]
These are not the only [b]L[/b]-states. I found (in java) that there are some other states $ L'$ (and [u]I think[/u] that there are not any other)

They are the states with the properties $ \{F_1$ odd, $ S$ odd, $ F_0\neq 0\}$

But they have the property $ L' \to \text{ opponent's move } \to W \to \text{your move} \to L$

where $ L$ is the set of the main losing states that described in this post[/color]

$ \star$ According to the theorem above, if there is a unique pile with $ 2008^{2008}$ coins, the 1st player wins
$ \star$ The player A wins if the initial state is not $ L$ and not $ L'$


geometry
circumcircle
geometric transformation
reflection
incenter
geometry solved

Let $(I)(O)$ be the incircle, circumcircle of the triagle $ABC$, respectively. $(I)$ tangent $ BC,CA,AB$ in $X,Y,Z$
Let $T$ is the reflection of  $X$ in $YZ$.
Suppose $AT$ meet $BC$ at $E$ . Prove that $E,I,O$ are collinear (where $O,I$ are incenter, circumcenter of the triangle $ABC$)

Discussed before on http://www.mathlinks.ro/Forum/viewtopic.php?t=20666 .

  darij


find the trigonometric form of  $(1+itan(\alpha))(1-itan(\theta)$

[hide="the Ansewr"]
$\displaystyle \frac 1 {\cos\alpha\cos \theta} e^{i(\alpha-\theta)}$[/hide]

[hide="Solution"]
$(1 + i\tan\alpha)(1 - i\tan\theta)$
$\left(1 + i\frac{\sin\alpha}{\cos\alpha}\right)\left(1 - \frac{\sin\theta}{\cos\theta}\right)$
$\frac{1}{\cos\alpha\cos\theta}(\cos\alpha + i\sin\alpha)(\cos\theta - i\sin\theta)$
$\frac{1}{\cos\alpha\cos\theta}(\cos\alpha + i\sin\alpha)(\cos(-\theta) + i\sin(-\theta))$
$\frac{1}{\cos\alpha\cos\theta}e^{\alpha i}\cdot e^{-\theta i}$
$\frac{1}{\cos\alpha\cos\theta}e^{(\alpha - \theta)i}$
[/hide]

[hide]$z= (1 + i \cdot \tan \alpha)(1 - i \cdot \tan \theta) = (1+\tan \alpha \tan \theta) + i \cdot (\tan \alpha - \tan \theta)$

But

$1 + \tan \alpha \tan \theta = \frac{\cos (\alpha - \theta)}{\cos \alpha \cos \theta}$

and

$\tan \alpha - \tan \theta = \frac{\sin (\alpha - \theta)}{\cos \alpha \cos \theta}$

Then

$z=\sec \alpha \sec \theta \cdot [\cos (\alpha - \theta) + i \cdot \sin (\alpha - \theta)]$.[/hide]


[color=darkblue] Consider the sequence $ (x_n)_{n \in \mathbb{N}}$ of natural numbers which satisfy
$ (i)$ $ x_{2n + 1} = 4x_n + 2n + 2$, $ \forall n \in \mathbb{N}$
$ (ii)$ $ x_{3n + 2} = 3x_{n + 1} + 6x_n$, $ \forall n \in \mathbb{N}$
If $ x_0 = 0$, prove that $ x_{3n - 1} = x_{n + 2} - 2x_{n + 1} + 10x_n$, $ \forall n \in \mathbb{N}^*$
[/color]

It is equivalent to prove that:$ x_{n\plus{}3}\minus{}2x_{n\plus{}2}\plus{}7x_{n\plus{}1}\minus{}6x_n\equal{}0$(*)
We will prove this lemma:((*) follow easily from this lemma)
For all $ n \in N$ we have:
$ x_{2n\plus{}1}\minus{}x_{2n}\equal{}4n\plus{}2$ (1)
$ x_{2n\plus{}2}\minus{}x_{2n\plus{}1}\equal{}4n\plus{}4$ (2)
Solution for (1):
From a problem's statement we have:
$ x_{6n\plus{}5}\equal{}4x_{3n\plus{}2}\plus{}6(n\plus{}1)\equal{}12x_{n\plus{}1}\plus{}24x_n\plus{}6(n\plus{}1)$
$ x_{6n\plus{}5}\equal{}3x_{2n\plus{}2}\plus{}6x_{2n\plus{}1}\equal{}3x_{2n\plus{}2}\plus{}12(n\plus{}1)\plus{}24x_n$
Subtract the second one to the first one we have:
->$ x_{2n\plus{}2}\equal{}4x_{n\plus{}1}\minus{}2(n\plus{}1)$(n>0)
Then
$ x_{2n\plus{}1}\minus{}x_{2n}\equal{}4x_n\plus{}2n\plus{}2\minus{}4x_n\plus{}2n\equal{}4n\plus{}2$
To prove (2) we will put $ n\plus{}1\equal{}2^p.q$(q odd)
From $ x_{q}\minus{}x_{q\minus{}1}\equal{}4\frac{q\minus{}1}{2}\plus{}2\equal{}2q$ we have:
$ x_{2q}\minus{}x_{2q\minus{}1}\equal{}4(x_{q}\minus{}x_{q\minus{}1})\minus{}4q\equal{}8q\minus{}4q\equal{}4q$
Then do the same by induction (2) was prove.


function
conics
ellipse
geometry
geometric transformation
reflection
invariant

Hello, in order to going on the same kind of exercise i submit you this one which i was completely unable to solve  :( .

Let $ \mathcal{E}$ an ellipse of equation $ a x^2 \plus{} b y^2 \equal{} 1$, we have to find the entire maps $ f$ which stabilize the ellipse.

That is to say :
\[ z \in \mathcal{E} \Rightarrow f(z) \in \mathcal{E}
\]
I wish someone could help me on this stuff, i surmise that if the ellipse is not a circle, only trivial maps stabilize it, but i may be wrong.

[color=green][Split by moderator from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=277820]this topic[/url].][/color]

I can't solve this problem, but:

In the Riemann mapping theorem we have quite a bit of freedom in choosing a biconformal self-map of the interior of the ellipse, for example we can choose which point to send to zero. What keeps us from repeatedly using the Schwarz reflection principle to extend this to an entire function? Maybe there is something I don't get about ellipses.

I have wasted quite some time trying to formalize (and making stupid mistakes, sorry) using maps similar to $ \xi \plus{} \frac {1}{\xi}$, but I give up.

Hi kalle, i dont understand your idea, this exercise is connected to the recent post of finding entire maps g such that $ |g(z)| \equal{} 1$ when $ |z| \equal{} 1$

In the solution of the precedent exercise, we used min-max principle and automorphisms of the circle, none of theses tools can be used there.

Well the other problem has another solution using reflection. Suppose $ g$ is analytic in the disc and $ |g(z)| \to 1$ as $ |z| \to 1$ in some sense. If we let $ z^* \equal{} \frac {1}{\overline{z}}$ be the reflection of $ z$ in the unit circle, then we can use the reflection principle to extend $ g$ to a meromorphic function by letting it be defined by $ g(z^*) \equal{} g(z)^*$. 0 and $ \infty$ are conjugate points with respect to the circle, so if we assume $ g(0) \neq 0$ then we get that $ g$ is bounded at infinity. And this leads us to the general form of $ g$.

So let us suppose that $ f$ is analytic in the interior of an ellipse and is invariant on the boundary. Now, ellipses are certainly nice curves, and in fact the map $ \xi \to \xi \plus{} \frac {1}{\xi}$ ($ \xi \to \frac {a \plus{} b}{2}\xi \plus{} \frac {a \minus{} b}{2}\frac {1}{\xi}$) maps the inside the circles to the exterior of ellipses. So we can use such maps to reflect in ellipses. Since they don't map to the entire interior of ellipses, we can not extend $ f$ to the entire plane immediately, but we can extend it to a bigger domain $ D_1$ that is "quite a lot" bigger than the original ellipse.

In particular, I considered the case when $ f$ is a non-trivial automorphism of the ellipse (which exist by the proof of the riemann mapping thm), then it looks intuitively to me like if we choose $ D_1$ as a new ellipse that the extension of $ f$ will map $ \partial D_1$ also to an ellipse (or something very close to it), because the values of $ f$ were also just gotten by reflection in the original ellipse. So we can use the reflection principle again to extend $ f$ to an even bigger domain $ D_2$, and it "looks to me like" the sizes of the domains increases fast, so by just repeatedly reflecting we should be able to extend $ f$ to an entire function. So this would "prove" that there are non-trivial maps that have the properties you are looking for.

But my intuition may be off, so I don't know.

The values of $ f$ on the inner ellipse may no longer lie on any reasonable curve, so I do not see how to make the reflection infinitely many times and keep things under control. Actually, I believe that I can prove that if $ f$ has a fixed point inside the ellipse, then it must be linear. This shows that quite many automorphisms cannot be extended to entire functions. I still cannot figure out what to do with the no fixed point case (of course, you do not need a fixed point of $ f$: one of something like $ \minus{}f(f(\minus{}f))$ will suffice as well, but I cannot get even that :( )

Seriously?I have proved that if $ f$ is not constant, one of the focuses of the ellipse is a fixed point of the map $ f$, or a fixed point of the map $ \minus{}f$. If you have proved that having a fixed point  implies that $ f$ is linear, we may have answered the question  :D
Could you please explain how you do that?

OK, let's try to prove that $ f$ must be linear. I'll present a proof independent on the fixed point assumption but, as you will see, it can be shortened a bit if you know in advance that there is a fixed point. So, I would really like to see your proof of the statement that one of the foci is fixed :)

First of all, note that if $ f$ maps $ \mathcal E$ to itself and is not constant, then $ f$ maps the interior $ E$ of the ellipse $ \mathcal E$ to itself.

Let $ \psi$ be any conformal mapping from $ \mathbb D$ to $ E$. Note that since the ellipse is an analytic curve, $ \psi$ can be analytically extended into some disk $ R\mathbb D$ with $ R > 1$. Consider the function $ \psi^{ \minus{} 1}\circ f\circ \psi$ in $ \mathbb D$. It is a continuous mapping from the unit disk to itself preserving the boundary. Thus, it is just some Blachke product $ B$. The Blashke product $ B$ is well-defined and analytic as a mapping from the Riemann sphere $ S$ to itself. Thus, we get the equation
\[ f\circ\psi \equal{} \psi\circ B\]
which is valid at least in $ \mathbb D$.

Now, take any point $ z_0\in S$ with $ |z_0| > 1$ and take $ r_0\in(0,1)$ so close to $ 1$ that $ S\setminus D(z_0,r_0)\subset R\mathbb D$ where $ D(z,r)$ denotes the hyperbolic distance in the exterior of the unit disk. Define the sequence $ z_n$ recursively by $ z_{n} \equal{} B(z_{n \minus{} 1})$ for $ n\ge 1$. Also define $ r_n$ recursively as the radius of the least hyperbolic disk centered at $ z_n$ containing $ B(D(z_{n \minus{} 1},r_{n \minus{} 1}))$. Note that, since $ B$ is non-expanding in $ S\setminus \mathbb D$, we have $ r_n\le r_{n \minus{} 1}$ for all $ n$.

Now we construct a sequence of $ \mathbb C$-valued analytic extensions $ \psi_n$ of  $ \psi$ from $ \mathbb D$ to $ S_n \equal{} S\setminus D(z_n,r_n)$. The extension $ \psi_0$ is just the restriction of $ \psi$ from $ R\mathbb D$ to $ S_0$. Suppose that $ \psi_{n \minus{} 1}$ is already constructed. Note that for every $ z\in S_n$ the full preimage $ B^{ \minus{} 1}(z)$ has cardinality  $ N \equal{} \text{deg\,}B < \plus{} \infty$ (counting with multiplicities, of course) and is contained in $ S_{n \minus{} 1}$. Now put
\[ \psi_n(z) \equal{} \frac 1N\sum_{w\in B^{ \minus{} 1}(z)}f(\psi_{n \minus{} 1}(w)),\qquad z\in S_n\]
Note that the sum over the full pre-image (again, we count with multiplicities) is an analytic function despite each term in it may (and almost certainly does) have branching points. Also, when $ z\in \mathbb D$, then all $ w\in\mathbb D$ and we have $ \psi_{n \minus{} 1}(w) \equal{} \psi(z)$ due to the commutation relation $ f\circ \psi \equal{} \psi\circ B$. Thus, $ \psi_n \equal{} \psi$ in $ \mathbb D$.

It is worth noticing here that, since $ S_n\cap S_m$ is connected, we actually have $ \psi_n \equal{} \psi_m$ on $ S_n\cap S_m$, so, as a matter of fact, we have just one analytic extension of $ \psi$ to the union of $ S_n$, $ n\ge 0$.

Now comes the casework.

[b]Case 1: $ N\ge 2$.[/b]
In this case, $ B$ is a strict contraction on any compact subset of the exterior of the unit disk with respect to the hyperbolic metric. Thus either $ r_n\to 0$, or $ |z_n|\to 1$. If the latter possibility holds, then the union of $ S_n$ is the entire $ S$ and $ \psi$ maps $ S$ to $ \mathbb C$, i.e., $ \psi$ is constant, which is impossible. It the former possibility takes place, then $ \psi$ extends to a function from the punctured Riemann sphere to $ \mathbb C$. We shall see later that it is not the case for the ellipse.

[b]Case 2: $ N \equal{} 1$.[/b]
Here we have to consider $ 2$ subcases:

[b]Subcase 2.1[/b] $ B$ has no fixed points in the unit disk.
In this case $ B$ is a single Blashke factor with two fixed points on the boundary, one attracting and one repelling. Choosing $ z_0$ near the attracting point, we see that $ |z_n|\to 1$ and we can finish as above.

[b]Subcase 2.2[/b] $ B$ has a fixed point in the unit disk.
This means that $ f$ has a fixed point in $ E$. So far we haven't specified any particular $ \psi$. Let us now choose one that maps $ 0$ to that fixed point $ \zeta$. Then $ B \equal{} cz$ with $ |c| \equal{} 1$. The commutation relation thus becomes $ \psi(cz) \equal{} f(\psi(z))$. Hence, $ \psi(c^mz) \equal{} f^{\circ m}(\psi(z))$. If $ z\mapsto cz$ is a rational rotation, i.e., if $ c^m \equal{} 1$ for some integer $ m$, we conclude that $ f^{\circ m}(z) \equal{} z$, which is possible for an entire function $ f$ only if $ f$ is linear. If the rotation is irrational, we immediately conclude that the largest star domain of analyticity of $ \psi$ is a circle (possibly, of infinite radius). Again we will show that this is not the case for the ellipse.

Now it remains to prove the above claims about $ \psi$. They follow from the known explicit formulae for $ \psi$ in terms of elliptic functions, of course, but we do not need so much here. Let $ \mathcal E$ be the image of $ \{z \equal{} \rho\}$ under the mapping $ z\mapsto z \plus{} \frac 1z$ with some $ \rho > 1$. Let $ \Gamma$ be any sector with the vertex at the origin not containing the $ \psi$-preimages of $ 2$ and $ \minus{} 2$. Then there exists an analytic solution $ w(z)$ of the equation $ w(z) \plus{} \frac 1{w(z)} \equal{} \psi(z)$ in $ \Gamma\cap\mathbb D$ continuous up to the boundary and such that $ |w(z)| \equal{} \rho$ when $ z\in\Gamma$, $ z \equal{} 1$. Note that $ w$ is separated from both $ 0$ and $ \infty$ in $ \Gamma\cap\mathbb D$. Now, by the Schwarz symmetry principle, $ w$ can be extended to the entire $ \Gamma$ as $ w(z) \equal{} \frac {\rho^2}{\overline{w(1/\bar z)}}$, thus giving a bounded extension of $ \psi$ to $ \Gamma$. Moreover, the bound involved does not depend on $ \Gamma$. Thus, we get a bounded analytic extension of $ \psi$ to the entire complex plane with 2 rays removed. Now both claims about $ \psi$ immediately follow from Liouville's theorem.

This argument proves the statement for many other analytic curves as well but not for all of them. So, the question now gets upgraded to whether there exists any simple closed analytic curve on the plane that is stabilized by an entire function that is not a polynomial. ;)

I have not understood all of the proof even on a technical level yet. If you don't mind, I have a couple of questions:

Why is $ \phi_n$ analytic on $ S_n$? Even better, what is the idea behind the construction of it?

I haven't read this part carefully yet, but does the $ n$ in the casework refer to the number of factors in $ B$?

Grrr... I used the same letter $ n$ for both the index of $ \psi$ and the degree of $ B$.  :o  I'll edit.

Yes, the casework refers to the degree of $ B$.

As to why $ \psi_n$ is analytic, the easiest way to see it is to write it as a Cauchy integral. If $ B$ is any mapping from a domain $ \omega$ bounded by a nice curve $ \gamma$ to $ \mathbb C$ and $ \Omega$ is another open set such that for every $ z\in\Omega$, the function $ B$ does not attain the value $ z$ on $ \gamma$, and $ F: \omega\to\mathbb C$ is any analytic function, then
\[ \sum_{w\in\omega: B(w)\equal{}z}F(w)\equal{}\frac 1{2\pi i}\int_\gamma \frac{B'(\zeta)F(\zeta)}{B(\zeta)\minus{}z}\]
by the residue theorem. The right hand side is, obviously, analytic in $ z$.

As to the idea, you want to use the identity $ \psi(B)\equal{}f(\psi)$ to extend $ \psi$. So, if $ z\equal{}B(w)$, you want to define $ \psi(z)\equal{}f(\psi(w))$. The unpleasant question is which $ w$ to take if there are several of them. To circumvent it, you just take all and average (this is similar to the fact that the roots of a polynomial do not depend analytically on the coefficients but all their symmetric sums do).


farz konid $ p$ yek adade avvale fard bashad.. agar $ g_{1},g_{2},\dots,g_{\varphi(p\minus{}1)}$ rishe avalliehayi be hange $ p$ bashad..betorike baraye har $ i$: $ 1\geq g_{i}\geq p\minus{}1$ sabet konid:
$ p|\sum_{i \equal{} 1}^{\varphi(p\minus{}1)}g_{i}\minus{}\mu(p\minus{}1)$

ide kollish estefade az chanjomleie zire:

farz konid $ S_{n}$ majmooe rishehaye $ n$ome vahed bashe oonvagh tarif mikonim:

$ \Phi_{n}(x)\equal{}\prod_{g\in S_{n}}(x\minus{}g)$

hala kafie azin nokteha estefade konim:

$ i)$rishehaye avalie adade $ p$,dar vaghe rishehaye chanjomleie $ \Phi_{p\minus{}1}(x)$ hastand (be hange $ p$)

$ ii)$darim $ \Phi_{n}(x)\equal{}\prod_{d\mid n}(x^{d}\minus{}1)^{\mu (\frac{n}{d})}$

hala ba tavajoh be nokte aval kafie jame rishehaye chanjomleie $ \Phi_{p\minus{}1}(x)$ ro hesab konim ke tebghe nokte dovom barabare jame rishehaye chanjomleie zire:

$ \prod_{d\mid n}(x^{d}\minus{}1)^{\mu (\frac{n}{d})}$

yeni kafie sabet konim dovomin zaribe in chanjomlei (ke adadi sahihe) barabare ba $ \minus{}\mu (p\minus{}1)$e...


number theory
relatively prime
number theory proposed

Let $ n,m\in\mathbb{Z}^ + ^}$. Let $ a_1,a_2,...a_n,b_1,b_2,...b_m$ be positive integers satisfying that $ \max\{a_i\}\ne \max\{b_i\}$. Prove that there exists infinitely many different $ m + n$-tuples $ (x_1,x_2,...,x_n,y_1,y_2,...,y_m)$ of positive integers such that
\[ x_1^{a_1} + x_2^{a_2} + ... + x_n^{a_n} = y_1^{b_1} + y_2^{b_2} + ... + y_m^{b_m}
\]
.

:)

I think it is correct now...

It's false.
Try $ n\equal{}2$, $ m\equal{}1$, $ a_1\equal{}a_2\equal{}b_1\equal{}4$. You will have $ x^4\plus{}y^4\equal{}z^4$ which hasn't integer solutions (without $ x\equal{}y\equal{}z\equal{}0$).

hmmm... you are right... let's see

I think the edition I made makes it truth...

:)

It's false, too. Before posting and editing some incorrect things, try to give counterexample for yourself - it's very easy. I'm really bored to give CE to this trivial-incorrect task.

$ n \equal{} 2$, $ m \equal{} 1$, $ a_1 \equal{} a_2 \equal{} 4$,  $ b_1 \equal{} 8$. You will have $ x^4 \plus{} y^4 \equal{} z^8$ which hasn't integer solutions (without $ x \equal{} y \equal{} z \equal{} 0$), too (because after substitution $ t \equal{} z^2$ you will have $ x^4 \plus{} y^4 \equal{} t^4$).

jiji you are complete right in your thought... I am sorry

the truth is I "discovered" this some time ago and it seems that I had a mistake in the solution...

I am sure it is ok now

sorry for the lost time...

 :oops:

Let $ m, n$ be positive integers with $ m<n$. Let $ a_1,a_2,...,a_n,b_1,b_2,...,b_m$ such that there existis $ k\in\mathbb{Z}\cap[1,m]$ such that $ b_k$ is relatively prime to at least $ n\minus{}m\plus{}1$ numbers among $ a_1,...,a_n$.
Prove that there are infinetely many $ m\plus{}n$-tuples of positive integers $ (x_1,...,x_n,y_1,...,y_m)$ such that $ x_1^{a_1} \plus{} x_2^{a_2} \plus{} ... \plus{} x_n^{a_n} \equal{} y_1^{b_1} \plus{} y_2^{b_2} \plus{} ... \plus{} y_m^{b_m}$

:)

[quote="ElChapin"]
Let $ m, n$ be positive integers with $ m < n$. Let $ a_1,a_2,...,a_n,b_1,b_2,...,b_m$ such that there existis $ k\in\mathbb{Z}\cap[1,m]$ such that $ b_k$ is relatively prime to at least $ n \minus{} m \plus{} 1$ numbers among $ a_1,...,a_n$.
Prove that there are infinetely many $ m \plus{} n$-tuples of positive integers $ (x_1,...,x_n,y_1,...,y_m)$ such that $ x_1^{a_1} \plus{} x_2^{a_2} \plus{} ... \plus{} x_n^{a_n} \equal{} y_1^{b_1} \plus{} y_2^{b_2} \plus{} ... \plus{} y_m^{b_m}$
[/quote]
Yes, it's correct now, thank you.
Solution...
[hide]
Without loss of generalization, we can think that $ b_1$ is relatively prime to $ a_1, a_2,...,a_{n \minus{} m \plus{} 1}$.
Taking $ y_2 \equal{} y_3 \equal{} ... \equal{} y_m \equal{} x_{n \minus{} m \plus{} 2} \equal{} ... \equal{} x_n \equal{} 1$, we wiil have:

$ y_1^{b_1} \equal{} x_1^{a_1} \plus{} x_2^{a_2} \plus{} ... \plus{} x_{n \minus{} m \plus{} 1}^{a_{n \minus{} m \plus{} 1}}$, or

$ y_1^{b_1} \equal{} x_1^{a_1} \plus{} x_2^{a_2} \plus{} ... \plus{} x_{k}^{a_k}$
, where $ k \equal{} n \minus{} m \plus{} 1$.

Let $ d \equal{} a_1a_2...a_k$, so $ b_1$ relatively prime to $ d$.
We can try to find solution of equation in form $ x_i \equal{} z_i ^{d/a_i}$, so, we get:
$ y_1^{b_1} \equal{} z_1^d \plus{} z_2^d \plus{} ... \plus{} z_k ^ d$.

Let try to take $ z_1 \equal{} z_2 \equal{} ... \equal{} z_k \equal{} k^{\alpha}$, $ y_1 \equal{} k^{\beta}$.

We will get:
$ k^{\beta b_1} \equal{} k^{d\alpha \plus{} 1}$.
This equation has infinitely many solutions, because there are exist infinitely many solutions of $ \beta b_1 \equal{} d\alpha \plus{} 1$ in unknown parametres $ \alpha$ and $ \beta$ (note, that it's true, because $ d$ relatively prime to $ b_1$). In other hand, $ k > 1$, so we get infinity serie for $ z_i$ and for $ y_1$.
[/hide]


mean
inequalities
IMO Shortlist
algebra
n-variable inequality

Let ${a_1,a_2,\dots,a_n}$ be positive real numbers, ${n>1}$. Denote by $g_n$ their geometric mean, and by $A_1,A_2,\dots,A_n$ the sequence of arithmetic means defined by \[ A_k=\frac{a_1+a_2+\cdots+a_k}{k},\qquad k=1,2,\dots,n. \]  Let $G_n$ be the geometric mean of $A_1,A_2,\dots,A_n$. Prove the inequality \[
n \root n\of{\frac{G_n}{A_n}}+ \frac{g_n}{G_n}\le n+1 \] and establish the cases of equality.

[i]Proposed by Finbarr Holland, Ireland[/i]

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in $\LaTeX$. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[hide="Solution by Bojan Basic"]

[quote="Bojan Basic"]Let \[m_k=\frac {A_{k-1}} {A_k}\] (suppose that $A_0=0$). Now it is obvious that \[\frac {a_k}{A_k}=\frac{kA_k-(k-1)A_{k-1}}{A_k}=k-(k-1)m_k\] We have:
\[\sqrt[n]{\frac {G_n}{A_n}}=\sqrt[n^2]{m_2m_3^2\dots m_n^{n-1}}\]
\[\frac {g_n}{G_n}=\sqrt[n]{\prod_{k=1}^n(k-(k-1)m_k)}\]
Using AM-GM we obtain:
\[n\sqrt[n^2]{1^{\frac{n(n+1)}2}m_2m_3^2\dots m_n^{n-1}}\leqslant \frac{\frac{n(n+1)} 2+\sum_{k=2}^n(k-1)m_k}n=\frac{n+1}2+\frac {\sum_{k=1}^n(k-1)m_k}n\]
\[\sqrt[n]{\prod_{k=1}^n(k-(k-1)m_k)}\leqslant \frac{n+1} 2-\frac {\sum_{k=1}^n(k-1)m_k}n\]

Summing the last two relations we get the desired inequality. Equality holds iff $a_1=a_2=\dots=a_n$.

[/quote]

[/hide]

Then the quality of the inequalities proposed that year wasn't very extraordinary if this is the most challenging. Write it in the form \[ n\cdot\sqrt[n^2]{\frac{A_1...A_{n-1}}{A_n^{n-1}}}+\sqrt[n]{(2-A_1/A_2)...(n-(n-1)A_{n-1}/A_n)}\leq n+1. \] Use the AM-GM inequality to deduce that the second term of the LHS is at most \[ \frac{1+(2-A_1/A_2)+...+(n-(n-1)A_{n-1}/A_n)}{n}. \]
Thus the inequality reduces to  \[  \frac{n+1}{2}+\frac{A_1/A_2+...+(n-1)A_{n-1}/A_n}{n} \geq n\cdot\sqrt[n^2]{\frac{A_1...A_{n-1}}{A_n^{n-1}}}. \] Now, apply generalized AM-GM inequality to deduce that the LHS is at least $ (n+1)/2+(n-1)/2\cdot\sqrt[(n^2-n)/2]{\frac{A_1...A_{n-1}}{A_n^{n-1}}}$.  Finally, kill the last beast with another AM-GM inequality.

This problem was used as problem 4 of the final exam of the 3rd TST of Taiwan 2005. (Only 1 person got it right.)

Solved with Amol Rama, Raymond Feng, Samuel Wang, and William Wang.

Let $c_i=A_i/A_{i+1}$. The inequality we wish to prove reads \[n\left(c_{n-1}^{\frac{n-1}{n^2}}c_{n-2}^{\frac{n-2}{n^2}}\cdots c_1^{\frac{1}{n^2}}\right)^{1/n^2}+\left(1\cdot(2-c_1)\cdot(3-2c_2)\cdots(n-(n-1)c_{n-1})\right)^{1/n}\le n+1.\] We see that \[n\left(c_{n-1}^{\frac{n-1}{n^2}}c_{n-2}^{\frac{n-2}{n^2}}\cdots c_1^{\frac{1}{n^2}}\right)^{1/n^2}\le \frac{n-1}{n}c_{n-1}+\cdots+\frac{1}{n}c_1+\frac{n+1}{2},\] and \[\left(1\cdot(2-c_1)\cdot(3-2c_2)\cdots(n-(n-1)c_{n-1})\right)^{1/n}\le \frac{n+1}{2}-\frac{1}{n}c_1-\cdots-\frac{n-1}{n}c_{n-1},\] so adding gives the desired estimate.

[b]Remark:[/b]  The solution is motivated by the $n=2$ case, which was solved with a similar AM-GM as above.


integration
calculus
function
algebra
partial fractions

Is it possible to simplify below sum by telescoping it?

$ \frac{1}{n\plus{}1} \plus{} \frac{1}{(n\plus{}1)(n\plus{}2)}\plus{}\frac{1}{(n\plus{}1)(n\plus{}2)(n\plus{}3)}\plus{}...$

That seems unlikely.  For example, for $ n \equal{} 0$, the sum becomes
\[ \frac{1}{1!} \plus{} \frac{1}{2!} \plus{} \frac{1}{3!} \plus{} \dots \equal{} e \minus{} 1.\]

[hide="Solution"] Let $ f_n(x) \equal{} \sum_{k \equal{} 1}^{\infty} \frac {x^{n \plus{} k}}{(n \plus{} 1)...(n \plus{} k)}$.  One has

$ f'_n(x) \equal{} x^n \plus{} f_n(x) \Leftrightarrow$
$ f'_n(x) \minus{} f_n(x) \equal{} x^n \Leftrightarrow$
$ e^{ \minus{} x} f'_n(x) \minus{} e^{ \minus{} x} f'_n(x) \equal{} x^n e^{ \minus{} x} \implies$
$ e^{ \minus{} x} f_n(x) \equal{} \int_{0}^{x} t^n e^{ \minus{} t} \, dt \implies$
$ f_n(1) \equal{} e \int_{0}^{1} t^n e^{ \minus{} t} \, dt$.

When $ n \equal{} 0$ the integral gives $ 1 \minus{} \frac {1}{e}$, which agrees with nsato's answer.  Of course, the integral is really only interesting when $ n$ is not an integer, and then what we're looking at is the [url=http://en.wikipedia.org/wiki/Incomplete_gamma_function#Properties]lower incomplete Gamma function[/url] evaluated at $ 1$. [/hide]

Let's break this into partial fractions and see what happens:

$ \frac {1}{(n \plus{} 1) \ldots (n \plus{} k)} \equal{} \sum _{j \equal{} 1}^k \frac {\frac {( \minus{} 1)^{j \minus{} 1}}{(k \minus{} 1)!} \binom{k \minus{} 1}{j \minus{} 1}}{n \plus{} j} \equal{} \sum_{j \equal{} 1}^k \frac {( \minus{} 1)^{j \minus{} 1}}{(k \minus{} j)! (j \minus{} 1)!} \cdot \frac {1}{n \plus{} j}$

So, to find the overall coefficient for a term $ \frac {1}{n \plus{} j}$, we need to compute:

$ \sum_{k \equal{} j}^{\infty} \frac {( \minus{} 1)^{j \minus{} 1}}{(k \minus{} j)! (j \minus{} 1)!} \equal{} \frac {( \minus{} 1)^{j \minus{} 1}}{(j \minus{} 1)!} \sum_{k \equal{} j}^{\infty} \frac {1}{(k \minus{} j)!}$

Since the above is never zero, no term of the form $ \frac {1}{n \plus{} j}$ will cancel out.

[i]
In an example such as $ \frac {1}{x(x \plus{} 1)} \equal{} \frac {1}{x} \minus{} \frac {1}{x \plus{} 1}$, each type of fraction shows up at most twice, with one coefficient of $ 1$ and one coefficient of $ \minus{} 1$. Once those two terms are added, the fraction type is out of the picture. That doesn't happen here; each fraction type shows up again and again, because any factor shows up in an infinite number of denominators. What's worse is that the number of factors in the denominators of each term keeps growing![/i]


In an increasing infinite sequence of positive integers, every term starting from the $2002$-th term divides the sum of all preceding terms. Prove that every term starting from some term is equal to the sum of all preceding terms.

This seems suspiciously similar to USAMO 2007 #1. I think a similar argument works.

Well, if it can be solved similarly, why don't you post a solution?

Let $ x_n$ is suth sequence. Then \[ x_n|\sum_{i\equal{}1}^{n\minus{}1}x_i\] for big n, or \[ \sum_{i\equal{}1}^nx_i\equal{}y_ny_n\] for big n ($ y_n$ is natural). It give \[ \sum_{i\equal{}1}^nx_n\equal{}y_nx_n\plus{}x_n\equal{}y_{n\plus{}1}x_{n\plus{}1}\] for all big n. Because $ x_{n\plus{}1}>x_n$ we get $ y_{n\plus{}1}<y_n\plus{}1$ or $ y_{n\plus{}1}\le y_n$. Because $ y_n$ is natural, exist N, suth that $ y_n\equal{}a \ \forall n>N$ and
$ (a\plus{}1)x_n\equal{}ax_{n\plus{}1} \ \forall n>N$.
If $ a>1$, let $ z_n\equal{}ord_a(x_n),n>N \to z_{n\plus{}1}\equal{}z_n\minus{}1$ give contradition. Therefore $ a\equal{}1,x_{n\plus{}1}\equal{}2x_{n} \ \forall n>N$.
It give $ x_n\equal{}2^{n\minus{}N}d, \ d\equal{}\sum_{i\equal{}1}^{N\plus{}1}x_n \ \forall n>N$.


Let $ f(x) \in Z[x]$ be an irreducible monic polynomial with integer coefficients
Suppose that $ \left|f(0) \right|$is not a perfect square
Show that $ f(x^2)$is also irreducible

I want to know how to prove problem.
I think the first step is to show if $ f(x^{2})$ is reducible over integer, then we can set $ f(x^{2})\equal{}p(x)q(x)$
Then degree of p, q are n. But I bumped into some difficulties, so it can't be worked well......

Please post your solution.

Let $ \deg f \equal{} n$. Assume that there exists a $ p(x)$ which is a divisor of $ f(x^2)$, and wlog assume that $ \deg p \le n$ and that $ p$ is irreducible.
Let $ \gcd(p(x),p( \minus{} x)) \equal{} q(x)$. Since $ p$ is irreducible we have $ q(x) \equal{} c \in \mathbb{Z}$ or $ q(x) \equal{} p(x)$. If $ q(x) \equal{} p(x)$, then $ p(x) \equal{} p( \minus{} x)$. Hence $ p(x)$ is even and $ p(\sqrt {x})$ is a divisor of $ f(x)$. Contradiction.Hence $ \gcd(p(x),p( \minus{} x)) \le c \in \mathbb{Z}$. So $ p(x)p( \minus{} x) \mid c \cdot f(x^2)$ and since $ p(x),f(x^2),p( \minus{} x)$ must be monic: $ p(x)p( \minus{} x) \mid f(x^2)$. If $ \deg p \equal{} n$ then $ f(x^2) \equal{} \pm p(x)p( \minus{} x)$ and therefore $ f(0) \equal{} \pm p(0)^2$, but $ |f(0)|$ is not a square, so $ \deg p < n$. Hence $ p(x)p( \minus{} x)$ is even and $ p(\sqrt {x})p( \minus{} \sqrt {x})$ is a divisor of $ f(x)$. Contradiction. Hence we are done :)


[quote="apollo"]find all $ f: R \rightarrow R$,
$ i) f(xy) \equal{} f(x)f(y)$
$ ii) f(x \plus{} 1) \equal{} f(x) \plus{} 1$[/quote]

$ i)$ implies $ f(0) \equal{} 0$ or $ f(x) \equal{} 1$ $ \forall x\neq 0$. This second point is impossible according to $ ii)$. So $ f(0) \equal{} 0$
$ ii)$ then implies $ f(n) \equal{} n$ $ \forall n\in\mathbb{Z}$
$ i)$ then implies $ f(x) \equal{} x$ $ \forall x\in\mathbb{Q}$
$ i)$ implies $ f(x)\geq 0$ $ \forall x\geq 0$ and $ f(x)\leq 0$ $ \forall x\leq 0$
$ ii)$ then implies $ f(x)\geq 1$ $ \forall x\geq 1$
$ i)$ then implies $ f(x)$ is non decreasing

Then $ f(x) \equal{} x$ $ \forall x\in\mathbb{Q}$ and $ f(x)$ non decreasing imply $ f(x) \equal{} x$ $ \forall x\in\mathbb{R}$

Posted before generally statment:
If $ f: R\to R$ satisfyed
1) $ f(x^2)\equal{}(f(x))^2$,
2) $ f(x\plus{}1)\equal{}f(x)\plus{}1$,
then $ f(x)\equiv x$.


geometry
incenter
circumcircle
cyclic quadrilateral
national olympiad

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58514]1.[/url] Let $ABCD$ be a cyclic quadrilateral. Show that the orthocenters of the triangles $ABC,BCD,CDA,DAB$ are vertices of a quadrilateral congruent to $ABCD$ and show that the centroids of the same triangles are the vertices of a cyclic quadrilateral.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58516]2.[/url] $ABCD$ is a cyclic quadrilateral with $BC=CD$. The diagonals $AC$ and $BD$ intersect at $E$. Let $X,Y,Z,W$ be the incenters of triangles $ABE,ADE,ABC,ADC$ respectively. Show that $X,Y,Z,W$ are concyclic iff $AB=AD$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58517]3.[/url] $A$ and $B$ are fixed points on a plane and $L$ is a line passing through $A$ and not $B$. $C$ is a variable point moving from A toward infinity along a half-line of $L$. The incircle of $\triangle ABC$ touches $BC$ at $D$ and $AC$ at $E$. Show that the line $DE$ passes through a fixed point.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58518]4.[/url] Let $\triangle ABC$ be given with circumradius $R$. Let $K_{1}$ and $K_{2}$ be the circles which pass through $C$ and are tangent to $AB$ at $A$ and $B$ respectively. Let $K$ be the circle of radius $r$ which is tangent to $K_{1},K_{2}$ and $AB$.
(a) Express $r$ in terms of $a,b,c$ and $R$.
(b) Suppose $r=\frac{R}{4}$, find $\angle C$.

Remark: Problems 1 and 4(a) are well known results.

[quote="mecrazywong"][quote="Valentin Vornicu"]Please also post each problem into it's own topic and supply the links in the post above ;) :D[/quote]
Sure :)
but would I know why the topic name cannot be capitalized?(IMO 2006 Hong Kong TST 2)[/quote]Topic name must be something meaningfull from the problem. It is automatically decapitalized to prevent users that posted titles with only capital letters to attract attention.

You can use the source field with capital letters though ;)

@Arne: sometimes team leaders want to give specific field tests. For example Romania also had a full geometry TST in 2005: http://www.mathlinks.ro/Forum/resources.php?c=142&cid=26&year=2005 (test 3).

[quote="Arne"]This test consists of geometry problems only, why?  :huh:[/quote]
In fact this test is also known as the geometry test. :)
In q.4 I thought $K_1$ and $K_2$ are the centres :oops: , so I gave up understanding the question... :( Only solved 1, 2, 3... You guys (refering to mecrazy) got them all correct...

[quote="silouan"]You say that problems $1,4$ were known. From where? Please reply thanks[/quote]
I said [i]well[/i] konwn. I think this makes sense to your question?

[quote="cezar lupu"]For example, first problem is not that hard, since we can use succesfully complex geometry here. :D[/quote]Comments about the problem's solution should be made in the problem's own topic ;)

I did very poor in this test,simply because I gave up it at the very beginning :(
(leepakhin should know that how weak is my geometry :? )
When I finish the second part of Problem 1,I even didn't have the "mood" to finish the first part :(
By the way,leepakhin,I think you are quite good among many F.4 student(except mecrazy) already.

In the title you said TST 2. Could you post the TST 1? Thanks.

It is already [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=49917]here[/url].


Is it allowed for multiple people using the same account to access a class? Also, is it possible to access a class with multiple different accounts?

No, you should not have multiple people using your account to access a class.  We will ban people that we find doing this.

You can register multiple users to access a class, but you'd have to enroll each of them separately.


geometry
circumcircle
geometric transformation
reflection
trapezoid
projective geometry
power of a point

Dear all,
Let ABC be a triangle, 0 the circumcircle of ABC, O the center of 0, P and Q two points on 0, Q1 the symmetric of Q wrt BC, Q2 the symmetric of Q wrt CA, S1 the point of intersection of PQ1 and BC, S2 the point of intersection of PQ2 and CA.
Prouve that the orthopole of ABC wrt the mediator of PQ, is on S1S2.
Any synthetic proof?
Sorry, I am not good with latex.
Jean-Louis

[b]Lemma:[/b] Let a line through the circumcenter $ O$ cut the triangle sidelines $ BC, CA, AB$ at $ X, Y, Z.$ Circles $ (P_a), (P_b), (P_c)$ with diameters $ AX, BY, CZ$ are coaxal with 2 common points $ K, L.$ The 1st point $ K$ lies on the triangle circumcircle $ (O)$ and the 2nd point $ L$ is orthopole of the line $ XYZ$ through $ O.$

The circle centers $ P_a, P_b, P_c$ lie on the triangle midlines $ B'C', C'A', A'B',$ where $ A', B', C'$ are midpoints of $ (BC), (CA), (AB).$ By Menelaus theorem for the $ \triangle ABC$ and its medial $ \triangle A'B'C',$

$ \frac {\overline{P_aB'}}{\overline{P_aC'}} \cdot \frac {\overline{P_bC'}}{\overline{P_bA'}} \cdot \frac {\overline{P_cA'}}{\overline{P_cB'}} = \frac {\overline{XC}}{\overline{XB}} \cdot \frac {\overline{YA}}{\overline{YC}} \cdot \frac {\overline{ZB}}{\overline{ZA}} = 1,$

which means that $ P_a, P_b, P_c$ are collinear. Since $ AX, BY, CZ$ are diameters of $ (P_a), (P_b), (P_c),$ these circles pass through the altitude feet $ D \in BC, E \in CA, F \in AB$ of the $ \triangle ABC.$ Therefore, powers of the orthocenter $ H$ to the circles $ (P_a), (P_b), (P_c)$ are all equal to half the power of $ H$ to the circumcircle $ (O)$:

$ \overline{HA} \cdot {\overline{HD} = \overline{HB} \cdot \overline{HE} = \overline{HC} \cdot \overline{HF} = \frac {_1}{^2}\ p(H, (O)).}$

Together with the collinear centers, this means that the circles $ (P_a), (P_b), (P_c)$ are coaxal. (The coaxality does not depend on the line $ XYZ$ passing through the circumcenter $ O.$) Let $ (P_a), (P_b), (P_c)$ meet $ XYZ$ again at $ X', Y', Z'.$ The reflection $ L$ of $ X', Y', Z'$ in $ B'C', C'A', A'B'$ is the orthopole of $ XYZ$ through $ O$ and it lies on $ (P_a), (P_b), (P_c)$ (well known, due to $ O$ being the orthocenter of the medial $ \triangle A'B'C',$ for example, see the last paragraph of [url]http://www.mathlinks.ro/viewtopic.php?t=173756[/url], or [url]http://www.mathlinks.ro/viewtopic.php?t=184885[/url], etc.) Let the ray $ (HL$ of the common radical axis $ KHL$ of $ (P_a), (P_b), (P_c)$ cut $ (O)$ at $ M.$ Since the orthopole $ L$ lies on the 9-point circle of the $ \triangle ABC,$ similar to the circumcircle $ (O)$ with center $ H$ and coefficient $ \frac {_1}{^2},$ it is the midpoint of $ (HM).$ Power of $ H$ to $ (P_a), (P_b), (P_c)$ is

$ \frac {_1}{^2}\ p(H, (O)) = \overline {HK} \cdot \overline {HL} = \frac {_1}{^2} \overline {HK} \cdot \overline {HM}.$

Since $ M \in (O),$ it follows that $ K \in (O).$ (This last fact will not be used any more.) This proves the lemma.
_____________________________________________

Let $ (O_a)$ be an arbitrary circle centered of the line $ XYZ$ through $ O$ and passing through $ X$ (intersecting the circumcircle $ (O)$ or not). The circle $ (O_a)$ meets the lines $ BC, XYZ$ and the circle $ (P_a)$ at $ X.$ Let $ (O_a)$ meet the line $ BC$ again at $ S_a,$ the line $ XYZ$ again at $ M_a$ and the circle $ (P_a)$ again at $ U.$ Since $ AX, M_aX$ are diameters of $ (P_a), (O_a),$ $ XU \perp UA, UM_a$ and $ A, U, M_a$ are collinear. From the cyclic hexagon $ AULX'XD$ and cyclic (isosceles) trapezoid $ ALX'D$ (because $ LX' \parallel AD,$ both perpendicular to $ B'C' \parallel BC$),

$ \measuredangle LUM_a = \measuredangle LUA = \measuredangle LX'A = \measuredangle LDA = \measuredangle DAX' \mod 180^\circ.$

Since $ AD \perp (BC \equiv X_aD)$ and $ AX' \perp (M_aX' \equiv XYZ),$

$ \measuredangle DAX' = \measuredangle S_aXM_a \mod 180^\circ,$

and from the cyclic quadrilateral $ XS_aM_aU,$

$ \measuredangle S_aXM_a = \measuredangle S_aUM_a \mod 180^\circ.$

Combining, $ \measuredangle LUM_a = \measuredangle S_aUM_a \mod 180^\circ,$ which means that the points $ S_a, L, U$ are collinear. Using the result of [url]http://www.mathlinks.ro/viewtopic.php?t=186477[/url], the posted problem is a particular case of the circle $ (O_a)$ intersecting the circumcircle $ (O)$ at two arbitrary points $ P, Q$ and passing through the intersection $ X$ of the perpendicular bisector $ XYZ$ of $ PQ$ with $ BC.$

The lemma used in the proof has the following generalization:

Let $ H$ be ortocenter of a $ \triangle ABC$ and $ P$ an arbitrary point in its plane. Let $ XYZ$ be the orthotransversal of $ P$ with respect to the $ \triangle ABC.$ The orthopole $ L$ of the line $ XYZ$ lies on the line $ PH.$

Any synthetic proof ?

[quote="yetti"]The lemma used in the proof has the following generalization:

Let $ H$ be ortocenter of a $ \triangle ABC$ and $ P$ an arbitrary point in its plane. Let $ XYZ$ be the orthotransversal of $ P$ with respect to the $ \triangle ABC.$ The orthopole $ L$ of the line $ XYZ$ lies on the line $ PH.$[/quote]

I am not feeling quite well right now, so please excuse me if the following has some mistakes:

[color=blue][b]Theorem 1.[/b] Let $ P$ be a point in the plane of a triangle $ ABC$, and let the perpendiculars to the lines $ AP$, $ BP$, $ CP$ at the point $ P$ meet the lines $ BC$, $ CA$, $ AB$ at the points $ X$, $ Y$, $ Z$. Then, the point $ P$, the orthocenter of triangle $ ABC$, and the orthopole of the line $ XYZ$ with respect to triangle $ ABC$ are collinear.[/color]

[i]Proof.[/i] This turns out to be pretty easy when considered from the right viewpoint, namely that of complete quadrilaterals:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=51844]It is known that[/url] the points $ X$, $ Y$, $ Z$ lie on one line. Consider the complete quadrilateral $ q$ formed by the lines $ XYZ$, $ BC$, $ CA$, $ AB$. After Theorem 2 of [1], the orthocenters of triangles $ ABC$, $ AYZ$, $ BZX$, $ CXY$ lie on the Steiner line of this complete quadrilateral $ q$, and after Corollary 4 of [1], the orthopoles of the lines $ XYZ$, $ BC$, $ CA$, $ AB$ with respect to the triangles $ ABC$, $ AYZ$, $ BZX$, $ CXY$ also lie on the Steiner line of this complete quadrilateral $ q$.

According to the proofs at http://www.mathlinks.ro/Forum/viewtopic.php?t=842 , the Steiner line of a complete quadrilateral coincides with the pairwise radical axis of the circles with the diagonals of the complete quadrilateral as diameters. Hence, the Steiner line of our complete quadrilateral $ q$ also happens to be the pairwise radical axis of the circles with diameters $ AX$, $ BY$, $ CZ$ (because $ AX$, $ BY$, $ CZ$ are the diagonals of the complete quadrilateral $ q$), so that the point $ P$ must lie on this Steiner line (because $ P$ has equal powers with respect to the circles with diameters $ AX$, $ BY$, $ CZ$ (in fact, $ P$ has the power $ 0$ with respect to all of these circles, because $ P$ lies on each of these circles, since $ \measuredangle APX \equal{} 90^{\circ}$, $ \measuredangle BPY \equal{} 90^{\circ}$ and $ \measuredangle CPZ \equal{} 90^{\circ}$)).

Altogether, we have seen that the Steiner line of the complete quadrilateral $ q$ passes through:
- the orthocenters of triangles $ ABC$, $ AYZ$, $ BZX$, $ CXY$;
- the orthopoles of the lines $ XYZ$, $ BC$, $ CA$, $ AB$ with respect to the triangles $ ABC$, $ AYZ$, $ BZX$, $ CXY$;
- the point $ P$.

This proves even more than Theorem 1 asked us to show.

[b]References[/b]

[1] [url=http://forumgeom.fau.edu/FG2004volume4/FG200406index.html]Atul Dixit and Darij Grinberg, [i]Orthopoles and the Pappus Theorem[/i], Forum Geometricorum, 4 (2004) pp. 53-59[/url]

  darij


The arithmetric progression series $\{a_n\}$ and $\{b_n\}$ has sum of first $n$ term as $S_n$ and $T_n$ respectively. If $\frac{S_n}{T_n}=\frac{2005n}{2006n+1}$ , find $\frac{a_n}{b_n}$ .

 :)

[quote="Boy Soprano II"][hide="Solution"] Clearly $S_{n}=k\cdot 2005n$, $k$ a constant, so for $n\neq 1$, $a_{n}=S_{n}-S_{n-1}=k\cdot 2005$.  However, this also holds for $n=1$.  Likewise, for $n\neq 1$, $T_{n}=k\cdot(2006n+1)$ and $b_{n}=T_{n}-T_{n-1}=k\cdot 2006$.  However, for $n=1$, we clearly must have $b_{n}=k(2006+1)$.  Thus for $n\neq 1$, $\frac{a_{n}}{b_{n}}=\frac{2005}{2006}$, and for $n=1$, we have $\frac{a_{n}}{b_{n}}=\frac{2005}{2007}$.[/hide][/quote]

sorry  you are wrong :(  . Logically we know that the sum of arithmetric progression is in second degree ....


geometry AMC AIME USA(J)MO USAMO blogs algorithm

sooo now that RSI apps have been turned in... just curious as to how many AoPSers applied and stuff or we could just generally discuss who we think will make it blah blah

ahh i just sent mine in today.  i wonder how many people around the nation apply to RSi annually.  They probably have to take what area you're from into account, and possibly gender?

no from the earlier RSI discussion they said that ethnicity/gender/race/etc are not taken into consideration XP

but there is a quota from the states so eeks

My understanding is there is explicit regional balance, but no consideration of ethnicity at all.

Can you be more specific about being explicitly balanced in terms of locations?

I applied! But it overlaps with MOP... hopefully I can get into one of them at least...

Yeah me too.  It does overlap with MOP but the three juniors in the Black group last year just left MOP a week early for RSI... so it's not that big of deal I guess.

yea the regionals thing sucks haha

i heard it ends up being like 1-2 per state T_T

I applied for it -- not really confident of going, but eh

I applied. However, my #Q@%#%@ school secretary forgot the envelope on her desk and as a result, it was postmarked on the 2nd! :(
Do you think that will significantly hurt my chances? Should I contact them and explain?

I think if you are postmarked late then you would be on the waiting list... or if only that one envelope was late then i dunno but definitely call them or something.

I think that would be getting too worked up over nothing. *shrugs*

The people at CEE are reasonable. If your envelope is one day late, I doubt they'd think it's because you're lately or incapable or something like that. A technical difficulty that could not be resolved the day of is much more likely.

surely more than about seven people from AOPS applied to RSI this year?

I applied, but I was in a bit of a rush to finish my applications (read: I was still working the day it was due... I had an early dismissal that day which basically saved me).

My stats are decent:
PSAT: 235 (77 math, 78 reading, 80 writing - 1 question wrong on each section, and I got the stupidest math question wrong)
AP Calc BC (as a sophomore): 5 (but that's my only AP b/c it's nearly impossible to take APs as anything under a junior at my school  :mad: )
AP classs: this year Stat, Bio, and Physics C
SAT II Chinese w/ Listening:  800
I took Chem and Math IIC on the 28th, so if the scores are good, I may send them as a supplement
GPA:  100.2 (only AP classes are weighted, .05)
Class rank:  Because everyone is equal, we don't have it.  I'm pretty sure I'm the only person in my class of 280-ish that has maintained a 100+ GPA for 3 years though (we had .04 weights for honors curriculum my freshman year, but they got rid of all weights other than AP my sophomore year), and I'm also pretty sure that I'm in the top 1 or 2%
SAT:  1420 (710 math 710 verbal) in 8th grade.  I wrote it down anyways b/c it was the last time I took them  :rotfl:
AMC 10 score of 126.5 last year (not too hot this year 101.5 on the 12  :( [size=59](3 stupid mistakes on questions that I knew how to do and 1 answer I was pretty sure was wrong but wrote in anyway).  Should easily have had a 118+... so mad at myself[/size]

btw... what did you guys put down as field and subfield of interest?  I had financial engineering (really cool stuff but I don't know if that was one of our options :lol:) and biomedical engineering  :)

Number Theory and Calculus/Geometry.

Don't worry about stats; they aren't heavily weighted, and your stats are especially fine. Mine are fairly mediocre; 4.49GPA, 230PSAT, 35ACT, 1470SAT. I mostly tried to emphasize my performance and enthusiasm in problem solving competitions.

I got in!  I am soooooo excited and look forward to meeting all of you this summer!

Yep. Rejected.  :mad:

Congrats to all of you who got in!  :)

ThAzn are you lying? From what I've heard about you, it's absolutely ridonculous that you'd be rejected.

[quote="agolsme"]waitlisted :([/quote]

Anyone know how many people are waitlisted? And if they ever even use the waitlist?

Haha I'm flattered, but no, I'm not kidding (really, even though it IS April fool's day  :? )

But then again (two years ago) they did reject two people from my school who ended up being Intel finalists (one of them got 2nd too!) and a few years back, rejected someone who ended up getting a gold medal at the IPhO (he was a MOPper as well)... Oh well, life goes on...

I know someone who got waitlisted last year and she said they didn't use it last year.

Yea, I know someone who got waitlisted and wasn't even notified about his status...they just never used him.

As for me, rejected. Bummer.

[quote="arjunlandes"]Does anyone have a suggestion about whether to do SEAP, which for me is 8 weeks of summer research at the Naval Observatory, or SSP? Anyone know how prestigious SSP is?[/quote]

Although I don't know much about SEAP, it doesn't hurt to apply there.  If it's not too late to round up application materials, then yeah...go for it.

When you mention SSP, I assume you mean the Summer Science Program.  If you haven't applied yet...you won't be able to download the application off the website anymore.

SSP is somewhat prestigious, though definitely not on par with RSI.  Then again, you shouldn't apply to summer programs based solely on prestige.  For example, the Simons + Garcia Research Programs at SUNYSB have excellent mentorships (~35 STS Semifinalists annually...consistently).  Though underadvertised and promoted usually just within the state of New York...this program produces results.

SSP advertises heavily, that's why it's somewhat prestigious.  RSI too, advertises heavily.  Does it produce as much as say...Simons?  Probably not.

Read the program descriptions and decide if SSP, SEAP, and whatever are right for you.  Don't give a **** about the prestige, but figure out for yourself whether you'll be happy at your placement.

I made it!  I'll see you all this summer...

I was rejected  :(

I'm going to Gov School for the Sciences at CMU instead.  Oh well, at least I got in somewhere  :|


hi

I want to have this result:

[u]blabla[/u]___________________

[u]blabla bal[/u]________________
[u]bla[/u]_______________[u]bla blah[/u]

several times for any given string.. what I have in mind is something like that:

\newcommand{\division}[1]{\Large \textsc{\underline{#1 \hspace{\textwidth} }}\normalsize}

but I need to substract the length of #1 to \textwidth and I don't know how...

thank

ledurt

Use \hfill instead.  \hfill is a rubber length which will fill the line.  In some cases (if your added space it at the beginning or the end of a line), you'll have to use \hspace*{\fill} rather than \hfill.

If this is not clear, let me know and I can elaborate.

well I ve tried with \hfill too but I didn't work either.

I found the answer in a book by Marie-Paule Kluth called "FAQ latex". The problem is the command \underline. Instead of this command, use the package

\usepackage{ulem}

and then

\newcommand{\division}[1]{\ULforem \Large \textsc{\emph{#1 \hfill }}\normalsize\\}

it redefines the command \emph to underline text even on several lines.  To use that mode of emphize, you need to use \ULforem before and to go back to the normal emph, use \normalem.

thanks

ledurt


The product of a set of distinct whole numbers is 120. What is the least possible
sum of the members of the set?

[hide]
You want a lot of small numbers and not big numbers. 120 is 5!.
5+4+3+2=14[/hide]


I downloaded miktex 2.7 and am currently using miktex 2.5. Can I get TexnicCenter to use 2.7 without reinstalling it?

No problem - all you have to do is tell TexnicCenter where to find MiKTeX. See [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Help#I.27ve_upgraded_MiKTeX_and_now_TeXnicCenter_won.27t_work]I've upgraded MiKTeX and now TeXnicCenter won't work[/url] for details.


linear algebra
matrix
search
linear algebra unsolved

I just got reminded of this beautiful problem (it's not hard, but I hadn't solved it):

Let $ \left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ and $ \left(b_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ be two nonnegatively definite $ n\times n$ matrices (what, in particular, means that they are symmetric with my notations). Prove that the matrix $ \left(a_{i,j}b_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ is also nonnegatively definite.

  darij

Thanks, the solution given there is the same as I knew. But I am wondering why on earth people define positive definiteness for asymmetric matrices. I would be surprised if one of the well-known properties of positive definite matrices would hold for such ones...

Another remark: Entrywise multiplication has nothing to do with the Kronecker product...
EDIT: Wow, it does! See the other topic.

  darij


Find the sum, expressed as a mixed number, of the first eight terms of the geometric sequence $ 2,1,\frac{1}{2},...$.

Actually, it's $ 3 \frac {63}{64}$, I think.  Basically, we add $ 2 \plus{} 1$ and $ \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \ldots \plus{} \frac {1}{64}$.

why can't you use the sum of terms in a geometric sequence formula?

[quote="kabi"]Hello
a=2
 r=$ \frac {1}{2}$
 n=8
Sum=$ 2\frac {1 - \frac {1}{2^8}}{\frac {1}{2}}$
=$ 4\frac {255}{256}$
Thank u[/quote]

The problem is $ 2\frac {1 - \frac {1}{2^8}}{\frac {1}{2}} \neq 4\frac {255}{256}$.

The error becomes apparent when we add parentheses:
\begin{align}
Sum &= 2\times \left( 2\frac {1 - \frac {1}{2^8}}{\frac {1}{2}} \right) \\
&= 4\times \left( \frac{255}{256} \right) \\
&= \frac{1020}{256} \\
&= 3+\frac{252}{256} \\
&= \boxed{3+\frac{63}{64} }
\end{align}

So the geometric seq formula is correct, but the evaluation was wrong.  The correct answer is indeed $ 3\frac{63}{64}$.

[quote="v_Enhance"]

The problem is $ 2\frac {1 - \frac {1}{2^8}}{\frac {1}{2}} \neq 4\frac {255}{256}$.

The error becomes apparent when we add parentheses:
\begin{align} Sum & = 2\times \left( 2\frac {1 - \frac {1}{2^8}}{\frac {1}{2}} \right) \\
& = 4\times \left( \frac {255}{256} \right) \\
& = \frac {1020}{256} \\
& = 3 + \frac {252}{256} \\
& = \boxed{3 + \frac {63}{64} } \end{align}

So the geometric seq formula is correct, but the evaluation was wrong.  The correct answer is indeed $ 3\frac {63}{64}$.[/quote]
thank u for correcting my solution .Sorry for d mistake.

[quote="kabi"]
thank u for correcting my solution .Sorry for d mistake.[/quote]
No problem  :P


search
linear algebra
linear algebra unsolved

If $ A\in M_3(\mathbb{R})$ and $ Tr(A) \equal{} Tr(A^2) \equal{} Tr(A^3) \equal{} 0$ then show that $ A^3 \equal{} 0$.

Let $ \lambda_1,\lambda_2,\lambda_3$ are the eigenvalues of  $ A$
Then we have :
$ Tr(A) \equal{} \lambda_1 \plus{} \lambda_2 \plus{} \lambda_3 \equal{} 0,Tr(A^2) \equal{} \lambda^2_1 \plus{} \lambda^2_2 \plus{} \lambda^2_3 \equal{} 0,Tr(A^3) \equal{} \lambda^3_1 \plus{} \lambda^3_2 \plus{} \lambda^3_3 \equal{} 0$
 Put $ \delta_1 \equal{} \lambda_1 \plus{} \lambda_2 \plus{} \lambda_3,\delta_2 \equal{} \lambda_1\lambda_2 \plus{} \lambda_2\lambda_3 \plus{} \lambda_3\lambda_1,\delta_3 \equal{} \lambda_1\lambda_2\lambda_3$
 easy calculate $ \delta_1 \equal{} 0,\delta_2 \equal{} 0,\delta_3 \equal{} 0$
 Characteristic equation of $ A$ is  : $ A^3 \minus{} \delta_1A^2 \plus{} \delta_2A \minus{} \delta_3I_3 \equal{} 0$ Then $ A^3 \equal{} 0$
General $ A \in M_n{\mathbb{(C)}}$ Such that : $ Tr(A^k) \equal{} 0 ,\forall k\in${1,2,...,n} then $ A^n \equal{} 0$
[color=blue]You can use Newton formula to prove that [/color]$ A^n \equal{} 0$
Or Click : http://www.mathlinks.ro/viewtopic.php?highlight=nilpotent+trace&t=64870
              http://www.mathlinks.ro/viewtopic.php?search_id=990017138&t=176564


For a $ \Delta ABC$ let $ F$, $ F_A$, $ F_B$, $ F_C$  be Feuerbach points, $ I$  - incenter.
Prove, that lines $ AF_A$, $ BF_B$, $ CF_C$,  $ IF$ concur.

Dear Edward and Mathlinkers,
this nice problem is an application of the Monge or d'Alembert's line (theorem).
Sincerely
Jean-Louis


function
number theory unsolved
number theory

Let there $ M$ a set of functions $ f : \mathbb{N} \to \mathbb{N}$ , such that $ \forall k \in \{ 1 , 2 , \dots , 2007\}$ we have $ f^{[k]} \neq \mathrm{Id}_{\mathbb{N}}$ , and $ f^{[2008]} \equiv \mathrm{Id}_{\mathbb{N}}$ . Where $ \mathrm{Id}_{\mathbb{N}}$ is the identity function ( $ f(x) \equal{} x$ ) . Has $ M$ an infinite number of elements ?

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=185415

Pierre.


I've put up all the problems on the forum of the Tournament of Towns, Fall 2004, although some of the problems are trivial.

Junior B-level
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=21840
2. http://www.mathlinks.ro/Forum/viewtopic.php?t=21841
3. http://www.mathlinks.ro/Forum/viewtopic.php?t=21842
4. http://www.mathlinks.ro/Forum/viewtopic.php?t=21843
5. http://www.mathlinks.ro/Forum/viewtopic.php?t=21844

Senior B-level
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=21845
2. Idea of the problem is the same as Junior B-level 2
3. http://www.mathlinks.ro/Forum/viewtopic.php?t=21846
4. http://www.mathlinks.ro/Forum/viewtopic.php?t=21844
5. http://www.mathlinks.ro/Forum/viewtopic.php?t=21847

Junior A-level
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=21848
2. http://www.mathlinks.ro/Forum/viewtopic.php?t=21849
3. http://www.mathlinks.ro/Forum/viewtopic.php?t=21850
4. http://www.mathlinks.ro/Forum/viewtopic.php?t=21851
5. http://www.mathlinks.ro/Forum/viewtopic.php?t=21852
6. http://www.mathlinks.ro/Forum/viewtopic.php?t=21391
7. http://www.mathlinks.ro/Forum/viewtopic.php?t=21390

Senior A-level
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=21853
2. http://www.mathlinks.ro/Forum/viewtopic.php?t=21855
3. http://www.mathlinks.ro/Forum/viewtopic.php?t=21851
4. http://www.mathlinks.ro/Forum/viewtopic.php?t=21856
5. http://www.mathlinks.ro/Forum/viewtopic.php?t=21390
6. http://www.mathlinks.ro/Forum/viewtopic.php?t=21857
7. http://www.mathlinks.ro/Forum/viewtopic.php?t=21858

Enjoy!

[quote="J"]Maybe someone could compile a .pdf file ?
I would be very grateful ;)[/quote]All the problems are there. You should not wait for others to do the work you want. If you want to contribute, you can create the .pdf yourself and attach it here ;)

[quote="Xixas"]I've put up all the problems on the forum of the Tournament of Towns, Fall 2004, although some of the problems are trivial.[/quote]
Many of them (maybe most of them) are far from not trivial.


Compare the electrical force between two protons in free space with the gravitational force between them: Express result as a ratio.

[hide="Hint"]
What are the laws governing the coloumbic and the gravitational interaction between particles?[/hide]

[hide="Solution"]

BTW, this is IP not AP.

Let the mass of protons be $ m$, charges $ e$, distance between them $ r$.

Gravitational force: $ F_1\equal{}\frac{\gamma m^2}{r^2}$
Electrical force: $ F_2\equal{}\frac{ke^2}{r^2}$

The ratio

$ \frac{F_2}{F_1}\equal{}\frac{ke^2}{\gamma m^2}$. These are all constants, I leave the computation to you

[/hide]


calculus
integration
limit
trigonometry
calculus computations

Evaluate

\[\lim_{n\to\infty} n\int_0^\pi e^{-nx} \sin ^ 2 nx\ dx\]

Integration by part, we have
$ \int e^{ax}\cos bx\ dx = \frac{a\cos bx+b\sin bx}{a^2+b^2}e^{ax} $

thus
$ \int e^{-nx} \sin ^ 2 nx\ dx =\frac{1}{2}\int e^{-nx} (1-\cos 2nx)\ dx = \frac{1}{2}\left(-\frac{e^{-nx}}{n}-e^{-nx}\frac{-n\cos 2nx+2n\sin 2nx}{5n^2}\right) = e^{-nx}\frac{-5+\cos 2nx-2\sin 2nx}{10n} $

$ \lim_{n\to\infty} n\int_0^\pi e^{-nx} \sin ^ 2 nx\ dx =\lim_{n\to\infty} n\cdot \frac{2}{5n}(1-e^{-n\pi}) = \frac{2}{5} $


search
induction
inequalities unsolved
inequalities

That's certainly false. I think you mean $ a_1^3+...+a_k^3=(a_1+...+a_k)^2$ for all $k=1,2,...,2000$ and then please try to search the forum in algebra since it has been solved several times. If I remember correctly (in case you cannot find with search button take it as a hint) the idea is to prove by induction that at each stage $a_1,...,a_i$ are integers and their sum if of the form $\frac{n(n+1)}{2}$ with $n$ integer.

oh sorry I've editd it.
and thanx for your help


linear algebra
matrix
vector
topology
group theory
abstract algebra

If $ v_1,v_2,...,v_n$ form an orthonormal basis for $ R^n$, prove that $ \displaystyle\sum_{k\equal{}1}^{n}{(v_k*v^T_k)}\equal{}I_n$, where $ I_n$ is the identity matrix with $ n$ columns.

There's some awkward notation there - what exactly do you mean by $ *$?

I'm going to assume that you simply mean matrix multiplication, giving the $ n\times n$ rank-1 matrix $ v_kv_k^T.$

Let $ Q$ be the $ n\times n$ matrix whose $ k$th column is $ v_k.$ Then $ Q^Tv_k$ is the column whose $ i$th element is $ v_i^Tv_k,$ also known as the inner product of $ v_i$ and $ v_k.$ That means that $ Q^Tv_k$ is $ e_k,$ the vector whose $ k$th element is one and all other elements are zero. Similarly, $ v_k^TQ \equal{} e_k^T.$

Let $ A \equal{} \sum_{k \equal{} 1}^nv_kv_k^T.$

Then $ Q^TAQ \equal{} \sum_{k \equal{} 1}^ne_kv_kQ \equal{} \sum_{k \equal{} 1}^ne_ke_k^T.$

But $ \sum_{k \equal{} 1}^ne_ke_k^T \equal{} I.$ So we have $ Q^TAQ \equal{} I.$

Since $ Q^TQ \equal{} I,$ (another characterization of an orthonormal basis), we can left-multiply that by $ Q$ and right-multiply it by $ Q^T$ to get that $ A \equal{} I.$

Sorry for the poor notation, but yes, i meant $ *$ to represent matrix multiplication.

I understand your proof, but I have some brief general questions:

If the $ v_k$ are orthonormal column vectors, then how do you know that $ v_kv^T_k$ is an n by n matrix with rank 1?

Also, can you explain how you go from:
$ Q^TA\equal{}\sum_{k\equal{}1}^ne_kv_kQ\equal{}\sum_{k\equal{}1}^ne_ke_k^T\equal{}I_n$ to $ Q^TAQ\equal{}I?$

One final question:
Are the orthonormal bases for $ R^n$ unique (standard orthonormal bases $ e_k$?)

Thanks

If $ a$ and $ b$ are any nonzero $ n\times 1$ column vectors, then $ ab^T$ will be an $ n\times n$ matrix of rank $ 1.$ That has nothing to do with any special properties of $ a$ and $ b.$ Furthermore, any rank $ 1$ matrix can be written in this form.

As for your second question: I had a typo which I have now fixed; I always intended to right-multiply by $ Q$ all along.

Orthonormal bases for $ \mathbb{R}^n$ are very, very far from unique. An orthonormal basis forms the columns of a matrix $ Q$ satisfying $ Q^TQ\equal{}I.$ Such a matrix $ Q$ is called an orthogonal matrix and belongs to the set $ O(n).$ How many members does $ O(n)$ have? Well, it's a Lie group, meaning that it's a group that is also a smooth manifold (is locally like $ \mathbb{R}^k$ for some $ k$). Its dimension as a manifold - the $ k$ of the $ \mathbb{R}^k$ that it looks like - is $ \frac{n(n\minus{}1)}2.$ That's how many.

The standard basis forms the identity matrix, but as an orthogonal matrix, the identity matrix is one among many.


Does anyone have idea how to show this question??

Let $\{ v_{1}, ..., v_{n}\}$ be a basis of a vecote space over $K$ and let $j=1,...,n$, $w_{j}= \sum_{i=1}^{n}\alpha_{ij}v_{i}$
Show that $\{w_{i},...,w_{n}\}$ is a basia of $V$ if and only if $det(\alpha_{ij}) \neq 0$


quadratics
modular arithmetic
algebra
quadratic formula
special factorizations

[quote="guile"]Prove that $ ax^2 \plus{} bx \plus{} c \equiv 0 (mod p)$, where $ p$ is odd prime and does not divide $ a$, is solvable if and only if, $ b^2 \minus{} 4ac \equiv 0 (mod p)$.[/quote]
[hide]
hmm, i recognize discriminant

and i know the radical in the quad formula is 0 (in other words, 1 soln) iff the discriminant is 0

hmm, nope i got nothing

[/hide]

I'm not sure where you're getting your result.  The correct statement is that $ b^2 \minus{} 4ac$ is a quadratic residue $ \bmod p$; in other words, $ (b^2 \minus{} 4ac)^{ \frac{p\minus{}1}{2} } \equiv 1 \bmod p$.  The proof is essentially a run-through of the proof of the quadratic formula, but with all calculations done $ \bmod p$.

[quote="t0rajir0u"]I'm not sure where you're getting your result.  The correct statement is that $ b^2 \minus{} 4ac$ is a quadratic residue $ \bmod p$; in other words, $ (b^2 \minus{} 4ac)^{\frac {p \minus{} 1}{2} } \equiv 1 \bmod p$.  The proof is essentially a run-through of the proof of the quadratic formula, but with all calculations done $ \bmod p$.[/quote]

Can  you show the proof?

[quote="t0rajir0u"]I'm not sure where you're getting your result.  The correct statement is that $ b^2 \minus{} 4ac$ is a quadratic residue $ \bmod p$; in other words, $ (b^2 \minus{} 4ac)^{\frac {p \minus{} 1}{2} } \equiv 1 \bmod p$.  The proof is essentially a run-through of the proof of the quadratic formula, but with all calculations done $ \bmod p$.[/quote]

Or maybe you mean to prove that there is only one solution $ \pmod{p}$ to the quadratic if $ b^2 \minus{} 4ac \equiv 0$.  Either way, the proof just involves completing the square in $ \pmod{p}$, so you should try it yourself.