Determine all the strictly increasing functions $ f\colon \mathbb{R}\to \mathbb{R}$ which satisfy

$ f(x\plus{}f(y))\equal{}f(x\plus{}y)\plus{}2009$,
 for any reals $ x,y$.

put in x such that x = -f(y), we get f(0) - 2009 = const = f(-f(y) + y), f is injection since f if strictly monotonic, so there exist a constant a such that a = -f(y) + y (*)

now we get f(y) = y - a

putting in x = y = 0, we get f(f(0)) = f(0) + 2009, now using (*) , -2a = -a + 2009, a = -2009,
so f(x) = x + 2009, putting back in starting relation we see that this function is the soluton

[quote="felixx"]put in x such that x = -f(y) [/quote]

This step throws out any potentially non-surjective solutions.  In other words you've only proven that $ f(x) \equal{} x \plus{} 2009$ for $ x\in f(\mathbb{R})$.  In order to make the argument solid, you'd need to show $ f$ is surjective.

[quote="Inequalities Master"]Determine all the strictly increasing functions $ f\colon \mathbb{R}\to \mathbb{R}$ which satisfy

$ f(x \plus{} f(y)) \equal{} f(x \plus{} y) \plus{} 2009$,
 for any reals $ x,y$.[/quote]
From hypothesis we have $ f(x\plus{}f(y))\equal{}f(y\plus{}f(x))\forall x,y\in\mathbb{R}$ therefore $ x\plus{}f(y)\equal{}y\plus{}f(x)\forall x,y\in\mathbb{R}$ because $ f$ is strictly increasing, so $ f(x)\equal{}x\plus{}C\forall x\in\mathbb{R}$. Now, check original identity.  :)

[quote="polskimisiek"][quote="JoeBlow"][quote="felixx"]put in x such that x = -f(y) [/quote]

This step throws out any potentially non-surjective solutions.[/quote]

Can you explain it? I think putting that into an equation does not need surjectivity, because here x is arbitrary.  :huh:[/quote]
Yes, he is wrong! I think so, too.  :lol:


Juanita called her friend long-distance and forgot the number, except for the area code. She did remember that when she dialed the last seven digits, she dialed the digits successively from columns #$ 1$, #$ 2$, #$ 3$, #$ 1$, #$ 2$, #$ 3$, and #$ 1$. Find the number of possible phone numbers Juanita could have dialed with this dialing pattern.
[asy]draw((0,0)--(2.6,0)--(2.6,4)--(0,4)--(0,0));
label("  1",(0,3.5),E);
label("  2",(1,3.5),E);
label("  3",(2,3.5),E);
label("  4",(0,2.5),E);
label("  5",(1,2.5),E);
label("  6",(2,2.5),E);
label("  7",(0,1.5),E);
label("  8",(1,1.5),E);
label("  9",(2,1.5),E);
label(" 0",(1,0.5),E);[/asy]

There are 3 ways to pick the first number,4 the 2nd,3,3,4,3,3

3x4x3x3x4x3x3=36x36x3=3x1296=3888


calculus
integration
search
number theory unsolved
number theory

Consquaring the equation $x^2+y^2+z^2+t^2-Nxyzt-N=0(*)$ where $N$ is known natural number .
a)Show that:there exists many infinitly natural number $N$ such that (*) has root.
b)Supposing $N=4^k(8m+7)$ where $k,m$ be non-negative integers.Show that: for which (*) has no positive integeral root .

Markov style again yea
Hmmm you should search for this, it's post by Mr [b]namdung[/b]


pigeonhole principle
combinatorics unsolved
combinatorics

Prove that if there are 101 people of different heights standing in a line, it is possible to find 11 (not necessarily consecutive) people in the order they are standing in the line with heights that are either increasing or decreasing.

This is Erdos-Szekeres with $ m \equal{} n \equal{} 10$.

I have found this page on the theorem you mentioned

http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Szekeres_theorem

but how did you determine that r = s = 10 (or m = n = 10)?  Thank you.

Actually, we have $ r \equal{} s \equal{} 11$.  Note that $ rs \minus{} r \minus{} s \plus{} 2 \equal{} (r \minus{} 1)(s \minus{} 1) \plus{} 1 \equal{} 101$ in this case.

(The statement that JBL is thinking of has $ m \equal{} r \minus{} 1, n \equal{} s \minus{} 1$, in which case it can be stated suggestively as follows:  a sequence of length $ mn \plus{} 1$ contains either a monotonically increasing subsequence of length $ m \plus{} 1$ or a monotonically decreasing subsequence of length $ n \plus{} 1$.  I say suggestive because the connection to the Pigeonhole principle is more obvious this way.)

As t0r noted, I use the formulation at [url=http://mathworld.wolfram.com/Erdos-SzekeresTheorem.html]mathworld[/url], not the wikipedia one, and that explains the off-by-one error.


function
calculus
derivative
calculus computations

The coefficient $ \beta$ of thermal expansion of a liquid relates the change in the volume V (in $ m^3$) of a fixed quantity of a liquid to an increase in its temperature T (in ° C) as shown below.

$ dV \equal{} \beta V dT$

Let $ \rho$ be the density (in kg/$ m^3$) of water as a function of temperature. Write an expression for $ d\rho$ in terms of $ \rho$ and dT

hello
$ \rho \equal{} \frac {M}{V}$
$ \frac {d\rho}{dt} \equal{} \minus{} \frac {M}{V^2}\beta V$
$ \frac {d\rho}{dt} \equal{} \minus{} \rho\beta$
$ {d\rho} \equal{} \minus{} \rho\beta dt$
thank u

Ok thanks for the reply. What if I have a graph that shows the density of water as a function of temperature.

How can I use it to estimate $ \beta$ when T = 40

[quote="kabi"]hello
if u have graph i.e. u have the derivative .
from the relation u can find $ \rho$ at the given $ T$.
put the values u will get $ \beta$
thank u[/quote]

Ok at T = 40 I got $ \rho$ to be somewhere near 992, so what do I do now?

you said to put the value, but to which equation?


We are given all the eigenvalues and eigenvectors of a $n\times n$ matrix $A$ , and we are also given another $n\times n$ matrix $Q$ . If matrix $B$ is defined by

$B=QAQ^{-1}$

how can we find the eigenvalues of $B$ without evaluating the matrix $B$ ?

[quote="Omid Hatami"]Eigenvalues of of similiar matrices are the same. :)[/quote]

so how we can prove that similar matrices have same eigenvalue ?

First of all look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=87254]this topic[/url].
Then you see that $\det(QAQ^{-1}-xI)=\det(QAQ^{-1}-Q(xI)Q^{-1})$ that is equal to $\det(Q)\det(Q^{-1})\det(A-xI)=\det(A-xI)$

More directly, if $Av=\lambda v$, $QAQ^{-1}(Qv)=QAv=\lambda (Qv)$. That matches every eigenvector of $A$ with an eigenvector of $QAQ^{-1}$ with the same eigenvalue.

Yes but it does not prove that the multiciplity of eigenvalues are equal.


[quote="Snailboto"]I'm taking the state mathcount test next week and I would like to know the scores from previous years to get a feel on what I need to attain to try and reach nationals. Thanks.[/quote]

okay, there are already like 5 forums asking about scores. look around before you post.  and maybe you should post this in the CT forum.

can a mod merge some of these or something.


geometry
circumcircle
trigonometry
LaTeX
angle bisector

(8) AB and CD are chords of a circle radius 4 on opposite sides of the centre. CB and DA produced meet at P outside the circle. THe length of arc CD is four times the length of arc AB. THe length of arc CD is 8(pi)/5. Find <APB.

(9) The side AB of a triangle ABC is diameter of a circle radius R and C lies on this circle. The angle bisector of <BAC meets BC at E and the circle at D. When produced AC meets the circumcircle of CED at F. GIven BC=a, express CF in terms of R and a.

(8) $R \widehat{CD} = \frac{8}{5}\pi \Longrightarrow \widehat{CD} = \frac{2}{5}\pi$
join $AC$, thus $\angle CAD = \frac{1}{2}\widehat{CD}=\frac{1}{5}\pi$ and $\angle BCA= \frac{1}{2}\widehat{AB} = \frac{1}{2}\cdot\frac{1}{4}\widehat{CD} = \frac{1}{20}\pi$
$\angle BPA = \angle CAD - \angle PCA = \frac{3}{20}\pi$

(9)
$AC\cdot AF = AE\cdot AD \quad (*)$,
suppose $\angle A = 2\alpha$
thus $AC=2R\cos 2\alpha$, $AE = AC/\cos\alpha$ and $AD = 2R\cos \alpha$.
hence from (*) we have
$AF = 2R$
thus $CF = AF-AC= 2R-\sqrt{4R^2-a^2}$

About (9), here is a synthetic solution:

Since the line AD is the angle bisector of the angle BAC, we have < FAD = < BAD.

Since the point C lies on the circle with diameter AB, we have < ACB = 90. Since the points C, E, D and F lie on one circle, we have < EDF = < 180 - < ECF = < ECA = < ACB = 90. In other words, < ADF = 90.

On the other hand, since the point D lies on the circle with diameter AB, we also have < ADB = 90.

From < ADF = 90 = < ADB, AD = AD and < FAD = < BAD, we infer that the triangles ADF and ADB are congruent, so that AF = AB, and CF = AF - AC = AB - AC. Now, since the segment AB is the diameter of a circle with radius R, we have AB = 2R, and since < ACB = 90, the triangle ABC is right-angled, and Pythagoras yields $AC = \sqrt{AB^2 - BC^2} = \sqrt{\left( 2R\right) ^2 - a^2} = \sqrt{4R^2 - a^2}$. Thus,

  $CF = AB - AC = 2R - \sqrt{4R^2 - a^2}$.

  Darij

[quote="darij grinberg"]Thus,

[unparseable or potentially dangerous latex formula].

Darij[/quote]

That's the best conclusion I've ever heard! :D

I think it's best to wait up. Seems like the LaTeX --> GIF compiler doesn't work today, so a formula can be viewed only if it was already compiled into a GIF before.

  Darij

NightFlarer wrote:For number 8 I got a different answer:
[hide]The question asked for CB and AD to be produced at P right?  That means to extend them to where they intersect, which is P right?  Well, if that's true, the answer would be 27 degrees I think.  We can find arc CD has measure of 72 degrees, and arc AB is 18 degrees, so (72-18)/2 gives 27 degrees for < APB[/hide]

[tex]\frac{3}{20}\pi[/tex] is 27 degrees


数列{bn}中，b1=0，b2=2，b3=3，b n+1=b n-1+b n-2，求证：当p为质数时，p|bp.而且，个人意见觉得，当p|bp时，p为质数。

[quote="shobber"]怎么求通项? :?[/quote]

To solve for the closed form of $a_n$ given $a_0, a_1, a_2$ and $a_n=a_{n-2}+a_{n-3}$, we first set up the characteristic polynomial of the recursion, which is $x^3=x+1$ (If the recursion was instead $2a_n=3a_{n-1}-7a_{n-2}+9a_{n-3}$, the char. poly. would be $2x^3=3x^2-7x+9$). Let the roots be $r_1,r_2,r_3$. (In this case, we can't really find the roots).

Then the closed form of $a_n$ will be $ur_1^n+vr_2^n+wr_3^n$ for some appropriate constants $u,v,w$. We can find the constants using undetermined coefficients. Namely, setting up $a_n=ur_1^n+vr_2^n+wr_3^n$ and then substituting $n=0,1,2$.

Finding the closed form was the first approach that came to my mind.

This method can be used to find, for example, the formular for the $n$th Fibonacci. Its char. polynomial is $x^2=x+1$. This is why the formula for Fibonacci has the golden ratio in it.

Joejia, maybe I know your math background? It is really impressive......


modular arithmetic
number theory
prime numbers

For starters it is divisible by 53.  :)

(And also 79 and 239 and 547 and  4649 and 14197 and 17837 and 4262077 and   43442141653
and  265371653 not to mention     316877365766624209  and
 110742186470530054291318013 )

Don't believe just   check it your self   :.D

Since 91=13*7  it is divisible by 1111111 and also 1111111111111  I just facotred those too ..  Easy to see that  (10^91-1) is divsible by 10^7-1)
 :)  :

BTW I have  not written a 10 digit prime factor above - (See if any one can find that...) :D

More generally, it is not difficult to prove that if $a_n = 11 \ldots 1$, with $n$ ones,  is prime then $n$ is prime.
Obviously the converse does not holds.

Pierre.

[quote="firefox"]Let N be the number which when expressed in decimal notation consists of 91 ones: 111...1. Show that N is a composite number.[/quote]

I already posted this problem before on the board... but then with 1991 :)

You can find it [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15174]here[/url]. It was succesfully solved in intermediate :)

Okay here is a little more  chalanging problem:

Prove that any prime factor of 1111111111111    (=((10^13-1)/9)  is  1 mod 26.

(That's how I got to try 53 -   ...first tryout as 27 is not prime,,,, and speaking of luck .. next one 79 was also a factor)

(or  in general  a  prime factor of ( 111.... p times)  where p is prime is 1 mod 2p. (Edited later - True for p>3)

(So if you want to find  a factor of 1111111 you need  to try only primes like 29,..43 ..71..  .... (only 14k+1 types)..).. (before you reach 239 which indeed is a factor of 1111111)

Proof is not hard .

[quote="Gyan"]Okay here is a little more  chalanging problem:
Prove that any prime factor of 1111111111111    (=((10^13-1)/9)  is  1 mod 26.
(or  in general  a  prime factor of ( 111.... p times)  where p is prime is 1 mod 2p.
[/quote]

Hi,

I think your statement doesn't hold true for all primes $p$.
For example,
$p = 2$, $11 \neq 1 \pmod{4}$, and
$p = 3$, $111 \neq 1 \pmod{6}$.

But for other primes $p$, this holds.

Proof for primes $p \neq 2,3$ :

$10^p - 1 = (10 - 1) (10^{p-1} + 10^{p-2} + ........ + 10^2 + 10 + 1)$ ---------- (1)

We know, for prime $p$, $10^p \equiv 10 \pmod{p}$, and so, $10^p - 1 \equiv 9 \pmod{p}$.

And in RHS of (1), (10-1)=9, so (1) reduced modulo prime $p$ gives
$9 \equiv 9 * x \pmod{p}$, ----- (2)
where $(10^{p-1} + 10^{p-2} + ........ + 10^2 + 10 + 1) \equiv x \pmod{p}$.

The only solution to (2) is $x = 1$ ($x<p$). (If $p$ was $3$, then there exist many solutions, but here $p \neq 3$ as $p = 2,3$ cases have been shown not to hold previously).

So, $(10^{p-1} + 10^{p-2} + ........ + 10^2 + 10 + 1) \equiv 1 \pmod{p}$

Thus, $\forall$ prime $p \neq 2, 3$,
$\frac{10^p-1}{9} = (10^{p-1} + 10^{p-2} + ........ + 10^2 + 10 + 1) \equiv 1 \pmod{p}$

Also,
$\frac{10^p-1}{9} = (10^{p-1} + 10^{p-2} + ........ + 10^2 + 10 + 1) \equiv 1 \pmod{2}$
since it is odd.

As $2$ and $p$ are coprime, $\frac{10^p-1}{9} \equiv 1 \pmod{2p}$.

*******************************************************************
Sambit

[quote="Gyan"]Another slight  variation of the problem:
Prove that  (1111.....1       (  2003 times) base 16  is composite.[/quote]

Hi,

111...1 (2003 times) in base 16 = (16^2003-1)/15 in base 10

Similar to the previous problem, one can show that, (16^2003-1)/(16-1) = 1 (mod 2p)
where p is a prime (not equal to 3 or 5)

Since, 2003 is prime, (16^2003-1)/15 = 1 (mod 4006).
Thus, we only need to check divisibility by numbers of the form 4006k+1.
Luckily, we 'see' that 4007 divides the number , and hence it is composite.

PS : But, how do we check this divisibility without multiprecision calculators?

********************************************************************
Sambit

Sambit: Yes you are right, I  should have said  [b]prime p > 3.[/b]

But I could not follow your proof.   I don't see that you have  proved it  for [b]all[/b] factors  of the number.

that is, for example 11111 = 41*271 .. both 41 and 271 are 1 mod 10 (=2*5)
and factors of 1111111 =  239*4649 both 239 and 4649 (and of course 1111111) are 1 mod 14.

For the second part a hint:  All such numbers (1111.. 'n'  times)  based 16 , except  n= 2, ( the number 17)  in decimal notation)  are not prime... :D

Hi Gyan,

Sorry about the carelessness. I guess I didn't read the question carefully, and hence thought that its asked to prove "1111...11(p times) is 1 mod 2p".

Now, I will try to prove that all prime factors of 111...11(p times) are 1 mod 2p.

[hide]

Let q be a prime factor of

111...11(p times) = (10^p - 1) / 9 = 10^(p-1) + 10^(p-2) + ....... + 10 + 1

Since, RHS is not divisible by 2 or 5, so q isn't 2 or 5. Hence, gcd(q,10)=1.

Further, p isn't 3 as 111 is not 1 (mod 6); this implies 111..11(p times) is not a multiple of 3, and hence q is not 3.

Now, q divides (10^p - 1). But we know 10^(q-1) = 1 (mod q) (Fermat's Little Theorem).

So, the order of 10 mod q is a common factor of both p and (q-1). Since, only factors of p are 1 and p, so order of 10 mod q is either 1 or p.

Now, 10^1 = 1 ( mod q) implies q = 3 (which is disproved before)

So, order of 10 mod q = p, implies p divides (q-1).

So, we conclude that p divides (q-1). Since, q isn't 2, so, 2 divides (q-1) too.

Since, p and 2 are coprime, and as both divide (q-1),

therefore 2p divides (q-1), or in other words, q is 1 mod 2p.

[/hide]

******************

Sambit

Gyan wrote:Another slight  variation of the problem:

Prove that  (1111.....1       (  2003 times) base 16  is composite.

Hint if any one is interested

[hide] 16 is 4^2.

and 16^n -1  =  (4^n+1)(4^n-1)

and if  n>2, the  required number  (10^n-1)/15  will prduce a number >1, which is not prime .  [/hide]

gyan, that doesn't matter.

16^2004 = 1 mod 15, that solves the problem on it's own: we got (15k+1-1)/15 is always a natural number...

it doesn't even matter base what it is, if it's just sufficiently high. (so that k is not 0 or 1)

Huh????

(Are you sure you are seeing the problem correctly?)

it is not about   (16^n-1)  but about (16^n-1)/15 )

Point is   to prove  [ (16^n-1)/15], itself,    is  composite  as  long  as n>2

In other words, what the above hint showed was:
 the number ((16^n-1)/15)  is divisible by 17 (if n is even) or  ((4^n+1)/5 ) (if n is odd)  - note that  ((4^n+1)/5 is an iteger if n is odd)).

For other bases, one can not  prove similar results  for ALL n's...
(If base is a not a  perfect square  (eg 16) .. you may find  prime numbers of the  type 1111111.... Trick was to realize that 16 was a perfect square)

[ eg  in base 10, not only 11,  but  (10^23)/9    is also  prime)

Hope that helps.


If $ \displaystyle\sum_{n\equal{}1}^\infty 1/n^2 \equal{} s$. ($ s\equal{}\frac{\pi^2}{6}$), prove that
$ \displaystyle\sum_{n\equal{}1}^\infty (\minus{}1)^{n\minus{}1}/n^2 \equal{}\frac{1}{2} s$

$ \displaystyle\sum_{n\equal{}1}^\infty (\minus{}1)^{n\minus{}1}/n^2 \equal{} \sum_{n\equal{}1}^\infty 1/n^2 \minus{} 2\sum_{n\equal{}1}^\infty 1/(2n)^2  \equal{} \sum_{n\equal{}1}^\infty 1/n^2 \minus{} \frac{1}{2}\sum_{n\equal{}1}^\infty 1/n^2 \equal{} s \minus{}\frac{1}{2} s \equal{} s/2$


[quote="pohoatza"]Let $a,b,c$ be sides of a triangle. Prove the following inequality:

$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq 6(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$[/quote]
Nice! :lol:
It's not true for all $a>0,$ $b>0$ and $c>0.$ :wink:

It works when $a=b=c=1$
the sign should be opposite


The cache seems to have corrupted again, or whatever the cause was.  :o

[quote="Temperal"]It's gone now, but it was there nevertheless.[/quote]We usually fix it when we see it, but sometimes none of us (me or Richard) are online so it takes a while to get fixed. Nevertheless please report it every time you see it.

I didn't see it at the time of the first post, but I do now.

[quote="Valentin Vornicu"][quote="i_like_pie"]OH GOD, IT'S BACK.

Can't you permanently do something about it? It's quite annoying not being able to use the "view new posts" button.[/quote]So it appears when you guys click on the view new posts / ego search buttons? Might be a clue as where the bug starts occurring if that is the case.[/quote]

Yes; it's a blank page with nothing but it on it, and then when we return to Mathlinks, then we find we're logged out.

Back again...
(just opened http://www.mathlinks.ro/Forum/index.php )

I see it too. AoPS skin, Firefox 2, first noticed at 3:21 PM EST.

edit: by the way, it appears at the top of the Topic Review when I'm posting replies, as well.

I've managed to trace back the problem to a single point, however it's still not clear what is causing it. A chain of events that I have yet been able to reproduce. Keep reporting when you see it.

[quote="Xevarion"]
edit: by the way, it appears at the top of the Topic Review when I'm posting replies, as well.[/quote]

This is because topic review is just an iframe of another page on the AoPS forums server. To be specific, posting.php?mode=topicreview&t=TOPIC_NUMBER


function
topology
real analysis
real analysis unsolved

Let $U$ be an open subset of $C$. Prove that there exists a closed discrete subset $S$ of $U$ whose closure in $C$ covers the frontier of $U$. From here it is easy to see that there is an analytic map on $U$ which cannot be extended in a larger open set. Is there any other approach?

Consider all sets $E\subset U$ with the following property: $|x-y|\ge \max\{\mathrm{dist}(x,\partial U),\mathrm{dist}(y,\partial U)\}$ for all $x,y\in E$. Order them by inclusion. Let $S$ be any maximal element with respect to this order.

I don't know of any other approach. We need an analytic function on $U$ which behaves in some "non-analytic" way near every point of the boundary. Accumulation of singularities of some kind (like zeros) is the most natural way to achieve this.

[quote="harazi"] From here it is easy to see that there is an analytic map on $U$ which cannot be extended in a larger open set. Is there any other approach?[/quote] Actually, there are a few alternative approaches but most of them require more effort. For instance, you can consider a dense countable set $\{z_{j}\}$ in $\mathbb C\setminus U$ and take $f(z)=\sum_{j}\frac{c_{j}}{z-z_{j}}$. This, obviously, will be a nice analytic function in $U$, provided that $\sum_{j}|c_{j}|<+\infty$. What is much less obvious is that you can always choose the coefficients $c_{k}$ in such a way that $f$ will not be extendable to any larger open set. Can you see how to do it or why not every choice of $c_{j}\ne 0$ with $\sum_{j}|c_{j}|<+\infty$ will work? ;)

I didn't make any progress with the construction of such a sequence, even though I guess that $c_{n}=1/5^{n}$ should work. Can you give some details, Fedja? I tried to use Landau's trick for series with nonnegative coefficients, but it didn't work.

Here are two possible approaches that you may want to explore:
a) Random choice (the idea is that, for fixed $c_{k}$ with $k\ne j$, there are not too many values of $c_{j}$ such that your function is analytically extendable to an open set containing $c_{j}$)
b) Very fast decaying coefficients (the idea is that the sum of the series will belong to some quasi-analytic class on almost every line and, therefore, if an analytic expansion exists, it must coincide with the sum of the series almost everywhere where it is defined)
I believe that the idea of taking all coefficients positive should work too but I do not see right away how to proceed with it. If I come up with something in this line, I'll post it.


i am too much confused betwwen fundamental principal of counting ,permutation and combination.WHAT IS MAIN DIFFERENCCE BETWWEN THEM.
pl explain a logical way.

In a permutation, order matters. In a combination, order doesn't matter. For example, $ _5C_2$ is different from $ _5P_2.$ The first is $ 10$ while the second  is $ 20.$


Is there anyway to determine when sqrt of a polynomial will give you a perfect square?
For example, for what values does $\sqrt{32x^2+16x+1}$ give an integer?
By trial and error the largest I could find was $x=7$

$x=42$ gives $\sqrt{32x^2+16x+1}=239$ also, according to my calculator :P

Anyway, I've seen a couple of methods to approach a problem like this, but I don't think there's any method that works for any polynomial...

I'll show you one way that is often useful (but it doesn't work for this probem.)
[hide]
[b]Problem[/b]: Find all positive integers $n$ such that $n^2+n+34$ is a perfect square.

[b]Solution[/b]:
Let $n^2+n+34=(n+k)^2$.
Solving for $n$, we obtain $n=\frac{34-k^2}{2k-1}$. ($n^2$ is gone! :) )

Now the problem becomes an integer problem!

Since $n$ has to be positive, $n=\frac{34-k^2}{2k-1}>0$
For positive values of $k$, $2k-1>0$ and thus $34-k^2>0$
So $k=1,2,3,4,5$

By trial and error, we see $k=1,2,3,5$ work, and these give us $n=1,5,10,33$

This method is only useful when the coefficient of $n^2$ is a perfect square. Also, if $n$ can be negative, it's often hard to restrict the possible values of $n$ and $k$.
[/hide]


porve Goldbach's theorem:
l=1/3+1/7+1/8+1/15+1/24+1/26+1/31+1/35+...= \sum  1/k-1
k  \in p
where P is the set of "perfect power" defined recursively as follows:
p={m   ^3   |m \geq 2,  ^3   \geq   ^3 }
m is not in p.

I think, author meant following:

Prove Goldbach's equality
\[1=\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\frac{1}{31}+\frac{1}{36}+...=\sum_{k\in PP}\frac{1}{k-1},\]
where $PP$ is a set of perfect powers $\{a^b\mid a,b\geq 2\}$.

Ok! Since problem was unsolved for 9 months, I will write solution.

Suppose $k\in PP$, then
\[\frac{1}{k-1}=\frac{1}{k}+\frac{1}{k^2}+\frac{1}{k^3}+...,\]
so
\[\sum_{k\in PP}\frac{1}{k-1}=\sum_{k\in PP}\left( \frac{1}{k}+\frac{1}{k^2}+\frac{1}{k^3}+...\right).\]
Note that summand $\frac{1}{a}$ appears in this sum iff $a\in PP$, moreover we will meet $\frac{1}{a}$ so much times how much times $a$ can be represented as a perfect power. Fox example, $\frac{1}{16}$ appears in sum twice, since $16=2^4=4^2$.
It follows that whole sum is
\[\sum_{n=2}^{\infty}\left(\frac{1}{n^2}+\frac{1}{n^3}+...\right)=\sum_{n=2}^{\infty}\frac{1}{n(n-1)}=1.\]
QED

fantastic Myth,now i read the proof it is beautiful :)


$\bigstar$
Let $a,b,c$ be non-negative real numbers with sum $3$. Prove that
\[a^{2}+b^{2}+c^{2}-\frac{8}{3}\max\{|a-b|,|b-c|,|c-a|\}\le 1+2abc.\]

Yes, you are really clever to realize the case of the equalities. However, my inequality can be proved with a very nice solution.

[quote="toanhocmuonmau"]This is a very tight inequality, Hung. Equality occur for $a=b=c=1$ and $a=3,b=c=0$, we can prove it by the same method with the following
Problem (Vo Quoc Ba Can).
\[\sum\sqrt{a+\frac{(b-c)^{2}}{4}}\le \sqrt{3}+\left( \frac{2}{\sqrt{3}}-1\right) (|a-b|+|b-c|+|c-a|) \]
if $a,b,c \ge 0$ and $a+b+c=1$.
:)[/quote]

How do you prove it? Can? I think it is mixing variable?


function
limit
real analysis
real analysis unsolved

Prove that if $ f: \mathbb{R} - > \mathbb{R}$ is non-decreasing function, continuous from the right, and if $ x > 0$ it's $ f(x) < x$, then for every ${ x > 0 \lim_{n\to\infty}\ {f^{n}(x)}} = 0$
where we define $ f^{n+1}$ as $ f(f^{n})$

What about $ f(x)\equal{}x\minus{}1$?

But if you assume in addition $ f(0)\equal{}0$, then the proof goes like this: any decreasing sequence of positive numbers has a limit; the limit $ L$ satisfies $ f(L)\equal{}L$ by right-continuity; hence $ L\equal{}0$.


Let a,b,c>0. Prove that:
$ \sum\frac{x\plus{}y}{\sqrt{2z}}\geq \sum \sqrt{x\plus{}y}$

We have
$ \sum\frac{x+y}{\sqrt{2z}}\geq \frac{(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})^2}{\sqrt{2a}+\sqrt{2b}+\sqrt{2c}}$
and we should to prove
$ \sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})\geq \sqrt{2a}+\sqrt{2b}+\sqrt{2c}$
 to finish solution.
So we have to prove:
   $ (\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})^2\geq (\sqrt{2a}+\sqrt{2b}+\sqrt{2c})^2$
or
 $ \sqrt{(a+b)(a+c)}+\sqrt{(b+c)(a+c)}+\sqrt{(a+b)(b+c)}\geq 2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}$
 which is right, becouse
  ${ \sqrt{(a+b)(a+c)}\geq 2\sqrt{ab}}+2\sqrt{ac}$
  ${ \sqrt{(b+c)(a+c)}\geq 2\sqrt{bc}}+2\sqrt{ac}$
  ${ \sqrt{(a+b)(b+c)}\geq 2\sqrt{ab}}+2\sqrt{bc}$
is right.


Let $p\ge 7$ be a prime number and $a\neq 0, b, c$ are natural numbers from $1$ to $9$. Prove that the number $\overline{\underbrace{aa\cdot\cdot\cdot a}_{p-1}\underbrace{bb\cdot\cdot\cdot b}_{p-1}\underbrace{cc\cdot\cdot\cdot c}_{p-1}}$ is devisible by $p\cdot\overline{a\underbrace{00\cdot\cdot\cdot 0}_{p-2}b\underbrace{00\cdot\cdot\cdot 0}_{p-2}}$.

huh? if 1<=c<=9, how can this happen? The first number is not even divisible by 10...huh???


inequalities
geometry
incenter
circumcircle
rectangle
trapezoid
3D geometry

Does somebody have any JBMO shortlists from past years? If somebody has any, please share them.

Yes, I want also to see the Shortlist pf JBMO 2007

If nobody has got any shortlists, can anyone tell me which books are good for prepairing for the JBMO and where can i find them?

[size=150][color=darkblue]Shortlist[/color] [color=darkred][b]JBMO 2002[/b][/color][/size]

[color=darkred][b]1.[/b][/color] [color=darkblue]A student is playing computer. Computer shows randomly $ 2002$ positive numbers. Game's rules let do the following operations
- to take $ 2$ numbers from these, to double first one, to add the second one and to save the sum.
- to take another $ 2$ numbers from the remainder numbers, to double the first one, to add the second one, to multiply this sum with previous and to save the result.
- to repeat this procedure, until all the $ 2002$ numbers won't be used.
Student wins the game if final product is maximum possible.
Find the winning strategy and prove it.[/color]

[b][color=darkred]2.[/color][/b] [color=darkblue]Positive real numbers are arranged in the form:
$ 1 \ \ \ 3 \ \ \ 6 \ \ \ 10 \ \ \ 15 \ \ \ ...$
$ 2 \ \ \ 5 \ \ \ 9 \ \ \ 14 \ \ \ ...$
$ 4 \ \ \ 8 \ \ 13 \ \ \ ...$
$ 7 \ \ 12 \ \ \ ...$
$ 11 \ \ \ ...$
Find the number of the line and column where the number $ 2002$ stays.[/color]

[b][color=darkred]3.[/color][/b]  [color=darkblue]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} \frac {9}{4}$. Prove the inequality:
\[ a^3 \plus{} b^3 \plus{} c^3 > a\sqrt {b \plus{} c} \plus{} b\sqrt {c \plus{} a} \plus{} c\sqrt {a \plus{} b}
\]
[/color]

[b][color=darkred]4.[/color][/b] [color=darkblue](Jury's variant) Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 2$. Prove the inequality:
\[ a^3 \plus{} b^3 \plus{} c^3 > a\sqrt {b \plus{} c} \plus{} b\sqrt {c \plus{} a} \plus{} c\sqrt {a \plus{} b}
\]
[/color]

[b][color=darkred]5.[/color][/b] [color=darkblue]Let $ a,b,c$ be positive real numbers. Prove the inequality:
\[ \frac {a^3}{b^2} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^3}{a^2}\ge \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}
\]
[/color]

[b][color=darkred]6.[/color][/b]  [color=darkblue]Let $ a_1,a_2,...,a_6$ be real numbers such that:
\[ a_1 \not \equal{} 0, a_1a_6 \plus{} a_3 \plus{} a_4 \equal{} 2a_2a_5 \ \mathrm{and}\ a_1a_3 \ge a_2^2
\]
Prove that $ a_4a_6\le a_5^2$. When does equality holds?[/color]

[b][color=darkred]7.[/color][/b] [color=darkblue]Consider integers $ a_i, i \equal{} \overline{1,2002}$ such that
\[ a_1^{ \minus{} 3} \plus{} a_2^{ \minus{} 3} \plus{} \ldots \plus{} a_{2002}^{ \minus{} 3} \equal{} \frac {1}{2}
\]
[/color]

[b][color=darkred]8.[/color][/b] [color=darkblue]Let $ ABC$ be a triangle with centroid $ G$ and $ A_1,B_1,C_1$ midpoints of the sides $ BC,CA,AB$. A paralel through $ A_1$ to $ BB_1$ intersects $ B_1C_1$ at $ F$ prove that triangles $ ABC$ and $ FA_1A$ are similiar if and only if quadrilateral $ AB_1GC_1$ is cyclic.[/color]

[b][color=darkred]9.[/color][/b] [color=darkblue]In triangle $ ABC$ $ H,I,O$ are orthocenter, incenter and circumcenter, respectively. $ CI$ cuts circumcircle at $ L$. IF $ AB \equal{} IL$ and $ AH \equal{} OH$, find angles of triangle $ ABC$.[/color]

I'll finish later...

[quote="Ahiles"]I'll finish later...[/quote]

It's been a month past since you said this. Can you please finish?

[quote="Bugi"]
It's been a month past since you said this. Can you please finish?[/quote]
OK..
 [b][color=darkred]10.[/color][/b] [color=darkblue]Let $ ABC$ be a triangle with area $ S$ and points $ D,E,F$ points on the sides $ BC,CA,AB$. Perpendiculars at points $ D,E,F$ to the $ BC,CA,AB$ cut circumcircle of the triangle $ ABC$ at points $ (D_1,D_2), (E_1,E_2), (F_1,F_2)$. Prove that:
\[ |D_1B\cdot D_1C \minus{} D_2B\cdot D_2C| \plus{} |E_1A\cdot E_1C \minus{} E_2A\cdot E_2C| \plus{} |F_1B\cdot F_1A \minus{} F_2B\cdot F_2A| > 4S
\]
[/color] [b][color=darkred]11.[/color][/b] [color=darkblue]Let $ ABC$ be an isosceles triangle with $ AB \equal{} AC$ and $ \angle{A} \equal{} 20^\circ$. On the side $ AC$ consider point $ D$ such that $ AD \equal{} BC$. Find $ \angle{BDC}$.[/color]
 [b][color=darkred]12.[/color][/b] [color=darkblue]Let $ ABCD$ be a convex quadrilateral with $ AB \equal{} AD$ and $ BC \equal{} CD$. On the sides $ AB,BC,CD,CA$ we consider points $ K,L,L_1,K_1$ such that quadrilateral $ KLL_1K$ is rectangle. Then consider rectangles $ MNPQ$ inscribed in the triangle $ BLK$, where $ M\in KB, N \in BL, P,Q \in LK$ and $ M_1N_1P_1Q_1$ inscribed in triangle $ DK_1L_1$ where $ P_1$ and $ Q_1$ are situated on the $ L_1K_1$, $ M_1$ on the $ DK_1$ and $ N_1$ on the $ DL_1$. Let $ S,S_1,S_2,S_3$ be the areas of the $ ABCD,KLL_1K_1,MNPQ,M_1N_1P_1Q_1$ respectively. Find the maximum possible valu of the expression:
\[ \frac {S_1 \plus{} S_2 \plus{} S_3}{S}
\]
[/color] [b][color=darkred]13.[/color][/b] [color=darkblue]Let $ A_1,A_2,...A_{2002}$ be the arbitrary points in the plane. Prove that for every circle of the radius $ 1$ and for every rectangle inscribed in this circle, exist 3 vertexes $ M,N,P$ of the rectangle such that:
\[ MA_1 \plus{} ... \plus{} MA_{2002} \plus{} NA_1 \plus{} ... \plus{} NA_{2002} \plus{} PA_1 \plus{} ... \plus{} PA_{2002}\ge 6006
\]
[/color]

That's all... :)

Problem 9 has appeared at a Brazilian's MO second round.  :P

Dear Ahiles,

What is the question for problem 7?

Is it JBMO Shortlist 2002 or 2007. BTW, could you plz post another jbmo shortlists ?

It's the shortlist for 2002, Ahiles wrote that. I have seen problem 7 and the question is: Prove that at least 3 of the numbers are equal.I would also like to see some more shortlists.

Okey thank you, and here is PDF file of JBMO 2002 :).

[size=150][color=darkred]JBMO Shortlist 2006[/color][/size]

 [b][color=darkred]1.[/color][/b] [color=darkblue] For an acute triangle $ ABC$ prove the inequality
\[ \frac {m_a^2}{ - a^2 + b^2 + c^2} + \frac {m_b^2}{a^2 - b^2 + c^2} + \frac {m_c^2}{a^2 + b^2 - c^2}\ge \frac {9}{4}
\]
where $ m_a,m_b,m_c$ represent the lengths of the corresponding medians.
[/color]

 [b][color=darkred]2.[/color][/b] [color=darkblue] Let $ x,y,z$ be positive real numbers such that $ x + 2y + 3z = \frac {11}{12}$. Prove the inequality
\[ 6(3xy + 4xz + 2yz) + 6x + 3y + 4z + 72xyz\le \frac {107}{18}
\]
When does equality occur?
[/color]

 [b][color=darkred]3.[/color][/b] [color=darkblue] Let $ n\ge3$ be a natural number. A set of real numbers $ \{x_1,x_2,...x_n\}$ is called $ summable$ if
\[ \frac {1}{x_1} + \frac {1}{x_2} + ... + \frac {1}{x_n} = 1.
\]
Prove that for every $ n\ge 3$ there always exists a $ summable$ set, which consits of $ n$ elements, such that the biggest element is

$ a)$ bigger that $ 2{}^2{}^{n - 2}$;
$ b)$ smaller that $ n^2$.
[/color]

[b][color=darkred]4.[/color][/b] [color=darkblue] Determine the biggest possible value of $ m$ for which equation $ 2005x + 2007y = m$ has unique solution in matural numbers.
[/color]

[b][color=darkred]5.[/color][/b] [color=darkblue] Determine all pairs $ (m,n)$ of real numbers for which $ m^2 = nk + 2$, where $ k = \overline{n1}$.
[/color]

[b][color=darkred]6.[/color][/b] [color=darkblue] Prove that for every composite number $ n > 4$, numbers $ kn$ divides $ (n - 1)!$ for every integer $ k$, such that $ 1\le k \le [\sqrt {n - 1}]$.
[/color]

[b][color=darkred]7.[/color][/b] [color=darkblue] Determine all numbers $ \overline{abcd}$, such that $ \overline{abcd} = 11(a + b + c + d)^2$.
[/color]

[b][color=darkred]8.[/color][/b] [color=darkblue] Prove that there do not exist natural numbers $ n\ge 10$, such that every $ n$'s digit is non-zero, and all numbers which are obtained by permutating of its digits are perfect squares
[/color]

[b][color=darkred]9.[/color][/b] [color=darkblue] Let $ ABCD$ be a trapezoid with $ AB\parallel CD, AB > CD$ and $ \angle{A} + \angle{B} = 90^\circ$. Prove that the distance between the midpoints of the bases is equal to the semidifference of the bases.[/color]

[b][color=darkred]10.[/color][/b] [color=darkblue] Let $ ABCD$ be an isosceles trapezoid inscribed in a circle $ \mathcal{C}$ with $ AB\parallel CD, AB = 2CD$. Let $ Q \in AD \cap BC$ and let $ P$ be intersection of the tangents to $ \mathcal{C}$ at $ B$ and $ D$. Calculate the area of the quadrilateral $ ABPQ$ in terms of the area of the triangle $ PDQ$.
[/color]

[b][color=darkred]10 (bis).[/color][/b] [color=darkblue] Let $ ABCD$ be atrapezoid inscribed in a circle $ \mathcal{C}$ of diameter $ AB$ with $ AB\parallel CD, AB = 2CD$. Let $ Q \in AD \cap BC$ and let $ P$ be intersection of the tangents to $ \mathcal{C}$ at $ B$ and $ D$. Calculate the area of the quadrilateral $ ABPQ$ in terms of the area of the triangle $ PDQ$.
[/color]

[b][color=darkred]11.[/color][/b] [color=darkblue] Circles $ \mathcal{C}_1$ and $ \mathcal{C}_2$ intersect at $ A$ and $ B$. Let $ M \in AB$. A line through $ M$ (different form $ AB$) cuts circles $ \mathcal{C}_1$ and $ \mathcal{C}_2$ in $ Z,D$ and $ E,C$ respectively, such that $ D,E \in ZC$. Perpendiculars at $ B$ to the lines $ EB$, $ ZB$ and $ AD$ respectively, cut circle $ \mathcal{C}_2$ at points $ F,K$ and $ N$ respectively. Prove that $ KF = NC$.[/color]

[b][color=darkred]12.[/color] [/b] [color=darkblue] Let $ ABC$ be  an equilateral triangle of center $ O$, and $ M \in BC$. Let $ K,L$ be proiections of $ M$ onto the sides $ AB$ and $ AC$ respectively. Prove that line $ OM$ passes through the midpoint of the segment $ KL$.
.[/color]

[b][color=darkred]13.[/color][/b] [color=darkblue] Let $ A$ be a subset of the set $ \{1,2,3,..,2006\}$, consisting of $ 1004$ elements. Prove that exists three distinct numbers $ a,b,c \in A$ such that $ gcd(a,b)$
$ a)$ divides $ c$.
$ b)$ doesn't divide $ c$.
[/color]

[b][color=darkred]14.[/color][/b] [color=darkblue] Let $ n\ge 5$ be a positive integer. Prove that the set $ \{1,2,3,...,n\}$ can not be partitioned into two non-zero subsets $ S_n$ and $ P_n$ such that the sum of the elements of $ S_n$ is equal to the product of the elements of $ P_n$.
[/color]

Thank you!
How did you find this (if it's internet please tell us!)?
Do you have 2007 shortlist?
And what's 10. bis, and are you sure that you've written 9. well?
I've made it in PDF:

[quote="Bugi"][quote="Ahiles"]I've got a book by Dan Branzei. But it is in romanian. There are shortlist 2001-2006. When I have time, I will post others.[/quote]

Can you tell us what's in that book (just shortlists or more)?[/quote]

Well, I have this book!
 It contains :
 [b]a) [/b]The problems for all JBMO 1997-2006
 [b]b)[/b] The SL - problems from 2000-2006
[b] c)[/b] The tst for Romany
[b] d) [/b]Solutions to these problems !

 A co-autor is a friend of me in Romany and he has sent it to me.
 I have the SL - 2006 in pdf format. But please find SL 2007 (and 2008, if possible) and we 'll give the list to each other with e mails.
  Babis Sterigou - Greece

SL 2008 is revealed at next JBMO, because participating countries can take these problems to the competitions in those countries (JBkTST-s).
Ahiles, can you tell us the full name of the book?

It's called "10 ani de Olimpiade Balcanice ale Juniorilor" and authors are Dan Brânzei, Dinu Șerbănescu, Gabriel Popa, Ioan Șerdean, Dan Ștefan Marinescu, Vasile Șerdean.
Pitești : Paralela 45, 2007.

Thanks. I've posted the most of these problems in Olympiad section, but are you sure about 9. and 14? In 14, someone proved that it always can be partitioned.
And, can you write the name of that book in English (exact please)?

[quote="Ahiles"]
[b][color=darkred]14.[/color][/b] [color=darkblue] Let $ n\ge 5$ be a positive integer. Prove that the set $ \{1,2,3,...,n\}$ can not be partitioned into two non-zero subsets $ S_n$ and $ P_n$ such that the sum of the elements of $ S_n$ is equal to the product of the elements of $ P_n$.
[/color][/quote]
For future reference, it's actually the other way around. See http://www.mathlinks.ro/viewtopic.php?t=238634.

Yep, I did a mistake there, but I can't edit it already.

Can you write the name of that Romanian book in English?

[size=150][color=darkred]JBMO Shortlist 2000[/color][/size]

 [b][color=darkred]1.[/color][/b] [color=darkblue] Let $ M$ be the set of positive integers which consist of $ 2006$ digits, one of which is $ 7$, but others are $ 1$. Prove that there are at least 666 composite numbers in $ M$.
[/color]

 [b][color=darkred]2.[/color][/b] [color=darkblue] Find all natural numbers, not divisible by 10, which are perfect cubes, and after deleting the last three digits, they are also nonzero perfect cubes.
[/color]

 [b][color=darkred]3.[/color][/b] [color=darkblue]  Find all numbers which are written as $ \overline{abcd}$ in the base $ 10$ and $ \overline{dcba}$ in the base 7.
[/color]

[b][color=darkred]4.[/color][/b] [color=darkblue] Find all pairs of natural numbers $ (m,n)$, for which numbers
$ A=3m^2+2mn+n^2+3n$
$ B=m^2+3mn+2n^2$
$ C=2m^2+mn+3n^2$
are consecutive in some order.
[/color]

[b][color=darkred]5.[/color][/b] [color=darkblue] Find all numbers $ n=\overline{abcd}$, $ n=\prod_{i=1}^k p_i^{\alpha_i}$, such that $ \sum p_i=\sum \alpha_i$.
[/color]

[b][color=darkred]6.[/color][/b] [color=darkblue] Find all positive integers $ a$ and $ b$, for which $ a^4+4b^4$ is prime.
[/color]

[b][color=darkred]7.[/color][/b] [color=darkblue] Solve in $ \mathbb{N}^3$ the equation
  \[ xy+yz+zy-xyz=2 \][/color]
[b][color=darkred]8.[/color][/b] [color=darkblue] Prove that there are no $ x,y,z\in \mathbb{Z}$ which satisfy the equation
\[ x^4+y^4+z^4-2x^2y^2-2y^2z^2-2y^2z^2=2000\][/color]
[b][color=darkred]9.[/color][/b] [color=darkblue]Prove that for every $ n \in \mathbb{Z}$ we can find $ a,b \in \mathbb{Z}$, such that
\[ n=[a\sqrt{2}]+[b\sqrt{3}]\][/color]

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

[b][color=darkred]11.[/color][/b] [color=darkblue] Let $ m,n \in \mathbb{N}$, $ m\le 2000$, and $ k=3-\dfrac{m}{n}$. Determine the minimum possible value for $ k$[/color]

[b][color=darkred]12.[/color] [/b] [color=darkblue] Let $ x,y,a,b\in \mathbb{R}_+$, $ x \not= y$, $ x \not = 2y$, $ y \not =2x$, $ a \not= 3b$ and $ \dfrac{2x-y}{2y-x}=\dfrac{a+3b}{a-3b}$. Prove that $ \dfrac{x^2+y^2}{x^2-y^2}\ge 1$[/color]

[b][color=darkred]13.[/color][/b] [color=darkblue] Find all $ (x,y,z) \in R^3$, for which
\[ 2x\sqrt{y-1}+2y\sqrt{z-1}+2z\sqrt{x-1}\ge xy+xz+yz\][/color]
[b][color=darkred]14.[/color][/b] [color=darkblue] Let $ ABC$ be a triangle. Find all segments $ XY$, which are situated in the interior of the triangle, such that $ XY$ and five from the segments $ XA,XB,XC$, $ YA,$ $ YB$, $ YC$ divide area of $ ABC$ in five equivalent parts. Prove that all found segments are concurrent.
[/color]

[b][color=darkred]14.[/color][/b] [color=darkblue] Let $ ABC$ be a triangle and $ a,b,c$ lengths of the corresponding sides. Consider another triangle $ DEF$ with sides $ EF=\sqrt{au}$, $ DF=\sqrt{bu}$, $ DE=\sqrt{cu}$. Denote by $ A,B,C,D,E,F$ measures of the angles of this triangles. Suppose that $ A>B>C$. Prove that $ A>D>E>F>C$.
[/color]

[b][color=darkred]15.[/color][/b] [color=darkblue] Let $ ABCDE$ be a hexagon with all angles equal. Prove that $ AB-DE=EF-BC=CD-FA$.
[/color]

[b][color=darkred]15.[/color][/b] [color=darkblue] Let $ ABCD$ be a quadrilateral with $ \angle{DAB}=60^\circ$, $ \angle{ABC}=90^\circ$, $ \angle{BCD}=120^\circ$. Diagonals of the quadrilateral intersect at $ M$, such that $ MB=1$, $ MD=2$. Calculate tha area of $ ABCD$.
[/color]

You have to post each problem in its own thread, otherwise this is not very useful to us.

Could someone please post the solution threads to these problems (if any)? Thanks!

I posted 2002 and 2006 a long time ago, Ahiles posted 2000. Search for ''JBMO Shortlist''...

[url=http://www.artofproblemsolving.com/community/c6h502939p2825631]JBMO Shortlist 2010.pdf (119kb)[/url]
[url=http://www.artofproblemsolving.com/community/c3240_jbmo_shortlists]JBMO Shortlist[/url]


find all integer n and k satisfaies

 $n|k^2+1 ,n^2|k^3+1 ,n^3|k^4+1$

Let $n \neq \pm 1$ and $p \in \mathfrak{P}$ such that $p \mid n$. It has to be $k^3 + 1 \equiv k^2 + 1 \equiv 0 \bmod p$. It follows $k^2(k-1) \equiv -k(k+1) \equiv 0 \bmod p$, and subsequently $k - 1 \equiv k + 1 \equiv 0 \bmod p$, since obviously $\gcd(k,p) = 1$. So $p = 2$, $k \equiv 1 \bmod 2$ and $n = 2^t$, for a suitable $t \in \mathbb{Z}^+$. Anyway $k^2 + 1 \equiv 2 \bmod 4$, if $k \equiv 1 \bmod 2$. Therefore $n = 2$ and you need $k^4 + 1 \equiv 0 \bmod 8$, which is impossibile, because $k^4 +1 \equiv 2 \bmod 4$, if $k \equiv 1 \bmod 2$. So necessarily $n = \pm 1$ and any $k\in\mathbb{Z}$ sounds good.


algebra
polynomial
quadratics
algebra unsolved

Let p(x) be the polynomial $x^3 + 14x^2 - 2x + 1$. Let $p^n(x)$ denote $p(p^(n-1)(x))$. Show that there is an integer N such that $p^N(x) - x$ is divisible by 101 for all integers x.

Cute...isn't it ;)
First of all $101$ is a prime number. This will be very,very usefull. Now for every polynomial $Q$ in $Z[x]$ ,value $Q(x) \; mod \; 101$ depends only on residue of $x$ modulo $101$ thus we have to prove our statement only for $0\leq x \leq 100$.
All the time we will operate on $Z_{101}$ so every equality will be taken modulo $101$
 It's easy  to see that all we have to prove now is that \[ p(0),p(1),...,p(100) \] is a permutation of numbers \[ 0,1,2,3,...100 \] With it, our statement will be obvious, cos' then for every $n$ \[ p^n(0),p^n(1),...,p^n(100) \]  will be such permutation. As the number of permutation is finite, the same permutation must appear again somewhere. So there exist $n_1<n_2$ such that for every $0\leq x \leq 100$ : $p^{n_1}(x)=p^{n_2}(x)$ but as the set of $p^{n_1}(x)$ is our permutation, we can permute our equalities such that for every $0 \leq x \leq 100$: \[ p(x)=x \].

Now all we have to do is to prove that $p(0),p(1),...,p(100)$ is a permutation of $0,1,2,3,..,100$. Well assume that there exist two numbers $x\neq y$ such that $p(x)=p(y)$ This means that \[ (x-y)(x^2+y^2+xy+14x+14y-2)=0 \] and as $x-y\neq 0$ we obtain \[ x^2+y^2+xy+14x+14y-2=0 \] Let's write it in a quadratic equation form: \[ x^2+x(y+14)+(y^2+14y-2) \] $x$ has to be root of this equation in $Z_{101}$. Operating on $Z_{101}$ we may solve this equation in a standard way by computing $\delta$. If we have a root, then $\sqrt{\delta}$ must exist in $Z_{101}$. Let's compute: \[ \delta=(y+14)^2-4(y^2+14y-2)=-3y^2-28y+14^2+8=-3y^2-28y+2 \] so there must exist $k$ for which \[ k^2=-3y^2-28y+2 \] One more time we get quadratic equation \[ -3y^2-28y+(2-k^2)=0 \] and again we may compute $\delta$ and say that there must exist $\sqrt{\delta}$, which means that there exists $t$ such that \[ t^2=28^2+12(2-k^2)=-12k^2 \], so finally we've obtained that there exist two numbers $t$ and $k$ for which \[ t^2=-12k^2 \] but this means that $-12$ is a quadratic residue modulo $101$ which is contradiction cos' \[ (-12)^{50}=-1 \] and thus $-12$ is not quadratic residue modulo $101$. We got a contradiction, which shows that our statement is true .

Well, I have another good manner to show that -12 is not a QR modulo 101. Well, the idea of computing $(-12)^{50}$ is not bad but there exists another manner to do it :P

Well,
$\left(\frac{-12}{101}\right)=\left(\frac{-3}{101}\right)\left(\frac{4}{101}\right)=\left(\frac{-3}{101}\right)=(-1)^{\frac{(-3-1)(101-1)}{4}}\left(\frac{101}{3}\right)=\left(\frac{2}{3}\right)$
by Quadratic Reciprocity. And this is the end :P

[b][color=#f00]Claim 1:-[/color][/b] $P(x)$ is bijective in the interval $[1,2,3,.......,100] \pmod {101}$
[b][color=#f00]Proof:-[/color][/b] It suffices to show $P(x)$ is injective so assume that $P(x)=P(y)$ for some $x \neq y$
This means that $\[ (x-y)(x^2+y^2+xy+14x+14y-2)=0 \] $ and as $x-y\neq 0$ we obtain $\[ x^2+y^2+xy+14x+14y-2=0 \]$
It can be rearranged to $\[ x^2+x(y+14)+(y^2+14y-2) \]$
It's discriminant $\alpha=(y+14)^2-4(y^2+14y-2)=-3y^2-28y+14^2+8=-3y^2-28y+2$ so there must exist $k$ for which $\[ k^2=-3y^2-28y+2 \]$
Again $\[ -3y^2-28y+(2-k^2)=0 \]$ whose discriminant is $\beta=28^2+12(2-k^2)=-12k^2 \]$
Or $X^2=-12k^2$,which isn't possible since $\left(\frac{-12}{101}\right)=\left(\frac{-3}{101}\right)\left(\frac{4}{101}\right)=\left(\frac{-3}{101}\right)=(-1)^{\frac{(-3-1)(101-1)}{4}}\left(\frac{101}{3}\right)=\left(\frac{2}{3}\right)=-1$ which isn't a quadratic residue,a contradiction.
Now since $P(x)$ is bijective we can find $\mathbb{X}$ such that $P(\mathbb{X}) \equiv \mathbb{X} \pmod {101}$,as required.

This is one of those problems where the numbers are kind of contrived to work. Let $p=101$.

Define $f(i) = P(i) \bmod p$ for each $0 \leq i \leq p-1$. It is equivalent to show that $f$ is bijective.

Notice if $f(i) = f(j)$ then $$p \mid i^2+ij+j^2+14(i+j) - 2.$$ Pick an $\alpha$ with $3\alpha \equiv 14 \pmod p$ (to complete the square), wherence $$(i+\alpha^2)+(i+\alpha)(j+\alpha)^2 = 3\alpha^2 + 2 \equiv 0 \pmod p.$$ But clearly $p \equiv 2 \pmod 3$, hence $p \mid i-j$, contradiction.


The figure below is composed of congruent squares. How many rectangles can be traced in the figure?

[asy]path A = (0,0)--(3,0)--(3,1)--(0,1)--(0,0)--(1,0)--(1,1)--(2,1)--(2,0);
draw(shift((1,-1))*A); draw(shift((2,-2))*A); draw(A);[/asy]

what are the ___  rectangles?  I keep on getting 24.  I won't say how many there are without your permision.


If $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}3$, then

$ \frac{a\minus{}1}{a^2\plus{}2bc}\plus{}\frac{b\minus{}1}{b^2\plus{}2ca}\plus{}\frac{c\minus{}1}{c^2\plus{}2ab}\leq\frac{1}{ab\plus{}bc\plus{}ca}$

\[ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab}\geq \frac {2}{ab \plus{} bc \plus{} ca}\]
it's enough to show
\[ \frac {a}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab}\leq \frac {3}{ab \plus{} bc \plus{} ca} \equal{} \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}\]

\[ \Longleftrightarrow \frac {a}{a^2 \plus{} 2bc}(a \minus{} b)(a \minus{} c) \plus{} \frac {b}{b^2 \plus{} 2ca}(b \minus{} c)(b \minus{} a) \plus{} \frac {c}{c^2 \plus{} 2ca}(c \minus{} a)(c \minus{} b)\geq 0\]
which is true by V-V Schur. :lol:


$AB$ is a fixed diameter of a circle whose center is $O$.  From $C$, any point on the circle, a chord $CD$ is drawn perpendicular to $AB$.  Then, as $C$ moves over a semicircle, the bisector of angle $OCD$ cuts the circle in a point that always:
\[ \text{(A) bisects the arc}\ AB \qquad \text{(B) trisects the arc}\ AB \qquad \text{(C) varies} \]
\[{ \text{(D) is as far from}\ AB\ \text{as from}\ D} \qquad \text{(E) is equidistant from}\ B\ \text{and}\ C \]

[hide="Answer"] (A) bisects the arc AB.[/hide]

[hide="Justification"]
For any segment $CD$, extend the segment so that it intersects $AB$ at $Q$. Let $\angle QCO = y$ and $\angle QOC = x$.

Let the bisector cut the circle at point $P$.

The external angle $\angle DCO = x + 90$. Thus, $\angle DCP = \frac{x+90}{2}$. Since $CO$ and $PO$ are radii, $\triangle CPO$ is isosceles. So, the angles $\angle OCP$ and $\angle OPC$ must sum to $x + 90$ and $\angle COP=y$.

Here, we see that $\angle AOP = x+y = 90^o$. Thus, the bisector must bisect arc $AB$.

 :lol:

[/hide]


[quote="#H34N1"]What is the area in square centimeters of the shaded region below?

[img]7208[/img][/quote]
[hide]So call the square thats 3x3 ABCD starting from the top left. Then call the little square DEFG starting form the the bottom left and going to the counter clockwise. So we have a 3-4-5 triangle BCE. We want to find the the area of quadrilateral BCDG. To do this we need to subtract the area of triangle DGE from triangle BCE. Triangle DGE is similiar to triangle CBE. And the ratio of the sides are 1:4. So the side length of DG will be 3/4. SO the area of triangle DGE is 3/8. Subtracting we get that the area of quafrilateral BCDG is 45/8. Then subtracting this from the area of the square which is 9 we get that the area of the shaded region is 27/8.[/hide]
Hope this isn't too confusing.

[hide]Using similar triangles, it is easily found that the length of the segment shared by the two squares (minus the part which touches the shaded region) is $\frac34$. This means the second side of the shaded region is $\frac94$.

$\frac{3\cdot\frac94}2=\boxed{\frac{27}8}$[/hide]


[quote]What is the sum of the numbers that are two more than their reciprocals?[/quote][hide]
Let $x$ be such a number.

Then $x = \frac{1}{x} + 2$

Then $x$ is a root of the quadratic: $x^2 - 2x - 1 = 0$

For a quadratic equation: $x^2 + bx + c = 0$, the sum of the roots is $-b$.

Therefore, the sum is $2$.[/hide]

[quote="h_s_potter2002"][hide]$x = \frac{1}{x} + 2\\
x^2 = 1+2x\\
x^2-1-2x = 0
(x-2)(x+1) = 0\\
{x = 2} or {x = -1}$

The answer is $1$.[/hide][/quote]

You made a little mistake after $x^2-2x-1=0$. It wasn't factored correctly. Close, though.

[hide]What is the sum of the numbers that are two more than their reciprocals?

x=(1/x)+2
x2=2x+1
x2-2x-1=0

if ax2+bx+c,  the roots of this equation is -b[/hide]

[hide][b]therefore the solution is 2[/b][/hide]


trigonometry
Pythagorean Theorem
geometry
AMC 10
power of a point

Square $ ABCD$ has side length $ 2$. A semicircle with diameter $ \overline{AB}$ is constructed inside the square, and the tangent to the semicricle from $ C$ intersects side $ \overline{AD}$ at $ E$. What is the length of $ \overline{CE}$?

[asy]
defaultpen(linewidth(0.8));
pair A=origin, B=(1,0), C=(1,1), D=(0,1), X=tangent(C, (0.5,0), 0.5, 1), F=C+2*dir(C--X), E=intersectionpoint(C--F, A--D);
draw(C--D--A--B--C--E);
draw(Arc((0.5,0), 0.5, 0, 180));
pair point=(0.5,0.5);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));[/asy]

$ \textbf{(A)}\  \frac {2 \plus{} \sqrt5}{2} \qquad \textbf{(B)}\  \sqrt 5 \qquad \textbf{(C)}\  \sqrt 6 \qquad \textbf{(D)}\  \frac52 \qquad \textbf{(E)}\ 5 \minus{} \sqrt5$

[hide] Call the point of tangency of CE to the semicircle R.  By POP Theorem, CR=CB=2.  Also ER=EA=x.  By Pythagorean Theorem on triangle EDC, 2^2+(2-x)^2=(2+x)^2, so x=1/2, so CE=5/2. D. [/hide]

[hide="Here is an another solution."][b]Denote:[/b] the middlepoint $O$ of the side $AB,$

the tangent point $T\in CE\cap C(O,1)$, i.e. $TC=BC=2$

and $TE=x$, i.e. $CE=2+x$. Therefore,

$OT\perp CE,\ OC\perp OE\Longrightarrow OT^2=TC\cdot TE\Longrightarrow$

$x=\frac 12\Longrightarrow \boxed {\ CE=\frac 52\ }\ ,$ i.e. the answer is $\boxed {\ D\ }$.[/hide]

Probably not rigorous, but who cares.

Let O be the midpoint of AB
$ \angle BOC \plus{} \angle AOE \equal{} 90$
$ tan ^{ \minus{} 1} {2} \plus{} tan^{ \minus{} 1} {x} \equal{} 90$
$ \frac {2 \plus{} x}{1 \minus{} 2x} \equal{} tan 90$
Sine $ tan 90$ is undefined, we have $ 1 \minus{} 2x \equal{} 0$, so $ x \equal{} \frac {1}{2}$
Now Pythag finishes it.

[quote="mathemagician1729"]So simply put you first use pythagoras to get 2 and then power of a point to find the other part is 1/2. Is this from an AMC 10?[/quote]
Yes, specifically [url=http://www.artofproblemsolving.com/Wiki/index.php/2004_AMC_12A_Problems/Problem_18]here.[/url]

alternate solution:  let AE=x, then ED=2-2, CD=2, CE=(2+x) by the tangent to a circle theorem. Pythagorean theorem on CDE gives x=1/2, hence the answer is 5/2  (D).

I go like : let tangent point be T. let center of circle be O. Draw CO. Note that triangle CBO is congruent to triangle CTO by HL congruence(radius length). So CT = 2 and TO =1. Draw EO. See how OT is the altitude of this triangle and forms 3 similar triangles. TOE is COT scaled down by 1/2. Then find ET and add to get D

There is a much easier solution.
Mark the point of contact of the tangent with the semicircle at $X$
$BC=CX$(TANGENTS FROM AN EXTERNAL POINT).
$CX=2$
ALSO $EA=EX=x$
$DE=2-x$
$CE=2+x$
$DC=2$.Apply Pythagoras in $DCE$ to get $2+x$

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20V.pdf  p. 46...

Sincerely
Jean-Louis

Why did people post in this thread after 6 years?

[b]Solution:[/b] Let $\overline{CE}$ be tangent to the semicircle at $P$. Notice that $PC=CB=2$. Now, let $EA=EP=x$. This makes $DE=2-x$ and we already know that $DC=2$. Now we can do the Pythagorean theorem to get that $(2-x)^2 + 2^2 = (2+x)^2 \Rightarrow x = \frac{1}{2}$. This makes $CE = 2+x = \frac{5}{2} \Rightarrow \boxed{D}$.


Determine a closed expression for S_(m,n) with m,n \in N:

S_(m,n) = \sum^{n}_{k=0} ((k+1)^2*m^k).

I do not call that a closed expression.  :)

I think it is much shoerter than your "closed" expresion ;)

[quote="orl"]I would like to see a fraction involving polynomials. Probably I have a different idea of closed expression than you seem to have...  :D[/quote]

No, I don't have a different idea of closed expresion. I've got smth much more fearsome. I call it a laziness attack :D. I think that the battle may be long lost.
---------------------------------------------------------------------------------
Here is a shot:
[(n+1)2mn+3-(2n 2 +6n+3)mn+2+(n+2) 2 mn+1-m-1]/(m-1)3.
-------------------------------------------------------------------------------
Happy?
This was an awful waste of an A4 sheet.

Hey well done, amfulger.  :)  I am proud you got it. But please do not forget to consider the case

m = 1: 1/6((n+1)(n+2)(2n+3)).


you are probably familiar with the equation for the surface are of a closed right cylinder.

S=2 pi r^2 + 2 pi r h

I need to isolate r on one side of the equation. i have been staring at this for hours and just cant figure it out. please help me! i just want to move on to the next section in by book!!!!

[hide="Hint"]$ ax^2\plus{}bx\plus{}c\equal{}0 \Rightarrow x\equal{}\frac{\minus{}b\pm\sqrt{b^2\minus{}4ac}}{2a}$[/hide]

It's just a quadratic equation in $ r$. Solve it using the quadratic formula and disregard the negative root because $ r>0$.

i dont know the quadratic formula yet, havnt got that far...
why did my book give me a problem that it has not yet taught me the solution to yet....
perhaps i left out important information....
find r when S=208m^2 and h= 4m
can you solve this without the quadratic formula now?

i can isolate S.....
S=2pi r^2+2pi r h   ......
h= [u]S-2pi r^2[/u]
  ------ 2pi r

If $ \pi \approx \frac {13}{4}$...

$ S \equal{} 2\pi r^2 \plus{} 2\pi r h$
So, $ 208 \equal{} 2 \times \frac {13}{4} \times r^2 \plus{} 2 \times \frac {13}{4} \times r \times 4$
$ 208 \times {1}{2} \times \frac {13}{4} \equal{} r^2 \plus{} 4r$
$ 32 \equal{} r^2 \plus{} 4r$

So, $ 32 \plus{} 4 \equal{} r^2 \plus{} 4r \plus{} 4 \equal{} (r \plus{} 2)^2$  i.e.  $ (r \plus{} 2)^2 \equal{} 6^2$  i.e.  $ r \plus{} 2 \equal{} \pm6$
Thus, $ r \equal{} 6 \minus{} 2 \equal{} \boxed{4}$.  (Because $ r > 0$)

EDIT:I fixed a typo.

[quote="davisdandrew"]oh, ooooopppss!!! im so sorry, but i wrote it wrong ......
S is actually 208 Pi m^2
 i just cant figure out how to isolate r.......

and.... i though the fraction for pi was 22/7....
please forgive me for being such a newbie!!!!![/quote]

Ok. So similarly my post,

$ 208 \pi \equal{} 2 \pi r^2 \plus{} 2 \pi r \times 4$
$ 208 \times \frac{1}{2} \equal{} r^2 \plus{} 4r$
$ 104 \equal{} r^2 \plus{} 4r$

So, $ 104 \plus{} 4 \equal{} r^2 \plus{} 4r \plus{} 4 \equal{} (r \plus{} 2)^2$  i.e.  $ (r \plus{} 2)^2 \equal{} 6^2 \times3$  i.e.  $ r \plus{} 2 \equal{} \pm 6\sqrt {3}$
Thus, $ r \equal{} 6\sqrt {3} \minus{} 2 \equal{} \boxed{6\sqrt {3} \minus{} 2}$.  (Because $ r > 0$)

hmmmm, the answer in the book says r= 8.....
im so lost....

If $ r \equal{} 8$,
$ 2\times8^2 \times \pi \plus{} (2\times8 \times \pi)\times4 \equal{} 128\pi \plus{} 64\pi \equal{} 192\pi \neq 208\pi$.

is there a way you can simply put the equation in r= form?


hi friends,

I cam across through a problem of this kind.

Solve for $n$ such that $n=2^{\frac{n}{3}}$

regards

bhupala

Considering the growth, you will have only a finite number of cases to check ...

If $n$ means any integer, then I suggest this problem be moved to the Pre-Olympiad section where, I am sure, there are plenty of people who will be able solve it.

They're two monotonically increasing functions, so they should intersect at most once. Actually, I'm not sure that they do.

[quote="kunny"]$n=2^{\frac{n}{3}}\Longleftrightarrow n^{3}=2^{n}.$[/quote]

If $n \in \mathbb{Z}$, it has to be in $\mathbb{N}$, cause if $n<0 \implies n^{3}<0$ but $2^{n}$ remains positive. So $n$ has to be a power of 2, call it $n=2^{k}$ for some positive integer $k$. So $2^{3k}=2^{2^{k}}\Longleftrightarrow 3k=2^{k}$, that is clearly impossible in $\mathbb{N}$

If $n$ is allowed to be any real:

They will intersect  between $n=1$ and $n=2$.
$2^{1}>1^{3}$ but $2^{2}<2^{3}$. The actual value at which they intersect is probably some really ugly number, which you could maybe find using logs.
They will also intersect at some other value $9<n<10$ since $2^{10}>10^{3}$.

But clearly no solutions for $n \in \mathbb{Z}$

$x=2^{x/3}\Leftrightarrow xe^{-x\frac{\ln{2}}{3}}=1\Leftrightarrow-\frac{\ln{2}}{3}xe^{-x\frac{\ln{2}}{3}}=-\frac{\ln{2}}{3}$
$\Leftrightarrow-x\frac{\ln{2}}{3}=W\left(-\frac{\ln{2}}{3}\right)\Leftrightarrow x=-W\left(-\frac{\ln{2}}{3}\right)\cdot\frac{3}{\ln{2}}$
Here W is the Lambert function (see http://en.wikipedia.org/wiki/Lambert's_W_function).
I get (with http://functions.wolfram.com) $x=1.373465876$. This is the value mentioned above that is between 1 and 2. The other value (almost 10) is due to the fact that the Lambert W-function is not single valued in the interval [$-\frac{1}{e}$,0[ and can be computed in e.g. Mathematica using the ProductLog[k,n] function which allows one to caluculate the values on other branches than the principle branch.


$\displaystyle 1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$ is equivalent to

$\displaystyle \frac{ab+bc+ca}{3} + \frac{ab+bc+ca}{a+b+c} \geq 2$.

By AM-GM, it suffices to prove $\displaystyle \frac{ab+bc+ca}{3} \cdot \frac{ab+bc+ca}{a+b+c} \geq 1$.

That is, $\displaystyle (ab+bc+ca)^2 \geq 3(a+b+c)$.

Let $x=1/a, y=1/b, z=1/c$, then $xyz=1$ and it reduces to prove $(x+y+z)^2\geq 3(xy+yz+zx)$ which is obvious.

[quote="Maverick"]Let $ a,b,c$ be three positive real numbers such that $ abc \equal{} 1$. Prove that:
\[ 1 \plus{} \frac {3}{a \plus{} b \plus{} c}\ge{\frac {6}{ab \plus{} bc \plus{} ca}} .
\]
[/quote]
We set \[ a\equal{}\frac{1}{x} ; b\equal{}\frac{1}{y} , c\equal{}\frac{1}{z}\] and observe xyz=1. The inequality is equivalent to:
\[ 1\plus{}\frac{3}{xy\plus{}yz\plus{}xz}\geq \frac{6}{x\plus{}y\plus{}z}\]
Clearly \[ 1\plus{}\frac{3}{xy\plus{}yz\plus{}xz}\geq 1\plus{}\frac{9}{(x\plus{}y\plus{}z)^{2}}\geq \frac{6}{x\plus{}y\plus{}z}\]

[quote="Maverick"]Let $ a,b,c$ be three positive real numbers such that $ abc = 1$. Prove that:
\[ 1 + \frac {3}{a + b + c}\ge{\frac {6}{ab + bc + ca}} .
\]
[/quote]
Let $ p = a + b + c$, $ q = ab + bc + ca$ and $ r = abc$.
Then we have $ 1 + \frac {3}{p} \ge \frac {6}{q}$. By AM-GM:
$ 1 + \frac {3}{p} \ge 2\sqrt {\frac {3}{p}}$. So we have to prove:
$ 2\sqrt {\frac {3}{p}} \ge \frac {6}{q}$, which is equivalent to $ q^2 \ge 3pr$. This is obviously true upon expanding.
Edit:
An one liner:
$ 1 + \frac {3}{a + b + c} - {\frac {6}{ab + bc + ca}} =$ $ \left(1 - \sqrt {\frac {3}{a + b + c} \right)^2 +}$ $ \frac {\sqrt {3}}{\sqrt {a + b + c}(ab + bc + ca)(ab + bc + ca + \sqrt {3a + 3b + 3c})} \cdot$ $ ( (ab - bc)^2 + (ab - ac)^2 + (ca - cb)^2 ) \ge 0$
(If it can fit in one line :P)

first ,we use $ {abc \equal{} 1}$,so the inequality $ {\Leftrightarrow}$ $ {\frac {6}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}} < \equal{} 1 \plus{} \frac {3}{a \plus{} b \plus{} c}}$;we know that :$ {\frac {3}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}} \le \frac {a \plus{} b \plus{} c}{3}}$ so  we only need to prove that :$ {\frac {2}{3}(a\plus{}b\plus{}c) \le 1 \plus{} \frac {3}{a \plus{} b \plus{} c}}$_______[#];let $ {a \plus{} b \plus{} c \equal{} x}$  by $ {a \plus{} b \plus{} c \le 3*(abc)^{\frac {1}{3}}}$ ,we know $ {x \in (0,3] }$; so [#]  $ {\Longleftrightarrow}$ $ {\frac {2}{3}x^2 \minus{} x \minus{} 3 > \equal{} 0}$ ; then consider the function: $ {\phi (x) \equal{} \frac {2}{3}(x \minus{} \frac {3}{4})^2 \minus{} \frac {27}{8}}$ ; we know that : $ {\phi (x) \ge 0}$ on $ {x \in (0,3]}$. then # is true ,so we have completed!

[quote="Maverick"]Let $ a,b,c$ be three positive real numbers such that $ abc \equal{} 1$. Prove that:
\[ 1 \plus{} \frac {3}{a \plus{} b \plus{} c}\ge{\frac {6}{ab \plus{} bc \plus{} ca}} .\]
[/quote]
$ LHS \equal{} 1 \plus{} \frac{9}{{3abc(a \plus{} b \plus{} c)}} \ge 1 \plus{} \frac{9}{{{{(ab \plus{} bc \plus{} ca)}^2}}} \ge 2\sqrt {\frac{9}{{{{(ab \plus{} bc \plus{} ca)}^2}}}}  \equal{} \frac{6}{{ab \plus{} bc \plus{} ca}}$
^^!

[quote=sqing][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=288827]Maranna SortList - 2009 - Italy:[/url]
Let $ a, b, c$ be positive numbers and $ abc\equal{}1$. Prove that\[ \frac{2}{a\plus{}b\plus{}c}\plus{}\frac{4}{3}\geq1\plus{}\frac{3}{ab\plus{}bc\plus{}ca}\geq\frac{6}{a\plus{}b\plus{}c}.\][/quote]

I think Buffalo Way helps us here (but looking for something quicker). Any ideas arqady sir ?

Let $a_1,a_2,\cdots,a_n  (n\ge 2)$ be positive real numbers . Prove that
$$1+\frac{n(n-1)}{2\sum_{1\le i<j\le n}a_ia_j}\geq\frac{2n}{a_1+a_2+\cdots+a_n}$$
[size=75]SXJX, 2(2020),Q1082[/size]
[url=https://artofproblemsolving.com/community/c6h1217573p6069200]h[/url]

Let $a,b,c$ be three positive real numbers such that $a+b+c=1$. Prove that$$ 1+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge 4(ab+bc+ca).$$
[size=50](Crux)[/size]
$$ 1+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge1+\frac{9abc}{(a+b+c)^2}=(a+b+c)^3+9abc\ge 4(ab+bc+ca).$$

[quote=sqing]Let $ a , b, c$ be positive real numbers such that $abc= 1 .$ Show that
$$ ab+bc+ca+\frac{9}{10(a^2+b^2+c^2)}\geq \frac{33}{10}. $$[/quote]
We use the well known inequality, that is
$$81abc(a^2+b^2+c^2)\leq (a+b+c)^5\implies 10(a^2+b^2+c^2)\leq\frac{10}{81}(a+b+c)^5$$
[hide=the proof]Write the inequality as
$$81abc(a+b+c)(a^2+b^2+c^2)\leq (a+b+c)^6$$
And since $3abc(a+b+c)\leq (ab+bc+ca)^2$ we have
$$81abc(a+b+c)(a^2+b^2+c^2)\leq 27(ab+bc+ca)(ab+bc+ca)(a^2+b^2+c^2)\leq (ab+bc+ca+ab+bc+ca+a^2+b^2+c^2)^3=(a+b+c)^6\quad\blacksquare$$[/hide]
and use $(ab+bc+ca)^2\geq 3abc(a+b+c)\implies ab+bc+ca\geq\sqrt{3(a+b+c)}$
And so by AM-GM
$$ ab+bc+ca+\frac{9}{10(a^2+b^2+c^2)}\geq \sqrt{3(a+b+c)}+\frac{9\cdot 81}{10(a+b+c)^5}=10\cdot\frac{\sqrt{3(a+b+c)}}{10}+\frac{9\cdot 81}{10(a+b+c)^5}\geq 11\sqrt[11]{\left(\frac{\sqrt{3(a+b+c)}}{10}\right)^{10}\cdot\frac{9\cdot 81}{10(a+b+c)^5}}=\frac{33}{10}.\quad\blacksquare$$

[quote=Maverick]Let $a,b,c$ be three positive real numbers such that $abc=1$. Prove that: \[ 1+\frac{3}{a+b+c}\ge{\frac{6}{ab+bc+ca}} . \][/quote]
Let $a,b,c$ be three positive real numbers such that $abc=1$. [url=https://artofproblemsolving.com/community/c6h2568906p22041884]Prove that:[/url]\[ 1+\frac{8}{a+b+c}\ge{\frac{11}{ab+bc+ca}} . \]

[quote=sqing]Let $a,b,c$ be three positive real numbers such that $a+b+c=1$. Prove that$$ 1+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge 4(ab+bc+ca).$$
Let $a,b,c$ be three positive real numbers such that $1+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=4(ab+bc+ca) $. Prove that$$ a+b+c\ge 1$$
Let $a,b,c$ be three positive real numbers such that $\frac{5}{2}+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=4(ab+bc+ca) $. Prove that$$ a+b+c\ge \frac{3}{2}$$
Let $a,b,c$ be three positive real numbers such that $\frac{5}{3}+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=4(ab+bc+ca) $. Prove that$$ a+b+c\ge \frac{5}{4}$$[url=https://artofproblemsolving.com/community/c6h3022986p27182277]here[/url][/quote]

[quote=Maverick]Let $a,b,c$ be three positive real numbers such that $abc=1$. Prove that: \[ 1+\frac{3}{a+b+c}\ge{\frac{6}{ab+bc+ca}} . \][/quote]
We have
\[\frac{1}{abc}+\frac{3}{a+b+c} \geqslant {\frac{6}{ab+bc+ca}},\]
for all $a,b,c$ are positive real numbers.

[quote=asdrojas]$ 1 \plus{} \frac {3}{a \plus{} b \plus{} c}\ge{\frac {6}{ab \plus{} bc \plus{} ca}}$ is equivalent to $ (3\plus{}a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)\ge 6(a \plus{} b \plus{} c)$.

Now lets consider two cases. First $ ab \plus{} bc \plus{} ca\ge a \plus{} b \plus{} c$ and then $ ab \plus{} bc \plus{} ca\le a \plus{} b \plus{} c$.

If $ ab \plus{} bc \plus{} ca\ge a \plus{} b \plus{} c$ then using AM-GM $ (3\plus{}a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)\ge (3\plus{}3)(ab \plus{} bc \plus{} ca))\ge 6(ab \plus{} bc \plus{} ca)\ge 6(a \plus{} b \plus{} c)$.

If $ ab \plus{} bc \plus{} ca\le a \plus{} b \plus{} c$ then let $ x\equal{} a \plus{} b \plus{} c$.

$ (3\plus{}a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)\ge (3\plus{}a \plus{} b \plus{} c)(a \plus{} b \plus{} c) \equal{} (3\plus{}x)x$. Now we net that $ (3\plus{}x)x\ge 6x$ but this equivalent to $ x(x\minus{}3)\ge 0$ and this is true becouse $ a\plus{}b\plus{}c\ge 3$ by AM-GM.[/quote]

You can prove that $ab+bc+ca\leq a+b+c$:
$$ab+bc+ca\leq a+b+c \Longleftrightarrow ab+bc+ca\leq \sqrt[3]{abc}(a+b+c) \Longleftrightarrow \sum_{sym} ab\leq\sum_{sym} a^{4/3}b^{1/3}c^{1/3}$$
And that is true by Muirhead since $\left[\frac{4}{3},\frac{1}{3},\frac{1}{3}\right]\succ[1,1,0]$


$f:=M_n(\mathbb{C})\to M_n(\mathbb C)$, $f(M)=M^k$, where $k>1$ is an integer.
Find all matrices $M$ such that $f$ is locally bijective around $M$
(i.e the differential is an isomorphism )

Here for $k=2$

$f(M)=M^2$

the differential at $M$ is $d_{M}f(H)=HM+MH$

If all eigenvalues of $M$ has real parts  $<0$ the
$HM+MH=0$ has a unique solution $H=0$
Then the differential is bijective in this case

If $M=aI_n$ an homotetie $a\neq 0$ gives also a differential bijective


Duke
college
AMC
USA(J)MO
USAMO
geometry
summer program

Palmer Mebane, from near Richmond, Virginia (born there also), going to be a senior next year. I'm probably the person most disconnected from immediate reality you'll ever meet. My biggest interests are math and logic-oriented puzzles, though I detest word puzzles and riddles. On the side, I play lots of classic and old-school electronic games, program some, and dabble into personality theories.

Jeremiah Siochi, from Iowa (Iowa City - home of University of Iowa), I am 15 and next year will be a sophomore.  I was born in Ohio and my dad is half Filipino and half Brazilian.  I am also homeschooled.  My interests are math, Latin, logic, computer programming, piano and double bass, classical music (especially Brahms), and jazz.  I spent the past two summers at Duke University's TIP summer studies.

I am Ofir, born in Israel, but living in Worcester, MA.  I am 15 yrs old.  I, too, enjoy classical music, and even play the clarinet (and the tenor sax but I'm not good at it).

I am Mark Chen (currently a junior) from Hsinchu, Taiwan.  My interests include math competitions, mathematics research, piano (Chopin, Liszt, Prokoffief), debate (National Debate Champion), British literature, chess, and Chinese.  I'm also resonably good at basketball, swimming, and ping-pong.  I'm not particularly good at chemistry, sudoku, go, organizing my room, and waking up for school.

I attend National Experimental High School, a superbly competitive high school where it's normal to study through lunchtimes in silence (so my GPA is nowhere near my post rating).  Needless to say, I can't wait for a summer back in the states :P .

I've taken the USAMO twice (from 12:30 AM-5:00 AM), won a silver medal in the Chinese Mathematical Olympiad, and trained as an alternate for the Taiwanese IMO team.  My AMC scores are indicated in my profile.  My school has a highly active Olympiad culture; the top IPhO scorer (absolute winner) in 2005 went to my school, as did a 2005 IChO gold medallist.

I'm looking forward to meeting everyone!  To all pingpong tournament contestants: bring it on.

Mark Chen (MC 04 05 x 07)

Laura Zehender, also from near Richmond, VA (I have lived here my whole life). I am 15 and will be a sophomore next year. I've been homeschooled since 2nd grade. My interests include math (especially combinatorics and geometry), dance (ballet, tap, jazz, pointe, and have done some other kinds of dancing), Latin, piano (especially Tschaikowsky), listening to country music, ancient Greek literature and myths, and art (especially modeling clay and drawing with pencil). I also am a bookworm and chocoholic. This is my first year at Mathcamp, but last summer I went to MathPath, and my older brother went to Mathcamp twice.

I'm David Corwin, from Acton, MA. I go to Acton-Boxborough Regional High School, one of the best public high schools in MA. My interests include math, foreign languages (French/Latin so far), science (I guess?), politics, classical music, tennis, school newspaper, something like that.

[quote="calc rulz"]I'm David Corwin, from Acton, MA. I go to Acton-Boxborough Regional High School, one of the best public high schools in MA. My interests include math, foreign languages (French/Latin so far), science (I guess?), politics, classical music, tennis, something like that.[/quote]

Algonquin Regional High School is way better.

(even though you beat us at the state meet  :mad: )

I'm Tom James from Louisville, KY, I'm 17 years old and will be a senior next year.  I'm a pretty cynical fellow with a dry sense of humor.  I like to make fun of myself and I don't mind criticism (based on logic, of course :-)).

At school, I play three sports: cross country, wrestling (my favorite), and track.  I'm active in yearbook, theatre, quick recall, and other assorted extra-curriculars.  I also really like French.

Outside of school, I'm very interested in philosophy and politics and love a good discussion or debate.  I also am interested in personality theory such as Myers-Briggs (I'm an INTJ).  I do a fair bit of independent reading and programming too.

My favorite area of mathematics would probably be logic/foundations such as Godel's theorems and theoretical computer science such as algorithmic information theory.  I think these fields penetrate into the deepest truths of our existence.  I hope to be a pure mathematician one day.

Although I'm normally slow to get comfortable around new people, I'm a good deal more extroverted at camps for kids like us (I went to Duke TIP for three years prior to Mathcamp).  I hope that I can get to know most kids during the five weeks because I'm normally such an introvert :P

I am looking forward to this very, very, very much  :D

[quote="tjames"]I'm a pretty cynical fellow with a dry sense of humor.[/quote]

Tom, I think you and I are going to get along!  :)

I guess I should also mention that I am taking Latin, and am enjoying it very much. I cannot play music, but I enjoy listening to it. I play Ultimate Frisbee for a sport, and participate in the Computer Club and Math Team as general extra-curriculars. I greatly enjoy debate and discussion in general, but have never participated in any team, and have no idea how competitive debating works. I go to Belmont High School, which, although far from the best school, is quite a nice, relaxing place to attend.

I am also something of a grammar stickler.

@tjames: Those are both extremely fascinating subjects! (Information theory + logical foundations) In fact, I have tried to delve into them but have found that my mathematical talents apparently lie elsewhere as I cannot seem to get the hang of the many bizarre and counterintuitive results that creep up. Perhaps at Mathcamp I'll have the proper time and environment to understand it better, as I'm sure you will too.

Though I may eschew that area entirely, as (since we're supposed to talk about ourselves) my primary interest is combinatorics, whatever that is supposed to mean. In the same vein, I do a lot of algorithmic style programming and also play chess, soccer, and frisbee, though I am not so great at the last of these (maybe some of you that are good can teach me while we are there).

Hey, I'm an INTJ too!

Anyway, might as well explain that I'm going to MathCamp as well.

Brian Hamrick, rising sophomore, from VA. I go to TJHSST. In math, I enjoy number theory, as well as anything else decently far away from geometry. This will be my second MathCamp of (hopefully) many. When we get to camp. You'll probably find me playing bughouse in the lounge during free time. I look forward to meeting all of you!

After discovering I've recieved a full ride, I'm most definitely going. Thanks MATHCAMP.

My name is Adrian from Alhambra, CA located in the LA Valley. I picked up Ultimate [Frisbee] from HCSSiM '06 and was like the worst player there, but when I returned home everyone was praising me, but I'm ready to be the worst again because it's so awesome to play.

I really like music and dabble in a bunch of instruments. My main instrument is guitar and I used to practice very technical music (you know those crazy driving rock songs with fast guitar solos), but now I became more mellow and acoustic. I'd  LOVE to start a technical (or nontechnical) rock band with anyone, if interested... that is... if Colby college happens to have an electric guitar and basses and drums and keyboards and trumpets for ska and amps sitting somewhere nearby.  I hate 4/4 time and I love playing around with time signatures. Dream Theater is my favorite band.

I've recently feel in love with singing, even though I'm not good, but I swear I soon will be. I think RENT is absolutely the best musical ever created (even though the only other one I've watched is Phantom of the Opera -_-). If you see some weirdo randomly bursting into song, there's a chance it'd be me.

And then I tend to be overly romantic and overly analyze things and dream of transcendence and connect my inner soul through the fleeting, tranquil depths of nature and thus I love hiking and things of the sort.

Hmn, CARPE DIEM is my life philosophy. Tchao.

[quote="Eternica"]
I've recently feel in love with singing, even though I'm not good, but I swear I soon will be. I think RENT is absolutely the best musical ever created (even though the only other one I've watched is Phantom of the Opera -_-). If you see some retard randomly bursting into song, there's a chance it'd be me.[/quote]

There's also a very good chance it won't be you. A lot of people sing :).

Wait. Then that'd be awesome. We can turn MATHCAMP '07 into a musical!

Thanks, I did it, and now I have the email!

[quote="hahafaha"]
So, I have obviously never been to Mathcamp before (can't wait though), but I do have friends who have. I am also not a Mormon, but I also have friends (one friend, actually) who is. And from everything I know about the two, I'd probably think that Mathcamp would be far stranger to the random person than Mormonism would be.[/quote]

It wasn't so much mormonism itself that we found strange, it was more the idea of mormon camp (and their crazy clap-chant-thingies).  Then again, it's true that with "games" like bunny-bunny-toki-toki (which I'm sure you'll all learn by the end of camp), we were probably the stranger of the two.

And, while we had the most amusing inter-camp relations with the mormons (barring, perhaps, the football players that Greg B. (who will be a Junior Counselor this year) explained the P vs NP problem to), I think the strangest camp there was, without a doubt, Yearbook camp.

Anyway, a bit about me.  I'm Youlian Simidjiyski, I enjoy math (surprise!) and philosophy (another surprise!).  I'm originally from Bulgaria, and I probably spend more time trying to convince random people that Math is awesome than I spend actually doing Math.  This will be my second and final MathCamp (since I'm a senior), and, as much as I look forward to seeing all of my friends from MC '06 again (*waves at Alex, Brian, and Dan*), I'm much more excited about meeting all of the new campers, and seeing all of your reactions to the craziness ;).  Simply put, people aren't lying when they say MathCamp is like a whole new world.  My two pieces of advice to new Campers:

1.  Go to TAU (Time: Academic, Un...scheduled?), do your homework (whoa!  Amazing, right?  Homework... a good thing... *headpop*), watch for signs of burnout, and resist the urge to take mid-day naps.  You're going to have the other 47 weeks of the year to do math at a sluggish pace, make sure to make the most of it while you're there.

2.  Balance your math time and your social time carefully.  It'll probably surprise most of you to hear that camp is a very social place.  It's generally full of amazing people, almost all of whom would be among my best friends were they to attend my school.  Thus, even though plenty of us are introverts and shun normal social behavior, at camp, the social life starts to detract first from sleep, and then, obviously, from Math.  Find a good balance.

Anyway, can't wait to meet all of you, and political discussions at camp would be quite fun, though, knowing how MC discussions end up, it's likely that two people originally debating free market vs communistic systems would end up talking about which one is better in a community with countably infinite, uncountably infinite, or negative amounts of capital...

Im going to have to agree with what the alumni are saying here. Mathcamp is pure awesomeness, and here are a few points:

A) Political discussion is good. Other stuff at Mathcamp always seems to seem more fun, so political discussion is not very common.

B) Mathcamp is VERY social. Someone once told me a story of how they did not take a single class their second year at Mathcamp because he was too busy enjoying the social life. Now, that does not mean you should not take any classes. In fact, you should take many classes, but Mathcamp is very fun socially as well.

C) I am also a news Junkie outside of Mathcamp. I read digg ~5 times a day, Google News ~2, Slashdot~1. Last year at Mathcamp, throughout the five weeks, I read digg once, Slashdot once, and Google News never. Mathcamp is really too much of its own awesome world to think about this "news" stuff.

Now, a little about me: I am Sergei Bernstein from Belmont, MA. This is my third year, and hopefully my best year. I enjoy math, frisbee, piano, video games, non-video games, living impulsively, and trying to make every problem a math problem. I am Xtremely excited about meeting new campers. In fact, I have already met some, and I can already tell that this year will be a great one. Also, I have already made some plans for Mathcamp.  Let's see... what else... I believe Dan still owes me some pizza or equivalent. Here is an awesome quote: "Your induction is weak, old man"

[quote="MellowMelon"][quote="TheBernstinator"]"Your induction is weak, old man"[/quote]
Oh lord, this is probably going to be all over AoPS now that MOP is over. For those attending Mathcamp, a Star Wars parody is already in the works; most of the progress is a lot of spoofed quotes like this one.

Other good ones:
"His diagram's gone. Luke, are you okay?"
"Leave your compass and straightedge. You will not need them here."
"If you invert me now I shall become more powerful than you can possibly imagine." (think power of a point)
(as you can see most of these were thought up during a Geometry lecture)

Evidently Mathcamp has already produced a matrix parody, so this is nothing too new.[/quote]

Also:

"We shall invert him to the dark side of the circle."
Keep in mind the compass and straightedge part was pertinent to one's entrance into the complex cave: straightedge and compass are unneeded when one complex-bashes a geo problem.

[quote]Oh lord, this is probably going to be all over AoPS now that MOP is over. For those attending Mathcamp, a Star Wars parody is already in the works; most of the progress is a lot of spoofed quotes like this one.[/quote]
Actually, Serge and I have several chunks of scenes written by now.

EDIT: To respond to the original subject. I'm "definitely going". 4 more days!!! Yay! :D

In response to the original topic, I might end up at mathcamp [b]next year[/b], so not me...

hey, i'm Prateek from Toronto, Canada.. I seem to be outnumbered by all you left-wing americans
anyhow, i'm 16, going into grade 11, and among other things i'm interested in video games, wrestling, food, sleep, and math.  this is my first year, and i hope i don't get lost on my way to camp.

[quote="Eternica"]
I've recently feel in love with singing, even though I'm not good, but I swear I soon will be. I think RENT is absolutely the best musical ever created (even though the only other one I've watched is Phantom of the Opera -_-). If you see some retard randomly bursting into song, there's a chance it'd be me.[/quote]

Hahaha. I'm a few weeks too late to warn you about Phantom of the Lecture Hall.

Is anyone doing another math musical this year?

(MC'06, not '07)

i love how this thread is literally from 2007 but we're all staring at it bc awesomeguy posted a link

Lol, yeah, cuz we don't know if Eternica is a boy or girl.

[quote=hahafaha]So, I realize it's a bit early for some to know whether or not they are going to camp this year, but for those that do, let's get to meet each other!

Post your name, age/grade, location, interests, and anything else you're are comfortable sharing.

My name is Leonid Grinberg from Boston, MA (a suburb, technically, of Boston, but whatever). I was born in Russia, and speak Russian fluently. I am 15 and an incoming sophomore (though I am in the class of 2009, for which there is somewhat of a long and complicated explanation). I like math (obviously), computers (IT and programming), physics, logic, economics, psychology, law and language.[/quote]

You’re 29 now

[quote=Jwenslawski]I was like eternica is a 49% boy 51% girl and now I know. Anyways these are his most personal posts that I have ever seen.[/quote]

I thought he was a girl until up to now o.o


This problem has been bugging me. I found a couple very different answers. But none worked. I really need this.  :wallbash: But I don't know if there is a solution.  :lol:  If there isn't I at least want a method to approximate it.

I have $ 5$ input values: $ Y_{1}$, $ Y_{2}$, $ Y_{3}$, $ C$, and $ M$.
And I need $ 1$ output  value: $ O$

$ Y_{1}$, $ Y_{2}$, and $ Y_{3}$ are three $ y$ values on the sine curve $ y\equal{}a \cdot sin(b\cdot x)$
The corresponding $ x$ values ($ x_{1}$, $ x_{2}$, and $ x_{3}$) have a distance of $ d$. (This means $ x_{1}\equal{}x_{2}\plus{}d\equal{}x_{3}\plus{}2\cdot d$)

$ C$, and $ O$ are two $ y$ values on the sine curve $ y\equal{}a \cdot sin(M\cdot b\cdot x)$
The corresponding $ x$ values ($ x_{c}$, and $ x_{o}$) have a distance of $ d$. (This means $ x_{c}\equal{}x_{o}\plus{}d$)

So pretty much solve for O:
$ Y_{1}\equal{}a \cdot sin(b\cdot x_{1})$
$ Y_{2}\equal{}a \cdot sin(b\cdot x_{2})$
$ Y_{3}\equal{}a \cdot sin(b\cdot x_{3})$
$ C\equal{}a \cdot sin(M\cdot b\cdot x_{c}))$
$ O\equal{}a \cdot sin(M\cdot b\cdot x_{o}))$
$ x_{1}\equal{}x_{2}\plus{}d\equal{}x_{3}\plus{}2\cdot d$
and,
$ x_{c}\equal{}x_{o}\plus{}d$

If this isn't the correct place to put this topic then move it.

Offhand, it looks like generically you need to know the value of $ a$ in advance; that lets you work with angle addition formulas and the Pythagorean theorem.  And that may not even be enough; consider what happens if $ d \equal{} \frac{2 \pi k}{b}$ for some unknown integer $ k$; then the value of $ O$ depends on the value of $ Mk \bmod 2 \pi$, which you can't compute.

Well, the value of $ b$, I think, can vary, it can be any multiple of a certain number.
Also, there will also be special cases where there is no solution. Such as where $ Y1$,$ Y2$,and $ Y3$ form a line, but I can consider those separatly.

Note that $ C$ and $ O$ correspond to points on the curve $ y \equal{} a \sin (bx)$, as well, if you just adjust the $ x$ components appropriately.

The question now becomes: can four points determine a sinusoid?

Can't three points determine a and b?
$ y \equal{} a \sin (bx)$
That's what (x1,y1),(x2,y2),(x3,y3) can do, right?
If that can't happen I can always add another point (x4,y4).


calculus
integration
trigonometry
logarithms
complex numbers
calculus computations

1)$ {\int_{0}^{\frac{\pi}{2}}\frac{x}{\sin x\plus{}\cos x\plus{}1}dx}$
2)$ {\int_{0}^{\frac{\pi}{2}}\frac{x^{2}}{\sin x\plus{}\cos x\plus{}1}dx}$
I can solve only the first one. :(

$ P\equal{}\int_{0}^{\frac{\pi }{2}}\frac{x}{\cos (x)\plus{}\sin (x)\plus{}1}\, dx$

using $ x\to\frac{\pi }{2}\minus{}x$

$ P\equal{}\int_{0}^{\frac{\pi }{2}}\frac{\pi }{2 (\cos (x)\plus{}\sin (x)\plus{}1)}\, dx\minus{}P$

$ x\equal{}i\log\left(\left(\frac{1}{2}\minus{}\frac{i}{2}\right) (y\minus{}1)\right)$

I will use complex numbers here.

$ \int_{0}^{\frac{\pi }{2}}\frac{\pi }{2 (\cos (x)\plus{}\sin (x)\plus{}1)}\, dx\equal{}\frac{1}{2}\pi\int_{2\plus{}i}^{2\minus{}i}\left(\frac{1}{\minus{}i\plus{}y}\minus{}\frac{1}{i\plus{}y}\right)\, dy\equal{}\frac{1}{2}\pi\log (2)$

So $ P\equal{}\frac{1}{4}\pi\log (2)$


If P(x) is a polynomial of degree n such that P(x) = 2^x for x = 1,2,3,...,n+1, find P(n+2).

There is an unique polynomial P with degree n and such that P(k) = 2 k  for k=1,2,...,n+1..

Using that C(k,0)+C(k,1)+...+C(k,k) = 2 k , it is easy to verify that P(x) = 2[C(x-1,0)+C(x-1,1)+...+C(x-1,n)], where C(x,j) = x(x-1)...(x-j+1)/j! .

It follows that
P(n+2) = 2[C(n+1,0)+...+C(n+1,n)] = 2[2 n+1 -1] = 2 n+2  - 2.

Pierre.

An other solution without having to guess the explicit form of Pn :

Let P(n) such defined polynom.

Q(n)(x)=P(n)(x+1) - P(n)(x)
d(Q(n)) = n-1 and Q(n)(k) = 2^k (for k=1..n) so Q(n) = P(n-1) and P(n)(x+1) = P(n)(x) + P(n-1)(x)  :)

After that we just have to sum : Pn(n+2) = 2^n+2 + 2^(n+1) + .... + 2^3 + P(1)(3) = 2^(n+2) - 2 (since P(1)(x) = 2x)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $D_{n}$ denote the dihedral group of order $2n$. Count the total number of homomorphisms from $S_{4}$ to $D_{6}$.

Is there any general method to count homomorphisms of finite groups?

Hi.

I don't think that there is a general method for counting homomorphisms, but in this case it's relative easy:
$D_{6}$ is isomorphic to $D_{3}\times\mathbb{Z}_{2}\cong S_{3}\times\mathbb{Z}_{2}$ (little exercise within the exercise ;-) )
Now $Hom(S_{4},S_{3}\times\mathbb{Z}_{2})$ is isomorphic to $Hom(S_{4},S_{3})\times\ Hom(S_{4},\mathbb{Z}_{2})$. The right-hand factor is easy since there is only one normal subgroup of index 2. The other factor can be calculated similarly: $S_{4}/V_{4}$ is isomorphic to $S_{3}$ which yields 6 homomorphisms $S_{4}\to S_{3}$ (the one from this isomorphism times the six automorphisms of $S_{3}$). Since there is no normal subgroup of index 3, there can't be an homomorphism from $S_{4}$ onto $A_{3}$. There are three elements of order 2 in $S_{3}$ so there are 3 additional homomorphisms onto the 2-elements-subgroups. This means that there are 10 Homomorphism $S_{4}\to S_{3}$ (6 onto + 3 with smaller image +1 trivial)


advanced fields
advanced fields unsolved

Can anyone help me with the problem found here - http://zgall1.t35.com/question2.jpg?

I am having a lot of trouble even getting started on it.

Thanks
Zack

I have managed to solve most of the question but I still need some help proving that X(jack) = X(bar). Can anyone help me out with that?


calculus
integration
function
limit
real analysis
real analysis unsolved

Given a sequence of measurable positive functions $(f_{n})$ defined on a measure space $(X,m,\Sigma)$ which converges pointwise to a function $f$, such that $lim \int_{X}{f_{n}dm}=\int_{X}{fdm}<+\infty$, show that $lim \int_{E}{f_{n}dm}=\int_{E}{fdm}$ for every measurable subset $E \subseteq X$.

N.B: this does not hold if we drop the integrability of $f$. So far, I could find a solution only if assuming $m$ to be $\sigma$-additive (and using some "complicated" stuff, like Egoroff's theorem), so I feel I must be missing something easy.

Would you please check your assertion. Aren't there some superfluous $\lim$'s?
And what do you mean by "converges to a function $f$"? Pointwise convergence? Or convergence in some weak sense?

Yep, pointwise convergence. I edited the lim's, thank you for pointing it out.

Fatou's Lemma states \[\int_{X}\liminf f_{n}dm\le\liminf\int_{X}f_{n}dm.\] We apply this for $E$ and $X\setminus E$ in the place of $X$ and notice that in the present situation $\liminf f_{n}=\lim f_{n}=f$. So \[\int_{X}f dm=\int_{E}f dm+\int_{X\setminus E}fdm\le\liminf\int_{E}f_{n}dm+\liminf\int_{X\setminus E}f_{n}dm\le\liminf\int_{X}f_{n}dm=\int_{X}fdm<\infty.\] Hence every $\le$ must be $=$ and we get $\int_{E}f dm=\liminf\int_{E}f_{n}dm,\ \int_{X\setminus E}fdm=\liminf\int_{X\setminus E}f_{n}dm.$
But since $\limsup\int_{E}f_{n}dm=\int_{X}fdm-\liminf\int_{X\setminus E}f_{n}dm=\int_{E}fdm,$ we conclude \[\int_{E}f dm=\lim\int_{E}f_{n}dm.\]

[quote="Kolumbus"]we get $\int_{E}f dm=\liminf\int_{E}f_{n}dm,\ \int_{X\setminus E}fdm=\liminf\int_{X\setminus E}f_{n}dm.$
[/quote]

Everything's looking fine but I cannot catch immediately how you infer the two independent equalities above.

If $a\le b$ and $c\le d$ and $a+c=b+d$, then "every $\le$ must be $=$", that is $a=b,c=d$.
Here we have $a=\int_{E}fdm, b=\liminf\int_{E}f_{n}dm, c=\int_{X\setminus E}fdm, d=\liminf\int_{X\setminus E}f_{n}dm$.
Note that this need not be true, if $a,b,c,d$ could be $\infty$.

Ok, I missed the fact that $a \leq b$ by Fatou's lemma. Thank you.


[hide](a+1)^4 can be expanded to a^4 + 4a^3 + 6a^2 + 4a + 1

In this case the a is (x-1), so we can replace a with (x-1) to get:

(a+1)^4

= [(x-1) + 1)]^4

= x^4

So the answer is C) x^4[/hide]

[hide](x+1)^4=x^4+4x^3+6x^2+4x+1. In this case the x is x-1, so subsitute it in. [(x-1)+1]^4=x^4. Therefore, the answer is (C) c^4[/hide]

[quote="balloo"][quote="MCrawford"]Let $S = (x-1)^4 + 4(x-1)^3 + 6(x-1)^2 + 4(x-1) + 1$.  Then $S$ equals
(A) $(x-2)^4$
(B) $(x-1)^4$
(C) $x^4$
(D) $(x+1)^4$
(E) $x^4+1$[/quote]

The answer is C. But I dont know how to reply. Am I doing it correctly?[/quote]

That is the right answer but I can't tell you if you are doing it correctly unless you post what you are doing.

[hide]
Just let $a=x-1$
Then u will get the answer in a minute.[/hide]

If  you have  already tried it, and want to look at something which might be new, here is something  which may be interesting ...
[hide]
For a moment, assume that you did not recognize that it is the  bionomial expansion , or right away saw  that  it is equal to $x^4$ .. Here is  an interesting[b] fast[/b] method:

[size=150]Check when  $x=1$ !!!
[/size]
What happens when $x = 1$ , you  see right away  that  $S=1$ and  option C can be correct.
You can also  have chosen some other values  such as $x=0$ (which makes the calculation easy) to rule out  some or all of the wrong  options.

Now can you be sure that  $x^4$ is correct, if you checked it for say a few   specific values.
The answer is Yes, if you checked it for 5 (or more different)  values of $x$ .

For example, if  particular values of , say $x$  for $0 , -1, +1, -2,+2$    $S$ gives values (here) (0,1,1,16,16) you can be sure that the identity will hold for all values for x.

Trick is  find the degree of the polynomial - obviously it is 4 here. and just try if for one more.
This one of not that well known trick, but often very easy to use!
[/hide]

[hide]
I thought multiping it out would be a lot of work, so I tried to plug in a number.
I tried to plug in 1.
You get:
0+0+0+0+1=1
The answer is very obviously "C".
[/hide]

[quote="TripleM"]Note to Gyan and asinha:
[hide] Answer A would also not be eliminated after substituting x=1 - you'll need to try at least one more value to eliminate A.
[/hide][/quote]
Yes ... but the point was to give  a tool to  students  to use it to solve such  problems quickly.. and the main point (that  if the substitution works for sufficient  number of values,- eg 5  in this case)  may ne new to many of the sutdents here.

Yep, I know that, but you both apparently made the same mistake which could have been costly :)

Not to nit-pick  but there is much more than "just pluggig for x=1 in my original  post ... see in the original:
[quote][b]x=0 [/b](which makes the calculation easy) to rule out [b]some or all of the wrong [/b]options. [/quote]  you are not depending on x=1 only. Yes  option A might have been overlooked  (BTW I did not even look that carefully at all the options  except x^4, as the expression, to  put it mildly, was found to  be extremely simple)  ..Okay so option A  could be a possibility for x=1   but not when x=0 and y[b]ou  need only one value to rule it out. [/b] ...  Again the important point here, you can  afford to be sloppy  (as to the concern as " the error could  be costly") if you are going to check it for another number where it is easy to see :) -

BTW, what you pointed out is really appriciated - and yes there was an error of adding extra "only"   - and yes I am VERY glad that  we have bright people like you  who would see that  and take time to point it out ..That is very helpful for the  forum.  (I have edited out the  extra "only" from the original  post :)

Anyway the  takeaway for the people here should be:

To rule out wrong option - Only one exception/contradiction  is needed.
To  'prove' the  option is correct: - You need  (n+1)  cases where n is the degree of the polynomial .

For others : Read the original message by  me, if you have not done so,   and you may have another  method in your tool box.  Enjoy.:)

Just a little remark for starters who read this topic, except TripleM and Gyan, which seems to be in an argument/concern type of debate: no matter what method/technique/formula you use, as long as you can get the answer fast and be done with it quick is what matters more because those tests/competitions are timed and there are no time remaining for you to think which method works faster.


$I=\int^{\frac{\pi}{4}}_{0}ln(1+tg t)dt$
Now with $u=\frac{\pi}{4}-t$ we have $I=\int^{\frac{\pi}{4}}_{0}ln(1+\frac{1-tg\: u}{1+tg\: u})du$
=$\int^{\frac{\pi}{4}}_{0}ln 2 \: du-\int^{\frac{\pi}{4}}_{0}ln(1+tg u)du$=$\frac{\pi}{4}ln2-I \rightarrow I$

...but i appreciate your hard-working :)


limit
integration
calculus
calculus computations

Let $ a_n\equal{}\frac{1}{1 \cdot n}\plus{}\frac{1}{2 \cdot (n\minus{}1)}\plus{}\cdots\plus{}\frac{1}{n \cdot 1}$
find the limit:
$ \lim_{n \to \plus{} \infty}a_n$ :)

zhaobin,
We'll use $ \dfrac{a\plus{}b}{ab}\equal{}\dfrac{1}{a}\plus{}\dfrac{1}{b}$, and that limit is equal to $ 0$ . :P

Hey,Thank you for your answer.That's also my way :lol: But maybe the following will be harder:
Let $ a_n \equal{} \frac {1}{\sqrt {1} \cdot \sqrt { n}} \plus{} \frac {1}{\sqrt {2} \cdot \sqrt {n \minus{} 1}} \plus{} \cdots \plus{} \frac {1}{\sqrt {n} \cdot {\sqrt 1}}$
find the limit:
$ \lim_{n \to \plus{} \infty}a_n$  :P

You can try, :)

That equals $ \int_0^1 \frac1{\sqrt {x(1 \minus{} x)}} \, dx \equal{} \pi$. [Well, some arguments are needed to justify the equality between the "Riemann sums" and the integral, but I reckon they are not too hard.]

PS: I got your PM and I'm currently thinking about it.

Yes,That's correct.I think it is nice limit.
Although we can generalize it,but not nice.

PS(Only to perfect_radio):Thank you,please reply as you have make clear of the problem.
I don't also how harizi apply that identity to the original problem.


modular arithmetic
number theory
relatively prime

In the title, don't you mean $ 3^{100}$ divides 1988n+1?

Yeah, I'm sorry. Just ignore the title. :D

[hide]Since $ 1988$ is relatively prime to $ 3^100$, there will be a multiplicative inverse of $ 1988 \pmod {3^100}$. Take the negative of this inverse and put that as $ n$.[/hide]

I would guess $ n\equal{}1988^{3^{99} \minus{}1}$

[quote="Johan Gunardi"]Prove or disprove that there is an $ n$ such that $ 1988n\equiv\minus{}1\pmod{3^{100}}$.[/quote]

We have $\gcd(1988,3^{100})=1$, so $1988^{\phi( 3^{100})} \equiv 1 \pmod{3^{100}}$.

Also $\phi(p^k)=p^{k-1}(p-1)$ for every prime $p$ and positive integer $k$.

So $\phi( 3^{100})=2 \times 3^{99}$ and you can easily find $n=1988^{2 \times 3^{ 99} -1}$.

[quote="amparvardi"]...
So $\phi( 3^{100})=2 \times 3^{99}$ and you can easily find $n=1988^{2 \times 3^{ 99} -1}$.[/quote]
(Not that it matters - as it is easy to adjust n) but wouldn't $1988 n$   would be $\equiv +1 \pmod {3^{100}}$

(Question is little more interesting if it was to find $n$ such that  $1988^n \equiv -1 \pmod {3^{100}}$