let us make predictions on:
1)India's medals tally
2)score tally

please do not make any direct comments on any team member.

1 gold,1 silver,2 bronze,and may be 1 H.M

my tally is:

1 gold ,1 silver, 3 bronzes ,1 HM

order of marks at IMO:
29,24,20,17,15,11

this is what i think is a practicable and reasonable estimate.

i would be happy if they prove me wrong and get more glittering medals.

(all that glitters is not gold :P )

[url=http://www.mathlinks.ro/viewtopic.php?t=159844&start=80]here[/url] are some initial results

As of now (from what Valentin Vornicu said)
[code]
Bhargav   5 ? 0 ? ? ?
Manas     3 ? 3 ? ? ?
Atul      0 ? 0 ? ? ?
Prashant  5 ? 0 ? ? ?
Utkarsh   3 ? 0 ? ? ?
Abhishek  3 ? 0 ? ? ?[/code]

Further updates:
Bhargav         5 0 0 ? 0 0
Manas           3 7 3 ? 1 0
Atul              0 7 0 ? 7 0
Prashant        5 1 0 ? 0 0
Utkarsh          3 1 0 ? 1 0
Abhishek        3 7 0 ? 7 0
Let's keep our fingers crossed- this year seems weird and medal cut-offs are likely to change :)

not much though going by what the mathlinks discussion says it will be pretty the same as that for last year   :maybe:

I know but the results seem to disagree with that.

seems peter scholze has done gr8 just a miss in problem 3 that's understandable....
well just a wait for another 24 grs would make things clearer

so the final results are out . [url=http://www.imo2007.edu.vn/statics/result/result%20imo2007.htm]here[/url] is the list.
supposedly we have three silvers  and three HM's :lol:
our rank 25 is better than that in last two years(35,36 i suppose  :maybe: )


I typoed in my protest when I was sending which polynomial I put in my solution. It should be -100x^2+100x+100 (I think... I can't find my solutions  :mad:   :mad: )

[quote="DPatrick"]A couple of notes:

1. If you wish to contest a score, follow the instructions on the website. [b]Don't post on the forum[/b] and don't send email. (You may send email for something unusual, for example, a problem that you submitted is completely missing.) [/quote]

And IMHO, I don't think that they would give you points back for typoing, since technically, it's your fault that you typoed, not the grader's fault that they didn't see through it.

I did the same thing in my protest yesterday. After realizing that I had typoed, I emailed USAMTS. They said that they had to approve/reject the protest as it was, and so I wasn't able to submit the protest as it should have been.

Then again, I managed to get a point back.

I didn't typo in my solution, just in the protest...

BTW, when will I know when my protests are regonized?

In any case, that is the wrong polynomial. Sorry :(

How long does it usually take for the USAMTS staff to respond to a protest?
And does the USAMTS protest rule mean that you are only allowed to protest on scores of 2 problems for the whole year, or what?

I totally don't get it.

[quote="gauss1181"]How long does it usually take for the USAMTS staff to respond to a protest?
And does the USAMTS protest rule mean that you are only allowed to protest on scores of 2 problems for the whole year, or what?[/quote]Both of these questions are answered in the protest rules -- please read them carefully.

BOGTRO, do you realize that at $ x\equal{}1/2$ your polynomial is $ 125$?  So it does not meet the given constraints.  If you are trying to use that to show that the max value $ |a|\plus{}|b|\plus{}|c| \equal{} 300$ can be obtained, it is an invalid example.  Assuming you still proved that $ 300$ is the maximum value, you should probably still get a 4.


function
advanced fields
advanced fields unsolved

Show that there isn´t a surjective function of S = {(x,y) $ \in \mathbb{R ^2}$ :  $ x^2 \plus{} y^2 \equal{} 1$} in $ \mathbb{R}$ .

The English words are "surjective" and "function," respectively.  Do you want a continuous function $ S \to \mathbb{R}$?

Need to prove that such function doesn´t exist.

$ S$ is compact, so a continuous function attains a maximum.


Let $ a,b,c\ge 0, a\plus{}b\plus{}c\equal{}1$. Prove that:
$ \frac{3\plus{}2a\plus{}b}{a^{2}\plus{}b^{2}}\plus{}\frac{3\plus{}2b\plus{}c}{b^{2}\plus{}c^{2}}\plus{}\frac{3\plus{}2c\plus{}a}{c^{2}\plus{}a^{2}}\ge 39$

[quote="NguyenDungTN"]Let $ a,b,c\ge 0, a\plus{}b\plus{}c \equal{} 1$. Prove that:
$ \frac{3\plus{}2a\plus{}b}{a^{2}\plus{}b^{2}}\plus{}\frac{3\plus{}2b\plus{}c}{b^{2}\plus{}c^{2}}\plus{}\frac{3\plus{}2c\plus{}a}{c^{2}\plus{}a^{2}}\ge 39$[/quote]
I think, it's wrong. Try $ a \equal{} 0.49,$ $ b \equal{} 0.51$ and $ c \equal{} 0.$ :wink:


Let's face it: eventually, earth will no longer exist for some reason or another. Be it getting hit by a gigantic asteroid, or the sun expanding to red-giant size and exploding at the end of it's life span, or a world-wide nuclear war, or anything else that might happen -- the earth will not last forever.

So the question is: [b]How do you think it will end?[/b]

Any ideas are welcome. This is by no means intended to be a purely philosophical or purely scientific discussion.

Nuclear war. I'm a pessimistic when it comes to the end of the world.

A classic reference for this question: http://www.livescience.com/technology/destroy_earth_mp.html

haha yeah I love that site. My favorite is the one with the tiny black hole consuming the earth. ;)

Gotta love the von Neumann machines.  It seems so perfect an idea.

[quote="ROLEX4life"]I say the world will end because we will blow ourselves up. We will decide randomly one day that "these chinese sons of bitches are going down" and then we will nuke china and they will be like "- - who the f*** is shooting us" and then some fat chinese guy in a spa in a room with a random JAPANESE flag will be like "oh well fire missles!" and then the french ppl will get involved and come up with some - technology that can somehow turn cigarettes into missles and they will take a nap because they are le tired and then attack and then that will get the russians involved and Stalin will be like AH MOTHALANDS!!! and then everyone will blow up cept Austrailia....they will be in nuclear winter and they will be like "wtf mate" and then the kangaross will die out cuz they are g@y[/quote]
 :rotfl: I shouldn't be laughing at that, but I am because it was funny.

Personally, I think von Neumann machines are the way to go. ;)
Anyone think that an AoPSer (no one in particular) will have a hand in ending the world?

lol the whole china thing - YOU STOLE THAT FROM A FLASH MOVIE!! lol and I was about to post it.. so  :P ..

Anyway, the video (its hilarious) can be found at http://www.albinoblacksheep.com i dont have the link but its titled "The End of the World".

Several thousand years: Gliese 310, an otherwise insignificant red dwarf star, may pass within 1.3 light-years of the Solar System, disturbing the Oort cloud. I doubt it will cause the end of life however.

3 billion years: Collision of Andromeda and Milky Way. It could very well mean the end, though it does not guarantee it.

5 billion years: Sun expands into red giant. HOWEVER, some scientific models predict that only 1 billion years from now, the Sun will expand to the point in the Earth will be too hot - stars tend to expand as they age, even while they remain on the main sequence.

Even if we build spaceships and colonize space and space is hyperbolic rather than elliptic, life's still going to end. Why?

1: Hydrogen, the fuel for stars, will run out eventually, so there will be no source of nuclear fusion or energy for us. We could think about radioactive decay but that obviously can't work out.
2: Entropy, which is related to #1 to an extent. Heard of the heat death of the universe? Well, eventually, the universe will reach an extremely high level of entropy and all that will be left is radiation, for even protons decay. We can't extract work from it.
3: Even if we could go through a supermassive spinning black hole to another universe, the radiation around it would kill us, an so would the super-high temperatures. And well... it probably won't work. ;)

And by using the collective "we", I am obviously referring to our intrinsic terrestrial bias.

The universe and life both seem so pointless. *sigh* I've turned nihilistic due to this all - though also because life and the universe seem to be accidents as well. Sad....

Well, i think pollution will killl us all,

perhaps due to lack of food and water pollution.

The human civilisation has been progressing drastically and dramatically yet we can't even solve the basic problem of lacking water.

The most basic of all is the hardest of all. :|

Notice that nuclear war does not actually destroy the earth, merely most of humanity. It is humbling to consider just how little an effect we humans can have at the scale of the cosmos.

he's not busy being born is busy dying...

so is the universe.

i hope it will last for at least 100 years so i can live :P

since u said that it is not meant 2b philosophical pr scientifidc i will say this.

About 10000 years from now the human beings will try to save earth from natural destruction posssibilities here in the solar system(such as the red giant phase of sun).So mankind decides to drive earth to a safe destination.Now the earth starts moving at 50 times the speed of light towards another galaxy.But on the way driver falls asleep and eath collides with a star.
Bang!Destroyed.


What's the smallest possible value of $ k \plus{} m \plus{} n$ if k, m, and n are positive integers such that $ \frac{1}{k} \plus{} \frac{1}{m} \plus{} \frac{1}{n} < 1$?

[hide]
Use Cauchy
$ (\frac 1k \plus{}\frac 1m\plus{} \frac 1n)(k\plus{}m\plus{}n) \ge 9$
We want $ (\frac 1k \plus{}\frac 1m\plus{} \frac 1n)$ to be as large as possible, so it will be 1.
Thus $ k\plus{}m\plus{}n\equal{}9$ is the smallest, when $ m\equal{}n\equal{}k\equal{}3$

But in the equation, it says only <, and not $ \le$, thus $ 9$ needs one more, making 10.
[/hide]

[hide="Solution"]
By AM-HM, $ \frac{a\plus{}b\plus{}c}3\geq \frac{3}{\frac 1a\plus{}\frac 1b\plus{}\frac 1c}>3\implies a\plus{}b\plus{}c>9$, making the minimum $ \boxed{10}$. This is clearly attainable.
[/hide]


would you please help me with that? the cumulants kn of X are defined by
g(t)=log(E(exp(tX)))
The nth cumulant is given by kn=the nth derivative of g at 0
Find the first 5 cumulants of the normal one dimensional distribution..
I am not english so i am not familiar with these terms. what do i have to do? which is the normal one dimensional distribution?


1.Let $ x,y,z>0$ and $ xyz\equal{}1$. Prove:
$ \frac{x}{{{y^3} \plus{} 2}} \plus{} \frac{y}{{{z^3} \plus{} 2}} \plus{} \frac{z}{{{x^3} \plus{} 2}} \ge 1$

2.Let $ x,y,z > 0$such that:$ 2\sqrt {xy} \plus{} \sqrt {xz} \equal{} 1$. Find minimum:
$ S \equal{} \frac {{3yz}}{x} \plus{} \frac {{4zx}}{y} \plus{} \frac {{5xy}}{z}$

[quote="Gnaht"]1.Let $ x,y,z > 0$ and $ xyz \equal{} 1$. Prove:
$ \frac {x}{{{y^3} \plus{} 2}} \plus{} \frac {y}{{{z^3} \plus{} 2}} \plus{} \frac {z}{{{x^3} \plus{} 2}} \ge 1$

2.Let $ x,y,z > 0$such that:$ 2\sqrt {xy} \plus{} \sqrt {xz} \equal{} 1$. Find minimum:
$ S \equal{} \frac {{3yz}}{x} \plus{} \frac {{4zx}}{y} \plus{} \frac {{5xy}}{z}$[/quote]
From which contest your inequalities?  :wink:


I know some proofs for the Pythagorean Theorem. There's one where u square the sides, there's the one where u put a huge square aruond the right triangle, any others because i'm curious to know others... maybe they can prove other triangle theorems (?)

[quote="shobber"]Consruct the altitude. Then use similar triangles.[/quote]
that's very interesting

There is one where you draw a square and then break it in 2 congrunet right triangles and a smaller square is formed.Then Use Areas and come to Pythagoras thm .

I remember a site with a lot of pythagorean proof animations that I saw earlier. Try using google search and typing in [pythagorean proof animation]

i read in a book that there are about 400 proofs of Pythagorean theorem, is that true??

There is a ton of proofs here:

http://www.cut-the-knot.org/pythagoras/index.shtml

Some of them are just variations of each other, though. :)

i found that website pretty useful and cool.  i really liked it.

Wow! That is an awesome site.

I really liked the proof where a circle was inscribed in the triangle and you proved it with the radius of the circle.

I liked about the Tangent Secant theorem


AMC AIME USA(J)MO USAMO MIT college probability

hi,
I am Dimitar from Bulgaria,and I've been on two IMOs and most surely I will go next year too. Is anybody else sure to go next year?

I know its still too early but there may be few like me .
I ' d like to contact with them :!:

As Peter said there are many MathLinkers returning next year to the IMO, and (probably) many new ones coming for their first IMO :)

im working very hard!! and im now enrolled to woot!! i hope to be there with manuel too!!

yeah, marko and i are working hard so everybody must hang out with us in slovenia!

[quote="Peter Scholze"]Hey, there are lots of guys here who will participate at IMO 2006... e.g. me, if nothing unpredictable happens ;). And from the rest of the german team, darij grinberg, NoName and ZetaX will most likely be there too.[/quote]

well, i think it's totally impossible for you not to go to slovenia.... have you ever heard an imo perfect scorer not being able to go to the next imo? :rotfl:

yes, but only those who finish their school education :blush:

[quote="orl"][quote="Soarer"] well, i think it's totally impossible for you not to go to slovenia.... have you ever heard an imo perfect scorer not being able to go to the next imo? :rotfl:[/quote]

I guess it could happen in the biggest of your neighbor countries.  ;)[/quote]

but not in Germany, right?  :lol:

i think i will also participate at IMO2006.

I hope to make it. I'm really working hard.
   Ashwath

Thanks all of you and Good luck !

davron

[quote="Albanian Eagle"]I guess it's time to say good luck to every one now! :lol:[/quote]

So, good luck to no one, jajaja. Good luck to everyone, more specially to all MathLinkers and more more specially to the Argentinin team!!! Have a nice time in Slovenia, and enjoy a lot!!!

Ultimately, results are important but not that important. What is more important than your results is whether you did your best. This competition is more of a test of yourself, of what you know you can achieve. So good luck to everyone!!

Anyway, all the best to Mathlinkers also!! I'm an Observer from Singapore and I haven't really seen many of you, but I hope to see more Mathlinkers!!

IMO Day 1 in a few more hours..

Good luck to all IMO participants.

Take it easy!

Japanese Communities moderator

kunny

[quote="dblues"]Yup, I am Observer B. :)[/quote]I thought so, because you weren't in the computer room here as I was online and you also :)

I've added you to the [url=http://www.mathlinks.ro/Forum/weblog.php?w=512]IMO 2006 AoPS - MathLinks Blog[/url] contributors

So guys,it's already eight o'clock in Slovenia and the competition will start after an hour,and now it is really time to wishe [b][size=150]good luck[/size]  [/b] to all you,and me too :blush: .
Thank you for averithing!!!!!!!!!!!!
Tigran

25 minutes before the test! I wish all the participants good luck! And I hope that China will have more perfect scorers than last year!

I am so Sorry that i am so late posting this:
Firts of all i wish allthe participants good luck and th eleaders too
Second I know some of the members of My country team:
5 winners of National Olympiad(there is 1 girl there) and one guy whoes name is [b]Utkir Boltayev[/b] who is 10 th grade and that's why did not participated in Nationals but he got medal in Kazakstan olympiad. I think that My country did mistake not makin the TST and take the members as mentioned above.
My prediction is 2 bronze  :(  :( medals(the medalists will be from the Buhara special math school).

[quote="kunny"]Do your utmost, Japanese Team and Tajikistan Team! :pilot:[/quote]
Hi kunny, how are you?
sorry what you are talking about I can't understand, please can you explain it one more time.

Sincerely
Davron Latipov

Hi! Davron. Today you worked hard. Give my regards to abdurashidjon.

kunny

Hi, I wish all of you the best of luck, and I hope you do a lot socialising now.  It makes me think back of my days (my score wasn't that superb (three points) but what an experience it was).

IMO is quite different, when I went we didn't have internet access really, I went to a cyber cafe in Glasgow to tell my parents my score and make arrangements to pick me up at the airport .  On my own.

Just a dumb question, guys you are actually meeting each other there right, you are not continuing mathlinks business as usual over there, talking to each other via mathlinks?  :rotfl:  :lol:


geometry
geometric transformation
reflection
function
algebra
polynomial
ratio

There are 4 parallel windows (parallel planes, that extend into infinity). A beam of light is shot into that from the top, and bounces around $n$ times, find the number of distinct paths that the light can follow in terms of $n$ (Note: it does not just hit the top layer and go away, the light has to enter between the panes of glass for it to count

well, here is a hint since i am not getting any responses

[hide]set up a recursion so say f(n) is what we are looking for, once it bounces off various windows, we get f(n)=f(n-1)+... (like for the first one)

i have this recursion, but i think it is wrong, because to convert it to an explicit equation, i have to solve a general cubic, i am not sure if there is an elegant answer though, this is an extension of an easier problem that i have seen[/hide]

Ok once we have a recursive formula this can be solved relatively easy, in my opinion I think the hardest part is determining the recursion.

Picture of problem:

Pane 1------------------------...
Pane 2------------------------...
Pane 3------------------------...
Pane 4------------------------...

Just to confirm, the light can only reflect on those dashed lines, right?
Let $f(n) = \text{number of possible paths for n reflections}$

Let's start with an easy case:

Pane 1------------------------...
Pane 2------------------------...

Let's work out some numbers by hand:

$f(0) = 1$ This occurs when the ray of light passes through the whole system.
$f(1) = 1$ This occurs when the ray of light passes through pane 1 and reflects off pane 2.  The note claims that reflection does not occur on top pane.
$f(2) = 1$ This occurs when the ray of light passes through pane 1 and reflects off pane 2 and then off of pane 1.

For the case where there is only 2 panes, it is obvious the function is $f(n) = 1$,  because the ray has no choices.

Let's start with a slightly less easy case, but one that is more enlightening:

Pane 1-----------------------...
Pane 2-----------------------...
Pane 3-----------------------...

Let's work out some numbers by hand:
$f(0) = 1$ The light passes through all three panes.
$f(1) = 2$ The light either reflects off pane 2 or pane 3. The note does not allow reflection from the top of pane 1.
$f(2) = 3$ Reflection at 3,2 or 3,1 or 2,1.

Let's try to generalize this:

$f(n) = g_{\text{reflection at pane 2}}(n-1) + g_{\text{reflection at pane 3}}(n-1)$

Notice that $g_{\text{reflection at pane 3}}(n-1) = f(n-1)$ if you look at the system upside down.
Also notice that if the ray reflects at pane $2$, then it must also reflect at $1$ for sufficiently large $n$.

$f(n) = f(n-1) + f(n-2)$  Wow, fibonacci!  The only exception is that $f(0) = 1$ and $f(1) = 2$, as opposed to both $1$ in the fibonacci sequence.

This can then be solved using generating functions relatively easy, but it's not as difficult as determining the recursion.

Let me end this post at that, and then I will attempt the $4$ pane problem.

Alright, the $4$ pane problem.

Picture:
Pane 1----------------------------...
Pane 2----------------------------...
Pane 3----------------------------...
Pane 4----------------------------...

Let's work out some values by hand:
$f(0) = 1$
$f(1) = 3$
$f(2) = 6$
$f(3) = 14$

Let  $f(n) = \text{number of possible paths for n reflections}$

Then, $f(n) = g_{\text{reflection at pane 2}}(n-1) + g_{\text{reflection at pane 3}}(n-1)+g_{\text{reflection at pane 4}}(n-1)$

Notice that $g_{\text{reflection at pane 4}}(n-1) = f(n-1)$, when you look at the system upside down.

Also notice that for sufficiently large $n$, then  $g_{\text{reflection at pane 2}}(n-1)$ must reflect off of pane $1$.

These two steps were used in the last post.

$f(n) = f(n-1) + f(n-2) + g_{\text{reflection at pane 3}}(n-1)$

$g_{\text{reflection at pane 3}}(n-1) = g_{\text{reflection at pane 2}}(n-2) + g_{\text{reflection at pane 1}}(n-2)$

Notice that this is from a different direction, so the scenario is changed, but we can transform the system by looking at our picture upside down again.  As the directions are changing I'm going to use a different variable to remove the changes imposed by the direction.

Notice $g_{\text{reflection at pane 1}}(n-2) = f(n-2)$, then

$y(n-1) = y(n-2)+f(n-2), y(0) = 1$
$y(n) = 1+\sum_{k=0}^{n-1}f(k)$

$f(n) = f(n-1)+f(n-2)+1+\sum_{k=0}^{n-2}f(k)$ for $n \ge 2$ and $f(0) = 1, f(1) = 3$

This is still not a desirable form.
Even though it can by slightly shortened, by $f(n) = f(n-2)+1+\sum_{k=0}^{n-1} f(k)$

In order to kill the sum, we employ $f(n-1) = f(n-3)+1+\sum_{k=0}^{n-2} f(k)$

$f(n) = 2f(n-1)+f(n-2)-f(n-3)$ for $n \ge 3$ and $f(0) = 1, f(1) = 3, f(2) = 6$

The recursive step is finished.

$(E^3-2E^2-E+1) f(n) = 0$

Then if $a,b,c$ are the roots of $E^3-2E^2-E+1$, $f(x) = C_a a^x + C_b b^x + C_c c^x$, where $C_a, C_b, C_c$ can be solved for using the known values.

Unfortunately, I don't have access to a CAS so I'm having difficulty computing the roots of the cubic.

Edit: whoops I didn't realize, you wrote that it Xevarion, my bad.  For all who read, the equivalent statement written on my post was written first by Xevarion.

You can solve that recursion as follows:

Find the three roots $r_1, r_2, r_3$ of $x^3 = 2x^2 + x - 1$.

$f(n) = C_1r_1^n + C_2r_2^n + C_3r_3^n$

where you can find the constants from the first three terms of $f$.

i got exactly that far too, i.e. getting that polynomial to solve for the ratios for the geometric series, but perhaps this problem was only meant to be done so you get the idea, because that seems that the only way to solve it will with the general cubic formula, and yes i followed the same idea of definining a new function, and then getting rid of it


Here's the problem I have no idea how to do! :(
Prove that with every c>8/3, there exist a real number k in order to have [k^c^n] is a prime number with every positive integers n([] is the greatest integer function).
Any hints for this problem will be very appreciated!

[quote="Akashnil"]I think managed to prove it with the following assumption:

For all $ c > \frac {8}{3}$, There exists a prime $ p_0$ such that for all primes $ p\ge p_0$, there exists another prime in the interval $ (p^{c},(p \plus{} 1)^{c})$

It might have been proven, like the Bertrand postulate, but I don't know.

Proof:
Let $ k_0 \equal{} \log_c p_0$
by our assumption there is a prime in the interval: $ ((k_0^c)^c,(k_0^c \plus{} 1)^c)$, call it $ p_1$
Let $ k_1 \equal{} \log_{c^2} p_1$
Since $ (k_0^c)^c < k_1^{c^2} < (k_0^c \plus{} 1)^c\implies k_0^c < k_1^c < k_0^c \plus{} 1$
Hence $ \lfloor k_1^c\rfloor \equal{} p_0 , \lfloor k_2^{c^2}\rfloor \equal{} p_1$
Again
by our assumption there is a prime in the interval: $ ((k_1^{c^2})^c,(k_1^{c^2} \plus{} 1)^c)$, call it $ p_2$
Let $ k_2 \equal{} \log_{c^3} p_2$
Since $ (k_1^{c^2})^c < k_2^{c^3} < (k_1^{c^2} \plus{} 1)^c\implies k_1^{c^2} < k_2^{c^2} < k_1^{c^2} \plus{} 1$
Also $ k_1^{c^2} \plus{} 1 < (k_1^c \plus{} 1)^c\implies k_1^{c^2} < k_2^{c^2} < (k_1^c \plus{} 1)^c\implies k_1^c < k_2^c < k_1^c \plus{} 1$
So we now have $ \lfloor k_2^c\rfloor \equal{} p_0,\lfloor k_2^{c^2}\rfloor \equal{} p_1,\lfloor k_2^{c^3}\rfloor \equal{} p_2$

Likewise Define $ p_{t \plus{} 1}$ as a prime in the interval $ ((k_t^{c^{t \plus{} 1}})^c,(k_t^{c^{t \plus{} 1}} \plus{} 1)^c)$ and define $ k_{t \plus{} 1} \equal{} \log_{c^{t \plus{} 2}} p_{t \plus{} 1}$ where $ t$ is a positive integer.
The sequence $ \{k_t\}_{t\in \mathbb N}$ converges because it is increasing, and still bounded because $ \lfloor k_t^c\rfloor \equal{} p_0$ for all $ t$.

Take $ k \equal{} \lim _{t\to \infty} k_t$ which clearly has the required property.

[color=blue]@Mods: I think it could do in the Olympiad forum as well  :maybe:[/color][/quote]
I really don't think your rephrased problem is equivalent to mine since my problem aslo includes n, which is any positive integers, except your assumption is still true if you prove the similar thing for:p^c^n and(p+1)^c^n . Do you use the Chebychev's theorem? I'm trying to use it but it still doesn't work for me.

You misunderstood me. I didn't rephrase your problem. The assumption I took, was what I used to solve your problem as it is. I didn't even try to prove that assumption. The proof I gave was not to the assumption, but to your problem as it is.

And plz don't quote the entire post before you.


Do there exist numbers $ x,y,a,b\in\mathbb R$ with $ x>y>0$ and $ a>b>0$ such that
\[ x^n\plus{}y^n\equal{}a^n\plus{}b^n,\quad\forall n\in\mathbb N\]
but $ (x,y)\not\equal{}(a,b)$?

\begin{align}
n=1&: &{}x+y&=a+b\\
n=2&: &{}x^2+y^2&=a^2+b^2
\end{align}
Squaring (1) and substracting 2 times (2), we obtain $ -(x-y)^2=-(a-b)^2$ and $ x-y=a-b$ since $ x>y,a>b$. But together with (1) this means $ x=a, y=b$, so there obviously is no such quadruple.

[quote="timwu"][hide]
You can also use Viete's Theorem[/hide][/quote]

Do you mean using $ xy\equal{}ab$ and looking at two quadratic equations, or is there some other application I'm not seeing?

[quote="timwu"]Yes. That's just my intuitive reaction to this question at the first sight.[/quote]

in fact that was my approach too...but then I realised proving it from the equality of the differences is faster :P


Does anyone know Chinese? If so, can we chat in this thread with Chinese to enhance our knowledge? Just so you know it's preferable to use Han Yu Pin Yin for me because I don't have a Chinese writing program.

STARTS OUT:

Yao shi ni bu dong wo zai shuo shen me, ni shi yi ge ben dan  :)

[quote="junggi"]Lee may lan jiao.(NO idea at ALL if that's spelled right.it's supposed to mean 'you don't have any peppers')[/quote]

"Ni mei lai jiao" ;)

I think all operating systems already come with some sort of Input Method Editor that allows you to type in Chinese or many other languages. All you need to do is set it up and then you can type stuff like:

谁说中文话，朋友遍天下.

:)

[quote="junggi"]Lee may lan jiao.(NO idea at ALL if that's spelled right.it's supposed to mean 'you don't have any peppers')[/quote]

wow you still remember that.

its ni mei you la jiao i think

Wo bu dong "bubka" zai shuo shen me. Ni men kan de dong ma?

[quote="Bictor717"]Wo ye bu ming bai 'bubka' xie de zi. Dui bu qi.[/quote]

Ni guang hui pu tong hua ma?

Well that does it - there are Chinese people everywhere. :P Use phonetic "accents", guys. I have no idea what half of you are trying to say, aside from the peppers stuff. And bubka is apparently using very odd pin yin's. Canton perhaps?

Bubka 是印度人。他写的汉语都是瞎编地。

我觉得我用英文会更make sense. Basically, each letter on the qwerty keyboard represents 一种笔画。[img]https://upload.wikimedia.org/wikipedia/commons/thumb/6/60/5strokes.jpg/1200px-5strokes.jpg[/img]
Essentially, a character is broken down into components, which are usually (but not always) the same as radicals. These are typed in the order in which they would be written by hand. In order to ensure that extremely complex characters do not require an inordinate number of keystrokes, any character containing more than 4 components is entered by typing the first 3 components written, followed by the last. In this way, each character's data can be entered with no more than 4 keystrokes.

Wubi distributes its characters very evenly and as such the vast majority of characters are uniquely defined by the 4 keystrokes discussed above. One then types a space to move the character from the input buffer onto the screen. In the event that the 4 letter representation of the character is not unique, one would type a digit to select the relevant character (for example, if two characters have the same representation, typing 1 would select the first, and 2 the second). In most implementations, a space can always be typed and simply means 1 in an ambiguous setting. Intelligent software will try to make sure that the character in the default position is the one desired.

Many characters have more than one representation. This sometimes is for ease of use, in case there is more than one obvious way to break down a character. More often though, it is because certain characters have a short representation that is less than 4 letters, as well as a "full" representation.

For characters with fewer than 4 components that do not have a short form representation, one types each component and then "fills up" the representation (that is, types enough extra keystrokes to make the representation 4 keystrokes) by manually typing the strokes of the last component, in the order they would be written. If there are too many strokes, one should write as many as possible, but put the last stroke last (this mirrors the component rule for characters with more than 4 components outlined above).

Once the algorithm is understood, one can type almost any character with a little practice, even if one hasn't typed it before. Muscle memory ensures that frequent typists using this method don't have to think about how the characters are actually constructed, just as the vast majority of English typists don't think very much about the spelling of words when they write. (taken from wikipedia)

你好厉害呀！我看起来觉得特别复杂  :laugh:

五笔我一个中国人都看不懂 :rotfl:

[quote=dhaops]你好厉害呀！我看起来觉得特别复杂  :laugh:[/quote]

(taken from wikipedia)

[quote=Felicityhe]我在学写作文，很无聊。回打拼音的人，你用哪个keyboard setting? 是用拼音还是五笔？[/quote]

"回打" y e s

天哪！大家的语言能力比我好太多了！有一点伤心 :(
(Idk... fresh from google translate)

[quote=Marcus_Zhang]its another da.[/quote]

no

it's 会打

I think it's supposed to be 回答 as in 回答问题.
Edit: never mind read the context

i can't write chinese to save my life so i just use qwerty han yu pin yin :P

[quote=mm999aops]ummmm
umm
how boring tell me what this translates too
画蛇添足
and this
孔子[/quote]

the first one: gild the lily
second: Confucius

[quote=g1zq][quote=azure123456]你们是要考中文AP吗？[/quote]

中文也有AP吗？[疑问][/quote]

有的啊，有APMandarin 好像，我有同学要考的


If I want to display a set of numbers like $\left\{1,2,3...52,000,000,000\right\}$ it does not carry over between lines like words do.  If the set is too long to fit in the line it is in, instead of carrying over to the next line, it just keeps going off the page.  I can edit the words around it to make it fit temporarily, but as soon as I change something, the problem reoccurs.  Do you know if there is a something that will allow the set to carry over to the next line?

Thanks

LaTeX doesn't normally do word wrap in math mode as it would require an understanding of the context which it isn't capable of. Perhaps you should display your set using \ [ ... \ ] so you get something like:

[quote]The required set is:
\[\left\{1,2,3...52,000,000,000\right\}\]
which has the property that etc etc[/quote]

If the set itself is too long to fit on one line then amsmath's split environment allows line breaks that you put in yourself. So
[code]\begin{equation*}\begin{split}
\{1,2,3,4,5,6,7,8,9,10,\dots,\\ 100,20000000000000\}
\end{split}\end{equation*}[/code] gives
\[\par\begin{equation*}\begin{split}\{1,2,3,4,5,6,7,8,9,10,\dots,\\ 100,20000000000000\}\end{split}\end{equation*}\]


algebra
polynomial
number theory unsolved
number theory

Find all the triples (x,y,z) of positive rational numbers such that x<=y<=z and x+y+z, 1/x + 1/y + 1/z, xyz are integers.

Let $x+y+z=a, \frac 1 x + \frac 1 y + \frac 1 z = b, xyz =c$, where $a,b,c$ are positive integers.
Then $xy+yz+zx = bc$ is an integer too. It follows that $x,y,z$ are roots of the polynomial $P(t)=t^3 - at^2 + bct - c$, which is normalized and has integer coefficients. It is well-known that any root of such a polynomial is either an integer or an irrational number. Since $x,y,z$ are supposed to be rational, it follows that they are integers.

And the problem is now to solve $\frac 1 x + \frac 1 y + \frac 1 z = b$ in positive integers.
Note that $b \in \{1,2,3 \}$. And $\frac 3 x \geq b$ so that $x \leq \frac 3 b.$

- For $b=3$ we easily have $x=y=z=1$.
- For $b=2$, we then have $x \leq \frac 3 2$ so that $x=1$ and $\frac 1 y + \frac 1 z = 1$.
Thus, $ \frac 2 y \geq 1$, from which we deduce that $y = 2$, and then $z=2$.
Conversely, $(1,2,2)$ is clearly a solution.
- For $b=1$, following the same lines as above we deduce that the solutions are $(2,3,6),(2,4,4),(3,3,3)$.

Thus there are five solutions which are $(1,1,1),(1,2,2),(2,3,6),(2,4,4),(3,3,3)$.

Pierre.


integration
calculus
calculus computations

put it into partial fractions, and then use laplace transformation. :wink:

Could you show how you mean more explicit please?

[quote="lehungvietbao"]By Laplace transform calculate the following sums

1)  $ \sum_{n=1}^{+\infty}\arctan \frac{2}{n^2}$

2) $ \sum_{n=1}^{+\infty}\arctan \frac{3}{n^2+3n+1}$

3) $ \sum_{n=1}^{+\infty} \frac{n}{\left ( n^2-\frac{1}{4} \right )^2}$

[hide]Apply the following formulas
 $ \sum_{n=k}^{\infty}F(n)=\int_{0}^ {+\infty}\frac{e^{-kt}f(t)}{1-e^{-t}}dt$

$ \sum_{n=k}^{\infty}(-1)^nF(n)=(-1)^k\int_{0}^ {+\infty}\frac{e^{-kt}f(t)}{1+e^{-t}}dt$
[/hide][/quote]

Apply MAZ Summation will get:
[img]https://sites.google.com/site/atzahrani/example-artofproblemsolving21.png[/img]


We have the set $A = \{ a\in R | a = \sqrt{\overline{xy1}-\overline{1xy}}\}$. Find $xy$ if $a \in (2\sqrt{5},12]$.

$ A \equal{}\{ a\in R | a \equal{}\sqrt{\overline{xy1}\minus{}\overline{1xy}}\}$. Find $ \overline{xy}$ if $ a\in (2\sqrt{5},12]$.
$ \overline{xy1}\minus{}\overline{1xy}\equal{}9\overline{xy}\minus{}99$. It is sufficient that $ x\geq1$ and $ y\geq1$for the radical to exist.
$ a\in (2\sqrt{5},12]\rightarrow 9\overline{xy}\minus{}99\in (20; 144]\leftrightarrow\overline{xy}\in [14;27]$
So we have fourteen pairs of numbers $ (x,y)$ which satisfy $ a\in (2\sqrt{5},12]$


abstract algebra
symmetry
group theory
superior algebra
superior algebra solved

G is non-abelian group of order 15. Prove that Z(G)=1. use the fact that <g> < Cg(g) for g in G to show that there is at most one possible class equation for G

There is an other way:
$G$ contains $5$ $3$-Sylows and $1$ $5$-Sylow. We just have to see why the $5$-Sylow is not $Z(G)$ ?

We does it have $5$ $3$-sylows¿ The only group with $15$ elements is $\mathbb{Z}/ 15 \mathbb{Z}$ and thus abelian.
This follows directly from the sylow-theorems (they give that there is just one $3$-sylow and one $5$-sylow). And then $G$ is isomorphic to the direct product of the sylows, meaning that $G$ is the cyclic group with $15$ elements.

If you start with: Let $G$ be a nonabelian group of order $15$, you will be able to deduce anything. There is no such group; the only type of group of order $15$ is the cyclic one.

[quote="Soarer"][quote="ZetaX"]$G$ is isomorphic to the direct product of the sylows[/quote]

May I know if this is true for arbitrary group or only abelian group?[/quote]
It holds for nilpotent groups
[url]http://en.wikipedia.org/wiki/Nilpotent_group[/url]

[quote="ZetaX"][quote="amfulger"]Are there non-abelian groups of order 15?[/quote]
What is this question intended to mean¿[/quote]

I don't really understand what has happened. I saw this post in the $10^{th}$ page or so in this forum, but I only saw the first reply. I don't know why though.
Anyway, given what I saw, I wanted to point out that there are no abelian groups with $15$ elements.
My mistake. Or maybe not.

the question itself is weird (but still valuable) as there actually aren't any nonabelian groups of that order (as many have pointed out), actually you only have the cyclic :

but still, it can be proven without sylow or anything, and can be done so in a much more general case

let $G$ be a group order $p q$ with p and q prime (they don't have to be different)
suppose the group is not abelian, show that the center of the group is trivial

solution : if the order is p, the you can take $G/Z(G)$ , and its order will be q, prime
as it has prime order, it is cyclic
but in any group, the quotient $G/Z(G)$ can only be cyclic when trivial

you can see this easily : it $G/Z(G)$ is cyclic, there is an element a such that every g in G can be written as
$z a^m$ with z in the center
this means $g_1 g_2=z1 a^m z_2 a^n = z_1 z_2 a^m a^n=z_2 z_1 a^n a^m=z_2 a^n z_1 a^m=g_2 g_1$ thus the group is abelian and thus the center is the full group

thus we have established that the order cannot be p, and by symmetry it can't be q either

[quote="amfulger"][quote="puuhikki"][quote="amfulger"]I wanted to point out that there are no abelian groups with $15$ elements.[/quote]

What about a cyclic group with 15 elements?[/quote]

$cyclic\Rightarrow commutative=abelian$[/quote]
@amfulger: read the part quoted by puuhikki again and you will understand ;)

[quote="ZetaX"][/quote][quote="amfulger"][quote="puuhikki"][quote="amfulger"]I wanted to point out that there are no abelian groups with $15$ elements.[/quote]

What about a cyclic group with 15 elements?[/quote]

$cyclic\Rightarrow commutative=abelian$[/quote][quote="ZetaX"]
@amfulger: read the part quoted by puuhikki again and you will understand ;)[/quote]

nope. I still don't understand.

[edit] oups. now I get it


Billy has three times as many nickels as dimes. If he had 45 more dimes he would have twice as much money. How many nickels and dimes does he have altogether now?

[hide]
solve the system

n=3d
5n+10d=450  [you notice the amount of money trivially b/c when u add 450 cents, you've doubled your quantity].

d=18,n=54

[/hide]


linear algebra
matrix
limit
algebra
polynomial
quadratics
quadratic formula

Let $ A\equal{}\begin{bmatrix}a&b\\c&d\end{bmatrix},\forall a,b,c,d\in\mathbb{R}$ find the relationships between $ a,b,c,d$ which make $ \lim_{n\rightarrow\infty}A^{n}\equal{}\begin{bmatrix}0&0\\0&0\end{bmatrix}$ :D

The condition is that the absolute values of both eigenvalues are less than $ 1.$ The eigenvalues are roots of the characteristic polynomial:

$ x^2 \minus{} (a \plus{} b)x \plus{} (ad \minus{} bc) \equal{} 0.$

The quadratic formula gives us $ \frac {a \plus{} b\pm\sqrt {(a \plus{} b)^2 \minus{} 4(ad \minus{} bc)}}2.$

If the discriminant is negative - that is, the case of complex roots - we would have roots smaller than $ 1$ precisely when $ ad \minus{} bc < 1.$

For the case of two real roots, $ |ad \minus{} bc| < 1$ is a necessary condition but not a sufficient condition. More could be teased out by algebra.


Let $n$ be an [b]odd [/b]positive integer.

In each square of the $n \times n$ grid, there is a lamp and a button. The pressing a button switches all the lamps in that row of column (i.e. pressing $(i, j)$ switches the state of all the lamps $(i, k)$, $(k, j)$, $k = 1, \ldots , n$, including $(i, j)$ itself.)

Determine all the states of the grid for which we can turn on all the lamps by pressing some set of buttons.

I think this has been discussed before. Look for MindFlyer's posts in the "Combinatorics" section (Proposed Problems, I think).

Found it: http://www.mathlinks.ro/Forum/viewtopic.php?t=5754

Great solution by the way.


#2 on part A of Irish olympiad:

The Fibonacci sequence is defined by F0 = 1, F1 = 1, Fn+2 = Fn+1 + Fn. Show that some Fibonacci number is divisible by 1000.

I'll post a solution soon. Just thought I would give people problems from a not so well published olympiad.

Just glancing at it, could you use mods to find the period mod some divisors of 1000, and then use the CRT to prove such a number exists?

By the way, this is a nice problem, thank you for posting it.

I think that's the right idea. This is a bit of an awkward definition- if we define $F_{1}=1, F_{2}= 1$, etc, then we can use binet's to show that every 6th number is divisible by 8, and do something similar for 125.

Yup, that is the way to go about it.

Is it just me, or does the Irish Olympiad seem much easier than the USAMO and Chinese olympiad. I was actually able to get 5 of 6 questions completely right on the Irish one, which I would never be able to do within the time limit for a USAMO.

Well...I think it's pretty logical that the Irish olympiads would be easier. Usually, the difficulty of the national olympiad is proportional to the country's strength on the IMO.

[quote="neelnal"]
Is it just me, or does the Irish Olympiad seem much easier than the USAMO and Chinese olympiad. I was actually able to get 5 of 6 questions completely right on the Irish one, which I would never be able to do within the time limit for a USAMO.[/quote]

If Chinese competitors were given something of IrMO difficulty, there would almost certainly be more than 6 perfect scorers, defeating the purpose of the competition. Whereas, a good majority of Irish competitors would have a hard time getting more than a few marks on a USAMO paper (indeed, a lot of IrMO competitors have difficulty with the current level of challenge). As Phelpedo said, nationals should be proportional to performance in the IMO (and our highest imedal is silver, which we've won only once; twice if we're allowed to count Martin Orr).

As regards the problem, a stronger statement: for all numbers there exists a Fibonacci number (in fact, an infinite number of Fibonacci numbers) that is divisible by it.

If your trying to prove $n$ divides some fib number, look at the sequence mod n.
to prove that 1,1 comes up again in the sequence mod n, notice that there are finitely many possible pairs a,b mod n of consecutive fib numbers that can occur. Thus some a,b mod n must be repeated. But if we have two fib numbers mod n, we can also work back to the previous term mod n. So if 1,1 didnt occur again, 1,1 could never have occured in the first place. Which is wrong because the first two terms are 1,1 ! So we know 1,1 occurs again in the sequence. The number before this must by equivlent to 0 mod n.

Sorry if my explanation sucked, im tired.


Its pretty easy I know, but my teacher solved it and ended up with n^5 with some constant whereas I ended up with n^6 with some constant. Who is correct?

http://img397.imageshack.us/img397/3014/summation4vk.jpg

Thanks in advanced!


A circular hatch in the hull of a submarine has a radius of 40.0 cm (0.40 m). The submarine is 850.0 m under water. If atmospheric pressure above the ocean is 1.01*10^5 Pa and the air pressure inside the submarine is 1.31*10^5 Pa, what net force is exerted on the hatch?

I have no clue how to work this problem. Please help me.

Basically what you have to do is figure out the total pressure being exerted by the air and water on the hatch, minus the pressure being exerted upward, and then rely on $P = F/A$.  I believe that pressure underwater increases at a constant rate per meter you travel downward, but you'd have to look that up.  I hope that helps.

Yes, the pressure increases lineraly:

$p=\varrho gh$,

where $h$ is the depth.

[quote="RHSLilSweetie07"]The density of sea water is 1025 kg/m^3.[/quote]
Then you can use that and the equation given above to figure out the pressure exerted by the seawater.  Add that to the pressure exerted by the air, and you have total downward pressure.  Subtract from that the pressure exerted upware by the air in the submarine, and you have net pressure.  Then, multiply that by the area of the door, and you have the net force exerted on the door.

[quote="RHSLilSweetie07"]How do you find the pressure exerted by the seawater?  I thought you used the formula

atmospheric pressure + density*height*gravity

I'm confused.  Please help.[/quote]
That IS what you do.

so, to find the pressure exerted by the air, it's

1.01*10^5 + (1025)(9.81)(850) = 8.6*10^6 Pa

To find the pressure exerted by the seawater, it's
(1025)(9.81)(850) = 8.5*10^6, right?

Now, how do I find the pressure exerted upward?  That's the buoyant force.  I know that buoyant force = density of the fluid*Volume displaced by the sub.*gravity.  How do I calculate the volume displaced by the submarine?

you don't need to calculate volume occupied by the submarine itself..
it's not that you have to use archimede's law: just pascal one, that states that pressure is the same at same heights..
so, pressure on the hatch is actually $p_0 + \ro gd - p_i$, where $d$ is depth, $p_0$ and $p_i$ are respectively amosferic pressure and pressure inside submarine.
hence, force on the hatch is $\pi r^2 \cdot (p_0 + \rho gd - p_i) = 41,5 N$

"so, pressure on the hatch is actually , where  is depth,  and  are respectively amosferic pressure and pressure inside submarine.
hence, force on the hatch is ...."

When I used that equation, I came up with -1.09*10^4:

(1.01*10^5)+(9.81)(850)-(1.31*10^5) = -21661.5

Then that number times pi*r^2, -10888.26.

Did I do something wrong?

actually, _i_ did it wrong..
but now magically a $\rho$ (density of salt water) has appeared in my post.. with that little correction it should be ok ;)
sorry :)

Thanks...so the pressure on the hatch is 4281092 or 4.3*10^6 Pa.  So that's the net force exerted on the hatch, or am I leaving out something?

uhm.. i've always known that pressure is totally different from force..
so i think net force (as someone already said here) is $F = pA$, where $A$ is the surface over where pressure is made..
in this case, pressure is $p = p_0 + \rho gd - p_i$ and force is $F = \pi r^2 p$...

Thanks for all the help ma_go!  You made it really simple.


can anyone help with dis prob:
With n, a, b as positive integers, find a/b using the equation below with n as the smallest possible number.

the above statement was pretty confusing since i didn't kno how to use latex,and i also screwed up the equation
here it is again,
With n, a, b as positive integers, find a/b using the equation below with n as the smallest possible number:
$(a+bi)^{n}=(a-bi)^{n}$
if anyone can answer, it'd be appreciated

Sorry about not seeing this earlier.  As for a solution the easiest way to start is to note that the real part of the power of a complex number and the same power of its conjugate are equal because they must contain only terms which have an even number of imaginary terms and therefore the sign of b will be taken to an even power and must be a positive factor.  Similarly the imaginary parts of the must be opposites of each other and are as a result only equal when the imaginary part is 0.  The case where n=1 is out of the question because b must be positive.  The case where n=2 has an imaginary part equal to $\pm(3a^{2}b-b^{3})i$.  In order for this to happen we need $3a^{2}=b^{2}$ which is impossible on the integers because b would need a factor of 3 which would require a to have a factor of 3, and so on forever.  So there is no solution for n=2.  Finally, consider n=3 which has imaginary part $\pm(4a^{3}b-4ab^{3})i$.  Thus $a^{2}=b^{2}$ so $\frac{a}{b}=1$.


calculus
integration
algebra
function
domain
abstract algebra
vector

$\mathbb R[X,Y]/(X^2+Y^2-1)$ is an integrally closed domain.

Let $A=\frac{\mathbb R[X,Y]}{(X^2+Y^2-1)}$. We also denote it by $\mathbb R[x,y]$ with $x^2+y^2=1$.
Let $B=\frac{\mathbb C[X,Y]}{(X^2+Y^2-1)}=\mathbb C[x,y]$ with $x^2+y^2=1$.
It is easy to see that $A$ and $B$ are domains.
Their fields of fractions are $Q(A)=\mathbb R(x,y)$ with $x^2+y^2=1$ and $Q(B)=\mathbb C(x,y)$.

[b]Lemma:[/b] $B$ is integrally closed.

Since the proof of the lemma is very technical, let's see first how it is applied.
Let $\frac pq$ with $p,q\in A$ be an element of $Q(A)$ which is integral over $A$. Since $A\subset B$, $\frac pq$ is also an element of $Q(B)$ integral over $B$. But by the lemma, $B$ is integrally closed, so $\frac pq\in B$. It is easy to see that $\mathbb C[x,y]\cap \mathbb R(x,y)=\mathbb R[x,y]$. From these it follows that $\frac pq\in A$, so $A$ is integrally closed.

[b]Proof of the lemma:[/b]: Integral closure is a local property so it suffices to prove that $B_m$ is integrally closed for every maximal ideal $m$ of $B$.
By the Nullenstellensatz, the maximal ideals of $B$ have the form $(x-u,y-v)$ for some $u,v\in\mathbb C$ such that $u^2+v^2=1$.
The Krull dimension of $B$ is $1$.

Then $B_m$ is a noetherial local domain of Krull dimension $1$. The maximal ideal of $B_m$ is $mB_m$.
It is not hard to see that $mB_m$ is the kernell of the ring homomorphism $B_m\to\mathbb C: \frac fg\to \frac{f(0)}{g(0)}$, $g\not\in m$. Hence $B_m/mB_m=\mathbb C$.

We use the following:
For a noetherian local domain $(R,m)$ of dimension $1$, the following are equivalent:
i) $R$ is a discrete valuation ring;
ii) $R$ is integrally closed;
iii) $R$ is a regular local ring i.e. $dim_{R/m}\frac m{m^2}=dim_{Krull}R$;
iv) $m$ is a principal ideal;

Therefore it suffices to prove that $dim_{\mathbb C}mB_m/m^2B_m=1$. The $\mathbb C$-vector space on the left is obviously generated by $x-u$ and $y-v$.
Up to a change of variables we can assume $uv\neq 0$. Then we have $(x+u)(x-u)+(y+v)(y-v)=x^2+y^2-u^2-v^2=1-1=0$.
Let $R=B_m$, $n=mB_n$. Then $R$ is a noetherian local domain with maximal ideal $n$, quotient field $\mathbb C$. We want to prove that $dim_{\mathbb C}\frac n{n^2}=1$. From $(x+u)(x-u)+(y+v)(y-v)=0$ we get $(x+u+n)(x-u+n^2)+(y+v+n^2)(y-u+n^2)=n^2$ in $n/n^2$ as an $R/n$ vector space. By the assumption $uv\neq 0$, we see that $x+u+n$ and $y+v+n$ are nonzero elements of $R/n$ i.e. they are not $n$ i.e. $x+u,y+v\not\in n$. This proves that $x-u+n^2$ and $y-v+n^2$ are linearly dependent over $R/n$. Since $B$ is not a field, $R$ is not a field either, hence $n\neq n^2$. This proves that $n/n^2$ is nonzero. Since it is generated by $2$ linearly dependent elements, it must have dimension $1$.


Here's a cool online edition of Euclid's [u]Elements[/u]: http://babbage.clarku.edu/~djoyce/java/elements/Euclid.html

And here is a neat page all about Triangle Geometry: http://www.xtec.es/~qcastell/ttw/ttweng/portada.html

For those with too much time on their hands, 43 Ways to Prove the Pythagorean Theorem: http://www.cut-the-knot.org/pythagoras/index.shtml


induction
real analysis
real analysis unsolved

[quote="jarod"]Let $(a_n)$ be defined :
 $a_1 =\frac{1}{2006}$
 $a_{n+1}=a_n+\frac{a_n^2}{n^2}$
 Prove that : $(a_n)$ is converge .[/quote]

Hmm.  $\frac{a_{n+1}}{a_n} =1 + \frac{a_{n}}{n^2}$. Since ${a_{1}} < 1$, by induction we see,
that ${a_{n}} < n$.


[quote="Interval"]After a 6% increase Adam earned $\$318$ per week, what was his weekly salary before the increase? This is my work: Adam now makes $\$318$ each week.

This amount came AFTER a 6% increase.

Whatever he made before the increase is LESS than $\$318$. My proportion set up: (increase/original amount) = (percent/100)$\$318/x = 0.06/100$

Solving for x, I got $\$53$. Now,$\$318-\$53 =\$265$.

He made $\$$265 before the raise. 6 times 53 = 318. QUESTION: I am told that the answer is $\$300$ BEFORE the increase.

Who is right?

If I am wrong, please show me why I am wrong.

If the book is right, please explain why.

interval

[size=59][color=red]preview your posts before posting and use dollarsign\dollarsign dollarsign for $\$$.
b-flat[/color][/size][/quote]

[hide]You aren't allowed to do that.

Let x be the initial price.

Then this means the new price is 1.06x.

So $1.06x=318$, so $x=300$

$300$[/hide]

[quote="Interval"]After a 6% increase Adam earned $\$318$ per week, what was his weekly salary before the increase? This is my work: Adam now makes $\$318$ each week.

This amount came AFTER a 6% increase.

Whatever he made before the increase is LESS than $\$318$. My proportion set up: (increase/original amount) = (percent/100)$\$318/x = 0.06/100$

Solving for x, I got $\$53$. Now,$\$318-\$53 =\$265$.

He made $\$$265 before the raise. 6 times 53 = 318. QUESTION: I am told that the answer is $\$300$ BEFORE the increase.

Who is right?

If I am wrong, please show me why I am wrong.

If the book is right, please explain why.

interval

[size=59][color=red]preview your posts before posting and use dollarsign\dollarsign dollarsign for $\$$.
b-flat[/color][/size][/quote]
in ur[b] (increase/original amount=percent/100)[/b] formula...u don't know by how much it's increased in dollars...u only know it as a percentage(6%), but u've taken increase to be 318. 318 is the final amount and not the increase so u can't do it like that.
So u take the original amount to be x.
x plus 6% of itself is equal to 318...u can continue from there

I want to thank bothof you for your help.

mathgeniuse^ln(x):

Where you get 1.06? Where did the whole number 1 come from? Where in the question does the 1 appear?

You cannot just use .06 because it is "one AND .06 more".  :)

you could think about it this way.
before the raise, he made x dollars.
let's call this 100% percent.
now he receives a 6% raise.
that means he now makes 106%
which is 106/100, which is 1.06
so 1.06x=318
x=300.


The distance light travels in one year is approximately $ 5,870,000,000,000$ miles. The distance light travels in $ 100$ years is:

$ \textbf{(A)}\ 587 * 10^8 \text{ miles} \qquad\textbf{(B)}\ 587 * 10^{10} \text{ miles} \qquad\textbf{(C)}\ 587*10^{ \minus{} 10} \text{ miles}$
$ \textbf{(D)}\ 587 * 10^{12} \text{ miles} \qquad\textbf{(E)}\ 587* 10^{ \minus{} 12} \text{ miles}$

This is nothing more than an exercise in scientific notation.  Since we're multiplying by $ 10^{2}$, we just need to determine the form of the original, which is just $ 5.87\times 10^{12}\equal{}587\times 10^{10}$, so the final answer is $ 587\times 10^{12}$, which is (D).


Find the greatest real number $a$ such that for any $x, y, z>0$  the following inequality holds:

       $\frac{x+1}{y}+\frac{y+1}{z}+\frac{z+1}{x} > a.$ ;)

Is this a joke? Obviously, $a = 3$ works since (AM-GM) \[ \frac{x + 1}{y} + \frac{y + 1}{z} + \frac{z + 1}{x} > \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq 3 \] and taking $x = y = z \rightarrow + \infty$ we see that that is the best constant we can get.


dar edameye poste kheili ghadimiyam aya kesi "abel summation formula" ro mitoone tozih bedeh? :oops:

akhe mishe azash estefadeh kard ta in soal ro hal kard vali nemi doonam chetori!!

Let \[ a1+a2+a3+....  \] be a convergant series (seriye hamgara). Prove that $a1+a2/2+a3/3+a4/4+...$ is also convergant.

Abel summation formula, dar asl jam zadan be vasileye ghestmat hast (summation by parts)...
mathematically:
$\sum\limits_{k=1}^na_kb_k=\sum\limits_{k=1}^n(\Sigma a_k)\cdot(\delta b_k)+(\Sigma a_n)\cdot b_{n+1}-(\Sigma a_0)\cdot b_1$, where $\Sigma a_k=c+\sum\limits_{i=1}^ka_i$ and $\delta b_k=b_k-b_{k+1}$.

dar asl dare mige:
$a_1 b_1 + a_2 b_2 + ... + a_n b_n$

$= (a_1-a_2)(b_1) + (a_2-a_3)(b_1+b_2) + ... + (a_{n-1}-a_n)(b_1+b_2+...+b_{n-1})+a_n(b_1+b_2+...+b_n)$

in formul, dar asl as hamoon Integration by parts miyad:

$\int{u}{dv}=uv-\int{v}{du}$


algebra
polynomial
inequalities
algebra proposed

Let the polynomial $ P(x) \equal{} \sum_{k \equal{} 0}^n\frac {1}{n \plus{} k \plus{} 1}x^k$. Prove that there doesn't exist a real number $ a$ satisfying $ P(a^2) \equal{}(P(a))^2$

You mean $ P(a^2) \equal{} P^2(a)$?  And does $ P^2(a)$ denote $ P(P(a))$?

I edited my first post in clearness. Let's solve this interesting problem.

By the Cauchy Schwartz inequality:
\[ \left(\sum_{k \equal{} 0}^n \frac {x^{2k}}{n \plus{} k \plus{} 1}\right)\left(\sum_{k \equal{} 0}^n \frac {1}{n \plus{} k \plus{} 1}\right)\ge \left(\sum_{k \equal{} 0}^n \frac {x^{k}}{n \plus{} k \plus{} 1}\right)^2
\]

\[ \iff (P(x^2))\left(\sum_{k \equal{} 0}^n \frac {1}{n \plus{} k \plus{} 1}\right)\ge (P(x))^2\cdots (*)
\]

\[ \sum_{k \equal{} 0}^n \left(\frac {1}{n \plus{} 1} \minus{} \frac {1}{n \plus{} k \plus{} 1}\right) > 0
\]

\[ \iff 1 \equal{} \sum_{k \equal{} 0}^n \frac {1}{n \plus{} 1} > \sum_{k \equal{} 0}^n \frac {1}{n \plus{} k \plus{} 1}\cdots (\#)
\]

\[ (*)\text{ and }(\#)\implies
\]

\[ P(x^2) > (P(x))^2\ \forall x
\]


Michael baked a chocolate cake. He left $ \frac{1}{2}$ of the cake for his family and took the rest to a friend's house. Of the amount he left for his family, his brother ate $ \frac{1}{4}$ of it, and his sister ate $ \frac{1}{3}$ of it. Later that night, his dad ate $ \frac{2}{5}$ of what was then remaining. When Michael got home the next morning, how much of the original cake was left? Express your answer as a common fraction.

There are the fractions 1/4,1/3, and 2/5. We want to add these together to see what fraction of 1/2 of the cake his family ate. First, find the LCM (least common multiple) of 4,3, and 5. You find that that it is 60. Now make the three fractions have denominators of 60. Now the fractions are 15/60, 20/60, and 24/60. Add these together to get 59/60. That means that his family ate 59/60 of 1/2 of the cake, so multiply 1/60 by 1/2 to get how much of the original cake is left. So ans is 1/120, I think.

When I solved it, I got a different answer.

His brother and sister ate 1/4 and 1/3 of the half cake respectively. That means that they ate 1/8 and 1/6 of the total cake. When you add these fractions, you get 7/24. So they ate 7/24 of the total cake, meaning that there was 17/24 cake left. Michael's dad ate 2/5 of the [b]remaining[/b] cake, so he ate 2/5 of 17/24, or 17/60 of the cake. Now we add 17/60, 1/8, and 1/6 to get 23/40. We want to find out how much is left, so there was 17/40 of the cake left.

It said how much of the [b]original[/b] cake is left, so I think the answer is 17/40....I might be wrong, though.

lol i got another answer...

michael left $ \frac{1}{2}$ of the cake, right?

and his brother ate $ \frac{1}{4}$ of it, which is $ \frac{3}{12}$.

and his sister ate $ \frac{1}{3}$ of it, which is $ \frac{4}{12}$.

so his brother and sister ate $ \frac{7}{12}$ of remaining cake, so it makes $ \frac{5}{12}$ of the half or the cake.

and his dad ate $ \frac{2}{5}$ of the remaining, so that gives us..

$ \frac{1}{2}*\frac{5}{12}*\frac{3}{5}\equal{}\frac{1}{8}$

i got $ \frac{1}{8}$ lol

Yes, I agree with Fantasy.

Isabella, your calculations are right expect when you mentioned that Michael's dad ate 2/5 of the remaining 17/24 cake. In fact, Michael took half to his friend's house, so the total remaining cake after his brother's ate some, was only 5/24 (or 17/24-12/24=5/24). Then, his dad ate

2/5*5/24 = 2/24

Which leaves 5/24 - 2/24 = 3/24 = 1/8 of the total cake.


[quote="ShOoTiNg_SpElLs"]How did you all do the area of the triangle question on the CNML? I did it by finding the ratio of the area of the smaller triangle to the larger triangle, by subtracting the area around the smaller triangle, but I don't think that solution was very elegant.  I can't seem to word the question very well right now, and will try to get the exact wording tomorrow, but I was wondering how you guys solved it (if any of you remember it).[/quote]

We can't discuss this contest yet, I believe.

at my school our math club is on fridays, and last friday was a PD day so I am not doing the contest until this friday..

 :wink:

I think we are allowed to talk about it know.

The key was to using the fact that the midline is half the base and the midpoint divides teh base into two equal segments.

from the answer sheet..

if you noticed, the triangle formed was similar to the big one, and its side lengths were $\frac14$ the size.. thus the area is $(\frac{1}{4})^{2}$ smaller

something I did not know  :ninja:

lol...didnt take it...went to join the knitting club instead... :rotfl:


[hide]That would simplify to 5/2 + 5/6 -3, or 1/3. [/hide]


[hide]
56*7=392
392+x/8=60
Multiply by 8 on each side,
392+x=480
Subtract 392 on each side,
x=88%[/hide]

[hide][quote="math92"]Flora had an average of 56% on her first 7 exams. What would she have to make on her eighth exam to obtain an average of 60% on 8 exams?[/quote]

56*7=392
60*8=480
480-392=88

88%[/hide]

[quote="math92"]Flora had an average of 56% on her first 7 exams. What would she have to make on her eighth exam to obtain an average of 60% on 8 exams?[/quote]
[hide]
56x7=392
60x8=480
480-392=88.

She has to get 88 on her eight exam.
[/hide]


[quote="arkhammedos"]hi all
where can i find some lecon of olympiads
thanks all  :blush:  :o[/quote]

What kind of olympiad do you denote by "lecon"?

I couldn't resist being away from mathlinks :P  By lecon, he/she means some lessons.  I believe he/she is French.  He/she is asking for some Olympiad level training documents perhaps.  I would recommend Problem-Solving Strategies or Problem-Solving Through Problems.  Visit [url=http://amazon.com]Amazon[/url] for more information. :)

Masoud Zargar[/url]

There are a few olympiad articles/lessons [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Articles.php]here[/url].


:oops: ok, was wondering if there was some sort of trick

now that i think about it, there is a closed form for $\sum k*a^k$

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=summation&t=50853[/url]

[quote="Hexaditidom"]Ok,[i] imagine[/i] this problem was on the AMC or AIME... how would you do it quickly?

$\frac{123456789}{9}$ has two factors. Find their sum.[/quote]

[hide]$123456789/9=9*3607*3803/9=3607*3803. 3607+3803=7410$

I don´t know if I´m answering what you asked.[/hide]

That's correct, but how did you find those factors? Computer/calculator program, or did you know some other way to do it?


ratio
projective geometry
geometry unsolved
geometry

Is possibile to have an equilater triangle such as an internal point P has the following distances from the vertexs:
PA = 2;
PB = 3
PC = 4
?

enjoy!

The answer is yes. We can define two points $ P,$ $ P'$ in the plane, with the property as the problem states, as the edges of the common cord of the three [b][size=100]Apollonian circles[/size][/b], with respect to the vertices of a given triangle $ \bigtriangleup ABC,$ taken per two of them.

So, we define the point $ D,$ between $ B,$ $ C,$ and similarly the points $ E,$ $ F,$ on $ AC,$ $ AB$ respectively, such that $ \frac{DB}{DC}= \frac{3}{4}$ and $ \frac{EA}{EC}=\frac{2}{4}$ and $ \frac{FA}{FB}= \frac{2}{3}.$

We define the point $ D',$ as the harmonic conjugate of $ D,$ with respect to $ B,$ $ C$ and similarly the points $ E',$ $ F',$ on the sidelines $ AC,$ $ AB$ respectively.

We draw the circle $ (A_{1}),$ with diameter the segment $ DD'$ and we have that this circle is the locus of the points in the plane, of which their distances from the vertices $ B,$ $ C,$ of $ \bigtriangleup ABC,$ have ratio, [b][size=100]$ r_{1}= 3 : 4$[/size][/b].

Similarly, about the circles $ (A_{2}),$ with diameter $ EE',$ as the locus of the points of which their distances from $ A,$ $ C,$ have ratio, [b][size=100]$ r_{2}= 2 : 4$[/size][/b] and $ (A_{3}),$ with diameter $ FF',$ as the locus of the points of which their distances from $ A,$ $ B,$ have ratio, [b][size=100]$ r_{3}= 2 : 3$[/size][/b] and we will prove that these three circles are coaxal.

$ \bullet$ We denote as $ P$ ( inwardly to $ %Error. "bigtriangle" is a bad command.
ABC$ ) and $ P',$ ( outwardly to $ \bigtriangleup ABC$ ), the intersection points of the circles $ (A_{2}),$ $ (A_{3})$ and it is easy to prove that these points also lie on the circle $ (A_{1}).$

We have $ \frac{PA}{PB}= \frac{2}{3}$ $ ,(1)$ and $ \frac{PA}{PC}= \frac{2}{4}$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac{PB}{PC}= \frac{3}{4}$ $ ,(3)$

From $ (3)$ we conclude that the point $ P,$ also lies on the circle $ (A_{1}).$

The same we can say about the point $ P'\equiv (A_{2})\cap (A_{3})$ and the proof is completed.

[b][size=100]REMARK.[/size][/b] – It is clear that we can generalize the problem about the point $ P,$ such that $ PA = m,$ $ PB = n,$ $ PC = q,$ where $ (m+n),$ $ (n+q),$ $ (q+m),$ are the numbers of equal segments, in which the side segments $ AB,$ $ BC,$ $ AC$ respectively, of the given triangle ( not necessary equilateral ) $ \bigtriangleup ABC,$ are divided.

$ ($ We define the points $ D,$ $ E,$ $ F,$ on $ BC,$ $ AC,$ $ AB$ respectively, such that $ \frac{FA}{FB}= \frac{m}{n}$ and $ \frac{DB}{DC}= \frac{n}{q}$ and $ \frac{EA}{EC}= \frac{m}{q}$ $ ).$

Kostas Vittas.


1. In how many ways can you split 10 people into 5 pairs?

2. A bag contains 10 balls, of which 3 are indistinguishable and the other 7 are distinct. In how many ways can 3 balls be chosen from the bag?

1.

[hide]

Consider the sequence of people abcdefghij. We want to examine the permutations of this such that we get unique pairs. However, each pair can be switched and our 'groups' are still the same. Thus, our answer is .

[/hide]

[hide]You forgot to divide by $5!$. Here's how I did it, BTW... \[ \frac{1}{5!}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{10!}{2^5\cdot 5!}. \][/hide]

[quote]Why do you divide by 5!?[/quote]

The order of the pairs does not matter; there are 5 pairs, so dividing $10!$ by $2^5$ will count each set of 5 pairs $5!$ times.[/quote]

[quote="Adunakhor"][hide="3"]

Dealing with separate cases:

All three are the indistinguishable ones $\Longrightarrow 1$ possibility

Two of the three are the indistinguishable ones $\Longrightarrow 7$ possibilities

One of the three is an indistinguishable one $\Longrightarrow {7\choose 2}=21$ possibilities

None of the three $\Longrightarrow {7\choose 3}=35$ possibilities

Sum them all up to get $\boxed{64}$

[/hide][/quote]

Correct. (I believe)

Yes, my bad, the order of the pairs is irrelevant.


For positive interger a, b, c,   prove that

the number of partition of (a-c) that have exactly (b-1) parts and each part not greater that c is equal to

                                                         the number of partition of (a-b) that have exactly (b-1) parts and each part not greater that b.

Thanks.

I think you meant to say $ c\minus{}1$ parts in the second part of the problem...?

make a $ b\times c$ table ($ b$ rows, $ c$ columns). let the parts of the partition of $ a\minus{}c$ be $ p_1\leq \cdots \leq p_{b\minus{}1}$. for each $ i$, $ 1\leq i\leq b\minus{}1$, put a $ 1$ in each of the first $ p_i$ cells of row $ i$, and fill in the entire row $ b$ with $ 1$'s. so the rows of this table give us a partition of $ a$ into exactly $ b$ parts, each at most $ c$, and the greatest of size $ c$. reading the columns, on the other hand, gives us a partition of $ a$ into exactly $ c$ parts, each at most $ b$, and the greatest of size $ b$. deleting the first column gives a partition of $ a\minus{}b$ into exactly $ c\minus{}1$ parts, each at most $ b$. so we have a bijection between the two sets of partitions, and that's it.


Any square is congruent $ 0,1,4,7$ modulus $ 9$.  But the given number for $ n\geq3$ is congruent $ (1!)^2\plus{}(2!)^2\equal{}5$ modulus $ 9$, hence cannot be a square.  $ 5$ is not a square either, and we are done.

For n=2,3,4   \[ \sum_{k\equal{}1}^n (k!)^2  \] is \[ 5,41,617 \]   which are not squares

If n>=5 then The sum will ended with digit 7, which can't be squares too.


analytic geometry
geometry
rectangle
LaTeX
graphing lines
slope

1. two vertices of a triangle are (5,-1) and (-2,3). If the orthocenter of the triangle is the origin, what will be the coordinates of the third point?

2.One side of a rectangle lies along the line 4x + 7y + 5 = 0. two of its vertices are (-3,1) & (1,1). Which of the following may be an equation of one of the other three straight lines?

a.	7x – 4y = 3
b.	7x – 4y + 3 = 0
c.	y + 1 = 0
d.	4x + 7y = 3

How to solve these???

#1
It's too hard $LaTeX$ing everything, so here's the method:
1) Determine the slopes of the altitudes
2) Determine the perpendicular slopes
3) Find the point using the slope (you have 2 equations)

#2
There is no answer...

1) ok i find the slopes of the altitude by using both the points and the origin....now how do i find the perpendicular slopes??? and what 2 equations are you talking about??? ironically i solved this question using the options but want to use this to actually learn....

[quote="fizsle"]2.One side of a rectangle lies along the line 4x + 7y + 5 = 0. two of its vertices are (-3,1) & (1,1). Which of the following may be an equation of one of the other three straight lines?

a.	7x – 4y = 3
b.	7x – 4y + 3 = 0
c.	y + 1 = 0
d.	4x + 7y = 3
[/quote]Hello, fizsle.
The way I would solve this is to notice that only the point $P(-3,1)$ satisfies the equation $4x+7y+5 = 0$, meaning that the $P$ is on the line. This means that the line segment $PQ$, where $Q(1,1)$, can either be a side length of the rectangle or a diagonal of the rectangle. Notice then that the slope of $PQ$ isn't the negative reciprocal of the slope of the line, which is $m_{l}=-\frac{4}{7}$. Therefore $PQ$ must be a diagonal.

Therefore you could find the line of the equation that has:
-- slope $m_{l}$ and goes through $(1,1)$,
-- slope $-\frac{1}{m_{l}}$ and goes through $(1,1)$, or
-- slope $-\frac{1}{m_{l}}$ and goes through $(-3,1)$

thanks...that helped with the 2nd one...anything about the first one???

[hide="On #1"]Let the given points be $A$ and $B$. Find the line $AO$ (the altitude). Then find the perpendicular from $B$ to $AO$, call it $a$ (the side). Then find the line $BO$ and the perpendicular from $A$ to $BO$, call it $b$ (the side). Now find the intersection of $a$ and $b$.

Note: if a line $y=kx+m$ is given, then the perpendiculars have the equation $y=-{1\over k}x+s$[/hide]


What an open ended question, it seems way too ambigious

What is your defintion of culture? Cultural diversity?

I live in GA, but in a county that's 60% white, 30% black, and 10% other. Cultural diversity hasn't really affected me. I'm Asian, and there isn't a lot of racism where I live.

by cultural diversity I mean differences in race, ethnicity, language, nationality, or religion among various groups within a community, organization, or nation.

Knowing there are black people in the world: good.

Making half of the people in my math textbook be in wheelchairs: whatever.

Extolling the virtues of the simplicity of African civilizations while forgetting to mention female genital mutilation: bad.

[quote="mdk"]How has cultural diversity or the lack of it, impacted your life so far? What is your view or opinion of how diversity fits into your education?[/quote]

Well, cultural diversity brings the world into one classroom :lol: .

Negative impact: some teachers are really hard to understand because of their accent.  :ninja:

i know this is reeeeally idealistic, but in my opinion it would be better if, instead of making sure that people of different ethnicities, genders, backgrounds, etc got equal opportunities we just simply ignored these during job interviews, etc
kinda like we (ignoring hitler) dont care about hair color, eye color, etc
the world would be a looooot better off

[quote="fredbel6"][quote="JBL"]Your question is of the sort that is often not conducive to discussion -- "how does X relate to you personally" tends to end up with a lot of disjoint answers and not very much discussion (if it gets responses at all), and then I usually send it off to the G&FF.[/quote]
This could be an interesting discussion, if the question was a bit more clear... :maybe:  what kind of cultural diversity are we talking about?[/quote]

difference in races and culture

okay, respond based on your take of what I'm trying to say.

Well my opinion is quite pessimistic then.

On a personal level, I think one should be "culturally diverse", meaning that one should be open to different religions, languages,....
I myself had to learn several other languages and I think that was great.
I consider myself an atheist but I am interested in "other" religions.

However I do not think it is good for nations or towns for that matter to be very hetergeneous... living in a nation like that and observing local politics on a daily basis, I am quite certain of that.
It is a phenomenen in Europe and beyond that small homogeneous nations do better.
And yes, linguistic differences are of utmost importance to me...I think when religions are practised in a secular way, differences can be much less of a problem.  An example of that is the Netherlands, a country I do happen to admire a bit, where people are catholic or protestant or atheist, but without splitting society up.

[quote="fredbel6"]Well my opinion is quite pessimistic then.

I think when religions are practised in a secular way, differences can be much less of a problem.  An example of that is the Netherlands, a country I do happen to admire a bit, where people are catholic or protestant or atheist, but without splitting society up.[/quote]
Nice said and good example.
Well I think religion is the only and the easiest way to win a War.
We have Moses,Jesus, Mohammand and others has examples.
They were great politicians,just by using one question " ًwho is the creator ?" brought peace to their region.
It is only my opinion I have lots of respects for all religions.

[quote="Treething"]People segregate themselves if given the opportunity, and differences (racial, social, theological, any kind really) breed mistrust. This is how life works; you can observe this anywhere, from schools to workplaces to cities.[/quote]

obviously true.

Interesting remarks Fredbel.

But on the other hand the difference in culture can be really beneficial, because the other person can have a different take on things than you, and it might make you think otherwise than you would otherwise have.

Further, if you don't really feel the same way as your native "culture", going and finding many of your friends in a different culture can make you be more self-confident or even keep your beliefs. This can be a good thing when, say, you want to be good at math but your culture frowns on mathematical excellence :D.

As for me, I find that many times Asians are more enjoyable friends than Whites. I do still have a great number of White friends and some Black friends too. [I'm white.] Accents are sometimes annoying, but then if you've talked to me I don't quite have a standard accent, so who am I to complain about accents?

[quote="solafidefarms"]But on the other hand the difference in culture can be really beneficial, because the other person can have a different take on things than you, and it might make you think otherwise than you would otherwise have.

Further, if you don't really feel the same way as your native "culture", going and finding many of your friends in a different culture can make you be more self-confident or even keep your beliefs. This can be a good thing when, say, you want to be good at math but your culture frowns on mathematical excellence :D.
[/quote]
I fear we have a different interpretation on the original question.
I am not talking about "setting out" and making some friends, I am talking about places where different cultures happen to coincide and... friends or not, they are there and at least one of them will stay.

[quote]As for me, I find that many times Asians are more enjoyable friends than Whites. I do still have a great number of White friends and some Black friends too. [I'm white.] Accents are sometimes annoying, but then if you've talked to me I don't quite have a standard accent, so who am I to complain about accents?[/quote]
What is wrong with an accent, be happy that you can speak English?


inequalities
quadratics
inequalities proposed

[quote="can_hang2007"]For $ a,b,c \ge 0$, prove that
\[ \frac {a \plus{} 2b}{a \plus{} 2c} \plus{} \frac {b \plus{} 2c}{b \plus{} 2a} \plus{} \frac {c \plus{} 2a}{c \plus{} 2b} \ge \sqrt {\frac {5(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} 4}\][/quote]
Put $ a\equal{}2x\plus{}y$ , $ y\equal{}2y\plus{}z$ , $ c\equal{}2z\plus{}x$ . Now , The inequality is symmetric , which is easy :)
The following is true , too :
$ \frac {a \plus{} 2b}{a \plus{} 2c} \plus{} \frac {b \plus{} 2c}{b \plus{} 2a} \plus{} \frac {c \plus{} 2a}{c \plus{} 2b} \ge \sqrt {\frac {11(a^2 \plus{} b^2 \plus{} c^2)}{2(ab \plus{} bc \plus{} ca)} \plus{} \frac{7}{2}}$

[quote="tanpham90"][quote="can_hang2007"]For $ a,b,c \ge 0$, prove that
\[ \frac {a \plus{} 2b}{a \plus{} 2c} \plus{} \frac {b \plus{} 2c}{b \plus{} 2a} \plus{} \frac {c \plus{} 2a}{c \plus{} 2b} \ge \sqrt {\frac {5(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} 4}
\]
[/quote]
Put $ a \equal{} 2x \plus{} y$ , $ y \equal{} 2y \plus{} z$ , $ c \equal{} 2z \plus{} x$ . Now , The inequality is symmetric , which is easy :)
The following is true , too :
$ \frac {a \plus{} 2b}{a \plus{} 2c} \plus{} \frac {b \plus{} 2c}{b \plus{} 2a} \plus{} \frac {c \plus{} 2a}{c \plus{} 2b} \ge \sqrt {\frac {11(a^2 \plus{} b^2 \plus{} c^2)}{2(ab \plus{} bc \plus{} ca)} \plus{} \frac {7}{2}}$[/quote]
Are you sure? Note that, we must have the condition $ 2x\plus{}y \ge 0,2y\plus{}z \ge 0,2z\plus{}x \ge 0$ which is not symmetric for $ x,y,z \in \mathbb R$.
Actually, this is not the strongest result. :)

[quote="can_hang2007"]For $ a,b,c \ge 0$, prove that
\[ \frac {a \plus{} 2b}{a \plus{} 2c} \plus{} \frac {b \plus{} 2c}{b \plus{} 2a} \plus{} \frac {c \plus{} 2a}{c \plus{} 2b} \ge \sqrt {\frac {5(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} 4}
\]
[/quote]
If we prove this problem then this one follows $ \sum \frac{1}{(2a\plus{}b)^2} \ge \frac{1}{ab\plus{}bc\plus{}ca}.$ (1) Since
\[ \sum (b\plus{}2c)^2 \cdot \sum \frac{1}{(2a\plus{}b)^2} \ge (\sum \frac{b\plus{}2c}{b\plus{}2a})^2 \ge \frac{5\sum a^2}{\sum ab} \plus{}4 \equal{}\frac{\sum (b\plus{}2c)^2}{\sum ab}
\]
:)
(1) was posted by hungkhtn. :)

Dear can_hang2007,

this your inequality is very nice.

[quote="can_hang2007"]For $ a,b,c \ge 0$, prove that
\[ \frac {a \plus{} 2b}{a \plus{} 2c} \plus{} \frac {b \plus{} 2c}{b \plus{} 2a} \plus{} \frac {c \plus{} 2a}{c \plus{} 2b} \ge \sqrt {\frac {5(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} 4}
\]
[/quote]

Can you please give a hint?

Thank you very much.

A hint please, even if I suspect it is very hard to solve. :(

Thank you very much. :)

Using BCS in Engel form we have

$ (\sum (a\plus{}2c)(a\plus{}2b)) (\sum \frac {a \plus{} 2b}{a \plus{} 2c}) \geq (3\sum a)^2$

so it remains to prove

$ \frac {(3\sum a)^2}{\sum a^2 \plus{}6\sum ab}  \ge \sqrt {\frac {5(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} 4}$

I would like to know if this step is right?  After that do I use p,q,r method or shur ineq ?

Thank you very much

[quote="manlio"]Using BCS in Engel form we have

$ (\sum (a \plus{} 2c)(a \plus{} 2b)) (\sum \frac {a \plus{} 2b}{a \plus{} 2c}) \geq (3\sum a)^2$

so it remains to prove

$ \frac {(3\sum a)^2}{\sum a^2 \plus{} 6\sum ab} \ge \sqrt {\frac {5(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} 4}$

I would like to know if this step is right?  After that do I use p,q,r method or shur ineq ?

Thank you very much[/quote]

Your inequality should be typed in

$ \frac {(3\sum a)^2}{\sum a^2 \plus{} 8\sum ab} \ge \sqrt {\frac {5(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} 4}$

Which is false for $ a \equal{} b \equal{} 1,c \equal{} 0$

[quote="can_hang2007"][quote="can_hang2007"]For $ a,b,c \ge 0$, prove that
\[ \frac {a \plus{} 2b}{a \plus{} 2c} \plus{} \frac {b \plus{} 2c}{b \plus{} 2a} \plus{} \frac {c \plus{} 2a}{c \plus{} 2b} \ge \sqrt {\frac {5(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \plus{} 4}
\]
[/quote]
If we prove this problem then this one follows $ \sum \frac {1}{(2a \plus{} b)^2} \ge \frac {1}{ab \plus{} bc \plus{} ca}.$ (1) Since
\[ \sum (b \plus{} 2c)^2 \cdot \sum \frac {1}{(2a \plus{} b)^2} \ge (\sum \frac {b \plus{} 2c}{b \plus{} 2a})^2 \ge \frac {5\sum a^2}{\sum ab} \plus{} 4 \equal{} \frac {\sum (b \plus{} 2c)^2}{\sum ab}
\]
:)
(1) was posted by hungkhtn. :)[/quote]
Similarly, if we prove that

$ \sum \frac{a\plus{}b}{b\plus{}c} \ge 3\sqrt{\frac 1{2}(\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\plus{}1)}$,

then from this follows Iran's Inequality

$ \sum \frac 1{(a\plus{}b)^2}\ge \frac {9}{4(ab\plus{}bc\plus{}ca)}$.

[quote="Vasc"]Similarly, if we prove that

$ \sum \frac {a \plus{} b}{b \plus{} c} \ge 3\sqrt {\frac 1{2}(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} 1)}$,

then from this follows Iran's Inequality

$ \sum \frac 1{(a \plus{} b)^2}\ge \frac {9}{4(ab \plus{} bc \plus{} ca)}$.[/quote]
I don't think so, VasC.
Since the Iran's Inequality have equality for $ a \equal{} b,c \equal{} 0$. But for the above inequality, the equality does not hold for $ a \equal{} b,c \equal{} 0$.

You are right. The claimed inequality doesn't hold. :oops:

Look at

http://reflections.awesomemath.org/2007_2/3variable.pdf  :wink:

[quote="manlio"]A hint, please.  :wink:[/quote]
We can use $ p,q,r$ technique here Manlio. :)
It is similar to http://www.mathlinks.ro/viewtopic.php?t=193004 :)

@To Manlio. I'm sorry since I can't sent you the solution as I promised. :(

Before using $ p,q,r$ mehod do you apply Cauchy-Schwarz at LHS or do you use squaring and full expansion ?

Thank you very much :)

[quote="can_hang2007"]For $ a,b,c \ge 0$, prove that
\[ \frac{a\plus{}2b}{a\plus{}2c}\plus{}\frac{b\plus{}2c}{b\plus{}2a}\plus{}\frac{c\plus{}2a}{c\plus{}2b} \ge \sqrt{\frac{5(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca}\plus{}4}\][/quote]
We can prove this inequality and we can find the best estimation by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=278791]uvw[/url]
because it's quadratic inequality of $w^3$. :wink:


VietMath 12.6.2006 có những tính năng sau đây:

1) Chuyển đổi độ C sang độ F và ngược lại
2) Tính xem bạn tuổi con gì
3) Xem thông tin máy tính của bạn
4) Làm tròn giá trị
5) Giải phương trình bậc 1
6) Giải phương trình bậc 2
7) Giải phương trình bậc 3
8) Giải hệ phương trình đường thẳng với các hệ số x, y
9) Giải hệ phương trình đường thẳng với các hệ số x, y, z
10) Phân tách 7 hằng đẳng thức đáng nhớ
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VietMath 21.7.2006 xin kính chào các bạn!

VietMath 21.7.2006 có thêm 2 tính năng mới. Đó là:
1) Làm việc với Hỗn số
2) Làm việc với Số nguyên tố

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VietMath 28.11.2006 ra lò với những tính năng nóng hổi sau đây:

1)  Giải phương trình bậc 4
2)  Chia đa thức
     2a) Chia phương trình bậc 2 cho phương trình bậc 1
     2b) Chia phương trình bậc 3 cho phương trình bậc 1
     2c) Chia phương trình bậc 4 cho phương trình bậc 1

 Tổng cộng những tính năng của VietMath cho đến phiên bản này là:
1) Chuyển đổi độ C sang độ F và ngược lại
2) Tính xem bạn tuổi con gì
3) Xem thông tin máy tính của bạn
4) Làm tròn giá trị
5) Giải phương trình bậc 1
6) Giải phương trình bậc 2
7) Giải phương trình bậc 3
8) Giải phương trình bậc 4
9) Chia đa thức
10) Giải hệ phương trình đường thẳng với các hệ số x, y
11) Giải hệ phương trình đường thẳng với các hệ số x, y, z
12) Phân tách 7 hằng đẳng thức đáng nhớ
13) Hệ thống đơn vị đo lường
14) Tính chu vi, diện tích và thể tích các hình
15) Làm việc với Hỗn số
16) Làm việc với Số nguyên tố
17) Cờ vây
18) Làm việc với Phân số
19) Cơ thể của bạn có cân đối không

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VietMath được viết bằng ngôn ngữ lập trình C# của Microsoft nên để có thể chạy được chương trình này thì bạn nhất thiết phải install .NET Framework 1.1 trở lên. Các bạn có thể tải về tại địa chỉ sau:

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Mình rất hi vọng chương trình phần mềm toán học đầu tay này của mình có thể giúp ích được cho các bạn một phần nhỏ nào đó trong học tập. Rất mong nhận được nhiều ý kiến đóng góp từ phía các bạn. Mình xin chân thành cảm ơn!

VietMath 5.5.2007 chính thức ra mắt với thêm 2 tính năng mới:
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2)	Chuyển đổi đơn vị đo lường

Trong phiên bản lần này, mình tạm thời bỏ phần “Giải phương trình bậc 4” do phần này đã đưa ra một vài kết quả sai. Mình sẽ tiếp tục nghiên cứu, khắc phục và phát triển để đưa phần “Giải phương trình bậc 4” và các tính năng khác vào VietMath trong phiên bản tiếp theo. Mình rất mong sẽ tiếp tục nhận được những ý kiến đóng góp từ phía các bạn. Mình xin chân thành cảm ơn!

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Tại sao không ai phát triển thêm để nó giống Mathlab hay Mathematica nhỉ? Chắc là khó lắm.
Mà giới hạn tính toán của chưong trình có lớn không vậy bạn? Ví dụ như tính $2^{1000}$ ấy?
P/s: Tại sao phần mềm toán lại có cả cái món cờ vây vậy?

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Em thấy cái này chẳng hay tí nào, trông thủ công quá :| thà dùng easy calc còn hơn  :D

Nếu máy tính của các bạn đang cài Windows Vista hoặc Windows 7 thì các bạn không cần phải cài .NET Framework nữa, bởi vì 2 hệ điều hành đó đã cài sẵn .NET Framework rồi  :)


analytic geometry
Asymptote
number theory unsolved
number theory

if both Asymptotes of a hiperbola pass through, each of them, by two points of integer coordinates, show that the hiperbola has a finite number of points of integer coordinates.


function
superior algebra
superior algebra unsolved

How many tridempotents does $Z_n$ have? A tridempotent verifies $x^3=x$.  :D

Oh, sorry, it is $Z/nZ$ the ring of classes mod n.

Let $f(n)$ denote the number of solutions.
If I'm not taken wrong, it suffices by the chinese remainder theorem (which implies that the function is multiplicative) to find $f(p^n)$ for primes $p$ and positive integers $n$.

$p \neq 2$:
whenever $p^n | x^3-x=(x-1)x(x+1)$ it divides (because $gcd(x-1,x)=1=gcd(x,x+1), gcd(x-1,x+1) \in \{ 1,2\}$) already one of the factors, so $x \equiv -1 , 0 , 1 \mod p^n$ and by this there are $f(p^n)=3$ solutions.

$p=2$:
when $2^n| x^3-x = (x-1)x(x+1)$ there are two cases:
a) $x$ even: then $x-1,x+1$ are odd, so $x\equiv 0 \mod 2^n$
b) $x$ odd: then $gcd(x-1,x+1)=2$, so $2^{n-1}$ divides at least one of the numbers $x \pm 1$, so $x\equiv 1,2^{n-1}-1,2^{n-1}+1,-1 \mod 2^n$
So there are $f(2^n)=5$ solutions except when some fall together which happens only for $n \in \{ 1,2\}$.


(1) $ x^4\minus{}ax^3\plus{}2x^2\minus{}bx\plus{}1$ has at least one real root. Prove $ a^2\plus{}b^2\ge 8.$
(2) $ x^4\minus{}ax^3\plus{}6x^2\minus{}bx\plus{}1$ has four real roots. Prove $ a^2\plus{}b^2\ge 32.$

1:

[hide]
Because it has one real root, we can write it as:

$ (x\minus{}u)(x^3 \plus{} px^2 \plus{} qx \plus{} r) \equal{} x^4 \plus{}(p\minus{}u)x^3 \plus{} (q\minus{}pu)x^2 \plus{} (r\minus{}qu)x \minus{} ru$.

From the original equation we get:

$ \minus{}ru \equal{} 1 \Rightarrow r \equal{} \frac{\minus{}1}{u}$

$ q\minus{}pu\equal{}2 \Rightarrow p \equal{} \frac{q\minus{}2}{u}$

$ a\equal{}p\minus{}u\equal{}\frac{q\minus{}2}{u}\minus{}u\equal{}\frac{q\minus{}2\minus{}u^2}{u}$

$ b\equal{}r\minus{}qu\equal{}\frac{\minus{}1}{u}\minus{}qu\equal{}\frac{\minus{}1\minus{}qu^2}{u}$.

The claim was: $ a^2 \plus{} b^2 \ge 8$.

$ \Leftrightarrow (q\minus{}2\minus{}u^2)^2\plus{}(1\plus{}qu^2)^2 \ge 8u^2$

Expanding and treating this as a polynomial in q we get:

$ (u^4 \plus{} 1)q^2 \minus{} 4q \plus{} 5 \plus{} u^4 \minus{} 4u^2 \ge 0$

$ \Leftrightarrow 16 \minus{} 4(u^4\plus{}1)(5\plus{}u^4\minus{}4u^2) \le 0$

$ \Leftrightarrow (u^4\plus{}1)(5\plus{}u^4\minus{}4u^2) \minus{} 4 \ge 0$

Substituting $ u^2 \equal{} v$ and expanding we get:

$ v^4 \minus{} 4v^2 \plus{} 6v^2 \minus{} 4v \plus{} 1 \equal{} (v\minus{}1)^4 \ge 0$, which holds.

[/hide]

[quote="hipsishopsis"]1:

[hide]
Because it has one real root, we can write it as:

$ (x \minus{} u)(x^3 \plus{} px^2 \plus{} qx \plus{} r) \equal{} x^4 \plus{} (p \minus{} u)x^3 \plus{} (q \minus{} pu)x^2 \plus{} (r \minus{} qu)x \minus{} ru$.

From the original equation we get:

$ \minus{} ru \equal{} 1 \Rightarrow r \equal{} \frac { \minus{} 1}{u}$

$ q \minus{} pu \equal{} 2 \Rightarrow p \equal{} \frac {q \minus{} 2}{u}$

$ a \equal{} p \minus{} u \equal{} \frac {q \minus{} 2}{u} \minus{} u \equal{} \frac {q \minus{} 2 \minus{} u^2}{u}$

$ b \equal{} r \minus{} qu \equal{} \frac { \minus{} 1}{u} \minus{} qu \equal{} \frac { \minus{} 1 \minus{} qu^2}{u}$.

The claim was: $ a^2 \plus{} b^2 \ge 8$.

$ \Leftrightarrow (q \minus{} 2 \minus{} u^2)^2 \plus{} (1 \plus{} qu^2)^2 \ge 8u^2$

Expanding and treating this as a polynomial in q we get:

$ (u^4 \plus{} 1)q^2 \minus{} 4q \plus{} 5 \plus{} u^4 \minus{} 4u^2 \ge 0$

$ \Leftrightarrow 16 \minus{} 4(u^4 \plus{} 1)(5 \plus{} u^4 \minus{} 4u^2) \le 0$

$ \Leftrightarrow (u^4 \plus{} 1)(5 \plus{} u^4 \minus{} 4u^2) \minus{} 4 \ge 0$

Substituting $ u^2 \equal{} v$ and expanding we get:

$ v^4 \minus{} 4v^2 \plus{} 6v^2 \minus{} 4v \plus{} 1 \equal{} (v \minus{} 1)^4 \ge 0$, which holds.

[/hide][/quote]

Well done! :)

[hide]A slightly shorter solution:
$ x^4\minus{}ax^3\plus{}2x^2\minus{}bx\plus{}1\equal{}\left(x^2\minus{}\frac{a}{2}x\right)^2\plus{}\left(\frac{b}{2}x\minus{}1\right)^2\plus{}2\minus{}\frac{a^2}{4}\minus{}\frac{b^2}{4}.$
Denote the root by $ x_0;$ then $ x_0\neq 0$ and
$ 0\equal{}\left(x_0^2\minus{}\frac{a}{2}x_0\right)^2\plus{}\left(\frac{b}{2}x_0\minus{}1\right)^2\plus{}\frac{8\minus{}a^2\minus{}b^2}{4}x_0^2\ge \frac{8\minus{}a^2\minus{}b^2}{4}x_0^2$

Dividing by $ x_0^2>0$ yields $ \frac{8\minus{}a^2\minus{}b^2}{4}\le 0 \iff a^2\plus{}b^2\ge8.$[/hide]

Nice!

I solved the second too :)
2:

[hide]
Let the roots be $ p,q,r,s$. By Vieta's formulas we know:

$ pqrs\equal{}1$
$ pq\plus{}pr\plus{}ps\plus{}qr\plus{}qs\plus{}rs\equal{}6$
$ a^2\equal{}(p\plus{}q\plus{}r\plus{}s)^2\equal{}p^2\plus{}q^2\plus{}r^2\plus{}s^2\plus{}12$
$ b^2\equal{}(pqr\plus{}pqs\plus{}prs\plus{}qrs)^2$

We only need prove:

$ p^2\plus{}q^2\plus{}r^2\plus{}s^2\plus{}(pqr\plus{}pqs\plus{}prs\plus{}qrs)^2 \ge 20$
$ \Leftrightarrow p^2\plus{}q^2\plus{}r^2\plus{}s^2\plus{}(\frac{1}{p}\plus{}\frac{1}{q}\plus{}\frac{1}{r}\plus{}\frac{1}{s})^2 \ge 20$

$ \Leftrightarrow p^2\plus{}q^2\plus{}r^2\plus{}s^2\plus{}\frac{1}{p^2}\plus{}\frac{1}{q^2}\plus{}\frac{1}{r^2}\plus{}\frac{1}{s^2}\plus{}2(\frac{1}{pq}\plus{}\frac{1}{pr}\plus{}\frac{1}{ps}\plus{}\frac{1}{qr}\plus{}\frac{1}{qs}\plus{}\frac{1}{rs})\ge 20$

Because $ \frac{1}{pq} \plus{} \frac{1}{pr} \plus{} \frac{1}{ps} \plus{} \frac{1}{qr} \plus{} \frac{1}{qs} \plus{} \frac{1}{rs} \equal{} pq \plus{} pr \plus{} ps \plus{} qr \plus{} qs \plus{} rs$ we have:

$ p^2\plus{}q^2\plus{}r^2\plus{}s^2\plus{}\frac{1}{p^2}\plus{}\frac{1}{q^2}\plus{}\frac{1}{r^2}\plus{}\frac{1}{s^2} \ge 8$.

Which holds since for each root $ x^2 \plus{} \frac{1}{x^2} \ge 2$ (By AM-GM)

[/hide]

[quote="bertram"](2) $ x^4 - ax^3 + 6x^2 - bx + 1$ has four real roots. Prove $ a^2 + b^2\ge 32.$[/quote]

[hide]Let the roots be $ r_1,r_2,r_3,r_4$. By condition,
\begin{eqnarray*}\sum r_i&=&a\\ \sum r_ir_j&=&6\\ \sum r_ir_jr_k&=&b\\ r_1r_2r_3r_4&=&1\end{eqnarray*}
Hence we need to show that
\[ \left(\sum r_i\right)^2+\left(\sum r_ir_jr_k\right)^2\ge 32\]
Or equivalently
\[ \left(\sum r_i\right)^2+\left(\sum \frac{1}{r_i}\right)^2\ge 32\]
This can be written as
\[ \sum r_i^2+2\sum r_ir_j+\sum\frac{1}{r_i^2}+2\sum\frac{1}{r_ir_j}\ge 32\]
Using $ r_1r_2r_3r_4=1$ and $ \sum r_ir_j=6=\sum\frac{r_ir_j}{r_1r_2r_3r_4}=\sum\frac{1}{r_ir_j}$ we can rewrite this as
\[ \sum r_i^2+\sum\frac{1}{r_i^2}\ge 8\]
Which is just AM-GM since $ \sum r_i^2+\sum\frac{1}{r_i^2}\ge \sum 2=8$.[/hide]

[quote="nayel"][quote="bertram"](2) $ x^4 - ax^3 + 6x^2 - bx + 1$ has four real roots. Prove $ a^2 + b^2\ge 32.$[/quote]

[hide]Let the roots be $ r_1,r_2,r_3,r_4$. By condition,
\begin{eqnarray*}\sum r_i & = & a \\
\sum r_ir_j & = & 6 \\
\sum r_ir_jr_k & = & b \\
r_1r_2r_3r_4 & = & 1\end{eqnarray*}
Hence we need to show that
\[ \left(\sum r_i\right)^2 + \left(\sum r_ir_jr_k\right)^2\ge 32
\]
Or equivalently
\[ \left(\sum r_i\right)^2 + \left(\sum \frac {1}{r_i}\right)^2\ge 32
\]
This can be written as
\[ \sum r_i^2 + 2\sum r_ir_j + \sum\frac {1}{r_i^2} + 2\sum\frac {1}{r_ir_j}\ge 32
\]
Using $ r_1r_2r_3r_4 = 1$ and $ \sum r_ir_j = 6 = \sum\frac {r_ir_j}{r_1r_2r_3r_4} = \sum\frac {1}{r_ir_j}$ we can rewrite this as
\[ \sum r_i^2 + \sum\frac {1}{r_i^2}\ge 8
\]
Which is just AM-GM since $ \sum r_i^2 + \sum\frac {1}{r_i^2}\ge \sum 2 = 8$.[/hide][/quote]

both your and hipsishopsis's solution is correct :) those two probelms were posted on a German MO for grade 11 and 12 respectively, and while everyone solved the second one hardly anyone could solve the first ^^

[hide="another solution for the first"]
Given a root $ x$ of that equation (note that $ x$ is nonzero), we have
\[ ax^3\plus{}bx\equal{}x^4\plus{}2x^2\plus{}1\]
\[ (a^2\plus{}b^2)(x^6\plus{}x^2)\geq (ax^3\plus{}bx)^2\equal{}(x^4\plus{}2x^2\plus{}1)^2\]
by Cauchy. Finally, we have $ (x^4\plus{}2x^2\plus{}1)^2\geq 8(x^6\plus{}x^2)$: If we substitute $ u\equal{}x^2\plus{}\frac{1}{x^2}$, it reduces to $ (u\plus{}2)^2\geq 8u$, or $ (u\minus{}2)^2\geq 0$.

So $ (a^2\plus{}b^2)(x^6\plus{}x^2)\geq 8(x^6\plus{}x^2)$ and $ x^6\plus{}x^2$ is positive $ \Rightarrow a^2\plus{}b^2\geq 8$.
[/hide]


From the top of a rough sphere of radius $R$ a small ball of radius $r$ and of mass $m$ rolls down with no initial speed. Where will the ball begin to slide and what is the frictional force at the moment when it begins to slide if $R=1$ metre, $r=0.1$ metre, $m=15$ kilograms, and the coefficient of static friction is $\mu =\mu_{0}=0.2$?

if $N$ is the normal force between the sphere and the ball, then at the moment of sliding the friction force will be $\mu N$. also until sliding the mechanical energy doesnt change, so you can find the speed of the ball at this point. you write the equations of motion using this, and will get the wanted angle. :D

Why the mechanical energy doesnt change? Doesn't friction act along the way?

[quote="Thaakisfox"]the ball is rolling until, then so the friction force doesnt have work on the ball, only the gravity force.[/quote]

I don't think so. The ball is trying to move downward, so there must be friction force to counter it.

Anyway, without friction, only gravity acts through the c.g., the ball can't roll?!

[quote="Thaakisfox"]but you are mixing the frictions together. Before sliding there is only static friction, no kinetic friction. And static friction force doesnt make work on the ball. Think about it, when the ball is rolling, the velocity of the point on which the friction force is exerted is zero. :wink:[/quote]

OK. Totally agreed.   :)

But now, do we need to take the rolling energy into account?

Yes, of course its part of the total kinetic energy :D


How is Zaimi(Albanian Eagle)-Marku(Son of Skanderbeg) ineq? :maybe:

Yeah I also wonder what is this inequality. :D

it's just the name I call the inequality [url=http://www.mathlinks.ro/Forum/viewtopic.php?&t=89757]here[/url]
post #5[I discovered it in collaboration with Keler Marku]

it's a generalisation to schur for n-variables.


Rock breaks scissors
Scissors cut paper
Paper covers rock.

Each poster must add one more item to the list, as well as documenting whether the item wins or loses against all of the items already listed. Each item must win against exactly half of the pre-existing items if possible; if the number of pre-existing items is odd, just round up or down  :wink:

Oh yeah, for clarity, if an item beats another, use the [b]active voice[/b], if it loses to another, use the [b]passive voice[/b]. Or switch it around.

Ready? Go!

Bomb blows up rock
Bomb blows up paper
scissors cut wires and defuse bomb.

Brick smashes rock
smashes scissors
gets covered by paper
crumbles by bomb.

Fire can't be cut by scissors
Fire burns paper
Brick stops fire
Rock gets red hot by fire
Bomb explodes in the presence of fire

Boulder smashes all but paper, fire.
fire makes boulder hot.
paper is too small to cover up boulder.

1) Rock
2) Scissors
3) Paper
4) Bomb
5) Brick
6) Fire
7) Boulder
8) Water

Antimatter blows up rock, scissors, paper, bomb, brick, fire, boulder, and water.

[quote="bubka"]List of items:

1) Rock
2) Scissors
3) Paper
4) Bomb
5) Brick
6) Fire
7) Boulder

How on earth is a boulder different from a rock and indeed, for the present purpose, a brick?

Anyway, I leave that upto random. Let's add water. It can put out a fire, defuse a bomb, soak a paper, but can't do anything to the rest.[/quote]
Well, it can weather rock, boulder, and brick... over millions of years.

and btw, nutz...
[quote="MithsApprentice"]3) Every post should be productive in some way, shape, or form. No one wants to scroll through pages of "HAHAHAHA..."

4) Except in cases where the theme of the thread allows (e.g. Mafia "aye" or "nay"), posts should always be longer than 5 words and 1 sentence.[/quote]

sledgehammer crushes rock.
sledgehammer crushes paper.
sledgehammer crushes scissors.
sledgehammer crushes bomb.
sledgehammer crushes brick.
sledgehammer crushes fire.
sledgehammer crushes boulder.
sledgehammer crushes water.
sledgehammer crushes antimatter.

why? because i said so. :P

sledgehammer crushes rock.
sledgehammer can't write on paper.
sledgehammer gets cut by scissors.
sledgehammer crushes bomb.
sledgehammer crushes brick.
sledgehammer burns in fire.
sledgehammer crushes boulder.
sledgehammer gets rusty by water.
sledgehammer crushes antimatter.

fixed.

[quote="randomdragoon"]
sledgehammer gets cut by scissors.
[/quote]
hmm??  :wink:
Howsabout...

 :spider:
Spider gets squished by sledgehammer
Spider entangles rock.
Spider gets squished by boulder (too big)
Spider blows up by bomb
Spider puts out fire ( :?: )
Spider floats in water...
Spider blows up by antimatter
Spider cuts up paper with sharp "claws"
Spider gets squished by brick.
Spider entangles scissors.

Pencil

writes on paper
squishes spider
smashed by rock
bashed by brick
smushed by boulder
Too small to be blown up by bomb
Burnd in fire
Wood breaks up in water
blows up by antimatter
it teams up with scissors and destroys paper. :P

Come on--there's got to be something that beats antimatter. :P

I forgot sledgehammer.
Smashed by sledgehammer

the sledgy smashed antimatter.


trigonometry
real analysis
real analysis unsolved

Let $f : \mathbb{R} \rightarrow \mathbb{R}, \;\; f \in C^2(\mathbb{R}), \;\; f \geq 0$
Let $g = \sqrt{f}$
Let $h: [a,b] \subset \mathbb{R} \rightarrow \mathbb{R}$ such that $h = g | _{[a,b]}$

1) $g \in C^n(\mathbb{R})$... $n$ = ?

2) is $h$ Lipschitz?

1. $n=0$. The best you can do is continuous; you can't have $C^1$ as is shown by the example $f(x)=x^2,\,g(x)=|x|.$

2. Still working on it. Have to consider odd examples like $x^a\sin^2(x^{-b})$ for exponents yet to be determined.