quadratics
algebra
number theory proposed
number theory

$ x,y,z$ are natural numbers.
Solve for
$ (x \plus{} y \plus{} z)^2 \equal{} xyz$
 $ (x \plus{} y \plus{} z)^2 \equal{} 2xyz$
 $ (x \plus{} y \plus{} z)^2 \equal{} 3xyz$
  $ (x \plus{} y \plus{} z)^2 \equal{} 4xyz$ ................
$ (x \plus{} y \plus{} z)^2 \equal{} kxyz$, $ k$ is a natuaral number.

It is a nice and easy problem.

Come on!! No takers??? THis is a very simple  problem but quite nice.
So post the solution....................Let us see how different people get different  solutions.

Someone please post the solution to the problem. Infact, I had not solved it properly so I am in need of a solution..........So someone even if it is kind of low level please post a solution.........Thanks........................

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

You sure this problem is correct? if $ x,y,z \in \mathbb{N}$ you can not say $ xyz\equal{}2xyz$ because you can delete each letter in two side of equality(NOTE: $ x,y,z \in \mathbb{N}$).
please talk me that how may this equality.???!!! :o

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

[quote="vandermond"][color=#666666]Hi..........[/color]

You sure this problem is correct? if $ x,y,z \in \mathbb{N}$ you can not say $ xyz \equal{} 2xyz$ because you can delete each letter in two side of equality(NOTE: $ x,y,z \in \mathbb{N}$).
please talk me that how may this equality.???!!! :o

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

Actually the question is not clear.I guess Agr_94_Math means solve $ (x\plus{}y\plus{}z)^2\equal{}kxyz$ for given $ k$ and not means to solve a system of equation.But I am not sure whether I guess it right.

May be that's why nobody posts solution, I saw several posts having the same problem before.

Sorry for all. It is not a system of equations. It is what stephenchang meant {Guessed in the last post}. Please post ur solutions now.
Thanks.

Well, I have almost finshed it. But I am trying using quadratics.
Someone else please suggest a nice way to solve the problem.

Thanks.

Can some one please post a solution?
ie Solve for $ x,y,z$ in natural numbers
$ (x\plus{}y\plus{}z)^2\equal{} kxyz$ where $ k$ is a antural number.
If u have not  got a solution for $ k$ as a natural no. in genral,
atleast please post a solution for $ k\equal{}1,2,3,...5$ atleast.

Let $ A_k \equal{} \{(x,y,z)\in \mathbb {N}^3| x\le y\le z,(x \plus{} y \plus{} z)^2 \equal{} kxyz\}$.
$ (x,y,z)\in A_k$ is called minimal solution in $ A_k$ if there does not exist $ (a,b,c)\in A_k$ with $ (a,b,c)\ne(x,y,z)$ and $ a\le x,b\le y,c\le z$.
Now we try to find all the minimal solutions $ (x,y,z)$ for different $ k$.

Suppose $ z > x \plus{} y$.
We can treat $ (x \plus{} y \plus{} z)^2 \equal{} kxyz$ as a quadratic equation in $ z$. If we let $ (a,b,c)$ be the ordered triple $ (x,y,\frac {(x \plus{} y)^2}z)$,
we see that $ (a,b,c)\in A_k$ and and $ a\le x, b \le y,c < z$. Contradiction with definition of minimal solution.

Case 1: $ y < z < x \plus{} y$
Let $ t \equal{} x \plus{} y \minus{} z$, we have $ 0 < t < x$.
$ kxy \equal{} \frac {(2z \plus{} t)^2}z \equal{} 4z \plus{} 4t \plus{} \frac {t^2}z$
Hence,
$ kxy > 4z \plus{} 4t \equal{} 4x \plus{} 4y\Rightarrow (kx \minus{} 4)(ky \minus{} 4) > 16$, or
$ (kx \minus{} 4)(ky \minus{} 4)\ge17$ (*)
Also,
$ z > y\ge\frac {x \plus{} y}2\ge t\Rightarrow kxy < 4z \plus{} 5t < 5x \plus{} 4y\Rightarrow (kx \minus{} 4)(ky \minus{} 5) < 20$, or
$ (kx \minus{} 4)(ky \minus{} 5)\le 19$ (**)
$ kx\le ky\Rightarrow kx\le 8$.
1a) $ kx \equal{} 5$,
Then $ 21\le ky\le24$, $ k\le gcd(kx,ky)\Rightarrow k \equal{} 1$.
Since $ 5|y \plus{} z,y < z < y \plus{} 5$, the only possible solutions are $ (5,21,24),(5,22,23),(5,23,27),(5,24,26)$. None of them are valid.
1b) $ kx \equal{} 6$. Then $ ky \equal{} 13$ or $ ky \equal{} 14$.
$ k \equal{} 1$ when $ ky \equal{} 13$;$ k \equal{} 1,2$ when $ ky \equal{} 14$. Similar to 1a), there is no valid solution.
1c) $ kx \equal{} 7$. Then $ ky \equal{} 10$ or $ ky \equal{} 11$.
$ k \equal{} 1$. No valid solution.
1d) $ kx \equal{} 8$. Then $ ky \equal{} 9$.
$ k \equal{} 1$. No valid solutions.

Case 2) $ z \equal{} y \plus{} x$.
So $ kxy \equal{} 4x \plus{} 4y$, or $ (kx \minus{} 4)(ky \minus{} 4) \equal{} 16$.
Then it is easy to get all possible $ (kx,ky,kz)$
$ (5,20,25), (6,12,18), (8,8,16)$
Then the possible $ (x,y,z,k)$ are
$ (5,20,25,1),(1,4,5,5),(6,12,18,1),(3,6,9,2),(2,4,6,3),(1,2,3,6),(8,8,16,1),(4,4,8,2),(2,2,4,4),(1,1,2,8)$.

Case 3) $ z \equal{} y$
$ kxy^2 \equal{} (x \plus{} 2y)^2$. We see $ 4 < kx\le 9$ and $ kx$ is a perfect square. So $ kx \equal{} 9$ and $ x \equal{} y \equal{} z$.
The possible $ (x,y,z,k)$ are
$ (9,9,9,1),(3,3,3,3),(1,1,1,9)$.

List of all minimal solutions:
$ k \equal{} 1$: $ (9,9,9),(5,20,25),(6,12,18),(8,8,16)$
$ k \equal{} 2$: $ (3,6,9),(4,4,8)$
$ k \equal{} 3$: $ (3,3,3),(2,4,6)$
$ k \equal{} 4$: $ (2,2,4)$.
$ k \equal{} 5$: $ (1,4,5)$
$ k \equal{} 6$: $ (1,2,3)$
$ k \equal{} 8$: $ (1,1,2)$
$ k \equal{} 9$: $ (1,1,1)$.

For each $ k$, the set $ A_k$ can be generated from its minimal solutions by repeating following operations:
Replace any number say $ x$, with $ \frac {(s \minus{} x)^2}x$, where $ s \equal{} x \plus{} y \plus{} z$.
Then put the new three numbers in ascending order.

Note if $ x \plus{} y \equal{} z$, replacing $ z$ will not give a different solution.


Let $ x,y,z>0,x\plus{}y\plus{}z\equal{}3,x^2\plus{}y^2\plus{}z^2\equal{}7$.Prove that:
$ 1\plus{}\frac 6{xyz}\ge \frac 1{3}(\frac x{z}\plus{}\frac z{y}\plus{}\frac y{x})$

Noone solve it for me?I think it's quite hard but you are very well at Inequality.


Let $f(n)$ be the number of those $n$-digit positive whole numbers in which the $1$ and $2$ digits both appear. Prove that $f(n)$ cant be a square number if $n \geq 2$.

I think $8 \cdot 9^{n-1}+8 \cdot 9^{n-1}-f(n) = 9 \cdot 10^{n-1}-7 \cdot 8^{n-1}$.
So, $f(n) = 16 \cdot 9^{n-1}+7 \cdot 8^{n-1}-9 \cdot 10^{n-1}$

But I can't continue the proof..  :(
Can anyone help ?


geometry
circumcircle
ratio
geometry unsolved

Let $ ABC$ be an acute triangle. Points $ D,E,F$ lie on segments $ BC,CA,AB$ respectively, and each of the three segments $ AD,BE,CF$ contains the circumcenter of $ ABC$. Prove that if any two of the ratios $ \frac{BD}{DC},\frac{CE}{EA}, \frac{AF}{FB}, \frac{BF}{FA},\frac{AE}{EC}, \frac{CD}{DB}$ are integers, the triangle $ ABC$ is isosceles

Please,  public all problems from Chinese Girl's MO 2007 in Resourses. Thanks.

They're posted here by zhaoli: http://www.mathlinks.ro/Forum/viewtopic.php?t=165337  :)


hi, i'm new. UK contestant 4 this IMO.

abc=1 a,b,c>0

(2a-1)(2b-1)(2c-1) + (a+b)(b+c)(c+a) >= k

what is the largest possible k?

Let $A=a+b+c, B=ab+bc+ca, C=abc=1$ with $B\geq 3$

Then we want to prove:

$9\leq 8C-4B+2A-1+AB-C=(A-4)(B+2)+14$

Because $A^2\geq 3B$,

Let $x=\sqrt{B}$ with $x\geq \sqrt{3}$

We need to prove:

$f(x)=(\sqrt{3}x-4)(x^2+2)+14=\sqrt{3}x^3-4x^2+2\sqrt{3}x+6\geq 9$

Because $f'(x)>0$, So $f(x)\geq f(\sqrt{3})=9$

At last with $a=b=c=1$, we have $k_{max}=9$.


greatest common divisor
number theory proposed
number theory

Let $a_{i}$, $i = 1,2, \dots ,n$, $n \geq 3$, be positive integers, having the greatest common divisor 1, such that \[a_{j}\textrm{ divide }\sum_{i = 1}^{n}a_{i}\]
for all $j = 1,2, \dots ,n$. Prove that \[\prod_{i = 1}^{n}a_{i}\textrm{ divides }\Big{(}\sum_{i = 1}^{n}a_{i}\Big{)}^{n-2}.\]

Let $p$ be a prime that divides some of the $a_{i}$'s.

Now, can all the $a_{i}$'s be multiples of p? Obviously not, because they are coprime.
Can all the $a_{i}$'s but one be multiples of p? Not, because is so, their sum would not be multiple of p, which is impossible.

So, if k is the maximum exponent of p that divides some of the $a_{i}$'s, we have $p^{k}\mid a_{1}+a_{2}+\ldots+a_{n}$, and $p^{(n-2)k}\mid (a_{1}+a_{2}+\ldots+a_{n})^{n-2}$. But the maximum exponent of p that divides $a_{1}a_{2}\cdots a_{n}$ is $\le (n-2)k$, because at least two of the $a_{i}$'s are coprime to p, and the others have exponent $\le k$.

If we to this reasoning for all the primes that divide some of the $a_{i}$'s, we get the result.

My solution is the same of yours, edriv!

Too easy for a romanian tst, no?  :P


I've got no idea why you asked, but I like the Flash. Because supposedly the same thing thats make him go super fast makes him think faster because when he is runnign he has to react faster. Plus he has got an attitude :D .

Yeah, the Flash is cool; gotta love his sense of humor.  Green Arrow's got a similar sense of humor, but a little less fun-loving and a little more cynical.

where do you guys know these things? cartoons?

these must be boring comic books. i only heard of superman. i gues this comic book isn't very popular.

They also have a show on Cartoon Network called Justice League...

i heard that kids WB is better than cartoon network.

Technically, the Cartoon Network show is Justice League Unlimited.  And what makes you think that comic books are boring?  The fact that you haven't heard even of Batman?

Green Arrow: The Greatest Archer That Ever Lived

I vote for Wonder Woman because she's hot!! jk  :P

What about this question: Which member do you not like? And to that, I would say Batman because he does not have any powers whatsoever. He's pretty much a person who has these cool gadgets and a cool car.

But back to the poll question. I like either Flash because he's so quick or Superman because he has the x-ray vision thing. Good for some things if you know what I'm talking about.......  :D I don't really know who Green Lantern is.....

I like superman becuase he's super and he's a man.

It used to be called Justice League, they changed it though.

I prefer Martian Manhunter because he has the best name.  But some of the reasoning in this thread is quite off.  H_s_potter2002 used Tarelike logic, complete with homosexuality (mind you, that's not necessarily a bad thing); gameworld reasoned that since he hadn't heard of something, it was boring, then used hearsay to argue that WB was superior to Cartoon Network for no apparent reason; and then our beloved administrator bumped Batman out of the running.

[quote="Valentin Vornicu"][quote="RobinHood3000"]Well, who's your favorite member of the Justice League?  Sorry, I'm something of a comic book geek.

Me, I'm a Green Arrow obsessor.

Sorry, I forgot the "Other" option, so just post here if you've got a different favorite.[/quote]There is a slight mistake here. Batman is NOT a member of the justice league :P He however tends to be an un-official member, having the same goals, but not the same means, as the Justice League (ie he does not have any special powers, just a lot of high-tech gadgets and ambition :) ).[/quote]

In the show on Cartoon Network, Batman was an official member...

Actually, Batman's intelligence rivals that of any other member of the Justice League, hence his membership, official or otherwise.  It's an aspect emphasized much more in the comics than in the television shows.  However, I am impressed with the Justice League Unlimited show remembering Green Arrow's, aka Oliver Queen's, relationship with Black Canary, aka Dinah Lance.

as i read somewhere Batman has one of the most analytical minds in the world and is one of the best detectives ever
when i first saw Batman and Superman i thought Batman was way cooler than Superman. I also liked a lot of the villains that were against Batman.
Now i like The Flash cause he's really cool and funny. I remember in an old Superman tv cartoon the Flash and Superman had a race around the world and Superman was like tied with Flash. It looked like Flash was worthless because not only was Superman as fast as Flash he could also do other stuff too.  But i still like Flash the best.

Yeah, I find Flash and Green Arrow have the best senses of humor, with Batman and Martian Manhunter as the deadpan straight men.  Lots of comic chemistry to be found.

Oh, yeah, and Batman is a martial arts expert equal to even the Birds of Prey.

what about robin? u know batman and robin.

Which Robin?  There have been at least three I can think of.

Robin was before the Justice League, I think

True, but he/she (there has been at least one girl Robin that I remember; isn't the unisex name "Robin" so convenient?) leads the Teen Titans, sort of a mini-Justice League.

Before anyone starts talking about the Cartoon Network Series (which is still pretty good), I'd like to point out that the Teen Titans in their original incarnation in the comic books was intended to be an association of sidekicks.  I believe that Robin (Dick Grayson/Robin), Speedy, Kid Flash, and Aqualad were among the founding members.

May be it is because I use to see the Justice League, the version of the seventies and eighties (and in spanish), but there are a lot of nice caracters that are not in the list and they were very nice. Assuming that the names were well translated, they should be

* The Wonder Twins
* Apache Chief
* Thunder
* Atomic Men

Well, and many other that I don't remember any more. I suppose I should see the new series to see the characters  :P .

[quote="GoBraves"]Wonder Woman[/quote]

that's my nickname. :)

um.. in ur profilhingy on the side it says ur a male.... I'M SCARED lol.  I think the green lantern is cool because his ring can do so many things XD

elvenchamp777


Fie $ABCD$ un tetraedru in care $AB\perp CD$ si $AD\perp BC$. Demonstrati ca proiectiile centrului de greutate ale tetraedrului pe fetele acestuia sunt centrele cercurilor lui Euler ale fetelor.

Desi poate ca unora o sa li se para usoara, cred ca pentru clasa a VIII-a este o problema foarte frumoasa.

[quote="mychrom"]Desi poate ca unora o sa li se para usoara, cred ca pentru clasa a VIII-a este o problema foarte frumoasa.[/quote]

Reluand aceste randuri, sper sa va determin sa publicati solutii cat mai frumoase.

[hide="indicatii"]
$ABCD$ e ortocentric (inaltimile sunt concurente), si proiectia lui $A$ pe $(BCD)$ este ortocentrul triunghiului $BCD$.
fie $O$ centrul cercului circumscris triunghiului $BCD$ si $X$ proiectia lui $G$ pe $BCD$. de asemenea, fie $G_{a}$ centrul de greutate al fetei $BCD$ si $H$ ortocentrul aceleasi fete.
atunci, $G_{a}X/G_{a}H=G_{a}G/G_{a}A$, si de aici se lucreaza cu rapoartele pana cand se ajunge la $OH=2OX$.
am omis multi pasi dar cred ca asta este ideea
[/hide]


A lock has 5 buttons numbered 1-5. The lock is opened by pushing two buttons simultaneously and then pushing one button alone. How many combinations are possible?

This is clearly $ \binom52\cdot5\equal{}\boxed{50}$.


inequalities
function
three variable inequality

Let $ a,b,c$ be real numbers such that $ a\plus{}b\plus{}c\equal{}3$. Prove that
\[\frac{1}{5a^2\minus{}4a\plus{}11}\plus{}\frac{1}{5b^2\minus{}4b\plus{}11}\plus{}\frac{1}{5c^2\minus{}4c\plus{}11}\leq\frac{1}{4}\]

Let $ f(x) \equal{} \frac {1}{5x^2 \minus{} 4x \plus{} 11}.$ Then $ f'(x) \equal{} \frac {4 \minus{} 10x}{(5x^2 \minus{} 4x \plus{} 11)^2}.$
Hence, $ f(x)\leq f\left(\frac {2}{5}\right)$ for all real $ x$ and $ f$ is a decreasing function on $ [\frac {2}{5}, \plus{} \infty].$
But $ \sum_{cyc}f(a)\leq\frac {1}{4}\Leftrightarrow\sum_{cyc}\left(\frac {1}{12} \minus{} \frac {a \minus{} 1}{24} \minus{} \frac {1}{5a^2 \minus{} 4a \plus{} 11}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(9 \minus{} 5a)(a \minus{} 1)^2}{5a^2 \minus{} 4a \plus{} 11}\geq0.$
If $ \max\{a,b,c\}\leq\frac {9}{5}$ then our inequality is true.
If $ \max\{a,b,c\} > \frac {9}{5}$ then $ \sum_{cyc}f(a) < f\left(\frac {9}{5}\right) \plus{} 2f\left(\frac {2}{5}\right) < \frac {1}{4}.$

It is the official solution,very nice!Thank you.

[quote="Ji Chen"]A stronger inequality is :

if $ a,b,c$ be real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ \frac {1}{9a^2 \minus{} 2a \plus{} 25} \plus{} \frac {1}{9b^2 \minus{} 2b \plus{} 25} \plus{} \frac {1}{9c^2 \minus{} 2c \plus{} 25}\leq\frac {3}{32},$

with equality if  $ a \equal{} \frac {17}{9},b \equal{} c \equal{} \frac {5}{9}.$

See also here : http://www.mathlinks.ro/viewtopic.php?t=223910[/quote]

Denote $ q\equal{}ab\plus{}bc\plus{}ca$ and $ r\equal{}abc$ then
\[ \frac {1}{{9a^2 \minus{} 2a \plus{} 25}} \plus{} \frac {1}{{9b^2 \minus{} 2b \plus{} 21}} \plus{} \frac {1}{{9c^2 \minus{} 2c \plus{} 21}} \le \frac {3}{{32}}
\]

\[ \Leftrightarrow f\left( r \right) \equal{} 2187r^2 \minus{} \left( {486q \plus{} 18276} \right)r \plus{} 3483q^2 \minus{} 7100q \plus{} 7500 \ge 0
\]
We have $ f'\left( r \right) \equal{} 4374r \minus{} 486q \minus{} 18276 \le 0$ thus it suffices us to prove the inequality in the case $ a \equal{} b$.
Hence the inequality becomes
\[ \frac {2}{{9a^2 \minus{} 2a \plus{} 25}} \plus{} \frac {1}{{9\left( {3 \minus{} 2a} \right)^2 \minus{} 2\left( {3 \minus{} 2a} \right) \plus{} 21}} \le \frac {3}{{32}}
\]

\[ \Leftrightarrow 12\left( {a \minus{} 1} \right)^2 \left( {9a \minus{} 5} \right)^2 \ge 0
\]
which is clearly true. Equality holds if and only if $ a \equal{} b \equal{} c \equal{} 1$ or $ a \equal{} b \equal{} \frac {5}{9}$ and $ c \equal{} \frac {{17}}{9}$ or any cyclic permutations, :).

[quote="sqing"]The following inequality is also true.
Let $ a,b,c$ be real numbers such that $ a\plus{}b\plus{}c\equal{}3$. Prove that
\[\frac{a}{5a^2\minus{}4a\plus{}11}\plus{}\frac{b}{5b^2\minus{}4b\plus{}11}\plus{}\frac{c}{5c^2\minus{}4c\plus{}11}\leq\frac{1}{4}.\][/quote]
Because $\frac{1}{4}-\sum_{cyc}\frac{a}{5a^2-4a+11}\geq\frac{5\left(\sum\limits_{cyc}(8a^3-15a^2b-15a^2c+22abc)\right)^2}{2916\prod\limits_{cyc}(5a^2-4a+11)}\geq0$.

[quote=Ji Chen]A stronger inequality is :

if $ a,b,c$ be real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ \frac {1}{9a^2 \minus{} 2a \plus{} 25} \plus{} \frac {1}{9b^2 \minus{} 2b \plus{} 25} \plus{} \frac {1}{9c^2 \minus{} 2c \plus{} 25}\leq\frac {3}{32},$

with equality if  $ a\equal{}\frac{17}{9},b\equal{}c\equal{}\frac{5}{9}.$

See also here : http://www.mathlinks.ro/viewtopic.php?t=223910[/quote]
A weaker inequality：
$$ \frac {1}{9a^2 \minus{} 6a \plus{} 21} \plus{} \frac {1}{9b^2 \minus{} 6b \plus{} 21} \plus{} \frac {1}{9c^2 \minus{} 6c \plus{} 21}\leq\frac {1}{8}$$
[quote=sqing]The following inequality is also true.
Let $ a,b,c$ be real numbers such that $ a\plus{}b\plus{}c\equal{}3$. Prove that
\[\frac{a}{5a^2\minus{}4a\plus{}11}\plus{}\frac{b}{5b^2\minus{}4b\plus{}11}\plus{}\frac{c}{5c^2\minus{}4c\plus{}11}\leq\frac{1}{4}.\][/quote]
 Let $ a,b,c$ be  non-negative real numbers such that $ a+b+c=3$. Prove that
\[\frac{3}{44}\leq\frac{a}{5a^2-4a+11}+\frac{b}{5b^2-4b+11}+\frac{c}{5c^2-4c+11}\leq\frac{1}{4}.\]


A zoo has a menagerie containing four pairs of different animals, one male and one female for each. The zookeeper wishes to feed the animals in a specific pattern: each time he feeds a single animal, the next one he feeds must be a different gender. If he starts by feeding the male giraffe, how many ways can he feed all the animals?

[quote="akhil0422"]You have giraffes $ AB,CD,EF,GH$, and the zoo keeper has $ 4$ choices for the first one, $ 4$ for the second one, $ 3$ for the third one and so on \[ 4\cdot4\cdot3\cdot3\cdot2\cdot2\Rightarrow\boxed{576}\][/quote]
Except the first animal he feeds is already chosen so there is only one choice for the first animal.  Therefore:

$1\cdot4\cdot3\cdot3\cdot2\cdot2\cdot1\cdot1=\boxed{144}$


inequalities
function
Asymptote
limit
algebra unsolved
algebra

$ f,g: R \to R$ satisfy $ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} 2f(x)g(y)$. If $ f$ is not identically 0 and $ |f(x)| \le 1$ for all $ x \in R$, prove that $ |g(y)| \le 1$ for all $ y \in R$.

[hide]
It's driving me crazy... I posted a related topic in the open problems subforum, and it is clear that $ f(x) \equal{} C\cos(h(x)), g(x)\equal{}\cos(h(x))$ are solutions where $ h$ satisfies $ h(x \plus{} y) \equal{} h(x) \plus{} h(y)$, and $ |C| \le 1$. Found this problem in an inequality collection, no source.
[/hide]

Let $ |sup f(x)|\equal{}m$ ($ m\leq 1$). Then $ 2|m|\geq |f(x \plus{} y) \plus{} f(x \minus{} y)| \equal{} |2f(x)g(y)|\to 2|m\parallel{}g(x)|$. Hence $ 1\geq |g(x)|$

[quote="Flakky"]Let $ |sup f(x)| \equal{} m$ ($ m\leq 1$). Then $ 2|m|\geq |f(x \plus{} y) \plus{} f(x \minus{} y)| \equal{} |2f(x)g(y)|\to 2|m\parallel{}g(x)|$. Hence $ 1\geq |g(x)|$[/quote]
Cool! Really cool.
More solutions plz :P although it looks like this solution may be the official one... It uses the boundness nicely.

[quote="FelixD"]It's problem 5 from the IMO 1972  :wink:

At home I've got a book about functional equations, there this problem appears too. The solution in the book is the same as Flakky's solution, so I won't post it here.[/quote]
Nice to know. Turns out it is good to check old IMOs, some nice problems there. I still wonder whether it is possible to solve this equation completely. Anyhow, I looked at the correspending thread of this IMO problem, and it is empty (perhaps some mod would merge these threads?)

[quote="hmm"]Sorry if this is a stupid question, but how do we know that this supremum can actually be attained? For example, a logistic function approaches two horizontal asymptotes (its "supremum") but never actually hits them; therefore, it's max/min cannot be attained. This would make the substitution f(x)=m invalid, which Flakky's solution seems to depend on...[/quote]
It wasn't a substition, but a limit. Flakky showed that $ |g(y)| \le \frac {|m|}{|f(x)|}$ when $ f(x) \neq 0$ and took a sequence $ x_k$ such that $ |f(x_k)| \to |m|$ when $ k \to \infty$ (must exist from the definition of supremum), and thus $ |g(y)| \le \lim_{k \to \infty} \frac {|m|}{|f(x_k)|} \equal{} 1$


x(1)=1  x(2)=4  x(n+2)=4*x(n+1)-x(n) find all natural m such that 3*x(n)*x(n)+m is perfect square(the numbers in braces after x mean indices)

Hi Makaveli,
[quote="Makaveli"]all natural m such that x(n)+(m*m)[/quote]
What is $n$ here ? Is it $x_m+ m \cdot m$ ?

Well, well,

I think solving the recurrence might be a great step on the solution of this problem

$q^2-4q+1=0 \Leftrightarrow (q-2)^2-3=0 \Rightarrow q=2\pm \sqrt{3}$

We know that a sequence that satisfies is:

$x_n=aq_1^n+bq_2^n$

Now its just solving the system

$1=aq_1+bq_2, 4=aq_1^2+bq_2^2$

where $q_1=2+\sqrt{3},q_2=2-\sqrt{3}$

After solving that, you find the general term which satisfies the recurrence...

$x_n=\frac{\sqrt{3}}{6}((2+\sqrt{3})^n-(2-\sqrt{3})^n)$

and after that I'll see if the problem is really written correct  ;)


geometry
3D geometry
sphere
combinatorics proposed
combinatorics

Given five radii $l_1,\ \cdots l_5$ of a sphere $S,$ no three of these radii are on a same plane.
Choose a pair to endpoint from each radius $l_1,\ \cdots l_5.$ Find the number of choices such that five points are in a hemisphere among 32 choices of an endpoint.

I think the answer is [hide]maybe 22?[/hide] (Thinking from the special condition,not the proof)

There's a typo in the statement. Change "radii" to "diameters".


I voted 1-3 months i never *completely* finished it cause it was the schools and school took it back cause of summer but i was a chapter off after doing it for 3 or 4 weeks

Lol, we're twin teletubbies :D

And how long did it take YOU to finish the c and p book?
I thought it was great mathcounts prep :)

I forgot, but I THINK it was somwhere between 3-6 months. or, more like, 1-6 months. I completely forgot, although I never COMPLETELY finished it in the sense that finish=do all the problems.

2 months
that was wayyy shorter than the alg one haha

[quote="isabella2296"][quote="Math Geek"]1 chapter a day= less than a month.  Thus, I cannot vote on your poll. :P[/quote]

Wow...a chapter a day? No way I could absorb all that information in just one day...I did more like 2 or 3 sections a day...plus some days I only did problems...wow...[/quote]

Well, intro books don't have that much info that I couldn't absorb a chapter of it in one day.  It's just the problems that took me some time, and I had a lot of time on my hands last summer. :roll:

I'm working on that book right now. Hopefully I finish by the end of the summer.

im working on it as fast as the class is which is 12 weeks

[quote="RussianRocket"]are you serious?[/quote]

that isn't very surprising becoz the book is just like 250 pages or something like that....

but (just curious) why are you taking the class if you read the whole freakin book?

[quote]

but (just curious) why are you taking the class if you read the whole freakin book? [/quote]

I haven't read the whole book actually but I am working on it [u][i][b]WITH[/b][/i][/u] the class meaning I am taking the class right now.

i think intro to c&p is one of the easiest in the AoPS series

[quote="RussianRocket"][quote="isabella2296"]I disagree. The last chapter was a killer.[/quote]
This is an Introductary book. Can you come up with a book that is easier?[/quote]
Nope, otherwise it would be made for elementary students who don't do Mathcounts :D

im not sure how long it took me but it is hard to work on it during a shcool year and also, next year,
 i heard there is atleast 4 hours of math hw a night for people in my mathclass next year.

!!!YAY A FULL STAR!!!

It's not so hard to work during the school year for me, because my teachers don't give enough homework. This is actually good for me, because everything is pretty easy, so I can spend more time devoted to other things, like AoPS and learning "real" math.

Introduction to C and P was very helpful and it was my first AoPS book. I really enjoyed it and I learned a lot of basic things that I didn't know before, like combinations. It took me about 3 months because I couldn't get the last few chapters.

I got the last few chapters, including Binomial Theorem, Hockey Stick Identity, and Pascal's Triangle!

w00t!!! :D

No; I understood them, it just took some extra time.

Dang i wish i still had the books (they are the school libraries) cause i totally forgot about the whole 2 or 3 chapters on probability but aaaw well its in the mail!!!! [size=200]w00t! w00t![/size]

I am almost finished and I started about a month ago

No not really in terms of understanding i seem to have a lot of problems with Aops Vol 1 excerpts they're going too fast for me  :(  :o  :rotfl:

Ugh I skipped it and went to Interm C&P =(


geometry
perpendicular bisector
geometry proposed

Let$ ABC$ is a triangle. $ (I_B)$ , $ (I_C)$ are the escribed circles. $ (I_B)$ meets $ BC, AC$ at$ M,N$. $ (I_C)$ meet $ BC, AB$ at $ P, Q$. $ I_BI_C$ meet$ PQ, MN$at $ H,K$.
$ D$ is the mid point of$ BC$.
Prove that$ DH\equal{}DK$ :)

[quote="TRAN THAI HUNG"]Let$ ABC$ is a triangle. $ (I_B)$ , $ (I_C)$ are the escribed circles. $ (I_B)$ meets $ BC, AC$ at$ M,N$. $ (I_C)$ meet $ BC, AB$ at $ P, Q$. $ I_BI_C$ meet$ PQ, MN$at $ H,K$.
$ D$ is the mid point of$ BC$.
Prove that$ DH \equal{} DK$ :)[/quote]

Very nice and quite easy problem.

It is well- known that $ CK\perp AI_b$ and $ BH\perp AI_c$ $ \Longrightarrow$ $ K,H$ respectively are the projections of $ C,$ and $ B$ onto $ I_bI_c$.  Let $ (I_a)$ is the center of the excircle of $ \triangle ABC$ wrt $ A$. Then it is followed that $ \triangle ABC$ is the orthic triangle wrt $ \triangle I_aI_bI_c$. Then if we call $ M$ is the midpoint of $ BC$ $ \Longrightarrow$ $ MD$ is the perpendicular bisector of $ BC$. Thus $ \angle MCD \equal{} \angle MBD$. But note that $ CDMK$ and $ MDBH$ are two concyclic quadrilaterals, then $ \angle MCD \equal{} \angle MKD$, $ \angle MBD \equal{} \angle MHD$, which implies $ \angle MHD \equal{} \angle MKD$. As the result, $ \triangle DKH$ is $ D \minus{}$ isoleces triangle $ \Longrightarrow$ $ DK \equal{} DH$. $ \square$

Let $ E$ be the midpoint of $ AC$ and $ H' \equiv DE \cap PQ.$ Since $ \triangle PBQ$ is isosceles and $ DE \parallel AB,$ then $ \triangle PDH'$ is also isosceles $\Longrightarrow$ $ DH' \equal{} DP \equal{} \frac {_1}{^2}(AB \plus{} AC)$ $ \Longrightarrow$ $ EH' \equal{} \frac {_1}{^2}AC.$ On the other hand, denote $ H'' \equiv I_bI_c \cap DE.$ From the isosceles $\triangle EAH''$ we get $ EH'' \equal{} EA \equal{} EH' \equal{} \frac {_1}{^2}AC$ $ \Longrightarrow$ $ H \equiv H' \equiv H''.$ Therefore $ DH$ and $ DK$ are the midlines of $AB$ and $AC$ $ \Longrightarrow$ Lines $ DK,DH$ and $ I_bI_c$ bound an isosceles triangle.


function
geometry
geometric transformation
reflection
complex analysis
complex analysis unsolved

Hello guys. I'm pretty lost on the following problem:

Let S = {(x + iy) in C| x,y are in [0,1]} be the unit square in the complex plane. Let f be analytic on a region R that contains S. Suppose the following is true:

(i) for all z with 0[u]<[/u] Re(z) [u]<[/u] 1, and Im(z)=0 or 1,
                                 Re f(z)=0;
(ii) for all z with Re(z)=0 or 1, and 0[u]<[/u]Im(z)[u]<[/u]1,
                                 Im f(z)=0.

Show that f is identically the zero function on R.

I don't even know where to start :/

It's tricky.

Here's my approach: Extend $ f$ by the reflection principle across each side; we can tile the plane, and we get that $ f(z)\equal{}f(z\plus{}2)\equal{}f(z\plus{}2i)$. The only doubly periodic entire functions are the constants.

This would break down if we allowed meromorphic functions.


[quote="canada"]Let $ n$ be positive integer. Prove that:

$ [\sqrt {n} \plus{} \sqrt {n \plus{} 2\sqrt [3]{n} \plus{} 1}] \equal{} [\sqrt {4n \plus{} 4\sqrt [3]{n} \plus{} 2} ]$,
 where $ [x]$ denotes integer part of $ x$[/quote]

It seems wrong for some $ n$ : 45, 95, .... .


$ x, y, z, w$ are intergers that are not negative.

Find all solutions of $ 2^x3^y \minus{} 5^z7^w \equal{} 1$.

This problem has been posted a few times before. http://www.mathlinks.ro/viewtopic.php?t=24692


Consider polynomial
$ p(x)\equal{}x^3\minus{}2nx\plus{}n$
Prove that $ P(x)$ has three different root. call $ x_1<x_2<x_3$
Prove that $ \{x_2(n)\}$ converge and find out it's limit.

Obviosly $ x_1(n) \equal{} \minus{} \sqrt {2n} \plus{} O(1),x_3(n) \equal{} \sqrt {2n} \plus{} O(1)$, therefore $ x_2(n) \equal{} \frac {1}{2} \plus{} O(n^{ \minus{} 1/2}).$
Exactly $ x_{1,3}\equal{}\mp \sqrt{2n}\minus{}\frac 14 \plus{}O(n^{\minus{}1/2}).$


Strange one  :? I think there's a flaw in a statement somewhere

we can get $ab^n(b^c-1)\equiv -c^2 \mod m$ for every $n$ so $b \equiv 0$ or $b \equiv 1 \mod m$. In both cases $m |c^2$

[quote="Melissa"]Prove that if $m\vert ab^n+cn+d$ for every integer $n\ge 0$ then $m\vert c^2$.[/quote]
Hi Melissa ,about this problem I can say that it is interesting :)  but not difficult :( :D
$ab^n+cn+d\equiv(ab^0+c*0+d)(\mod{m}) \Rightarrow  a(b^n-1)\equiv(-cn)(\mod{m})$for every $n\geq 0$=>$a(b^{2n}-1)\equiv(-2cn)(\mod{m})$but
$a(b^{2n}-1)\equiv(b^n+1)(-cn)(\mod{m})$ =>
$(b^n+1)(-cn)\equiv(-2cn) (\mod{m})$ =>
$(cn)(b^n-1)\equiv{0}(\mod{m})$
But we also know that $a(b^n-1)\equiv(-cn)(\mod{m})$=>$a(cn)(b^n-1)\equiv-(cn)^2(\mod{m}) \Rightarrow m|(cn)^2$ for every $n$ => $m|c^2$ ;)


Let f be a function from the positive integers into the positive integers and satisfying
f(n + 1) > f(n) and f(f(n)) = 3n for all n. Find f(100).

Here's a dumb, mechanical way to do it.

[hide="Solution."]
=====
We have $ f: \mathbb{N}\to \mathbb{N}$, $ f(n \plus{} 1) > f(n)$, $ f(f(n)) \equal{} 3n$ $ \forall n\in \mathbb{N}$. We claim that $ f(100) \equal{} 181$.

[b]LEMMA ONE:[/b] If $ f(n \plus{} a) \equal{} f(n) \plus{} a$ for some $ n,a\in \mathbb{N}$, then $ f(n \plus{} k) \equal{} f(n) \plus{} k$ $ \forall k\in [0,a]$.
[i]Proof:[/i] This is a direct consequence of $ f(n \plus{} 1) > f(n)$. If we suppose the contrary, then either $ f(n \plus{} k) < f(n) \plus{} k$ or $ f(n \plus{} k) > f(n) \plus{} k$. In the former case, we have less than $ f(n) \plus{} k \minus{} f(n) \plus{} 1 \equal{} k \plus{} 1$ consecutive numbers mapped to by the $ (n \plus{} k) \minus{} n \plus{} 1 \equal{} k \plus{} 1$ numbers $ \{n,n \plus{} 1,\dots, n \plus{} k\}$, a contradiction. We use similar logic in the latter case. $ \square$

Now we show that $ f(1) \equal{} 2$. Suppose, for sake of contradiction, that $ f(1)\neq 2$. If $ f(1) < 2$, then $ f(1) \equal{} 1$ and $ f(f(1)) \equal{} f(1) \equal{} 1$, which contradicts $ f(f(1)) \equal{} 3$. If $ f(1) \equal{} c > 2$, then $ f(c) \equal{} f(f(1)) \equal{} 3$. But $ c > 1\implies f(c) \equal{} 3 > f(1)$, which contradicts $ f(1) > 2$. Thus $ f(1) \equal{} 2$.

Now we repeatedly apply $ f(f(n)) \equal{} 3n$ on $ f(1) \equal{} 2$. We have: $ f(f(1)) \equal{} f(2) \equal{} 3$, $ f(f(2)) \equal{} f(3) \equal{} 6$, $ f(f(3)) \equal{} f(6) \equal{} 9$, $ f(f(6)) \equal{} f(9) \equal{} 18$, $ f(f(9)) \equal{} f(18) \equal{} 27$, $ f(f(18)) \equal{} f(27) \equal{} 54$, $ f(f(27)) \equal{} f(54) \equal{} 81$, and $ f(f(54)) \equal{} f(81) \equal{} 162$.

We also have $ f(3 \plus{} 3) \equal{} f(6) \equal{} 9 \equal{} 6 \plus{} 3 \equal{} f(3) \plus{} 3$, so by LEMMA ONE, $ f(3 \plus{} k) \equal{} f(3) \plus{} k$ for $ k\in [0,3]$. In particular, $ f(5) \equal{} f(3 \plus{} 2) \equal{} f(3) \plus{} 2 \equal{} 6 \plus{} 2 \equal{} 8$. Now we repeatedly apply $ f(f(n)) \equal{} 3n$ on $ f(5) \equal{} 8$. We have: $ f(f(5)) \equal{} f(8) \equal{} 15$, $ f(f(8)) \equal{} f(15) \equal{} 24$, $ f(f(15)) \equal{} f(24) \equal{} 45$, $ f(f(24)) \equal{} f(45) \equal{} 72$, $ f(f(45)) \equal{} f(72) \equal{} 135$, and $ f(f(72)) \equal{} f(135) \equal{} 216$.

Finally, we see that $ f(81 \plus{} 54) \equal{} f(135) \equal{} 216 \equal{} 162 \plus{} 54 \equal{} f(81) \plus{} 54$, so by LEMMA ONE, $ f(81 \plus{} k) \equal{} f(81) \plus{} k$ for $ k\in [0,54]$. In particular, $ f(100) \equal{} f(81 \plus{} 19) \equal{} f(81) \plus{} 19 \equal{} 162 \plus{} 19 \equal{} \boxed{181}$, as desired. $ \blacksquare$
=====
[/hide]

[quote="1=2"][hide="solution"]If we have $ f(1)\geq 3$ then we have $ f(3) \equal{} f(f(1)) > f(1) \equal{} 3$, so we must have either $ f(1) \equal{} 1$ or $ f(2) \equal{} 2$. We can't have $ f(1) \equal{} 1$ because $ f(1(1)) \equal{} f(1) \equal{} 1\neq 3$. Hence $ f(1) \equal{} 2$. We have $ f(f(1)) \equal{} f(2) \equal{} 3$, so $ f(2) \equal{} 3$. Note that $ f(f(2)) \equal{} f(3) \equal{} 6$, and $ f(6) \equal{} f(f(3)) \equal{} 9$. Since $ f(n \plus{} 1) > f(n)$, we have $ f(4) \equal{} 7$ and $ f(5) \equal{} 8$. From this information, we can find $ f(7)$, $ f(8)$, and $ f(9)$:

$ f(1) \equal{} 2$
$ f(2) \equal{} 3$
$ f(3) \equal{} 6$
$ f(4) \equal{} 7$
$ f(5) \equal{} 8$
$ f(6) \equal{} 9$
$ f(7) \equal{} f(f(4)) \equal{} 12$
$ f(8) \equal{} f(f(5)) \equal{} 15$
$ f(9) \equal{} f(f(6)) \equal{} 18$
-----------------------------------------------------------------
Lemma: $ f(3^n) \equal{} 2*3^n$ for all non-negative integers $ n$.
Proof: It is true for the base case $ n \equal{} 0$.
If $ f(3^k) \equal{} 2*3^k$, then we have $ f(f(3^k)) \equal{} f(2*3^k) \equal{} 3^{k \plus{} 1}$, and $ f(f(2*3^k)) \equal{} f(3^{k \plus{} 1}) \equal{} 2*3^{k \plus{} 1}$. Hence by induction $ f(3^n) \equal{} 2*3^n$ for all non-negative integers $ n$.
-----------------------------------------------------------------

Since $ f(3^k) \equal{} 2*3^k$ and $ f(2*3^k) \equal{} 3^{k \plus{} 1}$ and $ 2*3^k \minus{} 3^k \equal{} 3^{k \plus{} 1} \minus{} 2*3^k \equal{} 3^k$, we have $ f(3^k \plus{} e) \equal{} 2*3^k \plus{} e$ for all $ 0 < e\leq 3^k$. Note that when $ e > 3^k$ we have $ f(3^k \plus{} e) \equal{} f(2*3^k \plus{} f) \equal{} f(f(3^k \plus{} f)) \equal{} 3^{k \plus{} 1} \plus{} 3f$ for some $ 0 < f\leq 3^k$. Note that $ 3^4 < 100 < 2*3^4$. Hence $ f(3^4 \plus{} 19) \equal{} 2*3^4 \plus{} 19 \equal{} \boxed{181}$.[/hide]

This is one of the fanciest-looking proofs I've made EVER.[/quote]

yeah, and all thanks to my problem.

good job guys.


geometry
parallelogram
inequalities
triangle inequality

Hello,

If we have a triangle and it has a median from each vertex how it can be proved that the three medians added together in smaller than the three triangle sides.  :(

Thanking you

Let the triangle be $ \triangle ABC$ with median AM, extend AM past M to D so that ABCD is a parallelogram, then by the triangle inequality we have AB+BD>AD=2AM, or b+c>2AM. Adding this cyclically and dividing by 2 gives the reverse of what you said. In other words, the sum of the sides is greater than the sum of the medians.

My point was like is there a range that medians in triangles can't get more than it or less than it in the term of sides :?:


modular arithmetic
number theory
prime numbers
number theory proposed

How to solve in elementary way(without using Dirichlet's theorem) that there is infinetely many prime numbers of the form $ 3k \plus{} 1$.

Suppose there are finitely many $ p_1, ... p_k$ and let $ P \equal{} 6p_1 p_2 ... p_k$.  What are the possible prime divisors of $ P^2 \plus{} P \plus{} 1$?

hello, here you can read something about this problem
http://www.physicsforums.com/showthread.php?t=60863
Sonnhard.

I will answer t0rajir0u's proposed question. Suppose q is a prime dividing $ P^2 \plus{} P \plus{} 1$. Then $ P^2 \plus{} P \plus{} 1 \equiv 0 \pmod q$. Multiplying both sides by $ P \minus{} 1$, we get $ P^3 \equiv 1 \pmod q$. So 3 is the order of P mod q. But by Fermat's little theorem, $ P^{q \minus{} 1} \equiv 1 \pmod q$, so $ 3|q \minus{} 1$, or $ q \equal{} 3r \plus{} 1$.


geometry
3D geometry
pigeonhole principle
geometry proposed

In the unit cube, given 75 points, no three of which are collinear. Prove that there exits a triangle whose vertices are among the given points and whose area is not greater than 7/72.

Lemma. If a triangle $ MNP$ is contained in a parallelepipe $ ABCDA'B'C'D'$ whose size is $ a\times b\times c$, the area of the triangle is at most $ \frac12\sqrt{a^2b^2\plus{}b^2c^2\plus{}c^2a^2}$

Proof of the lemma: Let $ S$ be the area of $ MNP$ and $ S_1,S_2,S_3$ be the areas of its projections to the sides of the parallelepipe. It is clear that $ S_1\le\frac12ab,S_2\le\frac12bc,S_3\le\frac12ca$. So using $ S^2\equal{}S_1^2\plus{}S_2^2\plus{}S_3^2$ the result follows.

Now divide the unit cube into 36 parallelepipe sized $ \frac16\times\frac13\times\frac12$. There are 75 points, and by pigeonhole principle there exist three points that is contained in the same parallelepipe. Use the lemma above, we are done.

Could you please clarify why $S^2=S_1^2+S_2^2+S_3^2$ is true?

This will help you.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=393609[/url]

I see, but we still have to prove that $MNP$ can always be adjusted so that its sides are parallel to the sides of the box, because that's what de Gua's theorem requires, right?

It shouldn't be dramatically hard to prove but in my opinion it should be included in the proof.

Or it might be that I missed something...  :roll:

it is better to divide the unit cube into 27 little cubes sized $ \frac13\times\frac13\times\frac13$
because
$S_i \le \frac{1}{2 \times 3^2}$ for $i=1,2,3$
and $S_i^2 \le \frac{1}{2^2 \times 3^{4}}$ for $i=1,2,3$
hence $S^2 \le \frac{3}{2^2 \times 3^{4}}$
or $S \le \frac{\sqrt 3}{2 \times 3^2}$
we must prove that:
$\frac{\sqrt 3}{2 \times 3^2} \le \frac{7}{72}$ $\iff$
$4\sqrt 3 \le  7$ $\iff$ $48<49$


There is an alley. In this alley there is a 20ft ladder and a 30ft ladder. These ladders cross over the alley from the ground to the opposite building. They cross each other 10ft above the ground. How wide is the alley?

Here's a simple drawing to represent this problem:

Did you make up that problem or did you find it in a well-known competition? I found the answer to be $ 30$ and since that is also the hypotenuse, the question is incorrect.

More likely, it means you screwed up somewhere :wink:

There is a very simple solution to this problem.

I dont know the answer. My friend asked me. tell me how the problem is solved?

Every way I tackled it, I ended up with a general quartic equation, so I highly doubt there's some elegant way to do it.

I continually got a quartic too...
And so did [url=http://mathforum.org/library/drmath/view/55279.html]everyone else[/url]


Help me with this probability quesition. Looks so simple,, but i can't get the answer!!.

A biased die, marked is such that: Pr(6)=Pr(1)=Pr(2),
Pr(6')=5*Pr(5)+Pr(1),     Pr(3)=Pr(4)+Pr(1)

What is the probabilityof each number?

Sorry, if the question was a bit confusing,,,

Pr(6') is not a typo(firstly,,),,
Pr(6')=1-Pr(6)=Pr(1)+Pr(2)+Pr(3)+Pr(4)+Pr(5)=5*Pr(5)+Pr(1)

What does "Pr(6[b]'[/b])" mean? Is that apostrophe a typo? Or it has some special meaning?

Pr(6) most likely means the probability of rolling a 6, and the 6' is probably a typo. It definitely doesn't mean 6 though, because then the problem has infinite solutions, so yeah, we need to wait for him to fix it.

[quote="wonki_noh"]Help me with this probability quesition. Looks so simple,, but i can't get the answer!!.

A biased die, marked is such that: Pr(6)=Pr(1)=Pr(2),
Pr(6')=5*Pr(5)+Pr(1),     Pr(3)=Pr(4)+Pr(1)

What is the probabilityof each number?[/quote]

If Pr(6)=Pr(1), then how is Pr(6)=5*Pr(5)+Pr(1) possible?

[quote="neelnal"]
If Pr(6)=Pr(1), then how is Pr(6)=5*Pr(5)+Pr(1) possible?[/quote]

[hide]
if Pr(5) = 0
[/hide]

[hide="full solution"]
Since Pr(5) = 0.

We can conlcude that P(1)=P(2)=P(6)=1/6 or 0 because if it were 2/6 or more then P(3)$\geq$2/6, but then the probabilities add up to more than one, contradiction.

If they are all zero, then P(3)=P(4)=1/2 and all the other probabilities are zero.

If they are all 1/6, then P(4) = 1/6, P(3)=2/6

So, there are two possible solutions:
0,0,1/2,1/2,0,0
or
1/6,1/6,2/6,1/6,0,1/6
[/hide]

I have fixed the problem~,,

can anyone now please solve this??


Given the numbers $ a_1, a_2, ..., a_n $ such that $ 0 \leq a_1 \leq a_2 \leq 2a_1 $, $ a_2 \leq a_3 \leq 2a_2 $, ..., $ a_{n-1} \leq a_n \leq 2a_{n-1} $.

Prove that in the sum $ s = \pm a_1 \pm a_2 \pm ... \pm a_n $ we can choose the signs appropriately to make $ 0 \leq s \leq a_1 $

By induction.
It is true for $n=2$: $0\leq a_2-a_1\leq a_1$.
Suppose it is true for $n$. Let's prove it for $n+1$. Indeed, apply it for numbers $a_2,...,a_{n+1}$. We will get some sum $0\leq S=\pm a_2\pm a_3\pm...\pma_{n+1}\leq a_2$. There are two cases:
Case 1. $S\geq a_1$. Take $S-a_1$.
Case 2. $S<a_1$. Take $a_1-S$.

yeah.  it's a fairly easy problem.  pretty though.


[hide="outline"]
Look at the denominators, express their form in function of $ n$ for the $ n$-th term.
[hide]
Use the formula for $ 1^2+2^2+...+n^2$.
[hide]
Rewrite as $ \sum{of something}$.

Almost done.

Remember the formula for $ 1+2+...+n$

[/hide]
[/hide]
[/hide]

[quote="ElChapin"][hide="outline"]
Look at the denominators, express their form in function of $ n$ for the $ n$-th term.
[hide]
Use the formula for $ 1^2 + 2^2 + ... + n^2$.
[hide]
Rewrite as $ \sum{of something}$.

Almost done.

Remember the formula for $ 1 + 2 + ... + n$

[/hide]
[/hide]
[/hide][/quote]

What's the formula for $ 1^2 + 2^2 + ... + n^2$?

$ 1^2\plus{}2^2\plus{}\cdots\plus{}n^2\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}6$.

Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1428326638&t=240562]this.[/url]

[hide] first we have some fraction multiplied by the sum of squares up until n
but upon closer notice, the fraction is actually 1/(n)(n+1)
so we can put this up into a neat sum

it is the sum of all numbers from one to sixty (sixty being the last square)      [1/(n)(n+1)] * [(n) (2n+1)(n+1)/6]

so we can simplify

and we get (2n+1)/6  from n=1 to n=60

and using my calculator i get

620[/hide]

sorry about my lack of latex, i'm working on it.


Here is an interesting problem:

For a binary sequence X of length n ( with terms x_1, x_1, ..., x_n ), define B(X) to be the sequence with terms b_1, b_2, ..., b_(n+1) such that:
b_i = Sum(k=i to n, (-1)^(1+x_k) )    (that is, add 1 for every 1 in position >= i and -1 for every 0)  for i<=n, and
b_(n+1) = 0.

Let p(X) be min(b_1, b_2, ..., b_(n+1) ).

Let S be the set of all binary sequences of length n for which p(X) = 0, and let T be the set of all binary sequences of length n for which p(X) = b_1.

Given n, find, with proof,
a)  #(S).
b)  #(T).
c)  #(S intersect T).

(the notation #(A) means the number of elements of the set A)

Doesn't anyone want to try this problem?  At least give it a shot! ;)

[color=cyan]Well, I wish I remembered the nice technique for this.

Another way to look at this problem, which I prefer, is to look at paths, beginning at (0,0).  From a point (a,b) you may either move to (a+1, b+1) or (a+1,b-1).

Part (a) is the number of paths which do not ever reach a point with negative y coordinate but do touch the axis at least once.

Part (b) is the number of paths for which the final point is the lowest point reached.

Part (c) is the number satisfying both, which should be 0 if n is odd and greater than 1, right?[/color]

Joel, you are correct in all your statements (although there is no need to stipulate in part (a) that the paths touch the axis at least once as you have already noted that the paths start at (0,0)).

Interestingly, this isn't all you can say about these sets, even without finding a formula for each of them.  [hide]For instance, #(S) = #(T) - can you find a simple bijection?  Using this, can you now easily compute #(S Intersect T)? [/hide] After someone posts a reasonable amount of solution I will post about an interesting connection these sets have with another problem.


calculus
integration
real analysis
real analysis unsolved

Consider $f(x)=\sum_{n\geq 2}{\frac{lnn\cdot sin(nx)}{n}}$.
   a) Find an equivalent of $f$ near 0.
   b) Is the integral of $f$ over $[0,\pi]$ absolutely convergent? Convergent? The same questions for $f^2$.

I've not checked the computation but i think that for 1/ one can do as follows:
$\sum_{n \geq 1} \frac{ln(n)sin(nx)}{n}=x.\sum_{n \geq 1} \frac{[ln(nx)-ln(x)]sin(nx)}{nx}$
$x \sum_{n \geq 1} \frac{ln(nx)sin(nx)}{nx} - ln(x) \sum_{n \geq 1} \frac{sin(nx)}{n}$

but it's here I've not carefully check the computations but it seems that there exists a constant $C$ such that:
$| (x \sum_{n \geq 1} \frac{ln(nx)sin(nx)}{nx}) - \int_0^{+\infty} \frac{ln(t)sin(t)}{t} | < C$ for $0<x<1$.
(and I'm convinced that $x \sum_{n \geq 1} \frac{ln(nx)sin(nx)}{nx})$ tends towards $\int_0^{+\infty} \frac{ln(t)sin(t)}{t}$)

If I'm right, it remains to study $\sum_{n \geq 1} \frac{sin(nx)}{n}$, wich is classical.

[quote="alekk"]
it remains to study $\sum_{n \geq 1} \frac{sin(nx)}{n}$, wich is classical.[/quote]

$\frac{\pi-x}{2}$ for $0<x<2\pi$

yes Moubi  :)
now, could you prove that the sum converges towards the integral ?

I think one can do a classical $3 \epsilon$ to prove that the sum converges towards the integral:
Let
$U(x)=x \sum_{1}^{\infty} \frac{ln(nx)sin(nx)}{nx}$
$V(x,p,q)=x \sum_{p}^{q} \frac{ln(nx)sin(nx)}{nx}$

1/
let $S_n=\sum_{1}^{n} sin(nx)$. One can check that $|S_n| < \frac{2}{x}$
Moreover,
$R_{a,x}=x \sum_{a}^{\infty} \frac{ln(nx)sin(nx)}{nx} = -xS_{a-1}\frac{ln(ax)}{ax}+x\sum_{a}^{\infty} S_n(\frac{ln(nx)}{nx}-\frac{ln((n+1)x)}{(n+1)x})$
$|R_{a,x}|<4|\frac{ln(ax)}{ax}|$

2/
let's choose one $\epsilon > 0$ and $Y>100$ such that $4\frac{ln(Y-1)}{Y-1} < \epsilon$ and $| \int_Y^{\infty}\frac{ln(x)sin(x)dx}{x}| < \epsilon$.
Then:
$V(x,1,[Y/x]-1)=x \sum_{1}^{[Y/x]-1} \frac{ln(nx)sin(nx)}{nx}$ converges towards $\int_{0}^{Y} \frac{ln(x)sin(x) dx}{x}$.
(Rieman sum)
And thanx to 1/ we know that:
$|V(x,[Y/x], \infty)|=|x \sum_{[Y/x]}^{\infty} \frac{ln(nx)sin(nx)}{nx}| < 4 |\frac{ln([Y/x]x)}{[Y/x]x}| < \epsilon$

Finally:
$|U(x)-\int_0^{\infty}\frac{ln(x)sin(x)dx}{x}|=..$
$..=|V(x,1,[Y/x]-1)+V(x,[Y/x], \infty)- \int_0^{Y}\frac{ln(x)sin(x)dx}{x} - \int_Y^{\infty}\frac{ln(x)sin(x)dx}{x}|<$

$<|V(x,1,[Y/x]-1) - \int_{0}^{Y} \frac{ln(x)sin(x) dx}{x}| + |V(x,[Y/x], \infty)|+|\int_Y^{\infty}\frac{ln(x)sin(x)dx}{x}|$.

This is why:
$|U(x)-\int_0^{\infty}\frac{ln(x)sin(x)dx}{x}|<|V(x,1,[Y/x]-1) - \int_{0}^{Y} \frac{ln(x)sin(x) dx}{x}|+2 \epsilon$

Hence for $x$ small enough, $|U(x)-\int_0^{\infty}\frac{ln(x)sin(x)dx}{x}|<3 \epsilon$.

[quote="alekk"]Then:
$V(x,1,[Y/x]-1)=x \sum_{1}^{[Y/x]-1} \frac{ln(nx)sin(nx)}{nx}$ converges towards $\int_{0}^{Y} \frac{ln(x)sin(x) dx}{x}$.
(Rieman sum)
[/quote]
I wonder that  if $\int_{0}^{Y} \frac{ln(x)sin(x) dx}{x}$ is finite? And therefore if $\int_0^{+infty}\frac{ln(t)sin(t)}{t}dt$ is convergent (finite)

near $0$ it is equivalent to $ln(x)$, so it's OK.
And $\int_1^{\infty} \frac{ln(x)sin(x)dx}{x}$ exists. (integrating by part .. )


I assume  $a,b,c,x,y,z$ are all positive. Then this directly follows from the generalized Holder.

Let $u=a^{(x+y+z)}, v=b^{(x+y+z)}, w=c^{(x+y+z)}$ and let

$\displaystyle p = \frac{x}{x+y+z}, q = \frac{y}{x+y+z}, r = \frac{z}{x+y+z}$, then $p+q+r=1$.

The inequality is equivalent to:

$u+v+w \geq u^p v^q w^r + w^p  u^q v^r + v^p w^q u^r$. ---------- (1)

By the generalized Holder, we have

RHS of (1) $\leq (u+w+v)^p (v+u+w)^q (w+v+u)^r = (u+v+w)^{p+q+r} = u+v+w$.

Q.E.D.


integration
calculus
calculus computations

Exactly. What I'm wondering is what I did wrong.

You didnt solved $f(x)=4$ where $f(x)=\int^{x}_{0}t^{3}\, dt$

but $f'(x)=0$ .

Your initial equation is only true for one (positive) value of $x$. If we differentiate something that's not generally true, it's not surprising we get nonsense.

[quote="1337h4x"]Solve for x:

$\int^{x}_{0}t^{3}\, dt=4$
[/quote]
when we actually integrate we get

$\left[\frac{t^{4}}{4}\right]_{t=0}^{x}=4$
or just
$\frac{x^{4}}{4}=4\Rightarrow x=2$

we choose 2 and not $-2$ since if $x=-2$ the integral would be negative.


wondering which forum i could post questions about dynamical systems in

I'd put them in "Advanced Fields" here within the College Playground.


[b][color=cyan]Here are problems from the 4-th Kyiv mathematical festival (May, 2005)
Any comments are welcome![/color][/b]

[b][size=200][color=orange]The fourth Kyiv mathematical festival[/color][/size][/b]

[b][size=150][color=indigo]8-th form[/color][/size][/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=199137]1.[/url] On Monday a school library was attended by 5 students, on Tuesday, by 6, on Wednesday, by 4, on Thursday, by 8, and on Friday, by 7. None of them have attended the library two days running. What is the least possible number of students who visited the library during a week?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=199138]2.[/url] Find the rightmost nonzero digit of $ \frac {100!}{5^{20}}$ (here $ n! \equal{} 1\cdot2\cdot3\cdot\ldots\cdot n$).

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=199132]3.[/url] $ \begin{picture}(40,50) \put(2,2){\put(0,0){\line(6,0){42}} \put(7,14){\line(6,0){28}} \put(14,28){\line(6,0){14}} \put(0,0){\line(1,2){21}} \put(14,0){\line(1,2){14}} \put(28,0){\line(1,2){7}} \put(14,28){\line(1,2){7}} \put(14,0){\line( \minus{} 1,2){7}} \put(28,0){\line( \minus{} 1,2){14}} \put(42,0){\line( \minus{} 1,2){21}} \put(0,0){\circle*{3}} \put(14,0){\circle*{3}} \put(28,0){\circle*{3}} \put(42,0){\circle*{3}} \put(7,14){\circle*{3}} \put(21,14){\circle*{3}} \put(35,14){\circle*{3}} \put(14,28){\circle*{3}} \put(28,28){\circle*{3}} \put(21,42){\circle*{3}}} \end{picture}$
Two players by turn paint the vertices of triangles on the given picture each with his colour. At the end, each of small triangles is painted by the colour of the majority of its vertices. The winner is one who gets at least 6 triangles of his colour. If both players get at most 5, then it is a draw. Does any of them have winning strategy? If yes, then who wins?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76163]4.[/url] Let $ M$ be the intersection point of medians of a triangle $ \triangle ABC.$ It is known that $ AC \equal{} 2BC$ and $ \angle ACM \equal{} \angle CBM.$ Find $ \angle ACB.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=199139]5.[/url] The first row of a table of size $ 2005\times5$ is filled with 1,2,3,4,5 so that every two neighbouring cells contain distinct numbers. Prove that it is possible to fill four other rows with 1,2,3,4,5 so that any neighbouring cells in them will contain distinct numbers as well as any cells of the same column will contain pairwise distinct numbers.

[b][size=150][color=indigo]9-th form[/color][/size][/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=199140]1.[/url] Prove that there exists a positive integer $ n$ such that for every $ x\ge0$ the inequality $ (x \minus{} 1)(x^{2005} \minus{} 2005x^{n \plus{} 1} \plus{} 2005x^n \minus{} 1)\ge0$ holds.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76164]2.[/url] The quadrangle $ ABCD$ is cyclic. Points $ E$ and $ F$  are chosen at the diagonals $ AC$ and $ BD$ in such a way that $ AF\bot CD$ and $ DE\bot AB.$ Prove that $ EF\|BC.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=199135]3.[/url] $ \begin{picture}(60,60) \put(5,3){\put(3,0){\line(6,0){8}} \put(17,0){\line(6,0){8}} \put(31,0){\line(6,0){8}} \put(45,0){\line(6,0){8}} \put(10,14){\line(6,0){8}} \put(24,14){\line(6,0){8}} \put(38,14){\line(6,0){8}} \put(17,28){\line(6,0){8}} \put(31,28){\line(6,0){8}} \put(24,42){\line(6,0){8}} \put(1,2){\line(1,2){5}} \put(15,2){\line(1,2){5}} \put(29,2){\line(1,2){5}} \put(43,2){\line(1,2){5}} \put(8,16){\line(1,2){5}} \put(22,16){\line(1,2){5}} \put(36,16){\line(1,2){5}} \put(15,30){\line(1,2){5}} \put(29,30){\line(1,2){5}} \put(22,44){\line(1,2){5}} \put(13,2){\line( \minus{} 1,2){5}} \put(27,2){\line( \minus{} 1,2){5}} \put(41,2){\line( \minus{} 1,2){5}} \put(55,2){\line( \minus{} 1,2){5}} \put(20,16){\line( \minus{} 1,2){5}} \put(34,16){\line( \minus{} 1,2){5}} \put(48,16){\line( \minus{} 1,2){5}} \put(27,30){\line( \minus{} 1,2){5}} \put(41,30){\line( \minus{} 1,2){5}} \put(34,44){\line( \minus{} 1,2){5}} \put(0,0){\circle{6}} \put(14,0){\circle{6}} \put(28,0){\circle{6}} \put(42,0){\circle{6}} \put(56,0){\circle{6}} \put(7,14){\circle{6}} \put(21,14){\circle{6}} \put(35,14){\circle{6}} \put(49,14){\circle{6}} \put(14,28){\circle{6}} \put(28,28){\circle{6}} \put(42,28){\circle{6}} \put(21,42){\circle{6}} \put(35,42){\circle{6}} \put(28,56){\circle{6}}} \end{picture}$
Two players by turn paint the circles on the given picture each with his colour. At the end, the rest of the area of each of small triangles is painted by the colour of the majority of vertices of this
triangle. The winner is one who gets larger area of his colour (the area of circles is taken into account). Does any of them have winning strategy? If yes, then who wins?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76166]4.[/url] Prove that there exist infinitely many collections of positive integers $ (a,b,c,d,e,f)$ such that $ a < b < c$ and the equalities $ ab \minus{} c \equal{} de,\;\; bc \minus{} a \equal{} ef\;\; \text{and}\;\; ac \minus{} b \equal{} df$ hold.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=199136]5.[/url] The plane is dissected by broken lines into some regions. It is possible to paint the map formed by these regions in three colours so that any neighbouring regions will have different colours. Call by knots the points which belong to at least two segments of broken lines. One of the segments connecting two knots is erased and replaced by arbitrary broken line connecting the same knots. Prove that it is possible to paint new map in three colours so that any neighbouring regions will have different colours.

[hide="[color=cyan"]authors[/color]]
The problems are [supposed to be] by:
V.Brayman (8.1, 8.3~9.3, 8.4, 9.2, 9.4), A.Dorogovtsev (8.5, 9.5), T.Malovichko (9.1), A.Pilipenko (8.2)[/hide]

To moderators:
The problems [url=http://www.mathlinks.ro/resources.php?c=180&cid=108&year=2005]here[/url] should be for Grade 8 and Grade 9 (not for Grade 9 and Grade 10). Please fix it


Prove that at least one of the following inequalities is true for positive reals $x_1, x_2, x_3,...,x_n$:

$\sum_{i=1}^{n}{{x_i \over x_{i+1}+x_{i+2}}}\geq {n \over 2}$
$\sum_{i=1}^{n}{x_i \over x_{i-1}+x_{i-2}}\geq {n \over 2}$

Ofcourse $x_{n+i}=x_i$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=46795 or http://www.mathlinks.ro/Forum/viewtopic.php?t=65232 .

  darij


Let $ S$ be the set of natural numbers $ n$ satisfying the following conditions:

$ (i)$ $ n$ has $ 1000$ digits,
$ (ii)$ all the digits of $ n$ are odd, and
$ (iii)$ any two adjacent digits of $ n$ differ by $ 2$.

Determine the number of elements of $ S$.

[hide="A straightforward solution by recurrences"]
We are essentially finding the number of words of length 1000 that use the alphabet $ (1,3,5,7,9)$, with the constraint that adjacent letters have to be consecutive in the alphabet.

Note that the alphabet is symmetric, that is, there is not much difference between 1 and 9, or between 3 and 7.

Define the following four sequences:
$ a_n$ is the number of $ n$-character words ending with either $ 1$ or $ 9$
$ b_n$ is the number of $ n$-character words ending with either $ 3$ or $ 7$
$ c_n$ is the number of $ n$-character words ending with $ 5$.
$ S_n = a_n + b_n + c_n$ is the total number of $ n$-character words.

It is pretty clear that the following recurrences hold for $ n \ge 2$:
\begin{align*}
a_n &= b_{n-1} \\
b_n &= a_{n-1} + 2c_{n-1} \\
c_n &= b_{n-1}
\end{align*}
Because $ a_n = b_{n-1} = c_n$, we know $ a_n=c_n$ for all $ n\ge2$.
This allows us to dramatically simplify our recurrences to, for $ n\ge3$:
\begin{align*}
a_n &= b_{n-1} \\
b_n &= 3a_{n-1}
\end{align*}
This can be further simplified to $ b_n = 3b_{n-2}$ for $ n\ge4$.

Since $ S_n = a_n + b_n + c_n = 2a_n + b_n$ for $ n\ge4$, we know that $ S_n = 3S_{n-2}$.

Simple casework gets us $ S_2 = 8$ (13,31,35,53,57,75,79,97).

Therefore, we can conclude that $ S_{1000} = \boxed{8\cdot3^{499}}$

[/hide]

My solution is the same as Anand.oza did.
But I want to see if there exists a nice bijective solution.  :D
Any idea?


linear algebra
matrix
superior algebra
superior algebra solved

Let $n\geq 2$
 $A\in Mn(R)-(0)$ not invertible

Prove that there exist $B\in Mn(R)-(0)$ not invertible
such that $AB\neq 0$, $AB\neq I_n$,  $(2AB-I_n)^{-1} = 2AB-I_n$

I'll prove this under the assumption that $A$ has at least one nonzero entry in the first row. The proof is essentially the same for a nonzero entry in some other row, but the notation is a bit more awkward.

The condition $(2AB-I)^{-1} = 2AB-I$ can be rewritten as $(2AB-I)^2 = I$, which can be expanded and rearranged to $ABAB = AB$. That is, we want $AB$ to be a nontrivial idempotent.

Do one row worth of column reduction on $A$, producing a matrix with a nonzero entry in the upper left corner and a row of zeros accross the rest of the top. That is, find an invertible matrix $P$ such that, in block-structured form,
$\displaystyle AP = \left(\begin{array}{cc}r&0\\**&*\end{array}\right)$, where $r\ne 0$. Now let $\displaystyle Q = \left(\begin{array}{cc}\frac1r&0\\0&0\end{array}\right)$. Then,
$\displaystyle APQ = \left(\begin{array}{cc}1&0\\u&0\end{array}\right)$, where $u$ is some column vector. By calculation, $(APQ)^2 = APQ$.
Thus, we can let $B = PQ$ to satisfy the problem. Since $B$ has rank 1, it is, as required, nonzero and noninvertible.

The problem was stated for real matrices. We could replace the real numbers by any other field, as long as the characteristic of the field is not 2. Also, it doesn't matter that $A$ is or isn't invertible. All that matters is that $A\ne 0$.


Hello everyone,

    The following two problems are from a different math teacher at our school and I am confused as to how the diagrams should look like. I have posted my intepretations.

Please only provide help to the diagrams; not to the final answer.

Thanks.

---

1. Three circles with radii 40 cm, 50 cm, and 60 cm are tangent to teach other without overlapping at any point. Determine the angles of the triangle formed when the centers are joined.

My interpretation: http://img526.imageshack.us/img526/8642/problem2rb3.jpg

2. A pendulum is 100 cm long. From one end of the swing to the other, the straight line distance is 20 cm. Find the angle through which the pendulum swings.

http://img265.imageshack.us/my.php?image=problem1yk7.jpg

Well, they want a triangle, and your interpretation has the circles collinear.
I think they want you to set up the circles so that the centers form a triangle.

As to the solution, think about the perpendicular lines formed by the tangency points, and think about using Power of a Point.

For the pendulum one, it needs to look a little different. From what I can see, you haven't drawn the other half of the swing.

What I think they want is a diagram that has a circular arc of N degrees, where the radius is 100 cm, and a chord is drawn that is of length 20.

Try solving that one.

1.
Actually, it means that each of the circles is tangent to both of the other circles (so their centers form a triangle.)

2.
Technically, it's correct, but it says
From one [b]end[/b] of the swing to the other.
So it forms an isosceles triangle with sides 100, 100, and 20.

Thank you very much for your assistance!


trigonometry
Trigonometric inequality
Inequality
algebra
IMO Shortlist
IMO Longlist

Find all $x$ for which, for all $n,$ \[\sum^n_{k=1} \sin {k x} \leq \frac{\sqrt{3}}{2}.\]

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]  :)


probability
quadratics
analytic geometry
conics
parabola
algebra

1. Michael is tossing Cheerios into the air and catching them in his mouth. Everytime that he throws the Cheerio, it stays in the air for a couple of seconds. Each second that the Cheerio stays in the air after Michael tossed it, the probability of him catching it is reduced by $\frac{1}{2}$ (starting with $1$, assuming he just pops it into his mouth). What is the probability that Michael catches the Cheerio $15 seconds$ after he threw it?

2. Michael also happens to be a math wiz, and calculates everything. He calculated that the path of his projectile Cheerios are constant that can be modelled by the quadratic equation of $-x^2+6x+9=0$. Assuming that the point the Cheerio starts (leaves his hand) is $(-2,6)$, what is the sum of the coordinates of the point that the Cheerio enters Michaels mouth?

For number one,
[hide="hint"]After 1 second, what is the power of two in the denominator? After 2 seconds, what is the power of 2 in the denominator? [/hide]

For number two, is the equation $y=x^2+6x+9$?

$x+y$ from the positive coordinates$(x,y)$. By the way, I kind of messed up, but i changed it...... :blush:.

This is my first problem, so I'm probably going to have many mistakes.

You almost got it, aznmonkey1992
[hide="1"]
The Cheerio is in the air for 15 seconds, but it takes him 1 second to just pop it in his mouth.
So the answer is $(\frac{1}{2})^{14} = \boxed{\frac{1}{16384}}$[/hide]

For 2, the parabola created from the equation would make the Cheerio go up forever if I graphed the equation correctly that 236 posted. :huh:

this is the $y = -x^2 + 6x +9$:

daaaaaaang it! i posted that first, pascal, but i stupidly changed it. :mad:  :mad:  :mad:  :mad: ironic :P

[quote="pascal_1623"]You almost got it, aznmonkey1992
[hide="1"]
The Cheerio is in the air for 15 seconds, but it takes him 1 second to just pop it in his mouth.
So the answer is $(\frac{1}{2})^{14} = \boxed{\frac{1}{16384}}$[/hide]
[/quote]

I don't think that it takes 1 second to pop it into his mouth, I think what they mean by starting with 1 is saying that the probability of his catching the Cheerio when he just pops it is one. After one second, the probability drops to 1/2, and so on.

If that's the case, aznmonkey is correct.

[quote="236factorial"]
If that's the case, aznmonkey is correct.[/quote]

that's the case :D

for #2, I changed it to $y=-x^2+6x+9$, that was the mistake I did earlier.


Fie $n \in \mathbb{N}^* $ si $X \in M_2 \left( \mathbb{R} \right)$

Cate solutii au ecuatiile:
1.
$X^{2n}  = \left( {\begin{array}{*{20}c}
   3 & 6  \\
   2 & 4  \\
\end{array}} \right)$

2.
$X^{2n + 1}  = \left( {\begin{array}{*{20}c}
   3 & 6  \\
   2 & 4  \\
\end{array}} \right)$

Cred ca gresesc, dar ar putea fi urmatoarea idee:
Din ecuatia caracteristica pentru matrici, avem $X^2-tr(A)X+det(A)=X^2-tr(A)X=0$, deoarece $det(X)=det(A)=0$, si astfel, daca $tr(X)=r => X^k=r^k X$, si de aici totul este clar?


Alberto and Barbara play the following game. Initially, there are some piles of coins on a table. Each player in turn, starting with Albert, performs one of the two following ways:
1) take a coin from an arbitrary pile;
2) select a pile and divide it into two non-empty piles.
The winner is the player who removes the last coin on the table. Determine which player has a winning strategy with respect to the initial state.

very good problem.i think its difficulty level is IMO-like.

yes, it's quite hard, expecially under pressure... It took me 2 hours and an half to solve it ... incompletely  :P

I think I solved this but there is just one thing. Is it enough to know just the number of coins and the number of piles? For example if you have two piles with two coins and one pile with one coin, then Alberto has a winning strategy, but if you have one pile with three coins and two piles with one coin then Barbara has a winning strategy. In both cases we have three piles and five coins, but different outcomes. Or am I missing something perhaps?

Well, knowing the theory behind that kind of games it is just an exercise to solve it:

Let the number of piles with exactly $k$ coins be $n_k$. Then Alberto wins iff at least one of $n_1$ and $n_2+n_4+n_6+...$ is odd.

Proof: It is easy to check if we define a "losing position" to be a position where both $n_1$ and $n_2+n_4+...$ are even, that you must go from a "losing position" to a "winning position" and that there is always a possibility to go from "winning position" to a "losing position". This immediately solves the problem.

Yes, Peter, you got it...

In fact we can also reconduce this game to NIM, i think a quite well-known "coin-game", in a simple way.

However we point out that the odd piles doesn't exists (if the other player removes a coin from ad odd pile i remove another coin, but if he leaves 2 coins then I split it into two "one-coin" piles that, soon or later will be removed with one move made by each player) and the even piles are like 2-coins piles. So we can only consider 2-coin piles and 1-coin piles. Now it's quite simple to check losing and winning position that obviously depends only on parity of these type of piles (in fact we can do "simmetric" moves to reduce piles modulo 2).

The unique point I've missed in my solution is the fact that the game is finite. But it's not difficult to check that 2N-P where N is the number of coins and P the number of piles is positive and strictly decreases each move.

If there are even number of odd piles, the second player have wining strategy, because after each move he can conserve the parity of the number of odd piles. Hence after the last move, this number will be 0.
If there are odd number of odd piles then, the first player has wining strategy by taking one coin from an odd pile and then acting like the second playr in the first case.
Since the product of the numbers of coins in each pile decreases after each move, the game will end.


Sue Perstition was born on Friday the 13th. She recently celebrated one of her birthdays below on Friday the 13th as well. Which one was it?

(A)10th
(B)20th
(C)30th
(D)40th
(E)50th
(F)60th

i don't get this question....she has a birthday on Friday the 13th and she celebrates it on a different date.... :?

What?
No, in fact, she's born on a friday 13th, and now, for her xth birthday, it's again a friday 13th, and you have to find the "x".

[quote="236factorial"]I've done this problem. Let me give a hint:
[hide]leap years. [/hide][/quote]
that can mean it can be B,D,F.

[quote="riddler"]Ah.. yes! friday the 13th doesnt come every year....[/quote]

You're incorrect about that.

This must be too hard for Getting Started if no one's solving it...

I think i got it...
[hide]When 1 year passes, we want to know how many days later her birthday finds place. Since there are 7 days in a week, we want to know $x$ in the equation $365\equiv x(mod 7)$
We find that $x=1$
Now, we know that every four years, we get a year with $366$ days. When we are living in such a year, her birthday will be $2$ days later than last year. (e.g. her birthday will be on a monday while last year, it was on a saturday)
So we know that with every $4$ years that pass, the birthday will take place $5$ days later.
So now we want to see which of the given possibilities in days is $0 (mod 7)$
50 is the only one fitting.[/hide]


geometry
circumcircle
inradius
trigonometry
inequalities
inequalities proposed

Given triangle $ABC$ with semiperimeter $s$ circumradius $R$, inradius $r$ prove that
$\sin^{4}\frac{A}{2}+\sin^{4}\frac{B}{2}+\sin^{4}\frac{C}{2}\le\frac{4R^{2}-8Rr+3r^{2}}{4R^{2}}$

Using the identity : $\boxed{\ \sum\, \sin^4\frac A2=\frac {8R^2+r^2-s^2}{8R^2}\ }$ we are left to prove that : $\frac {8R^2+r^2-s^2}{8R^2}\, \le\, \frac {4R^2-8Rr+3r^2}{4R^2}$

$\iff 8R^2+r^2-s^2\, \le\, 8R^2-16Rr+6r^2\iff\boxed{\ 16Rr-5r^2\, \le\, s^2\ }$ , which is just [u][b]Gerretsen[/b]'s inequality[/u]. $\square$

[hide="Remark."]The previous identity can be proved like this: Since $(\sum\, \sin^2\frac A2)^2=\sum\, \sin^4\frac A2+2\sum\, \sin^2\frac B2\sin^2\frac C2$

and $\{\begin{array}{ccc}
\sum\, \sin^2\frac A2=\frac {2R-r}{2R} \\ \\
\sum\, \sin^2\frac B2\sin^2\frac C2=\frac {s^2+r^2-8Rr}{16R^2}\end{array}$ our identity now follows.[/hide]


In triangle $\Delta ADC$ we got $AD=DC$ and $D=100^\circ$.
In triangle $\Delta CAB$ we got $CA=AB$ and $A=20^\circ$.

Prove that $AB=BC+CD$.

Let M be the point in AB such that ADM is an equilateral triangle. Thus we have $\measuredangle DMA=60^0$ and AD=DM=AM. But AD=CD so AM=DC. (1)
   => $\measuredangle CDM=100^0-60^0$
   => $\measuredangle CDM=40^0$
We have :
    DM=DC
  So triangle DCM is an isosceles triangle
   => $\measuredangle DMC=DCM$
   => $\measuredangle DMC+DCM=180^0-40^0=140^0$
   => $\measuredangle DMC=DCM=\frac{140^0}{2}=70^0$
 We have :
    $\measuredangle DMA=60^0$ and $\measuredangle DMC=70^0$
   => $\measuredangle CMB=50^0$
 We have :
    AC=AB so triangle ABC is an isosceles triangle.
   => $\measuredangle ACB=ABC=180^0-20^0=160^0$
   => $\measuredangle ACB=ABC=\frac{160^0}{2}=80^0$
    Now we return to triangle MCB
   => $\measuredangle MCD=180^0-(50^0+80^0)=50^0$
 Hence triangle MCB is an isosceles triangle.
   => MB=BC (2)
   From (1) and (2) we have :
       AB=CD+BC :cool:

Here's a nice easy way with the Law of Cosine

DB=AD^2+AB^2-2AD*AB cos(DAB)=AD^2+AB^2-AD*AB

DB=DC^2+CB^2-2DC*CB cos(DCB)=DC^2+CB^2+DC*CB

Set them equal to each other to receive
AD^2+AB^2-AD*AB=DC^2+CB^2+DC*CB
AD=DC so....

DC^2+AB^2-DC*AB=DC^2+CB^2+DC*CB
Cancel to get AB^2-DC*AB=CB^2+DC*CB

Move the CB^2 to the LHS and the DC*AB to the RHS to get (AB-CB)(AB+CB)=DC(AB+CB)

Divide by AB+CB to get (AB-CB)=DC
Which is also AB=DC+CB

I know this is the third solution, but it seems my way is different from the other two (and pretty short):

If we set AB = k, then by some simple trig BC = 2ksin10 and  CD = k/(2sin50). To prove AB = BC + CD, it suffices to prove 2sin10 + 1/(2sin50) = 1 (cancelling the k's out).

2sin10 + 1/(2sin50) = 2cos80 + 1/(2sin50) = (4cos80sin50+1)/(2sin50).

cosasinb = 1/2(sin(b-a)+sin(b+a)) --> 4cos80sin50 = 2sin(-30)+2sin(130).

Thus, (4cos80sin50+1)/(2sin50) = (2sin(-30)+2sin(130)+1)/(2sin(50)) = (-1+2sin(130)+1)/(2sin(50)) = sin(130)/sin(50). sin(x) = sin(180-x) --> sin(130)/sin(50) = 1. In conclusion, 2sin(10)+1/(2sin(50)) = 1 --> BC + CD = AB.

Sayoonara.

Love ur trig solution using sine law .But ur simplification could be done a bit fast.After taking lcm we get 4sin10sin50+1/2sin50.
Using product to sum formula we get 2(cos40-cos60)+1/2sin50=2cos40/2sin50=1.


topology
geometry
3D geometry
function
geometric transformation
calculus
integration

As I understood, no contestant (from THAT team) solved this problem.
So you have the chance to prove you are better than them :)

In each lattice point is written a real number between 0 and 1. For every point, the number written in it is the average of the 4 numbers written in the points at unit distance from it. Prove that all numbers are equal

As far as I know, there is a more general problem, concerning only positive numbers. But I read the solution and I didn't understand anything. It is highly non-elementary. But I believe the bulgarians have an elementary solution for this case, in which the numbers are smaller than 1.

I saw a similar problem in Engels book, but the corners were labeled by positive integers so one could choose the smallest element and it was easy to prove all other elemnts were equal to it, but in this problem we cant assume the existence of a minimal element. also as Harazi said the result works for  positive reals, and the prof is non elementary, I dont see how the extra assumption of the upper bound for the numbers may help to solve the problem :? However there must be a nice sol :lol:

Someone should write solution at last.
This theorem appeared on forum at least 4 times. No one solution was proposed!

Just link to the more general problem http://www.mathlinks.ro/Forum/viewtopic.php?t=22086

I remember having read the solution in the book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0387907653/qid=1105210939/sr=1-2/ref=sr_1_2/103-2168213-6089437?v=glance&s=books]A Problem Seminar by Donald Newman.[/url] The solution was really short. But it used an unknown approach to me.  :)

[quote="grobber"]I'd love to have more time to think about it, but I'd also love to see this short solution. Would you please post it, orl?[/quote]

I let you think for some more days and post the solution later. I don't want to spoil your fun.  :lol:

The problem is much easier for one dimensional case, if on a line, integer points are given real number in (0,1).

Then $a_0=\frac{a_{-1}+a_1}{2}\Rightarrow\ a_0+k=a_1; a_0-k=a_{-1}$ and it is easy to show that
$a_0+kn=a_n$ for integer n.  Hence there is some n such that $|a_0+kn|>1$ contradiction.  So k=0.

But extending this to the 2d case seems messy and requires some more insight...

That is the main feature of this problem: it is very hard to make step from 2D to 3D. Just remember how many time we were waiting for 3D accelerators...  :D

Hm...Myth!! Now you have the solution, you just want to tease people here. :mad:

[hide="Convexity"]

Convexity:

"The idea of sets is one with rather extensive generality. Although we tend to think of a convex set as a picture we can draw on a blackboard, the notion is not at all limited to two dimensions, or even finitely many dimensions. The general backdrop is a linear space with a topology attached, and then there is then the very important concept of EXTREME POINT, which means a point which is not convex combination of other points of the set.
The basic theorem(which will prove useful to us) is that a compact convex set always contains extreme points. Enough of them in fact, so that the whole set is the closure of their convex combinations.
Actually we will be involved with what are called convex cones rather than convex sets. These are defined as being closed under taking positive linear combinations rather than only such combinations with coefficients summing to 1. The concepts of extreme points, etc., go over lock, stock and barrel."

[/hide]

[hide="Hint"]

Hint:

Take an extreme point of such functions (assignments). Notice that it is a positive combination of its own translation. Conclude that it is equal to its translates. Conclusion: Every extreme poin is constant. Conclusion: Each one of them is constant.

[/hide]

[hide="Solution"]

Solution:

If we look at all functions $f(m,n)$ with this property then they form a locally compact convex cone under the topology of uniform pointwise convergence (because at each neighboring point the bound is at most 4 times this
point's value). We wish to show that there is only one such function (up to constant multiples), so it suffices to show that there is only ONE extreme point. But if $T$ denotes translation by $(1,0)$ and $S$ translation by $(0,1)$, we have $f = \frac{1}{4} T f + \frac{1}{4} T^{-1} f + \frac{1}{4} S f + \frac{1}{4} S^{-1} f$, so for an extreme point we must have $Tf = \lambda f, S f = \mu f$. But then

\[f = \frac{1}{4} \left ( \lambda + \frac{1}{\lambda} + \mu + \frac{1}{\mu} \right ) f\]

so

\[\lambda + \frac{1}{\lambda} + \mu + \frac{1}{\mu} = 4,\]

so

\[\lambda = \mu = 1,\]

and hence $f$ is constant.

[/hide]

Thanks, orl.  Quick question: is that the solution in Newman?

I do if there is something wrong in my understanding.If I understand precisely.
I think there is a point's number is larger than any others'.but it is the average of the 4 numbers written in the points at unit distance from it.so we can get these five numbers are equal.similarly we can prove all numbers are equal.
Is there something wrong in my proof?

I think there is, because you don't know that there is a maximal number. Where did you get this?:

"I think there is a point's number is larger than any others'" (I quoted you :))

Yes, you are right, I have edited it. The statement for all functions except polynomials in $x,y$, for which the process of passing to the difference function $f_{1},f_{2},f_{3},f_{4}$ will eventually lead to the constant void function, which does not satisfy the original conditions any more.

[quote="iura"]Coming back to the original problem, we can prove a generalisation, which is true for higher dimensions:

Let $f$ be a discrete harmonic function on $\mathbb{Z}^{2}$ i.e. $f(x,y)=\frac{f(x+1,y)+f(x,y+1)+f(x-1,y)+f(x,y-1)}4$. Assume that $f$ is not a polynomial in $\mathbb{R}[x,y]$. Denote $M_{n}=max\{|f(x,y)|, |x|, |y|\leq n\}$. Then for any sufficiently big $n$, we have $M_{n}>|p(n)|$, where $p$ is some polynomial. [/quote]

I do not fully understand that statement or your proof, but the fact that $M_{n}> Kn$ (K a constant) is clear from the probability interpretation of harmonic functions.

Suppose $f$ has two different values at given adjacent points $p$ and $p'$.  Then $f(p)$ and $f(p')$ are weighted (probabilistic) averages of the the value of $f$ on the boundary of the square of size $n$.  The boundary has perimeter $O(n)$ and the probabilistic weight of each point on the boundary is $O(1/n)$  (the weight will be much smaller, $o(1/n)$, as one approaches the corners, but this will only help our argument).  The weight distribution will be quite smooth and, in some rescaling, analytic or at least with a nice asymptotic expansion in powers of $1/n$.

There are two different probability distributions on the boundary, one for $p$ and one for $p'$.  The difference between the $p$ and $p'$ distributions will behave like the derivative of the weight, i.e. $O(1/n^{2})$.   Now, if $|f| \leq M$ on the boundary, then

$f(p)-f(p') = \Sigma (w-w') f(u)$

where  $u$ runs through the points on the boundary. There are $4n$ such points, and $|w-w'| < C/{n^{2}}$ for some constant C,  so that

$|f(p)-f(p')| \leq M C (4n) /{n^{2}}$ .

Thus $M$ grows at least linearly with $n$.    This argument works in higher dimensions as well (in fact better) than in 2 dimensions.

Exercise:  find a sufficient condition for $f$ to have growth of order at least $n^{2}$.

Well, please tell me if you don't understand some exact step in the proof. From the statement (if it's correct) it follows that  all harmonic functions except constants and linear functions have growth of order at least $n^{2}$ i.e. $M_{n}>Kn^{2}$ (did i understand correctly?)

[quote="iura"]Well, please tell me if you don't understand some exact step in the proof.[/quote]

First let me understand the statement.  Is the theorem "if $M_{n}\leq Cn^{k}$  (for large $n$) then $f$ is a polynomial of degree at most $k$"?

Second, I don't understand where in the proof do we use the fact that this harmonic function lives on $Z^{2}$ instead of an infinite tree of degree 4.
On a tree there is the freedom to construct a harmonic function that grows as fast or as slow as we want.

Indeed, that's iuras theorem.

And for the infinite tree of degree 4, $f_{1},...,f_{4}$ were not harmonic anymore, that's where he uses that we are on $\mathbb{Z}^{2}$.

Peter

there is a nice way to prove that a discrete harmonic function $f: \mathbb{Z}^{n}\rightarrow (0,1)$ is constant.

Let $g(x_{1},x_{2},...,x_{n})=f(x_{1}+1,x_{2},...,x_{n})-f(x_{1},x_{2},...,x_{n})$ and let
$M=\sup g(X)$ suppose $M> 0$ then we proceed as follows:
obviously $g(X)$ is also harmonic
easily by induction one shows that
$g(x_{1},x_{2},...,x_{n})\geq M-\epsilon$ implies $g(x_{1}+m,x_{2},...,x_{n})\geq M-(2n)^{m}\epsilon$
choose $\epsilon \in [0,\frac{M}{2 (2n)^{m}})$ and $m>\frac{2}{M}$
then
$1>f(x_{1}+m,x_{2},...,x_{n})-f(x_{1},x_{2},...,x_{n})=\sum_{i=1}^{m}g(x_{1}+i,x_{2},...,x_{n})\geq m \frac{M}{2}\geq 1$

Well, the proof looks beautiful, but I don't understand that at all.. Hm.. Can someone write with more details ?  :maybe:

we should see if any Romanian guy has the official solution here because this appeared as Romanian TST 1978 day 2 Prob 1 !!!  :o  :)

I want to add one more conjecture:

Let $ G$ be any subgraph of the $ \mathbb{Z}^2$ grid. Consider harmonic functions as defined on the previous page. Is the conclusion that any bounded harmonic function is constant still true?

Maybe we even have stronger versions using only boundedness from below or subharmonicity... A connected question is whether a random walk in that "labyrinth" will return to its starting point with probability 1(in fact, it would imply the question above; on the other hand, the question with subharmonicity would imply the random-walk-property).

Note: The infinite tree given on the previous page proves that this fails for dimension at least 3. However, that construction can not be used to disprove the 2-dimensional version(at least not without additional thoughts).

I assume you want the graph $ G$ to be connected.  If $ G$ is finite, then all harmonic functions are constant.  So we can assume that $ G$ is infinite.

There is a connection between random walks and electric networks.  For example, Doyle and Snell wrote a book called [i]Random Walks and Electric Networks[/i].  It is free to download on [url=http://math.dartmouth.edu/~doyle]Doyle's website[/url].

One result they show (page 102) is that an infinite, connected, bounded-degree graph is recurrent (meaning random walks return) if and only if the resistance of the graph is infinite.  Another result (page 76), called the Cutting Law, is that removing edges can only increase the resistance of the graph.

We know that $ \mathbb{Z}^2$ is recurrent, which means that $ \mathbb{Z}^2$ has infinite resistance.  Thus, by the Cutting Law, $ G$ also has infinite resistance, which means $ G$ is also recurrent.  As you point out, that means every harmonic function on $ G$ bounded from below is constant.

I hope I haven't made a huge blunder in my reasoning.

Maybe I am being silly, but I think I have a simple Proof

[b]Problem:[/b] If a function $f: \mathbb{Z}^n \to \mathbb{R}_{\ge 0}$ is harmonic, then it is constant.

If $f$ is non-zero then wlog we can assume $f(0,0,...,0)=M > 0$. Define a function $g_0$ such that $g_0(0,0,...,0)=M$ and $g_0(x_1,...,x_n)=0$ for all other points in $\mathbb{Z}^n$.

Define a sequence of functions $\{g_k\}_{k=0}^{\infty}$, so that for each $p\in \mathbb{Z}^n$, $g_k(p)$ is the maximum of $g_{k-1}(p)$ or the average of $g_{k-1}$ over the $2n$ neighbouring points of $p$. It is easily shown by induction that for every $k\in \mathbb{N}$ we have $f(p) \ge g_k(p)$.

For each point $p\in\mathbb{Z}^n$, the sequence $\{g_k(p)\}_{k=0}^{\infty}$ is monotonic increasing and bounded above by $M$, so it converges to a limit we call $g(p)$. Indeed the sequence of functions $\{g_k\}_{k=0}^{\infty}$ will therefore converge to a function $g$ that is harmonic and bounded above, and attains it's maximum of $M$ at $(0,0,...,0)$.

It follows by the extremal argument that $g$ is constant and equal to $M$ at all points. So for our original function $f$, the value at each point is at least $M$, and it attains this value at $(0,0,...,0)$. It follows directly from the extremal argument that $f$ is constant $\square$

Very interesting.  Can you say a little more about why $g$ must be harmonic?

[b]Edit:[/b] sorry, this is indeed wrong. It's not harmonic at the origin. $g(p)$ is actually the probability that a random walk from the origin passes through $p$. So it only works as a proof of the 2-D case.

Sure, for each $p\in\mathbb{Z}^n$ let $N(p)$ denote the set of $2n$ neighbours of $p$, Let $S(k,p) = \frac{1}{2n}\sum_{q \in N(p)}g_k(q)$

Then we know that $g_k(p)$ is monotonic increasing in $k$ and bounded above by $M$, therefore $S(k,p)$ is as well. So they both approach some limit. Now by the definition of $g_k(p)$ there exists some time $s \le k$ such that $g_k(p)=g_s(p) =S(s-1,p) \le S(k,p)$. Furthermore if the last inequality is strict then $g_{k+1}(p) = S(k,p)$ by definition. It follows that $S(k,p) \le g_{k+1}(p) \le S(k+1,p)$. So $g_k$ is sandwiched to the same limit as $S(k,p)$ which says that $g(p)$ is harmonic.

Btw, it was p.3 of the Bulgarian TST (03.03.2003) for BMO 2003,
[b]The official solution[/b]. Assume on the contrary that $f(x,y)$ is a non constant bounded harmonic function (assume $0\leq f \leq 1$). Then there exist two consecutive points (along x or y axes) at which $f$ takes different values. Rotating the plane, if necessary, WLOG we may assume that $f(x_0+1,y_0) > f(x_0,y_0)$. Let us denote
$$g(x,y)=f(x+1,y)-f(x,y)$$
Then $g$ is not identically zero and $g$ is bounded. Take $M= \sup_{(x,y)\in\mathbb{Z}^2} g(x,y)$. It follows $0< M < \infty$. It is easy to see $g$ is also a harmonic function. Let $\varepsilon >0$ and $(x,y)$ satisfies $g(x,y) > M-\varepsilon$. Because $g$ is harmonic and $M$ is its supremum, it follows:
$$g(x+1,y)  > M-4\varepsilon$$
Applying several times this argument it yields
$$g(x+n,y) > M-4^n \varepsilon$$
Using the definition of $g$ we get
$$f(x+n,y)-f(x,y) = \sum_{k=0}^{n-1} g(x+k,y)\quad (1)$$

Now, see that the LHS of $(1)$ is between $-1$ and $1$, while we can make the RHS as big as we want, in particular bigger than $1$. How ? Take first $n>\frac{2}{M},$ then sufficiently small $\varepsilon$ and after that $(x,y)$ such that $f(x,y)>M-\varepsilon$.


function
integration
logarithms
calculus
Euler
calculus computations

How do you find $ \int_{0}^{\infty}\ln xe^{ \minus{} x^2}dx$?
I tried to do it
$ \int_{0}^{\infty}\ln xe^{ \minus{} x^2}dx$

$ \int_{0}^{\infty}\ln xx^ne^{ \minus{} x^2}dx$

$ \int_{0}^{\infty}\frac {d}{dn}x^ne^{ \minus{} x^2}dx$

$ \frac {d}{dn}\int_{0}^{\infty}u^{\frac {n \minus{} 1}{2}}e^{ \minus{} u \minus{} ln2}du$

$ \frac {d}{dn}.5\Gamma\left(\frac {n \plus{} 1}{2}\right)$

$ .25\Gamma'(.5)$

but I'm not sure where to go from there

Using the substitution $ t \equal{} x^2$,

$ \int_{0}^{\infty} e^{\minus{}x^2} \ln x \, dx \equal{} \frac{1}{4} \int_{0}^{\infty} t^{\frac{1}{2}\minus{}1} e^{\minus{}t} \ln t \, dt$

Integrals of this kind is closely related to the [url=http://en.wikipedia.org/wiki/Digamma_function]digamma function[/url], defined by $ \psi(x) \equal{} \frac{d}{dx} \ln \Gamma(x) \equal{} \frac{1}{\Gamma(x)} \int_{0}^{\infty} t^{x\minus{}1}e^{\minus{}t} \ln t \, dt$. With this function, you can simply write

$ \int_{0}^{\infty} e^{\minus{}x^2} \ln x \, dx \equal{} \frac{1}{4}\Gamma\left(\frac{1}{2}\right)\psi\left(\frac{1}{2}\right) \equal{} \frac{\sqrt{\pi}}{4}\psi\left(\frac{1}{2}\right)$

Also, for $ s > \minus{}1$, $ \psi(s\plus{}1) \equal{} \minus{}\gamma \plus{} \int_{0}^{1} \frac{1\minus{}x^{s}}{1 \minus{} x} \, dx$, where $ \gamma$ is the Euler-Mascheroni constant. This enables us to evaluate $ \psi\left(\frac{1}{2}\right)$, and the result is $ \minus{}\gamma \minus{} 2\ln 2$. Therefore,

$ \int_{0}^{\infty} e^{\minus{}x^2} \ln x \, dx \equal{} \minus{}\frac{\sqrt{\pi}}{4}(\gamma \plus{} 2\ln2)$

[quote="sos440"]$ \psi(s \plus{} 1) \equal{} \minus{} \gamma \plus{} \int_{0}^{1} \frac {1 \minus{} x^{s}}{1 \minus{} x} \, dx$[/quote]

How do you prove that?


integration
logarithms
function
superior algebra
superior algebra unsolved

If i want to show that $ R(n) /n/log n$ tends to 1 as n tends to infinity. Here
$ R(n)  \equal{} \sum_{k\equal{}1}^{\infty} \frac {\mu(k)}{k} Li(n^{1/k})$  and $ Li(x) \equal{} \int_2^n \frac {dt}{\log t}$
here $ \mu$ is mobius function. Is it enough if I show that $ R(n) /n/log n$ is less then something which tends to 1 or should I also say that $ R(n) /n/log n$ is greater then something which tends to 1 as n tends to infinity, and then I can conclude that $ R(n) /n/log n$ tends to 1 as n tends to infty?

I hope you understand what I mean.

You should also say that the function is greater then something which tends to $ 1$ as $ n$ tends to infinity. Look at for example $ f(n) \equal{} \frac {1}{2}$. Now $ f(n) < 1$ but $ f(n)\to \frac {1}{2}$ as $ n\to \infty$. Note that in your notation $ Li(x)$ is a constant function.


function
real analysis
real analysis unsolved

Find all functions f defined on $(0,infinity)$ with values in the same set such that $ f(f(f(x)))+2x=f(3x)$ and $ f(x)-x$ tends to 0 when x tends to infinity.

I want to see solution.
Harazi! Is it possible that you forgot to say $f(x)$ is continuous? :?

No, f is not supposed to be continuous. I will post solution later, now I have to go. Sorry.

I don't need your solution anymore, since I have my own!  :D  :D  :D

[b]1)[/b] Easy task 1: prove that that $f(x)\geq x$.
[b]2)[/b] Easy task 2: using 1), prove that it is impossible $f(x)-x>0$.

Anyway, I wouldn't have written my solution... ;)  I knew you will find a solution in 5 minutes.


induction
trigonometry
real analysis
real analysis unsolved

Define $ x_1 \equal{} \sqrt{2}$, $ x_{n\plus{}1} \equal{} \sqrt{2 \plus{} x_n}$ for $ n\geq 1$, therefore $ x_{n\plus{}1}^2 \equal{} 2 \plus{} x_n$.
Notice that $ x_1 \equal{} 2\cos \left ( \frac{\pi}{2^{1\plus{}1}} \right )$. Assume $ x_n \equal{} 2\cos \left ( \frac {\pi}{2^{n\plus{}1}} \right )$. But $ 2 \plus{} 2\cos x \equal{} 4\cos^2 \frac {x} {2}$, and this yields $ x_{n\plus{}1} \equal{} 2\cos \left ( \frac {\pi}{2^{(n\plus{}1)\plus{}1}} \right )$, q.e.d.


Hi all! I want to ask question and discuss about this man: Prof. Jucelino Nobrega da Luz. Local media in my hometown said that he predicts there will be a very terrific earthquake on next Dec 23rd. I know that scientifically it is impossible to predict when an earthquake will occur. But, local media in my hometown says, he is a [b]scientist[/b] from Brazil (!). I'm very not sure he is so. Since I have red that in Portuguese-speaking country, an oracle is often called: "professor", I think he is just an oracle who want to give his prediction by his knowledge. Can you give me some information about this man and everything about him? It will be very valuable to me. Thanks. :)

NB: I have use google, but its links are in Portuguese, and unfortunately I don't understand it. :blush: I'.m sorry if you don't understand my English, because I think mine is not too good.

I don't know him, but perhaps you could use online translators for the websites you found in Portugease. There are many great websites where you just copy and paste paragraphs in a box and choose what language you want to translate it too, and it translates it for you. Try Babelfish.

I don't think that's very helpful, though, sorry.

According to http://news.yahoo.com/s/afp/20071119/od_afp/indonesiabrazilquakepsychicoffbeat_071119125913
he is just a psychic.

Ok, thanks for the informations. The earthquake didn't happen that day. It is very absurd to predict scientifically :rotfl: the time exactly when the quake happen.  :wink:

I think in every country there are such kind of peopele....who predict the world ending and earthquakes.... :) ....as far as I know...an earthquake can be prevent with a few minutes before it hapens...not in years...


modular arithmetic
number theory unsolved
number theory

Show that for any natural numbers $ a_k$, for $ k\equal{}1,2,...,5$, there exist numbers $ t_k\equal{}\minus{}1,0,1$,not all equal to zero,
s.t. $ \sum^5_{k\equal{}1}t_ka_{k}^2$ is divisible by $ 11$.

All perfect squares are $ \equiv0,1,3,4,5,9\pmod{11}$. If there is $ a_m\equiv0\pmod{11}$, take $ t_m\equal{}1$ and $ t_n\equal{}0$ for all $ n\ne m$. If there are two numbers $ a_n\equiv a_m\pmod{11}$, take $ t_n\equal{}1,t_m\equal{}\minus{}1$ and $ t_l\equal{}0$ for all $ l\ne m,n$. If all the numbers are distinct, and none of them is divisible by 11, just take $ t_n\equal{}1$ for all $ n$, hence $ \sum a_i^2\equiv22\pmod{11}$. :)

more general:
for any natural numbers $ a_k$, for $ k \equal{} 1,2,...,m$,and evrery natural number $ n$ with $ n<2^m$ , there exist numbers $ t_k \equal{} \minus{} 1,0,1$,not all equal to zero,
s.t. $ \sum^m_{k \equal{} 1}t_ka_{k}$ is divisible by $ n$


[quote="4everwise"]The length of the common chord of two intersecting circles is $16$ feet. If the radii are $10$ feet and $17$ feet, a possible value for the distance between the centers of teh circles, expressed in feet, is:

$\text{(A)} \ 27 \qquad \text{(B)} \ 21 \qquad \text{(C)} \ \sqrt{389} \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{undetermined}$[/quote]

[hide](B) 21
So the chord is the length of the segment of the two poitns where the circles intersect. Radii are drawn from each circle to these intersecting points so on the circle with radius 10, an isoscoles triangle of length 10, 10, and 16 form, while an isoscoles triangle of length 17, 17, and 16 form. A line bisects the chord and forms a 6-8-10 and a 17-15-8 triangle and $15+6=\boxed{21}$.[/hide]


algebra
polynomial
function
invariant
group theory

Hi, given a polynomial:

$ f(x) \equal{} (x \minus{} r_1)(x \minus{} r_2)(x \minus{} r_3)...(x \minus{} r_n) \equal{} y_n x^n \plus{} y_{n \minus{} 1}x^{n \minus{} 1} \plus{} ... \plus{} y_1x \plus{} y_0$

What is the formal notation to express each term:

$ y_n \equal{} 1$
$ y_{n \minus{} 1} \equal{}$ negative sum of the roots
$ y_{n \minus{} 2} \equal{}$ sum of product of every pair of roots
$ y_{n \minus{} 3} \equal{}$ negative sum of product of every triplet of roots

What is the correct way (correct terms) to write it in English and is math notation?

- Thanks

Thank you,

Does this topic has a name? What is the typical course title for a course that teaches this?

Notation:

So in my example above, there are n+1 elementary symmetric polynomials. The coefficients can be expressed using these elementary symmetric polynomials. For example:

$ y_n \equal{} e_0(r_1,...,r_n) \equal{} 1$

$ y_{n \minus{} 1} \equal{} \minus{} e_1(r_1,...,r_n) \equal{} \minus{} \Sigma_{1\leq j\leq n} r_j \equal{} \minus{} ( r_1 \plus{} r_2 \plus{} ... \plus{} r_n)$

The terms e_0(...) and e_1(...) are the elementary symmetric polynomials. They do not have the variable $ x$ in them but they are called polynomials.

Is what I say above correct?

- Thanks

[quote="Twinbee"]They do not have the variable $ x$ in them but they are called polynomials. [/quote]
They are polynomials in the roots.

Your other question is interesting.  You could say that this is just standard knowledge about polynomials, and it's certainly relevant to abstract algebra, so that might be one of the first places you see the symmetric polynomials.  On the other hand, the elementary symmetric functions belong to a larger subject called symmetric function theory, which you might learn about in a combinatorics class or a representation theory class because symmetric functions turn out to have interesting connections to Young tableaux and the representation theory of the symmetric group, as well as connections to classical invariant theory.  Symmetric functions pop up in a lot of places.


When I go to the Science Olympiad Camp in January we will be doing some laboratory exercises (plus a trial and an official laboaratory examination). These include:

-Calibration of Pipette
-Acid Base
-$ \ce{KMnO4}$
-Compleximetric Titration
-Qualitative Analysis
-Inorganic Synethesis
-Trial Lab Exam (2 hours)
-Gravimetric Analysis of Nickel
-Organic Synthesis
-Iodometric Titration
-Combined Exercises
-Lab Exam (4 hours)

Some/Most of them are ambiguous but what are some possibilities? Bear in mind this camp is an attempt to instruct first year University Chemistry in 2 weeks.

What are some laboratory techniques I should be familiar with?

Thanks.

titration should be easy .It's just how u solve titration problems.
organic synthesis-i haven't done that
inorganic synthesis???

[quote="BanishedTraitor"]These include:[/quote]

Those are tipical laboratory experiences that we do in university, and if I remember well,  think I have done all of them.

[quote="BanishedTraitor"]What are some laboratory techniques I should be familiar with?[/quote]

You should be familiarized with the following:

- the name of the material used in a laboratory (pipettes, burettes, etc);

- basic behaviour in a laboratory;

- the symbols and some R/S phrases that appear in solute's flasks;

- preparation of solutions, which includes proper handling of solid and liquid solutes, know how to make the relevant calculations and work with different concentration units;

- know when it is a good idea to work in the hotte;

- the proper disposal of waste products/solvents;

- know how to wash correctly the (glass) material;

- what to do in any titration: what we will determine, who is going to the burette, know how to "affer" the common solutions of acids and bases, what indicator will be used (in particular what is an indicator, its colors, its pH range for color changing, and its minimum detectable concentration), how to calculate the pH of equivalence points, the proper control of the burette when the equivalent point is approaching, etc;

- know how to determine melting points;

- in organic synthesis you will almost certainly use solvents that must be free of water, so it is a good idea to know how to dry some solvents;

- if you are to work with an inert atmosphere, you should know what this is, how to handle a bottle of gaseous nitrogen, how to work with vacuum pumps (this also with vacuum destilation), and so on.

Needless to say, you must study carefully the laboratory experience that you will do, organise what, when, and how you will do the steps of your work, what to do if something goes wrong, etc.

Thanks for the feedback.

Would it be wise to first have a theoretical background of the possible reactions that are employed?

Like how would you carry out the gravimetric analysis of nickel ina solution? Dimethylglyoxime? Or whatever etc.

Not only it would be very wise, but also it is absolutely necessary to understand as good as possible what you are doing in the laboratory and what is happening in your reactions.

Any last minute suggestions before I go to Canberra tomorrow? :D